INFLUENCE LINE DAIGRAMS

INFLUENCE LINES FOR STATICALLY INDETERMINATE BEAMS !

!

!

Comparison Between Indeterminate and Determinate Influence line for Statically Indeterminate Beams Qualitative Influence Lines for Frames

1

Comparison between Indeterminate and Determinate

Indeterminate

1 A

D

B

E

Determinate

C

RA

1 A

D

B

E

C

RA

2

Indeterminate D

1

Determinate

B

E

C

B

E

C

A

1

D

A

D A

B

E

1

RA

VD

1 A

A

D

B

E

C

1 D

B

E

C

RA

VD

1 C

ME

A

D

B

E

C

ME 3

Influence Lines for Reaction Redundant R1 applied

1

4

2

j

3

1 Compatibility equation: f1 j + f11 R1 = ∆1 = 0

= R1 = − f1 j (

∆´1 = f1j

1 R1 = (

fjj

f j1 f11

1 ) f11

)

+

f11 ×R1

1 fj1

4

Redundant R2 applied

1

4

2

j

3

1 Compatibility equation. ∆ '2 + f 22 R2 = ∆ 2 = 0

=

f 2 j + f 22 R2 = ∆ 2 = 0

1

+

f2j

f22

1

R2 = − f 2 j (

fjj

R2 = (

fj2

f j2 f 22

1 ) f 22

)

×R2 5

Influence Lines for Shear

1

4

2

j

3

1

VE = ( 1

1 ) f j4 f 44

1 f44 fj4 1

6

Influence Lines for Bending Moment

4

1

1

2

j

3

1

ME = ( 1

α44

1

α 44

) f j4

1

fj4

7

Using Equilibrium Condition for Shear and Bending Moment • Influence line of Reaction 4

1

2

j

3

f 41 f11

f11 =1 f11

R1

R1 = (

f j1 f11

)

f j1

1 f 22 =1 f 22

f 42 f 22

f11 f j2 f 22

R2

R2 = (

1

f j2 f 22

)

f 33 =1 f 33

f 43 f 33

f j3 f 33

R3 1

R3 = (

f j3 f 33

) 8

• Influence line of Shear

• Unit load to the left of x

1

1 4

1

2

j

M4

1

3

R1 R1

R2

4

V4

+ ↑ ΣFy = 0; R1 − 1 − V4 = 0

R3

V4 = R1 - 1 R1 1

• Unit load to the right of M4

1

V4 = R1

4 1 1 1

V4 = R1 - 1

R1 V4

4

V4

+ ↑ ΣFy = 0; R1 − V4 = 0

V4 = R1

9

• Influence line of Bending moment • Unit load to the left of

l

x 1 4

1

R1

2

l1

R2

j

l2

1

3

l R1

4

1 M4 V4

+ Σ M4 = 0: M4 - 1 (l-x) - l R1 = 0

R3

M4 = - l + x + l R1 R1

• Unit load to right of

4

1 1

1

4

M4 = - l + x + l R1 1 1 M4

l R1

M4 V4

+ Σ M4 = 0: M4 - l R1 = 0 M4 = l R1

M4 = lR1

10

Qualitative Influence Lines for Frames

1

I 1

Influence Line of VI

Maximum positive shear

Maximum negative shear

11

1

1 I

Influence Line of MI

Maximum positive moment

Maximum negative moment

12

Influence Line for MOF D

A

MA

Ay

15 m

Dy

G

15 m

Gy

1.0 Gy 1.0 Dy 1.0 Ay

13

D

A

15 m

H

G

15 m

MA

1

1 MH

14

G A

B

C

D

E

1.0 RA 1.0 RB

MB 1 1 MG

15

G A

B

C

D

1

E

VG

1

VF

1

VH 16

Example 1 Draw the influence line for - the vertical reaction at A and B - shear at C - bending moment at A and C EI is constant . Plot numerical values every 2 m.

A

C

2m

D

2m

B

2m

17

• Influence line of RB A

C

D

2m

2m

f AB f BB

f CB f BB

B

2m

f DB f BB

f BB =1 f BB

1

18

• Find fxB by conjugate beam A

6 kN•m

1

C

D

2m

2m

6 EI

18 EI

Real Beam

1

2m x

72 EI Conjugate Beam

V´x

18 EI

x2 2 EI

x EI

x3 72 18 x + − = M´x 6 EI EI EI

B

x 3

x

2x 3

72 EI

18 EI

19

x A

C

D

2m

2m

B

f xB

x3 72 18 x = M 'x = + − 6 EI EI EI

fxB

fxB / fBB

Point

x (m)

B

0

72 EI

1

D

2

37.33 EI

0.518

C

4

10.67 EI

0.148

A f BB 72 = =1 f BB 72

6

0

2m

72/EI 0

10.67/EI

37.33/EI

fBB 1

fxB

37.33 /72 = 0.518 10.67 /72 = 0.148 0

0

1 Influence line of RB 20

Influence line of RA A

C

D

2m

2m

1 kN f AA =1 f AA

f CA f AA

B

2m

f DA f AA

21

• Find fxA by conjugate beam 6 kN•m

1 kN A

C

B

D

2m

2m

Real Beam 1 kN

2m x

M 'A =

72 EI

Conjugate Beam B' y = 18 EI

6 EI

x 3

3

18 x x − = M´x EI 6 EI

2x 3

V´x x EI

18 EI

x x2 2 EI

B' y =

18 EI

22

x A

C

2m

1 kN fAA

72/EI

D

2m

61.33 /EI

B

f xA

2m

34.67 /EI

AA AA

1 kN =1

Point

x (m)

fxA

fxA / fAA

B

0

0

0

D

2

34.67 EI

0.482

C

4

61.33 EI

0.852

6

72 EI

1.0

fxA

72 /72 = 1.0 61.33 /72=0.852 f f

18 x x 3 = M 'x = − EI 6 EI

A

34.67 /72 = 0.482 Influence line of RA

23

Alternate Method: Use equilibrium conditions for the influence line of RA x A

MA

1 C

D

B + ↑ ΣFy = 0;

RA

2m

2m

2m

RB

RA + RB − 1 = 0 RB = 1 − RA

RB 0.148

1 .518 RB 1

RA = 1- RB 1.0 1 kN

0.852 0.482 RA 24

Using equilibrium conditions for the influence line of VC A

MA RA

• Unit load to the left of C

x

1 C

D

2m

2m

MC

B

2m

VC

RB

RB

+ ↑ ΣFy = 0; + RB + VC = 0

1 0.148

0.518 RB 1

VC = - RB • Unit load to the left of C 1

MC 0.852 1 -0.148

x

VC

0.482 VC

RB

+ ↑ ΣFy = 0; + VC − 1 + RB = 0

VC = 1 - RB

25

Using equilibrium conditions for the influence line of MA x

1 A

MA

C

D

B + ΣM A = 0;

RA

2m

2m

2m

RB

− M A − 1(6 − x) + 6 RB = 0 M A = −6 + x + 6 RB

1 0.148

0.518 RB 1

MA -1.112 1

-0.892 26

Using equilibrium conditions for the influence line of MC A

MA

• Unit load to the left of C

x

1 C

D

B

MC 4m

VC RA

2m

2m

2m

RB

ΣM C = 0;

+

− M C + 4 RB = 0

1

MC = 4RB

0.518

0.148

RB 1

• Unit load to the left of C x

1

C

0.592

RB

MC

1

VC 0.074 MC

4m

RB

+ ΣM C = 0; − M C − 1(4 − x) + 4 RB = 0 MC = -4 + x + 4RB

27

Example 2 Draw the influence line and plot numerical values every 2 m for - the vertical reaction at supports A, B and C - Shear at G and E - Bending moment at G and E EI is constant.

A

G

2@2=4 m

B

D

E

F

C

4@2 = 8 m

28

Influence line of RA A

G

2@2=4 m

B

D

E

F

C

4@2 = 8 m

f AA =1 f AA

1

f DA f AA

f EA f AA

f FA f AA

29

• Find fxA by conjugate beam A

B

1

D

E

2m

C

Real beam

0.5

1.5 4m

F

6m

4 /EI Conjugate beam 4/EI 0 10.67 EI

5.33 EI

4/EI

64/EI

0 18.67/EI

10.67 EI

f AA = M ' A =

64 EI

30

4/EI

x2

64/EI

x1 Conjugate beam

4m

18.67 EI

8m

5.33 EI

2

x1 4 EI

x1 2 EI

x1

3

x1 5.33 x1 − = 12 EI EI

M´ x1 V´ x1 2

64 EI

x2

x2 2 EI

x1 3

x2 EI

2 x2 3

x2 V´ x2 3

5.33 EI

3

M´ x2 18.67 EI

2 x1 3

x 64 18.67 = 2 + − 6 EI EI EI

31

x2 A

G

4m 64 EI

fAA

1

B

D

2m

E

F

C

f xA

6m

28 EI

Point

x (m)

fxA

fxA / fAA

C

0

0

0

−

10 EI

-0.1562

16 EI

-0.25

14 EI

-0.2188

F

2

fxA

1

f AA =1 f AA

f xA

x1

5.33x x3 = M ' x1 = − , for B to C EI 12 EI 3 18.67 x2 64 x2 = M 'x 2 = − + , for A to B EI EI 6 EI

− 14 EI

− 16 EI

− 10 EI

Influence line of RA 0.438 -0.219 -0.25 -0.156

E

4

−

D

6

−

B

8

0

G

10

28 EI

A

12

64 EI

0 0.4375 1

32

Using equilibrium conditions for the influence line of RB 1 A

G

x B

RA

D

E

F

RB 4m

+ ΣM C = 0;

C

− 8 RB + x − 12 RA = 0 RB =

RC

2m

6m

1 0.438 -0.219 -0.25 -0.156

1

1

0.59

1.078 0.875

RA

0.485 RB

1

x 12 − RA 8 8

Point

x (m)

RA

RB

C

0

0

0

F

2

-0.1562 0.485

E

4

-0.25

D

6

-0.2188 1.078

B

8

G

10

A

12

0 0.4375 1

0.875

1 0.5939 0 33

Using equilibrium conditions for the influence line of RC 1 A

G

RA

x B

D

E

F

C

RB 4m

+ ΣM B = 0;

4 RA − 1( x − 8) − 8 RC = 0

RC

2m

6m

1 0.438 RA

-0.219 -0.25 -0.156

1

0.141 0.375 -0.0312

RC = 0.5 RA −

1

0.672

RC 1

x +1 8

Point

x (m)

RA

RC

C

0

0

1

F

2

-0.1562

0.6719

E

4

-0.25

0.375

D

6

-0.2188

0.1406

B

8

G

10

A

12

0 0.4375 1

0 -0.0312 0 34

• Check ΣFy = 0 1 A

G

x B

RA

D

E

F

C

RB 4m

+ ↑ ΣFy = 0; RA + RB + RC = 1

RC

2m

6m

1 0.438 RA

-0.219 -0.25 -0.156

1 1

0.59

1.08

0.875

0.49

RB

1 0.141 0.375 -0.0312

1

0.672 1

Point

RA

RB

RC

ΣR

C

0

0

1

1

F

-0.1562 0.485

0.6719

1

E

-0.25

0.375

1

D

-0.2188 1.078

0.1406

1

0

1

B G A

RC

0

0.875

1

0.4375 0.5939 -0.0312 1

0

0

1 1 35

Using equilibrium conditions for the influence line of VG 1

• Unit load to the left of G

x A

G

B

D

E

F

x

C

1 MG

A 4m

2m

RA

6m

VG

+ ↑ ΣFy = 0; RA − 1 − VG = 0

1 VG = RA - 1

0.438 -0.219 -0.25 -0.156

1

RA • Unit load to the right of G

0.438 1 -0.562

MG

A VG -0.219 -0.25 -0.156

RA VG = RA

VG

36

Using equilibrium conditions for the influence line of VE 1 A

G

B

D

• Unit load to the left of E

x F

E

C

ME VE

4m

2m

RC

+ ↑ ΣFy = 0; RC + VE = 0

6m

VE = - RC 0.141 0.375

1

0.672 1

-0.0312

• Unit load to the right of E RC

1

x

ME 0.625 0.0312

1 -0.141

VE

0.328 VE

-0.375

RC

+ ↑ ΣFy = 0; VE − 1 + RC = 0

VE = 1 - RC 37

Using equilibrium conditions for the influence line of MG 1

• Unit load to the left of G

x A

G

B

D

E

F

A 4m

2m

1

x

C

RA

6m

2m

MG VG

ΣM G = 0;

+

M G + 1(2 − x) − 2 RA = 0

1

MG = -2 + x + 2RA

0.438 -0.219 -0.25 -0.156

1

RA

• Unit load to the right of G A

0.876

RA

1 MG -0.438 -0.5

-0.312

+

2m

MG VG

ΣM G = 0; M G − 2 RA = 0

MG = 2RA

38

Using equilibrium conditions for the influence line of ME 1 A

G

B

D

• Unit load to the left of E

x F

E

4m

C

ME VE

4m

2m

0.141

6m

0.375

0.564

1

ME = 4RC

1

1 1.5

-0.125

+ ΣME = 0;

0.672

-0.0312

RC

RC

• Unit load to the right of E x 1 ME VE

0.688

4m

RC

+ ΣM E = 0;

ME

− M E − 1(4 − x) + 4 RC = 0

ME = - 4 + x+ 4RC 39

Example 3 For the beam shown (a) Draw quantitative influence lines for the reaction at supports A and B, and bending moment at B. (b) Determine all the reactions at supports, and also draw its quantitative shear, bending moment diagrams, and qualitative deflected curve for - Only 10 kN downward at 6 m from A - Both 10 kN downward at 6 m from A and 20 kN downward at 4 m from A 20 kN

10 kN

2EI

3EI B

A 4m

C

2m

2m

40

Influence line of RA 3EI

2EI

B

A C

4m

fAA

fDA

2m

2m

fCA

fEA

fCA /fAA

fEA /f

1

fAA /fAA

1

fDA /f

AA

AA

41

• Find fxA by conjugate beam 8 kN•m

3EI

2EI

B

A 1

C

4m

1 V (kN)

1 kN

2m

2m

Real Beam

1 +

x (m)

8

4 +

M (kN•m)

2EI

1.33EI

x (m)

2.67 EI

60.44/EI 12/EI

fAA = M´A = 60.44/EI

Conjugate Beam 42

• Quantitative influence line of RA 2EI

2.67 EI

1.33EI

60.44/EI A

12/EI

60.44/EI

1

B

C

37.11/EI

0.614

17.77/EI

0.294

Conjugate Beam

4.88/EI

0

0.081

0

fxA

RA = fxA/fAA

43

Using equilibrium conditions for the influence line of RB and MB x

1

B

A RA 1

0

4m

0.614

0.386

2m

0.294

0.706

2m

0.081

MB

RB

0

0.919

RA

1 RB = 1 - RA

0

0 -1.088

-1.648

-1.352

MB = 8RA - (8-x)(1) 44

Using equilibrium conditions for the influence line of VB x

1

B

A RA 1

4m

0.614

0.294

VB = RA - 1

2m

2m

0.081

0

RA

0

VB = RA -1 -0.386

-0.706

-0.919

-1

45

Using equilibrium conditions for the influence line of VC and MC x

1

C

B

A RA 1

4m

2m

0.614

0.294

2m

MB

RB

0.081

0

0.081

0

RA

VC = RA - 1 VC = RA 1 -0.386 MC = 4RA - (4-x)(1) 0.456

0.294

VC

-0.706 MC = 4RA 1.176 0.324

MC

46

The quantitative shear and bending moment diagram and qualitative deflected curve 10 kN MB= 13.53 kN•m B A 4m

10(0.081)=0.81 kN

1

V (kN)

0.81

0.614

2m

0.294

2m

RB=9.19 kN

0.081

0

RA

0.81 -9.19

MA (kN•m)

+

4.86 -13.53

47

The quantitative shear and bending moment diagram and qualitative deflected curve 10 kN 20 kN MB=46.48 kN•m B A 4m

20(.294) +1(0.081) = 6.69 kN 1

V (kN)

6.69

0.614

2m

0.294

2m

RB=23.31 kN

0.081

0

RA

6.69 -13.31

MA (kN•m)

26.76 +

-23.31

-23.31

0.14 -46.48

48

APPENDIX •Muller-Breslau for the influence line of reaction, shear and moment •Influence lines for MDOF beams Example 1 Draw the influence line for - the vertical reaction at B

A

C

2m

D

2m

B

2m

49

Influence line of RA A

C

D

2m

2m

1 kN f AA =1 f AA

B

f CA f AA

2m

f DA f AA

50

• Find fxA by conjugate beam 6 kN•m

1 kN A

C

B

D

2m

2m

Real Beam 1 kN

2m x

M 'A =

72 EI

Conjugate Beam B' y =

6 /EI

18 EI

x 3

3

18 x x − = M´x EI 6 EI

2x 3

V´x x EI

18 EI

x x2 2 EI

B' y =

18 EI

51

x A

C

2m

1 kN fAA

72/EI

D

2m

61.33 /EI

B

f xA

2m

34.67 /EI

Point

x (m)

fxA

fxA / fAA

B

0

0

0

D

2

34.67 EI

0.482

4

61.33 EI

0.852

6

72 EI

1.0

fxA C 72/72 = 1.0 61.33 /72=0.852

1 kN f AA =1 f AA

18 x x 3 = M 'x = − EI 6 EI

A

34.67 /72 = 0.482 Influence line of RA

52

• Influence line of RB A

C

D

2m

2m

f AB f BB

f CB f BB

B

2m

f DB f BB

f BB =1 f BB

1

53

• Find fxB by conjugate beam A

6 kN•m

1

C

D

2m

2m

6 EI

B 1

2m

18 EI

Real Beam

x 72 EI 18 EI

x2 2 EI

x EI

Conjugate Beam

x 72 EI

x3 72 18 x + − = M´x 6 EI EI EI

V´x

x 3

2x 3

18 EI 54

x A

C

D

2m

2m

B

f xB

2m

x3 72 18 x = M 'x = + − 6 EI EI EI

Point

x (m)

fxB

fxB / fBB

B

0

72 EI

1

D

2

37.33 EI

0.518

C

4

10.67 EI

0.148

A

6

0

72/EI 0

10.67/EI

37.33/EI

fBB 1

37.33 /72 = 0.518 10.67 /72 = 0.148 0

fxB

72 /72 = 1 fBB =1 fBB

0

1 Influence line of RB 55

Example 2 For the beam shown (a) Draw the influence line for the shear at D for the beam (b) Draw the influence line for the bending moment at D for the beam EI is constant.Plot numerical values every 2 m.

A

D 2m

B 2m

E 2m

C 2m

56

The influence line for the shear at D 1 kN A

D

B

E

C

D 1 kN 2m

2m f DD =1 f DD

2m

2m

1 kN D 1 kN

f ED f DD

VD

57

• Using conjugate beam for find fxD 1 kN A

D

2m

1 kN

B

E

2m

2m

C

2m

2 kN•m 1 kN

1 kN 1 kN

2 kN•m 2k

1 kN

2 kN

1 kN

1 kN

58

1 kN A

D

B

1 kN 2m

E

C

2kN 1 kN

V( kN)

2m

2m

Real beam 1 kN

2m

1 x (m) -1 4

M (kN •m)

x (m) 4/EI Conjugate beam M´D

59

• Determine M´D at D 4/EI A

M´D D

B 2m

2m

C

E 2m

Conjugate beam

2m

8/EI 8 m 3

4/EI 0

8/EI 8 m 3

128/3EI 40 3EI

16 3EI

8 3EI

4/EI 16 3EI 60

4/EI A D 128/3EI

40 3EI

B 2m

2m

C

E 2m

2m

Conjugate beam

8 3EI

2/EI 2/EI M´DL 40 3EI

2 3

=(

2 2 40 76 )( ) − ( )(2) = − EI 3 3EI 3EI

V´DL

2/EI 128/3EI 40 3EI

2 3

M´DR = (

2 2 128 40 52 )( ) + −( )(2) = EI 3 3EI 3EI 3EI

V´DR

2/EI −

4 2 2 8 = ( )( ) − ( )(2) = M´E EI EI 3 3EI

2/EI

2 V´E 3

8 3EI 61

4/EI A 40 3EI

D 128/3EI

B 2m

2m

E 2m

V´

−

40 3EI

34 3EI 52 3EI

−

128/3EI = M´D = fDD

16 3EI

fxD = M ´ −

0.406 =

2m

2 3EI −

52 128

C

76 3EI

−

Conjugate beam

8 3EI 8 3EI x (m)

θ

x (m)

∆

4 EI

Influence line of VD = fxD/fDD 76 /128 = -0.594

4 /(128/3) = -0.094

62

The influence line for the bending moment at D αDD A

D

1 kN •m

B

E

C

1 kN •m 2m

2m

2m

2m

f DD

α DD

f ED

MD

α DD

63

• Using conjugate beam for find fxD

A

D

1 kN •m 2m

B

E

C

1 kN •m 2m

2m

2m

1 kN•m 0.5 kN

0.5 kN 0.5 kN

1 kN•m 1k

0.5 kN

1 kN

0.5 kN

0.5 kN

64

1 kN •m 1 kN •m A D B 0.5 kN 2m V (kN)

2m

1 kN 2m

E

C

Real beam 0.5 kN

2m

0.5 x (m) 1 2

M (kN •m)

x (m) 2/EI Conjugate beam 65

2/EI Conjugate beam

2m

2m

2m

2m

4/EI 8 m 3

2/EI 0

4/EI 8 m 3

4 3EI

8 3EI

4 3EI

2/EI

32 3EI

8 3EI 66

2/EI Conjugate beam 4 EI

4 3EI

32 3EI

2m

2m 1/EI

2m

1/EI M´D = (

4 EI

2m

1 2 4 26 )( ) + ( )(2) = EI 3 EI 3EI

2 V´D 3

1/EI −

2 1 2 4 = ( )( ) − ( )(2) = M´E EI EI 3 3EI

V´E

1/EI

2 3

4 3 EI 67

2/EI Conjugate beam 4 EI 2m

4 3EI

32 3EI 2m

2m

2m

5 1 EI αDD = 32/3EI 3EI

4 EI V´ 17 − 3EI

−

4 3EI x (m)

8 3EI

θ

26/3EI fxD = M ´

x (m) 26 = 0.813 32

f xD

α xD

∆

-2/EI θDL = 5/(32/3) = 0.469 rad. θDR = -17/32 = -0.531 rad.

θD = 0.469 + 0.531 = 1 rad

Influence line of MD 68 -2 /(32/3) = -0.188

Example 3 Draw the influence line for the reactions at supports for the beam shown in the figure below. EI is constant.

A 5m

B

C 5m

D 5m

E 5m

F 5m

G 5m

69

Influence line for RD A 5m

B

C 5m

fBD

D 5m

fCD

E

F

5m

5m

fDD

fED

G 5m

fFD

fXD

1

f BD f DD

f CD f DD

f DD = 1 f ED f DD f DD

f FD f DD

fXD/fDD = Influence line for RD

1

70

• Use the consistency deformation method A

G 3@5 =15 m

3@5 =15 m

=

1 ∆´G

+

1

∆´G + fGGRG = 0

15

15 EI

------(1)

x RG

1

- Use conjugate beam for find ∆´G and fGG 30 Real beam Real beam G A

A 1

fGG

15 m

112.5 EI

1

30 m

15 m

2812.5 ∆'C = M 'C = EI

Conjugate beam 15 + (2/3)(15)

G

15 EI

1

450 EI

1 f 'GG = M ' 'G =

Conjugate beam

112.5/EI

20 m

9000 EI

450 71 EI

Substitute ∆´G and fGG in (1) :

2812.5 9000 + RG = 0 EI EI RG = −0.3125 kN , ↓

5.625 A

G 1

0.3125

=

0.6875

15 1

+

1 30

x RG = -0.3125 kN

1

1 72

• Use the conjugate beam for find fXD fCD 5.625 fBD A 3@5 =15 m

fDD

fED

fFD

G

Real beam

3@5 =15 m

23.01 1 0.3125 EI 5.625 x2 6.818 m EI G Conjugate beam A 8.182 m 28.13 − 4.688 35.16 2 x1 0.6875 x1 EI 15 .

98 EI EI 2 EI 5.625 x1 − 0.6875 x12 x2 EI EI V´2 5.625 (5.625-0.6875x1)/EI M´2 G EI 2 2 A x − x x 5 .

625 0 .

6875 x x2 0 .

3125 1 1 1 2 M´ 0.3125x2 = ( ) − ( ) 1 28.13 x1 EI 2 2 3 2 2 V´1 EI 0 .

3125 x 2 .6875 x1 2 x1 28 .

13 + ( ) + x2 = M ' x 2 2 EI 2 EI 3 EI

0.6875

x1 = 5 m -----> fBD = M´1= 56/EI

x2 = 5 m -----> fFD = M´2= 134.1/EI

x1 = 10 m -----> fCD = M´1= 166.7/EI

x2 = 10 m -----> fED = M´2= 229.1/EI

x1 = 15 m -----> fDD = M´1= 246.1/EI

x2 = 15 m -----> fDD = M´2= 246.1/EI

73

• Influence Line for RD 56 EI

166.7 EI

246.1 229.2 134.1 EI EI EI

fXD

1 246.1 166.7 229.2 134.1 56 246.1 246.1 246.1 246.1 246.1

fXD/fDD

1

0.228 0.677

1.0 0.931 0.545

Influence Line for RD

74

Influence line for RG A 5m

B

C 5m

D 5m

E 5m

F 5m

fEG fBG

5m

fFG

f CG f GG

fGG fXG

fCG

1 f EG f GG

f BG f GG

G

f FG f GG

f GG =1 f GG

fXG/fGG 1

75

• Use consistency deformations fXG 1

3@5 =15 m

+

∆´D

=

3@5 =15 m

1

fDD

X RD

1

30

A

∆´D + fDDRD = 0

- Use conjugate beam for find ∆´D and fDD 15 Real beam Real beam G A 30 m

1 30 EI

------(2)

1

1

450 EI

9000 EI

Conjugate beam 20 m

15 EI

15 m

1

G 15 m

112.5 EI Conjugate beam

450/EI

15 + (2/3)(15)

76

112.5 EI

1.5 EI

M´

15 m

V´

∆'D = M ' =

9000 EI

15 EI

112.5 EI M´´

450/EI

V´´

112.5 15 9000 450 2812.5 ( )+ − (15) = EI 3 EI EI EI

f DD = M ' ' =

112.5 2 1125 ( × 15) = EI 3 EI

2812.5 1125 + RD = 0, RD = −2.5kN = 2.5kN , ↓ Substitute ∆´D and fDD in (2) : EI EI 7.5

1.5

1

=

30

2.5

15

1

+

1

x RD = -2.5 kN

1

1

77

• Use the conjugate beam for find fXG 7.5 fBG

fCG

1.5 3@5 =15 m

Real beam

2.5

15 EI

112.5 EI

fGG = M´G = 1968.56/EI G

A − 7.5 EI 18.75 EI

10 m 15 + (10/3) = 18.33 m

Conjugate beam

168.75/EI

25 + (2/3)(5) = 28.33 m

18.75 2 62.5 ( × 5) = − EI 3 EI 18.75 EI 6.67 m

M´1 = f BG = −

A − 7.5 EI 18.75 EI

1

3@5 =15 m

75 EI

5m

fGG

fFG

fEG

V´1

A − 7.5 EI

M´2 = f CG = −

− 18.75 EI

V´2

18.75 125.06 (6.67) = − EI EI 78

x2 2 x

2

x3 x3 1968.56 168.75 ( )+ x3 =M´ − 2 3 EI EI

V´2

fGG = M´G = 1968.56/EI G

x

168.75/EI

x = 5 m -----> fFG = M´= 1145.64/EI x = 10 m -----> fEG = M´ = 447.73/EI 1968.56 1145 .

64 447.73 EI EI EI fXG − 62.5 − 125 1 EI EI 1145.64 1968.56 447.73 1968.56 1968.56 1968.56 fXG/fGG − 125 − 62.5 1 1968.56 1968.56 1.0 0.582 0.227 Influence line for RG 79 -0.032 -0.064

Using equilibrium condition for the influence line for Ay 1 x A

MA

B

5m

C 5m

D 5m

Ay

E

5m RD

F 5m

G 5m RG

+ ↑ ΣFy = 0 : RA = 1 − RD − RC

1

1 Unit load 0.228 0.678

1.0 0.929 0.542 Influence Line for RD 0.227 0.582

-0.032 -0.064 1.0

0.804 0.386 -0.156 -0.124

1.0 Influence line for RG

Influence line for Ay 80

Using equilibrium condition for the influence line for MA 1 x A

MA

B

5m

C 5m

D 5m

Ay

E

5m RD

F 5m

G 5m RG

+ ΣM A = 0 : 1x − 15RD − 30 RC

5

10

15

20

25

30 1x

0.228 0.678

1.0 0.929 0.542 0.227 0.582

-0.032 -0.064 2.54 1.75

RD

x 15

RG

x 30

1.0

Influence line for MA -0.745 -0.59

81