Solution Manual for Dynamics of Structures, 4E Anil K Chopra by Ace Test Banks

Solution Manual for Dynamics of Structures, 4E By Anil K. Chopra

CHAPTER 13 m /2

Problem 13.1 m/2 h

0.604 m

0.104 m

0.854 m

m u

0.146 m

h 1

h s

Fig. P13.1a

LM 2 − 1OP N− 1 1Q

m = m

Part b

LM1 OP N 1 2Q

Substituting Γn and φ jn in Eq. (13.2.5) gives floor displacements due to each mode:

Vibration properties (from Problem 10.6): k

ω1 = 0. 765

φ1 = 0. 707 1

RSu (t ) UV = 1207 RS0.707UV D (t ) = RS0.854UV D (t ) . u t ( ) . T W T 1 W T1207 W . U RSu (t ) UV = − 0.207 RS− 0.707UV D (t ) = RS 0146 D (t ) Tu (t )W T 1 W T− 0.207VW 1

ω 2 = 1. 848

Fig. P13.1b

Stiffness and mass matrices (from Problem 9.5): k = k

k m

φ2 = − 0. 707 1

Part a The modal quantities given by Eq. (13.2.3) are M1 = m

M2 = m

K1 = 0. 586 k

K2 = 3. 414 k

L1h = 1. 207 m

Lh2 = − 0. 207 m

L1h

Lh2

Γ1 =

= 1. 207

Γ2 =

= − 0. 207

Substituting Γn , m, and φn in Eq. (13.2.4) gives

RS0.854UV T0.604W . U R 0146 s = Γ mφ = mS V . W T− 0104 s1 = Γ1 m φ 1 = m 2

The modal expansion of effective earthquake forces is shown in Fig. P13.1b.

Combining the contributions of the two modes gives u1 ( t ) = 0. 854 D1 ( t ) + 0.146 D2 ( t ) u2 ( t ) = 1. 207 D1 ( t ) − 0. 207 D2 ( t )

Part c st

The modal static responses Vjn for the story shears are determined in Fig. P13.1c. First mode

Second mode s22 = – 0.104 m

s21 = 0.604 m V21st = 0.604 m s11 = 0.854 m

V22st = – 0.104 m s12 = 0.146 m

V11st = 1.458 m

V12st = 0.042 m

Fig. P13.1c st

Substituting Vjn in Eq. (13.2.8) gives the modal responses: V11 (t ) = 1458 . m A1 (t ) V12 (t ) = 0.042 m A2 (t ) V21 (t ) = 0.604 m A1 (t ) V22 (t ) = − 0104 . m A2 (t ) © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 1 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Combining the modal responses gives the total responses: V1 ( t ) = V11 ( t ) + V12 ( t ) = 1. 458 m A1 ( t ) + 0. 042 m A2 ( t ) V2 ( t ) = V21 ( t ) + V22 ( t ) = 0. 604 m A1 ( t ) − 0.104 m A2 ( t )

Part d Static analysis of the structure for external floor forces st sn gives the modal static responses Mbn and M1stn for Mb and M1 , the overturning moments at the base and the first floor, respectively: st Mb1 = mh[ 0. 854 (1) + 0. 604 ( 2 )] = 2. 062 mh st Mb2 = mh[ 0.146 (1) − 0.104 ( 2 )] = − 0. 062 mh st M11 = 0. 604 mh st M12 = − 0.104 mh st and M1stn in Eq. (13.2.8) gives the modal Substituting Mbn responses:

Mb1 ( t ) = 2. 062 mh A1 ( t )

Mb2 ( t ) = − 0. 062 mh A2 ( t )

M11 ( t ) = 0. 604 mh A1 ( t )

M12 ( t ) = − 0.104 mh A2 ( t )

Combining the modal responses gives the total response: Mb ( t ) = Mb1 ( t ) + Mb 2 ( t ) = 2. 062 mh A1 ( t ) − 0. 062 mh A2 ( t )

M1 ( t ) = M11 ( t ) + M12 ( t ) = 0. 604 mh A1 ( t ) − 0.104 mh A2 ( t )

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 2 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.2

The modal static responses for the various response quantities are given in Table P13.2b.

System properties: w

m = k =

24 EI

100 386 =

Table P13.2b

= 0. 2591 kip - sec2 in.

Mode n 3

24 ( 727 ) ( 29 × 10 )

u1stn u2stn st Vbn m st V2 n m st Mbn mh st M1n mh

= 169. 5 kips in.

(12 × 12 )3

Vibration properties (from Problem 10.6):

ω1 = 0. 765 ω 2 = 1. 848

k m k m

= 19. 565 = 47. 263

T1 = 0. 321 sec

T2 = 0.133 sec

2. 23 × 10 3.15 × 10

−3

6. 529 × 10

−5

− 9. 237 × 10 −5

1.458

0.042

0.604

– 0.104

2.062

– 0.062

0.604

– 0.104

Step 5c of Section 13.2.4 is implemented to determine the contribution of the nth mode to selected response quantities — floor displacements, story shears, and story overturning moments:

. U RS0.707UV = RS1389 V . W m T 1 W T1965 . U R 1389 1 R0.707 U φ = = S S V V . W m T −1 W T−1965 1

φ1 =

rn ( t ) = rnst An ( t )

where rnst and An ( t ) are both known. These results for roof displacement u2 ( t ) , base shear Vb ( t ) , and base overturning moment Mb ( t ) are plotted in Figs. P13.2c-e, where their peak values are noted.

Modal properties: Table P13.2a

Part c

Mode

Lhn

Lθn h

1.0

0.614

0.869

1.0

0.105

– 0.149

The modal contributions to each response quantity are combined at each time instant to obtain the total responses shown in Figs. P13.2c-e. Table P13.2c summarizes the peak values of the total responses.

Part a The displacements Dn ( t ) and pseudo-accelerations An ( t ) of the two modal SDF systems are calculated using the procedure of Section 5.2 with Δt = 0. 01 sec and are shown in Figs. P13.2a and P13.2b. Part b

Table P13.2c Floor or Story

Displacement, in.

Shear, kips

Overturning Moment, kip-ft

2 1

0.964 0.679

49.56 115.11

594.65 1959.25

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 3 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Mode 1

1 0

Dn, in.

-1

0.797

Mode 2

1 0

0.118

-1 0

Time, sec Fig. P13.2a

Mode 1

1 0

An, g

-1

0.791 Mode 2

1 0 -1

0.684 0

Time, sec Fig. P13.2b

Mode 1

u2n, in.

0 -1

0.962

Mode 2

0.025

0 -1

un, in.

Total

1 0

0.964

-1 0

Time, sec

Fig. P13.2c © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 4 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Mode 1

150 0 -150

Vbn, kips

115.28

Mode 2

150 0

2.933

-150

Total

150

Vb, kips

115.11

-150 0

Time, sec

Fig. P13.2d

Mode 1

Mbn, 3

10 kip-ft

0 -2

1956.32

Mode 2

49.78

0 -2

Mb,

10 kip-ft

Total

1959.25

-2 0

Time, sec

Fig. P13.2e

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 5 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.3 2

ω1 = 0. 765

φ1 = 0. 707 1

ω 2 = 1. 848 T

j =1

L1h = ∑ m j φ j1 = m ( 0. 707) +

j =1

L1 = ∑ h j m j φ j1 = h m ( 0. 707) + 2 h j =1 2

L2 =

∑hmφ j

j =1

= 0.293mh 2

L2 =

∑hmφ j

j =1

n =1 2

∑ Mn = ∑ m j *

n =1

(1) = − 0. 207 m

∑ m = m + 2 = 15. m

(1) = 1. 207 m

Lh2 = ∑ m j φ j 2 = m ( − 0. 707) +

* n

n =1 2

φ2 = − 0. 707 1

U V| ⇒ |W

. m + 0.043m = 15 .m | ∑ M = 1457 |

From problem 13.2:

Verify Eq. (13.2.17): 2

(1) = 1. 707 mh

n =1

* n

n =1

∑ h m = h m + 2h FGH 2 IJK = 2mh

F mI = hm ( − 0.707) + 2h G J (1) H 2K

n =1

⇒

F mI = h m ( − 0.707) + 2h G J (1) H 2K

U |V || |W

. h (1457 . m) + ( − 1415 . h) (0.043m) = 2mh | ∑ h M = 1414 |

∑

n =1

hn* M n* =

∑hm j

n =1

= 0.293mh 2

j =1

M1 = ∑ m j φ2j1 = m ( 0. 707)2 + 2

(1)2 = m

M2 = ∑ m j φ2j 2 = m ( − 0. 707 )2 +

j =1

(1)2 = m

From Eq. (13.2.9a), the effective modal masses are M1* =

( L1h )2 M1

= 1. 457 m

( Lh2 )2

M2* =

= 0. 043m

From Eq. (13.2.9a), the effective modal heights are h1* =

L1θ L1h

h2* =

= 1. 414 h

Lθ2 Lh2

= − 1. 415h

m /2 h

1.457 m 0.043 m m

&& ug(t)

1.414 h Mode 1

1.415 h Mode 2

&& ug(t)

Verify Eq. (13.2.14):

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 6 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

. − 0.411O R|u (t ) U| LM− 0164 . − 0.411PP R0.647 U |u (t )|V = 1 M− 0164 u (t ) = S VW D (t ) . ||u (t ) || h MM 0.904 − 0.740PP ST1341 N 0.904 − 0.740Q Tu (t )W R|− 0.657U| 1 |− 0.657| (b) = S V D (t ) h |− 0.407| |T− 0.407|W

Problem 13.4

m/2

5 6

Fig. P13.4a

The floor displacements due to the second mode are Mass and lateral stiffness matrices (from Problem 9.6): m = m

LM1 OP N 1 2Q

LM N

37.15 − 1512 . EI k$ tt = 3 . 1019 . h − 1512

OP Q

Vibration properties (from Problem 10.10): EI

ω1 = 2. 407 φ1 =

ω 2 = 7.193

mh3

RS0.482UV T 1 W

φ2 = m 2

M2 = m ( − 1. 037) +

L1h = m ( 0. 482 ) +

Lh2 = m ( − 1. 037) +

Γ1 =

L1h M1

= 1. 341

mh3

. U RS− 1037 T 1 VW

(f)

Part b

= − 0. 341

RSu (t ) UV = 1341 RS0.482UV D (t ) = RS0.647UV D (t ) . . Tu (t )W T 1 W T1341 W 1

(d)

RSu (t ) UV = RS0.647 D (t ) + 0.353 D (t ) UV (e) . D (t ) − 0.341 D (t ) W Tu (t )W T 1341 R|u (t ) U| R|− 0.657 D (t ) + 0.082 D (t )U| |u (t )|V = 1 |S− 0.657 D (t ) + 0.082 D (t )|V u (t ) = S ||u (t ) || h ||− 0.407 D (t ) + 0.572 D (t ) || Tu (t )W T− 0.407 D (t ) + 0.572 D (t ) W

Part a From Eq. (13.2.5), the floor displacements due to the first mode are u1 (t ) =

u(t ) =

(1) = − 0. 537 m

The joint rotations associated with u2 , u0 2 = T u2 , can be computed following Eq. (b) to obtain

Combining the modal responses gives the total floor displacements u( t ) and total joint rotations u0 ( t ) :

(1) = 1. 575 m

Lh2

(c)

Γ2 =

(1) = 0. 982 m

(1)2 = 0. 732 m

. U RSu (t ) UV = − 0.341 RS− 1037 R 0.353UV D (t ) D (t ) = S V Tu (t )W T 1 W T− 0.341W

R|u (t ) U| R|0.082U| u t ( ) | |V = 1 |S0.082|V D (t ) u (t ) = S ||u (t ) || h ||0.572|| Tu ( t ) W T0.572W

Modal properties: M1 = m ( 0. 482 )2 +

u 2 (t ) =

(a) The joint rotations associated with u1 are

The bending moments at the ends of a flexural element are related to the nodal displacements by Ma =

Mb =

4 EI L

2 EI L

θa +

2 EI L

4 EI L

θb +

6 EI 2

ua −

6 EI L2

(g)

(h)

For a first story column, L = h and the nodal displacements are as shown in Fig. P13.4b

u01 = T u1

where T is available in the solution to Problem 10.10:

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 7 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

θa = u 3

Ma =

Mb =

u5 +

2 EI

4 EI

u6 +

θb = 0

(m)

2 EI

u5 2h 2h = mh − 0. 211 A1 ( t ) + 0. 0332 A2 ( t )

ub = 0

4 EI

u6 2h 2h = mh − 0. 211 A1 ( t ) + 0. 0332 A2 ( t )

ua = u1

(n)

Fig. P13.4b

Substituting these ua , ub , θa and θb and Eqs. (e)-(f) in Eqs. (g)-(h) gives: Ma =

4 EI h

2 EI

u3 +

6 EI

(0) +

u1 −

6 EI h2

1. 254 D1 ( t ) + 2. 446 D2 ( t )

(0)

(i)

Relate Dn ( t ) to An ( t ) : D1 ( t ) = D2 ( t ) =

A1 ( t )

ω12

A2 ( t )

ω 22

EI ( 2. 407 )

(j)

mh3 EI ( 7.193)

A1 ( t )

A2 ( t )

Substituting Eq. (j) in Eq. (i) gives Ma = mh 0. 216 A1 ( t ) + 0. 0473 A2 ( t )

(k)

Similarly, Mb =

2 EI h

u3 +

6 EI h2

= mh 0. 443 A1 ( t ) + 0. 0441 A2 ( t )

(l)

For the second floor beam, L = 2 h and the nodal displacements are as shown in Fig. P13.4c. ua = 0

ub = 0

a θa = u5

θb = u6

Fig. P13.4c

Substituting these ua , ub , θa and θb and Eqs. (e)-(f) in Eqs. (g)-(h) gives: © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 8 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

LM1 OP R|− 1U| R| 0.3333U| s = − 0.3333m M 1 S 0V = m S 0 V| MN 0.5PPQ |T 1|W |T− 01667 . W 1 0 . 5 0 . 0447 L U| OP R| U| R| s = 0.0893m MM 1 − 0 . 866 = m − 0 . 0773 S V S V MN 0.5PPQ |T 1 |W |T 0.0447|W

Problem 13.5

m /2 h

The modal expansion of m1 is shown next:

m/2

0.6220m

Mass and stiffness matrices (from Problem 9.7): 1 2 −1 0 m = m 1 k = k −1 2 −1 12 0 −1 1

LM MM N

OP PP Q

LM MM N

OP PP Q

where k = 24 EI h3 . Vibration properties (from Problem 10.11): k k k ω12 = ( 2 − 3 ) ; ω 22 = 2 ; ω 32 = ( 2 + 3 ) m m m 0.5 −1 0.5 0 φ 1 = 0.866 φ2 = φ 3 = − 0.866 1 1 1

R| S| T

U| V| W

R| U| S| V| T W

R| S| T

U| V| W

ι=1 3

m = m j φ j1 = m (0.5) + m (0.866) + (1) = 1866 . m 2 j =1

∑

M1 = ∑

m j φ2j1

= m ( 0. 5) + m ( 0. 866 ) +

j =1

Γ1 =

m 2

(1) = 1. 5 m

L1h = 1244 . M1 Lh3 = 0134 . m

M2 = 1. 5 m

M3 = 1. 5 m

Γ2 = − 0. 3333

Γ3 = 0. 0893

Part a

Substituting Γn , m, and φn in Eq. (13.2.4) gives

LM1 OP R| 0.5 U| R|0.6220U| s = 1244 . mM 1 . S0.866V = m S 10773 V MN 0.5PPQ |T 1 |W |T0.6220|W

0.0773m

+ 0.3333m

0.0447m

Part b

The floor displacements due to the nth mode are un = Γn φn Dn ( t ) Substituting for Γn and φn gives

R|u (t ) U| R| 0.5 U| R|0.6220U| . . S|u (t )V| = 1244 S|0.866V| D (t ) = S|10774 V| D (t ) ( ) . Tu t W T 1 W T12440 W R|u (t ) U| R|− 1U| R| 0.3333U| S|u (t )V| = − 0.3333 S| 0V| D (t ) = S| 0 V| D (t ) Tu ( t ) W T 1W T− 0.3333W R|u (t ) U| R| 0.5 U| R| 0.0447U| S|u (t )V| = 0.0893 S|− 0.866V| D (t ) = S|− 0.0774V| D (t ) Tu ( t ) W T 1 W T 0.0893W 1

Lh2 = − 0.5m

0.0447m

+ 0.6220m

Similar calculations for the second and third modes give:

The first-mode properties are computed from Eq. (13.2.3):

= m

The influence vector is

L1h

1.0773m

0.1667m

Combining the displacements:

modal

responses

gives

the

floor

u1 (t ) = 0.6220 D1 (t ) + 0.3333 D2 (t ) + 0.0447 D3 (t ) . u2 (t ) = 10774 D1 (t ) − 0.0774 D3 (t ) . u3 (t ) = 12440 D1 (t ) − 0.3333 D2 (t ) + 0.0893 D3 (t )

Part c

Static analysis of the frame for external floor forces sn gives Vinst , i = 1, 2, 3: V31st = 0. 6220 m

V32st = − 0.1667 m

V33st = 0. 0447 m

st V21 = 1. 6993 m

st V22 = − 0.1667 m

st V23 = − 0. 0326 m

V11st = 2. 3213m

V12st = 0.1667 m

V13st = 0. 0121m

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 9 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

The total story shears are 3

n =1

Vj ( t ) = ∑ Vjn ( t ) = ∑ Vjnst An ( t ) st

Substituting values of Vjn gives V1 ( t ) = m 2. 3213 A1 ( t ) + 0.1667 A2 ( t ) + 0. 0121 A3 ( t ) V2 ( t ) = m 1. 6993 A1 ( t ) − 0.1667 A2 ( t ) − 0. 0326 A3 ( t ) V3 ( t ) = m 0. 6220 A1 ( t ) − 0.1667 A2 ( t ) + 0. 0447 A3 ( t )

Part d

Static analysis of the frame for external floor forces sn st gives Mbn : Mbst1 = mh 0. 6220 (1) + 1. 0773 ( 2 ) + 0. 6220 ( 3) = 4. 6426 mh

Mbst2 = mh 0. 3333 (1) + 0 ( 2 ) − 0.1667 ( 3) = − 0.1667 mh Mbst3 = mh 0. 0447(1) − 0. 0773( 2 ) + 0. 0447( 3) = 0. 0242 mh

The base overturning moment response is 3

n =1

st Mb ( t ) = ∑ Mbn ( t ) = ∑ Mbn An ( t ) st Substituting values of Mbn gives

Mb ( t ) = mh 4. 6420 A1 ( t ) − 0.1667 A2 ( t ) + 0. 0242 A3 ( t )

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 10 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.6 m/2 EI/3

Rigid beams

M 3 =15.46m

12'

Lh3 = ∑ m j φ j 3 = 1503 . m

12'

Γ3 =

j =1

2EI/3

L2 =−0.5 M2

Γ2 =

u1 12'

L3 = 0.0972 M3

Part a

Substituting Γn , m and φ n in Eq. (13.2.4) gives ⎡ 0.44 ⎤ ⎡0.314⎤ ⎥ ⎢ s1 = Γ1mφ1 = 1.403m ⎢0.686⎥ = m ⎢⎢0.962⎥⎥ ⎢⎣ 0.701⎥⎦ ⎢⎣ 0.5 ⎥⎦

24'

Mass and stiffness matrices (from Problem 9.8) ⎡1 ⎤ ⎥ m = m ⎢⎢ 1 ⎥ ⎢⎣ 1 / 2⎥⎦

⎡ 0.25 ⎤ ⎡− 1 / 2⎤ ⎥ ⎢ s 2 = Γ2mφ2 = −0.5m ⎢− 1 / 2⎥ = m ⎢⎢ 0.25 ⎥⎥ ⎢⎣− 0.25⎥⎦ ⎢⎣ 1 / 2 ⎥⎦ ⎡ 0.31 ⎤ ⎡ 3.189 ⎤ ⎥ ⎢ s3 = Γ3mφ3 = 0.0972m ⎢− 2.186⎥ = m ⎢⎢− 0.212⎥⎥ ⎢⎣ 0.049 ⎥⎦ ⎢⎣ 0.5 ⎥⎦

⎡ 5 −2 0 ⎤ k = k ⎢⎢− 2 3 − 1⎥⎥ ⎢⎣ 0 − 1 1 ⎥⎦

The modal expansion of m1 is shown next:

where k = 8 EI / h and h = story height Vibration properties (from Problem 10.12):

ω 1 = 2.241

EI EI EI ; ω 2 = 4.899 ; ω 3 =7.14 mh3 mh3 mh3

⎡0.314⎤

⎡− 1 / 2⎤

⎡ 3.189 ⎤

⎢⎣ 1 ⎥⎦

⎢⎣

M1 =1.069m 3

L1h = ∑ m j φ j1 = 15 .m j =1

L Γ1 = 1 =1.403 M1

Similar calculations for the second and third modes give: M2 =m

0.44m

⎥⎦

The first mode properties are computed from Eq. (13.2.3):

0.962m

φ1 = ⎢⎢0.686⎥⎥ ; φ2 = ⎢⎢− 1 / 2⎥⎥ ; φ3 = ⎢⎢− 2.186⎥⎥ 1

0.701m

m/2

0.25m

0.049m

0.25m

0.212m

0.25m

0.31m

Part b

Equation (13.2.5) gives floor displacements due to each mode: u jn (t ) = Γnφ jn Dn (t )

Substituting for Γn and φ n gives: ⎡ 0.44 ⎤ ⎡ u1(t ) ⎤ ⎢u (t )⎥ = ⎢0.962⎥ D (t ) ⎥ 1 ⎢ ⎢ 2 ⎥ ⎢⎣1.403 ⎥⎦ ⎢⎣u3 (t ) ⎥⎦ 1

Lh2 = ∑ m j φ j 2 = −0.5m j =1

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 11 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

M b ( t ) = 4.467mhA1 ( t ) + 0.033mhA3 ( t )

⎡ 0.25 ⎤ ⎡ u1(t ) ⎤ ⎢u (t )⎥ = ⎢ 0.25 ⎥ D (t ) ⎥ 2 ⎢ ⎢ 2 ⎥ ⎢⎣− 0.25⎥⎦ ⎢⎣u3 (t ) ⎥⎦ 2

Static analysis of the frame for external floor forces s n st

gives M1n :

⎡ 0.31 ⎤ ⎡ u1(t ) ⎤ ⎢u (t )⎥ = ⎢− 0.212⎥ D (t ) ⎥ 3 ⎢ ⎢ 2 ⎥ ⎢⎣ 0.0972 ⎥⎦ ⎢⎣u3 (t ) ⎥⎦ 3

Combining the displacements:

modal

responses

M11 = mh( 0.962(1) + 0.701( 2)) = 2.364mh st

M12 = mh( 0.25(1) − 0.25( 2)) = −0.25mh

gives

the

floor

M13 = mh( −0.212(1) + 0.049( 2)) = −0.114mh

u1 ( t ) =0.44 D1 ( t ) + 0.25D2 ( t ) + 0.31D3 ( t )

The first floor overturning moment response is

u2 ( t ) =0.962 D1 ( t ) + 0.25D2 ( t ) − 0.212 D3 ( t )

M1 ( t ) = ∑ M1n ( t ) = ∑ M1stn An ( t )

u3 ( t ) =1.403D1 ( t ) −0.25D2 ( t ) + 0.0972 D3 ( t )

Part c

Static analysis of the frame for external floor forces s n gives Vinst , i = 1, 2, 3:

n =1

Substituting values of M1n gives

M1 ( t ) = 2.364mhA1 ( t ) − 0.25mhA2 ( t ) − 0.114mhA3 ( t )

Static analysis of the frame for external floor forces s n st

V32 = − 0.25m

V12 = 0.25m

V31 = 0.701m V21 = 1.663m V11 = 2.103m

V33 = 0.049m

V22 = 0

V23 = −0.163m

V13 = 0.147m

gives M 2 n : st

M 21 ( t ) = 0.701mh

M 22 = −0.25mh

st st

M 23 ( t ) = 0.049mh

The total story shears are: 3

n =1

The second floor overturning moment rseponse is

V j ( t ) = ∑ V jn ( t ) = ∑ V jnst An ( t )

n =1

M 2 ( t ) = ∑ M 2 n ( t ) = ∑ M 2stn An ( t )

Substituting values of V jnst gives

V1 ( t ) = 2.103mA1 ( t ) + 0.25mA2 ( t ) + 0.147mA3 ( t ) V2 ( t ) = 1.663mA1 ( t ) + 0

− 0.163mA3 ( t )

V3 ( t ) = 0.701mA1 ( t ) − 0.25mA2 ( t ) + 0.049mA3 ( t )

Substituting values of M 2 n gives M 2 ( t ) = 0.701mhA1 ( t ) − 0.25mhA2 ( t ) + 0.049mhA3 ( t ) 2

* n

n =1

Part d

∑ h m = h m + 2h FGH 2 IJK = 2mh 2

Static analysis of the frame for external floor forces

n =1

s n gives M bnst :

st M b1 = mh 0.44(1) +0.962( 2) + 0.701( 3)

= 4.467mh

U |V || |W

. h (1457 . m) + ( − 1415 . h) (0.043m) = 2mh | ∑ h M = 1414 |

⇒

∑

n =1

hn* M n* =

∑hm j

n =1

M b2 = mh 0.25(1) + 0.25( 2) −0.25( 3) = 0 st

M b3 = mh 0.31(1) −0.212( 2) + 0.049( 3) = 0.033mh

The base overturning moment response is: 3

n =1

st M b ( t ) = ∑ Mbn ( t ) = ∑ Mbn An ( t ) st

Substituting values of M bn gives

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 12 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.7

Step 5c of Section 13.2.4 is implemented to determine the contribution of the nth mode to selected response quantities:

System properties: m = k =

100 g

24 EI h3

100 386. 4

= 0. 2588 kip − sec2 in .

24 ( 29 × 103 ) (1400 )

(12 × 12 )3

rn ( t ) = rnst An ( t )

= 326. 32 kips in .

Vibration properties (from Problem 10.11):

ω12 = ( 2 − 3 )

k m

; ω 22 = 2

T1 = 0. 3418 sec

k m

; ω 32 = ( 2 + 3 )

Part c

T2 = 0.1251 sec T3 = 0. 0716 sec

R| 0.5 U| S| V| T 1 W

φ 1 = 0.866

φ2 =

R|− 1U| S| 0V| T 1W

φ3 =

R| 0.5 U| S|− 0.866V| T 1 W

Modal properties (from Problem 13.5): Γ1 = 1. 244

Γ2 = − 0. 3333

where rnst and An ( t ) are both known. These results for roof displacement u3 ( t ) , base shear Vb ( t ) , and base overturning moment Mb ( t ) are plotted in Figs. P13.7c-e where their peak values are noted.

Γ3 = 0. 0893

Part a

The displacements Dn ( t ) and pseudo-acceleration An ( t ) of the three modal SDF systems (with Tn given above and ζn = 0. 05 ) are calculated using the numerical procedure of Section 5.2 with Δt = 0. 01sec . The results are shown in Figs. P13.7a-b.

The modal contributions to each response quantity are combined at each time instant to obtain Figs. P13.7c-e. Table P13.7b summarizes the peak values of the total responses. Table P13.7b Floor or story

Displacement,

Shear,

in.

kips

Overturning moment, kip-ft

1.103

52.22

626.6

0.957

138.08

2267.5

0.580

189.29

4320.8

Part b

The modal static responses for the various response quantities are given in Table P13.7a (also see Problem 13.5). Table P13.7a Mode n

2 −3

3 −3

u3stn

3. 682 × 10

st Vbn m

2.3213

0.1667

0.0121

V2stn m

1.6993

-0.1667

-0.0326

V1stn m

0.6220

-0.1667

0.0447

st Mbn mh

4.6426

-0.1667

0.0242

0.132 × 10

0. 019 × 10

−3

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 13 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

0.8859

Mode 1

0 -1

Mode 2

Dn, in.

0.1096

-1

Mode 3

1 0

0.0498

-1 0

Time, sec Fig. P13.7a

Mode 1

0.7746

0 -1

Mode 2

Ag, g

0 -1

0.7153 Mode 3

1 0

0.6065

-1 0

Time, sec Fig. P13.7b

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the14 publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

1.102

Mode 1

0 -1 1

u3n, in.

Mode 2

0.0365

0 -1

Mode 3

1 0

0.0044

-1

u3, in.

1.103

Total

0 -1 0

Time, sec Fig. P13.7c

Mode 1

179.82

200 0 -200

Mode 2

200

Vbn, kips

11.92

-200

Mode 3

200 0

0.73

-200

Total

200

Vb, kips

189.29

-200 0

Time, sec Fig. P13.7d

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the15 publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

4315.6

Mode 1

0 -4 4

Mbn, 10 3 kip-ft

Mode 2

143.1

0 -4

Mode 3

17.5

0 -4

Mb, 3

10 kip-ft

4320.8

Total

0 -4 0

Time, sec Fig. P13.7e

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the16 publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.8

Step 5c of section 13.2.4 is implemented to determine the contribution of the n th mode to selected response quantities:

System properties: m=

100 100 = = 0.2588 kip − sec 2 / in. g 386.4

rn (t ) = rnst An (t )

8EI 8(29 × 103 )(1400) k= 3 = = 108.77 kips/in. h (12 × 12)3

Vibration properties (from Problem 10.12): k ; m T1 = 0.3868 sec

k k ω 32 = 6.372 ; m m T2 = 0.1769 sec T3 = 0.1214 sec

⎧0.314⎫ ⎪ ⎪ φ1 = ⎨0.686⎬ ⎪ 1 ⎪ ⎭ ⎩

⎧− 0.5⎫ ⎪ ⎪ φ 2 = ⎨− 0.5⎬ ⎪ 1 ⎪ ⎭ ⎩

ω 12 = 0.6277

ω 22 = 3

⎧ 3.186⎫ ⎪ ⎪ φ 3 = ⎨− 2.186⎬ ⎪ 1 ⎪ ⎭ ⎩

Modal properties from (from Problem 13.6): Γ1 = 1.403

Γ2 = −0.5

Γ3 = 0.0972

Part a

The displacements D n (t ) and pseudo-acceleration An (t ) of the three modal SDF systems (with Tn given above and ς n = 0.05 ) are calculated using the numerical procedure of Section 5.2 with Δt = 0.02 sec . The results are shown in Figs. P13.8a-b.

where rnst and An (t ) are both known. These results for roof displacement u 3 (t ) , base shear Vb (t ) , and base overturning moment M b (t ) are plotted in Figs. P13.8c-e where their peak values are noted. Part c

The modal contributions to each response quantity are combined at each time instant to obtain Figs. 13.8c-e. Table P13.8b summarizes the peak values of the total responses. Table P13.8b Floor or Story 3 2 1

Displacement, in.

Shear kips.

1.4332 1.1085 0.5281

54.85 126.17 172.23

Overturning moment. kip-ft. 658.2 2136.0 3965.9

Part b

The modal static responses for the various response quantities are given in Table P13.8a (also see Problem 13.6). Table P13.8a u 3stn

1 5.3710×10-3

2 -0.3965×10-

3 0.0363×10-3

Vbnst m

2.103

0.25

0.147

V 2stn m

1.663

-0.163

V3stn m

0.701

-0.25

0.049

st M bn mh

4.467

0.033

Mode n

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the17 publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

1 Mode 1 0

-1

1.0855

1 Mode 2 Dn, in. 0 0.2717 -1 1 Mode 3 0 0.1090 -1 0

Time, sec

Fig. P13.8a

1 Mode 1 0

0.7412 -1 1 Mode 2 A n, g 0

-1

0.8868

1 Mode 3 0

0.7556

-1 0

Time, sec

Fig.. P13.8b © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 18 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

2 Mode 1 0 1.5227

-2 2

Mode 2 0.1359 u3n, in. 0

-2 2 Mode 3 0 0.0106 -2 2

1.4332

Total

u3, in. 0 -2 0

Time, sec

Fig. P13.8c 200

Mode 1

0 155.87

-200 200

Mode 2 Vbn,

0 22.17

kips -200 200

Mode 3 0 11.11 -200 200 Vb ,

Total

kips 172.23

-200 0

Time, sec

Fig. P13.8d

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 19 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Mode 1

-4

3973.0

4 Mode 2 Mbn, 0 0.0

10 kip-ft. -4 4

Mode 3 0 29.9 -4 4

Total

Mb, 0 3

10 kip-ft. 3965.9

-4 0

Time, sec

Fig. P13.8e

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 20 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.9 The floor masses, and the height of each floor above the base are m2 = m

m3 =

h1 = h

h2 = 2 h

h3 = h

R| U| S| V| T 1 W

φ 1 = 0.866 M1 = 1. 5 m

R| U| S| 0V| T 1W −1

φ2 =

M2 = 1. 5 m

R| U| S|− 0.866V| T 1 W

j =1

Lh3

= 1. 866 m = ∑ m j φ j 2 = m 1 × ( − 1) + 0 + 0. 5 × 1 = − 0. 5 m

Lθ3

Substituting for m j , h j and φ jn in Eq. (13.2.9b) gives Lθn : 3

L1θ = ∑ h j m j φ j1 j =1

= mh 1 × 1 × 0. 5 + 2 × 1 × 0. 866 + 3 × 0. 5 × 1 = 3. 732 mh 3

Lθ2 = ∑ h j m j φ j 2 = mh 1 × 1 × ( − 1) + 0 + 3 × 0. 5 × 1 j =1

= 0. 5 mh 3

L3 = ∑ h j m j φ j 3 θ

j =1

0. 268

= h

The effective modal masses and effective modal heights are given by Eq. (13.2.9a):

= 2h

0.134

2.3213 m

0.0121 m 0.1667 m

&&ug(t)

Mode 1

Mode 2

Mode 3

&& ug(t)

Verify Eq. (13.2.14): 3

∑ Mn* = 2. 3213m + 0.1667 m + 0. 0121m = 2. 5m

n =1 3

∑ m j = m + m + 2 = 2. 5 m

n =1

∴

n =1

∑ Mn* = ∑ m j

Verify Eq. (13.2.17): 3

∑ hn* Mn* = 2 h ( 2. 3123m ) + ( − h )( 0.1667 m ) + 2 h ( 0. 0120 m )

n =1

= 4. 5 mh

= mh 1 × 1 × 0. 5 + 2 × 1 × ( − 0. 866 ) + 3 × 0. 5 × 1 = 0. 268 mh

= −h

− 0. 5

j =1

= 2h

0. 5

= h

= 0. 0121m

1. 5

1. 866

Lh3 = ∑ m j φ j 3 = m 1 × 0. 5 + 1 × ( − 0. 866 ) + 0. 5 × 1 = 0.134 m

m ( 0.134 )2

3. 732

= h

= 0.1667 m

m /2

L1h = ∑ m j φ j1 = m 1 × 0. 5 + 1 × 0. 866 + 0. 5 × 1

Lh2

h3* =

Substituting for m j and φ jn in Eq. (13.2.3) gives Lhn :

Lh2

Lθ2

h2* =

M3 = 1. 5 m

j =1

L1h

1. 5

M3 L1θ

h1* =

0.5

φ3 =

( Lh3 )2

M3* =

m ( − 0 . 5 )2

The natural modes and generalized masses (from Problem 13.5) are: 0.5

( Lh2 )2

M2* =

m1 = m

m (1866 ( L1h ) 2 . )2 = = 2.3213m M1 15 .

M 1* =

∑ hj m j = h ( m ) + 2 h ( m ) + 3h ( 0. 5m ) = 4. 5mh

n =1

∴

n =1

∑ hn* Mn* = ∑ hj m j

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 21 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.10 The floor masses, and the height of each floor above the base are:

h1 =

0 Lθ =0 h2* = h2 = L2 −0.5m

m 2

m1 =m

m2 =m

m3 =

h1 =h

h2 =2h

h3 =3h

h3 =

The natural modes and generalized masses (from Problem 13.6) are:

M1 =1.069m

3189 .

2.104m

M2 =m

M 3 =15.46m

Substituting for m j and φ jn in Eq. (13.2.3) gives

= 0.25m

2.124h

0.146m

0.211h

Lhn :

3 h L1 = ∑ m jφ j1 = m( 0.314 + 0.686 + 0.5) = 15 .m j =1

u&&() gt

Mode 1

Mode 2

Mode 3

u&&() gt

Verify Eq. (13.2.14):

L2 = ∑ m jφ j 2 = m( −0.5 − 0.5 + 0.5) = −0.5m

∑ Mn* = m( 2.104 + 0.25 + 0.146) = 2.5m

j =1 3

n =1

j =1

∑ m j = m(1 + 1 + 0.5) = 2.5m

− 2.186 + 0.5) = 1503 L3 = ∑ m jφ j 3 = m( 3189 . . m

j =1

Substituting for m j , h j , and φ jn in Eq. (13.2.9b) gives Ln : 3

L1 = ∑ h j m jφ j1 = 3186 . mh j =1

j =1

L3 = ∑ h j m j φ j 3 = 0.317mh j =1

The effective modal masses and effective modal heights are given by Eq. (13.2.9a):

* M1 =

n =1

∑ hn* M n = mh ( 2.104 × 2.124 + 0.25 × 0 + 0.146 × 0.211) = 4.5mh

n =1 3

Verify Eq. (13.2.17): 3

L2 = ∑ h j m jφ j 2 =0

∴ ∑ Mn* = ∑ m j

M 2* =

Lθ3 0.317mh = = 0.211h 1503 . m Lh3 m/2

L O L O L O φ = MM 0.686PP ; φ = MM −1/ 2 PP ; φ = MM −2.186PP MN 1 PQ MN 1 PQ MN 1 PQ −1/ 2

0.314

. mh Lθ1 3186 = 2.124h h = 15 .m L1

∑ h j m j = mh(1 + 2 + 3 × 0.5) = 4.5mh

j =1

n =1

∴ ∑ hn* Mn* = ∑ h j m j

e L j = b15. mg2 = 2.104m h 2 1

1.069m

e j = b−0.5mg2 =0.25m

* M3 =

2 Lh2

e L j = b1503 . mg2 = 0.146m h 2 3

15.46m

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 22 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Γ2 = − 0. 5083

Problem 13.11

Γ3 = 0.1569

Part a m /2 u8 h h

From Eq. (13.2.5) the floor displacements due to the first mode are

u9 m

u1 (t ) =

m u4

u01 ( t ) = T u1 ( t )

where T was determined in solving Problem 10.19. Thus:

Mass and lateral stiffness matrices (from Problem 9.9):

LM1 OP 1 MM PP 0 . 5 N Q . O L 40.85 − 23.26 511 $k = EI M− 23.26 − 14.25PP 3109 . M h MN 511 − 14.25 . 10.06PQ m = m

− 0.5342 . 0.0774O LM− 01084 − 0.5342 . 0.0774PP MM− 01084 R0.4625U| 0.5961 − 0.0619 − 0.4258P | 1 . u (t ) = P S10070 M V| D (t ) h M 0.5961 − 0.0619 − 0.4258P | . 13515 W MM − 01703 . 0.8748 − 0.7355P T P . 0.8748 − 0.7355PQ MN − 01703 01

R|− 0.4796U| |− 0.4796|| 1 |− 0.3836 = S V D (t ) h |− 0.3836| || − 01856 || . |T − 01856 |W .

Natural frequencies and modes (from Problem 10.19):

ω1 = 1. 4576 ω 3 = 8.1980

EI mh3

ω 2 = 4. 7682

(b)

EI mh

From Eq. (13.2.3), for the first mode:

FG mIJ = 15607 H 2K . m F mI M = (0.3156) m + (0.7451) m + (1) G J H 2K

L1h = 0.3156m + 0.7451m + 1 2

mh3

. R|0.3156U| R|− 0.7409U| R| 12546 U| φ = − φ = − 0 . 7451 0 . 3572 12024 . S| V| S| V| S| V| 1 1 1 T W T W T W

Similarly, the floor displacements due to the second and third modes are

R|u (t ) U| R|− 0.7409U| u (t ) = Su (t ) V = − 0.5083 S− 0.3572V D (t ) |Tu (t ) |W |T 1 |W R| 0.3766U| = S 01816V D (t ) (c) |T− 0.5083|W 1

. R|u (t ) U| R| 12546 U| u (t ) = Su (t ) V = 01569 . . S|− 12024 V| D (t ) |Tu (t ) |W T 1 W . U| R| 01968 . = S− 01887 V| D (t ) |T 01569 . W 1

Computed similarly, these quantities for the second and third modes are Lh3 = 0. 5522 m M3 = 3. 5198 m

= 1. 3515

Lh2 = − 0. 5981m M2 = 1.1765 m

= 11548 . m

The joint rotations associated with u1 are

Fig. P13.11a

Γ1 =

(a)

L1h

φ1 =

R|u (t ) U| R|0.3156U| R|0.4265U| u ( t ) = 13515 . 0 . 7451 D ( t ) = . S| V| S| V| S|10070 V| D (t ) . Tu ( t ) W T 1 W T13515 W 1

(d)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 23 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Substituting these ua , ub , θa and θb , and Eqs. (g)-(h) in Eqs. (i) and (j) gives

The joint rotations associated with u2 and u3 can be computed following Eq. (b): . U| R|− 01772 − . 01772 | | 1 | 0.4297| u (t ) = V D (t ) S h | 0.4297| || 0.4686|| |T 0.4686|W R| 0.0916U| || 0.0916|| 0.0622 1 u (t ) = V D (t ) S h | 0.0622| ||− 0.3140|| |T− 0.3140|W 02

Ma =

Mb =

(e)

EI h2 EI h2

Ma = mh 0. 7526 A1 ( t ) + 0. 08381 A2 ( t ) + 0. 02030 A3 ( t )

Mb = mh 0. 3014 A1 ( t ) + 0. 06823 A2 ( t ) + 0. 02302 A3 ( t )

For the second floor beam, L = 2 h and the nodal displacements are shown in Fig. P13.11c.

(f)

ua = 0

. u1 (t ) = 0.4265 D1 (t ) + 0.3766 D2 (t ) + 01968 D3 (t ) . . . u2 (t ) = 10070 D1 (t ) + 01816 D2 (t ) − 01887 D3 (t ) . . u3 (t ) = 13515 D1 (t ) − 0.5083 D2 (t ) + 01569 D3 (t )

(g) Combining modal contributions to joint rotations gives

u0 ( t ) = u01 ( t ) + u0 2 ( t ) + u03 ( t )

(h)

Part b

ub = 0

a θa = u6

θb = u7

U| V| W

Fig. P13.11c Substituting these ua , ub , θa and θb , and Eqs. (g)-(h) in Eqs. (i) and (j) gives Ma = Mb

The bending moments at the ends of a flexural element are related to the nodal displacements by

Mb =

0. 6408 D1 ( t ) + 1. 5511 D2 ( t ) + 1. 5475 D3 ( t )

Substituting Dn ( t ) = An ( t ) ω 2n and ω n in terms of E, I, m, and h gives

Combining modal responses gives the total floor displacements:

Ma =

1. 5999 D1 ( t ) + 1. 9054 D2 ( t ) + 1. 3643 D3 ( t )

4 EI L 2 EI L

θa + θa +

2 EI L 4 EI L

θb + θb +

6 EI 2

6 EI L2

ua − ua −

6 EI 2

6 EI L2

EI h2

− 1.1508 D1 ( t ) + 1. 2892 D2 ( t ) + 0.1867 D3 ( t )

(i)

Substituting Dn ( t ) = An ( t ) ω 2n and ω n in terms of E, I, m, and h gives

(j)

Ma = Mb

For a first story column, L = h and the nodal displacements are shown in Fig. P13.11b.

= mh − 0. 5414 A1 ( t ) + 0. 0567 A2 ( t ) + 0. 00278 A3 ( t )

θb = u4

ub = – u1

b EI

ua = 0

a θa = 0

Fig. P13.11b

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 24 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.12

. m M 2 = 1193

M 3 = 4.567m Lh3 =0.714m

Lh2 = −0.647m Lh Γ2 = 2 = −0.542 M2

u3 h

u2 u1 h

⎧0.377 ⎫ ⎧0.273⎫ ⎪ ⎪ ⎪ ⎪ u1 (t ) = 1.386 ⎨0.698⎬ D1 (t ) = ⎨0.966⎬ D1 (t ) ⎪1.386 ⎪ ⎪ 1 ⎪ ⎭ ⎩ ⎭ ⎩

u 01 ( t ) = Tu1 ( t )

Mass and lateral stiffness matrices (from Problem 9.10) ⎡1 ⎤ ⎢ ⎥ m = m⎢ 1 ⎥ ⎢⎣ 0.5⎥⎦ ⎡39.38 − 22.68 5.486 ⎤ ˆk = EI ⎢ 27.13 − 11.75⎥⎥ tt 3 ⎢ h ⎢ 7.418 ⎥⎦ ⎣

Natural frequencies and modes (from Problem 10.20): EI mh 3

ω 2 = 4.178

EI mh

ω 3 = 7.903

EI mh 3

⎡0.273⎤ ⎡− 0.706⎤ ⎡ 1.529 ⎤ ⎢ ⎢ ⎥ ⎥ φ1 = ⎢0.698⎥ φ 2 = ⎢ − 0.441⎥ φ3 = ⎢⎢− 1.315⎥⎥ ⎢⎣ 1 ⎥⎦ ⎢⎣ 1 ⎥⎦ ⎢⎣ 1 ⎥⎦

From Eq. (13.2.3), the modal properties are: First Mode: M 1 = (0.273) 2 m + (0.698) 2 m + (1) 2 (0.5m ) = 1.06m L1h = 0.273m + 0.698m + 1(0.5m ) = 1.47 m Γ1 =

(a)

The joint rotations associated with u1 are Fig. P13.12(a)

L1h

Lh3 = 0.156 M3

Part a From Eq. (13.2.5) the floor displacements due to the first mode are:

ω 1 = 1197 .

Γ3 =

= 1.386

Computed similarly, these quantities for the second and third modes are

where T was determined in solving Problem 10.20. Thus: ⎡− 0.1512 − 0.6084 ⎢− 0.1512 − 0.6084 ⎢ − 0.11 1 ⎢ 0.7184 u01(t ) = ⎢ − 0.11 h ⎢ 0.7184 ⎢− 0.2612 1.131 ⎢ 1.131 ⎣⎢− 0.2612

0.0962 ⎤ 0.0962 ⎥⎥ ⎧0.377 ⎫ − 0.457 ⎥ ⎪ ⎪ ⎥ ⎨0.966⎬ D1(t ) − 0.457 ⎥ ⎪ 1.386 ⎪⎭ − 0.925⎥ ⎩ ⎥ − 0.925⎦⎥

⎡ − 0.511⎤ ⎢ − 0.511⎥ ⎢ ⎥ 1 ⎢− 0.469⎥ = ⎢ ⎥ D1 (t ) h ⎢− 0.469⎥ ⎢ − 0.288⎥ ⎢ ⎥ ⎣⎢ − 0.288⎦⎥

(b)

Similarly, the floor displacements due to the second and third modes are ⎧− 0.706⎫ ⎧ 0.383 ⎫ ⎪ ⎪ ⎪ ⎪ u 2 (t ) = −0.542⎨ − 0.441⎬ D2 (t ) = ⎨ 0.239 ⎬ D2 (t ) ⎪ 1 ⎪ ⎪− 0.542⎪ ⎩ ⎭ ⎩ ⎭

(c)

⎧ 1.529 ⎫ ⎧ 0.238 ⎫ ⎪ ⎪ ⎪ ⎪ u 3 (t ) = 0.156 ⎨− 1.315⎬ D3 (t ) = ⎨− 0.205⎬ D3 (t ) ⎪ 1 ⎪ ⎪ 0.156 ⎪ ⎩ ⎭ ⎩ ⎭

(d)

The joint rotations associated with u 2 and u 3 can be computed following Eq. (b):

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 25 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

⎡− 0.256⎤ ⎢− 0.256⎥ ⎢ ⎥ 1 ⎢ 0.497 ⎥ u02 (t ) = ⎢ ⎥ D2 (t ) h ⎢ 0.497 ⎥ ⎢ 0.671 ⎥ ⎢ ⎥ ⎢⎣ 0.671 ⎥⎦

⎡ 0.104 ⎤ ⎢ 0.104 ⎥ ⎥ ⎢ 1 ⎢ 0.122 ⎥ u03 (t ) = ⎢ ⎥ D3 (t ) h ⎢ 0.122 ⎥ ⎢− 0.438⎥ ⎥ ⎢ ⎣⎢− 0.438⎦⎥

ub = −u1

(e)

ua =0

θa =0

Fig. P13.12(b) (f) Substituting ua , ub , θ a , and θ b , and Eqs. (g)-(h) in Eqs. (i) and (j) gives:

Combining modal responses gives the total floor displacements: u1 ( t ) = 0.377 D1 ( t ) + 0.383D2 ( t ) + 0.238 D3 ( t ) u2 ( t ) = 0.966 D1 ( t ) + 0.239 D2 ( t ) − 0.205D3 ( t )

θb =u4

(g)

u3 ( t ) = 1.386 D1 ( t ) − 0.542 D2 ( t ) + 0.156 D3 ( t )

EI 0.998 D1 ( t ) +1.787 D2 ( t ) +1.624 D3 ( t ) h2 EI . D3 ( t ) M b = 2 0.216 D1 ( t ) +1.274 D2 ( t ) +1811 h Ma =

Substituting Dn ( t ) = An ( t )/ω 2n and ω n in terms of E, I, m, h gives

Combining modal contributions to joint rotations gives

M a = mh 0.697 A1 ( t ) + 0.102 A2 ( t ) + 0.026 A3 ( t )

u 0 ( t ) =u 01 ( t ) +u 02 ( t ) + u03 ( t )

M b = mh 0.151 A1 ( t ) + 0.073 A2 ( t ) + 0.029 A3 ( t )

(h)

u4 ( t ) = u5 ( t ) = −0.511D1 ( t ) − 0.256 D2 ( t ) + 0.104 D3 ( t ) u6 ( t ) = u7 ( t ) = −0.469 D1 ( t ) + 0.239 D2 ( t ) − 0.205D3 ( t )

For the second floor beam, L =2h and the nodal displacements are shown in Fig. P13.12(c):

u8 ( t ) = u9 ( t ) = −0.288 D1 ( t ) + 0.671D2 ( t ) − 0.438 D3 ( t )

ua = 0 EI/2

Part b a

The bending moments at the ends of a flexural element are related to the nodal displacements by: Ma =

4 EI 2 EI 6 EI 6 EI θa + θb + ua − ub 2 L L L L2

(i)

2 EI 4 EI 6 EI 6 EI θ + θ + ua − ub 2 L a L b L L2

(j)

Mb =

For a first story column, L =h and the nodal displacements are shown in Fig. P13.12(b):

θa = u6

ub = 0 b

θ b = u7

Fig. P13.12(c) Substituting ua , ub , θ a , and θ b , and Eqs. (g)-(h) in Eqs. (i) and (j) gives: M a = Mb =

EI −1.407 D1 ( t ) + 0.717 D2 ( t ) − 0.615D3 ( t ) h2

Substituting Dn ( t ) = An ( t )/ω 2n and ω n in terms of E, I, m, h gives M a = M b = mh[ − 1.407 D1(t ) + 0.717 D2 (t ) − 0.615.026 D3 (t )]

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 26 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.13

M 3 = 15.678m

h L2 = −0.603m

Lh3 =1.792m

Γ2 =

Lh2 = −0.543 M2

Γ3 =

Lh3 = 0.114 M3

Part a h

From Eq. (13.2.5) the floor displacements due to the first mode are:

M 2 = 111 . m

⎧0.234⎫ ⎧0.333⎫ ⎪ ⎪ ⎪ ⎪ u1 (t ) = 1.426 ⎨0.639⎬ D1 (t ) = ⎨0.911⎬ D1 (t ) ⎪ 1 ⎪ ⎪1.426 ⎪ ⎩ ⎭ ⎩ ⎭

(a)

The joint rotations associated with u1 are

u 01 ( t ) = Tu1 ( t )

Fig. P13.13(a)

where T was determined in solving Problem 10.21. Thus: Mass and lateral stiffness matrices (from Problem 9.11) ⎡1 ⎤ ⎢ ⎥ m =m ⎢ 1 ⎥ ⎢⎣ 0.5⎥⎦ ⎡33.36 − 14.91 1.942 ⎤ ˆk = EI ⎢ 15.96 − 5.489⎥⎥ tt 3 ⎢ h ⎢ 3.923 ⎥⎦ ⎣

Natural frequencies and modes (from Problem 10.21):

ω 1 = 1.329

EI mh3

ω 2 = 3.514

EI mh3

ω 3 = 6.562

EI mh3

⎡0.234⎤ ⎡− 0.512⎤ ⎡ 3.324 ⎤ ⎢ ⎢ ⎥ ⎥ φ1 = ⎢0.639⎥ φ2 = ⎢ − 0.591⎥ φ3 = ⎢⎢− 2.032⎥⎥ ⎢⎣ 1 ⎥⎦ ⎢⎣ 1 ⎥⎦ ⎢⎣ 1 ⎥⎦

From Eq. (13.2.3), the modal properties are:

⎡− 0.3006 − 0.3695 0.0313 ⎤ ⎢− 0.3006 − 0.3695 0.0313 ⎥ ⎢ ⎥ ⎧0.333⎫ − 0.321 − 0.227 ⎥ ⎪ 1 ⎢ 0.6795 ⎪ u 01 (t ) = ⎢ ⎥ ⎨0.911⎬ D1 (t ) − 0.321 − 0.227 ⎥ ⎪ h ⎢ 0.6795 1.426 ⎪⎭ ⎢− 0.1942 0.9489 − 0.7923⎥ ⎩ ⎢ ⎥ ⎣⎢− 0.1942 0.9489 − 0.7923⎦⎥ ⎡− 0.392⎤ ⎢− ⎥ ⎢ 0.392⎥ 1 ⎢− 0.390⎥ = ⎢ ⎥ D1 (t ) h ⎢− 0.390⎥ ⎢ − 0.333⎥ ⎢ ⎥ ⎢⎣ − 0.333⎥⎦

(b)

Similarly, the floor displacements due to the second and third modes are ⎧ 0.278 ⎫ ⎧− 0.512⎫ ⎪ ⎪ ⎪ ⎪ u 2 (t ) = − 0.543 ⎨ − 0.591⎬ D2 (t ) = ⎨ 0.321 ⎬ D2 (t ) ⎪− 0.543⎪ ⎪ 1 ⎪ ⎩ ⎭ ⎩ ⎭

(c)

Lh Γ1 = 1 = 1.426 M1

⎧ 3.324 ⎫ ⎧ 0.379 ⎫ ⎪ ⎪ ⎪ ⎪ u 3 (t ) = 0.114 ⎨− 2.032⎬ D3 (t ) = ⎨− 0.232⎬ D3 (t ) ⎪ 1 ⎪ ⎪ 0.114 ⎪ ⎩ ⎭ ⎩ ⎭

(d)

Computed similarly, these quantities for the second and third modes are

The joint rotations associated with u 2 and u 3 can be computed following Eq. (b):

First mode: M 1 = (0.234) m + (0.639) m + (1) (0.5m ) = 0.963m 2

L1h = 0.234m + 0.639m + 1(0.5m ) = 1.373m

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 27 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

⎡− 0.219⎤ ⎢− 0.219⎥ ⎢ ⎥ 1 ⎢ 0.209 ⎥ u02 (t ) = ⎢ ⎥ D2 (t ) h ⎢ 0.209 ⎥ ⎢ 0.681 ⎥ ⎢ ⎥ ⎢⎣ 0.681 ⎥⎦ ⎡ − 0.025⎤ ⎢ − 0.025⎥ ⎢ ⎥ 1 ⎢ 0.306 ⎥ u03 (t ) = ⎢ ⎥ D3 (t ) h ⎢ 0.306 ⎥ ⎢− 0.384⎥ ⎢ ⎥ ⎣⎢− 0.384⎦⎥

ub = −u1

(e)

ua =0

θa =0

Fig.P13.13(b) (f) Substituting ua , ub , θ a , and θ b , and Eqs. (g)-(h) in Eqs. (i) and (j) gives:

Combining modal responses gives the total floor displacements: u1 ( t ) = 0.333D1 ( t ) + 0.278 D2 ( t ) + 0.379 D3 ( t ) u2 ( t ) = 0.911D1 ( t ) + 0.321D2 ( t ) − 0.232 D3 ( t )

θb =u4

(g)

u3 ( t ) = 1.426 D1 ( t ) − 0.543D2 ( t ) + 0.114 D3 ( t )

EI 2.75D1 ( t ) +1.235D2 ( t ) +1.765D3 ( t ) h2 EI M b = 2 0.43D1 ( t ) + 0.79 D2 ( t ) + 2.153D3 ( t ) h Ma =

Substituting Dn ( t ) = An ( t )/ω n2 and ω n in terms of E, I, m, h gives

Combining modal contributions to joint rotations gives

M a = mh 1555 . A1 ( t ) + 0.1 A2 ( t ) +0.041 A3 ( t )

u 0 ( t ) =u 01 ( t ) +u 02 ( t ) + u03 ( t )

M b = mh 0.243 A1 ( t ) + 0.064 A2 ( t ) +0.05 A3 ( t )

(h)

u4 ( t ) = u5 ( t ) = −0.392 D1 ( t ) − 0.219 D2 ( t ) − 0.025D3 ( t ) u6 ( t ) = u7 ( t ) = −0.39 D1 ( t ) + 0.209 D2 ( t ) + 0.306 D3 ( t )

For the second floor beam, L =2h and the nodal displacements are shown in Fig. P13.13(c):

u8 ( t ) = u9 ( t ) = −0.33D1 ( t ) + 0.681D2 ( t ) − 0.384 D3 ( t )

ua = 0 2EI/3

Part b

The bending moments at the ends of a flexural element are related to the nodal displacements by: Ma =

4 EI 2 EI 6 EI 6 EI θa + θb + ua − ub 2 L L L L2

2 EI 4 EI 6 EI 6 EI Mb = θ + θ + ua − ub 2 L a L b L L2

θa = u6

ub = 0 b

θ b = u7

(i) Fig. P13.13(c) (j)

For a first story column, L = h and the nodal displacements are shown in Fig. P13.13(b):

Substituting ua , ub , θ a , and θ b , and Eqs. (g)-(h) in Eqs. (i) and (j) gives: M a = Mb =

EI −117 . D1 ( t ) + 0.628 D2 ( t ) + 0.918 D3 ( t ) h2

Substituting Dn ( t ) = An ( t )/ω n2 and ω n in terms of E, I, m, h gives M a = M b = mh −0.66 A1 ( t ) + 0.05 A2 ( t ) +0.021 A3 ( t )

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 28 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

From Eq. (13.2.5) the floor displacements due to the first mode are:

Problem 13.14

u2 u1

⎧0.289⎫ ⎧0.200⎫ ⎪ ⎪ ⎪ ⎪ u1 (t ) = 1.447 ⎨0.597 ⎬ D1 (t ) = ⎨0.863⎬ D1 (t ) ⎪1.447 ⎪ ⎪ 1 ⎪ ⎭ ⎩ ⎭ ⎩

h h

The joint rotations associated with u1 are u 01 (t ) = Tu1 (t )

⎡ −0.4005 −0.4152 ⎢ ⎢− 0.4005 − 0.4152 1 ⎢ 0.9530 − 0.4569 u 01 (t ) = ⎢ h ⎢ 0.9530 − 0.4569 ⎢− 0.3465 1.2570 ⎢ ⎣⎢− 0.3465 1.2570

Fig. P13.14(a) Mass and lateral stiffness matrices (from Problem 9.12) ⎡1 ⎤ ⎢ ⎥ m=m ⎢ 1 ⎥ ⎢⎣ 0.5⎥⎦ 2.43 ⎤ ⎡ 30.77 − 14.01 ⎥ ˆk = EI ⎢ 13.82 − 4.80⎥ tt ⎢ h3 ⎢ 2.92 ⎥⎦ ⎣ Symm

Natural frequencies and modes (from Problem 10.22):

⎡0.200⎤ φ1 = ⎢⎢0.597 ⎥⎥ ⎢⎣ 1 ⎥⎦

EI mh

ω 2 = 3.081

EI mh

⎡ − 0.545⎤ φ 2 = ⎢⎢− 0.656⎥⎥ ⎢⎣ 1 ⎥⎦

ω 3 = 6.314

EI mh

(b)

where T was determined in solving Problem 10.22. Thus:

ω1 = 1.043

(a)

⎡ 3.220 ⎤ φ3 = ⎢⎢− 1.916⎥⎥ ⎢⎣ 1 ⎥⎦

0.0458 ⎤ ⎥ 0.0458 ⎥ ⎧0.289⎫ − 0.2803⎥⎪ ⎪ ⎥⎨0.863⎬ D1 (t ) − 0.2803⎥⎪ 1.447 ⎪⎭ − 0.9890⎥⎩ ⎥ − 0.9890⎦⎥

⎡ −0.408 ⎤ ⎥ ⎢ ⎢ − 0.408⎥ 1 ⎢ − 0.525⎥ = ⎢ ⎥ D1 (t ) h ⎢ − 0.525⎥ ⎢− 0.446⎥ ⎥ ⎢ ⎣⎢− 0.446⎦⎥

Similarly, the floor displacements due to the second and third modes are ⎧− 0.545⎫ ⎧ 0.312 ⎫ ⎪ ⎪ ⎪ ⎪ u 2 (t ) = −0.571⎨− 0.656⎬ D2 (t ) = ⎨ 0.374 ⎬ D2 (t ) ⎪ 1 ⎪ ⎪− 0.571⎪ ⎩ ⎭ ⎩ ⎭

(c)

⎧ 3.220 ⎫ ⎧ 0.399 ⎫ ⎪ ⎪ ⎪ ⎪ u 3 (t ) = 0.124⎨− 1.916⎬ D3 (t ) = ⎨− 0.238⎬D3 (t ) ⎪ 1 ⎪ ⎪ 0.124 ⎪ ⎭ ⎩ ⎭ ⎩

(d)

From Eq. (13.2.3), for the first mode: M 1 = (0.200) 2 m + (0.597) 2 m + (1) 2 (0.5m ) = 0.896m L1h = 0.200m + 0.597 m + 1(0.5m ) = 1.296m

Γ1 =

L1h = 1.447 M1

Computed similarly, these quantities for the second and third modes are M 2 = 1.227 m

M 3 = 14.541m

Lh2 = −0.701m

Lh3 = 1.804m

Γ2 = −0.571

Γ3 = 0.124

The joint rotations associated with u 2 and u 3 can be computed following Eq. (b): ⎡ −0.306 ⎤ ⎡ −0.056 ⎤ ⎢ ⎢ ⎥ ⎥ − 0 . 306 ⎢ ⎢− 0.056⎥ ⎥ 1 ⎢ 0.286 ⎥ 1 ⎢ 0.455 ⎥ u 02 (t ) = ⎢ ⎥ D2 (t ) u 03 (t ) = ⎢ ⎥ D3 (t ) h ⎢ 0.286 ⎥ h ⎢ 0.455 ⎥ ⎢ 0.928 ⎥ ⎢− 0.560⎥ ⎢ ⎢ ⎥ ⎥ ⎣⎢ 0.928 ⎦⎥ ⎣⎢− 0.560⎦⎥

(e)

Part a: © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 29 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Combining modal responses gives the total floor displacements: u1 (t ) = 0.289 D1 (t ) + 0.312 D2 (t ) + 0.399 D3 (t ) u 2 (t ) = 0.863D1 (t ) + 0.374 D2 (t ) − 0.238 D3 (t )

(f)

u3 (t ) = 1.447 D1 (t ) − 0.571D2 (t ) + 0.124 D3 (t )

M a = mh [ 0.844 A1 (t ) + 0.132 A2 (t ) + 0.057 A3 (t ) ] M b = mh [ 0.094 A1 (t ) + 0.068 A2 (t ) + 0.055 A3 (t ) ]

For the second floor beam, L = 2 h and the nodal displacements are shown in Fig. P13.14(c):

Combining modal contributions to joint rotations gives

1 u 4 (t ) = u5 (t ) = [− 0.408 D1 (t ) − 0.306 D2 (t ) − 0.056 D3 (t )] h 1 u6 (t ) = u 7 (t ) = [− 0.525D1 (t ) + 0.286 D2 (t ) + 0.455 D3 (t )] (g) h 1 u8 (t ) = u9 (t ) = [− 0.446 D1 (t ) + 0.928 D2 (t ) − 0.560 D3 (t )] h

Part b:

The bending moments at the ends of a flexural element are related to the nodal displacements by: Ma =

ub = 0

ua = 0

u 0 ( t ) =u 01 ( t ) +u 02 ( t ) + u03 ( t )

4EI/3

θa = u6

θb = u7

Fig. P13.14(c) Substituting ua , ub , θ a , and θ b , and Eqs. (f)-(g) in Eqs. (h) and (i) gives: Ma = Mb =

EI h2

[− 0.525D1 (t ) + 0.286D2 (t ) + 0.455D3 (t )]

4 EI 2 EI 6 EI 6 EI θa + θb + ua − ub 2 L L L L2

(h)

2 EI 4 EI 6 EI 6 EI θa + θb + ua − ub 2 L L L L2

Substituting Dn ( t ) = An ( t )/ω n2 and ω n in terms of E, I, m, and h gives

(i)

M a = M b = mh[− 0.482 A1 (t ) + 0.030 A2 (t ) + 0.011A3 (t )]

Mb =

For a first story column, L =h and the nodal displacements are shown in Fig. P13.14(b):

u b = − u1

θb =u4 b EI

ua =0

θa =0

Fig. P13.14(b) Substituting ua , ub , θ a , and θ b , and Eqs. (f)-(g) in Eqs. (h) and (i) gives: Ma = Mb =

EI h2 EI h2

[0.918D1 (t ) + 1.256 D2 (t ) + 2.286D3 (t )] [0.102 D1 (t ) + 0.644 D2 (t ) + 2.174 D3 (t )]

Substituting Dn ( t ) = An ( t )/ω n2 and ω n in terms of E, I, m, and h gives

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 30 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.15 2m

Substituting Γn , m and φn in Eq. (13.1.6) gives

EI L

s1 = Γ1 m φ 1 = 0.2834

LM3m OP RS 1 UV = RS0.849mUV N mQ T2.097W T0.594mW

s 2 = Γ2 m φ 2 = − 0.2834

LM3m OP RS 1 UV = RS− 0.849mUV . W N mQ T− 1431 T 0.406mW

EI L

Part a

From Example 9.6, the mass and stiffness matrices

The modal expansion of effective forces is shown in the following figure.

are:

6 EI ⎡ 8 − 3⎤ k= ⎢ 2⎥⎦ 7 L2 ⎣− 3

⎡3 ⎤ m= m⎢ ⎥ ⎣ 1⎦

0 L

ω 1 = 1874 .

⎧ 1 ⎫ ⎬ ⎩2.097 ⎭

⎧

EI m L3

0.849 m

φ2 =⎨

⎧0⎫ ⎩1⎭

= 2.097m

= 0. 2834

L3m OP RS0UV u&& (t ) p (t ) = − m ι u&& (t ) = − M N mQ T1W R0U = − S V u&& (t ) TmW g

Combining the modal displacements gives the total displacements:

= − 0. 2834

The effective earthquake forces are given by Eq. (13.1.2): eff

M2 = φ2T m φ2 = 5. 048 m M2

RSu (t ) UV = Γ φ D (t ) Tu (t )W R 1 UV D (t ) = RS0.283UV D (t ) = 0.2834 S T2.097W T0.594W Ru (t ) UV = Γ φ D (t ) u (t ) = S Tu (t )W R 1 UV D (t ) = RS− 0.283UV D (t ) = − 0.2834 S . W T− 1431 T 0.406W u1 (t ) =

L2 = φ 2T m ι = − 1431 . m

M b 2= – 0.443 mL

M1 = φ1T m φ1 = 7. 397 m

Γ2 =

M b 1= 1.443 mL

The modal displacements from Eq. (13.1.10) are

φ 1T m ι

0.406 m

Part b

Substituting for m , ι and φn in Eq. (13.1.5) gives the modal quantities:

0.594 m

ι =⎨ ⎬

Γ1 =

0.849 m

The equations of motion are given by Eqs. (13.1.1) and (13.1.2) where, for vertical ground motion taken positive downward, the influence vector is:

L1 =

⎫ ⎬ ⎩− 1.431⎭

φ1 = ⎨

The natural frequencies and modes of the system (from Example 10.3) are:

ω 1 = 0.6987

u1 ( t ) = 0. 283 D1 ( t ) − 0. 283 D2 ( t ) u2 ( t ) = 0. 594 D1 ( t ) + 0. 406 D2 ( t )

Part c st Using Mbn shown in the figure, the modal responses for Mb are

Mb1 ( t ) = Mbst1 A1 ( t ) = 1. 443 m L A1 ( t )

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 31 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Mb 2 ( t ) = Mbst2 A2 ( t ) = − 0. 443 m L A2 ( t )

The total bending moment is Mb ( t ) = Mb1 ( t ) + Mb 2 ( t ) = 1. 443 m L A1 ( t ) − 0. 443 m L A2 ( t )

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 32 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.16 2m

ug(t)

⎡3m ⎤ ⎪⎧ 1 p eff (t ) = −m ι u&&g (t ) = − ⎢ ⎨ m⎥⎦ ⎪⎩− 1 ⎣ − m ⎧ 3⎫ = ⎨ ⎬ u&&g (t ) 2 ⎩− 1⎭

2 ⎫⎪ ⎬ u&&g (t ) 2 ⎪⎭

Substituting Γn , m and φ n in Eq. (13.1.6) gives: ⎧0.259m⎫ s 1 = Γ1mφ1 = ⎨ ⎬ ⎩0.181m ⎭ ⎧ 1.862m⎫ s 2 = Γ2 mφ 2 = ⎨ ⎬ ⎩− 0.888m⎭

Part a From Example 9.6, the mass and stiffness matrices are:

⎡3 ⎤ m = m⎢ ⎥ ⎢⎣ 1⎥⎦

k =

6 EI ⎡ 8 − 3⎤ ⎢ ⎥ 7 L3 ⎢⎣− 3 2⎥⎦

The modal expansion of the spatial distribution m ι of the effective forces is shown in the following figure. EI

2m The natural frequencies and modes of the system (from Example 10.3) are:

ω 1 = 0.6987

mL3

⎧ 1 ⎫ ⎬ ⎩2.097 ⎭

φ1 = ⎨

ω 2 = 1.874 ⎧

mL3

The equations of motion are given by Eqs. (13.1.1) and (13.1.2) where, for the ground motion assumed to be positive in the direction shown above, the influence vector is:

ι=

⎫ ⎬ 1 . 431 − ⎩ ⎭

φ2 = ⎨

3 2

L EI

|RS 1 / 2 |UV |T−1 / 2 |W

Substituting for m , ι and φ n in Eq. (13.1.5) gives the modal quantities: L1 = φ1T mι = 0.639m

L 2 = φ 2T mι = 3.133m

M 1 = φ1T mφ1 = 7.397 m

M 2 = φ 2T mφ 2 = 5.048m

L Γ1 = 1 = 0.0863 M1

L Γ2 = 2 = 0.6206 M2

The effective earthquake forces are given by Eq. (13.1.2):

L 0.259 m

1.862 m

0.181 m

0.888 m

M bst1 = 0.440 mL

M bst2 = 0.974 mL

Part b The modal displacements from Eq. (13.1.10) are: ⎧0.086⎫ u 1 (t ) = Γ1φ1 D1 (t ) = ⎨ ⎬ D1 (t ) ⎩0.181⎭ ⎧ 0.621⎫ u 2 (t ) = Γ2φ 2 D 2 (t ) = ⎨ ⎬ D 2 (t ) ⎩− 0.888⎭

Combining the modal displacements gives the total displacements: u1 (t ) = 0.086 D1 (t ) + 0.621D2 (t ) u 2 (t ) = 0.181D1 (t ) − 0.888 D2 (t )

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 33 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Part c st Using M bn shown in the figure above, the modal responses for M b are:

M b1 (t ) = M bst1 A1 (t ) = 0.440 m L A1 (t ) M b 2 (t ) = M bst2 A2 (t ) = 0.974 m L A2 (t )

Thus, the total bending moment is: M b (t ) = M b1 (t ) + M b 2 (t ) = 0.440 m L A1 (t ) + 0.974 m L A2 (t )

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 34 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.17

Similar calculations for the second and third modes gives:

u2 u1

L2 = φ 2T m ι = 5m

L3 = φ T3 m ι = 0

M 2 = φ 2T m φ 2 = 8.292m

M3 = φ3T m φ3 = 2 m

Γ2 =

L2 M2

= 0. 603

Γ3 =

L3 M3

= 0

The effective earthquake forces are given by Eq. (13.1.4): Fig. P13.17a

OP R|1U| LM5m m p (t ) = − m ι u&& (t ) = − M PP S|0V| u&& (t ) MN mQ T0W R|5mU| = − S 0 V u&& (t ) |T 0 |W eff

Part a The equations of motion are given by Eqs. (13.1.1) and (13.1.2). The mass and stiffness matrices (from Problem 9.13) are

LM5 OP m =m M 1 P MN 1PQ

LM 28 6 − 6OP k = 6 7 3 10 L M MN− 6 3 7PPQ 3EI

The influence vector due to horizontal ground motion is (from Problem 9.13)

R|1U| S| V| T0W

ι = 0

The natural frequencies and modes of the system (from Problem 10.23) are

ω1 = 0. 526 ω 3 = 1. 732

ω 2 = 1. 614

m L3

EI m L3

EI mL3

R| 1 U| R| 1 U| R|0U| φ = S− 1949 . V φ = S 1283 . V φ = S 1V |T 1949 |T− 1283 |T1|W . |W . |W 1

Substituting for m , ι and φn in Eq. (13.1.5) gives the first-mode quantities: L1 =

φ 1T m ι

= 5m

LM5 OP R|1U| = 1 − 1949 . 1949 . m M 1 P S0V MN 1PQ |T0|W

M1 = φ 1T m φ 1 = 1 − 1949 . 1949 . m = 12.597m

Γ1 =

L1 M1

Substituting Γn , m and φn in Eq. (13.1.6) gives

LM5m OP R| 1 U| s = Γ m φ = 0.397 M m . V P S− 1949 MN mPQ |T 1949 . |W . mU R| 1985 | = S− 0.774mV |T 0.774m|W LM5m OP R| 1 U| s = Γ m φ = 0.603 M m . V S| 1283 P MN mPQ T− 1283 . |W R| 3.015mU| = S 0.774mV |T− 0.774m|W LM5m OP R|0U| R|0U| s = Γ mφ = 0M m P S1V = S|0V| MN mPQ |T 1|W T0W 1

The modal expansion of the spatial distribution of effective forces is shown in Fig. P13.17b. The effective forces in the third mode are all zero, implying that this mode will not be excited by horizontal ground motion. Part b

The modal displacements from Eq. (13.1.10) are

LM5 OP R| 1 U| . V MM 1 PP S|− 1949 . |W N 1Q T 1949

= 0. 397

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 35 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

R|u (t ) U| R| 1 U| u ( t ) = Su ( t ) V = Γ φ D (t ) = 0.397 S− 1949 . V D (t ) |Tu (t )|W |T 1949 . |W R| 0.397U| = S− 0.774 V D ( t ) |T 0.774|W

0.774 m

M a 1= – 0.774 mL 1.985 m

0.774 m

M b 1= 3.533 mL

R|u (t ) U| R| 1 U| u (t ) = Su (t ) V = Γ φ D (t ) = 0.603 S 1283 . V D (t ) |Tu (t ) |W |T− 1283 . |W R| 0.603U| = S 0.774 V D (t ) |T− 0.774|W 1

Ma2= 0.774 mL

3.015m

0.774m

R|u (t ) U| R|0U| u (t ) = Su (t ) V = Γ φ D (t ) = 0 S 1V D (t ) |Tu (t ) |W |T1|W R|0U| = S0V D (t ) |T0|W

0.774m

Mb2= 1.467mL

M a3= 0

Combining the modal displacements gives the total displacements:

bg bg u bt g = −0.774 D (t ) + 0.774 D bt g u bt g = 0.774 D (t ) − 0.774 D bt g u1 t = 0.397 D1 (t ) + 0.603D2 t 2

Mb3 = 0

Fig. P13.17b

Observe that the third mode does not contribute to the total displacements, and the total displacements u2 ( t ) and u3 ( t ) are antisymmetric.

m EI L

m EI

Part c

The bending moment at the base of the column due to the nth mode is st Mbn ( t ) = Mbn An ( t )

st , shown in Fig. Substituting the modal static responses Mbn P13.17b, and combining modal responses gives 3

Mb ( t ) = ∑ Mbn ( t ) = 3. 533 m L A1 ( t ) + 1. 467 m L A2 ( t )

n =1

The bending moment at location a of the beam due to the nth mode is st Man ( t ) = Man An ( t )

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 36 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

st , shown in Fig. Substituting the modal static responses Man P13.17b, and combining modal responses gives 3

Ma ( t ) = ∑ Man ( t ) = − 0. 774 m L A1 ( t ) + 0. 774 m L A2 ( t ) n =1

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 37 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.18 Part a

The properties of the structure, m and k, ω n and φn are given in Problem 13.17. The influence vector is (from Problem 9.13):

R|0U| ι = S1V |T1|W

L2 =

φ1T m φ1 L1 M1

= 12. 579 m

= 0

The first two modes are not excited. Due to the third mode, the bending moments at the base of the column and at section a of the beam are Mb3 ( t ) = Mbst3 A3 ( t )

= 0

M3 = φ3T m φ3 = 2 m L3 M3

Ma3 ( t ) = Mast3 A3 ( t )

st = 0 Static analysis of the system in Fig. P13.3 gives Mb3 st and Ma3 = m L . Thus the total responses are

L3 = φ 3T m ι = 2m

Γ3 =

Part c

φ 2T m ι L2

The modal expansion of effective forces is shown in Fig. P13.18. The effective forces in the first two modes are zero, implying that these modes will not be excited by vertical ground motion.

R|0U| u (t ) = S 1V D (t ) |T1|W

M2 = φT2 m φ2 = 8. 292 m Γ2 =

The response is only due to the third mode; from Eq. (13.1.10):

L1 = φ 1T m ι = 0

Γ1 =

Part b

The modal quantities, given by Eq. (13.1.5), are:

M1 =

R| 0U| s = Γ m φ = SmV |Tm|W

Mb ( t ) = 0

Ma ( t ) = m L A3 ( t )

= 1

The effective forces, Eq. (13.1.4), are

LM5m OP R|0U| m p (t ) = − m ι u&& (t ) = − M P S1V u&& (t ) MN mPQ |T1|W R| 0 U| = − SmV u&& (t ) |Tm|W eff

Substituting Γn , m and φn in Eq. (13.1.6) gives

R|0U| S|0V| T0W R|0U| s = Γ m φ = S0V |T0|W s1 = Γ1 m φ 1 =

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 38 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

m 3m

L L

Ma1= 0

0 0

= st

Mb1= 0

Ma2= 0

0 0

+ st

Mb2= 0 m

Ma3= mL

m 0

+ st

M b3= 0

Fig. P13.18

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 39 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.19 The properties of the structure, m and k , ω n and φ n are given in Problem 13.17. The influence vector due to ground motion in the direction b-d is (from Problem 9.13, Part c). ⎧1 ⎪ ι = ⎨1 ⎪1 ⎩

2⎫ ⎪ 2⎬ 2⎪ ⎭

⎧ 0 ⎫ ⎪ ⎪ s 3 = Γ3 mφ 3 = ⎨0.707 m⎬ ⎪0.707 m⎪ ⎩ ⎭

The modal expansion of the spatial distribution of effective forces is shown in Fig. P13.19. Part b The modal displacements, form Eq. (13.1.10) are ⎧ 0.281⎫ ⎪ ⎪ u1 = Γ1φ1 D1 (t ) = ⎨− 0.547 ⎬ D1 (t ) ⎪ 0.547 ⎪ ⎩ ⎭

Substituting for m , ι , and φ n in Eq. (13.1.5) gives the modal quantities for the three modes: L1 = φ1T mι = 3.536m

⎧ 0.426⎫ ⎪ ⎪ u 2 = Γ2φ 2 D2 (t ) = ⎨ 0.547 ⎬ D2 (t ) ⎪− 0.547 ⎪ ⎩ ⎭

M 1 = φ1T mφ1 = 12.597 m Γ1 =

L1 = 0.281 M1

⎧ 0 ⎫ ⎪ ⎪ u 3 = Γ3φ3 D3 (t ) = ⎨0.707 ⎬ D3 (t ) ⎪0.707 ⎪ ⎩ ⎭

L2 = φ 2T mι = 3.536m M 2 = φ 2T mφ 2 = 8.292m Γ2 =

L2 = 0.426 M2

Combining the modal displacements gives the total displacements:

L3 = φ3T mι = 1.414m

u1 (t ) =

M 3 = φ3T mφ3 = 2m Γ3 =

0 D3 (t )

u 2 (t ) = − 0.547 D1 (t ) + 0.547 D 2 (t ) + 0.707 D3 (t ) u 3 (t ) = 0.547 D1 (t ) − 0.547 D 2 (t ) + 0.707 D3 (t )

L3 = 0.707 M3

Part c

The effective earthquake forces are given by Eq. (13.1.4): ⎡5m ⎤ ⎧1 ⎢ ⎥ ⎪1 m p eff (t ) = −m ι u&&gbd (t ) = − ⎢ ⎥⎨ ⎢⎣ m⎥⎦ ⎪1 ⎩ ⎧ 5m ⎪ = −⎨ m ⎪ m ⎩

0.281D1 (t ) + 0.426 D 2 (t ) +

2⎫ ⎪ 2 ⎬u&&gbd (t ) 2⎪ ⎭

Substituting Γn , m and φ n in Eq. (13.1.6) gives

The bending moment at the base of the column due to the n th mode is st M bn = M bn An (t ) st Substituting the modal static responses M bn , shown in Fig. P13.19, and combining modal responses gives

M b (t ) =

∑ M (t ) bn

n =1

= 2.499 mL A1 (t ) + 1.036 mL A2 (t )

⎧ 1.405m ⎫ ⎪ ⎪ s1 = Γ1mφ1 = ⎨− 0.547 m ⎬ ⎪ 0.547 m⎪ ⎩ ⎭

The bending moment at location a of the beam due to the n th mode is

⎧ 2.130m⎫ ⎪ ⎪ s 2 = Γ2 mφ 2 = ⎨ 0.547 m⎬ ⎪− 0.547 m ⎪ ⎩ ⎭

st Substituting the modal static responses M an , shown in Fig. P13.19, and combining modal responses gives

st M an (t ) = M an An (t )

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 40 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

M a (t ) =

∑ M (t ) = −0.547 mL A (t ) + 0.547 mL A (t ) an

n =1

+ 0.707 mL A3 (t ) m

m 3m

EI L

M ast1 = −0.547mL

0.547m

1.405m 0.547 m

M bst1 = 2.499mL

M ast2 = 0.547 mL

0.547 m 2.130m

0.547 m

M bst2 = 1.036mL

0.707 m

M ast3 = 0.707 mL

0.707 m 0

M bst3 = 0 © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 41 Fig. P13.19 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.20 Part a The properties of the structure, m and k , ω n and φ n are given in Problem 13.17. The influence vector due to ground motion in the direction b-c is (from Problem 9.13, Part c). ⎧− 1 ⎪ ι=⎨ 1 ⎪ 1 ⎩

2⎫ ⎪ 2⎬ 2⎪ ⎭

M 1 = φ1T mφ1 = 12.597m L1 = −0.281 M1

M 2 = φ 2T mφ 2 = 8.292m

⎧ 0 ⎫ ⎪ ⎪ u 3 = Γ3φ3 D3 (t ) = ⎨0.707⎬ D3 (t ) ⎪0.707⎪ ⎭ ⎩

L2 = −0.426 M2

L3 = φ3T mι = 1.414m

Combining the modal displacements gives the total displacements:

M 3 = φ3T mφ3 = 2m L3 = 0.707 M3

u1 (t ) = −0.281D1 (t ) − 0.426 D2 (t ) +

The effective earthquake forces are given by Eq. (13.1.4): ⎡5m ⎤ ⎧− 1 ⎢ ⎥⎪ 1 m p eff (t ) = −mιu&&gbc (t ) = − ⎢ ⎥⎨ ⎢⎣ m⎥⎦ ⎪ 1 ⎩ ⎧ − 5m 2 ⎫ ⎪ ⎪ = −⎨ m 2 ⎬u&&gbc (t ) ⎪ m 2⎪ ⎭ ⎩

The modal displacements, form Eq. (13.1.10). are

⎧− 0.426⎫ ⎪ ⎪ u 2 = Γ2φ 2 D 2 (t ) = ⎨− 0.547 ⎬ D 2 (t ) ⎪ 0.547⎪ ⎭ ⎩

L2 = φ 2T mι = −3.536m

Γ3 =

Part b

⎧ − 0.281⎫ ⎪ ⎪ u 1 = Γ1φ1 D1 (t ) = ⎨ 0.547⎬ D1 (t ) ⎪− 0.547 ⎪ ⎭ ⎩

L1 = φ1T m ι = −3.536m

Γ2 =

⎧ 0 ⎫ ⎪ ⎪ s 3 = Γ3 mφ 3 = ⎨0.707 m⎬ ⎪0.707 m⎪ ⎭ ⎩

The modal expansion of the spatial distribution of effective forces is shown in Fig. P13.20.

Substituting for m , ι and φ n in Eq. (13.1.5) gives the modal quantities for the three modes:

Γ1 =

⎧− 2.130m ⎫ ⎪ ⎪ s 2 = Γ2 mφ 2 = ⎨− 0.547 m ⎬ ⎪ 0.547 m⎪ ⎭ ⎩

2⎫ ⎪ 2 ⎬u&&gbc (t ) 2⎪ ⎭

u 2 (t ) = 0.547 D1 (t ) − 0.547 D2 (t ) + 0.707 D3 (t ) u3 (t ) = −0.547 D1 (t ) + 0.547 D2 (t ) + 0.707 D3 (t )

Part c The bending moment at the base of the column due to the n th mode is st M bn = M bn An (t ) st Substituting the modal static responses M bn , shown in Fig. P13.20, and combining modal responses gives

Substituting Γn , m and φ n in Eq. (13.1.6) gives ⎧ − 1.405m ⎫ ⎪ ⎪ s1 = Γ1mφ1 = ⎨ 0.547m⎬ ⎪− 0.547 m ⎪ ⎭ ⎩

0 D3 (t )

M b (t ) =

∑ M (t ) bn

n =1

= −2.499 mL A1 (t ) − 1.036 mL A2 (t )

The bending moment at location a of the beam due to the n th mode is st M an (t ) = M an An (t )

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 42 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

st , shown in Substituting the modal static responses M an Fig. P13.20, and combining modal responses gives

m 3m

M a (t ) =

∑ M (t ) = 0.547 mL A (t ) − 0.547 mL A (t ) an

n =1

+ 0.707 mL A3 (t )

M ast1 = 0.547mL

0.547 m 1.405m

0.547 m

M bst1 = −2.499mL

0.547 m

M ast2 = −0.547 mL

2.130m 0.547 m

M bst2 = −1.036mL

0.707 m

M ast3 = 0.707 mL

0.707 m 0

M bst3 = 0

Fig. P13.20 © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 43 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.21 Part a The properties of the structure, m and k , ω n and φ n are given in Problem 13.17. The influence vector due to ground rocking u gθ (in radians) in the counter-clockwise direction is (from Problem 9.13, Part c). ⎧− L ⎫ ⎪ ⎪ ι = ⎨ L⎬ ⎪− L ⎪ ⎩ ⎭

Substituting for m , ι and φ n in Eq. (13.1.5) gives the modal quantities for the three modes: L1 = φ1T m ι = −8.898mL M 1 = φ1T mφ1 = 12.597m Γ1 =

L1 = −0.7064 L M1

L2 = φ 2T mι = −2.4348mL M 2 = φ 2T mφ 2 = 8.292m L Γ2 = 2 = −0.294 L M2 L3 = φ 3T mι = 0 M 3 = φ 3T mφ 3 = 2m L Γ3 = 3 = 0 M3

⎧ − 1.470mL ⎫ ⎪ ⎪ s 2 = Γ2 mφ 2 = ⎨− 0.377 mL ⎬ ⎪ 0.377 mL ⎪ ⎭ ⎩ ⎧0⎫ ⎪ ⎪ s 3 = Γ3 mφ 3 = ⎨0⎬ ⎪0⎪ ⎩ ⎭

The modal expansion of the spatial distribution of effective forces is shown in Fig. P13.21. Part b The modal displacements, from Eq. (13.1.10), are ⎧− 0.706 L ⎫ ⎪ ⎪ u 1 = Γ1φ1 D1 (t ) = ⎨ 1.377 L ⎬ D1 (t ) ⎪ − 1.377 L ⎪ ⎭ ⎩ ⎧− 0.294 L ⎫ ⎪ ⎪ u 2 = Γ2φ 2 D 2 (t ) = ⎨− 0.377 L ⎬ D 2 (t ) ⎪ 0.377 L ⎪ ⎭ ⎩ ⎧0⎫ ⎪ ⎪ u 3 = Γ3φ 3 D3 (t ) = ⎨0⎬ D3 (t ) ⎪0⎪ ⎩ ⎭

Note that, here, D1 (t ) , D 2 (t ) and D3 (t ) are in radians. Combining the modal displacements gives the total displacements: u1 (t ) = −0.706 L D1 (t ) − 0.294 L D 2 (t )

The effective earthquake forces are given by Eq. (13.1.4):

u 2 (t ) = 1.377 L D1 (t ) − 0.377 L D 2 (t )

⎡5m ⎤ ⎧− L ⎫ ⎥ ⎪ L ⎪u&& (t ) m p eff (t ) = −m ι u&&gθ (t ) = − ⎢⎢ ⎥ ⎨ ⎬ gθ ⎢⎣ m⎥⎦ ⎪⎩− L ⎪⎭ ⎧ − 5mL ⎫ ⎪ ⎪ = − ⎨ mL ⎬u&&gθ (t ) ⎪ − mL ⎪ ⎭ ⎩

u 3 (t ) = −1.377 L D1 (t ) + 0.377 L D 2 (t )

Substituting Γn , m and φ n in Eq. (13.1.6) gives ⎧− 3.532mL ⎫ ⎪ ⎪ s 1 = Γ1mφ1 = ⎨ 1.377 mL ⎬ ⎪ − 1.377 mL ⎪ ⎭ ⎩

Observe that the third mode does not contribute to the displacement response. Part c The bending moment at the base of the column due to the n th mode is st M bn = M bn An (t )

where An (t ) is in radians per second per second. st Substituting the modal static responses M bn , shown in Fig. P13.21, and combining modal responses gives

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 44 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

mL M b (t ) =

∑ M (t )

n =1

= −6.826 mL2 A1 (t ) + 1.470 mL2 A2 (t )

5mL

EI EI

The bending moment at location a of the beam due to the n th mode is

st M an (t ) = M an An (t ) st Substituting the modal static responses M an , shown in Fig. P13.21, and combining modal responses gives

M ast1 = 1.377 mL2

1.377 mL

M a (t ) =

3.532mL

∑ M (t ) an

n =1

= 1.377 mL2 A1 (t ) − 1.470 mL2 A2 (t )

1.377 mL 1.47 mL M bst1 = −6.286mL2

0.3771mL

M ast2 = −1.47 mL2

0.3771mL 1.47 mL

M bst2 = 1.47 mL2

M ast3 = 0

0 0

M bst3 = 0

Fig. P13.21

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 45 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.22

Solution of the eigenvalue problem gives the natural periods and modes: m /1000

T1 = 1. 027 sec

T3 = 0.1503 sec

. U R|− 0.3054U| R| 0.3139U| R| 1366 | φ = S − 0.9583V φ = S 0.9735V φ = S− 0.4379 V |T − 32.34|W |T− 3180 |T 0.0773|W . |W

10 '

10 ' u1

T2 = 0. 974 sec

The modes have been normalized so that

10 '

Mn = φTn m φn = 1

Data: 32.34

m = 0. 486 kip − sec2 in .

EI L3 EI ′ L3

31.80

0.0773

= 56. 26 kips in.

0.9583

= 0. 0064 kip in .

0.3054

0.3139

1.366

Mode 1 T1 = 1.027 sec

Mode 2 T2 = 0.974 sec

Mode 3 T3 = 0.1503 sec

0.9735

0.4379

ζn = 0. 05 , n = 1, 2, 3 Part a

First compute the flexibility matrix: f$31 f$21 1

f$32 f$22

f$11

f$33

1 f$23

f$12

f$13

Part b

Computing Eq. (13.2.3) gives the modal properties: M1 = 1

M2 = 1

M3 = 1

L1h = − 0. 6296

Lh2 = 0. 6104

Lh3 = 0. 4511

Γ1 = − 0. 6296

Γ2 = 0. 6104

Γ3 = 0. 4511

Substituting these data in Eq. (13.2.4) gives

LM L 3EI 5L 6EI OP 4 L 3EI 8 L 3EI 14 L 3EI MM PP sym 26 L 3 EI L 3 EI + ( ) ′ N Q 3

f$ =

Next determine the lateral stiffness matrix k$ : k$ = f$ − 1 =

. . 0.0329O − 24119 LM77162 96 . 55 0.0439PP kip in. − MM 0.0192PQ N (sym)

R|0.0934U| R| 0.0931U| R| 0.2995U| s = S 0.2931V s = S 0.2889 V s = S − 0.0960V |T0.0099|W |T− 0.0094|W |T17. × 10 |W 1

−5

Thus the modal expansion of m 1 is [Eq. (13.2.2)]:

R| 0.486 U| R|0.0934U| R| 0.0931U| R| 0.2995U| S| 0.486 V| = S| 0.2931V| + S| 0.2889V| + S| − 0.0960V| T0.000486W T0.0099W T− 0.0094W T17. × 10 W −5

(c)

This expansion is shown graphically:

The mass matrix is

LM1 OP m = 0.486 M 1 kip − sec in. P MN 0.001PQ 2

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 46 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

0.000486 0.486

0.486

0.0099 0.2931

1.7 × 10

0.0094 0.2889

0.0934

–5

0.096

0.0931

0.2995

Part c

We compute u

LM MM N

0.0051 0.0046 0.0004 −1 $ 0.0143 − 0.0001 = k m 1 = 0.0161 0.5437 − 0.4664 0

OP PP Q

The modal static responses of the displacement of the appendage mass are st u31 = 0. 5437

st u32 = − 0. 4664

st u33 = 2. 0 × 10 − 5 (a)

Static analysis of the system for forces s n (n = 1, 2 and 3) gives the shear force in the appendage: Vast1 = 0. 0099

Vast2 = − 0. 0094

Vast3 = 1. 7 × 10 − 5 (b)

and the shear force at the base of the tower: Vbst1 = 0. 3965

Vbst2 = 0. 3726

Vbst3 = 0. 2035

(c)

Part d

From Eq. (a) we expect that the first two modes should give the largest contribution to the displacement of the appendage mass. Similarly, these two modes will dominate the shear at the base of the appendage. In addition to important contributions from the first two modes, the third mode should have, according to Eq. (c), significant contribution to the shear at the base of the tower.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 47 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.23

frequency of the main tower and the appendage mass is small.

Part a

The three modal SDF systems have the following properties: T1 = 1. 027 sec

T2 = 0. 974 sec T3 = 0.1503 sec

ζ1 = 0. 05

ζ2 = 0. 05

ζ3 = 0. 05

The displacement responses Dn ( t ) of these three SDF systems to the El Centro ground motion are shown in Fig. P13.23a. The pseudo-acceleration responses are An ( t ) = ω 2n Dn ( t ) and these are shown in Fig. P13.23b. Part b

Step 5c of Section 13.2.4 is implemented to determine the contribution of the nth mode rn ( t ) = rnst An ( t )

to the desired response quantities: displacement u3 ( t ) of the appendage mass, shear Va ( t ) in the appendage, and shear force Vb ( t ) at the base of the tower. The modal static responses u3stn are available in Eq. (a) st st in Eq. (b), and Vbn in Eq. (c). These of Problem 13.22, Van modal static responses are multiplied by An ( t ) (Fig. P13.23b) at each time step to obtain the results presented in Figs. P13.23c, P13.23d, and P13.23e. The peak values are noted for each modal response. Part c

The modal responses are combined at each time instant to determine the total responses shown in Figs. P13.23c, P13.23d, and P13.23e. The peak values are noted: u3o = 44. 58 in . Vao = 0. 879 kips

Vbo = 159. 83 kips

Part d

The seismic coefficient for the appendage is Vao wa

0. 879 w 1000

0. 879 (1000 ) 0. 486 ( 386 )

= 4. 69

and that for the tower is Vbo w

159. 83 2. 001 w

159. 83 2. 001 ( 0. 486 ) ( 386 )

= 0. 426

The seismic coefficient for the appendage is large because its natural frequency is tuned to the fundamental natural

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 48 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

4.364

Mode 1

0 -5

Mode 2

Dn, in.

0 -5

4.531

Mode 3

5 0

0.1649

-5 0

Time, sec Fig. P13.23a

Mode 1

0.4232

0 -1

Mode 2

1 0

0.4879

-1

Mode 3

1 0

0.7452

-1 0

Time, sec Fig. P13.23b

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 49 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

100

88.90

Mode 1

0 -100

Van, u3n, in. kips

87.93

100

Mode 2

0 -100

Mode 3

100 0

0.0058

-100

Va, u3, in. kips

100

Total

44.58

-100 0

Time, sec Fig. P13.23c 1.619

Mode 1

0 -2

1.778

Mode 2

0 -2

Mode 3

2 0

0.0049

-2

Total

2 0

0.879

-2 0

Time, sec Fig. P13.23d

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 50 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

100

64.83

Mode 1

0 -100

Mode 2

100

Vbn, kips

0 -100

70.23 Mode 3

100 0 -100

Vb, kips

58.59 Total

200 0

159.83

-200 0

Time, sec Fig. P13.23e

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 51 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.24 Part b

Data (see Problem 9.14): m = 0. 2331 kip − sec2 in .

From Eq. (13.3.10),

k = 1. 5 kips in .

Mn* =

b = 25 ft

uθ

IOn = sθn

(b)

Substituting numerical data for Mn , Lhn and sθn gives Table P13.24.

ζn = 0. 05 , n = 1, 2 and 3 DOFs: u = ux

( Lhn )2

Table P13.24

Equations of motion: 0 ⎤ ⎧0⎫ ⎡6 0 ⎤ ⎡1 ⎪ ⎪ ⎥u && + k ⎢0 6 −b ⎥ u = −m⎨1⎬u&&gy (t ) m ⎢⎢ 1 ⎥ ⎢ ⎥ ⎪0⎪ ⎢⎣0 −b 3b 2 ⎥⎦ ⎢⎣ b 2 6⎥⎦ ⎩3⎭ 12 mι

Mode

Mn*

IO* n

0.2245

5.3966

0.0086

– 5.3966

0.2331 = m

∑

Natural frequencies and modes (from Problem 10.17):

Mn* or

∑

IO* n

The last row of the table verifies Eq. (13.3.9).

ω1 = 5. 96

ω 2 = 6. 21

Part c

ω 3 = 10. 90

The displacements due to the nth mode are

⎧ 0 ⎫ ⎧2.071⎫ ⎧ 0 ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ φ1 = ⎨ 2.032 ⎬ φ 2 = ⎨ 0 ⎬ φ 3 = ⎨−0.3988⎬ ⎪0.0033⎪ ⎪ 0 ⎪ ⎪ 0.0166 ⎪ ⎩ ⎭ ⎩ ⎭ ⎩ ⎭

u( t ) = ∑ Γn φn Dn ( t ) n =1

Substituting for Γn and φn gives

Part a Mn = φTn m φ n

Lhn = φTn m ι

Γn =

Lhn

(a)

Substituting for m, φn and ι in Eq. (a) gives Γ1 = 0. 4738

Γ2 = 0

Γ3 = − 0. 0930

Substituting for m, φn and Γn in Eq. (13.1.6) gives ⎧0 ⎫ ⎪ ⎪ s1 = ⎨0.2245⎬ ⎪5.3966⎪ ⎩ ⎭

⎧0⎫ ⎪ ⎪ s 2 = ⎨0⎬ ⎪0⎪ ⎩ ⎭

⎧ 0 ⎫ ⎪ ⎪ s 3 = ⎨ 0.0086⎬ ⎪−5.3966⎪ ⎩ ⎭

syn

sθn

0 ⎫ ⎧ ⎧u x (t ) ⎫ ⎧ 0 ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨u y (t )⎬ = ⎨ 0.9629 ⎬D1 (t ) + ⎨ 0.0371 ⎬D3 (t ) ⎪−0.00156⎪ ⎪u (t ) ⎪ ⎪0.00156⎪ ⎭ ⎩ ⎭ ⎩ θ ⎭ ⎩

Part d The modal static responses for base shear in the ydirection and base torque are

where sn = sxn

⎧u x (t ) ⎫ ⎧ 0 ⎫ ⎧ 0 ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨u y (t ) ⎬ = 0.4738⎨ 2.032 ⎬D1 (t ) − 0.0930⎨−0.3988⎬ D3 (t ) ⎪0.0033⎪ ⎪ 0.0166 ⎪ ⎪u (t ) ⎪ ⎭ ⎩ θ ⎭ ⎩ ⎭ ⎩

The modal expansion of m ι is shown in the following figure:

st Vbyn = syn

Substituting numerical values for syn and sθn gives st

5.3966

st Tbn = sθn

Vby1 = 0. 2245

Vby2 = 0

Vby3 = 0. 0086

Tb1st = 5. 3966

st Tb2 = 0

st Tb3 = − 5. 3966

0.0086 0.2331

0.2245

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 52 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

st Substituting Vbyn and Tbn in Eq. (13.3.15) gives the modal responses:

Vby1 ( t ) = 0. 2245 A1 ( t ) Vby 2 ( t ) = 0

Vby3 ( t ) = 0. 0086 A3 ( t ) Tb1 ( t ) = 5. 3966 A1 ( t ) Tb 2 ( t ) = 0 Tb3 = − 5. 3966 A3 ( t )

Combining the modal responses gives the total response: Vby ( t ) = 0. 2245 A1 ( t ) + 0. 0086 A3 ( t ) Tb ( t ) = 5. 3966 A1 ( t ) − 5. 3966 A3 ( t )

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 53 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Table P13.25

Problem 13.25 Equations of motion: ⎧1 2 ⎫ 0 ⎤ ⎡6 0 ⎤ ⎡1 ⎪ ⎪ ⎥ ⎢ ⎥ ⎢ && && m⎢ 1 ⎥ u + k ⎢0 6 −b ⎥ u = − m ⎨1 2 ⎬ u g (t ) ⎪ 0 ⎪ ⎢⎣0 −b 3b 2 ⎥⎦ ⎢⎣ b 2 6⎥⎦ ⎭ ⎩243 14 mι

Mode

Mn*

IO* n

0.1123

3.816

0.1166

0.0043

– 3.816

0.2332 = m

∑

Mn* or

∑

IO* n

where m, b, k are given in the solution to Problem 13.24. Also the ω n and φn are given in the solution to Problem 13.24.

The last row of the table verifies Eq. (13.3.9)—which is for a one-way symmetric syste—after it is generalized to an arbitrarily unsymmetric system.

Part a

Part c

Mn = φTn m φ n

Lhn = φTn m ι

Γn =

The displacement response is

Lhn

(a)

u ( t ) = ∑ Γn φn Dn ( t )

Substituting for m, φn and ι in Eq. (a) gives Γ1 = 0. 3350

Γ2 = 0. 3414

Substituting for Γn and φn gives

Γ3 = − 0. 0658

R|u (t )U| R| 0 U| R|2.071U| = 0 3350 2 032 + 0 3414 u ( t ) . . D ( t ) . S| V| S| V| S| 0 V| D (t ) T0.0033W T 0 W Tu ( t ) W R| 0 U| − 0.0658 S− 0.3988V D (t ) |T 0.0166 |W x

Substituting for m, φn and Γn in Eq. (13.1.6) gives

⎧ 0 ⎫ ⎧0.1649⎫ ⎧ 0 ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ s 1 = ⎨0.1588⎬ ; s 2 = ⎨ 0 ⎬ ; s 3 = ⎨ 0.0061⎬ ⎪ 3.816 ⎪ ⎪ 0 ⎪ ⎪−3.816⎪ ⎭ ⎭ ⎭ ⎩ ⎩ ⎩

syn

sθn

where sn = sxn

(c)

n =1

R|u (t )U| R| 0 U| R|0.7071U| S|u (t )V| = S|0.6809V| D (t ) + S| 0 V| D (t ) T 0 W Tu (t )W T0.0011W R| 0 U| + S 0.0262 V D (t ) |T− 0.0011|W x

The modal expansion of m ι is shown in the following figure: 0.1649 0.1649

0.1588

0.0061 0

3.816

0.1649

( Lhn )2 Mn

0 3.816

Part d

The modal static responses for the x-component Vbx ( t ) of base shear, the y-component Vby ( t ) of base shear, and the base torque Tb ( t ) are

Part b From Eq. (13.3.10), Mn* =

IO* n = sθn

(b)

Substituting numerical data for Mn , Lhn and sθn gives Table P13.25.

st Vbxn = sxn

Vbyn = syn

Tbyn = sθn

(d)

Substituting numerical values for sxn , syn and sθn gives Vbxst1 = 0 st

Vbxst 2 = 0.1649 Vbxst 3 = 0 st

Vby1 = 0.1588

Vby 2 = 0

Vby3 = 0. 0061

Tbst1 = 3. 816

Tbst2 = 0

Tbst3 = − 3. 816

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 54 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Substituting these modal static responses in Eq. (13.3.15) gives the modal responses: Vbx1 ( t ) = 0 Vbx 2 ( t ) = 0.1649 A2 ( t ) Vbx 3 ( t ) = 0 Vby1 ( t ) = 0.1588 A1 ( t ) Vby 2 ( t ) = 0

Vby3 ( t ) = 0. 0061 A3 ( t ) Tb1 ( t ) = 3. 816 A1 ( t ) Tb 2 ( t ) = 0 Tb3 = − 3. 816 A3 ( t )

Combining the modal responses gives the total response: Vbx ( t ) = 0.1649 A2 ( t ) Vby ( t ) = 0.1588 A1 ( t ) + 0. 0061 A3 ( t ) Tb ( t ) = 3. 816 A1 ( t ) − 3. 816 A3 ( t )

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 55 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.26 Part a Ground motion in the y-direction excites the first and third natural vibration modes of the system. The corresponding SDF systems have the following properties: T1 = 1. 054 sec T3 = 0. 5764 sec ζ1 = 0. 05 ζ3 = 0. 05 The displacement responses Dn ( t ) of the two SDF systems are shown in Fig. P13.26a. The pseudoacceleration responses are An ( t ) = ω 2n Dn ( t ) and these are shown in Fig. P13.26b. Parts b and c From Problem 13.24,

uy ( t ) = 0. 9629 D1 ( t ) + 0. 0371 D3 ( t )

(a)

In Eq. (a), the Dn values at each time instant are substituted from the data of Fig. P13.26a to obtain the modal contributions uyn ( t ) and the total y-displacement uy ( t ) . The results are shown in Fig. P13.26c. Similar results for ( b 2 ) uθn ( t ) and ( b 2 ) uθ ( t ) are presented in Fig. P13.26d. In Eq. (c), the An values at each time instant are substituted from the data of Fig. P13.26b to obtain the modal contributions Vbyn ( t ) and the total base shear Vby ( t ) . The results are shown in Fig. P13.26e. Similar results for Tbn ( t ) and the total base torque Tb ( t ) are presented in Fig. P13.26f. For various response quantities, their peak values are

( b 2 ) uθ ( t ) = ( b 2 ) 0. 00156 D1 ( t ) − 0. 00156 D3 ( t ) (b)

Vby ( t ) = 0. 2245 A1 ( t ) + 0. 0086 A3 ( t )

(c)

Tb ( t ) = 5. 3966 A1 ( t ) − 5. 3966 A3 ( t )

(d)

uyo = 4.182 in.

( b 2 ) uθ o = 1. 22 in.

Vbyo = 35. 6 kips

Tbo = 1899 kip - in.

4.291

Mode 1

Dn , in .

-5 5

Mode 3

2.604

0 -5 0

Time, sec Fig. P13.26a

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 56 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Mode 1

0.3943

An, g

-1

0.8012

Mode 3

0 -1 0

Time, sec Fig. P13.26b

4.132

Mode 1

0 -5

uyn, in.

Mode 3

0.0966

0 -5

4.182

uy, in.

Total

0 -5 0

Time, sec Fig. P13.26c

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 57 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

0.9931

Mode 1

b uθ n in. 2

-1

Mode 3

1 0 -1

b uθ in. 2

0.6027 Total

1 0 -1

1.2219 0

Time, sec Fig. P13.26d

34.21

Mode 1

Vyn, kips

-50 50

Mode 3

2.667

0 -50

35.60

Vy, kips

Total

0 -50 0

Time, sec Fig. P13.26e

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 58 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Tbn,

Mode 1

822.2

0 -2

10 kip-ft

Mode 3

2 0 -2

Tbn, 10 3 kip-ft

1671

Total

1899

-2 0

Time, sec Fig. P13.26f

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 59 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.27

⎧ 0.3532⎫ ⎪ ⎪ s 3 = Γ3 mφ 3 = m ⎨ 0.4632⎬ ⎪ 0.1179⎪ ⎭ ⎩

uy m

L b

⎧ 0.3532⎫ ⎧ 0.0434⎫ ⎧ 0.6034⎫ ⎧1 ⎫ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ m ⎨0⎬ = m ⎨− 0.3824⎬ + m ⎨− 0.0808⎬ + m ⎨ 0.4632⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩ 0.1179⎪⎭ ⎪⎩ 0.1872⎪⎭ ⎪⎩ − 0.3051⎪⎭ ⎪⎩0⎪⎭

y a

Graphically:

Part a From Problem 9.18: ⎡m ⎤ ⎢ ⎥ m=⎢ ι = [ 1 0 0] T m ⎥ ⎢⎣ m⎥⎦

From Problem 10.28: ⎧− 0.2084⎫ ⎧ 0.7767⎫ ⎪ ⎪ ⎪ ⎪ φ1 = ⎨− 0.4923⎬ φ 2 = ⎨ 0.3875⎬ ⎪− 0.8980⎪ ⎪− 0.3928⎪ ⎭ ⎩ ⎭ ⎩

Thus, Eq. (13.1.4) specializes to:

0.3051m 0.3824m m 0.6034m

⎧ 0.5943⎫ ⎪ ⎪ φ 3 = ⎨ 0.7794⎬ ⎪ 0.1984⎪ ⎭ ⎩

Substituting m, ι and φn in Eq. (13.1.5) gives:

0.1872m

0.0808m

L1 = φ1T m ι = 0.7767 m M 1 = φ1T mφ1 = m Γ1 =

0.0434m

L1 = 0.7767 M1

L 2 = φ 2T m ι = −0.2084m

0.1179m

0.4632m

M 2 = φ 2T mφ 2 = m Γ2 =

L2 = −0.2084 M2

0.3532m

L3 = φ 3T m ι = 0.5943m M 3 = φ 3T mφ 3 = m L3 = 0.5943 M3 Substituting , m and φ n in Eq. (13.1.6) gives the effective earthquake forces: ⎧ 0.6034⎫ ⎪ ⎪ s 1 = Γ1mφ1 = m ⎨− 0.3824⎬ ⎪ − 0.3051⎪ ⎭ ⎩ ⎧ 0.0434⎫ ⎪ ⎪ s 2 = Γ2 mφ 2 = m ⎨− 0.0808⎬ ⎪ 0.1872⎪ ⎭ ⎩ Γ3 =

Part b Substituting for Γn and φ n in Eq. (13.1.10) gives the modal displacements: ⎧ 0.6034⎫ ⎧u x (t ) ⎫ ⎪ ⎪ ⎪ ⎪ u 1 (t ) = ⎨u y (t )⎬ = Γ1φ1 D1 (t ) = ⎨− 0.3824⎬ D1 (t ) (a) ⎪ − 0.3051⎪ ⎪u (t ) ⎪ ⎭ ⎩ ⎩ z ⎭1 ⎧u x (t ) ⎫ ⎧ 0.0434⎫ ⎪ ⎪ ⎪ ⎪ u 2 (t ) = ⎨u y (t )⎬ = Γ2φ2 D2 (t ) = ⎨− 0.0808⎬ D2 (t ) ⎪u (t ) ⎪ ⎪ 0.1872⎪ ⎩ z ⎭2 ⎩ ⎭

(b)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 60 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

⎧ 0.3532⎫ ⎧u x (t ) ⎫ ⎪ ⎪ ⎪ ⎪ u3 (t ) = ⎨u y (t )⎬ = Γ3φ3 D3 (t ) = ⎨ 0.4632⎬ D3 (t ) ⎪ 0.1179⎪ ⎪u (t ) ⎪ ⎭ ⎩ ⎩ z ⎭3

(c)

Combining modal responses in Eqs. (a), (b) and (c) gives the total displacement response: ⎧ 0.6034⎫ ⎧ 0.0434⎫ ⎧0.3532⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ u(t ) = ⎨− 0.3824⎬D1(t ) + ⎨− 0.0808⎬D2 (t) + ⎨0.4632⎬D3 (t) (d) ⎪− 0.3051⎪ ⎪ 0.1872⎪ ⎪0.1179⎪ ⎩ ⎭ ⎩ ⎭ ⎩ ⎭ Part c From Eq. (13.1.13), the earthquake-induced bending moments and torques at the base due to the nth mode are: st M xn (t ) = M xn An (t ) st M yn (t ) = M yn An (t )

Tn (t ) = Tnst An (t ) st st , M yn and Tnst are expressed in terms of s n where M xn

as the scalar products: st M xn = [ 0 L − L] s n st M yn = [ − L 0 L] s n

Tnst = [ L − L 0] s n

and s n are given in Part a:

st M xn = [− 0.0773mL, − 0.2679mL, 0.3453mL]

st M yn = [− 0.9084mL, 0.1437 mL, − 0.2353mL]

Tnst = [ 0.9858mL, 0.1242mL, − 0.1100mL] Combining these modal contributions for each response gives the total response in terms of An (t ) : 3

M x (t ) =

∑

M xn (t ) =

n =1

∑ M A (t ) st xn

n =1

= −0.0773mLA1 (t ) − 0.2679mLA2 (t ) + 0.3453mLA3 (t ) 3

M y (t ) =

∑

M yn (t ) =

n =1

∑ M A (t ) st yn

n =1

= −0.9084mLA1 (t ) + 0.1437 mLA2 (t ) − 0.2353mLA3 (t ) 3

T (t ) =

∑ T (t ) = ∑ T A (t ) n

n =1

st n

n =1

= 0.9858mLA1 (t ) + 0.1242mLA2 (t ) − 0.1100mLA3 (t )

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 61 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.28

⎧ 0.4632⎫ ⎪ ⎪ s 3 = Γ3 mφ 3 = m ⎨ 0.6074⎬ ⎪ 0.1546⎪ ⎭ ⎩

uy L

m ux

d b

x a Part a From Problem 9.18: ⎡m ⎤ ⎢ ⎥ ⎥ m m=⎢ ι = [ 0 1 0] T ⎢ ⎥ ⎢⎣ m⎥⎦ From Problem 10.28: ⎧ 0.5943⎫ ⎧− 0.2084⎫ ⎧ 0.7767⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ φ1 = ⎨− 0.4923⎬ φ 2 = ⎨ 0.3875⎬ φ 3 = ⎨ 0.7794⎬ ⎪ 0.1984⎪ ⎪− 0.8980⎪ ⎪− 0.3928⎪ ⎭ ⎩ ⎭ ⎩ ⎭ ⎩ Substituting m, ι and φ n in Eq. (13.1.5) gives:

Thus, Eq. (13.1.4) specializes to: ⎧ 0.4632⎫ ⎧ − 0.0808⎫ ⎧− 0.3824⎫ ⎧0 ⎫ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ m ⎨1⎬ = m ⎨ 0.2424⎬ + m ⎨ 0.1502⎬ + m ⎨ 0.6074⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩ 0.1546⎪⎭ ⎪⎩− 0.3480⎪⎭ ⎪⎩ 0.1934⎪⎭ ⎪⎩0⎪⎭

Graphically: m

0.1934m

0.2424m

0.3824m

0.3480m

0.1502m

L1 = φ1T m ι = −0.4923m M 1 = φ1T mφ1 = m Γ1 =

L1 = −0.4923 M1

L 2 = φ 2T m ι = 0.3875 m M 2 = φ 2T mφ 2 = m Γ2 =

L2 = 0.3875 M2

L3 = φ 3T m ι = 0.7794 m M 3 = φ 3T mφ 3 = m Γ3 =

L3 = 0.7794 M3

Substituting Γn , m and φ n in Eq. (13.1.6) gives the effective earthquake forces: ⎧− 0.3824⎫ ⎪ ⎪ s 1 = Γ1mφ1 = m ⎨ 0.2424⎬ ⎪ 0.1934⎪ ⎭ ⎩ ⎧− 0.0808⎫ ⎪ ⎪ s 2 = Γ2 mφ 2 = m ⎨ 0.1502⎬ ⎪− 0.3480⎪ ⎭ ⎩

0.0808m

0.1546m

0.6074m

0.4632m Part b Substituting for Γn and φn in Eq. (13.1.10) gives the modal displacements: ⎧− 0.3824⎫ ⎧u x (t ) ⎫ ⎪ ⎪ ⎪ ⎪ u 1 (t ) = ⎨u y (t )⎬ = Γ1φ1 D1 (t ) = ⎨ 0.2424⎬ D1 (t ) (a) ⎪ 0.1934⎪ ⎪u (t ) ⎪ ⎭ ⎩ ⎩ z ⎭1 ⎧− 0.0808⎫ ⎧u x (t ) ⎫ ⎪ ⎪ ⎪ ⎪ u 2 (t ) = ⎨u y (t )⎬ = Γ2φ2 D2 (t ) = ⎨ 0.1502⎬ D2 (t ) ⎪− 0.3480⎪ ⎪u (t ) ⎪ ⎭ ⎩ ⎩ z ⎭2

(b)

⎧ 0.4632⎫ ⎧u x (t ) ⎫ ⎪ ⎪ ⎪ ⎪ u3 (t ) = ⎨u y (t )⎬ = Γ3φ3 D3 (t ) = ⎨ 0.6074⎬ D3 (t ) ⎪ 0.1546⎪ ⎪u (t ) ⎪ ⎭ ⎩ ⎩ z ⎭3

(c)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 62 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Combining modal responses in Eqs. (a), (b) and (c) gives the total displacement response: ⎧− 0.3824⎫ ⎧− 0.0808⎫ ⎧0.4632⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ u(t ) = ⎨ 0.2424⎬D1(t ) + ⎨ 0.1502⎬D2 (t) + ⎨0.6074⎬D3 (t ) ⎪ 0.1934⎪ ⎪− 0.3480⎪ ⎪0.1546⎪ ⎩ ⎭ ⎩ ⎭ ⎩ ⎭ (d) Part c From Eq. (13.1.13), the earthquake-induced bending moments and torques at the base due to the nth mode are: st M xn (t ) = M xn An (t ) st M yn (t ) = M yn An (t )

Tn (t ) = Tnst An (t ) st st , M yn and Tnst are expressed in terms of s n where M xn

as the scalar products: st M xn = [ 0 L − L] s n st M yn = [ − L 0 L] s n

Tnst = [ L − L 0] s n

and s n are given in Part a: st M xn = [ 0.0490mL,

0.4982mL,

0.4528mL]

st M yn = [ 0.5758mL, − 0.2672mL, − 0.3086mL]

Tnst = [ − 0.6248mL, − 0.2310mL, − 0.1442mL]

Combining these modal contributions for each response gives the total response in terms of An (t ) : 3

M x (t ) =

∑

M xn (t ) =

n =1

∑ M A (t ) st xn

n =1

= 0.0490mLA1 (t ) + 0.4982mLA2 (t ) + 0.4528mLA3 (t ) 3

M y (t ) =

∑

M yn (t ) =

n =1

∑ M A (t ) st yn

n =1

= 0.5758mLA1 (t ) − 0.2672mLA2 (t ) − 0.3086mLA3 (t ) 3

T (t ) =

∑ T (t ) = ∑ T A (t ) n

n =1

st n

n =1

= −0.6248mLA1 (t ) − 0.2310mLA2 (t ) − 0.1442mLA3 (t )

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 63 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.29

⎧ 0.1179⎫ ⎪ ⎪ s 3 = Γ3 mφ 3 = m ⎨ 0.1546⎬ ⎪ 0.0394⎪ ⎭ ⎩

uz z

uy m

L b

Thus, Eq. (13.1.4) specializes to: ux

y a

Graphically: m

Part a From Problem 9.18: ⎡m ⎤ ⎢ ⎥ m=⎢ ι = [ 0 0 1] T m ⎥ ⎢⎣ m⎥⎦

From Problem 10.28: ⎧− 0.2084⎫ ⎧ 0.7767⎫ ⎪ ⎪ ⎪ ⎪ φ1 = ⎨− 0.4923⎬ φ 2 = ⎨ 0.3875⎬ ⎪− 0.8980⎪ ⎪− 0.3928⎪ ⎭ ⎩ ⎭ ⎩

⎧ 0.1179⎫ ⎧ 0.1872⎫ ⎧− 0.3051⎫ ⎧0 ⎫ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ m ⎨0⎬ = m ⎨ 0.1934⎬ + m ⎨− 0.3480⎬ + m ⎨ 0.1546⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩ 0.0394⎪⎭ ⎪⎩ 0.8064⎪⎭ ⎪⎩ 0.1542⎪⎭ ⎪⎩1⎪⎭

0.1542m

0.1934m

0.3051m

⎧ 0.5943⎫ ⎪ ⎪

φ 3 = ⎨ 0.7794⎬

⎪ 0.1984⎪ ⎭ ⎩ Substituting m, ι and φ n in Eq. (13.1.5) gives:

0.8064m

0.3480m

L1 = φ1T m ι = −0.3928m M 1 = φ1T mφ1 = m Γ1 =

0.1872m

L1 = −0.3928 M1

L 2 = φ 2T m ι = −0.8980m

0.0394m

0.1546m

M 2 = φ 2T mφ 2 = m Γ2 =

L2 = −0.8980 M2

L3 = φ 3T m ι = 0.1984m M 3 = φ 3T mφ 3 = m L3 = 0.1984 M3 Substituting Γn , m and φ n in Eq. (13.1.6) gives the effective earthquake forces: ⎧− 0.3051⎫ ⎪ ⎪ s 1 = Γ1mφ1 = m ⎨ 0.1934⎬ ⎪ 0.1542⎪ ⎭ ⎩ Γ3 =

⎧ 0.1872⎫ ⎪ ⎪ s 2 = Γ2 mφ 2 = m ⎨− 0.3480⎬ ⎪ 0.8064⎪ ⎭ ⎩

0.1179m Part b Substituting for Γn and φ n in Eq. (13.1.10) gives the modal displacements: ⎧− 0.3051⎫ ⎧u x (t ) ⎫ ⎪ ⎪ ⎪ ⎪ u 1 (t ) = ⎨u y (t )⎬ = Γ1φ1 D1 (t ) = ⎨ 0.1934⎬ D1 (t ) ⎪ 0.1542⎪ ⎪u (t ) ⎪ ⎭ ⎩ ⎩ z ⎭1

(a)

⎧ 0.1872⎫ ⎧u x (t ) ⎫ ⎪ ⎪ ⎪ ⎪ u 2 (t ) = ⎨u y (t )⎬ = Γ2φ2 D2 (t ) = ⎨− 0.3480⎬ D2 (t ) ⎪ 0.8064⎪ ⎪u (t ) ⎪ ⎭ ⎩ ⎩ z ⎭2

(b)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 64 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

⎧ 0.1179⎫ ⎧u x (t ) ⎫ ⎪ ⎪ ⎪ ⎪ u3 (t ) = ⎨u y (t )⎬ = Γ3φ3 D3 (t ) = ⎨ 0.1546⎬ D3 (t ) ⎪ 0.0394⎪ ⎪u (t ) ⎪ ⎭ ⎩ ⎩ z ⎭3

(c)

Combining modal responses in Eqs. (a), (b) and (c) gives the total displacement response: ⎧− 0.3051⎫ ⎧ 0.1872⎫ ⎧0.1179⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ u(t ) = ⎨ 0.1934⎬ D1 (t ) + ⎨− 0.3480⎬ D2 (t ) + ⎨0.1546⎬ D3 (t ) ⎪ 0.1542⎪ ⎪ 0.8064⎪ ⎪0.0394⎪ ⎩ ⎭ ⎩ ⎭ ⎩ ⎭

(d) Part c From Eq. (13.1.13), the earthquake-induced bending moments and torques at the base due to the nth mode are: st M xn (t ) = M xn An (t ) st M yn (t ) = M yn An (t )

Tn (t ) = Tnst An (t ) st st and Tnst are expressed in terms of s n where M xn , M yn

as the scalar products: st M xn = [ 0 L − L] s n st M yn = [ − L 0 L] s n

Tnst = [ L − L 0] s n

and s n are given in Part a: st M xn = [ 0.0391mL, − 1.1544mL,

0.1153mL]

st M yn = [ 0.4594mL,

0.6192mL, − 0.0786mL]

Tnst = [ − 0.4985mL,

0.5352mL, − 0.0367mL]

Combining these modal contributions for each response gives the total response in terms of An (t ) : 3

M x (t ) =

∑

M xn (t ) =

n =1

∑ M A (t ) st xn

n =1

= 0.0391mLA1 (t ) − 1.1544mLA2 (t ) + 0.1153mLA3 (t ) 3

M y (t ) =

∑

M yn (t ) =

n =1

∑ M A (t ) st yn

n =1

= 0.4594mLA1 (t ) + 0.6192mLA2 (t ) − 0.0786mLA3 (t ) 3

T (t ) =

∑ n =1

Tn (t ) =

∑ T A (t ) st n

n =1

= −0.4985mLA1 (t ) + 0.5352mLA2 (t ) − 0.0367mLA3 (t ) © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 65 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.30

⎧ 0.5394⎫ ⎪ ⎪ s 3 = Γ3 mφ 3 = m ⎨ 0.7074⎬ ⎪ 0.1801⎪ ⎭ ⎩

uy m

L b

⎧ 0.5394⎫ ⎧ 0.0865⎫ ⎧− 0.0485⎫ ⎧1⎫ m ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 0 . 1608 + 0 . 0308 1 + − m m m = ⎨ 0.7074⎬ ⎬ ⎨ ⎬ ⎨ ⎨⎬ 3⎪⎪ ⎪ 0.1801⎪ ⎪ 0.3727⎪ ⎪ 0.0246⎪ ⎭ ⎩ ⎭ ⎩ ⎭ ⎩ ⎩1⎭

y a

Part a From Problem 9.18: ⎡m ⎤ 1 ⎢ ⎥ [ 1 1 1] T m=⎢ ι= m ⎥ 3 ⎢⎣ m⎥⎦

From Problem 10.28: ⎧− 0.2084⎫ ⎧ 0.7767⎫ ⎪ ⎪ ⎪ ⎪ φ1 = ⎨− 0.4923⎬ φ 2 = ⎨ 0.3875⎬ ⎪− 0.8980⎪ ⎪− 0.3928⎪ ⎭ ⎩ ⎭ ⎩

Thus, Eq. (13.1.4) specializes to:

Graphically: m 3

m 3

⎧ 0.5943⎫ ⎪ ⎪ φ 3 = ⎨ 0.7794⎬ ⎪ 0.1984⎪ ⎭ ⎩

0.0246m

0.0308m

m 3

0.0485m

Substituting m, ι and φ n in Eq. (13.1.5) gives: L1 = φ1T m ι = −0.0626m M 1 = φ1T mφ1 = m

0.3727m

0.1608m

L Γ1 = 1 = −0.0626 M1

0.0865m

L 2 = φ 2T mι = −0.4150m M 2 = φ 2T mφ 2 = m Γ2 =

L2 = −0.4150 M2

0.1801m

0.7074m

L3 = φ 3T m ι = 0.9077 m M 3 = φ 3T mφ 3 = m L Γ3 = 3 = 0.9077 M3 Substituting Γn , m and φ n in Eq. (13.1.6) gives the effective earthquake forces: ⎧− 0.0485⎫ ⎪ ⎪ s1 = Γ1mφ1 = m ⎨ 0.0308⎬ ⎪ 0.0246⎪ ⎭ ⎩ ⎧ 0.0865⎫ ⎪ ⎪ s 2 = Γ2 mφ 2 = m ⎨− 0.1608⎬ ⎪ 0.3727⎪ ⎭ ⎩

0.5394m

Part b Substituting for Γn and φ n in Eq. (13.1.10) gives the modal displacements: ⎧− 0.0485⎫ ⎧u x (t ) ⎫ ⎪ ⎪ ⎪ ⎪ u 1 (t ) = ⎨u y (t )⎬ = Γ1φ1 D1 (t ) = ⎨ 0.0308⎬ D1 (t ) (a) ⎪ 0.0246⎪ ⎪u (t ) ⎪ ⎭ ⎩ ⎩ z ⎭1

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 66 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

⎧ 0.0865⎫ ⎧u x (t ) ⎫ ⎪ ⎪ ⎪ ⎪ u 2 (t ) = ⎨u y (t )⎬ = Γ2φ2 D2 (t ) = ⎨− 0.1608⎬ D2 (t ) ⎪ 0.3727 ⎪ ⎪u (t ) ⎪ ⎭ ⎩ ⎩ z ⎭2 ⎧ 0.5394⎫ ⎧u x (t ) ⎫ ⎪ ⎪ ⎪ ⎪ u3 (t ) = ⎨u y (t )⎬ = Γ3φ3 D3 (t ) = ⎨ 0.7074⎬ D3 (t ) ⎪ 0.1801⎪ ⎪u (t ) ⎪ ⎩ ⎭ ⎩ z ⎭3

(b)

Combining these modal contributions for each response gives the total response in terms of An (t ) : 3

M x (t ) =

∑

M xn (t ) =

n =1

⎧− 0.0485⎫ ⎧ 0.0865⎫ ⎧0.5394⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ u(t ) = ⎨ 0.0308⎬D1(t) + ⎨− 0.1608⎬D2 (t ) + ⎨0.7074⎬D3 (t ) ⎪ 0.0246⎪ ⎪ 0.3727⎪ ⎪0.1801⎪ ⎩ ⎭ ⎩ ⎭ ⎩ ⎭ (d) Part c From Eq. (13.1.13), the earthquake-induced bending moments and torques at the base due to the nth mode are: st M xn (t ) = M xn An (t )

st xn

n =1

= 0.0062mLA1 (t ) − 0.5335mLA2 (t ) + 0.5273mLA3 (t )

(c)

Combining modal responses in Eqs. (a), (b), and (c) gives the total displacement response:

∑ M A (t )

M y (t ) =

∑ M (t ) = ∑ M A (t ) st yn

n =1

= 0.0732mLA1 (t ) + 0.2862mLA2 (t ) − 0.3594mLA3 (t ) 3

T (t ) =

∑ n =1

Tn (t ) =

∑ T A (t ) st n

n =1

= −0.0794mLA1 (t ) + 0.2474mLA2 (t ) − 0.1680mLA3 (t )

st M yn (t ) = M yn An (t )

Tn (t ) = Tnst An (t ) st st , M yn and Tnst are expressed in terms of s n where M xn

as the scalar products: st M xn = [ 0 L − L] s n st M yn = [ − L 0 L] s n

Tnst = [ L − L 0] s n

and s n are given in Part a:

st M xn = [ 0.0062mL, − 0.5335mL,

0.5273mL]

[ 0.0732mL, 0.0262mL, − 0.03594mL] Tnst = [ − 0.0794mL, 0.2474mL, − 0.1680mL]

st M yn =

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 67 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.31

6. Total displacements. ut ( t ) = u s ( t ) + u ( t )

1. Equations of motion (from Problem 9.19). m

LM1 OP RSu&& UV + k LM 2 − 1OP RSu UV N 1Q Tu&& W N− 1 2Q Tu W L1 OP LM2 3 1 3OP RSu&& UV = −mM N 1Q N1 3 2 3Q Tu&& W 1

where u( t ) and u s ( t ) are given by Eqs. (b) and (c), respectively.

7. Excitation a. ug1 ( t ) = − ug 2 ( t ) = ug ( t )

2. Natural frequencies and modes. k

ω1 =

φ1 =

R|S1 2 U|V T|1 2 W|

For these excitations, Dn1 ( t ) = Dn ( t ) Dn 2 ( t ) = − Dn ( t )

ω2 =

3k m

φ2 =

R|S− 1 2 U|V T| 1 2 W|

LM 1 3OP D (t ) N− 1 3Q L 1 3OP u (t ) u (t ) = M N− 1 3Q L u (t ) 3 + D (t ) 3OP u (t ) = M N− u (t ) 3 − D (t ) 3Q u(t ) =

u&&g1

LM 0.7071 0.7071OP ← mode 1 N− 0.2357 0.2357Q ← mode 2 A A

Γ = Γnl =

(e)

Substituting Eq. (e) in Eqs. (b), (c) and (d) gives

3. Determine Γnl = Lnl Mn from Eq. (13.5.3).

u&&g 2

Note that for this symmetric excitation only the second mode is excited.

4. Dynamic displacements. 2

(d)

u ( t ) = ∑ ∑ Γnl φn Dnl ( t )

8. Excitation b.

(a)

ug1 ( t ) = ug 2 ( t ) = ug ( t )

l =1 n =1

For these excitations, Dn1 ( t ) = Dn ( t ) Dn 2 ( t ) = Dn ( t )

Substituting for Γnl and φn gives

LM1 2 OP D (t ) + (− 0.2357) LM− 1 2 OP D (t ) MN1 2 PQ MN 1 2 PQ L1 2 OP D (t ) + 0.2357 LM− 1 2 OP D (t ) + 0.7071 M MN1 2 PQ MN 1 2 PQ L1 2O L 1 6OP D (t ) + LM1 2OP D (t ) = M P D (t ) + M 1 2 N Q N− 1 6Q N1 2Q L− 1 6OP D (t ) + M N 1 6Q

u(t ) = 0.7071

(b) 5. Quasistatic displacements. u s (t ) =

L2 3O

L1 3 O

∑ ι u (t ) = MN1 3PQ u (t ) + MN2 3PQ u (t ) l

(f)

Substituting Eq. (f) in Eqs. (b), (c) and (d) gives

LM1OP D (t ) N1Q L1O u (t ) = M P u (t ) N1Q Lu (t ) + D (t )OP u (t ) = M Nu ( t ) + D ( t ) Q u(t ) =

Note that (1) for this antisymmetric excitation only the first mode is excited;. and (2) because both supports undergo identical motion, the response analysis reduces to the standard case of a single support excitation.

l =1

(c) © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 68 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.32

From Problem 9.20,

From Problem 9.20, the equation of motion is:

Ru&& (t ) UV mu&& + ku = − m 1 2 1 2 S Tu&& (t )W g1

6 EI

k =

(a)

6 EI

kg =

6 EI

ωn =

m L3

6 ( 5 × 108 )

( 0. 2 ) ( 50 × 12 )3

−

(f.2)

LM N

OP Q

(f.3)

u(t )

(g)

From Eq. (13.5.9),

The natural vibration frequency of the beam is 6 EL

6 EI 1 4 1 4 L3 1 4 1 4

k gg =

−

where k =

(f.1)

= 8. 333 rad sec

6 EI

fS ( t ) = ku( t ) =

From Eq. (13.5.12),

Excitation (i) ug1 ( t ) = ugo sin ( 0. 8 ω nt )

ug 2 ( t ) = 0

f S g = k Tg u t + k gg u g

Substituting in Eq. (a) gives the equations of motion: mu&& + ku =

FG mIJ ω u sin ωt H 2K 2

(b)

= =

( m 2 ) ω ugo

1 − ( ω ω n )2

ugo

2 ( ω n ω )2 − 1 ugo

2 (1. 25)2 − 1

Therefore,

sin ωt

fS g = −

sin ωt

(c)

From Problem 9.20, the quasistatic displacement is ugo us (t ) = sin 6. 667t (d) 2 The total displacement is ut ( t ) = u s ( t ) + u ( t ) = 1. 3889 ugo sin 6. 667t (e)

fsg 2

R|S3EI L U|V u(t ) |T3EI L |W 3

(h)

Static analysis of the beam due to the forces in Fig. P13.32a gives the bending moment at mid-span: 3 EI M ( t ) = ( fS g1 ) L = − 2 u ( t ) (i) L

Alternative derivation Because the quasistatic displacements represent rigidbody motion, only the dynamic displacement u causes bending moment. For a simply supported beam,

fS =

6 EI

u fS

To compute the bending moment we determine the equivalent static forces fs , fsg1 and fsg 2 . fsg 1

RSu UV Tu W g2

sin ωt

= 0. 8889 ugo sin 6. 667t

us = 1 2 1 2

The steady state solution of Eq. (b) is u( t ) =

where

where ω = 0. 8 ω n = 6. 667 . 2

R|R− 1 2U (u + u) + L1 4 1 4O Ru UU| SS V MN1 4 1 4PQ STu VWV|W L |TT− 1 2W

6 EI

u L

Fig. P13.32b

L Fig. P13.32a © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 69 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

The bending moment at mid-span is

M (t ) = −

fS ( t ) 2

L = −

3 EI

u(t )

(j)

Substituting Eq. (c) in Eq. (i) or (j) gives M (t ) = −

3 ( 5 × 108 ) ( 50 × 12 )2

0. 8889 ugo sin 6. 667t

M ( t ) = − 3704. 2 sin 6. 667t

(k)

Excitation (ii) ug1 ( t ) = ug 2 ( t ) = ugo sin ( 0. 8 ω nt ) Substituting in Eq. (a) gives

mu&& + ku = m ω 2 ugo sin ωt

(l)

where ω = 0. 8ω n = 6. 667 . The steady state solution of Eq. (l) is u ( t ) = 1. 7778 ugo sin 6. 667t

(m)

The quasistatic displacement is

us (t ) =

1 2

ug1 ( t ) + ug 2 ( t ) = ugo sin 6. 667t (n)

The total displacement is

ut ( t ) = u s ( t ) + u ( t ) = 2. 7778 ugo sin 6. 667t (o) Substituting Eq. (m) in Eq. (i) or (j) gives M (t ) = −7408.3sin u go 6.667t

(p)

Comments (1) If the beam is subjected to motion at only one support, the dynamic response (displacement and bending moment) and quasistatic displacement at the mid-point is one-half of the corresponding values if both supports are excited.

(2) Because the structure is statically determinate, the quasistatic displacements do not cause any bending moment.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 70 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.33

Solve k φ = ω 2 m φ for natural frequencies and modes:

L/2

ω1 = 10.4745

L/2

φ1 =

u2 m EI

L/2 L/2

m EI

ug2

1 ⎧1⎫ ⎨⎬ 2 ⎩1⎭

Γ nl =

ω 2 =13.8564

φ2 =

EI mL3

1 ⎧− 1⎫ ⎨ ⎬ 2 ⎩ 1⎭

Determine Γnl :

EI mL3

ug1

Lnl Mn

for mode n and ground motion at support l, where Lnl = φnT mι and M n = φnT mφn

Preliminaries From Problem 9.21

⎡ 0.4861 0.2210 ⎤ Γ = [ Γ nl ] = ⎢ ⎥ ⎣ −0.3536 −0.3536 ⎦

u2 u6 u5

Then D n1 (t ) = D n (t ) and D n 2 (t ) = D n (t − t ′) .

u1 u3 = ug1 u1 u2

⎡ k tt ⎢ ⎣k θ t

Part a Given u&&g1 (t ) = u&&g (t ) and u&&g 2 (t ) = u&&g (t − t ′) .

u4=ug2

Also, An1 (t ) = An (t ) and An 2 (t ) = An (t − t ′) .

0 − 24 − 24 0 − 6L 0⎤ ⎡ 48 ⎢ ⎥ 0 0 0 − 6L 0⎥ ⎢ 0 48 ⎢ − 24 0 24 0 − 6L 0 0⎥ k tθ ⎤ 4 EI ⎢ ⎥ = − 24 0 0 24 6 L 6 L 0⎥ ⎥ k θθ ⎦ L3 ⎢ ⎢ 0 0 − 6 L 6 L 4 L2 L2 0⎥ ⎢ 2 2 2⎥ 0 6L L 4L L ⎥ ⎢− 6 L − 6 L 2 ⎢ 0 0 0 0 0 L 4 L2 ⎥⎦ ⎣

Condense out the rotational DOF [u 5 , u 6 , u 7 ]T

⎡m ⎤ m=⎢ m⎥⎦ ⎣

u1 u2

ι=

⎧0.3438⎫ ⎧0.1563⎫ +⎨ ⎬ D1 (t ) + ⎨ ⎬ D1 (t − t ′) ⎩0.3438⎭ ⎩0.1563⎭

(a)

2. Determine the bending moments at a, b, c, d and e.

⎡ 176 − 48 − 100 − 76⎤ ⎢ 36 ⎥⎥ 6 EI − 48 176 12 = 3 ⎢ 67 33 ⎥ 7 L ⎢− 100 12 ⎢ ⎥ 33 43 ⎦ ⎣ − 76 36

⎧⎪u1t (t ) ⎫⎪ ⎧0.5938⎫ ⎧ 0.4062⎫ ⎨ t ⎬=⎨ ⎬u g (t ) + ⎨ ⎬u g (t − t ′) 0 . 0938 ⎪⎩u 2 (t )⎪⎭ ⎩ ⎭ ⎩− 0.0938⎭

⎧ 0.2500⎫ ⎧ 0.2500⎫ +⎨ ⎬ D2 (t ) + ⎨ ⎬ D2 (t − t ′) ⎩− 0.2500⎭ ⎩− 0.2500⎭

⎡ k kg ⎤ −1 ⎢ T ⎥ = k tt − k tθ k θθ k θ t k k ⎥ gg ⎦ ⎣⎢ g

1. Determine the total displacements, u1 and u2 of the valves. Substituting for Γ nl , φ n , Dnl(t) and ι l in Eq. (13.5.6) with N = 2 and Ng = 2 gives:

ug1

Recover the rotational DOF [u 5 , u 6 , u 7 ]T ⎧ u1t (t ) ⎫ ⎧u5 (t ) ⎫ ⎪ t ⎪ ⎪ ⎪ u 2 (t ) ⎪ ⎪ −1 ⎬ ⎨u6 (t ) ⎬ = −k θθ k θ t ⎨ ⎪u (t )⎪ ⎪ u g1 (t ) ⎪ ⎩ 7 ⎭ ⎪u g 2 (t )⎪ ⎩ ⎭

ug2

1 ⎡ 266 182 ⎤ ⎢ ⎥ = [ι 1 448 ⎣ 42 − 42⎦

ι2 ]

⎧ u t (t ) ⎫ 4 − 15 11⎤ ⎪ 1t ⎡ 4 ⎪ 3 ⎢ ⎪ u 2 (t ) ⎪ ⎥ =− − 16 − 16 4 12⎥ ⎨ ⎬ (b) 28 L ⎢ u g (t ) ⎪ ⎪ ⎢⎣ 4 4 − 1 − 3⎥⎦ ⎪u g (t − t ′)⎪ ⎩ ⎭

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 71 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Substitute Eq. (a) into Eq. (b) and collect terms

M d (t ) =

⎡⎧ 1.3125⎫ ⎧u 5 (t ) ⎫ ⎧ − 1.3125⎫ ⎪ ⎪ 1 ⎢⎪ ⎪ ⎪ ⎪ ⎨u 6 (t )⎬ = ⎢⎨ 0.7500⎬u g (t ) + ⎨− 0.7500⎬u g (t − t ′) L ⎪u (t )⎪ ⎪ 0.1875⎪ ⎢⎪⎩− 0.1875⎪⎭ ⎩ 7 ⎭ ⎩ ⎭ ⎣ ⎤ ⎧− 0.2946⎫ ⎧− 0.1339⎫ ⎥ ⎪ ⎪ ⎪ ⎪ ′ + ⎨ 1.1786⎬ D1 (t ) + ⎨ 0.5357⎬ D1 (t − t )⎥ ⎪− 0.1339⎪ ⎪− 0.2946⎪ ⎥ ⎩ ⎭ ⎩ ⎭ ⎦

+6.0000 D2 (t ) + 6.0000 D2 (t − t ′) ]

(c)

Compute bending moments in terms of nodal DOF and support displacements

[

4 EI M b (t ) = 2 6u 2t (t ) + Lu 7 (t ) L M c (t ) =

]

(e)

[

4 EI − 6u1t (t ) + 6u g 2 (t ) + Lu5 (t ) + 2 Lu 6 (t ) 2 L

[

4 EI M d (t ) = 2 6u1t (t ) − 6u g1 (t ) + 2 Lu5 (t ) L M e (t ) =

(d)

]

[

4 EI 6u 2t (t ) + 2 Lu 7 (t ) L2

]

(f) (g) (h)

Substituting D n (t ) = An (t ) / ω n2 : EI ⎡ −18u g (t ) + 18u g (t − t ′) ⎤ M a (t ) = ⎦ 4 L2 ⎣ mL + [0.258 A1 (t ) + 0.117 A1 (t − t ′) 4 +0.125 A2 (t ) + 0.125 A2 (t − t ′) ]

EI ⎡ −4.5u g (t ) + 4.5u g (t − t ′) L2 ⎣ +7.0714 D1 (t ) + 3.2143D1 (t − t ′)

+6.0000 D2 (t ) + 6.0000 D2 (t − t ′) ]

M b (t ) =

EI ⎡1.5u g (t ) − 1.5u g (t − t ′) L2 ⎣ +7.0714 D1 (t ) + 3.2143D1 (t − t ′)

−6.0000 D2 (t ) − 6.0000 D2 (t − t ′) ]

M c (t ) =

EI ⎡ −3.0u g (t ) + 3.0u g (t − t ′) L2 ⎣ −6.0000 D2 (t ) − 6.0000 D2 (t − t ′) ]

(i)

EI ⎡6.0u g (t ) − 6.0u g (t − t ′) ⎤ ⎦ 4 L2 ⎣ mL + [ 0.258 A1 (t ) + 0.117 A1 (t − t ′) 4 −0.125 A2 (t ) − 0.125 A2 (t − t ′) ]

(j)

EI ⎡ −12.0u g (t ) + 12.0u g (t − t ′) ⎤ ⎦ 4 L2 ⎣ mL + [ −0.125 A2 (t ) − 0.125 A2 (t − t ′)]⎤⎥ 4 ⎦

(k)

EI ⎡3.0u g (t ) − 3.0u g (t − t ′) ⎤ ⎦ 4 L2 ⎣ mL + [0.215 A1 (t ) + 0.098 A1 (t − t ′) 4 + 0.125 A2 (t ) + 0.125 A2 (t − t ′) ]

(l)

EI ⎡ −3.0u g (t ) + 3.0u g (t − t ′) ⎤ ⎦ 4 L2 ⎣ mL + [ −0.215 A1 (t ) − 0.215 A1 (t − t ′) 4 +0.125 A2 (t ) + 0.125 A2 (t − t ′) ]

(m)

M b (t ) =

M c (t ) =

M d (t ) =

Substitute Eqs. (a) and (c) into Eqs. (d) through (h) and collect terms M a (t ) =

EI ⎡ −0.75u g (t ) + 0.75u g (t − t ′) L2 ⎣ −5.8929 D1 (t ) − 2.6786 D1 (t − t ′)

M e (t ) =

+6.0000 D2 (t ) + 6.0000 D2 (t − t ′) ]

Note that there is no contribution from the second mode to the rotational displacements, an observation that could have been predicted from the second mode shape presented above.

4 EI M a (t ) = 2 6u1t (t ) − 6u g1 (t ) + Lu5 (t ) L

EI ⎡0.75u g (t ) − 0.75u g (t − t ′) L2 ⎣ +5.8929 D1 (t ) + 2.6786 D1 (t − t ′)

M e (t ) =

Notes: (1) When the supports undergo different ground motions, the resulting bending moments in the structure depend, in part, on the relative displacements of the supports. (2) The first mode of vibration does not contribute to the bending moment at c. This result could have been predicted by plotting the first mode shape.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 72 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Part b For identical support motions, u&&g1 (t ) = u&&g 2 (t ) = u&&g (t )

(n)

D n1 (t ) = D n 2 (t ) = D n (t )

(o)

An1 (t ) = An 2 (t ) = An (t )

(p)

Substituting Eqs. (n), (o) and (p) into Eqs. (a) and (i) (m) and collecting terms gives the displacements and bending moments: 1. Total displacements, u1 and u2 of the valves. ⎧⎪u1t (t ) ⎫⎪ ⎧ 1⎫ ⎧0.5⎫ ⎧ 0.5⎫ ⎬ D2 (t ) ⎨ t ⎬ = ⎨ ⎬u g (t ) + ⎨ ⎬ D1 (t ) + ⎨ ⎪⎩u 2 (t )⎪⎭ ⎩0⎭ ⎩0.5⎭ ⎩− 0.5⎭

The first term in Eq. (q) corresponds to the rigid body motion of the structure when the supports are subjected to identical ground motions. The displacements of the valves relative to the ground are given by the last two terms in Eq. (q). 2. Bending moments at a, b, c, d, and e. M a (t ) = ( mL 4 ) [ 0.375 A1 (t ) + 0.25 A2 (t ) ] M b (t ) = ( mL 4 ) [ 0.375 A1 (t ) − 0.25 A2 (t ) ] M c (t ) = ( mL 4 ) [ −0.25 A2 (t ) ] M d (t ) = ( mL 4 ) [ 0.3125 A1 (t ) + 0.25 A2 (t ) ] M e (t ) = ( mL 4 ) [ −0.3125 A1 (t ) + 0.25 A2 (t ) ]

As we might expect, when the supports undergo identical motions, the resulting bending moments in the structure are functions of the modal responses only; they are not affected by the support displacements (c.f. with the results obtained in Part a).

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 73 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.34

Solve k φ = ω 2 m φ for natural frequencies and modes: L/2

L/2

ω1 = 10.4745

u2 m/4 c

m/4 e

m/2 d

L/2

⎧ 1⎫ ⎩0⎭

φ1 = ⎨ ⎬

EI mL3

ω 2 = 15.4919

EI mL3

⎧0⎫ ⎩ 1⎭

φ2 = ⎨ ⎬

EI a

Determine Γnl :

g1 uug1

ug2

L Γ nl = nl Mn

Preliminaries From Problem 9.22:

for support mode n and ground motion at support l, where

u2 u5

L nl = φ nT m ι l and M n = φ nT mφ n

⎡ 0.50 0.50 ⎤ Γ = [ Γ nl ] = ⎢ ⎥ ⎣ 0.30 −0.30 ⎦

Part a

⎡ k tt ⎢k ⎣ θt

Given u&&g1 (t ) = u&&g (t ) and u&&g 2 (t ) = u&&g (t − t ′) . Then

u4 = ug2

u3 = ug1 u3

0 − 24 − 24 6 L 0 6L⎤ ⎡ 48 ⎢ 0 48 0 0 − 6L 0 6 L ⎥⎥ ⎢ ⎢− 24 0 24 0 − 6L 0 0⎥ k tθ ⎤ 4 EI ⎢ ⎥ = 3 ⎢− 24 0 0 24 0 0 − 6L⎥ ⎥ k θθ ⎦ L ⎢ 6L − 6L − 6L 0 4 L2 L2 0⎥ ⎢ 2 2 2⎥ 0 0 0 L 4L L ⎥ ⎢ 0 2 ⎢ 6L 6L 0 − 6L 0 L 4 L2 ⎥⎦ ⎣

Condense out the rotational DOF [u 5 , u 6 , u 7 ]T ⎡ k kg ⎤ −1 ⎢ T ⎥ = k tt − k tθ k θθ k θ t ⎢⎣k g k gg ⎥⎦

and D n 2 (t ) = D n (t − t ′) . An1 (t ) = An (t ) and An 2 (t ) = An (t − t ′) .

D n1 (t ) = D n (t )

Also,

1. Determine the total displacements, u1 and u2. Substituting for Γnl , φn , Dnl(t) and ιl in Eq. (13.5.6) with N = 2 and Ng = 2 gives: ⎧⎪ u1t (t ) ⎫⎪ ⎧0.5⎫ ⎧ 0.5⎫ ⎨ t ⎬ = ⎨ ⎬ u g (t ) + ⎨ ⎬ u g (t − t ′) ⎩−0.3⎭ ⎩⎪u2 (t ) ⎭⎪ ⎩0.3⎭ ⎧0.5⎫ ⎧0.5⎫ + ⎨ ⎬ D1 (t ) + ⎨ ⎬ D1 (t − t ′) 0 ⎩ ⎭ ⎩0 ⎭ ⎧ 0 ⎫ ⎧ 0 ⎫ +⎨ ⎬ D2 (t ) + ⎨ ⎬ D2 (t − t ′) ⎩0.30 ⎭ ⎩−0.30 ⎭

(a)

0 − 64 − 64⎤ ⎡ 128 ⎢ 0 140 − 42 42⎥ 6 EI ⎥ = 3⎢ 7 L ⎢− 64 − 42 67 − 3⎥ ⎥ ⎢ ⎣− 64 42 − 3 67 ⎦ u2 ug1 ug2 u1

⎡m ⎤ m=⎢ ⎥ m 2 ⎣ ⎦ u1

0.5⎤ ⎡ 0.5 ⎥ = [ι1 ⎣ 0.3 − 0.3⎦

ι=⎢

ι2 ]

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 74 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

2. Determine the bending moments at a, b, c, d and e. Recover the rotational DOF [u5 , u 6 , u 7 ]T

M a (t ) =

+1.8 D2 (t ) − 1.8 D2 (t − t ′) ]

⎧ u1t (t ) ⎫ ⎧u5 (t ) ⎫ ⎪ t ⎪ ⎪ ⎪ ⎪ u 2 (t ) ⎪ −1 = − k k u ( t ) ⎨ 6 ⎬ ⎬ θθ θ t ⎨ ⎪u (t )⎪ ⎪ u g1 (t ) ⎪ ⎩ 7 ⎭ ⎪u g 2 (t )⎪ ⎩ ⎭

EI ⎡7.2u g (t ) − 7.2u g (t − t ′) L2 ⎣ +8.5714 D1 (t ) + 8.5714 D1 (t − t ′)

M b (t ) =

−1.8 D2 (t ) + 1.8 D2 (t − t ′)]

⎧ u1t (t ) ⎫ ⎡ 16 − 14 − 15 − 1⎤ ⎪ t ⎪ 3 ⎢ ⎪ u 21 (t ) ⎪ ⎥ =− −8 0 4 4⎥ ⎨ ⎬ 28 L ⎢ u (t ) ⎪ ⎢⎣ 16 − 1 − 15⎥⎦ ⎪ g 14 ⎪u g (t − t ′)⎪ ⎩ ⎭

EI ⎡ 2.4u g (t ) − 2.4u g (t − t ′) L2 ⎣ −5.1429 D1 (t ) − 5.1429 D1 (t − t ′)

M c (t ) =

−3.6 D2 (t ) + 3.6 D2 (t − t ′) ]

(b) Substitute Eq. (a) into Eq. (b) and collect terms

M d (t ) =

⎡ ⎧ 1.2 ⎫ ⎧u5 (t ) ⎫ ⎧−1.2 ⎫ ⎪ ⎪ 1 ⎢⎪ ⎪ ⎪ ⎪ ⎨u6 (t ) ⎬ = ⎢ ⎨ 0 ⎬ u g (t ) + ⎨ 0 ⎬ u g (t − t ′) ⎪ ⎪ L ⎢ ⎪−1.2 ⎪ ⎪ 1.2 ⎪ ⎭ ⎩ ⎭ ⎩u7 (t ) ⎭ ⎣⎩

M e (t ) =

⎧−0.8571⎫ ⎧−0.8571⎫ ⎪ ⎪ ⎪ ⎪ + ⎨ 0.4286 ⎬ D1 (t ) + ⎨ 0.4286 ⎬ D1 (t − t ′) (c) ⎪−0.8571⎪ ⎪−0.8571⎪ ⎩ ⎭ ⎩ ⎭

]

(d)

[

]

(e)

M b (t ) =

4 EI 6u1t (t ) − 6u g 2 (t ) + Lu7 (t ) 2 L

M c (t ) =

4 EI − 6u 2t (t ) + 2 Lu5 (t ) + Lu6 (t ) L2

M d (t ) =

4 EI − 6u 2t (t ) + Lu5 (t ) + 2 Lu6 (t ) L2

M e (t ) =

4 EI 6u 2t (t ) + Lu6 (t ) + 2 Lu7 (t ) L2

[

]

(f)

[

]

(g)

]

EI ⎡ −7.2u g (t ) + 7.2u g (t − t ′) ⎤ ⎦ L2 ⎣ + mL [ 0.0781A1 (t ) + 0.0781A1 (t − t ′)

(i)

EI ⎡7.2u g (t ) − 7.2u g (t − t ′) ⎤ ⎦ L2 ⎣ + mL [ 0.0781A1 (t ) + 0.0781A1 (t − t ′)

(j)

EI ⎡ 2.4u g (t ) − 2.4u g (t − t ′) ⎤ ⎦ L2 ⎣ + mL [ −0.0469 A1 (t ) − 0.0469 A1 (t − t ′)

(k)

M a (t ) =

[

4 EI 6u1t (t ) − 6u g1 (t ) + Lu5 (t ) L2

EI ⎡ −2.4u g (t ) + 2.4u g (t − t ′) L2 ⎣ −5.1429 D1 (t ) − 5.1429 D1 (t − t ′)

Substituting D n (t ) = An (t ) / ω n2 :

Compute bending moments in terms of nodal DOF and support displacements M a (t ) =

EI ⎡ −2.4u g (t ) + 2.4u g (t − t ′) L2 ⎣ −5.4 D2 (t ) + 5.4 D2 (t − t ′) ]

+3.6 D2 (t ) − 3.6 D2 (t − t ′) ]

⎤ ⎧ 0.45 ⎫ ⎧−0.45⎫ ⎪ ⎪ ⎪ ⎪ ⎥ + ⎨ 0 ⎬ D2 (t ) + ⎨ 0 ⎬ D2 (t − t ′) ⎥ ⎪−0.45⎪ ⎪ 0.45 ⎪ ⎥⎦ ⎩ ⎭ ⎩ ⎭

[

EI ⎡ −7.2u g (t ) + 7.2u g (t − t ′) L2 ⎣ +8.5714 D1 (t ) + 8.5714 D1 (t − t ′)

(h)

Substitute Eqs. (a) and (c) into Eqs. (d) through (h) and collect terms

+0.0075 A2 (t ) − 0.0075 A2 (t − t ′) ] M b (t ) =

−0.0075 A2 (t ) + 0.0075 A2 (t − t ′) ]

M c (t ) =

−0.015 A2 (t ) + 0.015 A2 (t − t ′) ]

EI ⎡ −2.4u g (t ) + 2.4u g (t − t ′) ⎤ ⎦ L2 ⎣ + mL [ −0.0225 A2 (t ) + 0.0225 A2 (t − t ′)]

M d (t ) =

EI ⎡ −2.4u g (t ) + 2.4u g (t − t ′) ⎤ ⎦ L2 ⎣ + mL [ −0.0469 A1 (t ) − 0.0469 A1 (t − t ′)

(l)

M e (t ) =

(m)

+0.015 A2 (t ) − 0.015 A2 (t − t ′)]

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 75 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Part b For identical support motions, u&&g1 (t ) = u&&g 2 (t ) = u&&g (t )

(n)

D n1 (t ) = D n 2 (t ) = D n (t )

(o)

An1 (t ) = An 2 (t ) = An (t )

(p)

Substituting Eqs. (n), (o), and (p) into Eqs. (a) and (i) (m) and collecting terms: 1. Total displacements, u 1 and u2. ⎧1⎫ ⎪⎧u1t (t ) ⎪⎫ ⎧ 1⎫ ⎨ t ⎬ = ⎨ ⎬u g (t ) + ⎨ ⎬ D1 (t ) ⎪⎩u 2 (t )⎪⎭ ⎩0⎭ ⎩0⎭ 2. Bending moments at a, b, c, d, and e. M a (t ) = 0.1563mLA1 (t ) M b (t ) = 0.1563mLA1 (t ) M c (t ) = −0.0938mLA1 (t ) M d (t ) = 0 M e (t ) = −0.0938mLA1 (t ) Observe that when the supports undergo identical ground motions, there is no contribution from the second mode to the response of the bridge. Because the masses do not displace horizontally in this mode, it is not excited by identical horizontal motion of the supports.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 76 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Part a Given: u&&ga (t ) = u&&gb (t ) = u&&g (t )

Problem 13.35 a k uθ

uy ux

uga(t)

k ugb(t)

D na (t ) = D nb (t ) = D n (t )

ugd(t)

u&&gc (t ) = u&&gd (t ) = u&&g (t − t ′) ,

then

and

ugc(t)

D nc (t ) = D nd (t ) = D n (t − t ′) ,

also, Ana (t ) = Anb (t ) = An (t )

and

Preliminaries From Problem 9.23:

⎡k ⎢ T ⎣⎢k g

⎡4 ⎢0 ⎢ ⎢0 kg ⎤ ⎢ ⎥ = k ⎢− 1 k gg ⎦⎥ ⎢− 1 ⎢ ⎢− 1 ⎢− 1 ⎣

0 4 0 0 0 0 0

−1 −1 −1 −1 ⎤ 0 0 0 0 ⎥⎥ b 2 b 2 − b 2 − b 2⎥ ⎥ 1 0 0 0 ⎥ b 2 0 1 0 0 ⎥ b 2 ⎥ 0 1 0 ⎥ −b 2 0 0 0 1 ⎥⎦ −b 2 0 0

0 2b 2

⎡ ⎤ ⎢1 ⎥ ⎢ ⎥ m=m⎢ 1 ⎥ ⎢ b2 ⎥ ⎢ 6 ⎥⎦ ⎣ 1 1 1 ⎤ ⎡ 1 1 0 0 0 ⎥⎥ = [ι a ι = ⎢⎢ 0 4 ⎢⎣− 1 b − 1 b − 1 b − 1 b ⎥⎦ k m

⎧ 1⎫ ⎪ ⎪ φ1 = ⎨0⎬ ⎪0⎪ ⎩ ⎭

⎧0⎫ ⎪ ⎪ φ 2 = ⎨ 1⎬ ⎪0⎪ ⎩ ⎭

ω 3= 2

1. Determine the total displacements, u, uy and uθ. Substituting for Γnl , φn , Dnl(t) and ι l in Eq. (13.5.6) with N = 3 and Ng = 4 gives: ⎧u tx (t ) ⎫ ⎡⎧ 1 ⎫ ⎧1⎫ ⎪ t ⎪ 1 ⎢⎪ ⎪ ⎪ ⎪ ⎨u y (t )⎬ = ⎢⎨ 0 ⎬u g (t ) + ⎨ 0 ⎬u g (t − t ′) ⎪1 b⎪ ⎪u t (t )⎪ 2 ⎢⎪− 1 b⎪ ⎩ ⎭ ⎭ ⎣⎩ ⎩ θ ⎭ ⎧1⎫ ⎧1⎫ ⎪ ⎪ ⎪ ⎪ + ⎨0⎬ D1 (t ) + ⎨0⎬ D1 (t − t ′) ⎪0⎪ ⎪0⎪ ⎩ ⎭ ⎩ ⎭

ιb ι c ι d ]

Solve kφ = ω 2 m φ for natural frequencies and mode:

ω1 = ω 2 = 2

Anc (t ) = And (t ) = An (t − t ′) .

3k m

⎧0⎫ ⎪ ⎪ φ 3 = ⎨0⎬ ⎪ 1⎪ ⎩ ⎭

⎤ ⎧0⎫ ⎧ 0 ⎫ ⎥ ⎪ ⎪ ⎪ ⎪ + ⎨ 0 ⎬ D3 (t ) + ⎨ 0 ⎬ D3 (t − t ′)⎥ ⎪1 b⎪ ⎪− 1 b ⎪ ⎥ ⎭ ⎩ ⎭ ⎩ ⎦ Note that there is no response in the y-direction and no contribution to the response from the second mode. 2. Determine the shear forces in the columns. Define nodal displacements at the top of the columns as shown:

u ax

u ay

uby

ubx

Determine Γnl : Γnl =

(a)

L nl Mn

uθ

for support mode n and ground motion at support l, where L nl = φ nT m ι l and M n = φnT mφn

Γ = [Γ nl ] =

1 ⎡ 1⎢ 0 4⎢ ⎢⎣ − 1 b

ucx

u dx

0 −1 b

1⎤ 0 ⎥⎥ − 1 b ⎥⎦

u dy

ucy

Relate these displacements to u xt , u ty and uθt :

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 77 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

b t b uθ uay = u ty − uθt 2 2 b t b t t t ubx = u x − uθ uby = u y + uθ 2 2 b b ucx = u xt + uθt ucy = u ty + uθt 2 2 b b u dx = u tx + uθt udy = u ty − uθt 2 2 The x and y components of the column drifts are: b t Δ ax (t ) = u ax (t ) − u g (t ) = u tx (t ) − uθt (t ) − u g (t ) 2 b t Δ ay (t ) = u ay (t ) = u y (t ) − uθt (t ) 2 b t Δ bx (t ) = ubx (t ) − u g (t ) = u x (t ) − uθt (t ) − u g (t ) 2 b t t Δ by (t ) = uby (t ) = u y (t ) + uθ (t ) 2 b t Δ cx (t ) = ucx (t ) − u g (t − t ′) = u x (t ) + uθt (t ) − u g (t − t ′) 2 b t t Δ cy (t ) = ucy (t ) = u y (t ) + uθ (t ) 2 b t Δ dx (t ) = udy (t ) − u g (t − t ′) = u x (t ) + uθ (t ) − u g (t − t ′) 2 b t Δ dy (t ) = udx (t ) = u y (t ) − uθ (t ) 2 The x and y components of the column shears are: b ⎞ ⎛ Vax (t ) = kΔ ax (t ) = k ⎜ u xt (t ) − uθt (t ) − u g (t ) ⎟ 2 ⎠ ⎝ b ⎞ ⎛ Vay (t ) = kΔ ay (t ) = k ⎜ u ty (t ) − uθt (t ) ⎟ 2 ⎠ ⎝ b t ⎞ ⎛ t Vbx (t ) = kΔ bx (t ) = k ⎜ u x (t ) − uθ (t ) − u g (t ) ⎟ 2 ⎠ ⎝ b ⎞ ⎛ Vby (t ) = kΔ by (t ) = k ⎜ u ty (t ) + uθt (t ) ⎟ 2 ⎠ ⎝ (b) b t ⎞ ⎛ t ′ Vcx (t ) = kΔ cx (t ) = k ⎜ u x (t ) + uθ (t ) − u g (t − t ) ⎟ 2 ⎠ ⎝ b ⎞ ⎛ Vcy (t ) = kΔ cy (t ) = k ⎜ u ty (t ) + uθt (t ) ⎟ 2 ⎠ ⎝ b ⎞ ⎛ Vdx (t ) = kΔ dx (t ) = k ⎜ u xt (t ) + uθt (t ) − u g (t − t ′) ⎟ 2 ⎠ ⎝ b ⎞ ⎛ Vdy (t ) = kΔ dy (t ) = k ⎜ u ty (t ) − uθt (t ) ⎟ 2 ⎠ ⎝ Substitute Eq. (a) into Eq. (b) and collect terms u ax = u tx −

1 1 1 ⎡ 1 Vax (t ) = k ⎢− u g (t ) + u g (t − t ′) + D1 (t ) + D1 (t − t ′) 2 2 4 ⎣ 4 +

1 1 ⎤ D3 (t ) − D3 (t − t ′)⎥ 4 4 ⎦

1 1 1 ⎡1 ⎤ Vay (t ) = k ⎢ u g (t ) − u g (t − t ′) + D3 (t ) − D3 (t − t ′)⎥ 4 4 4 ⎣4 ⎦ 1 1 1 ⎡ 1 Vbx (t ) = k ⎢− u g (t ) + u g (t − t ′) + D1 (t ) + D1 (t − t ′) 4 4 2 2 ⎣ +

1 1 ⎤ D3 (t ) − D3 (t − t ′)⎥ 4 4 ⎦

1 1 1 ⎡ 1 ⎤ Vby (t ) = k ⎢− u g (t ) + u g (t − t ′) − D3 (t ) + D3 (t − t ′)⎥ 4 4 4 4 ⎣ ⎦ 1 1 1 ⎡1 Vcx (t ) = k ⎢ u g (t ) − u g (t − t ′) + D1 (t ) + D1 (t − t ′) 4 2 2 ⎣4 −

1 1 ⎤ D3 (t ) + D3 (t − t ′)⎥ 4 4 ⎦

1 1 1 ⎡ 1 ⎤ Vcy (t ) = k ⎢− u g (t ) + u g (t − t ′) − D3 (t ) + D3 (t − t ′)⎥ 4 4 4 ⎣ 4 ⎦ 1 1 1 ⎡1 Vdx (t ) = k ⎢ u g (t ) − u g (t − t ′) + D1 (t ) + D1 (t − t ′) 4 4 2 2 ⎣ −

1 1 ⎤ D3 (t ) + D3 (t − t ′)⎥ 4 4 ⎦

1 1 1 ⎡1 ⎤ Vdy (t ) = k ⎢ u g (t ) − u g (t − t ′) + D3 (t ) − D3 (t − t ′)⎥ 4 4 4 4 ⎣ ⎦

Substituting D n (t ) = An (t ) / ω n2 : k Vax (t ) = − u g (t ) + u g (t − t ′) 4

[

]

1 1 m⎡ ⎤ A1 (t ) + A1 (t − t ′) + A3 (t ) − A3 (t − t ′)⎥ ⎢ 8⎣ 6 6 ⎦ k m [A3 (t ) − A3 (t − t ′)] Vay (t ) = u g (t ) − u g (t − t ′) + 4 48 k Vbx (t ) = − u g (t ) + u g (t − t ′) 4 +

[

]

[

]

m⎡ 1 1 ⎤ A1 (t ) + A1 (t − t ′) + A3 (t ) − A3 (t − t ′)⎥ ⎢ 8⎣ 6 6 ⎦ k m [− A3 (t ) + A3 (t − t ′)] Vby (t ) = − u g (t ) + u g (t − t ′) + 4 48 k Vcx (t ) = u g (t ) − u g (t − t ′) 4 +

[

]

m⎡ 1 1 ⎤ A1 (t ) + A1 (t − t ′) − A3 (t ) + A3 (t − t ′)⎥ 8 ⎢⎣ 6 6 ⎦

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 78 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

[

]

k m [− A3 (t ) + A3 (t − t ′)] − u g (t ) + u g (t − t ′) + 4 48 k Vdx (t ) = u g (t ) − u g (t − t ′) 4

Vcy (t ) =

[

]

m⎡ 1 1 ⎤ A1 (t ) + A1 (t − t ′) − A3 (t ) + A3 (t − t ′)⎥ 8 ⎢⎣ 6 6 ⎦ k m [A3 (t ) − A3 (t − t ′)] Vdy (t ) = u g (t ) − u g (t − t ′) + 4 48 Part b For identical support motions, u&&ga (t ) = u&&gb (t ) = u&&gc (t ) = u&&gd (t ) = u&&g (t ) +

[

]

(c)

Dna (t ) = Dnb (t ) = Dnc (t ) = Dnd (t ) = Dn (t ) Ana (t ) = Anb (t ) = Anc (t ) = And (t ) = An (t )

(d)

Substituting Eqs. (d) into Eqs. (a) and (c) and collecting terms: 1. The total displacements, ux, uy and uθ. ⎧u tx (t ) ⎫ ⎧1⎫ ⎧1⎫ ⎪ t ⎪ ⎪ ⎪ ⎪ ⎪ ⎨u y (t )⎬ = ⎨0⎬u g (t ) + ⎨0⎬ D1 (t ) ⎪u t (t )⎪ ⎪0⎪ ⎪0⎪ ⎩ ⎭ ⎩ θ ⎭ ⎩ ⎭ 2. The column shear forces. m Vax (t ) = Vbx (t ) = Vcx (t ) = Vdx (t ) = A1 (t ) 4 Vay (t ) = Vby (t ) = Vcy (t ) = Vdy (t ) = 0

If the support motions in the x-direction are identical, only the first mode of vibration is excited, causing slab displacements and column shears in the xdirection only. However, if the supports are subjected to spatially varying ground motions, the third mode of vibration (a purely torsional mode) is also excited and contributes to the response of the structure.; i.e., the structure undergoes coupled lateral-torsional motion. Note however that due to the symmetry of the structure and the direction of the applied ground motions, the second mode of vibration is not excited for either excitation case considered. Consequently, as evidenced by the above results, there is no contribution to the response of the structure from the second mode.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 79 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.36

u ga ( t )

uθ

u gd ( t ) 2k

⎡ ⎢ 0.3333 ⎢ ι = −k −1k g = ⎢ − 0.0588 ⎢ ⎢ − 0.3529 ⎢⎣ b

u gb ( t ) b

ω1 = 5.96

b = 25 ft

ζ n = 0.05, n = 1, 2 and 3

DOFs: 2.

u = u

t θ

Lnl = φ Tn m ι n = 1, 2, 3

M n = φ Tn mφ n = 1

(f)

l = 1, 2

OP PP Q

u&&ga (t ) = u&&g (t ) , u&&gb (t ) = u&&g (t ) , u&&gc (t ) = u&&g (t − t ′)

then

0⎤ −1 − 2⎤ ⎡− 2 − 1 0⎥ ⎥, kg = k ⎢ 0 0 0 0 ⎥⎥ ⎢ 0⎥ ⎢⎣ b 0.5b − 0.5b − b⎥⎦ ⎥ 2⎦

and

Given:

(b)

⎡2 ⎢0 k gg = k ⎢ ⎢0 ⎢ ⎣0

and the influence matrix:

0 ⎧ ⎫ ⎪ ⎪ φ 3 = ⎨− 0.3988⎬ ⎪ 0.0166 ⎪ ⎩ ⎭

5. Determine the response of the nth -mode SDF system to u&&gl (t ) .

and

0 0 1 0

(d)

−0.0413 −0.0206 0.0206 0.0413 . . 0.0805 0.0805 01609 Γ = Γnl = 01609 −0.0628 −0.0314 0.0314 0.0628

The stiffness matrix associated with support DOFs, and the coupling submatrix between structural DOFs and support DOFs (from Problem 9.24).are 0 1 0 0

⎧2.071⎫ ⎪ ⎪ φ2 = ⎨ 0 ⎬ ⎪ 0 ⎪ ⎩ ⎭

LM MM N

(a)

The structural stiffness matrix (from Problem 9.14) is 0 ⎤ ⎡6 0 k = k ⎢⎢0 6 − b ⎥⎥ ⎢⎣0 − b 3b 2 ⎥⎦

ω 3 = 10.90

4. Determine Γnl = Lnl M n .

The mass matrix (from Problem 9.14) is ⎤ ⎥ ⎥ ⎥ 2 b ⎥ 6 ⎥⎦

0.1765 b

(e) T

Set up mass and stiffness matrices.

⎡ ⎢1 m = m⎢ 1 ⎢ ⎢ ⎢⎣

0.1765 b

−

ω 2 = 6.21

⎧ 0 ⎫ ⎪ ⎪ φ1 = ⎨ 2.032 ⎬ ⎪0.0033⎪ ⎩ ⎭

k = 1.5 kip / in

t y

0.0294

From Problem 10.24 ω n in rad/sec are:

m = 0.2331 kip - sec 2 / in

t x

− 0.0294

⎤ 0.3333 ⎥ ⎥ 0.0588 ⎥ ⎥ 0.3529 ⎥ b ⎥⎦

3. Determine the natural vibration frequencies and modes.

d c Establish data (from Problem 9.14).

0.1667

(c)

ugc (t )

0.1667

u&&gd (t ) = u&&g (t − t ′) ,

D na (t ) = D nb (t ) = D n (t )

(h)

D nc (t ) = Dnd (t ) = Dn (t − t ′)

Determine the displacement response. In Eq. (13.5.6) with N = 3 and N g = 4 , substituting

for Γnl , φ n , and D nl (t ) from Eqs. (g), (e), and (h), respectively, and for ι l from Eq. (e) of Problem 9.31 give © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 80 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

⎧u xt (t ) ⎫ ⎧ 0.3333 ⎫ ⎪ t ⎪ ⎪ ⎪ ⎨u y (t ) ⎬ = ⎨− 0.0588⎬u g (t ) + ⎪u t (t ) ⎪ ⎪− 0.0012 ⎪ ⎭ ⎩ θ ⎭ ⎩

⎧ 0.1667 ⎫ ⎪ ⎪ ⎨− 0.0294 ⎬u g (t ) + ⎪− 0.0006 ⎪ ⎩ ⎭

⎧0.1667 ⎫ ⎪ ⎪ ⎨0.0294 ⎬u g (t − t ′) ⎪0.0006 ⎪ ⎩ ⎭

⎧0.3333⎫ ⎪ ⎪ ⎨0.0588⎬u g (t − t ′) ⎪0.0012 ⎪ ⎩ ⎭

⎧ 0 ⎫ ⎪ ⎪ − 0.0413⎨ 2.032 ⎬D1 (t ) ⎪0.0033⎪ ⎩ ⎭

⎧ 0 ⎫ ⎪ ⎪ − 0.0206⎨ 2.032 ⎬ D1 (t ) ⎪0.0033⎪ ⎩ ⎭

⎧ 0 ⎫ ⎪ ⎪ + 0.0206⎨ 2.032 ⎬ D1 (t − t ′) ⎪0.0033⎪ ⎩ ⎭

⎧ 0 ⎫ ⎪ ⎪ + 0.0413⎨ 2.032 ⎬ D1 (t − t ′) ⎪0.0033⎪ ⎩ ⎭

⎧2.071⎫ ⎪ ⎪ + 0.1609 ⎨ 0 ⎬D 2 (t ) ⎪ 0 ⎪ ⎩ ⎭

⎧2.071⎫ ⎪ ⎪ + 0.0805⎨ 0 ⎬ D 2 (t ) ⎪ 0 ⎪ ⎩ ⎭

⎧2.071⎫ ⎪ ⎪ + 0.0805⎨ 0 ⎬ D 2 (t − t ′) ⎪ 0 ⎪ ⎩ ⎭

⎧2.071⎫ ⎪ ⎪ + 0.1609⎨ 0 ⎬ D 2 (t − t ′) ⎪ 0 ⎪ ⎩ ⎭

0 0 0 0 ⎧ ⎧ ⎧ ⎧ ⎫ ⎫ ⎫ ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ′ − + + − 0.0628⎨− 0.3988⎬ D3 (t ) 0.0314 ⎨− 0.3988⎬ D3 (t ) 0.0314⎨− 0.3988⎬ D3 (t − t ) 0.0628⎨− 0.3988⎬ D3 (t − t ′) ⎪ 0.0166 ⎪ ⎪ 0.0166 ⎪ ⎪ 0.0166 ⎪ ⎪ 0.0166 ⎪ ⎩ ⎩ ⎩ ⎩ ⎭ ⎭ ⎭ ⎭

or ⎧u xt (t ) ⎫ ⎧ 0.5 ⎫ ⎧ 0.5 ⎫ 0 ⎧ ⎫ ⎧ 0 ⎫ ⎧0.5⎫ ⎪ t ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨u y (t ) ⎬ = ⎨− 0.0882 ⎬u g (t ) + ⎨0.0882 ⎬u g (t − t ′) + ⎨− 0.1258 ⎬ D1 (t ) + ⎨0.1258⎬ D1 (t − t ′) + ⎨ 0 ⎬D 2 (t ) ⎪u t (t ) ⎪ ⎪− 0.0018⎪ ⎪0.0018⎪ ⎪− 0.0002 ⎪ ⎪0.0002 ⎪ ⎪0⎪ ⎭ ⎩ ⎭ ⎩ ⎭ ⎩ ⎭ ⎩ ⎭ ⎩ θ ⎭ ⎩ 0 ⎧ ⎫ 0 ⎫ ⎧0.5⎫ ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ + ⎨ 0 ⎬D2 (t − t ′) + ⎨ 0.0376 ⎬ D3 (t ) + ⎨− 0.0376⎬ D3 (t − t ′) ⎪ 0.0015 ⎪ ⎪0⎪ ⎪− 0.0015⎪ ⎩ ⎭ ⎩ ⎭ ⎩ ⎭

(i) 7.

Determine shears in columns.

Define nodal displacements at the top of the columns as shown in the accompanying figure uay

ubx

uy uθ

udy d

b t uθ 2 b u byt = u yt + u θt 2 b t t t u cy = u y + u θ 2 b u dyt = u ty − u θt 2 t u ay = u ty −

(h)

The x- and y- components of deformations in columns a, b, c, and d are:

ucy udx

b t uθ 2 b u bxt = u xt − u θt 2 b t t t u cx = u x + u θ 2 b u dxt = u xt + u θt 2 t u ax = u xt −

uby uax

Relate these displacements to u x , u y and u θ :

ucx

b t u θ (t ) − u g (t ) , 2 b Δ ay (t ) = u ayt (t ) = u yt (t ) − u θt (t ) 2

Δ ax (t ) = u axt (t ) − u g (t ) = u xt (t ) −

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 81 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

b t u θ (t ) − u g (t ) , 2 b Δ cx (t ) = u cxt (t ) − u g (t − t ′) = u xt (t ) + u θt (t ) − u g (t − t ′) , 2 b t t Δ dx (t ) = u dx (t ) − u g (t − t ′) = u x (t ) + u θt (t ) − u g (t − t ′) , 2

b t u θ (t ) 2 b Δ cy (t ) = u cyt (t ) = u yt (t ) + u θt (t ) 2 b t t Δ dy (t ) = u dy (t ) = u y (t ) − u θt (t ) 2

Δ bx (t ) = u bxt (t ) − u g (t ) = u xt (t ) −

Δ by (t ) = u byt (t ) = u yt (t ) +

(k)

The x and y components of shear in columns a, b, c and d are: Vax = 2 k Δ ax

Vay = 2 k Δ ay

Vbx = k Δ bx

Vby = k Δ by

Vcx = k Δ cx

V cy = k Δ cy

Vdx = 2 k Δ dx

(l)

Vdy = 2 k Δ dy

Substituting Eqs (i) and (k) into ( l ) we obtain: Vax = −0.69 u g (t ) + 0.69 u g (t − t ′)

Vdy = 0.5454 u g (t ) − 0.5454 u g (t − t ′)

+ 0.09 D1 (t ) + 1.5 D2 (t ) + 0.675 D3 (t ) − 0.09 D1 (t − t ′) + 1.5 D2 (t − t ′) − 0.675 D3 (t − t ′)

− 0.2874 D1 (t ) + 0.7878 D3 (t ) + 0.2874 D1 (t − t ′) − 0.7878 D3 (t − t ′)

Vay = 0.5454 u g (t ) − 0.5454 u g (t − t ′)

or, using the fact that D n (t ) = An (t ) / ω 2n , we can write: Vax = −0.69 u g (t ) + 0.69 u g (t − t ′)

− 0.2874 D1 (t ) + 0.7878 D3 (t ) + 0.2874 D1 (t − t ′) − 0.7878 D3 (t − t ′) Vbx = −0.345 u g (t ) + 0.345 u g (t − t ′) + 0.045 D1 (t ) + 0.75 D 2 (t ) + 0.3375 D3 (t ) − 0.045 D1 (t − t ′) + 0.75 D2 (t − t ′) − 0.3375 D3 (t − t ′) Vby = −0.5373 u g (t ) + 0.5373 u g (t − t ′) − 0.2337 D1 (t ) − 0.2811 D3 (t ) + 0.2337 D1 (t − t ′) + 0.2811 D3 (t − t ′)

+ 0.0025 A1 (t ) + 0.0389 A2 (t ) + 0.0057 A3 (t ) − 0.0025 A1 (t − t ′) + 0.0389 A2 (t − t ′) − 0.0057 A3 (t − t ′) Vay = 0.5454 u g (t ) − 0.5454 u g (t − t ′) − 0.0081 A1 (t ) + 0.0066 A3 (t ) + 0.0081 A1 (t − t ′) − 0.0066 A3 (t − t ′) Vbx = −0.345 u g (t ) + 0.345 u g (t − t ′)

(m)

Vcx = 0.345 u g (t ) − 0.345 u g (t − t ′)

+ 0.0013 A1 (t ) + 0.0194 A2 (t ) + 0.0028 A3 (t ) − 0.0013 A1 (t − t ′) + 0.0194 A2 (t − t ′) − 0.0028 A3 (t − t ′)

− 0.045 D1 (t ) + 0.75 D 2 (t ) − 0.3375 D3 (t ) + 0.045 D1 (t − t ′) + 0.75 D2 (t − t ′) + 0.3375 D3 (t − t ′) Vcy = −0.5373 u g (t ) + 0.5373 u g (t − t ′) − 0.2337 D1 (t ) − 0.2811 D3 (t ) + 0.2337 D1 (t − t ′) + 0.2811 D3 (t − t ′) Vdx = 0.69 u g (t ) − 0.69 u g (t − t ′) − 0.09 D1 (t ) + 1.5 D 2 (t ) − 0.675 D3 (t ) + 0.09 D1 (t − t ′) + 1.5 D2 (t − t ′) + 0.675 D3 (t − t ′)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 82 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Vby = −0.5373 u g (t ) + 0.5373 u g (t − t ′) − 0.0066 A1 (t ) − 0.0024 A3 (t ) + 0.0066 A1 (t − t ′) + 0.0024 A3 (t − t ′)

Comparison

(n)

If the support motions in the x-direction are identical, only the second mode is excited, causing floor displacements and column shears in the xVcx = 0.345 u g (t ) − 0.345 u g (t − t ′) direction only. However, this symmetric plan structure responds to spatially varying ground − 0.0013 A1 (t ) + 0.0194 A2 (t ) − 0.0028 A3 (t ) + 0.0013 A1 (t − t ′) + 0.0194 A2 (t − t ′) + 0.0028 A3 (t − t ′) motion in a much more complicated manner: (1) the three vibration modes are excited; (2) the roof Vcy = −0.5373 u g (t ) + 0.5373 u g (t − t ′) slab displaces in x and y directions and rotates about the vertical axis (implying that the building − 0.0066 A1 (t ) − 0.0024 A3 (t ) undergoes coupled lateral-torsional motion); and + 0.0066 A1 (t − t ′) + 0.0024 A3 (t − t ′) (3) shears in columns develop in both x and y directions. Vdx = 0.69 u g (t ) − 0.69 u g (t − t ′) − 0.0025 A1 (t ) + 0.0389 A2 (t ) − 0.0057 A3 (t ) + 0.0025 A1 (t − t ′) + 0.0389 A2 (t − t ′) + 0.0057 A3 (t − t ′) Vdy = 0.5454 u g (t ) − 0.5454 u g (t − t ′) − 0.0081 A1 (t ) + 0.0066 A3 (t ) + 0.0081 A1 (t − t ′) − 0.0066 A3 (t − t ′)

Identical support motions.

If all the supports undergo identical motions, the motion at the structure is given by Eq.(13.1.15), where Γn is defined by Eq.(13.1.5) with

ι= 1 0 0

For this case: Γ1 = 0 , Γ2 = 0.4828 , Γ3 = 0

Then Eq.(13.1.15) gives ⎧1⎫ ⎪ ⎪ u(t ) = Γ2 φ 2 D2 (t ) = ⎨0⎬ D2 (t ) ⎪0⎪ ⎩ ⎭

(o)

Observe that the first and third modes are not excited by identical support motions in the x-direction; all the response is due to the second mode. Because u y = u θ = 0 , column shears in the ydirection are zero:

Vay = V by = Vcy = V dy = 0 and the shears in the x-direction in the four columns are: Vax (t ) = Vdx (t ) = 2k D2 (t ) = 3.0 D2 (t ) Vbx (t ) = Vcx (t ) =k D2 (t ) = 1.5D2 (t )

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 83 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.37 In the solution of Problem 13.36 the displacements u , u ty and u θt were expressed in terms of u g (t ) and t x

D n (t ) , Eq. (i); and column shears in terms of u g (t )

and An (t ) , Eq. (n). 1. Determine ground displacement. Double integration of u&&g (t ) gives the ground displacement u g (t ) shown in Fig. P13.37a. 2. Determine D n (t ) . The properties of the three, nth-mode SDF systems are ( n = 1, 2, 3 ) ω1 = 5.96 rad/sec ζ1 = 5 % ω2 = 6.21 rad/sec ζ2 = 5 % ω3 = 10.90 rad/sec ζ3 = 5 % The deformation responses D n (t ) of these systems are shown in Fig. P13.37b.

6. Comparison. Identical support motions in the x-direction cause floor displacements and column shears in the x-direction only. However, this symmetric plan structure responds to spatially varying ground motion in a much more complicated manner. The roof slab displaces in both x and y directions and also rotates about the vertical axis (implying that the building undergoes coupled lateral-torsional motion). The peak floor displacements and the peak column shear forces are given in Tables P13.37a and P13.37b, respectively. Table P13.37a Displacement Non-uniform Uniform excitation excitation 9.941 u t , in x0

3. Determine displacement response. Substituting u g (t ) , D n (t ) and t′ in Eq. ( i ) of

u ty 0 , in

0.324

u θt 0 , in

0.379

t y

t x

Problem 13.36 gives the displacements u (t ) , u (t ) and u θt (t ) shown in Fig. P13.37c.

column

4. Determine column shears.

a b c d

From known u g (t ) and D n (t ) , the x- and ycomponents of shear in each column are determined from Eq. (m). The results are shown in Fig. P13.37d.

Table P13.37b column shear, kips non-uniform uniform excitation excitation x y x y 13.684 2.661 13.078 0 6.842 1.214 6.539 0 6.892 1,214 6.539 0 13.783 2.661 13.078 0

5. Identical support motions. Substituting D2 (t ) from Fig. P13.37b into Eq. (o) gives the displacement response of the structure relative to the ground (Fig. P13.37e), and Eq. (p) gives the column shears (Fig. P13.37g). The total displacements (relative displacement plus the ground displacement) is shown in Fig. P13.37f.

Observe that, in the case of different support motions the shear forces in the y-direction are about 20% of those in the x-direction, although the total displacement in the y-direction is negligible compared to the one in x-direction. This is due to the coupled lateral- torsional response of the structure.

0.4

u&&g (t ) g

0.319g

-0.4

2 0

u g (t ) in 8.40

-10 0

Time Sec © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. publication Fig.This P13.37a is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 84 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

D1(t), in

4.297

0 -6

D2(t), in

4.359

0 -6

D3(t), in

6 2.617 0 -6 0

Time, sec Fig. P13.37b utx(t), in

4 0

9.941

-10

uty(t), in

0.324

-10

(b/2) utθ (t), i

4 0

0.379

-10 0

Time, Sec Fig. P13.37c

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 85 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Vax(t), kips

13.684

-16 Vay(t), kips

16 0

2.661

-16

Vbx(t), kips

6.842 0

-16

Vby(t), kips

16 0

1.214

-16

Vcx(t), kips

16 0 6.892

-16

Vcy(t), kips

16 0

1.214

-16

Vdx(t), kips

16 0

13.783

-16

Vdy(t), kips

16 0

2.661

-16 0

Time, sec Fig. P13.37d © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 86 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

ux(t), in

10 4.359 0 -10

uy(t), in

10 0 -10

(b/2) uθ (t), i

10 0 -10 0

Time, sec Fig. P13.37e

utx(t), in

4 0

10.077

-10

uty(t), in

4 0

-10

(b/2) utθ (t), i

4 0

-10 0

Time, Sec Fig. P13.37f

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 87 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Vax(t), kips

13.078

0 -16

Vay(t), kips

16 0 -16

Vbx(t), kips

6.539 0 -16

Vby(t), kips

16 0 -16

Vcx(t), kips

6.539 0

Vcy(t), kips

-16

16 0 -16

Vdx(t), kips

13.078

0 -16

Vdy(t), kips

16 0 -16 0

Time, sec Fig. 13.37g © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 88 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.38

4. Determine displacements.

Part a From Problem 9.25,

l =1

l =1 n =1

ut ( t ) = ∑ ι l ugl ( t ) + ∑ ∑ Γnl φn Dnl ( t )

LM0.9359 0.7701OP kips in. . N 0.7701 15088 Q L3.624 OP kip − sec in. m = M 1812 . Q N

Substituting Eqs. (a), (b) and (c) in Eq. (d) gives

k = 104

|RSu (t ) |UV = RS 0.6035UV u (t ) + RS0.3965UV u (t − t ′) . |Tu (t )|W T− 0.2143W T12143 W RS 0.4830UV D (t ) + 11698 . T− 0.2920W R 0.4830UV D (t − t ′) +0.0515 S T− 0.2920W RS0.2065UV D (t ) + 01864 . T0.6831W RS0.2065UV D (t − t ′) + 17997 . (e) T0.6831W t 1 t 2

ι 1 = 0.6035 − 0.2143 ι 2 = 0.3965 12143 .

(d)

1. Evaluate natural frequencies and modes.

ω1 = 36. 02

ω 2 = 98. 04

R 0.4830UV φ = S T− 0.2920W

R0.2065UV φ = S T0.6831W

0.2920

The displacement at the top of the tower is

0.6831

u2t ( t ) = − 0. 2143 ug ( t ) + 1. 2143 ug ( t − t ′ ) 0.4830

− 0. 3416 D1 ( t ) − 0. 0150 D1 ( t − t ′ )

0.2065

+ 0.1273 D2 ( t ) + 1. 2294 D2 ( t − t ′ )

(f)

5. Compute equivalent static forces. 2

2. Determine Γnl = Lnl Mn .

(g)

l =1 n =1

Substituting for m and φn , Eq. (b) for Γnl , and Eq. (c) for Anl ( t ) gives

Lnl = φ Tn m ι l Mn = 1

ι 1 = 0.6035 − 0.2143 T . ι 2 = 0.3965 12143 Γ = Γnl

fS ( t ) = ∑ ∑ Γnl m φ n Anl ( t )

(a)

. 0.0515O ← mode 1 L 11698 = M PQ ← mode 2 . 17997 . N01864 (b) A A u&&g1

RS 2.0476UV A (t ) + RS 0.0901UV A (t − t ′) T− 0.6189W T− 0.0272W . . R 01395 UV A (t ) + RS 13469 UV A (t − t ′) + S T0.2307W T2.2277W

f S (t ) = −

u&&g 2

(h)

6. Compute equivalent static support forces.

f S g (t ) = k Tg u(t ) + k gg + k Tg ι u g (t )

(i)

3. Determine response of the nth-mode SDF system to u&&gl ( t ) . ug1 ( t ) = ug ( t )

ug 2 ( t ) = ug ( t − t ′ )

Dn1 ( t ) = Dn ( t )

Dn 2 ( t ) = Dn ( t − t ′ ) (c)

An1 ( t ) = An ( t )

An 2 ( t ) = An ( t − t ′ )

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 88 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Substituting for k g and k gg from Problem 9.25 and u( t ) from the second half of Eq. (d), and ι from Eq. (a) gives . RS13686 UV A (t ) − RS0.0603UV A (t − t ′) T0.0603W T0.0027W R0.0348UV A (t ) − RS0.3355UV A (t − t ′) − S T0.3355W T3.2388W L 1 − 1OP RS u (t ) UV + 159.46 M N− 1 1Q Tu (t − t ′)W

f S g (t ) = −

Substituting φn and Γn in Eq. (l) gives u(t ) = 12212 .

RS 0.4830UV D (t ) + 19861 RS0.2065UV D (t ) . T− 0.2920W T0.6831W 1

(m)

The total displacements are

u t (t ) = u g (t )

(j)

RS1UV + u(t ) T1W

(n)

In particular, the displacement at top of the tower is

7. Compute internal forces.

200'

u2t ( t ) = ug ( t ) − 0. 3567 D1 ( t ) + 1. 3566 D2 ( t ) (o)

fs2(t) fsg2(t)

The equivalent static forces are

fs1(t)

fS ( t ) = ∑ Γn m φn An ( t )

175'

(p)

n =1

100'

Substituting for Γn , m , and φn gives

fsg1(t)

f S (t ) =

Observe that at each time instant, the equivalent static forces shown in the accompanying figure and defined by Eqs. (h) and (j) are in equilibrium. By statics, the shear at the base is Vb ( t ) = fS g1 ( t ) kips

(k.1)

The moment at the base is Mb ( t ) = 200 fS 2 ( t ) + 175 fS g 2 ( t ) + 100 fS1 ( t ) kip - ft

(k.2) and the

. RS 2.1375UV A (t ) + RS 14863 UV A (t ) T− 0.6461W T2.4583W 1

(q)

The equivalent static support forces are obtained by substituting An ( t ) = An ( t − t ′ ) and ug ( t ) = ug ( t − t ′ ) in Eq. (j): f S g (t ) = −

. RS14289 UV A (t ) − RS0.3703UV A (t ) (r) . T0.0630W T35743 W 1

These support forces could also be obtained by static analysis of the structure subjected to fS ( t ) . With fS and fS g given by Eqs. (q) and (r), the internal forces are given by Eqs. (k).

axial force in the bridge = fS g2 ( t ) kips (k.3) Part b If both supports undergo identical motion ug ( t ) , the relative motion of the structure is

Comments

If the support motions are different, quasistatic support forces are developed. These disappear if the support motions are identical.

u( t ) = ∑ Γn φn Dn ( t )

(l)

n =1

where Γn = φTn m 1

Substituting φn and m gives Γ1 = 1. 2212

Γ2 = 1. 9861

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 89 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.39

Mb1 = 0. 443 mh A1 = 480. 2 kip − ft

Initial calculations

Second mode: Ma2 = 0. 0473 mh A2 = 51. 27 kip - ft Mb2 = 0. 0441 mh A2 = 47. 80 kip − ft

Substituting values of E = 29, 000 ksi , I = 727 in. , m = 100 kips g , and h = 12 ft in the solution for Problem 13.4 gives: 4

ω1 = 2. 407

ω 2 = 7.193 2π

T1 =

EI mh3

= 12. 57 rad sec

= 37. 55 rad sec 2π

= 0.167 sec 12. 57 37. 55 Corresponding to these periods, the spectral ordinates are 1 A1 = ( 2. 71g ) = 348. 7 in. sec2 3 A2 =

1 3

T2 =

( 2. 71g ) = 348. 7 in. sec2

D1 =

= 0. 5 sec

ω12

= 2. 208 in.

D2 =

ω 22

= 0. 247 in.

Part a Substituting the above D1 and D2 for D1 ( t ) and D2 ( t ) in Eqs. (a) and (c) of the solution to Problem 13.4 gives the peak values of floor displacements due to each of the two modes: . RSu UV = RS0.647UV (2.208) = RS1429 UV in. u 1341 . 2 . 961 T W T W T W RSu UV = RS 0.353UV (0.247) = RS 0.087UV in. Tu W T− 0.341W T− 0.084W 1

2 1

Using the SRSS rule gives an estimate of the total bending moments: Ma ≈

( 234.1)2 + ( 51. 27)2 = 239. 69 kip − ft

Mb ≈

( 480. 2 )2 + ( 47. 80 )2 = 482. 58 kip − ft

Bending moments in a second-floor beam [from Eqs. (m) and (n) in the solution to Problem 13.4] are determined as follows.

First mode: Ma1 = − 0. 211 mh A1 = − 228.7 kip - ft Mb1 = − 0. 211 mh A1 = − 228. 7 kip − ft Second mode: Ma2 = 0. 0332 mh A2 = 35. 99 kip - ft Mb2 = 0. 0332 mh A2 = 35. 99 kip - ft Using the SRSS rule gives the total values: Ma ≈

( − 228. 7)2 + ( 35. 99 )2 = 231. 53 kip − ft

Mb ≈

( − 228. 7)2 + ( 35. 99 )2 = 231. 53 kip − ft

Table P13.39 summarizes the bending moments in all elements which were calculated similarly. Table P13.39: Peak bending moments, kip-ft

Element

Node

Mode 1

Mode 2

Total

2 2

Beam 1

3 4 5 6 3 1 5 3

– 369 – 369 229 229 234 480 229 136

5 5 36 36 51 48 – 36 – 56

369 369 232 232 240 483 232 147

Using the SRSS rule gives an estimate of the total displacements: (1. 429 )2 + ( 0. 087 )2 = 1. 43 in.

u1 ≈

u2 ≈

( 2. 961) + ( − 0. 084 )

Beam 2 Column 3

= 2. 96 in.

Part b Bending moments in a first-story column [from Eqs. (k) and (l) in the solution to Problem 13.4] are determined as follows.

Column 5

First mode: Ma1 = 0. 216 mh A1 = 0. 216 (100 386 ) (12 × 12 ) 348. 7 = 2810 kip − in. = 234.1 kip - ft © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 90 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

5 5

6 1

3 3 1

4 4 2

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 91 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.40

2. Peak responses due to mode 2.

Initial calculations From Problem 13.2, the modal properties are

φ1 =

. U RS1389 V . W T1965

φ2 =

Γ1 = 0. 614

. U RS 1389 V . W T− 1965

u 2 = Γ2 φ 2 D2 = 0105 .

Equivalent static forces: . )U RS m φ UV A = 0105 100 R 1 (1389 . V (0.684g) S . )W g T0.5 ( − 1965 Tm φ W R 9.98UV kips = S T− 7.06W

f 2 = Γ2

Γ2 = 0.105

Part a: Spectral ordinates

From Problem 13.2, the peak values of Dn ( t ) and An ( t ) are D1 = 0. 797 in.

D2 = 0.118 in.

A1 = 0. 791g

A2 = 0. 684 g

2 21

12' 9.98 12'

. U RS1389 R0.680UV in. (0.797) = S V . W T1965 T0.962W

. )U R m φ UV A = 0.614 100 RS 1 (1389 f = Γ S V (0.791g) . )W g T0.5 (1965 Tm φ W R67.46UV kips = S T47.72W 1

2 22

7.06

Equivalent static forces: 1 11

1 12

Story shears:

Part b 1. Peak responses due to mode 1 Displacements: u1 = Γ1 φ 1 D1 = 0.614

. U RS 1389 R 0.0172UV in. (0118 . ) = S V . W T− 1965 T− 0.0243W

Story shears: 47.72 12' 67.46 12'

V22 = − 7. 06 kips V12 = 9. 98 − 7. 06 = 2. 92 kips

Overturning moments: M12 = − 7. 06 (12 ) = − 84. 72 kip - ft Mb 2 = 9. 98 (12 ) + ( − 7. 06 ) ( 24 ) = − 49. 68 kip - ft

The above-determined responses due to each mode should be identical to those determined in Problem 13.2 by RHA; the slight differences are due to numerical truncation errors in the computations for this problem whereas more significant digits were used in the computer work for Problem 13.2. Part c

The peak modal responses are combined by the SRSS rule: V21 = 47.72 kips V11 = 67.46 + 47.72 = 11518 . kips

r =

r12 + r22

The resulting RSA estimate and the RHA results (from Problem 13.2) are presented in Table P13.40.

Overturning moments: These moments are denoted by M1 at the first floor level and Mb at the base. M11 = 47. 72 (12 ) = 572. 64 kip - ft Mb1 = 67. 46 (12 ) + 47. 72 ( 24 ) = 1954. 8 kip - ft

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 92 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Table P13.40 Response

RSA

RHA

Error, %

u1 , in.

0.680

0.679

0.15

u2 , in.

0.962

0.964

0.21

Vb , kips

115.22

115.11

0.10

V2 , kips

48.24

49.56

2.66

Mb , kip-ft

1955.43

1959.25

0.20

M1 , kip-ft

578.87

594.65

2.65

Part d: Comments For this problem, the RSA method gives results that are very close to the RHA results. The errors are small for all response quantities because the first mode response is dominant.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 93 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.41

The bending moments in a first-story column due to the first mode are

Initial calculations Substituting values of E = 29, 000 ksi , I = 1400 in.4 , m = 100 kips g and h = 12 ft in the solution for Problem 13.11 gives:

ω1 = 10. 57

ω 2 = 34. 56

ω 3 = 59. 42

T1 = 0. 595 sec T2 = 0.182 sec T3 = 0.106 sec

From the design spectrum of Fig. 6.9.5, scaled to ugo = (1 3) g , the spectral ordinates are D1 = 3.13 in.

D2 = 0. 292 in. D3 = 0. 088 in.

= 0. 903

A3 g

= 0. 803

Part a: Floor displacements

Substituting numerical values for Γn and φn from Problem 13.11 and Dn values above in Eq. (13.8.1a) gives the peak displacements due to each of the three modes: . U| R|0.3156U| R|1335 u = Γ φ D = 13513 . 0 . 7451 ( 313 . ) = 3152 . S| V| S| V| in. T 1 W T4.230W 1

. U R|− 0.7409U| R| 0110 | u = Γ φ D = − 0.5083 S− 0.3572V (0.292) = S 0.053V in. |T 1 |W |T− 0148 . |W . R| 12546 U| R| 0.017U| u = Γ φ D = 01569 . . V| (0.088) = S|− 0.017V| in. S|− 12024 1 T W T 0.014W 2

Combining modal displacements by the SRSS rule gives u1 ≈

(1. 335)2 + ( 0.110 )2 + ( 0. 017)2 = 1. 340 in.

u2 ≈

( 3.152 )2 + ( 0. 053)2 + ( − 0. 017 )2 = 3.152 in.

u3 ≈

( 4. 230 )2 + ( − 0.148)2 + ( 0. 014 )2 = 4. 233 in.

Observe that the first mode contributes essentially the entire floor displacements.

M a1 = 0.7526 mh A1 = 0.7526 (100 g) (12) 0.903g = 815.52 kip - ft Mb1 = 0. 3014 mh A1 = 0. 3014 (100 g ) (12 ) 0. 903g = 326. 60 kip - ft

These bending moments due to the second mode are Ma2 = 0. 08381 mh A2 = 0. 08381 (100 g ) (12 ) 0. 903g = 90. 82 kip - ft Mb2 = 0. 06823 mh A2 = 0. 06823 (100 g ) (12 ) 0. 903g = 73. 93 kip - ft

Due to the third mode the bending moments are Ma3 = 0. 02030 mh A3 = 0. 02030 (100 g ) (12 ) 0. 803g = 19. 56 kip - ft Mb3 = 0. 02302 mh A3 = 0. 02302 (100 g ) (12 ) 0. 803g = 22.18 kip - ft

Combining modal responses by the SRSS rule gives Ma ≈

( 815. 52 )2 + ( 90. 82 )2 + (19. 56 )2

= 820. 80 kip − ft Mb ≈

( 326. 60 )2 + ( 73. 93)2 + ( 22.18)2

= 335. 60 kip − ft

The bending moments in the second-story beam due to each of the three modes are computed similarly to obtain Ma1 = Mb1 = − 586. 66 kip − ft Ma 2 = Mb 2 =

61. 44 kip − ft

Ma3 = Mb3 =

2. 68 kip − ft

Combining modal responses by the SRSS rule gives Ma = Mb ≈

( − 586. 66 )2 + ( 61. 44 )2 + ( 2. 68)2

= 589. 87 kip − ft

Part b: Element forces

To compute the element forces use the results from Problem 13.11 and replace An ( t ) by the spectral values An .

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 94 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.42

The bending moments in a first-story column due to the first mode are

Initial calculations 4

Substituting values of E = 29,000 ksi, I = 1400 in , m = 100 kips/g and h = 12 ft in the solution for Problem 13.12 gives ω 1 = 8.67 rad / sec ω 2 = 30.27 rad / sec ω 3 = 57.25 rad / sec T2 = 0.207 sec

T1 = 0.725 sec

T3 = 0.109 sec

From the design spectrum of Fig. 6.9.5, scaled to u&&g 0 = (1 / 3) g , the spectral ordinates are D1 = 4.25 in.

D2 = 0.381 in.

D3 = 0.096 in.

A1 = 0.828 g

A2 = 0.903 g

A3 = 0.819 g

Part a: Floor displacements Substituting numerical values for Γn and φ n from Problem 13.12 and Dn values above in Eq. (13.8.1a) gives the peak displacements due to each of the three modes: ⎧ 0.273 ⎫ ⎧ 1.608 ⎫ ⎪ ⎪ ⎪ ⎪ u1 = Γ1φ1 D1 =1.386⎨ 0.698 ⎬ (4.25) = ⎨ 4.112 ⎬ in. ⎪ 1 ⎪ ⎪ 5.89 ⎪ ⎩ ⎭ ⎩ ⎭

⎧− 0.706⎫ ⎧ 0.146 ⎫ ⎪ ⎪ ⎪ ⎪ u 2 = Γ2φ2 D2 = −0.542 ⎨ − 0.441⎬ (0.381) = ⎨ 0.091 ⎬ in. ⎪ 1 ⎪ ⎪− 0.206⎪ ⎩ ⎭ ⎩ ⎭ ⎧ 1.529 ⎫ ⎧ 0.022 ⎫ ⎪ ⎪ ⎪ ⎪ u 3 = Γ3φ 3 D3 = 0.156 ⎨− 1.315 ⎬ (0.096) = ⎨− 0.019⎬ in. ⎪ 1 ⎪ ⎪ 0.015 ⎪ ⎩ ⎭ ⎩ ⎭

Combining modal displacements by the SRSS rule gives 2

u1 ≈ (1608 . ) + (0146 . ) + (0.022) = 1615 . u2 ≈ (4.112) 2 + (0.091) 2 + ( −0.019) 2 = 4.117

M a 1 = 0.697 mhA1 = 0.697 (100 / g )(12) 0.828 g = 692.5 kip − ft M b1 = 0.151mhA1 = 0.151(100 / g )(12 ) 0.828 g = 150.0 kip − ft

The bending moments due to the second mode are M

= 0.102mhA2 = 0.102(100 / g )(12 ) 0.903 g = 110.5 kip − ft

M b 2 = 0.073mhA2 = 0.073(100 / g )(12 ) 0.903 g = 79.10 kip − ft

Due to the third mode the bending moments are M a 3 = 0.026mhA3 = 0.026(100 / g )(12) 0.819 g = 25.55 kip − ft M b 3 = 0.029mhA3 = 0.029(100 / g )(12) 0.819 g = 28.50 kip − ft

Combining modal responses by the SRSS rule gives M a ≈ ( 692.5) 2 + (110.5) 2 + ( 25.55) 2 = 7018 . kip − ft

M b ≈ (150.0) 2 + ( 79.10) 2 + ( 28.5) 2 = 169.6 kip − ft The bending moments in the second-story beam due to each of the three modes are computed similarly to obtain M a1 = M b1 = −973.7 kip − ft M a 2 = M b 2 = 44.43 kip − ft M a 3 = M b 3 = −9.63 kip − ft

Combining modal displacements by the SRSS rule gives M a = M b ≈ ( −973.7 )2 + ( 44.43)2 + ( −9.63)2 = 974.8 kip − ft

u3 ≈ (589 . ) 2 + ( −0.206) 2 + (0.015) 2 = 5893 .

Observe that the first mode contributes essentially the entire floor displacements. Part b: Element forces

To compute the element forces use the results from Problem 13.12 and replace An (t ) by the spectral values An . © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 95 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.43

The bending moments in a first-story column due to the first mode are

Initial calculations 4

Substituting values of E = 29,000 ksi, I = 1400 in , m = 100 kips/g and h = 12 ft in the solution for Problem 13.13 gives ω 1 = 9.63 rad / sec ω 2 = 25.46 rad / sec ω 3 = 47.54 rad / sec T1 = 0.653 sec

T2 = 0.247 sec

T3 = 0.132 sec

From the design spectrum of Fig. 6.9.5, scaled to ug 0 =(1/ 3) g , the spectral ordinates are D1 = 3.761 in.

D2 = 0.538 in.

D3 = 0.154 in.

A1 = 0.903 g

A2 = 0.903 g

A3 = 0.903 g

Substituting numerical values for Γn and φ n from Problem 13.13 and Dn values above in Eq. (13.8.1a) gives the peak displacements due to each of the three modes:

R| 0.234 U| R| 1.255 U| u = Γ φ D = 1.426S 0.639 V( 3.761) = S 3.427 V in. |T 1 |W |T 5.363 |W R|−0.512U| R| 0.15 U| u = Γ φ D = −0.543S −0.591V( 0.538) = S 0.173 V in. |T 1 |W |T−0.292|W R| 3.324 U| R| 0.058 U| u = Γ φ D = 0.114 S−2.032 V( 0.154) = S−0.036V in. |T 1 |W |T 0.018 |W 2

1 1 1

2 2

3 3

Combining modal displacements by the SRSS rule gives 2

= 1685 kip − ft M b1 = 0.243mhA1 = 0.243(100 / g )(12 ) 0.903 g = 263.3 kip − ft

The bending moments due to the second mode are M

= 0.1mhA2 = 0.1(100 / g )(12) 0.903g = 108.4 kip − ft

M b 2 = 0.064 mhA2 = 0.064 (100 / g )(12 ) 0.903 g = 69.35 kip − ft

Due to the third mode the bending moments are

Part a: Floor displacements

M a1 = 1555 . mhA1 = 1555 . (100 / g )(12 ) 0.903g

u1 ≈ (1.255) + ( 0.15) +( 0.058) = 1.265 u2 ≈ ( 3.427) 2 +( 0.173)2 +( −0.036) 2 = 3.432 u3 ≈ (5.363) 2 +( −0.292) 2 + ( 0.018) 2 = 5.371

M a 3 = 0.041mhA3 = 0.041(100 / g )(12) 0.903g = 44.43 kip − ft M b 3 = 0.050mhA3 = 0.050(100 / g )(12) 0.903g = 54.18 kip − ft

Combining modal responses by the SRSS rule gives M a ≈ (1685) 2 + (108.4) 2 + ( 44.43) 2 = 1689.1 kip − ft

M b ≈ ( 263.3) 2 + ( 69.35) 2 + (54.18) 2 = 277.6 kip − ft The bending moments in the second-story beam due to each of the three modes are computed similarly to obtain M a1 = M b1 = −715.2 kip − ft M a 2 = M b 2 = 54.18 kip − ft M a 3 = M b 3 = −22.76 kip − ft

Combining modal displacements by the SRSS rule gives M a = M b ≈ ( −715.2 )2 + ( 5418 . )2 + ( −22.76 )2 = 717.6 kip − ft

Observe that the first mode contributes essentially the entire floor displacements. Part b: Element forces

To compute the element forces use the results from Problem 13.13 and replace An (t ) by the spectral values An . © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 96 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.44

The bending moments in a first-story column due to the first mode are

Initial calculations Substituting values of E = 29,000 ksi, I = 1400 in 4 , m = 100 kips/g and h = 12 ft in the solution for Problem 13.14 gives ω n in rad/sec and Tn in sec: ω1 = 7.558 ω 2 = 22.320 ω 3 = 45.740 T1 = 0.831

T2 = 0.282

T3 = 0.137

From the design spectrum of Fig. 6.9.5, scaled to u&&g 0 = (1 / 3) g , the spectral ordinates are D1 = 4.877 in.

D2 = 0.700 in.

D3 = 0.167 in.

A1 = 0.722 g

A2 = 0.903 g

A3 = 0.903 g

Part a: Floor displacements.

Substituting numerical values for Γn and φn from Problem 13.14 and Dn values above in Eq. (13.8.1a) gives the peak displacements due to each of the three modes:

u 1 = Γ1φ1 D1 =

⎧ 0.200 ⎫ ⎧ 1.410 ⎫ ⎪ ⎪ ⎪ ⎪ 1.447⎨ 0.597 ⎬(4.877) = ⎨ 4.211⎬ in. ⎪ 1 ⎪ ⎪ 7.058 ⎪ ⎩ ⎭ ⎩ ⎭

⎧− 0.545⎫ ⎧ 0.218 ⎫ ⎪ ⎪ ⎪ ⎪ u 2 = Γ2φ 2 D 2 = −0.571⎨− 0.656⎬(0.700) = ⎨ 0.262 ⎬ in. ⎪ 1 ⎪ ⎪− 0.400⎪ ⎩ ⎭ ⎩ ⎭ u 3 = Γ3φ 3 D3

⎧ 3.220 ⎫ ⎧ 0.067 ⎫ ⎪ ⎪ ⎪ ⎪ = 0.124⎨− 1.916 ⎬(0.167) = ⎨− 0.040⎬ in. ⎪ 1 ⎪ ⎪ 0.021 ⎪ ⎩ ⎭ ⎩ ⎭

Combining modal displacements by the SRSS rule gives

M a1 = 0.844mh A1 = 0.844(100 / g )(12)0.722 g = 730.81 kip − ft M b1 = 0.094mh A1 = 0.094(100 / g )(12)0.722 g = 81.51 kip − ft

Bending moments due to the second mode are M a 2 = 0.132mh A 2 = 0.132(100 / g )(12)0.903 g = 143.47 kip − ft M b 2 = 0.068mh A 2 = 0.068(100 / g )(12)0.903 g = 73.50 kip − ft

Bending moments due to the third mode are M a 3 = 0.057 mh A 3 = 0.057(100 / g )(12)0.903 g = 62.15 kip − ft M b3 = 0.055mh A 3 = 0.055(100 / g )(12)0.903 g = 59.13 kip − ft

Combining modal responses by the SRSS rule gives Ma ≈

(730.81)2 + (143.47 )2 + (62.15)2 = 747.35 kip − ft

Mb ≈

(81.51)2 + (73.50)2 + (59.13)2

= 124.67 kip − ft

The bending moments in the second-story beam due to each of the three modes are computed similarly to obtain M a1 = M b1 = −417.50 kip − ft M a 2 = M b2 =

32.65 kip − ft

M a3 = M b3 =

12.36 kip − ft

Combining modal displacements by the SRSS rule gives Ma ≈

(− 417.50)2 + (32.65)2 + (12.36)2 = 418.96 kip − ft

u1 ≈ (1.410) 2 + ( 0.218) 2 + ( 0.067) 2 = 1.428 u 2 ≈ (4.211) 2 + ( 0.262) 2 + (−0.040) 2 = 4.219 u 3 ≈ (7.058) 2 + (−0.400) 2 + ( 0.021) 2 = 7.069

Observe that the first mode contributes essentially the entire floor displacements. Part b: Element forces To compute the element forces use the results from Problem 13.14 and replace An (t ) by the spectral values An . © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 97 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.45

These forces are shown in the following figure.

Initial calculations From Problem 13.7, the modal properties are

R| 0.5 U| φ = S0.866V |T 1 |W

R|−1U| φ = S 0V |T 1|W

Γ1 = 1. 244

Γ2 = − 0. 3333 Γ3 = 0. 0893

R| 0.5 U| φ = S− 0.866V |T 1 |W 3

Part a: Spectral ordinates From Problem 13.7, the peak values of Dn ( t ) and An ( t ) are D1 = 0. 8859 in.

D2 = 0.1096 in.

D3 = 0. 0498 in.

A1 = 0. 7746 g

A2 = 0. 7153g A3 = 0. 6065g

Part b: Peak modal responses Substituting numerical values for Γn , φn , and Dn in Eq. (13.8.1a) gives the peak displacements due to each of the three modes:

R| 0.5 U| R|0.551U| u = 1244 . S|0.866V| (0.8859) = S|0.954V| in. . W T 1 W T1102 R|− 1U| R| 0.037 U| u = − 0.3333 S 0V (01096 . ) = S 0 V in. |T 1|W |T− 0.037|W R| 0.5 U| R| 0.0022U| u = 0.0893 S− 0.866V (0.0498) = S− 0.0039 V in. |T 1 |W |T 0.0044|W 1

Substituting Γn , m j , φn and An in Eq. (13.8.2) gives the equivalent static forces due to each of the three modes: . U R|4818 R 1 (0.5) U| F 100 I | f = 1244 . GH g JK S|1 (0.866)V| (0.7746g) = S|83.45|V| kips . W T4818 T 0.5 (1) W 1

. U FG 100IJ R|S11((−01))U|V (0.7153g) = R|S 2384 | 0 V kips H g K |T0.5 (1) |W |T− 1192 . |W 5) U R| 2.71U| F 100 IJ R|S1 (1−(00..866 | f = 0.0893 G ) V (0.6065g) = S− 4.69 V kips H g K |T 0.5 (1) |W |T 2.71|W f 2 = − 0.3333

m/2

48.18

11.92

2.71

4.69

23.84

2.71

12 ' m

83.45

12 ' m

48.18

12 ' f1

Static analysis of the structure subjected to forces fn gives the responses due to each mode. These computations give Table P13.45a for story shears (in kips) and Table P13.45b for story overturning moments (in kip-ft). Table P13.45a Story shear

Mode 1

Mode 2

Mode 3

48.18

– 11.92

2.71

131.63

– 11.92

– 1.98

Vb = V1

179.81

11.92

0.73

Table P13.45b Floor moments

Mode 1

Mode 2

Mode 3

578.2

– 143.0

32.52

2157.7

– 286.1

8.76

4315.5

– 143.0

17.52

The above-determined responses due to each mode should be identical to those determined in Problem 13.7 by RHA; the slight differences are due to numerical truncation errors in the computations for this problem whereas more significant digits were used in the computer work for Problem 13.7. Part c The peak modal responses are combined by the SRSS rule: r =

∑ rn2

n =1

For each response quantity rn available in Part b are substituted to obtain the total response (Tables P13.45c-d).

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 98 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Table P13.45c Floor or story, j

u j , in.

Vj , kips

1.103

49.71

0.954

132.15

0.552

180.21

Table P13.45d Floor or story, j

Overturning moment, M j or Mb (kip-ft)

596.5

2176.6

Base, b

4317.9

Part d: Comments Comparison of the RSA results in Tables P13.23cd with RHA results from Problem 13.9 is summarized in Tables P13.45e-f.

Floor or story, j

Table P13.45e uj Shear Vj (in.) (kips) RSA

RHA

RSA

RHA

1.103

49.71

52.22

0.954

0.957

132.15

138.08

0.552

0.580

180.21

189.29

Table P13.45f Floor or story, j 2

Overturning moment M j or Mb (kip-ft) RSA

RHA

596.5

626.6

2176.6

2267.5

Base, b

4317.9

4320.8

For this particular problem, the RSA method gives results that are very close to the RHA results, in part, because most of the response is due to one mode, the first mode.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 99 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.46

These forces are shown in the following figure.

Initial calculations

m/2 51.99

From Problem 13.6, the modal properties are

12'

Γ2 =−0.5

Γ3 = 0.0972

Part a: Spectral ordinates

From Problem 13.8, the peak values of Dn ( t ) and An ( t ) are D1 = 1.086 in.

D2 = 0.272 in.

D3 = 0.109 in.

A1 = 0.741g

A2 =0.887 g

A3 =0.756 g

Part b: Peak modal responses

Substituting numerical values for Γn , φ n , and Dn in Eq. (13.8.1a) gives the peak displacements due to each of the three modes: ⎧0.314⎫ ⎧0.478⎫ ⎪ ⎪ ⎪ ⎪ u1 = 1.403⎨0.686⎬(1.086) = ⎨1.045 ⎬ in ⎪ 1 ⎪ ⎪1.523⎪ ⎭ ⎩ ⎭ ⎩

⎧− 1 / 2 ⎫ ⎧ 0.068⎫ ⎪ ⎪ ⎪ ⎪ u 2 = −0.5⎨− 1 / 2⎬(0.272) = ⎨ 0.068⎬ in ⎪ 1 ⎪ ⎪− 0.136⎪ ⎩ ⎭ ⎩ ⎭ ⎧0.034⎫ ⎧ 3.189 ⎫ ⎪ ⎪ ⎪ ⎪ u3 = 0.0972⎨− 2.186⎬(0.109) = ⎨0.023⎬ in ⎪0.011⎪ ⎪ 1 ⎪ ⎭ ⎭ ⎩ ⎩ Substituting Γn , m j , φ n and An in Eq. (13.8.2) gives the equivalent static forces due to each of the three modes: ⎧1(0.314)⎫ ⎧32.63⎫ ⎛ 100 ⎞⎪ ⎪ ⎪ ⎪ ⎟⎟⎨1(0.686)⎬(0.741g ) = ⎨71.34⎬ kips f1 = 1.403⎜⎜ g ⎠⎪ 0.5(1) ⎪ ⎝ ⎪51.99⎪ ⎩ ⎭ ⎩ ⎭ ⎧ 22.17⎫ ⎧1(−0.5)⎫ ⎛ 100 ⎞⎪ ⎪ ⎪ ⎪ ⎟⎟⎨1(−0.5)⎬(0.887 g ) = ⎨ 22.17⎬ kips f 2 = −0.5⎜⎜ g ⎠⎪ 0.5(1) ⎪ ⎝ ⎪− 22.17 ⎪ ⎭ ⎩ ⎭ ⎩ ⎧ 1(3.189) ⎫ ⎧ 23.40⎫ ⎛ 100 ⎞ ⎪ ⎪ ⎪ ⎪ ⎟⎟ ⎨1( −2.186) ⎬(0.756 g ) = ⎨ − 16.06 ⎬ kips f = 0.0972⎜⎜ 3 g ⎝ ⎠ ⎪ 0.5(1) ⎪ ⎪ 3.67 ⎪ ⎩ ⎭ ⎩ ⎭

71.34

3.67

22.17

16.06

22.17

23.40

12'

⎡− 1 / 2⎤ ⎡ 3.189 ⎤ ⎡0.314⎤ ⎢ ⎢ ⎥ ⎥ φ1 = ⎢0.686⎥ ; φ2 = ⎢− 1 / 2⎥ ; φ3 = ⎢⎢− 2.186⎥⎥ ⎢⎣ 1 ⎥⎦ ⎢⎣ 1 ⎥⎦ ⎢⎣ 1 ⎥⎦ Γ1 =1.403

22.17

32.63

12'

Static analysis of the structure subjected to forces f n gives the responses due to each mode. These computations give Table 13.46a for story shears (in kips) and Table 13.46b for story overturning moments (in kip-ft). Table 13.46a Story Shear

Mode 1

Mode 2

Mode 3

V3 V2 Vb = V1

51.99 123.33 155.96

-22.17 0 22.17

3.67 -12.38 11.02

Table 13.46b Floor Moments

Mode 1

Mode 2

Mode 3

M3 M2 M b = M1

623.8 2103.8 3975.3

-266.0 -266.0 0

44.1 -104.5 27.7

The above-determined responses due to each mode should be identical to those determined in Problem 13.8 by RHA; the slight differences are due to numerical truncation errors in the computations for this problem whereas more significant digits were used in the computer work for Problem 13.8. Part c

The peak modal reponses are combined by the SRSS rule: r=

∑ rn

n =1

For each response quantity rn available in Part b are substituted to obtain the total response (Tables 13.46c-d)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 100 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Table 13.46c Floor or story, j

u j , in.

V j , kips

3 2 1

1.529 1.047 0.484

56.64 123.95 157.92

Table 13.46d Overturning moment,

Floor j

M j or Mb (kip-ft) 3 2 Base, b

679.6 2123.1 3975.4

Part d: Comments

Comparison of the RSA results with RHA results from Problem 13.8 is summarized in Tables 13.46e-f. Table 13.46e Floor or

u j (in.)

Shear V j (kips)

story, j 3 2 1

RSA 1.529 1.047 0.484

Floor or

RHA 1.4332 1.1085 0.5281

RSA 56.64 123.95 157.92

RHA 54.85 126.17 172.23

Table 13.46f Overturning moment,

M j or Mb (kip-ft)

story, j 2 1 Base, b

RSA 679.6 2123.1 3975.4

RHA 658.2 2136.0 3965.9

For this problem, the RSA method gives results that are close to the RHA results, in part, because most of the response is due to one mode, the first mode.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 101 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.47

4. Determine spectral ordinates. From Fig. 6.9.5: m

T1 = 0.815 s

Mode 1:

in ⎛ 1.80 ⎞ A1 = 0.20 ⎜ ⎟ = 0.44 g = 171 2 0 . 815 ⎝ ⎠ s D1 =

E = 29000 ksi

I = 28.1 in4

m = 1.5 kips/g

D2 =

mL3

kips 1.5 k ⋅s2 = = 0.00388 g 386.4 in

( 29000) ( 28.1)

( 0.00388) (120) 3

= 11.02 s −1

From Example 13.1:

ω1 = 0.6987

EI 3

mL EI

ω 2 = 1.874

mL3

= 7.70

rad ⇒ T1 = 0.815 s s

rad = 20.65 ⇒ T2 = 0.304 s s

Γ1 = 0.406

Γ2 = 0.594

⎧ 1 ⎫ ⎬ ⎩2.097⎭

φ1 = ⎨

1 ⎫ ⎬ ⎩− 1.431⎭ ⎧

φ2 = ⎨

3. Determine correlation coefficient. Use both the SRSS and CQC modal combination rules. For CQC, we require ρ12:

ζ = 0.05 ω 7.70 = 0.373 β12 = 1 = ω 2 20.65 ∴ From Eq. (13.7.10):

ρ 12 =

171

= 2.88 in

(7.70 )2

A2 = 0.20 (2.71) = 0.54 g = 209

2. Natural frequencies and modes. m = 1.5

ω 12

T2 = 0.304 s

Mode 2:

1. Data. L = 120 in.

ω 22

209

(20.65)2

in s2

= 0.49 in

5. Determine peak displacements. u n = u nst An

u nst =

Γn

ω n2

φn

∴ u n = Γn φ n D n ⎧ 1 ⎫ ⎧1.169 ⎫ u 1 = 0.406 ⎨ ⎬ 2.88 = ⎨ ⎬ 2 . 097 ⎩ ⎭ ⎩2.452⎭ ⎧ 1 ⎫ ⎧ 0.291⎫ u 2 = 0.594 ⎨ ⎬ 0.49 = ⎨ ⎬ ⎩− 1.431⎭ ⎩− 0.417 ⎭

SRSS estimates of the peak displacements: u1 =

(1.169) 2 + (0.291) 2

u2 =

(2.452) 2 + (−0.417) 2 = 2.487 in

= 1.205 in

CQC estimates of the peak displacements: u1 =

(1.169) 2 + 2 (0.00836) (1.169) (0.291) + (0.291) 2

= 1.207 in u2 =

(2.452) 2 + 2 (0.00836) (2.452) (−0.417) + (−0.417) 2

= 2.484 in

8 (0.05) 2 (1 + 0.373) (0.373)1.5 [ 1 − (0.373) 2 ] 2 + 4 (0.05) 2 (0.373) (1 + 0.373) 2

= 0.00836

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 102 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

6. Determine peak bending moments. Peak modal responses: st M bn = M bn An

From Figure E13.1: M bst1 = 2.069 mL = 2.069 (0.00388) (120) = 0.964 k ⋅ s 2 M bst2 = 0.931 mL = 0.931 (0.00388) (120) = 0.434 k ⋅ s 2

Hence, M b1 = 0.964 (171) = 164.8 k ⋅ in M b 2 = 0.434 (209) = 90.8 k ⋅ in

SRSS estimate of the peak bending moment: Mb =

(164.8) 2 + (90.8) 2 = 188.2 k ⋅ in

CQC estimate of the peak bending moment: Mb =

(164.8) 2 + 2 (0.00836)(164.8)(90.8) + (90.8) 2

= 188.8 k ⋅ in 7. Comments

The CQC estimate of the peak response is close to the SRSS result because ω1 and ω2 are well separated resulting in ρ12 close to zero.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 103 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.48 2m

ρ 12 =

8 (0.05) 2 (1 + 0.373) (0.373)1.5 [ 1 − (0.373) 2 ] 2 + 4 (0.05) 2 (0.373) (1 + 0.373) 2

= 0.00836

4. Determine spectral ordinates. From Fig. 6.9.5:

in ⎛ 1.80 ⎞ A1 = 0.20 ⎜ ⎟ = 0.44 g = 171 2 s ⎝ 0.815 ⎠

1. Data. L = 120 in.

E = 29000 ksi

I = 28.1 in4

m = 1.5 kips/g

T1 = 0.815 s

Mode 1:

D1 =

kips 1.5 k ⋅s2 m = 1.5 = = 0.00388 g 386.4 in =

( 29000) ( 28.1) ( 0.00388) (120) 3

= 11.02 s −1

ω 2 = 1.874

EI 3

mL EI mL3

= 7.70

rad ⇒ T1 = 0.815 s s

rad = 20.65 ⇒ T2 = 0.304 s s

Γ1 = 0.2834

Γ2 = −0.2834

⎧ 1 ⎫ φ1 = ⎨ ⎬ ⎩2.097 ⎭

⎧ 1 ⎫ φ2 = ⎨ ⎬ ⎩− 1.431⎭

3. Determine correlation coefficient. Use both the SRSS and CQC modal combination rules. For CQC, we require ρ12:

ς = 0.05 β12 =

ω1 7.70 = = 0.373 ω 2 20.65

∴ From Eq. (13.7.10):

171

= 2.88 in

(7.70 )2

D2 =

ω 22

209

(20.65)2

in s2

= 0.49 in

5. Determine peak displacements: u n = u nst An

From Problem 13.15:

ω 1 = 0.6987

A2 = 0.20 (2.71) = 0.54 g = 209

2. Natural frequencies and modes.

ω 12

T2 = 0.304 s

Mode 2:

u nst =

Γn

ω n2

φn

∴ u n = Γn φ n D n ⎧ 1 ⎫ ⎧0.816⎫ u 1 = 0.2834 ⎨ ⎬ 2.88 = ⎨ ⎬ 2 . 097 ⎩ ⎭ ⎩1.712 ⎭ ⎧ 1 ⎫ ⎧− 0.139⎫ u 2 = − 0.2834 ⎨ ⎬ 0.49 = ⎨ ⎬ ⎩− 1.431⎭ ⎩ 0.199⎭

SRSS estimates of the peak displacements: u1 =

(0.816) 2 + (−0.139) 2

u2 =

(1.712) 2 + (0.199) 2 = 1.724 in

= 0.828 in

CQC estimates of the peak displacements: u1 =

(0.816) 2 + 2 (0.00836) (0.816) (−0.139) + (−0.139) 2

= 0.827 in u2 =

(1.712) 2 + 2 (0.00836) (1.712) (0.199) + (0.199) 2

= 1.725 in

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 104 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

6. Determine peak bending moments. Peak modal responses: st M bn = M bn An

From the figure in the solution to Problem 13.15: M bst1 = 1.443 mL = 1.443 (0.00388) (120)

= 0.672 k ⋅ s 2 M bst2 = − 0.443 mL = − 0.443 (0.00388) (120)

= − 0.206 k ⋅ s 2

Hence, M b1 = 0.672 (171) = 114.9 k ⋅ in M b 2 = − 0.206 (209) = − 43.1 k ⋅ in

SRSS estimate of the peak bending moment: Mb =

(114.9) 2 + (−43.1) 2 = 122.7 k ⋅ in

CQC estimate of the peak bending moment: Mb =

(114.9) 2 + 2 (0.00836)(114.9)(−43.1) + (−43.1) 2

= 122.4 k ⋅ in

7. Comments: The CQC estimate of the peak response is close to the SRSS result because ω1 and ω2 are well separated resulting in ρ12 close to zero.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 105 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

4. Determine spectral ordinates.

Problem 13.49 2m

From Fig. 6.9.5:

T1 = 0.815 s

Mode 1:

in ⎛ 1.80 ⎞ A1 = 0.20 ⎜ ⎟ = 0.44 g = 171 2 ⎝ 0.815 ⎠ s

D1 =

E = 29000 ksi

I = 28.1 in4

m = 1.5 kips/g

2. Natural frequencies and modes.

EI mL

D2 =

kips 1.5 k ⋅s = = 0.00388 g 386.4 in ( 29000) ( 28.1)

( 0.00388) (120)

= 11.02 s

−1

ω1 = 0.6987

mL3 EI

ω 2 = 1.874

mL3

rad = 7.70 ⇒ T1 = 0.815 s s = 20.65

Γ1 = 0.0863

rad ⇒ T2 = 0.304 s s Γ2 = 0.6206

⎧ 1 ⎫ ⎬ ⎩2.097⎭

⎧

1 ⎫ ⎬ ⎩− 1.431⎭

φ1 = ⎨

171

= 2.88 in

(7.70 )2

ω 22

209

(20.65)2

in s2

= 0.49 in

5. Determine peak displacements. u n = u nst An

u nst =

Γn

ω n2

φn

∴ u n = Γn φ n D n

From Problem 13.16: EI

A2 = 0.20 (2.71) = 0.54 g = 209

L = 120 in.

m = 1.5

ω 12

T2 = 0.304 s

Mode 2:

1. Data.

φ2 = ⎨

⎧ 1 ⎫ ⎧0.249⎫ u 1 = 0.0863 ⎨ ⎬ 2.88 = ⎨ ⎬ ⎩2.097⎭ ⎩0.521⎭ ⎧ 1 ⎫ ⎧ 0.304⎫ u 2 = 0.6206 ⎨ ⎬ 0.49 = ⎨ ⎬ 1 . 431 − ⎩ ⎭ ⎩− 0.435⎭

SRSS estimates of the peak displacements: u1 =

(0.249) 2 + (0.304) 2

u2 =

(0.521) 2 + (−0.435) 2 = 0.679 in

= 0.393 in

3. Determine correlation coefficient.

CQC estimates of the peak displacements:

Use both the SRSS and CQC modal combination rules. For CQC, we require ρ12:

u1 =

ς = 0.05 ω 7.70 = 0.373 β12 = 1 = ω 2 20.65 ∴ From Eq. (13.7.10):

ρ 12 =

8 (0.05) 2 (1 + 0.373) (0.373)1.5 [ 1 − (0.373) 2 ] 2 + 4 (0.05) 2 (0.373) (1 + 0.373) 2

= 0.00836

(0.249) 2 + 2 (0.00836) (0.249) (0.304) + (0.304) 2

= 0.395 in u2 =

(0.521) 2 + 2 (0.00836) (0.521) (−0.435) + (−0.435) 2

= 0.676 in

6. Determine peak bending moments. Peak modal responses: st M bn = M bn An

From the figure in the solution to Problem 13.16:

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 106 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

M bst1 = 0.440 mL = 0.440 (0.00388) (120)

= 0.205 k ⋅ s 2 M bst2 = 0.974 mL = 0.974 (0.00388) (120)

= 0.453 k ⋅ s 2

Hence, M b1 = 0.205 (171) = 35.1 k ⋅ in M b 2 = 0.453 (209) = 94.7 k ⋅ in

SRSS estimate of the peak bending moment: Mb =

(35.1) 2 + (94.7) 2 = 101.0 k ⋅ in

CQC estimate of the peak bending moment: Mb =

(35.1) 2 + 2 (0.00836)(35.1)(94.7) + (94.7) 2

= 101.3 k ⋅ in

7. Comments. The CQC estimate of the peak response is close to the SRSS result because ω1 and ω2 are well separated resulting in ρ12 close to zero.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 107 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.50 1. Determine effect of weight of pipe. Pipe weight equals 3 × 10 × 18.97 = 569 lbs. This is negligible compared to the lumped weights equals 5 mg = 7.5 kips. 2. Determine design spectrum ordinates. From Problem 10.23,

ω1 = 0.5259 ω 2 = 1.6135

ω 3 = 1.7321

⎧0⎫ ⎪ ⎪ u 3 (t ) = ⎨0⎬ D3 (t ) ⎪0⎪ ⎩ ⎭

Substituting numerical values for D1 and D 2 , the peak values of D1 (t ) and D 2 (t ) , gives the peak values of the modal responses:

EI mL3

⎧ 1.517 ⎫ ⎪ ⎪ u1 = ⎨− 2.970⎬ in. ⎪ 2.970⎪ ⎩ ⎭

EI 3

mL EI

mL3

Substituting the given values of E , I , m, and L :

ω1 = 0.5259

⎧ 0.603⎫ ⎪ ⎪ u 2 (t ) = ⎨ 0.774⎬ D 2 (t ) ⎪− 0.774 ⎪ ⎩ ⎭

29 × 10 3 × 28.1

(1.5 386.4)× (10 × 12)3

= 0.5259 × 11.022 = 5.798 rad / s

T1 = 1.084 secs

ω 2 = 17.79 rad / s

⎧ 0.399⎫ ⎪ ⎪ u 2 = ⎨ 0.512⎬ in. ⎪− 0.512 ⎪ ⎩ ⎭

Substituting numerical values for Γn , m , and φ n (available from Problem 13.17) in Eq. (13.8.2) gives the equivalent static forces (in kips) shown in the accompanying figure. 0.386

T2 = 0.353 secs

0.989

ω 3 = 19.09 rad / s 0.386

T3 = 0.329 secs

f1 (kips)

For these Tn , the design spectrum of Fig. 6.9.5 gives 1.80g = 0.332g 1.084

D1 = 3.82 in.

A2 = 0.2 × 2.71g = 0.542g

D 2 = 0.661 in.

A3 = 0.2 × 2.71g = 0.542g

D3 = 0.574 in.

A1 = 0.2 ×

0.629

3. Determine peak modal responses. The modal displacements, available from Problem 13.17, are ⎧ 0.397 ⎫ ⎪ ⎪ u1 (t ) = ⎨− 0.774 ⎬ D1 (t ) ⎪ 0.774⎪ ⎩ ⎭

2.451 0.629

f 2 (kips)

0 0

f 3 (kips)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 108 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Static analysis of the structure subjected to forces f n gives the peak values of the bending moments (kip-ft.) due to each mode (Table P13.50a). Table P13.50a: Bending Moments (kip-ft.) Mode 1

Mode 2

Mode 3

−3.85

6.29

17.59

11.93

4. Combine modal responses. The correlation coefficients, ρ in , in the CQC rule depend on the frequency ratios β in = ω i ω n , computed from the known natural frequencies. Table P13.50b: Natural Frequency Ratios, β in Mode, i

n =1

n=2

n=3

ω i (rad/sec)

1 2 3

1.0 3.0683 3.2925

0.3259 1.0 1.0731

0.3037 0.9319 1.0

5.798 17.79 19.09

Table P13.50e: Bending Moments (kip-ft.) SRSS CQC

7.375 7.395

21.25 21.31

5. Comments. The cross-correlation coefficients ρ12 and ρ13 are negligible, implying that the 1-2 and 1-3 cross-terms in Eq. (13.7.5) are insignificant. In contrast, ρ 23 = 0.66731 , which is large. Thus the 2-3 cross-term in Eq. (13.7.5) may be significant. However, this term turns out to be zero because the response due to the third mode is zero. Thus all the cross-terms are either zero or negligible, which explains why the CQC estimate is very close to the SRSS estimate.

Correlation coefficients, ρ in , are calculated from Eq. (13.7.10): Table P13.50c: Correlation Coefficients, ρ in Mode, i

n =1

n=2

n=3

1 2 3

1.0 0.00613 0.00526

0.00613 1.0 0.66731

0.00526 0.66731 1.0

Substituting the peak modal responses in Eq. (13.7.3) with N = 3 gives the SRSS estimate of the total response, and, in addition, using the ρ in values in Eq. (13.7.5) with N = 3 gives the CQC estimate of the total response. Tables P13.50d and P13.50e summarize the results for u1 , u 2 and u 3 , and M a and M b . Table P13.50d: Displacements (in.) SRSS CQC

1.569 1.571

3.014 3.014

3.014 3.011

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 109 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.51

1. Determine design spectrum ordinates.

From the values obtained in Problem 13.50, we have A1 = 0.332g

D1 = 3.82 in.

A2 = 0.542g

D 2 = 0.661 in.

A3 = 0.542g

D3 = 0.574 in.

f 2 (kips)

2. Determine peak modal responses. (a) The modal displacements, available from Problem 13.18, are

0.813

⎧0⎫ ⎪ ⎪ u 1 (t ) = u 2 (t ) = ⎨0⎬ ⎪0⎪ ⎩ ⎭

f 3 (kips)

⎧0⎫ ⎪ ⎪ u 3 (t ) = ⎨1⎬ D3 (t ) ⎪1⎪ ⎩ ⎭

Substituting numerical values for D3 , the peak value of D3 (t ) , gives the peak values of the modal responses: ⎧0⎫ ⎪ ⎪ u 1 = ⎨0⎬ ⎪0⎪ ⎩ ⎭

⎧0⎫ ⎪ ⎪ u 2 = ⎨0⎬ ⎪0⎪ ⎩ ⎭

⎧ 0 ⎫ ⎪ ⎪ u 3 = ⎨0.574⎬ in. ⎪0.574⎪ ⎩ ⎭

Table P13.51a: Bending Moments (kip-ft.)

(b) Substituting numerical values for Γn , m , and φ n (available from Problem 13.18) in Eq. (13.8.2) gives the equivalent static forces (in kips) shown in the accompanying figure. 0

Static analysis of the structure subjected to forces f n gives the peak values of the bending moments (kip-ft.) due to each mode (Table P13.51a)

Mode 1

Mode 2

Mode 3

8.13

3. Combine modal responses. Because the entire response is due to the third mode, we do not need to combine modal responses. The total response is he same as the third mode response:

u1 = 0 in.

u 2 = 0.574 in. u 3 = 0.574 in.

f1 (kips)

M a = 8.13 kip - ft. M b = 0 kip - ft.

4. Comments. In this case, since only one mode contributes to the response, the peak value of the total response is identical to the peak response in the third mode. Consequently, SRSS and CQC rules give identical results. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 110 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.52 1. Determine design spectrum ordinates.

0.272

From the values obtained in Problem 13.50, we have A1 = 0.332g

D1 = 3.82 in.

A2 = 0.542g

D 2 = 0.661 in.

A3 = 0.542g

D3 = 0.574 in.

1.050 0.272

f1 (kips)

2. Determine peak modal responses. (a) The modal displacements, available from Problem 13.19, are ⎧ 0.281⎫ ⎪ ⎪ u 1 (t ) = ⎨− 0.547 ⎬ D1 (t ) ⎪ 0.547 ⎪ ⎩ ⎭ ⎧ 0.426⎫ ⎪ ⎪ u 2 (t ) = ⎨ 0.547 ⎬ D 2 (t ) ⎪− 0.547 ⎪ ⎩ ⎭ ⎧ 0 ⎫ ⎪ ⎪ u 3 (t ) = ⎨0.707 ⎬ D3 (t ) ⎪0.707 ⎪ ⎩ ⎭

0.445 1.732 0.445

f 2 (kips)

0.575

0.575 0

Substituting numerical values for D1 , D 2 and D3 , the peak values of D1 (t ) , D 2 (t ) and D3 (t ) , gives the peak values of the modal responses: ⎧ 1.073 ⎫ ⎪ ⎪ u 1 = ⎨− 2.090 ⎬ in. ⎪ 2.090⎪ ⎩ ⎭ ⎧ 0.282⎫ ⎪ ⎪ u 2 = ⎨ 0.362⎬ in. ⎪− 0.362 ⎪ ⎩ ⎭ ⎧ 0 ⎫ ⎪ ⎪ u 3 = ⎨0.403⎬ in. ⎪0.403⎪ ⎩ ⎭

(b) Substituting numerical values for Γn , m , and φ n (available from Problem 13.19) in Eq. (13.8.2) gives the equivalent static forces (in kips) shown in the accompanying figure.

f 3 (kips)

Static analysis of the structure subjected to forces f n gives the peak values of the bending moments (kip-ft.) due to each mode (Table P13.52a)

Table P13.52a: Bending Moments (kip-ft.) Ma

Mode 1 −2.724

Mode 2 4.447

Mode 3 5.748

12.445

8.423

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 111 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

3. Combine modal responses. The correlation coefficients, ρ in , in the CQC rule depend on the frequency ratios β in = ω i ω n , computed from the known natural frequencies. Table P13.52b: Natural Frequency Ratios, β in Mode, i

n =1

n=2

n=3

ω i (rad/sec)

1 2 3

1.0 3.0683 3.2925

0.3259 1.0 1.0731

0.3037 0.9319 1.0

5.798 17.79 19.09

Because the cross-term is positive, the CQC estimate for M a is significantly larger than its SRSS estimate. For u1 and M b the 2-3 cross-term is zero because the third mode response is zero; hence the SRSS and CQC estimates of peak response are close. For u 2 and u 3 the 2-3 cross-term is negligible because the response due to the second and third modes are much smaller than due to the first mode; hence the SRSS and CQC estimates of peak response are close.

Correlation coefficients, ρ in , are calculated from Eq. (13.7.10): Table P1352: Correlation Coefficients, ρ in Mode, i 1 2 3

n =1

n=2

1.0 0.00613 0.00526

0.00613 1.0 0.66731

n=3 0.00526 0.66731 1.0

Substituting the peak modal responses in Eq. (13.7.3) with N = 3 gives the SRSS estimate of the total response, and, in addition, using the ρ in values in Eq. (13.7.5) with N = 3 gives the CQC estimate of the total response. Tables P13.52d and P13.52e summarize the results for u1 , u 2 and u 3 , and M a and M b . Table P13.52d: Displacements (in.) SRSS CQC

1.110 1.112

2.159 2.1996

2.159 2.113

Table P13.52e: Bending Moments (kip-ft.) SRSS CQC

7.763 9.699

15.028 15.070

Problem 13.53 0.272

1. Determine design spectrum ordinates. From the values obtained in Problem 13.50, we have A1 = 0.332g

D1 = 3.82 in.

A2 = 0.542g

D 2 = 0.661 in.

A3 = 0.542g

D3 = 0.574 in.

1.050 0.272

f1 (kips)

2. Determine peak modal responses. (a) The modal displacements, available from Problem 13.20, are

0.445

⎧ − 0.281⎫ ⎪ ⎪ u 1 (t ) = ⎨ 0.547 ⎬ D1 (t ) ⎪− 0.547 ⎪ ⎩ ⎭

1.732 0.445

⎧− 0.426 ⎫ ⎪ ⎪ u 2 (t ) = ⎨− 0.547 ⎬ D 2 (t ) ⎪ 0.547 ⎪ ⎩ ⎭ ⎧ 0 ⎫ ⎪ ⎪ u 3 (t ) = ⎨0.707 ⎬ D3 (t ) ⎪0.707 ⎪ ⎩ ⎭

f 2 (kips)

0.575

0.575 0

Substituting numerical values for D1 , D 2 and D3 , the peak values of D1 (t ) , D 2 (t ) and D3 (t ) , gives the peak values of the modal responses: ⎧ − 1.073 ⎫ ⎪ ⎪ u 1 = ⎨ 2.090⎬ in. ⎪− 2.090 ⎪ ⎩ ⎭ ⎧− 0.282⎫ ⎪ ⎪ u 2 = ⎨− 0.362⎬ in. ⎪ 0.362⎪ ⎩ ⎭ ⎧ 0 ⎫ ⎪ ⎪ u 3 = ⎨0.403⎬ in. ⎪0.403⎪ ⎩ ⎭

(b) Substituting numerical values for Γn , m , and φ n (available from Problem 13.20) in Eq. (13.8.2) gives the equivalent static forces (in kips) shown in the accompanying figure.

f 3 (kips)

Static analysis of the structure subjected to forces f n gives the peak values of the bending moments (kip-ft.) due to each mode (Table P13.53a) Table P13.53a: Bending Moments (kip-ft.) Ma

Mode 1 2.724

Mode 2 −4.447

Mode 3 5.748

−12.445

−8.423

3. Combine modal responses. The correlation coefficients, ρ in , in the CQC rule depend on the frequency ratios β in = ω i ω n , computed from the known natural frequencies.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 113 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Table P13.53b: Natural Frequency Ratios, β in Mode, i

n =1

n=2

n=3

1 2 3

1.0 3.0683 3.2925

0.3259 1.0 1.0731

0.3037 0.9319 1.0

ω i (rad/sec) 5.798 17.79 19.09

hence the SRSS and CQC estimates of peak response are close.

Correlation coefficients, ρ in , are calculated from Eq. (13.7.10): Table P13.53c: Correlation Coefficients, ρ in n =1 1.0 0.00613 0.00526

Mode, i 1 2 3

n=2 0.00613 1.0 0.66731

n=3 0.00526 0.66731 1.0

Substituting the peak modal responses in Eq. (13.7.3) with N = 3 gives the SRSS estimate of the total response, and, in addition, using the ρ in values in Eq. (13.7.5) with N = 3 gives the CQC estimate of the total response. Tables P13.53d and P13.53e summarize the results for u1 , u 2 and u 3 , and M a and M b . Table P13.53d: Displacements (in.) SRSS CQC

1.110 1.112

2.159 2.113

2.159 2.199

Table P13.53e: Bending Moments (kip-ft.) SRSS CQC

7.763 5.114

15.028 15.070

4. Comments. The cross-correlation coefficients ρ12 and ρ13 are negligible, implying that the 1-2 and 1-3 cross-terms in Eq. (13.7.5) are insignificant. In contrast, ρ 23 = 0.66731 , which is large. Thus the 2-3 cross-term in Eq. (13.7.5) may be significant. This is indeed the case for M a because the second and third mode contributions are significant. Because the cross-term is negative, the CQC estimate for M a is smaller than its SRSS estimate. For u1 and M b the 2-3 cross-term is zero because the third mode response is zero; hence the SRSS and CQC estimates of peak response are close. For u 2 and u 3 the 2-3 cross-term is negligible because the response due to the second and third modes are much smaller than due to the first mode; © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 114 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.54

Table P13.54a: Correlation Coefficients ρin

1. Determine design spectrum ordinates.

Mode, i

From Problem 13.22, T1 = 1. 027 sec

T2 = 0. 974 sec T3 = 0.1503 sec

For these Tn , the design spectrum of Fig. 6.9.5 gives 1 1. 80 g

A1 =

= 0. 584 g

D1 = 6. 02 in.

= 0. 616 g

D2 = 5. 70 in.

2. 71g = 0. 903g

D3 = 0. 2 in.

3 1. 027

A2 = A3 =

1 1. 80 g 3 0. 974 1 3

n=1

n=2

n=3

1.0

0.7847

0.0013

0.7847

1.0

0.0015

0.0013

0.0015

1.0

Substituting the peak modal responses in Eq. (13.7.3) with N = 3 gives the SRSS estimate of the total response and, in addition, using the ρin values in Eq. (13.7.5) with N = 3 gives the CQC estimate of the total response. Table P13.54b summarizes the results for u3 , Va and Vb . Table P13.54b

2. Determine peak modal responses.

u3 , in.

Va , kips

Vb , kips

The appendage displacement is

SRSS

165.0

3.158

144.3

u3n = Γn φ3n Dn

CQC

77.27

1.465

182.3

Substituting numerical values for Γn and φ3n available from Problem 13.22 gives u31 = ( − 0. 6296 ) ( − 32. 34 ) 6. 02 = 122. 57 u32 = ( 0. 6104 ) ( − 31. 80 ) 5. 70 = − 110. 6 u33 = ( 0. 4511) ( 0. 0773) 0. 2 = 0. 007

Substituting numerical values for Γn , m, and φn (available from Problem 13.22) in Eq. (13.8.2) gives the equivalent static forces shown in the accompanying figure.

2.2295

2.2366

66.015

68.500

21.036

22.086

4. Comments. The cross correlation coefficient for the first two modes is 0.7847 which is significant. Therefore, neglecting the 1-2 cross term in the SRSS procedure introduces significant error. The CQC method is more accurate.

0.0059 33.537 104.62

Static analysis of the structure subjected to forces fn gives the peak value of modal response rn . These results for the appendage shear and base shear are (in kips): Va1 = 2. 2295 Va 2 = − 2. 2366 Va3 = 0. 0059 Vb1 = 89. 281 Vb 2 = 88. 349

Vb3 = 71. 089

(a)

3. Combine modal responses. Correlation coefficients, ρin , are calculated from Eq. (13.7.10) using known values of βin = ω i ω n : © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 115 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.55

Part d

Part a: Response spectrum ordinates

Substituting the peak modal responses (from Table P13.55a) in Eq. (13.7.3) with N = 3 gives the SRSS estimate of peak response:

From Problem 13.23, the peak values of Dn ( t ) and An ( t ) are D1 = 4. 3641 in.

A1 = 0. 4237 g

D2 = 4. 5312 in.

A2 = 0. 4884 g

D3 = 0.1649 in.

A3 = 0. 7460 g

Vao = 2. 405 kips Vbo = 112.1 kips

Part e: Comments

Part b: Peak modal responses

The peak values of the appendage displacement u3 , appendage shear Va and tower base shear Vb are u3n = u3stn An

st Van = Van An

u3o = 125. 04 in.

st Vbn = Vbn An

Summarized in Table P13.55b are the CQC and SRSS results from Parts c and d and the RHA results from Problem 13.23. Table P13.55b

(a)

where the modal static responses rnst are available from the solution to Problem 13.22. Substituting for rnst and An in Eq. (a) gives the results in Table P13.55a.

u3o , in.

Vao , kips

Vbo , kips

SRSS

125.04

2.405

112.1

CQC

58.02

1.125

140.5

RHA

44.58

0.879

159.8

Table P13.55a Mode, n

u3 , in.

Va , kips

Vb , kips

88.90

1.619

64.83

– 87.93

– 1.778

70.23

0.0058

0.0049

58.61

Clearly the peak response from the CQC method is much closer to that from RHA because the CQC method considers the correlation between modes. The SRSS method should not be used when frequencies are closely spaced.

These peak modal responses are essentially identical to those determined in Problem 13.23 by RHA. Part c

For each response quantity, the peak modal responses (from Table P13.55a) and the correlation coefficients (from Table P13.54a) are substituted in Eq. (13.7.4) with N = 3 to obtain peak values of the total response: u3o = 58. 02 in. Vao = 1.125 kips Vbo = 140. 5 kips

The cross-correlation coefficient for the first two modes is 0.7847 (see Problem 13.54), which is significant and contributes importantly to the response. The correlation coefficient between modes 1 and 3 and between 2 and 3 is insignificant because the frequencies are well separated.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 116 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Table P13.56c

Problem 13.56 uy ,

( b 2 ) uθ ,

Vb ,

Tb ,

in.

kips

kip-ft

SRSS

4.133

1.162

34.3

1862

CQC

4.135

1.149

34.4

1844

RHA

4.182

1.222

35.6

1899

From Problem 13.24, m, φ n , and Γn are available. Part a: Spectral ordinates

From Problem 13.26, D1 = 4. 291 in.

A1 = 0. 3047 g

D2 = 4. 354 in.

A2 = 0. 4354 g

D3 = 2. 604 in.

A3 = 0. 8020 g

Part d: Comments

Part b: Peak modal responses Displacements: u = Γn φn Dn

(a)

Equivalent static forces: fn = Γn m φn An

(b)

The accuracy of both combination rules (SRSS and CQC) is good in this case because the two modes that contribute to the response have well separated frequencies. Both methods are similarly accurate in this case.

Base shear and torque: Vbn = fyn

Tbn = fθn

(c)

Substituting m, φn , and Γn from Problem 13.24 and Dn and An from Part a in Eqs. (a)-(c) gives the peak modal responses in Table P13.56a. Table P13.56a ( b 2 ) uθ

0.9931

34.21

822.2

0.0966

– 0.6027

2.677

– 1671

Mode, n

4.132

2 3 Part c

The correlation coefficients ρin are calculated from Eq. (13.7.10) using known values of βin = ω i ω n : Table P13.56b: Correlation Coefficients ρin n=1

n=2

n=3

1.0

0.8513

0.0247

0.8513

1.0

0.0287

0.0247

0.0287

1.0

Mode, i

The modal peaks rn (or rno ) and ρin are substituted in SRSS and CQC formulas [Eqs. (13.7.3) and (13.7.4) with N = 3], specialized for each response quantity, to obtain estimates of the peak value of the total response. The exact values are available from RHA in Problem 13.26.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 117 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.57 From Problem 13.24, m, ω n , φ n , and Γn are available.

(13.7.3) and (13.7.4) for each response quantity to obtain estimates of the peak value of the total response. The results using SRSS and CQC methods are presented in Table P13.57a.

1. Determine spectral ordinates. The spectral ordinates are determined from Fig. 6.9.4 scaled by 0.5:

4. Determine bending moments in columns.

ω1 = 5. 96 , T1 = 1. 054 sec ⇒

Δi

D1 = 0. 5 × 110. 4 5. 96 = 9. 26 in.

ω 2 = 6. 21 , T2 = 1. 012 sec ⇒

Vi h

D2 = 0. 5 × 110. 4 6. 21 = 8. 89 in.

ω 3 = 10. 90 , T3 = 0. 576 sec ⇒ D3 = 0. 5 × 2. 71g (10. 90 )

2. Determine peak modal responses. Due to mode n, the peak displacements in the x, y, and

θ directions are uxn = 0

uyn = Γn φ yn Dn

uθn = Γn φθn Dn

(a)

Vbxn = 0

Vbyn = fyn = Γn m φ yn An

(b)

h 2

kh 2

Δi

(c)

where the two components of Δ i — Δ ix and Δ iy — cause moments My and Mx about the y and x axes, respectively. We illustrate the procedure for column a in Fig. P13.24. The peak modal values of Δ ax and Δ ay are

Tbn = fθn = Γn ( m r ) φθn An

In Eqs. (a) and (b), we substitute for m, r, φ yn , φθn and Γn from Problem 13.24; Dn from Part a; and An = ω n2 Dn to obtain the peak modal responses in Table P13.57a. Table P13.57a

Δ axn = Γn ( − ya φθn ) Dn = − ya uθn Δ ayn = Γn ( φ yn + xa φθn ) Dn = uyn + xa uθn

(d)

For mode 1 (n = 1), substituting the peak values of ux and uθ from Table P13.57a in Eqs. (d) and (c) gives Δ ax1 = − ( b 2 ) uθ1 = − 2.143 in.

Peak total response

Peak modal responses

The bending moment Mi at the ends of a clampedclamped column i is related to the lateral displacement Δ i of the column as follows: Mi = Vi

The base shear (x and y components) and base torque due to mode n are

Response

= 4. 40 in.

n=1

n=2

n=3

CQC

SRSS

ux , in.

uy , in.

8.917

0.163

8.922

8.918

( b 2) uθ , in.

2.143

-1.018

0.0157

0.0158

Vbx , kips

Vby , kips

73.8

4.5

74.1

74.0

Tb , kip-in.

1774.7

-2823.1

3297.1

3334.4

3. Combine peak modal responses. The modal peaks (from Part b) and correlation coefficients ρin from Table P13.56b are substituted in Eqs.

Δ ay1 = uy1 − ( b 2 ) uθ1 = 8. 917 − 2.143 = 6. 774 in.

May1 =

( 2 × 1. 5) (12 × 12 )

Δ ax1 = 216 Δ ax1 2 = 216 ( − 2.143) = − 462. 9 kip - in.

Max1 = 216 Δ ay1 = 216 ( 6. 774 ) = 1463 kip - in.

The computations for mode 3 proceed similarly: Δ ax 3 = − ( b 2 ) uθ3 = 1. 018 in. Δ ay3 = uy 3 − ( b 2 ) uθ3 = 0.163 − ( − 1. 018) = 1.181 in.

May3 = 216 Δ ax 3 = 216 (1. 018) = 220 kip - in.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 118 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Max 3 = 216 Δ ay3 = 216 (1.181) = 255. 2 kip - in.

Specializing Eqs. (13.7.3) and (13.7.4) for the SRSS and CQC methods to response May , and substituting for May1 and May3 (from above) and ρ13 = 0. 0247 (from Problem 13.56) gives May =

2 2 May 1 + May 3 = 512. 5 kip − in.

May =

2 2 May 1 + May 3 + 2 ρ13 May1 May 3

= 507. 6 kip − in. Similar computations lead to Table P13.57b for Max and bending moments in other columns.

Table P13.57b: Column Moments in kip-in. Response

Peak modal responses

Peak total response

n=1

n=2

n=3

CQC

SRSS

May

– 462.9

220

507.6

512.6

Max

1463

255.2

1491.3

1485.1

Mby

– 231.5

110.0

253.8

256.3

Mbx

1194.5

– 92.4

1195.7

1198.0

253.8

256.3

Mcy

231.5

– 110.0

Mcx

1194.5

– 92.4

1195.7

1198.0

Mdy

462.9

– 220

507.6

512.6

Mdx

1463

255.2

1491.3

1485.1

In this case both CQC and SRSS methods render similar results because modes 1 and 3, which are the only contributions to the response, have well separated frequencies.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 119 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.58 Solution to this problem closely follows that to Problem 13.57; therefore many details are omitted.

Table P13.58b: Column Moments in kip-in.

1. Spectral ordinates (from Problem 13.57). D1 = 9. 26 in.

D2 = 8. 89 in.

peak values of modal responses and the peak value of total response are calculated. The results are summarized in Table P13.58b.

D3 = 4. 40 in.

(a)

Response

2. Determine peak modal responses. From Problem 13.25, Γ1 = 0. 3350

Γ2 = 0. 3414

Γ3 = − 0. 0658 (b)

Displacements:

R|u U| R|φ U| S|u V| = Γ S|φ V| D Tu W Tφ W xn

θn

(c)

θn

Base forces: Vbxn = fxn = Γn m φ xn An

Vbyn = fyn = Γn m φ yn An

(d)

Tbn = fθn = Γn ( m r 2 ) φθn An

In Eqs. (c) and (d), substituting for m, r, φ xn , φ yn and φθn from Problem 13.24, Γn from Eq. (b), Dn from Eq. (a), and An = ω 2n Dn gives the results in Table P13.58a. Table P13.58a Response

Peak modal responses

Peak total response

n=1

n=2

n=3

CQC

SRSS

ux , in.

6.286

uy , in.

6.305

0.115

6.309

6.306

( b 2 )uθ ,

1.516

– 0.720

1.662

1.678

Vx , kips

56.6

Vy , kips

52.2

3.2

52.4

52.3

T , kip-in.

1254.7

– 1996.2

2331.4

2357.8

Peak modal responses

Peak total response

n=1

n=2

n=3

CQC

SRSS

May

– 327.4

1357.8

155.5

1108.1

1405.4

Max

1034.5

180.5

1054.5

1050.1

Mby

– 163.7

678.9

77.8

554.0

702.7

Mbx

844.6

– 65.3

845.5

847.1

Mcy

163.7

678.9

– 77.8

824.2

702.7

Mcx

844.6

– 65.3

845.5

847.1

Mdy

327.4

1357.8

– 155.5

1648.4

1405.4

Mdx

1034.5

180.5

1054.5

1050.1

It is interesting to note that the differences between the SRSS and CQC methods only show up in those bending moments that have contributions of the first and second modes. These modes have closely spaced frequencies and, hence, the cross correlation term for these two modes is significant. Thus, all moments about the yaxis will be in significant error if estimated using the SRSS method. Therefore, we conclude that, depending on the response quantity selected, the cross-correlation terms may or may not be important.

in.

3. Combine peak modal responses. The modal peaks (from Table P13.58a) and correlation coefficients ρin (from Table P13.56b) are substituted in Eqs. (13.7.3) and (13.7.4) for each response quantity to obtain estimates of the peak value of the total response. The results obtained by SRSS and CQC methods are presented in Table P13.58a. 4. Determine bending moments in columns. Following the solution of Problem 13.57 for each (x and y) component of bending moment in each column, the © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 120 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.59

3. Determine correlation coefficients. Use Eq. (13.7.10) to compute ρ ij for ζ = 0.05:

uz z

uy m

L b

ω1 = 0.969 ω2

∴ ρ 12 = 0.9089

β 13 =

ω1 = 0.326 ω3

∴ ρ 13 = 0.0061

β 23 =

ω2 = 0.336 ω3

∴ ρ 23 = 0.0066

d L

β 12 =

y x

4. Determine spectral ordinates. 1. Data.

From Fig. 6.9.5:

L = 36 in.

m = 1.0 kips/g

E = 30000 ksi

I = 3.017 in4

G = 12000 ksi

J = 6.034 in4

D1 =

2. Natural frequencies and modes. kips 1.0 k ⋅s2 m = 1.0 = = 0.00259 g 386.4 in

mL3

(30000)(3.017)

(0.00259)(36) 3

= 27.38 s −1

EI mL3

rad s

= 13.24

⎧ 0.7767⎫ ⎪ ⎪ φ1 = ⎨− 0.4923⎬ ⎪− 0.3928⎪ ⎩ ⎭ EI 3

= 13.66

⎧− 0.2084⎫ ⎪ ⎪ φ 2 = ⎨ 0.3875⎬ ⎪− 0.8980⎪ ⎩ ⎭

ω 3 = 1.4827 ⎧ 0.5943⎫ ⎪ ⎪ φ 3 = ⎨ 0.7794⎬ ⎪ 0.1984⎪ ⎩ ⎭

rad s

= 40.59

rad s

209 (13.24) 2

= 1.20 in

A2 = 0.20 (2.71g) = 0.54 g = 209 D2 =

ω 22

209 (13.66) 2

in s2

= 1.12 in

A3 = 0.20 (2.71g) = 0.54 g = 209

⇒ T1 = 0.475 s

⇒ T2 = 0.460 s

D3 =

ω 32

⇒ T3 = 0.155 s

209 (40.59) 2

in s2

= 0.127 in

5. Determine peak modal displacements. u n = u nst An

Γ2 = −0.2084

ω 12

Mode 2: T2 = 0.460 s

Γ1 = 0.7767

ω 2 = 0.4990

Mode 3: T3 = 0.155 s

From Problems 10.28 and 13.27:

ω 1 = 0.4834

A1 = 0.20 (2.71g ) = 0.54 g = 209

4 EI 5

Note: GJ =

Mode 1: T1 = 0.475 s

u nst =

Γn

ω n2

φn

∴ u n = Γn φ n D n

u1 =

⎧ 0.7767⎫ ⎪ ⎪ 0.7767 ⎨− 0.4923⎬ 1.20 = ⎪− 0.3928⎪ ⎩ ⎭

(a) ⎧ 0.7212⎫ ⎪ ⎪ ⎨− 0.4572⎬ ⎪− 0.3647 ⎪ ⎩ ⎭

Γ3 = 0.5943

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 121 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

⎧− 0.2084⎫ ⎪ ⎪ = − 0.2084 ⎨ 0.3975⎬ 1.12 = ⎪− 0.8980⎪ ⎩ ⎭

⎧ 0.5943⎫ ⎪ ⎪ u 3 = 0.5943 ⎨ 0.7794⎬ 0.127 ⎪ 0.1984⎪ ⎩ ⎭

⎧ 0.0488⎫ ⎪ ⎪ ⎨− 0.0906⎬ ⎪ 0.2100⎪ ⎩ ⎭

⎧ 0.0449⎫ ⎪ ⎪ ⎨ 0.0589⎬ ⎪ 0.0150⎪ ⎩ ⎭

calculated. These estimates are summarized in the following table. Response Mx My T

SRSS rule (k ⋅ in) 8.66 18.52 19.50

CQC rule (k ⋅ in) 9.42 15.92 21.55

9. Comments. 6. Combine peak modal displacements. Using Eqs. (13.7.3) and (13.7.4), the SRSS and CQC estimates for the peak displacements of the mass can be calculated. These estimates are summarized in the following table. Displacement ux uy uz

SRSS rule (inches) 0.724 0.470 0.421

CQC rule (inches) 0.767 0.544 0.195

Modes 1 and 2 are strongly correlated with each other (ρ12=0.9089), but weakly correlated with mode 3. Consequently, significant differences between the SRSS and CQC estimates should result only when both modes 1 and 2 contribute significantly to the total response. Examining the above results, we see that this is the case. Only the SRSS and CQC estimates for the displacement uz differ significantly. The effect of modal correlation is less pronounced for the other displacements and responses either because (1) the contribution from mode 3 is larger than that of modes 1 and 2 (e.g. Mx) or (2) the contributions from modes 1 and 2 are not comparable (e.g. ux, uy, My and T).

7. Determine peak modal responses. st M xn = M xn An

st M yn = M yn An

Tn = Tnst An (b)

st st where M xn , M yn and Tnst are given in the solution to

Problem 13.27 as: st M xn = [− 0.0773mL, − 0.2679mL, 0.3453mL] st M yn = [− 0.9084mL, 0.1437 mL, − 0.2353mL]

Tnst = [ 0.9858mL, 0.1242mL, − 0.1100mL] st st Substituting numerical values for M xn , M yn , Tnst and An

in Eq. (b) gives: M x1 = −1.509 k ⋅ in

M y1 = −17.725 k ⋅ in

M x 2 = −5.228 k ⋅ in

M y2 =

M x 3 = 6.737 k ⋅ in

M y 3 = − 4.591 k ⋅ in

= 19.234 k ⋅ in

= 2.424 k ⋅ in

= −2.146 k ⋅ in

2.804 k ⋅ in

8. Combine peak modal responses. Using Eqs. (13.7.3) and (13.7.4), the SRSS and CQC estimates for the peak values of these responses can be © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 122 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.60

ω 3 = 1.4827

uy m

L b

1. Data. L = 36 in.

m = 1.0 kips/g

E = 30000 ksi

I = 3.017 in4

G = 12000 ksi

J = 6.034 in4

4 Note: GJ = EI 5

mL3

(30000)(3.017) (0.00259)(36) 3

= 27.38 s −1

EI mL3

rad = 13.24 s

⎧ 0.7767 ⎫ ⎪ ⎪

φ1 = ⎨− 0.4923⎬

⎧− 0.2084⎫ ⎪ ⎪ φ 2 = ⎨ 0.3875⎬ ⎪− 0.8980⎪ ⎩ ⎭

3. Determine correlation coefficients.

β 12 =

ω1 = 0.969 ω2

∴ ρ 12 = 0.9089

β 13 =

ω1 = 0.326 ω3

∴ ρ 13 = 0.0061

β 23 =

ω2 = 0.336 ω3

∴ ρ 23 = 0.0066

4. Determine spectral ordinates.

A1 = 0.20 (2.71g ) = 0.54 g = 209 D1 =

ω 12

209 (13.24) 2

⇒ T1 = 0.475 s

D2 =

ω 22

209 (13.66) 2

= 13.66

rad s

⇒ T2 = 0.460 s

Γ2 = 0.3875

in s2

= 1.12 in

Mode 3: T3 = 0.155 s A3 = 0.20 (2.71g ) = 0.54 g = 209

= 1.20 in

A2 = 0.20 (2.71g) = 0.54 g = 209

Γ1 = −0.4923

⎪− 0.3928⎪ ⎩ ⎭

ω 2 = 0.4990

Γ3 = 0.7794

Mode 2: T2 = 0.460 s

From Problems 10.28 and 13.28:

ω 1 = 0.4834

⎧ 0.5943⎫ ⎪ ⎪ φ 3 = ⎨ 0.7794⎬ ⎪ 0.1984⎪ ⎩ ⎭

⇒ T3 = 0.155 s

Mode 1: T1 = 0.475 s

kips 1.0 k ⋅s2 m = 1.0 = = 0.00259 g 386.4 in =

rad s

From Fig. 6.9.5:

2. Natural frequencies and modes.

= 40.59

Use Eq. (13.7.10) to compute ρ ij for ζ = 0.05: x

D3 =

ω 32

209 (40.59) 2

in s2

= 0.127 in

5. Determine peak modal displacements. u n = u nst An

u nst =

Γn

ω n2

φn

(a)

∴ u n = Γn φ n D n

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 123 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

u1 =

⎧− 0.4572⎫ ⎪ ⎪ ⎨ 0.2898⎬ ⎪ 0.2312⎪ ⎩ ⎭

⎧ 0.7767⎫ ⎪ ⎪ − 0.4923 ⎨− 0.4923⎬ 1.20 = ⎪− 0.3928⎪ ⎩ ⎭

⎧− 0.2084⎫ ⎪ ⎪ = 0.3875 ⎨ 0.3975⎬ 1.12 = ⎪− 0.8980⎪ ⎩ ⎭

⎧ 0.5943⎫ ⎪ ⎪ u 3 = 0.7794 ⎨ 0.7794⎬ 0.127 ⎪ 0.1984⎪ ⎩ ⎭

⎧− 0.0906⎫ ⎪ ⎪ ⎨ 0.1685⎬ ⎪− 0.3905⎪ ⎩ ⎭

⎧ 0.0589⎫ ⎪ ⎪ ⎨ 0.0772⎬ ⎪ 0.0197 ⎪ ⎩ ⎭

6. Combine peak modal displacements. Using Eqs. (13.7.3) and (13.7.4), the SRSS and CQC estimates for the peak displacements of the mass can be calculated. These estimates are summarized in the following table. SRSS rule (inches) 0.470 0.344 0.454

Displacement ux uy uz

CQC rule (inches) 0.544 0.456 0.205

7. Determine peak modal responses. st M xn = M xn An

st M yn = M yn An

= − 12.191 k ⋅ in

= − 4.507 k ⋅ in

= − 2.814 k ⋅ in

8. Combine peak modal responses. Using Eqs. (13.7.3) and (13.7.4), the SRSS and CQC estimates for the peak values of these responses can be calculated. These estimates are summarized in the following table. Response Mx My T

SRSS rule (k ⋅ in) 13.17 13.77 13.30

CQC rule (k ⋅ in) 13.84 9.10 16.65

9. Comments. Modes 1 and 2 are strongly correlated with each other (ρ12=0.9089), but weakly correlated with mode 3. Consequently, significant differences between the SRSS and CQC estimates should result only when both modes 1 and 2 contribute significantly to the total response. Examining the above results, we see that this is the case. The SRSS and CQC estimates for the displacements uy and uz and responses My and T differ significantly. The effect of modal correlation is less pronounced for ux and Mx because the contributions from modes 1 and 2 are not comparable.

Tn = Tnst An (b)

st st where M xn and Tnst are given in the solution to , M yn

Problem 13.28 as: st = [ 0.0490mL, M xn

0.4982mL,

0.4528mL]

st = [ 0.5758mL, − 0.2672mL, − 0.3086mL ] M yn

Tnst = [ − 0.6248mL, − 0.2310mL, − 0.1442mL ] st st Substituting numerical values for M xn , M yn , Tnst and An

in Eq. (b) gives: M x1 = 0.956 k ⋅ in

M y1 = 11.235 k ⋅ in

M x 2 = 9.721 k ⋅ in

M y 2 = −5.214 k ⋅ in

M x 3 = 8.835 k ⋅ in

M y 3 = − 6.021 k ⋅ in

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 124 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.61

ω 3 = 1.4827

uz z

uy m

L b

1. Data. L = 36 in.

m = 1.0 kips/g

E = 30000 ksi

I = 3.017 in4

G = 12000 ksi

J = 6.034 in4

4 EI 5

Note: GJ =

kips k ⋅s2 1.0 m = 1.0 = = 0.00259 in g 386.4

mL3

(30000)(3.017) (0.00259)(36) 3

= 27.38 s −1

EI 3

= 13.24

⎧ 0.7767⎫ ⎪ ⎪

φ1 = ⎨− 0.4923⎬

⎧− 0.2084⎫ ⎪ ⎪ φ 2 = ⎨ 0.3875⎬ ⎪− 0.8980⎪ ⎭ ⎩

rad s

⇒ T1 = 0.475 s

Γ1 = −0.3928

⎪− 0.3928⎪ ⎭ ⎩

ω 2 = 0.4990

⎧ 0.5943⎫ ⎪ ⎪ φ 3 = ⎨ 0.7794⎬ ⎪ 0.1984⎪ ⎭ ⎩

⇒ T3 = 0.155 s

Γ3 = 0.1984

3. Determine correlation coefficients.

β 12 =

ω1 = 0.969 ω2

∴ ρ 12 = 0.9089

β 13 =

ω1 = 0.326 ω3

∴ ρ 13 = 0.0061

ω2 = 0.336 ∴ ρ 23 = 0.0066 ω3 4. Determine spectral ordinates. From Fig. 6.9.5: β 23 =

A1 = 0.20 (2.71g ) = 0.54 g = 209 D1 =

ω 12

209 (13.24) 2

in s2

= 1.20 in

Mode 2: T2 = 0.460 s

From Problems 10.28 and 13.29:

ω 1 = 0.4834

rad s

Mode 1: T1 = 0.475 s

2. Natural frequencies and modes.

= 40.59

Use Eq. (13.7.10) to compute ρ ij for ζ = 0.05: x

A2 = 0.20 (2.71g) = 0.54 g = 209 D2 =

ω 22

209 (13.66) 2

in s2

= 1.12 in

Mode 3: T3 = 0.155 s EI mL3

= 13.66

rad s

A3 = 0.20 (2.71g ) = 0.54 g = 209 ⇒ T2 = 0.460 s

Γ2 = −0.8980

D3 =

ω 32

209 (40.59) 2

in s2

= 0.127 in

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 125 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

5. Determine peak modal displacements. u n = u nst An

u nst =

Γn

ω n2

st st Substituting numerical values for M xn , M yn , Tnst and An

in Eq. (b) gives:

φn

(a)

∴ u n = Γn φ n D n

u1 =

⎧ 0.7767 ⎫ ⎪ ⎪ − 0.3928 ⎨− 0.4923⎬ 1.20 = ⎪− 0.3928⎪ ⎭ ⎩

⎧− 0.3647 ⎫ ⎪ ⎪ ⎨ 0.2312⎬ ⎪ 0.1844⎪ ⎭ ⎩

⎧ 0.2100⎫ ⎧− 0.2084⎫ ⎪ ⎪ ⎪ ⎪ u 2 = − 0.8980 ⎨ 0.3975⎬ 1.12 = ⎨− 0.3905⎬ ⎪ 0.9049⎪ ⎪− 0.8980⎪ ⎭ ⎩ ⎭ ⎩

M x1 =

M y1 =

0.763 k ⋅ in

8.963 k ⋅ in

M x 2 = −22.524 k ⋅ in

M y 2 = 12.082 k ⋅ in

M x3 =

M y 3 = − 1.533 k ⋅ in

2.249 k ⋅ in

= − 9.726 k ⋅ in

= 10.442 k ⋅ in

= − 0.716 k ⋅ in

8. Combine peak modal responses. ⎧ 0.5943⎫ ⎪ ⎪ u 3 = 0.1984 ⎨ 0.7794⎬ 0.127 ⎪ 0.1984⎪ ⎭ ⎩

⎧ 0.0150⎫ ⎪ ⎪ = ⎨ 0.0197 ⎬ ⎪ 0.0050⎪ ⎭ ⎩

6. Combine peak modal displacements.

Response

Using Eqs. (13.7.3) and (13.7.4), the SRSS and CQC estimates for the peak displacements of the mass can be calculated. These estimates are summarized in the following table. Displacement ux uy uz

SRSS rule (inches) 0.421 0.454 0.923

CQC rule (inches) 0.195 0.205 1.075

7. Determine peak modal responses. st M xn = M xn An

st M yn = M yn An

Tn = Tnst An (b)

st st where M xn , M yn and Tnst are given in the solution to

Problem 13.29 as: st M xn = [ 0.0391mL, − 1.1544mL,

Using Eqs. (13.7.3) and (13.7.4), the SRSS and CQC estimates for the peak values of these responses can be calculated. These estimates are summarized in the following table.

Mx My T

SRSS rule (k ⋅ in) 22.65 15.12 14.29

CQC rule (k ⋅ in) 21.93 20.62 4.42

9. Comments. Modes 1 and 2 are strongly correlated with each other (ρ12=0.9089), but weakly correlated with mode 3. Consequently, significant differences between the SRSS and CQC estimates should result only when both modes 1 and 2 contribute significantly to the total response. Examining the above results, we see that this is the case. The SRSS and CQC estimates for the displacements ux and uy and responses My and T differ significantly. The effect of modal correlation is less pronounced for uz and Mx because the contributions from modes 1 and 2 are not comparable.

0.1153mL]

st M yn = [ 0.4594mL,

0.6192mL, − 0.0786mL]

Tnst = [ − 0.4985mL,

0.5352mL, − 0.0367mL]

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 126 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Use Eq. (13.7.10) to compute ρ ij for ζ = 0.05:

Problem 13.62 uz z

uy m

L b

∴ ρ 12 = 0.9089

β 13 =

ω1 = 0.326 ω3

∴ ρ 13 = 0.0061

β 23 =

ω2 = 0.336 ω3

∴ ρ 23 = 0.0066

4. Determine spectral ordinates.

ω1 = 0.969 ω2

β 12 =

From Fig. 6.9.5: 1. Data. L = 36 in.

m = 1.0 kips/g

E = 30000 ksi

I = 3.017 in4

G = 12000 ksi

J = 6.034 in

Mode 1: T1 = 0.475 s A1 = 0.20 (2.71g ) = 0.54 g = 209 D1 =

4 Note: GJ = EI 5

kips 1.0 k ⋅s2 m = 1.0 = = 0.00259 g 386.4 in

mL3

(30000)(3.017) (0.00259)(36) 3

= 27.38 s

−1

ω 1 = 0.4834

rad = 13.24 s

(13.24) 2

= 1.20 in

ω 22

209 (13.66) 2

in s2

= 1.12 in

⇒ T1 = 0.475 s

A3 = 0.20 (2.71g ) = 0.54 g = 209 A3

ω 32

209 (40.59) 2

in s2

= 0.127 in

5. Determine peak modal displacements. EI 3

⎧− 0.2084⎫ ⎪ ⎪ φ 2 = ⎨ 0.3875⎬ ⎪− 0.8980⎪ ⎩ ⎭

⎧ 0.5943⎫ ⎪ ⎪ φ 3 = ⎨ 0.7794⎬ ⎪ 0.1984⎪ ⎩ ⎭

D2 =

Γ1 = −0.0626

ω 2 = 0.4990

ω 3 = 1.4827

209

A2 = 0.20 (2.71g) = 0.54 g = 209

D3 =

⎧ 0.7767⎫ ⎪ ⎪ φ1 = ⎨− 0.4923⎬ ⎪− 0.3928⎪ ⎩ ⎭

Mode 3: T3 = 0.155 s

From Problems 10.28 and 13.30: EI

ω 12

Mode 2: T2 = 0.460 s

2. Natural frequencies and modes.

= 13.66

rad s

⇒ T2 = 0.460 s

u n = u nst An

mL3

rad = 40.59 s

Γn

ω n2

φn

(a)

∴ u n = Γn φ n D n

Γ2 = −0.4150

u nst =

⎧ 0.7767⎫ ⎪ ⎪ − 0.0626 ⎨− 0.4923⎬ 1.20 = ⎪− 0.3928⎪ ⎩ ⎭

⎧− 0.0581⎫ ⎪ ⎪ ⎨ 0.0368⎬ ⎪ 0.0294⎪ ⎩ ⎭

⎧− 0.2084⎫ ⎪ ⎪ u 2 = − 0.4150 ⎨ 0.3975⎬ 1.12 = ⎪− 0.8980⎪ ⎩ ⎭

⎧ 0.0971⎫ ⎪ ⎪ ⎨− 0.1805⎬ ⎪ 0.4182⎪ ⎩ ⎭

u1 =

⇒ T3 = 0.155 s

Γ3 = 0.9077

3. Determine correlation coefficients. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 127 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

⎧ 0.5943⎫ ⎪ ⎪ u 3 = 0.9077 ⎨ 0.7794⎬ 0.127 ⎪ 0.1984⎪ ⎩ ⎭

⎧ 0.0686⎫ ⎪ ⎪ ⎨ 0.0899⎬ ⎪ 0.0229⎪ ⎩ ⎭

Displacement ux uy uz

CQC rule (inches) 0.085 0.173 0.446

7. Determine peak modal responses. st M xn = M xn An

st M yn = M yn An

Tn = Tnst An (b)

Response Mx My T

SRSS rule (k ⋅ in) 14.64 9.08 6.04

CQC rule (k ⋅ in) 14.51 9.81 4.76

9. Comments. Modes 1 and 2 are strongly correlated with each other (ρ12=0.9089), but weakly correlated with mode 3. Consequently, significant differences between the SRSS and CQC estimates should result only when both modes 1 and 2 contribute significantly to the total response. Examining the above results, we see that this is the case. The SRSS and CQC estimates for the displacement ux and response T differ significantly. The effect of modal correlation is less pronounced for the other displacements and responses either because (1) the contribution from mode 3 is larger than that of modes 1 and 2 (e.g. My) or (2) the contributions from modes 1 and 2 are not comparable (e.g. uy, uz and Mx ).

st st where M xn , M yn and Tnst are given in the solution to

Problem 13.30 as: st M xn = [ 0.0062mL, − 0.5335mL,

0.5273mL]

[ 0.0732mL, 0.0262mL, − 0.03594mL] Tnst = [ − 0.0794mL, 0.2474mL, − 0.1680mL]

st M yn =

st st Substituting numerical values for M xn , M yn , Tnst and An

in Eq. (b) gives: M x1 =

0.122 k ⋅ in

M x 2 = − 10.410 k ⋅ in M x 3 = 10.289 k ⋅ in T1

= − 1.549 k ⋅ in

= − 3.277 k ⋅ in

M y1 =

1.428 k ⋅ in

M y 2 = 5.584 k ⋅ in M y 3 = − 7.012 k ⋅ in

4.826 k ⋅ in

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 128 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.63

3. Determine correlation coefficients. Use Eq. (13.7.10) to compute ρ ij for ζ = 0.05:

uz z

uy m

L b

ω1 = 0.969 ω2

∴ ρ 12 = 0.9089

β 13 =

ω1 = 0.326 ω3

∴ ρ 13 = 0.0061

β 23 =

ω2 = 0.336 ω3

∴ ρ 23 = 0.0066

d L

β 12 =

y x

4. Determine spectral ordinates. 1. Data.

From Fig. 6.9.5:

L = 36 in.

m = 1.0 kips/g

E = 30000 ksi

I = 3.017 in4

G = 12000 ksi

J = 6.034 in4

A1 = 0.20 (2.71g ) = 0.54 g = 209

4 Note: GJ = EI 5

D1 =

2. Natural frequencies and modes. m = 1.0 EI mL3

kips 1.0 k ⋅s2 = = 0.00259 g 386.4 in

Mode 1: T1 = 0.475 s

(30000)(3.017) (0.00259)(36) 3

ω 12

EI 3

D2 =

rad s

⎧ 0.7767⎫ ⎪ ⎪ φ1 = ⎨− 0.4923⎬ ⎪− 0.3928⎪ ⎩ ⎭ EI 3

= 13.66

⎧− 0.2084⎫ ⎪ ⎪

φ 2 = ⎨ 0.3875⎬

ω 22

209 (13.66) 2

in s2

= 1.12 in

Mode 3: T3 = 0.155 s

D3 =

ω 32

209 (40.59) 2

in s2

= 0.127 in

⇒ T2 = 0.460 s

Γ2 = −0.2084

⎪− 0.8980⎪ ⎩ ⎭

⎧ 0.5943⎫ ⎪ ⎪ φ 3 = ⎨ 0.7794⎬ ⎪ 0.1984⎪ ⎩ ⎭

rad s

= 1.20 in

A3 = 0.20 (2.71g ) = 0.54 g = 209

Γ1 = 0.7767

ω 2 = 0.4990

ω 3 = 1.4827

⇒ T1 = 0.475 s

(13.24) 2

A2 = 0.20 (2.71g) = 0.54 g = 209

= 27.38 s −1

= 13.24

209

Mode 2: T2 = 0.460 s

From Problems 10.28 and 13.27:

ω 1 = 0.4834

EI 3

= 40.59

rad s

⇒ T3 = 0.155 s

Γ3 = 0.5943

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 129 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

5. Determine modal static responses for Mα . The bending moment, Mα , about an axis oriented at an angle α =30° counterclockwise from the x-axis can be expressed in terms of the bending moments, Mx and My , about the x and y axes as:

7. Combine peak modal responses. (a) SRSS rule Using Eq. (13.7.3), the SRSS estimate for the peak value of Mα can be calculated:

(

M α = M α21 + M α2 2 + M α23

)

= 11.21 k ⋅ in

My Mα

(b) CQC rule Using Eq. (13.7.4), the CQC estimate for the peak value of Mα can be calculated :

α = 30°

M α = ( M α21 + M α2 2 + M α23 + 2 ρ 12 M α 1 M α 2

+ 2 ρ 13 M α1 M α 3 + 2 ρ 23 M α 2 M α 3 ) 2

M α = M x cos α + M y sin α

= 13.52 k ⋅ in

= 0.866M x + 0.500 M y

8. Comments.

Therefore, M αn = 0.866 M xn + 0.500M yn st st = (0.866M xn + 0.500M yn ) An

(a)

Because modes 1 and 2 are strongly correlated (ρ12 = 0.9089) and both contribute significantly to the total response, the SRSS and CQC estimates differ significantly.

= M αstn An

Thus, st st st st M αstn = 0.866 M xn + 0.500 M yn where M xn and M yn

are given in the solution to Problem 13.27: M xst1 = −0.0773mL

M yst1 = −0.9085mL

M xst2 = −0.2680mL

M yst2 = 0.1438mL

M xst3 =

M yst3 = −0.2353mL

0.3453mL

(b)

Equation (a), after substituting Eq. (b) becomes M αst1 = −0.5212mL M αst2 = −0.1602mL M αst3 = 0.1814mL

6. Determine peak modal responses Mαn. The peak values, Mαn, of the modal contributions Mαn(t) to response Mα are determined next M α1 = −0.5212 mLA1 = −10.17 k ⋅ in M α 2 = −0.1602 mLA2 = −3.13 k ⋅ in M α 3 = 0.1814 mLA3 = 3.54 k ⋅ in © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 130 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.64

ω 3 = 1.4827

uz z

uy m

L b

1. Data. L = 36 in.

m = 1.0 kips/g

E = 30000 ksi

I = 3.017 in4

G = 12000 ksi

J = 6.034 in

4 EI 5

Note: GJ =

kips 1.0 k ⋅s2 m = 1.0 = = 0.00259 g 386.4 in

(30000)(3.017) (0.00259)(36) 3

= 27.38 s −1

From Problems 10.28 and 13.28:

ω 1 = 0.4834

EI mL3

⎧ 0.5943⎫ ⎪ ⎪ φ 3 = ⎨ 0.7794⎬ ⎪ 0.1984⎪ ⎩ ⎭

rad s

⇒ T3 = 0.155 s

Γ3 = 0.7794

3. Determine correlation coefficients.

rad = 13.24 s

⎧ 0.7767⎫ ⎪ ⎪ φ1 = ⎨− 0.4923⎬ ⎪− 0.3928⎪ ⎩ ⎭

β 12 =

ω1 = 0.969 ω2

∴ ρ 12 = 0.9089

β 13 =

ω1 = 0.326 ω3

∴ ρ 13 = 0.0061

β 23 =

ω2 = 0.336 ω3

∴ ρ 23 = 0.0066

4. Determine spectral ordinates. From Fig. 6.9.5:

2. Natural frequencies and modes.

= 40.59

Use Eq. (13.7.10) to compute ρ ij for ζ = 0.05: x

Mode 1: T1 = 0.475 s A1 = 0.20 (2.71g ) = 0.54 g = 209 D1 =

ω 12

209 (13.24) 2

in s2

= 1.20 in

Mode 2: T2 = 0.460 s ⇒ T1 = 0.475 s

Γ1 = −0.4923

A2 = 0.20 (2.71g) = 0.54 g = 209 D2 =

ω 22

209 (13.66) 2

in s2

= 1.12 in

Mode 3: T3 = 0.155 s

ω 2 = 0.4990 ⎧− 0.2084⎫ ⎪ ⎪ φ 2 = ⎨ 0.3875⎬ ⎪− 0.8980⎪ ⎩ ⎭

EI 3

= 13.66

rad s

⇒ T2 = 0.460 s

Γ2 = 0.3875

A3 = 0.20 (2.71g ) = 0.54 g = 209 D3 =

ω 32

209 (40.59) 2

in s2

= 0.127 in

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 131 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

5. Determine modal static responses for Mα .

7. Combine peak modal responses.

The bending moment, Mα , about an axis oriented at an angle α =30° counterclockwise from the x-axis can be expressed in terms of the bending moments, Mx and My , about the x and y axes as:

(a) SRSS rule Using Eq. (13.7.3), the SRSS estimate for the peak value of Mα can be calculated:

(

M α = M α21 + M α2 2 + M α23

My Mα

)

= 9.84 k ⋅ in

(b) CQC rule

α = 30°

Using Eq. (13.7.4), the CQC estimate for the peak value of Mα can be calculated :

M α = ( M α21 + M α2 2 + M α23 + 2 ρ 12 M α 1 M α 2

M α = M x cos α + M y sin α

+ 2 ρ 13 M α1 M α 3 + 2 ρ 23 M α 2 M α 3 ) 2

= 0.866 M x + 0.500M y

= 12.87 k ⋅ in

Therefore,

8. Comments.

M αn = 0.866 M xn + 0.500 M yn

Because modes 1 and 2 are strongly correlated (ρ12 = 0.9089) and both contribute significantly to the total response, the SRSS and CQC estimates differ significantly.

st st = (0.866 M xn + 0.500 M yn ) An

= M αstn An

Thus, st st M αstn = 0.866 M xn + 0.500 M yn

(a)

st st where M xn and M yn are given in the solution to Problem

13.28: M xst1 = 0.0490mL

M yst1 = 0.5758mL

M xst2 = 0.4982mL

M yst2 = − 0.2672mL

M xst3 = 0.4528mL

M yst3 = −0.3086mL

(b)

Equation (a), after substituting Eq. (b) becomes M αst1 = 0.3303mL M αst2 = 0.2978mL M αst3 = 0.2378mL

6. Determine peak modal responses Mαn. The peak values, Mαn, of the modal contributions Mαn(t) to response Mα are determined next M α1 = 0.3303 mLA1 = 6.45 k ⋅ in M α 2 = 0.2978 mLA2 = 5.81 k ⋅ in M α 3 = 0.2378 mLA3 = 4.64 k ⋅ in © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 132 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.65

ω 3 = 1.4827

uz z

uy m

L b

1. Data. L = 36 in.

m = 1.0 kips/g

E = 30000 ksi

I = 3.017 in4

G = 12000 ksi

J = 6.034 in

4 EI 5

Note: GJ =

= 40.59

⎧ 0.5943⎫ ⎪ ⎪ φ 3 = ⎨ 0.7794⎬ ⎪ 0.1984⎪ ⎩ ⎭

rad s

⇒ T3 = 0.155 s

Γ3 = 0.1984

3. Determine correlation coefficients. Use Eq. (13.7.10) to compute ρ ij for ζ = 0.05:

β 12 =

ω1 = 0.969 ω2

∴ ρ 12 = 0.9089

β 13 =

ω1 = 0.326 ω3

∴ ρ 13 = 0.0061

β 23 =

ω2 = 0.336 ω3

∴ ρ 23 = 0.0066

4. Determine spectral ordinates. From Fig. 6.9.5: Mode 1: T1 = 0.475 s A1 = 0.20 (2.71g ) = 0.54 g = 209

2. Natural frequencies and modes. m = 1.0 EI 3

kips 1.0 k ⋅s2 = = 0.00259 g 386.4 in (30000)(3.017)

(0.00259)(36) 3

= 27.38 s −1

EI 3

= 13.24

⎧ 0.7767⎫ ⎪ ⎪

φ1 = ⎨− 0.4923⎬

⎧− 0.2084⎫ ⎪ ⎪ φ 2 = ⎨ 0.3875⎬ ⎪− 0.8980⎪ ⎩ ⎭

⇒ T1 = 0.475 s

Γ1 = −0.3928

⎪− 0.3928⎪ ⎩ ⎭

ω 2 = 0.4990

rad s

ω 12

209

(13.24) 2

= 1.20 in

Mode 2: T2 = 0.460 s A2 = 0.20 (2.71g) = 0.54 g = 209

From Problems 10.28 and 13.29:

ω 1 = 0.4834

D1 =

D2 =

ω 22

209 (13.66) 2

= 13.66

rad s

⇒ T2 = 0.460 s

in s2

= 1.12 in

Mode 3: T3 = 0.155 s A3 = 0.20 (2.71g ) = 0.54 g = 209

in s2

D3 =

ω 32

209 (40.59) 2

in s2

= 0.127 in

Γ2 = −0.8980

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 133 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

0.2635 mLA1 =

5.14 k ⋅ in

5. Determine modal static responses for Mα .

M α1 =

The bending moment, Mα , about an axis oriented at an angle α =30° counterclockwise from the x-axis can be expressed in terms of the bending moments, Mx and My, about the x and y axes as:

M α 2 = − 0.6901 mLA2 = − 13.47 k ⋅ in Mα3 =

0.0605 mLA3 =

1.18 k ⋅ in

7. Combine peak modal responses. (a) SRSS rule

Using Eq. (13.7.3), the SRSS estimate for the peak value of Mα can be calculated:

Mα

(

α = 30°

M α = M α21 + M α2 2 + M α23

)

= 14.46 k ⋅ in

(b) CQC rule Using Eq. (13.7.4), the CQC estimate for the peak value of Mα can be calculated :

M α = M x cos α + M y sin α = 0.866 M x + 0.500M y

M α = ( M α21 + M α2 2 + M α23 + 2 ρ 12 M α 1 M α 2

Therefore,

+ 2 ρ 13 M α1 M α 3 + 2 ρ 23 M α 2 M α 3 ) 2

M αn = 0.866 M xn + 0.500 M yn

= 9.12 k ⋅ in

st st = (0.866 M xn + 0.500 M yn ) An

8. Comments.

= M αstn An

Thus, st st M αstn = 0.866 M xn + 0.500 M yn

(a)

Because modes 1 and 2 are strongly correlated (ρ12 = 0.9089) and both contribute significantly to the total response, the SRSS and CQC estimates differ significantly.

st st where M xn and M yn are given in the solution to Problem

13.29: M xst1 =

0.0392mL

M yst1 = 0.4593mL

M xst2 = − 1.1544mL

M yst2 = 0.6192mL

M xst3 =

M yst3 = −0.0785mL

0.1152mL

(b)

Equation (a), after substituting Eq. (b) becomes M αst1 =

0.2635mL

M αst2 = − 0.6901mL M αst3 =

0.0605mL

6. Determine peak modal responses Mαn. The peak values, Mαn, of the modal contributions Mαn(t) to response Mα are determined next

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 134 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.66 3. Determine correlation coefficients.

uz z

Use Eq. (13.7.10) to compute ρ ij for ζ = 0.05:

uy m

L b

d L

y x

1. Data.

β 12 =

ω1 = 0.969 ω2

∴ ρ 12 = 0.9089

β 13 =

ω1 = 0.326 ω3

∴ ρ 13 = 0.0061

β 23 =

ω2 = 0.336 ω3

∴ ρ 23 = 0.0066

L = 36 in.

m = 1.0 kips/g

4. Determine spectral ordinates. From Fig. 6.9.5:

E = 30000 ksi

I = 3.017 in4

Mode 1: T1 = 0.475 s

G = 12000 ksi

J = 6.034 in

4 EI 5

Note: GJ =

D1 =

2. Natural frequencies and modes. m = 1.0 EI 3

kips 1.0 k ⋅s2 = = 0.00259 g 386.4 in

(30000)(3.017) (0.00259)(36)

A1 = 0.20 (2.71g ) = 0.54 g = 209

= 27.38 s −1

ω 12

rad s

= 13.24

ω 22

209 (13.66) 2

= 1.12 in

Mode 3: T3 = 0.155 s

Γ1 = −0.0626 D3 = EI

ω 2 = 0.4990

= 13.66

⎧− 0.2084⎫ ⎪ ⎪

φ 2 = ⎨ 0.3875⎬

rad s

ω 32

209 (40.59) 2

in s2

= 0.127 in

⇒ T2 = 0.460 s

Γ2 = −0.4150

⎪− 0.8980⎪ ⎩ ⎭

⎧ 0.5943⎫ ⎪ ⎪ φ 3 = ⎨ 0.7794⎬ ⎪ 0.1984⎪ ⎩ ⎭

A3 = 0.20 (2.71g) = 0.54 g = 209

⎧ 0.7767⎫ ⎪ ⎪ φ1 = ⎨− 0.4923⎬ ⎪− 0.3928⎪ ⎩ ⎭

ω 3 = 1.4827

⇒ T1 = 0.475 s

= 1.20 in

A2 = 0.20 (2.71g) = 0.54 g = 209

From Problems 10.28 and 13:30: EI

(13.24) 2

Mode 2: T2 = 0.460 s

D2 =

ω 1 = 0.4834

209

EI 3

= 40.59

rad s

⇒ T3 = 0.155 s

Γ3 = 0.9077

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 135 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

5. Determine modal static responses for Mα

7. Combine peak modal responses.

The bending moment, Mα , about an axis oriented at an angle α =30° counterclockwise from the x-axis can be expressed in terms of the bending moments, Mx and My, about the x and y axes as:

(a) SRSS rule

Using Eq. (13.7.3), the SRSS estimate for the peak value of Mα can be calculated:

(

M α = M α21 + M α2 2 + M α23

Mα

)

= 8.28 k ⋅ in

(b) CQC rule

α = 30°

Using Eq. (13.7.4), the CQC estimate for the peak value of Mα can be calculated :

M α = ( M α21 + M α2 2 + M α23 + 2 ρ 12 M α 1 M α 2

M α = M x cos α + M y sin α

+ 2 ρ 13 M α1 M α 3 + 2 ρ 23 M α 2 M α 3 ) 2

= 0.866 M x + 0.500M y

= 7.68 k ⋅ in

Therefore,

8. Comments.

M αn = 0.866 M xn + 0.500 M yn

The contributions to the total response from modes 2 and 3 are much larger than the contribution from mode 1. Because modes 2 and 3 are only weakly correlated (ρ23 = 0.0066), the SRSS and CQC estimates do not differ significantly.

st st = (0.866 M xn + 0.500 M yn ) An

= M αstn An

Thus, st st M αstn = 0.866 M xn + 0.500 M yn

(a)

st st where M xn and M yn are given in the solution to Problem

13.30: M xst1 =

0.0062mL

M yst1 = 0.0731mL

M xst2 = − 0.5335mL

M yst2 = 0.2862mL

M xst3 = 0.5273mL

M yst3 = −0.3593mL

(b)

Equation (a), after substituting Eq. (b) becomes M αst1 =

0.0420mL

M αst2 = − 0.3190mL M αst3 =

0.2770mL

6. Determine peak modal responses Mαn. The peak values, Mαn, of the modal contributions Mαn(t) to response Mα are determined next M α1 = 0.0420 mLA1 =

0.82 k ⋅ in

M α 2 = − 0.3190 mLA2 = − 6.22 k ⋅ in M α 3 = 0.2770 mLA3 =

5.40 k ⋅ in

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 136 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.67

3. Determine correlation coefficients. Use Eq. (13.7.10) to compute ρ ij for ζ = 0.05:

uz z

uy m

L b

ω1 = 0.969 ω2

∴ ρ 12 = 0.9089

β 13 =

ω1 = 0.326 ω3

∴ ρ 13 = 0.0061

β 23 =

ω2 = 0.336 ω3

∴ ρ 23 = 0.0066

d L

β 12 =

y x

4. Determine spectral ordinates. 1. Data.

From Fig. 6.9.5:

L = 36 in.

m = 1.0 kips/g

E = 30000 ksi

I = 3.017 in4

G = 12000 ksi

J = 6.034 in4

Mode 1: T1 = 0.475 s A1 = 0.20 (2.71g ) = 0.54 g = 209

4 Note: GJ = EI 5

D1 =

2. Natural frequencies and modes. m = 1.0 EI mL3

kips 1.0 k ⋅s2 = = 0.00259 g 386.4 in

(30000)(3.017) (0.00259)(36) 3

= 27.38 s

EI mL3

rad s

= 13.24

EI mL3

= 13.66

⎧− 0.2084⎫ ⎪ ⎪ φ 2 = ⎨ 0.3875⎬ ⎪− 0.8980⎪ ⎩ ⎭

ω 3 = 1.4827 ⎧ 0.5943⎫ ⎪ ⎪ φ 3 = ⎨ 0.7794⎬ ⎪ 0.1984⎪ ⎩ ⎭

rad s

209 (13.24) 2

= 1.20 in

D2 =

ω 22

209 (13.66) 2

in s2

= 1.12 in

Mode 3: T3 = 0.155 s ⇒ T1 = 0.475 s

Γ1 = 0.7767

ω 2 = 0.4990

A2 = 0.20 (2.71g) = 0.54 g = 209

A3 = 0.20 (2.71g ) = 0.54 g = 209 D3 =

⎧ 0.7767⎫ ⎪ ⎪ φ1 = ⎨− 0.4923⎬ ⎪− 0.3928⎪ ⎩ ⎭

ω 12

Mode 2: T2 = 0.460 s

−1

From Problems 10.28 and 13.27:

ω 1 = 0.4834

ω 32

209 (40.59) 2

in s2

= 0.127 in

⇒ T2 = 0.460 s

Γ2 = −0.2084

EI 3

= 40.59

rad s

⇒ T3 = 0.155 s

Γ3 = 0.5943

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 137 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

M α1 = [(−0.0773 cos α − 0.9085 sin α ) mL] A1

5. Determine modal static responses for Mα . The bending moment, Mα , about an axis oriented at an angle α counterclockwise from the x-axis can be expressed in terms of the bending moments, Mx and My, about the x and y axes as:

= −1.51 cos α − 17.73 sin α M α 2 = [(−0.2680 cos α + 0.1438 sin α ) mL] A2 = −5.23 cos α + 2.80 sin α M α 3 = [(0.3453 cos α − 0.2353 sin α ) mL] A3

= 6.74 cos α − 4.59 sin α

Mα

7. Combine peak modal responses.

(a) SRSS rule Mx

Using Eq. (13.7.3), the SRSS estimate for the peak value of Mα can be calculated:

(

M α = M x cos α + M y sin α

M α = M α21 + M α2 2 + M α2 3

)

= [(− 1.51 cos α − 17.73 sin α )2

Therefore,

+ (− 5.23 cos α + 2.80 sin α )2

M αn = M xn cos α + M yn sin α

+ (6.74 cos α − 4.59 sin α )2

st st = ( M xn cos α + M yn sin α ) An

(

]

= A sin 2 α + B sin α cos α + C cos 2 α

= M αstn An

)

where

Thus, st st M αstn = M xn cos α + M yn sin α

(a)

st st where M xn and M yn are given in the solution to Problem

13.27: M xst1 = −0.0773mL

M yst1 = −0.9085mL

M xst2 = −0.2680mL

M yst2 = 0.1438mL

M xst3 =

M yst3 = −0.2353mL

0.3453mL

A = 343.12 k ⋅ in

B = −37.70 k ⋅ in C = 75.00 k ⋅ in

(b) CQC rule Using Eq. (13.7.4), the CQC estimate for the peak value of Mα can be calculated : M α = ( M α21 + M α2 2 + M α2 3 + 2 ρ12 M α 1 M α 2

(b)

Equation (a), after substituting Eq. (b) becomes M αst1 = (−0.0773 cos α − 0.9085 sin α )mL

+ 2 ρ 13 M α 1 M α 3 + 2 ρ 23 M α 2 M α 3 ) 2

(d)

= ( A sin 2 α + B sin α cos α + C cos 2 α ) 2

where A = 253.59 k ⋅ in

B = 122.25 k ⋅ in C = 88.75 k ⋅ in

M αst2 = (−0.2680 cos α + 0.1438 sin α )mL

8. Determine maximum of Mα over α.

M αst3 = (0.3453 cos α − 0.2353 sin α )mL

To find the value of α corresponding to the maximum value of Mα , solve

6. Determine peak modal responses Mαn. The peak values, Mαn, of the modal contributions Mαn(t) to response Mα are determined next

dM α = 0 for α dα dM α 1 2( A − C ) sin α cos α + B (cos 2 α − sin 2 α ) = dα 2 A sin 2 α + B sin α cos α + C cos 2 α =

1 2

( A − C ) sin 2α + B cos 2α A sin 2 α + B sin α cos α + C cos 2 α

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 138 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Therefore, the maximum value of Mα occurs for α that satisfies ( A − C ) sin 2α + B cos 2α = 0

Hence, tan 2α =

−B A−C

(e)

(a) SRSS rule Substituting A, B and C from Eq. (c) into Eq. (e) gives tan 2α = 0.1406

α=

1 tan −1 (0.1406) = 0.0698 rad , − 1.5010 rad 2

The two values of α obtained correspond to the maximum and minimum values of Mα. The value of α corresponding to the maximum value of Mα can be ascertained by substituting each value of α into Eq. (c) for α = 0.0698 rad,

M α = 8.58 k ⋅ in

for α = −1.5010 rad,

M α = 18.56 k ⋅ in

Thus, the peak value of Mα estimated by the SRSS combination rule is Mα = 18.56 k⋅in about an axis oriented 1.5010 rad (86.0°) clockwise from the x-axis.

(b) CQC rule Substituting A, B and C from Eq. (d) into Eq. (e) gives tan 2α = −0.7416

α=

1 tan −1 (−0.7416) = −0.3191 rad , 1.2517 rad 2 for α = − 0.3191 rad,

M α = 8.28 k ⋅ in

for α = 1.2517 rad,

M α = 16.55 k ⋅ in

Thus, the peak value of Mα estimated by the CQC combination rule is Mα = 16.55 k⋅in about an axis oriented 1.2517 rad (71.7°) counterclockwise from the x-axis. 9. Comments. Modes 1 and 2 are strongly correlated (ρ12 = 0.9089); hence, the CQC estimate, which accounts for this correlation, is expected to be more accurate than the SRSS estimate. Note that the SRSS and CQC estimates for the peak value of Mα differ by 11.4%. The values of α predicted by the two combination rules differ by 14.3°.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 139 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.68 3. Determine correlation coefficients.

uz z

Use Eq. (13.7.10) to compute ρ ij for ζ = 0.05:

uy m

L b

d L

y x

1. Data. L = 36 in.

m = 1.0 kips/g

E = 30000 ksi

I = 3.017 in4

G = 12000 ksi

J = 6.034 in4

EI mL3

kips 1.0 k ⋅s2 = = 0.00259 g 386.4 in

(30000)(3.017) (0.00259)(36) 3

= 27.38 s −1

EI 3

rad s

= 13.24

⎧ 0.7767 ⎫ ⎪ ⎪ φ1 = ⎨− 0.4923⎬ ⎪− 0.3928⎪ ⎭ ⎩

∴ ρ 13 = 0.0061

β 23 =

ω2 = 0.336 ω3

∴ ρ 23 = 0.0066

4. Determine spectral ordinates. From Fig. 6.9.5: Mode 1: T1 = 0.475 s

⇒ T1 = 0.475 s

ω 12

209

(13.24) 2

in s2

= 1.20 in

Mode 2: T2 = 0.460 s A2 = 0.20 (2.71g) = 0.54 g = 209 A2

ω 22

209 (13.66) 2

in s2

= 1.12 in

Mode 3: T3 = 0.155 s

Γ1 = −0.4923 D3 = EI 3

= 13.66

⎧− 0.2084⎫ ⎪ ⎪ φ 2 = ⎨ 0.3875⎬ ⎪− 0.8980⎪ ⎭ ⎩

⎧ 0.5943⎫ ⎪ ⎪ φ 3 = ⎨ 0.7794⎬ ⎪ 0.1984⎪ ⎭ ⎩

ω1 = 0.326 ω3

A3 = 0.20 (2.71g ) = 0.54 g = 209

ω 2 = 0.4990

ω 3 = 1.4827

β 13 =

D2 =

From Problems 10.28 and 13.28:

ω 1 = 0.4834

∴ ρ 12 = 0.9089

D1 =

2. Natural frequencies and modes. m = 1.0

ω1 = 0.969 ω2

A1 = 0.20 (2.71g ) = 0.54 g = 209

4 EI 5

Note: GJ =

β 12 =

rad s

ω 32

209 (40.59) 2

in s2

= 0.127 in

⇒ T2 = 0.460 s

Γ2 = 0.3875

EI 3

= 40.59

rad s

⇒ T3 = 0.155 s

Γ3 = 0.7794

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 140 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

M α 1 = [(0.0490 cos α + 0.5758 sin α ) mL] A1

= 0.96 cos α + 11.23 sin α M α 2 = [(0.4982 cos α − 0.2672 sin α ) mL] A2 = 9.72 cos α − 5.21 sin α M α 3 = [(0.4528 cos α − 0.3086 sin α ) mL] A3 = 8.83 cos α − 6.02 sin α

My Mα

7. Combine peak modal responses.

(a) SRSS rule Using Eq. (13.7.3), the SRSS estimate for the peak value of Mα can be calculated:

(

M α = M α21 + M α2 2 + M α2 3

M α = M x cos α + M y sin α

= [(0.96 cos α + 11.23 sin α )2

Therefore,

+ (9.72 cos α − 5.21 sin α )2

M αn = M xn cos α + M yn sin α

+ (8.83 cos α − 6.02 sin α )2

(

st st = ( M xn cos α + M yn sin α ) An

]

= A sin 2 α + B sin α cos α + C cos 2 α

= M αstn An Thus,

)

where A = 189.66 k ⋅ in

st st M αstn = M xn cos α + M yn sin α st where M xn

)

(a)

st and M yn are given in the solution to Problem

13.28: M xst1 = 0.0490mL

M yst1 = 0.5758mL

M xst2 = 0.4982mL

M yst2 = − 0.2672mL

M xst3 = 0.4528mL

M yst3 = −0.3086mL

B = −186.27 k ⋅ in

C = 173.46 k ⋅ in

(b) CQC rule Using Eq. (13.7.4), the CQC estimate for the peak value of Mα can be calculated : M α = ( M α21 + M α2 2 + M α2 3 + 2 ρ12 M α 1M α 2 1

+ 2 ρ13M α 1M α 3 + 2 ρ 23 M α 2 M α 3 ) 2

(b)

Equation (a) after substituting Eq. (b) becomes M αst1 = (0.0490 cos α + 0.5758 sin α )mL

(d) 1

= ( A sin 2 α + B sin α cosα + C cos 2 α ) 2

where A = 82.75 k ⋅ in

B = 2.97 k ⋅ in

C = 191.60 k ⋅ in

M αst2 = (0.4982 cos α − 0.2672 sin α )mL

8. Determine maximum of Mα over α.

M αst3 = (0.4528 cos α − 0.3086 sin α )mL

To find the value of α corresponding to the maximum value of Mα , solve

6. Determine peak modal responses Mαn. The peak values, Mαn, of the modal contributions Mαn(t) to response Mα are determined next

dM α = 0 for α dα dM α 1 2( A − C ) sin α cos α + B (cos 2 α − sin 2 α ) = dα 2 A sin 2 α + B sin α cos α + C cos 2 α =

1 2

( A − C ) sin 2α + B cos 2α

A sin 2 α + B sin α cos α + C cos 2 α

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 141 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Therefore, the maximum value of Mα occurs for α that satisfies ( A − C ) sin 2α + B cos 2α = 0

Hence, tan 2α =

−B A−C

(e)

(a) SRSS rule Substituting A, B and C from Eq. (c) into Eq. (e) gives tan 2α = 11.5000 1 α = tan −1 (11.5000) = 0.7420 rad , − 0.8288 rad 2

M α = 9.38 k ⋅ in

for α = −0.8288 rad,

M α = 16.58 k ⋅ in

Thus, the peak value of Mα estimated by the SRSS combination rule is Mα = 16.58 k⋅in about an axis oriented 0.8288 rad (47.5°) clockwise from the x-axis. (b) CQC rule Substituting A, B and C from Eq. (d) into Eq. (e) gives tan 2α = 0.0273

α=

1 tan −1 (0.0273) = 0.0136 rad , − 1.5572 rad 2 for α = 0.0136 rad,

M α = 13.84 k ⋅ in

for α = −1.5572 rad,

M α = 9.10 k ⋅ in

Thus, the peak value of Mα estimated by the CQC combination rule is Mα = 13.84 k⋅in about an axis oriented 0.0136 rad (0.78°) counterclockwise from the x-axis. 9. Comments. Modes 1 and 2 are strongly correlated (ρ12 = 0.9089); hence, the CQC estimate, which accounts for this correlation, is expected to be more accurate than the SRSS estimate. Note that the SRSS and CQC estimates for the peak value of Mα differ by 18.0%. The values of α predicted by the two combination rules differ by 46.7°.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 142 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 13.69

Use Eq. (13.7.10) to compute ρ ij for ζ = 0.05:

d L

y x

ω1 = 0.969 ω2

∴ ρ 12 = 0.9089

β 13 =

ω1 = 0.326 ω3

∴ ρ 13 = 0.0061

β 23 =

ω2 = 0.336 ω3

∴ ρ 23 = 0.0066

L b

β 12 =

4. Determine spectral ordinates.

1. Data. L = 36 in.

m = 1.0 kips/g

E = 30000 ksi

I = 3.017 in4

G = 12000 ksi

J = 6.034 in4

4 Note: GJ = EI 5

From Fig. 6.9.5: Mode 1: T1 = 0.475 s A1 = 0.20 (2.71g ) = 0.54 g = 209 D1 =

2. Natural frequencies and modes. kips 1.0 k ⋅s2 = = 0.00259 m = 1.0 g 386.4 in

EI 3

(30000)(3.017) (0.00259)(36)

= 27.38 s −1

From Problems 10.28 and 13.29:

ω 1 = 0.4834

EI mL3

rad = 13.24 s

⎧ 0.7767 ⎫ ⎪ ⎪ φ1 = ⎨− 0.4923⎬ ⎪− 0.3928⎪ ⎭ ⎩

EI mL3

= 13.66

⎧− 0.2084⎫ ⎪ ⎪ φ 2 = ⎨ 0.3875⎬ ⎪− 0.8980⎪ ⎭ ⎩

ω 3 = 1.4827 ⎧ 0.5943⎫ ⎪ ⎪ φ 3 = ⎨ 0.7794⎬ ⎪ 0.1984⎪ ⎭ ⎩

rad s

⇒ T1 = 0.475 s

mL3

= 40.59

rad s

209 (13.24) 2

= 1.20 in

Mode 2: T2 = 0.460 s A2 = 0.20 (2.71g) = 0.54 g = 209 D2 =

ω 22

209 (13.66) 2

in s2

= 1.12 in

A3 = 0.20 (2.71g ) = 0.54 g = 209 D3 =

ω 32

209 (40.59) 2

in s2

= 0.127 in

5. Determine modal static responses for Mα. ⇒ T2 = 0.460 s

Γ2 = −0.8980

ω 12

Mode 3: T3 = 0.155 s

Γ1 = −0.3928

ω 2 = 0.4990

The bending moment, Mα , about an axis oriented at an angle α counterclockwise from the x-axis can be expressed in terms of the bending moments, Mx and My, about the x and y axes as: My

⇒ T3 = 0.155 s

Mα

α Γ3 = 0.1984

3. Determine correlation coefficients.

Mx M α = M x cos α + M y sin α

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 143 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Therefore,

where

M αn = M xn cos α + M yn sin α

A = 228.65 k ⋅ in

B = −537.47 k ⋅ in

C = 512.97 k ⋅ in

st st cos α + M yn sin α ) An = ( M xn

(b) CQC rule

= M αstn An

Using Eq. (13.7.4), the CQC estimate for the peak value of Mα can be calculated:

Thus,

M α = ( M α21 + M α2 2 + M α2 3 + 2 ρ12 M α 1 M α 2

st st M αstn = M xn cos α + M yn sin α

(a)

+ 2 ρ 13 M α 1 M α 3 + 2 ρ 23 M α 2 M α 3 ) 2 = ( A sin 2 α + B sin α cos α + C cos 2 α ) 2

st st where M xn and M yn are given in the solution to Problem

13.29:

(d)

where

M xst1 =

0.0392mL

M yst1 =

0.4593mL

M xst2 = − 1.1544mL

M yst2 =

0.6192mL

M xst3 = 0.1152mL

M yst3 = −0.0785mL

(b)

To find the value of α corresponding to the maximum value of Mα , solve

Equation (a) after substituting Eq. (b) becomes

dM α = 0 for α dα

M αst1 = (0.0392 cos α + 0.4593 sin α )mL M αst2 = (−1.1544 cos α + 0.6192 sin α )mL

dM α 1 2( A − C ) sin α cos α + B (cos 2 α − sin 2 α ) = dα 2 A sin 2 α + B sin α cos α + C cos 2 α

M αst3 = (0.1152 cos α − 0.0785 sin α )mL

6. Determine peak modal responses Mαn. The peak values, Mαn, of the modal contributions Mαn(t) to response Mα are determined next M α1 = [(0.0392 cos α + 0.4593 sin α ) mL] A1 = 0.76 cos α + 8.96 sin α

7. Combine peak modal responses. (a) SRSS rule Using Eq. (13.7.3), the SRSS estimate for the peak value of Mα can be calculated: 2

= [(0.76 cos α + 8.96 sin α )2 + (− 22.52 cos α + 12.08 sin α )2 + (2.25 cos α − 1.53 sin α )2

(

Therefore, the maximum value of Mα occurs for α that satisfies

]

(c)

= A sin 2 α + B sin α cos α + C cos 2 α

)

−B A−C

(e)

(a) SRSS rule

= 2.25 cos α − 1.53 sin α

A sin 2 α + B sin α cos α + C cos 2 α

tan 2α =

M α 3 = [(0.1152 cos α − 0.0785 sin α ) mL] A3

)

( A − C ) sin 2α + B cos 2α

Hence,

= −22.52 cos α + 12.08 sin α

(

1 2

( A − C ) sin 2α + B cos 2α = 0

M α 2 = [(−1.1544 cos α + 0.6192 sin α ) mL] A2

M α = M α21 + M α2 2 + M α2 3

A = 425.09 k ⋅ in B = −886.66 k ⋅ in C = 481.09 k ⋅ in 8. Determine maximum of Mα over α.

Substituting A, B and C from Eq. (c) into Eq. (e) gives tan 2α = −1.8904

α=

1 tan −1 (−1.8904) = −0.5421 rad , 1.0287 rad 2

M α = 25.98 k ⋅ in

for α =

M α = 8.17 k ⋅ in

1.0287 rad,

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 144 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Thus, the peak value of Mα estimated by the SRSS combination rule is Mα = 25.98 k⋅in about an axis oriented 0.5421 rad (31.1°) clockwise from the x-axis. (b) CQC rule Substituting A, B and C from Eq. (d) into Eq. (e) gives tan 2α = −15.8335 1 tan −1(−1.8335) = −0.7539 rad , 0.8169 rad 2 for α = − 0.7539 rad, Mα = 29.96 k ⋅ in

α=

for α =

0.8169 rad,

Mα = 2.98 k ⋅ in

Thus, the peak value of Mα estimated by the CQC combination rule is Mα = 29.96 k⋅in about an axis oriented 0.7539 rad (43.2°) clockwise from the x-axis. 9. Comments. Modes 1 and 2 are strongly correlated (ρ12 = 0.9089); hence, the CQC estimate, which accounts for this correlation, is expected to be more accurate than the SRSS estimate. Note that the SRSS and CQC estimates for the peak value of Mα differ by 14.2%. The values of α predicted by the two combination rules differ by 12.1°.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 145 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Use Eq. (13.7.10) to compute ρ ij for ζ = 0.05:

Problem 13.70 uz z

uy m

L b

∴ ρ 12 = 0.9089

β 13 =

ω1 = 0.326 ω3

∴ ρ 13 = 0.0061

β 23 =

ω2 = 0.336 ω3

∴ ρ 23 = 0.0066

4. Determine spectral ordinates.

ω1 = 0.969 ω2

1. Data.

β 12 =

L = 36 in.

m = 1.0 kips/g

E = 30000 ksi

I = 3.017 in4

G = 12000 ksi

J = 6.034 in4

4 Note: GJ = EI 5

From Fig. 6.9.5: Mode 1: T1 = 0.475 s A1 = 0.20 (2.71g ) = 0.54 g = 209 A1

D1 =

ω 12

2. Natural frequencies and modes. kips 1.0 k ⋅s2 m = 1.0 = = 0.00259 g 386.4 in EI 3

(30000)(3.017) (0.00259)(36)

= 27.38 s −1

From Problems 10.28 and 13.30:

ω 1 = 0.4834

EI mL3

rad = 13.24 s

⎧ 0.7767 ⎫ ⎪ ⎪ φ1 = ⎨− 0.4923⎬ ⎪− 0.3928⎪ ⎩ ⎭

209 (13.24) 2

in s2

= 1.20 in

Mode 2: T2 = 0.460 s A2 = 0.20 (2.71g) = 0.54 g = 209 D2 =

ω 22

209 (13.66) 2

in s2

= 1.12 in

Mode 3: T3 = 0.155 s ⇒ T1 = 0.475 s

Γ1 = −0.0626

A3 = 0.20 (2.71g ) = 0.54 g = 209 D3 =

ω 32

209 (40.59) 2

in s2

= 0.127 in

5. Determine modal static responses for Mα . EI

ω 2 = 0.4990

mL3

⎧− 0.2084⎫ ⎪ ⎪ φ 2 = ⎨ 0.3875⎬ ⎪− 0.8980⎪ ⎩ ⎭

ω 3 = 1.4827 ⎧ 0.5943⎫ ⎪ ⎪ φ 3 = ⎨ 0.7794⎬ ⎪ 0.1984⎪ ⎩ ⎭

= 13.66

rad s

⇒ T2 = 0.460 s

The bending moment, Mα , about an axis oriented at an angle α counterclockwise from the x-axis can be expressed in terms of the bending moments, Mx and My, about the x and y axes as:

Γ2 = −0.4150

My EI

mL3

rad = 40.59 s

Mα ⇒ T3 = 0.155 s

Γ3 = 0.9077

3. Determine correlation coefficients.

Mx M α = M x cos α + M y sin α

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 146 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Therefore,

where

M αn = M xn cos α + M yn sin α

A = 82.38 k ⋅ in

B = −260.20 k ⋅ in

C = 214.25 k ⋅ in

st st = ( M xn cos α + M yn sin α ) An

(b) CQC rule

= M αstn An

Using Eq. (13.7.4), the CQC estimate for the peak value of Mα can be calculated :

Thus,

M α = ( M α21 + M α2 2 + M α2 3 + 2 ρ12 M α 1 M α 2

st st M αstn = M xn cos α + M yn sin α

(a)

+ 2 ρ 13 M α 1 M α 3 + 2 ρ 23 M α 2 M α 3 ) 2 = ( A sin 2 α + B sin α cos α + C cos 2 α ) 2

st st where M xn and M yn are given in the solution to Problem

13.30:

(d)

where

M xst1 =

0.0062mL

M yst1 =

0.0731mL

M xst2 = − 0.5335mL

M yst2 =

0.2862mL

M xst3 = 0.5273mL

M yst3 = −0.3593mL

A = 96.24 k ⋅ in

B = −284.10 k ⋅ in

C = 210.56 k ⋅ in

(b) 8. Determine maximum of Mα over α. To find the value of α corresponding to the maximum value of Mα , solve

Equation (a) after substituting Eq. (b) becomes M αst1 = (0.0062 cos α + 0.0731 sin α )mL

dM α = 0 for α dα

M αst2 = (−0.5335 cos α + 0.2862 sin α )mL M αst3 = (0.5273 cos α − 0.3593 sin α )mL

6. Determine peak modal responses Mαn. The peak values, Mαn, of the modal contributions Mαn(t) to response Mα are determined next

dM α 1 2( A − C ) sin α cos α + B (cos 2 α − sin 2 α ) = dα 2 A sin 2 α + B sin α cos α + C cos 2 α =

M α 1 = [(0.0062 cos α + 0.0731 sin α ) mL] A1 = 0.12 cos α + 1.43 sin α

1 2

( A − C ) sin 2α + B cos 2α A sin 2 α + B sin α cos α + C cos 2 α

Therefore, the maximum value of Mα occurs for α that satisfies

M α 2 = [(−0.5335 cos α + 0.2862 sin α ) mL] A2 = −10.41 cos α + 5.58 sin α

( A − C ) sin 2α + B cos 2α = 0

M α 3 = [(0.5273 cos α − 0.3593 sin α ) mL] A3 = 10.29 cos α − 7.01 sin α

Hence, tan 2α =

7. Combine peak modal responses.

−B A−C

(e)

(a) SRSS rule Using Eq. (13.7.3), the SRSS estimate for the peak value of Mα can be calculated:

(

M α = M α21 + M α2 2 + M α2 3

)

= [(0.12 cos α + 1.43 sin α )2

α=

+ (− 10.41 cos α + 5.58 sin α )2

(

Substituting A, B and C from Eq. (c) into Eq. (e) gives tan 2α = −1.9732

+ (10.29 cos α − 7.01 sin α )2

(a) SRSS rule

]

(c)

1 tan −1 (−1.9732) = −0.5509 rad , 1.0199 rad 2

= A sin 2 α + B sin α cos α + C cos 2 α

)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 147 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

M α = 17.15 k ⋅ in

for α =

M α = 1.57 k ⋅ in

1.0199 rad,

Thus, the peak value of Mα estimated by the SRSS combination rule is Mα = 17.15 k⋅in about an axis oriented 0.5509 rad (31.6°) clockwise from the x-axis. (b) CQC rule Substituting A, B and C from Eq. (d) into Eq. (e) gives tan 2α = −2.4851

α=

1 tan −1 (−2.4851) = −0.5941 rad , 0.9767 rad 2

for α = − 0.5941 rad,

M α = 17.51 k ⋅ in

for α =

M α = 0.53 k ⋅ in

0.9767 rad,

Thus, the peak value of Mα estimated by the CQC combination rule is Mα = 17.51 k⋅in about an axis oriented 0.5941 rad (34.0°) clockwise from the x-axis. 9. Comments. Modes 1 and 2 are strongly correlated (ρ12 = 0.9089); hence, the CQC estimate, which accounts for this correlation, is expected to be more accurate than the SRSS estimate. Note that the SRSS and CQC estimates for the peak value of Mα differ by 2.1%. The values of α predicted by the two combination rules differ by 2.4°.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 148 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

CHAPTER 14 Problem 14.1

From these eigenvalues, the damped frequencies are determined from their definition in Eq. (14.5.4):

The mass and stiffness matrices of the system are: ⎡m m=⎢ ⎣

⎤

− k⎤ k ⎥⎦

⎡ 2k k=⎢ ⎣− k

1 m⎥ 2 ⎦

24 EI h3

the damping matrix corresponding to given modal damping ratios ζ n = 5% is determined by the Rayleigh damping procedure (Section 11.4.1). a0 = 0.0541

k m

a1 = 0.0383

0.1307

1m 2

−0.0383

0 0 ⎡ −m ⎢ 1 ⎡ −m 0 ⎤ ⎢ 0 − 2 m 0 b=⎢ ⎥=⎢ 0 2k ⎣ 0 k⎦ ⎢ 0 ⎢⎣ 0 0 −k

k (− 0.0383 ± 0.7062 i ) m

λ2 , λ 2 =

k (− 0.0924 ± 1.8454 i ) m

k m

Solution of the eigenvalue problem provides also the eigenvectors: ⎧− 0.7071⎫ ⎬ ⎩ 1.0000⎭

ψ2 = ⎨

Note that the eigenvectors are real valued as expected for a classically damped system, and identical to the natural modes of the associated undamped system determined by solving Problem 10.6.

To verify that the eigenvectors ψ n are orthogonal, we compute the individual terms in the left side of Eqs. (14.6.1) and (14.6.2).

1m 2

ψ 1T mψ 2T = 0

⎤ ⎥ ⎥ ⎥ −0.0383 ⎥ 0.0653⎥⎦

ψ 1T kψ 2T = 0

ψ 1T cψ 2T = 0

0⎤ ⎥ 0⎥ ⎥ −k ⎥ k ⎥⎦

Substituting individual terms in the left side of Eqs. (14.6.1) and (14.6.2).

(λ1 + λ 2 )ψ 1T mψ 2T + ψ 1T cψ 2T = 0

The eigenvalue problem can be solved numerically using an appropriate algorithm, e.g., the Matlab function eig ( b, −a ) , resulting in the eigenvalues

λ1 , λ 1 =

ω 2 D = Im(λ2 ) = 1.8454

ψ1 = ⎨

The eigenvalue problem to be solved is defined by Eq. (A14.2.8) with the matrices a and b , defined in Eq. (A14.2.5). 0

k m

⎧0.7071⎫ ⎬ ⎩1.0000⎭

m k

⎡ 0.1307 − 0.0383⎤ k c = a0m + a1k = ⎢ 0.0653⎥⎦ m ⎣ − 0.0383

⎡0 ⎢ ⎡ 0 m⎤ ⎢ 0 a=⎢ ⎥=⎢ ⎣m c ⎦ ⎢ m ⎢0 ⎣

ω1D = Im(λ1 ) = 0.7644

ψ 1T kψ 2T − λ1λ 2ψ 1T mψ 2T = 0 Thus, orthogonality of modes is demonstrated.

From these eigenvalues, ω n and ζ n can be determined from Eq. (14.5.6):

ω1 = λ1 = 0.7654

ζ1 = −

Re(λ1 )

λ1

k m

= 0.05

ω 2 = λ2 = 1.8478

ζ2 = −

Re(λ2 )

λ2

k m

= 0.05

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 1 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 14.2 The mass, stiffness and damping matrices of the system are: ⎡m m=⎢ ⎣ k=

⎤

1 m⎥ 2 ⎦

− k⎤ k ⎥⎦

⎡ 2k k=⎢ ⎣− k

⎡0 . 4 0 ⎤ c=⎢ ⎥ km ⎣ 0 0⎦

24 EI h3

The eigenvalue problem to be solved is defined by Eq. (A14.2.8) with the matrices a and b , defined in Eq. (A14.2.5). ⎡0 ⎢ ⎡ 0 m⎤ ⎢ 0 a=⎢ ⎥=⎢ ⎣m c ⎦ ⎢ m ⎢ ⎣⎢ 0

0.4 km

1m 2

0 0 ⎡ −m ⎢ ⎡ −m 0 ⎤ ⎢ 0 − 12 m 0 b=⎢ ⎥=⎢ 0 2k ⎣ 0 k⎦ ⎢ 0 ⎢⎣ 0 −k 0

Solution of the eigenvalue problem provides also the eigenvectors: ⎧0.7132 − 0.0775 i ⎫ ⎬ 1.0000 ⎩ ⎭

ψ1 = ⎨

⎧− 0.6732 − 0.1806 i ⎫ ⎬ 1.0000 ⎩ ⎭

ψ2 = ⎨

Note that the eigenvectors are complex valued, as expected for a nonclassically-damped system. To verify that the eigenvectors ψ n are orthogonal, we compute the individual terms in the left side of Eqs. (14.6.1) and (14.6.2).

ψ 1T mψ 2T = m (0.0059 − 0.0766 i ) ψ 1T kψ 2T = k (− 0.0282 + 0.1049 i )

0 ⎤ 1 m⎥ 2 ⎥ ⎥ 0 ⎥ ⎥ 0 ⎦⎥

ψ 1T cψ 2T = km (− 0.1976 − 0.0306 i ) Substituting individual terms in the left side of Eqs. (14.6.1) and (14.6.2)

(λ1 + λ 2 )ψ 1T mψ 2T + ψ 1T cψ 2T = 0

0⎤ ⎥ 0⎥ ⎥ −k ⎥ k ⎥⎦

ψ 1T kψ 2T − λ1λ 2ψ 1T mψ 2T = 0 Thus, orthogonality of modes is demonstrated.

The eigenvalue problem can be solved numerically using an appropriate algorithm, e.g., the Matlab function eig ( b, −a ) , resulting in the eigenvalues

λ1 , λ 1 =

k (− 0.1014 ± 0.7641 i ) m

λ2 , λ 2 =

k (− 0.0986 ± 1.8320 i ) m

From these eigenvalues, ω n and ζ n can be determined from Eq. (14.5.6):

ω1 = λ1 = 0.7709

ζ1 = −

Re(λ1 )

λ1

k m

ω 2 = λ2 = 1.8346

ζ2 = −

= 0.1316

Re(λ2 )

λ2

k m

= 0.0537

From these eigenvalues, the damped frequencies are determined from their definition in Eq. (14.5.4):

ω1D = Im(λ1 ) = 0.7641

k m

ω 2 D = Im(λ2 ) = 1.8320

k m

Problem 14.3 The initial displacement and velocity vectors are: ⎧1⎫ u(0) = ⎨ ⎬ ⎩2⎭

⎧0⎫ u& (0) = ⎨ ⎬ ⎩0⎭

Substituting them in Eq. (14.7.4) together with m , c , λn , and ψ n determined in Problem 14.1 gives:

λ ψ mu ( 0 ) + ψ 1 cu ( 0 ) + ψ 1 mu& ( 0 ) B1 = 1 1 = 0.8536 − 0.0427i 2λ1ψ 1T mψ 1 + ψ 1T cψ 1 T

B2 =

λ2ψ 2T mu ( 0 ) + ψ 2T cu ( 0 ) + ψ 2T mu& ( 0 ) 2λ2ψ 2T mψ 2 + ψ 2T cψ 2

= 0.1464 − 0.0073i

Using Bn and ψ n determined in solving Problem 14.1, β n and γ n are determined from Eq. (14.7.2) as follows: ⎧1.2071⎫ ⎬ ⎩1.7071⎭

β 1 = Re(2 B1ψ 1 ) = ⎨

⎧− 0.0604⎫ ⎬ ⎩− 0.0855⎭

γ 1 = Im(2 B1ψ 1 ) = ⎨

⎧ − 0.2071⎫ ⎬ ⎩ 0.2929⎭

β 2 = Re(2 B2ψ 2 ) = ⎨

⎧ 0.0104⎫ ⎬ ⎩ − 0.0147 ⎭

γ 2 = Im(2 B2ψ 2 ) = ⎨

Substituting β n and γ n into Eq. (14.7.6) gives the freevibration response: ⎡⎧1.2071⎫ ⎤ ⎧0.0604⎫ u(t ) = e − 0.05 ω1t ⎢⎨ ⎬ cos ω1D t + ⎨ ⎬ sin ω1D t ⎥ ⎩0.0855⎭ ⎣⎢⎩1.7071⎭ ⎦⎥ ⎡⎧ − 0.2071⎫ ⎤ ⎧ 0.0104⎫ + e − 0.05 ω 2 t ⎢⎨ ⎬ cos ω2 D t − ⎨ ⎬ sin ω2 D t ⎥ ⎩− 0.0147 ⎭ ⎣⎢⎩ 0.2929⎭ ⎦⎥

This result agrees with the solution of Problem 10.9.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 3 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 14.4 The initial displacement and velocity vectors are: ⎧1⎫ u(0) = ⎨ ⎬ ⎩2⎭

⎧0⎫ u& (0) = ⎨ ⎬ ⎩0⎭

Substituting them in Eq. (14.7.4) together with m , c , λn , and ψ n determined in Problem 14.2 gives:

λ ψ mu ( 0 ) + ψ 1 cu ( 0 ) + ψ 1 mu& ( 0 ) B1 = 1 1 = 0.8534 − 0.0189 i 2λ1ψ 1T mψ 1 + ψ 1T cψ 1 T

B2 =

λ2ψ 2T mu ( 0 ) + ψ 2T cu ( 0 ) + ψ 2T mu& ( 0 ) 2λ2ψ 2T mψ 2 + ψ 2T cψ 2

= 0.1466 − 0.0473 i

Using Bn and ψ n determined in solving Problem 14.2, β n and γ n are determined from Eq. (14.7.2) as follows: ⎧1.2144⎫ ⎬ ⎩1.7069⎭

β 1 = Re(2 B1ψ 1 ) = ⎨

⎧− 0.1592⎫ ⎬ ⎩− 0.0378⎭

γ 1 = Im(2 B1ψ 1 ) = ⎨

⎧− 0.2144⎫ ⎬ ⎩ 0.2931⎭

β 2 = Re(2 B2ψ 2 ) = ⎨

⎧ 0.0107⎫ ⎬ ⎩ − 0.0945 ⎭

γ 2 = Im(2 B2ψ 2 ) = ⎨

Substituting β n and γ n into Eq. (14.7.6) gives the freevibration response: ⎡⎧1.2144⎫ ⎤ ⎧0.1592⎫ u(t ) = e − 0.1316 ω1t ⎢⎨ ⎬ cos ω1D t + ⎨ ⎬ sin ω1D t ⎥ ⎢⎣⎩1.7069⎭ ⎥⎦ ⎩0.0378⎭ ⎡⎧− 0.2144⎫ ⎤ ⎧ 0.0107⎫ + e − 0.0537 ω 2 t ⎢⎨ ⎬ cos ω2 D t − ⎨ ⎬ sin ω 2 D t ⎥ ⎢⎣⎩ 0.2931⎭ ⎥⎦ ⎩ − 0.0945 ⎭

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 4 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 14.5 Bng are determined by substituting m , c , and ψ n from Problem 14.1 in Eq. (14.8.1) B1g =

B2g =

−ψ 1T mι 2λ1ψ 1T mψ 1 + ψ 1T cψ 1

= 0.7896

−ψ 2T mι 2λ2ψ 2T mψ 2 + ψ 2T cψ 2

m i k

= −0.0561

m i k

Using the Bng and ψ n from the solution to Problem 14.1,

β ng and γ ng are determined from Eq. (14.8.2) as follows:

(

) ⎧00⎫

(

⎫ m ) ⎧11..1166 ⎬ 5791 k

(

) ⎧00⎫

(

0794⎫ m ) ⎧− 00..1122 ⎬ k

β1g = Re 2 B1gψ 1 = ⎨ ⎬ ⎩ ⎭

γ 1g = Im 2 B1gψ 1 = ⎨ ⎩

⎭

β 2g = Re 2 B2gψ 2 = ⎨ ⎬ ⎩ ⎭

γ 2g = Im 2 B2gψ 2 = ⎨ ⎩

⎭

Substituting the β ng and γ ng into Eq. (14.8.3) gives the desired response: ⎧− 1.1166⎫ m u(t ) = e − 0.05ω1t ⎨ sin ω1D t ⎬ ⎩ − 1.5791⎭ k ⎧− 0.0794⎫ m sin ω2 D t + e − 0.05ω 2 t ⎨ ⎬ ⎩ 0.1122⎭ k

The response can also be expressed in terms of the unit impulse response functions hn (t ) . For this purpose Eq. (14.8.8) is used to obtain: ⎧− 1.1152⎫ m ⎬ ⎩− 1.5772⎭ k

α 1g = ⎨

⎧− 0.0793⎫ m ⎬ ⎩ 0.1121⎭ k

α 2g = ⎨

which are substituted in Eq. (14.8.7), to obtain ⎧ 0.1464⎫ ⎧0.8536⎫ u(t ) = ⎨ ⎬h2 (t ) ⎬h1 (t ) + ⎨ ⎩ − 0.2071⎭ ⎩1.2071⎭

where hn ( t ) Eq. (14.8.4).

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 5 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 14.6 Bng are determined by substituting m , c , and ψ n from Problem 14.2 in Eq. (14.8.1) B1g = B2g =

−ψ 1T mι 2λ1ψ 1T mψ 1 + ψ 1T cψ 1

m ( −0.0368 + 0.7991 i ) k

m ( 0.0368 − 0.0603 i ) k

−ψ 2T mι 2λ2ψ 2T mψ 2 + ψ 2T cψ 2

Using these Bng and ψ n from the soluation to Problem 14.2, β ng and γ ng are determined from Eq. (14.8.2) as follows:

(

⎫ m ) ⎧− 00..0714 ⎬ 0736 k

β 1g = Re 2 B1gψ 1 = ⎨

⎧− 0.0714 ⎫ m & ⎧0.8681⎫ h1 (t ) u(t ) = ⎨ ⎬ ⎬h1 (t ) + ⎨ ⎩ 0.0736⎭ k ⎩1.2287⎭

⎧ 0.1315⎫ ⎧ 0.0714⎫ m & +⎨ h2 (t ) ⎬h2 (t ) + ⎨ ⎬ − 0 . 2283 ⎩ ⎭ ⎩− 0.0736⎭ k where hn ( t ) and h&n ( t ) are given by Eq. (14.8.4) and (14.8.5). The vectors α ng and β ng satisfy Eq. (14.8.9):

β 1g + β 2g = 2

∑ ωn ⎡⎣α ng − 2ζ n β ng ⎤⎦ =

n =1

⎭

⎩

(

1455⎫ m ) ⎧11..5982 ⎬ k

ω1

(

)

+ ω2

(

0679⎫ m ) ⎧− 00..1207 ⎬ k

γ 1g = Im 2 B1gψ 1 = ⎨ ⎩

β 2g = Re 2 B2gψ 2

⎭

⎧− 0.0714⎫ m =⎨ ⎬ ⎩ 0.0736⎭ k

γ 2g = Im 2 B2gψ 2 = ⎨ ⎩

m ⎡⎧ 0.0714⎫ ⎧− 0.0714⎫⎤ ⎧0⎫ ⎢⎨ ⎬⎥ = ⎨ ⎬ ⎬+⎨ k ⎢⎣⎩− 0.0736⎭ ⎩ 0.0736⎭⎥⎦ ⎩0⎭

⎧ 0.0714⎫⎤ m ⎡⎧− 1.1262⎫ ⎢⎨ ⎬⎥ ⎬ − 2ζ 1 ⎨ k ⎣⎢⎩− 1.5940⎭ ⎩− 0.0736⎭⎦⎥ ⎧− 0.0714⎫⎤ ⎧− 1⎫ m ⎡⎧− 0.0717⎫ ⎢⎨ ⎬⎥ = ⎨ ⎬ ⎬ − 2ζ 2 ⎨ k ⎢⎣⎩ 0.1244⎭ ⎩ 0.0736⎭⎥⎦ ⎩− 1⎭

⎭

Substituting the β ng and γ ng into Eq. (14.8.3) gives the desired response: ⎤ m ⎡⎧ 0.0714⎫ ⎧1.1455⎫ u(t ) = e − 0.1316 ω1t ⎢⎨ ⎬ sin ω1D t ⎥ ⎬ cos ω1D t − ⎨ ⎩1.5982⎭ ⎦⎥ k ⎣⎢⎩− 0.0736 ⎭ ⎤ m ⎡⎧ − 0.0714 ⎫ ⎧ 0.0679⎫ + e − 0.0537 ω 2 t ⎢⎨ ⎬ sin ω2 D t ⎥ ⎬ cos ω 2 D t + ⎨ ⎩− 0.1207 ⎭ ⎦⎥ k ⎣⎢⎩ 0.0736⎭

The response can also be addressed in terms of the unit impulse response functions hn (t ) . For this purpose Eq. (14.8.8) is used to obtain: ⎧− 1.1262⎫ m ⎬ ⎩− 1.5940⎭ k

α 1g = ⎨

⎧− 0.0717 ⎫ m ⎬ ⎩ 0.1244 ⎭ k

α 2g = ⎨

which are substituted together with β ng in Eq. (14.8.7), leading to:

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 6 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 14.7 Substituting the α ng and β ng determined in Problem 14.5 into Eq. (14.9.2) provides equations for the floor displacements: ⎧ 0.1464⎫ ⎧0.8536⎫ u(t ) = ⎨ ⎬ D2 (t ) ⎬ D1 (t ) + ⎨ ⎩ − 0.2071⎭ ⎩1.2071⎭

This result matches that obtained by solving Problem 13.1.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 7 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 14.8 Substituting the α ng and β ng determined in solving Problem 14.6 into Eq. (14.9.2) provides equations for the floor displacements: ⎧ − 0.0714 ⎫ m & ⎧0.8681⎫ D1 (t ) u(t ) = ⎨ ⎬ ⎬ D1 (t ) + ⎨ ⎩ 0.0736⎭ k ⎩1.2287 ⎭ ⎧ 0.0714⎫ m & ⎧ 0.1315⎫ D2 (t ) +⎨ ⎬ ⎬ D2 (t ) + ⎨ ⎩ − 0.0736 ⎭ k ⎩ − 0.2283⎭

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 8 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 14.9 The mass, stiffness and damping matrices of the system are: ⎡m m=⎢ ⎣ k=

⎤

⎡ 2k k=⎢ ⎣−k

1 m⎥ 2 ⎦

⎡ 3 −2 ⎤ 0.6 km c=⎢ 2 ⎥⎦ ⎣ −2

−k ⎤ k ⎥⎦

24 EI h3

The eigenvalue problem to be solved is defined by Eq. (A14.2.8) with the matrices a and b , defined in Eq. (A14.2.5). ⎡0 ⎢ ⎡ 0 m⎤ ⎢ 0 a=⎢ ⎥=⎢ ⎣m c ⎦ ⎢ m ⎢ ⎣⎢ 0

1.8 km

1m 2

−1.2 km

0 0 ⎡ −m ⎢ ⎡ −m 0 ⎤ ⎢ 0 − 12 m 0 b=⎢ ⎥=⎢ 0 2k ⎣ 0 k⎦ ⎢ 0 ⎢⎣ 0 −k 0

⎤ ⎥ 1m ⎥ 2 ⎥ −1.2 km ⎥ ⎥ 1.2 km ⎦⎥ 0

k m

λ3 = −2.4339

ζ1 = −

λ2 − λ3 2

= 0.5316

k m

Solution of the eigenvalue problem provides also the 4 × 1 eigenvectors, but only the third and fourth components [see Eq. (A14.2.7)] are shown below: ⎧0.7611 + 0.0873i ⎫ ⎬ 1.0000 ⎩ ⎭

ψ1 = ⎨

⎧− 0.4568⎫ ⎬ ⎩ 1.0000 ⎭

ψ2 = ⎨

⎧ − 0.5421⎫ ⎬ ⎩ 1.0000⎭

ψ3 = ⎨

Note that ψ 1 is complex-valued, but ψ 2 and ψ 3 are real-valued.

ψ 1T kψ 2T = k ( −0.0003 − 0.1671 i )

ψ 1T cψ 2T = km ( 0.2090 − 0.1766 i )

(λ1 + λ 2 )ψ 1T mψ 2T + ψ 1T cψ 2T = 0

k m

ψ 1T kψ 2T − λ1λ 2ψ 1T mψ 2T = 0

From the eigenvalues λ1 and λ1 , ω 1 and ζ 1 can be determined from Eq. (14.5.6): k m

ω2 D =

Substituting these individual terms in the left side of Eqs. (14.6.1) and (14.6.2)

Note that two of the eigenvalues of the system are complex conjugates, whereas two are real and negative valued, with λ3 ≥ λ2 .

ω1 = λ1 = 0.7743

However, ω 2 D should be determined from Eq. (14.10.5):

ψ 1T mψ 2T = m ( 0.1523 − 0.0399 i )

k (− 0.1977 ± 0.7486 i ) m

λ2 = −1.3707

k m

To verify that the eigenvectors ψ 1 and ψ 2 are orthogonal, we compute the individual terms in the left side of Eqs. (14.6.1) and (14.6.2).

0⎤ ⎥ 0⎥ ⎥ −k ⎥ k ⎥⎦

The eigenvalue problem can be solved numerically using an appropriate algorithm, e.g., the Matlab function eig ( b, −a ) , resulting in the eigenvalues

λ1 , λ 1 =

ω1D = Im(λ1 ) = 0.7486

Re(λ1 )

λ1

This verifies that the eigenvectors ψ 1 and ψ 2 computed for the system are orthogonal; similar calculations show that other pairs of modes are orthogonal.

= 0.2553

Substituting eigenvales λ2 and λ3 in Eqs. (14.10.3) and (14.10.4) gives

ω 2 = λ2 λ3 = 1.8265

k m

ζ2 = −

λ2 + λ3 = 1.0415 2ω 2

From the eigenvalue λ1 , the corresponding frequency ω1D of the damped system is determined from its definition in Eq. (14.5.4):

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 9 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 14.10 The initial displacement and velocity vectors are: ⎧1⎫ u(0) = ⎨ ⎬ ⎩2⎭

⎧0⎫ u& (0) = ⎨ ⎬ ⎩0⎭

Substituting them in Eq. (14.7.4) together with m , c , λn , and ψ n determined in the solution to Problem 14.9 gives: B1 = B2 =

B3 =

λ1ψ 1T mu ( 0 ) + ψ 1T cu ( 0 ) + ψ 1T mu& ( 0 ) 2λ1ψ 1T mψ 1 + ψ 1T cψ 1

λ2ψ 2T mu ( 0 ) + ψ 2T cu ( 0 ) + ψ 2T mu& ( 0 ) 2λ2ψ 2T mψ 2 + ψ 2T cψ 2

λ3ψ 3T mu ( 0 ) + ψ 3T cu ( 0 ) + ψ 3T mu& ( 0 ) 2λ3ψ 3T mψ 3 + ψ 3T cψ 3

= 0.7460 − 0.3125 i = 1.0005

= −0.4924

Using Bn and ψ n from Problem 14.1, β1 and γ 1 are determined from Eq. (14.7.5) and β 2 and γ 2 from Eqs. (14.10.8) and (14.10.9) as follows ⎧1.1901⎫ ⎬ ⎩1.4919⎭

β 1 = Re(2 B1ψ 1 ) = ⎨

⎧− 0.3454⎫ ⎬ ⎩− 0.6250⎭

γ 1 = Im(2 B1ψ 1 ) = ⎨

⎧ − 0.1901⎫ ⎬ ⎩ 0.5081⎭

β 2 = B2ψ 2 + B3ψ 3 = ⎨

⎧ 0.7240⎫ ⎬ ⎩ − 1.4928 ⎭

γ 2 = B2ψ 2 − B3ψ 3 = ⎨

The free-vibration response is given by u(t ) = u1 (t ) + u 2 (t )

where u1 (t ) is determined by substituting β1 and γ 1 into the n = 1 term on the right side of Eq. (14.7.6):

⎤ ⎡⎧1.1901⎫ ⎧0.3454⎫ u1 (t ) = e − 0.2553ω1t ⎢⎨ ⎬ cos ω1Dt + ⎨ ⎬ sin ω1Dt ⎥ ⎥⎦ ⎢⎣⎩1.4919⎭ ⎩0.6250⎭ and u 2 (t ) is determined by substituting β 2 and γ 2 into Eq. (14.10.7):

⎡⎧ −0.1901 ⎫ ⎬ cosh ω2 D t − ⎣⎩ 0.5081⎭

u 2 ( t ) = e −1.0415 ω t ⎢ ⎨ 2

{

}

0.7240 −1.4928

⎤ ⎦

sinh ω 2 D t ⎥

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 10 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 14.11 B1g are determined by substituting m , c , and ψ n from the solution of Problem 14.9 in Eq. (14.8.1): B1g =

−ψ 1T mι 2λ1ψ 1T mψ 1 + ψ 1T cψ 1

m (0.0549 + 0.7897 i ) k

−ψ 2T mι

m B2g = = −0.0592 T T k 2λ2ψ 2 mψ 2 + ψ 2 cψ 2 B3g =

−ψ 3T mι 2λ3ψ 3T mψ 3 + ψ 3T cψ 3

= −0.0505

m k

Using the Bng and ψ n , β1g and γ 1g are determined from Eq. (14.8.2), and β 2g and γ 2g are determined from Eqs. (14.10.11) and (14.10.12) as follows:

(

⎫ m ) ⎧− 00..0544 ⎬ 1097 k

(

⎫ m ) ⎧11..2117 ⎬ 5794 k

β 1g = Re 2 B1gψ 1 = ⎨

⎭

⎩

γ 1g = Im 2 B1gψ 1 = ⎨ ⎩

⎭ 0 . 0544⎫ m ⎧ β 2g = B2gψ 2 + B3gψ 3 = ⎨ ⎬ ⎩− 0.1097 ⎭ k ⎧0.0003⎫ m ⎬ ⎩0.0087⎭ k

γ 2g = B2gψ 2 − B3gψ 3 = ⎨

The free-vibration response is given by u(t ) = u1 (t ) + u 2 (t )

⎧ 0.0566⎫ m ⎬ ⎩− 0.1168⎭ k

α 2g = ζ 2 β 2g − ζ 22 − 1 γ 2g = ⎨

The response of the system can be determined from u(t ) = u1 (t ) + u 2 (t )

The first term u1 ( t ) is determined by substituting α1g and β1g into the n = 1 term on the right side of Eq. (14.8.7): ⎧ 0.0544⎫ m & ⎧0.9178⎫ h1 (t ) u1 (t ) = ⎨ ⎬ ⎬h1 (t ) + ⎨ 1 . 1607 ⎩− 0.1097⎭ k ⎭ ⎩

where h1 (t ) and h&1 (t ) are given by Eqs. (14.8.4) and (14.8.5), respectively. The second term u 2 (t ) is determined by substituting α 2g and β 2g (14.10.16): ⎧− 0.0544⎫ m & ⎧− 0.1034⎫ h2 (t ) u 2 (t ) = ⎨ ⎬ ⎬h2 (t ) + ⎨ 0 . 2133 ⎩ 0.1097⎭ k ⎭ ⎩

where h2 (t ) and h&2 (t ) are given by Eq. (14.10.13) and (14.10.14), respectively. The vectors β ng and γ ng satisfy Eq. (14.8.9):

β 1g + β 2g =

⎡ ⎧ −0.0544 ⎫ ⎤ m ⎧1.2117 ⎫ u1 ( t ) = e−0.2553ω1t ⎢ ⎨ ⎬ cos ω1D t − ⎨ ⎬ sin ω1D t ⎥ 0.1097 1.5794 ⎭ ⎩ ⎭ ⎣⎩ ⎦ k

m ⎡⎧− 0.0544⎫ ⎧ 0.0544⎫⎤ ⎧0⎫ ⎢⎨ ⎬⎥ = ⎨ ⎬ ⎬+⎨ k ⎣⎢⎩ 0.1097 ⎭ ⎩− 0.1097 ⎭⎦⎥ ⎩0⎭

∑ω [α − 2ζ β ] = ω 2

where u1 (t ) is determined by substituting β 1g and γ 1g into the n = 1 term on the right side of Eq. (14.8.3):

into Eq.

n =1

ω2

g n

⎧− 0.0544⎫⎤ m ⎡⎧− 1.1854⎫ ⎢⎨ ⎬⎥ + ⎬ − 2ζ 1 ⎨ k ⎣⎢⎩− 1.4990⎭ ⎩ 0.1097 ⎭⎦⎥

⎧ 0.0544⎫⎤ ⎧− 1⎫ m ⎡⎧ 0.0566⎫ ⎢⎨ ⎬⎥ = ⎨ ⎬ ⎬ − 2ζ 2 ⎨ k ⎢⎣⎩− 0.1168⎭ ⎩− 0.1097⎭⎥⎦ ⎩− 1⎭

and u 2 (t ) is determined by substituting β 2g and γ 2g into Eq. (14.10.10):

⎡⎧ 0.0544 ⎫ ⎤ ⎧0.0003⎫ u 2 (t ) = e −1.0415 ω2t ⎢⎨ ⎬ cosh ω 2 D t − ⎨ ⎬ sinh ω 2 D t ⎥ − 0 . 1097 0 . 0087 ⎭ ⎩ ⎭ ⎣⎢⎩ ⎦⎥ The response can also be expressed in terms of the unit impulse response functions hn (t ) . For this purpose Eqs. (14.8.8) and (14.10.17) give ⎧− 1.1854⎫ m ⎬ ⎩− 1.4990⎭ k

α1g = ζ 1β1g − 1 − ζ 12 γ 1g = ⎨

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 11 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 14.12 Substituting the α ng and β ng determined in the solution of Problem 14.11 into Eq. (14.9.2) provides equations for the floor displacements: ⎧0.9178⎫ ⎧ 0.0544 ⎫ m & u (t ) = ⎨ D1 ( t ) + ⎬ D1 ( t ) + ⎨ ⎬ ⎩1.1607 ⎭ ⎩ −0.1097 ⎭ k ⎧ −0.1034 ⎫ ⎧ −0.0544 ⎫ m & D2 ( t ) ⎨ ⎬ D2 ( t ) + ⎨ ⎬ ⎩ 0.2133⎭ ⎩ 0.1097 ⎭ k

where Dn (t ) and D& n (t ) represent the deformation and relative velocity response of the nth-mode SDF system; see Eqs. (14.9.3) and (14.9.4). Recall that in these equations, h1 (t ) and h&1 (t ) are given by Eq. (14.8.4) and (14.8.5), respectively, whereas h (t ) and h& (t ) are 2

given by Eq. (14.10.13) and (14.10.14), respectively

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 12 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 14.13

Part a

φ2 = − 1.5610 1.0000 , which give Φ (see Section 10.3). Substituting c and Φ gives

This a 2-DOF system; u1=displacement of the base slab; u2 displacement of the mass m. The differential equations governing the nodal displacements u are:

⎡ 0.9680 − 1.5080⎤ C=⎢ m 6.6724⎥⎦ ⎣ − 1.5080

⎡mb ⎢ ⎣

⎤ ⎧u&&1 ⎫ ⎡ cb + c − c ⎤ ⎧u&1 ⎫ ⎡kb + k − k ⎤ ⎧u1 ⎫ ⎨ ⎬+ ⎨ ⎬+ ⎨ ⎬= m ⎥⎦ ⎩u&&2 ⎭ ⎢⎣ − c c ⎥⎦ ⎩u&2 ⎭ ⎢⎣ − k k ⎥⎦ ⎩u2 ⎭ ⎡m ⎤ ⎧1⎫ −⎢ b ⎨ ⎬u&&g (t ) m⎥⎦ ⎩1⎭ ⎣

where:

kb = (m + mb )ωb2 = 5π 2 m / 3

c = 2mω f ζ f = π m / 5

cb = 2(m + mb )ωbζ b = π m / 3

⎡0.9680 ⎤ C=⎢ ⎥m 6 . 6724 ⎣ ⎦ The corresponding modal damping ratios are ζ 1 = 9.651% and ζ 2 = 5.0571%

mb = 2m / 3 k = mω 2f = 25π 2 m

which is not a diagonal matrix. Thus, the equations of motion in modal coordinates are coupled. It is this coupling we are neglecting in classical modal analysis to obtain the diagonal C matrix.

Displacements are given by Eq. (14.4.2):

u(t ) = u1 (t ) + u 2 (t )

Solving these coupled differential equations numerically by the Runge-Kutta method, using the Matlab function ode45, gives u1(t) shown in the figure. The peak value of the deformation in the isolation system (displacement of the base slab) = 4.7714 in. Part b

Modal analysis of the nonclassically damped system following Examples 14.13 and 14.14, gives the displacement response:

u(t ) = u1 (t ) + u 2 (t )

⎧0.9758⎫ u1 (t ) = ⎨ ⎬ D1 (t ) ⎩1.0155 ⎭

⎧ 0.0242⎫ u 2 (t ) = ⎨ ⎬ D2 (t ) ⎩ − 0.0155 ⎭ where Dn (t ) represent the deformation of the nth-mode SDF system. Dn (t ) was computed by numerical integration of Eq. (14.4.3). The resulting u1(t) is shown in the figure with a peak value of 4.7767 in. For this example, the off-diagonal terms of C is not small (compared to diagonal terms) and one would expect that neglecting it would produce significant error. However, the error is small because the response is dominated by the first mode; see relative values of u1 ( t ) and u 2 ( t ) .

⎧ − 0.0015 ⎫ & ⎧0.9766⎫ u1 (t ) = ⎨ ⎬ D1 (t ) ⎬ D1 (t ) + ⎨ ⎩ 0.0009⎭ ⎩1.0164 ⎭

⎧ 0.0263⎫ ⎧ 0.0015⎫ & u 2 (t ) = ⎨ ⎬ D2 (t ) + ⎨ ⎬ D2 (t ) − 0 . 0182 ⎩ ⎭ ⎩ − 0.0009 ⎭ where Dn (t ) and D& n (t ) represent the deformation and relative velocity response, respectively, of the nth-mode SDF system; these quantities were obtained by numerical integration of Eqs. (14.9.3) and (14.9.4). The resulting u1(t) is shown in the figure; its peak value is 4.7732 in. As expected, the results are essentially identical to those in part (a). Part c

The damping matrix in modal coordinates is given by Eq. (10.9.5): C = Φ Tc Φ , where the natural vibration modes of the undamped system are φ1 = 0.9610 1.0000

and

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 13 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

4.7714 (a) 4.7732 (b) 4.7767 (c)

4 u1, in.

part (a) part (b) part (c)

0 -2 -4 -6 0

15 time, sec

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 14 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

CHAPTER 14 Problem 14.1

From these eigenvalues, the damped frequencies are determined from their definition in Eq. (14.5.4):

The mass and stiffness matrices of the system are: ⎡m m=⎢ ⎣

⎤

− k⎤ k ⎥⎦

⎡ 2k k=⎢ ⎣− k

1 m⎥ 2 ⎦

24 EI h3

the damping matrix corresponding to given modal damping ratios ζ n = 5% is determined by the Rayleigh damping procedure (Section 11.4.1). a0 = 0.0541

k m

a1 = 0.0383

0.1307

1m 2

−0.0383

0 0 ⎡ −m ⎢ 1 ⎡ −m 0 ⎤ ⎢ 0 − 2 m 0 b=⎢ ⎥=⎢ 0 2k ⎣ 0 k⎦ ⎢ 0 ⎢⎣ 0 0 −k

k (− 0.0383 ± 0.7062 i ) m

λ2 , λ 2 =

k (− 0.0924 ± 1.8454 i ) m

k m

Solution of the eigenvalue problem provides also the eigenvectors: ⎧− 0.7071⎫ ⎬ ⎩ 1.0000⎭

ψ2 = ⎨

Note that the eigenvectors are real valued as expected for a classically damped system, and identical to the natural modes of the associated undamped system determined by solving Problem 10.6.

To verify that the eigenvectors ψ n are orthogonal, we compute the individual terms in the left side of Eqs. (14.6.1) and (14.6.2).

1m 2

ψ 1T mψ 2T = 0

⎤ ⎥ ⎥ ⎥ −0.0383 ⎥ 0.0653⎥⎦

ψ 1T kψ 2T = 0

ψ 1T cψ 2T = 0

0⎤ ⎥ 0⎥ ⎥ −k ⎥ k ⎥⎦

Substituting individual terms in the left side of Eqs. (14.6.1) and (14.6.2).

(λ1 + λ 2 )ψ 1T mψ 2T + ψ 1T cψ 2T = 0

The eigenvalue problem can be solved numerically using an appropriate algorithm, e.g., the Matlab function eig ( b, −a ) , resulting in the eigenvalues

λ1 , λ 1 =

ω 2 D = Im(λ2 ) = 1.8454

ψ1 = ⎨

The eigenvalue problem to be solved is defined by Eq. (A14.2.8) with the matrices a and b , defined in Eq. (A14.2.5). 0

k m

⎧0.7071⎫ ⎬ ⎩1.0000⎭

m k

⎡ 0.1307 − 0.0383⎤ k c = a0m + a1k = ⎢ 0.0653⎥⎦ m ⎣ − 0.0383

⎡0 ⎢ ⎡ 0 m⎤ ⎢ 0 a=⎢ ⎥=⎢ ⎣m c ⎦ ⎢ m ⎢0 ⎣

ω1D = Im(λ1 ) = 0.7644

ψ 1T kψ 2T − λ1λ 2ψ 1T mψ 2T = 0 Thus, orthogonality of modes is demonstrated.

From these eigenvalues, ω n and ζ n can be determined from Eq. (14.5.6):

ω1 = λ1 = 0.7654

ζ1 = −

Re(λ1 )

λ1

k m

= 0.05

ω 2 = λ2 = 1.8478

ζ2 = −

Re(λ2 )

λ2

k m

= 0.05

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 1 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 14.2 The mass, stiffness and damping matrices of the system are: ⎡m m=⎢ ⎣ k=

⎤

1 m⎥ 2 ⎦

− k⎤ k ⎥⎦

⎡ 2k k=⎢ ⎣− k

⎡0 . 4 0 ⎤ c=⎢ ⎥ km ⎣ 0 0⎦

24 EI h3

The eigenvalue problem to be solved is defined by Eq. (A14.2.8) with the matrices a and b , defined in Eq. (A14.2.5). ⎡0 ⎢ ⎡ 0 m⎤ ⎢ 0 a=⎢ ⎥=⎢ ⎣m c ⎦ ⎢ m ⎢ ⎣⎢ 0

0.4 km

1m 2

0 0 ⎡ −m ⎢ ⎡ −m 0 ⎤ ⎢ 0 − 12 m 0 b=⎢ ⎥=⎢ 0 2k ⎣ 0 k⎦ ⎢ 0 ⎢⎣ 0 −k 0

Solution of the eigenvalue problem provides also the eigenvectors: ⎧0.7132 − 0.0775 i ⎫ ⎬ 1.0000 ⎩ ⎭

ψ1 = ⎨

⎧− 0.6732 − 0.1806 i ⎫ ⎬ 1.0000 ⎩ ⎭

ψ2 = ⎨

ψ 1T mψ 2T = m (0.0059 − 0.0766 i ) ψ 1T kψ 2T = k (− 0.0282 + 0.1049 i )

0 ⎤ 1 m⎥ 2 ⎥ ⎥ 0 ⎥ ⎥ 0 ⎦⎥

ψ 1T cψ 2T = km (− 0.1976 − 0.0306 i ) Substituting individual terms in the left side of Eqs. (14.6.1) and (14.6.2)

(λ1 + λ 2 )ψ 1T mψ 2T + ψ 1T cψ 2T = 0

0⎤ ⎥ 0⎥ ⎥ −k ⎥ k ⎥⎦

ψ 1T kψ 2T − λ1λ 2ψ 1T mψ 2T = 0 Thus, orthogonality of modes is demonstrated.

The eigenvalue problem can be solved numerically using an appropriate algorithm, e.g., the Matlab function eig ( b, −a ) , resulting in the eigenvalues

λ1 , λ 1 =

k (− 0.1014 ± 0.7641 i ) m

λ2 , λ 2 =

k (− 0.0986 ± 1.8320 i ) m

From these eigenvalues, ω n and ζ n can be determined from Eq. (14.5.6):

ω1 = λ1 = 0.7709

ζ1 = −

Re(λ1 )

λ1

k m

ω 2 = λ2 = 1.8346

ζ2 = −

= 0.1316

Re(λ2 )

λ2

k m

= 0.0537

From these eigenvalues, the damped frequencies are determined from their definition in Eq. (14.5.4):

ω1D = Im(λ1 ) = 0.7641

k m

ω 2 D = Im(λ2 ) = 1.8320

k m

Problem 14.3 The initial displacement and velocity vectors are: ⎧1⎫ u(0) = ⎨ ⎬ ⎩2⎭

⎧0⎫ u& (0) = ⎨ ⎬ ⎩0⎭

Substituting them in Eq. (14.7.4) together with m , c , λn , and ψ n determined in Problem 14.1 gives:

λ ψ mu ( 0 ) + ψ 1 cu ( 0 ) + ψ 1 mu& ( 0 ) B1 = 1 1 = 0.8536 − 0.0427i 2λ1ψ 1T mψ 1 + ψ 1T cψ 1 T

B2 =

λ2ψ 2T mu ( 0 ) + ψ 2T cu ( 0 ) + ψ 2T mu& ( 0 ) 2λ2ψ 2T mψ 2 + ψ 2T cψ 2

= 0.1464 − 0.0073i

Using Bn and ψ n determined in solving Problem 14.1, β n and γ n are determined from Eq. (14.7.2) as follows: ⎧1.2071⎫ ⎬ ⎩1.7071⎭

β 1 = Re(2 B1ψ 1 ) = ⎨

⎧− 0.0604⎫ ⎬ ⎩− 0.0855⎭

γ 1 = Im(2 B1ψ 1 ) = ⎨

⎧ − 0.2071⎫ ⎬ ⎩ 0.2929⎭

β 2 = Re(2 B2ψ 2 ) = ⎨

⎧ 0.0104⎫ ⎬ ⎩ − 0.0147 ⎭

γ 2 = Im(2 B2ψ 2 ) = ⎨

This result agrees with the solution of Problem 10.9.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 3 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 14.4 The initial displacement and velocity vectors are: ⎧1⎫ u(0) = ⎨ ⎬ ⎩2⎭

⎧0⎫ u& (0) = ⎨ ⎬ ⎩0⎭

Substituting them in Eq. (14.7.4) together with m , c , λn , and ψ n determined in Problem 14.2 gives:

λ ψ mu ( 0 ) + ψ 1 cu ( 0 ) + ψ 1 mu& ( 0 ) B1 = 1 1 = 0.8534 − 0.0189 i 2λ1ψ 1T mψ 1 + ψ 1T cψ 1 T

B2 =

λ2ψ 2T mu ( 0 ) + ψ 2T cu ( 0 ) + ψ 2T mu& ( 0 ) 2λ2ψ 2T mψ 2 + ψ 2T cψ 2

= 0.1466 − 0.0473 i

Using Bn and ψ n determined in solving Problem 14.2, β n and γ n are determined from Eq. (14.7.2) as follows: ⎧1.2144⎫ ⎬ ⎩1.7069⎭

β 1 = Re(2 B1ψ 1 ) = ⎨

⎧− 0.1592⎫ ⎬ ⎩− 0.0378⎭

γ 1 = Im(2 B1ψ 1 ) = ⎨

⎧− 0.2144⎫ ⎬ ⎩ 0.2931⎭

β 2 = Re(2 B2ψ 2 ) = ⎨

⎧ 0.0107⎫ ⎬ ⎩ − 0.0945 ⎭

γ 2 = Im(2 B2ψ 2 ) = ⎨

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 4 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 14.5 Bng are determined by substituting m , c , and ψ n from Problem 14.1 in Eq. (14.8.1) B1g =

B2g =

−ψ 1T mι 2λ1ψ 1T mψ 1 + ψ 1T cψ 1

= 0.7896

−ψ 2T mι 2λ2ψ 2T mψ 2 + ψ 2T cψ 2

m i k

= −0.0561

m i k

Using the Bng and ψ n from the solution to Problem 14.1,

β ng and γ ng are determined from Eq. (14.8.2) as follows:

(

) ⎧00⎫

(

⎫ m ) ⎧11..1166 ⎬ 5791 k

(

) ⎧00⎫

(

0794⎫ m ) ⎧− 00..1122 ⎬ k

β1g = Re 2 B1gψ 1 = ⎨ ⎬ ⎩ ⎭

γ 1g = Im 2 B1gψ 1 = ⎨ ⎩

⎭

β 2g = Re 2 B2gψ 2 = ⎨ ⎬ ⎩ ⎭

γ 2g = Im 2 B2gψ 2 = ⎨ ⎩

⎭

The response can also be expressed in terms of the unit impulse response functions hn (t ) . For this purpose Eq. (14.8.8) is used to obtain: ⎧− 1.1152⎫ m ⎬ ⎩− 1.5772⎭ k

α 1g = ⎨

⎧− 0.0793⎫ m ⎬ ⎩ 0.1121⎭ k

α 2g = ⎨

which are substituted in Eq. (14.8.7), to obtain ⎧ 0.1464⎫ ⎧0.8536⎫ u(t ) = ⎨ ⎬h2 (t ) ⎬h1 (t ) + ⎨ ⎩ − 0.2071⎭ ⎩1.2071⎭

where hn ( t ) Eq. (14.8.4).

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 5 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 14.6 Bng are determined by substituting m , c , and ψ n from Problem 14.2 in Eq. (14.8.1) B1g = B2g =

−ψ 1T mι 2λ1ψ 1T mψ 1 + ψ 1T cψ 1

m ( −0.0368 + 0.7991 i ) k

m ( 0.0368 − 0.0603 i ) k

−ψ 2T mι 2λ2ψ 2T mψ 2 + ψ 2T cψ 2

Using these Bng and ψ n from the soluation to Problem 14.2, β ng and γ ng are determined from Eq. (14.8.2) as follows:

(

⎫ m ) ⎧− 00..0714 ⎬ 0736 k

β 1g = Re 2 B1gψ 1 = ⎨

⎧− 0.0714 ⎫ m & ⎧0.8681⎫ h1 (t ) u(t ) = ⎨ ⎬ ⎬h1 (t ) + ⎨ ⎩ 0.0736⎭ k ⎩1.2287⎭

β 1g + β 2g = 2

∑ ωn ⎡⎣α ng − 2ζ n β ng ⎤⎦ =

n =1

⎭

⎩

(

1455⎫ m ) ⎧11..5982 ⎬ k

ω1

(

)

+ ω2

(

0679⎫ m ) ⎧− 00..1207 ⎬ k

γ 1g = Im 2 B1gψ 1 = ⎨ ⎩

β 2g = Re 2 B2gψ 2

⎭

⎧− 0.0714⎫ m =⎨ ⎬ ⎩ 0.0736⎭ k

γ 2g = Im 2 B2gψ 2 = ⎨ ⎩

m ⎡⎧ 0.0714⎫ ⎧− 0.0714⎫⎤ ⎧0⎫ ⎢⎨ ⎬⎥ = ⎨ ⎬ ⎬+⎨ k ⎢⎣⎩− 0.0736⎭ ⎩ 0.0736⎭⎥⎦ ⎩0⎭

⎭

The response can also be addressed in terms of the unit impulse response functions hn (t ) . For this purpose Eq. (14.8.8) is used to obtain: ⎧− 1.1262⎫ m ⎬ ⎩− 1.5940⎭ k

α 1g = ⎨

⎧− 0.0717 ⎫ m ⎬ ⎩ 0.1244 ⎭ k

α 2g = ⎨

which are substituted together with β ng in Eq. (14.8.7), leading to:

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 6 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

This result matches that obtained by solving Problem 13.1.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 7 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 8 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 14.9 The mass, stiffness and damping matrices of the system are: ⎡m m=⎢ ⎣ k=

⎤

⎡ 2k k=⎢ ⎣−k

1 m⎥ 2 ⎦

⎡ 3 −2 ⎤ 0.6 km c=⎢ 2 ⎥⎦ ⎣ −2

−k ⎤ k ⎥⎦

24 EI h3

The eigenvalue problem to be solved is defined by Eq. (A14.2.8) with the matrices a and b , defined in Eq. (A14.2.5). ⎡0 ⎢ ⎡ 0 m⎤ ⎢ 0 a=⎢ ⎥=⎢ ⎣m c ⎦ ⎢ m ⎢ ⎣⎢ 0

1.8 km

1m 2

−1.2 km

0 0 ⎡ −m ⎢ ⎡ −m 0 ⎤ ⎢ 0 − 12 m 0 b=⎢ ⎥=⎢ 0 2k ⎣ 0 k⎦ ⎢ 0 ⎢⎣ 0 −k 0

⎤ ⎥ 1m ⎥ 2 ⎥ −1.2 km ⎥ ⎥ 1.2 km ⎦⎥ 0

k m

λ3 = −2.4339

ζ1 = −

λ2 − λ3 2

= 0.5316

k m

Solution of the eigenvalue problem provides also the 4 × 1 eigenvectors, but only the third and fourth components [see Eq. (A14.2.7)] are shown below: ⎧0.7611 + 0.0873i ⎫ ⎬ 1.0000 ⎩ ⎭

ψ1 = ⎨

⎧− 0.4568⎫ ⎬ ⎩ 1.0000 ⎭

ψ2 = ⎨

⎧ − 0.5421⎫ ⎬ ⎩ 1.0000⎭

ψ3 = ⎨

Note that ψ 1 is complex-valued, but ψ 2 and ψ 3 are real-valued.

ψ 1T kψ 2T = k ( −0.0003 − 0.1671 i )

ψ 1T cψ 2T = km ( 0.2090 − 0.1766 i )

(λ1 + λ 2 )ψ 1T mψ 2T + ψ 1T cψ 2T = 0

k m

ψ 1T kψ 2T − λ1λ 2ψ 1T mψ 2T = 0

From the eigenvalues λ1 and λ1 , ω 1 and ζ 1 can be determined from Eq. (14.5.6): k m

ω2 D =

Substituting these individual terms in the left side of Eqs. (14.6.1) and (14.6.2)

Note that two of the eigenvalues of the system are complex conjugates, whereas two are real and negative valued, with λ3 ≥ λ2 .

ω1 = λ1 = 0.7743

However, ω 2 D should be determined from Eq. (14.10.5):

ψ 1T mψ 2T = m ( 0.1523 − 0.0399 i )

k (− 0.1977 ± 0.7486 i ) m

λ2 = −1.3707

k m

To verify that the eigenvectors ψ 1 and ψ 2 are orthogonal, we compute the individual terms in the left side of Eqs. (14.6.1) and (14.6.2).

0⎤ ⎥ 0⎥ ⎥ −k ⎥ k ⎥⎦

The eigenvalue problem can be solved numerically using an appropriate algorithm, e.g., the Matlab function eig ( b, −a ) , resulting in the eigenvalues

λ1 , λ 1 =

ω1D = Im(λ1 ) = 0.7486

Re(λ1 )

λ1

This verifies that the eigenvectors ψ 1 and ψ 2 computed for the system are orthogonal; similar calculations show that other pairs of modes are orthogonal.

= 0.2553

Substituting eigenvales λ2 and λ3 in Eqs. (14.10.3) and (14.10.4) gives

ω 2 = λ2 λ3 = 1.8265

k m

ζ2 = −

λ2 + λ3 = 1.0415 2ω 2

From the eigenvalue λ1 , the corresponding frequency ω1D of the damped system is determined from its definition in Eq. (14.5.4):

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 9 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 14.10 The initial displacement and velocity vectors are: ⎧1⎫ u(0) = ⎨ ⎬ ⎩2⎭

⎧0⎫ u& (0) = ⎨ ⎬ ⎩0⎭

Substituting them in Eq. (14.7.4) together with m , c , λn , and ψ n determined in the solution to Problem 14.9 gives: B1 = B2 =

B3 =

λ1ψ 1T mu ( 0 ) + ψ 1T cu ( 0 ) + ψ 1T mu& ( 0 ) 2λ1ψ 1T mψ 1 + ψ 1T cψ 1

λ2ψ 2T mu ( 0 ) + ψ 2T cu ( 0 ) + ψ 2T mu& ( 0 ) 2λ2ψ 2T mψ 2 + ψ 2T cψ 2

λ3ψ 3T mu ( 0 ) + ψ 3T cu ( 0 ) + ψ 3T mu& ( 0 ) 2λ3ψ 3T mψ 3 + ψ 3T cψ 3

= 0.7460 − 0.3125 i = 1.0005

= −0.4924

Using Bn and ψ n from Problem 14.1, β1 and γ 1 are determined from Eq. (14.7.5) and β 2 and γ 2 from Eqs. (14.10.8) and (14.10.9) as follows ⎧1.1901⎫ ⎬ ⎩1.4919⎭

β 1 = Re(2 B1ψ 1 ) = ⎨

⎧− 0.3454⎫ ⎬ ⎩− 0.6250⎭

γ 1 = Im(2 B1ψ 1 ) = ⎨

⎧ − 0.1901⎫ ⎬ ⎩ 0.5081⎭

β 2 = B2ψ 2 + B3ψ 3 = ⎨

⎧ 0.7240⎫ ⎬ ⎩ − 1.4928 ⎭

γ 2 = B2ψ 2 − B3ψ 3 = ⎨

The free-vibration response is given by u(t ) = u1 (t ) + u 2 (t )

where u1 (t ) is determined by substituting β1 and γ 1 into the n = 1 term on the right side of Eq. (14.7.6):

⎡⎧ −0.1901 ⎫ ⎬ cosh ω2 D t − ⎣⎩ 0.5081⎭

u 2 ( t ) = e −1.0415 ω t ⎢ ⎨ 2

{

}

0.7240 −1.4928

⎤ ⎦

sinh ω 2 D t ⎥

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 10 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 14.11 B1g are determined by substituting m , c , and ψ n from the solution of Problem 14.9 in Eq. (14.8.1): B1g =

−ψ 1T mι 2λ1ψ 1T mψ 1 + ψ 1T cψ 1

m (0.0549 + 0.7897 i ) k

−ψ 2T mι

m B2g = = −0.0592 T T k 2λ2ψ 2 mψ 2 + ψ 2 cψ 2 B3g =

−ψ 3T mι 2λ3ψ 3T mψ 3 + ψ 3T cψ 3

= −0.0505

m k

Using the Bng and ψ n , β1g and γ 1g are determined from Eq. (14.8.2), and β 2g and γ 2g are determined from Eqs. (14.10.11) and (14.10.12) as follows:

(

⎫ m ) ⎧− 00..0544 ⎬ 1097 k

(

⎫ m ) ⎧11..2117 ⎬ 5794 k

β 1g = Re 2 B1gψ 1 = ⎨

⎭

⎩

γ 1g = Im 2 B1gψ 1 = ⎨ ⎩

⎭ 0 . 0544⎫ m ⎧ β 2g = B2gψ 2 + B3gψ 3 = ⎨ ⎬ ⎩− 0.1097 ⎭ k ⎧0.0003⎫ m ⎬ ⎩0.0087⎭ k

γ 2g = B2gψ 2 − B3gψ 3 = ⎨

The free-vibration response is given by u(t ) = u1 (t ) + u 2 (t )

⎧ 0.0566⎫ m ⎬ ⎩− 0.1168⎭ k

α 2g = ζ 2 β 2g − ζ 22 − 1 γ 2g = ⎨

The response of the system can be determined from u(t ) = u1 (t ) + u 2 (t )

where h2 (t ) and h&2 (t ) are given by Eq. (14.10.13) and (14.10.14), respectively. The vectors β ng and γ ng satisfy Eq. (14.8.9):

β 1g + β 2g =

⎡ ⎧ −0.0544 ⎫ ⎤ m ⎧1.2117 ⎫ u1 ( t ) = e−0.2553ω1t ⎢ ⎨ ⎬ cos ω1D t − ⎨ ⎬ sin ω1D t ⎥ 0.1097 1.5794 ⎭ ⎩ ⎭ ⎣⎩ ⎦ k

m ⎡⎧− 0.0544⎫ ⎧ 0.0544⎫⎤ ⎧0⎫ ⎢⎨ ⎬⎥ = ⎨ ⎬ ⎬+⎨ k ⎣⎢⎩ 0.1097 ⎭ ⎩− 0.1097 ⎭⎦⎥ ⎩0⎭

∑ω [α − 2ζ β ] = ω 2

where u1 (t ) is determined by substituting β 1g and γ 1g into the n = 1 term on the right side of Eq. (14.8.3):

into Eq.

n =1

ω2

g n

⎧− 0.0544⎫⎤ m ⎡⎧− 1.1854⎫ ⎢⎨ ⎬⎥ + ⎬ − 2ζ 1 ⎨ k ⎣⎢⎩− 1.4990⎭ ⎩ 0.1097 ⎭⎦⎥

⎧ 0.0544⎫⎤ ⎧− 1⎫ m ⎡⎧ 0.0566⎫ ⎢⎨ ⎬⎥ = ⎨ ⎬ ⎬ − 2ζ 2 ⎨ k ⎢⎣⎩− 0.1168⎭ ⎩− 0.1097⎭⎥⎦ ⎩− 1⎭

and u 2 (t ) is determined by substituting β 2g and γ 2g into Eq. (14.10.10):

α1g = ζ 1β1g − 1 − ζ 12 γ 1g = ⎨

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 11 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

given by Eq. (14.10.13) and (14.10.14), respectively

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 12 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 14.13

Part a

φ2 = − 1.5610 1.0000 , which give Φ (see Section 10.3). Substituting c and Φ gives

This a 2-DOF system; u1=displacement of the base slab; u2 displacement of the mass m. The differential equations governing the nodal displacements u are:

⎡ 0.9680 − 1.5080⎤ C=⎢ m 6.6724⎥⎦ ⎣ − 1.5080

⎡mb ⎢ ⎣

where:

kb = (m + mb )ωb2 = 5π 2 m / 3

c = 2mω f ζ f = π m / 5

cb = 2(m + mb )ωbζ b = π m / 3

⎡0.9680 ⎤ C=⎢ ⎥m 6 . 6724 ⎣ ⎦ The corresponding modal damping ratios are ζ 1 = 9.651% and ζ 2 = 5.0571%

mb = 2m / 3 k = mω 2f = 25π 2 m

which is not a diagonal matrix. Thus, the equations of motion in modal coordinates are coupled. It is this coupling we are neglecting in classical modal analysis to obtain the diagonal C matrix.

Displacements are given by Eq. (14.4.2):

u(t ) = u1 (t ) + u 2 (t )

Modal analysis of the nonclassically damped system following Examples 14.13 and 14.14, gives the displacement response:

u(t ) = u1 (t ) + u 2 (t )

⎧0.9758⎫ u1 (t ) = ⎨ ⎬ D1 (t ) ⎩1.0155 ⎭

⎧ − 0.0015 ⎫ & ⎧0.9766⎫ u1 (t ) = ⎨ ⎬ D1 (t ) ⎬ D1 (t ) + ⎨ ⎩ 0.0009⎭ ⎩1.0164 ⎭

The damping matrix in modal coordinates is given by Eq. (10.9.5): C = Φ Tc Φ , where the natural vibration modes of the undamped system are φ1 = 0.9610 1.0000

and

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 13 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

4.7714 (a) 4.7732 (b) 4.7767 (c)

4 u1, in.

part (a) part (b) part (c)

0 -2 -4 -6 0

15 time, sec

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 14 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

CHAPTER 15 Problem 15.1 Story Stiffness k

m m

0.9

0.8 – 0.5

0.6

0.3

1 0.5

–1

. OP LM 0.3101 − 12091 . P MM0.6524 − 12394 ~ Φ = Ψ z = 0.8966 − 0.6702 P P MM 10428 . 0.4986P PQ MN 11730 . . 10780 % n and φ% n with the approximate Comparing these ω results of Example 15.1 and the exact results in Section 12.8, we note that the selection of Ritz vectors has little influence on the two vibration frequencies or the first mode but affects considerably the second mode.

–1

k ψ

LM 0.3 − 1OP MM0.6 − 1PP = 0.8 − 0.5 MM0.9 0.5PP MN 10. 1PQ

Ψ = ψ1 ψ2

1. Determine mass and stiffness matrices. Available in Example 15.1. % 2. Compute k% and m.

LM 0.24 − 0.05OP N− 0.05 2.5Q . O ~ = Ψ m Ψ = m L 2.9 015 m MN015 35 . . PQ

~ k = ΨT k Ψ = k

3. Determine approximate frequencies and modes.

ek~ − ω~ m~ j z = 0 2

ω% 1 = 0. 2866 z1 =

RS0.9996UV T0.0282W

k m

ω% 2 = 0. 8474 z2 =

k m

RS− 0.0853UV T 0.9964W

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 1 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 15.2

5. Compute natural vibration modes.

The k and m are available in Example 15.1 and s = 0 0 0 0 1

1. Determine the first Ritz vector. Solve

R|0.2649U| ||0.5298|| y ky = s ⇒ ψ = = S0.7948V || ey m y j ||10597 . |W |T 13246 . 1

T 1

LM0.330 0.560OP M0.636 0.910PP ~ 0.838 Φ = Ψ z = M 0.895 MM1082 P . . P 0134 MN 1173 . − 1414 . PQ We note that this selection of s gives essentially the same values for ω1 and φ1 as from s = m1 (Example 15.2) and both are close to the exact results. However, this selection of s gives less accurate values for ω 2 and φ2 than from s = m1 (Example 15.2).

2. Determine the second Ritz vector. Solve

R| 0.0061U| || 0.0100|| k y = m ψ ⇒ y = S 0.0096V || 0.0026|| |T− 0.0130|W ψ$ = y − eψ m y j ψ ⇒ R| 0.5939U| || 0.9757|| ψ$ ψ = = S 0.9333V || 0.2545|| eψ$ m ψ$ j . W| T|− 12726 2

T 1

T 2

% 3. Compute k% and m. Ψ = ψ1 ψ 2 − 10.64O . LM 11075 − 10 64 103 . .932PQ N . 0 ~ = Ψ mΨ = L10 m MN 0 10. OPQ ~ ~ m ~ z = 0. 4. Solve ek − ω j ~ k = ΨT k Ψ =

ω% 1 = 3.1418 rads sec z1 =

RS0.9937UV . T01124 W

ω% 2 = 10. 2536 rads sec z2 =

. RS− 01124 UV 0 . 9937 T W

Problem 15.3 Story Stiffness k/2

3k/4 3k/4 k

m/2

m m m

z1 = u5

0.8

0.6

0.4

0.2

1 0 – 0.5 –1

RS 0.9979UV T− 0.0647W

4. Determine natural vibration modes.

LM0.2319 − 0.4366OP MM0.4639 − 0.8732PP ~ Φ = Ψ z = 0.6311 − 0.3396 PP MM 0.7983 01940 . PQ MN0.9332 12126 . ω1 = 8. 262 rads sec

RS0.2425UV T0.9702W

5. Compare with exact values.

– 0.5

k ψ

z2 =

k = 200 kips in. m = 100 386 = 0. 259 kip - sec2 in.

LM0.2 − 0.5OP M0.4 − 1P Ψ = M 0.6 − 0.5P MM 0.8 0PP MN 10. 1PQ

ω 2 = 22. 347 rads sec

LM0.2319 − 0.4366OP MM 0.4433 − 0.5908PP Φ = 0.6728 − 0.2868 MM 0.8231 0.2646PP MN0.9029 0.7493PQ

1. Determine mass and stiffness matrices.

LM400 − 200 0 0 0OP 350 − 150 0 0P MM 0P k = P MM (sym) 300 − 150 250 − 100P MN 100PQ LM0.259 OP 0.259 M PP 0.259 m = M MM PP 0.259 MN PQ 01295 . %. 2. Compute k% and m

LM32 10OP N10 275Q ~ = Ψ m Ψ = L 0.4404 − 0.0777O m MN− 0.0777 0.5181PQ ~ k = ΨT k Ψ =

3. Solve eigenvalue problem. %2 m % z k% z = ω

ω% 1 = 8. 389 rads sec

ω% 2 = 23. 59 rads sec

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 3 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 15.4 The k and m are available in the solution to Problem 15.3 and s = 1 1 1 1 0.5

1. Determine the first Ritz vector.

6. Comment on accuracy.

Solve

R|0.3982U| ||0.7078|| y . = S 10027 ky = s ⇒ ψ = V| | y m y e j | 11797 . || . T|12682 W 1

LM0.2319 − 0.4366OP MM0.4449 − 0.4200PP ~ Φ = Ψ z = 0.6753 − 01042 MM0.8249 0..2095PP MN0.9038 0.4108PQ The results using force dependent Ritz vectors are better than those obtained using the vectors guessed in Problem 15.3.

T 1

2. Determine the second Ritz vector. Solve ky 2 = mψ 1 ⇒ y 2

R|− 0.7396U| ||− 0.7016|| . ψ$ = y − eψ m y j ψ = S − 01490 || 0.3965V|| T| 0.7456W| . R| − 11879 U| − 11269 . || || ψ$ = S− 0.2393V ψ = eψ$ m ψ$ j || 0.6368|| |T 11976 |W . 2

T 1

T 2

%. 3. Compute k% and m Ψ = ψ1 ψ 2

LM 69.413 − 23.333OP N− 23.333 547.737Q ~ = Ψ m Ψ = L 1 0O m MN0 1PQ ~ ~ m ~ z = 0. 4. Solve ek − ω j ~ k = ΨT k Ψ =

ω% 1 = 8. 263 rads sec z1 =

RS 0.9988UV T0.04861W

ω% 2 = 23. 428 rads sec z2 =

RS− 0.04861UV T 0.9988W

5. Compute natural vibration modes.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 4 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

~ 5. Determine the natural modes, φ n = Ψ z n , n = 1,2.

Problem 15.5 The k and m are given in the solution to Problem 15.3 and s= 0 0 0 0 1

φ 2 = 0.4467 0.7459 0.7516 0.1019 − 2.2489

1. Determine the first Ritz vector, ψ 1 .

The exact values of the first two frequencies and modes are:

y 1 = 0.0050 0.0100 0.0167 0.0233 0.0333

ω 1 = 8.2619

( y m y ) = 0.0197 to obtain the 1/ 2

T 1

6. Compare with exact results:

Solve k y 1 = s to obtain

Divide y 1 by

φ1 = 0.3338 0.6422 0.9860 1.2173 1.3281

φ1 = 0.3407 0.6513 0.9886 1.2093 1.3266

normalized vector:

ψ 1 = 0.2534 0.5068 0.8447 11826 . 1.6894

φ 2 = 0.8698 11769 . 0.5713 − 0.5271 − 1.4927

The approximate values of the first frequency and mode from Problems 15.4 and 15.5 are similarly accurate, whereas the use of s = 1 1 1 1 0.5 T in Problem

2. Determine the second Ritz vector, ψ 2 . Solve k y 2 = mψ 1 to obtain: y 2 = 0.0047 0.0091 0.0140 0.0175 0.0197

ω 2 = 22.3473

15.4 provides better accuracy for the second frequency and mode than from Problem 15.5.

Orthogonalize y 2 with respect to ψ 1 : a12 = ψ 1T m y 2 = 0.0143

ψ$ 2 = y 2 − 0.0143 ψ 1 = 10 − 2 0.1089 0.1851 0.1991 0.0672 0.4370

Divide

ψ$ 2

(ψ$ m ψ$ ) =0.0022 to get the T 2

1/ 2

normalized vector:

ψ 2 = 0.4967 0.8438 0.9076 0.3064 − 1.9918

~ ~ . 3. Compute k and m

Ψ = [ψ 1 ψ 2 ]

− 100.948⎤ ⎡ 85. 619 ~ k = Ψ T kΨ = ⎢ ⎥ ⎢⎣ − 100.948 656.450 ⎥⎦ ⎡1.0 0 ⎤ ~ = ΨT mΨ = ⎢ m ⎥ ⎢⎣ 0 1.0⎥⎦

4. Solve the reduced eigenvalue problem, Eq. (15.3.11). ~ = 8.2640 ~ = 25.9572 ω ω 1

⎡0.9856⎤ z1 = ⎢ ⎥ ⎢⎣ 0.1692 ⎥⎦

⎡− 0.1692⎤ ⎥ z2 = ⎢ ⎢⎣ 0.9856 ⎥⎦

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 5 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 15.6

φ2 = − 01367 . − 01403 . 0.2100 11921 . − 2.2137

The stiffness and mass matrices , k and m, are given in Problem 15.3 and s = 0 0 0 −2 1

6. Compare with exact results. The first two exact modes and frequencies of the structure are:

ω 1 = 8.2619

1. Determine the first Ritz vector, ψ 1 . Solve k y 1 = s to obtain T

1/ 2

T 1

Problems 15.4, 15.5 and 15.6 lead to almost the same accuracy for the first frequency and mode. The second frequency and mode differ among the three solutions, being the best from Problem 15.4 and the worst from Problem 15.6.

normalized vector:

ψ 1 = − 0.3052 − 0.6104 − 1.0173 − 1.4242 − 0.8138

φ 2 = 0.8698 11769 . 0.5713 − 0.5271 − 1.4927

( y m y ) = 0.0164 to obtain the

y 1 by

ω 2 = 22.3473

φ1 = 0.3407 0.6513 0.9886 1.2093 1.3266

y1 = − 0.0050 − 0.0100 − 0.0167 − 0.0233 − 0.0133 Divide

2. Determine the second Ritz vector, ψ 2 . Solve k y 2 = mψ 1 to obtain: y 2 = − 0.0049 − 0.0094 − 0.0143 − 0.0174 − 0.0185

Orthogonalize y 2 with respect to ψ 1 : a12 = ψ 1T m y 2 = 0.0140

ψ 2 = y2 − 0.0140ψ1 = 10−2 − 0.0600 − 0.0804 − 0.0023 0.2515 − 0.7090

Divide ψ$ 2 by

(ψ$ m ψ$ ) T 2

1/ 2

= 4.7518 ⋅ 10 −5 to get the

normalized vector:

ψ 2 = − 0.2068 − 0.2774 − 0.0079 0.8673 − 2.4446

~ ~ . 3. Compute k and m Ψ = [ψ 1 ψ 2 ]

− 100.948⎤ ⎡ 85. 619 ~ k = Ψ T kΨ = ⎢ ⎥ ⎢⎣ − 100.948 656.450 ⎥⎦ ⎡1.0 0 ⎤ ~ = ΨT mΨ = ⎢ m ⎥ ⎢⎣ 0 1.0⎥⎦

4. Solve the reduced eigenvalue problem, Eq. (15.3.11). ~ = 8.2630 ~ = 35.8902 ω ω rad/sec 1

⎡0.9768⎤ ⎥ z1 = ⎢ ⎢⎣ 0.2141⎥⎦

⎡− 0.2141⎤ z2 = ⎢ ⎥ ⎢⎣ 0.9768 ⎥⎦

~ 5. Determine the natural modes, φ n = Ψ z n , n=1,2.

φ1 = 0.3424 0.6556 0.9954 1.2055 1.3183

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 6 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 15.7

•

The stiffness and mass matrices , k and m, are given in Problem 15.3.

Ψ = [ψ 1 ψ 2 ψ 3 ψ 4 ψ 5 ]

1. Compute the Ritz vectors following the steps in Table 15.4.1 •

For the force distribution s a :

Ψ = [ψ 1 ψ 2 ψ 3 ψ 4 ψ 5 ] 0.4967 0.8386 0.9932 1.3636⎤ ⎡0.2534 ⎥ ⎢ ⎢ 0.5068 0.8438 1.0229 0.3766 − 1.3050⎥ ⎥ ⎢ 0.9076 0.0358 − 1.4282 0.5307⎥ = ⎢ 0.8447 ⎥ ⎢ ⎢ 11826 . 0.3064 − 1.2992 0.8148 − 0.1258⎥ ⎥ ⎢ ⎢ 1.6894 − 1.9918 0.9178 − 0.2364 0.0194⎥⎦ ⎣

For the force distribution s c :

1.2697 − 0.796 0.2122⎤ ⎡ 0.3981 − 1.1878 ⎢0.7077 − 1.1268 − 0.3903 1.2493 − 0.6136⎥ ⎢ ⎥ = ⎢1.0026 − 0.2393 − 1.0788 − 0.5555 1.1511⎥ ⎢ ⎥ 0.6367 − 0.0110 − 0.7105 − 1.2483⎥ ⎢1.1796 ⎢⎣1.2680 1.1974 1.3650 1.3056 1.0538⎥⎦

2. Compute e J . The error norm, e J , is defined as: eJ =

sT e J sT s

where J ~ e J = s − ∑ Γn mψ n

•

n =1

For the force distribution s b :

3. Comments.

Ψ = [ψ 1 ψ 2 ψ 3 ψ 4 ψ 5 ] ⎡ − 0.3052 ⎢ ⎢− 0.6104 ⎢ = ⎢ − 1.0173 ⎢ ⎢ − 1.4242 ⎢ ⎢ − 0.8138 ⎣

~ Γn = ψ Tn s

− 0.2068

0.3007

1.2005

− 0.2774

0.7990

11183 .

− 0.0079

1.2630

− 1.0426

0.8673

− 1.0313

0.1138

− 2.4446 − 0.9721 − 0.3694

1.4807⎤ ⎥ − 1.2335⎥ ⎥ 0.3780⎥ ⎥ − 0.0552⎥ ⎥ − 0.0123⎥⎦

The error norm, e J , computed using the above formula is plotted versus the number of Ritz vectors in Fig. P15.7. For a given force distribution, the error decreases as more Ritz vectors are included, and is zero when all five Ritz vectors are included. For a fixed number of Ritz vectors, the error is smallest for the force distribution s c , and is larger for force distributions s a and sb.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 7 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

1.0

Error eJ

0.8 0.6 0.4 0.2 0.0 0

1.0

Error eJJ

0.8

0.6 0.4 0.2 0.0 0

1.0

0.5

Error eJ

0.8

1 1 1

0.6 0.4

0.2 0.0 0

J = number of Ritz vectors Fig. P15.7

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 8 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 15.8 The stiffness and mass matrices, k and m, are given in Problem 15.3. 1. Determine the natural frequencies and modes.

ω1 = 8.2619 ω 4 = 41.7423

which have several advantages. In particular: (1) these uncoupled equations are easier to solve in response history analysis; (2) they permit estimation of the peak value of the earthquake response of a structure by response spectrum analysis.

ω 2 = 22.3473 ω 3 = 32.9203 ω 5 = 48.9507 rad /sec

Φ = [φ1 φ 2

φ3 φ 4 φ5 ]

. 0.8698 0.9108 0.8706 11832 ⎤ ⎡ 0.3407 ⎥ ⎢ ⎢ 0.6513 . 11769 0.5430 − 0.2238 − 1.3061⎥ ⎥ ⎢ 0.5713 − 0.9638 − 1.0095 0.7800⎥ = ⎢ 0.9886 ⎥ ⎢ ⎢ 1.2093 − 0.5271 − 0.6666 1.2427 − 0.3620⎥ ⎥ ⎢ ⎢ 1.3266 − 1.4927 1.6507 − 0.9886 0.1721⎥⎦ ⎣

2. Compute e J . The error e J is defined as: eJ =

sT e J sT s

where J

e J = s − ∑ Γn m φ n n =1

Γn = φnT s

The error norm, e J , computed using the above formula is plotted versus the number of natural modes in Fig. P15.8. 3. Comments. For a given force distribution, the error decreases as more modes are included, and is zero then all five modes are included. For a fixed number of modes, the error is smallest for the force distribution s c , largest for s b , and has an intermediate value for s a . Figure P15.8 also shows the error norm, e J , versus the number of Ritz vectors (from Problem 15.7). The error is smaller when Ritz vectors are used, because they are derived from the force distribution. Ritz vectors are useful for dynamic analysis of large systems with classical damping, since the vibration properties of the system can be obtained by solving, a smaller eigenvalue problem of order J, instead of original eigenvalue problem of size N. It must be noted that the resulting frequencies and mode shapes are approximations to the first J natural vibration frequencies and modes of the system. While this property indicates that Ritz vectors are preferable to natural modes, the latter leads to uncoupled modal equations of motion, © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 9 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Error eJ

1.0

Ritz vectors

0.8

Natural Modes

0.6 0.4 0.2 0.0 0

1.0

Error eJ

0.8

0.6 0.4 0.2 0.0 0

1.0

0.5

Error eJ

0.8

1 1 1

0.6 0.4

0.2 0.0 0

J = number of natural modes or Ritz vectors Fig. P15.8 © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 10 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Substituting for Ψ and z 0 gives

Problem 15.9 The stiffness and mass matrices , k and m, are given in Problem 15.3. From Problem 15.7 the first two Ritz vectors and approximate natural frequencies using the force distribution s = 1 1 1 1 0.5

Ψ = [ψ 1

are:

From Problem 15.8 the natural frequencies and modes of the structure are:

ω 1 = 8.2619 ω 2 = 22.3473 ω 3 = 32.9203 ω 4 = 417423 ω 5 = 48.9507 rad / sec . Φ = [φ1

⎡ 69.3929 − 23.3226⎤ ~ k = ΨT k Ψ = ⎢ ⎥ ⎣− 23.3226 547.6113 ⎦

(a)

⎡1.0 0 ⎤ ~ = ΨT mΨ = ⎢ m ⎥ ⎢⎣ 0 1.0⎥⎦

(b)

⎧ 1.0161⎫ ~ T L = Ψ m1 = ⎨ ⎬ ⎩− 0.3416 ⎭

(c)

φ4

φ5 ]

⎡0.3407 0.8698 0.9108 0.8706 1.1832⎤ ⎢0.6513 1.1769 0.5430 − 0.2238 − 1.3061⎥ ⎥ ⎢ = ⎢0.9886 0.5713 − 0.9638 − 1.0095 0.7800⎥ ⎥ ⎢ ⎢1.2093 − 0.5271 − 0.6666 1.2427 − 0.3620⎥ ⎢⎣1.3266 − 1.4927 1.6507 − 0.9886 0.1721⎥⎦

5. Determine the steady-state response.

Eq.(13.1.). J

3. Determine steady-state response. Equation (d) is specialized for the steady-state part of the response. By substituting z(t ) = z 0 sin ω t and u&&g (t ) = 0.2 g sin ω t and canceling the sin ω t terms on

both sides: (e)

Solving the two coupled algebraic equations gives (f)

The peak values of the displacements are u0 = Ψ z0

φ3

The response of the structure to ground motion u&&g ( t ) using first J modes of vibration is given by

~ ~ ~ , k , and L are given by Eqs. (a)-(c), and where m u&&g (t ) = 0.2 g sin ω t , where ω = rad/sec.

⎧0.4866 ⎫ z0 = ⎨ ⎬ ⎩0.1169 ⎭

φ2

(i)

(j)

2. Set up equations in generalized coordinates, Eq. (15.3.3) is specialized for this problem. ~ ~ ~ &z& + k m z = −L u&&g (t ) (d)

b g

(h)

4. Determine frequencies and modes.

Part a: Rayleigh-Ritz analysis ~ ~ ~ 1. Compute, k , m and L .

~ k − 15 m z0 = −0.2 g L

Part b: Modal analysis

. ⎡ 0.3981 − 11878 ⎤ ⎢ ⎥ ⎢0.7077 − 11268 ⎥ . ⎢ ⎥ ψ 2 ] = ⎢ 1.0026 − 0.2393⎥ ⎢ ⎥ ⎢ 11796 . 0.6367⎥ ⎢ ⎥ ⎢ 1.2680 ⎥ 11974 . ⎣ ⎦

2 ~

u 0 = 0.0548 0.2126 0.4599 0.6484 0.7570

u(t ) = ∑ Γn φn Dn ( t )

(k)

n =1

where Γn =

φ nT m 1 φ nT m φ n

(l)

and Dn (t ) is governed by && + ω2 D = −u&& (t ) D n n n g

(m)

= −0.2 g sinω t

The steady state solution is Dn ( t ) = −

0.2 g

ω n2 − ω 2

sin ω t

(n)

where ω = rad/sec and ω n are given by Eq. ( i ). Substituting φ n and m into Eq. ( l ) gives Γ1 = 0.9982,

Γ2 = 0.3483,

Γ4 = 0.0999,

Γ5 = 0.0988

Γ3 = 0.1681,

(o)

(g)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 11 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Substituting Γn from Eq. (o), φ n from Eq. ( j ) and Dn (t ) from Eq. (n) into Eq. (k) gives the displacement response: u(t ) = u 0 sin ω t

(p)

where u 0 is given as follows. Considering two modes only, i.e., J = 2: u 0 = 0.0823 0.2049 0.4301 0.6462 0.7985

(q) Considering all five modes, i.e., J = 5: u 0 = 0.0599 0.2024 0.4470 0.6130 0.7780

(r) 6. Comments. Comparison of Eqs. (h) and (q) with the exact solutions, Eq. (r), indicates that Rayleigh-Ritz method using two force-dependent Ritz vetors gives a roughly similar accuracy as modal analysis with two modes for responses u2, u3, u4, and u5. However, the error in u1 determined by modal analysis is large. This becomes obvious from the following table: Percentage Error Response

Reyleigh-Ritz Method, J = 2

Modal Analysis, J = 2

-2.7 %

2.6 %

5.8 %

5.4 %

u3 u2

2.9 %

-3.8 %

5.0 %

1.2 %

8.5 %

37.4 %

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 12 recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

CHAPTER 16

Problem 16.1 First we set up modal equations. These are given by Eq. (16.2.3) with M , C , K and P(t ) available from Example 16.1: ⎡1 ⎤ M=⎢ ⎥ ⎣ 1⎦

⎡0.559 ⎤ C=⎢ 1.632⎥⎦ ⎣

⎡31.27 ⎤ K=⎢ ⎥ 266 . 4 ⎣ ⎦

⎡−1.067 ⎤ P (t ) = ⎢ ⎥ u&&g (t ) ⎣ 0.336 ⎦

The first two natural frequencies and modes are given by Eq. (b) of Example 16.1. We now implement the procedure of Table 16.2.1. 1.0 Initial calculations 1.1 Since the system starts from rest, u 0 = u& 0 = 0 ; therefore, q 0 = q& 0 = 0 . 1.2

p 0 = 0 ; therefore, P0 = 0 .

1.3

&&0 = 0 . q

1.4

Δt = 0.1 sec.

1.5

q −1 = 0 .

2.3 Solve: ⎡ Pˆ ⎤ ⎡102.8 ⎤ ⎡ q1 ⎤ = ⎢ 1⎥ ⎥ ⎢ ⎢ ⎥ 108.2⎦ ⎣q 2 ⎦ i +1 ⎣ Pˆ2 ⎦ i ⎣

⎡q ⎤ ⇒ ⎢ 1⎥ ⎣q2 ⎦ i +1

The resulting modal displacements qi are given in Table P16.1.

2.5

⎡ u1 ⎤ ⎡0.334 −0.895 ⎤ ⎢u ⎥ ⎢ 0.641 − 1.173⎥ ⎢ 2⎥ ⎢ ⎥ ⎡q ⎤ ⎢u3 ⎥ = ⎢0.895 − 0.641⎥ ⎢ 1 ⎥ ⎢ ⎥ ⎢ ⎥ ⎣q 2 ⎦ i +1 ⎢u 4 ⎥ ⎢1.078 0.334 ⎥ ⎢⎣u5 ⎥⎦ i +1 ⎢⎣1.173 1.078 ⎥⎦

These displacements are also presented in Table P16.1.

1.6 Substituting M , C , and Δt in step 1.6 gives ⎡102.8 ⎤ K̂ = ⎢ 108.2⎥⎦ ⎣

1.7 Substituting M , C , K , and Δt in step 1.7 gives ⎡97.20 ⎤ a=⎢ ⎥ 91 . 84 ⎣ ⎦

⎡−168.7 ⎤ b=⎢ ⎥ 66 . 45 ⎣ ⎦

2.0 Calculations for each time step i 2.1

⎡ P1 ⎤ ⎡−1.067⎤ ⎢ P ⎥ = ⎢ 0.336 ⎥ (u&&g ) i ⎦ ⎣ 2 ⎦i ⎣

2.2

⎡ Pˆ1 ⎤ ⎡ P1 ⎤ ⎡ 97.20q1 ⎤ ⎡− 168.7q1 ⎤ ⎢ ˆ ⎥ = ⎢ ⎥ −⎢ ⎥ − ⎢ 66.45q ⎥ P 91 . 84 q 2 ⎦ i −1 ⎣ 2 ⎦i ⎣ P2 ⎦ i ⎣ 2 ⎦ i ⎣

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 1 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Table P16.1: Numerical solution of modal equations by the central difference method ti

0.00 0.10

0.0000 0.0000

0.20

-1.1779

-0.7087

-1.1678

-1.2804

-1.1518

-1.0015

0.30

-3.8393

-1.5983

-2.8740

-3.6642

-4.0195

-4.1211

0.40

-7.0939

-2.4159

-4.6068

-6.3861

-7.6271

-8.2612

0.50

-9.1914

-3.0850

-5.9097

-8.2420

-9.8991

-10.7581

0.60

-8.3786

-2.7453

-5.2995

-7.4653

-9.0487

-9.8873

0.70

-3.8833

-0.9976

-2.0963

-3.2634

-4.2963

-4.9133

0.80

3.4547

1.4367

2.5843

3.2962

3.6174

3.7100

0.90

11.2485

3.8374

7.3135

10.1309

12.0915

13.0913

1.00

16.3744

5.4893

10.5194

14.6783

17.6376

19.1734

1.10

16.2401

5.3357

10.2909

14.4802

17.5335

19.1470

1.20

11.1728

3.7613

7.1984

10.0268

12.0289

13.0638

1.30

2.9823

1.0480

1.9791

2.7079

3.1944

3.4341

1.40

-5.6699

-1.9519

-3.7095

-5.1192

-6.0882

-6.5775

1.50

-12.1266

-4.0557

-7.7780

-10.8636

-13.0656

-14.2110

1.60

-14.5430

-4.7988

-9.2426

-12.9818

-15.6935

-17.1212

1.70

-12.4039

-4.1674

-7.9806

-11.1256

-13.3574

-14.5133

1.80

-6.6078

-2.2359

-4.2722

-5.9382

-7.1098

-7.7123

1.90

0.8832

0.3365

0.6204

0.8206

0.9362

0.9855

2.00

7.6980

2.5696

4.9309

6.8927

8.2959

9.0272

Problem 16.2 The relevant data are available from the solution to Problem 16.1. The same procedure is implemented. 1.0 Initial calculations 1.1 Since the system starts from rest, u 0 = u& 0 = 0 ; therefore, q 0 = q& 0 = 0 . 1.2

p 0 = 0 ; therefore, P0 = 0 .

1.3

&&0 = 0 . q

1.4

Δt = 0.05 sec.

1.5

q −1 = 0 .

⎡ Pˆ1 ⎤ ⎡ P1 ⎤ ⎡ 394.4q1 ⎤ ⎡ − 768.7 q1 ⎤ −⎢ ⎢ˆ ⎥ =⎢ ⎥ −⎢ ⎥ ⎥ ⎣ P2 ⎦ i ⎣ P2 ⎦ i ⎣383.7q2 ⎦ i −1 ⎣− 533.6q2 ⎦ i

2.3 Solve: ⎡ Pˆ1 ⎤ ⎤ ⎡ q1 ⎤ ⎡405.6 = ⎢ ⎥ ⎢ 416.3⎥⎦ ⎢⎣q2 ⎥⎦ i +1 ⎣ Pˆ2 ⎦ i ⎣

⎡q ⎤ ⇒ ⎢ 1⎥ ⎣q2 ⎦ i +1

The resulting modal displacements qi are given in Table P16.2a.

1.6 Substituting M , C , and Δt in step 1.6 gives ⎡405.6 ⎤ K̂ = ⎢ 416.3⎥⎦ ⎣

1.7 Substituting M , C , K , and Δt in step 1.7 gives ⎡394.4 ⎤ a=⎢ 383.7⎥⎦ ⎣

2.2

⎡−768.7 ⎤ b=⎢ − 533.6⎥⎦ ⎣

2.5

These displacements are also presented in Table P16.2a. 3.0 Comments The solutions to Problems 16.1 and 16.2a are compared in Table P16.2b which shows that the smaller time step leads to more accurate results.

2.0 Calculations for each time step i 2.1

⎡ P1 ⎤ ⎡−1.067⎤ ⎢ P ⎥ = ⎢ 0.336 ⎥ (u&&g ) i ⎦ ⎣ 2 ⎦i ⎣

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 3 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Table P16.2a: Numerical solution of modal equations by the central difference method ti

0.00 0.05

0.0000 0.0000

0.10

-0.1570

-0.0955

-0.1570

-0.1714

-0.1531

-0.1322

0.15

-0.5960

-0.3362

-0.5615

-0.6318

-0.5912

-0.5338

0.20

-1.3879

-0.7123

-1.2151

-1.4208

-1.4029

-1.3279

0.25

-2.5340

-1.1711

-2.0492

-2.5016

-2.6096

-2.5801

0.30

-3.9611

-1.6490

-2.9653

-3.7805

-4.1469

-4.2516

0.35

-5.5264

-2.0965

-3.8696

-5.1284

-5.8619

-6.1777

0.40

-7.0334

-2.4819

-4.6810

-6.3936

-7.5297

-8.0863

0.45

-8.2551

-2.7774

-5.3165

-7.4072

-8.8882

-9.6542

0.50

-8.9636

-2.9395

-5.6727

-7.9883

-9.6795

-10.5746

0.55

-8.9614

-2.9037

-5.6254

-7.9612

-9.6902

-10.6143

0.60

-8.1114

-2.6005

-5.0555

-7.1863

-8.7815

-9.6409

0.65

-6.3610

-1.9848

-3.8931

-5.5965

-6.9067

-7.6260

0.70

-3.7574

-1.0625

-2.1557

-3.2272

-4.1208

-4.6370

0.75

-0.4529

0.0996

0.0383

-0.2262

-0.5816

-0.8329

0.80

3.3033

1.3868

2.4881

3.1610

3.4540

3.5317

0.85

7.1843

2.6638

4.9499

6.6228

7.6433

8.1050

0.90

10.8153

3.8015

7.1784

9.8208

11.5840

12.4524

0.95

13.8108

4.6924

8.9543

12.4249

14.8529

16.0964

1.00

15.8158

5.2514

10.0942

14.1413

17.0548

18.5804

1.05

16.5462

5.4113

10.4522

14.7353

17.8732

19.5378

1.10

15.9807

5.2166

10.0822

14.2247

17.2660

18.8818

1.15

14.1986

4.6915

9.0321

12.6790

15.3195

16.7081

1.20

11.3710

3.8425

7.3451

10.2160

12.2369

13.2781

1.25

7.7447

2.6893

5.0973

7.0090

8.3074

8.9566

1.30

3.6212

1.2990

2.4377

3.3069

3.8688

4.1378

1.35

-0.6678

-0.2033

-0.4021

-0.5839

-0.7270

-0.8067

1.40

-4.7870

-1.6559

-3.1423

-4.3277

-5.1373

-5.5438

1.45

-8.4236

-2.9042

-5.5167

-7.6084

-9.0435

-9.7669

1.50

-11.3104

-3.8402

-7.3298

-10.1736

-12.1648

-13.1853

1.55

-13.2455

-4.4192

-8.4816

-11.8584

-14.2753

-15.5353

1.60

-14.1061

-4.6460

-8.9536

-12.5856

-16.2252

-16.6172

1.65

-13.8554

-4.5465

-8.7723

-12.3498

-14.9609

-16.3422

1.70

-12.5433

-4.1440

-7.9783

-11.2004

-13.5336

-14.7608

1.75

-10.3002

-3.4554

-6.6203

-9.2350

-11.0940

-12.0582

1.80

-7.3249

-2.5065

-4.7724

-6.6026

-7.8710

-8.5158

1.85

-3.8669

-1.3536

-2.5593

-3.5074

-4.1439

-4.4589

1.90

-0.2061

-0.0927

-0.1633

-0.2016

-0.2132

-0.2129

1.95

3.3697

1.1521

2.1942

3.0367

3.6213

3.9187

2.00

6.5870

2.2557

4.2939

5.9387

7.0775

7.6558

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 4 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Table P16.2b ti

u5 (Δt = 0.1)

u5 (Δt = 0.05)

0.00

0.0000

0.10

0.0000

-0.1322

-0.1804

0.20

-1.0015

-1.3279

-1.4153

0.30

-4.1211

-4.2516

-4.2998

0.40

-8.2612

-8.0863

-8.0443

0.50

-10.7581

-10.5746

-10.4925

0.60

-9.8873

-9.6409

-9.5676

0.70

-4.9133

-4.6370

-4.5850

0.80

3.7100

3.5317

3.4995

0.90

13.0913

12.4524

12.2796

1.00

19.1734

18.5804

18.3511

1.10

19.1470

18.8818

18.7924

1.20

13.0638

13.2781

13.3779

1.30

3.4341

4.1378

4.3529

1.40

-6.5775

-5.5438

-5.2533

1.50

-14.2110

-13.1853

-12.8543

1.60

-17.1212

-16.6172

-16.4102

1.70

-14.5133

-14.7608

-14.8285

1.80

-7.7123

-8.5158

-8.7981

1.90

0.9855

-0.2129

-0.5740

2.00

9.0272

7.6558

7.2622

u5 (Theoretical)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 5 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 16.3 First we set up modal equations. These are given by Eq. (16.2.3) with M , C , K and P(t ) available from Example 16.1: ⎡1 ⎤ M=⎢ ⎥ ⎣ 1⎦

⎡0.559 ⎤ C=⎢ 1.632⎥⎦ ⎣

⎡31.27 ⎤ K=⎢ 266.4⎥⎦ ⎣

⎡−1.067 ⎤ P (t ) = ⎢ ⎥ u&&g (t ) ⎣ 0.336 ⎦

The first two natural frequencies and modes are given by Eq. (b) of Example 16.1. We now implement the procedure of Table 16.2.2. 1.0 Initial calculations 1.1 Since the system starts from rest, u 0 = u& 0 = 0 ; therefore, q 0 = q& 0 = 0 . 1.2

p 0 = 0 ; therefore, P0 = 0 .

1.3

&&0 = 0 . q

1.4

Δt = 0.10 sec.

1.5 Substituting M , C , Δt , γ = 12 , and β = 14 in

2.0 Calculations for each time step i For the parameters of this problem, computational steps 2.1 through 2.5 are specialized and implemented for each time step i as follows. 2.1

⎡ 411.2 q1 + 40.56 q&1 + q&&1 ⎤ ⎢432.6 q + 41.63 q& + q&& ⎥ ⎣ 2 2 2 ⎦i

⎡ Pˆ ⎤ ⎤ ⎡ q1 ⎤ ⎡442.5 = ⎢ 1 ⎥ ⇒ q i +1 2.2 Solve ⎢ ⎥ ⎢ ⎥ 699.1⎦ ⎣q 2 ⎦ i +1 ⎣ Pˆ2 ⎦ i +1 ⎣ 2.3

⎛ ⎡ q1 ⎤ ⎡ q1 ⎤ ⎞ ⎡ q&1 ⎤ ⎡ q&1 ⎤ ⎢q& ⎥ = 20 ⎜⎜ ⎢q ⎥ − ⎢q ⎥ ⎟⎟ − ⎢q& ⎥ ⎣ 2 ⎦ i +1 ⎝ ⎣ 2 ⎦ i +1 ⎣ 2 ⎦ i ⎠ ⎣ 2 ⎦ i

2.4

⎛ ⎡ q1 ⎤ ⎡ q1 ⎤ ⎞ ⎡ q&1 ⎤ ⎡ q&&1 ⎤ ⎡ q&&1 ⎤ ⎢q&& ⎥ = 400 ⎜⎜ ⎢q ⎥ − ⎢q ⎥ ⎟⎟ − 40 ⎢q& ⎥ − ⎢q&& ⎥ ⎣ 2 ⎦i ⎣ 2 ⎦i ⎣ 2 ⎦ i +1 ⎝ ⎣ 2 ⎦ i +1 ⎣ 2 ⎦ i ⎠

The resulting modal displacements qi are given in Table P16.3a.

step 1.5 gives ⎡411.2 ⎤ a1 = ⎢ 432.6⎥⎦ ⎣

⎡40.56 ⎤ a2 = ⎢ 41.63⎥⎦ ⎣

⎡1 ⎤ a3 = ⎢ ⎥ ⎣ 1⎦

1.6 Substituting K and a1 in step 1.6 gives ⎡442.5 ⎤ K̂ = ⎢ ⎥ 699 . 1 ⎣ ⎦

⎡ Pˆ1 ⎤ ⎡− 1.067⎤ ⎢ˆ ⎥ =⎢ ⎥ (u&&g ) i +1 + ⎣ P2 ⎦ i +1 ⎣ 0.336 ⎦

2.5

⎡ u1 ⎤ ⎡0.334 −0.895 ⎤ ⎢u ⎥ ⎢ 0.641 − 1.173⎥ ⎢ 2⎥ ⎥ ⎡ q1 ⎤ ⎢ ⎢u3 ⎥ = ⎢0.895 − 0.641⎥ ⎢ ⎥ ⎢ ⎥ ⎥ ⎣q 2 ⎦ i +1 ⎢ ⎢u 4 ⎥ ⎢1.078 0.334 ⎥ ⎢⎣u5 ⎥⎦ i +1 ⎢⎣1.173 1.078 ⎥⎦

These displacements are also presented in Table P16.3a. 3.0 Comments Because of different period elongations in the two numerical methods, their comparison at each time instant is not especially meaningful.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 6 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Table 16.3a: Numerical Solution of modal equations by the average acceleration method ti

0.00 0.10

0.0000 -0.2737

0.0000 -0.1402

0.0000 -0.2392

0.0000 -0.2800

0.0000 -0.2767

0.0000 -0.2622

0.20

-1.4463

-0.6780

-1.1822

-1.4347

-1.4858

-1.4610

0.30

-3.7528

-1.5688

-2.8179

-3.5863

-3.9265

-4.0203

0.40

-6.4838

-2.3938

-4.4537

-5.9697

-6.9020

-7.3272

0.50

-8.2339

-2.7261

-5.2448

-7.3565

-8.8819

-9.6827

0.60

-7.5677

-2.3093

-4.5636

-6.6210

-8.2364

-9.1354

0.70

-3.8056

-1.0310

-2.1243

-3.2363

-4.1905

-4.7508

0.80

2.4442

0.9967

1.8024

2.3179

2.5667

2.6487

0.90

9.2892

3.2386

6.1308

8.4161

9.9594

10.7273

1.00

14.1522

4.7892

9.1505

12.7183

15.2272

16.5173

1.10

15.0420

4.9695

9.5676

13.4315

16.2297

17.7014

1.20

11.6343

3.8044

7.3487

10.3606

12.5676

13.7385

1.30

5.1099

1.7219

3.2944

4.5869

5.5007

5.9726

1.40

-2.5294

-0.7664

-1.5182

-2.2091

-2.7549

-3.0599

1.50

-9.0673

-3.0116

-5.7883

-8.1080

-9.7773

-10.6512

1.60

-12.7114

-4.3077

-8.2269

-11.4279

-13.6747

-14.8285

1.70

-12.5776

-4.2365

-8.1064

-11.2890

-13.5404

-14.7035

1.80

-8.8947

-2.9247

-5.6395

-7.9325

-9.6022

-10.4840

1.90

-2.8835

-0.9109

-1.7794

-2.5449

-3.1267

-3.4436

2.00

3.6390

1.1943

2.3043

3.2438

3.9293

4.2919

Table P16.3b: Comparison of central difference and average acceleration methods ti

u5 (Δt = 0.1)

central diff.

avg. accel.

u5 (Theoretical)

0.00

0.0000

0.10

0.0000

-0.2622

-0.1804

0.20

-1.0015

-1.4610

-1.4153

0.30

-4.1211

-4.0203

-4.2998

0.40

-8.2612

-7.3272

-8.0443

0.50

-10.7581

-9.6827

-10.4925

0.60

-9.8873

-9.1354

-9.5676

0.70

-4.9133

-4.7508

-4.5850

0.80

3.7100

2.6487

3.4995

0.90

13.0913

10.7273

12.2796

1.00

19.1734

16.5173

18.3511

1.10

19.1470

17.7014

18.7924

1.20

13.0638

13.7385

13.3779

1.30

3.4341

5.9726

4.3529

1.40

-6.5775

-3.0599

-5.2533

1.50

-14.2110

-10.6512

-12.8543

1.60

-17.1212

-14.8285

-16.4102

1.70

-14.5133

-14.7035

-14.8285

1.80

-7.7123

-10.4840

-8.7981

1.90

0.9855

-3.4436

-0.5740

2.00

9.0272

4.2919

7.2622

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 7 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 16.4 Equation (16.2.3) gives the modal equations with M , C , K and P(t ) defined in the solution of Problem 16.3. We now implement the procedure of Table 16.2.2 as follows. 1.0 Initial calculations 1.1 Since the system starts from rest, u 0 = u& 0 = 0 ; therefore, q 0 = q& 0 = 0 .

The resulting modal displacements qi are given in Table P16.4a.

2.5

1.2

p 0 = 0 ; therefore, P0 = 0 .

These displacements are also presented in Table P16.4a.

1.3

&&0 = 0 . q

3.0 Comments

1.4

Δt = 0.05 sec.

The solutions to Problems 16.3 and 16.4 are compared in Table P16.4b, which shows that the smaller time step leads to more accurate results.

1.5 Substituting M , C , Δt , γ = 12 , and β = 14 in step 1.5 gives ⎤ ⎡1622 a1 = ⎢ 1665⎥⎦ ⎣

⎤ ⎡80.56 a2 = ⎢ 81.63⎥⎦ ⎣

⎡1 ⎤ a3 = ⎢ ⎥ ⎣ 1⎦

1.6 Substituting K and a1 in step 1.6 gives ⎤ ⎡1654 K̂ = ⎢ ⎥ 1932 ⎣ ⎦

2.0 Calculations for each time step i For the parameters of this problem, computational steps 2.1 through 2.5 are specialized and implemented for each time step i as follows. 2.1

⎡ Pˆ1 ⎤ ⎡− 1.067⎤ ⎢ˆ ⎥ =⎢ ⎥ (u&&g ) i +1 + ⎣ P2 ⎦ i +1 ⎣ 0.336 ⎦ ⎡ 1622 q1 + 80.56 q&1 + q&&1 ⎤ ⎢1665 q + 81.63 q& + q&& ⎥ ⎣ 2 2 2 ⎦i

⎡ Pˆ ⎤ ⎡1654 ⎤ ⎡ q1 ⎤ = ⎢ 1 ⎥ ⇒ q i +1 2.2 Solve ⎢ ⎢ ⎥ ⎥ 1932⎦ ⎣q2 ⎦ i +1 ⎣ Pˆ2 ⎦ i +1 ⎣

⎛ ⎡q ⎤ ⎡ q& ⎤ ⎡ q ⎤ ⎞ ⎡ q& ⎤ 2.3 ⎢ 1 ⎥ = 40 ⎜⎜ ⎢ 1 ⎥ − ⎢ 1 ⎥ ⎟⎟ − ⎢ 1 ⎥ ⎣q& 2 ⎦ i +1 ⎝ ⎣q2 ⎦ i +1 ⎣q2 ⎦ i ⎠ ⎣q& 2 ⎦ i ⎛ ⎡q ⎤ ⎡q ⎤ ⎞ ⎡ q&& ⎤ 2.4 ⎢ 1 ⎥ = 1600 ⎜⎜ ⎢ 1 ⎥ − ⎢ 1 ⎥ ⎟⎟ − & & q q ⎣ 2 ⎦ i +1 ⎝ ⎣ 2 ⎦ i +1 ⎣q2 ⎦ i ⎠ ⎡ q& ⎤ ⎡ q&& ⎤ 80 ⎢ 1 ⎥ − ⎢ 1 ⎥ ⎣q& 2 ⎦ i ⎣q&&2 ⎦ i

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 8 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Table P16.4a: Numerical Solution of modal equations by the average acceleration method ti

0.00 0.05

0.0000 -0.0385

0.0000 -0.0221

0.0000 -0.0368

0.0000 -0.0411

0.0000 -0.0380

0.0000 -0.0340

0.10

-0.2233

-0.1236

-0.2073

-0.2350

-0.2223

-0.2027

0.15

-0.6719

-0.3523

-0.5980

-0.6932

-0.6763

-0.6338

0.20

-1.4508

-0.7102

-1.2252

-1.4606

-1.4793

-1.4294

0.25

-2.5612

-1.1590

-2.0388

-2.5108

-2.6468

-2.6375

0.30

-3.9339

-1.6396

-2.9474

-3.7559

-4.1178

-4.2203

0.35

-5.4343

-2.0927

-3.8460

-5.0652

-5.7525

-6.0371

0.40

-6.8763

-2.4706

-4.6342

-6.2823

-7.3451

-7.8526

0.45

-8.0448

-2.7364

-5.2199

-7.2397

-8.6507

-9.3725

0.50

-8.7234

-2.8551

-5.5134

-7.7702

-9.4223

-10.2981

0.55

-8.7238

-2.7859

-5.4229

-7.7210

-9.4485

-10.3820

0.60

-7.9141

-2.4843

-4.8632

-6.9736

-8.5876

-9.4702

0.65

-6.2419

-1.9147

-3.7771

-5.4682

-6.7897

-7.5229

0.70

-3.7495

-1.0704

-2.1644

-3.2277

-4.1084

-4.6151

0.75

-0.5787

0.0112

-0.1030

-0.3719

-0.6998

-0.9246

0.80

3.0363

1.2440

2.2468

2.8836

3.1863

3.2832

0.85

6.7862

2.5044

4.6602

6.2474

7.2241

7.6700

0.90

10.3145

3.6532

6.8822

9.3858

11.0374

11.8425

0.95

13.2529

4.5598

8.6672

11.9638

14.2317

15.3776

1.00

15.2595

5.1216

9.8110

13.6832

16.4345

17.8608

1.05

16.0961

5.2938

10.2067

14.3556

17.3759

18.9706

1.10

15.6925

5.1028

9.8745

13.9540

16.9619

18.5650

1.15

14.1127

4.5975

8.8915

12.5553

15.2513

16.6860

1.20

11.5082

3.8111

7.3319

10.2826

12.4135

13.5319

1.25

8.1037

2.7687

5.2742

7.3015

8.7095

9.4263

1.30

4.1783

1.5108

2.8285

3.8242

4.4595

4.7599

1.35

0.0430

0.1153

0.1597

0.1107

0.0087

-0.0711

1.40

-3.9837

-1.2987

-2.5111

-3.5447

-4.3047

-4.7090

1.45

-7.6001

-2.5881

-4.9352

-6.8416

-8.1714

-8.8508

1.50

-10.5438

-3.6191

-6.8842

-9.5120

-11.3257

-12.2445

1.55

-12.6102

-4.2990

-8.1949

-11.3552

-13.5564

-14.6798

1.60

-13.6669

-4.5929

-8.7947

-12.2593

-14.7171

-15.9897

1.65

-13.6612

-4.5190

-8.6968

-12.2027

-14.7378

-16.0696

1.70

-12.6223

-4.1277

-7.9730

-11.2406

-13.6348

-14.9049

1.75

-10.6567

-3.4772

-6.7213

-9.4847

-11.5144

-12.5931

1.80

-7.9382

-2.6191

-5.0446

-7.0858

-8.5663

-9.3459

1.85

-4.6928

-1.5986

-3.0480

-4.2248

-5.0453

-5.4644

1.90

-1.1802

-0.4669

-0.8515

-1.1089

-1.2446

-1.2961

1.95

2.3267

0.7059

1.3977

2.0327

2.5338

2.8136

2.00

5.5627

1.8271

3.5243

4.9595

6.0059

6.5590

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 9 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Table P16.4b ti

u5 (Δt = 0.1)

u5 (Δt = 0.05)

u5 (Theoretical)

0.00

0.0000

0.10

-0.2622

-0.2027

-0.1804

0.20

-1.4610

-1.4294

-1.4153

0.30

-4.0203

-4.2203

-4.2998

0.40

-7.3272

-7.8526

-8.0443

0.50

-9.6827

-10.2981

-10.4925

0.60

-9.1354

-9.4702

-9.5676

0.70

-4.7508

-4.6151

-4.5850

0.80

2.6487

3.2832

3.4995

0.90

10.7273

11.8425

12.2796

1.00

16.5173

17.8608

18.3511

1.10

17.7014

18.5650

18.7924

1.20

13.7385

13.5319

13.3779

1.30

5.9726

4.7599

4.3529

1.40

-3.0599

-4.7090

-5.2533

1.50

-10.6512

-12.2445

-12.8543

1.60

-14.8285

-15.9897

-16.4102

1.70

-14.7035

-14.9049

-14.8285

1.80

-10.4840

-9.3459

-8.7981

1.90

-3.4436

-1.2961

-0.5740

2.00

4.2919

6.5590

7.2622

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 10 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 16.5 Solution to this problem is available as Example 16.1 the book. Table P16.5 shows these results using linear acceleration method, those from the average acceleration method (Problem 16.3), and theoretical results. Table P16.5 ti

u5 (Δt = 0.1)

avg. accel.

linear accel.

u5 (Theoretical)

0.00

0.0000

0.10

-0.2622

-0.1742

-0.1804

0.20

-1.4610

-1.3357

-1.4153

0.30

-4.0203

-4.0229

-4.2998

0.40

-7.3272

-7.5893

-8.0443

0.50

-9.6827

-10.0877

-10.4925

0.60

-9.1354

-9.4087

-9.5676

0.70

-4.7508

-4.7301

-4.5850

0.80

2.6487

2.9758

3.4995

0.90

10.7273

11.3579

12.2796

1.00

16.5173

17.3602

18.3511

1.10

17.7014

18.2966

18.7924

1.20

13.7385

13.6304

13.3779

1.30

5.9726

5.1327

4.3529

1.40

-3.0599

-4.2003

-5.2533

1.50

-10.6512

-11.6901

-12.8543

1.60

-14.8285

-15.5745

-16.4102

1.70

-14.7035

-14.9016

-14.8285

1.80

-10.4840

-9.8223

-8.7981

1.90

-3.4436

-2.0069

-0.5740

2.00

4.2919

5.8944

7.2622

Comments Table P16.5 shows that the linear acceleration method gives a more accurate value for the peak response.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 11 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 16.6 Equation (16.2.3) gives the modal equations with M , C , K and P(t ) defined in the solution of Problem 16.3. We now implement the procedure of Table 16.2.2 as follows.

2.4

⎛ ⎡ q1 ⎤ ⎡ q1 ⎤ ⎞ ⎡ q&&1 ⎤ ⎢q&& ⎥ = 2400 ⎜⎜ ⎢q ⎥ − ⎢q ⎥ ⎟⎟ − ⎣ 2 ⎦ i +1 ⎝ ⎣ 2 ⎦ i +1 ⎣ 2 ⎦ i ⎠

⎡ q& ⎤ ⎡ q&& ⎤ 120 ⎢ 1 ⎥ − 2 ⎢ 1 ⎥ ⎣q& 2 ⎦ i ⎣q&&2 ⎦ i

1.0 Initial calculations 1.1 Since the system starts from rest, u 0 = u& 0 = 0 ; therefore, q 0 = q& 0 = 0 . 1.2

p 0 = 0 ; therefore, P0 = 0 .

1.3

&&0 = 0 . q

1.4

Δt = 0.05 sec.

1.5 Substituting M , C , Δt ,

2.5

These displacements are also presented in Table P16.6a.

γ = 12 , and

β = 16

step 1.5 gives ⎤ ⎡2434 a1 = ⎢ 2498⎥⎦ ⎣

⎤ ⎡121.1 a2 = ⎢ 123.3⎥⎦ ⎣

3.0 Comments The solutions to Problems 16.5 and 16.6 are compared in Table P16.6b, which shows that the smaller time step leads to more accurate results.

⎡2.014 ⎤ a3 = ⎢ ⎥ 2 . 041 ⎣ ⎦

1.6 Substituting K and a1 in step 1.6 gives ⎤ ⎡2465 K̂ = ⎢ ⎥ 2764 ⎣ ⎦

2.0 Calculations for each time step i For the parameters of this problem, computational steps 2.1 through 2.5 are specialized and implemented for each time step i as follows. 2.1

⎡ Pˆ1 ⎤ ⎡− 1.067⎤ ⎢ˆ ⎥ =⎢ ⎥ (u&&g ) i +1 + ⎣ P2 ⎦ i +1 ⎣ 0.336 ⎦ ⎡ 2434 q1 + 121.1q&1 + 2.014 q&&1 ⎤ ⎢2498 q + 123.3 q& + 2.041q&& ⎥ ⎣ 2 2 2 ⎦i

⎡ Pˆ ⎤ ⎡2465 ⎤ ⎡ q1 ⎤ = ⎢ 1 ⎥ ⇒ q i +1 2.2 Solve ⎢ ⎢ ⎥ ⎥ 2764⎦ ⎣q2 ⎦ i +1 ⎣ Pˆ2 ⎦ i +1 ⎣

The resulting modal displacements qi are given in Table P16.6a. 2.3

⎛ ⎡ q1 ⎤ ⎡ q&1 ⎤ ⎡ q1 ⎤ ⎞ ⎢q& ⎥ = 60 ⎜⎜ ⎢q ⎥ − ⎢q ⎥ ⎟⎟ − ⎣ 2 ⎦ i +1 ⎝ ⎣ 2 ⎦ i +1 ⎣ 2 ⎦ i ⎠ ⎡ q&& ⎤ ⎡ q& ⎤ 2 ⎢ 1 ⎥ − 0.025 ⎢ 1 ⎥ ⎣q&&2 ⎦ i ⎣q& 2 ⎦ i

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 12 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Table 16.6a: Numerical Solution of modal equations by the linear acceleration method ti

0.00 0.05

0.0000 -0.0258

0.0000 -0.0151

0.0000 -0.0250

0.0000 -0.0278

0.0000 -0.0254

0.0000 -0.0225

0.10

-0.2014

-0.1143

-0.1907

-0.2140

-0.1995

-0.1795

0.15

-0.6468

-0.3463

-0.5852

-0.6725

-0.6485

-0.6015

0.20

-1.4300

-0.7101

-1.2208

-1.4469

-1.4543

-1.3968

0.25

-2.5521

-1.1630

-2.0422

-2.5076

-2.6343

-2.6183

0.30

-3.9426

-1.6440

-2.9550

-3.7648

-4.1266

-4.2287

0.35

-5.4643

-2.0958

-3.8562

-5.0872

-5.7875

-6.0807

0.40

-6.9277

-2.4752

-4.6507

-6.3194

-7.4052

-7.9281

0.45

-8.1139

-2.7485

-5.2498

-7.2937

-8.7293

-9.4667

0.50

-8.8025

-2.8795

-5.5614

-7.8396

-9.5083

-10.3933

0.55

-8.8024

-2.8215

-5.4854

-7.7980

-9.5297

-10.4628

0.60

-7.9797

-2.5222

-4.9261

-7.0437

-8.6523

-9.5278

0.65

-6.2820

-1.9414

-3.8202

-5.5136

-6.8280

-7.5540

0.70

-3.7533

-1.0739

-2.1697

-3.2326

-4.1116

-4.6169

0.75

-0.5386

0.0354

-0.0631

-0.3283

-0.6607

-0.8907

0.80

3.1229

1.2908

2.3257

2.9739

3.2728

3.3631

0.85

6.9161

2.5620

4.7620

6.3739

7.3588

7.8052

0.90

10.4786

3.7096

6.9894

9.5339

11.2137

12.0331

0.95

13.4365

4.6085

8.7683

12.1191

14.4342

15.6081

1.00

15.4434

5.1631

9.9028

13.8336

16.6400

18.1003

1.05

16.2459

5.3255

10.2788

14.4767

17.5442

19.1683

1.10

15.7899

5.1315

9.9319

14.0385

17.0684

18.6839

1.15

14.1444

4.6237

8.9322

12.5948

15.2795

16.7043

1.20

11.4669

3.8249

7.3415

10.2654

12.3587

13.4502

1.25

7.9895

2.7538

5.2315

7.2159

8.5777

9.2644

1.30

3.9985

1.4546

2.7183

3.6660

4.2644

4.5446

1.35

-0.1882

0.0184

-0.0142

-0.1104

-0.2331

-0.3184

1.40

-4.2465

-1.4193

-2.7224

-3.8036

-4.5758

-4.9777

1.45

-7.8712

-2.7052

-5.1437

-7.1034

-8.4536

-9.1367

1.50

-10.7979

-3.7077

-7.0520

-9.7423

-11.5982

-12.5380

1.55

-12.8231

-4.3476

-8.3019

-11.5297

-13.7941

-14.9563

1.60

-13.8169

-4.6068

-8.8435

-12.3677

-14.8922

-16.2091

1.65

-13.7320

-4.5140

-8.7046

-12.2455

-14.8247

-16.1870

1.70

-12.6040

-4.1178

-7.9563

-11.2215

-13.6164

-14.8880

1.75

-10.5473

-3.4646

-6.6826

-9.4038

-11.3875

-12.4360

1.80

-7.7437

-2.5921

-4.9697

-6.9388

-8.3425

-9.0721

1.85

-4.4269

-1.5395

-2.9166

-4.0081

-4.7478

-5.1169

1.90

-0.8635

-0.3650

-0.6537

-0.8281

-0.9020

-0.9202

1.95

2.6686

0.8445

1.6488

2.3564

2.8932

3.1851

2.00

5.9015

1.9787

3.7917

5.2904

6.3566

6.9099

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 13 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Table P16.6b ti

u5 (Δt = 0.1)

u5 (Δt = 0.05)

u5 (Theoretical)

0.00

0.0000

0.10

-0.1742

-0.1795

-0.1804

0.20

-1.3357

-1.3968

-1.4153

0.30

-4.0229

-4.2287

-4.2998

0.40

-7.5893

-7.9281

-8.0443

0.50

-10.0877

-10.3933

-10.4925

0.60

-9.4087

-9.5278

-9.5676

0.70

-4.7301

-4.6169

-4.5850

0.80

2.9758

3.3631

3.4995

0.90

11.3579

12.0331

12.2796

1.00

17.3602

18.1003

18.3511

1.10

18.2966

18.6839

18.7924

1.20

13.6304

13.4502

13.3779

1.30

5.1327

4.5446

4.3529

1.40

-4.2003

-4.9777

-5.2533

1.50

-11.6901

-12.5380

-12.8543

1.60

-15.5745

-16.2091

-16.4102

1.70

-14.9016

-14.8880

-14.8285

1.80

-9.8223

-9.0721

-8.7981

1.90

-2.0069

-0.9202

-0.5740

2.00

5.8944

6.9099

7.2622

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 14 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 16.7 The 5× 5 mass, damping and initial stiffness matrices were defined in Example 16.4. We now implement the procedure of Table 16.3.1 as follows. 1.0 Initial calculations 1.1 Since the system starts from rest, u 0 = u& 0 = 0 ; therefore, (f S )0 = 0 .

Table P16.7: Numerical solution by the central difference method ti

0.00 0.05

0.0000 0.0000

0.10

-0.1432

-0.1465

-0.1473

-0.1476

-0.1477

0.15

-0.4208

-0.5618

-0.5702

-0.5727

-0.5735

-0.8020

-1.2167

-1.3588

-1.3728

-1.3766

1.2

&& 0 = 0 . u

0.20

1.3

Δt = 0.05 sec.

0.25

-1.2518

-2.0406

-2.4520

-2.5954

-2.6128

0.30

-1.7246

-2.9563

-3.7338

-4.1396

-4.2779

u −1 = 0 .

0.35

-2.5784

-3.8765

-5.0819

-5.8396

-6.2157

0.40

-3.6875

-5.1181

-6.3669

-7.5168

-8.1226

0.45

-4.8851

-6.5821

-7.8163

-8.9214

-9.6809

0.50

-5.9959

-8.0892

-9.2473

-10.1205

-10.6374

0.55

-6.8531

-9.3900

-10.3592

-10.9546

-11.1065

0.60

-7.3119

-10.1437

-10.9290

-11.2071

-11.2937

0.65

-7.2581

-10.0618

-10.7194

-10.9997

-11.1181

0.70

-6.6398

-8.9990

-9.7769

-10.2531

-10.4411

0.75

-5.5877

-7.4676

-8.1300

-8.7881

-9.1320

0.80

-4.6349

-5.7953

-6.0604

-6.5835

-7.0331

0.85

-3.8460

-4.2993

-3.8473

-3.9174

-4.1022

0.90

-2.6658

-3.0051

-1.7986

-1.0502

-0.7024

0.95

-1.1925

-1.4014

0.0688

1.6356

2.5320

1.00

0.3644

0.3874

2.0666

4.0417

4.9676

1.05

1.7963

2.1930

4.0292

6.0310

6.4831

1.10

3.0682

3.9888

5.9220

7.1856

7.5612

1.15

4.1607

5.7289

7.0520

8.1679

8.2798

1.20

5.0590

6.8074

7.9022

8.8572

8.8938

1.25

5.7262

7.0683

8.5265

9.3454

9.4638

1.30

5.7746

7.2737

8.4733

9.8272

9.9055

1.35

5.4096

7.0208

8.5179

9.7603

10.2484

1.40

5.1000

6.5184

8.2479

9.6609

10.1044

1.45

4.9397

6.1701

7.6325

9.3028

9.5438

1.50

4.8866

5.8819

7.1719

8.2805

8.7658

1.55

4.7642

5.6994

6.4835

7.3875

7.5490

1.60

4.6329

5.2003

5.8631

6.4840

6.2303

2.0 Calculations for each time step i

1.65

4.2730

4.6411

5.1459

5.4529

5.1979

Computational steps 2.0 and 3.0 are implemented for i = 0, 1, 2,K to obtain the displacements u1 , u 2 , u3 , u 4 , and u5 presented in Table P16.7.

1.70

3.8564

4.0565

4.2031

4.5989

4.4291

1.75

3.6118

3.2808

3.4723

3.9072

3.8311

1.80

3.2637

2.8706

2.9356

3.4235

3.3147

1.85

3.0881

2.7373

2.7534

3.0641

2.9156

1.90

3.1203

2.7849

2.7864

2.9598

2.6743

1.95

3.3270

2.9803

2.9176

3.0997

2.7192

2.00

3.4953

3.2683

3.2178

3.3828

3.1263

1.4 1.5

1 kˆ = m+ c 2 2Δ t (Δ t )

⎡110.5 −1.981 −0.339 −0.137 −0.087 ⎤ ⎢ 110.2 − 2.118 − 0.427 − 0.224⎥ ⎥ ⎢ k̂ = ⎢ 110.1 − 2.205 − 0.564⎥ ⎥ ⎢ 110.0 − 2.545⎥ ⎢ ⎢⎣(sym) 108.0 ⎥⎦

1.6 Substituting m , c , and Δt in step 1.6 gives ⎡96.72 1.981 0.339 ⎢ 97.06 2.118 ⎢ a=⎢ 97.14 ⎢ ⎢ ⎣⎢(sym)

0.137 0.427 2.205 97.28

0.087⎤ 0.224⎥ ⎥ 0.564⎥ ⎥ 2.545⎥ 99.26⎥⎦

⎤ ⎡1 ⎥ ⎢ 1 ⎥ ⎢ b = (− 207.3) ⎢ 1 ⎥ ⎥ ⎢ 1 ⎥ ⎢ ⎢⎣ 1⎥⎦

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 15 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 16.8 Solution to this problem is available as Example 16.2 in the book.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 16 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 16.9 Steps 1.0 to 4.0 of the procedure of Table 16.3.2 are implemented to obtain the floor displacements presented in Table P16.9 and plotted in Fig. P16.9a. The base shear is plotted as a function of the roof displacement in Fig. P16.9b. Story shear-story drift relationships are presented in Figs. E16.9c-g for the five stories of the building. Table P16.9: Results of nonlinear static analysis for a uniform distribution of lateral forces λi

0 1.0

0.00 1.25

0.00 2.25

0.00 3.00

0.00 3.50

0.00 3.75

1.1

3.75

4.85

5.68

6.23

6.50

1.2

6.25

7.45

8.35

8.95

9.25

1.3

8.75

11.00

11.98

12.63

12.95

1.4

11.25

15.50

16.55

17.25

17.60

1.5

13.75

20.00

21.13

21.88

22.25

1.6

16.25

24.50

25.70

26.50

26.90

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 17 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

λ=0

1 1.11.2 1.3 1.4 1.5 1.6

0.4 Base shear/weight

Floor

4 3 2 1

(a)

200

1 2 3 Displacement/height, %

0.1

(b) 0

150 100 50 5

150 100

(d) 0

δ1 , in. 200

Story 3

150 100 50 0

150 100

(f) 0

δ3 , in.

V5 , kips

200

Story 4

(e) 0

δ2 , in.

V4 , kips

V3 , kips

200

Story 2

1 2 3 Displacement/height, %

200

Story 1 V2 , kips

V1 , kips

0.2

0 0

0.3

δ4 , in. Story 5

150 100 50 0

(g) 0

δ5 , in.

Fig. P16.9

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 18 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 16.10 Solution to this problem is available as Example 16.3 in the book.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 19 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 16.11 Solution to this problem is available as Example 16.4 in the book.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 20 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 16.12 The 5× 5 mass, damping and initial stiffness matrices were defined in Example 16.4. We now implement the procedure of Table 16.3.3 as follows. 1.0 Initial calculations 1.1 State determination for u 0 = 0 (f S ) 0 = 0

⎤ ⎡ 2 −1 ⎥ ⎢− 1 2 − 1 ⎥ ⎢ ⎥ (k T ) 0 = k ⎢ −1 2 −1 ⎥ ⎢ − − 1 2 1 ⎥ ⎢ ⎢⎣ − 1 1 ⎥⎦

1.2

&& 0 = 0 . u

1.3

Δt = 0.10 sec.

1.4 Matrices a1 and a 2 ⎡117.4 −3.962 −0.679 −0.274 −0.175 ⎤ ⎢ 116.8 − 4.236 − 0.854 − 0.449⎥ ⎥ ⎢ a1 = ⎢ 116.6 − 4.411 − 1.128 ⎥ ⎥ ⎢ 116.3 − 5.090⎥ ⎢ ⎢⎣(sym) 112.4 ⎥⎦ ⎡110.5 −1.981 −0.340 −0.137 −0.087 ⎤ ⎢ 110.2 − 2.118 − 0.427 − 0.224⎥ ⎥ ⎢ a2 = ⎢ 110.1 − 2.205 − 0.564⎥ × 10 −1 ⎥ ⎢ 110.0 − 2.545⎥ ⎢ ⎢⎣(sym) 108.0 ⎥⎦

2.0 Calculations for each time step i Computational steps 2.0 and 3.0 are implemented for i = 0, 1, 2,K to obtain the displacements u1 , u 2 , u3 , u 4 , and u5 presented in Table P16.12.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 21 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Table P16.12: Numerical solution by constant average acceleration method with modified Newton-Raphson iteration ti

0.00 0.05

0.0000 -0.0302

0.0000 -0.0355

0.0000 -0.0366

0.0000 -0.0368

0.0000 -0.0369

0.10

-0.1599

-0.2027

-0.2134

-0.2162

-0.2170

0.15

-0.4277

-0.5919

-0.6452

-0.6614

-0.6659

0.20

-0.8085

-1.2221

-1.3939

-1.4561

-1.4753

0.25

-1.2534

-2.0385

-2.4452

-2.6220

-2.6838

0.30

-1.8010

-2.9592

-3.7088

-4.1001

-4.2559

0.35

-2.5766

-3.9741

-5.0609

-5.7505

-6.0639

0.40

-3.5903

-5.1570

-6.4189

-7.3881

-7.8912

0.45

-4.7051

-6.5128

-7.7597

-8.8434

-9.4778

0.50

-5.7500

-7.8894

-9.0346

-10.0035

-10.6082

0.55

-6.5600

-9.0435

-10.0805

-10.7903

-11.1921

0.60

-6.9909

-9.7228

-10.6480

-11.1123

-11.2668

0.65

-6.9424

-9.7065

-10.5213

-10.8622

-10.9019

0.70

-6.4173

-8.8925

-9.6124

-9.9684

-10.0854

0.75

-5.5559

-7.4268

-7.9902

-8.4289

-8.7022

0.80

-4.5725

-5.6692

-5.8777

-6.3031

-6.6348

0.85

-3.5279

-3.9638

-3.5938

-3.7188

-3.9028

0.90

-2.2746

-2.3902

-1.4073

-0.9122

-0.7441

0.95

-0.7813

-0.7892

0.6306

1.7917

2.4076

1.00

0.7791

0.9285

2.6105

4.1492

5.0590

1.05

2.2373

2.6985

4.5483

6.0849

6.9087

1.10

3.5179

4.4570

6.3342

7.5641

8.0206

1.15

4.6007

6.1143

7.7698

8.5751

8.6969

1.20

5.4733

7.4444

8.7198

9.2110

9.2146

1.25

6.1201

8.1676

9.2235

9.6286

9.6854

1.30

6.4583

8.2244

9.3834

9.9268

10.0786

1.35

6.3411

7.8896

9.2664

10.0696

10.3110

1.40

5.8352

7.4688

8.9357

9.9435

10.2928

1.45

5.3511

7.0371

8.4709

9.4932

9.9357

1.50

5.2204

6.5992

7.9125

8.7781

9.1911

1.55

5.3184

6.2569

7.2706

7.8960

8.1164

1.60

5.2974

6.0435

6.6032

6.9179

6.8865

1.65

5.0327

5.7728

5.9850

5.9319

5.7154

1.70

4.6731

5.2528

5.3892

5.0787

4.7669

1.75

4.3591

4.5658

4.7298

4.4549

4.1222

1.80

4.0926

3.9866

4.0614

4.0236

3.7778

1.85

3.8581

3.6847

3.6145

3.7094

3.6421

1.90

3.7094

3.6345

3.5589

3.5522

3.5834

1.95

3.7308

3.7660

3.8210

3.6585

3.5546

2.00

3.9438

4.0617

4.2047

4.0169

3.6657

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 22 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

CHAPTER 17 Problem 17.1

where C1 is an arbitrary constant. The first three natural vibration modes are shown in the accompanying figure. The natural vibration frequencies of the clamped beam are higher than for the simply supported beam.

φ (x )

m , EI

φ (x)

m, EI

φ( x ) = C1 sin β x + C2 cos β x + C3 sinh β x + C4 cosh β x (a) u( 0 ) = 0 ⇒ φ ( 0 ) = 0 ⇒ C2 + C4 = 0 ⇒ C4 = − C2

(b)

u′ ( 0 ) = 0 ⇒ φ′ ( 0 ) = 0 ⇒ β ( C1 + C3 ) = 0 ⇒ C3 = − C1

φ (x) 1

ω 1 = 22.37

EI m

φ 2 (x)

ω 2 = 61.67

EI m

φ (x)

ω 3 = 120.9

EI m

(c)

u ( L ) = 0 ⇒ C1 (sin β L − sinh β L ) + C2 (cos β L − cosh β L ) = 0

(d)

u′ ( L ) = 0 ⇒ C1 (cos β L − cosh β L ) + C2 ( − sin β L − sinh β L ) = 0 (e)

Rewrite Eqs. (d) and (e) in matrix form:

LM bsin β L − sinh β Lg bcos β L − cosh β Lg OP RSC UV = RS0UV MNbcos β L − cosh β Lg b− sin β L − sinh β LgPQ TC W T0W 1

(f)

Setting the determinant of the coefficient matrix to zero gives 1 − cos β L cosh β L = 0

(g)

Equation (g) is solved numerically to obtain

βn L = 4. 730, 7. 853, and 10. 996 for n = 1, 2, and 3. Eq. (17.3.8) then gives the natural frequencies: ω1 =

22. 37

; ω2 =

61. 67

; ω3 =

120. 9

To determine the natural vibration modes corresponding to each value of βn L , we express C2 in terms of C1 from Eq. (e) and substitute in Eq. (a) together with Eqs. (b) and (c). The result is φ n ( x) =

LM MN

C1 sin β n x − sinh β n x +

FG cos β L − cosh β L IJ bcos β x − cosh β xgOP H sin β L + sinh β L K PQ n

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 1 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

φ (x)

Problem 17.2 φ (x )

m , EI

m, EI L

φ ( x ) = C1 sin β x + C2 cos β x + C3 sinh β x + C4 cosh β x (a) u ( 0 ) = 0 ⇒ φ ( 0 ) = 0 ⇒ C2 + C4 = 0 (b)

M (0) = 0 ⇒ EI φ ′′(0) = 0 ⇒ β2 ( − C2 + C4 ) = 0

(c)

φ 1(x)

ω 1 = 15.42

EI m

φ 2(x)

ω 2 = 49.97

EI m

φ 3(x)

ω 3 = 104.2 2

EI m

These two equations gives C2 = C4 = 0 and the general solution reduces to

φ( x ) = C1 sin β x + C3 sinh β x

(d)

u( L ) = 0 ⇒ φ( L ) = 0 ⇒ C1 sin β L + C3 sinh β L = 0

(e)

u′ ( L ) = 0 ⇒ φ′ ( L ) = 0 ⇒ β ( C1 cos β L + C3 cosh β L ) = 0

(f)

Rewrite Eqs. (e) and (f) in matrix form:

LM sin β L sinh β L OP RSC UV = RS0UV Ncos β L cosh β LQ TC W T0W 1

(g)

For a nontrivial solution, the determinant of the coefficient matrix must be zero. Thus sin β L cosh β L − cos β L sinh β L = 0 or tan β L = tanh β L

(h)

Equation (g) is solved numerically to obtain

βn L = 3. 927, 7. 069, and 10. 210 for n = 1, 2, and 3. Eq. (17.3.8) then gives the natural frequencies: ω1 =

15. 42 2

EI m

; ω2 =

49. 97 2

EI m

; ω3 =

104 . 2

Corresponding to each value of βn L , the natural vibration mode is obtained by expressing C3 in terms of C1 from Eq. (g) and substituting in Eq. (d):

FG H

φ n ( x ) = C1 sin β n x −

sin β n L sinh β n x sinh β n L

IJ K

where C1 is an arbitrary constant. The first three natural vibration modes are shown in the accompanying figure. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 2 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

where D1 is an arbitrary constant.

Problem 17.3

The non-zero roots of Eq. (f) can be obtained numerically:

φ (x )

m , EI

βn L = 4. 730, 7. 853, and 10. 996

for n = 3, 4, and 5. Thus

φ( x ) = C1 sin β x + C2 cos β x + C3 sinh β x + C4 cosh β x (a) M (0) = 0 ⇒ EI φ ′′ (0) = 0 ⇒ β2 ( − C2 + C4 ) = 0 ⇒ C4 = C2

(b)

V (0) = 0 ⇒ EI φ ′′′(0) = 0 ⇒ β3 ( − C1 + C3 ) = 0 ⇒ C3 = C1

V ( L) = 0 ⇒ C1 (cosh β L − cos β L) + C2 (sinh β L + sin β L ) = 0

(e)

For a nontrivial solution of C1 or C2 , the determinant of the coefficients must be zero. Thus (f)

Equation (f) is identical to the frequency equation for the clamped beam, but now it also possesses a zero root. Substituting β = 0 in Eq. (17.3.7) gives

φiv ( x ) = 0

EI m

; ω4 =

61. 67 2

EI m

; ω5 =

120. 9

To determine the natural vibration mode corresponding to each non-zero value of βn L , we express C2 in terms of C1 from Eq. (e) and substituted in Eq. (a) together with Eqs. (b) and (c) to obtain: φ n ( x) =

LM N

C1 sin β n x + sinh β n x +

cos β n L − cosh β n L cos β n x + cosh β n x sin β n L + sinh β n L

where C1 is an arbitrary constant.

M ( L) = 0 ⇒ C1 (sinh β L − sin β L) + C2 (cosh β L − cos β L ) = 0

1 − cos β L cosh β L = 0

22. 37

ω3 =

gOPQ

The natural vibration modes are shown in the accompanying figure. The first two modes, with zero natural frequencies, are rigid body translation and rigid body rotation, respectively. Modes 3 to 5 have the same vibration frequencies as the first three modes of the beam clamped at both ends, although the mode shapes are not identical. φ (x)

m, EI

(g)

φ (x) 1

ω1 = 0

(h)

φ (x)

ω2 = 0

φ (x) 3

ω 3 = 22.37

EI m

φ (x) 4

ω 4 = 61.67

EI m

φ (x)

ω 5 = 120.9

EI m

The general solution of Eq. (g) is

φ ( x ) = D1 + D2 x + D3 x 2 + D4 x 3

Applying the boundary conditions at x = 0 and x = L , we get D3 = D4 = 0 , and

φ ( x ) = D1 + D2 x

(i)

Equation (i) in fact represents two eigenfunctions corresponding to two zero eigenvalues. We can select the two eigenfunctions as, say, D1 and D2 x , but any linear combination of these is also an eigenfunction. Let D1 be the first eigenfunction and D1 + D2 x be the second. In order to satisfy the condition of modal othogonality, D2 = − 2 D1 L . Therefore, the two functions corresponding to β1 = 0 and β2 = 0 are

φ1 ( x ) = D1

(j)

FG H

φ 2 ( x ) = D1 1 −

x 2L

IJ K

(k)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 3 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 17.4

The integral in Eq. (g) vanishes for n even because ϕ n ( x ) is antisymmetric about x = L 2 but u ( x , 0 ) is symmetric. For n odd, Eq. (g) is

1. Natural vibration frequencies and modes.

ωn =

n2 π 2

φn ( x ) = sin

nπ x

(a)

2m W qn (0) = M n 48 EI

2. Set up modal equations. q&&n ( t ) + ω 2n qn ( t ) = 0

W L u( x ,0) = 4 x 3 − 3 L2 x ; 0 ≤ x ≤ 48 EI 2

nπ x L

4 x 3 − 3 L2 x dx

where qn ( 0 ) and q&n ( 0 ) are determined from u ( x , 0 ) and u& ( x , 0 ) , respectively. The initial displacements due to the weight W are

sin

R| − 2WL 1 n = 1, 5, 9, ⋅⋅⋅ | = S π EI n || 2WL 1 n = 3, 7, 11, ⋅⋅⋅ T π EI n

(b)

3. Initial conditions

z FGH IJK e L

(h)

Because the initial velocity is zero, q&n ( 0 ) = 0

(i)

(c)

The initial velocities are 4. Modal responses.

u& ( x , 0 ) = 0

Determine qn ( 0 ) from u ( x , 0 ) :

q&n ( 0 )

qn ( t ) = qn ( 0 ) cos ω nt +

ωn

u ( x , 0 ) = ∑ φr ( x ) qr ( 0 )

(d)

r =1

Multiplying both sides of Eq. (d) by m ( x )ϕ n ( x ) and integrating over the length of the beam:

z L

m( x ) ϕ n ( x ) u( x ,0) dx =

z L

qn ( t ) = qn ( 0 ) cos ω nt

m( x ) ϕ n ( x ) u( x ,0) dx = qn (0)

∞ u ( x , t ) = ∑ ϕ n ( x ) qn ( t )

z L

(j)

5. Total response.

Because of the modal othogonality relation of Eq. (17.4.6a), all terms in the summation on the right side vanish except the one for which r = n . Thus

where qn ( 0 ) and q&n ( 0 ) are given by Eqs. (h) and (i), respectively. Therefore,

∑ q (0) m( x) ϕ ( x) ϕ ( x) dx r =1

sin ω nt

n =1

Substituting Eqs. (a) and (j) gives u( x , t ) =

m( x ) [ϕ n ( x )]2 dx

−

FG − sin π x cos ω t + 1 sin 3π x cos ω t L L 81 π EI H 2WL3

IJ K

5π x 7π x 1 1 sin cos ω 5t + sin cos ω 7 t − ⋅⋅⋅ L L 625 2401

(k)

6. Specialize for mid-span deflection.

qn (0) =

m( x ) ϕ n ( x ) u( x ,0) dx

(e)

z L

IJ (l) K

m sin

qn (0) =

FG nπ x IJ dx = mL H LK 2

1 1 + cosω 5t + cos ω 7 t + ⋅⋅⋅ 625 2401

Substituting ϕ n ( x ) in Eqs. (17.5.3a) and (e) gives Mn =

FG L , tIJ = − 2WL FG cos ω t + 1 cos ω t H 2 K π EI H 81 3

1 Mn

z L

m sin

FG nπ x IJ u( x,0) dx H LK

(f)

(g)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 4 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 17.5 1. Natural vibration frequencies and modes.

ωn =

n2 π 2

φn ( x ) = sin

nπx L

(a)

2. Set up modal equations. Mn =

Kn =

Pn (t ) =

p(t ) φ n ( x ) dx =

n 4 π 4 EI 2 L3

R| 0 n = 2, 4, 6, ⋅⋅⋅ S| 2 p L n = 1, 3, 5, ⋅⋅⋅ T nπ

(b) The nth modal equation (where n is an odd number) is Mn q&&n + Kn qn =

2 pL

(c)

nπ

3. Determine dynamic response. The solution to Eq. (c) is: qn ( t ) =

2 pL

(1 − cos ω nt ) =

nπKn

4 pL4 1

π 5 EI n5

(1 − cos ω nt )

(d) ∞

∑ φn ( x ) qn (t )

u( x , t ) =

(e)

n =1 n odd

where φn ( x ) and qn ( t ) are given in Eqs. (a) and (d). Note that the modes with antisymmetric mode shape ( n = 2 , 4, 6, ⋅⋅⋅ ) do not contribute to the response. 4. Specialize for x = L 2 .

φn

FG L IJ = RS 1 n = 1, 5, 9, ⋅⋅⋅ H 2 K T−1 n = 3, 7, 11, ⋅⋅⋅

Substituting these in Eq. (e) and using Eqs. (a) and (d) gives

FG L , tIJ = 4 p L FG 1 − cos ω t − 1 − cos ω t + H 2 K π EI H 1 243 4

IJ K

1 − cos ω 5t 1 − cos ω 7 t − + ⋅⋅⋅ 3125 16807

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 5 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 17.6 1. Natural vibration frequencies and modes.

ωn =

FG L , tIJ = 8 p L FG 1 − cos ω t − 1 − cos ω t + H 4 K π EI H 32 7776 1 − cos ω t I − ⋅⋅⋅J K 100,000 4

n2 π 2

φn ( x ) = sin

n πx L

(a)

2. Set up modal equations. Mn =

Kn =

z z

n 4 π 4 EI 2 L3

Pn (t ) =

p( x , t ) φ n ( x ) dx

0 L2

p sin

nπ x dx + L

z L

( − p) sin

R| 0 n = 1, 3, 4, 5, 7, 8, ⋅⋅⋅ S| 4 pL n = 2, 6, 10, ⋅⋅⋅ T nπ

nπ x dx L

(b)

For n = 2, 6,10, ⋅⋅⋅ , the modal equation is Mn q&&n + Kn qn =

4 pL

(c)

nπ

3. Determine dynamic response. The solution to Eq. (c) is qn ( t ) =

4 pL nπ Kn

(1 − cos ω nt ) =

8 pL4 1

π 5 EI n5

(1 − cos ω nt )

(d) u( x , t ) =

∞

∑ φn ( x ) qn (t )

(e)

n = 2, 6, 10, ⋅⋅⋅

where φn ( x ) and qn ( t ) are given in Eqs. (a) and (d), respectively. Note that only modes n = 2, 6,10, ⋅⋅⋅ contribute to the response. Modes n = 1, 3, 5, ⋅⋅⋅ , which are symmetric about x = L 2 , do not respond because the applied force is antisymmetric about x = L 2 . Antisymmetric modes n = 4, 8,12, ⋅⋅⋅ do not respond because each half of the beam with constant force contains one or more complete sine waves, and thus provides zero contribution to Pn ( t ) . 4. Specialize for x = L 4 .

φn

FG L IJ = RS 1 n = 2, 10, 18, ⋅⋅⋅ H 4 K T−1 n = 6, 14, 22, ⋅⋅⋅

Substituting these in Eq. (e) and using Eqs. (a) and (d) gives © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 6 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 17.7 1. Determine the natural vibration frequencies and modes.

ωn =

n 2π 2 L2

2 po L

nπx φ n ( x) = sin L

EI m

(a)

2. Setup the modal equations. Substituting φ n (x) in Eq. (17.5.3a) gives

Mn ,

which is substituted in Eq. (17.5.5) together with ω n2 to get K n : mL Mn = 2

Kn =

n 4π 4 EI

(b)

2 L3

From Problem 8.25, the applied force is ⎧ po ⎪ p ( x, t ) = ⎨ 0 ⎪p ⎩ o

0 ≤ x ≤ vt

2 po L

0 ≤ t ≤ td

vt < x ≤ L 0 ≤ t ≤ t d 0 ≤ x ≤ L t ≥ td

(c)

Substituting for p( x, t ) in Eq. (17.5.3c) gives L

Pn (t ) =

∫

The nth modal equation of motion is

p ( x, t ) φ n ( x) dx

M n q&&n (t ) + K n q n (t ) = Pn (t )

⎧ vt ⎪ p o sin (nπx L) dx 0 ≤ t ≤ t d = ⎨ 0L ⎪ p o sin (nπx L) dx t ≥ t d ⎩0 ⎧ po L ⎡ nπv ⎤ 1 − cos t 0 ≤ t ≤ td ⎪⎪ nπ ⎢⎣ L ⎥⎦ =⎨ ⎪ p o L 1 − (−1) n t ≥ td ⎪⎩ nπ

∫ ∫

[

(e)

with M n , K n , and Pn (t ) as given above. (d)

3. Solve the modal equations. Response for 0 ≤ t ≤ t d .The particular solution to Eq. (e) can be obtained by superposing the steady-state responses to the constant and the cosine term on the right-hand side of Eq. (d). The steady-state response to the constant term is ( Pn ) o K n , where ( Pn ) o = p o L nπ , and that for the cosine term is adapted from Eqs. (3.2.3) and (3.2.26), noting that ς = 0 and replacing p o and k by ( Pn ) o and K n respectively. The complete solution is obtained by adding the complementary solution, given by Eq. (3.1.4), to the particular solution, and determining the constants A and B by imposing zero initial conditions.

]

Equation (d) is plotted next for n = 1, 2, and 3: P1 2 po L

The result is Lv

qn (t ) =

2 po 1 ⎡ ω n2 nπv − t 1 cos ⎢ nπm ω n2 ⎣⎢ ω n2 − (nπv L )2 L +

(nπv L )2 cosω t ⎤ n ⎥ ω n2 − (nπv L )2 ⎦⎥ t≤L v

(f)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 7 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Equation (g) is valid if ω n ≠ nπv L ; otherwise the particular solution to the cosine term should formulated similar to that in Eq. (3.1.12), noting that the forcing function is a cosine function instead of a sine function.

4. Determine the displacement response. From Eq. (17.5.7) the displacement response is ∞

u ( x, t ) =

∑

φ n ( x)qn (t ) =

n =1

Response for t ≥ t d . The motion is described by Eq. (4.5.3) with q n instead of u , t d instead of t r , and q n (t d ) and q& n (t d ) determined from Eq. (f) : qn (t d ) =

2 po 1 ⎡ ω n2 − 1 (−1) n ⎢ nπm ω n2 ⎢⎣ ω n2 − (nπv L )2 +

q& n (t d ) = −

(nπv L )2 cosω t ⎤ n d⎥ ω n2 − (nπv L )2 ⎥⎦

(g)

nπx

∑ q (t ) sin L n

(k)

n =1

where q n (t ) is given by Eqs. (f), (i), and (j). 5. Specialize for x = L 2 . At midspan, x = L 2 and ⎧ 0 nπ (L 2 ) nπ ⎪ sin = sin =⎨ 1 L 2 ⎪ −1 ⎩

n = 2,4,6, K

n = 1,5,9, K n = 3,7,11, K

(l)

Substituting Eq. (l) into Eq. (k) gives the deflection at midspan:

2 po 1 (nπv L )2 sinω t n d nπm ω n ω n2 − (nπv L )2

(h)

Noting that Pn (t ) = 2 p o L nπ for n = 1,3,5, K and Pn (t ) = 0 for n = 2,4,6, K , substituting these in Eq. (4.5.3), using trigonometric identities and manipulating the mathematical quantities, we obtain qn (t ) =

∞

⎛L ⎞ u⎜ , t ⎟ = q1 (t ) − q 3 (t ) + q 5 (t ) − q 7 (t ) + q 9 (t ) − L (m) ⎝2 ⎠

where q n (t ) is given by Eq. (f) for t ≤ L v and by Eq. (i) for t ≥ L v .

⎫ 2 po 1 ⎡ ⎧⎪ ω n2 n⎪ ⎢ 2 1 ( 1 ) − + − ⎨ ⎬ cosω n (t − t d ) nπm ω n2 ⎢ ⎪⎩ ω n2 − (nπv L )2 ⎪⎭ ⎣ (nπv L )2 cosω t ⎤ + 2 n ⎥ ω n − (nπv L )2 ⎥⎦ n = 1,3,5, K

t≥L v

(i)

and q n (t ) =

⎫⎪ 2 po 1 ⎡⎪⎧ ω n2 ⎢ 1 (−1) n ⎬ cosω n (t − t d ) − ⎨ 2 2 2 nπm ω n ⎢⎪⎩ ω n − (nπv L ) ⎪⎭ ⎣ +

(nπv L )2 cosω t ⎤ n ⎥ ( ) ⎦⎥

ω n2 − nπv L 2

n = 2,4,6, K

t≥L v

(j)

Thus, the modal response q n (t ) is given by Eq. (f) while the moving load is on the bridge span and by Eqs. (i) and (j) after the load has crossed the span.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 8 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 17.8 1. Determine the natural vibration frequencies and modes.

ωn =

n 2π 2

EI m

φ n ( x) = sin

nπx L

individual responses are adapted from Eq. (3.1.6b) by changing the notation from u (t ) to q n (t ) , substituting for ω with ω + nπt t d in one case and with ω − nπt t d in the other, and noting that

(a)

td =

2. Setup the modal equations. Substituting φ n (x) in Eq. (17.5.3a) gives

Mn ,

The result is

which is substituted in Eq. (17.5.5) together with ω n2 to get K n : mL Mn = 2

Kn =

n 4π 4 EI 2 L3

q n (t ) =

(b) −

∫ p( x, t ) φ ( x) dx n

⎧ ⎪ p cosωt δ ( x − vt ) sin (nπx L) dx 0 ≤ t ≤ t d =⎨ 0 o ⎪⎩ 0 t ≥ td

∫

⎧ p cosωt sin (nπx L) 0 ≤ t ≤ t d =⎨ o t ≥ td ⎩0 ⎧ p cosωt sin (nπt t d ) 0 ≤ t ≤ t d =⎨ o t ≥ td ⎩0

⎧ ⎛ nπv ⎞ sin ⎜ ω − ⎟t L ⎠ ) ⎝

⎨ ω n2 − ω − nπv L 2 ⎩

(

⎫⎤ ⎛ ω − nπv L ⎞ ⎟ sinω n t ⎪⎬⎥ − ⎜⎜ ⎟ ωn ⎪⎭⎥⎦ ⎝ ⎠

(c)

where δ ( x − vt ) is the Dirac delta function centered at x = vt . Substituting for p( x, t ) in Eq. (17.5.3c) gives Pn (t ) =

⎧ ⎛ po ⎡ nπ v ⎞ 1 sin ⎜ ω + ⎟t ⎢ 2 2 ⎨ mL ⎢⎣ ω n − (ω + nπv L ) ⎩ ⎝ L ⎠ ⎫ ⎛ ω + nπ v L ⎞ ⎟ sinω n t ⎪⎬ − ⎜⎜ ⎟ ωn ⎪⎭ ⎝ ⎠

From Problem 8.26, the applied force is ⎧ p cosωt δ ( x − vt ) 0 ≤ t ≤ t d p ( x, t ) = ⎨ o t ≥ td ⎩0

p po (u st ) o = ~o = 2k mLω n2

L v

t≤L v

(f)

Based on Eq. (3.1.6b), Eq. (g) is valid if ω n ≠ ω + nπv L and ω n ≠ ω − nπv L ; otherwise Eq. (3.1.13a) should be used instead of Eq. (3.1.6b). Free vibration phase. The motion is described by Eq. (4.7.3) with q n (t ) instead of u (t ) , and q n (t d ) and q& n (t d ) determined from Eq. (f) : qn (t d ) =

⎧⎪ ⎫⎪ po ⎡ 1 1 − 2 ⎢(−1) n ⎨ 2 2 2⎬ mL ⎢⎣ ⎪⎩ω n − (ω + nπv L) ω n − (ω − nπv L) ⎪⎭ × sin ωL v

−

This can be re-written as

⎤ 1 ⎧⎪ (ω + nπv L) (ω − nπv L) ⎫⎪ − 2 sinω n L v ⎥ ⎨ 2 2 2⎬ ω n ⎪⎩ω n − (ω + nπv L) ω n − (ω − nπv L) ⎪⎭ ⎥⎦

⎧ po ⎪ [sin (ω + nπt t d ) − sin(ω − nπt t d )] 0 ≤ t ≤ t d Pn (t ) = ⎨ 2 ⎪⎩ 0 t ≥ td

(d) The nth modal equation of motion is M n q&&n (t ) + K n qn (t ) = Pn (t )

(e)

with M n , K n , and Pn (t ) as given above. 3. Solve the modal equations. Forced vibration phase. The response q n (t ) can be obtained by superposing the responses due to the two sine terms in the right-hand side of Eq. (d). The

(g)

⎡⎧⎪ (ω + nπv L) (ω − nπv L) ⎫⎪ − 2 q& n (t d ) = ⎢⎨ 2 ⎬ 2 ω n − (ω − nπv L) 2 ⎪⎭ ⎢⎣⎪⎩ω n − (ω + nπv L)

ω L⎫ ⎤ ωL ⎧ × ⎨(−1) n cos − cos n ⎬ ⎥ v v ⎭⎦ ⎩ (h) Substituting these in Eq. (4.7.3), using trigonometric identities, and manipulating the mathematical quantities, we obtain

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 9 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

q n (t ) =

⎧⎪ ⎫⎪ po ⎡ 1 1 − 2 ⎢(−1) n ⎨ 2 2 2⎬ mL ⎢⎣ ⎪⎩ ω n − (ω + nπv L) ω n − (ω − nπv L) ⎪⎭ × sin

ωL v

⎛ L⎞ cosω n ⎜ t − ⎟ ⎝ v⎠

1 ⎧⎪ (ω + nπv L) (ω − nπv L) ⎫⎪ − 2 ⎨ 2 ⎬ 2 ω n ⎪⎩ω n − (ω + nπv L) ω n − (ω − nπv L) 2 ⎪⎭ ⎧ ⎫⎤ ωL ⎛ L⎞ sinω n ⎜ t − ⎟ − sinω n t ⎬⎥ × ⎨(−1) n cos v ⎝ v⎠ ⎩ ⎭⎥⎦ t≥L v

(i)

Thus, the modal response q n (t ) is given by Eq. (f) while the moving load is on the bridge span and by Eq. (i) after the load has crossed the span. 4. Determine the displacement response. From Eq. (17.5.7), the displacement response is ∞

u ( x, t ) =

∞

nπx

∑φ ( x)q (t ) = ∑ q (t ) sin L n

n =1

(j)

n =1

where q n (t ) is given by Eqs. (f) and (i). 5. Specialize for x = L 2 At midspan, x = L 2 and ⎧ 0 nπ (L 2 ) nπ ⎪ = sin =⎨ 1 sin L 2 ⎪ ⎩ −1

n = 2,4,6, K n = 1,5,9, K n = 3,7,11, K

(k)

Substituting Eq. (k) into Eq. (j) gives the deflection at midspan: ⎛L ⎞ u⎜ , t ⎟ = q1 (t ) − q 3 (t ) + q 5 (t ) − q 7 (t ) + q 9 (t ) − L (l) ⎝2 ⎠

where q n (t ) is given by Eq. (f) for t ≤ L v and by Eq. (i) for t ≥ L v .

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 10 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 17.9 Equation (17.6.3) ⇒ ∞ m ( x ) = ∑ Γn m ( x ) φn ( x )

(a)

n =1

Integrating over the length of the cantilever beam gives

∞

m( x ) dx =

∑

n =1

∞

Γn

m( x ) φ n ( x ) dx =

∑Γ L n

h n

n =1

(b) By definition, Γn Lhn = Mn* and Eq. (b) becomes ∞

∑

n =1

z L

M n* =

m( x ) dx

(c)

which is the same as Eq. (17.6.19).

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 11 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 17.10

Table P17.11a

Equation (17.6.3) ⇒ ∞ m ( x ) = ∑ Γn m ( x ) φn ( x )

(a)

n =1

Multiplying both side by x and integrating over the length of the cantilever gives

∞

x m( x ) dx =

∑

n =1

∞

Γn

x m( x ) φ n ( x ) dx =

∑

Γn Lθn

(b) By definition, (b) becomes ∞

∑

Γn Lhn

hn*

and

Ln Lhn

and Eq.

hn* M n* =

n =1

unst ( L )

st V bn

M stbn

0.0258

53.274

7,740

− 3. 66 × 10 − 4

16.365

684.77

2. 73 × 10 − 5

5.625

143.27

− 4. 87 × 10 − 6

2.873

51.83

n =1

Mn*

Mode

(c)

x m( x ) dx

4. Determine spectrum ordinates. From Fig. 6.9.5, scaled to u&&go = 1 3g , the design spectrum ordinates are A1 g A2

which is the same as Eq. (17.6.20).

Problem 17.11

1. Tower properties.

L = 200 ft

A4 2

m = π (12. 5) − (11. 25)

0.15

= 0. 4345 kip - sec ft

= 3. 4184 × 10 kip - ft

= = =

1. 80 T1− 1 3 2. 71 3

= 0. 744

= 0. 903

11. 70 T30.704 3 1 3

= 0. 446

= 0. 333

5. Determine peak responses.

32. 2

EI = ( 3600 × 144 ) ( π 4 ) (12. 5)

− (11. 25)

For each response quantity r, substituting rnst and An in rno = rnst An

gives the peak modal responses for the first four modes. 2. Natural vibration periods and modes.

Table P17.11b un ( L ) (in.)

Vbn (kips)

(kip-ft)

7.409

1,276.2

185,425

T2 = 0.129

– 0.128

475.8

19,911

T3 = 0. 046 , T4 = 0. 023 , etc.

0.0047

80.8

2,057.5

3. Determine the modal static responses.

0.0006

30.8

555.8

Equation (17.3.22) gives the natural vibration frequencies and the corresponding periods in seconds are T1 = 0. 806

The modal properties given in Table E17.2 are valid for a uniform cantilever. Substituting these in the equations of Table 17.6.1 gives Table P17.9a for the modal static responses.

Mode

M bn

These peak modal responses are combined according to the SRSS rule to obtain: Top displacement

uo ( L ) = 7. 410 in.

Base shear

V bo = 1,365 kips

Base overturning moment

M bo = 186,503 kip - ft

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 12 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

CHAPTER 18 Problem 18.1

~ = m 12 I(x) = I ( 1 –

Thus

x ) 2L

LM OP Q N . 0.363296O ~ = mL L 013376 m MN0.363296 12978 PQ .

EI 2.59145 2.7758 ~ k = 3 2.7758 187.7 L

The shape functions selected are

ψ 2 ( x ) = 1 − cos

2. Solve eigenvalue problem.

πx

%2 m % z k% z = ω

2L 3π x

ω% 1 = 4. 3178

% 1. Set up k% and m.

For the selected ψ j ( x ) , the stiffness coefficients are computed from Eq. (18.1.4a): ~ k11 =

z FGH L

EI 1 −

~ k 22 =

z z L

IJ LMFG π IJ cos π x OP dx = 2.59145 EI K MNH 2 L K 2 L PQ L 2

F x IJ LMFG 3π IJ cos 3π x OP dx = 187.7 EI EI G 1 − H 2 L K MNH 2 L K 2 L PQ L

F x IJ LMFG π IJ cos π x OP LMFG 3π IJ cos 3π x OP dx EI G 1 − H 2 L K MNH 2 L K 2 L PQ MNH 2 L K 2 L PQ 2

= 2.7758

z1 =

ω% 2 = 24. 8415

m L4

LM0.9997OP N0.0244Q

z2 =

m L4

LM− 0.9406OP N 0.3396Q

3. Determine natural modes from Eq. (18.1.8). ~

FG H

φ 1 ( x ) = 0.9997 1 − cos

πx 2L

IJ + 0.0244 FG1 − cos 3π x IJ K H 2L K

. = 10241 − 0.9997 cos

~ k12 =

x 2L

IJ FG1 − cos π x IJ FG1 − cos 3π x IJ dx KH 2LK H 2L K

= 0.363296 mL

ψ1 ( x ) = 1 − cos

x 2L

m 1 −

x ) 2L

m(x) = m ( 1 –

z FGH L

FG H

φ 2 ( x ) = − 0.9406 1 − cos

πx 2L

− 0.0244 cos

3π x 2L

IJ + 0.3396 FG1 − cos 3π x IJ K H 2L K

= − 0.6010 + 0.9406 cos

πx 2L

− 0.3396 cos

3π x 2L

EI L3

Similarly the mass coefficients are determined from Eq. (18.1.4b): ~ = m 11

m 1 −

FG H

x 2L

m 1 −

FG H

x 2L

~ = m 22

IJ FG1 − cos π x IJ dx = 013376 . mL KH 2LK 2

IJ FG1 − cos 3π x IJ dx = 12978 mL . KH 2L K 2

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 1 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 18.2

Thus

x L

m(x)=2mo

m(x)=2mo (1 –

LM N

0 EI 48.705 ~ k = 3 0 3945.07 L

x ) L

%. 3. Set up m

mass distribution EI

x L/2

~ = m 11

ψ1′ ( x ) =

cos

ψ 2 ( x ) = sin

πx

ψ ′2 ( x ) =

3π L

3π x

cos

3π x

FG π IJ sin π x ψ ′′( x) = − FG 3π IJ sin 3π x H LK L H LK L

FG π IJ EI sin π x dx H LK L F π I L π x L − sin bπ x Lg cos bπ x Lg OP = EI G J M H L K MN 2 2 PQ F π I π = π EI = 48.705 EI = EI G J H LK 2 2 L L 4 L

= 0

sin 2

z L

sin 2

FG π x IJ d FG π x IJ + H LK H LK

FG π x IJ d FG π x IJ − H LK H LK πx L

LM MN

sin 2

FG π x IJ d FG π x IJ H LK H LK

2mo L (π x L) 2 (π x L) sin (2 π x L) − − 2 4 4 π

OP Q

cos (2 π x L) 8 0

π π

πx

1 L

2mo

π π

2mo π L

2. Set up k% . ~ k11 =

2mo L

ψ 1′′( x ) = −

FG IJ H K F x I F π xI 2m G 1 − J sin G J dx H LK H L K πx x dx + sin 2 L L

2mo

0 L

1. Select shape functions.

ψ1 ( x ) = sin

z z

L/2

πx

OP Q

LM OP N Q 2m L L (π x L) (π x L) sin (2 π x L) − − M 4 4 π MN cos (2 π x L) O PQ 8 L

FG π IJ FG 3π IJ EI sin π x sin 3π x dx H LK H L K L L F π I F 3π I L sin b− 2π x Lg − sin b4 π x Lg OP = EI G J G J M H L K H L K MN 2 b− 2π Lg 2 b4π Lg PQ 2 L

~ k12 =

sin (2 π x L) 2mo π x L − − π L 2 4 L2

k%21

= 0 = k%

= 0

FG 3π IJ EI sin 3π x dx H LK L F 3π I L 3π x L − sin b6 π x Lg OP = EI G J M H L K MN 2 4 PQ F 3π I 3π = (3π ) EI = EI G J H LK 2 2 L

~ k 22 =

o 2

4 L

OP + 2m L L π − π O − LM MN PQ π MN 2 4 PQ (π 2) 2m L L π 1 1O − − − P M 8 4 8 PQ π MN 4

2mo L (π 2) 2 1 1 + + 2 4 8 8 π o 2

= 3945. 07

EI L3

LM MN LM MN

(π 2) 2 1 + 4 4

2mo L π 2 1 = + 2 8 2 π

OP PQ

~ = m 22

OP PQ

z z

2mo

2mo L = 3π 3π

0 L

R| S| T

2mo − L

z L

sin (2π x L) dx + 2 (2π L)

z L

sin (4π x L) dx 2 (4π L)

U| V| W

FG IJ FG IJ + H K H K

3π x 3π x 3π x d sin 2 L L L

sin 2

FG IJ FG IJ H K H K

3π x 3π x 3π x d sin 2 L L L

LM MN

OP Q

= − 0.1013 mo L

% 21 = m % 12 = − 0.1013 mo L m

cos (6 π x L) 8 0

LM 3π x L − sin (6π x L) OP − 4 Q N 2 (3π x L) sin(6 π x L) 2m L L (3π x L) − − M 4 4 (3π ) MN cos(6 π x L) O PQ 8 2m L L (3π 2) 1 1O = + + P + M 8 8 PQ (3π ) MN 4 2m L L 3π 3π O − − M 3π N 2 4 PQ (3π 2) 2m L L (3π ) 1 1O − − − P M 8 4 8 PQ (3π ) MN 4 2m L L (3π 2) (3π ) (3π ) 1 = + + − + M 4 4 4 (3π ) MN 4 (3π 2) O 1 + P 4 4 PQ 2m L L 9π 1O = + P M 2 PQ (3π ) MN 8 L

2mo 3π L

FG 3π x IJ d FG 3π x IJ − H LK H LK

(3π x L) sin (6 π x L) 2mo L (3π x L) 2 − − 2 4 4 (3π )

2mo L 3π 3π

2mo 3π L

πx 3π x x sin sin dx + L L L

3π x x sin 2 dx + L L

FG IJ FG IJ H K H K F x I F π x I F 3π x IJ dx 2m G 1 − J sin G J sin G H LK H L K H L K 2m R| L sin (2π x L) sin (4π x L) O = − SxM P − L | N 2 (2π L) 2 (4π L) Q T U| sin (2π x L) sin (4π x L) dx + dx V + 2 (2π L) 2 (4π L) |W L sin (2π x L) − sin (4π x L) OP − 2m M 2 (4π L) Q N 2 (2π L)

~ = m 12

FG IJ H K F x I F 3π x IJ dx 2m G 1 − J sin G H LK H L K

2mo

0 L

= 0. 3513mo L

z z

2mo L (π 2) 2 π2 π2 1 + + − + 4 4 4 4 π2

= 0. 2613 mo L

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 3 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Thus ~ = m L m o

. OP LM 0.3513 − 01013 . 0.2613Q N− 01013

4. Solve reduced eigenvalue problem. %2 m % z k% z = ω

ω% 1 = z1 =

11. 765 2

EI mo

RS 10. UV T−0.0036W

ω% 2 = z2 =

130. 467

RS0.2790UV T0.9603W

5. Determine natural vibration modes from Eq. (18.1.8).

FG π x IJ − 0.0036 sin FG 3π x IJ H LK H LK ~ F π xI F 3π x IJ φ ( x ) = 0.2790 sin G J + 0.9603 sin G H LK H LK ~

φ 1 ( x ) = 10 . sin

Note that the first mode of this beam with nonuniform mass is only slightly different from that for a uniform beam; the second mode differs more between the two beams.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 4 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

0 0O − 0.0386 LM 0.2196 − 0.6898 L − 0.5774 L − 0.7066 L − 0.5774 L PP Φ = M MM 0 L 0.5774 L 0 L − 0.5774 L P N 0.6898 L − 0.5774 L 0.7066 L − 0.5774 LPQ

Problem 18.3 1. Stiffness and mass matrices. 5

L/2

3. Solution using lumped mass matrix

OP LM0.5 0 P m = mL M MM 0 P P 0Q N

L/2

Fig. P18.3a 1 4

Eliminate the three rotational DOFs by static condensation to obtain

48 EI −1 k$lat = ktt − kt 0 k00 k0 t = L3

Fig. P18.3b The 6 × 6 stiffness and mass matrices with reference to the DOFs in Fig. P18.3a are given in Example 18.2. Impose boundary conditions u1 = u5 = 0 by eliminating the corresponding rows and columns to obtain k and m with reference to the DOFs in Fig. P18.3b:

LM 24 3L 0 − 3LOP L L 2 0 P Lk 8 EI M 3 L k = = M L M 0 MN− 3L L 0 2 L2 L 2 LL 2PPQ Nk 2

tt 0t

6.5 L O LM 312 − 6.5L 0 PP − 0.75 L L 0 m L M− 6.5 L m = 840 M 0 MN 6.5L − 0.750 L − 02.75L L − 0.L75L PPQ 2

2. Solve eigenvalue problem. 2

( k − ω m )φ = 0

ω 3 = 110.14

EI m L4

EI 4

ω1 =

k$lat 0. 5 m L

Natural mode equations:

OP Q

ω 2 = 43. 818

ω 4 = 200. 80

EI m L4

= 9. 798

after

m L4

using

static

condensation

φ1T = 0. 2196 − 0. 6588 L 0 L 0. 6588 L 4. Compare with exact frequencies. For a simply-supported beam, the exact values of the first four natural frequencies are

ω1 = 9. 8696

ω1 = 9. 9086

kt0 k 00

Natural frequency:

ω 3 = 88. 826

EI 4

mL EI

ω 2 = 39. 478 ω 4 = 157. 914

EI m L4 EI m L4

The finite element method using consistent mass provides an excellent result for the fundamental frequency, but the accuracy deteriorates for higher modes. The lumped mass approximation provides only the fundamental frequency and the result is less accurate. Note that the finite element method overestimates the fundamental frequency whereas the lumped mass approximation provides an under estimate.

EI m L4

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 5 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 18.4

4. Compare with exact frequencies.

1. Stiffness and mass matrices.

For a beam clamped at both ends, the exact values of the first two natural frequencies are

ω1 = 22. 37

Fig. P18.4 The 6 × 6 stiffness and mass matrices are given in Example 18.2. Impose boundary conditions u1 = u2 = u5 = u6 = 0 by eliminating the corresponding rows and columns to obtain k and m with reference to the two DOFs in Fig. P18.4: k =

LM N

8 EI 24 0 2 L3 0 2 L

OP Q

m =

LM N

mL 312 0 840 0 2 L2

EI m L4

ω 2 = 61. 67

EI m L4

The finite element method using consistent mass provides an excellent result for the fundamental frequency, but the accuracy deteriorates for the second mode. The lumped mass approximation provides only the fundamental frequency, but the result is less accurate and lower than the exact value.

OP Q

2. Solve eigenvalue problem. 2

( k − ω m )φ = 0

ω1 = 22. 74 φ1 =

EI 4

RS1UV T0W

ω 2 = 81. 98 φ2 =

EI m L4

RS 0 UV T1 LW

3. Solution using lumped mass matrix. m = mL

LM0.5 OP N 0Q

Eliminate the DOF 2 by static condensation to obtain 8 EI 192 EI k$lat = 3 ( 24 ) = L L3

Natural frequency:

ω1 =

k$lat 0. 5 m L

= 19. 60

EI m L4

Natural mode after calculating rotations using static condensation equations:

φ1T = 1 0

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 6 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 18.5

Natural mode after calculating rotation using static condensation equations:

1. Stiffness and mass matrices.

φ1T = 0. 2375 − 0. 8143 L 0. 2036 L 4. Compare with exact frequencies.

1 3

For a beam clamped at one end and simply supported at the other, the exact values of the first three natural frequencies are

Fig. P18.5

ω1 = 15. 42 The 6 × 6 stiffness and mass matrices are given in Example 18.2. Impose boundary conditions u1 = u5 = u6 = 0 by eliminating the corresponding rows and columns to obtain k and m with reference to the three DOFs in Fig. 18.5:

LM 24 3L 0 OP L k k O L 2P = M 3L L P L M MN 0 L 2 2 L PQ Nk k Q 0O L 312 − 6.5L mL M m = − 6.5 L L − 0.75 L PP M 840 MN 0 − 0.75L 2 L PQ k =

8 EI

ω 3 = 104. 2

EI m L4

ω 2 = 49. 97

EI m L4

The finite element method using consistent mass provides an excellent result for the fundamental frequency, but the accuracy deteriorates increasingly for the second and third modes. The lumped mass approximation provides only the fundamental frequency; the result is less accurate and lower than the exact value.

2. Solve eigenvalue problem. 2

( k − ω m )φ = 0

ω1 = 15. 56

EI mL

ω 3 = 155. 64

Φ =

ω 2 = 58. 41

EI m L4

LM 0.2375 − 0.0349 − 0.0205OP MM− 0.9370 L − 0.7449 L − 0.8516 LPP N 0.2561 L 0.6662 L − 0.5237 LQ

3. Solution using lumped mass matrix.

LM0.5 OP m = mL M 0 P 0PQ NM Eliminate the two rotational DOFs by static condensation to obtain 109. 71EI −1 k$lat = ktt − kt 0 k00 k0 t = L3 Natural frequency:

ω1 =

k$lat 0. 5 m L

= 14. 81

EI m L4

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 7 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 18.6 2 h

2. Mass matrix. 2

22 mh /420

2m , I/2 m, I

2m (2 h) + 2(156 mh /420)

m, I

&u&1 = 1

4mh 3/420 + 4(2 m )(2 h)3/420

1. Stiffness matrix.

– 3(2 m )(2 h ) /420 2

22 mh /420

6EI h2

24 EI h3

&u&2 = 1 3

4mh /420 +

u 1= 1

4(2 m )(2 h)3/420

4EI 4E(I/2) + 2h h

2E(I/2) 2h

– 3(2 m )(2 h) /420 2

22 mh /420

6EI h2

&u&3 = 1

u 2= 1

2E(I/2)

4EI h

Thus +

4E(I/2)

mh m = 420

2h 6EI h2

LM 1992 22h 22h OP 68h − 48h P MM 68h PQ N(sym) 2

3. Solve eigenvalue problem. 2

( k − ω m )φ = 0 u 3= 1

ω1 = 1. 5354

Thus

LM 24 6h 6h OP k = 5h h 2P h M MN(sym) 5h PQ EI 3

ω 3 = 10. 7471

EI mh

ω 2 = 4. 0365

EI mh 4

0 − 0.0001O LM 0.5440 Φ = M − 0.5933 h − 1 2 h 1 2 hPP MN− 0.5933 h 1 2h 1 2hPQ

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 8 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

4. Plot modes.

– 0.5933/ h

0.5440

– 0.5933/ h

– 1/ 2 h

– 0.001

1/ 2 h

1/ 2 h 1/ 2 h

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 9 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 18.7 1. Determine lateral stiffness. Statically condense joint rotations in the stiffness matrix of Problem 18.6: −1 k 0t k$lat = k tt − k t 0 k 00

F L 5 1 2OP LM6OPIJ 24 − 6 6 M G h GH N1 2 5 Q N6QJK −1

EI 3

= 10.9091

2. Calculate lumped mass. mlat = 2 m ( 2 h ) + (1 2 ) ( mh + mh ) = 5 mh 1 424 3 144 42444 3 beam columns

3. Calculate natural frequency.

ω1 =

k$lat mlat

= 1. 477

EI mh 4

The exact value is

ω1 = 1. 5354

EI mh 4

LM 0.544 OP 0.544 O L φ = M = − 0.594 h N− k k (0.544)PQ MMN− 0.594 hPPQ −1 00

4. Comment on accuracy. The accuracy of the lumped mass procedure is satisfactory; the fundamental frequency is estimated with approximately 4% error and the first mode is very accurate (less than 0.03% error). Using the lumped mass procedure we have reduced the order of the system to obtain only the fundamental frequency and mode.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 10 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 18.8

This is the same as determined in Problem 18.6.

2m, I/2

m, I

m, I 2

4. Element mass matrix

LM 156 22 L 54 − 13LOP 1 0 1 4 L − 6 L − 3L P 2 2 3 mL M m = M 156 − 22 L P 0 0 0 420 MN(sym) P 4L Q 0 3 0

1. Identify DOFs. The frame (assemblage) DOFs that correspond to the element DOFs are identified for each element in Table P18.8.

Table P18.8 Element 1

1 2 0 0

0 2 0 3

1 3 0 0

2. Element stiffness matrix

LM 12 6 L − 12 6 LOP 1 0 1 EI M 6 L 4 L − 6 L 2 L P 2 2 3 k = L M− 12 − 6 L MN 6 L 2 L − 612L −46LLPPQ 00 03 00 e

For elements L = 2h.

0 1

2 3

0 0

3 0

and

2 ,

L = h ; for element

3 ,

3. Assemble element stiffness matrices. 6 EI 6 EI LM12 EI + 12 EI OP M h 6EI h 4 EI h4 E ( I 2) 2 E h( I 2) PP k = M + h h 2h MM 6hEI P 2 E ( I 2) 4 E ( I 2) 4 EI P + MN h 2h h h PQ 3

2 2

LM24 6h 6h OP 6h 5h h 2P h M NM6h h 2 5h PQ EI 3

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 11 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

5. Assemble element mass matrices.

LM156mh + 156mh + 2m (2h) 22mh 420 420 MM 420 4mh 4 ( 2m) ( 2h) m = M + 420 420 MM ( sym) MN L 1992 22h 22h OP mh M = 68h − 48h P 420 M MN(sym) 68h PQ 2

22mh 2 420 3 (2m) ( 2h) 3 − 420 4 ( 2m) ( 2h) 3 4mh 3 + 420 420

OP PP PP PP Q

where 2 m ( 2 h ) in m11 is due to rigid translation of the beam. This mass matrix is the same as determined in Problem 18.6.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 12 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

CHAPTER 1 Problem 1.1 If ke is the effective stiffness, fS = keu k1

u fS

k1 u k2 u

Equilibrium of forces: Effective stiffness: Equation of motion:

fS = ( k1 + k2 ) u ke = fS u = k1 + k2 mu&& + keu = p( t )

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 1 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 1.2 If ke is the effective stiffness, fS = keu

(a)

u fS

If the elongations of the two springs are u1 and u2 , u = u1 + u2

(b)

Because the force in each spring is fS , fS = k1u1

fS = k2u2

(c)

Solving for u1 and u2 and substituting in Eq. (b) gives fS f f 1 1 1 = S + S ⇒ = + ⇒ ke k1 k2 ke k1 k2 ke =

k1 k2 k1 + k2

Equation of motion: mu&& + keu = p( t )

Problem 1.3 k1 k2

k3 m

Fig. 1.3(a)

Fig. 1.3(b)

⇓

k 1+ k 2 ⇓

ke m

Fig. 1.3(c)

This problem can be solved either by starting from the definition of stiffness or by using the results of Problems P1.1 and P1.2. We adopt the latter approach to illustrate the procedure of reducing a system with several springs to a single equivalent spring. First, using Problem 1.1, the parallel arrangement of k1 and k2 is replaced by a single spring, as shown in Fig. 1.3(b). Second, using the result of Problem 1.2, the series arrangement of springs in Fig. 1.3(b) is replaced by a single spring, as shown in Fig. 1.3(c): 1 1 1 = + ke k1 + k2 k3

Therefore the effective stiffness is ke =

( k1 + k2 ) k3 k1 + k2 + k3

The equation of motion is mu&& + keu = p( t ) .

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 3 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 1.4 1. Draw a free body diagram of the mass. O T

L θ m

mgsinθ

mgcosθ

2. Write equation of motion in tangential direction. Method 1: By Newton’s law. −mg sin θ = ma − mg sin θ = mLθ&& mLθ&& + mg sin θ = 0

(a)

This nonlinear differential equation governs the motion for any rotation θ. Method 2: Equilibrium of moments about O yields mL2θ&& = −mgL sin θ

or mLθ&& + mg sin θ = 0

3. Linearize for small θ. For small θ, sin θ ≈ θ , and Eq. (a) becomes mLθ&& + mgθ = 0 ⎛g⎞ ⎝L⎠

θ&& + ⎜ ⎟θ = 0

(b)

4. Determine natural frequency.

ωn =

g L

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 4 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 1.5 1. Find the moment of inertia about O. From Appendix 8, 2

I0 =

1 1 ⎛ L⎞ mL2 + m⎜ ⎟ = mL2 ⎝ 2⎠ 12 3

2. Draw a free body diagram of the body in an arbitrary displaced position.

L/2

3. Write the equation of motion using Newton’s second law of motion.

∑ M = I θ&& 0

− mg

L 1 sin θ = mL2θ&& 2 3

mL2 && mgL θ+ sin θ = 0 3 2

(a)

4. Specialize for small θ. For small θ, sinθ ≅ θ and Eq. (a) becomes mL2 && mgL θ+ θ =0 3 2

θ&& +

3 g =0 2 L

(b)

5. Determine natural frequency.

ωn =

3 g 2 L

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 5 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 1.6 1.

5. Determine natural frequency.

Find the moment of inertia about about O.

ωn =

In each case the system is equivalent to the springmass system shown for which the equation of motion is

∫ r xd A 2

I0 = ρ

⎛ w⎞ ⎜⎜ ⎟⎟ u&& + ku = 0 ⎝g⎠

∫

= ρ r 2 ( r α dr )

4 g 3 L

L4 α 4 1 = mL 2 2 =

w u

2. Draw a free body diagram of the body in an arbitrary displaced position.

2L/3

The spring stiffness is determined from the deflection u under a vertical force fS applied at the location of the lumped weight: Simply-supported beam: u =

fS L3 48 EI

⇒ k =

Cantilever beam: u = mg

Clamped beam: u =

fS L

3 EI fS L3 192 EI

⇒ k = ⇒ k =

48 EI L3

3 EI L3 192 EI L3

3. Write the equation of motion using Newton’s second law of motion.

∑ M = I θ&& 0

− mg

2L 1 sin θ = mL2θ&& 3 2

mL2 && 2mgL θ+ sin θ = 0 2 3

(a)

4. Specialize for small θ. For small θ, sinθ ≅ θ, and Eq. (a) becomes mL2 && 2mgL θ+ θ =0 2 3

θ&& +

4 g =0 3 L

(b)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 6 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 1.7 Draw a free body diagram of the mass: fS

mü u p(t)

Write equation of dynamic equilibrium: mu&& + fS = p( t ) Write the force-displacement relation: AE fS = u L

FG IJ H K

(a)

(b)

Substitute Eq. (b) into Eq. (a) to obtain the equation of motion: AE mu&& + u = p( t ) L

FG IJ H K

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 7 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 1.8 Show forces on the disk: fS O θ

Write the equation of motion using Newton's second law of motion: − fS = IO&& θ

where

IO =

m R2 2

(a)

Write the torque-twist relation: fS =

FG GJ IJ θ where J = π d H LK 32

(b)

Substitute Eq. (b) into Eq. (a): GJ I Oθ&& + θ = 0 L

FG IJ H K

or,

F mR I θ&& + F π d G I θ = 0 GH 2 JK GH 32 L JK 2

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 8 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problems 1.9 through 1.11 In each case the system is equivalent to the springmass system shown for which the equation of motion is

FG w IJ u&& + ku = 0 H gK k

w u

The spring stiffness is determined from the deflection u under a vertical force fS applied at the location of the lumped weight: 48 EI f L3 Simply-supported beam: u = S ⇒ k = 48 EI L3 3 f L 3EI Cantilever beam: u = S ⇒ k = 3EI L3 3 192 EI f L Clamped beam: u = S ⇒ k = 192 EI L3

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 9 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 1.12

2. Determine the effective stiffness. f s = k eu L

(d)

where

simply supported

u = δ spring + δ beam

(e)

f s = kδ spring = k beam δ beam

(f)

Substitute for the δ’s from Eq. (f) and for u from Eq. (d): Fig. 1.12a

fs f f = s + s ke k k beam ke = Undeformed position

ke =

δst Static Equlibrium

kk beam k + k beam

(

k 48 EI / L3 48 EI k+ 3 L

)

3. Determine the natural frequency.

u Deformed position

ωn =

ke m

Fig. 1.12b

fs ..

mu p(t) w

Fig. 1.12c 1. Write the equation of motion. Equilibrium of forces in Fig. 1.12c gives m u&& + f s = w + p (t )

(a)

where fs = k e u

(b)

The equation of motion is: mu&& + k e u = w + p(t )

(c)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 10 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 1.13 Compute lateral stiffness: 1 3EI c /h

k = 2 × kcolumn = 2 ×

3 EIc 6 EIc = 3 h h3

Equation of motion: mu&& + ku = p( t )

Base fixity increases k by a factor of 4.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 11 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 1.14

k 23

k 22

1. Define degrees of freedom (DOF). 3

3EI c 4 EI c 5EI c + = ( 2h ) h h 2 EI c EI c = k 32 = ( 2h ) h 3EI c k12 = 2 h k 22 =

2. Reduced stiffness coefficients. Since there are no external moments applied at the pinned supports, the following reduced stiffness coefficients are used for the columns.

u3 = 1, u1 = u2 = 0

k23

k33 k13

Joint rotation: 3EI L

EI 1

3EI

3 EI

3EI c 4 EI c 5EI c + = ( 2h ) h h 2 EI c EI c = k 23 = ( 2h ) h 3EI c k13 = 2 h k33 =

Joint translation: EI

3EI 1

3EI L3

L2 3EI L3

3. Form structural stiffness matrix.

Hence ⎡ 6 3h EI c ⎢ k = 3 ⎢3h 5h 2 h ⎢⎣3h h 2

3h ⎤ ⎥ h2 ⎥ 5h 2 ⎥⎦

u1 = 1, u2 = u3 = 0

k 31

k21

k11 = 2

4. Determine lateral stiffness. The lateral stiffness k of the frame can be obtained by static condensation since there is no force acting on DOF 2 and 3:

3EI c h

k 21 = k31 =

6 EI c h3

3EI c h2

u2 = 1, u1 = u3 = 0

⎡ 6 3h EI c ⎢ 3h 5h 2 h3 ⎢ ⎢⎣3h h 2

3h ⎤ ⎧ u1 ⎫ ⎧ f S ⎫ ⎥⎪ ⎪ ⎪ ⎪ h 2 ⎥ ⎨u2 ⎬ = ⎨ 0 ⎬ 5h 2 ⎥⎦ ⎪⎩u3 ⎪⎭ ⎪⎩ 0 ⎪⎭

First partition k as ⎡ 6 3h EI c ⎢ k = 3 ⎢3h 5h 2 h ⎢⎣3h h 2

3h ⎤ ⎥ ⎡ k tt h2 ⎥ = ⎢ k t0 5h 2 ⎥⎦ ⎣

k t0 ⎤ k 00 ⎥⎦

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 12 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

where EI c

k tt = k t0 = k 00 =

h3 EI c h3

[6] [3h 3h]

EI c ⎡5h 2 ⎢ h 3 ⎣⎢ h 2

h2 ⎤ ⎥ 5h 2 ⎦⎥

Then compute the lateral stiffness k from −1 T k = k tt − k t 0 k 00 k t0

Since −1 k 00 =

h ⎡ 5 − 1⎤ ⎢ ⎥ 24 EI c ⎣− 1 5 ⎦

we get k= k= k=

6 EI c 3

h EI c

−

EI c h

⎡5

[3h 3h] ⋅ 24 EI ⎢− 1 c ⎣

− 1⎤ EI c ⎡3h ⎤ ⋅ 5 ⎥⎦ h 3 ⎢⎣3h ⎥⎦

[6 − 3]

h3 3EI c h3

5. Equation of motion. m u&& +

3EIc u = p(t ) h3

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 13 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

u3 = 1, u1 = u2 = 0

Problem 1.15 1

Ib = I c / 2 Ic

k23

k33

k13

Define degrees of freedom (DOF): 2

k31

k11

k13 =

6 EIc h2

⎡24 EI c ⎢ k = 3 ⎢6 h h ⎢ ⎣6 h

⎡ 24 EI c ⎢ ⎢6 h h3 ⎢ ⎣6 h

6 EIc h2

u2 = 1, u1 = u3 = 0 k22

2 EIb EIc = (2 h) 2h

6h ⎤ 1 2⎥ h ⎥ 2 5h 2 ⎥⎦

6h 5h

1 2 h 2

The lateral stiffness k of the frame can be obtained by static condensation since there is no force acting on DOF 2 and 3:

12 EIc 24 EIc = h3 h3

k21 = k31 =

k23 =

Hence

u1 = 1, u2 = u3 = 0

k11 = 2

EIc 4 EIc 4 EIb 4 EIc 5 EIc + = + = h h h h (2 h)

Form structural stiffness matrix: k21

k33 =

6h 5h

1 2 h 2

⎤

6 h ⎧ u1 ⎫ ⎧ f S ⎫ ⎪ ⎪ ⎪ 1 2⎥ ⎪ h ⎥ ⎨u 2 ⎬ = ⎨ 0 ⎬ 2 5h 2 ⎥⎦ ⎪⎩u 3 ⎪⎭ ⎪⎩ 0 ⎪⎭

First partition k as k32

k12

k22 =

4 EIc 4 EIb 4 EIc 5 EIc EIc + = + = (2h) h h h h

k32 =

2 EIb EIc = (2h) 2h

k12 =

6 EIc h2

⎡ 24 EI c ⎢ k = 3 ⎢6 h h ⎢ ⎣6 h

6h 5h

1 2 h 2

6h ⎤

1 2⎥ h ⎥ = 2 2 ⎥

5h ⎦

⎡k tt ⎢k T ⎣ t0

k t0 ⎤ k 00 ⎥⎦

where ktt =

EIc 24 h3

kt 0 =

EIc 6h 6h h3

k 00 =

EI c ⎡ 5h 2 ⎢ 2 h 3 ⎢⎣ 12 h

1 2⎤ h 2 2 ⎥

5h ⎥⎦

Then compute the lateral stiffness k from −1 T k = ktt − kt 0 k00 kt 0

Since −1 = k 00

4h ⎡ 5 − 12 ⎤ ⎢ ⎥ 5⎥⎦ 99 EI c ⎢⎣− 12

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 14 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

we get k =

24 EI c EI c 4h ⎡ 5 − 12 ⎤ EI c ⎡6h ⎤ − 3 [6h 6h] ⋅ ⎢ ⎥⋅ 3 5⎦⎥ h3 ⎢⎣6h ⎥⎦ 99 EI c ⎣⎢− 12 h h

EI c 144 (24 − ) 3 11 h 120 EI c = 11 h3 =

This result can be checked against Eq. 1.3.5: k =

24 EI c ⎛ 12 ρ +1 ⎞ ⎜⎜ ⎟⎟ h 3 ⎝ 12 ρ + 4 ⎠

Substituting ρ = Ib 4 Ic = 1 8 gives 24 EI c ⎛⎜ 12 8 +1 ⎞⎟ 24 EI c ⎛ 5 ⎞ 120 EI c = ⎜ ⎟= h 3 ⎜⎝ 12 18 + 4 ⎟⎠ h 3 ⎝ 11 ⎠ 11 h 3 1

k =

Equation of motion: ⎛ 120 EI c ⎞ m u&& + ⎜⎜ ⎟ u = p (t ) 3 ⎟ ⎝ 11 h ⎠

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 15 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

u3 = 1 , u1 = u2 = u4 = u5 = 0

Problem 1.16 1. Define degrees of freedom (DOF). u3

u3 =1

EIc /2 h

EIc

EIc u2

u5 2h

4EI c 4EIc 5 EIc + = h 2(2h ) h 6EI 2 EIc k13 = 2 c k 23 = h h 2 EIc EI k 43 = = c k53 = 0 2(2h) 2h k 33 =

2. Form the structural stiffness matrix. u1 = 1 , u2 = u3 = u4 = u5 = 0 u1 =1

u4 = 1 , u1 = u2 = u3 = u5 = 0

u4 =1

k11 = 2

12EI c 24EIc = 3 3 h h

k 21 = k 31 = k 41 = k 51 =

6 EIc 2 h

u2 = 1 , u1 = u3 = u4 = u5 = 0

u2 =1

4 EIc 4EI c 5EI c + = h 2(2h) h 6 EIc k14 = 2 k 24 = 0 h 2 EIc EI 2EI c k 34 = = c k54 = 2(2h) 2 h h k 44 =

u5 = 1 , u1 = u2 = u3 = u4 = 0

4 EIc h 2EI c k 32 = h k 22 =

k12 =

6 EI c 2 h

u5 =1

k 42 = k52 = 0 4 EIc h 2 EIc k 45 = h k 55 =

k15 =

6EI c 2 h

k 25 = k 35 = 0

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 16 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Assemble the stiffness coefficients: ⎡24 6h ⎢6h 4h 2 EI c ⎢ 2 k = 3 ⎢6h 2h h ⎢ 6h 0 ⎢ ⎢⎣6h 0

2h 5h 1 2

0 1 2

5h 2

2h 2

6h ⎤ 0 ⎥ ⎥ 0 ⎥ 2h2 ⎥ ⎥ 4h2 ⎥⎦

3. Determine the lateral stiffness of the frame. First partition k. ⎡24 6h ⎢6 h 4 h 2 EIc ⎢ k = 3 ⎢6 h 2 h 2 h ⎢ ⎢6 h 0 ⎢⎣6h 0

6h 2h 2 5h 2 1 2

h2 0

6h 1 2

0 h2

5h 2 2h 2

6h ⎤ 0 ⎥ ⎥ ⎡ k tt 0 ⎥=⎢ T ⎥ kt 0 2h 2 ⎥ ⎣ 4h 2 ⎥⎦

kt 0 ⎤ k 00 ⎥⎦

Compute the lateral stiffness. k = k tt − k t 0k −1 kT 00 t 0 k=

24EIc 22EI c 2 EI c − = 3 h3 h3 h

4. Write the equation of motion. mu&& + ku = p (t ) ⎛ 2 EI ⎞ mu&& + ⎜⎜ 3 c ⎟⎟ u = p (t ) ⎝ h ⎠

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 17 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 1.17 (a) Equation of motion in the x-direction. The lateral stiffness of each wire is the same as the lateral stiffness of a brace derived in Eq. (c) of Example 1.2: ⎛ AE ⎞ 2 kw = ⎜ ⎟cos θ ⎝ L ⎠ ⎛ AE ⎞ 2 o 1 AE ⎟cos 45 = = ⎜⎜ ⎟ 2 2 h ⎝h 2⎠

Each of the four sides of the structure includes two wires. If they were not pretensioned, under lateral displacement, only the wire in tension will provide lateral resistance and the one in compression will go slack and will not contribute to the lateral stiffness. However, the wires are pretensioned to a high stress; therefore, under lateral displacement the tension will increase in one wire, but decrease in the other; and both wires will contribute to the lateral direction. Consequently, four wires contribute to the stiffness in the x-direction: kx = 4 kw =

AE h

Then the equation of motion in the x-direction is mu&&x + kx ux = 0

(b) Equation of motion in the y-direction. The lateral stiffness in the y direction, ky = kx , and the same equation applies for motion in the y-direction: mu&&y + kyuy = 0

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 18 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 1.18 z h h O

kw h 2

uθ = 1

Fig. 1.18(a)

Fig. 1.18(b)

1. Set up equation of motion. The elastic resisting torque fS and inertia force fI are shown in Fig. 1.18(a). The equation of dynamic equilibrium is fI + fS = 0 or IOu&&θ + fS = 0

(a)

where IO = m

h2 + h2 mh2 = 12 6

(b)

2. Determine torsional stiffness, kθ . fS = kθuθ

(c)

Introduce uθ = 1 in Fig. 1.18(b) and identify the resisting forces due to each wire. All the eight forces are the same; each is kw h 2 , where, from Problem 1.17, kw =

AE 2 2 h

The torque required to equilibrate these resisting forces is kθ = 8 kw =

h h

2 2 AEh

= 2 kw h2 =

2 2 2

(

AE h

) h2

(d)

3. Set up equation of motion. Substituting Eq. (d) in (c) and then Eqs. (c) and (b) in (a) gives the equation of motion: mh2 AEh u&&θ + uθ = 0 6 2

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 19 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 1.19 ut m k

ut - u

ug Fig. 1.19(a)

Fig. 1.19(b)

Displacement u t is measured from the static equilibrium position under the weight mg . From the free-body diagram in Fig. 1.19(b) fI + fD + fS = 0

(a)

where

fI = mu&&t fD = c ( u& t − u&g )

(b)

fS = k ( u − ug ) Substituting Eqs. (b) in Eq. (a) gives mu&&t + c ( u& t − u&g ) + k ( ut − ug ) = 0

Noting that x = vt and transferring the excitation terms to the right side gives the equation of motion: mu&&t + cu& t + ku t = cu&g ( vt ) + kug ( vt )

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 20 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

CHAPTER 2 Problem 2.1 Given: Tn = 2 π

m = 0. 5 sec k

(a)

Tn′ = 2 π

m + 50 g = 0. 75 sec k

(b)

1. Determine the weight of the table. Taking the ratio of Eq. (b) to Eq. (a) and squaring the result gives 2

⎛ Tn′ ⎞ m + 50 g ⎜⎜ ⎟⎟ = m ⎝ Tn ⎠

⇒ 1+

50 ⎛ 0.75 ⎞ =⎜ ⎟ = 2.25 mg ⎝ 0.5 ⎠

or mg =

50 = 40 lbs 1. 25

2. Determine the lateral stiffness of the table. Substitute for m in Eq. (a) and solve for k: ⎛ 40 ⎞ k =16π 2 m =16π 2 ⎜ ⎟ =16.4lbs in. ⎝ 386 ⎠

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 1 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 2.2 1. Determine the natural frequency. m =

k = 100 lb in.

ωn =

k m

100 400 386

400 386

lb − sec2 in.

= 9. 82 rads sec

2. Determine initial deflection. Static deflection due to weight of the iron scrap u( 0 ) =

200 = 2 in. 100

3. Determine free vibration. u ( t ) = u ( 0 ) cos ω nt = 2 cos ( 9. 82 t )

Problem 2.3 1. Set up equation of motion. ku+mg/2

mü u mg

mu&& + ku =

mg 2

2. Solve equation of motion. u ( t ) = A cos ω nt + B sin ω nt +

mg 2k

At t = 0 , u( 0 ) = 0 and u& ( 0 ) = 0 ∴ A = − u(t ) =

mg , B = 0 2k

mg (1 − cos ω nt ) 2k

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 3 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 2.4 k

u m

v0 m0

m =

10 = 0. 0259 lb − sec2 in. 386

m0 =

0. 5 = 1. 3 × 10 −3 lb − sec2 in. 386

k = 100 lb in.

Conservation of momentum implies m0 v0 = ( m + m0 ) u& ( 0 ) u& ( 0 ) =

m0 v0 = 2. 857 ft sec = 34.29 in. sec m + m0

After the impact the system properties and initial conditions are Mass = m + m0 = 0. 0272 lb − sec2 in. Stiffness = k = 100 lb in. Natural frequency:

ωn =

k = 60. 63 rads sec m + m0

Initial conditions: u ( 0 ) = 0, u& ( 0 ) = 34. 29 in. sec The resulting motion is u( t ) =

u& ( 0 )

ωn

sin ω nt = 0. 565 sin ( 60. 63t ) in.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 4 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 2.5

f S = ku

m2 m1

m1 u

m 2g

With u measured from the static equilibrium position of m1 and k, the equation of motion after impact is ( m1 + m2 ) u&& + ku = m2g

(a)

The general solution is u ( t ) = A cos ω nt + B sin ω nt +

ωn =

m2 g k

k m1 + m2

(b) (c)

The initial conditions are u( 0 ) = 0

u& (0) =

m2 m1 + m 2

2gh

(d)

The initial velocity in Eq. (d) was determined by conservation of momentum during impact: m2u&2 = ( m1 + m2 ) u& ( 0 )

where u&2 =

2 gh

Impose initial conditions to determine A and B: u( 0 ) = 0 ⇒ A = −

m2 g k

u& ( 0 ) = ω n B ⇒ B =

m2 m1 + m2

(e) 2 gh

ωn

(f)

Substituting Eqs. (e) and (f) in Eq. (b) gives

u(t ) =

m2 g (1 − cos ω nt ) + k

2 gh

ωn

m2 sin ω nt m1 + m2

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 5 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 2.6 1. Determine deformation and velocity at impact. u( 0) =

mg 10 = = 0.2 in. k 50

u& ( 0 ) = − 2 gh = − 2( 386 )( 36 ) = − 166.7 in./sec

2. Determine the natural frequency.

ωn =

kg (50)(386) = = 4393 . rad/sec w 10

3. Compute the maximum deformation. u(t ) = u(0) cos ω n t +

u& (0)

ωn

sin ω n t

⎛ 166.7 ⎞ = ( 0.2) cos 316.8t − ⎜ ⎟ sin 316.8t ⎝ 4393 . ⎠

⎡ u&( 0) ⎤ uo = [u( 0)]2 + ⎢ ⎥ ⎣ ωn ⎦

= 0.2 2 + ( −3.795) 2 = 38 . in.

4. Compute the maximum acceleration.

u&&o = ω n 2 uo = ( 4393 . )2 (38 . ) = 7334 in./sec2 = 18.98g

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 6 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 2.7 Given: m =

200 = 6. 211 lb − sec2 ft 32. 2

fn = 2 Hz

Determine EI: k = fn =

3 EI 3 EI EI = lb ft 3 = 3 3 9 L 1

2π

⇒ 2 =

2π

55. 90

⇒

EI = ( 4 π )2 55. 90 = 8827 lb − ft 2

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 7 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 2.8 Equation of motion: mu&& + cu& + ku = 0

(a)

Dividing Eq. (a) through by m gives u&& + 2 ζω n u& + ω 2n u = 0

(b)

where ζ = 1.

Equation (b) thus reads u&& + 2 ω n u& + ω 2n u = 0

(c)

Assume a solution of the form u ( t ) = e st . Substituting this solution into Eq. (c) yields ( s 2 + 2 ω n s + ω 2n ) e st = 0

Because e st is never zero, the quantity within parentheses must be zero: s 2 + 2 ω n s + ω 2n = 0

or s =

− 2ω n ±

( 2 ω n ) 2 − 4 ω 2n 2

= − ωn

(double root) The general solution has the following form: u ( t ) = A1 e − ω n t + A2 t e − ω n t

(d)

where the constants A1 and A2 are to be determined from the initial conditions: u( 0 ) and u& ( 0 ) . Evaluate Eq. (d) at t = 0 : u (0) = A1 ⇒ A1 = u (0)

(e)

Differentiating Eq. (d) with respect to t gives u& ( t ) = − ω n A1 e − ω n t + A2 (1 − ω n t ) e − ω n t

(f)

Evaluate Eq. (f) at t = 0 : u& ( 0 ) = − ω n A1 + A2 (1 − 0 ) ∴ A2 = u& ( 0 ) + ω n A1 = u& ( 0 ) + ω n u ( 0 )

(g)

Substituting Eqs. (e) and (g) for A1 and A2 in Eq. (d) gives u (t ) = { u (0) + [u& (0) + ω n u (0) ] t} e −ω nt

(h)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 8 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 2.9

A2 ω n ⎡−ζ + ζ 2 −1 + ζ + ζ 2 −1 ⎤ = ⎢⎣ ⎥⎦

Equation of motion: mu&& + cu& + ku = 0

Dividing Eq. (a) through by m gives u&& + 2 ζω n u& +

ω n2 u

= 0

or (b)

where ζ > 1.

A2 =

Assume a solution of the form u ( t ) = e st . Substituting this solution into Eq. (b) yields ( s 2 + 2ζω n s + ω n2 ) e st = 0

Because e st is never zero, the quantity within parentheses must be zero: s

+ 2 ζω n s +

ω 2n

−2ζω n ±

(2ζω n ) 2 − 4ω n2

= 0

u& (0) + ⎛⎜ ζ + ⎝

⎠

2 ζ −1 ω n

(f)

Substituting Eq. (f) in Eq. (d) gives u& (0) + ⎛⎜ ζ + ζ 2 − 1 ⎞⎟ω nu (0) ⎝ ⎠ A1 = u (0) − 2 2 ζ −1ω n 2 ζ 2 − 1 ω nu (0) − u& (0) − ⎛⎜ ζ + ζ 2 − 1 ⎞⎟ ω nu (0) ⎝ ⎠ = −u& (0) + ⎛⎜ −ζ + ζ 2 − 1 ⎞⎟ ω nu (0) ⎝ ⎠ = 2 ζ 2 − 1ω n

= ⎛⎜ − ζ ± ⎝

ζ 2 −1 ⎞⎟ ω n u (0)

2 ζ 2 −1ω n

or s =

u& (0) + ⎛⎜ ζ + ζ 2 −1 ⎞⎟ ω n u (0) ⎝ ⎠

(a)

(g)

ζ 2 −1 ⎞⎟ ω n ⎠

The solution, Eq. (c), now reads:

The general solution has the following form: ⎡ ⎤ u (t ) = A1 exp ⎢⎛⎜ −ζ − ζ 2 −1 ⎞⎟ω n t ⎥ ⎠ ⎣⎝ ⎦ ⎡ ⎤ + A2 exp ⎢⎛⎜ −ζ + ζ 2 −1 ⎞⎟ω n t ⎥ ⎠ ⎣⎝ ⎦

u (t ) = e −ζω nt

ω ′D =

(d)

Differentiating Eq. (c) with respect to t gives

⎡ ⎤ + A2 ⎛⎜ −ζ + ζ 2 −1 ⎞⎟ ω n exp⎢⎛⎜ −ζ + ζ 2 −1 ⎞⎟ ω nt ⎥ ⎝ ⎠ ⎠ ⎣⎝ ⎦

−ω ′D t

′

+ A2 e ω D t

)

ζ2 − 1 ω n

−u& (0) + ⎛⎜ −ζ + ζ 2 − 1 ⎞⎟ω n u (0) ⎝ ⎠ A1 = 2ω ′D

Evaluate Eq. (c) at t = 0 :

⎡ ⎤ u& (t ) = A1 ⎛⎜ −ζ − ζ 2 −1 ⎞⎟ ω n exp⎢⎛⎜ −ζ − ζ 2 −1 ⎞⎟ ω nt ⎥ ⎝ ⎠ ⎝ ⎠ ⎣ ⎦

where (c)

where the constants A1 and A2 are to be determined from the initial conditions: u( 0 ) and u& ( 0 ) . u (0) = A1 + A2 ⇒ A1 + A2 =u (0)

(A e

A2 =

u& (0) + ⎛⎜ ζ + ⎝

ζ 2 −1 ⎞⎟ ω n u (0) ⎠

2ω ′D

(e)

Evaluate Eq. (e) at t = 0 : u& (0) = A 1 ⎛⎜ −ζ − ζ 2 −1 ⎞⎟ ω n + A 2 ⎛⎜ −ζ + ζ 2 −1 ⎞⎟ ω n ⎠ ⎠ ⎝ ⎝ = [u (0) − A2 ] ⎛⎜ −ζ − ζ 2 −1 ⎞⎟ ω n + A2 ⎛⎜ −ζ + ζ 2 −1 ⎞⎟ ω n ⎝ ⎠ ⎝ ⎠

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 9 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 2.10

The general solution is u(t) = A1 e − ωn t + A2 t e −ω nt

Equation of motion: u&& + 2ζω n u& + ω n2 u = 0

(a)

u(t) = e

A2 = u& (0)

(j)

Substituting in Eq. (i) gives

Substituting this solution into Eq. (a) yields:

u(t ) = u& (0) t e −ω n t

FH s2 + 2ζω n s + ω 2n IK e st = 0

The roots of the characteristic equation [Eq. (b)] are:

s 2 + 2ζω n s + ω 2n = 0

(b)

The roots of this characteristic equation depend on ζ.

s1,2 = ω n −ζ ± ζ 2 − 1

(l)

The general solution is:

(a) Underdamped Systems, ζ<1

u(t ) = A1e s1t + A2 e s2 t

The two roots of Eq. (b) are

(m)

which after substituting Eq. (l) becomes

s = ω F −ζ ± i 1 − ζ I H K 2

(k)

Because e is never zero

1,2

Determined from the initial conditions u(0) = 0 and u& (0) : A1 = 0

Assume a solution of the form

(i)

(c)

F −ζ + ζ −1IK ω t

u(t ) = A1e H

F −ζ − ζ −1IK ω t

+ A 2 eH

(n)

Hence the general solution is st

Determined from the initial conditions u(0) = 0 and u& (0) :

s t

u(t) = A1e 1 + A2e 2

which after substituting in Eq. (c) becomes

u(t ) = e −ζω n t A1e iω D t + A2 e −iω D t

− A1 = A2 =

(d)

(o)

ζ 2 −1

2ω n

Substituting in Eq. (n) gives

where 1− ζ 2

(e)

u(t) = e −ζω nt (A cos ω D t + Bsin ω D t)

ω n 1−ζ

0.8

(g)

(b) Critically Damped Systems, ζ = 1 The roots of the characteristic equation [Eq. (b)] are: s2 = −ω n

− ω n t ζ 2 −1

IJ K

(p)

ζ = 0.1

ωD

e −ζω n t sin ω n 1 − ζ 2 t

−e

Plot Eq. (g) with ζ = 0.1; Eq. (k), which is for ζ = 1; and Eq. (p) with ζ = 2.

u& (0)

Substituting A and B into Eq. (f) gives u& (0)

ω n t ζ 2 −1

(d) Response Plots (f)

Determine A and B from initial conditions u(0) = 0 and u& (0) :

s1 = −ω n

FG e 2ω ζ − 1 H u& (0) e −ζω n t n

Rewrite Eq. (d) in terms of trigonometric functions:

A=0

u( t ) =

. u(t) ÷ (u(0)) / ωn )

ω D = ωn

u( t ) =

u& (0)

ζ = 1.0

ζ = 2.0

0.4

0 0.25

0.5

0.75

1.25

1.5

t/Tn

-0.4

-0.8

(h)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 10 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 2.11

F GH

u1 1 ln j uj +1

I ≈ 2πζ ⇒ 1 ln F 1 I ≈ 2πζ GH 01. JK JK j 10%

∴ j10% ≈ ln (10 ) 2 πζ ≈ 0. 366 ζ

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 11 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 2.12 ⎛ ⎞ ui ⎜ 2πζ ⎟ = exp ⎜ u i +1 ⎜ 1−ζ 2 ⎟⎟ ⎝ ⎠

(a) ζ = 0. 01: (b) ζ = 0. 05 : (c) ζ = 0. 25 :

ui ui + 1 ui ui + 1 ui ui + 1

= 1. 065 = 1. 37 = 5. 06

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 12 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 2.13 Given: w = 20.03 kips (empty); m = 0.0519 kip-sec2/in. k = 2 (8.2) = 16.4 kips/in. c = 0.0359 kip-sec/in. (a) Tn = 2 π (b) ζ =

m 0. 0519 = 2π = 0. 353 sec 16. 4 k

c 2 km

0. 0359 2 (16. 4 ) ( 0. 0519 )

= 0. 0194

= 1. 94%

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 13 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 2.14 (a) The stiffness coefficient is k=

3000 = 1500 lb/in. 2

The damping coefficient is c = ccr = 2 km c = 2 1500

3000 = 215.9 lb - sec / in. 386

(b) With passengers the weight is w = 3640 lb. The damping ratio is

ζ=

c 2 km

215.9 3640 2 1500 386

= 0.908

ω D = ω n 1− ζ 2 1500 1 − (0.908) 2 3640 / 386 = 12.61 × 0.419 = 5.28 rads / sec =

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 14 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 2.15 1. Determine ζ and ω n . ζ ≈

⎛ u ⎞ 1 ⎛ 1 ⎞ ln⎜ 1 ⎟ = ln⎜ ⎟ = 0.0128 = 1.28% ⎜ ⎟ 2π j ⎝ u j +1 ⎠ 2π (20) ⎝ 0.2 ⎠ 1

Therefore the assumption of small damping implicit in the above equation is valid. TD =

3 = 0.15 sec ; Tn ≈ TD = 0.15 sec ; 20

ωn =

2π = 41. 89 rads sec 0.15

2. Determine stiffness coefficient. k = ω 2n m = ( 41. 89 )2 0.1 = 175. 5 lbs in.

3. Determine damping coefficient. ccr = 2 mω n = 2 ( 0.1) ( 41. 89 ) = 8. 377 lb − sec in. c = ζ ccr = 0. 0128 ( 8. 377 ) = 0.107 lb − sec in.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 15 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 2.16 250 = 312. 5 lbs in. 0. 8

(a) k =

m =

250 386

= 0. 647 lb − sec2 in.

k = 21. 98 rads sec m

ωn =

(b) Assuming small damping,

F u I ≈ 2 jπζ ⇒ GH u JK F u IJ = ln (8) ≈ 2 (2) π ζ ⇒ ζ = 0.165 ln G H u 8K 1

j +1

This value of ζ may be too large for small damping assumption; therefore we use the exact equation: ln

F u I = 2 jπ ζ GH u JK 1 − ζ 1

j +1

or, ln ( 8) =

2 (2) π ζ 1 − ζ

⇒

ζ 1 − ζ2

= 0.165 ⇒

ζ2 = 0. 027 (1 − ζ2 ) ⇒ ζ =

0. 0267 = 0.163

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 16 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 2.17 Reading values directly from Fig. 1.1.4b: Peak 1 31

Time, t i (sec) 0.80 7.84

Peak, u&&i (g) 0.78 0.50

7.84 − 0.80 = 0.235 sec 30 ⎛ 0.78g ⎞ 1 ζ= ln⎜ ⎟ = 0.00236 = 0.236% 2π (30) ⎝ 0.50g ⎠ TD =

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 17 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 2.18 1. Determine buckling load.

wcr θ

L k

wcr ( L θ ) = k θ wcr =

k L

2. Draw free-body diagram and set up equilibrium equation. fI w θ

L fS

where

∑ MO = 0 ⇒ fI L + fS = w Lθ

fI =

w 2 && L θ g

fS = k θ

Substituting Eq. (b) in Eq. (a) gives w 2 && L θ + (k − w L) θ = 0 g

(a) (b)

(c)

3. Compute natural frequency.

ω n′ =

k − wL = ( w g) L2

k wL ⎞ ⎛ ⎜1 − ⎟ 2 k ⎠ ( w g) L ⎝

ω ′n = ω n 1 −

w wcr

(d)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 18 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 2.19 For motion of the building from left to right, the governing equation is mu&& + ku = − F

(a)

for which the solution is u ( t ) = A2 cos ω nt + B2 sin ω nt − uF

(b)

With initial velocity of u& ( 0 ) and initial displacement u( 0 ) = 0 , the solution of Eq. (b) is u(t ) =

u& ( 0 )

ωn

sin ω nt + uF (cos ω nt − 1)

u& ( t ) = u& ( 0 ) cos ω nt − uFω n sin ω nt

At the extreme right, u& ( t ) = 0 ; hence from Eq. (d) tan ω nt =

u& ( 0 ) 1 ω n uF

Substituting ω n = 4 π , uF = 0.15 in. 20 in. sec in Eq. (e) gives tan ω nt =

(e) and

u& ( 0 ) =

20 1 = 10. 61 4 π 0.15

or sin ω nt = 0. 9956; cos ω nt = 0. 0938

Substituting in Eq. (c) gives the displacement to the right: u =

20 ( 0. 9956 ) + 0.15 ( 0. 0938 − 1) = 1. 449 in. 4π

After half a cycle of motion the amplitude decreases by 2 uF = 2 × 0.15 = 0. 3 in.

Maximum displacement on the return swing is u = 1. 449 − 0. 3 = 1.149 in.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 19 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 2.20 Given: F = 0.1w , Tn = 0. 25 sec 0.1w 0.1mg 0.1g 0.1g F = = = = ( 2 π Tn )2 k k k ω 2n 0.1g = = 0. 061 in . ( 8 π )2

uF =

The reduction in displacement amplitude per cycle is 4uF = 0. 244 in.

The displacement amplitude after 6 cycles is 2.0 − 6 (0.244) = 2.0 − 1.464 = 0.536 in.

Motion stops at the end of the half cycle for which the displacement amplitude is less than uF . Displacement amplitude at the end of the 7th cycle is 0.536 – 0.244 = 0.292 in.; at the end of the 8th cycle it is 0.292 – 0.244 = 0.048 in.; which is less than uF . Therefore, the motion stops after 8 cycles.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 20 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

CHAPTER 3 Problem 3.1 From the given data, k = ω 2n = ( 2 π fn ) 2 = ( 8 π ) 2 = 64 π 2 m

(a)

k = ( ω ′n ) 2 = ( 2 π fn′ ) 2 = ( 6 π ) 2 = 36 π 2 m + Δm (b)

Dividing Eq. (a) by Eq. (b) gives 1 +

Δm 16 = m 9

m =

9 9 5 6. 43 Δm = = lbs g 7 7 g g

(c)

From Eq. (a), k = 64 π 2 m = 64 π 2

6. 43 = 10. 52 lbs in. g

(d)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 1 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 3.2 At ω = ω n , from Eq. (3.2.15), uo = ( ust ) o

1 = 2 2ζ

(a)

At ω = 0.1ω n , from Eq. (3.2.13), uo ≈ ( ust ) o = 0. 2

Substituting ( ust ) o = 0. 2 in Eq. (a) gives

ζ = 0. 05

Problem 3.3 Assuming that damping is small enough to justify the approximation that the resonant frequency is ω n and the resonant amplitude of Rd is 1 2ζ , then the given data implies: ( uo ) ω = ω n = ( ust ) o ( uo ) ω = 1.2ω n = ( ust ) o

= ( ust ) o

1 2ζ

(a) 1

1 − (ω ω n )2

+ 2ζ ω ω n

1 2 2

[1 − (1. 2 ) ]

+ [ 2ζ (1. 2 )]2

(b)

Combining Eq. (a) and Eq. (b): 2

⎛ (u o ) ω2 = 1.2ω ⎞ 1 n ⎟ ⎜ = 2 2 2 ⎜ ⎟ (2ζ ) ⎝ (u o ) ω = ω n ⎠ (−0.44) + (2.4ζ ) 2 1

1 4

Equation (c) gives 64ζ 2 = 0.1935 + 5. 76ζ 2 ⇒ ζ = 0. 0576

Assumption of small damping implied in Eq. (a) is reasonable; otherwise we would have to use the exact resonant frequency = ω n 1 − 2ζ 2 and exact resonant amplitude = (u st ) o

⎡2ζ ⎢⎣

1− 2ζ 2 ⎤ . ⎥⎦

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 3 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 3.4

(d) Summarizing these results together with given data:

(a) Machine running at 20 rpm.

ω 20 = = 0 .1 ωn 200 uo =

(u st ) o

(a)

1− (ω ω n )2

( ust ) o 1 − ( 0.1) 2

( uo ) ζ = 0

( uo )ζ = 0.25

0.1

0.2

0.1997

0.9

1.042

0.4053

3.0

0.0248

0.0243

The isolator is effective at ω ω n = 0. 9 ; it reduces the deformation amplitude to 39% of the response without isolators. At ω ω n = 0.1 or 3, the isolator has essentially no influence on reducing the deformation.

or 0. 2 =

ω ωn

⇒ ( ust ) o = 0.1980 in.

For ζ = 0. 25 and ω ω n = 0.1 , 1

uo = ( ust )o

1 − (ω ωn )

2 2

+ 2ζ ω ω n

(b)

or 1

uo = 0.1980

2 2

[1 − ( 0.1) ] + [ 2 ( 0. 25) ( 0.1)]2

= 0.1997 in.

(b) Machine running at 180 rpm.

ω 180 = = 0. 9 ωn 200 From Eq. (a), 1. 042 =

( ust )o 1 − ( 0 . 9 )2

⇒ ( ust )o = 0.1980 in .

For ζ = 0. 25 and ω ω n = 0. 9 , Eq. (b) reads 1

uo = 0.1980

1 − ( 0 . 9 )2

+ 2 ( 0. 25) ( 0. 9 )

= 0. 4053 in.

ω 600 = = 3 ωn 200 From Eq. (a), 0. 0248 =

( ust )o 1 − ( 3 )2

⇒ ( ust )o = 0.1980 in .

For ζ = 0. 25 and ω ω n = 3 , Eq. (b) reads uo = 0.1980

1 2 2

[1 − ( 3) ]

+ [ 2 ( 0. 25) ( 3)]2

= 0. 0243 in. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 4 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 3.5 Given: w = 1200 lbs ,

E = 30 × 10 6 psi ,

I = 10 in. 4 ,

L = 8 ft ;

po = 60 lbs ;

ω =⎜

ζ = 1%

⎛ 300 ⎞ ⎟ 2π = 10π rads sec ⎝ 60 ⎠

Stiffness of two beams:

⎛ 48EI ⎞ k = 2 ⎜ 3 ⎟ = 32,552 lbs in. ⎝ L ⎠ Natural frequency:

ωn =

k m

32, 552

1200 386

= 102. 3 rads sec

Steady state response: Rd =

1 1 − (ω ω n )

2 2

+ 2ζ ω ω n

where ω ω n = 10 π 102. 3 = 0. 3071 . Therefore, Rd =

1 2

[1 − 0. 0943] + [ 2 × 0. 01 × 0. 3071]2

= 1.104

Displacement: uo = ( ust )o Rd = =

60 32, 552

po k

× 1.104 = 2. 035 × 10 −3 in.

Acceleration amplitude: u&&o = ω 2uo = (10 π )2 2. 035 × 10 −3 = 2. 009 in. sec2 = 0. 0052 g

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 5 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 3.6 In Eq. (3.2.1) replacing the applied force by po cosω t and dividing by m we get u&& + 2ζω n u& + ω n2 u =

po cos ω t m

(a)

(a) The particular solution is of the form: u p (t ) = C sin ω t + D cos ω t

(b)

Differentiating once and then twice gives u& p (t ) = Cω cos ω t − Dω sin ω t

(c)

u&&p (t ) = − Cω 2 sin ω t − Dω 2 cos ω t

(d)

Substituting Eqs. (b)-(d) in Eq. (a) and collecting terms: (ω 2n − ω 2 ) C − 2ζω nω D sin ω t + 2ζω nω C + (ω n2 − ω 2 ) D cos ω t =

po cos ω t m

Equating coefficients of sin ωt and of cos ω t on the two sides of the equation: ( ω 2n − ω 2 ) C − ( 2ζω n ω ) D = 0

(e)

po m

(f)

( 2ζω n ω ) C + ( ω 2n − ω 2 ) D =

Solving Eqs. (e) and (f) for C and D gives C = D =

2ζω n ω po 2 2 2 m ( ω n − ω ) + ( 2ζω n ω ) 2

(g)

ω n2 − ω 2 m ( ω n2 − ω 2 )2 + ( 2ζω nω )2

(h)

Substituting Eqs. (g) and (h) in Eq. (b) gives u p (t ) =

2 po 1 − ( ω ω n ) cos ωt + 2ζ ω ω n sin ωt 2 2 k 1 − (ω ω n )2 + 2ζ ω ω n

(b) Maximum deformation Substituting for C and D gives uo =

po k

uo =

C 2 + D2 .

1 1 − (ω ω n )

2 2

+ 2ζ ω ω n

This is same as Eq. (3.2.11) for the amplitude of deformation due to sinusoidal force.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 6 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 3.7 (a) The displacement amplitude in given by Eq. (3.2.11): uo = (ust ) o (1 − β 2 ) 2 + (2ζ β ) 2

−1 2

(a)

where β = ω ω n . Resonance occurs at β when uo is maximum, i.e., duo dβ = 0 . Differentiating Eq. (a) with respect to β gives −

−3 2 1 (1 − β 2 ) 2 + (2ζ β ) 2 2

× 2(1 − β 2 ) ( −2 β ) + 2 (2ζβ ) 2ζ = 0

or −4 (1 − β 2 ) β + 8ζ 2 β = 0

1 − β 2 = 2ζ 2 ⇒ β =

1 − 2ζ 2

(b)

Resonant frequency:

ω r = ω n 1 − 2ζ 2 (b) Substituting Eq. (b) in Eq. (a) gives the resonant amplitude: uo = ( ust ) o

1 2 2

( 2ζ )

+ 4ζ 2 (1 − 2ζ 2 )

or uo = (ust ) o

1 2ζ 1 − ζ 2

(c)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 7 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 3.8

u&&o = −

(a) From Eq. (3.2.19) the acceleration amplitude is

po 1 m 2ζ 1 − ζ 2

u&&o = − = −

po p ⎛ ω ⎞ ⎟ Rd R a = − o ⎜⎜ m m ⎝ ω n ⎟⎠ po β2 m (1 − β 2 ) 2 + (2ζβ ) 2 1 2

(a)

where β = ω ω n . Resonance occurs at β where u&&o is maximum, i.e., du&&o dβ = 0 . Differentiating Eq. (a) with respect to β and setting the result equal to zero gives 1 ⎧ 2β ⎨ (1 − β 2 ) 2 + (2ζβ ) 2 ⎩

[(1− β ) + (2ζβ ) ] − 2 12

2 2

[

] [

]

⎫⎪ ⎛ β2 ⎞ −1 2 ⎜ ⎟ (1 − β 2 ) 2 + ( 2ζβ ) 2 − 4 β (1 − β 2 ) + 8ζ 2 β ⎬ = 0 ⎜ 2 ⎟ ⎪⎭ ⎝ ⎠

Multiplying the numerator by [(1 − β 2 ) 2 + (2ζβ ) 2 ] 1 2 and dividing it by β gives 2 (1 − β 2 ) 2 + 4ζ 2 β 2 −

β2

− 4 (1 − β 2 ) + 8ζ 2 = 0

or 1 − 2 β 2 + β 4 + 4ζ 2 β 2 + β 2 − β 4 − 2ζ 2 β 2 = 0

or 1 = β 2 (1 − 2ζ 2 ) ⇒ β =

1 1 − 2ζ

(b)

Resonant frequency:

ωr =

ωn

(c)

1 − 2ζ 2

(b) Dividing both the numerator and denominator of Eq. (a) by β 2 gives: u&&o = −

po 1 12 m ⎡ 2 2⎤ ⎛ 1 ⎞ ζ 4 ⎢⎜ ⎥ −1⎟ + ⎢⎜⎝ β 2 ⎟⎠ β2 ⎥ ⎣ ⎦

(d)

Substituting Eq. (b) in Eq. (d) gives the resonant amplitude: u&&o = −

m ( − 2ζ ) + 4ζ2 (1 − 2ζ2 ) 2 2

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 8 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 3.9 (a) From Eq. (3.2.17), the velocity amplitude is u&0 = =

p0 km

p0 ⎛ ω ⎞ ⎜ ⎟ Rd km ⎜⎝ ω n ⎟⎠

Rv =

p0 2 2

km [(1 − β ) + (2ζβ ) 2 ]1 / 2

(a)

where β = ω / ω n . Resonance occurs at β where u& is maximum, i.e., du& 0 / dβ = 0 . Differentiating Eq. (a) with respect to β and setting the derivative equal to zero gives [(1 − β 2 ) 2 + ( 2ζβ ) 2 ]−1/ 2 −

β 2

[(1− β 2)2 + (2ζ β )2]−3 / 2 [ 2 (1 − β 2) ( − 2 β ) + 2( 2ζ β ) 2ζ ] = 0

Multiplying the numerator by [(1 − β 2 ) 2 + ( 2ζβ ) 2 ]3/ 2 gives

[(1− β 2)2 + (2ζ β )2] − [ 2(1− β 2 ) ( − 2 β ) + 2 ( 2ζ β ) ( 2ζ ) ] = 0 2

or 1 − 2 β 2 + β 4 + 4ζ 2 β 2 − β (−2 β + 2 β 3 + 4ζ 2 β ) = 0

β 4 = 1⇒ β = 1

(b)

Resonant frequency:

ωr = ωn

(c)

(b) Substituding Eq. (b) into Eq. (a) gives u& 0 =

km 2ζ

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 9 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 3.10 We assume that the structure has no mass other than the roof mass. From Eq. (3.3.4), u&&o =

FG IJ R H K

me e ω 2n ω ωn m

(a)

At ω = ω n , Ra = 1 2ζ and Eq. (a) gives

ζ =

1 me e ω n2 2 u&&o m

(b)

Given: me = 2 × 50 g = 100 lbs g = 0.1 kips g m = 500 kips g e = 12 in.

ω n = 2 π fn = 2 π × 4 = 8 π u&&o = 0. 02g = 7. 72 in. sec2

Substituting the above data in Eq. (b) gives

ζ =

. 1 01 (12) (8π ) 2 = 0.0982 = 9.82% 2 (7.72) 500

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 10 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 3.11 The given data is plotted in the form of the frequency response curve shown in the accompanying figure:

(a) Natural frequency The frequency response curve peaks at fn = 1. 487 Hz

Assuming small damping, this value is the natural frequency of the system. (b) Damping ratio The acceleration at the peak is rpeak = 8. 6 × 10 −3 g .

Draw a horizontal line at rpeak ÷ obtain fa and fb in Hz: fa = 1. 473 Hz

2 = 6. 08 × 10 −3 g to

fb = 1. 507 Hz

Then,

ζ =

fb − fa 2 fn

1. 507 − 1. 473 2 (1. 487)

= 0. 0114

= 1.14%

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 11 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 3.12 (a) Transmissibility is given by Eq. (3.5.3) with ζ = 0. Thus the force transmitted is ( f T ) o = po

where ω n =

1 1 − (ω ω n ) 2

g δ st and ω = 2 π f . Therefore

( f T ) o = po

(a)

4π 2 2 f δ st 1 − g

(b) For ( fT )o = 0.1 po and f = 20 , Eq. (a) gives 0 .1 =

1 − 4 π2 386

1 2

1 − ( 4 π 386 ) ( 20 ) δ st

4 π2 386

⇒

( 20 )2 δ st = ± 10 ⇒

( 20 )2 δ st = − 9 or 11

The negative value is invalid. Therefore,

δ st =

11 2

( 4 π 386 ) ( 20 )2

= 0. 269 in.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 12 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 3.13 The equation governing the deformation u ( t ) in the suspension system is mu&& + cu& + ku = − mu&&g ( t )

(a)

where u g (t ) = u go sin ω t and ω = 2 π v L . Substituting in Eq. (a) gives mu&& + cu& + ku = mω 2 u go sin ω t

(b)

The amplitude of deformation is m ω 2ugo

uo = ( ust )o Rd =

The amplitude of the spring force is 2

fSo = kuo = m ω ugo Rd

(c)

where Rd =

1 1 − (ω ω n )2

+ 2ζ ω ω n

(d)

Numerical calculations: k = 800 lbs in. ; m = 4000 386 lb - sec 2 in. ugo = 3 in.

ω n = 8. 768 rads sec ; ω = 3. 686 rads sec ω ω n = 0. 420 Thus Eq. (d) gives Rd =

1 2 2

[1 − ( 0. 42 ) ]

+ [ 2 ( 0. 4 ) ( 0. 42 )]2

= 1.124

Substituting ω n and Rd in Eq. (c) gives fSo =

4000 ( 3. 686 ) 2 3 (1.124 ) = 474. 8 lbs 386

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 13 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 3.14

bg bg p bt g = − mu&& bt g = mω u sin ωt ≡ p sin ω t mω u F ω IJ R = u R u = bu g R = R =u G k Hω K u g t = u go sin ω t ⇒ u&&g t = −ω 2 u go sin ω t eff

st o

Spring force: f S = kuo The resonant frequency for f S is the same as that for uo , which is the resonsnat frequency for Ra (from section 3.2.5):

ω res = =

ωn 1 − 2ζ 2 8.786 1 − 2(0.4) 2

= 10.655 rads / sec

Resonance occurs at this forcing frequency, which implies a speed of v=

ωL 10.655 100 = = 169.6 ft / sec = 116 mph 2π 2π

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 14 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 3.15 Given: w = 2000 lbs , f = 1500 cpm = 25 Hz , and TR = 0.10 For an undamped system, TR =

1 1 − ( ω ω n )2

For TR < 1, ω ω n > 1 ( ω ω n )2 − 1

= 0 .1

2 and the equation becomes

= 0 .1 ⇒

( ω ω n )2 = 11 ⇒ ω ω n = 3. 32

ωn = k =

ω 3. 32

ω 2n m

2π f 2 π ( 25) = = 47. 31 rads sec 3. 32 3. 32

= ω 2n w g = ( 47. 31)2 2000 386

= 11. 6 kips in.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 15 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 3.16 The excitation is u g (t ) = u go sin ω t

u&&g (t ) = − ω 2 u go sin ω t (a)

The equation of motion is mu&& + cu& + ku = peff ( t )

(b)

where peff (t ) = − m u&&g (t ) = ω 2 m u go sin ω t

ω 2 m u go

Rd sin (ω t − φ )

(d)

where Rd and φ are defined by Eqs. (3.2.11) and (3.2.12), respectively. The total displacement is u t ( t ) = u ( t ) + ug ( t ) Substituting Eqs. (a), (d), (3.2.11) and (3.2.12) gives

aω ω f aω ω f a ζ ω ω f 2

u (t ) = u go sin ω n t +

× =

1−

n aω ω f 1−

u go

+ 2

sin ω n t − 2ζ ω ω n cos ω t

u go

a f a ζω ω f { 1 − aω ω f + 4ζ aω ω f 2 2

1− ω ωn

2 2

+ 2

sin ω n t − 2ζ ω ω n

cos ω t

}

(e) Equation (e) is of the form u t (t ) = C sin ω t + D cos ω t ; hence uot =

C 2 + D 2 . Substituting for C and D gives

R| 1 + 2ζ ω ω b g u =u S |T 1 − bω ω g + 2ζ ω ω t o

2 2

U| V| W

1/ 2

2 n

(f)

Equation (f) is the same as Eq. (3.6.5).

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 16 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 3.17 The accelerometer properties are fn = 50 cps and

ζ = 0. 7 ; and the excitation is

u&&g ( t ) = 0.1g sin ( 2 π f t ) ; f = 10 , 20 and 40 Hz

(a)

Compare the excitation with the measured relative displacement u ( t ) = ( − 1 ω 2n ) Rd u&&g ( t − φ ω )

(b)

where Rd =

1 1 − (ω ω n )

2 2

+ 2ζ ω ω n

(c)

and ⎡

⎤ ⎥ 2 ⎢⎣ 1 − (ω ω n ) ⎥⎦

φ = tan −1 ⎢

2ζ ω ω n

(d)

Because the instrument has been calibrated at low excitation frequencies, after separating the instrument constant − 1 ω 2n , the recorded acceleration is u&&g ( t ) = Rd u&&g ( t − φ ω )

(e)

For a given f (or ω ), ω ω n is computed and substituted in Eqs. (c) and (d) to calculate Rd and φ . With Rd and φ known for a given excitation frequency, the recorded acceleration is given by Eq. (e). The computed amplitude Rd and time lag φ ω agrees with Fig. 3.7.3. The difference between Rd and 1 is the error in measured acceleration (Table P3.17). Table P3.17 f (Hz)

ω ωn

φ ω , sec

% error in Rd

0.2

0.0045

0.4

0.991

0.0047

0.9

0.8

0.850

0.0050

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 17 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

f ≤ 4. 86 Hz

Problem 3.18 From Eq. (3.7.1),

ω 2n u ( t ) = − Rd u&&g ( t − φ ω ) = − Rd u&&go sin ( 2 π f t − φ )

and therefore,

ω 2n uo u&&go

= Rd =

1 1 − (ω ω n )

2 2

+ 2ζ ω ω n

(a)

Equation (a) is plotted for ζ = 0. 6 , the accelerometer damping ratio: 1.5 1.01 1

ω n2 uo

&u&go

0.5

0.194

0.723

0.5

f fn

Error < 1%

1.5

or ω ω n

We want to bound Rd as follows: 0. 99 ≤ Rd ≤ 1. 01

(b)

The relevant condition is Rd ≤ 1. 01 because we are interested in a continuous range of frequencies over which the error is less than 1%. Therefore, impose 1 1 − (ω ω n )2

+ 2ζ ω ω n

= 1. 01

(c)

Defining β ≡ ω ω n , Eq. (c) can be rewritten as ⎛ 1 ⎞ ( 1 − β 2 ) 2 + (2ζβ ) 2 = ⎜ ⎟ ⎝ 1.01 ⎠

⇒

β 4 − 0. 56β 2 + 1 = 0. 9803 ⇒ β 4 − 0. 56β 2 + 0. 0197 = 0 ⇒ β2 = 0. 0377, 0. 5223 ⇒ β = 0.194, 0. 723 Choose β = 0.194 (see figure) which gives the desired frequency range: f ≤ 0.194 fn = 0.194 ( 25) ⇒ © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 18 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 3.19

which gives the desired frequency range:

From Eq. (3.7.1),

f ≤ 0. 405 fn = 0. 405 ( 50 ) ⇒

ω 2n u ( t ) = − Rd u&&g ( t − φ ω )

f ≤ 20. 25 Hz

= − Rd u&&go sin ( 2 π f t − φ )

and therefore,

ω 2n uo u&&go

= Rd =

1 1 − (ω ω n )

2 2

+ 2ζ ω ω n

(a)

Equation (a) is plotted for ζ = 0. 7 , the accelerometer damping ratio: 1.5

ωn2 u o

&u&go

0.99 0.5 0.405

0.5

1.5

f f n or ω ω n

Error < 1%

We want to bound Rd as follows 0. 99 ≤ Rd ≤ 1. 01

(b)

The relevant condition is 0. 99 ≤ Rd because Rd is always smaller than 1.01 in this case. Therefore, impose 1 1 − (ω ω n )

2 2

+ 2ζ ω ω n

= 0. 99

(c)

Defining β ≡ ω ω n , Eq. (c) can be rewritten as 2

⎛ 1 ⎞ ( 1 − β 2 ) 2 + (2ζ β ) 2 = ⎜ ⎟ ⇒ ⎝ 0.99 ⎠

β 4 − 0.04 β 2 + 1 = 10203 . ⇒ β 4 − 0.04 β 2 − 0.0203 = 0 ⇒ β 2 = − 01239 . , 01639 . Take only the positive value:

β 2 = 01639 . ⇒ β = 0.405 © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 19 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 3.20 From Eq. (3.7.3), uo = Ra ugo

(a)

For maximum accuracy, uo ugo = 1; this condition after using Eqs. (3.2.20) and (3.2.11) for Ra gives (ω ω n )2 1 − (ω ω n )2

+ 2ζ ω ω n

= 1

(b)

Squaring, simplifying and defining β = ω ω n : (1 − β 2 ) 2 + (2ζβ ) 2 = β 4 ⇒ 1 − 2 β 2 + 4ζ 2 β 2 = 0 ⇒

ζ2 =

1 1 − 2 4β 2

For ω ω n >> 1 (β >> 1) ,

ζ2 =

1 ⇒ ζ = 0. 707 2

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 20 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 3.21 From Eq. (3.7.3), (ω ω n )2

uo = Ra = ugo

1 − (ω ω n )2

+ 2ζ ω ω n

(a)

Equation (a) is plotted for ζ = 0. 6 , the instrument damping ratio: 1.5 1.01

Ra 0.5 1.383 0

Error < 1%

5.149

f fn

or ω ω n

We want to bound Ra as follows (see figure) Ra ≤ 1. 01

(b)

Therefore imposing Ra = 1. 01 which, after defining β ≡ ω ω n , gives

β2 (1 − β 2 ) 2 + (2ζβ ) 2

= 101 .

Squaring this equation gives 2

⎛ β2 ⎞ ⎟ ⇒ (1 − β 2 ) 2 + ( 2ζβ ) 2 = ⎜ ⎜ 1.01 ⎟ ⎝ ⎠ 1 − 2 β 2 + β 4 + 144 . β 2 = 0.9803β 4 ⇒ 0.0197 β 4 − 0.56β 2 + 1 = 0 ⇒

β 2 = 1914 . , 26.51 ⇒ β = 1383 . , 5149 . Choose β = 5.149 (see figure) which gives the desired frequency range: f ≥ 5.149 fn = 5.149 ( 0. 5) ⇒ f ≥ 2. 575 Hz

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 21 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 3.22 From Eq. (3.7.3), (ω ω n )2

uo = Ra = ugo

1 − (ω ω n )2

+ 2ζ ω ω n

(a)

Equation (a) is plotted for ζ = 0. 7 , the instrument damping ratio: 1.5

1 0.99

Ra 0.5

Error < 1% 2.470

f fn

ω ωn

Since Ra is always smaller than 1.01, we want to bound Ra as follows Ra ≥ 0. 99

(b)

Therefore imposing Ra = 0. 99 which, after defining β ≡ ω ω n , gives

β2 (1 − β 2 ) 2 + (2ζβ ) 2

= 0.99

Squaring this equation gives 2 2

(1 − β )

+ (2ζβ )

⎛ β2 ⎞ ⎟ =⎜ ⎜ 0.99 ⎟ ⎝ ⎠

⇒

1 − 2 β 2 + β 4 + 196 . β 2 = 10203 . β4 ⇒ 0.0203β 4 + 0.04 β 2 − 1 = 0 ⇒

β 2 = − 8.073, 6102 . Take only the positive value:

β 2 = 6.102 ⇒ β = 2. 470 which gives the desired frequency range: f ≥ 2. 470 fn = 2. 470 ( 0. 5) ⇒ f ≥ 1. 235 Hz © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 22 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 3.23 From Eq. (3.8.1), E D = 2 πζ ( ω ω n ) kuo2

(a)

where uo =

po Rd k

(b)

In Eq. (a) substituting Eq. (b) and Eq. (3.2.11) for Rd gives ED =

π po2 k

2ζ ω ω n 1 − ( ω ω n )2

+ 2ζ ω ω n

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 23 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 3.24 From Eq. (3.8.9) the loss factor is

ξ =

1 ED 2 π ES o

(a)

where E D = π cω uo2 and ES o = k uo2 2. Substituting these in Eq. (a) gives

ξ =

cω k

(b)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 24 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 3.25 Based on equivalent viscous displacement amplitude is given by

{

uo 1 − [ ( 4 π ) ( F p o )]2 = (u st ) o 1 − (ω ω n ) 2

damping,

}

the

(a)

From the given data 50 1 F = = 100 2 po T 0. 25 ω = n = = 0. 833 0. 3 ωn T

Substituting these data in Eq. (a) gives

{

uo 1 − [ ( 4 π ) (1 2)] 2 = (u st ) o 1 − (0.833) 2

}

= 2.52

Now, ( ust ) o =

po k

where po = 100 kips

⎛ 2π ⎞ ⎟ k = ω n2 m = ⎜⎜ ⎟ ⎝ Tn ⎠

w 500 = (8π ) 2 g 386

= 818kips in.

Thus ( ust ) o =

100 = 0.1222 in. 818

and uo = 2. 52 ( 0.1222 ) = 0. 308 in.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 25 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

where ( ust ) o = po k and β j = j ω 0 ω n . Equation (f) is indeterminate when β j = 1; corresponding values of T0 are Tn , 3Tn , 5Tn , etc.

Problem 3.26 (a) p( t ) is an even function: ⎛ 2 ⎞ p(t ) = p o ⎜⎜1 − t⎟ T0 ⎟⎠ ⎝ a0 =

T0 2

1 T0

∫

−T0 2

2 T0

T0 2

(c) For T0 Tn = 2 , β j = j ω 0 ω n = j Tn T0 = j 2 and Eq. (f) becomes ∞ u( t ) 1 16 1 2 π jt ) = + 2 ∑ 2 2 cos ( ( ust )o 2 T0 π j = 1, 3, 5, L j ( 4 − j )

(a)

⎛ 2 ⎞ p o ⎜⎜1 − t ⎟ dt T0 ⎟⎠ ⎝

∫ 0

po 2

(b)

z bg b g z FGH IJK b g

T0 2

2 aj = T0 =

p(t )dt =

0 ≤ t ≤ T0 2

p t cos jω 0 t dt

− T0 2

4 p0 T0

T0 2

1−

two terms

0.8

2 t cos jω 0 t dt T0

u (t )

0.6

( u st ) o

0.4

three terms

0.2

U| 4 p R| 1 2 0 t cos b jω t gdt V sin b jω t g = − S 0 0.2 0.4 0.6 0.8 1 T | jω T | T W t T F 8p 2π jt I t cos G =− H T JK dt T of the j in the denominator of the series, two R F 2π j I O T L F 2π j I O U| Because 4 p |L terms are enough to obtain reasonable convergence of the =− SMt sin GH T tJK PPQ + 2π j MMNcos GH T tJK PPQ V| series solution. π jT |MN T W F 2π j tIJ OP 2p L =− cos G M π j MN H T K PQ 2p cos bπ j g − 1 =− π j T0 2

T0 2

0 2 0

T0 2

0 2 2

⎧ 4 po ⎪ ∴a j = ⎨ π 2 j 2 ⎪ 0 ⎩

j = 1,3,5,L

(c)

j = 2,4,6,L

bn = 0 because p( t ) is an even function

(d)

Thus the Fourier series representation of p ( t ) is p( t ) =

po 2

4 po

∞

∑

j = 1, 3, 5, L

1 j

2 cos ( j ω 0 t )

(e)

(b) The steady-state response of an undamped system is obtained by substituting Eqs. (b), (c) and (d) in Eq. (3.13.6) to obtain ∞ u( t ) 1 4 1 cos ( j ω 0t ) = + 2 ∑ π j = 1, 3, 5, L j 2 (1 − β2j ) ( ust )o 2 (f) © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 26 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

CHAPTER 4 Problem 4.1

Equation (4.1.7) with τ = 0 gives the response to p( t ) = δ ( t ) : u(t ) =

1 mω D

− ζω n t

sin ω Dt

= =

1 mω D 1 mω D

attaining a maximum is

− ζω n t

( − ζω n ) sin ω Dt + e

− ζω n t

− ζω n sin ω Dt + ω D cos ω Dt = 0

m ωn u o

0.6 0.4 0.2

The condition for u ( t ) du dt = 0 : du

(a)

0.8

− ζω n t

ω D cos ω Dt

0.2

0.4

0.6

0.8

Damping Ratio, ζ

or tan ω Dt =

1 ωD

ζ ωn

1 − ζ2

or t =

⎛ 1− ζ 2 ⎞ ⎜ ⎟ tan −1 ⎜ ⎟⎟ ωD ⎜ ζ ⎝ ⎠ 1

(b)

The maximum displacement is given by Eq. (a) evaluated at t given by Eq. (b). For this t, sin ω Dt =

1 − ζ

ωD t

1− ζ 2

Substituting Eqs. (b) and (c) in Eq. (a) gives ⎡ ⎛ ζ ⎜ exp ⎢ − tan −1 ⎜ uo = ⎢ 2 mω n ⎜ 1− ζ ⎢⎣ ⎝ 1

⎤ 1−ζ 2 ⎞⎟ ⎥ ⎟⎟ ⎥ ζ ⎠ ⎥⎦

(d)

From Eq. (d) mω nuo is plotted against ζ . The effect of damping is small; e.g., 10% damping reduces the response by about 15%.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 1 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 4.2 Response to a step function is given by Eq. (4.3.5); for po = 1 it becomes g (t ) =

⎧ ⎡ 1 ⎪ −ζω n t ⎢ − 1 e cos ω D t + ⎨ ⎢ k ⎪ ⎣ ⎩

⎤⎫ ⎪ sin ω D t ⎥ ⎬ ⎥ 2 1 −ζ ⎦ ⎪⎭ (a)

The response to a unit impulse force is given by Eq. (4.1.7) with τ = 0 : h(t ) =

1 e −ζω n t sin ω Dt mω D

(b)

From Eq. (a), g& (t ) = e −ζω n t

1 ⎡⎢ ζω n cos ω Dt + k ⎢ ⎣

+ ω D sin ω Dt −

ζ 1− ζ 2

ζ2 1− ζ 2

ω n sin ω Dt ⎤

ω D cos ω Dt ⎥ ⎥ ⎦

Canceling the cos ω Dt terms gives ⎡ ⎤ 1 ⎢ ζ2 ω n + ω D ⎥ sin ω D t ⎥ k ⎢ 1−ζ 2 ⎣ ⎦ ⎡ ⎤ ζ2 1 = e −ζω n t ω n ⎢ + 1 −ζ 2 ⎥ sin ω D t ⎢ 1 −ζ 2 ⎥ k ⎣ ⎦

g& (t ) = e −ζω n t

= e −ζω n t

= e

− ζω n t

= h( t )

1 mω n

1 mω D

1 1 −ζ 2

sin ω D t

sin ω Dt Q. E.D.

Problem 4.3 1. Determine response to the first impulse. The response of the system to the first impulse is the unit impulse response of Eq. (4.1.6) times I : ⎤ ⎡ 1 sin ω n t ⎥ u1 (t ) = I ⎢ ⎦ ⎣ mω n

(a)

2. Determine response to the second impulse. ⎤ ⎡ 1 u 2 (t ) = − I ⎢ sin ω n (t − t d )⎥ t ≥ t d ⎦ ⎣ mω n

(b)

If t d is shorter than Tn / 4 no peak will develop during 0 ≤ t ≤ t d and the response simply builds up from zero to u(td), where 2π t d u (t d ) = sin I / mω n Tn

(f)

The maximum deformation during 0 ≤ t ≤ t d is ⎧sin 2π t d / Tn uo ⎪ =⎨ I / mω n ⎪ ⎩1

t d / Tn ≤ 1 / 4

(g) t d / Tn ≥ 1 / 4

Equation (g) is plotted in Fig. P4.3e.

3. Determine response to both impulses. For 0 ≤ t ≤ t d :

6. Determine maximum response during t ≥ t d .

⎤ ⎡ 1 sin ω n t ⎥ u (t ) = I ⎢ ⎦ ⎣ mω n

From Eq. (d), the maximum deformation during t ≥ t d is

2π t = sin ω m n Tn I

(c)

(h)

Equation (h) is plotted in Fig. P4.3e

For t ≥ t d : u( t ) =

uo = 2 sin (π t d / Tn ) I / mω n

I [sin ω n t − sin ω n ( t − t d )] mω n

ω t ω ( 2t − t d ) I 2 sin n d cos n 2 2 mω n

⎡ ⎛ t 2I ⎛ π t d ⎞ 1 t d ⎞⎤ ⎟⎥ ⎟ cos ⎢2π ⎜ ⎜ sin − ⎟ ⎜ ⎟ ⎜ mω n ⎝ Tn ⎠ ⎣⎢ ⎝ Tn 2 Tn ⎠⎦⎥

(d) 4. Plot displacement response. Equations (c) and (d) are plotted for t d / Tn = 1/8, 1/4, 1/2, and 1 in Figs. P4.1a, b, c, and d, respectively. 5. Determine maximum response during 0 ≤ t ≤ t d . The number of peaks in u(t) depend on t d / Tn ; the longer the time t d between the pulses, more such peaks occur. The first peak occurs at t o = Tn / 4 with the deformation: uo =1 I / mω n

(e)

Thus t d must be longer than Tn / 4 for at least one peak to develop during 0 ≤ t ≤ t d . © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 3 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

t d Tn = 1 4

t d Tn = 1 8 2

First impulse

u1 ( t ) ÷ ( I / m ω n )

First impulse

-1

-2

-1

-2 0

0.5

1.5

0.5

1.5

Second impulse

u2( t ) ÷ ( I / mω n )

u2 ( t ) ÷ ( I / m ω n )

Second impulse

-1

-2

-1

-2 0

0.5

1.5

0.5

Both impulses

-1

-2

1.5

Both impulses

u ( t ) ÷ ( I / mω n )

-1

-2 0

0.5

1.5

0.5

t Tn

Fig. P4.3a

Fig. P4.3 b

1.5

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 4 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

t d Tn = 1 2

-1

-2 0.5

1.5

-1

Second impulse

-1

0.5

u2( t ) ÷ ( I / mω n )

-2 0

-2

1.5

Second impulse

-1

-2 0

0.5

1.5

-1

-2

0.5

1.5

Both impulses u ( t ) ÷ ( I / mω n )

u ( t ) ÷ ( I / mω n )

First impulse

u1 ( t ) ÷ ( I / m ω n )

t d Tn = 1

Both impulses

-1

-2 0

0.5

1.5

0.5

t Tn

Fig. P4.3c

Fig. P4.3d

1.5

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 5 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Eq. (h)

u o ÷ ( I / mω n )

1.5

Eq. (g) 1

0.5

0 0

0.5

1.5

t d Tn Fig. P4.3e 7. Determine the overall maximum response. From Eqs. (g) and (h), the overall maximum response is given by ⎧ 2 sin(π td / Tn ) uo ⎪ 1 ⎧ =⎨ I / mωn ⎪max. of ⎨ ⎩2 s in (π td / Tn ) ⎩

td / Tn ≤ 1/ 4 td / Tn ≥ 1/ 4

(i) Equation (i) is plotted t d / Tn in Fig. P4.3f to obtain the response spectrum. Overall maximum 2

u o ÷ ( I / mω n )

1.5

0.5

0 0

0.5

1.5

t d Tn Fig. P4.3f

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 6 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 4.4 1. Determine response to the first impulse. The response of the system to the first impulse is the unit impulse response of Eq. (4.1.6) times I : ⎡ 1 ⎤ u1 ( t ) = I ⎢ sin ω n t ⎥ ⎣ mω n ⎦

(a)

If t d is shorter than Tn / 4 no peak will develop during 0 ≤ t ≤ t d and the maximum displacement will occur at t d : u (t d ) 2π td = sin I / mω n Tn

(f)

The maximum deformation during 0 ≤ t ≤ t d is

2. Determine response to the second impulse. ⎡ 1 ⎤ u2 ( t ) = I ⎢ sin ω n ( t − t d )⎥ t ≥ t d ⎣ mω n ⎦

(b)

⎧sin 2 π t d / Tn uo ⎪ =⎨ I / mω n ⎪ ⎩1

t d / Tn ≤ 1 / 4

(g) t d / Tn ≥ 1 / 4

3. Determine response to the both impulses. Equation (g) is plotted in Fig. P4.4e.

For 0 ≤ t ≤ t d :

6. Determine maximum response during t ≥ t d .

⎡ 1 ⎤ u( t ) = I ⎢ sin ω n t ⎥ ⎣ mω n ⎦ =

I mω n

sin

From Eq. (d), t ≥ t d is

2π t Tn

uo = 2 cos (π t d / Tn ) I / mω n

I mω n

the maximum deformation during

(h)

Equation (h) is plotted in Fig. P4.4e [sin ω n t + sin ω n ( t − t d )]

I mω n

2 cos

ω ntd 2

sin

ω n ( 2t − t d ) 2

πt ⎞ ⎡ ⎛ t 1 t d ⎞ ⎤ 2I ⎛ ⎜ cos d ⎟ sin ⎢2π ⎜ − ⎟⎥ mω n ⎝ Tn ⎠ ⎢⎣ ⎝ Tn 2 Tn ⎠ ⎥⎦

(d)

4. Plot displacement response. Equations (c) and (d) are plotted for t d / Tn = 1/8, 1/4, 1/2, and 1 in Figs. P4.4a, b, c, and d, respectively. 5. Determine maximum response during 0 ≤ t ≤ t d . The number of peaks in u(t) depend on t d / Tn ; the longer the time t d between the pulses, more such peaks occur. The first peak occurs at t o = Tn / 4 with the deformation: uo =1 I / mω n

(e)

Thus t d must be longer than Tn / 4 for at least one peak to develop during 0 ≤ t ≤ t d .

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 7 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

t d Tn = 1 4

t d Tn = 1 8 2

First impulse u1 ( t ) ÷ ( I / m ω n )

u1 ( t ) ÷ ( I / m ω n )

First impulse 1

-1

-2

-1

-2 0

0.5

1.5

Second impulse

0.5

1.5

u2( t ) ÷ ( I / mω n )

Second impulse

-1

-2

-1

-2 0

0.5

1.5

Both impulses

-1

-2

0.5

u ( t ) ÷ ( I / mω n )

1.5

Both impulses

-1

-2 0

0.5

1.5

0.5

t Tn

Fig. P4.4a

Fig. P4.4 b

1.5

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 8 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

t d Tn = 1 2

t d Tn = 1

First impulse u1 ( t ) ÷ ( I / m ω n )

u1 ( t ) ÷ ( I / m ω n )

First impulse 1

-1

-2

-1

-2 0

0.5

1.5

0.5

Second impulse

Second impulse 1

u2( t ) ÷ ( I / mω n )

u2( t ) ÷ ( I / m ω n )

1.5

-1

-2

-1

-2 0

0.5

1.5

0.5

1.5

Both impulses

u ( t ) ÷ ( I / mω n )

Both impulses 1

-1

-2

-1

-2 0

0.5

t Tn

1.5

0.5

1.5

t Tn

Fig. P4.4c Fig. P4.4d

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 9 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Eq. (h)

uo ÷ ( I / mω n )

1.5

Eq. (g)

0.5

0 0

0.5

1.5

t d Tn Fig. P4.4e 7. Determine the overall maximum response. From Eqs. (g) and (h), the overall maximum response is given by ⎧ 2 cos (π td / Tn ) uo ⎪ 1 ⎧ =⎨ I / mωn ⎪max. of ⎨ ⎩2 cos (π td / Tn ) ⎩

td / Tn ≤ 1/ 4 td / Tn ≥ 1/ 4

(i) Eq. (i) is plotted t d / Tn response spectrum.

in Fig. P4.4f to obtain the

Overall maximum 2

u o ÷ ( I / mω n )

1.5

0.5

0 0

0.5

1.5

t d Tn Fig. P4.4f

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 10 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 4.5

a ω n = 0.01 2

(a) The equation of motion is mu&& + ku = p o e − at

(a)

Using Duhamel’s integral the solution is

p u( t ) = o mω n

Integrate

−aτ

dτ :

dy = e

(ust )

b g

e − aτ sin ω n t − τ dτ

parts

o 0

(b)

b g

v = sin ω n t − τ

letting

u( t ) 1

-1

and

∫

u(t) = vy − ydv

t/Tn

a ω n = 0.1

RSL−e u( t ) = mω a TMN po

− aτ

b g

sin ω n t − τ

u( t )

(ust )

b g UVW

− ω n e − aτ cos ω n t − τ dτ 0

Integrating again by parts, this −aτ v = cos ω n t − τ and dy = e dτ gives

b g

u( t ) =

po mω n (a 2 + ω n2 )

-1

time

with

a ω n = 1.0

a sin ω n t − ω n cos ω n t + ω n e − at u( t )

⎤ u( t ) 1 ⎡ a = sin ω nt − cos ω nt + e − at ⎥ 2 ⎢ (ust )o 1 + a ⎣ω n ⎦

(ust )

-1

a/ω

0.5

a/ω 0

t/Tn

− at

bu g = 1 + a ω

= 0.01

becomes very small and the As t increases, term e system attains a steady state harmonic motion whose amplitude is given by usteady

(c)

(b) The force p( t) = po e −at is plotted for three values of a ωn : p/po

ωn2

a/ω

t/Tn

Written in terms of (ust )o , the displacement response is

st o

2 n

= 0.1

= 1.0 t/Tn

The motion given by Eq. (c) is plotted next.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 11 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

a/ω

Problem 4.6 0.4

(a) The equation of motion is

(

mu&& + ku = p o e − at − e −bt

)

u( t ) (ust )

o -0.2

t/Tn

-0.4 1

LM OP b g N Q LM b sin ω t − cosω t + e OP 1 − 1 + b ω Nω Q u( t ) 1 a = sin ω n t − cos ω n t + e − at ust o 1 + a 2 ω 2n ω n 2

2 n

= 0.1

0.2

(a)

The response to each exponential function is given by an expression of the form in Eq. (c) of Problem 4.5. Combine the responses to the two components of p(t) to obtain the total response:

(b)

− bt

a / ω n = 0.5 0.4

(b) The force p(t) is plotted for b = 2a and three values of a ωn .

0.2

u( t )

p/po

(ust )

o -0.2

0.3

a / ω n = 0.05

t/Tn

-0.4 1

0.2

a / ω n = 0.1 a / ω n = 0.5

0.1

t/Tn

The motion given by Eq. (b) is plotted next for b = 2a and three values of a ω n = 0.05, 0.1, 0.5. a / ω n = 0.05

0.4

0.2

u( t ) (ust )

o -0.2

t/Tn

-0.4 1

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 12 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 4.7 The differential equation to be solved is t mu&& + ku = p( t ) = po tr

(a)

The complimentary and particular solutions are uc ( t ) = A cos ω nt + B sin ω nt up (t ) =

po t k tr

= ( ust )o

t tr

The complete solution is u ( t ) = A cos ω nt + B sin ω nt + ( ust )o

t tr

Differentiating Eq. (b) gives the velocity ( ust )o u& ( t ) = − ω n A sin ω nt + ω n B cos ω nt + tr

(b)

(c)

The constants A and B are determined from the initial conditions u( 0 ) = 0 ⇒ A = 0 u& ( 0 ) = 0 ⇒ ω n B +

( ust )o tr

= 0; B = −

( ust )o

ω ntr

Substituting A and B in Eq. (b) gives u(t ) = (ust ) o

FG t − sin ω t IJ Ht ω t K n

n r

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 13 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 4.8 u s = vt

Free body diagram k ( u s - ut) = k( vt- u)t

k ut m

m && ut

Dynamic equilibrium gives mu&&t − k ( vt − ut ) = 0

or u&&t + ω 2nut = ω 2nvt

(a)

The general solution of the differential equation is

ut ( t ) = uc ( t ) + u p ( t ) = A cos ω nt + B sin ω nt + vt (b) Impose initial conditions: ut ( 0 ) = 0 and u& t ( 0 ) = 0 : ut ( 0 ) = 0 ⇒ A = 0 u& t ( 0 ) = Bω n + v = 0 ⇒ B = −

ωn

Substituting for A and B in Eq. (b) gives u t (t ) = vt −

FG v IJ sin ω t Hω K

(c)

Equation (c) is written as

FG IJ H K

u t (t ) t 1 t = − sin 2π vTn Tn 2π Tn

(d)

and plotted in the accompanying diagram. 4 3

ut (t ) 2 vTn 1 0 0

t Tn

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 14 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 4.9

The equation of motion is

1.5

mu&& + cu& + ku = p(t )

(a)

(u ) st

The complementary solution is given by Eq. (f) in Derivation 2.2 in the book and the particular solution is u p = po k . Then the general solution is

g pk

u(t ) = e −ζω n t A cos ω D t + B sin ω D t +

1 o

0.5 0

0.2

0.4

0.6

0.8

(b)

When the system starts from rest, i.e. u(0) = u& (0) = 0 , the constants A and B are A=−

po k

B=−

po k

F ζ I GG J H 1 − ζ JK

(c)

Then Eq. (b) becomes

F GG H

u( t ) ζ sin ω D t = 1 − e −ζω n t cos ω D t + ust o 1−ζ 2

b g

I JJ K

The time t p when u(t) attains its maximum (or peak) value is determined by setting the velocity equal to zero: u& (t ) = e −ζω n t ust o

b g

Fζω I + ω J sin ω t = 0 GG JK H 1− ζ 2

or tp =

nπ

ωD

(e)

n = 0,1,2,3....

The maximum response occurs at t p given by Eq. (e) with n=1, i.e. t p = π ω D . Substituting this t p in Eq. (d) gives u F I bu g = 1 + expH −πζ 1 − ζ K o

st o

The relation between uo and ζ is shown in the following plot.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 15 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 4.10

The first integral in Eq. (c) is

z tr

Equation of motion:

R| p t 0 ≤ t ≤ t mu&& + ku = p(t ) = S t |T p t ≥ t o

(a)

mω n

po mω n t r

[ − (t − τ ) sin ω n (t − τ )

−

LM O χ sin ω χ dχ − t sin ω χ dχ P MN PQ 1 p R| 1 = S − χ cos ω χ + ω1 sin ω χ mω t |T ω U t + cos ω χ V ω W I p sin ω t 1 F t cos ω t − = + t − t cos ω t J G ω K mω t H F t − sin ω t IJ = (u ) G Ht ω t K po mω n t r

2 n

0 t

2 n

0 t

sin ω n t

ω 2n

n r

(b) (b) Solution for t ≥ tr Substituting p( τ ) in Eq. (4.2.4) and separating into two integrals gives

LM p u( t ) = τ sin ω (t − τ ) dτ mω M t N n

z t

po sin ω n (t − τ ) dτ

OP PQ

ωn

−

2 n

t cos ω n t

ωn

n r

0 t

(d) The second integral in Eq. (c) is

z t

po sin ω n (t − τ ) dτ =

ωn po

ωn

cos ω n χ t − t

1 − cos ω n (t − t r )

(e) Substituting Eqs. (d) and (e) in Eq. (c) gives po u( t ) = cos ω n ( t − t r ) mω n2 sin ω n ( t − t r ) − sin ω n t + + 1 − cos ω n (t − t r ) ω ntr

RS T

po t cos ω n (t − t r ) sin ω n (t − t r ) sin ω n t − − r + 2 ωn tr ωn ω 2n

= (ust ) o 1 −

t cos ω n ( t − t r )

2 n

st o

t − tr

OP LM PQ N p L sin ω (t − t ) − sin ω t O cos ω (t − t ) + = M PQ ω N ω t

Introducing χ ≡ t − τ and integrating the first term by parts gives 1

+ t sin ω n (t − τ )] dτ

u( t ) =

t − tr

po τ sin ω n (t − τ ) dτ tr

( χ − t ) sin ω n χ dχ

t − tr

O p LM = χ sin ω χ dχ − t sin ω χ dχ P PQ t M N R L cos ω χ OP U|V sin ω χ O p |L − χ cos ω χ = + + tM M P S t |MN ω PQ N ω Q |W T ω sin ω (t − t ) p L ( t − t ) cos ω (t − t ) t cos ω t = + + M− ω ω t MN ω t − tr

(a) Solution for 0 ≤ t ≤ tr Substituting p( τ ) in Eq. (4.2.4) gives u( t ) =

po p τ sin ω n (t − τ ) dτ = o tr tr

ω ntr

sin ω n t − sin ω n ( t − t r )

UV W

(f)

(c)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 16 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 4.11 p (t)

p (t)

p1 (t) = po t t r

= tr

p2 (t) = - po ( t - t r) t r

Equation (4.5.2) gives the response to a ramp function p( t ) = po ( t tr ) : u(t ) = (ust ) o

FG t − sin ω t IJ Ht ω t K n

(a)

n r

(a) Response for 0 ≤ t ≤ tr Because p2 ( t ) = 0 , the response is only due to p1 ( t ) , which is given by Eq. (a): u1 (t ) = (ust ) o

FG t − sin ω t IJ Ht ω t K n

(b)

n r

(b) Response for t ≥ tr The response to p1 ( t ) is given by Eq. (b) and that due to p2 ( t ) is obtained by replacing t in Eq. (a) by t − tr and po by − po : u2 (t ) = − (ust ) o

FG t − t − sin ω (t − t ) IJ (c) ω t H t K r

n r

The total response is the sum of Eqs. (b) and (c): u(t ) = u1 (t ) + u2 (t )

FG t − sin ω t − t + 1 + sin ω (t − t ) IJ ω t Ht ω t t K F 1 sin ω t − sin ω (t − t ) IJ = (u ) G1 − H ωt K n

= (ust ) o

st o

n r

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 17 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 4.12 System properties: m = w g = 100. 03 386 = 0. 2591 kip − sec2 in. k = 8. 2 kips in. Tn = 2 π

m = 1.12 sec k

Applied force: po = 50 kips ; (a) tr = 0. 2 sec , (b) tr = 4 sec

(a) tr Tn = 0. 2 1.12 = 0.179 The rise time of the force is relatively short, and the structure will “see” this excitation as a suddenly applied force (Fig. 4.5.3); therefore ⎛p ⎞ ⎛ 50 ⎞ u o ≈ 2(u st ) o = 2 ⎜ o ⎟ = 2 ⎜ ⎟ = 2(6.1) = 12.2in. ⎝ 8.2 ⎠ ⎝ k ⎠

(b) tr Tn = 4 1.12 = 3. 57 The rise time of the force is relatively long, and it will affect the structure like a static force (Fig. 4.5.3); therefore uo ≈ ( ust )o = 6.1 in.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 18 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 4.13

(a) Response results.

We have from Eq. (4.3.2)

u( t )

b gb gU| 0 ≤ t ≤ T 2 V u& (t ) = bu g ω sin ω t |W At t = T 2 , u = 2bu g and u& = 0 . u(t ) = ust o 1 − cos ω n t st o

(u ) st

(a)

-5

-10 0

st o

0.5

1.5

2.5 t/Tn

For Tn 2 ≤ t ≤ Tn , u(t ) = A1 cos ω n t + A2 sin ω n t − t = t − Tn 2 . Substituting

where

po k

b g

u(0) = 2 ust o

and

The displacement peaks un at the end of n half cycles of applied force are

u& (0) = 0 gives A1 = 3(ust )o and A2 = 0 . Hence

b g

g At t = T , i.e. at t = T 2 , u = -4bu g and u& = 0.

u(t ) = ust o −1 + 3 cos ω n t − Tn 2 , Tn 2 ≤ t ≤ Tn (b) n

st o

b g

st o

-4

-8

b g b g

u(0) = −4 ust o

Substituting

un n −1 = −1 2 n ust o

p u(t ) = A1 cos ω n t + A2 sin ω n t + o k t = t - Tn .

b g

In general,

For Tn ≤ t ≤ 3Tn 2

where

and

& 0) = 0 gives A1 = 5 ust o and A2 = 0 . Hence u(

b g

u(t ) = ust o 1 − 5 cos ω n t − Tn

g , T ≤ t ≤ 3T 2 (c) n

b g

At t = 3Tn 2 , i.e. at t = Tn 2 , u = 6 ust o and u& = 0.

In a similar manner the following results can be obtained:

b g

g , 3T 2 ≤ t ≤ 2T (d)

b g

g , 2T < t < 5T 2 (e)

u( t ) = ust o −1 + 7 cos ω n t − 3Tn 2

u(t ) = ust o 1 − 9 cos ω n t − 2Tn

(b) Response plot.

b g

From Eqs. (a), (b), (c), (d), and (e), u(t ) ust o is plotted

against

f a

t Tn .

f a f a

Note

that

ω n t - Tn 2 = 2π t Tn − 1 2 , ω n t - Tn = 2π t Tn − 1 , etc.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 19 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 4.14 p (t)

p (t)

+ t

- po

The response to p1 ( t ) is given by Eq. (4.7.2): po

(1 − cos ω nt ) t ≥ 0 (a) k To obtain the response to p2 ( t ) , this equation can be modified as follows: replace po by − po and t by t − td : u1 ( t ) =

u2 ( t ) = −

po k

[1 − cos ω n ( t − td )]

t ≥ td (b)

During 0 ≤ t ≤ td , p2 ( t ) = 0 and the response is given by Eq. (a): u(t ) =

po (1 − cos ω nt ) k

0 ≤ t ≤ td

(c)

For t ≥ td , the response is the sum of Eqs. (a) and (b): u( t ) =

po [cos ω n ( t − td ) − cos ω nt ] k

t ≥ td

(d)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 20 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 4.15

RS p T0

0 ≤ t ≤ td t ≥ td

p( t ) =

z t

u( t ) =

(a)

mω n

p(τ ) sin ω n (t − τ ) dτ

(b) For 0 ≤ t ≤ td , Eq. (b) after substituting for p ( t ) gives u( t ) =

z t

1 mω n

po sin ω n (t − τ ) dτ

LM N

OP Q

(1 − cos ω n t ) =

po (1 − cos ω n t ) k (c)

po 1 cos ω n (t − τ ) mω n ω n 0

mω n2

For t ≥ td , Eq. (b) after substituting for p( t ) gives

LM mω M N 1

u( t ) =

po sin ω n (t − τ ) dτ

0 ⋅ sin ω n (t − τ ) dτ

po = mω n

LM 1 cos ω (t − τ )OP Nω Q

OP PQ

p = o [cos ω n (t − t d ) − cos ω n t ] k

(d) The complete response is:

R| p (1 − cos ω t ) 0 ≤ t ≤ t u( t ) = S k |T pk [cos ω (t − t ) − cos ω t ] t ≥ t o o

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 21 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 4.16 The half-cycle sine pulse p( t ) is the sum of two sinusoidal excitations: p1 (t ) = po sin ω t starting at t = 0 and p2 (t ) = po sin ω t starting at t = td , where td = π td (Fig. 4.6.2b).

The response to p1 ( t ) is given by Eq. (3.1.6b): u1 (t ) =

po 1 sin ω t − (ω ω n ) sin ω n t k 1 − (ω ω n ) 2 (a)

To obtain the response to p2 ( t ) , in Eq. (a) replace t by t − td : u2 ( t ) =

k 1 − ( ω ω n )2

sin ω ( t − td ) − ( ω ω n ) sin ω n ( t − td )

(b) The response to the half cycle sine pulse is 0≤t ≤t d ⎧ u (t ) u (t ) = ⎨ 1 ⎩ u1 (t ) + u 2 (t ) t ≥ t d

These two equations can be shown to be equivalent to Eqs. (4.8.2) and (4.8.3), respectively.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 22 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 4.17 1. Determine the natural vibration period. With the base of the columns clamped, the lateral stiffness will be four times of the value computed in Example 4.1: k = 14. 93 kips in.

The natural period will be halved, i.e., Tn = 0. 25 sec

2. Determine Rd . td Tn

0. 2

= 0. 8 >

0. 25

1 2

Equation (4.7.12) gives Rd =

uo ( ust )o

= 2

3. Determine ( ust ) o . po 4 = = 0. 268 in. k 14. 92

( ust )o =

4. Determine the maximum dynamic deformation. uo = ( ust )o Rd = ( 0. 268) ( 2 ) = 0. 536 in.

5. Determine the maximum bending stress. uo M

M =

6 EI 2

uo =

LM 6 (30,000) (619. ) OP 0.536 MN (12 × 12) PQ 2

= 287.9 kip − in.

The bending stress is largest at the outside of the flange at the top and bottom of columns:

σ =

M S

287. 9 15. 2

= 18. 9 ksi

6. Effect of base fixity. For this excitation, the deformation as well as bending stress is reduced by clamping the columns at their base.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 23 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 4.18 1. Determine Rd . 0. 25 1 td = = Tn 0. 5 2

For this td tn , Eq. (4.8.12) gives Rd =

π 2

2. Determine the maximum dynamic deformation. uo = ( ust )o Rd =

po 5 π Rd = = 2.105 in. k 3. 73 2

3. Determine the maximumbending stress. uo M

M =

3EI 2

uo =

LM 3 (30,000) (619. ) OP 2.105 MN (12 × 12) PQ 2

= 565.7 kip − in.

σ =

565. 7 M = = 37. 2 ksi S 15. 2

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 24 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 4.19

The equation of motion to be solved is ⎧ po sin( 2πt / t d ) ⎪ mu&& + ku = p( t ) = ⎨ ⎪0 ⎩

t ≤ td

(a) t ≥ td

⎡ 2π ⎛ 2π ⎞ ⎤ t t t⎟ ⎥ sin ( − ) − sin ⎜ ⎢ d ⎝ Tn ⎠ ⎥⎦ 1 − (Tn / t d ) 2 ⎢⎣ Tn (Tn / t d )

2(Tn / t d ) sin(π t d / Tn ) 2

(Tn / t d ) − 1

with at rest initial conditions.

t ≥ td

1. Determine response u(t).

(d)

Case 2: td / Tn = 1

Case 1: td / Tn ≠ 1

Forced Vibration Phase.

The forced response is now given by Eq. (3.1.13b)

The response is given by Eq. (3.1.6b). Substituting ω = 2π / t d ,

2π t ⎤ u (t ) 1 ⎡ 2π t 2π t = ⎢sin − cos ⎥ Tn Tn Tn ⎦ (u st ) o 2 ⎣

ω n = 2π / Tn , and ( ust ) o = po / k , Eq. (3.1.6b) becomes

−

t ≤ td

(e)

⎡ ⎛ 1 u( t ) t ⎞ = ⎢sin ⎜ 2π ⎟ 2 ( ust ) o 1 − (Tn / t d ) ⎢⎣ ⎝ t d ⎠

Free Vibration Phase

⎛ Tn t ⎞⎤ sin ⎜ 2π ⎟ ⎥ t ≤ td td ⎝ Tn ⎠ ⎥⎦

(b)

From Eq. (e) determine u( t d ) = −π ( ust ) o

Free Vibration Phase. The motion is described by Eq. (4.7.3) with u(t d ) and u& (t d ) , determined from Eq. (b) : u( t d ) = ( ust ) o

⎡ ⎛ t 1 t d ⎞⎤ ⎟⎥ cos ⎢2π ⎜⎜ − ⎟ ⎢⎣ ⎝ Tn 2 Tn ⎠⎥⎦

and

u& (t d ) = 0

The second equation implies that the displacement in the forced vibration phase reaches its maximum at the end of this phase. Substituting Eq. (f) in Eq. (4.7.3) gives ⎛ t ⎞ u (t ) = −π cos 2π ⎜⎜ − 1⎟⎟ (u st ) o ⎝ Tn ⎠

⎛ Tn t ⎞ sin ⎜ 2π d ⎟ 2 t ⎝ Tn ⎠ 1 − (Tn / t d ) d −1

(c.1) 2π ⎡ ⎛ td ⎞ ⎤ u&( t d ) = ( ust ) o ⎢1 − cos⎜ 2π ⎟ ⎥ 2 ⎝ Tn ⎠ ⎥⎦ 1 − (Tn / t d ) t d ⎢⎣ 1

(c.2) Substituting Eq. (c) in Eq. (4.7.3) gives

(f)

t = −π cos 2π Tn

(g) t ≥ td

2. Plot response history. The time variation of the normalized deformation, u( t ) / ( ust ) o , given by Eqs. (b) and (d), is plotted in Fig. P4.19a for several values of td / Tn . For the special case of td / Tn = 1, Eqs. (e) and (g) describe the response of the system and these are also plotted in Fig. P4.19a. The static solution is included in these figures.

⎛ 2π ⎞ Tn ⎡ 1 2π u( t ) = t d ⎟ cos (t − t d ) ⎢− sin ⎜ 2 ( ust ) o 1 − (Tn / t d ) t d ⎢⎣ Tn ⎝ Tn ⎠ + sin

⎤ ⎛ 2π ⎞ 2π 2π ( t − t d ) − cos⎜ ( t − t d )⎥ t d ⎟ sin Tn Tn ⎝ Tn ⎠ ⎥⎦

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 25 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

t d Tn = 1 2

u( t ) / ( ust ) o

t d Tn = 1 8

1 0

-1

-2

-3

-3 0

0.05

0.1

0.15

0.2

0.25

0 -1

-2

-3

-3 0

0.5

t Tn

1.5

t d Tn = 3 4

0.1

0.2

0.3

0.4

0.5

t d Tn = 1.5

u( t ) / ( ust ) o

0.75

t d Tn = 1 4

u( t ) / ( ust ) o

t d Tn = 1

0.5

-1

-2

-3

-3 0

0.3

0.6

0.9

1.2

1.5

t Tn

Fig. P4.19a

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 26 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

t d Tn = 2.5

u( t ) / ( ust ) o

t d Tn = 2

-1

-2

-3

-3 0

t d Tn =3.5

t d Tn = 3

3 2 1

u( t ) / ( ust ) o

2 1 0

0 -1

-1 -2 -2 -3 0

-3 0

t d Tn = 5

t d Tn = 4

3 2

u( t ) / ( ust ) o

2 1 0

1 0 -1

-1 -2

-2

-3 0

t Tn

Fig. P4.19a (continued) © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 27 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

3. Determine maximum response. During the forced vibration phase, the number of local maxima and minima depends on td / Tn; the longer the pulse duration, more such peaks occur. These peaks occur at time instant t0 when u& (t ) = 0 . This condition applied to Eq. (b) gives 2π t 0 2π t 0 = cos cos td Tn (t 0 ) l =

l 1 + (t d / Tn )

l=1, 2, 3......

(h)

Only those l for which (t0)l < td Substituting Eq. (h) into Eq. (b) gives Rd =

are relevant.

uo 1 = * ( ust ) o 1 − (Tn / t d ) 2

⎡ ⎤ T 2π l 2π l − n sin ⎢sin ⎥ + 1 + ( / ) t T t T t 1 ( / ) d n d n d ⎦ ⎣

Similarly, the maximum response during free vibration can be determined from Eq. (g): Rd = π

(l)

The overall maximum response is the larger of the two maxima determined separately for the forced and free vibration phases. Fig. P4.19c shows that if td > Tn, the overall maximum is the largest peak that develops during the force pulse. On the other hand, if td < Tn, the overall maximum is given by the peak response during the free vibration phase. For the special case of td = Tn, as mentioned earlier, the two individual maxima are equal. The overall maximum response is plotted against td / Tn in Fig. P4.19d; for each td / Tn it is the larger of the two plots in Fig. P4.19c. This is the shock spectrum for the fullcycle sine pulse force.

(i)

At least one local maximum occurs during the force pulse, irrespective of the td / Tn value. If td / Tn > 1/2 the displacement reverses in sign during the excitation and has a negative value at the end of the excitation. If td / Tn > 1 a local minimum develops during the force pulse. If td / Tn > 2 more than one local maximum and/or more than one local minimum may develop. We define umax = max u( t ) t

umin = min u( t ) t

Figure P4.19b shows umax/(ust)o and −umin/(ust)o plotted as a function of td / Tn. The response spectrum for the larger of the two values during the force pulse is shown as ‘Forced Response’ in Fig. P4.19c. During the free vibration phase, the response is given by Eq. (d) and its amplitude is

Rd =

uo 2(Tn / t d ) sin (π t d / Tn ) = (u st ) o (Tn / t d ) 2 − 1

(j)

This equation is plotted in Fig. P4.19c. For the special case of td / Tn = 1, the maximum response during the forced vibration can be determined from Eq. (e): u& (t ) = 0 ⇒ t 0 = Tn and u( t0 ) = −π ⇒ Rd = π ( ust )o

(k)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 28 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

umax for t ≤ t d (ust )o

R d = u o / ( ust )o

−− 2

umin for t ≤ t d (ust )o

0 0

t d Tn Fig. 4.19b

R d = u o / ( ust )o

F o rc e d R e sp o n se 1

F re e R e sp o n se

0 0

t d Tn Fig. 4.19c

R d = u o / ( ust )o

2 O v e ra ll M a x im u m 1

0 0

t d Tn Fig. 4.19d © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 29 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 4.20

For t ≥ td , the total response is the sum of either Eqs. (a), (b) and (c) or Eqs. (c) and (d):

A symmetrical triangular pulse can be expressed as the superposition of three linear functions as shown.

u( t ) =

t t p3 (t) = 2 po - d td

LM N

2 sin ω n ( t − t d ) po 2(t − t d ) − ω ntd k td

p1 (t) = 2 po t td

p (t)

LM N

2 po 1 t 1 − 2 sin ω n ( t − t d 2) − sin ω n t + ω ntd k td

2 po k

OP Q

gOPQ

RS 1 [2 sin ω (t − t 2) − sin ω (t − t ) − sin ω t ]UV Tω t W n

n d

(e)

The total response can be summarized as

R| || t − 1 sin ω t 0 ≤t ≤t 2 t ω t | 2p | u( t ) = S1 − tt + ω 1t LMN2 sin ω FGH t − t2 IJK − sin ω t OPQ t 2 ≤ t ≤ t k | || 1 L2 sin ω FG t − t IJ − sin ω (t − t ) − sin ω t O PQ H 2K ||ω t MN t ≥t T n

t t p (t) = – 4 po - d 2 2 td

The response to p1 ( t ) is given by Eq. (4.4.2) with t r replaced by t d 2 : po k

u1 (t ) =

FG 2t − 2 sin ω t IJ ω t K Ht n

n d

t ≥ 0

n d

(a)

(f)

n d

which is the same as Eq. (4.9.1) because po k = ( ust )o and ω n = 2 π Tn .

The response to p2 ( t ) is given by Eq. (a) with t replaced by t − ( td 2 ) and po replaced by − 2 po :

LM N

n d

OP Q

2 po 2(t − t d 2) 2 sin ω n (t − t d 2) − k td ω ntd t t ≥ d (b) 2 The response to p3 ( t ) is obtained by replacing t by t − td in Eq. (a): u2 (t ) = −

LM N

po 2(t − t d ) 2 sin ω n (t − t d ) − k td ω n td

u3 (t ) =

OP t ≥ t Q

(c)

For 0 ≤ t ≤ td 2 , the total response is given by Eq. (a). For td 2 ≤ t ≤ td , the total response is the sum of Eqs. (a) and (b): u( t ) = u1 (t ) + u2 ( t ) =

LM N

po 2t 2 sin ω n t − k td ω n td −

LM N

OP Q

2 po 2(t − t d 2) 2 sin ω n ( t − t d 2) − k td ω ntd

RS T

OP Q

UV W

2 po 1 t 1 − + [ 2 sin ω n (t − t d 2) − sin ω n t ] k td ω ntd

(d)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 30 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 4.21

Rewriting in terms of t d Tn gives u(t )

au f

= cos

2π Tn

(t − t d ) +

FG T IJ sin 2π (t − t ) T 2π H t K 1

FG T IJ sin 2π t − T 2π H t K 1

(f)

t ≥ td

3. Response Plots.

b g

t td

The normalized deformation u(t ) ust o given by

Fig. P.4.21a

Eqs. (b) and (f) is plotted as a function of t Tn for t d Tn = 1/2 and 2 (Fig. P4.21 b-c).

Equation of motion:

t d Tn = 1 2

R| p bt t g 0 ≤ t ≤ t mu&& + ku = S |T0 t≥t o

ust ( t ) (ust )o d

(a)

u( t )

1. Forced Vibration Phase. The response is given by Eq. (4.4.2) with t r replaced by td : u( t )

au f st

t td

−

1 sin ω n t

0 o

-1

t Tn

t ≤ td

ωn

(u )

b g

FG T IJ sin 2π t Ht K T n

0 ≤ t ≤ td

(b)

The free vibration resulting from u(td ) and u& (t d ) is u(t ) = u(t d ) cos ω n (t − t d ) +

t d Tn = 2

ust ( t ) (ust )o

u& (t d )

ωn

sin ω n (t − t d )

(u ) st

0 o

-1

(c)

t Tn 1

From Eq. (b), u(td ) and u& (t d ) are determined:

F sin ω t IJ u ( t ) = bu g G 1 − H ωt K 1 u& (t ) = bu g b1 − cosω t g t st o

au f st

Fig. P4.21c

n d

(d)

= cos ω n ( t − t d ) −

sin ω n t

ω ntd

4. Response Spectrum. During the forced vibration phase u is a nondecreasing function of t. Thus, the maximum value of u during this phase can be found by evaluating Eq. (b) at t = td :

n d

Substituting Eq. (d) in Eq. (c) gives u( t )

n d

st o

u( t )

2. Free Vibration Phase.

Fig. P4.21b

Rewritting in terms of td / Tn gives 1 u( t ) t = − ust o t d 2π

sin ω n ( t − t d )

ω ntd t ≥ td

(e)

Rd =

u( t d ) T 2π t d = 1 − n sin ust o 2π t d Tn

b g

(g)

This equation is plotted in Fig. P4.21d. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 31 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Rd Overall maximum

Forced Response 1

t d Tn 1

t d Tn

Fig. P4.21f

Fig. P4.21d In the free vibration phase the response of the system is given by Eq. (c) with the amplitude

L u&(t ) OP u = u( t ) + M Nω Q 2

(h)

Substituting Eq. (d) and manipulating gives

⎡ ⎛ 2π t ⎞⎤ T T 2 ⎡ ⎛ π t ⎞⎤ d ⎟ ⎢1 − n sin ⎜ ⎥ + n ⎢sin ⎜ d ⎟⎥ ⎜ T ⎟⎥ ⎜ T ⎟ 2 2 ⎢ 2π t d ⎝ n ⎠⎦ π t d ⎢⎣ ⎝ n ⎠⎥⎦ ⎣

Rd =

(i) This equation is plotted in Fig. P4.21e. Rd

Free Response

Forced Response

0 1

t d Tn

Fig. P4.21e The overall maximum response is the larger of the maxima shown in Figs. P4.21d and e. This maximum is always given by Eq. (i) and is plotted in Fig. P4.21f to obtain the shock spectrum.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 32 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 4.22 The given excitation is expressed as the superposition of four linear functions shown. p (t)

u( t ) =

p4 (t) = po t - 3t 1 t1

UV W

po k

n 1

The response to p1 ( t ) is given by Eq. (4.2.2) with t r replaced by t1 : n

t ≥ 0

(a)

n 1

po k

FG t − t − sin ω (t − t ) IJ ω t H t K 1

n 1

t ≥ t1

or u( t ) =

RS T

po t 1 − sin ω n t − sin ω n (t − t1 ) 3 − k t1 ω n t1 − sin ω n (t − 2t1 )

For t ≥ 3t1, the response is the sum of Eqs. (a), (b), (c), and (d):

FG t − sin ω t IJ Ht ω t K p Ft − t sin ω (t − t ) I − − G JK ω t k H t p F t − 2t sin ω (t − 2t ) I − − G JK ω t k H t p F t − 3t sin ω (t − 3t ) I + − G JK ω t k H t

n 1

FG t − 3t − sin ω (t − 3t ) IJ ω t H t K 1

n 1

FG t − sin ω t IJ Ht ω t K n

n 1

t ≤ t1

t ≥ 3t1

n 1

For t ≤ t1, the response is given by Eq. (a): po k

n 1

(d)

u( t ) =

po k

u( t ) =

FG t − 2t − sin ω (t − 2t ) IJ t ≥ 2t ω t H t K 1

2 t1 ≤ t ≤ 3t1 (g)

(b) p u3 (t ) = − o k

n 1

This equation can be adapted to write the responses u2 ( t ) , u3 ( t ) and u4 ( t ) to p2 ( t ) , p3 ( t ) and p4 ( t ) , respectively: u2 (t ) = −

n 1

FG t − sin ω t IJ Ht ω t K

n 1

RS T

po k

FG t − sin ω t IJ Ht ω t K p Ft − t sin ω (t − t ) I − − G JK ω t k H t sin ω (t − 2t ) I p F t − 2t − − G JK ω t k H t

Fig. P4.22

u1 (t ) =

n 1

po 1 1 − sin ω n t − sin ω n (t − t1 ) ω n t1 k t1 ≤ t ≤ 2 t1 (f)

u( t ) =

p3 (t) = – po t - 2t 1 t1

p2 (t) = – po t - t 1 t1

For 2 t1 ≤ t ≤ 3t1 , the response is the sum of Eqs. (a), (b) and (c):

3t 1

FG t − sin ω t IJ − p FG t − t − sin ω (t − t ) IJ ω t Ht ω t K k H t K

or u( t ) =

p1 (t) = po t t1

po k

(e)

For t1 ≤ t ≤ 2 t1 , the response is the sum of Eqs. (a) and (b):

u(t ) =

FG H

po 1 − k ω n t1

IJ sin ω t − sin ω (t − t ) K n

− sin ω n (t − 2t1 ) + sin ω n (t − 3t1 )

t ≥ 3t1 (h)

The desired solution is given by Eqs. (e), (f), (g) and (h). © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 33 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 4.23 The response for t ≥ 3t1 is given by Eq. (h) of Problem 4.22. After some trigonometric and algebraic manipulation this can be rewritten as u( t ) =

FG H

ω t 3t po 4 sin ω n t1 sin n 1 sin ω n t − 1 2 2 k ω n t1

IJ K (a)

The response attains a maximum when

FG H

3t du = 0 ⇒ cos ω n t − 1 dt 2

IJ = 0 K

The first peak occurs at t = to given by

FG H

ω n to −

3t1 2

IJ = π ⇒ t = T + 3t K 2 4 2 n

(b)

where Tn = 2 π ω n . The maximum response is obtained by evaluating Eq. (a) at t o given by Eq. (b): uo =

k ω nt1

sin ω nt1 sin

ω nt1 2

(c)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 34 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 4.24 If td << Tn , the maximum deformation can be estimated by assuming that the force of Fig. P4.22 is a pure impulse. Its magnitude is

I =

p(t ) dt = 2 po

td 3

(a)

From Eq. (4.10.3) the maximum deformation is uo =

2 po t d 3mω n

mω n

(b)

Defining ( ust )o = po k , Eq. (b) becomes uo 4π = 3 (ust ) o

FG t IJ HT K d

(c)

A plot of Eq. (c) gives the shock spectrum: 4 3

uo (u st)o

2 1 0

0.2

0.4

0.6

0.8

t d Tn

For td Tn = 1 4 , Eq. (c) gives the approximate result: uo π = = 1. 047 ( ust )o 3

(d)

The exact value of the maximum response is given by Eq. (c) of Problem 4.23: uo ωt 4 = sin ω nt1 sin n 1 ( ust )o ω nt1 2

where

ω nt1 =

ω n td 3

π 2 π td 2π 1 = = 3 Tn 3 4 6

Therefore, uo π π 4 = sin sin = 0. 9885 ( ust )o π6 6 12

Error =

(e)

1. 047 − 0. 9885 × 100% = 5. 9% 0. 9885

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 35 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 4.25

u(t ) = (ust ) o

(a) During the first half of the excitation, p( t ) = po and the response is given by Eq. (4.3.2) where ( ust )o = po k : u( t ) = 1 − cos ω nt ( ust )o

0 ≤ t ≤ td 2

(a)

For the second half of the excitation, p( t ) = − po and the general solution is u ( t ) = A cos ω nt + B sin ω nt − ( ust )o td 2 ≤ t ≤ td

(b)

where t = t − td 2 ; A and B are to be determined from u ( t d 2 ) and u& ( t d 2 ) . Equation (a) gives u ( td 2 ) = ( ust )o 1 − cos ( ω n td 2 )

(c)

u& ( td 2 ) = ( ust )o ω n sin ( ω n td 2 )

(d)

Differentiating Eq. (b) gives u& ( t ) = − Aω n sin ω nt + Bω n cos ω nt

(e)

From Eqs. (b) and (c),

n d

t ≥ td

(l)

Equations (a), (h) and (l) give the desired results. (b) The maximum displacement during the free vibration phase is uo =

u( t d ) 2 +

LM u&(t ) OP Nω Q

d n

which after substituting Eqs. (j) and (k) and much trigonometric and algebraic manipulation becomes uo = 4 (ust ) o sin 2

FG ω t IJ = 4 (u ) sin FG π t IJ H 4 K H 2T K n d

st o

The corresponding deformation response factor is Rd ≡

FG IJ H K

π td uo = 4 sin 2 (ust ) o 2Tn

(m)

A plot of Eq. (m) gives the shock spectrum:

u ( t = 0 ) = A − ( ust )o = ( ust )o [1 − cos ( ω n td 2 )] ⇒ A = ( ust )o [ 2 − cos ω n td 2 ]

(f)

4 3

From Eqs. (d) and (e),

Rd 2

u& ( t = 0 ) = ω n B = ( ust )o ω n sin ( ω n td 2 ) ⇒ B = ( ust )o ω n sin ( ω n td 2 )

(g)

Substituting Eqs. (f) and (g) in Eq. (b) gives u(t ) = (ust ) o

m 2 cos (ω t 2) − cos ω t − 1 cos ω (t − t ) + sin ω t − 2 sin (ω t 2) sin ω (t − t ) r

m 2 − cos (ω t 2) cos ω (t − t 2) + sin (ω t 2) sin ω (t − t 2) − 1q n d

n d

td 2 ≤ t ≤ td

(h)

After the force ends at t d , the system vibrates freely and the response is u ( t ) = u ( td ) cos ω n ( t − td ) +

u& ( td )

ωn

1 0 0

t d Tn (c) If the excitation is considered as a pure impulse, its magnitude I = 0, implying zero response, a meaningless result. The pure impulse approximation is valid only for excitations which are a single pulse.

sin ω n ( t − td )

t ≥ td

(i)

This free vibration is initiated by u ( t d ) and u& ( t d ) determined from Eq. (h): u ( td ) = ( ust )o 2 cos ( ω n td 2 ) − cos ω ntd − 1

(j)

u& ( td ) = ( ust )o ω n sin ω ntd − 2 sin ( ω n td 2 )

(k)

Substituting Eqs. (j) and (k) in Eq. (i) gives

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 36 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 4.26 (a) From Example 2.7, weight w = 20. 03 kips (empty tank), Tn = 0. 5 sec , and ζ = 2. 75%. Therefore, td Tn = 0. 08 0. 5 = 0.16

Because td Tn < 0. 25 , the force may be treated as a pure impulse of magnitude I = 12 . kip − sec (see Example 4.2). Neglecting the effect of damping,

uo =

(12 . ) 2π I 2π . = = 184 k Tn 8.2 (0.5)

The equivalent static force is fSo = kuo = ( 8. 2 ) 1. 83 = 15. 08 kips

The resulting shear and bending moment diagrams for the tower are as shown: 15.08 kips

80 '

15.08 kips

1206 kip-ft

Shear

Moment

(b) The maximum responses of the tank for empty and full conditions are summarized next: Empty

Full

uo , in.

1.84

0.821

Vb , kip s

15.08

6.73

Mb , kip − ft

1206

538

Increasing the mass has the effect of reducing the dynamic response. This can be explained by observing that a pure impulse of magnitude I imparts an initial velocity u& (0) = I m which is smaller for the larger mass.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 37 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

CHAPTER 5 Problem 5.1 Over the time interval ti ≤ t ≤ ti +1 the excitation function is p( τ ) = p% i

0 ≤ τ ≤ Δti

(a)

where Δti ≡ ti +1 − ti and the equation to be solved is mu&& + ku = p% i

(b)

subject to the initial conditions u ( τ = 0 ) = ui and u& ( τ = 0 ) = u&i . The response u( τ ) over 0 ≤ τ ≤ Δti is the sum of two parts: (1) free vibration due to initial displacement ui and velocity u&i at τ = 0 ; and (2) response to step force p% i with zero initial conditions: u ( τ ) = ui cos ω n τ + u& ( τ )

u&i

ωn

= − ui sin ω n τ +

ωn

sin ω n τ +

u&i

ωn

p% i (1 − cos ω n τ ) k (c.1) p% i sin ω n τ (c.2) k

cos ω n τ +

Evaluate Eqs. (c) at τ = Δti or t = ti + Δti : ui +1 = ui cos (ω n Δti ) + u&i

LM sin (ω Δt ) OP N ω Q

~ p + i 1 − cos (ω n Δti ) k

u&i +1 = ui − ω n sin ( ω n Δti ) + u&i cos ( ω n Δti ) p% + i ω n sin ( ω n Δti ) k

(d.1)

(d.2)

Substituting p% i = ( pi + pi +1 ) 2 gives ui +1 = Aui + Bu&i + Cpi + Dpi +1

(e.1)

u&i +1 = A′ui + B′ u&i + C′pi + D′ pi +1

(e.2)

where A = cos ( ω n Δti ) B =

sin ( ω n Δti )

ωn

A′ = − ω n sin ( ω n Δti )

(f.1)

B′ = cos ( ω n Δti )

(f.2)

C =

1 − cos ( ω n Δti ) 2k

C′ =

ω n sin ( ω n Δti )

D =

1 − cos ( ω n Δti ) 2k

D′ =

ω n sin ( ω n Δti )

2k 2k

(f.3) (f.4)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 1 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 5.2 1. Initial calculations. Substituting ω n = 6. 283 , k = 10 and Δti = 0.1 in the equations for A, B , C , L, D′ in Problem 5.1 gives A = 0. 8090 B = 0. 09355 C = 0. 009550 D = 0. 009550

A′ = − 3. 6932

B′ = 0. 8090 C ′ = 0.1847 D′ = 0.1847

2. Apply the recurrence Eqs. (e) of Problem 5.1. The resulting computations are summarized in Table P5.2a and P5.2b, wherein the theoretical result is calculated from Eqs. (3.1.6b) and (4.7.3).

Table P5.2a: Numerical solution using piecewise constant interpolation of excitation ti

Cpi

Dpi +1

Bu&i

u&i

Aui

Theoretical ui

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

0.0000 5.0000 8.6602 10.0000 8.6603 5.0000 0.0000 0.0000 0.0000 0.0000 0.0000

0.0000 0.0477 0.0827 0.0955 0.0827 0.0477 0.0000 0.0000 0.0000 0.0000

0.0477 0.0827 0.0955 0.0827 0.0477 0.0000 0.0000 0.0000 0.0000 0.0000

0.0000 0.0864 0.2894 0.4682 0.4682 0.2030 – 0.2894 – 0.7575 – 0.9364 – 0.7575

0.0000 0.9233 3.0931 5.0047 5.0047 2.1698 – 3.0931 – 8.0978 – 10.0095 – 8.0979 – 3.0931

0.0000 0.0386 0.2067 0.5454 0.9642 1.2644 1.2257 0.7575 0.0000 – 0.7575

0.0000 0.0477 0.2554 0.6742 1.1918 1.5628 1.5151 0.9364 0.0000 – 0.9364 – 1.5151

0.0000 0.0333 0.2405 0.6789 1.2312 1.6364 1.6031 0.9907 0.0000 – 0.9907 – 1.6031

Table P5.2b: Numerical solution using piecewise constant interpolation of excitation ti

C ′pi

D′pi +1

A′ui

B′u&i

u&i

Theoretical u&i

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

0.0000 5.0000 8.6602 10.0000 8.6603 5.0000 0.0000 0.0000 0.0000 0.0000 0.0000

0.0000 0.9233 1.5992 1.8466 1.5992 0.9233 0.0000 0.0000 0.0000 0.0000

0.9233 1.5992 1.8466 1.5992 0.9233 0.0000 0.0000 0.0000 0.0000 0.0000

0.0000 – 0.1763 – 0.9434 – 2.4899 – 4.4016 – 5.7718 – 5.5955 – 3.4582 0.0000 3.4582

0.0000 0.0477 0.2554 0.6742 1.1918 1.5628 1.5151 0.9364 0.0000 – 0.9364 – 1.5151

0.0000 0.7470 2.5024 4.0489 4.0489 1.7554 – 2.5024 – 6.5513 – 8.0979 – 6.5513

0.0000 0.9233 3.0931 5.0047 5.0047 2.1698 – 3.0931 – 8.0978 – 10.0095 – 8.0979 – 3.0931

0.0000 0.9769 3.2727 5.2953 5.2953 2.2958 – 3.2727 – 8.5680 – 10.5906 – 8.5680 – 3.2727

Problem 5.3 Solution to this problem is available as Example 5.2 in the textbook.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 3 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 5.4 1.0 Initial calculations

Comparison with Problem 5.3 and theoretical solution:

m = 0. 2533 k = 10 c = 0.1592

Table P5.4b shows this comparison, where the smaller Δt is seen to give more accurate results.

u0 = 0 u&0 = 0 Δt = 0. 05 sec

1.1 u&&0 = ( p0 − cu&0 − ku0 ) m = 0 1.2 u−1 = u0 − u&0 Δt + u&&0 ( Δt )2 2 = 0 1.3 k$ = m ( Δt )2 + c 2 Δt = 102. 91 1.4 a = m ( Δt )2 − c 2 Δt = 99. 73 1.5 b = k − 2 m ( Δt )2 = − 192. 64 2.0 Calculation for each time step 2.1

p$ i = pi − aui −1 − bui

= pi − 99. 73ui −1 + 192. 64 ui

2.2 ui +1 = p$ i k$ = p$ i 102. 91 3.0 Repetition for next time step. Computational steps 2.1 and 2.2 are repeated for i = 0, 1, 2, 3, ... leading to Table P5.4a, wherein the theoretical result is also included.

Table P5.4a: Numerical solution by central difference method ti

ui−1

p$ i (Eq. 2.1)

0.00 0.0000 0.0000 0.0000 0.0000 0.05 2.5882 0.0000 0.0000 2.5882 0.10 5.0000 0.0000 0.0251 9.8448 0.15 7.0711 0.0251 0.0957 22.9915 0.20 8.6602 0.0957 0.2234 42.1576 0.25 9.6593 0.2234 0.4096 66.2938 0.30 10.0000 0.4096 0.6442 93.2417 0.35 9.6593 0.6442 0.9060 119.9551 0.40 8.6603 0.9060 1.1656 142.8465 0.45 7.0711 1.1656 1.3880 158.2206 0.50 5.0000 1.3880 1.5374 162.7449 0.55 2.5882 1.5374 1.5814 153.9036 0.60 0.0000 1.5814 1.4955 130.3811 0.65 0.0000 1.4955 1.2669 94.9169 0.70 0.0000 1.2669 0.9223 51.3267 0.75 0.0000 0.9223 0.4987 4.0973 0.80 0.0000 0.4987 0.0398 – 42.0695 0.85 0.0000 0.0398 – 0.4088 – 82.7204 0.90 0.0000 – 0.4088 – 0.8038 – 114.0761 0.95 0.0000 – 0.8038 – 1.1085 – 133.3771 1.00 0.0000 – 1.1085 – 1.2960 139.1210 © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This–publication

ui+1 (Eq. 2.2)

Theoretical ui+1

0.0000 0.0251 0.0957 0.2234 0.4096 0.6442 0.9060 1.1656 1.3880 1.5374 1.5814 1.4955 1.2669 0.9223 0.4987 0.0398 – 0.4088 – 0.8038 – 1.1085 – 1.2960 – 1.3518

0.0042 0.0328 0.1053 0.2332 0.4176 0.6487 0.9060 1.1605 1.3782 1.5241 1.5665 1.4814 1.2602 0.9245 0.5101 0.0593 – 0.3832 – 0.7751 – 1.0802 – 1.2718 – 1.3349

is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 4 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Table P5.4b ti

ui ( Δt = 0.1)

ui ( Δt = 0. 05)

ui (Theoretical)

0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00

0.0000 0.0000 0.1914 0.6293 1.1825 1.5808 1.5412 0.9141 – 0.0247 – 0.8968 – 1.3726

0.0000 0.0251 0.2234 0.6442 1.1656 1.5374 1.4955 0.9223 0.0398 – 0.8038 – 1.2960

0.0000 0.0328 0.2332 0.6487 1.1605 1.5241 1.4814 0.9245 0.0593 – 0.7751 – 1.2718

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 5 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 5.5 — and from Eq. (2.2.4) modified appropriately — valid for t ≥ 0. 6 sec .

1.0 Initial calculations m = 0. 2533 k = 10 c = 0. 6366

The response of systems with ζ = 0. 05 and ζ = 0. 20 is compared in the accompanying figure; numerical solution gives the peak displacement uo = 1. 5814 in. for ζ = 0. 05 and 1. 2947 in. for ζ = 0. 20 .

u0 = 0 u&0 = 0 Δt = 0. 05 sec

1.1 u&&0 = ( p0 − cu&0 − ku0 ) m = 0 1.2 u−1 = u0 − u&0 Δt + u&&0 ( Δt )2 2 = 0 1.3 k$ = m ( Δt )2 + c 2 Δt = 107. 69 1.4 a = m ( Δt )2 − c 2 Δt = 94. 95

1.5 b = k − 2 m ( Δt )2 = − 192. 64

2.0 Calculations for each time step 2.1

p$ i = pi − aui −1 − bui = pi − 94. 95ui −1 + 192. 64 ui

-1

ζ = 0.05 (Problem 5.4) ζ = 0.20 (Problem 5.5)

2.2 ui +1 = p$ i k$ = p$ i 107. 69 3.0 Repetition for the next time step. Computational steps 2.1 and 2.2 are repeated for i = 0, 1, 2, 3, ... leading to Table P5.5; also included is the theoretical result, calculated from Eq. (3.2.5) — valid for t ≤ 0. 6 sec

-2 0.0

0.2

0.4

0.6

0.8

1.0

Time (sec)

Table P5.5: Numerical solution by central difference method ti

ui−1

pi (Eq. 2.1)

ui+1 (Eq. 2.2)

Theoretical ui+1

0.00 0.0000 0.0000 0.0000 0.0000 0.0000 0.0041 0.05 2.5882 0.0000 0.0000 2.5882 0.0240 0.0313 0.10 5.0000 0.0000 0.0240 9.6300 0.0894 0.0984 0.15 7.0711 0.0240 0.0894 22.0161 0.2044 0.2133 0.20 8.6602 0.0894 0.2044 39.5535 0.3673 0.3744 0.25 9.6593 0.2044 0.3673 61.0036 0.5665 0.5704 0.30 10.0000 0.3673 0.5665 84.2527 0.7824 0.7822 0.35 9.6593 0.5665 0.7824 106.5884 0.9898 0.9854 0.40 8.6603 0.7824 0.9898 125.0455 1.1612 1.1530 0.45 7.0711 0.9898 1.1612 136.7796 1.2702 1.2593 0.50 5.0000 1.1612 1.2702 139.4247 1.2947 1.2826 0.55 2.5882 1.2702 1.2947 131.3981 1.2202 1.2087 1.2947 0.60 0.0000 1.2202 112.1188 1.0412 1.0362 1.2202 0.65 0.0000 1.0412 84.7074 0.7866 0.7888 1.0412 0.70 0.0000 0.7866 52.6710 0.4891 0.4980 0.7866 0.75 0.0000 0.4891 19.5312 0.1814 0.1957 0.4891 0.80 0.0000 0.1814 – 11.5039 – 0.1068 – 0.0889 0.1814 0.85 0.0000 – 0.1068 – 37.8014 – 0.3510 – 0.3316 – 0.1068 0.90 0.0000 – 0.3510 – 57.4792 – 0.5338 – 0.5152 – 0.3510 0.95 0.0000 – 0.5338 – 69.4929 – 0.6453 – 0.6298 © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication – 0.5338 1.00 0.0000 – 0.6453 – 73.6329 – 0.6838 0.6729 in a retrieval system, is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction,– storage

or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 6 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 5.6 2.2 ui +1 = p$ i k$ = p$ i 2. 518

1.0 Initial calculations m = 0. 2533 k = 10 c = 0.1592

3.0 Repetition for the next time step. Computational steps 2.1 and 2.2 are repeated for i = 0, 1, 2, 3, ... leading to Table P5.6; also included is the theoretical result, calculated from Eq. (3.2.5) — valid for t ≤ 0. 6 sec — and from Eq. (2.2.4) modified appropriately — valid for t ≥ 0. 6 sec .

u0 = 0 u&0 = 0 Δt = 1 3 sec

1.1 u&&0 = ( p0 − cu&0 − ku0 ) m = 0 1.2 u−1 = u0 − u&0 Δt + u&&0 ( Δt )2 2 = 0 2 1.3 k$ = m ( Δt ) + c 2 Δt = 2. 518

The central difference method gives meaningless results because Δt Tn = 1 3 exceeds the stability limit of 1 π.

1.4 a = m ( Δt ) − c 2 Δt = 2. 041 1.5 b = k − 2 m ( Δt )2 = 5. 441 2.0 Calculations for each time step 2.1

p$ i = pi − aui −1 − bui = pi − 2. 041ui −1 − 5. 441ui

Table P5.6: Numerical solution by central difference method ti

ui−1

pi (Eq. 2.1)

ui+1 (Eq. 2.2)

Theoretical ui+1

0.00 0 . 33 0 . 66 1.00 1. 33 1. 66 2.00

0.0000 9.8481 0.0000 0.0000 0.0000 0.0000 0.0000

0.0000 0.0000 0.0000 3.9104 – 8.4474 15.0796 – 25.7299

0.0000 0.0000 3.9104 – 8.4474 15.0796 – 25.7299 43.3626

0.0000 9.8481 – 21.2744 37.9773 – 64.7998 109.2072 – 183.4013

0.0000 3.9104 – 8.4474 15.0796 – 25.7299 43.3626 – 72.8227

0.8190 1.1595 – 1.2718 0.1998 0.8524 – 0.9262 0.1390

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 7 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 5.7 The solution to this problem is available as Example 5.3 in the book. Table P5.7 shows the results obtained by the average and central difference methods. For this problem, both methods give the peak response with roughly the same accuracy. Table P5.7 ti

0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00

u i ( Ex. 5.3)

ui (Ex. 5.2)

(ave. accel.)

(cent. diff.)

0.0000 0.0437 0.2326 0.6121 1.0825 1.4309 1.4230 0.9622 0.1908 -0.6043 -1.1441

0.0000 0.0000 0.1914 0.6293 1.1825 1.5808 1.5412 0.9141 -0.0247 -0.8968 -1.3726

u i (Theoretical)

0.0000 0.0328 0.2332 0.6487 1.1605 1.5241 1.4814 0.9245 0.0593 -0.7751 -1.2718

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 8 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 5.8 1.0 Initial calculations

2.4

m = 0.2533 k = 10 c = 0.1592 u0 = 0 u&0 = 0

]

3.0 Repetition for the next time step. Steps 2.1 through 2.4 are repeated for successive time steps and are summarized in Table P5.8a, wherein the theoretical result is also included.

p0 = 0

1.1

u&&0 = ( p0 − cu&0 − ku0 ) m = 0

1.2

Δt = 0.05 sec

1.3

a1 = 4 (Δt ) 2 m + (2 Δt ) c = 411.7

[

u&&i = 4 (Δt ) 2 (ui +1 − ui ) − (4 Δt ) u&i − u&&i

Table P5.8b shows that the smaller Δt gives a more accurate value of the peak response.

]

a 2 = (2 Δt ) m + c = 20.42

1.4

Table P5.8b

a3 = m = 0.2533

u i (Δt = 0.1)

ui (Δt = 0.05)

u i (Theoretical)

kˆ = k + a1 = 421.7

0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00

0.0000 0.0437 0.2326 0.6121 1.0825 1.4309 1.4230 0.9622 0.1908 -0.6043 -1.1441

0.0000 0.0356 0.2329 0.6389 1.1400 1.5002 1.4673 0.9360 0.0950 -0.7306 -1.2406

0.0000 0.0328 0.2332 0.6487 1.1605 1.5241 1.4814 0.9245 0.0593 -0.7751 -1.2718

2.0 Calculations for each time step 2.1

pˆ i +1 = pi +1 + a1ui + a2u&i + a3u&&i = pi +1 + 411.7u i + 20.42u& i + 0.2533u&&i

2.2

ui +1 = pˆ i +1 kˆ = pˆ i +1 421.7

2.3

u& i +1 = (2 Δt ) (u i +1 − u i ) − u& i

Table P5.8a: Numerical solution by average acceleration method ti

p̂i

u&&i

u&i

Theoretical u i

0.00 0.05

0.0000 2.5882

0.0000 9.8212

0.0000 0.2455

0.0000 0.0061

0.0000 0.0042

0.10

5.0000

15.0291

17.7448

0.9347

0.0356

0.0328

0.15

7.0711

45.3277

22.4527

1.9396

0.1075

0.1053

0.20

8.6603

98.2134

23.0598

3.0774

0.2329

0.2332

0.25

9.6593

174.2355

19.2215

4.1345

0.4132

0.4176

0.30

10.0000

269.4111

11.1778

4.8944

0.6389

0.6487

0.35

9.6593

375.4724

-0.2694

5.1672

0.8905

0.9060

0.40

8.6603

480.6894

-13.8430

4.8143

1.1400

1.1605

0.45

7.0711

571.1782

-27.9330

3.7699

1.3546

1.3782

0.50

5.0000

632.5507

-40.7761

2.0522

1.5002

1.5241

0.55

2.5882

651.7213

-50.6559

-0.2336

1.5457

1.5665

0.60

0.0000

618.6631

-56.1012

-2.9025

1.4673

1.4814

0.65

0.0000

530.5016

-46.2386

-5.4610

1.2582

1.2602

0.70

0.0000

394.6766

-32.2874

-7.4242

0.9360

0.9245

0.75

0.0000

225.5129

-15.6947

-8.6237

0.5348

0.5101

0.80

0.0000

40.0653

1.8857

-8.9689

0.0950

0.0593

0.85

0.0000

-143.5816

18.7562

-8.4529

-0.3405

-0.3832

0.90

0.0000

-308.0604

33.3378

-7.1505

-0.7306

-0.7751

0.95

0.0000

-438.3466

44.3164

-5.2092

-1.0396

-1.0802

1.00

0.0000

-523.1134

50.7592

-2.8323

-1.2406

-1.2718

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 9 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 5.9 1.0 Initial calculations m = 0. 2533 k = 10 c = 0.1592 p0 = 0 Δt = 0.05 sec

u0 = 0 u&0 = 0

1.1

u&&0 = ( p0 − cu&0 − ku0 ) m = 0

1.2

Δt = 1 / 3 sec

1.3

a1 = 4 (Δt ) 2 m + (2 Δt ) c = 10.07

[

ui +1 = pˆ i +1 kˆ = pˆ i +1 20.07

2.3

u&i +1 = (2 Δt ) (ui +1 − ui ) − u&i

2.4

Δu&&i = [4 (Δt ) 2 ] (ui +1 − ui ) − (4 Δt ) u&i − u&&i

3.0 Repetition for the next time step. Steps 2.1 through 2.4 are repeated for successive time steps and are summarized in Table P5.9; also included is the theoretical result, calculated from Eq. (3.2.5) —valid for t ≤ 0.6 sec—and from Eq. (2.2.4) modified appropriately—valid for t ≥ 0.6 sec.

]

a2 = (2 Δt ) m + c = 3.199 a3 = m = 0.2533

1.4

2.2

The numerical results are in large error but the solution is stable, unlike Problem 5.6.

kˆ = k + a1 = 20.07

2.0 Calculations for each time step 2.1

pˆ i +1 = pi +1 + a1ui + a2u&i + a3u&&i = pi +1 + 10.07ui + 3.199u&i + 0.2533u&&i

Table P5.9: Numerical solution by average acceleration method ti

p̂i

u&&i

u& i

Theoretical u i

0.00 0.3 3 0.6 6 1.00 1.3 3 1.6 6 2.00

0.0000 9.8481 0.0000 0.0000 0.0000 0.0000 0.0000

0.0000 9.8481 18.8315 -0.7161 -16.9765 2.1384 15.1732

0.0000 17.6612 -36.8730 4.9180 32.9321 -7.3413 -29.1534

0.0000 2.9435 -0.2584 -5.5842 0.7241 4.9892 -1.0932

0.0000 0.4906 0.9381 -0.0357 -0.8457 0.1065 0.7559

0.0000 0.8189 1.1595 -1.2718 0.1997 0.8525 -0.9262

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 10 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 5.10 The solution to this problem is available as Example 5.4 in the book. Table P5.10 compares the numerical results obtained from the average acceleration method (Example 5.3) and from the linear acceleration method (Example 5.4) with the theoretical results. The linear acceleration method gives more accurate results. Table P5.10 ti

0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00

u i ( Ex. 5.3)

u i (Ex. 5.4)

(ave. accel.)

(lin. Accel.)

0.0000 0.0437 0.2326 0.6121 1.0825 1.4309 1.4230 0.9622 0.1908 -0.6043 -1.1441

0.0000 0.0300 0.2193 0.6166 1.1130 1.4782 1.4625 0.9514 0.1273 -0.6954 -1.2208

u i (Theoretical)

0.0000 0.0328 0.2332 0.6487 1.1605 1.5241 1.4814 0.9245 0.0593 -0.7751 -1.2718

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 11 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 5.11 1.0 Initial calculations

3.0 Repetition for the next time step. Steps 2.1 through 2.4 are repeated for successive time steps and are summarized in Table P5.11a, wherein the theoretical result is also included.

m = 0. 2533 k = 10 c = 0.1592 u0 = 0 u&0 = 0

p0 = 0

Table P5.11b compares the numerical results using Δt = 0.05 and 0.1 sec, and the theoretical results. The smaller Δt gives more accurate results.

1.1

u&&0 = ( p0 − cu&0 − ku0 ) m = 0

1.2

Δt = 0.05 sec

1.3

a1 = 6 (Δt ) 2 m + (3 Δt ) c = 617.5

[

]

Table P5.11b

a2 = (6 Δt ) m + 2c = 30.71 a3 = 2m + (Δt 2 ) c = 0.5106

1.4

kˆ = k + a1 = 627.5

2.0 Calculations for each time step 2.1

pˆ i +1 = pi +1 + a1ui + a2u&i + a3u&&i = pi +1 + 617.5ui + 30.71u&i + 0.5106u&&i

2.2

ui +1 = pˆ i +1 kˆ = pˆ i +1 627.5

2.3

u&i +1 = (3 Δt ) (ui +1 − ui ) − 2u&i − (Δt 2 )u&&i

2.4

u&&i +1 = 6 ( Δt )2 (ui +1 − ui ) − (6 Δt ) u&i − 2u&&i

[

u i (Δt = 0.1)

u i (Δt = 0.05)

u i (Theoretical)

0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00

0.0000 0.0437 0.2326 0.6121 1.0825 1.4309 1.4230 0.9622 0.1908 -0.6043 -1.1441

0.0000 0.0322 0.2298 0.6406 1.1484 1.5124 1.4767 0.9318 0.0771 -0.7546 -1.2592

0.0000 0.0328 0.2332 0.6487 1.1605 1.5241 1.4814 0.9245 0.0593 -0.7751 -1.2718

]

Table P5.11a: Numerical solution by average acceleration method ti

p̂i

u&&i

u& i

Theoretical u i

0.00 0.05

0.0000 2.5882

0.0000 9.8995

0.0000 0.2475

0.0000 0.0041

0.0000 0.0042

0.10

5.0000

20.2029

17.8764

0.9419

0.0322

0.0328

0.15

7.0711

65.0086

22.5977

1.9537

0.1036

0.1053

0.20

8.6603

144.1786

23.1713

3.0980

0.2298

0.2332

0.25

9.6593

258.5228

19.2544

4.1586

0.4120

0.4176

0.30

10.0000

401.9626

11.0978

4.9174

0.6406

0.6487

0.35

9.6593

561.9174

-0.4782

5.1829

0.8955

0.9060

0.40

8.6603

720.5679

-14.1738

4.8166

1.1484

1.1605

0.45

7.0711

856.8575

-28.3544

3.7534

1.3656

1.3782

0.50

5.0000

949.0079

-41.2352

2.0137

1.5124

1.5241

0.55

2.5882

977.2661

-51.0841

-0.2943

1.5575

1.5665

0.60

0.0000

926.5688

-56.4231

-2.9820

1.4767

1.4814

0.65

0.0000

791.4030

-46.3047

-5.5502

1.2613

1.2602

0.70

0.0000

584.6771

-32.0666

-7.5095

0.9318

0.9245

0.75

0.0000

328.3373

-15.1958

-8.6910

0.5233

0.5101

0.80

0.0000

48.4058

2.6145

-9.0056

0.0771

0.0593

0.85

0.0000

-227.6315

19.6325

-8.4494

-0.3628

-0.3832

0.90

0.0000

-473.4981

34.2550

-7.1022

-0.7546

-0.7751

0.95

0.0000

-666.6023

45.1569

-5.1169

-1.0624

-1.0802

1.00

0.0000

-790.0855

51.4088

-2.7028

-1.2592

-1.2718

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 12 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 5.12 The governing equation is k$ ui +1 = p$ i

where k$ =

m c + = 102. 91 2 Δt ( Δt )2

and

LM m − c OP u − b f g + 2m u ( Δt ) N (Δt ) 2Δt Q = p − 99.730u − b f g + 202.64u

p$ i = pi −

i −1

S i

The solution steps are summarized in Table P5.12.

Table P5.12: Numerical solution by central difference method ti

( fS )i

ui−1

p$ i

u&i+1

ui+1

0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90

0.0000 2.5882 5.0000 7.0711 8.6602 9.6593 10.0000 9.6593 8.6603 7.0711 5.0000 2.5882 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000

0.0000 0.2515 0.9566 2.2341 4.0964 6.4418 7.5000 7.5000 7.5000 7.5000 7.5000 7.5000 7.5000 7.5000 7.5000 6.2816 4.4905 2.3184 – 0.0117

0.0000 0.0000 0.0000 0.0251 0.0957 0.2234 0.4096 0.6442 0.9060 1.1808 1.4583 1.7230 1.9553 2.1327 2.2317 2.2547 2.2042 2.0824 1.9033

0.0000 0.0000 0.0251 0.0957 0.2234 0.4096 0.6442 0.9060 1.1808 1.4583 1.7230 1.9553 2.1327 2.2317 2.2547 2.2042 2.0824 1.9033 1.6861

0.0000 2.5882 9.8448 22.9915 42.1576 66.2938 93.2417 121.5154 150.0748 177.3219 201.2263 219.4796 229.6683 232.0418 226.8420 214.3029 195.8701 173.5170 149.5368

0.0000 0.2515 0.9566 1.9826 3.1398 4.2077 4.9638 5.3659 5.5225 5.4227 4.9704 4.0964 2.7637 1.2207 – 0.2746 – 1.7237 – 3.0095 – 3.9632 – 4.5022

0.0000 0.0251 0.0957 0.2234 0.4096 0.6442 0.9060 1.1808 1.4583 1.7230 1.9553 2.1327 2.2317 2.2547 2.2042 2.0824 1.9033 1.6861 1.4530

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 13 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 5.13 The solution to this problem is available as Example 5.5 in the textbook.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 14 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 5.14 The solution to this problem is available as Example 5.6 in the textbook.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 15 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 5.15 Table P5.15 shows the results obtained by the linear acceleration method with Newton-Raphson iteration.

Table P5.15: Numerical solution by linear acceleration method with Newton-Raphson iteration R̂i( j )

(kT )(i j )

(k̂T )i or (k̂T ) (i j )

5.0000

166.756

R̂i or ti

0.00 0.10

0.0000 5.0000

(kT )i or

ui or ( j +1)

Δu ( j )

0.0300

0.0000 0.0300

( f S )(i j +1) 0.0000 0.2998

u&i

u&&i

0.0000 0.8995

0.0000 17.9904

0.20

8.6603

31.5748

166.756

0.1893

0.2193

2.1933

2.9819

23.6566

0.30

10.0000

66.2473

166.756

0.3973

0.6166

6.1660

4.7716

12.1372

0.40

8.6603

82.7769

166.756

0.4964

1.1130

7.5000

3.6300

156.756

0.0232

1.1362

7.5000

5.4366

1.1636

0.50

5.0000

82.4552

156.756

0.5260

1.6622

7.5000

4.8489

-12.9173

0.60

0.0000

61.0910

156.756

0.3897

2.0519

7.5000

2.6396

-31.2682

0.70

0.0000

17.3684

156.756

0.1108

2.1627

7.5000

-0.3919

-29.3628

0.80

0.0000

-28.6900

156.756

-0.1830

1.9797

5.6698

1.8302

166.756

0.0110

1.9906

5.7795

-2.9095

-20.9883

0.90

0.0000

-61.7237

166.756

-0.3701

1.6205

2.0781

-4.2360

-5.5417

1.00

0.0000

-70.6566

166.756

-0.4237

1.1968

-2.1590

-3.9624

11.0140

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 16 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 5.16 Table P5.16 shows the results obtained by the linear acceleration method with modified NewtonRaphson iteration.

Observe that more iterations are necessary copared to Problem 5.15.

Table 5.16: Numerical solution by linear acceleration method with modified Newton-Raphson iteration R̂i or

(kT )i or

R̂i( j )

(kT )i

(k̂T )i or (k̂T ) (i j )

( j)

ui or ( j)

( j +1)

0.00 0.10

0.0000 5.0000

5.0000

166.756

0.0300

0.0000 0.0300

0.20

8.6603

31.5748

166.756

0.1893

0.2193

0.30

10.0000

66.2473

166.756

0.3973

0.40

8.6603

82.7769

166.756

0.4964

Δu

( f S )(i j +1)

u&i

u&&i

0.0000 0.2998

0.0000 0.8995

0.0000 17.9904

2.1933

2.9819

23.6566

0.6166

6.1660

4.7716

12.1372

1.1130

7.5000

3.6300

0.0218

1.1348

7.5000

0.2177

1.305E-3

1.1361

7.5000

7.828E-5

1.1361

7.5000

5.4364

1.1606

0.50

5.0000

82.4513

0.0131 0

156.756

0.5260

1.6621

7.5000

4.8486

-12.9171

0.60

0.0000

61.0866

156.756

0.3897

2.0518

7.5000

2.6394

-31.2680

0.70

0.0000

17.3643

156.756

0.1108

2.1626

7.5000

-0.3922

-29.3627

0.80

0.0000

-28.6938

156.756

5.6695

-20.9856

-0.1830

1.9796

1.8305

0.0117

1.9912

5.7863

-0.1168

-7.449E-4

1.9905

5.7788

4.752E-5

1.9905

5.7793

-2.9096

0.90

0.0000

-61.7238

0.0074 10

166.756

-0.3701

1.6204

2.0779

-4.2359

-5.5409

1.00

0.0000

-70.6550

166.756

-0.4237

1.1967

-2.1592

-3.9622

11.0144

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 17 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

CHAPTER 6 Problem 6.1 We will solve this problem by the procedure described in Section 5.2 using piece-wise linear interpolation of u&&g (t ) with Δt = 0.02 sec . 1. Initial calculation. Δt = 0.02 sec .

ωD = ωn = π

0.0000 0.0200 0.0400 0.0600 0.0800 0.1000 0.1200 0.1400 0.1600 0.1800 0.2000

B ′ = 0.99803

C ′ = 9.9101× 10 −3

D ′ = 9.9967 × 10 −3 p i = − mu&&g (t i ) = −u&&g (t i )

and the resulting computations are summarized in Table P6.1a and P6.1b.

e −ζω n Δt = 1

3. Plot response history.

Substituting these in Table 5.2.1 of the book with m = 1 and k = ω n2 gives

A′ = −0.19726

In these equations,

sin ω D Δt = 0.062791 cos ω D Δt = 0.99803

A = 0.99803

D = 6.6654 × 10 −5

2. Apply recurrence Eq. (5.2.5).

Tn = 2 sec . ω n = π

ζ =0

C = 1.3328 × 10 −4

The deformation is plotted as a function of time in Fig. 6.4.1b for ζ = 0 .

B = 0.019987

Table P6.1a: Numerical solution using piece-wise linear interpolation of excitation pi Cp i Dp i +1 Bu& i u& i Au i 0.0000 -2.4318 -1.4050 -0.3821 -1.6521 -2.9259 -4.1958 -2.6325 -1.0692 0.4941 -1.4205

0.0000 -0.0003 -0.0002 -0.0001 -0.0002 -0.0004 -0.0006 -0.0004 -0.0001 0.0001 -0.0002

-0.0002 -0.0001 -0.0000 -0.0001 -0.0002 -0.0003 -0.0002 -0.0001 0.0000 -0.0001 -0.0002

0.0000 -0.0005 -0.0013 -0.0016 -0.0020 -0.0029 -0.0043 -0.0056 -0.0063 -0.0063 -0.0063

0.0000 -0.0243 -0.0626 -0.0801 -0.0998 -0.1445 -0.2141 -0.2798 -0.3133 -0.3142 -0.3174

0.0000 -0.0002 -0.0011 -0.0025 -0.0043 -0.0067 -0.0102 -0.0152 -0.0212 -0.0275 -0.0337

ui 0.0000 -0.0002 -0.0011 -0.0025 -0.0043 -0.0067 -0.0102 -0.0152 -0.0212 -0.0275 -0.0338

Table P6.1b: Numerical solution using piece-wise linear interpolation of excitation ti

C ′p i

D ′p i +1

B ′u& i

u& i

A′u i

0.0000 0.0200 0.0400 0.0600 0.0800 0.1000 0.1200 0.1400 0.1600 0.1800 0.2000

0.0000 -2.4318 -1.4050 -0.3821 -1.6521 -2.9259 -4.1958 -2.6325 -1.0692 0.4941 -1.4205

0.0000 -0.0243 -0.0140 -0.0038 -0.0165 -0.0292 -0.0419 -0.0263 -0.0107 0.0049 -0.0142

-0.0243 -0.0140 -0.0038 -0.0165 -0.0292 -0.0419 -0.0263 -0.0107 0.0049 -0.0142 -0.0333

0.0000 -0.0243 -0.0624 -0.0799 -0.0996 -0.1442 -0.2136 -0.2793 -0.3127 -0.3136 -0.3168

0.0000 -0.0243 -0.0626 -0.0801 -0.0998 -0.1445 -0.2141 -0.2798 -0.3133 -0.3142 -0.3174

0.0000 0.0000 0.0002 0.0005 0.0008 0.0013 0.0020 0.0030 0.0042 0.0054 0.0067

0.0000 -0.0002 -0.0011 -0.0025 -0.0043 -0.0067 -0.0102 -0.0152 -0.0212 -0.0275 -0.0338

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 1 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 6.2 A = 0.99803

We will solve this problem by the procedure described in Section 5.2 using piece-wise linear interpolation of u&&g (t ) with Δt = 0.02 sec .

C = 1.3297 × 10 −4

A′ = −0.19664

1. Initial calculation. Tn = 2 sec . ω n = π

B = 0.019924 D = 6.6549 × 10 −5

B ′ = 0.99177

C ′ = 9.9484 × 10 −3

Δt = 0.02 sec .

D ′ = 9.9758 × 10 −3

2. Apply recurrence Eq. (5.2.5).

ζ = 0.05

ω D = ω n 1 − ζ 2 = 3.1377

e −ζω n Δt = 0.99686

In these equations,

sin ω D Δt = 0.062712

cos ω D Δt = 0.99803

p i = − mu&&g (t i ) = −u&&g (t i )

and the resulting computations are summarized in Table P6.2a and P6.2b. 3. Plot response history.

Substituting these in Table 5.2.1 of the book with m = 1 and k = ω n2 gives

The deformation is plotted as a function of time in Fig. 6.4.1b for ζ = 5% .

Table P6.2a: Numerical solution using piece-wise linear interpolation of excitation ti

Cp i

Dp i +1

Bu& i

u& i

Au i

0.0000 0.0200 0.0400 0.0600 0.0800 0.1000 0.1200 0.1400 0.1600 0.1800 0.2000

0.0000 -2.4318 -1.4050 -0.3821 -1.6521 -2.9259 -4.1958 -2.6325 -1.0692 0.4941 -1.4205

0.0000 -0.0003 -0.0002 -0.0001 -0.0002 -0.0004 -0.0006 -0.0004 -0.0001 0.0001 -0.0002

-0.0002 -0.0001 -0.0000 -0.0001 -0.0002 -0.0003 -0.0002 -0.0001 0.0000 -0.0001 -0.0002

0.0000 -0.0005 -0.0012 -0.0016 -0.0020 -0.0028 -0.0042 -0.0055 -0.0061 -0.0061 -0.0061

0.0000 -0.0243 -0.0622 -0.0793 -0.0984 -0.1424 -0.2109 -0.2752 -0.3068 -0.3059 -0.3073

0.0000 -0.0002 -0.0011 -0.0025 -0.0042 -0.0066 -0.0101 -0.0150 -0.0209 -0.0270 -0.0331

0.0000 -0.0002 -0.0011 -0.0025 -0.0042 -0.0066 -0.0101 -0.0150 -0.0209 -0.0271 -0.0331

Table P6.2b: Numerical solution using piece-wise linear interpolation of excitation ti

C ′p i

D ′p i +1

B ′u& i

u& i

A′u i

0.0000 0.0200 0.0400 0.0600 0.0800 0.1000 0.1200 0.1400 0.1600 0.1800 0.2000

0.0000 -2.4318 -1.4050 -0.3821 -1.6521 -2.9259 -4.1958 -2.6325 -1.0692 0.4941 -1.4205

0.0000 -0.0242 -0.0140 -0.0038 -0.0164 -0.0291 -0.0417 -0.0262 -0.0106 0.0049 -0.0141

-0.0243 -0.0140 -0.0038 -0.0165 -0.0292 -0.0419 -0.0263 -0.0107 0.0049 -0.0142 -0.0333

0.0000 -0.0241 -0.0617 -0.0787 -0.0976 -0.1412 -0.2092 -0.2729 -0.3043 -0.3034 -0.3048

0.0000 -0.0243 -0.0622 -0.0793 -0.0984 -0.1424 -0.2109 -0.2752 -0.3068 -0.3059 -0.3073

0.0000 0.0000 0.0002 0.0005 0.0008 0.0013 0.0020 0.0030 0.0041 0.0053 0.0065

0.0000 -0.0002 -0.0011 -0.0025 -0.0042 -0.0066 -0.0101 -0.0150 -0.0209 -0.0271 -0.0331

u&&0 = 0 u −1 = 0 kˆ = 2507.9 a = 2942.1

Problem 6.3

b = −4990.1

We will solve this problem by the central difference method described in Section 5.3 with Δt = 0.02 sec .

We apply the recurrence Eq.s (2.1)-(2.3) in Table 5.3.1 of the book. In these equations, pi = −mu&&g (ti ) = −u&&g (ti ) and the resulting computations

Initial calculations.

Tn = 2 sec

ωn = π

ζ = 0.05

m =1

k = ω n2 = 9.8696

u0 = 0

u& 0 = 0

Δt = 0.02 sec

Calculations for each time step.

are summarized in Table P6.3.

c = 2ζω n = 0.3142

Plot response history The deformation is plotted as a function of time in Fig. P6.3. Note that it is essentially the same as that shown in Fig. 6.4.1b with ζ = 5% . 3.

Substituting these into Eq.s (1.1)-(1.5) in Table 5.3.1 of the book gives

Table P6.3: Numerical solution by central difference method ti

u i −1

p̂ i

u i +1

0 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20

0 -2.4343 -1.4065 -0.3825 -1.6538 -2.9289 -4.2002 -2.6352 -1.0703 0.4946 -1.4220

0 0 0 -0.0010 -0.0025 -0.0041 -0.0064 -0.0099 -0.0149 -0.0209 -0.0272

0 0 -0.0010 -0.0025 -0.0041 -0.0064 -0.0099 -0.0149 -0.0209 -0.0272 -0.0331

0 -2.4343 -6.2503 -10.4003 -16.1372 -24.7036 -37.3192 -52.3443 -68.1395 -83.0730 -99.0081

0 -0.0010 -0.0025 -0.0041 -0.0064 -0.0099 -0.0149 -0.0209 -0.0272 -0.0331 -0.0395

Deformation u, in

-10 0

Time, sec

Fig. P6.3 © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 3 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 6.4

Substituting Eqs. (f) and (g) in Eq. (e) and using

ω D = ω n 1 − ζ2 gives

The equation of motion to be solved is mu&& + cu& + ku = − mu&&g (t ); u&&g (t ) = u& goδ (t )

(a)

We have solved a related equation:

1 mω D

− ζω n t

sin ω Dt

= − = −

u&go

ωD u&go

ωD

− ζω n t

sin ω Dt

( − ζω n ) sin ω Dt + e

− ζω n t

(c)

− ζω n t

ω D cos ω Dt

ωD

− ζω n sin ω Dt + ω D cos ω Dt = 0

1 − ζ2

ωD = ζω n

tan −1

2π

u&goTn ;

V = 0. 8626 u&go ;

A =

2 π ( 0. 8626 ) Tn

u&go

(d)

u& go

1 0.5 0

exp − ζω nto sin ω Dto

1 − ζ

ζ=0 ζ = 0.1

1.5

(e)

1.5

∴ sin ω Dto =

0. 8626

F 1−ζ I ω t = tan G GH ζ JJK −1

2 π u&go

ωD

2π

A =

Tn ; V = u&go ;

F 1−ζ I GG ζ JJ H K

u&go

These spectra are plotted in the accompanying figure.

From Eq. (d) D o

FG 2π IJ D HT K

1 − ζ2

F 1 − ζ I OP GG ζ JJ P KQ H

For ζ = 0.1,

and the maximum response is uo = u ( to ) =

A =

D =

2π D Tn

tan −1

The maximum occurs at to =

V =

For ζ = 0 ,

or tan ω Dt =

LM MN

Tn exp −

u&go

2π

(b)

For maximum response, du dt = 0 : du

u& go

D =

Therefore solution to Eq. (a) is Eq. (b) multiplied by − mu&go : u(t ) = −

(h)

Response spectra:

and its solution is given by Eq. (4.1.7) specialized for τ = 0: u( t ) = h( t ) =

−1

p( t ) = δ ( t )

mu&& + cu& + ku = p( t );

LM ζ F 1 − ζ I OP u = exp − MN 1 − ζ tan GGH ζ JJK PQ ω u& go

(f)

u& go

0.8626 0.5 0

(g)

10 8

ωD t ζ

1− ζ

u& go

6 4 2 0

Tn © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 4 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 6.5

Thus t d must be longer than Tn / 4 for at least one peak to develop during 0 ≤ t ≤ t d .

1. Determine response to the first impulse. The ground motion impulse can be represented by the effective earthquake force, peff ( t ) = − mu&&g ( t ) = − mu& goδ ( t )

The response of the system to the first impulse is the unit impulse response of Eq. (4.1.6) times −mu& go : u1 ( t ) = −

u& go

sin ω n t

ωn

(a)

2. Determine response to second impulse. u2 ( t ) =

u& go

sin ω n ( t − t d )

ωn

t ≥ td

(b)

=−

u& go

ωn

sin

= sin

2π td Tn

The absolute 0 ≤ t ≤ t d is

(f)

maximum

⎧sin 2π t d / Tn uo ⎪ =⎨ u& go / ω n ⎪ ⎩1

deformation

during

t d / Tn ≤ 1 / 4

(g) t d / Tn ≥ 1 / 4

From Eq. (d), the peak deformation uo during t ≥ t d is given by

sin ω n t

ωn

u& go / ω n

6. Determine the peak response during t ≥ t d .

For 0 ≤ t ≤ t d : u( t ) = −

u (t d )

Equation (g) is plotted in Fig. P6.5e.

3. Determine response to both impulses.

u& go

If t d is shorter than Tn / 4 no peak will develop during 0 ≤ t ≤ t d and the response simply builds up from zero to u(td), where

2πt Tn

(c)

uo = 2 sin (π t d / Tn ) u& go / ω n

(h)

Equation (h) is plotted in Fig. P6.5e

For t ≥ t d : u( t ) = −

=−

u& go

ωn

u& go

ωn

[sin ω n t − sin ω n ( t − t d )]

2 sin

ω ntd 2

cos

ω n ( 2t − t d ) 2

2u& go ⎛ πt d ⎞ ⎡ ⎛ t 1 td ⎞ ⎤ ⎜ sin ⎟ cos ⎢2π ⎜ − ⎟⎥ Tn ⎠ ωn ⎝ ⎢⎣ ⎝ Tn 2 Tn ⎠ ⎥⎦

(d)

4. Plot displacement response. Equations (c) and (d) are plotted for t d / Tn = 1/8, 1/4, 1/2, and 1 in Figs. P6.5a, b, c, and d, respectively. 5. Determine the peak response during 0 ≤ t ≤ t d . The number of peaks in u(t) depend on t d / Tn ; the longer the time t d between the pulses, more such peaks occur. The first peak occurs at t o = Tn / 4 with the deformation uo given by uo =1 u& go / ω n

(e)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 5 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

t d Tn = 1 4

t d Tn = 1 8 2

First impulse

u1 ( t ) ÷ ( u& go ω n )

First impulse

-1

-2

-1

-2 0

0.5

1.5

0.5

1.5

Second impulse

u2 ( t ) ÷ ( u& go ω n )

Second impulse

-1

-2

-1

-2 0

0.5

1.5

0.5

Both impulses

-1

-2

1.5

Both impulses

u ( t ) ÷ ( u& go ω n )

-1

-2 0

0.5

1.5

0.5

t Tn

Fig. P6.5a

Fig. P6.5b

1.5

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 6 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

t d Tn = 1 2

-1

-2

-1

-2 0

0.5

1.5

Second impulse

-1

0.5

u2 ( t ) ÷ ( u& go ω n )

First impulse

u1 ( t ) ÷ ( u& go ω n )

t d Tn = 1

-2

1.5

Second impulse

-1

-2 0

0.5

1.5

0.5

1.5

u ( t ) ÷ ( u& go ω n )

Both impulses 2

-1

-2

Both impulses

-1

-2 0

0.5

1.5

0.5

t Tn

Fig. P6.5c

Fig. P6.5d

1.5

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 7 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Eq. (h)

1.5 2

Eq. (g) 1 1.5

V ÷ u& go

uo ÷ ( u& go ω n )

Thus, the pseudo-velocity response spectrum for td = 0.5 sec. is given by Eq. (i) with td = 0.5 sec. This is plotted in Fig. P6.5g

0.5

0 0

0.5

1.5

0.5

t d Tn 0 0

Fig. P6.5e

f n =1 Tn

7. Determine the overall maximum response. From Eqs. (g) and (h), the overall maximum response is given by ⎧ 2 sin(π td / Tn ) uo ⎪ 1 ⎧ =⎨ u&go / ωn ⎪max. of ⎨ ⎩2 sin(π td / Tn ) ⎩

Fig. P6.5.g

td / Tn ≤ 1/ 4 td / Tn ≥ 1/ 4

(i) Equation (i) is plotted t d / Tn in Fig. P6.5f to obtain the response spectrum. Overall maximum

uo ÷ ( u& go ω n )

1.5

0.5

0 0

0.5

1.5

t d Tn Fig. P6.5f 8. Determine the pseudo-velocity response spectrum. uo ω D ω u V = n = n o = u& go u& go u& go u& go ω n

(j)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 8 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 6.6

Thus t d must be longer than Tn / 4 for at least one peak to develop during 0 ≤ t ≤ t d .

1. Determine response to the first impulse. The ground motion impulse can be represented by the effective earthquake force, peff ( t ) = − mu&&g ( t ) = − mu& goδ ( t )

The response of the system to the first impulse is the unit impulse response of Eq. (4.1.6) times − mu& go : u1 ( t ) = −

u& go

ωn

sin ω n t

(a)

2. Determine response to second impulse. u2 ( t ) = −

u& go

ωn

sin ω n ( t − t d )

t ≥ td

(b)

=−

ωn

sin

ωn

= sin

2π t d Tn

(f)

The absolute deformation during 0 ≤ t ≤ t d is ⎧sin 2π t d / Tn uo ⎪ =⎨ u& go / ω n ⎪ ⎩1

t d / Tn ≤ 1 / 4

(g) t d / Tn ≥ 1 / 4

From Eq. (d), the peak deformation uo during t ≥ t d is given by

sin ω n t

u& go

u& go / ω n

6. Determine the peak response during t ≥ t d .

For 0 ≤ t ≤ t d : u( t ) = −

u (t d )

Equation (g) is plotted in Fig. P6.6e.

3. Determine response to both impulses.

u& go

If t d is shorter than Tn / 4 no peak will develop during 0 ≤ t ≤ t d and the response simply builds up from zero to u(td), where

2π t Tn

(c)

uo = 2 cos(π t d / Tn ) u& go / ω n

(h)

Equation (h) is plotted in Fig. P6.6e

For t ≥ t d : u( t ) = − =− =−

u& go

ωn u& go

ωn

[sin ω n t + sin ω n (t − t d ) 2 cos

ω ntd 2

sin

ω n ( 2t − t d ) 2

FG cos π t IJ sin LM2π FG t − 1 t IJ OP T K ω H MN H T 2 T K PQ

2u& go

(d)

4. Plot displacement response. Equations (c) and (d) are plotted for t d / Tn = 1/8, 1/4, 1/2, and 1 in Figs. P6.6a, b, c, and d, respectively. 5. Determine the peak response during 0 ≤ t ≤ t d . The number of peaks in u(t) depend on t d / Tn ; the longer the time t d between the pulses, more such peaks occur. The first peak occurs at t o = Tn / 4 with the deformation uo given by uo =1 &u go / ω n

(e)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 9 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

t d Tn = 1 4

t d Tn = 1 8 2

First impulse u1 ( t ) ÷ ( u& go ω n )

u1 ( t ) ÷ ( u& go ω n )

First impulse 1

-1

-2

-1

-2 0

0.5

1.5

Second impulse

0.5

1.5

u2 ( t ) ÷ ( u& go ω n )

Second impulse

-1

-2

-1

-2 0

0.5

1.5

Both impulses

-1

-2

0.5

u ( t ) ÷ ( u& go ω n )

1.5

Both impulses

-1

-2 0

0.5

1.5

0.5

t Tn

Fig. P6.6a

Fig. P6.6b

1.5

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 10 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

t d Tn = 1 2

t d Tn = 1

First impulse u1 ( t ) ÷ ( u& go ω n )

u1 ( t ) ÷ ( u& go ω n )

First impulse 1

-1

-2

-1

-2 0

0.5

1.5

0.5

Second impulse

Second impulse 1

u2 ( t ) ÷ ( u& go ω n )

1.5

-1

-2

-1

-2 0

0.5

1.5

0.5

1.5

Both impulses 2

u ( t ) ÷ ( u& go ω n )

Both impulses 1

-1

-2

-1

-2 0

0.5

t Tn

1.5

0.5

1.5

t Tn

Fig. P6.6c Fig. P6.6d

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 11 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Eq. (h)

1.5 2

Eq. (g)

1.5

V ÷ u& go

uo ÷ ( u& go ω n )

Thus, the pseudo-velocity response spectrum for td = 0.5 sec. is given by Eq. (i) with td = 0.5 sec. This is plotted in Fig. P6.6g.

0.5

0 0

0.5

1.5

0.5

t d Tn

0 0

Fig. P6.6e

f n =1 Tn

7. Determine the overall maximum response. From Eqs. (g) and (h), the overall maximum response is given by ⎧ 2 cos(π t d / Tn ) uo ⎪ 1 ⎧ =⎨ u& go / ωn ⎪max. of ⎨ ⎩2 cos(π t d / Tn ) ⎩

Fig. P6.6g

t d / Tn ≤ 1/ 4 t d / Tn ≥ 1/ 4

(i) Equation (i) is plotted t d / Tn in Fig. P6.6f to obtain the response spectrum. Overall maximum

uo ÷ ( u& go ω n )

1.5

0.5

0 0

0.5

1.5

t d Tn Fig. P6.6f 8. Determine the pseudo-velocity response spectrum.

ω D ω u uo V = n = n o = u& go u& go u& go u& go ω n

(j)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 12 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 6.7 (a) The equation of motion is ⎛ 2π t ⎞ ⎛ 2π t ⎞ mu&& + cu& + ku = − mu&&go sin ⎜ ⎟ = − ( peff )o sin ⎜ ⎟ ⎝ T ⎠ ⎝ T ⎠

(a) u&&go ω 2n

Substituting ( ust )o = ( peff )o k = and ω ω n = Tn T in Eq. (3.2.11) gives the peak deformation D =

u&&go

ω 2n

2 2

1 − ( Tn T )

+ 2ζ Tn T

(b)

The peak pseudo-acceleration A = ω 2n D is A = u&&go

1 1 − ( Tn T )

2 2

+ 2ζ Tn T

(c)

Substituting ω ω n = Tn T in Eq. (3.6.4) gives the true acceleration

(

)

2 ⎧ ⎫ 2 1 + 2ζ Tn T ⎪ ⎪ t u&&o = u&&go ⎨ ⎬ 2 2 2 ⎪ 1− (Tn T ) + [2ζ Tn T ] ⎪ ⎩ ⎭

[

]

(d)

(b) For ζ = 0, Eqs. (c) and (d) become identical: A = u&&go

1 1 − ( Tn T )

; u&&ot = u&&go

1 1 − ( Tn T )2

(e)

(c) Figure 3.2.6a with the abscissa ω ω n replaced by Tn T and the ordinate by A u&&go gives the pseudoacceleration response spectrum. Figure 3.5.1 with the abscissa ω ω n replaced by Tn T gives the trueacceleration response spectrum.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 13 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 6.8

1. Determine response u(t). Case 1: td / Tn ≠ 1 Forced Vibration Phase

The equation of motion to be solved is ⎧− mu&&go sin(2π t / t d ) t ≤ t d ⎪ mu&& + ku = peff (t ) = ⎨ ⎪0 t ≥ td ⎩

The response solution is adapted from Eq. (3.1.6b). Substituting ω = 2π / t d , ω n = 2π / Tn , po = mu&&go and including a minus sign in Eq. (3.1.6b) gives

(a)

u( t ) u&&go ω n2

with at rest initial conditions.

=−

⎡ ⎛ t ⎞ sin ⎜ 2π ⎟ t ⎝ 1 − (Tn / t d ) ⎢⎣ d ⎠ 1

2 ⎢

u&&g

−

u&&go

⎛ Tn t ⎞⎤ sin ⎜ 2π ⎟ ⎥ td ⎝ Tn ⎠ ⎥⎦

t ≤ td

(b)

Free Vibration Phase The motion is described by Eq. (4.7.3) with u(t d ) and u& (t d ) , determined from Eq. (b) : u( t d ) =

u& g 2u&&go

u&( t d ) =

td 2π

u&&go

ω n2 1 − (Tn / t d ) 2 u&&go

−1

ω n2 1 − (Tn / t d ) 2

⎛ Tn t ⎞ sin ⎜ 2π d ⎟ td ⎝ Tn ⎠

(c.1)

2π ⎡ ⎛ td ⎞ ⎤ ⎢1 − cos⎜ 2π ⎟ ⎥ t d ⎢⎣ ⎝ Tn ⎠ ⎥⎦

(c.2) Substituting Eq. (c) in Eq. (4.7.3) gives u( t )

u&&go ω n2 − sin

ug = u&&go

t d2 2π

Tn ⎡ ⎛ 2π ⎞ 2π t d ⎟ cos (t − t d ) ⎢sin ⎜ Tn 1 − (Tn / t d ) t d ⎢⎣ ⎝ Tn ⎠ 1

⎤ ⎛ 2π ⎞ 2π 2π ( t − t d ) + cos⎜ ( t − t d )⎥ t d ⎟ sin Tn Tn ⎝ Tn ⎠ ⎥⎦

⎛ 2π ⎞ ⎤ − (Tn / t d ) ⎡ 2π sin ( t − t d ) − sin ⎜ t⎟ ⎥ 2 ⎢ ⎝ Tn ⎠ ⎥⎦ 1 − (Tn / t d ) ⎢⎣ Tn 2(Tn / t d )sin(πt d / Tn ) 1 − (Tn / t d ) 2

⎡ ⎛ t 1 td ⎞ ⎤ cos ⎢2π ⎜ − ⎟⎥ ⎢⎣ ⎝ Tn 2 Tn ⎠ ⎥⎦

t ≥ td

(d)

Case 2: td / Tn = 1

Forced Vibration Phase The forced response is now given by Eq. (3.1.13b) u( t ) u&&go ω n2

1 ⎡ 2πt 2πt 2πt ⎤ = − ⎢sin − cos ⎥ t ≤ td 2⎣ Tn Tn Tn ⎦

(e)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 14 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Free Vibration Phase From Eq. (e) determine u( t d ) = π

u&&go

ω n2

and

u&(t d ) = 0

(f)

The second equation implies that the displacement in the forced vibration phase reaches its maximum at the end of this phase. Substituting Eq. (f) in Eq. (4.7.3) gives u( t ) u&&go ω n2

⎞ ⎛ t = π cos 2π ⎜ − 1⎟ ⎠ ⎝ Tn = π cos 2π

t Tn

t ≥ td

(g)

2. Plot response history. The time variation of the normalized deformation, u( t ) ( u&&go ω n2 ) , given by Eqs. (b) and (d), is plotted in Fig. P6.8a for several values of td / Tn . For the special case of td / Tn = 1, Eqs. (e) and (g) describe the response of the system and these are also plotted in Fig. P6.8a. The static solution is included in these figures.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 15 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

t d Tn = 1 4

ω n2 u ( t ) / u&& go

t d Tn = 1 8

1 0 -1

1 0 -1 -2

-2

-3

-3 0

0.05

0.1

0.15

0.2

0.25

0.1

ω n2 u ( t ) / u&& go

0.3

0.4

0.5

1.2

1.5

t d Tn = 3 4

t d Tn = 1 2

0 -1

0 -1 -2

-2

-3

-3 0

0.25

0.5

0.75

t d Tn = 1

0.3

0.6

0.9

t d Tn = 1.5

ω n2 u ( t ) / u&& go

0.2

0 -1 -2

-3

-3 0

0.5

t Tn

1.5

t Tn

Fig. P6.8a

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 16 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

t d Tn = 2.5

ω n2 u ( t ) / u&& go

t d Tn = 2

1 0 -1

-2

-3

-3 0

1 0 -1 -2

-3

-3 0

t d Tn = 4

t d Tn = 5

ω n2 u ( t ) / u&& go

t d Tn =3.5

ω n2 u ( t ) / u&& go

t d Tn = 3

1 0 -1

1 0 -1 -2

-2

-3

-3 0

t Tn

Fig. P6.8a (continued)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 17 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

During the free vibration phase, the response is given by Eq. (d) and its amplitude is uo; using Eq. (h) gives

3. Determine maximum response. The pseudo acceleration A = ω n2 uo

(h)

will be determined from u o ≡ max u (t ) . t

2π t0 2π t0 = cos td Tn

(t 0 ) l =

l 1 + (t d / Tn )

(i)

1 A * = u&&go 1 − (Tn / t d ) 2 ⎡ ⎤ T 2π l 2π l − n sin ⎢sin ⎥ 1 ( / ) + + t T t 1 ( T / t ) d n d n d ⎦ ⎣

(j)

At least one local maximum occurs during the ground acceleration pulse, irrespective of the td / Tn value. If td / Tn > 1/2 the displacement reverses in sign during the excitation and has a negative value at the end of the excitation. If td / Tn > 1 a local minimum develops during the ground acceleration pulse. If td / Tn > 2 more than one local maximum and/or more than one local minimum may develop. We define t

u& (t ) = 0 ⇒ t 0 = Tn and u&&go

Only those l for which (t0)l < td are relevant. Substituting Eq. (i) into Eq. (b) and using Eq. (h) gives

umax = max u( t )

For the special case of td / Tn = 1, the maximum response during the forced vibration can be determined from Eq. (e):

ω n2 u( t 0 )

l=1, 2, 3......

(k)

This equation is plotted in Fig. P6.8c.

During the forced vibration phase, the number of local maxima and minima depends on td / Tn ; the longer the pulse duration, more such peaks occur. These peaks occur at time instants t0 when u& (t ) = 0 . This condition applied to Eq. (b) gives cos

2(Tn / t d ) sin (π t d / Tn ) A = u&&go (Tn / t d ) 2 − 1

umin = min u( t ) t

Figure P6.8b shows umax/(ust)o and −umin/(ust)o plotted as a function of td / Tn. The response spectrum for the larger of the two values during the acceleration pulse is shown as ‘Forced Response’ in Figure P6.8c.

=π ⇒

A =π u&&go

(l)

Similarly, the maximum response during free vibration can be determined from Eq. (g): A =π u&&go

(m)

The overall maximum response is the larger of the two maxima determined separately for the forced and free vibration phases. Figure P6.8c shows that if td > Tn , the overall maximum is the largest peak that develops during the force pulse. On the other hand, if td < Tn , the overall maximum is given by the peak response during the free vibration phase. For the special case of td = Tn , as mentioned earlier, the two individual maxima are equal. The overall maximum response is plotted against td / Tn in Fig. P6.3d; for each td / Tn it is the larger of the two plots in Fig. P6.8c. This is the pseudo acceleration response spectrum for the full-cycle sine pulse ground motion. 4. True Acceleration Response Spectrum. As shown in Section 6.3, for undamped systems: u&&ot = ω n2 uo

Thus the true acceleration response spectrum is also given by Fig. P6.8b-d with the ordinate axis showing u&&ot u&&go .

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 18 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

u m ax for t ≤ t d ( u st ) o

A u&&go = u&&ot u&&g o

−− 2

u m in ( u st ) o

for t ≤ t d

0 0

td Tn

Fig. P6.8b

A u&&g o = u&&o u&&g o

Forced Response

Free Response 0 0

t d Tn

Fig. P6.8c

A u&&g o = u&&ot u&&g o

Overall Maximum 2

0 0

Fig. P6.8d

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 19 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

1. Determine response u(t).

Problem 6.9

Case 1: td / Tn ≠ 1

The equation of motion to be solved is ⎧− mu&&go cos(2π t / t d ) t ≤ t d ⎪ mu&& + ku = peff (t ) = ⎨ (a) ⎪0 ≥ t t d ⎩

Forced Vibration Phase

with at rest initial conditions.

u&&go ω n2

The response solution of Eq. (a) is u (t )

⎛ t ⎞ ⎟ cos⎜⎜ 2π 1 − (Tn / t d ) ⎢⎣ ⎝ Tn ⎟⎠

u&&go

(b)

Free Vibration Phase

The motion is described by Eq. (4.7.3) with u(t d ) and u&(t d ) determined from Eq. (b) : u (t d ) =

t u& (t d ) =

u& g td 2π

u&&go

ω n2

⎡

⎛ t ⎞ ⎤ cos⎜⎜ 2π d ⎟⎟ − 1⎥ 1 − (Tn / t d ) ⎣⎢ ⎝ Tn ⎠ ⎦⎥

2 ⎢

⎛ t ⎞ sin ⎜⎜ 2π d ⎟⎟ ω n 1 − (Tn / t d ) ⎝ Tn ⎠

u&&go

−1

(c.1)

(c.2)

Substituting Eq. (c) in Eq. (4.7.3) gives u (t )

= 2

u&&go ω n

⎧⎪⎡ ⎛ 2π ⎞ ⎤ 2π t d ⎟⎟ − 1⎥ cos cos⎜⎜ (t − t d ) 2 ⎨⎢ Tn 1 − (Tn / t d ) ⎪⎩⎢⎣ ⎝ Tn ⎠ ⎥⎦ 1

⎫⎪ ⎛ 2π ⎞ 2π − sin ⎜⎜ t d ⎟⎟ sin (t − t d )⎬ Tn ⎪⎭ ⎝ Tn ⎠

−1

⎡

2 ⎢

1 − (Tn / t d ) ⎣

=− ⎛t ⎞ 2u&&go ⎜⎜ d ⎟⎟ ⎝ 2π ⎠

2 ⎢

⎛ t ⎞⎤ ⎟⎥ t ≤ td − cos⎜⎜ 2π ⎟ ⎝ t d ⎠⎦⎥

u&&g

u&&go

⎡

− cos

⎤ 2π 2π t d + cos (t − t d )⎥ Tn Tn ⎦

⎡ ⎛ t 1 t ⎞⎤ sin ⎢2π ⎜⎜ − d ⎟⎟⎥ 1 − (Tn / t d ) ⎢⎣ ⎝ Tn 2 Tn ⎠⎥⎦

2 sin(π t d / Tn ) 2

t ≥ td

(d)

Case 2: td / Tn = 1

Forced Vibration Phase The forced response is

πt 2π t u (t ) = − sin Tn Tn &u&go ω n2

t ≤ td

(e)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 20 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Free Vibration Phase From Eq. (e) determine u (t d ) = 0

u& (t d ) = −

and

u&&go t d 2

(f)

The second equation implies that the displacement in the forced vibration phase reaches its maximum at the end of this phase. Substituting Eq. (f) in Eq. (4.7.3) gives u (t ) u&&go ω n2

⎛ t ⎞ = −π sin 2π ⎜⎜ − 1⎟⎟ T ⎝ n ⎠

= −π sin 2π

t Tn

t ≥ td

(g)

2. Plot response history. The time variation of the normalized deformation, u( t ) ( u&&go ω n2 ) , given by Eqs. (b) and (d), is plotted in Fig. P6.9a for several values of td / Tn. For the special case of td / Tn = 1, Eqs. (e) and (g) describe the response of the system and these are also plotted in Fig. P6.9a. The static solution is included in these figures.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 21 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

ω n2 u ( t ) / u&& go

t d Tn = 1 8

t d Tn = 1 4

−1

−2

−3 0

0.05

0.1

0.15

0.2

0.25

−3 0

ω n2 u ( t ) / u&& go

t d Tn = 1 2

0.1

0.2

0.3

0.4

0.5

t d Tn = 3 4

−1

−2

−2 −3

−3 0

0.2

0.4

0.6

0.8

0.5

t d Tn = 1

1.5

t d Tn = 1.5 4

ω n2 u ( t ) / u&& go

−1

−1 −2

−2

−3

−4 0

0.5

1.5

t Tn

Fig. 6.9a © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 22 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

ω n2 u ( t ) / u&& go

t d Tn = 2

t d Tn = 2.5

−1

−2

−3

−3 0

ω n2 u ( t ) / u&& go

t d Tn = 3

t d Tn = 3.5

−1

−2

−2 −3

−3 0

t d Tn = 4

ω n2 u ( t ) / u&& go

t d Tn = 5

−1

−2

−3

−3 0

t Tn

Fig. 6.9a (cont.) © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 23 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Thus the true acceleration response spectrum is also given by Figs. P6.9b-d with the ordinate axis showing u&&ot u&&go .

3. Determine maximum response. The pseudo acceleration A = ω n2 uo

(h)

will be determined from u o ≡ max u (t ) .We define t

u max = max u (t )

u min = min u (t ) t

Figure P6.9b shows umax/(ust)o and −umin/(ust)o plotted as a function of td / Tn. The response spectrum for the larger of the two values during the acceleration pulse is shown as “Forced Response” in Fig. P6.9c. During the free vibration phase, the response is given by Eq. (d) and its amplitude is uo; using Eq. (h) gives 2 sin(πt d / Tn ) A = u&&go (Tn / t d ) 2 − 1

(i)

This equation is plotted in Fig. P6.9c. For the special case of td / Tn = 1, the maximum response during the forced vibration can be determined from Eq. (e): u& (t ) = 0 ⇒ t 0 = 3Tn / 4 and

ω n2 u (t 0 ) u&&go

= 3π / 4 ⇒

A = 3π / 4 &u&go

(j)

Similarly, the maximum response during free vibration can be determined from Eq. (g): A =π u&&go

(k)

The overall maximum response is the larger of the two maxima determined separately for the forced and free vibration phases. The overall maximum response is plotted against td / Tn in Fig. P6.9d; for each td / Tn it is the larger of the two plots in Fig. P6.9c. This is the pseudo-acceleration response spectrum for the full-cycle cosine pulse ground motion. 4. True acceleration response spectrum. As shown in Section 6.3, for undamped systems: u&&ot = ω n2 uo

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 24 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

u max for t ≤ t d (u st ) o

t A /&u& go =&u& go /&u& go

−u min for t ≤ t d (u st ) o

3 2 1 0 0

t d Tn Fig. 6.9b

t A /&u& go =&u& go /&u& go

Forced response 3

Free response

2 1 0 0

t d Tn Fig. 6.9c

t A /&u& go =&u& go /&u& go

4 3

Overall maximum

2 1 0 0

t d Tn Fig. 6.9d

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 25 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 6.10 The lateral stiffness of the SDF system is k =

3 EI 3 ( 29 ⋅ 103 ) 28.1 = = 1. 415 kips in. L3 (10.12 )3

The total weight of the pipe is 18. 97 × 10 = 189. 7 lbs , which may be neglected relative to the lumped weight. Thus w 3 = = 0. 0078 kip − sec2 in. 386. 4 g

m =

The natural frequency and period are

ωn =

k m = 13. 47 rads sec

Tn = 2 π ω n = 0. 47 sec

At Tn = 0. 47 sec the response spectrum curve for ζ = 5% gives A = 0. 9 g . The peak value of deformation is uo =

ω n2

0. 9 ( 386 ) = 1. 91 in. (13. 47 )2

The peak value of the equivalent static force is

FG A IJ = 3 × 0.9 = 2.7 kips H gK

f So = mA = w

The bending moment diagram increases from zero at the top to the maximum moment at the base: M = 2. 7 × 10 = 27 kip − ft

The maximum bending stress is

σ =

Mc ( 27 × 12 ) ( 6. 625 2 ) = = 38. 2 ksi 28.1 I

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 26 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 6.11 For each system we compute ω n = k m and Tn = 2 π ω n . For the computed Tn and ζ = 5% we read one of D, V or A from the design spectrum of Fig. 6.9.4 multiplied by 0.50; and compute the other two among D, V and A. The peak deformation uo = D and the base shear is Vbo = ( A g ) w . These results are summarized in the accompanying table. Syste m

(a) (b) (c)

ωn

sec-1

sec

in./sec

in.

3.93 5.56 3.93

1.60 1.13 1.60

55.2 55.2 55.2

14.1 9.93 14.1

A/g

Vbo

kips 0.562 0.795 0.562

56.2 79.5 112.4

Comparing the response of systems (a) and (b), we observe that stiffening the tower shortens the natural period and reduces the design deformation, but increases the base shear; the latter is a disadvantage. Comparing systems (a) and (c), both have the same natural period and spectral ordinates. However, system (c) has twice the mass of system (a), which doubles the base shear. The preceding comments are restricted to systems in the constant-V region of the spectrum.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 27 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 6.12 For each system we compute ω n = k m and Tn = 2π ω n . For the computed Tn and ζ = 5% we read A from the design spectrum of Fig. 6.9.5 multiplied by 0.50 and compute D and V. The peak deformation uo = D and the base shear is Vbo = A g w . The results are summarized in the accompanying table.

b g

System

ωn

D in.

A/g

Vbo

sec-1

sec

in./sec

(a)

9.82

0.64

53.24

5.42

1.355

21.7

(b)

13.89

0.45

37.79

2.71

1.355

21.7

(c)

9.82

0.64

53.24

5.42

1.355

43.4

kips

Comparing the response of systems (a) and (b), we observe that stiffening the tower shortens the natural period and reduces the design deformation; however the base shear remains unchanged. Comparing systems (a) and (c), both have the same natural period and spectral ordinates. The deformations of the two systems are the same. However, system (c) has twice the mass of system (a), which doubles the base shear. The preceding comments are restricted to systems in the constant-A region of the spectrum.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 28 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 6.13 For each system we compute ω n = k m and Tn = 2π ω n . For the computed Tn and ζ = 5% we read A from the design spectrum of Fig. 6.9.5 multiplied by 0.50 and compute D and V. The peak deformation uo = D and the base shear is Vbo = A g w . The results are summarized in the accompanying table.

b g

ωn

sec-1

sec

in./se c

in.

(a) (b)

0.98 1.39

6.40 4.52

35.5 50.3

36.2 36.2

0.09 0.18

144.7 289.4

(c)

0.98

6.40

35.5

36.2

0.09

289.4

System

A/g

Vbo kips

Comparing the response of systems (a) and (b), we observe that stiffening the tower shortens the natural period and increases the base shear; however the deformations remain unchanged. Comparing systems (a) and (c), both have the same natural period and spectral ordinates. The deformations of the two systems are the same. However, system (c) has twice the mass of system (a), which doubles the base shear. The preceding comments are restricted to systems in the constant-D region of the spectrum.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 29 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 6.14

The bending moment at the base of columns is 1355 . × 12 1355 . M = h = = 8130 . kip − ft 2 2

FG H

(a) Frame with rigid beam ( EIb = ∞ ) k = 2

ωn = Tn =

LM12 EI OP = 24 (3 × 10 ) (10 12) = 20.09 kips in. (12 × 12) N h Q 3

IJ K

k = m

20. 09 = 27. 85 rads sec 10 386

13.55 kips 12' 24'

2π

= 0. 226 sec

ωn

13.55 kips 2

For Tn = 0. 226 sec and ζ = 5% , Fig. 6.9.5 scaled by 0.50 gives 1. 355 × 386 A A = 1. 355 ⇒ D = 2 = = 0. 674 in. ( 27. 85)2 g ωn

Design quantities

81.30 M, kip-ft

A rigid beam increases the lateral stiffness by a factor of 4 and shortens the natural period by a factor of 2. In the constant-A region of the spectrum, this change in Tn does not affect the lateral force; however, the maximum bending moment is halved. The design deformation is reduced by a factor of four.

Deformation, uo = 0. 674 in. Lateral force, fSo = ( A g ) w = 13. 55 kips Bending moments at top and bottom of columns:

1355 . × 12 . I h FG 1355 H 2 JK 2 = 4 = 40.65 kip − ft

M =

Bending moments in the columns are shown in the accompanying diagram. 13.55 kips

0.674 in.

40.65

13.55 kips 2

12' 24'

40.65 M, kip-ft

(b) Frame with flexible beam ( EI = 0 ) 3EI c k =2 = 5.02 kips in. h3

LM N

OP Q

k = 13. 92 rads sec; Tn = 0. 452 sec m For Tn = 0. 452 sec and ζ = 5% , Fig. 6.9.5 gives

ωn =

1. 355 × 386 A A = 1. 355 ⇒ D = 2 = = 2. 7 in. (13. 92 )2 g ωn

Design quantities uo = 2. 7 in. f So =

FG A IJ w = 1355 . kips H gK

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 30 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 6.15

h = 12'

EIc

24'

k = 2

LM 3EI OP = 6 (3 × 10 ) (10 12) = 5.02 kips in. (12 × 12) Nh Q

ωn =

k = m

Tn =

5. 02 = 13. 92 rads sec 10 386

2π = 0. 452 sec ωn

For Tn = 0. 452 sec and ζ = 5% , Fig. 6.9.5 scaled by 0.5 gives 1. 355 × 386 A A = 1. 355 ⇒ D = 2 = = 2. 7 in. (13. 92 )2 g ωn

Design quantities uo = 2. 7 in. f So =

FG A IJ w = 1355 . kips H gK

The bending moment diagram for each column is as shown; at the top M =

1355 . × 12 . I FG 1355 h = = 8130 . kip − ft J H 2 K 2

13.55 kips

2.7 in.

81.30

13.55 kips 2

12' 24'

M, kip-ft

Influence of base fixity If the columns are hinged at the base instead of clamped, the lateral stiffness is reduced by a factor of 4 and the vibration period is lengthened by a factor of 2. In the constant-A region of the spectrum, this change in Tn does not affect the lateral force; however, the maximum bending moment is doubled. The design deformation is quadrupled.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 31 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 6.16

Alternatively, from Eq. (A1.1) in the book with ua = uo, ub = θa = θb = 0,

From Example 1.2: w = 30 × 30 × 20 = 18,000 lbs=18.0 kips m = w g = 0.04663 kip - sec 2 in . k N-S = 38.58 kips in.

6 EI

6 × 29000 × 82.8

uo =

h (12 × 12 ) M = 18.3 kip - ft

0.316 12

The bending moment diagram drawn on the compression side is shown in the accompanying figure.

k E-W = 119.6 kips in. (a) North-South excitation 1. Determine the natural vibration period. . 004663 Tn = = 2π = 2π k N-S . ωn 3858

18.3

(

18.3

(

sec Tn = 0219 .

(

2π

18.3 kip-ft

18.3

(

2. Determine the pseudo-acceleration. From Fig. 6.9.5 scaled by 0.25, the pseudoacceleration for Tn = 0.219 sec is

A = 0.25( 2.71g ) = 0.6775 g

(b) East-West excitation

1. Determine the natural vibration period.

3. Compute peak responses. The peak lateral displacement uo is 2

18.3

⎛T ⎞ ⎛ 0.219 ⎞ uo ≡ D = ⎜ n ⎟ A = ⎜ ⎟ ( 0.6775 × 386) ⎝ 2π ⎠ ⎝ 2π ⎠ uo = 0.316 in. To determine bending moments in the columns, we first determine the equivalent static force:

A f So = w = 0.6775 × 18.0 = 1219 . kips g

Tn =

2π

ωn

= 2π

m kE-W

= 2π

2. Determine the pseudo-acceleration. From Fig. 6.9.5 scaled by 0.25 the pseudo-acceleration for Tn = 0.124 sec is

A = 0.25(1170 . × 0124 . 0.704 g ) = 0.6728 g 3. Compute peak responses. The peak lateral displacement uo is 2

The bending moments in the columns are determined from the static analysis of the frame subjected to the equivalent static force, fSo = 12.19 kips. Each column carries 1/4th of the force: M

f So 4

004663 . = 0124 . sec 1196 .

. ⎞ ⎛T ⎞ ⎛ 0124 uo ≡ D = ⎜ n ⎟ A = ⎜ ⎟ (0.6728 × 386) ⎝ 2π ⎠ ⎝ 2π ⎠ . in. uo = 01013 The equivalent static force is

f So =

A w = 0.6728 × 18.0 = 1211 . kips g

Neglecting the lateral resistance of the columns, the axial force in each brace is

12'

f So 4

p brace =

f So 4 1211 . 4 = = 353 . kips cos θ 0.8575

f So 1219 . × 12 = × 12 4 4 M = 18.3 kip - ft

2M =

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 32 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

p brace

12'

f So

θ 20'

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 33 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 6.17 1. Compute Tn . k=

∑

(

)

⎤ 3EI 3 3 × 103 104 ⎡ 1 1 = + ⎢ 3 3 3⎥ 12 L ⎣ (10 × 12 ) (20 × 12 ) ⎦ = 4.340 + 0.543 = 4.883 kip/in .

Tn = 2π

10 386 = 0.458 sec ; ω n = 13.73 rads/sec 4.883

2. Compute peak deformation uo . From spectrum: D=

ω n2

0.25 × 2.71 × 386

(13.73)2

= 1.39 in.

u o = D = 1.39 in.

3. Compute bending moments. 3EI Method 1: M = 2 uo L

(

)

(

)

M short =

3 3 × 10 3 10 4 1.39 = 724 k - in = 60.3 k - ft (10 × 12)2 12

M long =

3 3 × 10 3 10 4 1.39 = 181 k - in = 15.1 k - ft (20 × 12)2 12

Method 2: From uo calculate lateral force for each column and then the bending moment.

( f s )short = k short u o = 4.340 × 1.39 = 6.03 kip M short = 6.03 × 10 = 60.3 k - ft

( f s )long = k long u o = 0.543 × 1.39 = 0.755 kip M long = 0.755 × 20 = 15.1 k - ft

The bending moment diagrams for both columns are shown. 60.3 k-ft.

15.1 k-ft.

Fig. P6.17 © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 34 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

3. Determine spectral ordinate.

Problem 6.18 2

w = 100 kips

From the design spectrum of Fig. 6.9.5, scaled by

0.5 1

I b = I c /2

A = 0. 5 ( 2. 71g) = 1. 355g

D =

ω 2n

1. 355 ( 386 ) = 3. 86 in. (11. 64 )2

4. Determine peak responses. E = 30 × 103 ksi ; Ic = 320 in.4 ; h = 12 ft ; ζ = 5% 1. Determine lateral stiffness of the frame.

Data:

Following the procedure of Example 1.1 of the text, the 3 × 3 stiffness matrix is

OP P 5h PQ

LM24 6h k = M6h 5h h M N6h h EI c

6h 1 2 h 2

1 2

(a)

From the second and third equations, the joint rotations can be expressed as

RSu UV = − LM 5h Tu W MN h 1 2

1 2 h 2 2

c 3

120 EI c u1 11 h 3

11 h

ωn =

(12 × 12 )

= 35. 07 kips in.

Tn =

w 100 = = 0. 2591 kip - sec2 in. g 386 klat = m 2π

ωn

θa

EI c

L c = 12 '

(c)

2. Calculate natural period. m =

θb

θa

35. 08 = 11. 64 rads sec 0. 2591

2π = 0. 540 sec 11. 64

EI b

θb

L b = 24 '

column

120 ( 30 × 103 ) ( 320 ) 11

Thus the lateral stiffness of the frame is klat =

0.02924

F 24 EI − EI 12 6h 6h L1OI u MN1PQJK GH h h 11h

120 EIc

0.02924

−1

Substituting Eq. (b) into the first of three equations in Eq. (a) gives fS =

3.86 in.

OP R6hU u = − 12 R1U u (b) SV S V 11h T1W 5h PQ T6h W

RSu UV = − 12 RS1UV u = − 12 RS1UV (386 . ) 11 (12 × 12) T1W Tu W 11h T1W R1U = − 0.02924 S V T1W The deflected shape at peak response is

OP R|u U| R| f U| PP S|u V| = S| 0 V| 5h Q Tu W T0W

c 3

To determine bending moments we recover joint rotations from Eq. (b):

6h 1 2 h 2

u1o = 3. 86 in.

The equilibrium equations are

L24 6h EI M M6h 5h h M N6h h

The peak lateral displacement is

beam

Bending moments in columns: Ma = Mb =

4 EIc Lc

2 EIc Lc

θa +

2 EIc

θa +

4 EIc

Lc Lc

θb +

6 EIc

θb +

6 EIc

L2c

ua − ua −

6 EIc L2c

ub ub

Substituting θ a = 0 , θb = − 0. 02924 , ua = 0 , and ub = − 3. 86 and values for E, Ic and Lc gives M a = 6824 kip - in.

M b = 2925 kip - in.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 35 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Bending moments in beam: Ma = Mb =

4 EIb Lb

2 EIb Lb

θa +

2 EIb

θa +

4 EIb

Lb Lb

θb +

6 EIb

θb +

6 EIb

L2b

ua − ua −

6 EIb L2b

ub ub

Substituting θ a = θb = − 0. 02924 , ua = ub = 0 , and values for E, Ib and Lb gives M a = M b = − 2925 kip - in.

The bending moment diagram drawn on the compression side is shown in the accompanying figure; the units are kip-in. 2925 2925

6824

2925 2925

6824

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 36 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Moment diagram:

Data: E = 30 × 103 ksi; Ic = 320 in4; h = 12ft; ζ = 5%

428.2 kip-ft

1. Determine lateral stiffness of the frame. The lateral stiffness of this frame was computed in Problem 1.16: 2 EI c h3

2(30 × 10 3 )(320) (12 × 12) 3

428.2

(

k lat =

(

428.2

(

Problem 6.19

428.2

= 6.43 kips in .

2. Calculate natural period. 5. Influence of base fixity.

w 100 = = 0.2591 kip - sec 2 in . g 386

The above results together with those from Problem 6.18 are summarized:

k lat 6.43 = = 4.982 rad sec m 0.2591 2π 2π Tn = = = 1261 . sec ω n 4.982

ωn =

3. Determine spectral ordinate. From the design spectrum of Fig. 6.9.5, scaled by 0.5,

(

)

A = 0.5 1.80 × 1.261−1 g = 0.7136 g D=

ω n2

0.7136 × 386 4.982 2

= 11.10 in.

Base

uo (in)

Mbase (kip-ft)

Mtop (kip-ft)

Hinged

11.1

428.2

Clamped

3.86

568.7

243.8

Base fixity shortens the period, reduces the design deformation, creates a larger bending moment at the base, but reduces the bending moment at the beam-column joints.

4. Determine peak responses. The peak lateral displacement is uo ≡ D = 1110 . in.

To determine bending moments, we first determine the equivalent static force f So = k lat uo = 6.43 × 1110 . = 714 . kips

Using symmetry, the moment at the beam-column joint is M=

f So 714 . h= × 12 = 428.2 kip - ft 2 2

f So 12'

f So 2

M f So 2 f So 2

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 37 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 6.20

2Vo

1. Determine structural properties. Lateral Stiffness: k =

48 EI h3

= 118.6 kips in .

12'

w Mass: m = = 0.2591 kip - sec 2 in . g

Peq

20'

Peq

Natural Frequency: ω n = k m = 214 . rad sec Natural Period: Tn = 2π ω n = 0.294 sec 2. Determine peak lateral displacement. From the design spectrum of Fig. 6.9.5, scaled to (1/3)g, 1 × 2.71g = 0.90 g 3 0.90 × 386 A D= 2 = = 0.76 in. ωn 214 . 2 A=

Peq =

2Vo (12 + 6) = 40.5 kips 20

The axial force in each column due to gravity load is Pgrav =

w = 25 kips 4

The total axial stress is Peq + Pgrav

40.5 + 25 = 3.28 ksi 20 A 5. Determine total stress. fa =

uo = D = 0.76 in.

3. Compute bending stress.

f o = f b + f a = 19.06 + 3.28 = 22.34 ksi

For one

Mo Vo

uo Vo

Mo Column: k 118.6 × 0.76 = 22.5 kips uo = 4 4 h h 24 × 12 = 3240 kip - in. M o = Vo = 22.5 × 2 2 M 3240 = 19.06 ksi fb = o = S 170

Vo =

12 EI 3

uo =

4. Compute stress due to axial forces. The sketch represents half of the structure, i.e., one pair of columns with rigid platform and rigid column bases; there is a point of inflection at mid-height of each column. Hence, taking moments about one of the inflection points, we get the column axial forces due to earthquake to be © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 38 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 6.21 The equation of motion to be solved is I O u&&θ + ( k x d 2 + k y b 2 )uθ = − I O u&&gθ ; u&&gθ = δ ( t )

(a)

We have solved a related equation: mu&& + ku = p( t ); p( t ) = δ ( t )

(b)

and its solution is given by Eq. (4.1.7) specialized for τ = 0: u( t ) = h( t ) =

1 sin ω n t mω n

(c)

Therefore, solution to Eq. (a) is Eq. (c) multiplied by -IO with m replaced by IO: uθ ( t ) = −

I Oω n

sin ω n t

(d)

or uθ ( t ) = −

ωn

sin ω n t

where 1

⎛ k x d 2 + k y b2 ⎞ 2 ⎟ ω n = ⎜⎜ ⎟ IO ⎝ ⎠

(e)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 39 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

30 × 12 b uθ o = × 0.00323 = 0.582 in 2 2 20 × 12 d uθ o = × 0.00323 = 0.388 in 2 2

Problem 6.22 1. Define structural properties. From Example 2.4: b = 30 ft; d = 20 ft; h = 12 ft; w = 01 . kip ft 2

6. Determine base torque To.

k x = 15 . kips in.; k y = 10 . kip in.

To = kθ uθo = 18000 × 0.00323 = 58.2 kip - ft

I O = 20186 . kip - sec - ft kθ = 18000 kip - ft rad

ωn =

7. Determine bending moment at top and base of column A.

kθ = 9.44 rad sec IO

fsy

fsx

Tn = 0.67 sec h

2. Write equation of motion. I O u&&θ + kθ uθ = − I O u&&gθ (t )

u&&θ + ω n2 uθ = −u&&gθ (t )

⎛d ⎞ f sx = k x ⎜ uθo ⎟ = 0.388 kips ⎝2 ⎠

Including damping gives

⎛b ⎞ f sy = k y ⎜ uθo ⎟ = 0.873 kips ⎝2 ⎠

u&&θ + 2ζω n u&θ + ω n2 uθ = −u&&gθ (t )

3. Determine spectral ordinate. From Fig. 6.9.5 scaled by 0.05, for Tn = 0.67 sec. and ζ = 5%, A = 0.05(180 . × 0.67 −1 g ) = 01343 . g

4. Determine peak rotation.

1 1 f sx h = × 0.873 × 12 = 5.24 kip - ft 2 2 1 1 M x = f sy h = × 0.388 × 12 = 2.33 kip - ft 2 2 My =

Bending moments in other columns are the same; the relative direction can be determined from the direction of displacements.

The peak value of rotation is uθo =

For a column clamped at both ends

. 2 A 2 01343 × 386 = × = 0.00323 rad 2 b ω n 30 × 12 (9.44) 2

5. Determine displacement at each corner of the roof slab. The corner displacements accompanying figure, where

du 2 θo

are

shown

du 2 θo

the

bu 2 θo

uθ o x

bu 2 θo

du 2 θo

bu 2 θo

du 2 θo

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 40 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 6.23 (a) With reference to Fig. P6.23a, the design spectrum is determined by the following steps: 1. The peak parameters for the ground motion: u&&go = 0. 50g , u&go = 24 in. sec , and ugo = 18 in. are plotted. 2. From Table 6.9.1 the amplification factors for the 84.1 percentile and 2% damping are obtained:

α A = 3. 66 , α V = 2. 92 , and α D = 2. 42 . 3-5. The ordinates for three branches of the spectrum are:

FG 2π IJ D = FG 2π IJ 4356 . H T K g H T K g = 4.46 T 2

A(Tn ) = g

•

−2 n

Tn ≥ 33 sec

FG 2π IJ u = 184 HT K g . T 2

A(Tn ) = g

−2 n

• 1 33 sec < Tn ≤ 1 8 sec : Find equation to straight line on log-log paper connecting point b with coordinates A g = 1. 83 and Tn = 1 8 sec to point a with coordinates A g = 0. 50 and Tn = 1 33 sec . The equation is A( Tn ) = 12. 28 Tn0.916 g

b-c: A = 0. 50 g × 3. 66 = 1. 83g c-d: V = 24 × 2. 92 = 70. 08 in. sec d-e: D = 18 × 2. 42 = 43. 56 in. 6. The line A = 0. 50 g is plotted for Tn < 1 33 sec and D = 18 in. for Tn > 33 sec . 7. The transition line b-a is drawn to connect the point A = 1. 83g at Tn = 1 8 sec to A = 0. 50 g at Tn = 1 33 sec . Similarly, the transition line e-f is drawn to connect the point D = 43. 56 in. at Tn = 10 sec to ugo = 18 in. at Tn = 33 sec . The same procedure is used to determine the median spectrum, using the amplification factors α A = 2. 74 , α V = 2. 03 , and α D = 1. 63 .

• 10 sec < Tn ≤ 33 sec : Find equation to straight line on log-log paper connecting point e with coordinates D = 43. 56 in. and Tn = 10 sec to point f with coordinates D = 18 in. and Tn = 33 sec . The equation is A( Tn ) = 24. 49 Tn−2.74 g

Pseudo-acceleration spectrum The preceding equations are plotted in Fig. P6.23b using log-log scale. (c) The same spectrum is plotted using linear scales in Fig. P6.23c.

(b) Determine Tc and Td . Fig. P6.9a

At Tc , A = 1. 83g and V = 70. 08 in. sec

IJ = 0.623 sec K

•

1 8 sec ≤ Tn ≤ 0. 623 sec

A( Tn ) = 1. 83 g

•

0. 623 sec < Tn ≤ 3. 91 sec

0. 1

a 1

0.5

0.05

0.1

0.2

0. 0

V, in./sec

A( Tn ) = 0. 5 g

0.2 0.02

Tn ≤ 1 33 sec

•

01 0.

Determine equations for A( Tn ) g .

IJ = 3.91 sec K

FG H

. 2π D 4356 D ⇒ Tn = 2π = 2π Tn V 70.08

1 0.

V =

e b

At Td , V = 70. 08 in. sec and D = 43. 56 in.

2π V 70.08 V ⇒ Tn = 2π = 2π Tn . × 386 A 183

A =

FG H

100

0.5

1 Tn, sec

Fig. 6.23a

A( Tn ) 2π V 2 π ( 70. 08) = = = 1.14 Tn−1 g gTn gTn

•

3. 91 sec < Tn ≤ 10 sec

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 41 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Fig. P6.9b 5

A, g

0.1 e

0.01

0.1

1 Tn, sec

Fig. 6.23b

Fig. P6.9c 2.0

Pseudo-acceleration A, g

1.5

1.0

0.5

a d

0.0 0

2 3 4 Natural vibration period Tn, sec

Fig. 6.23c

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 42 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

CHAPTER 8 Problem 8.1

The relation between z and the rotation θ about fulcrum O is

1. Determine the shape function.

z =

x′

3. Determine natural frequency and damping ratio.

ωn =

2. Draw the free body diagram and write the equilibrium equation.

O z L) I 1 (2 && z k (3z 2 ) m 1 &&

c z&

m 2 ( && z 2)

m 2 ( 2 && z)

k% % m

ζ =

c% %% 2 km

(e)

which are the same as in Example 8.1. 4. Solve the equation of motion. p% ( t ) =

p p( t ) = o ≡ p% o 2 2

z(t ) =

p% o 4 po (1 − cos ω nt ) = (1 − cos ω nt ) 9k k%

p (t ) z L) I 2 (2 &&

(d)

Substituting Eq. (d) into Eq. (a) leads to the same equation of motion as in Example 8.1.

u ( x ′, t ) = ( 2 x ′ L ) z u (x , t ) = ( 2 x L ) z

FG L IJ θ H 2K

(f)

5. Determine displacements.

u( x , t ) =

2x z(t ) L

u( x ′, t ) =

2 x′ z(t ) L

(g)

which are identical to the results in Example 8.1. Thus the results are independent of the choice of generalized displacement.

∑ MO = 0 ⇒ && 2 && z 2 && z z L L + m1&& + I2 + m2 ( 2 && z z ) L + m2 2 2 4 L L L 3z 3 L L + cz& + k = p( t ) 2 2 4 2 I1

Substituting I1 = m1L2 12 and I2 = m2 L2 128 gives

FG 2m L + 137m L IJ &&z + FG c L IJ z& + FG 9k L IJ z = p(t ) L H 3 H2K H 8 K 64 K 2 1

(a) The equation of motion (after dividing by L) is

% = p% ( t ) % && + cz % & + kz mz

(b)

where m% =

c 2 m1 137 m2 9k + , c% = , k% = 2 3 64 8

and p% ( t ) =

p( t ) 2

(c)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 1 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 8.2

4. Solve the equation of motion. For p( t ) = δ ( t ) , the solution of Eq. (a) is

1. Determine the shape function.

θ( t ) =

u (x , t ) = x θ

9θ L 8

L 8

where ω D = ω n 1 − ζ2 .

L 8 L 8

5. Determine displacements.

2. Draw the free body diagram and write the equilibrium equation.

u ( x , t ) = x θ( t )

(e)

where θ ( t ) is given by Eq. (d).

p(t )

m ( Lθ&& 2 ) O

9 L 8 − ζω n t 72 e sin ω Dt = e− ζω n t sin ω Dt % mω D 103m Lω D (d)

c θ& I 2 θ&&

I 1 θ&& k ( Lθ 2) L2

m (9 L θ&& 8 ) L2

L 8

∑M = 0⇒ O

F m Lθ&& I L + I θ&& + F 9m Lθ&& I 9 L + cθ& GH 2 JK 2 GH 8 JK 8 F k Lθ IJ L = p(t ) 9 L + G H 2 K2 8 I1θ&& +

Substituting I1 = m L2 12 and I2 = ( m 12 ) [( L 4 )2 + ( L 4 )2 ] = m L2 96 gives % && θ + c% θ& + k%θ = p% ( t ) m

(a)

where m% =

103 m L2 9L k L2 p( t ) , c% = c , k% = , and p% ( t ) = 64 4 8 (b)

3. Determine natural frequency and damping ratio.

ωn =

k% = m%

16 k ; 103 m

ζ =

c% = %% 2 km

8c 103k m L4 (c)

Problem 8.3

5. Determine displacements.

1. Determine the shape function.

u( x , t ) =

u ( x , t ) = ( 8 x 9L ) z O x

8x 72 x z = e− ζω n t sin ω Dt 9L 103m Lω D

(e)

This result is identical to Eqs. (d) and (e) in Problem 8.2, i.e., the response is independent of the choice of generalized displacement.

2. Draw the free body diagram and write the equilibrium equation. p (t )

z 9) m ( 4 && c θ&

θ = 8z 9L

I 2 θ&&

I 1 θ&& k ( 4z 9 )

m && z

∑ MO = 0 ⇒ 8 && z 4 mz&& L 8&& z 9L 8 z& + + I2 + mz&& + c 9L 9 2 9L 8 9L 4 kz L 9L + = p( t ) 9 2 8 I1

Substituting I1 = m L2 12 dividing by L gives

and

I2 = m L2 96

% = p% ( t ) % && + cz % & + kz mz

and (a)

where % = m

2k 8c % 9 103m , c% = ,k = , and p% ( t ) = p( t ) 9 9 L2 8 72 (b)

3. Determine natural frequency and damping ratio.

ωn =

k% = % m

16 k ; 103m

ζ =

c% = %% 2 km

8c 103k m L4 (c)

4. Solve the equation of motion. For p( t ) = δ ( t ) , the solution of Eq. (a) is z(t ) =

9 8 − ζω n t 81 e sin ω Dt = e − ζω n t sin ω Dt % 103mω D mω D (d)

where ω D = ω n 1 − ζ2 .

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 3 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 8.4 1. Determine the shape function. u (x , t ) = x θ ( t )

O x

2. Draw the free body diagram and write the equilibrium equation. ( 2x L ) p ( t ) O x mx θ&&+ cx θ& + kx θ

∑ MO = 0 ⇒

z LMN

(mxθ&& + cx θ& + kx θ ) x dx =

−L 2

OP Q

2x p(t ) x dx L

or m L3 && c L3 & k L3 L2 θ + θ + θ = p( t ) 12 12 12 6

The equation of motion is m% && θ + c% θ& + k%θ = p% ( t )

where % = m

k L3 m L3 c L3 % , c% = ,k = , 12 12 12

and p% ( t ) =

L2 p( t ) 6

3. Determine natural frequency and damping ratio.

ωn =

k% = m%

c% c k ; ζ = = m 2 km % 2 km%

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 4 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 8.5 1. Determine the shape function. u (x ′ , t ) = ( x ′ L ) z ′ B

A x′

u(x , t ) = ( x L ) z

z′

2. Draw the free body diagram and write the equilibrium equation. p (t ) B

z 2 m && 2

k ( z- z ′ ) k z′ 2

z L ( mL 12 ) &&

C c z& 2

∑ MA = 0 for bar AB ⇒ kz ′ L L + p( t ) − k ( z − z′ ) L = 0 ⇒ 2 2 2 z′ =

4z 2 p( t ) − 5 5 k

(a)

The force in spring BC is fS = k ( z − z ′ ) =

kz 2 + p( t ) 5 5

(b)

∑ MD = 0 for bar CD ⇒ m L2 && z mz&& L cz& L + + + fS L = 0 12 L 2 2 2 2

(c)

Substituting Eq. (b) in Eq. (c) gives % = p% ( t ) % && + cz % & + kz mz

(d)

where m% =

m c 2 k , c% = , k% = , and p% ( t ) = − p( t ) 5 3 4 5

(e)

3. Determine natural frequency and damping ratio.

ωn =

c% k% k = 0. 6 ; ζ = = 0. 484 m% m %% 2 km

c km

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 5 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 8.6

7. Compute shear and bending moment at mid-height and at the base.

1. Determine the generalized properties.

~ = m

m( x ) ψ ( x )

M o ( L 2) =

EI ( x ) ψ ′′( x )

z FGH L

= EI

~ L =

ξ −

L 2

IJ f (ξ) dξ K o

= 0.3806 × 106 kip - ft

~ k =

f o (ξ ) dξ = 2,151 kips

V o ( L 2) =

F 3 x − 1 x I dx = 33 mL = m G 140 H 2 L 2 L JK L

z z FGH L

3 2

−

IJ dx = 3EI L K L

Mo =

= m

z 0

ξ f o (ξ ) dξ = 1108 . × 106 kip - ft

m( x ) ψ ( x ) dx L

f o (ξ ) dξ = 2,518 kips

z z L

V bo =

F 3 x − 1 x I dx = 3mL GH 2 L 2 L JK 8 2

2. Determine the natural period.

ωn =

k% 3. 57 = m% L2

EI = 1. 76 ; Tn = 3. 57 sec m

3. Formulate the equation of motion. && z + ω 2n z = − Γ% u&&g ( t ) ; Γ% = 1. 59

4. Determine the peak value of z ( t ) . For Tn = 3. 57 sec spectrum gives

ζ = 0. 05 , the design

and

A FG IJ = 0126 H K . ; D = ω = 15. 7 in.

A 180 . = 0.25 g 357 .

2 n

∴ zo = Γ% D = 25. 0 in.

5. Determine peak displacements of the tower. uo ( x ) = zo ψ ( x ) = 25.0

F 3 x − 1 x I in. GH 2 L 2 L JK 2

6. Determine equivalent static forces.

F GH

I JK

3 x2 1 x3 ~ f o ( x ) = Γ m( x ) ψ ( x ) A = 1119 . − kip ft 2 L2 2 L3

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 6 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 8.7

2π k% = 1. 896 rad sec ; Tn = = 3. 313 sec m% ωn

ωn = 25'

3. Determine the peak value of z ( t ) . and ζ = 0. 05 , the design

For Tn = 3. 313 sec spectrum gives

FG H

. 180 A = 0.25 3.313 g

2 n

∴ zo = Γ% D = 25.1 in.

600 '

R avg (x)

A IJ = 01358 K . ; D = ω = 14. 58 in.

4. Determine peak displacements at the top. uo ( L ) = ψ ( L ) zo = 25.1 in. x

t= 2.5'

5. Determine equivalent static forces. ~ f o ( x ) = Γ m( x ) ψ ( x ) A

LM1 − cos π x OP (01358 g) kip ft . 2L Q N

1. Determine properties of the chimney. L = 600 ft Ravg ( x ) = 23. 75 −

6. Determine shear and bending moment at mid-height and at the base.

12. 5 x 600

V o ( L 2) =

m( x ) =

150 A( x ) = 1738 . − 1523 . × 10 −3 x kip − sec 2 ft 2 32.2

I ( x) =

3 π Ravg ( x) t

M o ( L 2) =

z z z

f o (ξ ) dξ

z z L

j x − 36.82 x kip − ft 3

V bo =

f o (ξ )dξ = 1,739 kips

2. Determine m% , k% , Γ% and Tn . ~ = m

L 2

= 0.2399 × 106 kip - ft

EI ( x ) = 5.454 × 1010 − 1435 . × 108 x + 1259 . × 10

ξ−

− 7.102 × 10 −5 x 3 ft 4

f o (ξ ) dξ = 1,426 kips

= 105,216 − 276.9 x + 0.243x

z z FGH IJK L

A( x ) = 2 π Ravg ( x ) t = 373. 06 − 0. 327 x ft 2

1738 = 1722 . . − 1523 . × 10 −3 x ×

50'

M bo =

ξ f o (ξ ) dξ = 0.7368 × 106 kip - ft

m( x ) ψ ( x )

dx = 134.5 kip - sec

~ k = ~ L =

EI ( x ) ψ ′′( x )

dx = 4835 . kips ft

0 L

m( x ) ψ ( x ) dx = 2316 . kip − sec 2 ft

L% Γ% = = 1. 722 m% © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 7 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 8.8

~ L L = ∫ m( x)ψ ( x)dx = 240.27 kip − sec2 / ft 0 ~ ~ ~ Γ = L / m = 1.709

1. Determine properties of the chimney. L=600 ft. Ravg ( x ) = 23.75 −

12.5 x 600

~ = 1.915rad / sec ; T = 2π /ω = 3.28 sec ωn = k / m n n

A( x ) =2πRavg ( x ) t =373.06− 0.327 x ft

3. Determine the peak value of z(t).

For Tn = 3.28 sec and ζ =0.05 , the design spectrum gives:

150 m( x ) = A( x ) 32.2

= 1738 − (1523 × 10 −3 )x kip − sec 2 / ft 2 . .

A = 0.25(1.80 / 3.28) = 0.137 g

D = A / ω n = 14.45 in. 2

25'

~ zo = ΓD = 24.7 in.

4. Determine peak dispalcement at the top. uo ( L ) =ψ ( L ) zo = 24.7in

5. Determine equivalent static forces. ~ f o ( x ) = Γm( x )ψ ( x ) A

Ravg(x) 600'

1738 = 1709 . . − (1523 . ×10 −3 ) x ×

F 3 x2 1 x3 I . g GGH 2 L2 − 2 L3 JJK × 0137

6. Determine shear and bending moment at mid-height and at base.

50'

z L

V o ( L 2) =

f o (ξ ) dξ = 1465 kips

I ( x ) =πRavg 3 ( x ) t =105,216−276.9 x 3

+0.243x 2 −( 7.102×10−5 ) x 3 ft 4

+ (1.259×10 ) x −36.82 x

f o (ξ )dξ = 1811 kips

EI ( x ) =5.454×1010 − (1.435×108 ) x 5

z L

V bo =

z FGH IJK L

kip − ft

M o ( L 2) =

ξ−

L 2

f o (ξ ) dξ

= 0.244 × 106 kip - ft ~ ~ , k~ , Γ , and Tn . 2. Determine m

L ~ = m( x)[ψ ( x)] 2 dx = 140.61kip − sec 2 / ft m ∫ 0

z L

M bo =

ξ f o (ξ ) dξ = 0.759 × 106 kip - ft

~ L k = ∫ EI ( x)[ψ ′′( x )] 2 dx = 515.93kips/ft 0 © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 8 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 8.9 The excitation force over the height is x p(t ) p( x , t ) = L

FG IJ H K

1. Formulate equation of motion. % && + k% z = p% ( t ) mz m% = 134. 5 kip - sec2 ft ; k% = 483. 5 kips ft ; ~ p (t ) =

z z LMN L

p( x , t ) ψ ( x ) dx

0 L

x p( t ) L

OP FG1 − cos π x IJ dx = 1612. p(t ) 2LK QH

2. Solve the equation of motion. Note that td Tn = 0. 25 3. 313 = 0. 075 . Because t d Tn << 0.25, the excitation can be approximated as a pure impulse. Adapting Eq. (4.10.3) to this problem gives

zo = ~ mω n

~ p (t )dt =

. 1612 (134.5) (1896 . )

0.25

p(t )dt

= 0.316 ft = 3.79 in.

3. Determine peak responses. uo ( L ) = ψ ( L ) zo = 3. 79 in. fo ( x ) = ω 2n m ( x ) ψ ( x ) zo = 1.136 m ( x ) ψ ( x ) kip ft

z z FGH IJK L

V o ( L 2) =

f o (ξ ) dξ = 216 kips

L2 L

M o ( L 2) =

ξ−

z z

L 2

f o (ξ ) dξ = 36,290 kip - ft

V bo =

f o (ξ ) dξ = 263 kips ;

M bo =

ξ f o (ξ ) dξ = 111,485 kip - ft

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 9 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 8.10 5. Determine the peak value of z ( t ) .

1. Determine the shape function.

w/2

3 k

w 2

w 1

w w

Story shear

Story drift

Floor deflection 4.5 w/k

0.5 w 0.5 w/k

For Tn = 0. 3405 sec and ζ = 0. 05 , the design spectrum gives A = 0. 25 ( 2. 71g ) = 0. 6775g ; D =

4w/k

ω 2n

= 0. 768 in.

zo = Γ% D = 1. 216 ( 0. 768) = 0. 934 in.

1.5 w 1.5 w/k 2.5 w/k 2.5 w 2.5 w/k

6. Determine floor displacements. u jo = ψ j zo ⇒ u1o = 0. 519 in. , u2 o = 0. 830 in. , u3o = 0. 934 in.

Shape vector:

7. Determine equivalent static forces.

ψ = 59 89 1

f jo = Γ% m j ψ j A ⇒

2. Determine generalized properties.

f 1o = 1216 .

f 2o

f 3o

259 w % = ∑ m j ψ 2j = m 162 g j =1

35k k% = ∑ k j ( ψ j − ψ j −1 )2 = 81 j =1 35 w L% = ∑ m j ψ j = 18 g j =1

3. Determine the natural period.

ωn =

FG 100 IJ FG 5 IJ b0.6775gg = 458. kips H g K H 9K FG 100IJ FG 8 IJ (0.6775g) = 73.2 kips . = 1216 H g K H 9K F 50 I = 1216 . G J (1) (0.6775g) = 412 . kips H gK

k% kg = 0. 520 % m w

8. Determine story shears and overturning moments. Static analysis of the structure with lateral forces f jo gives the story shears in Fig. (c) and story overturning moments in Fig. (d).

where w = 100 kips

L12 (29000 × 1400) OP = 326.3 kips in. k = 2M MN (12 × 12) PQ

0.934 in.

41.2 kips

0.830

73.2

114.4

494 kip-ft

Thus

ω n = 0. 520 Tn =

2π

ωn

326. 3 × 386 = 18. 45 rads sec 100

0.519

45.8

160.2

1867 3789

= 0. 3405 sec

(a) u jo

(b) f jo

(d)

M jo

4. Determine the equation of motion. L% && z + ω 2n z = − Γ% u&&g ( t ) ; Γ% = = 1. 216 m% © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 10 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 8.11

6. Determine the peak value of z(t).

1. Determine deflections to applied forces. story shear

w/2 w/2 w

EI/3

w/2

floor story deflection drift 25w/12k w/2k 19w/12k

w 2EI/3

3w/2

10w/12k

5w/2

5w/6k

In this figure, the stiffness of the top story is k = 2×

12( EI / 3) 8 EI = 3 h3 h

ω n2

= 0.966 in.

~ zo = ΓD = 1297 . in. u jo =ψ j zo ⇒ u10 =0.519 ; u20 =0.985 ; u30 =1.297 ; all in inches (Fig. a).

8. Determine equivalent static forces. ~ f jo = Γm jψ j A= 350.69 m jψ j ⇒ f10 =36.34 ; b).

2. Determine shape vector:

ψ = 2 /5 19 / 25 1

A/ g =2.71×0.25=0.677 ; D =

7. Determine floor displacements.

3w/4k

For Tn =0.381sec and ζ =0.05 ,the design spectrum gives:

f 20 =69.04 ;

f 30 =45.42 ; all in kips (Fig.

9. Determine story forces.

3. Determine generalized properties. 3

Static analysis of the structure due to external forces f j 0 gives the story shears and floor overturning moments in Figs. (c) and (d).

~ = ∑ m ψ 2 =1.237 w / g m j j j =1

1.30

~ 3 k = ∑ k j (ψ j −ψ j −1 ) 2 = 0.797 k j =1

0.99

~ L = ∑ m jψ j =1.66w / g j =1

0.52

45.42

69.04

45.42 114.46 150.8

36.34

545 1919

4. Determine the natural period. ~ ~ ω n = k /m =0.803 k / m

3728 (a) uj0 ,in

(b) f j0 ,kips

(d) M j ,kips-ft

For E = 29000 ksi , I = 1400 in . , and h = 12 ft , k=108.77 kip/in For w = 100 kips, m = w/g = 0.259 kip- sec 2 /in.

ω n = 16.45 rad / sec Tn = 2π / ω n = 0.381 sec 5. Determine the equation of motion. ~ ~ ~ Γ= L / m =1.342 && z + ω n 2 z = −1342 . u&&g (t )

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 11 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 8.12

5. Determine peak responses.

1. Determine the shape function.

The floor displacements using Eq. (8.4.14) are given in Fig. (a). The equivalent static forces, calculated from Eq. (8.4.15) are shown in Fig. (b). Static analysis of the structure gives the story shears and story overturning moments in Figs. (c) and (d).

w/2

Story shear

Floor Story drift deflection 3p/k

p/k 2p/k

48.2 kips

0.624

64.2

p/k p/k

1 p

0.935 in.

p/k

0.312

32.1

48.2 kips 112.4

578 kip- ft

144.5

1927

Shape vector: ψ = 13 2 3 1 V jo

3661

(a) u jo

2. Determine generalized properties. 3

% = ∑ m j ψ 2j = m j =1

(b) f jo

(d)

M jo

19 w 18 g

k k% = ∑ k j ( ψ j − ψ j −1 )2 = 3 j =1 3

3 w L% = ∑ m j ψ j = g 2 j =1 L% Γ% = = 1. 421 % m

3. Determine the natural period. k% kg = 0. 5620 % m w

ωn =

where w = 100 kips Problem 8.9).

and

k = 326. 3 kips in.

(from

ω n = 19. 94 rads sec ; Tn = 0. 315 sec 4. Determine the peak value of z ( t ) . For Tn = 0. 315 sec and ζ = 0. 05 , the design spectrum gives A = 0. 25 ( 2. 71g ) = 0. 6775g ; D =

ω 2n

= 0. 6577 in.

zo = Γ% D = 0. 935 in.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 12 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 8.13

6. Determine the peak responses. u jo =ψ j zo ⇒

1. Determine deflections due to applied forces. story shear

w/2

floor story deflection drift

p EI/3

2EI/3

11p/6k

7. Determine equivalent static forces. ~ f jo = Γm jψ j A=401.65 m jψ j

5p/6k

⇒ f10 =18.91 ; f 20 =47.29 ; f 30 =52.02 ; all in kips (Fig. b).

p/k

p/2k

8. Determine story forces. 2p/6k

p/3k

u10 =0.236 ; u20 =0.591 ; u30 =1.3 ; all in inches. (Fig. a).

Static analysis of the structure due to external forces f j 0 gives the story shears and floor overturning moments in Figs. (c) and (d).

2. Determine shape vector:

ψ = 2 /11 5/11 1

1.30

0.59

3. Determine generalized properties. 3

~ = ∑ m ψ 2 =0.739 w / g m j j

0.27

52.02 47.29

52.0 99.3

18.91

118.2

624 1816

j =1

3235

~ 3 k = ∑ k j (ψ j −ψ j −1 ) 2 =0.545k

(a) uj0 ,in

j =1

(b) f j0 ,kips

(d) M j ,kips-ft

~ 3 L = ∑ m jψ j =1136 . m j =1

~ ~ ~ Γ= L / m =1537 .

4. Determine the natural period. ~

~ =0.858 k / m ω n = k /m

From Problem 8.11: k =108.77 kip/in.; m = 0.259 kip − sec 2 / in. Substituting in above equation:

ω n = 17.58 rad / sec Tn =2π /ω n =0.357 sec

5. Determine the peak value of z(t). For Tn = 0. 357 sec spectrum gives:

ξ = 0. 05 ,the design

and

A/ g =2.71×0.25= 0.677 ; D =

ω n2

= 0.846 in

~ zo = ΓD =1.3 in © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 13 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

5. Determine the peak values u j 0 of floor displacements.

Problem 8.14

u jo = ψ j zo

Rigid Beams

k j kips/ in

100

150

100

150

100

200

5@12'

50 kips

4/5

3/5

ψj =5

u10 = 0.95 ; u20 = 1.90 ; u30 = 2.85 ;

⇒

u40 =3.80 ; u50 = 4.77 ; all in inches. (Fig. a)

6. Determine equivalent static forces. (Fig. b). ~ f j 0 = Γm jψ j A= 346.6 m jψ j kips ⇒ f10 =17.94 ; f 20 = 35.88 ; f 30 =53.82 ;

2/5

f 40 =71.76 ; f50 =44.85 all in kips 1/5

1.90

~ = ∑ m ψ 2 = 0.44 kip − sec 2 / in m j j j =1

0.95

170.43

53.82

2.85

538

116.61

71.76

3.80

1. Determine generalized properties.

44.85

4.77

200

206.31

35.88

224.25

17.94

~ 2 k = ∑ k j (ψ j − ψ j −1 ) = 32.0 kips / in

1937

3982

6458

9149

j =1

(a) uj0 ,in

5 ~ 2 L = ∑ m jψ j = 0.647 kip − sec / in

j =1

(b) f j0 ,kips

(d) M j ,kips-ft

7. Determine story forces.

~ ~ ~ Γ= L / m =1.471

Static analysis of the structure subjected to external forces f j 0 gives the story shears and floor overturning

2. Formulate the equation of motion. 2 ~ && z + ω n z = − Γu&&g ( t ) = −1.471 u&&g ( t ) ,

moments in Figs. (c) and (d).

where z is the lateral displacement at the location where ψ = 1, in this case the top of the frame. j 3. Determine the natural vibration frequency and period. ~

~ = 8.528 rad/sec ω n = k /m

Tn = 2π / ω n = 0.737 sec Note that the exact value of ω n =8.262 rad/sec. 4. Determine the peak value of z(t). For Tn = 0.737 and ζ =0.05 , the design spectrum gives 2

A/g = 0.25(1.80/0.737) = 0.61; D = A / ω n = 3.244 in. zo =1.471 D =4.77 in. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 14 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 8.15

6. Determine the peak value of z(t). Story shear

k j kips/ in

w/2

100

Floor deflection 86w/1200

w/2

150

Story drift

w/2

w/200

3w/2

w/100

5w/2

5w/300

80w/1200

68w/1200

48w/1200

200 w

7w/2

7w/400

9w/2

9w/400

27w/1200

200

In the above figure, w = 100 kips 1. Determine the deflections due to applied forces. The static deflections are determined by calculating the story shears and the resulting story drifts,and adding these drifts from the bottom to the top to obtain u = ( w / 1200) 27 48 68 80 86

. A 18 A = × 0.25 = 0.597 ; D = = 3.32 in g Tn ω n2

7. Determine the peak values u j 0 of floor displacements.

150

For Tn =0.754 sec and ζ =0.05 ,the design spectrum gives:

u jo = ψ j zo ⇒u10 =1.34

u20 =2.389

u40 =3.98

u50 = 4.28 all in inches. (Fig. a).

8. Determine the equivalent static forces. (Fig. b). ~ f j 0 = Γm jψ j A=297.03 m jψ j ⇒ f10 = 24.16 f 40 =7156 .

ψ = 0.314 0.558 0.791 0.93 1

∑m ψ j

f 30 =60.87

f 50 =38.47 all in kips

9. Determine story forces. Static analysis of the structure subjected to external forces f j 0 gives the story shears and floor overturning

T 4.28

3. Determine generalized properties. ~= m

f 20 =42.94

moments in Figs. (c) and (d).

2. Determine the shape vector.

u30 = 3.385

3.98

= 0.622 kip − sec 2 / in.

3.39

38.47 71.56 60.87

38.47 461.6

110.0 170.9

j =1

~ K=

2.39

∑ k (ψ − ψ j

j −1 )

∑ m ψ = 0.801 kip − sec / in. j

213.8

3833

= 4315 . kip / in.

j =1

~ L=

42.94

1782

1.34

24.16

(a) uj0 ,in

(b) f j0 ,kips

6399

238.0

j =1

9255 (c) V j ,kips

(d) M j ,kips-ft

~ ~ ~ Γ = L /m = 1.289

4. Determine te natural vibration period.

ωn =

~ ~ k /m = 8.329 rad / sec

Tn = 2π / ω n = 0.754 sec 5. Formulate the equation of motion. ~ 2 && z + ω n z = − Γu&&g ( t ) = −1.289u&&g ( t )

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 15 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 8.16

7. Determine peak responses.

u jo = ψ j z o

1. Determine deflections due to applied force. u = ( p / 600) 3 6 10 14 20

⇒ u10 = 0.713

u40 = 3.32

2. Determine the shape vector.

ψ = 3 6 10 14 20 k j kips/in

50 kips

story shear

story drift

p/100

p 100

100

floor displacement 20p/600 14p/600

150

100

150

100

200

100

p/150

p/200

u50 = 4.75 all in inches.

⇒ f10 = 15.81

f 40 = 73.79

f 20 = 31.62

f 30 =52.71

f 50 =52.7 all in kips.

9. Determine story forces. 10p/600

Static analysis of the structure subjected to external forces f j 0 gives the story shears and floor overturning

6p/600

moments in Figs. (c) and (d). 4.75

p/200

3. Determine generalized properties. 5 ~ 5 ~= ∑ m m jψ j 2 = 140.2 ; k = ∑ k j (ψ j − ψ j −1 ) 2 = 12,000 j =1

j =1

u30 =2.37

8. Determine equivalent static forces. ~ f jo = Γm jψ j A= 20.345 m jψ j

3p/600 200

u20 =1.426

5 ~ ~ ~ ~ = 0.0795 L = ∑ m jψ j = 11.14 ; Γ = L / m j =1

52.70

3.32

73.79

2.37

52.71

1.43

52.7

179.2 210.8

31.62

0.71

15.81

(a) uj0 ,in

(b) f j0 ,kips

632

126.5

2150 4301

6831

226.6

9550 (c) V j ,kips

(d) M j ,kips-ft

4. Determine the natural vibration period. ~

~ = 9.25 rad / sec ωn = k / m

Tn = 2π / ω n = 0.679 sec

5. Formulate the equation of motion.

~ 2 && z + ω n z = − Γu&&g (t ) = −0.0795u&&g (t ) 6. Determine the peak value of z(t). For Tn =0.679 sec and gives:

ζ =0.05 ,the design spectrum

A A 1.8 = 2.989 in. = × 0.25 = 0.663 ; D = g Tn ωn2

~ z o = ΓD = 0.237 in.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 16 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 8.17 1. Determine the shape function.

zz x x

ψ 1 ( x) =

0 0

⇒ ψ1 ( x ) =

ψ 1′′(ξ ) d ξ d ξ ; ψ1 ( 0 ) = ψ1′ ( 0 ) = 0 L 2 x 2 EI

m C

ψ ′′ ( x ′ ) = (L − x ′ ) EI 2

x′

ψ ′′ ( x ) = L EI 1

x A

Moment diagram

Curvature

x′ x′

ψ 2 ( x ′) =

ψ ′′2 (ξ ) d ξ d ξ ;

0 0

L2 EI

ψ 2 ( 0 ) = 0 , ψ ′2 ( 0 ) = ψ1′ ( L ) = ⇒ ψ 2 ( x′) =

L2 L 1 ( x ′ )2 − ( x ′ )3 x′ + EI 2 EI 6 EI

u1 = ψ1 ( L ) =

L3 4 L3 ; u2 = ψ 2 ( L ) = 3 EI 2 EI

2. Determine the natural frequency. m% = ( 2 m + m ) u12 + mu22 = ~ k =

z L

EI ψ 1′′( x )

dx +

ωn =

91m L6 36 E 2 I 2

EI ψ ′′2 ( x ′)

dx ′ =

4 L3 3EI

k% EI = 0. 726 % m m L3

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 17 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 8.18 1. Determine a shape function.

ψ (x ) = 1 − x (1 − ψ r ) L

For ψ r = −1. 366 , ω n = shape is shown:

ψr

1 x

0.366

9. 464 k m and the vibration

1.366

2. Determine natural frequency. k% = k (1)2 + 2 k ψ 2r = k (1 + 2 ψ r2 ) ~ = m

z z L

0 L

ω 2n =

m( x ) ψ ( x )

m 2 1 − x (1 − ψ r ) L dx = m (1 + ψ r + ψ r2 ) L

k% 3k (1 + 2 ψ 2r ) = % m m (1 + ψ r + ψ 2r )

A plot of ω 2n vs. ψ r is shown. 10 8

ω n2

k m

4 2 -1.366

0 -4

-3

-2

0.366

-1

ψr

3. Find stationary values of ω 2n . dω 2n = 0 ⇒ dψ r 4 ψ r (1 + ψ r + ψ 2r ) − (1 + 2 ψ r2 ) (1 + 2 ψ r ) = 0 (1 + ψ r + ψ 2r )2 ⇒ 2 ψ 2r + 2 ψ r − 1 = 0 ⇒ ψ r = 0. 366 or − 1. 366

For ψ r = 0. 366 , ω n = shape is shown:

2. 536 k m and the vibration

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 18 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 8.19 Mass = m

1. Determine the shape function. Assume that the slab is rigid in flexure and column is axially rigid. The system has two degrees of freedom: lateral deflection u ( L ) and rotation θ of the slab. To represent the system as a generalized SDF system, we assume a suitable deflected shape for the column: πx ψ ( x ) = 1 − cos 2L Then u ( x , t ) = zo ψ ( x ) sin ω nt ′ ; uo ( x ) = zo ψ ( x ) u& ( x , t ) = ω n zo ψ ( x ) cos ω nt ′ ; u&o ( x ) = ω n zo ψ ( x )

2. Determine ESo and EKo .

z L

E So =

1 2 EI ( x ) uo′′( x ) dx 2

1 2 = zo 2 =

z L

EI ( x ) ψ ′′( x )

FG IJ H K

π 1 2 zo EI 2 2L

4 L

cos2

FG π x IJ dx = π EI z H 2L K 64 L 4

2 o

1 1 2 2 E Ko = m u&o ( L) + I O u&o′ ( L) 2 2 1 = ω n2 zo2 m ψ ( L) 2 + I O ψ ′( L) 2 2

F π I OP H 2 L K PQ L π F R I OP 1 = ω z m M1 + 2 MN 16 H L K PQ =

LM MN

m R2 1 2 2 ω n zo m + 2 4 2 2 n o

3. Determine ω n . EKo = ESo ⇒ ω n =

π 4 EI 32 L3 m 1 + ( π 2 16 ) ( R L )2

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 19 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 8.20 2L

Moment diagram

0.5

Curvature

1.5

L EI

1 E Ko = 2

1. Determine the shape function. Calculate the deflection due to a unit force applied at the free end: x 0 ≤ x ≤ 2L 2 EI u ′′( x ) = 3L − x 2 L ≤ x ≤ 3L EI Boundary conditions:

R| S| T

u ( 0 ) = u ( 2 L ) = 0 , u′ ( 2 L− ) = u′ ( 2 L+ ) ⇒

LM F 1 L I MN GH 12 x − 3 xJK dx F − 1 x + 3L x − 10 L x + 2 L I dxOP GH 6 2 JK P 3 Q

1 m ω 2n zo2 2 ( EI ) 2 3L

m( x ) u&o ( x )

51m L7 280 ( EI ) 2

ω 2n zo2

3. Determine ω n . EKo = ESo ⇒

ωn =

L3 2 EI 7

51m L 280 ( EI )

1. 657

R| 1 F 1 x − L xI 0 ≤ x ≤ 2L 3 JK | EI GH 12 u( x ) = S || 1 FG − 1 x + 3L x − 10 L x + 2 L IJ 2 L ≤ x ≤ 3L 2 3 K T EI H 6 2

Assumed shape function ψ ( x ) = u ( x ) . u ( x , t ) = zo ψ ( x ) sin ω nt ′ ; uo ( x ) = zo ψ ( x ) u& ( x , t ) = ω n zo ψ ( x ) cos ω nt ′ ; u&o ( x ) = ω n zo ψ ( x )

2. Determine ESo and EKo .

1 E So = 2

z2 = o 2 EI

EI ( x ) uo′′( x )

L3 2 EI

LM F x I MM GH 2 JK dx + N 2L

OP (3 L − x ) dx P PQ 2

zo2

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 20 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 8.21 The beam shown in Fig. P8.21 is statically indeterminate. We will first compute the deflection of the simply supported beam by releasing the two bents. Then, we’ll solve the unknown reactions in the two bents and the total deflection. This is the flexibility method.

m ( x) =

u1 ( x ) =

L2 L (−3x 2 + 3Lx − ) 18 EI y 9

(d)

L 3≤ x≤ L 2

Thus, at x = L / 3 the compatibility condition states u 0 ( L / 3) + Fu1 ( L / 3) + F / k bent = 0

(e)

where the first term is computed from Eq. (a) and the second term is from Eq. (c) or (d).

18 . 45 g

k bent

Substituting L = 375 ft, E = 3,000 ksi, Iy = 65,550 ft4, x = L/3, and k bent = 12,940 kip ft into Eqs. (a) and (c) and after some simplifications, the unknown reaction force F in the bent is

k bent

u 0 ( L / 3) u1 ( L / 3) + 1 / k bent

F =− p ( x) = 1

F = −58.65 kips

Fig. P8.21 1.3 Determine the total deflection. 1. Determine the total deflection of the beam by the flexibility method. We will make the structure statically determinate by releasing the two bents. The resulting primary structure is a simply supported beam.

The total deflection is u ( x) = u 0 ( x) + Fu1 ( x)

It is convenient to express F = −58.65 kips = −0.1564 L kips. Substituting Eqs. (a), (c), and (d) in Eq. (f) gives the total deflection:

1.1 Determine the deflection of the primary structure. The deflection u 0 ( x) of a simply supported beam under uniform force p( x) = 1 is u 0 ( x) =

x ( L3 − 2 Lx 2 + x 3 ) 24 EI y

(a)

u 0 ( L 2) =

5L4 384 EI y

0≤ x ≤ L/3 u ( x) =

(g)

10 −2 (4.167 x 4 − 8.333Lx 3 + 2.6065 L2 x 2 EI y L/3≤ x ≤ L/2

(h) 2. Compute natural frequency.

Since the structure is symmetric about x = L 2 , we can utilize the property of symmetry and thus reduce the problem to one-degree indeterminacy. It can be shown that the deflection of the beam due a unit force at the bent is given by x (−3x 2 + 2 L2 ) 18 EI y

x (0.04167 x 3 − 0.05727 Lx 2 + 0.02429 L3 ) EI y

(b)

1.2 Determine the unknown reaction force F at the bent.

u1 ( x ) =

u ( x) =

+ 1.5602 L3 x + 0.09654 L4 )

and the deflection at the midspan is

(f)

0≤ x≤L 3

(c)

From Eq. 8.6.1, specialized for p( x) = 1 m( x) = m , the natural vibration frequency is:

and

ω n2 =

∫ u( x)dx 0

∫

(i)

m[u ( x)] 2 dx

Substituting Eqs. (g) and (h) into Eq. (i) and evaluating the integrals using MathCAD gives

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 21 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

ωn = Tn =

1.255 = 20.612 rad/sec. 0.00296 2π

ωn

= 0.305 sec.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 22 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 8.22 1. Determine the mid-span deflection u o of the beam due to p( x) = 1 . Substituting L = 375 ft, E = 3,000 ksi, Iy = 65,550 ft4, and x = L / 2 into Eq. (h) of Problem 8.21, the mid-span deflection is u o = 5.217 × 10 −3 ft

(a)

2. Compute natural frequency. The deflection is given by u ( x) = u o sin(πx / L)

(b)

with u o given by Eq. (a). From Eq. 8.6.1, specialized for p( x) = 1 and m( x) = m , the natural vibration frequency is: L

ω n2 =

∫ u( x)dx 0

∫

(c)

m[u ( x)] 2 dx

Substituting Eq. (b) into Eq. (c) with m = 18.45 / g and evaluating the integrals gives L

ω n2 = uo

ωn = Tn =

πx

∫ sin L dx L

2 πx

∫ m sin L dx

π mu o

4g 18.45π ⋅ 5.217 × 10 −3

= 20.639 rad/sec.

2π = 0.304 sec. ωn

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 23 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 8.23 1. Determine the mid-span deflection u o of the beam due to p( x) = 1 . Substituting L = 375 ft, E = 3,000 ksi, Iy = 65,550 ft4, and x = L / 2 into Eq. (h) of Problem 8.21, the mid-span deflection is u o = 5.217 × 10 −3 ft

2. Determine the deflection u(x). The deflection is given by u ( x) = u oψ ( x)

(a)

where ψ (x) will be taken as the deflected shape of a simply supported beam without bents. Dividing Eq. (a) of Problem 8.21 by Eq. (b), also of Problem 8.21, gives

ψ ( x) =

3 4 16 ⎡ x ⎛x⎞ ⎛x⎞ ⎤ ⎢ − 2⎜ ⎟ + ⎜ ⎟ ⎥ 5 ⎢⎣ L ⎝ L ⎠ ⎝ L ⎠ ⎥⎦

(b)

Substituting ψ (x) in Eq. (a) gives u ( x) =

3 4 16u o ⎡ x ⎛x⎞ ⎛x⎞ ⎤ ⎢ − 2⎜ ⎟ + ⎜ ⎟ ⎥ 5 ⎢⎣ L ⎝ L ⎠ ⎝ L ⎠ ⎥⎦

(c)

3. Compute natural frequency. From Eq. 8.6.1, specialized for p( x) = 1 m( x) = m , the natural vibration frequency is:

and

ω n2 =

∫ u( x)dx 0

∫

(d)

m[u ( x)] 2 dx

Substituting Eq. (c) into Eq. (d) with m = 18.45 / g and evaluating the integrals gives

ωn = Tn =

1.252 = 20.614 rad/sec. 0.00295 2π = 0.305 sec. ωn

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 24 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 8.24

Δ 22 = u ' 2 ( L / 3) = 2.207 × 10 −9 rad

Including the torsional stiffness of the bents adds another degree of freedom to the system, namely, the rotational degree of freedom. Thus, the beam is now indeterminate to two degrees. We will use the classical flexibility method in solving this problem. 1. Determine the total deflection of the beam by the flexibility method. Note that the deflection u 0 ( x) of the simply support beam (the primary structure) as well as the deflection u1 ( x) of the beam due to unit lateral force at the bend (first released structure) were given by Eqs. (a), (c) and (d) of Problem 8.21, respectively. Next, we need to determine the deflection u 2 ( x) of the beam due to unit rotational moment at the bend. It can be easily shown that the deflection of the beam due to unit moment at the bent is given by u 2 ( x) =

u 2 ( x) =

Lx 6 EI y

0≤ x ≤ L/3

F2 = −40.98 kips-ft.

Finally, the total deflection is u ( x) = u 0 ( x) + F1u1 ( x) + F2 u 2 ( x)

2 1 ⎡ Lx ⎛ L⎞ ⎤ ⎢ −⎜x− ⎟ ⎥ 2 EI y ⎢⎣ 3 ⎝ 3 ⎠ ⎥⎦

2. Compute natural frequency. From Eq. 8.6.1, specialized for p( x) = 1 m( x) = m , the natural vibration frequency is:

Δ 10 + F1 (Δ 11 + 1 k bent ) + F2 Δ 12 = 0

(c)

Δ 20 + F1 Δ 21 + F2 (Δ 22 + 1 kθ ) = 0

(d)

and

∫ u( x)dx 0

∫

(b)

(e)

where u 0 ( x) is defined by Eq. (a) of Problem 8.21; u1 ( x) is defined by Eqs. (c) and (d) of Problem 8.21; and u 2 ( x) is given by Eqs. (a) and (b) above.

(a)

The compatibility conditions at the location of the bents ( x = L / 3 ) state

kθ

F1 = −58.56 kips.

ω n2 =

L/3< x ≤ L/2

where

Thus, solving Eqs. (c) and (d) simultaneously gives

(f)

m[u ( x)] 2 dx

Substituting Eq. (e) into Eq. (f) and evaluating the integrals using MathCAD gives

ωn = Tn =

1.254 = 20.628 rad/sec. 0.002946 2π

ωn

= 0.305 sec.

is the torsional stiffness of the bent,

kθ = 1.941 × 10 6 kips-ft assuming that the distance between the adjacent columns in the bent is 30 ft.

Substituting L = 375 ft, E = 3,000 ksi, Iy = 65,550 ft4 and x = L/3 into Eqs (a) and (c) of Problem 8.21 and Eqs. (a) and (b) above, we have Δ 10 = u 0 ( L / 3) = 7.903 × 10 −3 ft Δ 20 = u ' 0 ( L / 3) = 3.736 × 10 −5 rad Δ 11 = u1 ( L / 3) = 5.748 × 10 −5 ft Δ 21 = u '1 ( L / 3) = 2.759 × 10 −7 rad Δ 12 = u 2 ( L / 3) = 2.759 × 10 −7 ft © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 25 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 8.25

From Eq. (8.3.26) the generalized force is Infinitely long i po v

~ p (t ) =

∫ ∫

vt L Fig. P8.25

1. Determine the generalized mass, generalized stiffness, and natural frequency.

πx

ψ ′′( x) = −

π2 2

sin

(e)

πx L

This generalized force is plotted next:

~ = m sin 2 πx dx = mL m 2 L

~ p

∫

(a)

⎛π 2 ⎞ ~ πx π 4 EI k = EI⎜⎜ 2 ⎟⎟ sin 2 dx = L 2 L3 ⎝L ⎠ 0

∫

ωn =

2 po L

~ π2 k = ~ m L2

EI m

(b) Lv

(c)

2. Determine the generalized force. The front of the load p( x, t ) traveling with a velocity v takes time t d = L v to cross the bridge. At any time t its position is as shown in Fig. P8.25. Thus the moving load can be described mathematically as ⎧ po ⎪ p ( x, t ) = ⎨ 0 ⎪p ⎩ o

0 ≤ x ≤ vt

∫ p( x, t )ψ ( x) dx

⎧ vt p o sin (π x L) dx 0 ≤ t ≤ t d ⎪⎪ = ⎨ 0L ⎪ p sin (πx L) dx t ≥ t d ⎪⎩ 0 o ⎧ po L ⎡ πv ⎤ 1 − cos t 0 ≤ t ≤ td ⎪⎪ ⎢ L ⎥⎦ =⎨ π ⎣ 2 po L ⎪ t ≥ td π ⎩⎪

m, EI

ψ ( x) = sin

0 ≤ t ≤ td

vt < x ≤ L 0 ≤ t ≤ t d 0 ≤ x ≤ L t ≥ td

(d)

Observe that this excitation has some similarity to Fig. 4.5.1b. 3. Solve the equation of motion. ~&z& + k~z = ~ m p (t )

(f)

Response for 0 ≤ t ≤ t d . The particular solution to Eq. (f) can be obtained by superposing the steady-state responses to the constant and the cosine term on the right-hand side of Eq. (e). The ~ steady-state response to the constant term is ~ p o k , where ~ p o = p o L π , and that for the cosine term is adapted from Eqs. (3.2.3) and (3.2.26), noting that ς = 0 and replacing ~ p and k by ~ p and k respectively. The complete o

solution is obtained by adding the complementary solution, given by Eq. (3.1.4), to the particular solution, and determining the constants A and B by imposing zero initial conditions. The result is

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 26 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

z (t ) =

At midspan, x = L 2 and

2 po 1 ⎡ ω n2 πv − 1 cos t ⎢ 2 2 2 L πm ω n ⎢⎣ ω n − (πv L )

⎛L ⎞ u⎜ , t ⎟ = z (t ) ⎝2 ⎠

(πv L ) cosω t ⎤ + 2 n ⎥ ω n − (πv L )2 ⎦⎥ 2

t≤L v

(g)

Thus the deflection at midspan is also given by Eqs. (g) and (j).

Equation (g) is valid if ω n ≠ πv L ; otherwise the particular solution to the cosine term should formulated similar to that in Eq. (3.1.12), noting that the forcing function is a cosine function instead of a sine function. Response for t ≥ t d . The motion is described by Eq. (4.5.3) with z instead of u , t d instead of t r , and z (t d ) and z& (t d ) determined from Eq. (g): z (t d ) =

2 po 1 ⎡ ω n2 1 + ⎢ πm ω n2 ⎢⎣ ω n2 − (πv L )2 +

(πv L )2 cosω t ⎤ n d⎥ ω n2 − (πv L )2 ⎥⎦ (h)

z& (t d ) = −

2 po 1 (πv L )2 sinω t n d πm ω n ω n2 − (πv L )2

(i)

Substituting these in Eq. (4.5.3), using trigonometric identities, and manipulating the mathematical quantities, we obtain z (t ) =

⎫⎪ 2 p o 1 ⎡ ⎪⎧ ω n2 ⎢2 − ⎨1 − ⎬ cosω n (t − t d ) 2 2 2 πm ω n ⎢ ⎪⎩ ω n − (πv L ) ⎪⎭ ⎣ +

(πv L )2 cosω t ⎤ n ⎥ ω n2 − (πv L )2 ⎦⎥ t≥L v

(j)

The generalized coordinate response is given by Eq. (g) while the front of the load is on the bridge span and by Eq. (j) after the front has crossed the span. 4. Determine the deflection at midspan. u ( x, t ) = z (t )ψ ( x) = z (t ) sin

πx L

(k)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 27 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 8.26 ~ p (t ) =

p o cosωt

⎧ L ⎪ p cos ω t δ ( x − vt ) sin (π x L) dx 0 ≤ t ≤ t d =⎨ 0 o ⎪⎩ 0 t ≥ td

∫

m, EI

⎧ p cosω t sin (πx L) 0 ≤ t ≤ t d =⎨ o t ≥ td ⎩ 0 ⎧ p cosω t sin (π t t d ) 0 ≤ t ≤ t d =⎨ o t ≥ td ⎩ 0

Fig. P8.26 1. Determine the generalized mass, generalized stiffness, and natural frequency.

ψ ′′ x = −

z F I z GH JK

∫ p( x, t )ψ ( x) dx 0

ψ x = sin

⎧ po ⎪ [sin (ω + π t t d ) − sin (ω − π t t d )] 0 ≤ t ≤ t d ~ p (t ) = ⎨ 2 ⎪⎩ 0 t ≥ td

π2

πx sin 2 L L

~ = m sin 2 πx dx = mL m L 2

This can be re-written as

(e)

(a)

~ k =

ωn =

π2

~ π2 k = ~ m L2

sin 2

πx L

dx =

π 4 EI 2 L3

EI m

(b)

(f)

Forced vibration phase (c)

2. Determine the generalized force. A load p( x, t ) traveling with a velocity v takes time t d = L v to cross the bridge. At any time t its position is as shown in Fig. P8.26. Thus the moving load can be described mathematically as ⎧ p cosωt δ ( x − vt ) 0 ≤ t ≤ t d p ( x, t ) = ⎨ o t ≥ td ⎩0

3. Solve the equation of motion. ~&z& + k~z = ~ m p (t )

(d)

where δ ( x − vt ) is the Dirac delta function centered at x = vt ; it is a mathematical description of the travelling concentrated load. From Eq. (8.3.26) the generalized force is

The response z (t ) can be obtained by superposing the responses due to the two sine terms in the right-hand side of Eq. (e). The individual responses are adapted from Eq. (3.1.6b) by changing the notation from u (t ) to z (t ) , substituting for ω with ω + πt t d in one case and with ω − πt t d in the other, and noting that td =

p po ( zst ) o = ~o = 2k mLω n2

L v

The result is z (t ) =

⎧ ⎛ po ⎡ πv ⎞ 1 ⎢ 2 ⎨sin ⎜ ω + ⎟ t 2 mL ⎣⎢ ω n − (ω + πv L ) ⎩ ⎝ L⎠ ⎫ ⎛ ω + πv L ⎞ ⎟ sinω n t ⎪⎬ − ⎜⎜ ⎟ ⎪⎭ ⎝ ωn ⎠ −

⎧ ⎛ πv ⎞ sin ⎜ ω − ⎟ t L⎠ ) ⎝

⎨ ω n2 − ω − πv L 2 ⎩

(

⎫⎤ ⎛ ω − πv L ⎞ ⎟ sinω n t ⎪⎬⎥ − ⎜⎜ ⎟ ⎪⎭⎥⎦ ⎝ ωn ⎠ t≤L v

(g)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 28 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Based on Eq. (3.1.6b), Eq. (g) is valid if ω n ≠ ω + πv L and ω n ≠ ω − πv L ; otherwise Eq. (3.1.13a) should be used instead of Eq. (3.1.6b). Free vibration phase. The motion is described by Eq. (4.7.3) with z (t ) instead of u (t ) , and z (t d ) and z& (t d ) determined from Eq. (g) : z (t d ) = +

At midspan, x = L 2 and ⎛L ⎞ u ⎜ , t ⎟ = z (t ) ⎝2 ⎠

Thus the deflection at midspan is also given by Eqs. (g) and (j).

⎫⎪ po ⎡⎧⎪ 1 1 − 2 sin ωL v ⎢⎨ 2 2 2⎬ mL ⎢⎣⎪⎩ ω n − (ω − πv L) ω n − (ω + πv L) ⎪⎭

⎤ 1 ⎧⎪ (ω − πv L) (ω + πv L) ⎪⎫ − L v sin ω ⎥ ⎨ 2 ⎬ n ω n ⎪⎩ω n − (ω − πv L) 2 ω n2 − (ω + πv L) 2 ⎪⎭ ⎥⎦ (h)

z& (t d ) =

po ⎡⎧⎪ (ω − πv L) (ω + πv L) ⎫⎪ − 2 ⎢⎨ 2 ⎬ 2 mL ⎢⎣⎪⎩ω n − (ω − πv L) ω n − (ω + πv L) 2 ⎪⎭

ωL ⎫ ⎤ ⎧ ω L × ⎨cos n + cos ⎬⎥ v v ⎭⎦ ⎩ (i) Substituting these in Eq. (4.7.3), using trigonometric identities, and manipulating the mathematical quantities, we obtain z (t ) =

⎫⎪ po ⎡⎧⎪ 1 1 − 2 ⎢⎨ 2 ⎬ 2 2 mL ⎣⎢⎪⎩ ω n − (ω − πv L) ω n − (ω + πv L) ⎪⎭ × sin

ωL v

⎛ L⎞ cosω n ⎜ t − ⎟ ⎝ v⎠

1 ⎧⎪ (ω − πv L) (ω + πv L) ⎫⎪ − ⎨ 2 ⎬ ω n ⎪⎩ω n − (ω − πv L) 2 ω n2 − (ω + πv L) 2 ⎪⎭ ⎧ ωL ⎫⎤ ⎛ L⎞ × ⎨cos sinω n ⎜ t − ⎟ + sinω n t ⎬⎥ v ⎝ v⎠ ⎩ ⎭⎦⎥ t≥L v

(j)

The generalized coordinate response is given by Eq. (g) while the moving load is on the bridge span and by Eq. (j) after the load has crossed the span. 4. Determine the deflection at midspan. u ( x, t ) = z (t )ψ ( x) = z (t ) sin

πx L

(k)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 29 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

CHAPTER 9

Problem 9.1

2. Determine the stiffness matrix. Apply a unit displacement u1 = 1 with u2 = 0 and identify the resulting forces and the stiffness influence coefficients, k11 and k21 .

pt u1

rigid bar

pθ m

k1 L /2

k 11

L /2

k 21

u1= 1

1. Determine the force vector. Introduce a virtual displacements δu1 along DOF 1. The work done by the applied forces pt and pθ , and by the equivalent forces p1 and p2 is

δW = pt ( δu1 ) + pθ ( 0 )

(a)

= p1 ( δu1 ) + p2 ( 0 ) p1

By statics, k11 = k1 + k2

k21 = k2 L

Similarly, apply a unit displacement u2 = 1 with u1 = 0 and identify the resulting forces and the stiffness influence coefficients, k12 and k22 . k 12

δ u1

k 22

pθ

Similarly, introduce a virtual displacement δu2 along DOF 2. The work done by the applied and equivalent forces is

δW = pt

FG δu L IJ + p (δu ) H 2 K 2

(b)

δ u 2 L /2

k22 = k2 L2

LMk + k N kL 1

δ u2 L

k2 L k 2 L2

OP Q

(d)

3. Determine the mass matrix.

From the above equations, the vector of equivalent forces is

k2 L

k12 = k2 L

k =

pθ

R| p U| S| p L + p V| W T2

Thus the stiffness matrix is

pt δ u2

u 2= 1

By statics,

= p1 (0) + p2 (δu2 )

p =

(c)

Impart a unit acceleration u&&1 = 1 with u&&2 = 0 , determine the distribution of acceleration and the associated inertia forces, and identify the influence coefficients, m11 and m21 .

&&u 1= 1 m 11

Inertia forces = m / L

m 21

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 1 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

By statics, m11 = m

m21 =

mL 2

Similarly, impart a unit acceleration u&&2 = 1 with u&&1 = 0 , determine the distribution of acceleration and the associated inertia forces, and identify the influence coefficients, m12 and m22 .

&&u 2 = 1 m 12

Inertia forces = ( m / L )/x

m 22

By statics, m L2 mL m22 = 2 3 Thus the mass matrix is

m12 =

LM m m L OP 2 m=M P m L m L P MM N 2 3 QP

(e)

4. Write the equations of motion. Substituting Eqs. (c), (d) and (e) in Eq. (9.2.12), with c = 0 , gives

LM m m L OP && MM m L m2L PP RSTu&&u UVW + LMNk k+Lk NM 2 3 QP 2

k2 L k 2 L2

OP RSu UV = R|S p L p U|V Q Tu W |T 2 + p |W 1

(f) These two differential equations are coupled because of mass coupling and stiffness coupling.

Similarly to obtain the third column, apply u3 = 1 with all other u j = 0 and identify the resulting elactic forces and the stiffness influence coefficients.

Problem 9.2 p1 ( t )

p2 ( t )

mL /3 L /3

mL /3

L /3

k 13

k 53

k 33

L /3

54 EI / L

Part a The elastic properties of the beam (neglecting axial deformation) are represented by six DOFs: two translational displacements and four rotational displacements.

6EI / L

k 31

k 41

54 EI /L

54 EI / L

LM 648 − 324 0 54 L − 54 L 0OP MM − 324 648 − 54 L 0 0 54 LPP 0 − 54 L 24 L 6L 6L 0 EI k = P M 0 6 L 24 L 0 6L P L M 54 L MM− 54 L 0 6 L 0 12 L 0PP 0 12 L PQ MN 0 54 L 0 6 L 2

(a)

The stiffness matrix is partitioned: k =

LM k Nk

tt 0t

kt0 k 00

OP Q

(b)

where the subscript t identifies the translational displacements, u1 and u2 , and the subscript 0 identifies the rotational displacements, u3 , u4 , u5 and u6 .

k 61

Part b

324 EI / L 3 54 EI / L

54 EI / L

Other elements of the stiffness matrix are obtained similarly. Apply a unit displacement ui = 1 while u j = 0 , j ≠ i . Identify the resulting elastic forces and by statics obtain the stiffness coefficients kij .

elastic forces and the stiffness influence coefficients.

k 51

12 EI / L

6EI / L

The coefficients of the stiffness matrix corresponding to these DOF are computed following Example 9.4. For instance, to obtain the first column of the stiffness matrix, apply a unit displacement u1 = 1 while the other displacements u j = 0 , j = 2, 3, ···, 6. Identify the resulting

k 21

k 43 2

5 6

k 11

k 63

The complete stiffness matrix is

R|u U| Ru U Ru U |u | u = S V u = S V u = S V Tu W Tu W ||u || Tu W t

k 23

The DOF representing the inertial properties are the two translational displacements u1 and u2 associated with the concentrated masses.

324 EI / L 3

By statics, k11 = k41 =

648 EI L3 54 EI L2

k21 = k51 =

− 324 EI L3 − 54 EI L2

k31 = 0

k61 = 0

u = ut =

RSu UV Tu W 1

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 3 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

To obtain the coefficients of the mass matrix for these DOF, apply first a unit acceleration u&&1 = 1 , while u&&2 = 0 . Next apply a unit acceleration u&&2 = 1 , while u&&1 = 0 . Determine the associated inertia forces and identify the mass coefficients: u&&1 = 1, u&&2 = 0

u 1 = 0, && u2= 1 &&

u&&1 = 1

u2= 1 &&

m11

m 21

m12

m 22

mL /3

m11 =

mL 3

m12 = 0

m22 =

mL 3

m21 = 0

Thus the mass matrix is m =

LM OP N Q

mL 1 0 3 0 1

(c)

Part c The condensed stiffness matrix for the two vertical DOF is k$ = k − k k − 1 k ⇒ tt

LM N

8 −7 162 EI k$ tt = 3 8 −7 5L

OP Q

(d)

The force vector is given by p(t ) =

RS p (t ) UV T p (t ) W 1

(e)

Substituting Eqs. (c), (d) and (e) in Eq. (9.2.12) with c = 0 , gives the equation governing the translational motion of the beam:

LM OP RSu&& UV + 162 EI LM 8 − 7OP RSu UV = RS p (t ) UV N Q Tu&& W 5L N− 7 8Q Tu W T p (t )W

mL 1 0 3 0 1

(f)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 4 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 9.3

3. Write the equations of motion. p1 ( t )

p2 ( t )

mL /3

L /3

Using Eqs. (d) and (e), the equation governing the translational motion of the beam is the same as in Problem 9.2:

LM OP RSu&& UV + 162 EI LM 8 − 7OP RSu UV = RS p (t ) UV N Q Tu&& W 5L N− 7 8Q Tu W T p (t )W

mL 1 0 3 0 1

L /3

(f) 1. Determine the stiffness matrix. The flexibility matrix is calculated first and inverted to obtain the stiffness matrix. The flexibility influence coefficient f$ij is the displacement in DOF i due to unit load applied along DOF j. The deflections due to unit load at node 1 are computed by standard procedures of structural analysis to obtain two of the influence coefficients: f$11 =

4 L3 243 EI

f$21 =

7 L3 486 EI

(a)

The deflections due to unit load at node 2 are computed to obtain the other two influence coefficients: f$12 =

7 L3 486 EI

f$11

f$22 =

4 L3 243 EI

(b) f$12

f$21

f$22

Thus the flexibility matrix is f$ =

LM OP N Q

8 7 L3 486 EI 7 8

(c)

The stiffness matrix is obtained by inverting f$ : k = f$ − 1 =

LM 8 − 7OP 8Q 5 L N− 7

162 EI 3

(d)

2. Determine the mass matrix. The mass matrix for the translational DOF is same as in Problem 9.2: m =

LM OP N Q

mL 1 0 3 0 1

(e)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 5 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 9.4

By statics,

m =

Impart a unit acceleration u&&1 = 1 , with u&&2 = 0 , determine the acceleration distribution and the associated inertia forces.

m ⎡2 1⎤ ⎢ ⎥ 6 ⎣1 2 ⎦

(c)

f$11 =

L3 3EI

f$21 =

L3 + 3EI

F L I L = 5L GH 2 EI JK 6EI 2

Inertia forces = (m/L)(x/L)

5 L3 f$12 = 6 EI

m11

5 L3 + f$22 = 6 EI

f$21 = f$11+ θ L

p2 = 1

F 3L I L = 7 L GH 2 EI JK 3EI 2

By statics, m 3

p1 = 1

m 6

m21 =

(a)

(e)

f$22 = f$12+ θ L θ =L2/2EI + L2/EI

θ =L2/2EI

m11 =

(d)

Similarly, apply a unit force p2 = 1 , with p1 = 0 . The other two displacements or influence coefficients due to this force are

m 21

u 1= 1 &&

(b)

Apply a unit force p1 = 1 along DOF 1 with p2 = 0 along DOF 2. The first two displacements or influence coefficients due to this force are computed following standard procedures of structural analysis:

1. Determine the mass matrix.

m 3

2. Determine the flexibility matrix.

EI, massless

u2 = 0 &&

m22 =

Thus the mass matrix is

Rigid; total mass = m u1

m 6

m12 =

f$11 =L3/3EI

=3L /2EI f$12 =L3/3EI + (L2/2EI )L =5 L3/6EI

Similarly, impart a unit acceleration u&&2 = 1 , with u&&1 = 0 , determine the acceleration distribution and the associated inertia forces. m 22

u2= 1 && x

&&u 1 = 0

Inertia forces = (m/L)(x/L) m12

Thus the flexibility matrix is f$ =

LM N

OP Q

L3 2 5 6 EI 5 14

(f)

The stiffness matrix is computed by inverting the flexibility matrix f$ : k = f$ − 1 =

LM 28 − 10OP 4Q L N− 10

EI 3

(g)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 6 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Apply u&&2 = 1 , u&&1 = 0 ⇒

Problem 9.5 1. Determine the stiffness matrix. The story stiffness is k = 2

m12 = 0

m22 =

m/2

&& u2 = 1

FG 12 EI IJ = 24 EI Hh K h 3

m 2 m22

m/2

Apply u1 = 1, u2 = 0 and determine ki1 :

m12

k21 = – k u1= 1

k11 = 2k

Thus, the mass matrix is m = m

Apply u2 = 1, u1 = 0 and determine ki 2 :

u 2= 1

LM1 OP N 0.5Q

3. Write the equations of motion.

k22 = k

LM1 OP RSu&& UV + k LM 2 − 1OP RSu UV = RS p (t ) UV N 0.5Q Tu&& W N− 1 1Q Tu W T p (t )W 1

k12 = – k

Thus, the stiffness matrix is k = k

LM 2 − 1OP N− 1 1Q

2. Determine the mass matrix. Apply u&&1 = 1 , u&&2 = 0 ⇒ m11 = m

m21 = 0

m/2 m

m 21

&& u1 = 1 m

m 11

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 7 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 9.6

Part b

p (t)

m/2

p (t)

The DOFs representing the inertial properties are the two translational displacements u1 and u2 . h

m/2

L=2h

Part a

The elastic properties of the shear frame (neglecting axial deformation) are represented by six DOFs: two horizontal displacements and four rotational displacements.

The mass matrix is

The coefficients of the stiffness matrix corresponding to these DOF are computed following Example 9.7.

m = m

u = ut =

RSu UV Tu W 1

LM1 OP N 0.5Q

(c)

Part c u3

The condensed stiffness matrix is −1 k$ tt = ktt − kt 0 k00 k0t

. O LM 37.15 − 1512 . 1019 . PQ h N− 1512

Substituting Eqs. (c) and (d) in Eq. (9.2.12) gives

R|u U| Ru U Ru U |u | u = S V u = S V u = S V u u T W T W ||u || Tu W t

. O Ru U LM1 OP RSu&& UV + EI LM 37.15 − 1512 R p (t ) UV = S S V P . . Q Tu W 1019 N 0.5Q Tu&& W h N− 1512 T p (t ) W 1

The complete stiffness matrix is

(d)

LM 48 − 24 0 0 − 6h − 6hOP MM− 24 24 6h 6h 6h 6hPP 0 6h 10h 1h 2h 0 EI k = P (a) M 6h 1h 10h 0 2h P h M 0 MM− 6h 6h 2h 0 6h 1h PP MN− 6h 6h 0 2h 1h 6h PQ 3

This matrix can be written in partitioned form as follows: k =

LM k Nk

kt0

k 00

OP Q

(b)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 8 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

The stiffness matrix is

Problem 9.7 Since the beams are rigid in flexure and axial deformation is neglected in columns, three DOFs associated with each story represent the properties of this three-story shear building. The corresponding story masses and story stiffness are: m1 = m

m2 = m

k1 = k2 = k3 = 2

m3 =

m 2

12 EI 24 EI = ≡ k h3 h3

(a)

LM 2 − 1 0OP k = k M− 1 2 −1 MN 0 − 1 1PPQ

(c)

2. Determine the mass matrix. Apply u&&1 = 1 , u&&2 = u&&3 = 0 ⇒ m11 = m m21 = m31 = 0

(b) m/2

m 31

1. Determine the stiffness matrix. Apply u1 = 1 , u2 = u3 = 0 and determine ki1 :

m 21

k 31 = 0 u1 = 1 m && m

k 21 = – k u 1= 1

m 11

k 11 = 2k

Apply u&&2 = 1 , u&&1 = u&&3 = 0 ⇒ m22 = m m12 = m32 = 0 Apply u2 = 1 , u1 = u3 = 0 and determine ki 2 : k 32 = – k u 2= 1

m/2

m 32

u 2= 1 m && m

k 22 = 2k

m 22 m 12

k 12 = – k

Apply u3 = 1 , u1 = u2 = 0 and determine ki 3 : u 3= 1

k 33 = k

Apply u&&3 = 1 , u&&1 = u&&2 = 0 ⇒ m m33 = m13 = m23 = 0 2 m/2

&& u3 = 1 k 23 = – k

k 13 = 0

m/2

m 33 m 23 m 13

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 9 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

The mass matrix is

OP LM1 m = mM 1 MN 0.5PPQ

(d)

3. Write the equations of motion.

LM1 OP R|u&& U| LM 2 − 1 0OP R|u U| R| p (t ) U| mM 1 Su&& V + k − 1 2 − 1 Su V = S p (t )V MN 0.5PPQ |Tu&& |W MMN 0 − 1 1PPQ |Tu |W |T p (t ) |W 1

(e)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 10 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 9.8 Since the beams are rigid in flexure and axial deformation is neglected in columns, the three floor displacements are the three DOFs. The floor masses and story stiffnesses are: m1 = m k3 = 2

m2 = m

m3 = m / 2

(a)

k2 = 2k

⎡ 5 −2 0 ⎤ k = k ⎢⎢− 2 3 − 1⎥⎥ ⎢⎣ 0 − 1 1 ⎥⎦

(c)

2. Determine the mass matrix.

12( EI / 3) 8 EI = 3 ≡k h3 h

k1 = 3k

The stiffness matrix is

Apply u&&1 =1, u&&2 =u&&3 = 0⇒ (b)

m11 =m

m21 = m31 = 0 m/2

1. Determine the stiffness matrix.

m 31

m 21

Apply u1 =1, u2 = u3 = 0 and determine ki1 : u1 = 1 m &&

k 31 = 0

m 11

k 21 = –2 k u 1= 1

Apply u&&2 =1, u&&1 =u&&3 = 0⇒ k 11 = 5k

m22 =m

m12 =m32 =0 m/2

Apply u2 =1, u1 = u3 = 0 and determine ki2 : k 32 = – k u 2= 1

u 2= 1 m && m m

m 22 m 12

k 22 = 3k k 12 = –2 k

Apply u&&3 =1, u&&1 = u&&2 = 0⇒ m33 =m/ 2

m13 =m23 = 0

m/2

&& u3 = 1

Apply u3 =1, u1 =u2 = 0 and determine ki3 : u 3= 1

k 33 = k

m/2

m 23 m 13

k 23 = – k

k 13 = 0

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 11 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

The mass matrix is ⎡1 0 0 ⎤ m = m ⎢⎢0 1 0 ⎥⎥ ⎢⎣0 0 1 2⎥⎦

(d)

3. Write the equations of motion. ⎡1 0 0 ⎤ ⎧ u&&1 ⎫ ⎡ 5 - 2 0 ⎤ ⎧ u1 ⎫ ⎧ p1 (t ) ⎫ ⎢0 1 0 ⎥ ⎪ && ⎪ + k ⎢- 2 3 - 1⎥ ⎪ ⎪ ⎪ ( )⎪ m⎢ ⎥ ⎨u 2 ⎬ ⎢ ⎥ ⎨u 2 ⎬ = ⎨ p 2 t ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎣⎢0 0 1 2⎦⎥ ⎩u&&3 ⎭ ⎣⎢ 0 - 1 1 ⎥⎦ ⎩u 3 ⎭ ⎩ p 3 (t ) ⎭

(e)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 12 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 9.9

p (t)

m/2

EI p (t)

p (t) 1

h h

⎡ 48 ⎢ − 24 ⎢ ⎢ 0 ⎢ ⎢ 0 EI ⎢ k = 0 h3 ⎢ ⎢− 6h ⎢− 6h ⎢ ⎢ 0 ⎢ ⎣ 0

0 48 − 24

− 24

0 6h

− 6h 0

24 0 0 6h 2 2 0 10h 1h 2h 2 2 2 0 1h 10h 0 6h 2 h 2 0 10h 2 6h 0 2h 2 1h 2

6h 6h 0 0 − 6h − 6h

6h 6h

0 0

2h 2 0

− 6h

0 0⎤ 0 − 6h − 6h ⎥⎥ 6h 6h 6h ⎥ ⎥ 0 0 0⎥ 2h 2 0 0⎥ ⎥ 2 2 1h 2h 0⎥ 10h 2 0 2h 2 ⎥ ⎥ 0 6h 2 1h 2 ⎥ ⎥ 2h 2 1h 2 6h 2 ⎦

(a) The stiffness matrix can be written in partitioned form as follows: ⎡k k = ⎢ tt ⎣k 0t

L= 2h

− 24

k t0 ⎤ k 00 ⎥⎦

(b)

Part b Part a

The elastic properties of the frame (neglecting axial deformation) are represented by 9 DOFs: three horizontal displacements and six rotational displacements.

⎧ u1 ⎫ ⎧u t ⎫ ⎪ ⎪ u = ⎨ ⎬ u t = ⎨u 2 ⎬ ⎩u 0 ⎭ ⎪u ⎪ ⎩ 3⎭

The DOFs representing the inertial properties are the three translational displacements, u1 , u2 and u3 . m/2

u3 u2 u1

⎧u 4 ⎫ ⎪ ⎪ ⎪u 5 ⎪ ⎪⎪u ⎪⎪ u0 = ⎨ 6 ⎬ ⎪u 7 ⎪ ⎪u 8 ⎪ ⎪ ⎪ ⎪⎩u 9 ⎪⎭

The coefficients of the stiffness matrix corresponding to these DOFs are computed following Example 9.7. The complete stiffness matrix is

⎧ u1 ⎫ ⎪ ⎪ u = u t = ⎨u 2 ⎬ ⎪u ⎪ ⎩ 3⎭

The mass matrix is ⎡1 ⎤ ⎥ m = m ⎢⎢ 1 ⎥ ⎢⎣ 0.5⎥⎦

(c)

Part c

The condensed stiffness matrix for the three lateral DOFs is −1 k$ tt = ktt − kt 0 k00 k0t

5.11⎤ ⎡ 40.85 −23.26 EI ⎢ = 31.09 −14.25⎥⎥ ⎢− 23.26 h 3 ⎢ 5.11 −14.25 10.06⎥ ⎣ ⎦

(d)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 13 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

The equation governing the translational motion of the building is 5.11⎤ ⎧ u1 ⎫ ⎧ p1(t ) ⎫ ⎡ 40.85 − 23.26 ⎡1 ⎤ ⎧ u&&1 ⎫ ⎪ ⎪ ⎪ ⎪ ⎥ ⎪u&& ⎪ + EI ⎢− 23.26 − m ⎢⎢ 1 31 . 09 14 .25⎥⎥ ⎨u2 ⎬ = ⎨ p2 (t ) ⎬ ⎬ ⎨ 2 ⎢ ⎥ 3 h ⎢ 5.11 −14.25 10.06⎥ ⎪u ⎪ ⎪ p (t ) ⎪ ⎢⎣ 0.5⎥⎦ ⎪⎩u&&3 ⎪⎭ ⎣ ⎦ ⎩ 3⎭ ⎩ 3 ⎭

(e)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 14 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 9.10 EI/2 m/2 EI EI

EI/2 EI/2

p3 (t )

m p2 ( t )

LM48 −24 0 0 0 −6h −6h 0 0 OP MM 48 −2424 60h 60h 60h 60h −66hh −66hhPP MM P h 2h 9h 0 0 0 P EI M 9h 0 2h 0 0 P k= P h M h 2h 9h 0 P MM Symm 9h 0 2h P MM P h P 5h MN P 2h Q

p1 ( t ) h

1 2 2 2

(a) The stiffness matrix can be written in partitioned form as follows:

L=2h

⎡k k = ⎢ tt ⎣k ot

Part a

The elastic properties of the frame (neglecting axial deformation) are represented by 9 DOFs: three horizontal displacements and six rotational displacements.

k to ⎤ k oo ⎥⎦

(b)

Part b

The DOFs representing the inertial properties are the three translational displacements, u1 , u2 and u3 . m/2

u8 u6

u9 u7

u3 m

⎧u 4 ⎫ ⎪u ⎪ ⎪ 5⎪ ⎧ u1 ⎫ ⎧u t ⎫ ⎪ ⎪ ⎪⎪u 6 ⎪⎪ u = ⎨ ⎬ u = ⎨u 2 ⎬ u 0 = ⎨ ⎬ ⎩u 0 ⎭ ⎪u ⎪ ⎪u 7 ⎪ ⎩ 3⎭ ⎪u 8 ⎪ ⎪ ⎪ ⎪⎩u 9 ⎪⎭

The coefficients of the stiffness matrix corresponding to these DOFs are computed following Example 9.7. The complete stiffness matrix is

⎧ u1 ⎫ ⎪ ⎪ u = u t = ⎨u2 ⎬ ⎪u ⎪ ⎩ 3⎭

The mass matrix is ⎡ ⎢1 0 m = m ⎢0 1 ⎢ ⎢0 0 ⎣

⎤ 0⎥ 0⎥ 1⎥ ⎥ 2⎦

(c)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 15 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Part c

The condensed stiffness matrix for the three lateral DOFs is −1 k$ tt =k tt − k tok oo k ot

⎡ 39.38 − 22.68 5.486 ⎤ EI ⎢ 27.13 − 11.75⎥⎥ = ⎢ h3 ⎢ Symm 7.418 ⎥⎦ ⎣

(d)

The equations governing the translational motion of the building are ⎡ ⎢1 0 m ⎢0 1 ⎢ ⎢0 0 ⎣

⎤ 0 ⎥ ⎧ u&&1 ⎫ ⎡ 39.38 − 22.68 5.486 ⎤ ⎧ u1 ⎫ ⎧ p1(t ) ⎫ ⎪ ⎪ EI ⎢ ⎪ ⎪ ⎪ ⎪ 0 ⎥ ⎨u&&2 ⎬ + 27.13 − 11.75⎥⎥ ⎨u2 ⎬ = ⎨ p2 (t )⎬ 1 ⎥ ⎪ && ⎪ h3 ⎢⎢ 7.418 ⎥⎦ ⎪⎩u3 ⎪⎭ ⎪⎩ p3 (t ) ⎪⎭ ⎥ ⎩u3 ⎭ ⎣ Symm 2⎦

(e)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 16 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 9.11 EI/3 m/2 EI/3 2EI/3

2EI/3 EI

p3 (t )

m p2 ( t )

p1 ( t )

LM40 −16 0 MM 24 −88 MM EI M k= h M MM Symm MM MM N 3

2h 4h

26 h 3

h2 26 2 h 3

−4 h 2h 2h 4 2 h 3

0 16 2 h 3

The elastic properties of the frame (neglecting axial deformation) are represented by 9 DOFs: three horizontal displacements and six rotational displacements. u3

0 2 2 h 3

0 0 2

2 2 3 1 2 3 2

(a)

⎡k k = ⎢ tt ⎣k ot

Part a

0 4 2 h 3 4 2 h 3 16 2 h 3

The stiffness matrix can be written in partitioned form as follows:

L=2h

0 0 −2 h − 2 h 2h 2h

OP PP P 0 P 0 PP 0 P P h P P h P 2 h PQ

−4 h 2h 2h

u7 u5

k to ⎤ k oo ⎥⎦

(b)

Part b

The DOFs representing the inertial properties are the three translational displacements, u1 , u2 and u3 . m/2 u3

u2 m

u1 m

⎧u 4 ⎫ ⎪u ⎪ ⎪ 5⎪ ⎧ u1 ⎫ ⎪⎪u ⎪⎪ u ⎧ ⎫ ⎪ ⎪ u = ⎨ t ⎬ u = ⎨u 2 ⎬ u 0 = ⎨ 6 ⎬ ⎩u 0 ⎭ ⎪u 7 ⎪ ⎪u ⎪ ⎩ 3⎭ ⎪u 8 ⎪ ⎪ ⎪ ⎩⎪u 9 ⎭⎪

The coefficients of the stiffness matrix corresponding to these DOFs are computed following Example 9.7. The complete stiffness matrix is

⎧ u1 ⎫ ⎪ ⎪ u = u t = ⎨u2 ⎬ ⎪u ⎪ ⎩ 3⎭

The mass matrix is ⎡ ⎢1 0 m = m ⎢0 1 ⎢ ⎢0 0 ⎣

⎤ 0⎥ 0⎥ 1⎥ ⎥ 2⎦

(c)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 17 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Part c

The condensed stiffness matrix for the three lateral DOFs is −1 kˆ tt = k tt − k to k oo k ot

⎡ 33.36 − 14.91 1.942 ⎤ EI ⎢ 15.96 − 5.489⎥⎥ = 3 ⎢ h ⎢ 3.923 ⎥⎦ ⎣ Symm

(d)

The equation governing the translational motion of the building is

LM1 0 0 OPRu&& U L 33.36 −14.91 1.942 ORu U R p bt gU | | EI M PP| | | | mM0 1 0 P Su&& V + 15.96 −5.489 Su V = S p bt gV M MM0 0 1 PP|Tu&& |W h MNSymm 3.923 PQ | Tu |W |T p bt g|W N 2Q 1

(e)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 18 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 9.12 m/2

EI/6

p3( t )

EI/3 EI/3 2EI/3

p2( t ) h

EI/2

p1( t )

0 ⎡40 − 16 ⎢ 24 8 − ⎢ ⎢ 8 ⎢ ⎢ ⎢ EI ⎢ k= 3⎢ h ⎢ ⎢ Symm ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

− 4h

4h 0

2h 2h

− 2h 2h

23 2 h 3

1 2 h 2 23 2 h 3

4 2 h 3

4 2 h 3 1 2 h 3 14 2 h 3

2 2 h 3

0 14 2 h 3

0 0 5 2 h 3

0 ⎤ − 2h ⎥⎥ 2h ⎥ ⎥ 0 ⎥ ⎥ 0 ⎥ 0 ⎥ ⎥ 2 2⎥ h 3 ⎥ 1 2⎥ h 6 ⎥ 5 2⎥ h ⎥ 3 ⎦

(a) The stiffness matrix can be written in partitioned form as follows:

L=2h

⎡ k tt k =⎢ ⎣ k ot

Part a

The elastic properties of the frame (neglecting axial deformation) are represented by 9 DOFs: three horizontal displacements and six rotational displacements.

k to ⎤ k oo ⎥⎦

(b)

Part b

The DOFs representing the inertial properties are the three translational displacements, u1 , u2 and u3 .

m/2

u2 m

⎧⎪ u t ⎫⎪ u=⎨ ⎬ ⎪⎩ u 0 ⎪⎭

⎧u ⎫ ⎪⎪ 1 ⎪⎪ ut = ⎨u2 ⎬ ⎪ ⎪ ⎪⎩ u 3 ⎪⎭

⎧u4 ⎫ ⎪ ⎪ ⎪ u5 ⎪ ⎪ ⎪ ⎪⎪ u 6 ⎪⎪ u0 = ⎨ ⎬ ⎪u7 ⎪ ⎪ ⎪ ⎪u8 ⎪ ⎪ ⎪ ⎪⎩ u 9 ⎪⎭

The coefficients of the stiffness matrix corresponding to these DOFs are computed following Example 9.7. The complete stiffness matrix is

u2 u1

⎧ u1 ⎫ ⎪ ⎪ u = u t = ⎨u 2 ⎬ ⎪u ⎪ ⎩ 3⎭

The mass matrix is ⎡ ⎢1 0 m = m ⎢0 1 ⎢ ⎢0 0 ⎣

⎤ 0⎥ 0⎥ 1⎥ ⎥ 2⎦

(c)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 19 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Part c

The condensed stiffness matrix for the three lateral DOFs is −1k kˆ tt = k tt − k to k oo ot

2.43 ⎤ ⎡ 30.77 − 14.01 EI ⎢ = 13.82 − 4.80⎥⎥ 3 ⎢ h ⎢ Symm 2.92 ⎥⎦ ⎣

(d)

The equation governing the translational motion of the building is

⎡1 ⎢ m 0 ⎢ ⎢⎣0

0 1 0

0 ⎤ ⎧ u&&1 ⎫ ⎡ 30.77 − 14.01 2.43 ⎤ ⎧ u1 ⎫ ⎧ p1 (t ) ⎫ ⎪ ⎪ ⎪ ⎪ ⎥ ⎪ ⎪ EI ⎢ 0 ⎨u&&2 ⎬ + 13.82 − 4.80⎥ ⎨u 2 ⎬ = ⎨ p 2 (t ) ⎬ 3 ⎥ 1 ⎪ && ⎪ h ⎢ ⎥ u ⎪ ⎪ p (t ) ⎪ u 2.92 ⎦ ⎪ ⎣ Symm ⎩ 3⎭ ⎩ 3 ⎭ 2 ⎥⎦ ⎩ 3 ⎭

(e)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 20 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 9.13 m

k41 = EI

Part a

The elastic properties of the umbrella (neglecting axial deformation of the elements) are represented by six DOFs: three translational displacements and three rotations. u3 u4

⎧ u1 ⎫ ⎧u 4 ⎫ ⎧u ⎫ ⎪ ⎪ ⎪ ⎪ u = ⎨ t ⎬ u t = ⎨u 2 ⎬ u 0 = ⎨u5 ⎬ u ⎩ 0⎭ ⎪u ⎪ ⎪u ⎪ ⎩ 3⎭ ⎩ 6⎭

The coefficients of the stiffness matrix corresponding to these DOFs are computed following Example 9.4. For instance, to obtain the first column of the stiffness matrix, apply a unit displacement u1 = 1 while the other displacements are zero, i.e., u j = 0 , j = 2, 3, ···, 6; identify the resulting elastic forces, and by statics obtain the stiffness coefficients: k 31 k 51

Other columns of k are determined similarly. The complete stiffness matrix is 0 0 6L 0 0⎤ ⎡ 12 ⎢ 0 ⎥ − − 12 0 6 0 6 L L ⎢ ⎥ 0 12 6L 6L 0⎥ EI ⎢ 0 k = ⎢ ⎥ (a) L3 ⎢ 6 L − 6 L 6 L 12 L2 2 L2 2 L2 ⎥ ⎢ 0 0 6 L 2 L2 4 L2 0⎥ ⎢ ⎥ 0 2 L2 0 4 L2 ⎥⎦ ⎢⎣ 0 − 6 L

⎡k k = ⎢ tt ⎣k 0t

k t0 ⎤ k 00 ⎥⎦

k 41

k 61

(b)

where the subscript t identifies the translational displacements, u1 , u2 and u3 , and the subscript 0 identifies the rotational displacements u4 , u5 and u6 . Part b

The DOFs representing the inertial properties are the three translational displacements u1 , u2 and u3 . u3

u2 u1

k 21

u 1= 1

k61 = 0

The stiffness matrix can be written in partitioned form as follows:

k51 = 0

6 EI L2

⎧ u1 ⎫ ⎪ ⎪ u = u t = ⎨u 2 ⎬ ⎪u ⎪ ⎩ 3⎭

k 11

To obtain the coefficients of the mass matrix for these DOF, first apply a unit acceleration u&&1 = 1 , while u&&2 = 0 and u&&3 = 0 . 6EI / L

m31

k11 =

12 EI L3

k21 = 0

12 EI / L

k31 = 0

m21

m11 m

m11 = 5 m

m21 = 0

m31 = 0

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 21 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Next apply a unit acceleration u&&2 = 1 , while u&&1 = 0 and u&&3 = 0 . m32

m22 m12

⎧u t ⎫ ⎪ 1 ⎪ ⎧ u1 ⎫ ⎧1⎫ ⎪ t⎪ ⎪ ⎪ ⎪ ⎪ ⎨u 2 ⎬ = ⎨u 2 ⎬ + ⎨0⎬ u gx ⎪ t ⎪ ⎪u ⎪ ⎪0⎪ ⎪⎩u 3 ⎪⎭ ⎩ 3 ⎭ ⎩ ⎭

(f)

Substituting Eq. (f) in Eq. (e) gives m

m12 = 0

m22 = m

⎧5m⎫ ⎤ ⎧ u&&1 ⎫ ⎡ 28 6 −6⎤ ⎧ u1 ⎫ ⎡5 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 3EI ⎢ ⎥ ⎥ ⎢ 6 7 3⎥ ⎨u2 ⎬ = − ⎨ 0 ⎬ u&&gx (t ) m ⎢ 1 ⎥ ⎨u&&2 ⎬ + 3⎢ ⎪0⎪ ⎢⎣ 1⎥⎦ ⎪⎩u&&3 ⎪⎭ 10 L ⎢⎣−6 3 7⎥⎦ ⎪⎩u3 ⎪⎭ ⎩ ⎭

m32 = 0

Finally, apply a unit acceleration u&&3 = 1 , while u&&1 = 0 and u&&2 = 0 . m33

(ii) If the excitation is vertical ground motion ugy ( t ), ⎧u t ⎫ ⎪ 1 ⎪ ⎧ u1 ⎫ ⎧0⎫ ⎪ t⎪ ⎪ ⎪ ⎪ ⎪ ⎨u 2 ⎬ = ⎨u 2 ⎬ + ⎨1⎬ u gy ⎪ t ⎪ ⎪u ⎪ ⎪1⎪ ⎪⎩u 3 ⎪⎭ ⎩ 3 ⎭ ⎩ ⎭

m23 m13

m13 = 0

(g)

(h)

Substituting Eq. (h) in Eq. (e) gives a matrix equation with its left hand side same as Eq. (g) and the right side is m23 = 0

⎧0⎫ ⎪ ⎪ p eff (t ) = −⎨m⎬ u&&gy (t ) ⎪m ⎪ ⎩ ⎭

m33 = m

Thus, the mass matrix is ⎡5 ⎤ ⎢ m = m ⎢ 1 ⎥⎥ ⎢⎣ 1⎥⎦

(iii) If the excitation is ground motion u gbd (t ) in the (c)

Part c

The condensed stiffness matrix for the three translational DOFs is −1 k$ tt = ktt − kt 0 k00 k0t

⎡ 28 6 −6⎤ 3EI ⎢ = 6 7 3⎥⎥ 3 ⎢ 10 L ⎢−6 3 7 ⎥ ⎣ ⎦

(i)

direction b-d, ⎧u1t ⎫ ⎧ u1 ⎫ ⎧1 ⎪ t⎪ ⎪ ⎪ ⎪ ⎨u 2 ⎬ = ⎨u 2 ⎬ + ⎨1 ⎪u t ⎪ ⎪u ⎪ ⎪1 ⎩ 3⎭ ⎩ 3⎭ ⎩

2⎫ ⎪ 2 ⎬u gbd 2⎪ ⎭

(j)

Substituting Eq. (j) into Eq. (e) gives a matrix equation with its left hand side the same as Eq. (g) and the right hand side is (d)

The equations governing the translational DOFs are ⎧ t⎫ ⎡ 28 6 −6⎤ ⎧ u1 ⎫ ⎧0⎫ ⎡5 ⎤ ⎪ u&&1 ⎪ 3EI ⎢ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ m ⎢⎢ 1 ⎥⎥ ⎨u&&2t ⎬ + 3⎥⎥ ⎨u 2 ⎬ = ⎨0⎬ ⎢ 6 7 3 ⎢⎣ 1⎥⎦ ⎪u&&t ⎪ 10 L ⎢⎣−6 3 7⎥⎦ ⎪⎩u 3 ⎪⎭ ⎪⎩0⎪⎭ ⎩⎪ 3 ⎭⎪

(e) (i) If the excitation is horizontal ground motion ugx ( t ), the total and relative displacements are related as follows:

⎧5m ⎪ p eff (t ) = −⎨ m ⎪ m ⎩

2⎫ ⎪ 2 ⎬u&&gbd (t ) 2⎪ ⎭

(k)

(iv) If the excitation is ground motion u gbc (t ) in the direction b-c, ⎧u1t ⎫ ⎧ u1 ⎫ ⎧− 1 ⎪ t⎪ ⎪ ⎪ ⎪ ⎨u 2 ⎬ = ⎨u 2 ⎬ + ⎨ 1 ⎪u t ⎪ ⎪u ⎪ ⎪ 1 ⎩ 3⎭ ⎩ 3⎭ ⎩

2⎫ ⎪ 2 ⎬u gbc 2⎪ ⎭

(l)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 22 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Substituting Eq. (l) into Eq. (e) gives a matrix equation with its left hand side the same as Eq. (g) and the right hand side is ⎧ − 5m ⎪ p eff (t ) = −⎨ m ⎪ m ⎩

2⎫ ⎪ 2 ⎬u&&gbc (t ) 2⎪ ⎭

(m)

(v) If the excitation is rocking ground motion defined by counter-clockwise rotation u gθ (in radians) in the plane of the structure, ⎧u1t ⎫ ⎧ u1 ⎫ ⎧− L ⎫ ⎪ t⎪ ⎪ ⎪ ⎪ ⎪ ⎨u 2 ⎬ = ⎨u 2 ⎬ + ⎨ L ⎬u gθ ⎪u t ⎪ ⎪u ⎪ ⎪− L ⎪ ⎩ 3⎭ ⎩ 3⎭ ⎩ ⎭

(n)

Substituting Eq. (n) into Eq. (e) gives a matrix equation with its left hand side the same as Eq. (g) and the right hand side is ⎧ − 5mL ⎫ ⎪ ⎪ p eff (t ) = −⎨ mL ⎬u&&gθ (t ) ⎪ − mL ⎪ ⎩ ⎭

(o)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 23 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Apply uθ = 1 , ux = uy = 0 :

Problem 9.14 b a 2k

uθ

kb/2

2kb/2

kxθ

2kb/2

kb/2

2kb/2

Apply ux = 1, uy = uθ = 0 : u x= 1 k b

kb/2

By statics, kxθ = 0

FG b − 2k b IJ = − kb H 2 2K F b I b + FG 3k b IJ b + FG 2k b IJ b = G 3k J H 2K 2 H 2K 2 H 2K 2 F b I b = 3k b + G 4k J H 2K 2

k yθ = 2 k

k yx k xx

kθ x

uθ=1

kθθ

Part a 1. Formulate the stiffness matrix.

kb/2

kyθ

2kb/2

kθθ

Hence the stiffness matrix is

LM6 0 0 OP k = k M0 6 − b P MN0 − b 3b PQ

By statics, kxx = 6 k

kyx = kθx = 0

where

Apply uy = 1 , ux = uθ = 0 :

k = u y= 1 2k

b k yy kθ y

By statics,

2. Formulate the mass matrix. Apply u&&x = 1, u&&y = u&&θ = 0 ⇒ mxx = m

k xy

12 EI h3

myx = mθx = 0

m yx

d k

u x =1 && O

mθ x m

m xx

kxy = 0

kyy = 6 k

FG b − 2k b IJ = − kb H 2 2K

kθ y = 2 k

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 24 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Apply u&&y = 1 , u&&x = u&&θ = 0 ⇒ myy = m mxy = mθy = 0

(ii) Ground motion in the y-direction.

R|u&& U| Ru&& U R 0 U S|u&& V| = |S|u&& |V| + |S|u&& |V| Tu&& W Tu&& W T 0 W t x t y t

m xy

Substituting Eq. (b) in Eq. (a) gives an equation with the left side same as Eq. (b) but the right side is

m yy mθ y

&& uy= 1

R|0U| p (t ) = − m S1V u&& (t ) |T0|W gy

eff

(iii) Ground motion in the d-b direction. Apply u&&θ = 1 , u&&x = u&&y = 0 ⇒ mθθ = IO mxθ = myθ = 0 ug = 1 1 O

m yθ O

mθ θ

&& uθ = 1

mxθ IO

The influence vector is

R|1 2 U| ι = S1 2 V |T 0 |W

Hence the mass matrix is

OP LM1 m = mM 1 MN r PPQ

The equations of motion have the same left side as Eq. (b) but the right side is

R|1 2 U| && (t ) = − m S1 2 V u&& (t ) p (t ) = − m ι u |T 0 |W

where r2 =

IO b2 + b2 b2 1 m = = m m 12 6

eff

Part b: Formulate the equations of motion. && t + ku = 0 mu (i) Ground motion in the x-direction.

(a)

R|u&& U| Ru&& U Ru&& U S|u&& V| = |S|u&& |V| + |S| 0 |V| Tu&& W Tu&& W T 0 W LM1 OP LM6 0 0 OP R|1U| && + k 0 6 − b u = − m S0V u&& (t ) u mM 1 MM P |T0|W MN b 6PPQ N0 − b 3b PQ t x t y t

(b)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 25 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 9.15

k 33

a 2k

u 3= 1 a

Part a 1. Formulate the stiffness matrix. Apply u1 = 1 , u2 = u3 = 0 : k

k 23

k 31

b O'

k 13

By statics, k 21

k13 = 2 k

k23 = − 2 k

k33 = 6 k

Hence the stiffness matrix is

LM 5 − 2 2OP k = k M− 2 5 − 2P MN 2 − 2 6PQ

O' O u 1= 1 2k

d k

By statics, k11 = 5 k

k 11

k21 = − 2 k

2. Formulate the mass matrix. Apply u&&1 = 1 , u&&2 = u&&3 = 0 :

k31 = 2 k

m31

1/b

m 21

Apply u2 = 1 , u1 = u3 = 0 :

IO (1/b) = mb/6

k 32 2k

1/2

k a

m/2

k 22

m/2

O O'

m11

1/2

u 2= 1

&& u1 =1

k 12

By statics, m11 =

2m 3

m21 = −

m 6

m31 =

m 2

By statics, k12 = − 2 k

k22 = 5 k

k32 = − 2 k

Apply u3 = 1 , u1 = u2 = 0 : © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 26 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Apply u&&2 = 1 , u&&1 = u&&3 = 0 : 1/b

&& u2 =1

(i) Ground motion in the x-direction. m32

ι3 = 0

m 22

ι2 = 1

m/2 1/2

ugx = 1

m/2

mb/6 m12

1/2

ι1 = 1

R|1U| p (t ) = − m S1V u&& (t ) |T0|W R|1 2U| = − m S1 2V u&& (t ) |T 0 |W eff

By statics, m12 = −

m 6

m22 =

2m 3

m32 = −

m 2

Apply u&&3 = 1 , u&&1 = u&&2 = 0 : && u3 = 1

m 33

m 23

(ii) Ground motion in the y-direction. ι2 = 0

ι3 = 1

ugy = 1

O m m 13

By statics,

m m m23 = − 2 2 Hence the mass matrix is m13 =

LM MM N

R| U| S| V| TW R| 1 2U| = − m S− 1 2 V u&& (t ) |T 1|W

0 p eff (t ) = − m 0 u&&gy (t ) 1

m33 = m

OP PP Q

2 3 −1 6 12 m = m −1 6 2 3 −1 2 1 2 −1 2 1

Part b: Formulate the equations of motion. && + ku = − m ι u&&g ( t ) mu 14243 peff (t )

ι1 = 0

(a)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 27 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

(iii) Ground motion in the d-b direction. ι3 = 1

ι2 = 1

ug = 1

ι1 = 1

R|1 2 U| p (t ) = − m S1 2 V u&& (t ) |T1 2 |W R|1 2 U| = − m S 0 V u&& (t ) |T1 2 |W eff

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 28 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

kθy = 2kb

Problem 9.16

Apply uθ = 1, u x = u y = 0 :

b uy

a ux

uθ

b k

2k b 2

k yθ

uθ = 1

kθθ

k xθ

Part a

1. Formulate the stiffness matrix.

2k b 2 d

Apply u x = 1, u y = uθ = 0 : ux = 1

By statics,

k xθ = 0

k yθ = 2kb

k yx

⎛ b⎞b ⎛ b⎞b kθθ = ⎜ 3k ⎟ + ⎜ 3k ⎟ + (2kb )b ⎝ 2⎠ 2 ⎝ 2⎠ 2 7 = kb 2 2

k xx

kθx

Hence the stiffness matrix is

0 ⎤ ⎡6 0 ⎢ 2b ⎥⎥ k = k ⎢0 6 ⎢⎣0 2b 7b 2 / 2⎥⎦

By statics, k xx = 6k

k yx = kθx = 0

Apply u y = 1, u x = uθ = 0 :

where k = 12

uy = 1 2k

k yy

h3 2. Formulate the mass matrix. Apply u&&x = 1, u&& y = u&&θ = 0 ⇒

k xy

kθy

kb 2

0 2k

kb 2

mxx = m

m yx = mθx = 0

m yx

u&&x = 1 O

mθx

m xx

By statics, k xy = 0

k yy = 6k © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 29 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Apply u&&y = 1, u&&x = u&&θ = 0 ⇒ m xy = 0

m yy = m

mθy = m

0 ⎤ 0 ⎤ ⎡1 0 ⎡6 0 ⎧1⎫ ⎪ ⎪ ⎢ ⎥ ⎢ ⎥ && + k 0 6 m ⎢0 1 b / 2 ⎥u 2b ⎥u = −m⎨0⎬u&&gx (t ) ⎢ ⎪0⎪ ⎢⎣0 b / 2 5b 2 / 12⎥⎦ ⎢⎣0 2b 7b 2 / 2⎥⎦ ⎩ ⎭

b 2

(b) (ii) Ground motion in the y-direction. m yy

u&& y = 1

⎧u&&tx ⎫ ⎧u&&x ⎫ ⎧ 0 ⎫ ⎪ t⎪ ⎪ ⎪ ⎪ ⎪ ⎨u&&y ⎬ = ⎨u&&y ⎬ + ⎨u&&gy ⎬ ⎪u&&t ⎪ ⎪u&& ⎪ ⎪ 0 ⎪ ⎩ θ⎭ ⎩ θ⎭ ⎩ ⎭

mθy

m xy

(c)

Substituting Eq. (c) in Eq. (a) gives an equation with the left hand side the same as Eq. (b) but the right side is Apply u&&θ = 1, u&&x = u&&y = 0 ⇒

⎧ 0 ⎫ ⎪ ⎪ p eff (t ) = −m⎨ 1 ⎬u&&gy (t ) ⎪b / 2⎪ ⎭ ⎩

mxθ = 0 m yθ =

mb 2

(iii) Ground motion in the d-b direction.

mθθ = I O + m

2 2 ⎛ b2 + b2 ⎞ bb ⎟ + m b = 5mb = m⎜ ⎜ 12 ⎟ 22 4 12 ⎝ ⎠

m yθ

u&&θ = 1 O

mθθ

m xθ

O 1

mb 2

The influence vector is ⎧1 2 ⎫ ⎪ ⎪ ι = ⎨1 2 ⎬ ⎪ 0 ⎪ ⎭ ⎩

Hence the mass matrix is 0 ⎤ ⎡1 0 m = m ⎢⎢0 1 b 2 ⎥⎥ ⎢⎣0 b 2 5b 2 12⎥⎦

The equations of motion have the same left side as Eq. (b) but the right side is

Part b: Formulate the equations of motion. &&t + ku = 0 mu

(i) Ground motion in the x-direction.

(a)

⎧1 2 ⎫ ⎪ ⎪ p eff (t ) = −m ι u&&g (t ) = − m⎨ 1 2 ⎬u&&g (t ) ⎪b 2 2 ⎪ ⎭ ⎩

⎧u&&tx ⎫ ⎧u&&x ⎫ ⎧u&&gx ⎫ ⎪ t⎪ ⎪ ⎪ ⎪ ⎪ ⎨u&&y ⎬ = ⎨u&&y ⎬ + ⎨ 0 ⎬ ⎪u&&t ⎪ ⎪u&& ⎪ ⎪ 0 ⎪ ⎩ θ⎭ ⎩ θ⎭ ⎩ ⎭

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 30 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Apply u 3 = 1, u1 = u 2 = 0 :

Problem 9.17 u3

k 23

u3 = 1

2k d

k 33

Part a

k13

1. Formulate the stiffness matrix.

Apply u1 = 1, u 2 = u 3 = 0 :

By statics, k13 = 3k

k 21

k 31 k

k 23 = −3k

k 33 = 7k

Hence the stiffness matrix is

3⎤ ⎡ 6 −3 5 − 3⎥⎥ k = k ⎢⎢ − 3 ⎢⎣ 3 − 3 7 ⎥⎦

where d

k11

k u1 = 1

By statics, k11 = 6k

k 21 = −3k

k = 12

EI h3

2. Formulate the mass matrix. Apply u&&1 = 1, u&&2 = u&&3 = 0 ⇒

k 31 = 3k

m11 = m

Apply u 2 = 1, u1 = u 3 = 0 :

m 21 = −

m 2

k 22 2k

k 32

u2 = 1

m31 =

m 2 m 21

m31

k m

m11

u&&1 = 1

By statics, k12 = −3k

k12

k 22 = 5k

k 32 = −3k

Apply u&&1 = 1, u&&2 = u&&3 = 0 ⇒ m12 = −

m 2

m 22 =

2m 3

m32 = −

m 6

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 31 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

u&&2 = 1

m 22

m32

⎧1⎫ ⎪ ⎪ p eff (t ) = −m ⎨0⎬u&&gx (t ) ⎪0⎪ ⎩ ⎭

I 0 (1 b) = m b 6

⎧ 1⎫ ⎪ ⎪ = −m⎨− 1 2⎬u&&gx (t ) ⎪ 1 2⎪ ⎩ ⎭

m 2

12 m 2

m12

(ii) Ground motion in the y-direction.

Apply u&&3 = 1, u&&1 = u&&2 = 0 ⇒ m13 =

m 2

m 23 = −

m 6

ι2 = 1

ι3 = 1 m33 =

u&&3 = 1

2m 3

u gy = 1

m 23

m33 I 0 (1 b) = m b 6 12

ι1 = 0

m 2

⎧0⎫ ⎪ ⎪ p eff (t ) = −m ⎨1⎬u&&gy (t ) ⎪1⎪ ⎩ ⎭

m 2

m13

⎧0⎫ ⎪ ⎪ = −m⎨1 2⎬u&&gy (t ) ⎪1 2⎪ ⎩ ⎭

Hence the mass matrix is ⎡ 1 m = m ⎢⎢ − 1 2 ⎢⎣ 1 2

−1 2 1 2⎤ 2 3 − 1 6 ⎥⎥ −1 6 2 3⎥⎦

(iii) Ground motion in the d-b direction.

ι3 = 1

Part b: Formulate the equations of motion. && + ku = −m ι u&&g (t ) mu 14243 p eff (t )

(a)

ι2 = 1

ug =1

i) Ground motion in the x-direction.

ι2 = 0

ι3 = 0

ι1 = 1

u gx = 1

ι =1

1 All rights reserved. This publication © 2012 Pearson Education, Inc., Upper Saddle River, NJ. is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 32 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

⎧1 ⎪ p eff (t ) = −m ⎨1 ⎪1 ⎩

2⎫ ⎪ 2 ⎬u&&g (t ) 2⎪ ⎭

⎧1 2 ⎫ ⎪ ⎪ = −m⎨ 0 ⎬u&&g (t ) ⎪1 2 ⎪ ⎩ ⎭

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 33 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 9.18

ψz =

uy L b

Px ( z − L) EI

ψy =

m ux

Curvatures, ψ x

φz =

Px L GJ

Px ( L − y ) EI

ψz =

Px L EI

Rate of twist, φ

(ii) Apply real force Py Py

1. Sign Convention. The assumed positive sense of the internal bending moments and torques acting on each element are as shown:

ψz =

ψx =

Py ( x − L)

φz =

− Py L GJ

Py ( L − z ) EI

y Curvatures, ψ

2. Determine mass matrix.

Rate of twist, φ

(iii) Apply real force Pz

The mass matrix, m, for this structure is

⎡m ⎤ ⎢ ⎥ ⎥ m=⎢ m ⎢ ⎥ ⎢⎣ m⎥⎦

3. Determine flexibility matrix. Compute the flexibility matrix, f̂ , in terms of the DOF u x , u y and u z using the principle of virtual forces.

ψy =

3.1 Establish the curvatures and rates of twist in the elements due to real forces Px, Py and Pz applied at the mass in the x, y and z directions respectively:

ψx =

(i) Apply real force Px Px

Pz ( L − x ) EI

− Pz L EI

ψx =

ψy =

Curvatures, ψ

Pz ( y − L) EI

φx =

− Pz L GJ

Pz L EI

Rate of twist, φ

3.2 Determine deflection ux due to forces Px, Py and Pz. Apply virtual force δPx in the direction of DOF u x © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 34 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

δPx

δM x = δPy ( L − z )

δM z = δPx L

Virtual Bending Moments

δM z = δPx L

Virtual Torques

Equate external and internal virtual work due to δPy

δM z = δPx ( L − y )

⎡ Py ( x − L) + Px L ⎤ ⎥ dx EI ⎣⎢ ⎦⎥

∫

δPy u y = δPy ( x − L) ⎢

δM y = δPx (z − L)

Virtual Torques

Virtual Bending Moments

⎡ Py ( L − z ) − Pz L ⎤ + δPy ( L − z ) ⎢ ⎥ dz EI ⎢⎣ ⎥⎦ 0

∫

Equate external and internal virtual work due to δPx L

∫

δPx u x = δPx ( L − y ) 0

⎡ − Py L + Px L ⎤ + − δPy L ⎢ ⎥ dz GJ ⎣⎢ ⎦⎥ 0

∫

Px ( L − y ) dy EI

Hence,

∫

⎡ L3 ⎤ ⎡ 2 L3 L3 ⎤ ⎡ L3 L3 ⎤ + − u y = Px ⎢− ⎥ ⎥ + Pz ⎢− ⎥ + Py ⎢ ⎣⎢ 2 EI ⎦⎥ ⎣⎢ 3EI GJ ⎦⎥ ⎣⎢ 2 EI GJ ⎦⎥

and

⎡ Px L + Py ( x − L) ⎤ + δPx L ⎢ ⎥ dx EI ⎢⎣ ⎥⎦ 0 ⎡ P ( z − L) + Pz L ⎤ + δPx ( z − L) ⎢ x ⎥ dz EI ⎦ ⎣ 0

∫

f̂ yx = −

⎡ Px L − Py L ⎤ + δPx L ⎢ ⎥ dx GJ ⎢ ⎣ ⎦⎥ 0

L3 L3 − 2 EI GJ

f̂ yy =

2 L3 L3 + 3EI GJ

f̂ yz = −

L3 2 EI

∫

3.4 Determine deflection uz due to forces Px, Py and Pz. Apply virtual force δPz in the direction of DOF uz.

Hence, ⎡ L3 ⎤ ⎡ L3 ⎡ 5L3 L3 ⎤ L3 ⎤ − + u x = Px ⎢ ⎥ ⎥ + Pz ⎢− ⎥ + Py ⎢− ⎣⎢ 2 EI ⎦⎥ ⎣⎢ 2 EI GJ ⎦⎥ ⎣⎢ 3EI GJ ⎦⎥

δPz

and f̂ xx =

5L3 L3 + 3EI GJ

f̂ xy = −

L3 L3 − 2 EI GJ

f̂ xz = −

L3 2 EI

3.3 Determine deflection uy due to forces Px, Py and Pz. Apply virtual force δPy in the direction of DOF uy

δM y = δPz ( L − x)

δM x = δPz ( y − L)

δM x = −δPz L

δM y = δPz L

δPy δM x = −δPz L

Virtual Bending Moments

δM z = δPy ( x − L)

δM z = −δPy L

Virtual Torques

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 35 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Equate external and internal virtual work due to δPz L

&& + ku = − mι u && g mu

⎡ P ( y − L) ⎤ ⎥ dy EI ⎣ ⎦

∫

δPz u z = δPz ( y − L) ⎢ z 0

5. Formulate the equations of motion u = [ ux, uy, uz ]T

where m and k are given above and influence vector ι = static displacements of DOF due to ug = 1.

⎡ P ( L − x) ⎤ + δPz ( L − x) ⎢ z ⎥ dx ⎣ EI ⎦ 0

∫

(i) ground motion in x-direction : ι =[ 1 0 0 ]T (ii) ground motion in y-direction : ι = [ 0 1 0 ]T (iii) ground motion in z-direction : ι = [ 0 0 1 ]T

⎡ − Pz L + Py ( L − z ) ⎤ + − δPz L ⎢ ⎥ dz EI ⎣⎢ ⎦⎥ 0

∫

(iv) ground motion in direction a-d:

ι=

⎡ P L + Px ( z − L) ⎤ + δPz L ⎢ z ⎥ dz EI ⎣ ⎦ 0

∫

1 3

[1 1 1]T

⎡− P L⎤ + − δPz L ⎢ z ⎥ dx ⎣ GJ ⎦ 0

∫

Hence, ⎡ L3 ⎤ ⎡ L3 ⎤ ⎡ 8L3 L3 ⎤ u z = Px ⎢− + ⎥ ⎥ + Py ⎢− ⎥ + Pz ⎢ ⎣⎢ 2 EI ⎦⎥ ⎣⎢ 2 EI ⎦⎥ ⎣⎢ 3EI GJ ⎦⎥

and f̂ zx = −

L3 2 EI

f̂ zy = −

L3 2 EI

f̂ zz =

8 L3 L3 + 3EI GJ

Therefore the flexibility matrix, f̂ , is ⎡ 5L3 L3 + ⎢ ⎢ 3EI3 GJ3 L ⎢ L f̂ = ⎢− − 2 EI GJ ⎢ 3 ⎢ − L ⎢⎣ 2 EI

L3 L3 − 2 EI GJ 2 L3 L3 + 3EI GJ L3 − 2 EI

−

L3 ⎤ ⎥ 2 EI ⎥ 3 L ⎥ − 2 EI ⎥ ⎥ 8L3 L3 ⎥ + 3EI GJ ⎥⎦ −

4. Determine stiffness matrix. The stiffness matrix, k, is obtained by inverting f̂ . For the case when GJ = 4/5 EI: ⎡ 35 − 21 − 6⎤ ⎥ L ⎢ ⎢ − 21 fˆ = 23 − 6⎥ 12 EI ⎢ ⎥ ⎢⎣ − 6 − 6 47 ⎥⎦ 3

⎡0.9283 0.9088 0.2345⎤ ⎥ EI ⎢ k = 3 ⎢0.9088 1.4294 0.2985⎥ ⎥ L ⎢ ⎢⎣0.2345 0.2985 0.3234⎥⎦

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 36 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 9.19 ug 1 k

ug 2

u1 k

1. Determine the stiffness matrix.. u1 u2 u g1 u g 2

B B B B L 2 −1 −1 0O LM k k OP = k MM−1 2 0 −1PP Nk k Q MM−1 0 1 0PP N 0 −1 0 1Q g

T g

where

LM 2 − 1OP N− 1 2Q L− 1 0OP k = kM N 0 − 1Q L 1 0OP k = kM N0 1Q 2. Determine the mass matrix.. L1 OP m = mM N 1Q k = k g

3. Determine the influence matrix..

ug1 = 1

ug 2 = 0 2/3

1/3

ug 1 = 0

ug2 = 1 1/3

ι = − k −1 k g =

2/3

LM2 3 1 3OP N 1 3 2 3Q

4. Write the equations of motion.. && + ku = peff ( t ) mu where && g (t ) = − m p eff (t ) = − m ι u

LM2 3 1 3OP RSu&& (t ) UV N1 3 2 3Q Tu&& (t )W g1

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 37 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Stiffness matrix in DOFs u , ug1 and ug2 :

Problem 9.20 ug1

ug2

u EI

⎡k ⎢ T ⎢⎣k g

⎧ 1 ⎫ kg ⎤ ⎪ ⎪ 6 EI ⎥ = ⎨− 1 2 ⎬ 3 1 − 1 2 − 1 2 k gg ⎥⎦ ⎪− 1 2 ⎪ L ⎩ ⎭

1. Formulate the stiffness matrix..

1 − 1 2 − 1 2⎤ ⎡ 6 EI ⎢ − 1 2 14 1 4 ⎥⎥ L3 ⎢ ⎢⎣ − 1 2 14 1 4 ⎥⎦

⎡ 6EI ⎤ k = ⎢ ⎥ ⎣ L3 ⎦ 6 EI −1 2 −1 2 kg = L3 6 EI ⎡1 4 1 4 ⎤ k gg = L3 ⎢⎣1 4 1 4 ⎥⎦

( 2 L )3 L3 f$11 = = 48 EI 6 EI $k = 6 EI 11 L3

2. Write the mass matrix.. m = m 3. Determine the influence matrix.. ug1 =1

Coordinate transformation:

u s = 1/2

u 1′ u s = 1/2

ug1

u g2

u g2 =1

By kinematics ι = 12 12 Alternatively,

ug1 =1

u 1′ = – 1/2

ι = − k −1 k g = − u g2 =1

= 12 12

4. Write the equation of motion. mu&& +

u=1

L3 6 EI −1 2 −1 2 6 EI L3

⎧⎪ u&&g1 (t ) ⎫⎪ 6 EI u = −m 12 12 ⎨ ⎬ L3 ⎩⎪u&&g 2 (t ) ⎭⎪

u 1′ = 1

⎧ u ⎫ ⎪⎪ ⎪⎪ u1′ = 1 −1 2 −1 2 ⎨ u g1 ⎬ ⎪ ⎪ ⎪⎩u g 2 ⎪⎭

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 38 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 9.21 L/2

Element e-b

L/2

m EI

L/2 L/2

m EI

d a

u7 e

u1 ug1

u4= ug2

u7 u1

Construct the element stiffness matrices

u3 = ug1

u5 d

u4 = ug2

⎡ 24 − 24 − 6 L − 6 L ⎤ ⎢ 6L 6 L ⎥⎥ 4 EI − 24 24 kˆ d −c = 3 ⎢ 2 L2 ⎥ L ⎢− 6 L 6 L 2 L ⎢ ⎥ L2 2 L2 ⎦ ⎣− 6 L 6 L

u2 u2 u7

⎡ 176 − 48 − 100 − 76⎤ ⎢ 12 36⎥⎥ 6 EI − 48 176 = 3 ⎢ 67 33⎥ 7 L ⎢− 100 12 ⎢ ⎥ 33 43⎦ ⎣ − 76 36 u1 u2 ug1 ug2

3. Write the mass matrix. u1

Element c-e

u6 c

⎡ 24 − 24 6 L ⎤ 4 EI ⎢ ˆk − 24 24 − 6 L ⎥⎥ a−d = 3 ⎢ L ⎢⎣ 6 L − 6 L 2 L2 ⎥⎦

Element d-c

⎡ k kg ⎤ −1 ⎢ T ⎥ = k tt − k tθ k θθ k θ t k k ⎥ gg ⎦ ⎣⎢ g

Element a-d

Condense out the rotational DOF u5 , u6 , u7

2. Formulate stiffness matrix..

− 24 − 24 − 6L 0 ⎤ 0 0 ⎡ 48 ⎢ − 6 L 0 ⎥⎥ 48 0 0 0 ⎢ 0 ⎢ − 24 − 6L 0 24 0 0 0 ⎥ ⎡ k tt k tθ ⎤ 4 EI ⎢ ⎥ 0 0 24 6L 6L 0 ⎥ ⎥ = 3 ⎢ − 24 ⎢ L ⎢ ⎣k θ t k θθ ⎦ − 6 L 6 L 4 L2 0 0 0 ⎥ L2 ⎢ ⎥ 2 2 0 6L 4L L L2 ⎥ ⎢− 6 L − 6 L T Condense out⎢⎣the0 rotational 0 DOF 0 [u5,0u6, u7]0 L2 4 L2 ⎥⎦

u3 = ug1

Add element stiffness matrices into global stiffness matrix

4 EI ⎡ 24 6 L ⎤ kˆ e −b = 3 ⎢ 2⎥ L ⎣6 L 2 L ⎦

ug2

1. Define the degrees of freedom.

⎡ 24 − 6 L − 6 L ⎤ 4 EI kˆ c −e = 3 ⎢⎢− 6 L 2 L2 L2 ⎥⎥ L ⎢⎣− 6 L L2 2 L2 ⎥⎦

u1 u2 ⎡m ⎤ m=⎢ ⎥ m ⎣ ⎦

4. Determine the influence matrix.

ι = − k −1k g =

1 ⎡ 266 182 ⎤ 448 ⎢⎣ 42 −42 ⎥⎦

Note that each column of ι can be interpreted as the displacement {u1, u2}T due to a unit displacement of one of the supports. The first column corresponds to ug1 = 1 and the second column corresponds to ug2 = 1.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 39 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

5. Write the equations of motion. && + ku = −m ι u && g (t ) mu

(a)

⎧⎪ u&&g1 (t ) ⎫⎪ ⎧u ⎫ && g (t ) = ⎨ where u = ⎨ 1 ⎬ and u ⎬ ⎩u2 ⎭ ⎩⎪u&&g 2 (t ) ⎭⎪

6. For the case of identical ground motion. ⎧1⎫ && g (t ) = ⎨ ⎬ u&&g1 (t ) u&&g 2 (t ) = u&&g1 (t ), i.e. u ⎩1⎭ Therefore, the equations of motion become ⎧1⎫ && + ku = − mι ⎨ ⎬ u&&g ( t ) = − mιˆ u&&g1 mu ⎩1⎭

(b)

⎧1⎫ ⎧1 ⎫ where ιˆ = ι ⎨ ⎬ = ⎨ ⎬ ⎩1⎭ ⎩0 ⎭

Note that this influence vector can be interpreted as the displacement {u1, u2}T due to a simultaneous unit displacement of ug1 and ug2.

Observe that ⎧m ⎫ p eff (t ) = − ⎨ ⎬u&&g1 (t ) , ⎩0⎭ implying that the effective force in the vertical DOF is zero because the horizontal support motions are identical. When different, even horizontal support motions create effective force in the vertical DOF; see Eq. (a).

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 40 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 9.22

Element c-d u2 u5 u6 ⎡ 24 − 6 L − 6 L ⎤ 4 EI ⎢ kˆ c − d = 3 ⎢− 6 L 2 L2 L2 ⎥⎥ L ⎢⎣− 6 L L2 2 L2 ⎥⎦

L/2

L/2 u2 m/4 c

m/4 e

m/2 d

L/2

u5 c

u6 d

EI Element d-e a

ug1

ug2

1. Define the degrees of freedom.

u2 u6 d

⎡ 24 6 L 6 L ⎤ 4 EI ⎢ ˆk 6 L 2 L2 L2 ⎥⎥ d −e = 3 ⎢ L ⎢⎣6 L L2 2 L2 ⎥⎦

u7 e

u2 u5

Add element stiffness matrices into global stiffness matrix u1

u4 = ug2

u3 = ug1

⎡ k tt ⎢ ⎣k θ t

2. Formulate stiffness matrix. Construct the element stiffness matrices Element a-c u5

⎡ 24 − 24 6 L ⎤ 4 EI ⎢ ˆk − 24 24 − 6 L ⎥⎥ a −c = L3 ⎢ ⎢⎣ 6 L − 6 L 2 L2 ⎥⎦

u3 = ug1

0 − 24 − 24 6 L 0 6L⎤ ⎡ 48 ⎢ ⎥ 0 0 − 6L 0 6L⎥ ⎢ 0 48 ⎢− 24 0 24 0 − 6L 0 0⎥ k tθ ⎤ 4 EI ⎢ ⎥ − 24 0 0 24 0 0 − 6L⎥ ⎥= k θθ ⎦ L3 ⎢ ⎢ 6L − 6L − 6L 0 4 L2 L2 0⎥ ⎢ 2 2 2⎥ 0 0 0 L 4L L ⎥ ⎢ 0 2 ⎢ 6L 6L 0 − 6L 0 L 4 L2 ⎥⎦ ⎣

Condense out the rotational DOF [u5, u6, u7]T ⎡ k kg ⎤ −1 ⎢ T ⎥ = k tt − k tθ k θθ k θ t ⎢⎣k g k gg ⎥⎦ 0 − 64 − 64⎤ ⎡ 128 ⎢ 0 140 − 42 42⎥ 6 EI ⎥ = 3 ⎢ 7 L ⎢− 64 − 42 67 − 3⎥ ⎢ ⎥ ⎣− 64 42 − 3 67⎦

Element b-e u7

u4 = ug2

⎡ 24 − 24 6 L ⎤ 4 EI kˆ b − e = 3 ⎢⎢− 24 24 − 6 L ⎥⎥ L ⎢⎣ 6 L − 6 L 2 L2 ⎥⎦

u1 u2 3. Write the mass matrix.

ug1 ug2

u1 u2 ⎤ ⎡m m=⎢ ⎥ ⎣ m / 2⎦ 4. Determine the influence matrix.

ι = − k -1k g =

LM0.5 0.5 OP N0.3 −0.3Q

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 41 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Note that each column of ι can be interpreted as the displacement {u1, u2}T due to a unit displacement of one of the supports. The first column corresponds to u g1 = 1 and the second column corresponds to ug2 = 1. 5. Write the equations of motion.

&& + ku = − m ι u && g t mu

where u =

(a)

RSu UV and u&& bt g = R|Su bt gU|V |Tu bt g|W Tu W g1

6. For the case of identical ground motion. ⎧1⎫ && g (t ) = ⎨ ⎬ u&&g1 (t ) u&&g 2 (t ) = u&&g1 (t ), i.e. u ⎩1⎭

Therefore, the equations of motion become && + ku = −m ι mu

RS1UV u&& (t ) = −m ι$ u&& (t ) T1W g1

(b)

⎧1⎫ ⎧1⎫ where ˆι = ι ⎨ ⎬ = ⎨ ⎬ ⎩1⎭ ⎩0⎭ Note that this influence matrix can be interpreted as the displacement {u1, u2}T due to a simultaneous unit displacement of ug1 and ug2. Observe that ⎧m ⎫ p eff (t ) = − ⎨ ⎬u&&g1 (t ) , ⎩0⎭

implying that the effective force in the vertical DOF is zero because the horizontal support motions are identical. When different, even horizontal support motions create effective force in the vertical DOF; see Eq. (a).

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 42 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

[

Problem 9.23

]

Apply uT uTg = [0 1 0 0 0 0 0] a k uθ uga(t)

uy ux

k ugb(t)

uy = 1

d k

kyy

ugd(t)

ugc(t) k

b 1. Define the degrees of freedom.

[u u ] = [u u u u T

T g

u gb u gc u gd

k xy = 0

k yy = 4k

kθ y = 0

ka y = 0

kb y = 0

kc y = 0

[

]

Apply u T uTg = [0 0 1 0 0 0 0]

2. Formulate mass matrix.

kd y = 0

From Problem 9.10 kb 2

kθθ

kcθ

]

Apply u T uTg = [1 0 0 0 0 0 0] ux = 1

kax

kdθ

kb 2 kb 2

kbx

k xθ = 0

k yθ = 0

k aθ = kb 2 kbθ = kb 2

kxx

[

kθθ = 2kb 2 kcθ = − kb 2

]

Apply uT uTg = [0 0 0 1 0 0 0]

kcx

kdx

kbθ

kaθ

2. Formulate stiffness matrix.

[

uθ = 1

kb 2

⎡ ⎤ ⎢1 ⎥ ⎢ ⎥ m=m⎢ 1 ⎥ ⎢ b2 ⎥ ⎢ 6 ⎥⎦ ⎣

kθa k xx = 4k

k yx = 0

kθ x = 0

k ax = − k

kbx = − k

kcx = −k

kaa

k dx = −k

k dθ = − kb 2

kxa

uga = 1

k xa = −k

k ya = 0

kθ a = kb 2

k aa = k

kba = 0

k ca = 0

k da = 0

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 43 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

[

]

Apply uT uTg = [0 0 0 0 1 0 0] kθb

Assemble the stiffness influence coefficients

kxb kbb

k xb = −k

k yb = 0

kθ b = kb 2

k ab = 0

kbb = k

k cb = 0

ugb = 1

k db = 0

⎡k ⎢ T ⎢⎣k g

⎡4 ⎢0 ⎢ ⎢0 kg ⎤ ⎢ k = ⎥ ⎢− 1 k gg ⎥⎦ ⎢− 1 ⎢ ⎢− 1 ⎢− 1 ⎣

0 4 0 0 0 0 0

−1 −1 −1 −1 ⎤ 0 0 0 0 ⎥⎥ b 2 b 2 − b 2 − b 2⎥ ⎥ b 2 1 0 0 0 ⎥ b 2 0 1 0 0 ⎥ ⎥ 0 1 0 ⎥ −b 2 0 0 0 1 ⎥⎦ −b 2 0 0

0 2b 2

4. Determine the influence matrix. 1 1 1⎤ ⎡ 1 1⎢ 0 0 0 ⎥⎥ ι = −k k g = ⎢ 0 4 ⎢⎣− 1 b − 1 b 1 b 1 b ⎥⎦ −1

[

]

Apply u T u Tg = [0 0 0 0 0 1 0] kθc

(a)

Note that each column of ι can be interpreted as the displacement [ux uy uθ]T due to a unit displacement of one uga = 1, of the supports. The first column corresponds to the second column to ugb = 1, the third column to ugc = 1 and the fourth column to ugd = 1.

kxc

5. Write the equations of motion. && + ku = −m ι u && g (t ) mu

where

kcc

ugc = 1

u = ux

k xc = −k

k yc = 0

kθ c = − kb 2

k ac = 0

kbc = 0

k cc = k

k dc = 0

uθ

and && g (t ) = u&&ga (t ) u&&gb (t ) u&&gc (t ) u&&gd (t ) u

[

Apply u

u Tg

] = [0 0 0 0 0 0 1] kθd

Observe that the equations of motion are uncoupled (matrices k and m are diagonal). Therefore, because the second row of the influence matrix, which corresponds to displacement uy, consists of zeros, there is no response in the y-direction. However, although the ground motion is only applied in the x-direction, torsional motion will occur in general due to the non-zero terms in the third row of the influence matrix. See Eq. (a).

kxd

6. For the case of identical ground motion at all supports. kdd

u gd = 1

u&&ga (t ) = u&&gb (t ) = u&&gc (t ) = u&&gd (t ) = u&&g (t )

k xd = − k

k yd = 0

kθ d = − kb 2

k ad = 0

kbd = 0

k cd = 0

k dd = k

⎧1⎫ ⎪1⎪ ⎪⎪ && g (equations i . e ., the u t ) = ⎨ ⎬ u&&gof (t )motion becom Therefore, ⎪1⎪ ⎩⎪1⎪⎭

(e)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 44 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Therefore, the equations of motion become

R|1U| |1| && + ku = −mι S V u&& bt g = − m ι$ u&& bt g mu ||1|| T1W R|1U| R1U |1| | | where ι$ = ι S V = S0V ||1|| |T0|W T1W g

Note that this result implies that, for the case of identical ground motions, the structure responds in the x-direction only.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 45 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Apply u a = 1, u b = 0, u c = 0, u d = 0

Problem 9.24 uy

u ga ( t )

uθ

kya 2k kθa

u gb ( t ) b

u gd ( t ) 2k

ugc (t )

k kxa

Data (from Problem 9.10). m = 0.2331 kip - sec 2 / in

kaa

k aa = 2 k

k ba = 0

k ca = 0

k da = 0

k xa = −2k

k ya = 0

kθa = bk

k = 1.5 kip/in b = 25 ft

Apply u b = 1, u a = 0, u c = 0, u d = 0

ζ n = 5%,

n = 1, 2 and 3 kyb

1. Define DOFs. u = ux

uθ

kθb

k kxb

2. Formulate the mass matrix.

From Problem 9.14:

⎡ ⎤ ⎢1 ⎥ ⎢ ⎥ m = m⎢ 1 ⎥ ⎢ b2 ⎥ ⎢ ⎥ 6 ⎦ ⎣

(a)

3. Formulate the stiffness matrix.

kbb

k ab = 0

k bb = k

k cb = 0

k xb = − k

k yb = 0

kθb = 0.5bk

k db = 0

Apply u c = 1, u a = 0, u b = 0, u d = 0

From Problem 9.14: 0 ⎤ ⎡6 0 k = k ⎢⎢0 6 − b ⎥⎥ ⎢⎣0 − b 3b 2 ⎥⎦

kyc 2k

(b)

The stiffness matrix associated with support degrees of freedom, k gg , and the coupling stiffness matrix

kxc

a 2k

between the support degrees of freedom and the structural degrees of freedom, k g , are computed by giving a unit displacement at a support degree of freedom, and computing the reactions at the support and structural degrees of freedom.

kθc

kcc

k ac = 0

k bc = 0

k cc = k

k xc = − k

k yc = 0

kθc = −0.5bk

k dc = 0

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 46 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Apply u d = 1, u a = 0, u b = 0, u c = 0

where

kyc k

kθc

kxc

a 2k

0.3529 b

ι b = 0 .1667 − 0.0294 −

0.1765 b

ι a = 0 .3333 − 0.0588 −

kdd

ι c = 0 .1667 0.0294

k ad = 0

k bd = 0

k cd = 0

k dd = 2 K

k xd = −2 k

k yd = 0

kθd = − bk

Assembling the stiffness influence coefficients gives k gg and k g : ⎡2 ⎢0 k gg = k ⎢ ⎢0 ⎢ ⎣0 (c)

0.1765 b

ι d = 0 .3333 − 0.0588 −

0.3529 b

6. Special case: Identical support motions. Equation (f) applies with p eff (t ) = −m ι u&&g (t )

0 0 0⎤ 1 0 0⎥⎥ 0 1 0⎥ ⎥ 0 0 2⎦

where

ι= 1 0 0

−1 − 2⎤ ⎡− 2 − 1 ⎢ kg = k ⎢ 0 0 0 0 ⎥⎥ ⎢⎣ b 0.5b − 0.5b − b⎥⎦

(d)

4. Determine the influence matrix. ⎡ ⎢ 0.3333 ⎢ ι = −k −1k g = ⎢ − 0.0588 ⎢ ⎢ 0.3529 ⎢− b ⎣

0.1667

− 0.0294

0.0294

0.1765 b

−

⎤ 0.3333 ⎥ ⎥ 0.0588 ⎥ ⎥ 0.3529 ⎥ b ⎥⎦

(e) 5. Formulate the equations of motion. The dynamic components of the displacements are governed by && + k u = p eff (t ) mu

(f)

where m and k are defined by Eqs. (a) and (b), and

bg = mι u&& bt g + mι u&& bt g + mι u&& bt g + mι u&& bt g

p eff t = − Σ m ι u&&gι t ι

a ga

b gb

c gc

d gd

(g) © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 47 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 9.25 mL/4 25'

f$22 =

Bridge

ug 2

100'

L3 3 EI

L3 f$33 = 2 3 EI L2 ⎛ L ⎞ fˆ23 = 2 ⎜ L − 2 ⎟ 2 EI ⎝ 3 ⎠

mL/2 u1 100'

ug 1

A =

( 25)2 − ( 22. 5)2 = 13, 430. 31 in.2

w = 150 lb ft 3 × m2 =

13, 430. 31 144

= 13, 989. 9 lb ft

13. 99 × 200 1 = 1. 812 kip − sec2 in. 4 386

m1 = 2 m2 = 3. 624 kip − sec2 in. I =

π ( 25 2 )4

−

π ( 22. 5 2 )4 4

= 6594. 2 ft 4

EI = 4. 9225 × 1011 kip − in.2

1. Formulate the stiffness matrix. The stiffness matrix can be formulated by the direct stiffness method and condensing the rotational DOFs. Instead we use a different method. We use the flexibility approach to determine k$ for DOFs u1 , u2 and u3 :

⎡ L3 ⎢ 1 ⎢ 3EI ⎢ ˆf = ⎢ ⎢ ⎢ ⎢ ⎢( sym) ⎢⎣

L12 ( L − L1 3) 2 EI L3 3EI

⎡0.9359 0.7701 −1.3063⎤ ˆk = fˆ −1 = ⎢ 1.5088 −2.1375⎥⎥ ×10 4 ⎢ 3.1294⎦⎥ ⎣⎢ ( sym)

Define the following transformation: u2 u g2

u 2′ u 3′

–1 –1

u 1′

–1

u g1

u2 u3 L

L2 L1

f$11 =

L12 ( L2 − L1 3) ⎤ ⎥ ⎥ 2 EI L22 ( L − L2 3) ⎥ ⎥ ⎥ 2 EI ⎥ 3 L2 ⎥ ⎥ 3EI ⎥⎦

a 448 6447 ⎧ u1′ ⎫ ⎡1 0 −1 0⎤ ⎪ ⎪ ⎢ ⎥ ⎨u 2′ ⎬ = ⎢0 1 −1 0⎥ ⎪u ′ ⎪ ⎢0 0 −1 1⎥ ⎦ ⎩ 3⎭ ⎣

ug1 =1

⎧ u1 ⎫ ⎪u ⎪ ⎪ 2 ⎪ ⎨u ⎬ ⎪ g1 ⎪ ⎪u g 2 ⎪ ⎩ ⎭

L13 3 EI

L2 ⎛ L ⎞ fˆ21 = 1 ⎜ L − 1 ⎟ 2 EI ⎝ 3 ⎠ L2 ⎛ L ⎞ fˆ31 = 1 ⎜ L2 − 1 ⎟ 2 EI ⎝ 3 ⎠ © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 48 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

The stiffness matrix in DOFs u1 , u2 , ug1 and ug2 is kg ⎤ ⎡k k = aT kˆ a = ⎢ T ⎥ ⎢⎣k g k gg ⎥⎦ ⎡0.9359 0.7701 −0.3998 −1.3063⎤ ⎢ 1.5088 −0.1415 −2.1375⎥⎥ =⎢ ×104 kips in. ⎢ 0.2269 0.3143⎥ ⎢ ⎥ 3.1294⎦ ⎣ ( sym)

2. Formulate the equations of motion.

&& + ku = − m ι u && g t mu

where ⎧⎪ u g1 ⎫⎪ ug = ⎨ ⎬ ⎪⎩u g 2 ⎪⎭

⎧u ⎫ u = ⎨ 1⎬ ⎩u 2 ⎭

⎡3.624 ⎤ m = ⎢ ⎥ 1 . 812 ⎣ ⎦ ⎡0.9359 0.7701⎤ 4 k = ⎢ ⎥ × 10 ⎣ 0.7701 1.5088⎦

⎡ 0.6035 0.3965⎤ ⎥ ⎣−0.2143 1.2143⎦

ι = −k −1 k g = ⎢

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 49 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

CHAPTER 10 that relates the displacements in the two sets of DOFs:

Problem 10.1 The mass and stiffness matrices were determined in Problem 9.1 with reference to the indicated DOF: mL 2 ⎤ ⎡ m m=⎢ 2 ⎥ ⎣mL 2 mL 3⎦ ⎡k1 + k 2 k = ⎢ ⎣ k2 L

Φ E10.1 = aΦ P10.1

k 2 L ⎤ ⎡ 3k 2k L ⎤ ⎥=⎢ ⎥ 2 k 2 L ⎦ ⎣2k L 2k L2 ⎦

u1 m

First mode, ω 1 = 2.536 k/m φ 11 = 1

φ 21 = 0.366

φ 11 = 1

φ 21 = –0.634/ L

Then ⎡ ⎢ 3− λ k −ω 2 m = k ⎢ ⎢2 L − λ L ⎢⎣ 2

L ⎤ 2 ⎥ ⎥ L2 ⎥ 2 2L − λ ⎥ 2 ⎦ 2L − λ

Second mode, ω 2 = 9.464 k/m

(a)

φ 12 = 1

φ 22 = –2.366/ L

φ 22 = –1.366

where

λ = ω2

Example 10.1

m k

Problem 10.1

(b)

Substituting Eq. (a) in Eq. (10.2.6) gives the frequency equation:

λ2 − 12 λ + 24 = 0 which has two solutions λ = ( 6 − 2 3 ) = 2. 536 and λ = ( 6 + 2 3 ) = 9. 464 . The corresponding natural frequencies are

ω1 =

2. 536

k m

ω2 =

9. 464

k m

(c)

The natural modes are determined from Eq. (10.2.5) following the procedure shown in Example 10.1 to obtain 1 1 ⎫ ⎧ ⎫ ⎬ ⎬ φ2 = ⎨ ⎩−2.366 L ⎭ ⎩−0.634 L ⎭ ⎧

φ1 = ⎨

(d)

The natural frequencies are the same as those obtained in Example 10.1, and the mode shapes are equivalent to those obtained in Example 10.1. The mode shapes in Example 10.1 can be obtained from the mode shapes obtained in this problem, using the transformation matrix

⎡1 0 ⎤ a= ⎢ ⎥ ⎣1 L ⎦

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 1 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Part c: Normalize modes so that Mn = 1 .

Problem 10.2 Part a: Determine natural frequencies and modes. The mass and stiffness matrices were determined in Problem 9.2:

LM OP N Q 162 EI L 8 − 7O k = M 8PQ 5 L N− 7 mL 1 0 m = 3 0 1

162 EI

k − ω m =

5 L3

T 2

1 mL 1 1 1 3

Divide φ1 of Eq. (d) by

2 m L 3 and φ2 of Eq. (d) by

2 m L 3 to obtain normalized modes:

Then

LM OP RS1UV = 2m L N Q T1W 3 L1 OP RS 1UV = 2m L mL M = φ mφ = 1 −1 M 3 N 1Q T− 1W 3 M 1 = φ 1T mφ 1 =

φ1 =

LM8 − λ −7 OP N −7 8 − λ Q

(a) φ 11 = 3/2 mL

where

RS UV φ = 3 RS 1UV 2m L T− 1W TW

1 3 2m L 1

φ 21 = 3/2 mL

5 m L4 2 ω (b) 486 EI Substituting Eq. (a) in Eq. (10.2.6) gives the frequency equation:

(e)

φ 12 = 3/2 mL

λ =

λ2 − 16 λ + 15 = ( λ − 1) ( λ − 15) = 0

First mode

The solution of the frequency equation gives: λ1 = 1 and λ 2 = 15 . The corresponding natural frequencies are EI m L4

ω1 = 9. 859

ω 2 = 38.184

EI m L4

φ 22 = –

3/2 mL

Second mode

The modes of Eq. (e) are scalar multiples of the modes in Eq. (d); the shapes of the two sets of modes are the same.

(c)

The natural modes are determined from Eq. (10.2.5) following the procedure shown in Example 10.1 to obtain:

φ1 =

RS1UV φ = RS 1UV T1W T− 1W

(d)

φ 11 = 1 φ 21 = 1

φ 12 = 1

φ 22 = –1 ω 1 = 9.859

EI /mL 4

ω 2 = 38.184

EI /mL 4

Second mode (antisymmetric)

First mode (symmetric)

Part b: Verify orthogonality properties.

LM OP RS 1UV = 0 N Q T− 1W L 8 − 7OP RS 1UV = 0 162 EI 1 1 M φ kφ = 5L N− 7 8 Q T− 1W φ 1T mφ 2 = T 1

1 mL 1 1 1 3 3

Thus the natural modes satisfy the orthogonality properties. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the2publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

q1 (0) = 1 q2 (0) = 0 q&1 (0) = 0 q& 2 (0) = 0

Problem 10.3 The free vibration response of the system without damping is computed using Eq. (10.8.6), 2

u(t ) =

⎡

q& (0)

⎤

∑ φ n ⎢⎣qn (0)cosω nt + ωn n sinω n t ⎥⎦

(a)

n =1

where according to Eq. (10.8.5)

φ T mu (0)

q n ( 0) = n

φ nT mφ n

φ T mu& (0)

q& n (0) = n

(b)

φ nT mφ n

Using m and k from Problem 10.2, the values of q1 ( 0 ) and q2 ( 0 ) for the initial displacements u( 0 ) are ⎡1 ⎤ mL 11⎢ ⎥u(0) 3 ⎣ 1⎦ = 0.5 0.5 u(0) q1 (0) = ⎡1 ⎤ ⎧1⎫ mL 11⎢ ⎥⎨ ⎬ 3 ⎣ 1⎦ ⎩1⎭

(c)

⎡1 ⎤ mL 1 −1 ⎢ ⎥ u ( 0) 3 ⎣ 1⎦ = 0.5 − 0.5 u(0) q 2 (0) = ⎡1 ⎤ ⎧ 1⎫ mL 1 −1 ⎢ ⎥⎨ ⎬ 3 ⎣ 1⎦ ⎩−1⎭

(d)

(h)

q2 ( 0 ) = 1

q&1 ( 0 ) = 0

q&2 ( 0 ) = 0 (i)

⎧ ⎞⎫ ⎛ ⎪ cos⎜ 38.184 EI t ⎟⎪ ⎟ ⎜ m L4 ⎠⎪⎪ ⎧ u1 (t ) ⎫ ⎪⎪ ⎝ = ⎬ ⎬ ⎨ ⎨ ⎛ ⎩u 2 (t )⎭ ⎪ EI ⎞⎟⎪ ⎜ t ⎪ ⎪−cos⎜ 38.184 ⎟ m L4 ⎠⎪⎭ ⎪⎩ ⎝

(j)

In case (a) both modes contribute to the response because the initial displacement condition has components in both modes. In case (b) the initial displacement is proportional to the first mode and therefore only this mode is excited and contributes to the response. In case (c) the initial displacement is proportional to the second mode and therefore only this mode contributes to the response.

(a) For u1 ( 0 ) = 1 and u2 ( 0 ) = 0 , ⎧0⎫ ⎧1⎫ u(0) = ⎨ ⎬ u& (0) = ⎨ ⎬ 0 ⎩0⎭ ⎩ ⎭

Substituting in Eqs. (c)-(d) gives q&1 ( 0 ) = 0

⎧ ⎛ ⎞⎫ ⎪cos⎜ 9.859 EI t ⎟⎪ ⎟ ⎜ m L4 ⎠⎪⎪ ⎧ u1 (t ) ⎫ ⎪⎪ ⎝ = ⎬ ⎬ ⎨ ⎨ ⎩u 2 (t )⎭ ⎪ ⎛⎜ EI ⎞⎟⎪ t ⎪ ⎪cos⎜ 9.859 ⎟ m L4 ⎠⎪⎭ ⎪⎩ ⎝

Substituting Eq. (i) in Eq. (a) and using the modes from Problem 10.2 gives

q&1 ( 0 ) = 0. 5 0. 5 u& ( 0 ) q&2 ( 0 ) = 0. 5 − 0. 5 u& ( 0 )

q2 ( 0 ) = 0. 5

Substituting Eq. (g) in Eq. (a) and using the modes from Problem 10.2 gives

Substituting in Eqs. (c)-(d) gives

Similarly,

q1 ( 0 ) = 0. 5

(g)

q&2 ( 0 ) = 0

(e) Substituting Eq. (e) in Eq. (a) and using the modes from Problem 10.2 gives ⎧ ⎛ ⎞ ⎛ ⎞⎫ ⎪0.5cos⎜ 9.859 EI t ⎟ + 0.5cos⎜ 38.184 EI t ⎟⎪ ⎜ ⎟ ⎜ ⎟ m L4 ⎠ m L4 ⎠⎪⎪ ⎧ u1 (t ) ⎫ ⎪⎪ ⎝ ⎝ ⎨ ⎬ ⎬=⎨ ⎛ ⎞ ⎛ ⎞ ⎩u 2 (t )⎭ ⎪ ⎜ 9.859 EI t ⎟ − 0.5cos⎜ 38.184 EI t ⎟ ⎪ 0 . 5 cos ⎪ ⎜ ⎟ ⎜ ⎟⎪ m L4 ⎠ m L4 ⎠ ⎪⎭ ⎪⎩ ⎝ ⎝

(f)

(b) For u1 ( 0 ) = 1 and u2 ( 0 ) = 1 , ⎧0⎫ ⎧1⎫ u(0) = ⎨ ⎬ u& (0) = ⎨ ⎬ ⎩0⎭ ⎩1⎭

Substituting in Eqs. (c)-(d) gives © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the3publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 10.4 The free vibration response of the system of Problem 9.2 including damping is computed using Eq. (10.10.4): 2

u(t ) =

∑ φ n e −ζ nω n t [qn (0)cosω n D t

n =1

q& n (0) +ζ nω n q n (0)

ωnD

(a)

⎤ sinω n D t ⎥ ⎦⎥

where ω n D = ω n 1 − ζ 2 ; ω n are given by Eq. (c) of Problem 10.2 and ζ n = 0. 05 . Therefore,

ω 1 D = 9. 847

EI m L4

ω 2 D = 38.136

EI m L4

(b) For the given initial conditions u1 ( 0 ) = 1 and u2 ( 0 ) = 0 , Eq. (e) of Problem 10.3 gives q1 ( 0 ) = 0. 5

q2 ( 0 ) = 0. 5

q&1 ( 0 ) = 0

q& 2 ( 0 ) = 0

(d)

where the modal contributions are

⎧1⎫ u1 (t ) = e − 0.493t 0.5cosω1D t + 0.025sinω1D t ⎨ ⎬ ⎩1⎭

(

)

(e) ⎧ 1⎫ u 2 (t ) = e −1.909t 0.5cosω 2 D t + 0.025sinω 2 D t ⎨ ⎬ ⎩−1 ⎭

(

)

(f)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the4publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 10.5

u(t ) =

Part a

⎤

q& (0)

(e)

n =1

From Problem 9.4, the mass and stiffness matrices are. m=

⎡

∑ φ n ⎢⎣qn (0)cosω nt + ωn n sinω nt ⎥⎦

m ⎡2 1⎤ ⎢ ⎥ 6 ⎣1 2 ⎦

where qn ( 0 ) =

2 EI ⎡ 14 −5⎤ ⎢ ⎥ L3 ⎣−5 2⎦

Substituting u& ( 0 ) = 0 1

Thus 2 EI ⎡ 2 − 2λ k −ω 2m = ⎢ L3 ⎣− 5 − λ

− 5− λ ⎤ 14 − 2λ ⎥⎦

(a)

where 3

mL ω2 12 EI

λ =

(b)

Substituting Eq. (a) in Eq. (10.2.6) gives the frequency equation:

λ2 − 14 λ + 1 = 0 The solution of this equation gives: λ 1 = 0. 0718 and λ 2 = 13. 928 . The corresponding natural frequencies are

ω1 = 0. 928

φ Tn mu( 0 ) φ Tn mφ n

ω 2 = 12. 928

EI m L3

(c)

q&1 (0) =

q& n ( 0 ) = T

φ Tn mu& ( 0 ) φ Tn mφ n

and u( 0 ) = 0 0

⎡1 ⎤ ⎧0⎫ m 1 2.7321 ⎢ ⎥⎨ ⎬ 6 ⎣ 1⎦ ⎩1⎭ ⎡1 ⎤ ⎧ 1 ⎫ m 1 2.7321 ⎢ ⎬ ⎥⎨ 6 ⎣ 1⎦ ⎩2.7321⎭

(f) T

gives:

= 0.2882

(g.1)

⎡1 ⎤ ⎧0⎫ m 1 − 0.7321 ⎢ ⎥⎨ ⎬ 6 ⎣ 1⎦ ⎩1⎭ q&2 (0) = = −0.2882 (g.2) ⎡1 ⎤ ⎧ 1 ⎫ m 1 − 0.7321 ⎢ ⎬ ⎥⎨ 6 ⎣ 1⎦ ⎩−0.7321⎭

q1 ( 0 ) = 0

q2 ( 0 ) = 0

(h)

Substituting Eqs. (g), (h) and (d) in Eq. (e) gives the displacements: ⎧ u1 (t ) ⎫ m L3 ⎧0.3110sinω1t − 0.0223sinω 2 t ⎫ ⎨ ⎬= ⎨ ⎬ EI ⎩0.8947sinω1t + 0.0164sinω 2 t ⎭ ⎩u 2 (t )⎭

Following the procedure used in Example 10.1, the natural modes are determined from Eq. (10.2.5): ⎧

⎧

1 ⎫ ⎬ ⎩2.7321⎭

1 ⎫ ⎬ ⎩−0.7321⎭

φ1 = ⎨

φ 21 = 2.7321

φ2 =⎨

φ 22 = – 0.7321

φ 11 = 1

φ 12 = 1

Second mode

First mode

ω1 = 0.928

(d)

EI /mL 3

ω 2 = 12.928

EI /mL 3

Part b

The free vibration response of the system is computed using Eq. (10.8.6): © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the5publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 10.6

Part b: Verify orthogonality.

From Problem 9.5, the mass and stiffness matrices are ⎤ ⎡1 m =m⎢ ⎥ ⎣ 0.5⎦

φ1T mφ 2 = 1 = m 1

⎡ 2 −1⎤ k = k⎢ ⎥ ⎣−1 1⎦

φ1T kφ 2 = 1

Part a: Determine the natural frequencies and modes.

⎛ m2 ⎞ ⎜ ⎟ω 4 − 2k ⎜ 2 ⎟ n ⎝ ⎠

k (2 ± m

2) ⇒

k m

ω1 = 3. 750

mh3

ω n2 + k 2 = 0 ⇒

ω 2 = 1. 848

–1/ 2

k m ω 1 = 3.750

ω 2 = 9. 052

EI mh3

⎡ 2 −1 ⎤ ⎧φ11 ⎫ ⎧0⎫ k ⎢ ⎥⎨ ⎬= ⎨ ⎬ ⎣⎢ −1 1 2 ⎦⎥ ⎩φ 21 ⎭ ⎩0⎭ 2

⎧1 ⎫ ⎬ ⎩ 2⎭

φ1 = ⎨

Second mode: k − ω 22 m φ 2 = 0 −1 ⎤ ⎧φ12 ⎫ ⎧0⎫ ⎥⎨ ⎬= ⎨ ⎬ −1 2 ⎦⎥ ⎩φ 22 ⎭ ⎩0⎭

ω 2 = 9.052

EI /mh 3

Second mode

First mode

Part d: Normalize modes so that M n = 1 .

M 1 =φ1T mφ1 = 1

k − ω 12 m φ1 = 0

⎡− 2 k ⎢ ⎣⎢ −1

φ2 = ⎨

1/ 2

First mode:

Select φ11 = 1 ⇒ φ 21 =

⎧⎪−1 2 ⎫⎪ ⎬ ⎪⎩ 1 ⎪⎭

φ1 = ⎨

Substituting for k gives EI

Part c: Normalize modes to unit value at roof. ⎧⎪1 2 ⎫⎪ ⎬ ⎪⎩ 1 ⎪⎭

det k − ω 2n m = 0 ⇒

ω1 = 0. 765

⎧⎪ 2 + 2 ⎫⎪ 2 ⎨ ⎬=0 ⎪⎩−1− 2 ⎪⎭

= k 1

24 EI k ≡ h3

⎡ 2 −1⎤ ⎧ 1 ⎫ k ⎢ ⎬ ⎥⎨ ⎣−1 1⎦ ⎩− 2 ⎭

where

ω 2n =

⎡1 ⎤⎧ 1 ⎫ m⎢ ⎬ ⎥⎨ ⎣ 1 2 ⎦ ⎩− 2 ⎭ ⎧ 1 ⎫ 2 ⎨ ⎬=0 ⎩−1 2 ⎭

M 2 =φ 2T mφ 2 = 1 − 2

⎡1 ⎤⎧ 1 ⎫ m⎢ ⎥ ⎨ ⎬ = 2m ⎣ 1 2⎦ ⎩ 2 ⎭ ⎡1 ⎤⎧ 1 ⎫ m⎢ ⎥ ⎨− 2 ⎬ = 2 m 1 2 ⎭ ⎣ ⎦⎩

Divide φ1 from part (a) by 2m and φ 2 from part (a) by 2m to obtain the normalized modes:

φ1 =

1 ⎧⎪1 2 ⎫⎪ 1 ⎧⎪1 2 ⎫⎪ ⎨ ⎬ φ2 = ⎨ ⎬ m ⎪⎩ 1 ⎪⎭ m ⎪⎩ − 1 ⎪⎭

These modes differ from these obtained in Part (c), only by a scale factor; the shapes of the two sets of modes are the same.

Select φ12 = 1 ⇒ φ22 = − 2 ⎧ 1 ⎫ ⎬ ⎩− 2 ⎭

φ2 = ⎨

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the6publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 10.7

⎧0.919⎫ ⎬ ⎩ 1 ⎭

φ1 = ⎨

1. Determine the stiffness matrix. Define k ≡

Determine second mode:

24 EI h3

k − ω 22 m φ 2 = 0

Apply u1 = 1 and u2 = 0 and determine ki1 : k 21 = – k u 1= 1

k 11= k + k /4 = 5 k/4

−1 ⎤ ⎧φ12 ⎫ ⎧0⎫ ⎡−1.837 k ⎢ ⎨ ⎬= ⎨ ⎬ 1 0 .544⎥⎦ ⎩φ 22 ⎭ ⎩0⎭ − − ⎣

Select φ22 = 1 ⇒ φ12 = − 0. 544 ⎧−0.544⎫ ⎬ ⎩ 1 ⎭

φ2 = ⎨

4. Compare vibration properties for hinged and clamped cases. Apply u2 = 1 and u1 = 0 and determine ki 2 :

hinged

clamped

k 22= k

u 2= 1

k 12= – k ω 1 = 1.971

EI /mh 3

ω 2 = 8.609

EI /mh 3

ω 1 = 3.750

EI /mh 3

ω 2 = 9.052

EI /mh 3

First mode

Second mode

Thus The system with columns hinged at the base is more flexible, and thus has lower frequencies; the first frequency is reduced by a factor of almost two whereas the second frequency is affected much less.

⎡5 4 − 1⎤ k = k ⎢ ⎥ ⎣ − 1 1⎦

2. Determine the mass matrix. ⎡1 ⎤ m = m⎢ ⎥ 1 2 ⎣ ⎦ 3. Determine natural frequencies and modes. det k − ω 2n m = 0 ( 0. 5 m2 ) ω n4 − (1. 625 k m ) ω n2 + 0. 25 k 2 = 0

ω 1 = 1. 971

EI mh 3

ω 2 = 8. 609

EI mh 3

Determine first mode: k − ω 12 m φ1 = 0 −1 ⎤ ⎧φ11 ⎫ ⎧0⎫ ⎡1.088 k⎢ ⎬= ⎨ ⎬ ⎥⎨ ⎣ −1 0.919⎦ ⎩φ 21 ⎭ ⎩0⎭

Select φ 21 = 1 ⇒ φ11 = 0. 919 © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the7publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

The second mode contributes more to the response than the first mode because the second mode has the more significant component in the specified initial conditions.

Problem 10.8 From Problem 10.6, ⎡ 2 −1⎤ k = k ⎢ ⎥ ⎣−1 1⎦

⎡1 ⎤ m = m⎢ ⎥ ⎣ 0.5⎦ ⎧⎪1 2 ⎫⎪ ⎬ ⎪⎩ 1 ⎪⎭

⎧⎪−1 2 ⎫⎪ ⎬ ⎪⎩ 1 ⎪⎭

φ1 = ⎨

ω1 = 3. 750

φ2 = ⎨ EI mh

ω 2 = 9. 052

EI mh3

Case a

⎧1⎫ u(0) = ⎨ ⎬ ⎩2 ⎭

⎧0⎫ u& (0) = ⎨ ⎬ ⎩0⎭

q1 ( 0 ) =

φ1T mu( 0 ) = 1. 707 φ1T mφ1

q&1 ( 0 ) = 0

q2 ( 0 ) =

φ T2 mu( 0 ) = 0. 293 φ T2 mφ 2

q& 2 ( 0 ) = 0

Substituting in Eq. (10.8.8) gives q1 ( t ) = 1. 707 cos ω 1t

q2 ( t ) = 0. 293 cos ω 2 t

Substituting qn ( t ) and φ n in Eq. (10.8.7) gives ⎧⎪−1 2 ⎫⎪ ⎧ u1(t ) ⎫ ⎧⎪1 2 ⎫⎪ ⎨ ⎬=⎨ ⎬ 1.707cos ω1t + ⎨ ⎬ 0.293cos ω 2t ⎪⎩ 1 ⎪⎭ ⎩u2 (t )⎭ ⎪⎩ 1 ⎪⎭ ⎧1.207⎫ ⎧− 0.207 ⎫ = ⎨ ⎬ cos ω1t + ⎨ ⎬ cos ω 2t ⎩1.707⎭ ⎩ 0.293 ⎭

The first mode contributes more to the response than the second mode because the first mode has the more significant component in the specified initial conditions. Case b ⎧−1⎫ u(0) = ⎨ ⎬ ⎩ 1⎭

⎧0⎫ u& (0) = ⎨ ⎬ ⎩0⎭

Following the procedure of part (a) we obtain q1 ( 0 ) = − 0. 207

q&1 ( 0 ) = 0

q2 ( 0 ) = 1. 207

q& 2 ( 0 ) = 0

q1 ( t ) = − 0. 207 cos ω1t

q2 ( t ) = 1. 207 cos ω 2 t

⎧⎪−1 2 ⎫⎪ ⎧ u1(t ) ⎫ ⎧⎪1 2 ⎫⎪ ⎨ ⎬ =⎨ ⎬ (−0.207) cos ω1t + ⎨ ⎬ 1.207 cos ω 2t ⎪⎩ 1 ⎪⎭ ⎩u2 (t )⎭ ⎪⎩ 1 ⎪⎭ ⎧−0.146⎫ ⎧−0.853⎫ =⎨ ⎬ cos ω1t + ⎨ ⎬ cos ω 2t − 0 . 207 ⎩ ⎭ ⎩ 1.207 ⎭

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the8publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 10.9 From Problem 10.8 q1 ( 0 ) = 1. 707

q2 ( 0 ) = 0. 293

q&1 ( 0 ) = 0

q&2 ( 0 ) = 0

Substituting qn ( 0 ) and q& n ( 0 ) in Eq. (10.10.2) gives

F GG H

q1 ( t ) = e − ζ 1ω 1t 1707 . cos ω 1 D t +

1707 . ζ1 1 − ζ 12

sin ω 1 D t

I JJ K

= 1707 . e −0.05ω 1t cos 0.999ω 1t + 0.05 sin 0.999ω 1t

(a)

q2 (t ) = 0.293 e −0.05ω 2 t cos 0.999ω 2 t + 0.05 sin 0.999ω 2 t

(b) Substituting Eqs. (a) and (b) in Eq. (10.8.7) gives . U RSu (t ) UV = RS1207 e bcos 0.999ω t + 0.05 sin 0.999ω t g u ( t ) 1707 T W T . VW R− 0.207UV e + S bcos 0.999ω t + 0.05 sin 0.999ω t g T 0.293W 1

− 0.05ω 1t

− 0.05ω 2 t

where

ω 1 = 3. 750

EI mh 3

ω 2 = 9. 052

EI mh 3

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the9publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

. − 0.411O LM− 0164 . 0164 − − 0.411PP 1M T = h M 0.904 − 0.740P MN 0.904 − 0.740PQ

Problem 10.10 u5

u2 u1

The joint rotations associated with the first mode are obtained by substituting u t = φ1 in Eq. (a): − 0.411O . R|u U| LM− 0164 R|− 0.490U| P − 0.411P R0.482 U . 1 |− 0.490| |Su |V = 1 M− 0164 = S V ||u || h MM 0.904 − 0.740PP T 1 W h S||− 0.304V|| Tu W N 0.904 − 0.740Q T− 0.304W 3

With reference to the lateral floor displacements u1 and u2 , the mass matrix and the condensed stiffness matrix (from Problem 9.6) are

L1 OP m = mM N 0.5Q

LM N

37.15 − 1512 . EI k$ tt = 3 1512 1019 . . − h

OP Q

5 6

Similarly, ut = φ2 in Eq. (a) gives the joint rotations associated with the second mode: − 0.411O . R|u U| LM− 0164 R|− 0.241U| P . U − 0.411P R− 1037 . 1 |− 0.241| |Su |V = 1 M− 0164 = V S V M P . | ||u || h M 0.904 − 0.740P T 1 W h S||− 1677 . |W Tu W N 0.904 − 0.740Q T− 1677 3

1. Determine natural frequencies.

det k$ tt − ω 2n m tt

ω 1 = 2. 407

= 0

EI mh 3

ω 2 = 7.193

EI mh 3

5. Summary.

2. Determine first mode. k$ tt − ω 12 m tt φ1 = 0

. − 1512 . O Rφ U LM 3136 R0U = S V S V P − 1512 . 7 . 295 φ h N Q T W T0W R0.482UV φ = S T 1 W EI

0.304/ h

0.490/ h

0.482

ω 1 = 2.407

EI mh 3

ω 2 = 7.193

EI mh 3

3. Determine second mode. First mode

k$ tt − ω 22 m tt φ 2 = 0

LM N

OP RSφ UV = RS0UV Q Tφ W T0W

. EI − 14.59 − 1512 3 − 1512 . − 15.68 h

1.677/ h

. U R− 1037 φ = S T 1 VW

0.241/ h –1.037

4. Determine joint rotations. The joint rotations corresponding to the lateral displacements φ n are computed using Eq. (9.3.3): u0 = T ut

Second mode

(a)

where −1 T = − k 00 k 0t

(b)

Substituting k 00 and k 0t from Problem 9.6 in Eq. (b) gives © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the10 publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 10.11

Part b: Verify modal orthogonality.

From Problem 9.7,

LM1 OP m = mM 1 MN 0.5PPQ L 2 − 1 0OP 24 EI M 2 −1 k = −1 h M MN 0 − 1 1PPQ 3

Part a: Natural frequencies and modes. 24 EI

k − ω m =

0 LM2 − λ −1 OP 1 2 1 − − λ − MM PP (a) N 0 − 1 1 − 0.5λ Q

where

λ =

mh ω2 24 EI

Substituting Eq. (a) in Eq. (10.2.6) gives the frequency equation:

λ3 − 6 λ2 + 9 λ − 2 = 0 The solution gives λ 1 = 2 − 3 = 0. 2679 , λ 2 = 2 and λ 3 = 2 + 3 = 3. 7321 . The corresponding natural frequencies are ω 1 = 2. 5359

EI mh 3

ω 2 = 6. 9282

LM1 OP R|− 1U| 1 MM PP S| 0V| = 0 N 0.5Q T 1W L1 OP R| 0.5U| φ mφ = m 0.5 0.866 1 MM 1 S|− 0.866V| = 0 P MN 0.5PQ T 1W LM1 OP R| 0.5U| φ mφ = m − 1 0 1 M 1 S− 0.866V = 0 MN 0.5PPQ |T 1|W (e) 2 −1 0O R− 1U L 24 EI | | 0.5 0.866 1 MM− 1 2 − 1PP S 0V = 0 φ kφ = h MN 0 − 1 1PQ |T 1|W φ 1T mφ 2 = m 0.5 0.866 1

EI mh 3

T 1

T 2

T 1

24 EI

φ T2 kφ 3 =

24 EI

0.5

First mode

Second mode

OP R| 0.5U| PP S|− 0.866V| = 0 Q T 1W

LM 2 − 1 MM N

−1 0 1 −1 2 −1 0 −1 1

LM1 OP R| 0.5U| = m 0.5 0.866 1 M 1 S0.866V MN 0.5PPQ |T 1|W

.m = 15

M2 = φ 2T mφ 2 = 1. 5 m

(c)

M3 = φ 3T mφ 3 = 1. 5 m

Divide φ1 by 1. 5 m , φ 2 by 1. 5 m and φ 3 by 1. 5 m to obtain the normalized modes:

0.5

–1

OP R| 0.5U| PP S|− 0.866V| = 0 Q T 1W

M 1 = φ 1T mφ 1

– 0.866

0.866

Part c: Normalize modes so that M n = 1 .

Following the procedure used in Example 10.1, the mode shapes are determined from Eq. (10.2.5):

LM 2 − 1 MM N

0.5 0.866 1 − 1 2 −1 0 −1 1

(f) Thus the computed modes satisfy the orthogonality properties.

(b)

R| 0.5U| R|− 1U| R| 0.5U| φ = S0.866V φ = S 0V φ = S− 0.866V |T 1|W |T 1|W |T 1|W

φ 1T kφ 3 =

EI mh 3

ω 3 = 9. 4641

Third mode

φ1 =

R| S| T

U| R− 1U R 0.5U| 2 | | 2 | φ = φ = 0 V| S V S− 0.866V| 3m | | 3m | 1 W T W T 1W

0.5 2 0.866 3m 1

(g)

These modes are scalar multiples of the modes in (a); the shapes of the two sets of modes are the same.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the11 publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Let φ 13 = 1, then the third and first equations give:

Problem 10.12 From Problem 9.8,

−φ 12 + ( 0.686)1 = 0 ⇒ φ 12 = 0.686

⎡1 0 0 ⎤ m = m ⎢⎢0 1 0 ⎥⎥ ⎢⎣0 0 1 / 2⎥⎦

4.3723φ 11 − 2( 0.6862) = 0 ⇒ φ 11 = 0.314 ⎡0.314⎤ φ1 = ⎢⎢0.686⎥⎥ ⎢⎣ 1 ⎥⎦

⎡ 5 −2 0 ⎤ k = k ⎢⎢− 2 3 − 1⎥⎥ ⎢⎣ 0 − 1 1 ⎥⎦

Substituting λ 2 = 3 in Eq. (a) gives ⎡ ⎤ ⎢ 2 − 2 0 ⎥ ⎡φ21 ⎤ ⎡0⎤ − 1 ⎥ ⎢⎢φ22 ⎥⎥ = ⎢⎢0⎥⎥ k ⎢− 2 0 ⎢ 1⎥ ⎢ 0 − 1 − ⎥ ⎢⎣φ23 ⎥⎦ ⎢⎣0⎥⎦ 2⎦ ⎣

where k =8 EI / h3 Part a: Natural frequencies and modes. ⎤ ⎡ 0 ⎥ ⎢5 − λ − 2 k − ω 2m = k ⎢ − 2 3 − λ −1 ⎥ ⎢ λ⎥ −1 1− ⎥ ⎢ 0 2⎦ ⎣

(a)

Let φ 23 =1 , then the third and first equations give: −φ 22 − 0.5 = 0 ⇒ φ 22 = −0.5

where λ =mω 2 / k

−2φ 21 − 1 = 0 ⇒ φ 21 = −0.5

λ λ ⎤ ⎡ ⎤ ⎡ det (k − ω 2 m) = (5 − λ ) ⎢(3 − λ ) (1 − ) − 1⎥ + 2 ⎢− 2 (1 − )⎥ 2 2 ⎦ ⎣ ⎦ ⎣

⎡− 0.5⎤ φ 2 = ⎢⎢− 0.5⎥⎥ ⎣⎢ 1 ⎦⎥

=−

λ3 2

+ 5λ2 −

25λ +6 2

Substituting Eq. (a) in Eq. (10.2.6) gives the frequency equation:

λ3 −10λ2 + 25λ −12 =0

(b)

The solution gives:

Substituting λ 3 =6.372 in Eq. (a) gives −2 0 ⎤ ⎡φ31 ⎤ ⎡0⎤ ⎡− 1.372 − 3.372 − 1 ⎥⎥ ⎢⎢φ32 ⎥⎥ = ⎢⎢0⎥⎥ k ⎢⎢ − 2 ⎢⎣ 0 −1 − 2.186⎥⎦ ⎢⎣φ33 ⎥⎦ ⎢⎣0⎥⎦

Let φ 33 =1 , then the third and first equations give:

λ 1 = 0.6277 λ 2 =3

−φ 32 − ( 2.186)1 = 0 ⇒ φ 32 = −2.186

λ 3 = 6.372

−1.3723φ 31 − 2 (−2.186) = 0 ⇒ φ 31 = 3.186

The corresponding natural frequencies are:

ω 1 = 2.241

EI mh 3

ω 2 = 4.899

EI mh 3

ω 3 = 7.14

Substituting λ 1 =0.6277 in Eq. (a) gives −2 0 ⎤ ⎡φ11 ⎤ ⎡0⎤ ⎡4.3723 ⎢ −2 − 1 ⎥⎥ ⎢⎢φ12 ⎥⎥ = ⎢⎢0⎥⎥ 2.3723 k⎢ ⎢⎣ 0 −1 0.686⎥⎦ ⎢⎣φ13 ⎥⎦ ⎢⎣0⎥⎦

EI mh 3 (c)

⎡ 3.186 ⎤ φ3 = ⎢⎢− 2.186⎥⎥ ⎢⎣ 1 ⎥⎦

In summary, the modes are: ⎡0.314⎤

φ1 = ⎢⎢0.686⎥⎥ ⎢⎣ 1 ⎥⎦

⎡− 0.5⎤

φ 2 = ⎢⎢− 0.5⎥⎥ ⎢⎣ 1 ⎥⎦

⎡ 3.186 ⎤

φ 3 = ⎢⎢− 2.186⎥⎥ ⎢⎣

(d)

⎥⎦

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the12 publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

0.686

-0.5

0.314

-0.5

First mode

Part c: Normalize modes so that M n =1 .

-2.186

M 1 = φ 1 mφ 1

⎡ ⎤ ⎢1 0 0 ⎥ ⎡0.314⎤ = m 〈 0.314 0.686 1〉 ⎢0 1 0 ⎥ ⎢⎢0.686⎥⎥ ⎢ 1⎥ ⎢0 0 ⎥ ⎢⎣ 1 ⎥⎦ 2⎦ ⎣

3.189

Third mode

Second mode

=1.069m

Part b: Verify modal orthogonality. ⎡ ⎢1 0 φ1T mφ2 = m [0.314 0.686 1] ⎢0 1 ⎢ ⎢0 0 ⎣

⎤ 0 ⎥ ⎡− 0.5⎤ 0 ⎥ ⎢⎢− 0.5⎥⎥ = 0 1⎥ ⎢ ⎥ ⎣ 1 ⎥⎦ 2⎦

⎡ ⎢1 0 T φ1 mφ3 = m [0.314 0.686 1] ⎢0 1 ⎢ ⎢0 0 ⎣

⎤ 0 ⎥ ⎡ 3.186 ⎤ ⎢ ⎥ 0 ⎥ ⎢− 2.186⎥ = 0 1⎥ ⎢ ⎥ ⎣ 1 ⎥⎦ 2⎦

⎡ ⎢1 0 T φ2 mφ3 = m [− 0.5 − 0.5 1] ⎢0 1 ⎢ ⎢0 0 ⎣

⎤ 0 ⎥ ⎡ 3.186 ⎤ 0 ⎥ ⎢− 2.186⎥⎥ = 0 1 ⎥ ⎢⎢ ⎥ ⎣ 1 ⎥⎦ 2⎦

M 2 = φ 2 mφ 2 ⎡ ⎢1 0 = m [− 0.5 − 0.5 1] ⎢0 1 ⎢ ⎢0 0 ⎣

⎤ 0 ⎥ ⎡− 0.5⎤ 0 ⎥ ⎢− 0.5⎥⎥ 1 ⎥ ⎢⎢ ⎥ ⎣ 1 ⎥⎦ 2⎦

=m T

M 2 = φ 3 mφ 3 ⎡ ⎢1 0 = m [3.186 − 2.186 1] ⎢0 1 ⎢ ⎢0 0 ⎣

(e)

⎤ 0 ⎥ ⎡ 3.186 ⎤ 0 ⎥ ⎢− 2.186⎥⎥ 1 ⎥ ⎢⎢ ⎥ ⎣ 1 ⎥⎦ 2⎦

= 15.46 m

φ1T kφ2 = k [0.314

⎡ 5 − 2 0 ⎤ ⎡− 0.5⎤ ⎢ ⎥ ⎢ 0.686 1] ⎢− 2 3 − 1⎥ ⋅ ⎢− 0.5⎥⎥ = 0 ⎢⎣ 0 − 1 1 ⎥⎦ ⎢⎣ 1 ⎥⎦ ⎡5

φ1T kφ3 = k [0.314 0.686 1] ⎢⎢− 2 ⎢⎣ 0

φ 2T kφ 3 = k [− 0.5

Divide

⎡ 5 − 2 0 ⎤ ⎡ 3.186 ⎤ − 0.5 1] ⎢⎢− 2 3 − 1⎥⎥ ⎢⎢− 2.186⎥⎥ = 0 ⎢⎣ 0 − 1 1 ⎥⎦ ⎢⎣ 1 ⎥⎦

1.069m ,

φ2

m and

φ 3 by

15.46m to obtain the normalized modes:

−2

0 ⎤ ⎡ 3.186 ⎤ 3 − 1⎥⎥ ⎢⎢− 2.186⎥⎥ = 0 − 1 1 ⎥⎦ ⎢⎣ 1 ⎥⎦

φ1

φ1 =

⎡0.314⎤ ⎡− 0.5⎤ ⎡ 3.189 ⎤ 1 ⎢0.686⎥ φ = 1 ⎢− 0.5⎥ φ = ⎢− 2.186⎥ ⎥ 2 ⎥ 3 ⎥ 1.069m ⎢ 15.46m ⎢ m⎢ ⎢⎣ 1 ⎥⎦ ⎢⎣ 1 ⎥⎦ ⎢⎣ 1 ⎥⎦ 1

These modes are multiples of the modes in (a); the shapes of the two sets of modes are the same.

(f) Thus the computed modes satisfy the orthogonality properties.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the13 publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 10.13 When the columns are hinged at the base, the stiffness of the first story is k1 = 2

FG 3EI IJ = 6EI Hh K h 3

The stiffness of the second and third stories does not change. Following the procedure in Problem 9.7 gives k =

0O −1 . LM125 1 2 1PP − − MM N 0 − 1 1PQ

24 EI h3

The structure with columns hinged at the base is more flexible than the structure with clamped columns, and thus has lower natural frequencies. The fundamental frequency is less than half, whereas the higher frequencies are affected less. The modes are also affected by the column fixity. Notice that the fundamental mode of the structure with hinged columns indicates a flexible first story relative to the other stories.

The mass matrix is the same as in Problem 10.12:

LM1 OP m = mM 1 MN 0.5PPQ Then 2

k − ω m =

24 EI h3

0 −1 . − λ LM125 OP MM −1 2 − λ −1 PP 1 − 0.5λ Q −1 N 0

(a)

where

λ =

mh 3 2 ω 24 EI

Substituting Eq. (a) in Eq. (10.2.6) gives the frequency equation: 4 λ3 − 21λ2 + 24 λ − 2 = 0

Following the procedure of Problem 10.12 we obtain ω1 = 1. 4726

EI mh

ω 2 = 6. 0413

EI mh

ω 3 = 9. 3453

mh3

(b)

R|0.8234U| R|− 0.8851U| R| 0.3430 U| φ = S0.9548V φ = S 0.2396 V φ = S− 0.8195V |T 1 |W |T 1 |W |T 1 |W 1

(c)

First mode

Second mode

Third mode

Problem 10.14 When the columns are hinged at the base, the stiffness of the first story is ⎛ 3EI ⎞ 6 EI ⎟= k1 = 2 ⎜⎜ ⎟ ⎝ h3 ⎠ h3

The stiffness of the second and third stories does not change. Following the procedure in Problem 9.8 gives ⎡ 11 ⎤ ⎡ 22 − 16 0 ⎤ ⎢ 4 −2 0 ⎥ EI ⎢ ⎥ ⎢ ⎥ k= ⎢− 16 24 − 8⎥ = k ⎢− 2 3 − 1⎥ 3 h ⎢ 0 − 8 8 ⎥⎦ ⎢ 0 −1 1 ⎥ ⎣ ⎢⎣ ⎥⎦

The structure with columns hinged at the base is more flexible than the structure with clamped columns, and thus has lower natural frequencies. The fundamental frequency is less than half, whereas the higher frequencies are affected less. The modes are also affected by column fixity. Notice that the fundamental mode of the structure with hinged columns indicates a flexible first story relative to the other stories.

where k = 8 EI / h 3 . The mass matrix is the same as in Problem 10.12: ⎡1 0 0 ⎤ m = m ⎢⎢0 1 0 ⎥⎥ ⎢⎣0 0 1 / 2⎥⎦

Then ⎡11 ⎤ 0 ⎥ ⎢ 4 −λ −2 3−λ k − ω 2m = k ⎢ − 2 −1 ⎥ ⎢ 1 ⎥ ⎢ 0 −1 1− λ ⎥ 2 ⎥⎦ ⎣⎢

where λ =

(a)

m 2 ω k

Substituting Eq. (a) in Eq. (10.2.6) gives: −

λ3 2

31λ2 55λ 3 − + =0 8 8 2

Following the procedure of Problem 10.12 gives:

ω 2 = 4.257

EI , mh3

⎡ 0.7 ⎤

⎡− 0.549⎤

⎡ 1.301 ⎤

⎢⎣ 1 ⎥⎦

⎢⎣

ω 1 = 1.423

EI , mh3

ω 3 = 6.469

EI mh3

ϕ1 = ⎢⎢0.873⎥⎥ ϕ 2 = ⎢⎢− 0.133⎥⎥ ϕ 3 = ⎢⎢− 1.614⎥⎥

First mode

⎥⎦

⎢⎣

Second mode

⎥⎦

Third mode

. R| u (t )U| R| 12440 U| R|− 0.3333U| S|u (t )V| = S|2.1547V| cos ω t + S| 0 V| cos ω t Tu (t )W T2.4880W T 0.3333W (f) R| 0.0893U| . + S− 01547 V| cos ω t |T 01786 . W

Problem 10.15

2 1

0.25

–1

Case b Fig. P10.15a

Fig. P10.15b

The initial conditions (from Fig. P10.15b) are

Fig. P10.15c

R| − 1U| R|0U| u(0) = S0.25V u& (0) = S0V |T 1|W |T0|W

From Problem 10.11, ω1 = 2. 5359

ω 2 = 6. 9282

mh3

EI mh3

ω 3 = 9. 4641

EI mh3

Then from Eq. (d),

(a) and

R| 0.5U| R|− 1U| R| 0.5U| φ = S0.866V φ = S 0V φ = S− 0.866V |T 1|W |T 1|W |T 1|W 1

(b)

The response of the system to initial displacements is obtained from Eqs. (10.8.6) and (10.8.5):

L O q& (0) sin ω t P u(t ) = ∑ φ Mq (0) cos ω t + ω N Q n

The displacement response is

R| u (t )U| R|0.0717U| R|− 1U| R|− 0.0717U| . cos cos . ( ) 0 u t 01241 t t = ω + ω + V| cos ω t S| V| S| V| S| V| S| 01241 . . W W Tu (t )W T01433 T 1W T −01433 1

(h) Case c The initial conditions (from Fig. P10.15c) are

q& n ( 0 ) =

φ Tn mu& ( 0 ) φ Tn mφ n

Then from Eq. (d),

The initial conditions (from Fig. P10.15a) are

q1 ( 0 ) = 0. 0893 q2 ( 0 ) = − 0. 3333 q3 ( 0 ) = 1. 2440

R| 1U| R|0U| & u(0) = S2V u(0) = S0V |T3|W |T0|W

q&1 ( 0 ) = 0 q&2 ( 0 ) = 0

(i)

q&3 ( 0 ) = 0

The displacement response is

Then from Eq. (d): q&1 ( 0 ) = 0 q&2 ( 0 ) = 0

R| 1U| R|0U| & u(0) = S− 1V u(0) = S0V |T 1|W |T0|W

(d)

Case a

q1 ( 0 ) = 2. 4800 q2 ( 0 ) = 0. 3333 q3 ( 0 ) = 0.1786

q&3 ( 0 ) = 0

(c)

where

φ Tn mu( 0 ) φ Tn mφ n

(g)

n =1

qn ( 0 ) =

q&1 ( 0 ) = 0 q&2 ( 0 ) = 0

q1 ( 0 ) = 0.1433 q2 ( 0 ) = 1 q3 ( 0 ) = − 0.1433

R| u (t )U| R|0.0447U| R| 0.3333U| . cos ( ) u t 0 0774 t = ω + S| V| S| V| S| 0 V| cos ω t Tu (t )W T0.0893W T− 0.3333W (j) R| 0.6220U| . + S− 10774 V| cos ω t |T 12440 . W 1

(e)

q&3 ( 0 ) = 0

Substituting Eqs. (b) and (e) in Eq. (c) gives u ( t ) in inches:

Although all three modes contribute to the response in each case, in the three cases the dominant response is due to the first, second, and third modes, respectively. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the16 publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Case a 4 2 0 -2 -4 Case b 2 un(t) 0

u1 u2 u3

-2

Case c 2 0 -2 0

t/T1

Fig. P10.15d

Substituting Eqs. (b) and (e) in Eq. (c) gives u ( t ) in inches:

Problem 10.16 3

1 0.25

–1

2 1

⎡ 0.065 ⎤ ⎡ u1 (t )⎤ ⎡0.935⎤ ⎢u (t )⎥ = ⎢ 2.04 ⎥ cos ω t + ⎢− 0.0668⎥ cos ω t 1 3 ⎥ ⎥ ⎢ ⎢ 2 ⎥ ⎢ ⎢⎣ 0.0205 ⎥⎦ ⎢⎣u 3 (t )⎥⎦ ⎢⎣ 2.98 ⎥⎦

Case b

The initial conditions (from Fig. P10.16b) are Fig. P10.16a

⎧ −1 ⎫ ⎪ ⎪ u (0) = ⎨0.25⎬ , ⎪ 1 ⎪ ⎩ ⎭

EI EI EI ω 2 = 4.899 ω 3 =7.14 (a) 3 3 mh mh mh3

Then from Eq. (d):

Fig. P10.16a

Fig. P10.16b

From Problem 10.12,

ω 1 = 2.241

⎡0.314⎤ ⎡− 0.5⎤ ⎡ 3.186 ⎤ ⎥ ⎥ ⎢ ⎢ φ1 = ⎢0.686⎥ φ2 = ⎢− 0.5⎥ φ3 = ⎢⎢− 2.186⎥⎥ ⎣⎢ 1 ⎦⎥ ⎣⎢ 1 ⎦⎥ ⎣⎢ 1 ⎦⎥

(b)

(c)

q&1 ( 0) = 0

q2 ( 0) = 0.875

q& 2 ( 0) = 0

q3 ( 0) = −0.209

q& 3 ( 0) = 0

⎡ u1(t )⎤ ⎡0.1048⎤ ⎡− 0.4375⎤ ⎡ − 0.666 ⎤ ⎢u (t )⎥ = ⎢ 0.229 ⎥ cos ω t + ⎢− 0.4375⎥ cos ω t + ⎢ 0.456 ⎥ cos ω t 1 ⎢ 2 ⎢ 3 ⎢ 2 ⎥ ⎢ ⎥ ⎥ ⎥ ⎢⎣u3 (t )⎥⎦ ⎢⎣ 0.334 ⎥⎦ ⎢⎣ 0.875 ⎥⎦ ⎢⎣− 0.2089⎥⎦

Case c

The initial conditions (Fig. P10.16c) are:

where T

φ mu(0) q n ( 0) = n φ nT mφ n

φ mu& (0) q& n (0) = n φ nT mφ n

(d)

The initial conditions from (Fig. P10.16a) are ⎧1⎫ ⎪ ⎪ u(0) = ⎨2⎬ , ⎪3 ⎪ ⎩ ⎭

⎧0⎫ ⎪ ⎪ u& (0) = ⎨0⎬ ⎪0⎪ ⎩ ⎭

q3 ( 0) = 0.0205 q& 3 ( 0) = 0

q&1 ( 0) =0

q2 ( 0) =0.5

q&2 ( 0) = 0

q3 ( 0) = 0.38

q&3 ( 0) = 0

⎡ 1.21 ⎤ ⎡− 0.25⎤ ⎡ u1 (t )⎤ ⎡0.038⎤ ⎢u (t )⎥ = ⎢0.082⎥ cos ω t + ⎢− 0.25⎥ cos ω t + ⎢− 0.83⎥ cos ω t 1 2 3 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 2 ⎥ ⎢ ⎢⎣ 0.38 ⎥⎦ ⎢⎣ 0.5 ⎥⎦ ⎢⎣u3 (t )⎥⎦ ⎢⎣ 0.12 ⎥⎦

q&1 (0) = 0 q& 2 (0) = 0

q1 ( 0) =0.12

The displacement response is

Then from Eq. (d): q1 (0 ) = 2.98

⎧1⎫ ⎪ ⎪ u(0) = ⎨− 1⎬ , ⎪1⎪ ⎩ ⎭

Then from Eq. (d):

Case a

q 2 (0 ) = 0

q1 ( 0) = 0.334

The displacement response is

The response of the system to initial displacements is obtained from Eqs. (10.8.6) and (10.8.5): 3 ⎡ ⎤ q& (0) u(t ) = ∑ φn ⎢qn (0) cos ω nt + n sin ω nt ⎥ ωn n =1 ⎣ ⎦

⎡0 ⎤ u& (0) = ⎢⎢0⎥⎥ ⎢⎣0⎥⎦

(e) Although all three modes contribute to the response in each case, in the three cases the dominant response is due to the first, second, and third modes, respectively.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 18 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Case a 4 2 0 -2 -4 Case b 2 un(t) 0

u1 u2 u3

-2

Case c 2 0 -2 0

t/T1

Fig. P10.16d

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 19 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 10.17

EI mh 3 EI ω 2 D = 6. 9195 mh 3 EI ω 3 D = 9. 4523 mh 3

From Problem 10.11, EI

ω1 = 2. 5359

ω 2 = 6. 9282

EI mh

ω 3 = 9. 4641

EI mh

(a) The response of the system is given by Eqs. (10.8.7) and (10.10.2): 2

u(t ) =

∑φ e ζω n

−

q n (0) cos ω n D t

n =1

EI mh 3 EI ζ 2 ω 2 = 0. 3461 mh 3 EI ζ 3ω 3 = 0. 4732 mh 3

ω 1 D = 2. 5327

q& n (0) + ζ nω n qn (0)

ωnD

(b)

O sin ω t P PQ

ζ1ω 1 = 0.1268

(c)

Substituting Eq. (c) and values of qn ( 0 ) from Problem 10.15 in Eq. (b) gives the following response for the first set of initial conditions: u(t ) =

. R| 12440 U S|2.1547|V| e T2.4880W

− ζ 1ω 1t

R|− 0.3333U| S| 0 V| e T 0.3333W R| 0.0893U| + S− 01547 . V| e |T 01786 . W

where

− ζ 2ω 2 t

ω n D = ω n 1 − ζ n2 For ζ n = 5% ,

cos ω 1 D t +

− ζ 3ω 3t

R|0.0623U| . S|01079 V| e 01249 . T W

− ζ 1ω 1t

R|− 0.0167U| S| 0 V| e T 0.0167W R| 0.0045U| cos ω t + S− 0.0077Ve |T 0.0089|W

cos ω 2 D t +

sin ω 1 D t

− ζ 2ω 2 t

sin ω 2 D t

−ζ 3ω 3t

sin ω 3 D t

Case a 4 2 0 -2 -4 Case b 2 un(t) 0

u1 u2 u3

-2

Case c 2 0 -2 0

t/T1

Fig. P10.17

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 20 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 10.18 From Problem 10.12,

ω 1 = 2.241

EI mh

ω 2 = 4.899

EI mh

ω 3 = 7.14

EI mh 3

(a) The response of the system is given by Eqs. (10.8.7) and (10.10.2): 3

u(t ) =

∑ φ e ς ω [q (0) cos ω t − n

n =1

q& n (0) + ς n ω n q n (0)

ω nD

⎤ sin ω nD t ⎥ ⎦

(b)

where

ω nD = ω n 1 − ς n2 For ς n = 5% ,

ω1D = 2.2382 ω 2 D = 4.8929 ω 3 D = 7.1311

EI mh

EI 3

ς 2ω 2 = 0.2450

ς 3ω 3 = 0.3570

mh EI mh

ς 1ω1 = 0.1121

EI mh 3 EI mh 3

(c)

EI mh 3

Substituting Eq. (c) and values of q n (0) from Problem 10.16 gives the following response for the first set of initial conditions: ⎧0.0469⎫ ⎧0.9358⎫ ⎪ ⎪ ⎪ −ς 1ω1t ⎪ cos ω 1D t + ⎨0.1023⎬e −ς 1ω1t sin ω 1D t u(t ) = ⎨2.0443⎬e ⎪0.1492⎪ ⎪ 2.98 ⎪ ⎭ ⎩ ⎭ ⎩ ⎧ 0.0033⎫ ⎧0.0653⎫ ⎪ ⎪ ⎪ −ς 3ω 3t ⎪ cos ω 3D t + ⎨− 0.0022 ⎬e −ς 3ω 3t sin ω 3 D t + ⎨0.0448⎬e ⎪ 0.0010⎪ ⎪0.0205⎪ ⎭ ⎩ ⎭ ⎩

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the21 publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 10.19 u8

The joint rotations associated with each mode are obtained by substituting u t = φ n , n = 1, 2 and 3, in Eq. (c); the results are

R|− 0.3548U| R| 0.3486U| R| 0.5837U| 0 3548 . 0 3486 . − | | | 0.5837| | | 1 |− 0.2838| 1 | − 0.8453| 1 | 0.3966| (φ ) = S V (φ ) = h S| − 0.8453V| (φ ) = h S| 0.3966V| h |− 0.2838| ||− 01374 ||− 0.9218|| ||− 2.0011|| || . |T− 01374 |T− 0.9218|W |T− 2.0011|W |W .

1 0

2 0

3 0

With reference to the lateral floor displacements u1 , u2 and u3 , the mass matrix and the condensed stiffness matrix (from Problem 9.9) are

LM1 OP m = mM 1 MN 0.5PPQ . O L 40.85 − 23.26 511 $k = EI M− 23.26 − 14.25PP 3109 . M h MN 511 − 14.25 . 10.06PQ tt

First mode

Second mode

Third mode

Obtained by using the same procedure as in Problem 10.11, the natural frequencies and modes are ω1 = 1. 4576

EI mh 3

ω 2 = 4. 7682

EI mh 3

ω 3 = 8.1980

EI mh 3

(a)

φ1 =

. U| R|0.3156U| R|− 0.7409U| R| 12546 12024 0 3572 0 . 7451 . . φ = − φ = − V| V| S| V| S| S| 1 1 1 W W T W T T 2

(b) The joint rotations corresponding to the modes of Eq. (b) are computed using Eq. (9.3.3): u0 = T ut

(c)

where −1 T = − k 00 k 0t

(d)

Substituting k 00 and k 0t from Problem 9.9 in Eq. (d) gives . 0.0744O − 0.5342 LM− 01084 . 0.0744PP − 0.5342 MM− 01084 0.5961 − 0.0619 − 0.4258P 1 T = P M h M 0.5961 − 0.0619 − 0.4258P MM − 01703 . 0.8748 − 0.7355P P . 0.8748 − 0.7355PQ MN − 01703 © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the22 publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

⎡− 0.1512 − 0.6084 ⎢− 0.1512 − 0.6084 ⎢ − 0.11 1 ⎢ 0.7184 T= ⎢ − 0.11 h ⎢ 0.7184 ⎢− 0.2612 1.131 ⎢ 1.131 ⎢⎣− 0.2612

Problem 10.20

The joint rotations associated with each mode are obtained by substituting u t =φ n , n = 1, 2, 3, in Eq. (c): The results are:

0.0962 ⎤ 0.0962 ⎥⎥ − 0.457⎥ ⎥ − 0.457⎥ − 0.925⎥ ⎥ − 0.925⎥⎦

With reference to the lateral floor displacements u1 , u2 and u3 , the mass matrix and the condensed stiffness matrix (from Problem 9.10) are ⎡1 0 0 ⎤ m = m ⎢⎢0 1 0 ⎥⎥ ⎣⎢0 0 0.5⎦⎥

⎡ 0.667 ⎤ ⎡ 0.472 ⎤ ⎡− 0.369⎤ ⎢ 0.667 ⎥ ⎢ 0.472 ⎥ ⎢− 0.369⎥ ⎥ ⎢ ⎥ ⎥ ⎢ ⎢ 1 ⎢ − 0.338⎥ 1 ⎢− 0.917⎥ 1 ⎢ 0.782 ⎥ (φ1 )0 = ⎢ ⎥ ⎥ (φ 3 )0 = ⎢ ⎥ (φ 2 )0 = ⎢ h ⎢ − 0.338⎥ h ⎢− 0.917⎥ h ⎢ 0.782 ⎥ ⎢− 2.808⎥ ⎢ − 1.238 ⎥ ⎢ − 0.208⎥ ⎥ ⎢ ⎥ ⎥ ⎢ ⎢ ⎣⎢− 2.808⎦⎥ ⎣⎢ − 1.238 ⎦⎥ ⎣⎢ − 0.208⎦⎥

⎡ 39.38 − 22.68 5.486 ⎤ EI ⎢ 27.13 − 11.75⎥⎥ kˆ tt = 3 ⎢ h ⎢ Symm 7.418 ⎥⎦ ⎣

Obtained by using the same procedure as in Problem 10.11, the natural frequencies and modes are:

ω 1 = 1197 .

EI mh3

ω 2 = 4.178

EI mh3

ω 3 = 7.903

⎡0.273⎤ ⎡− 0.706⎤ ⎡ 1.529 ⎤ ⎢ ⎢ ⎥ ⎥ φ1 = ⎢0.698⎥ φ 2 = ⎢ − 0.441⎥ φ3 = ⎢⎢− 1.315⎥⎥ ⎢⎣ 1 ⎥⎦ ⎢⎣ 1 ⎥⎦ ⎢⎣ 1 ⎥⎦

EI mh3 (a)

First mode

Second mode

Third mode

(b)

The joint rotations corresponding to the modes of Eq. (b) are computed using Eq. (9.3.3): u0 =T ut

(c)

where −1 T = − k oo ⋅ k oT

(d)

Substituting k 00 and k 0t from Problem 9.10 in Eq. (d) gives

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the23 publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

⎡− 0.3006 − 0.3695 0.0313 ⎤ ⎢− 0.3006 − 0.3695 0.0313 ⎥ ⎥ ⎢ − 0.321 − 0.227 ⎥ 1 ⎢ 0.6795 T= ⎢ ⎥ − 0.321 − 0.227 ⎥ h ⎢ 0.6795 ⎢− 0.1942 0.9489 − 0.7923⎥ ⎥ ⎢ ⎢⎣− 0.1942 0.9489 − 0.7923⎥⎦

Problem 10.21

The joint rotations associated with each mode are obtained by substituting u t =φ n , n = 1, 2, 3, in Eq. (c): The results are:

With reference to the lateral floor displacements u1 , u2 and u3 , the mass matrix and the condensed stiffness matrix (from Problem 9.11) are ⎡1 0 0 ⎤ m = m ⎢⎢0 1 0 ⎥⎥ ⎢⎣0 0 0.5⎥⎦

⎡− 0.217 ⎤ ⎡ 0.404 ⎤ ⎡− 0.275⎤ ⎢− 0.217 ⎥ ⎢ 0.404 ⎥ ⎢− 0.275⎥ ⎥ ⎢ ⎥ ⎥ ⎢ ⎢ 1 ⎢ − 0.273⎥ 1 ⎢− 0.385⎥ 1 ⎢ 2.684 ⎥ (φ1 )0 = ⎢ ⎥ ⎥ (φ 3 )0 = ⎢ ⎥ (φ 2 )0 = ⎢ h ⎢ − 0.273⎥ h ⎢− 0.385⎥ h ⎢ 2.684 ⎥ ⎢ − 3.366 ⎥ ⎢ − 1.254 ⎥ ⎢ − 0.231⎥ ⎥ ⎢ ⎥ ⎥ ⎢ ⎢ ⎣⎢ − 3.366 ⎦⎥ ⎣⎢ − 1.254 ⎦⎥ ⎣⎢ − 0.231⎦⎥

⎡ 33.36 − 14.91 1.942 ⎤ ˆk = EI ⎢ 15.96 − 5.489⎥⎥ tt ⎢ 3 h ⎢ Symm 3.923 ⎥⎦ ⎣

Obtained by using the same procedure as in Problem 10.12, the natural frequencies and modes are:

ω 1 = 1.329

First mode

Second mode

Third mode

EI EI EI ω 2 =3.514 ω 3 = 6.562 3 3 mh mh mh3

(a) ⎡0.234⎤ ⎡− 0.512⎤ ⎡ 3.324 ⎤ ⎥ ⎥ ⎢ ⎢ φ1 = ⎢0.639⎥ φ 2 = ⎢ − 0.591⎥ φ 3 = ⎢⎢− 2.032⎥⎥ ⎢⎣ 1 ⎥⎦ ⎢⎣ 1 ⎥⎦ ⎢⎣ 1 ⎥⎦

(b)

The joint rotations corresponding to the modes of Eq. (b) are computed using Eq. (9.3.3): u0 =T ut

(c)

where −1 T = − k oo ⋅ k oT

(d) Substituting k 00 and k 0t from Problem 9.11 in Eq. (d) gives

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the24 publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 10.22

⎡ −0.4005 ⎢ ⎢ − 0.4005 1 ⎢ 0.9530 T= ⎢ h ⎢ 0.9530 ⎢ − 0.3465 ⎢ ⎢⎣ − 0.3465

0.0458 ⎤ ⎥ 0.0458 ⎥ − 0.2803⎥ ⎥ − 0.4569 − 0.2803⎥ 1.2570 − 0.9890⎥ ⎥ 1.2570 − 0.9890⎥⎦

The joint rotations associated with each mode are obtained by substituting u t =φ n , n = 1, 2, 3, in Eq. (c): The results are:

−0.4152 − 0.4152 − 0.4569

With reference to the lateral floor displacements u1 , u2 and u3 , the mass matrix and the condensed stiffness matrix (from Problem 9.12) are

⎡ −0.2819 ⎤ ⎡ 0.5364 ⎤ ⎡ −0.4482 ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ − 0 . 5364 0 . 2819 ⎢ ⎥ ⎢ ⎥ ⎢ − 0.4482 ⎥ ⎢ − 0.3626⎥ 1 ⎢ 3.6639 ⎥ 1 ⎢ − 0.5006⎥ (φ1 )0 = 1 ⎢ ⎥ (φ2 )0 = ⎢ ⎥ (φ3 )0 = ⎢ ⎥ h ⎢ − 0.3626⎥ h ⎢ − 0.5006⎥ h ⎢ 3.6639 ⎥ ⎢ − 0.3083⎥ ⎢ − 1.6240 ⎥ ⎢ − 4.5137 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣⎢ − 0.3083⎦⎥ ⎣⎢ − 1.6240 ⎦⎥ ⎣⎢ − 4.5137 ⎦⎥

⎡1 0 0 ⎤ m = m ⎢⎢0 1 0 ⎥⎥ ⎢⎣0 0 0.5⎥⎦ 2.43 ⎤ ⎡ 30.77 − 14.01 ˆk = EI ⎢ 13.82 − 4.80⎥⎥ tt h 3 ⎢⎢ 2.92 ⎥⎦ ⎣ Symm

First mode

Second mode

Third mode

Obtained by using the same procedure as in Problem 10.12, the natural frequencies and modes are: ω1 = 1.043

EI mh

ω 2 = 3.081

EI mh

ω 3 = 6.314

EI mh 3

(a) ⎡0.1997 ⎤ φ1 = ⎢⎢0.5966⎥⎥ ⎢⎣ 1 ⎥⎦

⎡ − 0.5454⎤ φ2 = ⎢⎢ − 0.6555⎥⎥ ⎢⎣ 1 ⎥⎦

⎡ 3.2201⎤ φ3 = ⎢⎢− 1.9163⎥⎥ ⎢⎣ 1 ⎥⎦

(b)

The joint rotations corresponding to the modes of Eq. (b) are computed using Eq. (9.3.3): u 0 = Tu t

(c)

where −1 T = −k oo k oT

(d)

Substituting k 00 and k 0t from Problem 9.12 in Eq. (d) gives

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the25 publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 10.23

φ 32 = –1.2826

Part a

φ 22 = 1.2826 φ 12 = 1

From Problem 9.13, the mass and stiffness matrices of the system are ⎡5 ⎤ ⎢ m = m ⎢ 1 ⎥⎥ ⎢⎣ 1⎥⎦

Second mode

φ 23 = 1

φ 33 = 1

⎡ 28 6 −6⎤ 3EI ⎢ 3⎥⎥ k = ⎢ 6 7 3 10 L ⎢−6 3 7 ⎥ ⎣ ⎦

φ 13 = 0

Then ⎡28 − 5λ 6 L ⎢ −6 ⎣

3EI ⎢ k −ω 2 m = ⎢ 10 3

6 −6 ⎤ 7−λ 3 ⎥⎥ 3 7 − λ ⎥⎦

Third mode

(a)

The vectors of initial displacements and velocities are

where 10 m L3 2 λ = ω 3 EI

(b)

Substituting Eq. (a) in Eq. (10.2.6) gives the frequency equation: 5 λ3 − 98λ2 + 520 λ − 400 = 0

EI m L3

ω 2 = 1. 6135

EI m L3

ω 3 = 1. 7321

EI m L3

(c) Following the procedure used in Example 10.1, the natural modes are determined from Eq. (10.2.5): 1⎫ 1⎫ ⎧ ⎧ ⎧0⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ φ1 = ⎨− 1.9492⎬ φ 2 = ⎨ 1.2826⎬ φ 3 = ⎨ 1⎬ ⎪ 1.9492⎪ ⎪− 1.2826⎪ ⎪ 1⎪ ⎩ ⎭ ⎩ ⎭ ⎩ ⎭

φ 11 = 1

φ 31 = 1.9492

⎧ 1⎫ ⎧0⎫ ⎪ ⎪ ⎪ ⎪ u(0) = ⎨0⎬ u& (0) = ⎨0⎬ ⎪0⎪ ⎪0⎪ ⎩ ⎭ ⎩ ⎭

⎫ q&1 (0) = 0 ⎪ ⎪ φ1T mφ1 ⎪ ⎪⎪ φ 2T mu (0) = 0.6031 q& 2 (0) = 0 ⎬ q 2 ( 0) = φ 2T mφ 2 ⎪ ⎪ T φ 3 mu (0) ⎪ & =0 q3 (0) = q3 (0) = 0 ⎪ T φ 3 mφ 3 ⎪⎭

φ 21 = –1.9492

q1 (0) = 1

= 0.3969

(f)

The free vibration response is given by Eq. (10.8.6) as 3

u(t ) =

(d)

(e)

Substituting m, k and Eqs. (d)-(e) in Eq. (10.8.5) gives

φ T mu (0)

The roots of this equation are λ 1 = 0. 9219 , λ 2 = 8. 6780 and λ 3 = 10 . The corresponding natural frequencies are ω1 = 0. 5259

Part b

⎡

q& (0)

⎤

∑ φ n ⎢⎣qn (0)cosω nt + ωn n sinω nt ⎥⎦

(g)

n =1

Substituting for φn , qn ( 0 ) and q&n ( 0 ) gives ⎧ u1 (t ) ⎫ ⎧ 0.3969⎫ ⎧ 0.6031⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨u 2 (t )⎬ = ⎨− 0.7736⎬cosω1t + ⎨ 0.7736⎬cosω 2 t ⎪u (t ) ⎪ ⎪ 0.7736⎪ ⎪− 0.7736⎪ ⎩ 3 ⎭ ⎩ ⎭ ⎩ ⎭

(h)

The third mode does not contribute to the free-vibration response because the initial conditions do not contain a component in this mode.

First mode © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the26 publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 10.24

2 n

b 2k

ek − ω m j φ = 0 ⇒ φ 2.071 0 O LM0 Φ = M2.0322 0 − 0.3988PP MN0.0033 0 0.0166PQ

uθ

k ux

The natural modes are sketched next.

k 0.0033 rad

1. Data.

2.032

m = 90 kips g k = 1. 5 kips in.

b = 25 ft

First mode

ω 1 = 5.96 rads/sec

2. Determine the mass and stiffness matrices. From Problem 9.140,

LM1 OP LM1 OP . = 1 0 2331 1 MM P MM P N (25 × 12) 6PQ N 15,000PQ 0 LM 6 0 OP k = 15 . M 6 − (25 × 12) P MN(sym) 3 × (25 × 12) PQ LM 6 0 0 OP . M 6 − 300 = 15 MN(sym) 270,000PPQ m =

90 386

2.071

Second mode

ω 2 = 6.21 rads/sec

3. Determine natural frequencies. det k − ω 2 m = 0 9 − 0. 2331ω 2

9 − 0. 2331ω − 450

0 0

− 450 = 0 2 405, 000 − 3495ω

e9 − 0.2331ω je9 − 0.2331ω j × e405,000 − 3495ω − 450 j = 0 2

0.0166 rad Third mode – 0.3988

ω 3 = 10.90 rads/sec

The roots of the above equation are

ω12 = 35. 55

ω 22 = 38. 63

ω 32 = 118. 85

Thus the natural frequencies are

ω1 = 5. 96

ω 2 = 6. 21

4. Determine natural modes.

ω 3 = 10. 90

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the27 publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 10.25 u3

k Second mode

u1 2k

ω 2 = 6.21 rads/sec

k 2.071

1. Determine the mass and stiffness matrices. From Problem 9.15,

– 2.489

LM 2 3 − 1 6 1 2OP m = 0.2331 M− 1 6 2 3 − 1 2P NM 1 2 − 1 2 1PQ LM 5 − 2 2OP k = 15 . M− 2 5 −2 MN 2 − 2 6PPQ

– 2.888

Third mode

ω 3 = 10.90 rads/sec 2.489

2. Determine natural frequencies. det k − ω 2 m = 0

(a) 4. Compare these modes with Problem 10.24.

The roots of Eq. (a) are

ω 1 = 5. 96

The two sets of DOF are related by

ω 2 = 6. 21

ω 3 = 10. 90

which are the same as in Problem 10.24. 3. Determine natural modes.

LM 0.4885 2.071 2.4889OP Φ = M− 0.4885 2.071 − 2.4889P . 0 − 2.8878 PQ NM 15437

R|u U| LM1 2 1 2 0OP R|u U| S|u V| = M1 2 − 1 2 1P S|u V| Tu W MN1 b − 1 b 0PQ Tu W

(c)

u = au

(d)

or (b)

where b = 25 × 12 = 300 in.

The natural modes are sketched next.

Substituting u = φn , n = 1, 2 and 3, from Eq. (b) in Eq. (d) gives 2.071 0 OP LM0 Φ = a Φ = M2.0332 0 − 0.3988P MN0.0033 0 0.0166PQ

– 0.489 1.544

These modes are the same as in Problem 10.24. First mode

ω 1 = 5.96 rads/sec 0.489 © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the28 publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 10.26 First mode

b uy

uθ

0.0033 rad

1.5437

ω 1 = 5.96 rad/sec

d 2k

Second mode

1. Data. m = 90 kips/g

k = 1.5 kips/in.

2.0710

b = 25 ft

ω 2 = 6.21 rad/sec

2. Determine mass and stiffness matrices. From Problem 9.16, 0 0 ⎤ ⎡1 90 ⎢ ⎥ 0 1 ( 25 12 ) 2 × ⎥ 386 ⎢ ⎢⎣0 (25 × 12) 2 5(25 × 12) 2 12⎥⎦ 0 ⎤ ⎡1 0 ⎢ = 0.2331 ⎢0 1 150 ⎥⎥ ⎢⎣0 150 37,500⎥⎦

⎡ 6 k = 1.5⎢⎢ ⎢⎣(sym) ⎡ 6 = 1.5⎢⎢ ⎢⎣(sym)

0 6 0 6

0.0166 rad

Third mode 2.8878

5. Compare these modes with Problem 10.24.

0 ⎤ 2(25 × 12) ⎥⎥ 7(25 × 12) 2 2⎥⎦

The two sets of DOF are related by

0 ⎤ 600 ⎥⎥ 315,000⎥⎦

]

det k − ω 2m = 0

(a)

ω 22 = 38.63

ω 2 = 6.21

u = au

(d)

where

In Eq. (d), u represents the degrees of freedom defined in Problem 9.14, while u represents the degrees of freedom defined in Problem 9.16.

ω 32 = 118.85

Thus the natural frequencies are

ω 1 = 5.96

(c)

b = 25 × 12 = 300 in.

The roots of Eq. (a) are

ω 12 = 35.55

⎧u x ⎫ ⎡1 ⎤ ⎧u x ⎫ ⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎨u y ⎬ = ⎢ 1 b 2⎥ ⎨u y ⎬ ⎪u ⎪ ⎢ 1 ⎥⎦ ⎪⎩uθ ⎪⎭ ⎩ θ⎭ ⎣

3. Determine natural frequencies.

[

ω3 = 10.90 rad/sec

ω 3 = 10.90

Substituting u = φ n , n = 1, 2 and 3 , from Eq. (b) in Eq. (d) gives

which are the same as in Problem 10.24. 4. Determine natural modes. 2.0710 0 ⎡ 0 ⎤ − 2.8878⎥⎥ 0 Φ = ⎢⎢1.5437 ⎢⎣0.0033 0 0.0166 ⎥⎦

The natural modes are sketched next.

(b)

2.071 0 ⎡0 ⎢ − 0.3988 Φ = a Φ = ⎢2.0322 0 ⎢⎣ 0.0033 0 0.0166

⎤ ⎥ ⎥ ⎥⎦

These modes are the same as in Problem 10.24.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the29 publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 10.27 u3

2k d

c b

2.5207

1.5437

First mode

ω 1 = 5.96 rad/sec

0.4885

1. Determine mass and stiffness matrices. From Problem 9.17, ⎡ 1 m = 0.2331⎢⎢ − 1 2 ⎢⎣ 1 2

−1 2

1 2⎤ 2 3 − 1 6 ⎥⎥ −1 6 2 3⎥⎦

Second mode

ω 2 = 6.21 rad/sec

3⎤ ⎡ 6 −3 ⎢ 5 − 3⎥⎥ k = 1.5⎢ − 3 ⎢⎣ 3 − 3 7 ⎥⎦

2.0710

2. Determine natural frequencies.

[

2.0901

]

det k − ω 2 m = 0

(a)

2.8878

Third mode

The roots of Eq. (a) are

ω 12 = 35.55

ω 22 = 38.63

ω 32 = 118.85

ω 3 = 10.90 rad/sec

Thus the natural frequencies are

ω1 = 5.96

ω 2 = 6.21

2.4889

ω3 = 10.90 4. Compare these modes with Problem 10.24.

which are the same as in Problem 10.24.

The two sets of DOF are related by

3. Determine natural modes. 2.4889⎤ ⎡0.4885 2.0710 ⎢ 0 2.0901⎥⎥ Φ = ⎢2.5207 ⎢⎣1.5437 − 2.8878⎥⎦ 0

The natural modes are sketched next.

(b)

1 2⎤ ⎧ u1 ⎫ ⎧u x ⎫ ⎡1 − 1 2 ⎪ ⎪ ⎢ ⎪ ⎪ = u 0 1 2 1 2⎥⎥ ⎨u 2 ⎬ ⎨ y⎬ ⎢ ⎪u ⎪ ⎢0 1 b − 1 b ⎥⎦ ⎪⎩u3 ⎪⎭ ⎩ θ⎭ ⎣

(c)

u = au

(d)

where b = 25 × 12 = 300 in.

In Eq. (d), u represents the degrees of freedom defined in Problem 9.14, while u represents the degrees of freedom defined in Problem 9.17.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the30 publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Substituting u = φ n , n = 1, 2 and 3 , from Eq. (b) in Eq. (d) gives 2.071 0 ⎡0 ⎢ Φ = a Φ = ⎢2.0322 0 − 0.3988 ⎢⎣ 0.0033 0 0.0166

⎤ ⎥ ⎥ ⎥⎦

These modes are the same as in Problem 10.24.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the31 publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 10.28

3. Determine the natural modes.

z m

L b

uz d

⎡ 0.7767 − 0.2084 0.5943⎤ ⎥ ⎢ Φ = ⎢ − 0.4923 0.3875 0.7794⎥ ⎥ ⎢ ⎢⎣ − 0.3928 − 0.8980 0.1984⎥⎦

(k − ω n2 m)φ n = 0 ⇒ φ n

Note: φ nT φ n = 1 .

1. Determine mass and stiffness matrices. From solution to Problem 9.18: ⎡m ⎤ ⎢ ⎥ ⎥ m=⎢ m ⎢ ⎥ ⎢⎣ m⎥⎦

and ⎡ 0.9283 0.9088 0.2345⎤ ⎥ EI ⎢ k = 3 ⎢0.9088 1.4294 0.2985⎥ ⎥ L ⎢ ⎢⎣0.2345 0.2985 0.3234⎥⎦

assuming GJ =

4 EI 5

2. Determine the natural frequencies. det[k − ω 2 m] = 0 ⎛ mL3 ⎞ ⎟ 0.9283 − ω 2 ⎜ ⎜ EI ⎟ ⎝ ⎠ EI L3

0.9088

0.2345

0.9088

⎛ mL3 ⎞ ⎟ 1.4294 − ω 2 ⎜ ⎜ EI ⎟ ⎠ ⎝

0.2985

0.2345

0.2985

= 0

⎛ mL3 ⎞ ⎟ 0.3234 − ω 2 ⎜ ⎜ EI ⎟ ⎝ ⎠

Solving for the roots of the characteristic equation yields the natural frequencies of the system.

ω 1 = 0.4834 ω 2 = 0.4990 ω 3 = 1.4827

EI mL3 EI mL3 EI mL3

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the32 publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 10.29

The fundamental natural frequency and mode are m/2

ω1 =

φ1 = 0. 8020 1. 3899 1. 6055

337. 86 = 18. 3809 T

u2 k

u1 k

1. Data: w = 100 kips ⇒ m = 0. 2588 kip - sec 2 in. k = 326. 32 kips in.

OP LM1 m = 0.2588 M 1 MN 0.5PPQ LM 2 − 1 0OP k = 326.32 M− 1 2 − 1P MN 0 − 1 1PQ Inverse iteration equations: k xj +1 = m x j

λ( j + 1) =

(a)

x Tj + 1k x j + 1

(b)

x j + 1m x j + 1

Implement Eqs. (a) and (b) to obtain Table P10.29. Table P10.29 Iteration

λ( j +1)

x j +1

R|0.0036U| S|0.0063V| 338.68 T0.0075W

R|0.7777U| . V| S|13826 . W T16418

R|0.7777U| R|0.0024U| . V| S|0.0041V| 337.87 S|13826 . W T0.0048W T16418

R|0.7989U| . V| S|13894 16094 . T W

R|0.7989U| R|0.0024U| . V| S|0.0041V| 337.86 S|13894 16094 . T W T0.0048W

R|0.8020U| . V| S|13899 16055 . T W

R|1U| S|2V| T 3W

x j +1

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the33 publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 10.30 Excitation frequency: f = 430 rpm =

430 = 7.17 Hz 60

ω = 2 π f = 45. 03 rads sec Shifted eigenvalue problem: ( ( kφ = λmφ ( ( k = k −μ m λ = λ − μ and m and k are given in Problem 10.29. Inverse iteration equations: ( k x j +1 = m x j ( x Tj + 1 k x j +1 ( j + 1) λ = T x j + 1 m x j +1 Use μ = ω 2 = ( 45. 03)2 = 2027. 7 as a shift and implement the inverse iteration equations to obtain Table P10.30.

Table P10.30 Iteration

R| 1 U| S| 0 V| T−0.5W

x j+1

λ( j+1)

x j+1

2027.7

R| 0.0017U| S|−0.0001V| 2517.3 T−0.0017W

. R| 15839 U| . S| −01323 V| . T−16357 W

. U| R| 15839 . V| 2027.7 S|−01323 − 16357 . W T

R| 0.0033U| S| 0.0000V| 2521.3 T−0.0032W

. U R| 16239 S| 0.0143|V| . W T−15663

. U| R| 16239 0 . 0143 V| 2027.7 S| − 15663 . W T

R| 0.0032U| S| 0.0000V| 2521.7 T−0.0033W

. U R| 16020 S|−0.0087|V| . W T −16110

. R| 16020 U 0 − S| .0087|V| 2027.7 . T −16110 W

R| 0.0033U| S| 0.0000V| T−0.0032W

. U R| 16063 0 S| .0017|V| . W T−16023

2521.8

The natural frequency of the structure closest to the machine frequency is

ωn =

2521. 8 = 50. 2174

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the34 publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 10.31 Shifted eigenvalue problem: ( ( kφ = λmφ ( ( k = k −μ m λ = λ − μ

Iteration

and m and k are given in Problem 10.30. Inverse iteration equations: ( k xj +1 = m x j ( x Tj + 1k x j +1 ( j + 1) λ = T x j + 1m x j +1

Table P10.31b: Second Mode μ xj x j+1 λ( j+1)

R| 1 U| S| 0 V| T−0.5W

x j+1

2000

R| 0.0016U| S|−0.0001V| 2516.5 T−0.0016W

. R| 15813 U| . S| −01405 V| . T−16394 W

2000

R| 0.0031U| S| 0.0000V| 2521.1 T−0.0030W

. U R| 16264 S| 0.0171|V| . W T−15611

. U R| 16264 S| 0.0171|V| 2000 . W T−15611

R| 0.0031U| S| 0.0000V| 2521.7 T−0.0031W

. R| 16011 U S|−0.0106|V| . T −16126 W

. R| 15813 U| . S|−01405 V| . T −16394 W

1. First mode. Select μ = 300 and implement the inverse iteration equations to obtain Table P10.31a.

The second natural frequency and mode are

ω2 =

2521. 7 = 50. 2167

φ2 = 1. 6011 − 0. 0106 − 1. 6126 Table P10.31a: First Mode Iteration 1

R|1U| S|2V| T 3W

300

R|0.7993U| . S|13891 V| . 16096 T W

300

x j+1

R|0.0327U| S|0.0569V| T0.0659W

R|0.0212U| S|0.0367V| T0.0424W

λ( j+1) 337.869

337.856

x j+1

R|0.7993U| . V| S|13891 . W T16096

R|0.8024U| . S|13900 V| 16051 . T W

The fundamental natural frequency and mode are

ω1 =

3. Third mode. Select μ = 4000 and implement the inverse iteration equations to obtain Table P10.31c. Table P10.31c: Third Mode Iteration

R| 1 U| S|−1 V| T 1W

x j+1

λ( j+1)

x j+1

R| 0.0006U| S|−0.0015V| 4669.7 T 0.0020W

R| 0.5813 U| . V| S|−13977 . W T 17735

R| 0.5813 U| . V| 4000 S|−13977 . W T 17735

R| 0.0013U| S|−0.0019V| 4697.6 T 0.0021W

R| 0.9010U| . V| S|−13837 . W T 15084

R| 0.9010U| . V| 4000 S|−13837 15084 . W T

R| 0.0011U| S|−0.0020V| 4703.9 T 0.0023W

R| 0.7550U| . V| S|−13901 16503 . W T

R| 0.7550U| . V| 4000 S|−13901 . 16503 W T

R| 0.0012U| S|−0.0020V| T 0.0022W

R| 0.8248 U| . V S|−13897 15827 T . |W

2. Second mode. Select μ = 2000 and implement the inverse iteration equations to obtain Table P10.31b.

4000

337. 856 = 18. 3809

φ1 = 0. 8024 1. 3900 1. 6051

4705.3

The third natural frequency and mode are

ω3 =

4705. 3 = 68. 5952

φ3 = 0. 8248 − 1. 3897 1. 5827

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the35 publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

4. Compare with exact results.

ω1 = 18. 4883

φ1 = 0. 8024 1. 3898 1. 6048

ω 2 = 50. 5111 φ2 = 1. 6011 0 − 1. 6011 ω 3 = 68. 9994

φ3 = 0. 8248 − 1. 4286 1. 6496

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the36 publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 10.32

3. Third mode.

Implement the iteration procedure of Eqs. (10.14.3) to (10.14.5) with the starting shift computed from Eq. (10.14.4) using the starting vector. The results are summarized below.

Table P10.32c: Third Mode Iteration

1. First mode. Table P10.32a: First Mode Iteration

R|1U| S|2V| T 3W

μ 398.18

R|−0.8078U| . V| 337.89 S| −13913 . W T −15973

x j+1

( j+1)

x j+1

R|−0.0208U| S|−0.0358V| 337.89 T−0.0410W

R|−0.8078U| . V| S| −13913 . W T −15973

R|22.2645U| S|38.5633V| 337.86 T44.5291W

R|0.8025U| . S|13900 V| . T16050 W

R| 1 U| S|−1 V| T 1W

x j +1

λ( j +1)

x j+1

4539.2

R| 0.0036U| S|−0.0065V| 4704.6 T 0.0076W

R| 0.7645U| . V| S|−13936 . W T 16355

R| 0.7645U| . V| 4704.6 S|−13936 16355 . W T

R| 0.7275U| . V| 4705.7 S|−12600 14549 . W T

R| 0.8025U| . V| S|−13900 16050 . W T

The third natural frequency and mode are

ω3 =

4705. 7 = 68. 5981

φ3 = 0. 8025 − 1. 3900 1. 6050

The fundamental natural frequency and mode are

ω1 =

337. 86 = 18. 3809

φ1 = 0. 8025 1. 3900 1. 6050

2. Second mode. Table P10.32b: Second Mode Iteration 1

R| 1 U| 2063.3 R| S| S| 0.5V| T T−0.5W

. R| 14524 U| . S|−01647 V| 2478.1 18587 − . T W

x j +1

λ( j +1)

x j +1

−0.0022

U| V 2478.1 −0.0021 |W

. R| 14524 U| . S|−01647 V| . T −18587 W

R| 0.0364U| S| 0.0001V| 2521.8 T−0.0363W

. U| R| 16070 0 . 0048 V| S| . W T−16010

0.0017

The second natural frequency and mode are

ω2 =

2521. 8 = 50. 2175

φ2 = 1. 6070 0. 0048 − 1. 6010

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the37 publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

CHAPTER 11 Problem 11.1 1. Set up mass and stiffness matrices.

LM100 OP MM 100 PP 50Q N L 16 − 7 0OP 168 M k = − 7 10 − 3P 9 M NM 0 − 3 3QP 1 m = 386

2. Determine a0 and a1 from Eq. (11.4.9).

LM 1 12.01OP Ra U 0.05 MM 121.01 38.90PP STa VW = 2 RST0.05UVW Q N 38.90 0 1

a0 = 0. 9177

a1 = 1. 964 × 10 −3

3. Evaluate the damping matrix. c = a0 m + a1k

0O LM 0.824 − 0.257 . PP 0.604 − 0110 = M MN(sym) 0.229PQ

4. Compute ζ2 from Eq. (11.4.8)

ζ2 =

a0 1 a + 1 ω 2 = 0. 0430 2 ω2 2

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 1 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 11.2 1. Caughey series for a 3-DOF system: c = a0m + a1k + a2 k m −1 k

(a)

(b) Determine a0 , a1 and a2 from Eq. (11.4.14).

LM 1 12.01 (12.01) OP MM 121.01 25.47 (25.47) PP R|aa U| = 2 R|00..0505U| PP S|Ta V|W S|T0.05V|W MM 251.47 MN 38.90 38.90 (38.90) PQ 3

3 3

a0 = 0. 7400 a1 = 3. 3137 × 10 −3

a2 = − 8.1417 × 10 −7

LM 0.848 − 0.234 − 0.023OP 0.628 − 0133 . P MM 0.252PQ N(sym)

Problem 11.3

LM 2ζ ω OP φ φ MN ∑ M PQ m N

c = m

n = 1

T n

(a)

1. Determine the individual terms in Eq. (a). 2 (0.05) (12.01) m φ 1 φ 1T m 1 0.0328 0.0655 0.0491 = 01310 0.0983 . 0.0737 ( sym)

c1 =

LM MM N

OP PP Q

2 (0.05) (38.90) mφ 3φ 3T m 1 0.6500 − 0.4643 0.0929 = 0.3316 − 0.0663 0.0133 ( sym)

c3 =

LM MM N

OP PP Q

2. Determine c. c = c1 + c 3

. O LM 0.683 − 0.399 0142 = M 0.463 0.032PP MN(sym) 0.087PQ

3. Check damping in the second mode. C2 = φT2 c φ2 = 0

Therefore the second mode is undamped.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 3 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 11.4 1. Determine individual terms in Eq. (a) of Problem 11.3. The first and third terms c1 and c3 are already computed in Problem 11.3 and c2 =

2 (0.05) (25.47) m φ 2 φ 2T m 1 01651 . 01651 . − 01650 .

LM OP = M 01652 . − 01650 . PP MN (sym) 01650 . Q 2. Determine c.

LM 0.848 − 0.234 − 0.023OP c = ∑ c = M 0.628 − 0133 . P MN(sym) 0.252PQ 3

n = 1

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 4 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

CHAPTER 12 Problem 12.1

1. Set up modal equations. Mn q&&n + Kn qn = Pno sin ωt ; Pno = po

Part a The equations of motion are

LMm OP RS UV + LM 2k − k OP RSu UV = RS 0 UV sin ωt (a) N m 2Q T W N− k k Q Tu W T p W u&&1 u&&2

(i) Direct Solution The steady-state solution is assumed as

RSu UV = RSu UV sin ωt Tu W Tu W 1

(b)

2. Solve modal equations. q1 (t ) =

po 1 C1 sin ωt ; C1 = 0.586k 1 − (ω ω 1 ) 2

(d)

q 2 (t ) =

po 1 C2 sin ωt ; C2 = 3.414 k 1 − (ω ω 2 ) 2

(e)

3. Determine modal responses. u1 ( t ) = φ1q1 ( t )

LM2k − mω OP RSu UV = RS 0 UV −k MN − k k − mω 2PQ Tu W T p W OP RS 0 UV RSu UV = LM2k − mω −k Tu W MN − k k − mω 2PQ T p W RS k UV p = (m 2) (ω − ω ) (ω − ω ) T2 k − mω W 1o

−1

2 1

u1 (t ) =

po (1207 . C1 − 0.207C2 ) sin ωt k

u2 (t ) =

po (1707 . C1 + 0.293C2 ) sin ωt k

2 2

where, from Problem 10.6, ω1 = 0. 765 k m

(f)

and

Part b

By algebraic manipulation it can be shown that these results are equivalent to those obtained by solving the coupled equations. Part c: Frequency Response Curves

ω 2 = 1. 848 k m , or u1o =

u2 ( t ) = φ2 q2 ( t )

4. Combine modal responses.

Substituting Eq. (b) into Eq. (a) gives 2

(c)

po k

1 − ( ω ω1 )2 1 − ( ω ω 2 )2

u2 o =

2 po ( 2 k − mω 2 )

u 1o

m 2ω12ω 22 1 − ( ω ω1 )2 1 − ( ω ω 2 )2

po k

(ii) Modal Analysis

1 ω1

ω2 ω1

ω ω1

1 ω1

ω2 ω1

ω ω1

ω1

-1 -2

From Problem 10.6, the following data are available:

ω1 = 0. 765 k m

ω 2 = 1. 848 k m

φ2 = − 0. 707 1

φ1 = 0. 707 1

M2 = m

K1 = 0. 586 k

K2 = 3. 414 k

P1 ( t ) = po sin ωt

P2 ( t ) = po sin ωt

From Eqs. (12.3.4) and (10.4.7), Mn , Kn and Pn ( t ) are obtained: M1 = m

-3

u 2o po k

0 -1

ω1

-2 -3

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 1 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 12.2

f 21 (t ) = ω 12

Part a: Story shears from displacements. V2 n ( t ) = k u2 n ( t ) − u1n ( t ) = k φ2 n − φ1n qn ( t )

From Problem 12.1,

ω1 = 0. 765 k m

ω 2 = 1. 848 k m

φ2 = − 0. 707 1

po C1 sin ωt k

q2 (t ) = 0.293

po C2 sin ωt k

f 12 (t ) = ω 22 m ( − 0.707) q2 (t ) = − 0.707 po C2 sin ωt f 22 (t ) = ω 22

q1 (t ) = 1707 .

For second mode ( n = 2 ),

V1n ( t ) = ku1n ( t ) = k φ1n qn ( t )

φ1 = 0. 707 1

FG mIJ (1) q (t ) = 0.5 p C sin ωt H 2K

FG m IJ (1) q (t ) = 0.5 p C sin ωt H 2K 2

Static analysis of the structure shown in the figure gives the story shears: V1n ( t ) = f1n ( t ) + f2 n ( t )

V2 n ( t ) = f2 n ( t )

Substituting for f jn ( t ) and combining responses, we get the story shears:

modal

V1 (t ) = V11 (t ) + V12 (t ) = (1207 . C1 − 0.207C2 ) po sin ωt

First mode responses:

V11 (t ) = 1207 . poC1 sin ωt

V2 (t ) = V21 (t ) + V22 (t )

= 0.5 (C1 + C2 ) po sin ωt

V21 (t ) = 0.5 poC1 sin ωt

These are identical to those obtained in part (a) directly from displacements.

Second mode responses: V12 (t ) = − 0.207 poC2 sin ωt V22 (t ) = 0.5 poC2 sin ωt

Total responses: V1 ( t ) = V11 ( t ) + V12 ( t ) = (1. 207C1 − 0. 207C2 ) po sin ωt V2 (t ) = V21 (t ) + V22 (t ) = 0.5 (C1 + C2 ) po sin ωt

Part b: Story shears from equivalent static forces. f 2n (t)

f 1n (t)

The equivalent static forces are f jn ( t ) = ω 2n m j φ jnqn ( t ) For first mode ( n = 1 ), f 11 (t ) = ω 12 m (0.707) q1 (t ) = 0.707 po C1 sin ωt

Problem 12.3 1. Set up modal equations. Mn q&&n + Cn q&n + Kn qn = Pno sin ωt

(a)

where Mn , Kn and Pno are available (Problem 12.1) and Cn is known in terms of ζ n . 2. Solve modal equations. qn (t ) =

Pno Cn sin ωt + D n cos ωt Kn

(b)

where 1 − (ω ω n ) 2

Cn =

Dn =

1 − (ω ω n ) 2 + 2ζ n ω ω n − 2ζ n ω ω n

1 − (ω ω n ) 2

+ 2ζ n ω ω n

(c)

Substituting for Pno and Kn into Eq. (b) gives po C1 sin ωt + D 1 cos ωt 0.586k po = 1707 . C1 sin ωt + D 1 cos ωt k

q1 (t ) =

po C2 sin ωt + D 2 cos ωt 3.414 k p = 0.293 o C2 sin ωt + D 2 cos ωt k

q2 (t ) =

3. Combine modal responses. u1 (t ) = φ 11q1 (t ) + φ 12 q2 (t ) p = o (1207 . C1 − 0.207C2 ) sin ωt k + (1207 . D 1 − 0.207D 2 ) cos ωt u2 (t ) = φ 21q1 (t ) + φ 22 q2 (t ) p = o (1707 . C1 + 0.293C2 ) sin ωt k + (1707 . D 1 + 0.293D 2 ) cos ωt

4. Determine displacement amplitudes. u1o =

po k

(1207 . C1 − 0.207C2 ) 2 + (1207 . D 1 − 0.207D 2 ) 2

u2 o =

po k

(1707 . C1 + 0.293C2 ) 2 + (1707 . D 1 + 0.293D 2 ) 2

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 3 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 12.4

4. Combine modal responses.

1. Set up equations of motion.

LMm 0 OP RSu&& UV + LM 2k − k OP RSu UV = RS p δ (t )UV (a) N 0 m 2Q Tu&& W N− k k Q Tu W T 0 W 1

u( t ) =

n = 1

2. Set up modal equations.

∑ φnqn (t )

u1 (t ) = φ 11q1 (t ) + φ 12 q 2 (t ) =

FG H

po sin ω 1t sin ω 2 t + ω1 ω2 2m

(f)

From Problem 12.1,

ω1 = 0. 765 k m

ω 2 = 1. 848 k m

φ2 = − 0. 707 1

φ1 = 0. 707 1

u2 (t ) = φ 21q1 (t ) + φ 22 q2 (t ) =

K1 = 0. 586 k

K2 = 3. 414 k

1. Determine initial velocities.

Ignoring the stiffness elements and using the impulse-momentum relationship of Eq. (4.1.3) for each mass gives

The generalized modal forces are P1 ( t ) = φ1T p( t ) = 0. 707 poδ ( t ) P2 ( t ) = φT2 p( t ) = − 0. 707 poδ ( t )

u&1 ( 0 ) =

Substituting these in Eq. (12.3.3) gives the modal equations: M1q&&1 + K1q1 = 0. 707 poδ ( t )

(b.1)

M2 q&&1 + K2 q1 = − 0. 707 poδ ( t )

(b.2)

po m

u&2 ( 0 ) = 0

q&n ( 0 ) =

φTn mu& ( 0 ) Mn 0.707 1 m

For an SDF system subjected to an impulse force, the governing equation is (c)

and its solution is (d)

Adapting this result, the solutions of Eqs. (b.1) and (b.2) are − 0. 707 po 0. 707 po sin ω 2t sin ω1t ; q2 ( t ) = M1ω1 M2ω 2

Substituting for Mn gives − 0. 707 po 0. 707 po sin ω 2t sin ω1t ; q2 ( t ) = mω1 mω 2 (e)

(h)

2. Determine initial velocities in modal coordinates.

3. Solve modal equations.

q1 ( t ) =

(g) Alternative method

q1 ( t ) =

M2 = m

po u( t ) = sin ω nt mω n

FG sin ω t − sin ω t IJ ω K H ω

0.707 po m

M1 = m

mu&& + ku = poδ ( t )

IJ K

LM1 OP RS p mUV N 1 2Q T 0 W = 0.707 p o

q&1 (0) =

Similarly, q&2 ( 0 ) = − 0. 707

po m

3. Determine free vibration response. qn ( t ) = q1 ( t ) =

q&n ( 0 )

ωn

sin ω nt

− 0. 707 po 0. 707 po sin ω 2t sin ω1t ; q2 ( t ) = mω1 mω 2 (i)

Note that these results are the same as Eq. (e).

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 4 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 12.5

4. Combine modal responses.

1. Set up equations of motion.

u ( t ) = φ1q1 ( t ) + φ2 q2 ( t )

LMm 0 OP RSu&& UV + LM 2k − k OP RSu UV = RS p (t )UV N 0 m 2Q Tu&& W N− k k Q Tu W T 0 W 1

(a)

po 0.853 po 0147 . 1 − cos ω 1t + 1 − cos ω 2 t k k p = o 1 − 0.853 cos ω 1t − 0147 . cos ω 2 t k

u1 (t ) =

2. Set up modal equations.

From Problem 12.1,

ω1 = 0. 765 k m

ω 2 = 1. 848 k m

φ2 = − 0. 707 1

φ1 = 0. 707 1 M1 = m

M2 = m

K1 = 0. 586 k

K2 = 3. 414 k

po 1207 . 0.207 po 1 − cos ω 1t − 1 − cos ω 2 t k k p = o 1 − 1207 . cos ω 1t + 0.207 cos ω 2 t k 5. Determine second story drift. u2 (t ) =

Δ 2 (t ) = u2 (t ) − u1 (t ) 0.354 po = − cos ω 1t + cos ω 2 t k

The generalized modal forces are P1 ( t ) = φ1T p( t ) = 0. 707 po

P2 ( t ) = φT2 p( t ) = − 0. 707 po

Substituting these in Eq. (12.3.3) gives the modal equations: M1q&&1 + K1q1 = 0. 707 po

(b.1)

M2 q&&2 + K2 q2 = − 0. 707 po

(b.2)

3. Solve modal equations. For an SDF system subjected to a suddenly applied force, the governing equation is mu&& + ku = po

(c)

and its solution is u(t ) =

po (1 − cos ω nt ) k

(d)

Adapting this result, the solutions of Eqs. (b.1) and (b.2) are q1 ( t ) =

0. 707 po (1 − cos ω1t ) K1

1. 207 po (1 − cos ω1t ) k

q2 ( t ) = =

− 0. 707 po K2 − 0. 207 po k

(1 − cos ω 2t ) (1 − cos ω 2t )

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 5 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 12.6

Adapting this result, the solution of Eq. (b) is

1. Set up equations of motion.

LMm 0 OP RSu&& UV + LM 2k − k OP RSu UV = RS p (t )UV N 0 m 2Q Tu&& W N− k k Q Tu W T 0 W 1

R| 0.707 p F1 − cos 2π t I 0 ≤ t ≤ t GH J K T K | q (t ) = S || 0.707 p FG 2 sin π t IJ sin LM2π FG t − t IJ OP t ≥ t T K H T K MN H T 2T K PQ 1

(a)

where p( t ) =

RS p t ≤ t ; T0 t ≥ t o

td =

(e)

T1 2

Substituting K1 = 0. 586 k and td = T1 2 gives

R| 1207 T 2π t I p F . 1 − cos 0 ≤ t ≤ G J T K 2 | kH q (t ) = S L O F I || 2.414 p sin M2π G t − 1 J P t ≥ T k 2 MN H T 4 K PQ T

2. Set up modal equations.

From Problem 12.1,

ω1 = 0. 765 k m

ω 2 = 1. 848 k m

φ2 = − 0. 707 1

φ1 = 0. 707 1

(f)

M1 = m

M2 = m

Similarly the solution of Eq. (c) is

K1 = 0. 586 k

K2 = 3. 414 k

R| − 0.207 p F1 − cos 2π t I 0 ≤ t ≤ T J G 2 k H T K | q (t ) = S O L F I || 0.252 p sin M2π G t − T J P t ≥ T 2 k MN H T 4T K PQ T o

The generalized modal forces are

R 0.707 p 0 ≤ t ≤ t P (t ) = φ p(t ) = S t ≥ t T0 R − 0.707 p 0 ≤ t ≤ t P (t ) = φ p(t ) = S t ≥ t T0 o

T 1

T 2

(g)

4. Combine modal responses.

u ( t ) = φ1q1 ( t ) + φ2 q2 ( t )

(h)

Substituting these in Eq. (12.3.3) gives the modal equations:

RS 0.707 p 0 ≤ t ≤ t t ≥ t T0 R − 0.707 p 0 ≤ t ≤ t M q&& + K q = S t ≥ t T0 o

M1q&&1 + K1q1 =

2 2

(b)

2 2

(c)

3. Solve modal equations. For an undamped SDF system subjected to a rectangular pulse force of amplitude po and duration td , the response is

R| p F1 − cos 2π t I 0 ≤ t ≤ t J T K | k GH u( t ) = S || p FG 2 sin π t IJ sin LM2π FG t − t IJ OP t ≥ t T k H T K MN H T 2T K PQ o

(d)

where Tn =

2π ωn

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 6 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 12.7

det k − ω 2 m = m1m2 m3 (ω 12 − ω 2 ) (ω 22 − ω 2 ) (ω 23 − ω 2 )

I F1 − ω I F1 − ω I ω ω ω JK GH ω JK GH ω JK F ω I F1 − ω I F1 − ω I = k G1 − H ω JK GH ω JK GH ω JK

Part a

From Problem 9.8, the mass and stiffness matrices are

LM1 OP m = mM 1 MN 0.5PPQ

LM 2 − 1 0OP k = k M−1 2 −1 MN 0 − 1 1PPQ

R| 0.5 U| φ = S0.866V |T 1 |W

k m

; ω 22 = 2

k m

; ω 32 = ( 2 +

k m

R|− 1U| R| 0.5 U| φ = S 0V φ = S− 0.866V |T 1|W |T 1 |W

2 2

2 3

2 1

2 2

2 3

2 2 2

Substituting Eqs. (d) and (e) in Eq. (c) gives 1o

LM1 OP R|u&& U| LM 2 − 1 0OP R|u U| R| 0 U| mM 1 Su&& V + k − 1 2 − 1 Su V = S 0 V sin ωt MN 0.5PPQ |Tu&& |W MMN 0 − 1 1PPQ |Tu |W |T p |W 2

2 2

(e)

2 1

2 2

2 3

The equations of motion are

2 3

U| R| 1 R| 0 U| | | adj k − ω m S 0 V = k p S 2 e1 − ω ω j V ||4 1 − ω ω − 1|| |T p |W j W Te

(i) Direct Solution

2 2

U| R| 1 R|u U| p S|u V| = k F ω I F 1 ω I F ω I |S| 2 e1 − ω ω j |V| Tu W GH1 − ω JK GH1 − ω JK GH1 − ω JK |T4 e1 − ω ω j − 1|W

2 1

Similarly,

From Problem 10.11, the natural frequencies ω n and modes φn are given by 3)

(d)

where k = 24 EI h3 .

ω12 = ( 2 −

F GH

1 3 ω2 m 1 − 2 2 ω1

(a)

2 2

2 2 2

(f) Substituting Eq. (f) in Eq. (b) gives u j ( t ) . (ii) Modal Analysis Using Eq. (12.3.4), the generalized modal mass, stiffness, and force are

The steady-state response is assumed as

R|u (t ) U| R|u U| S|u (t )V| = S|u V| sin ωt T u ( t ) W Tu W 1

(b)

Substituting Eq. (b) in Eq. (a) gives

R|u U| S|u V| = k − ω m Tu W 1o

2o 3o

−1

M1 = 1. 5 m

K1 = 1. 5 ( 2 −

M2 = 1. 5 m M3 = 1. 5 m

K2 = 3 k K3 = 1. 5 ( 2 +

3) k

P1 ( t ) = po sin ωt

3) k

P2 ( t ) = po sin ωt P3 ( t ) = po sin ωt (g)

The modal equations and their steady-state solution are

R| 0 U| S| 0 V| Tp W

Mn q&&n + Kn qn = Pno sin ωt

R| 0 U| 1 adj k − ω m S 0 V = det k − ω m |T p |W

qn (t ) =

Pno Cn sin ωt Kn

(h)

where

(c)

Cn =

1 1 − (ω ω n )2

(i)

The determinant can be expressed in terms of the natural frequencies:

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 7 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Substituting Eq. (i) in Eq. (h) gives q1 (t ) =

po 2 (2 + 3 k

q2 (t ) =

po 1 C2 sin ωt k 3

q3 (t ) =

po 2 (2 − 3 k

Part c

C1 sin ωt

(j)

C3 sin ωt

Substituting Eq. (j) and φn in Eq. (12.3.2) gives the floor displacements:

R|u (t ) U| p R| (2 + 3) C − C + (2 − 3) C U| S|u (t )V| = 3k S| (3 + 2 3) C + 0 + (3 − 2 3) C V| sin ωt Tu ( t ) W T(4 + 2 3) C + C + (4 − 2 3) C W 1

1st story 15

2nd story 3rd story

uo 10 po k

(k)

Part b

The direct solution Eq. (f) can be written in terms of Cn : u1o = C1C2C3 po k u2 o = 2C1C 3 po k u3o 4 = C 1C 2C 3 − 1 po k C22

F GH

The amplitude of displacements is given by Eq. (l) where Cn are defined by Eq. (i). These amplitudes are plotted against ω ω 1 in Fig. P12.7.

ω ω1 Fig. P12.7

(l)

I JK

This result is equivalent to Eq. (k) from modal analysis. This equivalence can be proven for u2 , for example, as follows: The direct solution gives u2 o ω 12 = 2C1C3 = 2 po k ω 12 − ω 2

ω 23

j eω − ω j 2 3

From modal analysis, u2 o 1 (3 + 2 3 ) C1 + (3 − 2 3 ) C3 = po k 3

LM OP MN PQ L O 1 6ω ω − 3ω ω ( 3 + 2) − ω ( 3 − 2) P = M PPQ 3M (ω − ω ) (ω − ω ) MN =

ω2 ω2 1 (3 + 2 3 ) 2 1 2 + (3 − 2 3 ) 2 3 2 3 ω1 − ω ω3 − ω 2 1

2 3

2 1

= 2

ω 12

ω 12 − ω

ω 23

2 3

ω 23 − ω2

2 3

j (m)

Hence, both methods give the same results. The same procedure can be applied to show such equivalence for displacements u1 and u3 . © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 8 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 12.8

Part b: Story shears from equivalent static forces.

Part a: Story shears from displacements.

Substituting qn ( t ) from Problem 12.7 in Eq. (12.6.2) gives the equivalent static forces:

V1n ( t ) = k1u1n ( t ) = k1φ1n qn ( t ) V2 n ( t ) = k2 u2 n ( t ) − u1n ( t )

(a)

= k2 ( φ2 n − φ1n ) qn ( t ) = k 3 (φ 3n − φ 2 n ) qn (t )

Substituting k1 = k2 = k3 = k , and the values of φ jn , and qn from Eq. (j) of Problem 12.7 in Eq. (a) gives the story shears for each mode:

Mode 1 po 2 + 3 C1 sin ωt 3 p V21 (t ) = o 1 + 3 C1 sin ωt 3 po (1) C1 sin ωt V31 (t ) = 3

e e

j j

(b.1)

po ( − 1) C2 sin ωt 3 p V22 (t ) = o (1) C2 sin ωt 3 po (1) C2 sin ωt V32 (t ) = 3

po C1 sin ωt 3 3 po f 21 (t ) = C1 sin ωt 3 p f 31 (t ) = o C1 sin ωt 3 f 11 (t ) =

(b.2)

po 2 − 3 C3 sin ωt 3 p V23 (t ) = o 1 − 3 C3 sin ωt 3 po (1) C3 sin ωt V33 (t ) = 3

j j

(b.3)

Mode 2 2 po C2 sin ωt 3

po C3 sin ωt 3 3 po C3 sin ωt f 23 (t ) = − 3 p f 33 (t ) = o C3 sin ωt 3 f 13 (t ) =

U| V| W

(g)

(c)

Substituting Eqs. (b) in Eq. (c) gives

f 3n(t)

po 2 + 3 C1 − C2 + 2 − 3 C3 sin ωt 3 p V2 (t ) = o 1 + 3 C1 + C2 + 1 − 3 C3 sin ωt 3 po V3 (t ) = C1 + C2 + C3 sin ωt 3 (d)

e e

(f.3)

Static analysis of the system gives the story shears due to the nth mode: V2 n (t ) = f 2 n (t ) + f 3n (t ) V1n (t ) = f 1n (t ) + f 2 n (t ) + f 3n (t )

Vj ( t ) = Vj1 ( t ) + Vj 2 ( t ) + Vj 3 ( t )

j j

(f.2)

V3n (t ) = f 3n (t )

The total story shears are

e e

(f.1)

Mode 3

V1 (t ) =

m j φ jn sin ωt (e)

f 22 (t ) = 0 p f 32 (t ) = o C2 sin ωt 3

V12 (t ) =

e e

Mode 1

f 12 (t ) = −

Mode 2

V13 (t ) =

ω 2n po Cn

Substituting ω 2n , Kn , m j and φ jn (from Problems 10.11 and 12.7) in Eq. (e) gives f jn for each mode:

V3n (t ) = k 3 u3n (t ) − u2 n (t )

V11 (t ) =

f jn (t ) = ω 2n m j φ jn qn (t ) =

j j

V3n(t) V2n(t)

f 2n(t)

f 1n(t)

V1n(t)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 9 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Substituting Eq. (f) in Eq. (g) gives Vjn ( t ) for each mode:

Mode 1 po (1) C1 sin ωt 3 p V21 (t ) = o 1 + 3 C1 sin ωt 3 po V11 (t ) = 2 + 3 C1 sin ωt 3 V31 (t ) =

e e

j j

(h.1)

Mode 2 po (1) C2 sin ωt 3 p V22 (t ) = o (1)C2 sin ωt 3 po V12 (t ) = ( − 1) C2 sin ωt 3 V32 (t ) =

(h.2)

Mode 3 po (1) C3 sin ωt 3 p V23 (t ) = o 1 − 3 C3 sin ωt 3 po V13 (t ) = 2 − 3 C3 sin ωt 3 V33 (t ) =

e e

j j

(h.3)

These modal contributions to the story shears are the same as in Eq. (b) determined from displacements.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 10 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 12.9

where

Using the results from Problem 10.11 and substituting m = 100 kips g and k = 24 EI h3 = 326. 32 kips in . gives ω n in rads sec and φn :

po = 1. 242 × 10 −3

ω1 = 18. 38

ω 2 = 50. 22

ω 3 = 68. 60

R| 0.5 U| φ = S0.866V |T 1 |W

R|− 1U| φ = S 0V |T 1|W

R| 0.5 U| φ = S− 0.866V (b) |T 1 |W

0 R| U| 0 p(t ) = S V| sin ω t |T1242 . × 10 ω W

(c)

Substituting qn ( t ) and φn in Eq. (12.5.2) gives the lateral displacements. In particular, the roof displacement, u3 ( t ) is

2o 3o

−3

(d) The modal equations are Mn q&&n + Cn q&n + Kn qn = Pno sin ωt

(e)

For ζn = 0. 05 and the values of ω n , φ jn , Kn , Mn and Pno defined in Eqs. (a), (b) and (d), the solution is po 2 ω Cn sin ωt + D n cos ωt Kn

(h)

FG C + C + C IJ + FG D + D + D IJ HK K K K H K K K K 2

M = 0.3882 K = 13116 . P = 1242 . × 10 ω U | M = 0.3882 K = 978.97 P = 1242 . × 10 ω V M = 0.3882 K = 1826.80 P = 1242 . × 10 ω |W 1

The amplitude of the displacement at the roof is

The generalized modal masses and stiffnesses have been computed in Eq. (g) of Problem 12.7. Substituting m =100 386.4 = 0.2588 kip − sec 2 in. and k = 326. 32 kips in. gives

qn (t ) =

+ 2ζ nω ω n

(g)

LMF C + C + C I sin ωt MNGH K K K JK F D + D + D IJ cos ωt OP + G H K K K K PQ

Thus

1 − (ω ω n ) 2

= ω 2 po

= 1242 × 10 −3 ω 2 sin ω t .

u3 (t ) = φ 31q1 (t ) + φ 32 q2 (t ) + φ 33q3 (t )

−3

Dn =

+ 2ζ nω ω n 1 − (ω ω n ) −2ζ nω ω n

FG m IJ e ω sin ω t H2K F 0.02 IJ (12) ω sin ω t = 2G H 386.4 K

p3 (t ) = 2

2 2

(a)

The excitation force due to the shaker is

U| | V| || W

1 − (ω ω n ) 2

Cn =

(f)

u3o = ω 2 po

(i) The amplitude of the roof acceleration is u&&3o = ω 2u3o

(j)

The frequency response curves for roof displacement and roof acceleration are shown in Figs. P12.9a-b. 0.04

u 3o , 0.03 in. 0.02 0.01 0 0

Exciting frequency, Hz Fig. P12.9a 0.5 0.4

u&&3o , 0.3 g 0.2 0.1 0 0

Exciting frequency, Hz Fig. P12.9b © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 11 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 12.10 Using the results from Problem 10.11 and substituting m = 100 kips g and k = 24 EI h3 = 326. 32 kips in . gives ω n in rads sec and φn :

ω1 = 18. 38

ω 2 = 50. 22

ω 3 = 68. 60

R| 0.5 U| φ = S0.866V |T 1 |W

R|− 1U| φ = S 0V |T 1|W

R| 0.5 U| φ = S− 0.866V |T 1 |W

(a) The generalized modal masses and stiffnesses have been computed in Eq. (g) of Problem 12.7. Substituting m = 400 386. 4 = 1. 0352 kip − sec2 in . and k = 326. 32 kips in. gives M1 = 0. 3882 M2 = 0. 3882 M3 = 0. 3882

K1 = 131.16 K2 = 978. 97 K3 = 1826. 80

(b)

The generalized modal forces due to the impulsive force at the second floor are P1 (t ) = φ 1T p(t ) = 0.5 0.886 1 = 17.32 δ (t )

R| 0 U| S|20 δ (t )V| (c.1) T 0 W

P2 ( t ) = φT2 p( t ) = 0

(c.2)

P3 ( t ) = φ3T p( t ) = − 17. 32 δ ( t )

(c.3)

The modal equations and their solutions are

Mn q&&n + Kn qn = Pno δ ( t ) qn ( t ) =

(d)

Pno sin ω nt Mnω n

(e)

Substituting Pno , Mn and ω n in Eq. (e) gives q1 (t ) = 2.427 sin ω 1t q2 (t ) = 0 q3 (t ) = − 0.650 sin ω 3t

U| V| W

(f)

Substituting φn and qn ( t ) in Eq. (12.5.2) gives the lateral displacements: . U| R|1214 R|− 0.325U| u(t ) = S2.102 V sin ω t + S 0.563V sin ω t (g) |T2.427|W |T− 0.650|W 1

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 12 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 12.11

Part b

The second-story drift is

Part a

Δ 2 ( t ) = u2 ( t ) − u1 ( t )

From Problem 12.10,

U| R| 0.5 U| R|− 1U| R| 0.5 U| | φ = S0.866V φ = S 0V φ = S− 0.866V | |T 1 |W |T 1|W |T 1 |W V| (a) M = 0.3882 M = 0.3882 M = 0.3882 | | K = 13116 . K = 978.97 K = 1826.80 |W

ω 1 = 13.38

ω 2 = 50.22

ω 3 = 68.60

= − 0. 2791 cos ω1t + 0. 2043 cos ω 2t + 0. 0748 cos ω 3t

The generalized modal forces are P1 (t ) =

φ 1T p(t )

R|200U| = 0.5 0.886 1 S 0 V = 100 |T 0 |W

P2 ( t ) = φT2 p( t ) = − 200

(b)

P3 ( t ) = φ3T p( t ) = 100

The modal equations and their solutions are Mn q&&n + Kn qn = Pno qn ( t ) =

(c)

Pno (1 − cos ω nt ) Kn

(d)

Substituting Pno , Mn and ω n in Eq. (d) gives q1 (t ) = 0.7624 (1 − cos ω 1t ) q2 (t ) = − 0.2043 (1 − cos ω 2 t ) q3 (t ) = 0.0547 (1 − cos ω 3t )

U| V| W

(e)

Substituting φn and qn ( t ) in Eq. (12.5.2) gives the lateral displacements: u(t ) =

R| 0.5 U| R− 1U S|0.866V| 0.7624 (1 − cos ω t ) + |S| 0|V| (− 0.2043) (1 − cos ω t ) T1 W T 1W R| 0.5 U| + S− 0.866V 0.0547 (1 − cos ω t ) |T 1 |W 1

or,

R|u (t ) U| R|0.6129U| R|0.3812U| S|u (t )V| = S|0.6129V| − S|0.6603V| cos ω t Tu (t ) W T0.6129W T0.7624W R| 0.2043 U| R| 0.0274U| − S 0 V cos ω t − S− 0.0474V cos ω t |T− 0.2043|W |T 0.0547|W 1

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 13 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

FG t − 0.250IJ U| HT K | F I| t q (t ) = − 0.3729 sin 2π G − 0.687J V t ≥ t HT K| | F I t q (t ) = − 0.0895 sin 2π G − 0.933J | HT K |W

Problem 12.12

q1 (t ) = 3.0498 sin 2π

From Problem 12.10,

ω1 = 18. 38

ω 2 = 50. 22

ω 3 = 68. 60

T1 = 0. 3418

T2 = 0.1244

T3 = 0. 0916

R| 0.5 U| φ = S0.866V |T 1 |W

R|− 1U| φ = S 0V |T 1|W

R| 0.5 U| φ = S− 0.866V |T 1 |W

M1 = 0. 3882

M2 = 0. 3882

M3 = 0. 3882

K1 = 131.16

K2 = 978. 97

K3 = 1826. 80

(f)

The story displacements are 3

u ( t ) = ∑ φn qn ( t )

(g)

n =1

where φn are known and qn ( t ) is given by Eq. (e) if 0 ≤ t ≤ td and by Eq. (f) if t ≥ td .

(a) The rectangular pulse force at the third floor is shown in the accompanying figure: p, kips 200 t, sec t d = T 1 /2 = 0.1709

The generalized modal forces are

R| P (t ) U| LM0.5 0.866 1OP R| 0 U| R|200U| S| P (t )V| = M− 1 0 1P S| 0 V| = S|200V| (b) T P (t ) W MN0.5 − 0.866 1PQ T200W T200W 1

The modal equations are M n q&&n + Kn qn =

RS P T0

0 ≤ t ≤ td t ≥ td

(c)

The solution of Eq. (c) is

R| P F1 − cos 2π t I 0≤ t ≤ t G J K H T K | q (t ) = S || P FG 2 sin π t IJ sin 2π FG t − 1 t IJ t ≥ t HT 2 T K TK H T K no

(d)

Substituting Pno , Kn and Tn in Eq. (d) gives

FG tI U J | TK | H F t I | q (t ) = 0.2043 G 1 − cos 2π J V 0 ≤ t ≤ t T K | H FG1 − cos 2π t IJ || q (t ) = 01095 . T K |W H q1 (t ) = 15249 . 1 − cos 2π

(e)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 14 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

For the given data, ω n in rads sec are

Problem 12.13 u1

ω1 = 10

mL/3

L/3

3. Set up modal equations. Mn q&&n + Kn qn = Pn ( t )

where Mn = φTn mφn

p1 (t) = poδ ( t)

E = 30, 000 ksi

ω 2 = 38. 73

I = 100 in.4

L = 150 in.

m L = 0. 864 kip − sec2 in .

po = 10 kips

Kn = ω 2n Mn

Pn ( t ) = φTn p( t )

For the given data, M1 = 0. 576

M2 = 0. 576

P1 ( t ) = P1oδ ( t ) = poδ ( t )

P2 ( t ) = P2 oδ ( t ) = poδ ( t )

1. Determine stiffness and mass matrices. 4. Solve modal equations.

From Problem 9.2,

LM 8 − 7OP k = 8Q 5 L N− 7

LM OP N Q

mL 1 0 m = 3 0 1

162 EI 3

For the given data,

L 230.4 − 2016. OP k = M N− 2016. 230.4Q

L0.288 0 OP m = M N 0 0.288Q

qn ( t ) = q1 ( t ) = q2 ( t ) =

R1U φ = SV T1W 1

sin ω nt

0. 576 (10 ) 10

sin 10 t = 1. 736 sin 10 t

0. 576 ( 38. 73)

sin 38. 73t = 0. 448 sin 38. 73t

u( t ) = ∑ φn qn ( t )

From Problem 10.2, EI m L4

Mnω n 10

5. Combine modal responses.

2. Determine natural frequencies and modes.

ω1 = 9. 859

Pno

n =1

ω 2 = 38.184

R 1U φ = S V T− 1W 2

EI m L4

RSu (t ) UV = RS1UV 1736 R 1U . sin 10t + S V 0.448 sin 38.73t Tu (t )W T1W T− 1W 1

The modal responses and total response are plotted in Figs. P12.13a-b.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 15 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Mode 1

u 11 0

u 21 0

-3

Mode 2

u 12 0

u22 0

-3

Total

Mode 2

Total

u1 0 in.

u2 0 in.

-3

Mode 1

-3

0.5

Time, sec

Fig. P12.13a

1.5

0.5

1.5

Time, sec

Fig. P12.13b

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 16 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 12.14

qn ( t ) = u1

q1 ( t ) =

mL/3

p1o Kn

(1 − cos ω nt )

100 (1 − cos 10 t ) 0. 576 (10 )2

= 1. 736 (1 − cos 10 t )

mL/3

100 (1 − cos 38. 73t ) 0. 576 ( 38. 73)2 = 0.116 (1 − cos 38. 73t )

q2 ( t ) = p1 (t ) =

{ 100 kips

t <0 t ≥0

4. Combine modal responses.

RSu (t ) UV = RS1UV 1736 . (1 − cos 10t ) Tu (t )W T1W R 1U . (1 − cos 38.73t ) + S V 0116 T− 1W

1. Determine natural frequencies and modes and Mn .

See Problem 12.13. 2. Set up modal equations. Mn q&&n + Kn qn = Pn ( t )

The modal responses and total response are plotted in Figs. P12.14a-b.

where Pn (t ) = φ Tn p(t ) = p1 (t ) =

t < 0 RS0 T p = 100 t ≥ 0 1o

3. Solve modal equations.

Mode 1

u 11 0

u 21 0

-4

Mode 2

u 12 0

u22 0

-4

Total

Mode 2

Total

u1 0 in.

u2 0 in.

-4

Mode 1

-4

0.5

1.5

0.5

Time, sec

Fig. P12.14a

Fig. P12.14b

1.5

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 17 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

R| p F t − sin ω t I 0 ≤ t ≤ t | k GH t ω t JK u( t ) = S || p RS1 − 1 sin ω t − sin ω (t − t ) UV t ≥ t W Tk T ω t

Problem 12.15

u1 mL/3

n r

(c)

mL/3

In Eq. (c), replacing po by 100 kips, and k by Kn = ω 2n Mn , and substituting for ω n and tr = 0.15 sec gives the solutions for qn ( t ) :

p (t) = p (t) 1

.748 R|641736 sin 10t I t 100 . sec − || 0.576 (10 ) FGH 015 J 0 ≤ t ≤ 015 15 . . K q (t ) = S RS1 − 1 sin 10t − sin 10 (t − 015 U ||1736 . . )V 15 . T W |T t ≥ 015 . sec

p , kips

100

(d)

t , sec

0.15

.744 R|6440116 8 sin 38.73t I 100 t − . sec || 0.576 (38.73 ) FGH 015 JK 0 ≤ t ≤ 015 . . 581 q (t ) = S RS1 − 1 sin 38.73t − sin 38.73 (t − 015 U ||0116 . . )V . 581 T W |T . sec t ≥ 015 2

1. Determine natural frequencies and modes and Mn .

See Problem 12.13. 2. Set up modal equations. Mn q&&n + Kn qn = Pn ( t )

(a)

where Pn (t ) = φ nT p(t ) = p1 (t ) =

R|100 t 0 ≤ t ≤ t S| t T100 t ≥ t r

(e) 4. Combine modal responses. The total displacements are 2

(b)

u( t ) = ∑ φn qn ( t )

3. Solve modal equations. The solution of the equation of motion for an SDF system mu&& + ku = p( t )

where p ( t ) is a step force po with rise time tr is

(f)

n =1

In Eq. (f),

φ1 =

RS1UV T1W

φ2 =

RS 1UV T− 1W

and qn ( t ) is given by Eq. (d) if 0 ≤ t ≤ 0.15 sec and by Eq. (e) if t ≥ 0.15 sec . The modal responses and total response are plotted in Figs. P12.15a-b.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 18 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Mode 1

u 11 0

u 21 0

-4

Mode 2

u 12 0

Mode 2

u22 0

-1

Total

u1 0 in.

u2 0 in.

-4

0 0.15

Mode 1

-4

0.5

1.5

0 0.15

0.5

Time, sec

Fig. P12.15a

Fig. P12.15b

1.5

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 19 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Substituting in Eq. (d) for t ≤ td gives

Problem 12.16 p, kips u1

q1 ( t ) =

100 t, sec

0.30

p2 (t) = p(t)

100 (1 − cos 10 t ) = 1. 736 (1 − cos 10 t ) 57. 6 0 ≤ t ≤ 0. 3 (e)

q1 ( td ) = 3. 455

q&1 ( td ) = 2. 450

Substituting q1 ( td ) and q&1 ( td ) for u ( td ) and u& ( td ) , respectively, in Eq. (d) gives

1. Determine natural frequencies and modes and Mn .

2. 45 sin 10 ( t − 0. 3) 10 t ≥ 0. 3 (f)

q1 ( t ) = 3. 455 cos 10 ( t − 0. 3) +

See Problem 12.13. 2. Set up modal equations. Mn q&&n + Kn qn = Pn ( t )

For mode 2,

(a)

po = − 100 ; ω 2 = 38. 73 ;

where

R100 0 ≤ t ≤ t U| = p (t ) = S P (t ) = t ≥ t |V T0 R− 100 0 ≤ t ≤ t | P (t ) = φ p(t ) = − p (t ) = S |W T0 t ≥ t 1

φ 1T p(t )

T 2

Substituting in Eq. (d) for t ≤ td gives

q2 ( t ) =

(b) The equation of motion for an SDF system is

(1 − cos 38. 73t ) = − 0.116 (1 − cos 38. 73t )

0 ≤ t ≤ 0. 3

(g)

Substituting q2 ( td ) and q&2 ( td ) for u ( td ) and u& ( td ) , respectively, in Eq. (d) gives

(c)

q2 ( t ) = − 0. 0483 cos 38. 73 ( t − 0. 3) 3. 648 sin 38. 73 ( t − 0. 3) + 38. 73

where p ( t ) is the rectangular pulse force shown: p po

t ≥ 0. 3

(h)

4. Combine modal responses.

RSu (t ) UV = RS1UV q (t ) + RS 1UV q (t ) 0 ≤ t ≤ 0.3 Tu (t )W T1W T− 1W (i) A A 1

t , sec

The solution of Eq. (c) is

R| p (1 − cosω t ) 0 ≤ t ≤ t k | u(t ) = Su(t ) cos ω (t − t ) + u& (t ) sin ω (t − t ) ω || t ≥ t T o

− 100 864. 0

q2 ( td ) = − 0. 0483 q&2 ( td ) = 3. 648

3. Solve modal equations. mu&& + ku = p( t )

k = K 2 = ω 2 M2 = ( 38. 73) ( 0. 576 ) = 864. 0

td = 0. 3 sec

(d)

Eq. (e)

Eq. (g)

RSu (t ) UV = RS1UV q (t ) + RS 1UV q (t ) t ≥ 0.3 Tu (t )W T1W T− 1W A A 1

Eq. (f)

(j)

Eq. (h)

The modal responses and total response are plotted in Figs. P12.16a-b.

For mode 1, po = 100 ; ω1 = 10 ;

td = 0. 3 sec

k = K1 = ω12 M1 = 10 2 ( 0. 576 ) = 57. 6 © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 20 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Mode 1

u 11 0

u 21 0

-4

Mode 2

u 12 0

u22 0

-1

Total

Mode 2

Total

u1 0 in.

u2 0 in.

-4 0

Mode 1

-4

0.3

0.5

1.5

0.3

0.5

Time, sec

Fig. P12.16a

Fig. P12.16b

1.5

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 21 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

0 R| V U| R| 0 U| R| U| − V u 0 3274 q 15274 q . . |S |V = k |S |V = k |S |V M 0 0 || || || || || || − M u 0 00948 q 0 00376 q . . T W T W T W R| − 6.928 q + 227.9 q U| | 6.928 q − 227.9 q |V = S ||− 666.03 q + 7019 q || T 250.37 q + 66555. q W

Problem 12.17

u 2( t )

u 1( t )

u 4( t )

1 d

u 3( t ) a

L /2

Substituting q1 = 1. 217 in. and q2 = 0. 0159 in. (from Example 12.6) gives

1. Determine displacements and rotations. From Example 12.6,

R| V U| R| − 4.8U| |S V |V = |S 4.8|V || M || ||− 699|| T M W T 411W d

u1 ( t ) = q1 ( t ) + q2 ( t )

(a)

u2 ( t ) = 0. 3274 q1 ( t ) − 1. 5274 q2 ( t )

(b)

Substituting Eq. (a) in Eq. (b) gives u3 ( t ) = 0. 01207 q1 ( t ) + 0. 06509 q2 ( t )

(c)

u4 ( t ) = 0. 00948 q1 ( t ) − 0. 00376 q2 ( t )

q U . R| V U| R|u U| R| 0.3274 q − 15274 || q + q |S V |V = k |Su |V = k |S || M || ||u || || 0.00948 q − 0.0376 q V|| TM W Tu W T0.01207 q + 0.06509 q W R| − 4.21 q − 110.8 q U| R|− 6.88U| | 4.21 q + 110.8 q |V = |S 6.88|V = S ||− 2513. q − 6650.7 q || || − 411|| T −0.97 q + 4.8 q W T 0W

In Eq. (d), substituting L 2 for L in ke , ua = u2 , ub = u1 , θ a = u4 , and θb = u3 , and using Eqs. (b) and (c) gives

2. Write force-displacement relations. ua

4. Determine forces in element 2.

RSu (t ) UV = LM0.0214 − 0.0286OP RSu (t ) UV Tu (t )W N0.0071 0.0071Q Tu (t )W 3

From Eq. (d) of Example 9.8 with L = 120 in. ,

θa

θb

a b

R| V U| LM 12 − 12 6 L 6 LOPR|u U| 12 − 6 L − 6 L P | u | |S V |V = EI M S V M 4L 2 L P |θ | || M || L M P ) 4 L Q |Tθ |W N(sym TM W 1 44444244444 3

L /2

2 cb

5. Draw shear and bending moment diagrams.

a b

V, kips 4.80

(d)

699

M, kip-in.

6.88 411

3. Determine forces in element 1. In Eq. (d), substituting L 2 for L in ke , ua = 0 , ub = u2 , θ a = 0 , and θb = u4 , and using Eqs. (b) and (c) gives

These results are identical to those obtained in Example 12.6.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 22 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 12.18

Static analysis gives the bending moment

1. Determine natural frequencies and modes.

Ma ( t ) = M f ( t ) = 0

L = 3 L Md ( t ) = Me ( t ) = RB ( t ) = 3

From Problem 12.13,

ω 1 = 10

Mb ( t ) = Mc ( t ) = RA ( t )

ω 2 = 38.73

R1U R 1U φ = SV φ = S V T1W T− 1W 1

Static analysis gives the shears:

2. Determine equivalent static forces for each mode.

ω 2nmφn qn ( t )

fn ( t ) = 1

n = 1, 2

(a)

RS f (t ) UV = (38.73) 0.288 LM1 OP RS 1UV q (t ) = RS 432UV q (t ) N 1Q T− 1W T f (t ) W T− 432W 1

where f1 ( t ) and f2 ( t ) are known from Eq. (c).

(b)

5. Determine modal coordinates at t = 0.1 sec.

3. Combine modal forces.

From Problem 12.14,

f1 ( t ) = 28. 8 q1 ( t ) + 432 q2 ( t ) f2 ( t ) = 28. 8 q1 ( t ) − 432 q2 ( t )

(c)

4. Determine internal forces.

A RA (t)

L/3

2 1 f 1 (t ) + f 2 (t ) 3 3 1 2 V f (t ) = Ve (t ) = f 1 (t ) + f 2 (t ) 3 3 (f) 1 1 Vc (t ) = R A (t ) − f 1 (t ) = − f 1 (t ) + f 2 (t ) 3 3 1 1 Vd (t ) = f 2 (t ) − RB (t ) = f 1 (t ) + f 2 (t ) 3 3 Va (t ) = Vb (t ) =

RS f (t ) UV = (10) 0.288 LM1 OP RS1UV q (t ) = RS28.8UV q (t ) N 1Q T1W T f (t ) W T28.8W 2

L 2 f1 ( t ) + f2 ( t ) (e) 9 L f1 ( t ) + 2 f2 ( t ) 9

q1 ( t ) = 1. 736 (1 − cos 10 t ) q2 ( t ) = 0.116 (1 − cos 38. 73t )

(g)

At t = 0.1 sec

f1(t)

f2(t)

L/3

q1 = 0. 798 in. f

L/3

q2 = − 0. 202 in.

(h)

6. Determine internal forces at t = 0.1 sec. Substituting Eq. (h) in Eq. (a) and Eq. (b) gives numerical values for the equivalent static forces shown in Figs. P12.18a-b where the shearing forces and bending moments due to each mode are plotted. The combined values are shown in Fig. P12.18c.

B RB (t)

By statics, the reactions are 2 1 f1 ( t ) + f2 ( t ) 3 3 1 2 RB ( t ) = f1 ( t ) + f2 ( t ) 3 3

RA ( t ) =

(d)

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 23 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

22.98

110.24

87.26

f S ( t), kips

f S ( t), kips 22.98

22.98

87.26

29.10

64.28

52.08

6.12

52.08 29.10

22.98

29.10 6.12

V , kips

22.98 58.16

58.16

1455 306

M , kip-in.

M , kip-in. 1149

1149

1455 2604

Fig. P12.18a

Fig. P12.18b

Fig. P12.18c

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 24 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 12.19 u1 ( t )

u2 ( t )

p1 ( t ) = 100 sin25 t

1. Determine natural frequencies and modes. See Problem 12.13. 2. Set up modal equations. Mn q&&n + Kn qn = Pno sin ωt

where Pno = φTn po = 100

3. Solve modal equations. qn ( t ) =

Pno Cn sin ωt Kn

Cn =

100

q1 (t ) =

1 1 − ( ω ω n )2

(10) 0.576 1 − (25 10) 2 = − 0.331 sin 25t

q2 ( t ) =

100

sin 25t

( 38. 73) 0. 576 1 − ( 25 38. 73)2

sin 25t

= 0.198 sin 25t

4. Determine displacements and accelerations. . sin 25t U RSq (t ) + q (t )UV = RS− 0133 Tq (t ) − q (t )W T− 0.529 sin 25t VW Rq&& (t ) + q&& (t )UV = (25) u && = Φ && u q = S Tq&& (t ) − q&& (t )W . sin 25t U R − 83125 = S V . sin 25t W − 330 63 T

u = Φq =

5. Determine accelerations. uo =

amplitudes

displacements

. U . U RS0133 R83125 && = S in. u in. sec V T0.529W T330.63VW o

and

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 25 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 12.20 1. Determine equivalent static forces. From Problem 12.18, f1 ( t ) = 28. 8 q1 ( t ) + 432 q2 ( t ) f2 ( t ) = 28. 8 q1 ( t ) − 432 q2 ( t )

(a)

2. Determine modal coordinate amplitudes. From Problem 12.19, q1 ( t ) = − 0. 331 sin 25t q2 ( t ) = 0.198 sin 25t

(b)

To determine the response amplitude we specialize Eq. (b) when sin 25t = 1; thus q1o = − 0. 331

q2 o = 0.198

Substituting qno for qn ( t ) in Eq. (a) gives f1o = 28. 8 ( − 0. 331) + 432 ( 0.198) = 76. 00 kips f2 o = 28. 8 ( − 0. 331) − 432 ( 0.198) = − 95. 07 kips

3. Determine bending moments. From Eq. (e) of Problem 12.18, Mbo = Mdo =

150 9 150 9

2 ( 76. 00 ) − 95. 07 = 948. 8 kip − in. ( 76. 00 ) − 2 ( 95. 07) = − 1902 kip − in.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 26 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 12.21

u = Φq = u1 ( t )

u2 ( t )

RSq (t ) + q (t )UV Tq (t ) − q (t )W 1

. sin 25t − 0.073 cos 25t U RS− 0139 V − 0 517 . sin 25t + 0.011 cos 25t W T

5. Determine displacement amplitudes.

p1 ( t ) = 100 sin25 t

. U . ) + ( − 0.073) U| RSu UV = R|S (− 0139 R0157 in. = S V 0 . u T W |T (− 0.517) + (0.011) |W T 517VW 2

1. Determine natural frequencies and modes.

6. Determine acceleration amplitudes.

See Problem 12.13.

&& o = ω 2 uo = ( 25)2 uo u

2. Set up modal equations. Mn q&&n + Cn q&n + Kn qn = Pno sin ωt

(a)

where

. U . U RSu&& UV = (25) RS0157 R98125 = S V V in. sec && . W Tu W T0.517W T32312 1o

Pno = φTn po = 100

(b)

3. Solve modal equations. qn ( t ) =

Pno Cn sin ωt + Dn cos ωt Kn

(c)

where Cn = Dn =

1 − (ω ω n ) 2 [1 − (ω ω n ) 2 ]2 + [2ζ n ω ω n ]2 −2ζ n ω ω n [1 − (ω ω n ) 2 ]2 + [2ζ n ω ω n ]2

(d) (e)

For the first mode,

ω 25 = = 2. 5 ω1 10

ζ1 = 0.10

P1o 100 = = 1. 736 K1 (10 )2 0. 576

Substituting these numerical values in Eqs. (c)-(e) gives q1 ( t ) = − 0. 328 sin 25t − 0. 031 cos 25t

For the second mode:

ω 25 = = 0. 645 ω2 38. 73

ζ2 = 0.10

P2 o 100 = = 0.116 K2 ( 38. 73)2 0. 576

Substituting these numerical values in Eqs. (c)-(e) gives q2 ( t ) = 0.189 sin 25t − 0. 042 cos 25t

4. Determine displacements. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 27 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

16137 11411 . . . LM 11411 OP Φ = M16137 0 − 16137 . . PP MN 11411 11411 . − 16137 . . Q

Problem 12.22 u1

u2 θ1

u3 θ2

L/4

The modes are normalized such that

θ3

Mn = φTn m φn = 1

L/4

1.1411

E = 30, 000 ksi

I = 100 in.

L = 150 in.

L′ =

1.6137

1.1411

ω 1 = 10.615

L = 37. 5 in. 4

1.6137

ω 2 = 42.164

m = 0.192 kip − sec2 in . -1.6137 1.1411

1. Define DOFs. ut = u1 u2

u0 = θ1 θ2

θ3

ω 3 = 89.523

-1.6137

2. Determine mass matrix in terms of ut .

6. Determine modal expansion of s.

. LMm OP LM0192 OP = m 0192 . MM P MM P m PQ 0192 . PQ N N 1

m =

n =1

s = ∑ sn = ∑ Γn m φn

where

3. Determine stiffness matrix.

LM7 L ′ MM EI k = M L′ M MM MN 3

OP PP PP PP PQ

2 L′ 2 8 L′ 2

0 3 L ′ −6 L ′ 0 2 2 L′ 6L′ 0 −6 L ′ 2 7 L′ 0 6 L ′ −3 L ′ 15 −12 0 ( sym) 24 −12 15

4. Determine lateral stiffness matrix.

LMk Nk

OP LMu OP = LM 0 OP k Q Nu Q Nf Q

k 0t

k$ tt = ktt −

−1 u0 = − k00 k 0 t ut

Γn =

φTn s Mn

5. Determine natural frequencies and modes.

ω n = 10. 615, 42.164, 89. 523 rads sec

= φTn s

The s n values for force distributions s a and s b are computed and summarized in Table P12.22a-b. Table P12.22a sa

R|1U| S|0V| T0W

0. 25 0. 3536 0. 25

0. 5 0 − 0. 5

0. 25 − 0. 3536 0. 25

−1 kt 0 k00 k0t

L630.86 − 603.43 246.86OP $k = EI M 877.71 − 603.43P L M MN (sym) 630.86PQ

1.1411

Table P12.22b sb

R| 2U| S| 0V| T− 1W

0. 25 0. 3536 0. 25

1. 5 0 − 1. 5

0. 25 − 0. 3536 0. 25

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 28 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

The modal expansions of s a and s b are shown graphically:

0.25 0.3536 0.25

0.5

M1stn due to s a

M1stn due to sb

1 2 3

– 0.1067L – 0.0625L – 0.0183L – 0.1875L = – 28.125 k-in.

– 0.1067L – 0.1875L – 0.0183L – 0.3125L = – 46.875 k-in.

sb 3

s2 +

0.25

0.5 0.25

0.25

0.3536

1.5 0.25

∑ M1stn

1.5

s2 s3

Mode, n 2

Table P12.22c

n = 1

Next we determine M1st : 2 1

0.3536

All three modes contribute similarly to force distribution s a , but the second mode has a predominant component in sb . Therefore the second mode will probably contribute more than other modes in response of the system to p( t ) = sb p( t ) ; however, this observation is tentative until we examine the dynamic response factor for p ( t ) . 7. Determine modal static responses Mnst . First determine M1 for following load cases:

0.25

0.75

M1 = −

3L 16

1.25

= − 0.1875 L

M1 = −

5L 16

= − 0. 3125 L

We have demonstrated that ∑ M1stn = M1st . n =1

8. Determine modal contribution factors, their cumulative values and error eJ . Table P12.22d

1 M 1 = – 3 L/16 0.75

0.25

M 1 = – L /8 0.5

0.5 M 1 = – L /16

0.25

Mode, n or no. of modes, J

Force distribution s a J

M1n

∑ M1n

n =1

0.5691

0.4309

0.3333

0.9024

0.0976

1.0

0.75 L

The values of Mnst due to forces sn is determined by linear combination of the above three load cases; the results are in Table P12.22c.

Table P12.22e Mode, n or no. of modes, J

Force distribution sb M1n

∑ M1n

n =1

0.3414

0.6586

0.6000

0.9414

0.0586

1.0

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 29 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

In case of s a , the modal contribution factor is largest for the first mode and progressively decreases for the second and third modes (Table P12.22d). In case of sb , the modal contribution factor is largest for the second mode (Table P12.22e). As a corollary, for the s b case e1 is larger but e2 is smaller, both relative to the s a case. 9. Determine response to half-cycle sine pulse. The peak modal response equation (12.11.2) is specialized for r ≡ M1 to obtain

(M1n )o = po M1st M1n Rdn Table P12.22f Spectral

Force Distribution

values

bM g

Rdn

Tn td

1.73

0.5691

0.985

0.3414

0.591

0.252

1.14

0.3333

0.380

0.6000

0.684

0.119

1.06

0.0976

0.103

0.0586

0.062

Mode,

M1n

1n o po M1st

M1n

1n o

po M1st

10. Comments. (a) For force distribution s a , the modal responses decrease for higher modes, as suggested by the modal contribution factors M1n ; the decrease is more rapid, however, because Rdn also decreases with mode number. For force distribution sb , the largest contribution is from the second mode, again as suggested by M1n . (b) No. The peak value of the total response can not be determined from the peak modal responses because the modal peaks occur at different time instants. Furthermore, the SRSS or CQC modal combination rules do not apply to pulse excitations.

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 30 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 12.23 From Problem 9.13 the mass and stiffness matrices are: ⎤ ⎡5 ⎥ ⎢ m = m⎢ 1 ⎥ ⎥ ⎢ ⎢⎣ 1⎥⎦

and

⎡ 28 6 − 6⎤ 3EI ⎢ k= 6 7 3 ⎥⎥ 3 ⎢ 10 L ⎢ ⎣− 6 3 7 ⎥⎦

From Problem 10.23 the natural frequencies and modes of the system are: ω 1 = 0.5259

EI EI , ω 2 = 1.6135 , 3 mL mL3

ω 3 = 1.7321

EI mL3

⎧ 1 ⎫ ⎪⎪ ⎪⎪ φ 1 = ⎨− 1.9492⎬ , ⎪ ⎪ ⎪⎩ 1.9492 ⎪⎭

⎧ 1 ⎫ ⎪⎪ ⎪⎪ φ 2 = ⎨ 1.2826 ⎬ , ⎪ ⎪ ⎪⎩− 1.2826⎪⎭

⎧0⎫ ⎪⎪ ⎪⎪ φ 3 = ⎨1⎬ ⎪ ⎪ ⎪⎩1⎪⎭

⎧ 1⎫ ⎪⎪ ⎪⎪ ⎨− 1⎬ ⎪ ⎪ ⎪⎩ 1⎪⎭

1.944

−0.944

− 0.758

− 0.242

0.758

0.242

⎧ 1⎫ ⎪⎪ ⎪⎪ ⎨ 1⎬ ⎪ ⎪ ⎪⎩− 1⎪⎭

−1150 .

2.150

0.448

0.552

− 0.448

− 0.552

⎧ 1⎫ ⎪⎪ ⎪⎪ ⎨ 2⎬ ⎪ ⎪ ⎪⎩2⎪⎭

0.397

0.603

− 0.155

0.155

− 0.155

The model expansions graphically

The generalized masses M n = φnT m φn

s a , s b and s c are shown

for the three modes are M 1 = 12.60,

M 2 = 8.29,

M 3 = 2.0

Part a

1 1

1. Determine modal expansion of s: s=

∑ s = ∑ Γ mφ n

n =1

0 758

1 944

n =1

0 758

where φT s Γn = n Mn

0 242

The s n values for force distributions s a , s b and s c are computed and summarized:

0 155

1 150

0 448

0 944

0 448

0 552

0 155

0 552 2 150

0 397

0 155

0 155 0 603

Note that because s a and s b are anti-symmetric they do not have any component in the symmetric third mode, whereas, the main component of s c is in the third mode. Part b

2a. Determine the modal static responses of M 1stn . First determine M1 for the following load cases: © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 31 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

The values of M 1stn due to forces s n is determined by linear combination of the above three load cases; the results are summarized below:

3b. Determine M 2stn : 1

M 2st 1

M 1stn due to

M 1stn due to s b M 1stn due to s c

3.460 L

-2.047 L

0.706 L

-0.460 L

1.047 L

0.294 L

M 2st

1 1

M 2st = L

M 2st = 2 L

We have demonstrated that: M 2st =

st 1n

−L

∑M

st 2n

4. Determine modal contribution factors, their cumulative values and error eJ .

n =1

2b. Determine M 1st : sa

Part c

∑M

M 2st = − L

1 1

Mode, n

M 2st 1

These results for M 1 are as follows: 1

1 1

M 1st

M 1st = L + L + L

M 1st

M 1st = L − L − L M 1st = L − 2 L + 2 L

= 3L

= −L

Mode, n or

Force distribution s a M1n

number of modes, J

∑M

n =1

0.883

0.117

2c. We have demonstrated that:

0.117

1.0

∑M

1.0

M = st 1

st 1n

st . 3a. Determine the modal static responses of M 2n

First determine M2 for the following load cases: M2 = 0

M2 = L

M2 = 0

Force distribution s b M1n

∑M

n =1

The values of M 2stn due to forces s n is determined by linear combination of the above three load cases; the results are summarized below:

Mode, n M 2stn due to s a M 2stn due to s b M 2stn due to s c

Mode, n ornumber of modes, J 1

0.662

0.338

1.0

Mode, n or number of modes, J

Force distribution s c M1n

∑M

-0.758 L

0.448 L

-0.155 L

-0.242 L

0.552 L

0.155 L

0.706

0.294

1.0

−L

1.0

∑M

st 1n

n =1

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 32 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

These results for M 2 are as follows: Mode, n or number of modes, J

( M1n ) o = p0 M1stn Rdn

Force distribution s a M2n

∑M

0.758

0.242

1.0

Force distribution s b M2n

∑M n =1

Mode,

0.448

0.552

1.0

Tn td

Rdn

M1n

( M 1n ) o p0 M1st

M1n

( M 1n ) o p0 M1st

M1n

( M 1n ) o p0 M1st

5.0

0.691

0.883

0.610

0.662

0.457

0.706

0.502

1.63

2.0

0.117

0.234

0.338

0.676

0.294

0.587

1.51

2.0

( M 2 n ) o = p0 M 2stn Rdn Force

Force

Distribution,

Spectral Values

Force distribution s c

∑M

Mode,

Values

Similarly for

0.448

M2n

Force Distribution,

Mode, n or number of modes, J

Force Distribution,

n =1

Mode, n or number of modes, J

Force Distribution,

Spectral

Tn td

Rdn

M1n

( M 1n ) o p0 M1st

M1n

( M 1n ) o p0 M1st

M1n

( M 1n ) o p0 M1st

n =1

0. 067

0.933

0. 067

0.134

0.866

1.0

5.0

0.691

0.758

0.524

0.448

0.310

0.067

0.046

1.63

2.0

0.242

0.484

0.552

1.104

0.067

0.046

1.51

2.0

0.866

1.732

Observe that:

Part e

1. For force distribution s a and s b , the third modal contribution factor is zero for responses M 1 and M 2 . It will be zero for every response quantity, because the third mode, being symmetric, does not contribute to the antisymmetric force distributions s a and s b (see Part a).

6. Comments.

2. For force distribution s c , the third modal contribution factor for response M 2 , is dominant relative to the other two modes; however, it is zero for response M 1 , because a symmetric mode will not contribute to M 1 .

The peak modal responses determined in Part (d) demonstrate that the relative response contributions of the three modes depend on the numerical values of the modal contribution factors ( M 1n and M 2n ) and on the dynamic response factors ( R dn ). As an early example for the force distribution s b , the second mode contribution to M 1 is larger than the first mode contribution, although M 11 >

M 12 , because R d 2 is almost three times R d1 .

Part d

Part f

5. Determine response to rectangular pulse.

7. No. The peak value of the total modal response can not be determined from the peak modal responses because the modal peaks occur at different time instants. Furthermore, the SRSS or CQC modal combination rules do not apply to pulse excitations.

The peak modal response Eq. (12.11.2) is specialized for r ≡ M 1 to obtain:

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 33 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problem 12.24 Part a: Classical modal analysis

The nth-mode response is given by Eq. (12.10.1), specialized for r ≡ M1 : M1n ( t ) = M1st M1n ω 2n Dn ( t )

(a)

where Dn ( t ) , the solution of Eq. (12.9.2), is Dn (t ) =

LMsin F π t I − F T I sin F 2π t I OP MN GH t JK GH 2t JK GH T JK PQ n

ω n2 1 − (Tn 2t d ) 2

t ≤ td Dn ( t ) = Dn ( td ) cos ω n ( t − td ) +

D& n ( td )

ωn

(b)

sin ω n ( t − td )

t ≥ td

(c)

For force distribution sb , from Problem 12.22, M1st = − 0. 3125 L M1n = 0. 3414, 0. 6000, and 0. 0586

In Eq. (a) we substitute these data, and numerical values for ω 2n Dn ( t ) from Eqs. (b) and (c) in which td = T1 =

2π = 0. 5919 sec (from Problem 12.22) 10. 615

The resulting bending moment due to each mode and the total response is plotted in Fig. P12.24a. Part b: Static correction method

Specializing Eq. (12.12.6) for M1 with Nd = 1 :

{

M 1 (t ) = M1st p(t ) + M11 ω 12 D1 (t ) − p(t )

}

(d) where M1st = − 0. 3125 L , M11 = 0. 3414 , D1 ( t ) was computed in Part a, and p( t ) =

R| p sin F π t I 0 ≤ t ≤ t S| GH t JK t ≥ t T0 o

(e)

Eq. (d) is plotted in Fig. P12.24b, where it is compared with the exact solution by classical modal analysis. The static correction method provides reasonable results; the error is primarily because the second mode has significant dynamic effects, as indicated by Rd2 = 1.14 being significantly larger than one. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 34 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

0.4

0.2

M1n / poL and

n= 3 0

M1 / poL

n= 1

-0.2

-0.4

0.2

n=2

Total

0.4

0.6

0.8

1.2

1.4

1.6

1.8

Time, sec Fig. P12.24a 0.4

Classical modal analysis Static correction method

0.2

M1 / poL -0.2

-0.4

0.2

0.4

0.6

0.8

1.2

1.4

1.6

Time, sec

Fig. P12.24b

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 35 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.