A FIRST COURSE IN THE FINITE ELEMENT METHOD ENHANCED 6TH EDITION BY DARYL L. LOGAN SOLUTIONS MANUAL by QuentinAxel

Solutions Manual For

A First Course in the Finite Element Method 6th EDITION DARYL L. LOGAN

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Contents Chapter 1:............................................................................................................................ 1 Chapter 2:............................................................................................................................ 3 Chapter 3:.......................................................................................................................... 25 Chapter 4:........................................................................................................................ 137 Chapter 5:........................................................................................................................ 203 Chapter 6:........................................................................................................................ 315 Chapter 7:........................................................................................................................ 363 Chapter 8:........................................................................................................................ 383 Chapter 9:........................................................................................................................ 397 Chapter 10:...................................................................................................................... 423 Chapter 11:...................................................................................................................... 449 Chapter 12:...................................................................................................................... 477 Chapter 13:...................................................................................................................... 499 Chapter 14:...................................................................................................................... 539 Chapter 15:...................................................................................................................... 561 Chapter 16:...................................................................................................................... 591 Appendix A: ..................................................................................................................... 629 Appendix B: ..................................................................................................................... 635 Appendix D:..................................................................................................................... 641

Chapter 1 1.1. A finite element is a small body or unit interconnected to other units to model a larger structure or system. 1.2. Discretization means dividing the body (system) into an equivalent system of finite elements with associated nodes and elements. 1.3. The modern development of the finite element method began in 1941 with the work of Hrennikoff in the field of structural engineering. 1.4. The direct stiffness method was introduced in 1941 by Hrennikoff. However, it was not commonly known as the direct stiffness method until 1956. 1.5. A matrix is a rectangular array of quantities arranged in rows and columns that is often used to aid in expressing and solving a system of algebraic equations. 1.6. As computer developed it made possible to solve thousands of equations in a matter of minutes. 1.7. The following are the general steps of the finite element method. Step 1 Divide the body into an equivalent system of finite elements with associated nodes and choose the most appropriate element type. Step 2 Choose a displacement function within each element. Step 3 Relate the stresses to the strains through the stress/strain law—generally called the constitutive law. Step 4 Derive the element stiffness matrix and equations. Use the direct equilibrium method, a work or energy method, or a method of weighted residuals to relate the nodal forces to nodal displacements. Step 5 Assemble the element equations to obtain the global or total equations and introduce boundary conditions. Step 6 Solve for the unknown degrees of freedom (or generalized displacements). Step 7 Solve for the element strains and stresses. Step 8 Interpret and analyze the results for use in the design/analysis process. 1.8. The displacement method assumes displacements of the nodes as the unknowns of the problem. The problem is formulated such that a set of simultaneous equations is solved for nodal displacements. 1.9. Four common types of elements are: simple line elements, simple two-dimensional elements, simple three-dimensional elements, and simple axisymmetric elements. 1.10 Three common methods used to derive the element stiffness matrix and equations are (1) direct equilibrium method (2) work or energy methods 1

(3) methods of weighted residuals 1.11. The term ‘degrees of freedom’ refers to rotations and displacements that are associated with each node. 1.12. Five typical areas where the finite element is applied are as follows. (1) Structural/stress analysis (2) Heat transfer analysis (3) Fluid flow analysis (4) Electric or magnetic potential distribution analysis (5) Biomechanical engineering 1.13. Five advantages of the finite element method are the ability to (1) Model irregularly shaped bodies quite easily (2) Handle general load conditions without difficulty (3) Model bodies composed of several different materials because element equations are evaluated individually (4) Handle unlimited numbers and kinds of boundary conditions (5) Vary the size of the elements to make it possible to use small elements where necessary

Chapter 2 2.1 (a)

[k(1)] =

k1 0 – k1 0

[k(2)] =

0 0 0 0

0 – k1 0 0 0 k1 0 0

0 0 0 0 0 k2 0 – k2

0 0 0 k3 [k 3(3)] = 0 0 0 – k3

0 0 0 0 0 0 – k2 k2

0 0 0 – k3 0 0 0 k3

[K] = [k(1)] + [k(2)] + [k(3)]

[K] =

k1 0 – k1 0

0 k3 0 – k3

– k1 0 k1 k2 – k2

0 – k3 – k2 k2 k3

(b) Nodes 1 and 2 are fixed so u1 = 0 and u2 = 0 and [K] becomes [K] =

k1 k2 – k2

– k2 k2 k3

{F} = [K] {d} F3 x F4 x



k1 k2 – k2

– k2 k2 k3

0 = P

k1 k2 – k2

– k2 k2 k3

u3 u4 u3 u4

 K] –1 {F} = {d} Using the adjoint method to find [K –1] C11 = k2 + k3

C21 = (– 1)3 (– k2)

C12 = (– 1)1 + 2 (– k2) = k2

C22 = k1 + k2

[C] =

and CT =

det [K] = | [K] | = (k1 + k2) (k2 + k3) – ( – k2) (– k2) 

| [K] | = (k1 + k2) (k2 + k3) – k22

[K –1] =

[C T ] det K k2

[K –1] =

u3 u4

 u3 =  u4 =

k3 k2

(k1

k2 ) (k2

k2 k2

k3 ) – k2

k2 k1 k2

k1 k2

k2 P k1 k3

k1 k2

k2 k1 k2 k1 k2 k1 k3 k2 k3

0 k1 k2 P k1 k3 k2 k3

(k1

k2 k3

k2 ) P k1 k3 k2 k3

F4 x

k1 0 k1 0

0 k3 0 k3

F1x = – k1 u3 = – k1  F1x =

k1 k2

0 k3 k2 k 2 k3

k1 k2

k2 P k1 k3 k2 k3

u1 u2 u3 u4

k1 k2 P k1 k3 k2 k3

F2x = – k3 u4 = – k3  F2x =

k1 0 k1 k2 k2

(k1 k1 k2

k2 ) P k1 k3 k2 k3

k3 (k1 k2 ) P k1 k2 k1 k3 k2 k3

2.2

lb in.

k1 = k2 = k3 = 1000

[k(1)] =

(1) (2) (2) (3) k k (1) k k (2) ; [k(2)] = k k (2) k k (3)

By the method of superposition the global stiffness matrix is constructed.

[K] =

(1)

(2)

(3)

k k 0

k k k

0 (1) k (2)  [K] = k (3)

k k 0

k 2k k

0 k k

Node 1 is fixed  u1 = 0 and u3 =  {F} = [K] {d} F1x

F2 x

0 =

F3 x



0 F3 x



= u2 =

k k 0

k 2k k

0 k k

u1  0    u2  ?  u     3 

2k u2

F3x

k u2

k 1 in. = =  u2 = 0.5 2 2 2k

F3x = – k (0.5) + k (1) F3x = (– 1000

lb lb ) (0.5) + (1000 ) (1) in. in.

F3x = 500 lbs Internal forces Element (1) f1x (1) f2 x



(2)

0.5

f1x (1) = (– 1000 f 2x (1) = (1000

lb ) (0.5)  f1x (1) = – 500 lb in.

lb ) (0.5)  f 2x (1) = 500 lb in.

k k

uf 2 x (2)0.5 = uf 33x (2)1

k k

0.5



f 2 x (2) f3x

– 500 lb

(2)

500 lb

2.3

k k k k By the method of superposition we construct the global [K] and knowing {F} = [K] {d} we have

(a) [k(1)] = [k(2)] = [k(3)] = [k(4)] =

(b)

F1x

F2 x

F3 x

P =

F4 x

F5 x

P =

k 0

2k k

k 2k

 u2 =

k k 0 0 0

k 2k k 0 0

0 k 2k k 0

0 0 k 2k k

0 0 0 k k

2ku2

P 0

ku2 2ku3 ku4 ku3 2ku4

u2 u3 u4 u5

0 ku3

(1) (2) (3)

u3 u ; u4 = 3 2 2

Substituting in the second equation above P = – k u2 + 2k u3 – k u4 

P= –k



P = ku3

 u3 = u2 =

u3 2

+ 2k u3 – k

P k P P ; u4 = 2k 2k

(c) In order to find the reactions at the fixed nodes 1 and 5 we go back to the global equation {F} = [K] {d} F1x = – ku2 = – k

P  F1x = 2k

P 2

F5x = – ku4 = – k

P  F5x = 2k

P 2



P 2

P + 2

+P=0

0=0 2.4

(a) [k(1)] = [k(2)] = [k(3)] = [k(4)] =

k k

By the method of superposition the global [K] is constructed. {F} = [K] {d} and u1 = 0 and u5 = 

Also F1x

F2 x

F3 x

0 =

F4 x

F5 x

k k 0 0 0

k 2k k 0 0

0 k 2k k 0

0 0 k 2k k

0 0 0 k k

(b) 0 = 2k u2 – k u3

(1)

0 = – ku2 + 2k u3 – k u4

(2)

0 = – k u3 + 2k u4 – k 

(3)

From (2) u3 = 2 u2 From (3) 2 u2 2

u4 =

Substituting in Equation (2)

  2 u2  – k (u2) + 2k (2u2) – k    2   – u2 + 4 u2 – u2 – u3 = 2 u4 =

 u3 = 2 4 2

= 0  u2 =

 u4 =

3 4

{F} = [K]{d} F1x = – k u2 = k

 F1x =

k 4

3 4

+k

F5x = – k u4 + k  = – k  F5x =

k 4

2.5

[k (1)] =

[k (3)] =

[k (5)] =

u1 1 1

u2 u2 1 2 ; [k (2)] = 1 2

u4 2 2

u2 3 3

u4 3 ; [k (4)] = 3

u4 4 4

u4 5 5

u3 5 5

u2 4 4

Assembling global [K] using direct stiffness method

[K] =

1 1 1 1 2 3 4 0 0 0 2 3 4

[K] =

1 1 1 10 0 0 0 9

0 0 0 2 3 4 5 5 5 2 3 4 5

Simplifying 0 0 5 5

0 9 kip 5 in. 14

2.6 Now apply + 3 kip at node 2 in spring assemblage of P 2.5.  F2x = 3 kip [K]{d} = {F} 8 © 2023 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

[K] from P 2.5 1 1 1 10

0 0

0 9

0 0

5 5

5 14

0 9

u2 u4

 F1  3    =   0  F3  0 

(A)

where u1 = 0, u3 = 0 as nodes 1 and 3 are fixed. Using Equations (1) and (3) of (A) 10

3  =    0

Solving u2 = 0.712 in., u4 = 0.458 in. 2.7

f1x = C, f2x = – C f = – k = – k(u2 – u1) 

f1x = – k(u2 – u1) f2x = – (– k) (u2 – u1) f1x

k –k

–k k

[K] =

k –k

–k same as for k tensile element

f2x



2.8

k1 = 1000

1 1

1 ; k2 = 1000 1

1 1

[K] = 1000

1 1 0

1 2 1

0 1 1

{F} = [K] {d}   F1  ?   1 1 0  u1 0    F2  0 = 1000  1 2 1 u2 ?      0 1 1 u3 ?  F3  500 



0 = 2000 u2 – 1000 u3

(1)

500 = – 1000 u2 + 1000 u3

(2)

From (1) u2 =

1000 u3  u2 = 0.5 u3 2000

(3)

Substituting (3) into (2) 

500 = – 1000 (0.5 u3) + 1000 u3



500 = 500 u3



u3 = 1 in.



u2 = (0.5) (1 in.)  u2 = 0.5 in.

Element 1–2 f1x (1) f 2 x (1)

 1 = 1000   1

f1x (1)   500lb 1  0 in.    1  0.5 in. f 2 x (1)  500lb

 1 = 1000   1

f 2 x (2)   500 lb 1  0.5 in.    1  1 in.  f3 x (2)  500 lb

Element 2–3 f 2 x (2) f 3 x (2)

0   F1x = 500 [1 –1 0] 0.5 in.  F1x   500 lb   1 in. 

2.9

(1) (2)  5000  5000 [k(1)] =  5000   5000 (2)

(3) 10

 5000 [k(2)] =    5000 (3) 5000  [k(3)] =    5000 (1)  5000   5000 [K] =   0   0

F1x F2 x F3 x F4 x

 5000 5000 (4)  5000 5000 (2) (3) (4)  5000 0 0  10000  5000 0    5000 10000  5000   0  5000 5000 

0 0  5000  5000    1000  5000 10000  5000 0  =  0 5000 10000  5000   0   0  5000 5000  4000  0



u2 u3 u4

u1 = 0 in. u2 = 0.6 in. u3 = 1.4 in. u4 = 2.2 in.

Reactions u1  u   F1x = [5000 – 5000 0 0]  2 u3  u4  Element forces Element (1) f1x (1) f2 x Element (2)

(1)

f 2 x (2) f3x Element (3)

(2)

f 3 x (3)

(3)

f4 x

0  0.6   F1x = – 3000 lb 1.4  2.2 

 5000 =    5000

f1x (1)  5000  0     5000  0.6 f 2 x (1)

3000lb

 5000 =    5000

f 2 x (2)  5000  0.6    5000  1.4  f 3 x (2)

4000lb

 5000 =    5000

f 3 x (3)  5000  1.4    5000  2.2 f 4 x (3)

3000lb

4000lb 4000lb 4000lb

2.10

[k(1)] =

1000 1000

[k(2)] =

500 500

[k(3)] =

500 500

{F} = [K] {d}  F1x  ?   F  – 8000   2x    =  F3 x  ?   F4 x  ? 



u2 =

1000 1000 0 0

1000 2000 500 500

0 500 500 0

0 500 0 500

8000 = – 4 in. 2000

Reactions F1x F2 x F3 x F4 x F1x



F2 x F3 x F4 x

0  1000 1000  1000 2000 500 =  500 500  0  500 0  0

0  0  500  4    0  0   500  0 

 4000   8000    =   lb  2000   2000 

Element (1) f1x (1) f2 x

(1)

 1000 =   1000

1000   0    1000   – 4

f1x (1)

 500 =   500

500    4    500   0

f 2 x (2)

f2 x

(1)

 4000  =   lb  4000 

Element (2) f 2 x (2) f3x

(2)

f3x

(2)

 – 2000  =   lb  2000 

f 2 x (3) f 4 x (3)

 500 =   500

500    4    500   0

f 2 x (3)  2000    lb f 4 x (3)  2000 

2.11

 1000 [k(1)] =   1000

1000   3000 ; [k(2)] =  1000   3000

3000  3000 

{F} = [K] {d} F1x

F2 x

F3 x



 1000 1000 0  u1 0  = 1000 4000 3000 u2 ?   3000 3000  u3 0.02 m  0

u2 = 0.015 m

Reactions F1x = (– 1000) (0.015)  F1x = – 15 N Element (1)  f1x   f1x  15  1000 1000  0    =          N   1000 1000  0.015  f2 x   f 2 x   15

Element (2)  f2 x   f 2 x  15  3000 –3000 0.015   =       =   N   –3000 3000   0.02  f3 x   f3 x   15

2.12

[k(1)] = [k(3)] = 10000

1 1

 3 3 [k(2)] = 10000   3 3

 F1x  ?   1 1 0 0   F  450 N   1 4 3 0   2x    = 10000    0 3 4 1  F3 x  0     F4 x  ?   0 0 1 1

0 = – 3 u2 + 4 u3  u2 =

4 u3  u2 = 1.33 u3 3

450 N = 40000 (1.33 u3) – 30000 u3  450 N = (23200  Element (1)

N ) u3  u3 = 1.93  10–2 m m

u2 = 1.5 (1.94  10–2)  u2 = 2.57  10–2 m

 f1x    = 10000  f2 x 

0 f1x (1)   257 N      2 f 2 x (1)  257 N 2.57  10 

Element (2)  f2 x    = 30000  f3 x 

2.57  102  f 2 x (2)  193 N    (2) f3 x  193 N 1.93  102 

Element (3)  f3 x   f3 x      = 10000  f4 x   f4 x  Reactions



f3 x (3)  193 N 1.93  102     0 f 4 x (3)  193 N  

{F1x} = (10000

0  N ) [1 – 1]  F1x = – 257 N 2  m 2.57  10 

{F4x} = (10000

1.93  102  N ) [–1 1]   m  0 

F4x = – 193 N

2.13

[k(1)] = [k(2)] = [k(3)] = [k(4)] = 60

1 1

{F} = [K]{d}

 F1x  ?  F  0   2 x   F3 x  5 kN  = 60 F  0   4x   F5 x  ?  0 2 u2 – u3 0



u2 u4

1 1 0 0 0

1 2 1 0 0

0.5 u3 0.5 u3

0 1 2 1 0

0 0 1 2 1

0 0 0 1 1

u2 u3

? ?

u4 u5

? 0

 u2 = u4

– u3

2 u4



5 kN = – 60 u2 + 120 (2 u2) – 60 u2



5 = 120 u2  u2 = 0.042 m



u4 = 0.042 m



u3 = 2(0.042)  u3 = 0.084 m

Element (1)  f1x    = 60  f2 x 

f1x (1)   2.5 kN  0     0.042  f 2 x (1)  2.5 kN

1 1

1 0.042  f 2 x (2)   2.5 kN    1 0.084  f3 x (2)  2.5 kN

1 = 60 f4 x 1 1

1 1

f 31x (3)0.084 2.5 kN f 3 x (3)     f14 x (3)0.042 2.5 kN f 4 x (3)

 f 4 x   f 4 x 1 1     = 60  f5 x   f5 x 1 1

1 1

f14 x (4)0.042  2.5 kN f 4 x (4)  2.5 kN     kN f5 x (4)   2.5 kN f15 x (4)  0 2.5

Element (2)  f 2 x 1 1   = 60  f3 x 1 1

Element (3) f 3x f4 x

f 3x 1

2.5 kN 2.5 kN

Element (4)

 0  F1x = 60 [1 –1]   F1x = – 2.5 kN 0.042  0.042  F5x = 60 [–1 1]   F5x = – 2.5 kN  0 

2.14

[k(1)] = [k(2)] = 4000

1 1

F1x ? = 4000 F2 x 100 F3 x 200

1 1 0

1 2 1

0 1 1

u3 ?

100 = 8000 u2 – 4000 u3 – 200 = – 4000 u2 + 4000 u3 – 100 = 4000 u2  u2 = – 0.025 m 100 = 8000 (– 0.025) – 4000 u3  u3 = – 0.075 m Element (1)  f1x    = 4000  f2 x  Element (2)

f1x (1)  100 N  0      0.025 f 2 x (1)  100 N

1 –1  0.025 f 2 x (2)  200 N  f2 x   f2 x      = 4000   –1 1   0.075   f3 x   f3 x  f3 x (2)   200 N

Reaction  0  {F1x} = 4000 [1 –1]    F1x = 100 N  0.025

2.15

[k(1)] =

500 500

500 ; [k(2)] = 500

 F1x  ?  F  ?   2x    =  F3 x  4 kN   F4 x  ? 

500 0 500 0

0 500 500 0

 Reactions

500 500 2000 1000

500 500

0 0 1000 1000

500 ; [k(3)] = 500

1000 1000

u1 0 u2 0 u3 ? u4 0

u3 = 0.002 m F1x = (– 500) (0.002)  F1x = – 1.0 kN F2x = (– 500) (0.002)  F2x = – 1.0 kN F4x = (– 1000) (0.002)  F4x = – 2.0 kN

Element (1)  f1x    =  f3 x 

500

 f1x   1.0 kN   0       =   0.002   f3 x   1.0 kN  16

Element (2)

500 f 2 x 500    = 500 f3 x 500  Element (3)

500 500

f 2 xkN    1.0 kN  0  1.0    =      f3 xkN 0.002   1.0    1.0 kN 

1000 0.001 1000  0.002  f3 x   2.0 kN        =   1000  1000 0  0   f 4 x   2.0 kN 

 f3 x    =  f4 x 

2.16

 F1x   200      =  200    F4 x   200    =  200 

100 100 100 100 100 0 100 0 0 200 100

100 200

0 100 100 100 100

0 0 100 100

0 u2 u3 0

u2 u3

2 in. 3 2 u3 = – in. 3

u2 =

2.17

F1x

500

0 1000 N F4 x ?

500

–500 400 300 500 300

–300 – 300

–400

–300 – 300

(300 300 400)

–400

– 400

400 400

u2 u3 u4

0 = 1500 u2 – 600 u3 1000 = – 600 u2 + 1000 u3

u3 =

15 0 0 u2 = 2.5 u2 60 0

1000 = – 600 u2 + 1000 (2.5 u2) 1000 = 1900 u2 u2 =

1000 1 = mm = 0.526 mm 1900 1.9

u3 = 2.5

1 1.9

mm = 1.316 mm

F1x = – 500

1 1.9

= – 263.16 N

F4x = – 400

1 1.9

– 400

= – 400

1 1.9

2.5 = –736.84 N 1.9

2.5

1 1.9

Fx = – 263.16 + 1000 – 736.84 = 0 2.18 (a)

As in Example 2.4 p = U +  1 U= k x2,  = – Fx 2 Set up table

p =

1 (2000) x2 – 1000 x = 1000 x2 – 1000 x 2

Deformation x, in.

p, lbin.

– 3.0 – 2.0 – 1.0 0.0 0.5 1.0

6000 3000 1000 0 – 125 0 18

2.0 p

1000

= 2000 x – 1000 = 0  x = 0.5 in. yields minimum p as table verifies.

(b)

p =



1 2 kx – Fx = 250 x2 – 1000 x 2

x, in.

p, lbin.

– 3.0 – 2.0 – 1.0 0 1.0 2.0 3.0

11250 3000 1250 0 – 750 – 1000 – 750

= 500 x – 1000 = 0

x = 2.0 in. yields p minimum

(c)

p = p

1 (2000) x2 – 3924 x = 1000 x2 – 3924 x 2

= 2000 x – 3924 = 0

x = 1.962 mm yields p minimum



p min =

1 (2000) (1.962)2 – 3924 (1.962) 2

 p min = – 3849.45 Nmm

p =

(d)



1 (400) x2 – 981 x 2

= 400 x – 981 = 0

x = 2.4525 mm yields p minimum

p min =

1 (400) (2.4525)2 – 981 (2.4525) 2

 p min = – 1202.95 Nmm 2.19



p =

1 2 kx – Fx 2

p =

1 (500) x2 – 1000 x 2

= 500 x – 1000 = 0

 x = 2.0 in.  2.20

F = k2

(x = )

dU = F dx U=

x 0

(kx2) dx

k x3 3

 = – Fx

p = p

1 3 kx – 500 x 3

= 0 = kx2 – 500

0 = 1000 x2 – 500 

x = 0.707 in. (equilibrium value of displacement)

p min =

1 (1000) (0.707)3 –500 (0.707) 3

p min = – 235.7 lbin. 2.21 Solve Problem 2.10 using P.E. approach

p =

p (e) =

e 1

1 1 1 k1 (u2 – u1)2 + k2 (u3 – u2)2 + k3 (u4 – u2)2 2 2 2

– f1x(1) u1 – f2x(1) u2 – f2x(2) u2 – f3x(2) u3 – f2x(3) u2 – f4x(3) u4 p

= – k1 u2 + k1 u1 – f1x(1) = 0

(1)

= k1 u2 – k1 u1 – k2 u3 + k2 u2 – k3 u4 u2 + k3 u2 – f2x(1) – f2x(2) – f2x(3) = 0

(2)

u1 p

u3 p

= k2 u3 – k2 u2 – f3x(2) = 0

(3)

= k3 u4 – k3 u2 – f4x(3) = 0

(4)

In matrix form (1) through (4) become k1 k1 k1 k1 k2 k3 0 k2 0 k3

0 k2 k2 0

0 k3 0 k3

u1 u2 u3 u4

f1x (1)

f 2 x (1)

f 2 x (2) f3 x (2)

f 2 x (3)

(5)

f 4 x (3)

or using numerical values 1000 1000 0 0

1000 2000 500 500

0 0 500 –500 500 0 0 500

u1 u2 u3 u4

 F1x   8000    =   0  F3 x   F4 x  0

(6)

Solution now follows as in Problem 2.10 Solve 2nd of Equations (6) for u2 = – 4 in. For reactions and element forces, see solution to Problem 2.10 2.22 Solve Problem 2.15 by P.E. approach

p =

1 1 k1 (u3 – u1)2 + k2 (u3 – u2)2 2 2

p (e) =

e 1

1 k3 (u4 – u3)2 – f1x(1) u1 2

– f3x(1) u3 – f2x(2) u2 – f3x(2) u3 – f3x(3) u3 – f3x(4) u4 p

= 0 = – k1 u3 + k1 u1 – f1x(1)

u1 p

= 0 = – k2 u3 + k2 u2 – f2x(2)

u2 p

= 0 = k1 u3 + k2 u3 – k2 u2 – k3 u4 + k3 u3 – f3x(2) – f3x(3) – f3x(1) – k1 u1

u3 p

= 0 = k3 u4 – k3 u3 – f3x(4)

In matrix form k1 0 k1 0

0 k2 k2 0

k1 k2 k2 k3 k3

0 0 k3 k3

u1 u2 u3 u4

F1x     F2 x   =    F3 x  4 kN    F4 x

For rest of solution, see solution of Problem 2.15. 2.23 I = a1 + a2 x I (0) = a1 = I1 I (L) = a1 + a2 L = I2 a2 =



I1 L

I = I1 +

I1 L

Now V = IR V = – V1 = R (I2 – I1) V = V2 = R (I2 – I1) V1 V2

1 1

I1 I2

Chapter 3 3.1

(a)  k (1)  =

A1 E1  1 1 L1 1 1 

1   1 1 

 k (2)   A2 E2  1 L2

1  1 1 

 k (3)  = A3 E3  1 L3

 A1E1  L1  – A1E1  L1 [K] =   0   0 (b)

– A1E1 L1

A1E1 AE  L2 2 L1 2

– A2 E2 L2

AE A2 E2  L3 3 L2 3

– A3 E3 L3

0   0   – A3 E3  L3  A3 E3  L3  

AE A1 E1 AE AE = 2 2 = 3 3 = L L3 L1 L2  1 1 0 0    AE  1 2 1 0  [K] = L  0 1 2 1    0 0 1 1 It is known that {F} = [K] {d}

 F1x  ?   1 1 0 0  u1  0  F  0    AE  1 2 1 0 u2  ?   2x     =   L  0 1 2 1  u3  ?   F3 x  P     F4 x  ?   0 0 1 1 u4  0 

2AE AE u2 – u3  u3 = 2 u2 L L

 AE 2AE u2 + u3 L L

 P=

 AE 2AE u2 + (2 u2) L L 25

 u2 = 

1 PL 3 AE

u3 = 2 

1 PL 3 AE

 u3 =

2 PL 3 AE

1 PL 1 (2000) (10) = 3 AE 3 (1)(10  106 )



u2 = 6.66  10–4 in. u3 =



2 PL = 2 u2 3 AE

u3 = 13.34  10–4 in.

(ii) Going back to {F} = [K] {d}



F1x =

 AE  AE 1 PL  1 u2 =  P L L 3 AE  3

F1x =

1 (2000)  F1x = – 666.6 lbs 3

F4x =

 AE  AE  2 PL  2 u3 = =– P 3 L L  3 AE 

F4x =

2 (2000)  F4x = – 1333.4 lbs 3

(iii) f =  A, where f = force,  = stress and A = area. Going back to the local system and substituting Element (1)

  1x  f1 x   AE  1 1 u1  0 A =    1 1 u  6.66  104  f2 x AL   2   2 x  A  (1)   1x = –

10  106 E u2 = – (6.66  10–4) 10 L

(1) = – 666.6 psi (C)   1x (1) =   2x

10  106 E u2 = (6.66  10–4) L 10

  2(1)x = 666.6 psi (T) Element (2) 4  2 x  f 2 x  AE  1 1 u2  6.66  10  A =    f3 x  AL  1 1 u3  13.34  10 –4   3 x  A 

(2)   2x =

E 10  106 (u2 – u3) =  10–4 (6.66 – 13.34) 10 L

(2)   2x = – 666.6 psi (C)

(2)   3x =

E 10  106 (u3 – u2) =  10–4 (13.34 – 6.66) 10 L

(2)   3x = 333.33 psi (T)

Element (3)  3 x  f3 x  AE  1 1 u3  13.34  104  A =     f AL  1 1 u4  0   4 x  A4 x    3x =

E 10  106 (u3 – u4) =  10–4 (13.34 – 0) 10 L

(3)   3x = 666.7 psi (T) (3)   4x =

E 10  106 (u4 – u3) =  10–4 (– 13.34) 10 L

(3)   4x = – 666.7 psi (C)

(1) = (2) = 666.6 psi (T) and (3) = 666.6 psi (C)

So 3.2

Element 1–2

 1 1 [k1–2] = 84  106   1 1 Element 2–3  1 1 [k2–3] = 84  106   1 1

{F} = [K] {d}

and

u1 = 0

 F1x  ?   1 1 0  u1  0      6   F2 x  0  = 84  10  1 2 1 u2  ?   F   10000     0 1 1  u3  ?   3x  

 2u2 – u3 = 0  u3 = 2u2

(1)



 – 10000 = 84  10 [– u2 + u3]

(2)

Substituting (1) in (2), we have – 10000 84  106 

= – u2 + 2 u2

 u2 = – 1.19  10–4 m  u3 = –2.38  10–4 m 27

Element 1–2

0 f1(1) 1 1    f1x  x  10000 N 6     = 84  10      4 (1) f  1 1  1.19  10   2x  f 2 x   10000 N Element 2–3 f 2(2) 1 1  1.19  104   f2 x  x  10000 N 6  = 84  10         1 1 2.38  104   f3x  f3(2) x   10000 N 0    4  6 F1x = 84  10 [1 –1 0]  1.19  10   F1x  10000 N  4   2.38  10  3.3

 1 1 [k1–2] = 3  106   1 1  1 1 [k2–3] = 1.2  106   1 1 0 3  3 [K] = 10 3 3  1.2 1.2     0 1.2 1.2  6 

–3 0 u1  0   F1x   3    6    F2 x  16000  = 10  –3 4.2 –1.2  u2  F     0 –1.2 1.2  u3  0  3x   u2 = 3.81 – in. 0  0  3  3  F1x     3   6   F2 x  = 10  3 4.2 1.2  3.81  10    F  0  0 –1.2 1.2     3x   F1x = – 11430 lb F2x = 16000 lb F3x = – 4570 lb

0 f1(1) 1 1    f1x  x   11430 lb 6  = 3  10        3   1 1 3.81  10   f2 x  f 2(1) x  11430 lb

f 2(2) 1 1 3.81  103   f2 x  x  4532 lb 6       = 1.2  10   (2) 0  1 1   f3x  f3 x   4532 lb  3.4

 1 1 [k1–2] = 4  106   1 1  1 1 [k2–3] = 4  106   1 1  1 1 [k3–4] = 4  106   1 1 {F} = [K] {d}  1 1 0 0 u1  0   F1x  0       F  5000  = 4  106  1 2 1 0 u2    2x  u   0 1 2 1      3  F 10000      3x   u  0  0 0 1 1   4  F4 x  0   u2 = 0 u3 = – 1.25  10–3 in.   F1x   1 1 0 0  u1  0  F   1 2 1 0  u  0  2x   6   2   = 4  10  –3  F 0 1 2 1     u 1.25 10    x 3  3       F4 x   0 0 1 1 u4  0 

 F1x = 0 F2x = 5000 lb F3x = – 10000 lb F4x = 5000 lb (1) f1(1) 1 1 0   f1x  x  0 lb 6      (1)  = 4  10    1 1 0  f 2(1)  f 2 x  x  0 lb

0  f 2(2) f 2(2) 1 1    x  x  5000 lb 6    (2)  = 4  10      3 (2)  1 1 1.25  10  f3 x   5000 lb  f3 x   f3(3)  f3(3) 1 1 1.25  103  x  x   5000 lb 6  = 4  10    (3)   1 1  (3) 0   f 4 x  5000 lb   f 4 x  3.5

Element 1–2  1 1 [k1–2] = 5  106   1 1 Element 2–3 29

1  2 1 2

 1 [k2–3] = 5  106  21 2

 1

1

Global [K] = 5  10  1  0 

3 2  12

6 

0

– 1 2  1 2

{F} = [K] {d} and u1 = 0  F1x  ?   1 1   6  3  F2 x  0  = 5  10  1 2  F  20000  1  3x   0 2

0 u1  0  1  u  2  2 1    u3  2

3 1 u2 – u3  u3 = 3 u2 2 2 1 1 u3]  20000 = 5  106 [ u2 + 2 2

 0=

(1) (2)

Substituting (1) in (2) (2  20000) 5  106 

= – u2 + 3 u2

u2 = 0.004 in.

u3 = 3 (0.004)  Element 1–2

 u3 = 0.012 in.

f1x   20000 lb 1 1  0   f1x  6    = 5  10      f 2 x  20000 lb  1 1 0.004   f2 x 

Element 2–3  1  f2 x  2 6   = 5  10   1  f3x   2

1  2 0.004     1  0.012  2

f 2 x   20000 lb f3 x  20000 lb

 0    F1x = 5  10 [1 –1 0] 0.004  0.012    6



F1x = – 20000 lb

3.6

 1.2 1.2  [k1–2] = 106    1.2 1.2  30

 0.667  0.667  [k2–3] = [k2–4] = 106  0.667    0.667 {F} = [K] {d} 0 0  1.2 1.2  u1  0   F1x  ?   1.2 2.533 0.667 0.667  u  ?    F  16000  = 106    2   2x   0 u 0  0.667 0.667 0      3  F  ?    3x  0 0.667  u4  0 0.667  0  F4 x  ?   u2 = 6.32  10–3 in. Reactions F1x = (– 1.2  106) (u2)  F1x = – 7579 lb F3x = (– 0.667  106) (u2)  F3x = – 4210 lb F4x = (– 0.667  106) (u2)  F4x = – 4210 lb

0 f1(1) 1.2 1.2     f1x  x   7579 lb 6     = 10      3 (1)  1.2 1.2  6.32  10   f2 x  f 2 x  7579 lb

f 2(2) 0.667 0.667  6.32  103   f2 x  x  4210 lb 6   = 10     0.667 0.667   (2) f 0    3x  f3 x   4210 lb  f3(3) 0.667 0.667  6.32  103   f3x  x  4210 lb 6     = 10     (2) 0  f4 x   0.667 0.667   f 4 x   4210 lb  3.7

 1 1 [k1–2] = [k3–4] = 1.5  106   1 1  1 1 [k2–3] = 20000   1 1 0 0  u1  0   1500 1500  F1x  0   1500 1505  u  ?  5 0   F  0  = 103    2   2x   ? u 0  5 1505  1500      3   F 10000    3x  0 0 1500 1500  u4  0   F4 x  0 

u2  8.70  105 in. 1505 5  u2   0  3     = 10      5 1505 u3  10000 u3  6.58  103 in. Reactions F1x = (– 1500  103) (u2)  F1x = – 130 lb F4x = (– 1500  103) (u3)  F4x = – 9870 lb 31

0 f1(1) 1 1    f1x  x   130 lb 6  = 1.5  10        5   1 1 8.70  10   f2 x  f 2(1) x  130 lb 5 f 2(2)  f2 x   1 1 8.70  10  x   130 lb = 20000         f3x   1 1 6.58  103  f3(2) x  130 lb

f3(3) 1 1 6.58  103   f3x  x  9870 lb 6  = 1.5  10        (3) 0  f4 x   1 1  f 4 x   9870 lb  3.8

[k(1)] =

(4  104 m 2 )(200  106 kN2 )  1 1 m 1 1 1m  

 1 1 kN [k(1)] = 800  102    1 1 m (2  104 m 2 )(70  106 kN2 )  1 1 m [k ] = 1 1 1m   (2)

 1 1 kN [k(2)] = 140  102    1 1 m 0  u1  0  F1x  0   800 800     2   F2 x  = 10  800 940 140 u2   F   40 kN   140 140  u3  0  3x   

0 = 102 (940 u2 – 140 u3)  u3 = 6.741 u2

 – 40000 = 102 (– 140 u2 + 140 u3) Substituting (1) into (2)  – 40000 = 102 (– 140 u2 + 140 (6.714) u2) 

u2 = – 0.50  10–3 m



u3 = – 3.356  10–3 m

F1x = 102 (– 800  (– 0.50  10–3))  F1x = 40 kN

(1) (2)

0 f1(1) 1 1    f1x  x  40 kN 2  = 800  10        3   1 1 0.50  10   f2 x  f 2(1) x   40 kN f 2(2) 1 1  0.50  103   f2 x  x  40 kN 2  = 140  10      1 1  (2) 3 f   3.356  10   3x  f3 x   40 kN 3.9

 1 1 kN [k1–2] = [k 2–3] = 4.2  104   1 1 m {F} = [K] {d}  F1x  0   1 1 0  u1  0       4   F2 x   5 kN  = 4.2  10  1 2 1 u2  ?  F  ?  u  0.025 m    0  1 1   3  3x   

5 kN 4.2  104 

= 2 u2 – 1(0.025)

u2 = 0.01244 m 0.01244  F3x = 4.2  104 [–1 1]    F3x = 527.5 kN 0.025   0  F1x = 4.2  104 [1 –1]    F1x = – 522.5 kN 0.01244 

Element forces (1) f1(1) 1 1  0   f1x  x   522.5 kN 4    (1)  = 4.2  10     (1)  1 1 0.01244  f 2 x  522.5 kN  f 2 x  (2) f 2(2) 1 1 0.01244   f 2 x  x   527.5 kN 4    (2)  = 4.2  10     (2)  1 1 0.025  f3 x  527.5 kN  f 3 x 

3.10

 1 1 [k(1)] = [k(2)] = 7000   1 1 33

 1 1 [k(2)] = 2000   1 1 {F} = [K] {d} 0 0 u1  0   7 7  F1x  ?   F  16kN   7 14 7 0 u2  ?   2x  3     = 10  9 2  u3  ?   0 7  F3 x  0     0 0 2 2  u4  0  F4 x  ?  

16 = 103 [14 u2 – 7 u3] 0 = 103 [– 7 u2 + 9 u3] 7  u3 = u2 9 Substituting (2) into (1) 

8 103

= 14 u2 – 7 

(1) (2)

7 u2 9



u2 = 1.870  10–3 m



u3 = 1.454  10–3 m

Element (1) 0 1 1    f1x  13.10   f1x  3        kN   = 7  10    3  1 1 1.870  10   f 2 x   13.10   f2 x  Element (2)

1 1 1.870  103   f 2 x   2.90   f2 x  3    = 7  10     kN   1 1 1.454  103   f3 x  2.90   f3x  Element (3) 1 1 1.454  103   f3 x   2.90   f3x  3    = 2  10     kN  0  1 1   f4 x    f 4 x  2.90  0   F1x = 103 [7 –7]   F1x   13.10 kN 3  1.87  10  1.454  103  F4x = 103 [–2 2]    F4 x   2.90 kN 0  

3.11

 1 1 [k1–2] = [k2–3] = [k2–4] = [k2–5] = 2.1  107   1 1 {F} = [K] {d}

 F1x  0   1  F  60 kN   1  2 x   7  F3 x  0  = 2.1  10  0  F  0   0  4x   0  F5 x  0  

1 0 0 0 u1  0   4 1 1 1 u2    1 1 0 0 u3  0   1 0 1 0 u4  0   1 0 0 1 u5  0 

u2 = 7.144  10–4 m

Reactions F1x = (2.1  107) (– 1) (u2)  F1x = – 15000 N F3x = (2.1  107) (– 1) (u2)  F3x = – 15000 N F4x = (2.1  107) (– 1) (u2)  F4x = – 15000 N F5x = (2.1  107) (– 1) (u2)  F5x = – 15000 N Element forces f1  2 =  f 2  3 =  f 2  4 =  f 2  5 = (2.1  107) (u2)



f1  2 = 15000 N f 2  3 = – 15000 N f 2  4 = – 15000 N

f 2  5 = – 15000 N 3.12

du P = E A( x) dx u= 

P dx A( x) E

u= 

P dx A0 1  Lx  E

= 

PL dx A0 L 1  Lx  E

= 

PL dx A0  L  x  E

= 

PL du A0 E u

PL 1 du A0 E  u

PL ln u A0 E

(Change variable u = L + x and du = dx)



u= u=

PL ln (L + x) A0 E ( 1000) (20) 2  10  106

ln (20 + x)

u = –10–3 ln (20 + x) u(x = 0) = (– ln 20)  10–3 = – 2.996 10–3 in.

u(x = 10) = (–ln (20 + 10)) (–10–3) = – 3.401  10–3 in. Two elements

  

A A

   

L   1 5 L = A0 1  4  = A0 1  = A0 4 4 L 4

3L   3 3 7 L = A0 1  4  = A0 1  = A0 4 4 L 4

[k(1)] =

5 A0 E  1 1 4 L2 1 1

[k(2)] =

7 A0 E  1 1 4 L2 1 1

 45 P  A E   0  5  0  = L 4 2  F   3x  0 

A0 E



A0 E



u2 =

L 2

5 4

5  74 4 7 4

0  u1   7   u  4  2 7   u 3 0    4 

(

5 5 u1 – u2) = – P 4 4

(

5 u1 + 3 u2) = 0 4

5 5 u1 = u1 12 43

Substituting (2) into (1)

  

A0 E L 2

(

5 5 5 u1 – ( u1)) = – P 4 4 12

  

12 5  5 5  u =  PL 12 4 4 12  1 2 A E 0  60  25  u = – PL 1  48  2A E 0



– PL u1 = 2 A0 E

24 48 35 36

(1)

(2)



u1 =

– PL 24 A 0 E 35



u2 =

5 24   PL  12 35  A0 E 



u2 = 

2 PL 7 A0 E

A0 = 2 in., L = 20 in., E = 10  106 psi P = 1000 lb (1000) (20) 24 u1 = –  2(10  106 ) 35

Now



u1 = – 0.6857  10–3 in.



u2 =



u2 = – 0.2857  10–3 in.

5 (– 0.6857  10–3) 12

One element L   1 3 A = A0 1  2  = A0 (1 + ) = A0 2 2 L

 P    = 0

1 1 u1   1 1 u  0   2 

3 2 A0 E 

 PL



u1 =



u1 = –



u1 = – 0.667  10–3 in.

3 2 A0 E

2 (1000)(20) 3 (2)(10  106 )

3.13 1

l 2

x l/2

u = a1 + a2x + a3x2

(A)

u(0) = u2 = a1

(1)

u(–

l l l ) = u1 = u2 + a2 (– ) + a3 (– )2 2 2 2

(2)

l l l ) = u 3 = u 2 + a 2 ( ) + a 3 ( )2 2 2 2

(3)

Solving for a2 and a3 from (2) and (3)

a2 =

u3  u1 2(u1  u3  2u2 ) , a3 = l l2

(4)

By (1) and (4) into (A) 2(u1  u3  2u2 ) 2  u  u1  u = u2 +  3 x+ x l  l2

(5)

u = [ N ] {d }

(6)

  x 2 x2  2 u=   l l

{} =

4 x2 1 2 l

u1  x 2 x2     2  u2  l l   u3 

(7)

u N = [B] {d} = {d } x x

(8)

Using (7) in (8) 1 4x {} =    2  l l

8 x l2

1 4x  [B] =    2  l l

8 x

l /2

[K] = A 

–l /2

u1  1 4x     u2  l l 2    u3  1 4x   l l 2 

[BT] E [B] dx

(9)

(10) (11)

A = cross sectional area of the bar E = Young’s Modulus of the bar

3.14 Given u = a + bx2 for 2 noded bar =

du = 2bx dx

u(0) = u1 = a u(L) = u2 = u1 + bL2 

u2  u1 L2

u u u = u1 +  2 2 1  x2  L  This displacement function allows for a rigid body displacement as the a = u1 term does this. Also should allow for constant strain, but have  = 2bx or a linear strain. Therefore, not complete. Need to complete 2nd degree polynomial and 3rd node for compatible function. Try

u = a1 + a2 x + a3 x2

du = a2 + 2a3x dx ‘a2’ allows for constant strain term. 3.15 (a) 38

1 2

,S=

C 2  EA  [K] = L    

1 2 CS

C 2

CS

–CS   S 2   CS  S 2 

 1 1 1 1   1 1 1 1 lb [K] = 2.25  106  1 1 1 1  in.   1 1 1 1 

(b)

C = 0, S = 1 0 0 15  10  1 0 1 [K] = 0 0 15  0 1 6

0 0 106 0 1 [K] = 4 0 0  0 1

0 0  0 1 0 0  0 1

0 0  0 1 lb 0 0  in.  0 1 39

(c)

3 , 2

S=–

 3  4 4   3 6 (210  10 )(4  10 )  4 [K] = 3  3  4  3  4

1 2  43

 34

1 4

3 4

 14

 43

   14    43   1  4  3 4

 3  3 3 3    3 3 1  kN 1 K = 7000   3  3 m 3 3    3 1  3 1  (d)

C = 0.9397 S = 0.3420

C2 = 0.883

CS = 0.321

S = 0.117

 0.883 0.321 0.883 0.321 (70  104 )(2  104 )  0.321 0.117 0.321 0.117  [K] =  0.883 0.321 0.883 0.321  1    0.321 0.117 0.321 0.117  40

 0.883 0.321 0.883 0.321   0.321 0.883 0.321 0.883 N [K] = 1.4  107   0.883 0.321 0.883 0.321  m    0.321 0.883 0.321 0.883 

3.16 (a)

C = 0.707

u1 = 0.5 in.

v1 = 0.0 in.

S = 0.707

u2 = 0.25 in.

v2 = 0.75 in.

u1 = u1 C + v1 S = 0.5 (0.707) + (0.0) (0.707) 

u1 = 0.3536 in. u 2 = u2 C + v2 S = (0.25) (0.707) + (0.75) (0.707) u 2 = 0.707 in.

(b)

3 , 2

1 2

S=–

u1 = u1 C + v1 S =



 

 1   3  + (0)  1    2  2  2

u1 = 0.433 in.

u 2 = u2 C + v2 S =



 1   3  +  3    4  2   4

  

1 2

u2 = – 0.1585 in.

3.17

u1 = 0.0

u2 = 5.0 mm

E = 210 GPa

v1 = 2.5 mm L=3m

v2 = 3.0 mm

A = 10  10–4 m2

(a) We know that {d} = [T] {d} C  S [T ] =  0  0

0 0

C S

0  0 S  C

C = cos 120° = – 0.5, S = sin 120° = 0.866

 0.5 0.866 0 0  0.0  u 1     0 0  0.0025  v   = 0.866 0.5      0 0.5 0.866  0.005  0     u 2   0 0 0.866 0.5  0.003  v    2.165  0.002165 u 1      1.25  0.00125    v   =   mm  m=  0.098 0.000098     u 2  5.830 0.00583  v   (b) C = cos (– 30°) = 0.866, S = – 0.5

0.866 0.5  0 0  0.0 u 1     0 0  0.0025  v   =  0.5 0.866      0 0 0.866 0.5  0.005     u    2  0 0 0.5 0.866  0.003  v   1.25  u 1     v   =  2.165 mm        3.03   u  2  5.098 v  

3.18

E [– C – S L

(a)  =

=

2 , 2

u1  v    C S]  1  ,  45 u2  v2  2 , E = 30  106 psi, L = 60 in. 2

30  10  2  60  2



2 2

 0  2   0    2  0.02  0.04 

  = 21200 psi (b) C =

3 1 , S = , E = 210 GPa, L = 3 m, θ = 30° 2 2

 =  23

 12

  = 45470

3 2

0.25   210  106 –3 1 0   10     2 3 1.00   0 

  = 45.47 MPa

3.19

10 lb

 f1x   0  f     1 y  10  f2 x   f2 x       f2 y   f2 y  {f} =     and f   f3 x   3 x   f 3 y   f3 y       f4 x   f4 x  f   f   4y   4y 

 u1  v   1 u2  v    {d} =  2  u3   v3    u4   v  4

(a) For element 1–3;  = 180°

 f1x   0  1 f      1y  10 0   = f  =k  f  1  3x   3x    f3 y   f3 y   0  

0 1 0 u1    0 0 0  v1    0 1 0  0   0 0 0  0 

For element 1–4;  = 225°

 f1x   0   1 1 1 1 u1  f        1y  10 k 1 1 1 1  v1      = f  =   4 x  2  1 1 1 1   0   f4x     f4 y   f 4 y   1 1 1 1   0    For element 1–2;  = 135°

 f1x   0   1 1 1 1 u1  f       10 k  1 1 1 1  v1   1y  = = f       2 x  2  1 1 1 1  0   f2x     f 2 y   f2 y   1 1 1 1  0   

Total [K]

1 1  2 0 2 2 1  1 1 2 0  0 1 2  1 1 1  12 0  2 2 2  1 1  12  12 0 2 [K] = k  2  0 0 0 1  1  0 0 0 0 0  1 1    2 0 0 0 2  1  1  2 0 0 0 2

1 0 2 1 0 2

0 0 0

0 0

1 2 1 2

 12    12   0  0  0 0 1 2 1 2

(b) Applying boundary conditions

u4 = v4 = u2 = v2 = u3 = v3 = 0 [K] is reduced to 2 0 [K] = k   0 1  f1x   2 0 u1   0  2 0  u1    = k   = k         0 1 v1   0 1  v1   10  f1 y   u1 = 0

v1 =

10 k

3.20

Element 1–2

2 3 ; S= ; L1–2 = 2 2

 12  1 A1 E1  2 [k1–2] =  L1 2  12  1  2

1 2 1 2  12  12

 12  12 1 2 1 2

2 L  12    12  1  2  1  2 

Element 2–3 45

2 2 ; S=– ; L2–3 = 2 2

 12  1 A2 E2  2 [k2–3] =  L2 3  12  1  2

 12

1 2 1 2  12

2 L

1  2   12    12  1  2 

Applying the boundary conditions, the global equations are:

1  10  103  12  12 0   =  2  100  12  12 10

1  12  u2  2  1  12   v2  2

 u2 = 0

v2 =

10  2  100 10  103

 v2 = 0.283 in.

{ f } = [k ]{d }  [k  ] [T *]{d } AE  1 1 C  f 1x    = L  1 1   0  f 2 x 

10  10  C  2  100  C

S 0 0 C

S S

C

 u1    0   v1    S  u2   v2   0    S   0    S  0  0.283

f 1x =

 10  103  2 (0.1415)  = – 7.07 kips    2 2  100

f 2 x =

 10  103  2 (0.1415)  = 7.07 kips   2  100 2

1–2 =

f 2 x 7.07 kips =  1–2 = 707 psi (T) A 10 in.2

2–3 =

f 3 x A

7.07 kips 10 in.2

 2–3 = 707 psi (T)

3.21

Element 1–2 L1–2 =

[k1–2] =

2 3

L;  = 60°

 1  4 3 3 AE  4  2L   1  4  3  4

3 4

 14

3 4

 43

1 4

 14

3 4

 43    34   3  4  3  4 

Element 1–3 L1–3 = L;  = 90° 0 0 0 0  AE 0 1 0 1 [k1–3] = L 0 0 0 0    0 1 0 1 Element 1–4 2 L1–4 = L;  = 120° 3  1  43  14  4 3 3  3 4 4 [k1–4] = 3 AE  4 3 1 2L   1 4 4  4 3 3  3    4 4 4

3 4

  43   3  4 

Applying the boundary conditions u2 = v2 = u3 = v3 = u4 = v4 = 0 [ K] =

3 1 3 1 AE  2 ( 4 )  0  2 ( 4 )  L  3 3 0 3  3 2 4 2 4

 



AE  4  = L  0

  1  3 4 3 



 

 34 

3 2

 43   0  23   43  

3 3 ( )  1  23 ( 43 ) 2 4

 

3  F1x  AE  4 0  u1      = 3 3 v L   F1 y   0 1  4   1  3 AE  4 100 0  u1      =   3 3 v L  100  0 1  4   1 

 u1 = v1 =

400 L 3 AE

 u1 =

400 L

231 L 231(100) 4 = = 7.7  10 in. 6 AE 30 10

 v1 =

(3 3  4) AE

43.5 L 43.5(100) 4 = = 1.45  10 in. AE 30 106

3.22

 C2 AE  CS [K] = [TT] [k  ] [T] = L  C 2  CS

C 2

CS C2 CS

S CS S 2

CS   S 2  CS   S2 

For element 1;  = 120°  1  4 3 AE  4  [k(1)] = 2L   1  4  3  4

 43

 14

3 4

1 4

 43

   34    43   3  4  3 4

For element 2;  = 180°  1 0 1 0  AE  0 0 0 0  [k(2)] = L 1 0 1 0     0 0 0 0 For element 3;  = 210° 3  3  34 4  4 3 1  43 3 AE  4 4 (3)  [k ] = 3 2L 3  3 4 4  4 3  3  1  4 4 4

 43    14   3  4  1  4 

Applying the boundary conditions 48

u2 = v2 = u3 = v3 = u4 = v4 = 0 3 3 1 AE  8  1  8 1000  =   L  3  0  3 1000  8 8

 83  0  83  u1    3 3   v1  0   8 8  AE 1.77 0.16  u1  1000  =   L 0.16 0.59   v1  1000  u1 =  v1 =

422 (100)

 u1 = 0.00422 in.

1  10  106 1570 (100)

 v1 = 0.0157 in.

1  10  106

Element (1)  1  f1x   4 f  3 AE   4  1y   =   L  1  f2x   4  f2 y     3  4

 43

 14

3 4

1 4

 34

 43

 L   422 AE     34  1570 L  AE    43   0    3  0   4 3 4

 f2x = 287 lb f2y = – 497 lb f (1) =

f 2 x 2  f 2 y 2  f (1) = 5741 lb (C)

(1) =

– 5741 f (1) psi = A A

 (1) = – 5741 psi (C) Element (2)  f1x  1 f  AE  0  1y    = L  1  f3 x    f3 y  0  

L 0   422 AE     L  0  1570 AE   0  0   0   0 

0 1 0 0 0 1 0 0

f3x = – 422 lb f3y = 0 lb f (2) =

f3 x 2  f3 y 2  f (2) = 422 lb (T)

(2) =

f (2) 422 psi  A A

 (2) = 422 psi (T)

Element (3)  f1x  f   1y    =  f4x   f4 y   

 3  4 3 3 AE  4  2L  3  4  3  4

3 4

 34

1 4

 43

3 4

 14

3 4

 43   422 L   AE    14  1570 L  AE   3   0  4   1  0  4 

 f4x = – 862.8 lb f4y = – 496 lb f (3) =

f 4 x 2  f 4 y 2  f (3) = 996 lb (T)

f (3) 996  psi (T) A A (3) = 996 psi (T)

(3) =

3.23

Element (1) C=

1 3 ; S= 2 2

3  14 4  3 3 AE  4 4  [k(1)] = L   – 

– 1 4 3 4

    3 4  3  4 

Element (2)  14 – 43  3 3 AE  – 4 4  [k(2)] = L   – 

– 1 4

– 43

    – 43   3  4 

{F} = [K]{d} 1 AE  2 0  u1  6000  =      L  0 23   v1   0  AE u1  6000 = L 2

 u1 =

6000  100  2 1  10  106

 u1 = 0.12 in. v1 = 0

(1) = [C]{d} =

 (1) =

E 1  L 2



3 2

1 2

u2  0    3  v2  0   u  0.12  2  1  v1  0 

10  106  1 (0.12)  2    2 10

 (1) = 6000 psi

3.24

L2–1 = 20

L2–3 = 15

L2–4 = 25

2–1 = 180°

sin 2–4 = 0.6

cos 2–1 = – 1

2–3 = 90° sin 2–3 = 1 cos 2–3 = 0

cos 2–4 = – 0.8

L1–4 = 15

L1–3 = 25

L3–4 = 20

1–4 = 90° sin 1–4 = 1 cos 1–4 = 0

1–3 = 36.87° sin 1–3 = 0.6 cos 1–3 = 0.8

3–4 = 180° sin 3–4 = 0 cos 3–4 = – 1

sin 2–1 = 0

2–4 = 143.13°

Boundary conditions u1 = v1 = u4 = v4 = 0 (2) x [k2–1] =

AE  1 20  0  1  0

AE 0 0 0 1 0  (2) ; [k2–3] = 15  0 0 0 0 1  0 0 0 1 0  (1)   0 1 0 0 0

[k3–4] =

AE  1 20  0   1  0

(3) y

AE 0 0 0 1 0   (3) ; [k1–4] = 15  0 0 0 0 1 0 0  0 1 0   (4) 0 1 0 0 0

(4) y

0  (2) 0 1 0 0  (3) 0 1

(1)

(4)

(3) x

(2)

(1) y

0  (1) 0 1 0 0  (4) 0 1

(2)

(4)

x [k2–4] =

AE  0.64 0.48 0.64 0.48  (2) 25  0.48 0.36 0.48 0.36     0.64 0.48 0.64 –0.48    (4)  0.48 0.36 –0.48 0.36  (1) x

[k1–3] =

(3) y

AE  0.64 0.48 0.64 0.48  (1) 25  0.48 0.36 0.48 0.36    0.64 0.48 0.64 0.48    (3)  0.48 0.36 0.48 0.36 

{F} = [K] {d}  F1x  ?   0.0756 0.0192 0.05 0 0.0256 F  ?   y 1 0.0811 0 0 0.0192    0.0192  F2 x  0   0.05 0 0.0756 0.0192 0     F2 y  1000 0.0192 0.0811 0 0  0   = AE   F 0   0.0256 0.0192 0 0 0.0756 3 x     F3 y  1000  0.0667 0.0192 0 0.0192 0.0144    0  F ? 0.0256 0.0192 0.05 0  4x   F  ?    0 0.0667 0.0172 0.0144 0  4y  0.0192

 u1  0    0.0144 0.0667  v1  0  0 0.0256 0.0192  u2  ?  0   0.0667 0.0192 0.0144  v2  ?    0.05 0.0192 0  u3  ?   0.0811 0 0  v3  ?    0 0.0756 0.0192  u4  0 0 0.0192 0.0811  v4  0  0

 0 = [0.0756 u2 – 0.0192 v2 + 0 u3 + 0 v3] AE  u2 = 0.254 v2

(1)

1000 = [– 0.0192 u2 + 0.0811 v2 + 0 u3 – 0.0667 v3] AE 0 = [0 u2 + 0 v2 + 0.0756 u3 + 0.0192 v3] AE  u3 = – 0.254 v3

(2)

1000 = [0 u2 – 0.0667 v2 + 0.0192 u3 + 0.0811 v3] AE 1000 = [–0.0192 (0.254 v2) + 0.0811 v2 – 0.0667 v3] AE  1000 = [0.0762 v2 – 0.0667 v3] AE

(3)

1000 = [ –0.0667 v2 + 0.0192 (– 0.254 v3) + 0.0811 v3] AE  1000 = [– 0.0667 v2 + 0.0762 v3] AE Multiplying (4) by 

0.0762 0.0667

 52

(4)

 1142.4 = [–0.0762 v2 + 0.0870 v3] AE

(5)

Adding (3) and (5) 2142.4 = [0 v2 + 0.204 v3] AE  v3 =

105021 AE

(6)

Substituting (6) into (3)





105021  1000 = 0.0762 v2  0.0667 AE  AE   v2 =

105021  v2 = v3 AE

Substituting in (1) and (2)



u2 = 0.254 v2 = 0.254  u2 =

105021 AE

26675 AE

u3 = – 0.254 v3 = – 0.254  u3 = –





105021 AE



26675 AE

Going back to the local stiffness matrices AE u2    f 2 x  1 u2 20  f2 x  AE 1 0 1 0  v2    =   26675 AE 20 0 0 0 0  u1  0   f2 y   f2 x  v1  0  20 AE  f2x = 1333 lb ; f2y = 0 lb f1–2 =

( f 2 x )2  ( f 2 y )2  f1–2 = 1333 lb (T)

Member 1–3 u1  0 f  3x  AE  0.64 0.48 0.64 0.48  v1  0      = 25  0.48 0.36 0.48 0.36  u3   f3 y   v3 













26675 105021  AE   0.48  f3x =  0.64  AE AE  25  f3x = – 1333 lb



26675 105021  AE  f3y =  0.48  0.36  AE AE  25 

 f3y = – 1000 lb 53

f1–3 =

( f3 x ) 2  ( f3 y )2  f1–3 = 1667 lb (T)

Member 2–4 u2    f v  2x  AE 0.64  0.48 0.64 0.48   2 =     u  0  f 25  0.48 0.36 0.48 0.36  4  2y   v4  0  

 f2x = 1333 lb (C) f2y = 1000 lb (C) f2–4 =

( f 2 x )2  ( f 2 y )2

 f2–4 = 1667 lb (C) Member 2–3 u2  f f2 x  0  2x  AE 0 0 0 0   v2    =     f2 y  0 15 0 1 0 1 u3   f2 y   v3   f2–3 = 0 lb Member 3–4 u3    f 3 x   1333 lb v  f3 x  AE 1 0 1 0  3  =      f  0 lb   20 0 0 0 0 u4  0 3y  f3 y  v4  0  f3–4 =

( f3 x ) 2  ( f3 y )2  f3–4 = 1333 lb (C)

Member 1-4

u1  0   f4 x  AE 0 0 0 0 v1  0    f  = 15 0 1 0 1  u4  0  4y  v4  0   f1–4 = 0 lb

3.25 The global stiffness matrix is changed since matrix [k2–4] is not incorporated in (1) (2)  F1x  0  0.0192 0.05 0 F  0   0. 0756  1y   0.0192 0.0811 0 0  F2 x  0     0 0.05 0   0  05  F2 y  1000  0 0 0 0.0667   = AE   F3 x  0   0.0256  0.0192 0 0  F3 y  1000   0  0.0667    0.0192  0.0144  F4 x  0   0 0 0 0 F  0    4y  0 0 0  0  0667 

0. 0256  0.0192 0 0 0.0756 0.0192  0.05 0 

(3)

(4)

0.0192

0.0144

 u =0 1  0.0667  v = 0   1  0 0 0  u   2  0 0   v2  0.0667    0.0192  0.05 0  u3   v  0.0811 0 0   3 u4 = 0   0 0.05 0  v = 0  0 0 0.0667   4

 0 = [005 u2 + 0 v2 + 0 u3 + 0 v3] AE u2 = 0 (1)



1000 = [0 u2 + 00667 v2 + 0 u3 – 00667 v3] AE

(2)

0 = [0 u3 + 0 v2 + 00756 u3 + 00192 v3] AE

(3)

 u3 = – 0254 v3

(3)

1000 = [0 u2 – 0.0667 v2 + 00192 u3 + 00811 v3] AE

(4)

Adding (2) and (4) 2000 = [00192 u3 + 00144 v3] AE Substituting (3) in (5) 2000 [00192 (– 0254 v3) + 00144 v3] AE 

 v3 =

210000 AE

 u3 = (– 0254)

 53340 (210000)  u3 = AE AE

Substituting in (2) 210000    1000 =  0.0667 v2  0.0667   AE  AE    

v2 =

224993 AE

Forces on members Member 1–2 u2  0   f2 x  0  f2x  AE 1 0 –1 0  v2  f  =    f  0   20 0 0 0 0  u1  0  2y  2y  v1  0   f1–2 = 0 Member 1–3

(5)

u1  0 f  1x  AE 0.64 0.48 –0.64 –0.48  v1  0    f  = 25 0.48 0.36 –0.48 –0.36  u3  1y   v3 

AE   53380   210000   – 0.64   – 0.48     25  AE AE  



f1x =



f1x = – 2666.5 lb f1y =



AE   53380  210000   – 0.48  – 0.36   AE   AE   25 

f1y = – 2000 lb

( f1x ) 2  ( f1 y ) 2

f1–3 =

 f1–3 = 3333 lb (T)

Member 2–3 u2  0   f  2x  AE 0 0 0 0 v2  f  =     u 0 1 0 –1 15    3  2y   v3   f2x = 0  f2y =

AE  (224993) (210000)  1 – 15  AE AE 

 f2y = – 1000 lb (C) f2–3 =

( f 2 x )2  ( f 2 y )2

 f2–3 = 1000 lb (C)

Member 3–4 u3    f3x  AE 1 0 – 1 0  v3    f  = 20 0 0 0 0  u4  0  3y  v4  0   f3x = 1

(  53340) AE  f3x = – 2666.5 lb (C) AE 20

 f3y = 0 f3–4 =

( f3 x ) 2  ( f3 y )2  f3–4 = 2667 lb (C)

Member 1–4 u1  0  f1x  0 0  v1  0   f1x  AE 0 0 0    f  f  =   0 15 0 1 0 – 1 u4  0 1y  1y  v4  0  56

 f1–4 = 0 lb

3.26

Since both elements 2–4 and 1–3 are removed, the global stiffness matrix will change (1) (2) (3) (4)  F1x  ?  F  ?  0 0 0 0 0 0  u1  0  0.05  0.05  1y   0 0.0667 0 0 0 0 0  0.0667 v1  0   F2 x  0       0 0.05 0 0 0 0 0  u2  ?  0.05   F2 y  1000  v ?  0 0 0.0667 0 0   0.0667 0    = AE  0   2  F 0  3 x   ? u  0 0 0 0 0.05 0 0.05 0    3   F3 y  1000   0 0 0 0 0.0667 0 0  v3  ?   0.0667       F4 x  ?  0 0 0 0 0.05 0  u4  0  0.05  0 F  ?   0 v4  0  0.0667 0 0 0 0 0 0.0667    4y  

0 = 005 u2

 u2 = 0

1000 = 00667 v2 – 00667 v3 0 = 005 u3

(1)

 u3 = 0

1000 = – 00067 v2 + 00667 v3

(2)

Adding (1) to (2) 2000 = 0v2 + 0v3 The matrix, therefore, is singular and we get an inconsistent equation. (The K is also 0). Of course this should have been expected since the truss is unstable.

3.27

(1)

(2)

(3)

1–3

2–3

3–4

cos 

50 ft 50 ft 20 ft 53.13 126.87 90  0.6 0.6 0

sin 

0.8



0.8

(1) (3)  0.48  0.36 0.48 0.36  (1)  AE 0.48 0.64 0.48 0.64  (1)   [k ] = 50     (3)   (2) (3)  0.36 0.48  0.36 0.48  (2) 0.48 0.64 AE 0.48 0.64  (2)   [k ] = 50     (3)  

(3)

(4)

0 0 0 0  (3) AE 0 1 0 1   [k(3)] = 20     (4)   Invoking boundary conditions. Therefore, need only 3–3 (0.36  0.36)  AE (0)  AE 50 20 [K] =  AE AE   (0.48 0.48) (0)  50 20

(0.72) AE 50  K=  0 

AE (0.48  0.48)  AE (0)  50 20  AE AE   (0.64 0.64) (1)  50 20

0  AE  (1.28) AE  50 20 

{F} = [K] {d}

 0.72  F3 x  5 K  50   = AE  F   20 K 3 y    0

 u3     v3 

1.28 1   20  50



5K=

(0.72) ( AE ) u3 50 (5000)  (50)  (12) (0.72) (3) (30  106 )



 u3 =



 u3 = 0.0463 in. 

1 – 20 K =  1.28  20  AE v3 50

 v3 = – 0.0352 in. Forces on the members (A = 3 in.2 , E = 30 psi) Member 1–3 (1) u1  0 f  0.8  0.6 0.8 v1  0  AE  0.6  1x      =  0.8 u3 600 0.6  0.8 0.6  f 3 x   v3   f 1 x (1) = 69.9 lb Member 2–3 (2) u2  0 0.8 0.6 0.8 v2  0  AE   0.6  f 2 x      =  0.8 u3 600  0.6  0.8  0.6  f 3 x   v3   f 2 x(2) = 8400 lb Member 3–4 (3)

u3    AE 0 1 0  1 v3   f 3 x      =   u  0 f  1  4 240 0 1 0  4x   v4  0   f 3 x =

AE (– 00352) 240

 f 3 x = – 13240 lb 3.28

0 0  C S  S C 0 0  [T ] =  C S  0 0    0 0  S C 0 C  S 0 S C 0 0  [TT] =  0 C  S 0   C 0 S 0 59

0 0  C  S 0 0  C S   S C  S C 0 0 0 0    [T]TT] =  C S  0 0 C  S  0 0     C 0 S  0 0  S C  0  C2  S2   CS  CS T  [TT ] =  0   0 

But

 SC  SC 2

S C

C2  S 2

 CS  CS

1 0 [T]TT] =  0  0

0 1 0 0

0 0 1 0

0 0 Identity   = [I] 0 matrix  1

1 0 [T]TT] =  0  0

0 1 0 0

0 0 1 0

0 0  =I  0  1



  0    SC  SC  S 2  C 2  0

0 C  S 0 S 0 0 C  [TT] = [T–1] =  0 C  S 0   C 0 S 0

3.29

AE (4  104 ) (210  106 ) kN = = 280  102 L 3 m

Element (1) C = 0, S = 1

(1)

(2)

0 0 1 [k(1)] = 280  102 0  0 0  0  1

0 0 0  1  0 0  0 1

Element (2)

2 2 ,S= 2 2 (1)

 12  [k(2)] = 280  102  12  1  12  2

(3)

1 2 1 2 – 12  12

 12 – 12 1 2 1 2

 12    12  1 2 1  2

Element (3)

C = 1, S = 0  1  0 [k(3)] = 280  102  1   0

0 1 0  0 0 0  0 1 0  0 0 1

Boundary conditions

u2 = v2 = u3 = v3 = u4 = v4 = 0 11  F1x   20 kN  2  2   = 280  10  1  F1 y   40 kN   2

1 u 2   1 1  1 2  v1 

 u1 = – 1.76  10–4 m

v1 = – 8.93  10–4 m Element stresses

(1) =

210  106 [0 – 1 3

  1.76  104     8.93 104  0 1]   0     0  

 (1) = 62.5 MPa (T)

(2) = 70  106  22

 22

2 2

  1.76  104    4 2    8.93  10   2   0     0  

 (2) = 52.9 MPa (T)

(3) = 70  106 [– 1 0

  1.76  104     8.93 104  0]   0     0  

 (3) = 12.3 MPa (T)

3.30

Element 1–2

C = 0, S = 1 (1)

0 0 1 –4 9 0 [k1–2] = (4  10 )(210  10 )  2 0 0  1 0  2

(2) 0 0  0 – 12   0 0   0 12 

Element 1–3

C = – 1, S = 0 (1) 1 3

 [k1–3] = 84  106  0   13   0

(3) 0

 13

0 0

1 3

0  0 0  0

Element 1–4

C = – 05, S = –

3 2 (1)

(4)

0.05  0.0866  0.0866  0.05 6  0.0866  0.15  0.15 [k1–4] = 84  10 0.0866   0.05 0.0866   0.05  0.0866   0.0866 0.15   0.0866  0.15

u2 = v2 = u3 = v3 = u4 = v4 = 0 62

0.3883 0.0866  u1   F1x  0  3    = 84  10     0.0866 0.65  v1   F1 y   80  

u1 = 3.37  10–4 m v1 = – 15.1  10–4 m

Element stresses

 3.37  104    210  10  15.1 104  (1) = [0 –1 0 1]   2 0     0   9

(1) = 158.5 MPa (T)  3.37  104    210  109  15.1 104  (2)  = [1 0 –1 0]   3 0     0  



 (2) = 23.59 MPa (T)

(3) =

210  109  1  5 2

3 2



1 2

 3.37  104    3   15.1  104     2   0    0  

 (3) = – 47.85 MPa (C)

3.31

[k1–3] and [k1–4] are the same as in Problem 3.30 u3 = v3 = u4 = v4 = 0  23  F1x  0  60 7   = 8.4  10  3 3 F – 40   y 1    60 

3 3 60  9  60 

u1    v1 

u1 = 16.5  10–4 m v1 = – 7.30  10–3 m 63

Element stresses  16.5  104    210  109  7.30  103  (1)  = [1 0 –1 0]   3 0     0    (1) = 115.5 MPa (T)

(2) =

210  10 5

 1   2



3 2

1 2

 16.5  104    3   7.30  103     2   0    0  

 (2) = – 231.0 MPa (C)

3.32

0 0 0 1 [k1–2] = 7000  0 0  0 1  

0 0  1   0 0 1  ; [k1–3] = 7000  0 0  1   0 1  0

0 1 0  0 0 0  0 1 0  0 0 0

1  1 1 1  1  1   0 1 1 1  ; [k2–4] = 7000  [k2–3] = 2475  1 1  1  1 1    1  1 1 1  0 1  1  1 1 [k1–4] = 2475   1  1   1  1

0 1 0 0 0 1 0 0

0  1  1 0 0 1  1  1  ; [k3–4] = 3500  1 1 0 0   1 1 0  1

0 0  0  0

0 0 0 1  0 0  0 1

u1 = v1 = u3 = u4 = 0  Global equations are 64

2475 0  u2   0  9475  2475  9475  2475 0  v2   – 50     =   5975  3500  v3   0     5975 v4   – 50   Solving simultaneously u3 = 0.135  10–2 m v2 = – 0.850  10–2 m v3 = – 0.137  10–1 m v4 = – 0.164  10–1 m

0     6 0 70  10   1–2 = 5–6 = [0 –1 0 1]  2  0.135 10  3    0.850  102    

 1–2 = 5–6 = – 198 MPa (C)

f x1 2 = f x5 6 = 1–2  A1–2 = – 198000  3  10–4  f x1 2 = f x5 6 = – 59.5 kN

1–3 = 5–3 =

70  10 3

0     0   [–1 0 1 0]   0    0.137  101 

 1–3 = 5–3 = 0  f x1 3 = f x5 3 = 0

Similarly

2–3 = 6–3 =

70  106  2  3 2  2

2 2

 0.135  102    2   0.850  102     2   0   0.137  101   

 2–3 = 6–3 = 44.6 MPa (T)

f x2  3 = f x6 3 = 13.39 kN

2–4 = 6–4 = – 31.6 MPa (C) f x2 4 = f x6 4 = – 9.47 kN

1–4 = 5–4 = – 191 MPa (C) f x1 4 = f x5 4 = – 57.32 kN

3–4 = – 63.1 MPa (C) 65

f x3 4 = – 18.93 kN Force equilibrium Node 2

Fy = – 50 – 13.39 sin 45° + 54.5  Fy = 0 Fx = – 9.47 + 13.39 cos 45° = – 0.001 kN Node 4

Fy = – 100 + 57.3 sin 45°  2 + 18.93 Fy = – 0.003 kN Fx = 9.47 + 57.3 cos 45° – 9.47 – 57.3 cos 45°  Fx = 0 3.33 (a)

Element 1–2 ;  = 135°

C2 = 05, CS = – 05, S2 = 05

(1) (2) 0.5  0.5 0.5  0.5 (210  109 ) (5.0  104 )  0.5 0.5 0.5  0.5 [k1–2] =   5 0.5  0.5 0.5 0.5   0.5  0.5 0.5 0.5 1  1 1 1  1 1 1  1  N/m  [k1–2] = 105  105  1 1  1  1   1  1 1 1 Element 1–3 ;  = 180°

C2 = 1.0, CS = 0, S2 = 0  1 (210  109 ) (5  104 )  0 [k1–3] =  10  1   0  1  0  [k1–3] = 105  105   1   0

(1) (3) 0  1 0 0 0 0  0 1 0  0 0 0

0  1 0 0 0 0  N/m 0 1 0  0 0 0

Element 1–4 ;  = 270°

C2 = 0, CS = 0, S2 = 1.0 (1) 0 0 1 [k1–4] = 20  10 0  0 0  0  1 5 

(4) 0 0 0 1 N/m  0 0  0 1

{F} = [K] {d} Boundary conditions are

u2 = v2 = u3 = v3 = u4 = v4 = 0 The final matrix (assembled)  F1x  0   210  105 u1  = 105   3     105 125  v1   F1 y   100  10  

0 = 210 u1 – 105 v1  v1 = 2 u1

– 100 103 = 105 [– 105 u1 + 125 (2 u1)]  u1 = – 6.897  10–3 m  v1 = – 14.0  10–3 m

 6.897  10 –3    210  109   14.0  10 –3  [0707 – 0707 – 0707 0707]  1–2 =  5m 0     0    1–2 = 210.9 MPa (T)

1–3 =

210  109 [1.0 0 10

–1.0

 6.897  103      14.0  103  0]   0     0  

 1–3 = – 144.8 MPa (C) Note: Can show equilibrium at node 1



Fs = 2000



kN (– 14.0  10–3 m) m

= 28 kN f1–3 = 72.4 kN f1–2 = 105.4 kN Fy = 0 – 100 + 28 + 74.8 = 0 (b)

A = 5  10–4 m2, E = 210  109

N m2

L1 = L2 = 5 m u1

 1  4 (1) 7 [k ] = 2.1  10   3  4   

 4

3 4

  N m   

 1  4 [k(2)] = 2.1  107  3  4   

3 4

  N m   

3 4

0 (3) 3  [k ] = 4  10 0    

0 1

  N m  

Boundary conditions u2 = v2 = u3 = v3 = 0 u4 = v4 = 0 {F} = [K] {d} 0 5.25  106  5.25  106  u1   F1x  0  =      5 7 7 0 1.58  10  1.58  10  4000  v1   F1 y   1  10   Solving 0 = 1.05  107 u1 

u1 = 0

– 1  105 = 3.15  107 v1 v1 = – 0.00317 m

 (1) =

E [– C – S C L

210  10 5

 (1) = 1.155  108

 (2) =

u1  v   1 S]   v2  v2 

210  10 5

1  2 N m2  1   2

3 2



–

1 2

0    3   0.00317     0 2      0

= 115 MPa



3 2

1 2

0      0.00317  3      0 2      0

 (2) = 115 MPa 3.34

 1 AE  0 [k1–2] = 8  1   0

0 1 0 0 0 1 0 0

0 0 0  0 1 0  ; [k1–4] = AE  0 4 0 0   0  0 1

0 0 0 1  0 0  0 1

0.40   0.80 0.40 0.80  0.20 0.40 0.20  AE 0.40  [k2–4] = 0.40 0.80 0.40  8.94 0.80   0.20   0.40 0.20 0.40 0 0 AE  0 1 [k2–3] = 4 0 0   0 1

0 0  1  0 0 1 AE  ; [k4–3] =  0 0 8  1   0 1  0

0 1 0 0 0 1 0 0

0 0  0  0

Boundary conditions u1 = v1 = u3 = v3 = u2 = 0, v2 = – 0.05 in. Applying the boundary conditions and superimposing the [k]s 0.0447  0.0223  v2   0.02  0  0.272      AE 0.0447 0.214  0.0447  u4  = 0    0 0.272  v4   0.0223 0.0447    Solving u4 = 3.97  10–3 in. v4 = – 0.987  10–3 in. Element stresses u2  0  v  0.02  30  10  2  (2) = [0 –1 0 1]   u 0  4  12  3  v3  0  3

 (2) = 12.5 ksi (T) 70

0      0.02 30  10   [ 0.894 – 0.447 – 0.894 0.447]  (5) = 3  8.94  12  3.97  10   – 0.987  103    3

 (5) = 1.384 ksi (T) 0     3 0 30  10  [0 –1 0 1]  (4) =  3.97  103  4  12    – 0.987  103   (4) = – 617 psi (C)  3.97  103    30  103 – 0.987  103   (3) [–1 0 1 0]  =   8  12 0   0    (3) = – 1240 psi (C)

(1) = 1–2 = 0 Note: This solution was also verified by a computer program.

3.35

Using symmetry Element (5)

 = 30°

(1)

(6)

AE 0.75 0.433 [k(5)] =  L 0.433 0.25 

   

Element (4)

 = 60° (1) [k(4)] =

(5)

AE  0.25 0.433  0.75 L  

   

Element (3)

 = 90° 0 2 AE  1   0 [k ] =   L  2   (3)

0 1

   

Applying the boundary conditions and superimposing the K’s 0.866 u1   0  AE  1  =    2  v1   1000 L 0.866 

v1 =

1000 2  106

 v1 = – 0.5  10–3 in.

u1 = 0

 0    0.0005 E   (1) = (5) = [0.866 – 0.5 …]   L  0   0   (1) = 250 psi (T)

3.36

{d } = [T*] {d} and [T*] = [Cx Cy Cz] Cx =

20  0 = 0.864 23.15

u1 = 0.1 in.

Cy =

10  0 = 0.432 23.15

v1 = 0.2 in.

Cz =

60 = 0.259 23.15

w1 = 0.25 in.

u 1 = 0.864 (0.1) + 0.432 (0.2) + 0.259 (0.25)

 u 1 = 0.2376 in.

3.37

Cx =

10 = 0.267 34.42

Cy =

20 = – 0.534 34.42

30 = 0.802 34.42 {d } = [T] {d} Cz =

u1    {d 1 x } = [Cx Cy Cz] v1  w   1 u 1 = (0.267) (0.1) – (0.534) (0.2) + (0.802) (0.25)



u 1 = 0.1205 in.

3.38

L1-2 = 3.0 in.

Cx =

1 = 0.333 3

Cy =

2 = 0.667 3

Cz =

2 = 0.667 3

6   {d 2 x } = [0.333 0.667 0.667] 12  18   

û2 = 22.0 mm

3.39

Cx =

3 = 0.515 5.831

Cy =

4 = 0.686 5.831

Cz = 

3 = – 0.515 5.831

{d 2 x } = [0.515 0.686

6   – 0.515] 12  18  

 û2 = 2.052 mm

3.40 From Figure P. 3.40

L1–2 =

(– 4) 2  0  (3) 2 = 5 m

Cx =

04 44 03 = – 0.8; Cy = = 0; Cz = = – 0.6 5 5 5

L1–3 =

( 4) 2  0  (3) 2 = 5 m

Cx =

04 44 63 = – 0.8; Cy = = 0; Cz = = 0.6 5 0 5

L1–4 =

0  ( 4) 2  0 = 4 m

Cx = 0; Cy = L1–5 = Cx =

04 = – 1; Cz = 0 4

42  ( 5)2  ( 2)2 = 3 5 m 4 3 5

= 0.596; Cy =

5 3 5

= – 0.745

Cz =

2 3 5

= – 0.298

Element 1–2  0.64 0 0.48 [  =  0 0 0     0.48 0 0.64 

 [ ] [   ] [k] = 42000   [   ] [ ] 

Element 1–3  0.64 0 0.48 0 0  [  =  0    0.48 0 0.64 

 [ ] [  ] [k] = 42000   [  ] [ ] 

Element 1–4 0 0 0 [  =  0 1 0     0 0 0 

 [ ] [  ] [k] = 52500   [  ] [ ] 

Element 1–5  0.356 0.444 0.178  [ =  0.444 0.556 0.222     0.178 0.222 0.0889   [ ] [   ] [k] = 31305   [   ] [ ]  Applying the boundary conditions where all deflections at node 2, 3, 4 and 5 are zero. The global equations are  64905 13899 5572  u1   0       6950  v1   10 =  13899 69906   0  6950 33023  w1   5572   0 = 64905 u1 – 13899 v1 – 5572 w1

(1)

– 10 = – 13899 u1 + 69906 v1 + 6950 w1

(2)

0 = – 5572 u1 + 6950 v1 + 33023 w1

(3)

From (1) and (3) 0 = 67058 v1 + 379094 w1

(4)

From (2) and (3) – 10 = 52570 v1 – 75424 w1 From (4) and (5), we get

w1 = 2.68374  10–5 m 76

(5)

v1 = –1.5171  10–4 m Substituting in (1)

u1 = –3.0183  10–5 m Element stresses

 3.0183  10 5    4  1.5171  10   5   1–2 = 42  106 [0.8 0 – 0.6 – 0.8 0 0.6]  2.6837  10  0     0     0  1–2 = – 337.846 Force 1–2 =

kN m2

(C)

337.846 = – 0.337 kN (C) 1000

 3.0183  10 5    4  1.5171  10   5   1–3 = 42  106 [0.8 0 – 0.6 – 0.8 0 0.6]  2.6837  10  0     0     0  1–3 = – 1690

Force 1–3 =

kN m2

(C)

1690 = – 1.69 kN (C) 1000

 3.0183  10 5    4  1.5171  10   5   1–4 = 52500  103 [0 1 0 0 –1 0]  2.68374  10  0     0   0    1–4 = – 7965 Force 1–4 = 

kN m2

(C)

7965 = – 7.695 kN (C) 1000

1–5 = – 2726

kN m2 77

Force 1–5 = – 2.726 kN Force equilibrium at node 1 x direction 0 = 0.388  0.8 + 1.69  0.8 – 2.726  0.596 0 = – 0.00307 y direction

5 15

– 10 = 0 + 0 – 7.965  1 – 2.726  – 10 = – 9.9968 z direction

0 = 0.338  0.6 – 1.69  0.6 + 0 + 2.726  0.298 0 = 0.0015

3.41

L1–2 =

(12  0) 2  (3  0) 2  (4  0) 2  L1–2 = 13 m

L1–3 =

(12  12) 2  (3  3) 2  (7  4) 2  L1–3 = 3 m

L1–4 =

(14  12) 2  (6  3) 2  (0  4) 2  L1–4 = 10.05 m

Element C x 

x j  xi Li  j

Cy 

y j  yi Li  j

Cz 

z j  zi Li  j

1 2 1 3

 12 13

3 13

1

1 4

2 10.05

9 10.05

4 10.05

4 13

Element

Cx2

1 2

0.852  0.213  0.284 0.053 0.071 0.095

C xC y

C xCz

1 3

1 4

0.040

0.178

0.079

C x 2  [  =   

C xC y C y2

Cy2

C yCz

C2 2 1

0.802 0.356 0.158

C xCz   C yCz   C z2 

 0.852 0.213 0.284  [1–2] =   0.213 0.053 0.071    0.284 0.071 0.095 0 0 0 [1–3] =  0 0 0     0 0 1  0.040 0.178 0.079  [1–4] =  0.178 0.802 0.356   0.079 0.356 0.158

u2 = v2 = w2 = u3 = v3 = w3 = 0 u 4 = v 4 = w4 = 0



k1(1) 2

]  [13 AE  [ ] [   ]  = L [ ] [ ] = AE  [ ]    13 1 2 

]   [13  [ ]  13 

 [ 3 ] AE  [ ] [   ] = AE  [ ] L13 [   ] [ ]   3

 [ 3 ]   [ ]  3 

k1(2) = 3 

[ ]  10.05 [ ] [   ] = AE  [ ] [ ] 1 4    10.05

 k1(3)4  = LAE [ ]

[ ]   10.05  [ ]  10.05 

{ F } = [ K ] {d }  F1x  40 kN  13.985  u1   69.519 1.327     3  40.885  v1   F1y = 0  = 210  10  1.327 83.879       13.985 40.885 356.363  w1   F1z = 0  

u1 = 2.766  10–3 m v1 = – 1.024  10–4 m w1 = 1.203  10–4 m

 2.766  103    4 1.024  10  E 12 3 4 12 3 4   1.203  104     (1) = (1)    13 13 13 13 13   L  13 0    0   0     (1) = 41.0 MPa (T)  2.766  103    4 1.024  10   4  E   (2) = (2) [0 0 1 0 0 –1]  1.203  10  L 0     0     0   (2) = 8.42 MPa (T)

 (3) =

E L(3)

 2.766  103    4 1.024  10  4 2 9 4   1.203  104   2  9    10.05 10.05 10.05 10.05 10.05 10.05   0    0     0

  (3) = – 11.58 MPa (C) 3.42

Element 1–5 L1–5 = 108 in. Cx =

x5  x1 0  ( 72) =  Cx = 0.667 108 L15

Cy =

y5  y1 0  ( 36)  Cy = 0.333 = 108 L15

Cz =

z5  z1 72 – 0  Cz = 0.667 = 108 L15

0.222 0.444  0.444 0.222  0.444   0.444  0.222 0.111 0.222  0.222 0.111  0.222    0.222 0.444  0.444 0.222  0.444  4  30  106  0.444 [K ] =   0.444 0.222 0.444  108 0.444  0.222 0.444 0.222  0.111 0.222 0.222 0.111 0.222    0.444 0.222 0.444  0.444  0.222 0.444 Element 2–5 (0  0)2  (0  ( 36)2 )  (72 – 0)2

L2–5 =

 L2–5 = 80.5 in. Cx =

00 =0 80.5

Cy =

0  ( 36)  Cy = 0.447 80.5

Cz =

72 – 0  Cz = 0.894 80.5

0 0 0 0 0.2 0.4  6 0.4 0.8 4  30  10  0 [K] =  0 0 0 80.5   0  0.2  0.4   0  0.4  0.8

0 0 0 0  0.2  0.4   0  0.4  0.8  0 0 0 0 0.2 0.4   0 0.4 0.8 Since the structure is symmetric to the x-z plane then we can assume v5 = 0 and a load of 500 lbs. Disregarding all rows and columns of zero displacement we form the new global stiffness matrix comprised only of the non-zero displacements. 0.0041 0.0041 [K] = (4)  30  106   0.0041 0.014  {F} = [K]{d} F5x = 500 = 492000 u5 + 492000 w5 F5z = 0 = 492000 u5 + 1685880 w5

 u5 = –

1685880 w5 492000

1685880   500 = 492000   w5 + 492000 w5  492000   w5 = – 0.00042in.  u5 = 0.0014in. Element stresses Element 1–5 81

1–5 =

E [– Cx – Cy – Cz L15

Cx Cy

u1  v   1  w1  Cz]   u5  v5     w5 

0     0   6 0   30  10 1–5 = [– 0.667 – 0.333 – 0.667 0.667 0.333 0.666]   108  0.00143    0   – 0.00042  1–5 = 185 psi (T) Element 2–5 0     0   0   30  106 2–5 = [0 – 0.447 – 0.894 0 0.447 0.894]   80.5  0.0014    0   – 0.00042 

2–5 = –140 psi (C)

From symmetry

3–5 = 140 psi (C) 4–1 = 180 psi (T) 3.43

Element 1–4

L1–4 =

(0  ( 36)2 )  0  (144 – 0)2  L1–4 = 148.4 in. 82

Cx =

x4  x1 36 = L1 4 148.4

 Cx = 0.2426

Cy = 0 Cz = 0.9704        AE  [k1–4] =  L1 4

     0.05885 0 0.2354   0 0 0  0.2354 0 0.9417 

Element 2–4

L2–4 =

(144) 2  ( 72) 2  1442 = 216 in.

Cx =

144 72 144 = 0.6667 = 0.667, Cy = = – 0.3333, Cz = 216 216 216

   AE  [k2–4] =  L2  4    

     0.4425 0.2222 0.4445  0.2222 0.1111  0.2222   0.4444 0.2222 0.4445

Element 3–4

L3–4 = 216 in., Cx = 0.6667, Cy = 0.3333, Cz = 0.6667    AE  [k3–4] =  L3 4    

[K] =

     0.4425 0.2222 0.4425  0.2222 0.1111 0.2222   0.4445 0.2222 0.4445 

0 0.2354  1.294  0.05885  1.294 AE   0 0.3234 0  L1 4  0 0.9417  1.294   0.2354  1.294

{F } = [ K ] {d } 0 1.5294  u4  1.3529  0  AE      = 0 0.3234 0  v4  0     148.4  4000 0 2.2357   w4  1.5294   0=

AE [0 u4 + 0.3234 v4 + 0 w4] 148.4 83

 v4 = 0 0=

AE [1.3529 u4 + 1.5294 w4] 148.4

 u4 = – 1.1305 w4



– 4000 =

AE [1.529 u4 + 2.2357 w4] 148.4

– 4000 =

AE [1.529 (– 1.1305 w4) + 2.2357 w4] 148.4

 w4 =

–1171501.87 6  30  106

 w4 = – 0.00683 in.  u4 = 0.00863 in. Stresses

1–4 =

E [– Cx – Cy – Cz L1 4

Cx Cy

u1  v   1   w1  Cz]   u4  v4     w4 

0     0   6 0   30  10  1–4 = [– 0.2426 0 – 0.9704 0.2426 0 0.9704]   148.4  0.00863    0    – 0.00683   1–4 = – 916 psi (C)

3.44 Derive Equation (3.7.21)

=

f 2 x A

f 2 x =

AE  u 1  [1 –1]   L  u  

=

E  u 1  [–1 1]   L  u  

Now in 3–D {d  = [T*]d} where by Equation (3.7.7) 84

Cx [T*] =  0

=

0   Cz 

Cy 0

Cz 0

0 Cx

0 Cy

u1  v   1   w1  Cz]   u2  v2     w2 

C x E [–1 1]  L 0

E [– Cx – Cy – Cz L

0  {d } C z 

3.46

E = 30  106 psi A(1) = A(2) = A(4) = A(5) = 10 in.2 A(2) = 20 in.2 Reduce the given figure by symmetry.

A(1) = A(2) = A(3) = 10 in.2 (Reducing given A(3) by half)

v1 = 0, u2 = u3 = 0 find u1, v2, v3 85

Data for reduced truss Element

°

36.9°

0.8

0.6

0.64

0.36

0.48

0°

1.0

90°

1.0

C 2

CS

S 2

CS   S 2   CS  S 2 

 C2   A E   CS () [k ] =    L   C 2  CS 

 0.64  0.48  0.48  0.36  0.64 0.48  0.48 0.36 

 0.64 0.48 6   30 10   1   1 ft   0.48 0.36 lbf    k(1)] = (10 in.2)     2  in.   10ft   12in.   0.64  0.48  0.48  0.36



u1 So

0.48 0.64 0.48 u1  0.64 [k(1)] = 2.5  106  0.48 0.36 0.48 0.36 v1   0.48  u3  0.64 0.48 0.64   0.36  v3  0.36 0.36 0.48

Finally

1.2 1.6 1.2  u1  1.6 [k(1)] = 106  1.2 0.9 1.2 0.9  v1   1.2  u3  1.6 1.2 1.6   0.9  v3  1.2 0.9 1.2  1.0  30  10 lbf   1   1 ft   0  [k(2)] = (10 in.2)    8ft   12 in.   1.0  in.2   0 6

0 1.0 0  0 0 0  0 1.0 0   0 0 0

and

u1  3.125 [k(2)] = 106  0   3.125   0

0 3.125 0  u1 0 0 0 v1  0 3.125 0  u3  0 0 0 v3

and 0 0   30  10 lbf   1   1 ft  0 1.0  [k(3)] = (10 in.2)    6 ft   12 in.  0 0  in.2  0 1.0 6

0 0  0 1.0  0 0   0 1.0 

0 0 [k ] = 10 0 4.167  0 0  0 4.167 6 

(3)

[K] = 10

 u2 0 4.167  v2  0 0  u3  0 4.167  v3 0

1.6  3.125

1.2

3.125

1.2  3.125

0.9 .

1.6

1.2

 0.9

3.125 4.167

1.6 1.2

1.2  0.9

 4.167

1.6 1.2

u1 v1 u2 v2 u3

 4.167 1.2 0.9  4.167 v3

[K] {d} = [F] gives

1.2  3.125 0  1.6  1.2   u1   0   4.725  1.2 0.9 0 0  1.2  0.9   0   0       0 3.125 0 0 0  0   0    3.125     =  0 0 4.167 0  4.167  v2   – 0.01  0   1.6  1.2 0 0 1.6 1.2   0   0       0 5.067   v3   0   4.167 1.2   1.2  0.9 and 0 1.200 u1   0   4.725      0 4.167 4.167  v2  =  – 0.01        1.2 4.167 5.067  v3   0 

u1 =

0 0 1.200 0.1 4.167 4.167 0 4.167 5.067 4.725 0 1.200 0 4.167 4.167 1.2

4.167

5.067

 0.01 4.167 0 4.167 u1 = 0 4.167 4.725 4.167 4.167  1.2 4.167 5.067 1.2 4.167 1.2

u1 =

0.025 = –0.00426 in. 17.72017  6.00048

4.725 0 1.2 0 0.01 4.167

v2 = =

1.2

0 5.067 11.71969

4.725   0.02534  1.2   0.006 11.71969

v2 = –0.0192 in. 4.725 0 1.2

v3 = v3 =

0 0 4.167 0.01 4.167 11.71969

4.725{(0.01)(4.167)} = – 0. 0168 in. 11.71969

now {} = [C {d} where [C =

E [– C – S C S] L

So, for element (1)

 0.00426   0  30  106 lbf    [C =  (–0.8 – 0.6 0.8 0.6)   0  120 in.2     0.0168   2.5  105 lbf  =   (– 0.0068 in.)  in.3

(1) = – 1668 psi (C) For element (2)

 30  106 lbf   1  (2) =    96 in.  [– 1.0  in.2

0 1.0

  0.00426    0   0]   0     0.0192 

 3.125  105 lbf   (0.00426 in.)  in.3

(2) = 

(2) = 1332 psi (T) For element (3)  0    0.0192   30  10 lbf   1    (3) =   72 in.  [0 –1.0 0 1.0]  0   2   in.     0.0168  6

 4.167  105 lbf   (0.0096 – 0. 0084)  in.3

(3) = 

= 1000 psi

(3) = 1000 psi (T) 3.47

Using symmetry

Boundary conditions

v1 = u2 = u3 = 0 (see solution to Problem 3.24 for individual [k]’s for each element) Global [K]  0.0256  0.0192 0 0 0   0.0756 0.0192  0.05  0.0192 0.0811  0.0192  0.0144  0.0667 0 0 0    0.0256 0.0192  0.0 0.0756  0.0192 0 0   0.05  0  0.0192 0.0811  0.0667 0.0192  0.0144 0 0  [K] = AE  0 0 0.0756 0.0192  0.05 0   0.0256 0.0192  0.0192 0.0144  0.0667 0.0192 0.0811 0 0  0   0  0.0256 0.0192  0.05 0 0.0756  0.0192  0  0 0.0667 0.0192  0.0144 0 0  0.0192 0.0811 

Applying the boundary conditions, we obtain

0  0.0192 0 0  0.0756  u1   0   0 0.0811  0.0667 0.0192  0.0144  v2   0        0 0 AE   0.0192  0.0667 0.0811  v3  =  2P    0 0.0192 0 0.0756  0.0192  u4   0       0  0.0144 0  0.0192 0.0811  v4   P   Using the simultaneous equation solver, we obtain

u1 = – 110

P P , v2 = – 405 AE AE

v3 = – 433

P P , u4 = 50 AE AE

v4 = – 208

P AE

where all displacements are now in units of inches. (Computer program TRUSS was also used to verify the above displacements (setting P = 1, A = 1, E = 1)) Stresses u4  v  E  4 (2) = [–1 0 1 0]   L u3  v3   50    208 P E   = [– 1 0 1 0]   0 L   AE   433 =

E  P   50  AE  240 in.

(2) = –0.208

P A

u1  110  v  0  E  1  P (3) = [0 –1 0 1]   L u4  50  AE v4  208 =

208 P 15  12 A

= –1.156

P A

(4) =

E [0.80 L

(4) =

(243  40  124.8) P A 25  12

(4) = 0.261

– 0.60

– 0.80

u2  0  v  405 P  2  0.60]   AE u4  50  v4  208

P A

Verify equilibrium at node 4

Fx = – 0.208 P + 0.261 P (0.8) = (–0.208 + 0.208)P = 0 Fy = 1.156 P – P + (0.261 P) (0.6) = (1.156 – 1 – 0.156)P  0

3.48

Element (1) (2) (3) (4) (5)

L  C S C 2 S 2 CS 4.47 26.565° 0.89443 0.44721 0.8 0.2 0.4 8 0 1 0 1 0 0 4.47 153.435° – 0.89443 0.44721 0.8 0.2 – 0.4 4.47 26.565° 0.89443 0.44721 0.8 0.2 0.4 4 90° 0 1 0 1 0 91

0.8 0.4 – 0.8 – 0.4  u1 0.2 – 0.4 – 0.2  v1 AE   [k(1)] = 0.8 0.4  u2 4.47  Symmetry  0.2  v2  0 1 1  0 0 AE  [k(2)] = 1 8   Symmetry 

0 u1 0 v1  0 u4  0 v4

0.8  0.4  0.8 0.2 0.4 AE  [k(3)] = 0.8 4.47   Symmetry 

0.4  u4  0.2  v4   0.4  u2  0.2  v2

0.8 0.4  0.8 0.2  0.4 AE  [k(4)] = 0.8 4.47   Symmetry 

 0.4  u2  0.2  v2  0.4  u3  0.2  v3

0 0 0 0  u4  1 0  12  v4 AE  2  [k(5)] = 0 0  u3 4   1  Symmetry  2  v3 Boundary conditions 

v1 = u4 = u3 = 0 u1 0.8  4.47 1   0.8 8  4.47 [K] = 210  106   0.4  4.47   0   0



F2y = – 20 kN

u2  0.8 4.47

v2  0.4 4.47

2.4 4.47

0.4 4.47

0.4 4.47  0.4 4.47

0.6 4.47  0.2 4.47  0.2 4.47

 0.4 4.47  0.2 4.47

0.4 4.47

0.2  0.5 4.47 4  0.5 4

v4  u1  0.4  u2 4.47  0.2  v  2 4.47  0.5  v  3 4 0.2 0.5  v  4 4 4.47 0

F3y = – 10 kN

F1x = F1y = F2x = F3x = F4x = F4y = ?



 2  18  5 5 F  1x   2    5 5  F2 x    1 3  20  10  =  5 5  3  0  10  10    F4 y   0 



 2 1  F   5 5 8 1x    2  F2 x   5 5    F = 0   4y    3 1   20  10   5 5  3   10  10  0 



2 5 5

1 5 5

6 5 5

1 5 5

1 5 5 1 5 5

3 10 5 1 10 5

1 5 5 1 10 5

1 5 5

1 10 5



1  81 10 5



–1 8

2 5 5

1 5 5

6 5 5

1 10 5

1 5 5 1 10 5

1 5 5 1 5 5 1 5 5

 K11 ?  6   = 210  10  K P  21



1  81 10 5



1 10 5 1 8



 u1    1  u  5 5  2 1  v   210  106  2 10 5   1  v3  8   1  1 v  8   4 10 5



 u1    1  u  5 5  2 1     210  106 8  v4    1  v2  10 5   1  1 v  5   3  10 5

3 10 5 1 10 5



K12  d1  K 22  d 2 

1.532 0.8074  2.602 1.3716    0.807 1.3716 6.6148   4.192

 k111  = 1.532

0.04759 0.02923 –1 [kc] = [k21]  k11  [k12] =  0.02923 0.0935    0.08657  0.073956 –1 [kc] = [k22] – [k21]  k11  [k12] =  0.073956 0.076213  

67.539 65.539  [kc] = [Nʹ ]–1 =   65.539 76.719  2 3 v2  67.539 65.539   21  10  =    65.539 76.719    1    21  103   v3 

  9.553  103  =   3  9.895  10  u1   0.20185  0.0361 3   1    9.553 10  u2  = –  k11  [k12] {d2} = –  0.15709  0.06134   3 v   9.895 10  0.34628 0.70417        4

  2.285  103  u1      u2  =  8.9381  104   v  2   4  1.0276  10  The stresses in each element u1  v  E  1  (1) = [– C – S C S]   L u2  v2 

 (1) =

210  109   2  5 2 5

 (1) = – 67.08  106

N mm 2

1

 2.285  103    0 1      5   8.9381  104   3    9.553  10 

= – 67.08 MPa (C)

  2.285  103    0 210  10    (2) = [– 1 0 1 0]   8 0    1.0276  102    9

 (2) = 59.99  106

(3) =

210  10 2 5

N mm 2

 2  5

= – 22.36  106

(4) =

1

2

0    2  1   1.0276  10     5   8.938  104   3    9.553  10 

= – 22.36 MPa (C)

mm 2

210  109   2  5 2 5

= – 44.72  106

= 60.0 MPa (T)

1

N mm 2

 8.938  104    1   9.553  103    5   0   3   9.895  10 

= – 44.72 MPa (C)

0    2  9 210  10 1.0276  10   (5) = [0 –1 0 1]   0 4   9.895  103    94

= 19.99  106

= 20.00 MPa (T)

mm 2

3.49

Element (1)

 = 90° (1)

(3) 0 0 0 1 0 1  0 0 0  1 0 1 

0 2  30  106  0 [k ] =  30 0  0 (1)

Element (2)

 = 0° (1) 1 2  30  106  0 [k ] =  30  1  0 (2)

0 0 0 0

(2) 1 0  0 0  1 0  0 0

Element (3)

 = 135° (2) (3) 0.5  0.5  0.5 0.5  2  30  106  (3)  0.5 0.5 0.5  0.5 [k ] =   30 2 0.5 0.5  0.5  0.5   0.5  0.5 0.5 0.5 Assembling the stiffness matrices 1 1 0 0 1 0   1 0 1  0.5 2 6  [K] = 2  10 0 0  0.5  2  0.5 0 0  2  0.5  0 1 2 95

0 0

 0.5

2 0.5 2 0.5 2  0.5 2

0  1   0.5  2   0.5  2   0.5  2  1  0.5  2

0 0 0 0  2  2  0 2 0 0 0 2     2 0 2.707  0.707  0.707 0.707  6 [K] = 10   0  0.707 0.707 0.707  0.707   0  0 0  0.707 0.707 0.707  0.707     0 2 0.707  0.707 0.707 2.707   0.707 0.707  0.707 0.707  0  0 [T] =  0 0   0 0  0  0

0 0 1 0 0 0

0 0 0 1 0 0

0 0 0 0 1 0

0 0  0  0 0  1

 0.707 0.707 0 0 0 0   0.707 0.707 0 0 0 0    0 1 0 0 0  0 T [T ] =   0 0 1 0 0  0  0 0 0 0 1 0   0 0 0 0 1  0 0  1.4 0 0 1.4   2.0  0 2.0  1.4 0 0  1.4     1.4  1.4 2.707  0.707  0.707 0.707   [T] [K]TT] = 106   0 0  0.707 0.707 0.707  0.707    0 0  0.707 0.707 0.707  0.707    0.707  0.707 0.707 2.707   1.4 1.4 Applying the boundary conditions v1 = u2 = u3 = v3 = 0 2.0 0  F1x  0   u '1  6    = 10   v  – 2000 F  0 0.707   2   2y   u1 = 0 v2 = – 0.00283 in.  F1x  0  0  2 F   0   0 y 1 2     F2 x  0   2 0 6   = 10  0  F2 y  2000  0 F  0   0 0   3x   0 2   F3 y  0 

0 0 0 0 2       0 0 0 0 2    2.707 0  0.707  0.707 0.707      0.707  0.707   – 0.00283  0.707 0.707   0.707 0.707 0.707  0.707   0    0.707  0.707 0.707 0.707   0 

 F1x = 0 F1y = 0 96

F2x = 2000 lb F2y = – 2000 lb F3x = – 2000 lb F3y = 2000 lb

Element stresses {} = [C {d} [C =

 (1) =

E [– C – S C S] L

30  10 30

[0 – 1

u1  0  v  0   1  0 1]   u3  0 v3  0 

  (1) = 0

 (2) =

30  10 [– 1 0 30

u1  0  v  0   1  1 0]   u2  0  v2  0.00283

  (2) = 0

 (3) =

30  10 30 2

0    – 0.00283   [0.707 – 0.707 – 0.707 0.707]   0     0

  (3) = 1414 psi (T) 3.50

0 0 0 1 [k(1)] = 2  106  0 0   0 1  1  0 [k(2)] = 2  106   1   0

0 0 0 1  0 0  0 1

0 1 0 0 0 1 0 0

0 0  0  0

 0.5 0.5 0.5 0.5  0.5 0.5 0.5 0.5  [k(3)] = 1.414  106   0.5 0.5 0.5 0.5    0.5 0.5 0.5 0.5 Assembling the stiffness matrices 0.707  0.707  0.707 0 0  0.707  0.707 2.707  0.707  0.707 0  2   2.707 0.707  2.0 0  0.707  0.707 6 [K] = 10   0.707 0.707 0 0  0.707  0.707  0 0  2.0 0  2.0 0   2.0 0 0 0 0 2.0   0.707 0.707 0.707 0.707  0 0  [T ] =  0 0   0 0  0 0   1.0 1.0   0 T 6 [T]K]T ] = 10   0  0   1.4

0 0 1 0 0 0

0 0 0 1 0 0

0 0 0 0 1 0

0 0  0  0 0  1

0 0 1.4  1.0 0 2.4 1.0 0 1.0 1.4  0 1.0 2.707 0.707 2  1.0 0.707 0.707 0 0 0 2.0 0 2.0 0  1.4 0 0 0 2.0  98

Applying the boundary conditions v1 = u2 = u3 = v3 = 0

 F1x  0   1, 000, 000   =    2000 F 0   2y 

0  u1  0    707,106.75 v2 

1–2 = –1414 psi

 u 1 = 0

1–3 = 0

v2 = – 0.00283 in.

3–2 = 0 3.51

Element (1) C = 1, S = 0,  = 0°

(1)  1 0 [k ] = 2  10  1   0 (1)

6 

(2)

0 1 0  0 0 0  0 1 0  0 0 0

Element (2) C= –

1 3 ,  = 120° ,S= 2 2 (2)

 0.25  [k(2)] = 2  106   4 3   0.25  3  4

(3)  3 4 3 4 3 4 3 4

0.25 3 4

0.25  3 4

3  4  3  4   3 4   3  4 

Element (3) 99

3 1 , S= 2 2 (1)

(3)

 3  0.25 3  0.25 4 4    3 [k(3)] = 2  106    0.75 0.75 4    0.25 43    0.75   Symmetry

Global [K]  3  3 1.25 1  0.25 0 4 4   3  3    0.75 0 0 0.75 4  4   3  1  3  0 1.25 0.25 4 4  [K] = 2  106   3  0  3  0 0.75 0.75 4 4     0.25  3 3  0.75 4 0.5 0  4  3  3  0.75 43 0 1.5   4 4

1 0  0 [T1] =  0  0  0

0 1 0 0 0 0

0 0 0 0 1 2 1 2

1 2 1 2

0 0 0 0

0 0 0 0  0 0  0 0  1 0  0 1

[K*] = [T1] [K]  T1T  with boundary conditions u1 = v1 = vʹ2 = u3 = v3 = 0

 F1x   u1  0  F  v 0  1y   1   F 2 x  2000  u2  F   = [K*] v  0   2   2y   u3  0   F3 x       v3  0   F3 y 

100

[T1] [K] = 6

 2.5  10  5  8.66  10  6  1.414  10  1.414  106   5  105   8.66  105

8.66  105

2  106

5  10 5

1.5  106

8.66  105

1.155  106 4.483  105

2.588  105

2.38  106

1.673  106

9.659  105

8.66  105

1.5  106

8.66  105

1  106

1.5  106

8.66  105 1.5  106

 2.5  106  5  8.66  10  6  1.414  10 T [K*] = [T1] [K]  T1  =  6  1.414  10  5  105  5  8.66  10

8.66  105   1.5  106   4.483  105  1.673  106   0  3  106 

8.66  105

1.414  106

1.5  106

1.134  106

5  105

8.66  105

4.483  105

1.5  106

4.483  105

1.414  106

5  105

8.66  105

5  105

2.588  105

2.866  106

9.659  105

1.673  106

1  106

1.673  106

8.66  105   1.5  106   4.483  105  1.673  106   0  3  106 

Solving the third equation of [K*] {d }= {F }yields

2000 u 2 = 1.134  106 = 1.764  10–3 in.

3.52 (a)

Using Equation (3.9.19), we get

p =

AL {d T } [BT] [D] [B] {d }  {d T }{ f } 2

For the bar above, Equation 3.9.26 yields

p =

AL 2 E u1 2 – u1 f 2 L 101

p =

AE 2 u1 – u1 f 2L

Putting in the numerical values

p =

(2)(30  106 ) 2 u1 – u1 (10000) (2)(30)

p = 106 u12 – 1  104 u1 u1, in.

p lbin.

–0.003 –0.001 0 0.001 0.003 0.005 0.007 0.009

39 11 0 –9 –21 –25 –21 –9

p min = – 25 lbin. Also by calculus  p u1

= 2  106 u1 – 1 104 = 0

 u1 = 0.005 in. for p minimum (b)

p =

AE 2 u1 – u1 f (see Problem 3.52 (a)) 2L

p =

2(30  106 ) 2 u1 – u1 (10000) 2(50)

p = 6  105 u12 – 0.1  105 u1

u1 in.

p lbin.

–0.004 –0.002 –0.001 0 0.002

49.6 44.0 10.6 0 4.0 102

0.004 0.006 0.008 0.010

–30.4 –38.4 –41.6  –40.6

p min = – 41.67 lbin. By calculus  p u1 

= 12  105 u1 – 0.1  105

u1 = 0.00833 in. yields p min.

3.53 du = x d x du U =   v

   d   dv

dv = A(x) dx



x

x A(x) = A0 1    L

1 L x  x x A0 1   dx  2 0 L

x 1 L T  L1  1 1 {d }  1 [ E ]   {d } A0 1   dx   0  L 2  L L  L

 1  x 1 L 1 1  u1   u1 u2   1L [ E ]      u  A0 1  L  dx L L 2 0  2  L

x 1 L EA0 2 (u1  2u1 u2  u22 ) 1   dx  2  0 2 L L

AL E A E U 1 = 0  2 (2 u1  2u2 )  + 0  2 (2 u1  2u2 )  2 L u1  2 L 2 3 A0 L  E U (2 u1  2u2 )   f1 = 4  L2 u1  Similarly 3 A0 L  E U = (2 u2  2u1 )   f 2 4  L2 u2   [K] =

3A0 E  1 1 2 L  1 1 103

3.54

(a) Two element solution

For element (1) force matrix is from Example 3.9

 f1x (1)   1500  (1)  =   3000   f 2 x 

f2x(2) =

1 1 (4500) + (30  300) 3 2

f3x(2) =

2 1 (4500) + 30  300  3 2

f2x(2) = 6000 lb

f3x(2) = 7500 lb

104



 f 2 x (2)  6000  (2)  =   7500  f3 x   F1x  1500     {F} = 3000  6000  7500    Stiffness matrices

[k(1)] = AE  1 1 30  1 1 2

[k ] = AE  1 1 30  1 1 (2)

Global [K]: [K] =

 1 1 0  AE  1 2 1  30   0 1 1

Global equations {F} = [K]{d}  F1x  1500  1 1 0 u1  0 (2)(30  106 )      1 2 1 u2 9000 =      30  7500  u    0  1 1      3  Solving equations 2 and 3 for u2 and u3, we obtain

u2 = 0.00825 in. u3 = 0.012 in. Element stresses u1  v  E  1 (1) = [C’] {d} = [– C – S C S]   L u2  v2 

30  10 30

 0   0    [– 1 0 1 0]   0.00825  0 

(1) = 106 (0.00825) = 8250 psi (T)

105



(2) =

30  10 30

u2  0.00825 v  0   2  [– 1 0 1 0]   u  0.012  3  v3  0 

(2) = 3750 psi (T) Four element solution

Basic triangular load

Total global forces at nodes

Global equations  F1x  375  1 1 0 0 0 u1  0  2250   1 2 1 0 0 u     2  AE   0 1 2 1 0 u3 4500 =  15    6750    0 0 1 2 1 u   4     0 0 0 1 1 u5  4120  Solve the last four equations for u2 through u5 106

2250 =

AE (2 u2 – u3) 15

(1)

4500 =

AE (– u2 + 2u3 – u4) 15

(2)

6750 =

AE (– u3 + 2u4 – u5) 15

(3)

4120 =

AE (– u4 + u5) 15

(4)

Using Gaussian Elimination, divide (1) by 2 1 AE 1125 =  u2  u3  2   15

(5)

Add (5) to (2) 1

5625 = (1 2 u3 – u4)

AE 15

6750 = (– u3 + 2u4 – u5) 4120 = (– u4 + u5) 1

1125 = (u2 – 2 u3)

(6)

AE 15

(7)

AE 15

(8)

AE 15

(9)

Divide (6) by 1.5 1 AE 3750 =  0  u3  u4   1.5  15

(10)

Add (10) to (7) 10500 = [(2 – 0.067) u4 – u5)]

AE 15

(11)

AE 15

(12)

1 AE 1125 =  u2  u3   2  15

(13)

1 AE u  3750 =    1.5 4  15

(14)

4120 = (– u4 + u5)

3 4  (11)

3 AE 7875 =  u4  u5    4 15

(15)

Add (15) to (12) 107

11995 =

1 AE u5 4 15

(16)

Solve (16) for u5

u5 =

11995  4  15 2  30  106

= 0.012 in.

By (15)

u4 = 7875  

15 2  30  10



3 (0.012 in.) 4

u4 = 0.01097 in.

By (10)

u3 =

3750 (15 ) 1  (0.01097) 6 2  30  10 1.5

 u3 = 0.00825 in. By (9)

u2 =

1125  15 2  30  10



1 (0.00825) 2

 u2 = 0.00441 in.

(1) =

30  106  0  [– 1 1]   15 0.00441

 (1) = 8812 psi (T)  (2) =

30  106  0.00441 [– 1 1]   15 0.00825

 (2) = 7688 psi (T)  (3) =

30  106 0.00825 [– 1 1]   15 0.01097 

 (3) = 5440 psi (T)  (4) =

30  106 0.01097  [– 1 1]   15  0.012 

 (4) = 2060 psi (T) Exact solution

108

F1x =

lb 1 (60) (600 ) in. 2

F1x = 18000 lb 1

P(x) = 18000 – 2 (10 x) x P(x) = 18000 – 5 x2 (18000  5 x 2 ) dx 0 AE

u(x) =  =

1  5 18000 x  x 3  + C  3  AE 

u(0) = 0 = C u=

18000  53 x 3 60  106

Analytical comparison with FEM Element stress Exact (x)

 (x) =

P( x) = 9000 – 2.5 x2 A

109

Analytical comparison with FEM

3.55

Analytical solution

F(x) at wall = – 300 (60) = – 18000 lb



f (x) = 18000 – 300 x

x = displacement (x) =

f ( x) 18000  300 x = A 2

  (x) = 9000 – 150 x

 (x) = 

x  ( x)

  (x) =

(1)

 1  150 x 2  9000 x  + C  E 2 

Applying the boundary conditions

 (x = 0) = 0 = C  (x) = – 2.5  10–6 x2 + 3  10–4 x

(2)

Finite element solutions (i) One element 1

f1x

f2x

110

Replace the distributed force with a concentrated force.

AE  1 1 u1  0  F1x     = L  1 1 u2   F2 x  F1x = F2x =

1 1  300  60 = 9000 lb F= 2 2

Solving for u2

u2 =

F2 x L 9000  60 = AE 2  30  106

u2 = 0.009 in. f2 x 9000 = = 4500 psi A 2

= (2) Two elements

f1x =

300  30 2

f2x =

300  30 (1  1) 2

f3x =

300  30 2

f1x = 4500 lb f2x = 9000 lb f3x = 4500 lb Global equation  F1x  4500   1 1 0 u1  0 AE      1 2 1 u2   F2 x  9000  =   L   F  4500   0 1 1 u3   3x  Solving the two equations

u2 = 6.75  10–3 in. u3 = 0.009 in.

(1) =

E u  [– 1 1]  1  L u2 

 (1) =

0   30  106 [– 1 1]  3  30 6.75  10 

  (1) = 6750 psi (T)

 (2) =

6.75  10 –2  30  106 [– 1 1]   30 0.009  111

  (2) = 2250 psi (T) Computer solutions One element NUMBER OF ELEMENTS (NELE) = 1 NUMBER OF NODES (KNODE) = 2 NODE POINTS K IFIX 1 111 2 011

XC(K) 0.000000E+00 6.000000E+01

YC(K) 0.000000E+00 0.000000E+00

ZC(K) 0.000000E+00 0.000000E+00

FORCE (1, K) 0.000000E+00 9.000000E+03

FORCE (2, K) 0.000000E+00 0.000000E+00

FORCE (3, K) 0.000000E+00 0.000000E+00

ELEMENTS K NODE (I, K) 1 1 2

E(K) 3.0000E+07

A(K) 2.0000E+00

NUMBER OF NONZERO UPPER CO-DIAGONALS (MUD) = 5 DISPLACEMENTS X Y Z NODE NUMBER 1 0.0000E+00 0.0000E+00 0.0000E+00 NODE NUMBER 2 0.9000E–02 0.0000E+00 0.0000E+00 STRESSES IN ELEMENTS (IN CURRENT UNITS) ELEMENT NUMBER 1= Two elements

STRESS 0.45000E+04

NUMBER OF ELEMENTS (NELE) = 2 NUMBER OF MODES (KNODE) = 3 NODE POINTS K IFIX 1 111 2 011 3 011

ELEMENTS K 1

XC(K) 0.000000E+00 3.000000E+01 6.000000E+01

YC(K) 0.000000E+00 0.000000E+00 0.000000E+00

ZC(K) 0.000000E+00 0.000000E+00 0.000000E+00

FORCE (1, K) 0.000000E+00 9.000000E+03 4.500000E+03

FORCE (2, K) 0.000000E+00 0.000000E+00 0.000000E+00

FORCE (3, K) 0.000000E+00 0.000000E+00 0.000000E+00

MODE (I, K) 1 2

K(K) 3.0000E+07 112

A(K) 2.0000E+00

3.0000E+07

2.0000E+00

NUMBER OF NONZERO UPPER CO-DIAGONALS (MUD) = 5 DISPLACEMENTS NODE NUMBER 1 NODE NUMBER 2 NODE NUMBER 3

X 0.0000E+00 0.6750E–02 0.9000E–02

Y 0.0000E+00 0.0000E+00 0.0000E+00

Z 0.0000E+00 0.0000E+00 0.0000E+00

STRESSES IN ELEMENT (IN CURRENT UNITS) ELEMENT NUMBER 1= 2=

STRESS 0.67500E+04 0.22500E+04

Four elements NUMBER OF ELEMENTS (NELE) = 4 NUMBER OF MODES (KNODE) = 5 NODE POINTS K IFIX 1 111 2 011 3 011 4 011 5 011

ELEMENTS K MODE (I, K) 1 1 2 2 2 3 3 3 4 4 4 5

XC(K) 0.000000E+00 1.500000E+01 3.000000E+01 4.500000E+01 6.000000E+01

YC(K) 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00

ZC(K) 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00

FORCE (1, K) 0.000000E+00 4.500000E+03 4.500000E+03 4.500000E+03 2.250000E+03

FORCE (2, K) 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00

FORCE (3, K) 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00

K(K) 3.0000E+07 3.0000E+07 3.0000E+07 3.0000E+07

A(K) 2.0000E+00 2.0000E+00 2.0000E+00 2.0000E+00

NUMBER OF NONZERO UPPER CO-DIAGONALS (MUD) = 5 DISPLACEMENTS

X 113

NODE NUMBER 1 NODE NUMBER 2 NODE NUMBER 3 NODE NUMBER 4 NODE NUMBER 5

0.0000E+00 0.3937E–02 0.6750E–02 0.8437E–02 0.9000E–02

0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00

STRESSES IN ELEMENTS (IN CURRENT UNITS) ELEMENT NUMBER 1= 2= 3= 4=

STRESS 0.78750E+04 0.56250E+04 0.33750E+04 0.11250E+04

Analytical comparison with FEM

3.56

114

 F1x  1500 lb    {F} = 1500  1500   F  1500   3x  [k(1)] =

AE  1 1 30  1 1

[k(2)] =

AE  1 1 30  1 1

Global equations 1 0 u1  0   F1x  1500   1 (2)(30  106 )      1 1  1 1 u2  =  3000    30  0 1 1 u3  0  F3 x  1500 Solving Equation (2) 2  106 (2 u2) = 3000

u2 = 0.75  10–3 in. Element stresses

E [– C L

–S

[– 1 0

0     0   1 0]  3  0.75 10      0

(1) = [C {d} =

30  10 30

u1  v   1 C S]   u2  v2 

  (1) = 750 psi (T)

F1x + 1500 = 2  106 (– 1) (0.75  10–3)  F1x = – 3000 lb

() 115

F3x = – 3000 lb

and

()

Total applied force = 60  100 = 6000 lb (

 = 0 (at node 2)  (x = 15) =

3000  1500 = 750 psi 2

3.57 Bar hanging under own weight Two element solution

 AL   4  1  1 0   u1    AE      AL  u =  1 2  1     2 2 L   2  0  1 1 u3  0  R   AL   4   3 x

u1 – u2 =

– u1 + 2u2 =

(A)

 AL2  1 

(1)

 AL2  1 

(2)

  2 AE  4    2 AE  2 

Adding (1) and (2) 2

u2 =  L  3  2 E  4 



u2 =

u1 =

3  L2   8E

 L2 8E



(3)

3  L2  L2 ( )  8E 2E

Analytical solution

116

Wx = V(x) = Ax = P(x)

= 



x P( x)

=



 x2 2E

 (L ) = 0 =

 =



dx = 

Ax dx AE

 L2

 x2

x

+CC=–

2E –

 L2 2E



 L2 2E

(x2 – L2)

u L  E  1 2E  (1)  = [1 –1]  2 L u2  38 EL    2

1 3 =  L =    2 8

(1) =

 L 8

(T)

u  3  L  E 8E = [1 –1]  2  L  u3  0  2



(2)

(2) =

3 L (T) 8

From Equation A.3 above, we obtain R3x:

117

R3x =

 AL 4



2 A E  3  L2    L  8E 

3 AL 4

R3x =  A L 3.58

Total Tx = 100  10 +

1  5(10)2 lb = 1250 lb 2

x u ( x)  (u2  u1 )( )  u1 L

f1x u1 + f2x u2 = 

= 

u1  u2   (100 + 5 x)   10  x  u1  dx

10 

10 (u2 – u1 ) x  100 u1 

= 500 (u2 – u1) + 1000 u1 +

(u2 – u1 ) 2 x  5 u1 x  dx  2

(u2  u1 ) 1000 + 250 u1 6

Let u1 = 1; u2 = 0 

f1x = – 500 + 1000 –



f1x = 583.33 lb

1000 + 250 = 583.33 6

Let u1 = 0, u2 = 1, f2x = 500 + 1000/6 = 666.7 lb (b)

118

u ( x)  (u2  u1 ) f1x u1 + f2x u2 = 

x  x1 L u  u1 (5 x 2 )  2 x  u1   4 

5 5 u1 (4)3 (u2 – u1) (44) + 16 3

Let u1 = 0; u2 = 1

f2x = (16) 5 = 80 kN Now let u2 =0; u1 = 1

f1x = – (16) 5 +

5(4)3 = 26.67 kN 3

f1x + f2x = 106.67 kN (Total force) Check parabolic yields

A= =

aR 3

(4) 5 (42 ) 

= 106.67 kN total force

3.59

Exact solution

P(x) = 18000 – 300 x u=

1 x P(x) dx AE 0

1 x (18000 – 300 x) dx AE 0

1  x2  18000 x  300  + C1  2 AE 

u (0) = 0  C1 = 0 u=

1 (18000 x – 150 x2) AE 119

1 2  30  10

(18000 x – 150 x2)

u = C0 + C1x + C2x2

Now choose

(1)

du = C1 + 2C2x dx Differential equation AE

du – P(x) = 0 dx

(2)

u(0) = 0  C0 = 0 Substituting (1) into (2) to yield R (error function)

AE [C1 + 2C2 x] – (18000 – 300 x) = R

(3)

By collocation: 2 unknowns so evaluate R at 2 points At x =

L and x = L 2

L L L R  C , x   0  AE C1  2C2  – 18000 + 300 =0   2 2 2  R(C, x = L) = 0 = AE [C1 + 2C2 L] – 18000 + 300 L = 0

(4)

Simplifying (4)

AE [C1 + C2 L] – 18000 + 150 L = 0

(5.1)

AE [C1 + 2C2 L] – 18000 + 300 L = 0

(5.2)

Subtracting (5.1) from (5.2)

AE C2 L + 0 + 150 L = 0 C2 =

 150 AE

(6)

Substituting (6) into (5.1)  150  AE C1  L – 18000 + 150 L = 0 AE  

AEC1 = 18000 – 150 L + 150 L

18000 AE Substituting (6) and (7) into (1), we get C1 =

(7)

18000 150  1 x– x = (18000 x – 150 x2) This gives the result which is same as the AE AE AE exact solution of differential equation. u=

Subdomain method: Use 2 subintervals as 2 C’s L/2

0

Rdx = 0 = 

L/2

{AE [C1 + 2C2 x] – 18000 + 300x}dx = 0

120

L/2 Rdx = 0 = L/2 {AE [C1 + 2C2 x] – 18000 + 300x}dx = 0

(8)

L /2

  x2  x2   AE C1x  2C2   18000x  300  2 20  

  x2  2 =0  AE C1x  2C2   18000x  150 x  2    L /2

(9)

 L L 2 L L 2 AE C1  C2    – 18000   + 150   = 0  2   2  2  2 2    L  L L 2 L AE C1  L    C2  L2     – 18000  L   + 150  L2     = 0 (10)    2  2  2 2    

and

Simplifying (10)  L L 2 L 2 AE C1  C2    – 9000 L + 150   = 0  2  2   2

(11.1)

 L  3L2  3L 2  =0 AE C1  C2  – 9000 L + 150  4   4   2

(11.2)

Subtracting (11.1) from (11.2)

AE C2 C2 =

L2 L2 + 0 + 150 =0 2 2

 150 AE

(12)

Substituting (12) into (11.1) 2  L  150   L  2   L AE C1    – 9000 L + 150   = 0     AE   2    2  2 

C1 =

18000 AE

(13)

Same values for C1 and C2 as previous solution. Same u as exact solution Least square method: 2 C’s need 2 integrals L

R

0 R Ci dx = 0 by (3.13.10) (i = 1 and 2) R R = AE 2x  AE (1) and  C2 C1 L

0 {AE [C1 + 2C2x] – 18000 + 300 x} AE dx = 0 121

(14) (15)

0 [AE (C1 + 2C2x] – 18000 + 300 x} AE 2x dx = 0

(16)

Integrating and simplifying

 300 x 2 x2 AE [C x + C ] – 18000 x + 2 1 2  2 2 

Subtract

 =0 

(171)

 x2 x2 x3 x3   2C2 AE C1 + 300 =0  – 18000 2 3 2 3 

(172)

x times (171) from (172) 2 2  x3 x3  x3  AE  x3   C2 + 0 + 300    = 0 2 4  3 3 AE x3

C2 =

1 1 C2 + 300 x3   = 0  6 12 

 300 6  150 = 12 AE AE

(18)

Substituting (18) into (171)

  150  2  2 AE C1 x    x  – 18,000 x + 150 x = 0  AE   C1 =

18000 AE

(19)

Galerkin’s method:

 R Wi dx = 0 by Equation (3.13.13)

(20)

Need 2 equations Let W1 = x and W2 = x2

(21)

0 {AE [C1 + 2C2x] – 18000 + 300 x] x dx = 0 L

0 {AE [C1 + 2C2x] – 18000 + 300 x} x2 dx = 0

(22)

  x2 x3  x2 x3   2C2   18000  300  = 0  AE C1 3 2 3 0  2  L

  x3 x4  x3 x4  2 18000 300    AE C C   =0  1  2 3 4 3 40    L2 2  AE C1  C2 L3  – 9000 L2 + 100 L3 = 0 2 3  

122

(231)

 L3 L4  300 4 3  C2 AE C1 L =0  –6000 L + 4 3 2   Subtract

(232)

2L times (231) from (232) 3  4 L4   300 – 200  L4 = 0 AE  C2 L4  C2  +0+  3  2 9  4   AE  

900  800  8  9  =0  C2 + 12 18  

1.5  100  18 C2 =    12  AE C2 =

 150 AE

(24)

(24) into (231)  L2 2   150 3   AE C1 L   – 9000 L3 + 100 L3 = 0     2 3 AE  C1 =

18000 AE

(25)

3.60

Exact solution

du = 0 dx

P – EA(x)

du =

P dx EA ( x)

P dx

P dx A –A  E  A1  2 L 1 x 

0 du = 0 EA ( x) dx u(x) = 

 u(x) =







PL   A  A1   ln  A1   2 x  ln A1  L   E ( A2  A1 )  x, in. 0



u(x), in. 0 123

6.66

3.642  10–4

13.32

8.099  10–4

19.98

1.384  10–3

Collocation method: Let u(x) = C1x + C2x2 + C3x3 u(0) = 0 satisfied

3 Ci’s. So need error to go to zero at 3 points. x= A(x) E

L 2L , ,L 3 3

du –P= 0 dx 

  A2  A1   2  A1   L  x  E(C1 + 2C2x + 3C3x ) – P = R (R = the error)



R(C , x)

x

L  0 , R (C , x ) 2L x 3 3

 0 , R(C , x)

xL

0

3 equations follow using Mathcad Given 2 1  20    20   3 C  20   – 1000 = 0 3  E C C   2 2      1 2 3   3  3   20   2 2   1   40   E C  2 C  40   3 C  40   – 1000 = 0          1 2 3   20   3    3  3   

[2 – (1)] E (C1 + 2 C2 20 + 3 C3 202) – 1000 = 0 11  200000   Find (C1, C2, C3)  0     3  80000000

C1 =

11 ; 200000

C2 = 0;

C3 =

3 80000000

uc(x) = C1 x + C2 x2 + C3 x3

Collocation solution uc(x) = 0

Exact solution u(x) = 0

6.66

3.774 10–4

3.642  10–4

13.32

8.212  10–4

8.099 10–4

19.98

1.398  10–3

1.384  10–3

124

(For use of other methods see P. 3.59 and P. 3.61) 3.61

Exact solution u=

x 1 P(x) dx AE 0

R1 =

1 (30  3) = 45 kN 2

P(x) = 45 – 10x

x = 45 – 5x2 2

x 1 (45 – 5x2) dx  AE 0

1  5 x3  x  +C 45 AE  3 

u(0) = 0  C = 0 1  5 x3   x 45 AE  3 

(A)

Let u = C1x + C2x2 + C3x3

(B)

 u= Collocation method:

du – P(x) = 0 AE dx

(1)

du = C1 + 2C2x + 3C3x2 dx

(2)

R = AE [C1 + 2C2 x + 3C3x2] – (45 – 5x2) = 0

(3)

3 C’s, therefore, need 3 equations L R  C , x   = 0  3 2L  R  C , x   = 0  3

(4)

R (C, x = L) = 0 125

Substituting for R using Equation (3) into (4)  L 2 L L 2 AE C1  2C2  3 C3    – 45 + 5   = 0  3  3  3   2L 2L 2  2L 2 AE C1  2C2    3C3     45  5   = 0  3  3   3  AE [C1 + 2 C2L +3 C3 L2 ] – 45 + 5 L2 = 0

(5)

Solving for C1 – C3 in Mathcad Given  L2 L L2 AE  3 C3  C1  2 C2  = 45 – 5 9 3 9   L2 L2 L AE  3 C3 4  C1  2 C2 2  = 45 – 5  4  9 9 3  AE (C1 + 2 C2 L + 3 C3 L2) = 45 – 5 L2  45   AE  Find (C1, C2, C3)   0   –5   

(6)

3 AE

Equation (6) into (B) u=

45 x  5 x3 +0+ AE 3 AE

(7)

Equation (7) is identical to exact solution given by Equation (A). Subdomain method: 3C’s, 3 intervals needed L/3

0 Rdx = 0 L/3

0= 

{AE [C1 + 2C2x + 3C3x2] – (45 – 5x2)}dx

2L/3

0= 

L/3

0= 

2L/3

{AE [C1 + 2C2x + 3C3x2] – (45 – 5x2)}dx {AE [C1 + 2C2 x + 3C3x2] – (45 – 5x2)}dx

(8)

Integrate and simplify (8)  L L 2 L 3 L 5 L 3 AE C1  C2    C3     45    = 0  3  3  3 3  3  3 L 1 7 3 L 5 7  L   45    L3 = 0 AE C1  C2  L2   C3   3 27  3 3 27  3 L 5 19 L 5 19 AE C1  C2 L2  C3   L3   45    L3 = 0   9 27 3 3  27   3  126

(9)

Solve using Mathcad for C1-C3 Given  L L 2 L 3 L 5 L 3 AE C1    C2    C3     45      = 0  3  3   3 3  3   3

1 1 7 35 3 AE C1 L + AE C2 L2 + AE C3 L3 – 15 L + L =0 3 3 27 81 1 5 19 95 3 AE C1 L + AE C2 L2 + AE C3 L3 – 15 L + L =0 3 9 27 81  45   AE  Find (C1, C2, C3)   0  Same C’s as in collocation method  –5    3 AE 

(10)

Least squares method: 3 C’s need 3 integrals R

0 R Ci dx = 0 by (3.13.10) R = AE, C1 R = AE3x2 C3

(11)

R = AE2x C2 (12)

0 {AE [C1 + 2C2x + 3C3x2] – 45 + 5x2] AE dx = 0 L

0 {AE [C1 + 2C2x + 3C3x2] – 45 + 5x2} AE 2x dx = 0 L

0 {AE [C1 + 2C2x + 3C3x2] – 45 + 5x2}AE 3x2 dx = 0

(13)

Simplifying (13) AE [C1L + C2L2 + C3L3 – 45 L +

5 3 L =0 3

 L2 2C2 L3 3C3 L4  L2 5 L4 AE C1    =0   45 2 3 4  2 4 

 L3  45 3 L4 3 AE C1 L  L5 = 0  C2  C3 L5   3 2 5 3   Solving (14) for C1 – C3 using Mathcad. Least squares method: Given 5 AE (C1 L + C2 L2 + C3 L3) – 45 L +   L3 = 0  3 127

(14)

  L2   L3   L4    L2   L4  AE C1    2 C2    3 C3     45    5   = 0  3  4   2  4   2   L3   L4   45 3 3 L  L5 = 0 AE C1    C2      C3 L5     3 2 5 3        45   AE  Find (C1, C2, C3)   0  Same C’s as in other methods  5     3 AE  Galerkin’s method: L

0 R Wi dx = 0

(3.13.13)

Need 3 equations Let W1 = x, W2 = x2, W3 = x3 L

0 {AE [C1 + 2C2x + 3C3x2] – 45 + 5x2} x dx = 0 L

0 {AE [C1 + 2C2x + 3C3x2] – 45 + 5x2} x2 dx = 0 L

0 {AE [C1 + 2C2x + 3C3x2] – 45 + 5x2} x3 dx = 0 L  x 2 2C2 x3 3C3 x 4  x2 5x4     =0 AE C1 45  2 3 4  2 4 0  L  x3 2C2 x 4 3C3 x5  x3 5 x5     =0 AE C1 45  3 4 5  4 5 0  L  x 4 2C2 x5 3C3 x6  x 4 5 x6 AE C1 =0      45 4 5 6  4 6 0 

Simplifying  L2 2C2 L3 3C3 L4  L2 5 4 AE C1    L =0   45 2 3 4  2 4 

  L3  C L4 3C L5  AE C1    2  3  – 15L3 + L5 = 0 2 5    3  L4 2C2 L5 C3 L6  45 4 5L6 L  AE C1   =0  4 5 2  4 6  Solve for C1 – C3 using Mathcad 128

Given   L2  L3 L4  L2 5 4  3 C3   45  L =0 AE C1    2 C2 3 4 2 4   2

  L3  L4 L5  AE C1    C2  3 C3  – 15 L3 + L5 = 0 2 5   3  L4 L5 L6  L4 L6 AE C1  2 C2  C3   45  5 = 0 4 5 2 4 6   45   AE  Find (C1, C2, C3)   0   –5    3 AE  3.64

Figure P3–64 Derrick truss (FS = 4.0) PRINT ELEMENT ***** POST 1 ELEMENT STAT CURRENT

TABLE ITEMS PER ELEMENT TABLE LISTING *** CURRENT

ELEM

SAXL

 3.3538E  12 50000 4.74 E  12  70000  70000 86023  98995

2 3 4 5 6 7 MINIMUM

MFORX 50000 4.74 E  12

 70000  70000 86023  98995

VALUES

ELEM VALUE 129

–98995

MAXIMUM

VALUES

ELEM

VALUES

86023

Figure 4 Free and deformed states Try W 21 22 A36 steel based on critical buckling member 7. Assumed Le = 2.1 L (conservative). 3.65

Figure P3–65 Truss bridge (FS = 3.0)

130

Load case: 1 of 1 Maximum value: 21627.6 lbf Minimum value: –28000 lbf Try S 8  18.4 A36 steel

(elements numbered)

or 4.5 4.5  165  square tube (assume pin-pin ends of members)

3.66

Figure P3–66 Tower (FS = 2.5)

131

Figure 4 Stress analysis model Try S 460  140 (mm 

kg ) A36 steel m

Pmax comp = – 466.35 kN in member CF

3.67

Figure P3–67 Boxcar lift (FS = 3.0) Try one of these cross sections – 1) a square solid bar, 3.25 in.  3.25 in., 2) a 6  6  1 in. structural square tube, 3) a W 8  35 wide flange section. Any of these made of A 2 36 steel. The critical force of – 120,000 lb is in element EG. So Johnson buckling formula dictates the section selected.

3.68

Figure P3–68 Howe scissors roof truss (FS = 2.0) Try an S 6  12.5 or a 3 in.  3 in.  63 in. structural square tube made of A 36 steel. The critical elements are AB and JL with force of –21,830 lb. Buckling dictates the cross section sizes recommended here.

3.69

132

Figure P3–69 Stadium roof truss (FS = 3.0) Try a W 6  25 of A 36 steel dictated by compressive force of –20,500 lb in elements AF, FJ, JL.

3.73

The recommended sqaure box section is 2 in. by 2 in. by ¼ in. thick. The maximum actual tensile stress is 360.4 psi < 46,000/3 psi So the truss is safe from yielding. The maximum actual compressive stress is 981.6 psi. Based on Euler buckling, the allowable compressive stress is 3,380 psi. So the truss is safe from buckling.

3.74

133

134

3.75

Nodal Displacements 135

The three largest displacements are shown on one side of the truss with the mirrored side of the truss having the same displacements

Principal Stress Plot The elements with the maximum stress of 1110 psi (high tensile stress) and minimum stress of −6450 psi (highest compressive stress) are shown. Those elements are also the same on the mirrored side. Of the truss.

136

Chapter 4 4.1 The degrees of freedom at each node of a truss element correspond to its axial displacement. It can be either 1, 2, or 3 DOF/s depending on how the element is oriented in 3D space. Trusses resist axial forces only. Beam elements have two DOF’s at each node, corresponding to transverse displacement and rotation. Beams resist shear forces and bending moments. 4.2 An element with axial force, shear force, and bending moment would be considered a beam column or frame element. Beam columns are commonly found in structures. For example, a building column that is also subjected to a relatively large bending moment. 4.3

Using Equation (4.1.7) plot N1,

dN 2 dN 4 , N3, dx dx

dN 2 dx

dN 4 dx

0 0.2L 0.4L 0.6L 0.8L 1.0L

1 0.896 0.648 0.352 0.104 0

1 0.32 –0.12 –0.32 –0.28 0

0 0.104 0.352 0.648 0.896 1.00

0 –0.28 –0.32 –0.12 0.32 1.00

where by Equation (4.1.7)

N1 =

1 (2 x 3  3x 2 L  L3 ) L3

N2 =

1 3 ( x L  2 x 2 L2  xL3 ) etc. L3

Plots

4.4

v (0) = v1 = a4 dv (0) = 1 = a3 dx

–1 v1 



v ( L) = v2 = a1L3 + a2L2 + a3 L  a4 dv ( L) = 2 = 3a1L2 + 2a2L + a3 dx





3 1 1 v2  2 ( L) = a1 L3  L3 + a2(0) + a3L – a3L + a4 2 2 2

v2 

2 L

         

1 3 1 L  ( 1 ) L  v1 = a1 2 2 2

a1 = v2 =

2 L L  v2  2  1  v1 3 2 2 L 2 3

 v2 



1L  v1 L3 + a2L2 – 1L  v1 2

a2L2 = v2  2v2  2 L  1L  2v1  1L  v1  a2 =

3 (21  2 ) ( v2  v1 )  L 2 L L2

3 1 1 2  v =  3  v1  v2  – 2 (1  2 )  x 3 +  2  v1  v2   (21  2 ) x 2 L  L  L L – 1 x  v1 Note: Terms with i s are opposite signs from v in Equation (4.1.4) f1y = V =

EI d 3v(0) EI  3 (12v1  12v2  6 L1  6 L2 ) L3 dx 3 L

m1 = m = EI

d 2 v(0) EI  3 ( 6 Lv1  6 Lv2  4 L21  2 L22 ) dx 2 L

f 2 y = −V =  EI

d 3v ( L) EI  3 ( 12v1  12v2  61L  62 L) dx 3 L

m2 = – m =  EI

d 2 v ( L) EI  3 ( 12 Lv1  12 Lv2  61L2  dx 2 L

62 L2  6 Lv1  6 Lv2  41 L2  22 L2 ) m2 =

EI L3

( 6 Lv1  6 Lv2  2 L21  4 L22 )

 f1 y   12 6 L 12 6 L   v1      EI  6 L 4 L2 6 L 2 L2  1  m1  =     L3  12 6 L 12 6 L   v2   f2 y    m2   6 L 2 L2 6 L 4 L2  2 

Note: All 6L terms have opposite signs from Equation (4.1.13), m1 and m2 are now negative of previous results.

4.5

L = l 2

Let Element 1–2

Element 2–3

 12 6l 12 6l   EI 6l 4l 2 6l 2l 2  ; [k1–2] = 3  l 12 6l 12 6l     6l 2l 2 6l 4l 2 

 12 6l 12 6l   EI 6l 4l 2 6l 2l 2  [k2–3] = 3  l 12 6l 12 6l     6l 2l 2 6l 4l 2 

 F1 y  ?   12 6l    2  M 0 1    6l 4l  F2 y   P  EI  12 6l   = 3  l  6l 2l 2  M2  0   F3 y  ?   0 0     0 0  M 3  ? 

12 6l 6l 2l 2

0   v1  0    0  1  ?  12 6l   v2  ?    6l 2l 2  2  ?  12 6l   v3  0    6l 4l 2  3  0

0   4l 2   EI   P  = 3  6l l  2   0   2l

2l 2  1    0   v2  8l 2  2 

6l 24 0

0 8l 2

24 0

12 6l 6l 2l 2

0 0

Rearrange





0   4l 2l   EI  2  0  = 3  2l 8l 2 l  P   6l 0   2



6l  1    0  2  24   v2 



Apply partition method N = [k – [kk–1 [k =

EI  4l 2   24 [ 6 l 0    2 l3  2l

d = N –1 F  v2 =

1

EI 2l 2   6l   ]  13.7148 3 2 l 8l   0 

l3 (– P) 13.7148 EI 140

7 Pl 3 7 PL3  v2  96 EI 768 EI

{d} = – [k–1 [k] {d}   4l 2 {d} =  1  =   2l 2 2  =

3 2l 2  6l   7 Pl       8l 2   0   96 EI 

1  0.2857 0.0714  6l   7 Pl 3      l 2 0.0714 0.1429   0   96 EI 

2  – Pl   – PL  1  32 EI EI 8    =  2  2    Pl  PL  2   32 EI   128 EI  2

Substituting back in the global matrix equation we have

 F1 y   12 6l    2 M 1    6l 4l  F2 y  EI 12 6l   = 3  l  6l 2l 2 M 2   F3 y   0 0     0 0  M 3  F1y =

12 6l 6l 2l 2 0 24 8l 2 12 6l 6l 2l 2 0

 0  0   – Pl 2    0   8 EI  3 –7 Pl 12 6l   96 EI     6l 2l 2   Pl 2  32 EI  12 6l     0  6l 4l 2     0  0 0

EI  6 Pl 3 7 Pl 3 6 Pl 3  EI 10 Pl 3 5P    F1 y    3 3   8 EI 8 EI 32 EI 32 EI 16 l l

Similarly M1  0

F3 y 

11P 16

F2 y   P

M3 

3PL 16

M2  0

4.6

12 6 L 12 6 L   EI  4 L2 6 L 2 L2  [K] = 3  L  12 6 L    Symmetry 4 L2  Boundary conditions v2 = 2 = 0 EI  12 6 L   v1  PL2  P   1      = 3  2 EI L 6 L 4 L2  1  0  v1 

 PL3 3EI

Matrix forces  PL  F1x  12 6 L 12 6 L   3 EI    M   2 EI   1 4 L2  6 L 2 L2   2PLEI      = 3 L  12 6 L   0   F2 y  Symmetry   M 2   4 L2   0  3

 F1y =

  PL3    PL2   EI    L 12 6   3 EI   2 EI   L3  

 F1y = – P Similarly

M1 = 0 F2y = P M2 = – PL

4.7

12 6 L 12 6 L  4 L2 6 L 2 L2  EI  0 24 [K] = 3  L  8L2   Symmetry E = 30  106,

0   0  12 6 L   6 L 2 L2  12 6 L   4 L2  0 0

I = 200 in.4,

L = 20 ft = 240 in.

 F1 y   10  v1       M1  0   1  F2 y  ?   v2  {F} = [K] {d}    = [K]   where v2 = v3 = 3 = 0  M2  0  2   F3 y  ?   v3      3   M 3  ?  500   12 6 L   30  106 (200)  2  0  = 6 L 4 L (240)3   6 L 2 L2  0 

6 L   v1    2 L2  1  8L2  2 

(1)

Solving for the displacements we have

1 = 0.0036 rad, 2 = 0.0012 rad, v1 = – 0.672 in. Substituting in the equation {F} = [K] {d} we have

 F1 y  12 6 L 12 6 L    4 L2 6 L 2 L2  M1    F2 y  30  106 (200)  0 24   =  (240) 8 L2 M 2    F3 y       Symmetry  M 3 

0  1.344 in.   0   0.0072   0 12 6 L     2 6 L 2 L   0.0024   0 12 6 L       0 4 L2   0 0

F1y = – 500, M1 = 0, F2y = 1250 lb, M2 = 0, F3y = – 750 lb, M3 = 60 kip-in. Element 1–2  f1 y  12 6 L   m EI   1 4 L2 =   3 L   f2 y    m2  

12 6 L 12

6 L  0.672    2 L2   0.0036    6 L   0   4 L2   0.0012 

f1y = – 500 lb

m1 = 0 f 2 y = 500 lb m2 = – 120,000 lb-in. Element 2–3  f2 y  12 6 L   EI   m2  4 L2 =   3 L   f3 y    m3  

12 6 L 12

f 2 y  750 lb 6L   0     2 L2  0.0024   m2  120, 000 lb-in.   f3 y   750 lb 6 L   0   m  60,000 lb-in. 4 L2   0  3

4.8

 F1 y   12 6 L 12 6 L 0 0   v1        2 2 M 0 0  1   1  6 L 4 L 6 L 2 L  F2 y  EI  12 6 L 24 0 12 6 L   v2    = 3    L  6 L 2 L2 0 8 L2 6 L 2 L2  2  M 2   F3 y   0 0 12 6 L 12 6 L   v3         0 0 6 L 2 L2 6 L 4 L2  3   M 3 

(1)

Boundary conditions v1 = 0, 1 = 0, v2 = – 0.5 in.  F2 y   24 0 12 6 L   0.5 in.      EI  0  0  8 L3 6 L 2 L2   2  =     L3  12 6 L 12 6 L   v3  500       6 L 2 L2 6 L 4 L2   3  0 

(2)

EI 3

(30  106 psi) (200 in.4 ) (240 in.)

= 434

lb in.

 – 0.5 in.  0  0  8 L2 6 L 2 L2         2  500  = 217  12 6 L 12 6 L    v3    2 2      0  6 L 2 L 6 L 4 L  3 

(3)

Solving (3) v3 = – 1.922 in.

2 = – 0.004325 rad 3 = – 0.006725 rad Back substituting into (1) F1y = – 99 lb M1 = – 96,250 lb  in. 

F2y = 599 lb Element 1 v1  0  f1 y    12 6 L 12 6 L         2 2 1  0 EI 6 L 4 L 6 L 2 L    m1       = 3 f   v 0.5 in. L  12 6 L 12 6 L   2   2y   2 2     0.004325  m2   6 L 2 L 6 L 4 L   2  99 lb    96, 250 lb  in.    =   99 lb   281,760 lb  in.

Element 2

 f 2(2) y   12 6 L 12 6 L   0.5 in.     m2(2)  EI 6 L 4 L2 6 L 2 L2  0.004325  (2)  = 3    L  12 6 L 12 6 L   1.922 in.   f3 y     (2)   6 L 2 L2 6 L 4 L2  0.006725 m  3  (2) f 2(2) y = 500 lb =  f 3 y

m2(2) = 120,000 lb  in. 

m3(2) = 0

4.9

Figure P4-9

 F1 y     M1    5000 EI   = 3 L  0    5000    0 

 24 12 L 24 12 L  2 12 L 4 L2 12 L 8L  24 12 L 6 L 36  2 6 L 12 L2 12 L 4 L  0 0 12 6 L   0 0 6L 2 L2

0  0    0 0  0  12 6 L   v2     (A) 6 L 2 L2  2  12 6 L   v3    6 L 4 L2  3  0

Solving the last four equations of (A) v2 = – 0.315 in.

2 = – 0.009 rad

(B)

3 = – 0.0135 rad (B) in (A) F1y =

(10  106 )(200) 603 (1000)

[– 24 (– 0.315) + 720 (– 0.009)]

= 10 kip M1 =

(10  106 )(200) 603 (1000  12)

[– 720 (– 0.315) + 14400 (– 0.009)]

= 75 kip  ft 

4.10

12 6 L 12 6 L   EI  4 L2 6 L 2 L2  [k1–2] = 3 L  12 6 L    Symmetry 4 L2  12 6 L 12 6 L   EI  4 L2 6 L 2 L2  [k2–3] = 3  L  12 6 L    Symmetry 4 L2  Boundary conditions v1 = 1 = v3 = 3 = 0 [K] =

EI 24 0    L3  0 8 L2 

 F2 y   20000 N  (210  109 )(4  104 )  24 0   v2    =    (3)3  0 8L2  2   M 2  20000 Nm  – 0.006428 = 24 v2  v2 = – 2.68  10–4 m 0.0064285 = 72 2  2 = 8.93  10–5 rad

  F1 y   12 6 L 12 6 L 0 0  0       2 2 0 0  0   M1   6 L 4 L 6 L 2 L 4  F2 y  (210  109 )(4  104 )  12 6 L 24 0 12 6 L  2.68  10    =    2 3 0 8 L2 6 L 2 L2  8.93  105  M2   6L 2L   F3 y   0 0 12 6 L 12 6 L  0        0 0 6 L 2 L2 6 L 4 L2  0 M3    F1y = 3.1  106 (– 12 (−2.68  10–4) + 6(3) (8.93  10–5))  F1y = 15000 N M1 = 3.1  106 (– 6(3) (−2.68  10–4) + 2(3)2 (8.93  10–5))  M1 = 20000 N  m 

Similarly F2y = – 20000 N M2 = 20000 N  m 

F3y = 5000 N M3 = – 10000 N  m 

Element 1–2

 0  f1 y   12 6 L 12 6 L       0 (210  109 )(4  104 )  6 L 4 L2 6 L 2 L2   m1   =    4  3 f 2.68 10     (3) 12 6 L 12 6 L    2y   5  2 2  m2   6 L 2 L 6 L 4 L   8.93  10   f1 y = 15000 N m1 = 20000 N  m 

f 2 y = – 15000 N

m2 = 25000 N  m 

Element 2–3 4  f2 y   12 6 L 12 6 L  1.34  10      (210  109 )(4  104 )  6 L 4 L2 6 L 2 L2   8.93  105  m2    =   12 6 L 12 6 L   (3)3   f3 y  0   m3    6 L 2 L2 6 L 4 L2    0

 f 2 y = – 5000 N m2 = −5000 N  m 

f 3 y = 5000 N

m3 = – 10000 N  m 

4.11

Using symmetry

 12 6  12  27 9 27 6 4  3 9 [k1–2] = EI  12  27  Symmetry

6  9



2  3   6 9  4  3 

 12 6  12 6  8 4 8 4   6 3  4  2 4 2  [k2–3] = EI   6 12  8 4   4   2  Applying the boundary conditions v1 = v2 = 3 = 0 we have

 43  M1  0  M  0  = EI  2 3  2    F   40,000 N   3y   0

2 3 10 3 –3 2

0  1   –3    2  2   3  v3   2 

which implies 0=

4 2 1 1 + 2  1 = – 2 3 3 2

2  1  10 1 3 v3  2 = (2) – v3   2  + 2 3 3 2 2

3 1 3 – 40,000 N = (70  109) (1  10–4)  –  v3   v3   2 2  2 

v3 = – 7.619  10–3 m  2 =

1 (– 7.619  10–3) m  2 = – 3.809  10–3 rad 2

 1 =

1 2 2

 1 = 1.904  10–3 rad 4 2 2  94 0 0   3 9 3 0    F1 y  2 4 2   2      3 0 0   1.904  10  3 3 3  3  M1     4  2 35 5  3 3    F2 y  0 3 18 6 2 2 9 –4  9   ) (1  10 ) = (70  10    3 5 2 4  2   3.809  103  M 2  3 3 6 3 1 2     F3 y  3  3 3  3 3   7.619  10    0 0 2 2 2 2     0  M 3  3 3    0 0 1 2 2 2 2 2 – 3.809  103  F1y = (7  106)  (1.904  10–3) + 3 3 





 F1y = – 8890 N Similarly F2y = 48890 N, M3 = 53330 N  m 

Element 1–2  4  f1 y  9    2  m1  9 –4  3   = (70  10 ) (1  10 )  4   f2 y  9  m2   2  3  f1 y = – 8890 N

2 3 4 3 2 3 2 3

4 9 2 3

4 9 2 3

2  3 





2  3   3  1.904  10    2 0   3  3.809  103   4  3 

m1 = 0 f 2 y = 8890 N m2 = – 26670 N  m 

Element 2–3 3  f2 y  2   3  m2  9 –4  2 = (70  10 ) (1  10 )   3  f3 y  2  m3  3 2  f 2 y = 40000 N = – f 3 y

3 2

2 3 2

3 2 3 2

3 2 3 2

3  2 

 0     3  1  3.809  10     3 7.619  103  2     0  2

m2 = 26670 N  m, m3 = 53330 N  m 



Elements 3–4 and 4–5 have same forces due to symmetry. Moments though will have opposite signs. 4.12

1

2

12 6 L 12  2 6 L 4L EI [K ]  3  3  12  KL L EI  Symmetry

6 L  v1  2 L2  1 6 L  v2  4 L2  2

Boundary conditions v1 = 1 = 0 Applying the boundary conditions on equation {F} = [K]{d} (29  106 )(200) 12  KL  F2   4000 lb  EI =    3 M  0 (20 12)   2   6 L

6 L   v 2    4 L2  2 

0 = – 6(240)v2 + 4(240)2 2  0 = – 6v2 + 4(240)2  v2 = 1602 – 4000 = 419.56 [(12 + (2.38) 1602 – 6(240)2] 

2 = – 0.01106 rad



v2 = 160(– 0.001106)  v2 = – 1.772 in.

Beam element  f1 y   12  6 L 12 6 L   0       6 2 2 (29  10 )(200)  6 L 4 L 6 L 2 L     m1  0     = (20  12)3  12 6 L 12 6 L   1.772   f2 y    m   6 L 2 L2 6 L 4 L2  0.01106   2 f1 y = 2230 lbs , m1 = 534 kip  in. 

f 2 y = – 2230 lbs , m2 = 0 The extra force at node 2 is resisted by the spring, ie., Fs = −kv2 = −1000 (−1.772) = 1770 lb 4.13

Applying symmetry

12 6 L 12  2 6 L 4L EI  [K] = 12  KL3 L3  EI  Symmetry

6L   2 L2  6 L   2 4L 

Applying the boundary conditions v1 = 0, 2 = 0 we have

2 6 L  1   M1  0  (70  109 )(2  104 )  4 L    = 12  KL3  v  43  F2 y   12000 N   6 L   2  EI

 0 = 4L21 – 6Lv2  1 =

6 6  v2  1 = v2 4L 16

 

6 – 12000 = 218750  24 v2  12.457 v2    16 



v2 = – 15.86  10–3 m

1 =

6 (– 0.014  10–3)  1 = – 5.95  10–3 rad 16

  F1 y  2  12 6L 6L   12      3  (70  109 )(2  104 )  6 L 4 L2  M1  6 L 2 L2   5.95  10    =    12 6 L 12.457 6 L  15.86  103  43  F2 y      M   6 L 2 L2 6 L 4 L2    2 0 F1y = 10.41 kN , M2 = 41.76 kN  m 

F2y = 0 kN  Fspring = (200

kN ) (15.86  10–3 m) m

Fspring = 3.18 kN From symmetry F3y = −10.41 kN  4.14

From symmetry

[k1–2] =

1

2

 23 EI  3 l 2 l3   3 2  3 l 2

3 l 2

3 2 3 l 2

3  l 2  v1 2  l  1 3 l  v2 2

[k2–3] =

2l 2 3 l 2

2

3 2 3 l 2

  2 2l 2 

3

 24 12 l 24 12 l  v2 12 l  8l 2 12 l 4l 2  2  24  12 l 24 12 l  v3   12 l 4l 2 12 l 8l 2  3

 wl    wl 2   1.5 1.5l 1.5 1.5l  3   2 1.5l l2   3wl   1.5l 2l  2  EI  1.5 15l 25.5 10.5l  wl 2  = 3  l  1.5l 10.5l 10l 2 l2  4    wl   0 0 24 12l   2   0 0 12l 4l 2  wl 2   12 

0 0

24 12l 24 12l

0   v1  0    0  1  0  12l   v2  ?    4l 2  2  ?  12l   v3  ?    8l 2  3  0 

Adding third row equation to fifth row equation we have





l 3 3wl wl = 25.5v2 + 10.5l2 – 24v3 – 24v2 – 12l2 + 24v3  EI 2 2 

2 wl 4 = 1.5v2 – 1.5l2 EI 155

 v2 =

4wl 4 + l2 3EI

(A)

Multiplying fourth row equation by –2 and third row equation by l and adding we have

l 3  3wl wl 2     = 25.5lv2 + 10.5l22 – 24lv3 – 21lv2 – 20l22 + 24lv3 EI  2 2  2 wl 5 = 4.5lv2 – 9.5l22 EI



(B)

Substituting (A) into (B)

 4 wl 4  2 wl 5  l2  – 9.5l22 = 4.5l   3EI  EI 

2 wl 5 18wl 5 = + 4.5l22 – 9.5l22 EI 3EI



4 wl 5 4 wl 3 = – 5l22  2  EI 5 EI

v2 =

 4wl 3  4 wl 4  l   3EI 5EI 



v2 

32 wl 4 15EI

Substituting in third row equation.  32 wl 4   4 wl 3  3wl 4 = 25.5   + 10.5l   – 24v3 2 EI 15 EI 5 EI  3wl 4 816 wl 4 42wl 4   = – 24v3  2 EI 15EI 5 EI

v3 

1839 wl 4 720 EI (Remember that l = L6 )

Now from symmetry

v4  v2 and 4   2  F1(ye)    M1( e)   (e)   F2 y  EI  (e)  = 3 l M 2   (e)   F3 y   (e)  M 3 

 1.5 1.5l 1.5 1.5l  2 1.5l l2 2l  1.5l  1.5 1.5l 25.5 10.5l  l2 10.5l 10l 2  1.5l  0 0 24 12l   0 0 12l 4l 2

0 0

24 12l 24 12l

 0  0    0  0  32 wl 4   12l   15 EI     4 wl 3  4l 2   5 EI  12l   1839 wl 4    720 EI  8l 2     0 

F1(ye) = 2 wl, M1( e) = 2.4 wl2, F2(ye) = –1.5 wl

M 2(e) = 0.25 wl 2, F3(ye) = – 0.5 wl, M 3(e) = 1.85 wl2 The equation {F} = [K] {d} – {F0} is now used to find the global nodal concentrated forces. L  F1 y  3wl  3w   6

  wl   F1 y   2 wl    wl 2  wL     F1 y  2   3  M wl 2.4 1 2      3   F2 y   1.5 wl   2 wl  2   =     wl 2   M  2.73 wl 2  3w  L  1 6  M 2   0.25 wl   4   0.5wl    wl   F3 y       2  wL2  M1  1.85 wl 2   wl 2   M 3  12  12  F2 y  0, M 2  0, F3 y  0 M3 =

21.5 wl 2 21.5  L 2 wL2  w   M3  12 12 6 24

In our case M3 is the maximum at midspan. From symmetry from elements 3–4 and 4–5

 M3 = So

– wL2 24

M3 = 0, F4y = 0, M4 = 0 F5 y 

wL 2

M5 

 wL2 12

Going back to the deflections we have

1839 w  L6  1839 wl 4 v3 = = EI 720 720 EI

 v3 

 wL4 507 EI

4.15

Applying the boundary conditions v1 = 1 = v3 = 3 = 0 EI EI  24 0   v2   wl  2    0 = 3 (8l 2)  2  0   = 3   2 l  0 8l  2  l  0 

– wl = F1(ye) = F3(ye) =

EI l3

(24v2) 

v2 

 wl 4 24 EI

wl wl 2 , M1(e)  , F2(ye ) = – wl, M 2(e) = 0, 2 4

wl  wl 2 ,M3= . These are obtained from the following matrix equation. 2 4 158

 F1(ye )    12 6l  M1(e)   4l 2  (e )   EI   F2 y   (e)  = 3  l  M 2   (e )    F3 y     (e)   M 3 

12 6l 6l 2l 2 0 24 8l 2

0  0    0  0   wl 4 12 6l   24 EI    6l 2l 2   0  12 6l   0    4l 2   0  0 0

{F} = [K] {d} – {F0}

 

F1y =

wl  wl wL  = wl = 2 2 2

M1 =

wl 2   wl 2  wl 2 wL2   =  = 4 12 3 12

F2y = – wl – (– wl) = 0, M2 = 0 – 0 = 0

 

F3y =

wl  wl wL wl 2  wl 2 wL2   , M3 =   2 2 2 4 12 12

v2 =

w( l ) 4 wl 4  wL4  2  v2  384 EI 24 EI 24 EI

4.16

After applying the boundary conditions v1 = v3 = 2 = 0 we have

1 

 wL3 24 EI

 – wL   L2 3L 0  1   48  5wL4 EI    v =  3 L 3 L v   24  wL    2   2 2 384 EI L3   2  2    0  3  3 L L wL  48  wL3 3  24 EI 2

Now

{F(e)}= [K] {d}  F1(ye)   12 3L 12 3L 0 0  0        wL3  2 (e) L M 1  2  3L 0 0   24 EI  3L 2 L  (e)    4  F2 y  0 12 3L    5 wL  EI 12 3L 24 2 2  384 EI   (e)  =  L2 3  3L L2 0 2L2 3L L2   0  M 2     (e)   0 0 12 3L 12 3L   0   F3 y    L2  wL3   (e)   0 3L L2   24 EI  0 3L 2 M 3  F1(ye) =

wL  wL2 wL , M1(e)  , F2(ye)  , M 2(e ) = 0, 4 48 2

F3(ye) =

wL wL2 , M3  4 48 160

{F} = {F(e)}– {F0} – wL  wL   4  4 wL  F1 y   2  2 , M1  0 F1 y  – wL – wL       2  M1   48   48    =  wL    – wL   wL  F3 y   4   4  , M3  0 F3 y   M   wL2   2  2 3 wL  48   48 

4.17

Total [K] for the whole beam  12 L 6 L2  2 4 L3  6L EI  12 L 6 L2 [K] = 4  L  6 L2 2 L3  0 0   0 0

12 L 6 L2

6 L2 3

2L 0 24 L 0 8L3 12 L 6 L2 6 L2 2 L3

0   0  12 L 6 L2   6 L2 2 L3  12 L 6 L2   6 L2 4 L3  0 0

v1 = 0, 1 = 0, v2 = 0 M   wL   8 L3 12   2   EI   wL  F3 y  2  = 4  6 L2 L  3   2  2L  M 3  wL  12  2

6 L2 12 L 6 L2

2 L3  2    6 L2   v2  4 L3  3 

Solving the 3 equations we get

2 

– wL3 or 8 EI

v3 

f1(1) y =

m1(1) =

f 2(1) y =

L4 4

– wL4  4 EI

(6L22) = (2L32) =

3 

3wL5 EI 4

4 EIL

– 7 wL3 or 24 EI

3wL or  4

2 wL6 EI  wL2 = or 8 EI L4 4

(–6L22) =

3wL  4

Reactions   3wL   3wL   F1 y   42  0   4    wL      wL2     M1  =  4    0  =  4   5 wL    wL   7 wL  F   2y   4   2   4 

using { f } = [k ] {d } – { f (0) }  f 2 y (2)   12 L 6 L2  (2)    m2  EI  6 L2 4 L3 =  (2)  L4  12 L 6 L2  f3 y   2  m (3)   6L 2 L3  3   m2(2) =

 0    2wL    12 L 6 L   wL3   2    wL  6 L2 2 L3   8 EI   12     wL4  –  12 L 6 L2   4 EI    2wL   6 L2 4 L3    7 wL3   wL2   24 EI   12  2

 wL2 2

4.18

Wdistributed =  w( x )v ( x )dx and 0

Wdiscrete = m11  m22  f1 y v1  f 2 y v2 and Wdistributed = Wdiscrete L

0 w( x ) v( x ) dx = m11  m22  f1 y v1  f 2 y v2  wx (since the L load is linear and increasing to the right), and in v(x) we substitute with a’s already evaluated, we get

Now evaluting the left side of the equation by substituting where w( x ) 

L  wx

0 w( x ) v( x )dx = 0 = 

L  wx

[a1x3 + a2x2 + a3x + a4]



 2 ( v  v )  1 (   )  x 3 – 1 2 1 2   L3  L2



 3 ( v  v )  1 (2   )  x 2   x  v dx 1 2 1 2  1 1 L  L2 

= 

L w  2

L  L3

( v1  v2 ) 

1 (1  2 )  x 4  2  L

 3 ( v  v )  1 (2   )  x 3  v x 2  v x  dx 1 2  1 1  L2 1 2 L 

w  2 1 w 3 1 (v1  v2 )  2 (1  2 )   2 (v1  v2 )   21  2     L L L 3 L  L L

x4 w x3 w x2  1  v1 4 0 L 3 0 L 2 0 =

 wL4  2 1 ( v1  v2 )  2 (1  2 )    5  L3 L 2 wL3  3 1   wL  wL v ( v v ) (2 )      1 2 1 2 1 4  L2 L 2  31

2 wL wL2 3wL ( v1  v2 )  (1  2 )  ( v1  v2 ) 5 5 4 

wL2 wL2 wL (21  2 )  1  v1 4 3 2

= m11  m22  f1 y v1  f 2 y v2 Now if we take the last equation and set 1  2  v1 = 0 and v1 = 1, we have 2 wL 3wL wL = f1y 1   5 4 2



8wL  15wL  10wL = 20

f1y 

f1 y 

3wL 20

2 = v2  v1 = 0 and

If 

 wL 2 wL2 wL2 = m1    5 4 3

m1 

1 = 1

 wL2 30

2  1  v1 = 0 and v2 = 1

If 

2 wL 3wL  = f2 y 5 4



f2 y 

7 wL 20

If 1  v1  v2 = 0 and 2 = 1 

 wL5 wL2  = m2 1  5 4

m2 

wL2 20

4.19

Work equivalent load system 12 6l 6l 12  2 6l 4l 2l 2  12  12 6l  6l EI  3  l  4l 2  4l 2   

F1 y  0   v1  0        0  M1 0   1    12 6l   v2   F2 y   1420wl  (3)     =  6l 2l 2   2   M 2  0  (4)   F3 y 12 6l   v3  0    M3  4l 2  3  0  0 0

Boundary Conditions v1 = 1 = v3 = 3 = 0 Use Equations (3) and (4)  – 14 wl  EI  24 0   v2   20  = 3    l  0 8l 2  2   0 

v2 =

7 wl 4 7 L4 w  240 EI 3840 EI

(L = 2l)

2 = 0 Reactions {F} = [K] {d} – {F0}

 F1 y  12 6l 12 6l    4l 2 6l 2l 2  M1    F2 y  EI  0 24   = 3  l  8l 2 M 2   F3 y    Symmetry     M 3 

F1y = F1y = M1 = M1 = F2y = M2 = F3y = M3 =

EI l3

3wl   20    0    0 0 2   0    wl  0 0    30  7 wl 4   7 wl   12 6l   240 EI     10   6l 2l 2   0   0  12 6l   0    3wl     20  4l 2   0   wl 2   30 

 7 wl 4   3wl  (– 12)   240 EI   20 

7 wl wL 3wl  10 wl +  wl = = =  20 20 20 2 4 EI l3

  L l 2

 7 wl 4    wl 2  1 7 (– 6l)      =    wl2 240 EI 30 40 30

5 2 5wL2 wl = 24 96 EI l3 EI l3 EI l3 EI l3

 7 wl 4   7 wl  (24)  =0  240 EI   10 

(0) – 0 = 0  7 wl 4   3wl  wl wL (– 12)  =   240 EI   20  2 4  7 wl 4   wl 2  5 2 5 (6l)  wl = wL4    = 240 EI   30  24 96

Note: Could use symmetry

v1 = 1 = 2 = 0

With Get 12 EI l

v2 =

7 wl 20

v2 =

7 wl 4 as before 240 EI 166

4.20

{F0} = [K]{d} wL  F01 y   320   12 6 L 12 6 L   v1  0    2 wL    M 01   30  EI  6 L 4 L2 6 L 2 L2  1  0  =     7 wL L3  12 6 L 12 6 L  v2  0  F02 y   20        6 L 2 L2 6 L 4 L2  2 wL2  M 02  20 



wL2 EI = 3 4L22  20 L

2 

wL3 80 EI

3wL 40  F1(ye)    0  12 6 L 12 6 L  wL2   (e)    M (e)   1 2 2 M 1  40 6 L 4 L 6 L 2 L   0   0    (e)  =  3wL   e ( ) 12 6 L 12 6 L    F2 y  F2 y   3   wL 2 2  (e)  40  6 L 2 L 6 L 4 L   80 EI  M 2  wL2 M 2( e)  20

F1(ye) 

wL  340    3wL   F1 y  9 wL  2   20 2  F1 y  wL   – wL   40   40 M 1  30    =   3wL  –    11 wL  F2 y   40   720wL  F2 y  M   wL2   wL2  40 2  20   20 

M1 

7 wL2 120



M2  0

   

4.21

 12 6 L 12 6 L 0 0    2 2 0 0   6 L 4 L 6 L 2 L EI  12 6 L 24 0 12 6 L  [ K] = 3   2 2 L  6L 2L 0 8L 6 L 2 L2   0 0 12 6 L 12 6 L     0 0 6 L 2 L2 6 L 4 L2  After imposing the boundary conditions and using work equivalence

v1 = 1 = v2 = 0, we have in {F} = [K]{d}   wL   8 L2 6 L 2 L2    12    2 EI   wL  =  6 L 12  6 L   v3   2  L3  2  wL2  2     2 L 6 L 4 L   3   12  2

(1) (2) (3)

From (1) and (3)

wL2 EI = 3 [– 8L22 + 6Lv3 – 2L23] 12 L wL2 EI = 3 [2L22 – 6Lv3 + 4L23] 12 L ------------------------------------------------wL3 EI = 3 [– 6L22 + 2L23] 6 L

(4)

From (2) and (3)  wL2 EI = 3 [– 3L22 + 6Lv3 – 3L23] 4 L

wL2 EI = 3 [2L22 – 6Lv3 + 4L23] 12 L ----------------------------------------------- wL2 EI = 3 [– L22 + L23] 6 L

(5) 168

Adding (4) and (5) we have wL2 EI = 3 [– 6L22 + 2L23] 6 L 2wL2 EI = 3 [2L22 – 2L23] 6 L ----------------------------------------------------------------------------EI wL2 = 3 [– 4L22]  2 L

2 

 wL3   1.524  103 rad 8 EI

  wL3   wL2 EI  2  = 3  L2    L 3   6 8 EI L 

Substituting in (5) 

 3 

7 wL3   3.556  103 rad 24 EI

Finally substituting in (1) 

v3 

 wL4   0.0122 m 4 EI

Reactions can be found from the global equation {F} = [K]{d} – {F0} F1y =

M1 =

F2y =

  wL3  EI 3wL [6L  ] – 0 = 6 L = – 12 kN   = 2 3 3 8 EI 4 L L

EI L3

[2L22] – 0 =

3  wL2 2   wL  2 L = = – 16 kN  m    8EI  4 L3



wL  7 wL  [–12v3 + 6L3]   = 28 kN 4  2  L 3

and M2 = 0 Element 1–2  f1 y   12 6 L 12 6 L   0  0      EI  6 L 4 L2 6 L 2 L2   0  0  m1  =      L3 12 6 L 12 6 L   0  0  f2 y  3       m2   6 L 2 L2 6 L 4 L2   8wL EI  0 f1 y =

3  wL2 wL = – 12 kN, m1 = = – 16 kN  m 4 4 

f2 y =

3 wL = 12 kN, m2 = – 32 kN  m 4 

Element 2–3  0    wL   f2 y   12 6 L 12 6 L   3   2 2   wL  wL   EI  6 L 4 L2 6 L 2 L2   8 EI   12   m2    wL4     wL    = 3 L 12 6 L 12 6 L   4 EI   2   f3 y     m3   6 L 2 L2 6 L 4 L2   7 wL3   wL2   24 EI   12   f 2 y = 16 kN

m2 = 32 kN  m 

f 3 y = m3 = 0

4.22

Global stiffness matrix 12 6 L 12  2 6 L 4L  3 24  kL EI  EI [K] = 3  L    Symmetry

6L 2

0 8L2

0  v1  0  1  12 6 L  v2 6 L 2 L2  2  12 6 L  v3 4 L2  3 0 0

{F} = [K]{d} 2  4 L2   wL  6 L 20    EI 6 L 24  kL3 6 wL EI  20  = 3  L   wL2   0 6L  20 

0    1 6 L  v  2  4 L2  3 

(1) (2) (3)

since 1 = – 3 we can ignore Equation (3)  Multiplying (1) by 3 and (2) by L and adding we have

– 3 wL2 EI = 3 [12L21 – 18Lv2] 20 L  – 6 wL2 EI  kL4   = 3   6 L21   24 L   v2  20 EI   L  

---------------------------------------------------------------–9 

wL2 6EI =  2  kL  v2 20  L 

 – 162000 = 140,060,000 v2  v2   0.0081 m

Substituting in (1) we have – 18000 = 5.6  107 1 + 11.34  104 

1   0.00235 rad

since 1 = – 3  3  0.00235 rad The reactions can be found by the global matrix {F} = [K] {d} – {F0}  F1 y   12 6L 12    2 6 L  M1   6L 4L 3  F2 y   EI 12 6 L 24  kL EI   = 3 L  6 L 2 L2 M 2  0   F3 y  0 12  0    0  M 3  0 6L

 720wL  0    2 0  wL  0   0.00235  20  2 L2     6 wL  0 12 6 L   0.0081  –  20  0   0  8L2 6 L 2 L2      7 wL  0 6 L 12 6 L     20  2 L2 6 L 4 L2   0.00235   wL2   20 

0 0

 F1y = 25.9 kN, M1 = 0

Element 1–2  7 wL   f1 y  12 6 L 12 6 L     20 2  0  wL     2 EI   m1  6 L 2 L2  0.00235  20  4 L   = 3    L  12 6 L   0.0081   320wL   f2 y  Symmetry    wL2   m2   0 4 L2    30 

 f1 y  25.9kN, m1  0, f 2 y  4.05 kN, m2  35.6 kN  m

Element 2–3  f2 y  12 6 L   EI   m2  4 L2   = 3 L   f3 y    m3  

12 6 L 12

 3wL  6 L    0.0081   20 2   wL     30  0 2 L2       7 wL  6 L   0   20  2    0.00235    wL2 4L    20 

 f 2 y  4.05, m3   35.6 kN  m, f3 y  25.9 kN, m3  0

Force in spring FS =

10 kN   0.0081 m  = – 8.1 kN m 

4.23

Global stiffness matrix of the beam  12 6 L 12 6 L 0 0    2 2 0 0   6 L 4 L 6 L 2 L EI 12 6 L 24 0 12 6 L  [K] = 3   2 2 L  6L 2L 0 8L 6 L 2 L2   0 0 12 6 L 12 6 L     0 0 6 L 2 L2 6 L 4 L2 

After imposing the boundary conditions v1 = 1 = v3 = 0 in {F} = [K] {d} 172 © 2023 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in

 24 0  wL  EI    2  0  = 3  0 8L L 2  wL  6 L 2 L2  12 

6 L   v3    2 L2  2  4 L2  3 

(1) (2) (3)

Multiplying (1) by L and (3) by – 4 and adding them EI – wL2 = 3 [24Lv2 + 02 + 6L23] L 2 EI  wL = 3 [– 24Lv2 – 8L22 – 16L23] 3 L --------------------------------------------------------

EI 4 wL2 = 3 [– 8L22 – 10L23] 3 L

(4)

Adding (2) and (4) we get 4 wL2 wL3 EI = 3 [– 8L23]  3  3 6 EI L Substituting in (2) 0=

EI  2 2 L2 wL3    8 L    2 6 EI  L3 

2 

 wL3 24 EI

Substituting in (1) – wL =

EI  EI 6 LwL3  24 v     – wL – wL = 24 3 v2 2 3  6 EI L L

  3 =

2 =

(1000 )(15  12)3 12





6 29  106 200



 wL4 12 EI

= 0.01396 rad

– ( 1000 )(15  12)3 12



v2 

24 29  106 200

= – 0.003491 rad

)(15  12)4 1 (1000 12 = – 1.2569 in.  12 (29  106 )  200 Substituting back in the global equation

v2 =

{F} = [K] {d} – {F0} we can find the reactions

 F1 y   12  6 L 12 6 L 0 0  0       2 2 0 0  0  M1   6 L 4 L 6 L 2 L   F2 y  EI  12 6 L 24 0 12 6 L   1.2569    = 3   L  6 L 2 L2 0 8 L2 6 L 2 L2  0.003491 M 2    F3 y   0 0 0 12 6 L 12 6 L          0 0 6 L 2 L2 6 L 4 L2   0.01396   M 3   wL   7500 lb   2  2  wL  12   225,000 lb  in.     wL   15,000 lb  –   0    wL     7500 lb 2    wL2  225,000 lb  in.   12 

F1y = 18750 lbs, M1 = 1350 K  in. 

F2y = 0, M2 = 0 F3y = 11250 lb, M3 = 0 Element 1–2  f1 y    EI  m1    = 3 L  f2 y   m2 

 12 6L 12 6L     7500  f1y  18750 lb 0       2 2  225000 m1  1350 k  in. 0  6L 4L 6L 2L     12 6L 12 6L  1.2569   7500  f2 y  – 3750 lb    6L 2L2 6L 4L2  0.003491  225000  m2  675 k  in.

Element 2–3  f2 y   12 6L 12 6L   1.2569   7500  f2 y  3750 lb   EI  6L 4L2 6L 2L2  0.003491 225000 m2   675 k  in.  m2       = 3  L 12 6L 12 6L  0   7500  f3 y  11250 lb  f3 y     m3   6L 2L2 6L 4L2   0.01396   225000  m3  0

4.24

12 6 L 12  4 L2 6  EI  24 [K] = 3  L    Symmetry

6L 2

2L 0 8L2

0   0  12 6 L   6 L 2 L2  6 L  12  4 L2  0 0

After applying the boundary conditions v1 = 1 = v2 = 0 in {F0} = [K] {d} 8L2 6 L 2 L2  2  (1)  –13333.33 ft  lb       6 L  v3  (2)  –17000 lb  = 30208.33  6 L 12   26666.67 ft  lb   2 L2 6 L 4 L2  3  (3)   Rewriting equations (1) (2) and (3) we get – 0.441379 = 8L22 – 6Lv3 + 2L23

(1)

– 0.562759 = – 6L2 + 12v3 – 6L3

(2)

0.882759 = 2L22 – 6Lv3 + 4L23

(3)

Adding (1) to – 4  (3) we get – 0.441379 = 8L22 – 6Lv3 + 2L23 – 3.53103 = – 8L22 + 24Lv3 – 16L23 -------------------------------------------------– 3.9724 = 18Lv3 – 14L23

(4)

Adding L  (2) to 3  (3) we get (where L = 10 ft) – 5.62759 = – 6L22 + 12Lv3 – 6L23 2.64827 = 6L22 – 18Lv3 + 12L23 -------------------------------------------------– 297931 = – 6Lv3 + 6L23

(5)

Adding Equation (4) to 3  (5) we have – 12.91034 = 4L23 

3   3.22758  10 –2 rad

Substituting in (4)  – 3.9724 = 180v3 – 1400 (– 3.22758  10–2)  v3   2.73103  10 –1 ft = – 3.27724 in. Substituting in (1) 175 © 2023 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in

– 0.441379 = 8L22 – 6L (– 2.73103  10–1) + 2L2 (– 3.22758) 

2   1.29655  10 –2 rad F1(ye) =

6 EI

2 = 2

6(29  106 )(150 in.4 ) (– 1.29655  10–2) = – 23500 lb (120)in.2

M1(e) =

2EI 2(29  106 )(150 in.4 ) (– 1.29655  10–2) = 78333 lb  ft 2 = 120  12 L

F2(ye) =

12 EI 6 EI v3  2 3 3 L L



6(29  106 ) 12(29  106 )(150) (– 3.27724) +  150 120  120  120 (120) 2

 (– 3.22758 10–1) = 40500 lb M 2(e) =

8L2 EI

2 

6 LEI L3

v2 

2 L2 EI L3

3 = – 13333.33 ft  lb 

F3(ye) =

6 LEI 12 LEI 6 LEI v3  3 3 = – 17000 lb 2  3 3 L L L

M 3(e) =

2 L2 EI 6 LEI 4 L2 EI  v   3 = 26,666.67 ft  lb 2 3 L3 L3 L3

Global forces



{F} = {F(e)} – {F0} F1y = – 23500 + 3000 = – 20500 lb M1 = – 78333.33 + 6666.67 = – 71,666.67 ft  lb 

F2y = 40,500 + 20,000 = 60500 lb M2 = – 13333.33 + 13333.33 = 0 F3y = – 17000 + 17000 = 0 M3 = 26666.67 – 26666.67 = 0 Element 1–2

Element 2–3

f1y = – 20500 lb

f2y = – 30,000 lb

m1 = –71666.67 ft  lb

m2 = 2000 kip  in.

f2y = 30,500 lb

f3y = 0



m2 = – 2000 kip  in. 



m3 = 0

4.25

After applying the boundary conditions on {F} = [K] {d} we have v1 = 1 = v3 = 3 = 2 = 0 So – wL =

 – 500  20 =

EI  KL3   24   v2 EI  L3 

1.6  106  100  4000(20  12)3   24   v2 (20  12)3  1.6  106 (100) 

 – 10000 = 277.78 v2  v2 = – 2.338 in. Reactions  F1 y   12 6L 6L 0 0   0   5000 lb  12     2 2 6L 2L 0 0   0  200000 in  l b  M1   6L 4L      F2 y   24  k 0 12 6L  2.338  10000   6 L  12    = 11.574     0  6L 2L2  M 2  0 8L2 6L 2L2   0         F3 y   5000 lb 0 12 6L 12 6L  0    0    200000 in  l b   0 0 6L 2L2 6L 4L2   0    M 3   F1y = 5325 lb M1 = 19,914 lb  ft 

F2y = 0 lb M2 = 0 F3y = 5325 lb M3 = – 19,914 lb  ft 

Element 1–2

 f1 y   12 6 L 12 6 L   0   5000         m  1 6 L 4 L2 6 L 2 L2   0  200000   = 11.574 –        12 6 L 12 6 L   2.338  5000   f2 y     m2   6 L 2 L2 6 L 4 L2   0   200000 

 f1 y = 5325 lb, m1 = 19914 lb  ft 

f 2 y = 4675 lb, m2 = – 13419 lb  ft 

Element 2–3 f 2 y = 4675 lb

m2 = 13419 lb  ft 

from symmetry f 3 y = 5325 lb

m3 = – 19914 lb  ft 

Note: Spring force is Fs = (4000

lb ) (2.338 in.) = 9352 lb in.

Equilibrium at node 2, Fy = 0  4675 lb from element 1  4675 lb from element 2  Fs = 9352 lb 4.26

v1 = 0 = v2 = v3  54 [k1–2] = EI   52

2  5 1  4   2 5

2 3 [k2–3] = EI  44 2  4

2  4  4  4

{F0} = [K]{d} 10416.667   0.8 0.4 0  1      9 –4  3750  = (210  10 ) (2  10 )  0.4 1.8 0.5  2         6666.67   0 0.5 1  3 

– 10416.667 = (210  109) (2  10–4) [0.81 + 0.42) 3750 = (210  109) (2  10–4) [0.41 + 1.82 + 0.53] 6666.67 = (210  109) (2  10–4) [0.52 + 3] Multiplying – 2 x (2) and adding it to (1) we have – 10416.667 = 4.2  107 [0.81 + 0.42] – 7500 = 4.2  107 [– 0.81 – 3.62 – 3] --------------------------------------------------------– 17916.667 = 4.2  107 [– 3.22 – 3] Adding (3) to (4) we have – 17916.667 = 4.2  107 [– 3.22 – 3] 6666.667 = 4.2  107 [0.52 + 3] -------------------------------------------------– 11250 = 4.2  107 (– 2.72)

(1) (2) (3)

(4)

 2  9.92  105 rad

Substituting into (4) we have – 17916.667 = 4.2  107 [– 3.2 (9.92  10–5) – 3]  3  1.091  104 rad

Substituting in (1)  – 10416.667 = 4.2  107 [0.8ϕ1 + 0.4(9.92  10–5)]  1   3.596  10 4 rad

Element 1–2  f1(ye )   0  12 6 L 12 6 L       (e )  4   m1  (210  109 )(2  104 )  6 L 4 L2 6 L 2 L2   3.596  10  =    (e )   12 6 L 12 6 L   53 0  f2 y    5  2 2   (2)   6 L 2 L 6 L 4 L   9.92  10  m2 

 fˆ1(ye ) = – 2625 N

m1( e) = – 10416.67 N  m 

m2( e) = – 2708.33 N  m 

 f1 y   2625   12500         m1  10416.667  10416.667      =   f2 y   2625   12500   m   2708.33  10416.667  2 f1y = 9875 N, m1 = 0

f2y = 15125 N, m2 = – 13125 N  m 

Element 2–3  f 2(ye)   0  12 6 L 12 6 L       (e)  5   9  4 2 2 (210  10 )(2  10 )  6 L 4 L 6 L 2 L   9.92  10  m2   (e)  =   12 6 L 12 6 L   43 0  f3 y    2 2   (e)    1.091  104   6 6 L L 2 4 L L m3 

 f2y = 13281.25 N

m2 = 13125 N  m 

f3 = 6718.5 N m3 = 0 Global F1y = f1y = 9875 N M1 = m1 = 0 F2y = 1525 + 13281.25 = 28406.25 N M2 = – 13125 + 13125 = 0 F3y = f3y = 6718.75 N M3 = m3 = 0 4.27

Figure P4–27

4.28

Figure P4–28

4.29

Figure P4–29

4.30

Figure P4–30 1 **** BEAM ELEMENTS number of beam elements =2 number of area property sets =1 number of fixed end force sets = 4 number of materials =1 number of intermediate load sets = 4

4.31

Figure P4–31

4.32

Figure P4–32

4.33 Design a beam of ASTM A36 steel with allowable bending stress of 160 MPa to support the load shown in Figure P4–33 Assume a standard wide flange beam from Appendix F or some other source can be used.

Figure P4–33

4.34 Select a standard steel pipe from Appendix F to support the load shown. The allowable L of bending stress must not exceed 24 ksi, and the allowable deflection must not exceed 360

any span.

Figure P4–34

1 **** BEAM ELEMENTS number of beam elements = number of area property sets = number of fixed end force sets = number of materials = number of intermediate load sets =

6 1 4 1 4

4.35 Select a rectangular structural tube from Appendix F to support the loads shown for the beam in Figure P4–33. The allowable bending stress should not exceed 24 ksi and allowable deflection ≤ 0.2 in. Answers for a 9” × 9” × ½ box section 187 © 2023 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in

Figure P4–35

4.36 Select a standard W section from Appendix F or some other source to support the loads shown for the beam in Figure P4–34. The bending stress must not exceed 160 MPa.

Figure P4 –36 Displacements/Rotations (degrees) of nodes NODE X– Y– Z– number translation translation translation 1 2 3 4 5 6

0.0000E+00 0.0000E+00 0.0000E+00 –4.2861E–03 0.0000E+00 0.0000E+00 0.0000E+00 –2.4492E–03 0.0000E+00 0.0000E+00 0.0000E+00 –9.7968E–03

0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00

X– rotation

Y– rotation

Z– rotation

0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00

0.0000E+00 –5.8470E–03 2.3388E–02 1.7541E–02 –9.3553E–02 –6.4317E–02

BENDING

1 **** BEAM ELEMENT FORCE AND MOMENTS ELEMENT

CASE

AXIAL

SHEAR

TORSION

NO.

(MODE)

FORCE R1

FORCE R2

FORCE R3

0.000E+00 –6.115E+04 0.000E+00 0.000E+00 –1.154E+03 0.000E+00 0.000E+00 –1.154E+03 0.000E+00 0.000E+00 5.885E+04 0.000E+00 0.000E+00 –5.654E+04 0.000E+00 0.000E+00 3.462E+03 0.000E+00 0.000E+00 3.462E+03 0.000E+00 0.000E+00 6.346E+04 0.000E+00 0.000E+00 –7.269E+04 0.000E+00 0.000E+00 –1.269E+04 0.000E+00 0.000E+00 –1.269E+04 0.000E+00 0.000E+00 4.731E+04 0.000E+00

MOMENT M1 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00

MOMENT M2 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00

MOMENT M3 –6.231E+04 3.115E+04 3.115E+04 –5.538E+04 –5.538E+04 2.423E+04 2.423E+04 –7.615E+04 –7.615E+04 5.192E+04 5.192E+04 0.000E+00

1**** BEAM ELEMENT STRESSES ELEMENT CASE P/A P/A + M2/S2 P/A–M2/S2 P/A+M3/S3 P/A-M3/S3 WORST SUM NO. (MODE) 1 1 0.000E+00 0.000E+00 0.000E+00 –1.139E+08 1.139E+08 1.139E+08 0.000E+00 0.000E+00 0.000E+00 5.695E+07 – 5.695E+07 5.695E+07 2 1 0.000E+00 0.000E+00 0.000E+00 5.695E+07 – 5.569E+07 5.695E+07 0.000E+00 0.000E+00 0.000E+00 –1.013E+08 1.013E+08 1.013E+08 3 1 0.000E+00 0.000E+00 0.000E+00 –1.013E+08 1.013E+08 1.013E+08 0.000E+00 0.000E+00 0.000E+00 4.430E+07 – 4.430E+07 4.430E+07 4 1 0.000E+00 0.000E+00 0.000E+00 4.430E+07 –4.430E+07 4.430E+07 0.000E+00 0.000E+00 0.000E+00 –1.392E+08 1.392E+08 1.392E+08 5 1 0.000E+00 0.000E+00 0.000E+00 –1.392E+08 1.392E+08 1.392E+08 0.000E+00 0.000E+00 0.000E+00 9.492E+07 –9.492E+07 9.492E+07 6 1 0.000E+00 0.000E+00 0.000E+00 9.492E+07 –9.492E+07 9.492E+07 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00

4.37 For the beam shown in Figure P4–37, determine a suitable sized W section from Appendix F or from another suitable source such that the bending stress does not exceed 150 MPa and L the maximum deflection does not exceed 360 of any span.

L 10 m = = 0.0278 m 360 360 Ymax = 0.0556 m

Bending stress max = 150 MPa = 1.50  108 Beam

I3 (m4)

S3(m3)

N m2

A(m2)

Bending stress ( N2 )

Ymax (m)

W310  143

0.000348

0.002150

0.018200

1.010 108

– 0.0269

W460  158

0.000796

0.00340

0.0201

6.389 10

– 0.0118

W760  257

0.003420

0.008850

0.0326

2.455 107

– 0.00275

W310  44.5

0.0000992

0.000634

0.00569

3.426 108

– 0.0944

W310  74

0.000165

0.001060

0.009480

2.049 108

– 0.0568

W310  107

0.000248

0.001590

0.013600

1.366 10

– 0.03776

For the problem given a W310  107 beam was chosen made of ASTM A36 steel. This made for a maximum deflection of 0.0378 m (see above table) which is less than the 0.0556 m maximum restraint.

4.38 For the stepped shaft shown in Figure P4–38, determine a solid circular cross section for each section shown such that the bending stress does not exceed 160 MPa and the maximum L deflection does not exceed 360 of the span.

Figure P4–38 Try small radius of 140 mm Large radius of 166 mm Yields max = 166 MPa close to

max allow = 160 MPa. Need to increase smaller diameter L 12 m 1 = = = 0.0333 m 360 360 30 Other trials shown below dAB and dDE

dBD, mm

(MPa)

(m)

279.6 290

332.5 340

166.3 MPa > 160 155.5

0.064 > 0.033 0.058

310

360

131

0.046

340

390

103

0.0330

 Too Large  Too Large  Finally at Limit of L 360

 Final dAB = 340 mm = dDE

dBD = 390 mm 4.39

Applying the boundary conditions v1 = 1 = 2 = v3 = 3 = 0 in the global equation {F} = [K] {d} we have



 P



wL EI = [24v2] 2 ( L2 )3

 P

wL 8 EI = 3 24v2  2 L

v2 

 PL3 wL4  192 EI 384 EI

6L 2

Reactions 6L  12 12 2 F  1y   2  62L 4  L2   26L   M 1    F2 y  8EI 12  26L 24  = 3  2 2L L  M2  6L 0 4  F3 y   0   0 12   M3  6L  0 0 2

 F1y =

2 L2 4

0 8L2 4 6 L 2 2 L2 2

  wL  0    4   0    wL2  0 0  0   48   6L    PL3  wL wL4  12 2 192 EI  384 EI   2   –      6L 2 L2     0  0 2 4     wL   6L  0    4  12 2     wL2  2 0 6L 4L   48   2 4

P  wL PL wL2 , M1 =  , F2y = 0 2 8 12

M2 = 0, F3y =

P  wL  PL wL2 , M3 =  2 8 12

4.40

After applying the boundary conditions v1 = 1 = 2 = v2 = 3 = 0 We have in the equation {F} = [K] {d} the following 40 P  40 Pl 3 EI = [24v ]  v = 2 2 27 24 EI (3l )3  v2 =  v2 =

L  5 Pl 3 since l = 3EI 6  5P ( L6 )3 3EI



v2 

5PL3 648 EI

4.41

Applying the boundary conditions v1 = 1 = 0   P  7 wL   2 EI  12 6 L   v2  20  = 3     2  L  6 L 4 L2  2   P L  wL   8 20 

(1) (2)

Multiplying (1) by L and (2) by 2 and then adding  PL 7 wL2 EI  = 3 [12Lv2 – 6L22] 2 20 L EI PL 2 wL2  = 3 [–12Lv2 + 8L22] 4 20 L ---------------------------------------------

( PL2  wL3 ) EI  PL wL2  = 3 (2L22)  2  4 4 8 EI L Substituting in (1)  

  PL2  wL3   P 7 wL EI   = 3 12v2  6 L     8EI 2 20 L 

5 P 22wL 12 EI (25P  22 wL) L3   = v v  2 2 240 EI 4 20 L3 3 wL P     2  20  0  F1 y   12 6 L 12 6 L        wL2 PL   0   8  30  EI  6 L 4 L2 6 L 2 L2    M1    (25 P  22 wL ) L3  –  P 7 wL    = 3 F   L 240 EI 12 6 L 12 6 L   2y     2  20   2 2    PL2 – wL3  M    2   6 L 2 L 6 L 4 L  2  wL    PL  EI 8 20 

wL PL 1 2 , M1   wL 2 2 3  F2 y  0, M 2  0 F1 y  P 

4.42

Assume the hinge as a part of the first element. Therefore, stiffness matrix for element 1 is

[k(1)] =

1 v2 2

 1 2 1 3EI  2 4 2 8   1 2 1   0 0 0

0  0 0  0

Stiffness matrix of element 2 is  12 12 12 12  v2 EI  12 16 12 8  2 [k(2)] = 8  12 12 12 12  v3    12 8 12 16  3

Adding the matrices by superposition v1

1

2

3

 3 6 3 0 0 0    0 0 0   6 12 6 EI [K] =  3 6 15 12 12 12    8  0 0 12 16 12 8   0 0 12 12 12 12     0 0 12 8 12 16 

Applying the boundary conditions v1 = 0, 1 = 0, v3 = 0, 3 = 0  

EI 15 12   v2   5 kN     =    8 12 16 2   0 

EI [15v2 + 122] = – 5000 N 8

(1)

12v2 + 162 = 0

(2)

From (2), v2 = –

4 2 3

(3)

Substituting (3) in (1) 5 5  103  8  4 15     2 + 122 =  3 210  109  2  104  – 82 = 

2 =

5  103  8 420  105 5000 420  105 195



2(2)  1.19  10 4 rad



v2 = – = –

Hence

4 2 3 4  1.19  10–4 3

v2   1.57  104 m

4.43

 1 2 1 3EI  2 4 2 [k(1)] = 8 1 2 1  0 0 0

0  0 0  0

 12 12 12 12  EI  12 16 12 8  [k(2)] = 8 12 12 12 12     12 8 12 16  By superposition

1

2

3

 3 6 3 0 0 0    6 12 –6 0 0 0  EI  [K] =  3 6 15 12 12 12  8   16 12 8   0 0 12  0 0 12 12 2 12     0 0 12 8 12 16  Boundary conditions v1 = 0, 1 = 0, v3 = 0 15 12 12   v3  5000   EI    12 16 8  2  =  0   8    12 8 16  3   0 

15 12 12 9.5  104    0 12 16 8  12 8 16  0 Select a11 = 15 as the pivot (a) Add the multiple

 a21 a11

 12 15

(b) Add the multiple

 a31 a11

4 5

of the first row to the third row.

4 5

of the first row to the second row

4 15 12 12 9.5  10    4  0 6.4 1.6 7.6  10   4   0 1.6 6.4 7.6  10 

Select a22 = 6.4 as the pivot Repeating the same procedure 15 12 12 9.5  104    4  0 6.4 1.6 7.6  10    9.5  104  0 0 6 

3  1.583  104 rad 2  1.583  104 rad v2   3.175  104 m

4.44

1

2

 32 3  12 12 12 12  2 EI   [K(1)] = 16 12 8  = EI  3 12  8 2  12 12 12 12  3   2  12 8 12 16 

3 2

2 3 2

3 2 3 2 3 2 3 2

3  2

 1 3  2   2

Assume the hinge as a + right end part of element (2)

[K(2)] =

2

v3 3

 1 1 1 0  3EI   1 1 1 0  (1)3   1 1 1 0     0 0 0 0 v3 3 v4  4

 12 6 12 6  EI   [K ] = 6 4 6 2  (1)3   12 6 12 6     6 2 6 4  (3)

By superposition

[ K] =

v1 1

2

 23 3 2  3 2 3 2

3 2 3 2 9 2 3 2

3 2

0  0  0  0

3 2

2 3 2

1 0 0 0 0

3

4

0  0 0 0 0 1  3 0 0 3 0 2  5 3 0 0 0 6 12 6  3 3 15  0 0 6 4 6 2   0 0 12 6 12 6 0 0 6 2 6 4  0

Applying the boundary conditions v1 = 1 = v2 = v4 = 4 = 0  5 3 0 2    0       EI  3 15 6  v3  =  10 kN      0 6 4  3   1.67 kN  m   2 =

3 v3 5

6v3 + 43 = −

(1)

1667 EI

– 416.75  3 =   1.5v3   EI 

(2)

EI(– 32 + 15v3 + 63) = – 10000

(3)

Substituting (2) and (1) in (3) –

9 10000 2500 v3 + 15v3 – – 9v3 = EI EI 5 

21 12500 v3 = – EI 5 198

v3   4.252  105 m

2   2.551  105 rad 3  5.386  105 rad 4.46

Figure P4–46 v  v  Y = AW  = AW G = Aw G  2 1  L

Y1 = – Y =

Y2 = Y =

 AW G  v2  v1  L

AW G  v2  v1  L

AW G  1 1  v1  Y1      = L  1 1 v2  Y2  [ k] =

4.47

AW G  1 1 L 1 1

From Equation (4.7.15)

p = 

L EI

 d   B T [ B ]  d  dx 

0 w  d   N  dx   d   P T

 v1    1 d =    v  2 2  EI  constant  p v1





L  2 EI L  T    =  B B dx  v1   N1wdx  f1 y = 0  0 0  2 

(1)

 p 1  p  v2  p  2

 2EI2  B B dx      N wdx  m = 0 L

 T  

(2)





L  2 EI L  T    =  B B dx  v2   N3 wdx 11  f 2 y = 0  0 0  2 

(3)

 2EI2  B B dx      N wdx  m = 0

(4)

 T  

Equations (1) – (4) in matrix form are  f1 y   v1   N1        L L N2  m  1  EI   B T  B  dx   –    wdx –  1  = 0 0 0 N  f2 y   v2   3  m2  2   N 4 

Simplifying L

EI   B T  B  dx  d     N T wdx   P = 0 4.49

p = 

EI ( v  )2 dx  

0 2

Lkfv

v = [N] {d} x =  y

dx   wvdx 0

d 2v =  yv  dx 2

x = – y [B] {d} [B] from Equation (4.7.10) L

p =   0

L1 L T 1   x  { x }dAdx   bTy vdx   k f {d }T [ N ]T [ N ]{d } dx 0 2 0 A2

bTy  w

L kf L EI L {d }T [ N ]T [ N ]{d } dx {d }T [ B ]T [ B ]{d } dx   w {d }T [N ]T dx +   0 0 0 2 2

 p d

= EI  [ B ]T [ B ] dx d   0

0 [ N ] wd x  0 k f [ N ] [ N ] dx d  T

 [k ] = EI  [ B ]T [ B ] dx  k f  [ N ]T [ N ] dx

 Equation.(4.7.19)

New part similar to convection part of heat transfer stiffness matrix.

4.79 Find the deflection at the mid-span using four beam elements, making the shear area zero and then making the shear area equal to 56 times the cross-sectional area (b times h). Then

make the beam have decreasing spans of 200 mm, 100 mm, and 50 mm with zero shear area and then 56 times the cross-sectional area. Compare the answers. Based on your program answers, can you conclude whether your program includes the effects of transverse shear deformation? Beam Span (m)

Shear Area

Displacement at center (m)

% difference

0.400

1.28E-03

4.61.E+00

0.400

0.001042

1.34E-03

0.200

1.60E-04

0.200

0.001042

1.91E-04

0.100

2.00E-05

0.100

0.001042

3.55E-05

0.0500

2.50E-06

0.0500

0.001042

1.02E-05

16.21 43.62 75.58

It would appear that the program DOES include the effects of transverse shear area which can be seen in the increasing per cent differences as the width of the beam approaches the span of the beam. As these width and span get closer and closer together the shear area becomes a larger factor, this would be the expected outcome if the program includes the effect of transverse shear area in the calculations. Note: For all of the beams the element definitions remained the same except the beam spans were changed. The figure below shows one example of when the 400mm beam was run with shear area included.

Chapter 5 5.1

Element (1) L(1) =

402  302 = 50 ft = 600 in.

cos  =

x2  x1

sin  =

y2  y1

(1)

30 – 0 = 0.6 50

40 – 0 = 0.8 50

E = 50000, L

12 I 2

= 0.0167,

6I = 5.0 L

 3.61  4.79  4   3.61 4.79  4  4.79 6.41 3  4.79  6.41 3     4 3 2000 4  3 1000   [k(1)] = 50000   3.61 4.79 4   3.61  4.79 4  4.79  6.41  3 4.79 6.41 – 3    3 1000 4 –3 2000  4

Element (2)

L(2) = 50 ft= 600 in. cos  =

30 – 0 40 – 0 = 0.8 = – 0.6 sin  = 50 50

203

 3.61  4.79   4 (2) [k ] = 50000   3.61 4.79   4

 4.79  4  3.61 4.79  4  6.41 3 4.79  6.41  3   3 2000 4 3 1000   4.79 4 3.61  4.79 4   6.41 3  4.79 6.41 3   3 1000 4 3 2000

After imposing the boundary conditions on each element stiffness matrix and assembling, we have  F2 x  5000  0 8  u2  7.22      0   v2   F2 y  0  = 50000  0 12.82    0 4000  2   0 M 2  0  Solving simultaneously, we obtain

u2 = 0.0139 in., v2 = 0, 2 = – 0.2775  10–4 rad The element forces are obtained using { f } = [k ] [T] {d} Element (1)

0 0  10 0  10 0  0 0.0167 5 0  0.0167 5    5 2000 0  5 1000   0 { f } = [k ] [T] {d} = 50000   0 10 0 0  10 0  0  0.0167  5 0 0.0167  5    5 0 0 5 2000  0    f 1x  4150 lb 0 0 0  u1  0  0.6 0.8 0      0.8 0.6 0 f   2.3 lb 0 0 0  v1  0    y         0 1387.5 lb in. m   0 1 0 0 0  1  0   1     =   0.0139 0 0 0.6 0.8 0     f 2 x  4150 lb   0    f 2 y  –2.3 lb  0 0 0 0 0.8 0.6 0        0 0 0 0 1  –0.2775  10 –4  m2  0  0  Element (2)  10 0 0 0 0  10   0  0.0167 5 0  0.0167 5   5 2000 0  5 1000   0 { f } =   0 10 0 0   10 0   0  0.0167  5 0 0.0167  0.5    5 1000 0  0.5 2000   0

204

0.8 0 0 0   0.6   0.8  0.6 0 0 0  0 1 0 0  0   0 0  0.6 0.8  0  0 0 0  0.8  0.6  0 0 0 0  0

0 0      0 0    0  0     0.0139 0    0 0    –4 1 0.2775  10 

 f 3 x  4150 lb   f   2.3 lb   3y  m3  1387.5 lb  in.    f 2 x  – 4150 lb   f 2 y  – 2.3 lb    m2  0  Equilibrium check

Fx = 0 5000 – 2(4150) cos 53.13° – 2 (2.3) cos 36.87°  0 Fy = 0 4150 (sin 53.13° – sin 53.13°) + 2.3 (sin 36.87° – sin 36.87°) = 0 5.2

205

E = 30 × 106 psi A = 10 in.2 I = 500 in.4

6I E 12 I = 0.104, = 12.5, = 125,000 2 L L L Element (1)

C (1) = 0,

S (1) = 1

0.104 0 12.5 0.104 0 12.5  10 0 0 0  10   2000 12.5 0 1000   [k (1)] = 125,000   0.104 0 12.5    10 0    2000   Symmetry Element (2)

C(2) = 1,

S(2) = 0

0 0 0 0  10  10  0.104 12.5 0  12.5    2000 0  12.5 1000   [k(2)] = 125,000   10 0 0    0.104 12.5   2000   Symmetry Element (3)

C (3) = 0,

S (3) = –1

10  0.104 0 10  0.104 0  10 0 0 0   10   2000  12.5 0 1000   (3) [k ] = 125,000   0.104 0  12.5   10 0    2000   Symmetry Boundary conditions are 206

u1 = v1 = 1 = 0, u4 = v4 = 4 = 0 Global equations {F} = [K] {d} are 0 12.5 10 0 0  u2  4000  10.104  10.104 12.5 0  12.5   v2   0       4000 0 –12.5 1000  2   0   125000     =  10.104 0 12.5  u3   0    10.104 12.5  v3   0   Symmetry    4000  3   0  

Solving simultaneously

u2 = 0.196 in., v2 = 0.000847 in., 2 = – 0.000103 rad u3 = 0.195 in., v3 = – 0.000847 in., 3 = – 0.000588 rad Local element forces { f } = [k ]{d } = [k  ] [T] {d} for each element Element (1)  0  1   0 (1) (1) [T ] {d} =   0  0   0

1 0 0 0 0 0

0 0 0 0 1 0 0 0 0 1 0 0

{ f }(1) = [k  ](1) [T ](1) {d }(1)

0 0 0 1 0 0

  0  0  0  0   0   0        0  0   0  =   0.196 –3  0   0.847  10  –3  0  0.847  10   –0.196     –3 –3 1 0.103  10  0.103  10   f 1 x (1)  1710 lb    (1)  f 1 y  2005 lb   (1)   m1  275,000 lb  in.  =   (1)  f 2 x  1710 lb    (1)  f 2 y  2005 lb   (1)   206, 000 lb  in.  m2

Similarly  f 2 x  1995 lb   f  1710 lb   2y  m2   206, 000 lb  in. { f }(2) = [k  ](2) [T ](2) {d }(2) =    f3 x  1995 lb   f3 y  1710 lb    m3   205,000 lb  in.  207

 f3 x  1710 lb   f  1995 lb   3y  m3  205,000 lb  in.  { f }(3) = [k ](3) [T ](3) {d }(3) =    f 4 x  1710 lb   f 4 y  1995 lb    m4  274, 000 lb  in. Free body diagram of frame (using local force results)

Check equation Fx = 0: 4000 – 2005 – 1995  0 Fy = 0: – 1710 + 1710 = 0 M1 = 275,000 + 274,000 + 1710 (20) (12

in. in. ) – 4000 (20) (12 )0 ft ft

5.3

208

bend < b =

2 y < 24000 psi 3 Mc I

Assume channel section, C6  8.2 Ix = 13.1 in.4, A = 2.40 in.2

Element (1) C= 1 12 I L2

S=0

= 2.3148  10–3 I = 0.03032

6I = 8.33  10–2 I = 1.092 L E 29  106 = = 4.028  105 (12) (6) L u1

1

2

0 0 0  2.4 0  2.4   0 0.0303 1.0917 0 0.303 1.0917    [k (1)] = (4.028  105)  0 1.0917 52.4 0  1.0917 26.2    0 2.4 0 0   2.4 0   0 0.0303  1.0917   0.0303 1.0917 0   1.0917 26.2 0  1.0917 52.4   0 Element (2) C = cos (−45°) = 0.707, S = sin (−45°) = −0.707 12 I 2

12 I 2

(8  82 ) 2 12

= 8.528  10–3

6I E = 0.5789, = 2.136  106 L L (2)  1.204 1.196   1.196 1.204  (2) 5 [k ] = 2.136  10     Symmetry  209

0.409 1.204

(3) 1.196

0.4094  1.204 0.409 1.196 0.4094   52.4 0.4094  0.4094 26.2   1.204 1.196 0.4094  1.204  0.4094   52.4 

Element (3) C = cos 0° = 1, S = sin 0° = 0 (3) (4) 0 0  2.4 0  2.4 0   0.0303 1.092 0  0.0303 1.092    52.4 0 1.092 26.2    [k(3)] = 4.028  105   2.4 0 0    0.0303 –1.092    52.4   Symmetry Boundary conditions u1 = v1 = 1 = 0 u4 = v4 = 4 = 0 Applying boundary conditions the reduced global equations become 2.554 0.8745 u2  12.24  2.554 0.8745  2.573   2.573 2.695  3.523 2.554 0.8745 v2      323 0.8745 0.8745 55.97   2  105    0.8745  u3  12.24 2.55   2.69 3.523  v3     323  3   Symmetry

 F2 x  0   F  2000  2y   M 2  0  =    F3 x  0   F3 y  2000     M 3  0  Solving simultaneously, we obtain u2 = –3.008  10–9 in. v2 = –0.402 in.

2 = –6.663  10–3 rad u3 = 3.30  10–9 in. v3 = –0.402 in.

3 = 6.663  10–3 rad Element forces Element (1)

210

 f 1x(1)   2.4 0   (1) f    1y  0.0303   (1)    m1  = 4.028  105    (1)  f 2 x     (1)   f 2 y    Symmetry  (1)   m2 

 0  2.4 0 0   0  0   1.092 0  0.0303 1.092     52.4 0 26.2   0 1.092   –9  2.4 0 0    3.008 ×10   0.0303 1.092    0.402    – 3 52.4    6.663×10 

 f 1 x (1)  0   (1)   f 1 y  2000 lb   (1)   106,900 lb  in.  m1 =   (1)  f 2 x  0   (1)   f 2 y  2000 lb   (1)   37060 lb  in.  m2

Element (2) –9  1.204 1.196  0.409  1.204  1.196  0.409  3.008  10     1.204 0.409  1.196  1.204 0.409   0.402    –3   52.4 0.409  0.409 26.2   6.66  10   2.316  105     1.204 1.196 0.409   3.3  10  9     1.204  0.409   0.402  Symmetry   52.4   6.66  10 –3    

211

 f 2 x(2)      0 (2) f    2y   0   (2)    m2    37060 lb  in. =    =  (2)   f 3 x   0  0   (2)    f 3 y    (2)   37060 lb  in.  m3 

Element (3) 2.4 0  0.0303   4.028  105      Symmetry

 2.4

1.092 52.4

0 0 2.4

–9   3.3  10     0.0303 1.092    0.402    1.092 26.2  6.66  10 –3     0 0 0     0.0303 1.092   0    52.4   0 

 f 3 x(3)    (3)   0   f 3 y    2000 lb   (3)   m3   – 37060 lb  in.  =   =   (3)  f 4 x   0    2000 lb  (3)   f 4 y     (3)   –106900 lb  in. m4 

212

Reactions

F1x = f 1x  (1) = 0 F1y = f 1y  (1) = 2000 lb M1 = m1(1) = 106900 lb  in. 

F4x = f 4 x(3) = 0 F4y = f 4 y(3) = 2000 lb M4 = m4(3) = – 106900 lb  in. 

b =

Mc 106900 (3 in.)  = 24,480 psi I 13.1 in.4

Close to allow = 24,000 psi 5.4

213

Element (1) C= 12 I 2

x4  x1 ~

= 0.0334,

20 40 = 0.447, S = = 0.895 44.7 44.7

6I E = 55.93 = 8.949, L L

Imposing boundary conditions u1 = v1 = 1 = 0, u2 = v2 = 2 = 0, u3 = v3 = 3 = 0 u4

4

447.9  90.87 178.2  358.8  223.7  [k ] =   Symmetry 179000 (1)

Element (2) 90.87  178.2 447.9  [k ] =  358.8 223.7    Symmetry 179000  (2)

0  400 0   1.334 400  [k ] =   Symmetry 160000  (3)

Equivalent forces (element 3) in. (20 kip) (50 ft) (12 ) PL ft = – 1500 kip  in. m04 = – = 8 8 f04y = – 10 kip 

214

Global equations 896 582 0  u4   0      v  719 400   10 kip  =    4   1500 kip  in.   Symmetry 517900  4    Solving u4 = 0.445  10–2 in., v4 = – 0.123  10–1 in. 4 = – 0.290  10–2 rad Element forces { f } = [k ] [T ]{d } – { f 0 } Element (1)

[T] {d}

 0  477 0 0 0 0  447   0      0  1.868 500.5   0 1.868 500.5 0  0   0   0  500.5 89490   500.5 179000 0  0    0.90193  10–2     0 447 0 0  0   447 0    0  1.868  500.5 0 1.868  500.5   0.94808  10–2  0        500.5 179000  0.2895  10–2  0 500.5 89490 0  0  f 1 x   4.04 kip  f    1.43 kip   1y    m1    254 kip  in.    =    4.04 kip   f 4 x   1.43 kip   f 4 y        513 kip  in.  m4  Element (2) Similarly  f 2 x   5.82 kip  f   –1.45 kip   2y    m2   – 260 kip  in.   =    f 4 x   – 5.82 kip   f 4 y   1.45 kip       – 519 kip  in. m4  Element (3) { f  }  f 4 x    –1.78 kip   0    1.78 kip f         8.83 kip 1.17 kip 10  4y        m4   – 468 kip  in.  –1500  1032 kip  in.    =      =   f 3 x    1.78 kip   0   1.78 kip  f 3 y   –1.17 kip   10    11.17 kip           – 236 kip  in.  1500    1736 kip  in. m3  Free-body diagrams of each element 215

Equilibrium at node 4

M4 = – 513 kip  in. – 519 kip  in. + 1032 kip  in.  0 



Fx = 1.43 cos 26.57° + 1.45 cos 26.57° – 1.78 + 4.04 cos 63.43° – 5.82 cos 63.43°  0 Fy = sin 26.57° (1.45 – 1.43) – 8.83 + sin 63.43° (4.04 + 5.82)  0

Reactions Support node 1

216

Fy = 4.04 sin 63.43° – 1.43 sin 26.57° = 2.96 kip  Fx = 4.04 cos 63.43° + 1.43 cos 26.57° = 31 kip 

M = 254 kip  in. 

Reactions support node 2

Fx = 1.45 cos 26.57° – 5.82 cos 63.43° = – 1.31 kip or  Fy = 5.82 sin 63.43° + 1.45 sin 26.57° = 58.86 kip  Mz = 260 kip  in. Reactions support node 3 Already in global x-y directions Fy = 11.17 kip, Fx = 1.78 kip , M = 1736 kip  in. 



5.5

217

Element 1–2 (1) 12 I 2

= 0.0178,

E 6I = 69.338, = 2.7735, C = 0.555, S = 0.832 L L

After imposing the boundary conditions u1 = v1 = 1 and u3 = v3 = 3 = we have  214.2 319.8 160  480.2  106.7  [k1–2] = 319.8    106.7 55470  160

Element 2 –3 (2) 12 I 2

= 0.0185,

6I E = 3.33, = 83.53, C = 1; S = 0 L L

0 0 853.3   [k2–3] = 0 1.54 277.49    277.49 66664  0

Equivalent forces

218

Assembled global equations. 160  0  1047.5 319.8  u2       481.74 170.8  v2   70  =    1200  122134.4  2    

Solving

u2 = 0.0562 in. v2 = – 0.1792 in.

2 = – 0.00965 rad Element forces {f } = [k ] {d } – {f0} = [k ] [T] {d} –{f 0} Element (1)  0   0     0  [T]{d} =     0.1179 in.    0.1462 in.      0.00965 rad  [k ] [T] {d} –{f 0} =

219

0  693.38 0 0  192 0  0.8875 192   55380 0 192 27690  693.33 0 0  0.8875  192   55380  f 1 x  90.07 kip  0   8.32   f 1x   0   5.55   f 1 y  f 1 y  3.83 kip        0   600   m1  m1  360.86 kip  in.  – =  = f 2 x  73.43 kip   0.1179   8.32   f 2 x    0.1462   5.55   f 2 y  f 2 y  7.27         0.00965  600   m2  m2  1106.35 kip  in.  693.38         Symmetry

0 0.8875

Reactions From the free body diagram of equivalent forces gives F1x = f 1 x (0.555) – f 1y  (0.832)  F1x = 46.8 kip

F1y = f 1 x (0.832) – f 1y  (0.555)  F1y = 77.06 kip M1 = 360.86 kip  in. 

Element (2)  833.3 0  1.5416        Symmetry

0  833.3 0 0   0.05618 277.488 0 1.5416 277.488   0.1792     66597 0  277.488 33299   0.00965    833.3 0 0    0   277.488  0 1.5416    66597   0 

 0   f 2 x   46.8 kip    160   f 2 y   17.1 kip          1800  m2   1108 kip  in.  –  =   =   0   f 3 x    46.8 kip    20   f 3 y   22.95 kip         1800   m3    2171 kip  in. Reactions

F3x = f 3x = – 46.8 kip  F3y = f 3 y = 22.95 kip  M3 = m3= – 2171 kip  in.

5.6

220

Replacement (Equivalent) force system

Element 1–2 (1)

C = S = 0.707, Since

12 I 2

= 0.0208,

E kip 6I = 88.3881 3 , = 3.536 L in. L

u1 = v1 = 1 = 0 u2

2

220.97   442.82 441.06 [k1–2] =  441.06 442.82  220.97     220.97  220.97 70710  Element 2 –3 (2)

C = 1, S = 0,

12 I 2

L u3 = v3 = 3 = 0 u2 v2

= 0.0289,

6I E = 4.166, = 104.167 L L

2

0 1041.67 0  [k 2–3] =  0 3.01 434.03    434.03 83333.28 0 Element 4 –2 (3) 221

C = – 0.60, S = 0.80,

12 I 2

= 0.0267,

6I E = 4, = 100 L L

L 361.71  478.72 320    [k4–2] =  478.72 641 240     320 240 80000  By direct stiffness method 540.97  0   1846.2  37.66  u2      v  1086.83 453.06   70  =   37.66  2   840 453.06 234043.28 2   540.97   Solving u2   0.000269 in.     v2  =  0.063 in.   – 0.00347 rad       2 Element 1–2 (1)

 0   0     0  [T]{d} =     0.0447 in.    0.0426 in.      0.00347 rad  { f }  f 1x   883.88 0   f  1.838   1y  6  m1  = (10 )      f 2 x    f 2 y       m2   Symmetry  { f 0 }

 k

[T ]{d } 0 0  883.88 0    0    0 311.92 0 311.92 1.838    70579.2 0  311.92 352.90   0     883.88 0 0    0.0447  1.838  311.92    0.0426     70579.2   0.00347 

f 1x  46.6 kip   7.07    7.07  f 1 y  6.07 kip     600  = m1  491.3 kip  in. –  f 2 x  32.4 kip   7.07    7.07  f 2 y  8.07 kip    600  m2  831.3 kip  in. From FBD of element (1) 1 2 (46.6 – 6.07) = 28.65 kip F1x = 2 1 2 (46.6 + 6.07) = 37.24 kip F1y = 2 M1 = m1= 491.3 kip in. Similarly Element 2–3 (2) Element 4 –2 (3) 222

f 2 x = – 0.28 kip

f 4x  = 50.2 kip

f 2 y = 58.31 kip

f 4 y = –1.49 kip

m2 = 1123.9 kip in.

m4 = – 154.2 kip in.

f 3x  = 0.28 kip

f 2x  = – 50.2 kip

f 3 y = 21.69 kip

f 2 y = 1.49 kip

m3 = – 1611 kip in.

m2 = – 293.2 kip in.

F3 x  0.28 kip

 F4 x  28.93 kip    F3 y  21.69 kip   Reaction   F4 y  41.05 kip   M 3  m3  1611.8 kip  in.  M 4  154.2 kip  in. 5.7

Equivalent force system

Boundary conditions

u1 = v1 = 1 = 0 u3 = v3 = 3 = 0 u4 = v4 = 4 = 0

Element (1) C = 0, S = 1 223

12 I 3

6I 2

= 2.344  10–6,

4I = 5  10–5 L

= 9.375  10–6,

A = 1.25  10–3 L

1.875  105

1.0  102

9.375  106   0  4  4.0  10 

0 210  10 [k ] = 1.875  105  8  Symmetry

  9.375  10   4.0  104 

[k(1)] =

6 

210  10  8   Symmetry

Element (2) C = 1, S = 0  1.0  102

6 

(2)

6

Element (3) C = 0.707, S = – 0.707 12 I 3

6I 2

= 8.286  10–7,

4I = 3.536  10–5 L

= 4.6875  10–6,

A = 8.839  10–4 L

[k(3)] =

210  10 8 2

5.005  103

 4.995  103

   Symmetry

5.005  103

6 

3.75  105   3.75  105   4.0  104 

Global equations 1.917  10 –2  20 6  210  10     40  = 8 2   40     Symmetry

 4.995  103 1.436  104  u2    1.917  102 1.436  104  v2    1.531  103  2 

Solving simultaneously u2 = 0.4308  10–4 m, v2 = – 0.9067  10–4 m

2 = – 0.1403  10–2 rad Local element forces Element (1) Effective forces = [k  ] {d } f1(xe ) = 23.8 kN, f1(ye ) = – 2.74 kN, m1( e ) = – 7.28 kN  m  

f 2(xe ) = – 23.8 kN, f 2(ye ) = 2.74 kN, m2( e ) = – 14.65 kN  m  

224

Actual forces { f } = [k  ] {d } – { f 0 } f 1x = 23.8 – 0 = 23.8 kN  f 1y = – 2.74 + 20.0 = 17.26 kN 

m1 = – 7.28 + 40 = 32.77 kN  m  

f 2x = – 23.8 kN  f 2 y = 2.74 + 20 = 22.74 kN 

m2 = – 14.65 – 40 = – 54.64 kN  m  

Element (2) Effective forces f 2(xe ) = 11.31 kN, f 2(ye ) = – 2.81 kN, m2( e ) = – 14.9 kN  m  

f 3(xe ) = – 11.31 kN, f 3(ye ) = 2.81 kN, m3( e ) = – 7.54 kN  m  

Actual forces { f } = [k ] {d } – { f 0 } f 2 x = 11.31 – 0 = 11.31 kN  f 2 y = – 2.81 – (– 40) = 37.19 kN 

m2 = – 14.91 – (– 80) = 65.09 kN  m f 3 x = – 11.31 – 0 = – 11.31 kN   

f 3 y = 2.81 – (– 40) = 42.81 kN 

m3 = – 7.54 – 80 = – 87.54 kN  m  

225

Element (3) { f 0 } = 0 f 2 x = 17.55 kN f 2 y = – 1.40 kN

m2 = – 10.51 N  m  

f 4 x = – 17.55 kN f 4 y = 1.40 kN

m4 = – 5.30 kN  m  

5.8

226

Calculate [k]’s based on node 2 and 3 contributions as u1 = v1 = 1 = 0, u4 = v4 = 4 = 0 Element (1) C = 0, S = 1 (3) 122I 0 L [k ] = E   A L Symmetry (1)

6I  L

 0  4I 

Element (2) C = 1, S = 0

227

(2) 0 A 0  12 I 16 I  L L2  E (2) [k ] = 4I  L     Symmetry

(3) A

 12 I

0 A

L2  6I L

0 12 I L2

0     2I  0   6I  L  4 I  6I L

Element (3) C = 0, S = – 1 (3)  122I 0 L [k ] = E   A L  Symmetry (3)

6I  L

 0  4I 

Assemble global [K] and Equations {F} = [K] {d} use numerical values for E, I, A, L 0 6.25  15 0 0  u2  15.05 2500    0  15.05 6.25 0 0.0521 6.25   v2        2000 0  6.25 500  2  100,000  E   =     15.05 0 6.25   u3  L  0   0   15.05  6.25  v3   Symmetry      2000  3   0   Solve simultaneously using an equation solver 5.9

Element [k]’s Element (1) C = 0, S = 1

228

12 I 2

12(2  10 4 ) 4

= 1.5  10–4 m2

6I 6(2  10 4 ) = 3.0  10–4 m3 = L 4

kN E 210  106 = 5.25  107 = 4 L m u2

1.5  104 [k(1)] = 5.25  107    

2

3  104   0   8  104 

2  102

Element (2) C = 1, S = 0 12 I

12(2  10 4 ) 52

= 9.6  10–5 m2

6I 6(2  10 4 ) = 2.4  10–4 m3 = L 5

kN E 210  106 = 4.2  107 = 5 L m (2)  2  102 0  [k(2)] = 4.2  107  9.6  105  Symmetry 

  2.4  10   8  104  0

4

Assemble global equations for node 2 8.48  102 0 1.58  103  u2  0       7 1.05  101 1.01  103  v2   0  = 10   Symmetry   600    7.56  103  2  

Solving simultaneously u2 = – 1.486  10–4 m, v2 = – 7.674  10–5 m

2 = 7.978  10–3 rad Element forces { f } = [k  ] [T] {d} 229

 0  1   0 [T] {d} =   0  0   0

1 0 0 0 0 0

0 0 0 0 1 0 0 0 0 1 0 0

 0  0  0   0    0   0   1.486  104  0   0   7.674  105    1  7.978  103 

0 0 0 1 0 0

 0   0     0  { f } = [k  ](1)  5  7.674  10   1.486  104     7.978  103  :  .  . 7  { f } = 5.25  10 .  .  :

: :  2  102

 0    0 3  10      4  104   0   7.674  105  0   4   1.486  104   3  10    3  4  7.978 10     8  10  0

4

. .

1.5  10

. .

 3  104

. .

2  102

. .

1.5  104

: :

 3  104

4

Multiplying matrices yields f 1 x (1) = 80.58 kN = – f 2 x (1) , f 1y  (1) = 124.48 kN =  f 2 y (1) m1(1) = 165.2 kN  m, m2(1) = 335.1 kN  m

Similarly for element (2) { f }(2) = [ K ](2) [T ](2) {d }(2) yields f 2 x (2) = 124.48 kN =  f 3 x (2) f 2 y (2) = 80.58 kN =  f 3 y (2)

m2(2) = 264.8 kN  m, m3(2) = 138.0 kN  m

230

5.10

Element (1) E 210  109 12 I = 70  109, 2 = 2.67  10–4 = L 3m L 6I = 4  10–4, C = 1, S = 0 L u1 = v1 = 1 = 0 and u4 = v4 = 4 = 0 u2

 1  102 0 [k(1)] = 70  109  2.67  104   Symmetry 231

2    4  10   8  10 –4  0

4

Element (2) E 210  109 = = 49.5  109, L 3 2

12 I L2

12(2  104 ) = 1.33  10–4 18

6I = 2.83  10–4, C = 0.707 = – S, 4I = 8  10–4 L u3 v3 3  v2 d2  u2 3.58  3.48 0.1415 3.58 3.48 0.1415      3.58 3.58 0.1415  3.48 0.1415    [k(2)] =  0.566  0.1415  0.1415 0.2828  70  106  3.48  0.1415 3.58    0.1415 3.58  Symmetry  0.566 

Element (3) E 6I 12 I = 70  109, 2 = 2.67  10–4, = 4  10–4 L L L C = 1, S = 0 u3 v3 3

1  102 0  2.67  104 [k(3)] = 70  109  0   Symmetry

  4  10   8  104  0

4

Global equations 0.0035 0.00014  u2   0  0.0135  0.0035 0.00014 0.0036      0.0038  0.0003  0.0035  0.0036 0.00014 v2 10000      0.0014  0.00014  0.00014 0.00028  2  5000   9 10 10    =   0.0035  0.00014 u3   0 0.0135    0.0038 0.0003  v3  10000      0.0014  3   5000   Symmetry

Using an equation solver u2 = 0.16  10–10 m

u3 = 0.85  10–11

v2 = – 0.1423  10–2 m

v3 = – 0.1423  10–2 m

2 = – 0.5917  10–3 rad

3 = 0.5917  10–3 rad

Element forces Element (1)

232

1 0  0 [T] {d} =  0 0  0

  0 0 0 0 0 0   0  1 0 0 0 0    ~ 0 1 0 0 0  0    0 0 1 0 0  0  0 0 0 1 0  0.1423  102    0 0 0 0 1  3    0.5917 10  

{ f }(1) = [k ](1) [T ](1) {d }(1) , C1 

AE EI , C2  3 L L

 0 0 0 0  0  C1 C1  0    12 C2 6 C2 L 0 12 C2 6 C2 L      0  2   4 C2 L 0  6 C2 L 2 C2 L   0  =  10  0 0   0.16 10 C1    12 C2  6 C2 L  0.1423 102      Symmetry 4 C2 L2   0.5917 103   

f 1 x (1) = 0 f 1 y (1) = 10028 N

m1(1) = 23276 N  m f 2 x (1) = 0 f 2 y (1) = – 10028 N

m2(1) = 6709 N  m

Element (2)  0.707  0.707 0 0 0 0  0.707 0.707 0 0 0 0   0 1 0 0 0  0 [T] =            

233

 0    4  0.387  10  0 11, 720  { f }(2) = [k  ](2) [T ](2) {d }(2) =    0   0.387  104  0     11, 720 

where [k  ](2) from Equation (6.1.8) text  0.16  1010  0    2  0.1423  10   3   0.5917  10  {d}(2) =   11  0.85  10  0   2   0.1423  10   0.5917  103    Similarly for element (3) f 3 x (3) = 0 f 3 y (3) = – 10,028 N

m3(3) = – 6709 N  m f 4 x (3) = 0 f 4 y (3) = 10,028 N

m4(3) = – 23276 N  m 5.11

Figure P5–11 This problem is done using symmetry and Mathcad E = 70  109 234

A = 3  10–2 I = 3  10–4

Element 1 x1 = 0

x2 = 3

y1 = 0

y2 = 4

L1 =

 x2  x1  2   y2  y1  2

x2  x1 L1

C = 0.6 N=

y2  y1 L1

S = 0.8

12I

L12

6I L1

 AC 2  NS 2   ( A  N )CS   MS  E  k1 =     L1   AC 2  NS 2    ( A  N )CS   MS 





( A  N )CS

 MS

AS 2  NC 2 MC

MC 4I

  A  N  CS

 AS 2  NC 2



  MC



 AC 2  NS 2



 ( A  N )CS



 ( A  N )CS

 AS 2  NC 2

 MC

( A  N )CS

( A  N )CS MS

AS 2  NC 2  MC

AC  NS



 MS   MC   2I   MS    MC  4 I 

Boundary conditions with symmetry u1 = 0 M2 = 0 u2 = 0 v1 = 0 2 = 0 1 = 0 Reduced set of equations Guess v2 = 1 Given F2y = – 30  103 E  AS 2  NC 2  (v2) (F2y) =   L1 





v2 = find (v2) 235

v2 = – 11.13  10–5 m Displacement of node 2  F1x   u1  F  v   1y   1  M1   1    = k1   u  F2 x   2 F   v2   2y     M   2  2

0    u1    v  0    1   0   1  Displacements and rotations u  =  0 2      11.13  105   v2       2  0  

 F1x   22331.7   F  =  30000  Reaction forces    1y   336.6   M    1 F1 =

F1x 2  F1 y 2

F1 = 37399 Reaction magnitude at node 1 same as node 3 by symmetry. Forces in elements

C S  0 T=  0  0  0 C1 =

AE L1

 C1  0   0 [k] =    C1  0   0

S C 0 0 0 0

0 0 1 0 0 0

C2 =

EI L13

0 12C2

0 6C2 L1

 C1 0

0  12C2

6C2 L1 0  12 C2

4C2 L12 0  6C2 L1

0 C1 0

 6C2 L1 0 12 C2

6C2 L1

2C2 L12

 6C2 L1

 f1x  f   1y   m1    = [k ] [T] {d}  f2x  f   2y   m  2

 u1  v   1  1  d=   u  2  v2     2 

0 0 0 0 0 0  0 0 0 C S 0   S C 0  0 0 1

0  6C2 L1   2C2 L12   0   6C2 L1   2  4C L  2

 f1x   37,399  f     1 y   135   m1   336    =   f 2 x   37,399  f   135   2y    m   336  2

Forces in element 1 and 2 equal by symmetry. 5.12 Determine displacements and rotations of the nodes, element forces, and reactions. Material properties & geometry 236

EE = 210  109 Pa

Modulus of elasticity

AA = 80  10–3 m2

Area of cross section of all elements

II = 1.2  10–4 m4

Area moment of inertia of all elements

LL1 = 3 m

Length of element 1

LL2 = 6 m

Length of element 2

1 = 0 deg

Angle between local x and global x for element 1.

2 = – 90 deg

Angle between local x and global x for element 2.

Figure P5.12 Applied loads

F1y = – 10000 N

Applied load (down at node 1)

Boundary conditions

u3= 0

x-displacement at node 3 is zero.

v3 = 0

y-displacement at node 3 is zero.

3 = 0

Angular displacement. at node 3 is zero.

Defining element properties in unitless format. (Mathcad does not allow elements with dissimilar units within the same matrix.)

E= A1 = A2 = I1 = I2 =

EE Pa AA m2 AA m2 II m4 II m4

Modulus of elasticity (Pa). Cross sectional area of element 1 (m^2) Cross sectional area of element 2 (m^2) Area moment of element 1 (m^4) Area moment of element 2 (m^4) 237

L1 =

LL1 m

Length of element 1(m)

L2 =

LL2 m

Length of element 2 (m)

C1 = cos (1)

Cosine of angle between local x and global x for element 1.

S1 = sin (1)

Sine of angle between local x and global x for element 1.

C2 = cos (2)

Cosine of angle between local x and global x for element 2.

S2 = sin (2)

Sine of angle between local x and global x for element 2.

Functional equations for the global stiffness matrix for a 2D beam/frame element with axial effects. Refer to text Equation (5.1.11) Functional expressions for large repeated terms in global stiffness matrix.

M1(A, C, S, I, L) = AC2 +

12 I 2

12 I M2 (A, C, S, I, L) =  A  2  CS  L 

M3(A, C, S, I, L) = AS2 +

12 I

Functional equation for global stiffness matrix of beam/frame element.  M 1 ( A, C , S , I , L) M 2 ( A, C , S , I , L)   M ( A, C , S , I , L) M ( A, C , S , I , L) 3  2 I  6 6I S C E  L L k(A, C, S, E, I, L) =  L  M ( A, C , S , I , L)  M ( A, C , S , I , L) 1 2    M 2 ( A, C , S , I , L)  M 3 ( A, C , S , I , L)   6I 6I S C  L L



– 6I S L



6I C L

4I 6I S L  6I C L

2I  6I S L  6I C L

    M 2 ( A, C , S , I , L)  M 3 ( A, C , S , I , L)   6I 6I S C 2I  L L 6I M1 ( A, C , S , I , L) M 2 ( A, C , S , I , L) S  L M 2 ( A, C , S , I , L) M 3 ( A, C , S , I , L)   6LI C    6I   6LI C  S 4I  L  M1 ( A, C , S , I , L)

Calculate global stiffness matrix for 1st element. 238

 M 2 ( A, C , S , I , L)

i=1

Set i = 1 so that the properties of element 1 are used in the functional expression.

k1 = k(Ai, Ci, Si, E, Ii, Li)  5.6  109  0   0 k1 =    5.6  109   0   0

 5.6  109

1.12  107

1.68  107

1.68  10

3.36  10

5.6  109

 1.12  107

 1.68  107

1.68  10

 1.12  10  1.68  10 0

1.12  107  1.68  107



1.68  10

7 

 1.68  107   0  7  1.68 10  3.36  107 

Augment element global k matrix with rows and columns of zeros to facilitate the assembly of the total global stiffness matrix. The global k matrix for each element needs to have the same number of rows and columns as there is degrees of freedom in the model. In this case, it is 9 (3 nodes, 3 DOF ). node

 0  0    0 ZeroCol =   0    0   0

ZeroRow = (0 0 0 0 0 0 0 0 0)

kIa = augment (k1, ZeroCol, ZeroCol, ZeroCol) kIb = stack(k1a, ZeroRow, ZeroRow, ZeroRow) kI b = 0

5.6 109

1.12107

1.68 107

3.36 107

4 –5.6109

– 5.6 109

5 0

6 0

–1.12 107

1.68 107

–1.68 107

1.68 107

5.6 109

–1.12 107

–1.68 107

1.12 107

–1.68 107

1.68 107

–1.68 107

3.36 107

239

Calculate global stiffness matrix for 2nd element.

i=2

Set i = 2 so that the properties of element 2 are used in the functional expression.

k2 = k (Ai, Ci, Si, E, Ii, Li)

 1.4106 0   1.714107  K2 =  4.2106   1.4106 0  7  1.71410   4.2106

1.714107

4.2106

2.8109

2.57210

1.4106

0 7

0 10

1.714107

4.2106

2.8109

2.57210



0 10

  1.68107 4.2106 2.572 1010 8.4 106  0  6 6 7 1.71410 4.210 1.410 1.714 10 4.2106  0 0 0 9 10 7 9 10 2.810 2.572 10 1.71410 2.810 2.57210  0 0  2.5721010 8.4106 4.2106 2.572 1010 1.68107  Augment element global k matrix with rows and columns of zeros to facilitate the assembly of the total global stiffness matrix. As before, we need 9 rows and columns. 0 2.5721010 0 7

1.71410

k2a = augment (ZeroCol,ZeroCol, ZeroCol, k2) k2b = stack(ZeroRow, ZeroRow, ZeroRow, k2a) k2b = 0

1 2

0 0

1.4 10

–7

–1.714 10

4.2 10

–1.4 10

0 0

0 –1.714 10–7

2.8 109

2.572 10–10

1.714 10–7

0 0

4.2 106 2.572 10–10

0 0

–1.4 106

0 0

1.714 

0 0

1.714 10–7

–7

4.2 106

– 2.8 109

2.572 10–10

1.68 107

– 4.2 106 – 2.572 10–10

8.4 106

– 4.2 106

1.4 106 –1.714 10–7

– 4.2 106

–2.8  –2.572  –1.714 

4.2 106 2.572 10–10

8.4 106

1.714 10

2.8  –2.572 

– 4.2 106 – 2.572 10–10

1.68 107

Calculate total global stiffness matrix by adding augmented matrices for each element. K = k1b + k2b K

5.610

2 0 1.1210

–5.610

1.6810

3.3610

–5.610 0

–7

4.2 106

1.68 10

–1.1210 0

–1.6810

1.68 10

–7

4.2 10

–1.4 10

5.60110 –1.71410

240

1.714 10

2.811109 –1.68 107 1.714 10–7

– 2.8 109 2.572 10–10

0 –1.12107 –1.68107 –1.71410–7

1.68107

4.2106

–1.68107

5.04107

–4.2 106 2.572 10–10

8.4 106

–1.4106

1.71410–7

–4.2 106

1.4 106 –1.714 10–7

– 4.2 106

1.71410–7

4.2 106 2.57210–10

–2.8109 2.572 10–10 –1.714 10–7

8.4106

2.8 109 2.57210–10

–4.2 106 2.57210–10

1.68107

Solve for displacements and rotations at node 1 and 2. First partition out rows and columns associated with homogenous boundary conditions (rows/columns 7, 8 and 9) Kpart = submatrix (K, 1, 6, 1, 6)

 5.6  109  5.6  109 0 0  0 1.12  107 1.68  107 0   7 7 0 1.68  10 3.36  10 0 Kpart =    5.6  109 0 0 5.601  109   1.12  107 1.68  107 1.714  107 0   0 1.68  107 1.68  107 4.2  106 0

 1.12  10

 1.68  10

 1.714  107 2.811  109  1.68  107

1.68  10



7 

 1.68  107   4.2  106    1.68  107   5.04  107 

Define partitioned vector of applied loads.  Fpart =  0 

F1 y N

 0 0 0 0 

Solve for displacements and rotations at node 1 and 2.  u1   v1     1  –1  u  = Kpart Fpart 2    v2     2

241

  0.214   u1  (m)    v1   0.25 (m)     (rad)   0.089   1  u  =   0.214  101   (m)  2   3.571105   v2  (m)      (rad) 0.071 2  

[Displacements are in meters, rotations in radians.] 0.0089 rad = 0.5099° 0.0071 rad = 0.4068° Solving for reactions.  F1x   u1   F1 y   v1       M1   1    u   F2 x   2  F2 y  = K  v2  Multiplying global stiffness matrix by displacement vector.      M2   2 F   u3   3x     F3 y   v3        M3  3   0  F1x     F1 y  3  10  10       M1  0     F   0  2x    F2 y  =  0       0  M2    F  0    3x    3  F3 y   10  10     3  104   M3   

Forces are in newtons, moment in N  m. Values agree with what one would expect given the simple nature of the problem. Vertical reaction at node 3 should be equal and opposite the applied load F1y and it is. Moment reaction at 3 (M3) should equal the applied load (F1y) times it’s moment arm (L1) and it does. Solve for element forces. Functional equations for local beam elements Functional equation for local stiffness matrix for a 2D beam/frame element with axial affects. Refer to text Equation (5.1.8). 242

 AE L  0   0 Klocal (A, C, S, E, I, L) =  AE   L  0   0

12 EI L3 6 EI L2

6 EI L2 4 EI L

 12 EI 3

6 EI L2 2 EI L

L 6 EI L2

AE L

0 0 AE L

0 0

0 12 EI L3 6 EI L2

0 

6 EI  L2  2 EI  L 

0  6 EI  L2  4 EI  L 

Functional equation for transformation matrix between local and global coordinates. C  S   0 T(C, S) =  0   0  0

0 0 0 0 0 0 0 0  1 0 0 0 0 C S 0  0  S C 0  0 0 0 1

S C 0 0

0 0

Solving for element forces in element 1 i= 1 klocal1 = klocal(Ai, Ci, Si, E, Ii, Li)  5.6  109  0   0 klocal1 =    5.6  109   0   0

Local k matrix for element 1 0

1.12  107

1.68  107

1.68  10

 1.12  107

 1.68  107

 1.68  107 0

1.12  107  1.68  107

5.6  109

0  1.12  10

3.36  10

0 1.68  10 7

 5.6  109

1.68  10

1.68  10 

7 

 1.68  107   0  7  1.68  10  3.36  107 

T1 = T(Ci, Si) Transformation matrix for element 1. 1 0  0 T1 =  0  0  0

0 1 0 0 0 0

0 0 1 0 0 0

0 0 0 1 0 0

0 0 0 0 1 0

0 0  0 0  0  1 243

 u1   v1     1  f1 = klocal T1   Calculate local forces/moments in element 1. u  2  v2     2

 0   f1x   10000   f     1y   0   m1  f1 =      0   f2x   10000   f 2 y       30000   m2 

Forces are in Newtons, moments in N  m. f1y comes back out of the equations as expected. f2y must be equal and opposite—and it is. m2 also checks (resists in CW direction the applied load of 1×104 N over 3m moment arm)

Solving for element forces in element 2 i= 2 klocal2 = klocal(Ai, Ci, Si, E, Ii, Li)  2.8  109  0   0 klocal2 =    2.8  109   0   0

Local k matrix for a element 2

 2.8  109

1.4  106

4.2  106

1.68  107

2.8  109

 1.4  106

 4.2  106

4.2  10

244

8.4  10

 1.4  10

 4.2  10 0

1.4  106  4.2  106



4.2  10

6 

 8.4  106   0  6  4.2  10  1.68  107 

T2 = T (Ci, Si) Transformation matrix for element 2.  0 1 1 0  0 0 T2 =  0 0  0 0  0 0

0 0 1 0 0 0

0 0 0 0 0 0 0 1 1 0 0 0

0 0  0 0  0  1

Calculate local forces/moments in element 2.  u2   v2      f2 = (klocal 2) T2  2  u  3  v3     3

Forces are in Newtons, moments in N  m. f2y and f3y again are of same magnitude as applied load at node 1 as expected. m2 and m3 also have correct magnitude. m3 must be equal and opposite of applied moment. m2 must be opposite of m3 for equilibrium.  10000   0     30000  f2 =    10000   0     30000 

 f2 x  f   2y   m2     f3 x  f   3y  m   3

245

FBD of element (2) 5.13

Boundary conditions

u1 = v1 = 1 = 0, u4 = v4 = 4 = 0

Element (1) by Equation (5.1.11) C = 0; S = 1 u2

v2  2

12 I

E  L2 [k ] = L  0  6I  L (1)

6I  L

 0  4I 

Element (2) C = 1, S = 0

246

2

0 A 0  6I 12 I  L L2 E  (2) [k ] = 4I  L      Symmetry

3

A

0 0

 12 I L2  6I L

0  6I  L   2I  0    6I  L  4 I 

12 I L2

Element (3) C = 0, S = – 1 u3 12 I  L2

E  [k ] = L    (3)

3

6I L 

 0 4 I  

Element (4) C=

 2  2 , S= , L = 20, 2 2

A  A  62I  62I 2 2 L L  E  (4) 6I A  2 [k ] = (4) 2 L L   Symmetry 

L(4) =

2 L

2 3LI   2 3LI   4I  

Assembling 5000  0     0  E   = L  0   0     0 

0 0  u2  10.75 – 0.707 5.0  10   0.0417 10.75 5.0 0 5   v2      5 1603 0 400  2       10.04 0 5   u3     5   v3  10.04     1600 3  

E lb 30  106 = = 0.125  106 3 in. L in. 20  ) ft

Solving simultaneously u2 = 0.055916 in.

u3 = 0.05576 in.

v2 = 0.003817 in.

v3 = – 0.000133 in.

2 = – 0.00015 rad

3 = – 0.000149 rad 247

PLANE FRAME PROBLEM 5.13 NUMBER OF ELEMENTS = 4 NUMBER OF NODES = 4 NODE POINTS K IFIX 1 111 2 000 3 000 4 111

XC(K) 0.000000 0.000000 240.000000 240.000000

YC(K) 0.000000 240.000000 240.000000 0.000000

ZC(K) 0.000000 0.000000 0.000000 0.000000

FORCE(1,K) 0.000000 5000.000000 0.000000 0.000000

ELEMENTS K NODE(I, K) E(K) 1 1 2 3.0000000E+07 2 2 3 3.0000000E+07 3 3 4 3.0000000E+07 4 2 4 3.0000000E+07

G(K) 0.0000000E+00 0.0000000E+00 0.0000000E+00 0.0000000E+00

A(K) 1.0000000E+01 1.0000000E+01 1.0000000E+01 2.0000000E+00

XI(K) 2.0000000E+02 2.0000000E+02 2.0000000E+02 1.0000000E+00

NODE 1 2 3 4

DISPLACEMENTS X Y 0.00000E+00 0.00000E+00 0.55918E–01 0.38170E–02 0.55760E–01 – 0.13304E–03 0.00000E+00 0.00000E+00

Z-ROTATION THETA 0.00000E+00 – 0.14987E–03 – 0.14913E–03 0.00000E+00

ELEMENTS K NODE(I, K) X-FORCE Y-FORCE Z-MOMENT X-FORCE Y-FORCE 1 1 2 – 0.4771E+04 0.1976E+03 0.2746E+05 0.4771E+04 – 0.1976E+03 2 2 3 0.1972E+03 – 0.1663E+03 – 0.1997E+05 – 0.1972E+03 0.1663E+03 3 3 4 0.1663E+03 0.1972E+03 0.1994E+05 – 0.1663E+03 – 0.1972E+03 4 2 4 0.6513E+04 0.1547E+00 0.1301E+02 – 0.6513E+04 – 0.1547E+00 Z–MOMENT 0.1996E+05 – 0.1994E+05 0.2739E+05 0.3950E+02

5.14

248

Element (1) C=

2 2 , S= 2 2

Use only node 2 part of [k(1)]

 A   CS

 AC 2  12 I S 2 L12   E [k(1)] =  L1   Symmetry 

u2 v2

12 I L12

AS 2  122I C 2 L1

6I S  L1 

  6I C L1  4I  

2

61 59  24 5 2 24   61  5 2  125 2 103 =  24   Symmetry 800 

Element (2) C = 1, S = 0  AC 2  12 I S 2 L 22  E  [k(2)] =  L2   Symmetry  5   =  Symmetry

 A   CS 12 I L 22

 6I  S L2 

AS 2  122I C 2

6I C L2

2 27

20 3

800



 4I  

  5 5   3  10 

[K] = [k(1)] + [k(2)] Then {F} = [K] {d} 249

Equivalent nodal forces f2x = 0

m2 =

f2y =

 wL 2

= – 22,500 lb

– wL22 = – 675,000 lb  in. 12

0 125000  u2    1282600 434580      461650 –138900   v2   22500  =  675000   27, 475,000 2    

Solving u2 = 0.0261 in. v2 = – 0.0721 in.

2 = – 0.0025 rad Element forces { f }(1) = [k ](1) [T ](1) {d }(1)

Element one (1) [T ]{d }

(1)

 4 2  f 1 x(1)    (1)    f 1y    (1)  6   m1  = 10   (1)    f 2x    (1)   f   2y    m(1)   2   5

1 48 2

5 4

100 2

5

  0  1 5   0 0 4 48 2     5 0  0 50 2 4   5 0 0    0.03250  4 2   0.06945  5 1   4 48 2   0.002475 Symmetry 100 2 

4 2

f 1 x (1) = 28740 lb = – f 2 x (1) f 2 y (1) = – 2077.7 lb = – f 2 y (1)

m1(1) = – 8.8578 × 104 lb  in. 

m2(1) = – 2.6403 × 105 lb  in. 

Element two (2) 250

{ f }(2) = [k ](2) [T ](2) {d }(2) – { f 0 }(2)

 0.0261   – 0.0721     – 0.0025 (2) (2) [T ] {d} =   0     0   0    f 2xe(2)    56 (2)  f 2 ye       m2(2)  e   6  (2) = 10    f 3xe    (2)   f 3 ye    (2)  m3e 

1 81

10 9 400 3

5 6

0 0 5 6

0 1 81 10 9

0 1 81

0   0.0261

10    0.0721 9    200   0.0025 3

 0 10   9  400   3 

0 0 0

    

21791      3647   5     4.1097 10   { f (2) } =     21791    3647     2.4556  105  Finally { f } = { f e} – { f 0}

 f 2x(2) 21791 0      21, 791 lb   (2)       f   2y  – 3647    – 22500   18,853 lb   (2)   – 4.1097  105   – 675000   264,030 lb  in.   m2  =  (2)   –  =  0   21791      21, 791 lb   f 3x      22500   26,147 lb   (2) 3647        f 3y   675, 000   675,000   915,980 lb  in.  m (2)   3 

251

5.15

Element (1)

C = 0, 12 I 2

S=1

12 (2  104 ) (4)

= 1.5  10–4 m2

6I = 3.0  10–4 m3 L

kN E 70  106 = 1.75  107 = L 4 m u2

v2 4

1.5  10 0  [k(1)] = 1.75  107  4  102 Symmetry 

2 3.0  104   0  4  8  10 

Element (2) S = 0,

C=1

252

2

 4  102 0 0  4 1.5  10 3  104   [k(2)] = 1.75  107  8  104   Symmetry

3

 4  102

  3  10   4  104  0  8  104  0

4

0 0 4  102

where boundary conditions u1 = v1 = 1 = 0 and v3 = 0

have been used in [k(1)] and [k(2)] The global equations are 4.015 102   7  1.75  10     Symmetry

3  104

0 4.015  10

2

3  10

 4  102

4

3

1.6  10

4  102

 u  2 3  10   v2     4  104  2    0  u3    8  104   3  0

4

 F2 x  – 30 F  0   2 y  =  M 2  30  F  0   3x   M 3  0  Solving simultaneously u2 = – 2.43  10–2 m, v2 0 m

2 = 6.4  10–3 rad u3 = – 2.43  10–2 m, 3 = – 3.2  10–3 rad

Element forces Element (1) { f }(1) = [k ](1) [T ](1) {d }(1)  0  1   0 [T ](1) {d }(1) =   0  0   0

1 0 0 0 0 0

0 0 0 0 1 0 0 0 0 1 0 0

0 0 0 1 0 0

 0 u1  0 v  0   0 1    0 1  0   2  0 u2   2.43  10   0 v2  0    1 2  6.4 103 

253

0     0   0   [T ](1) {d }(1) =   0   2.43  102    3  6.4  10  { f }(1) = [k  ](1) [T ](1) {d }(1) =  7  105   7 105 0 0 0 0   3 3 3 3  2.625  10 2.625  10 5.25  10 0 5.25  10     5.25  103 1.4  104 0 7  103     7  105 0 0    2.625  103  5.25  103    Symmetry 1.4  104  0     0   0     0    – 2.43  102    3  6.4  10  f 1 x = 16.87 kN f 1y  = – 30.0 kN m 1 = – 82.5 kN m

f 2 x = –16.87 kN

f 2 y = 30.0 kN m 2 = –37.5 kN m

Similarly for element (2) { f }(2) = [k  ](2) [T ](2) {d }(2) f 2 x = f 3 x = 0

 = – 16.87 kN f 2 y = 16.87 kN, f 3y m2 = 67.5 kN m, m3 = 0 5.16

254

Element (1) C = 0, S = 1

(2) (1) 0 0 0 0 200  109 (1  102 )  (1) [k ] = 0 1 0  1 1   0 0 0 0 L   0  1 0 1 E

Element (3) C = 0, S = 1

(4) 0 0 EA  (3) [k ] = 0 1 L  0 0  0  1

(3) 0 0 0  1  0 0  0 1

Element (2) C = 1, S = 0

255

(2) (4)  1 0  1 0 AE  (2) [k ] = 0 0 0 0  L  1 0  1 0   0 0  0 0 Element (4) C = 1, S = 0

(1) (3)  1 0  1 0 AE  (4) [k ] = 0 0 0 0  L  1 0  1 0   0 0  0 0 Element (5) C=

2 2 , S= 2 2 (2)

0.5  0.5  0.5 AE  [k ] = 0.5 0.5  0.5 2L  0.5  0.5  0.5  0.5  0.5  0.5 (5)

(3)  0.5  0.5  0.5  0.5

Boundary conditions u1 = v1 = u2 = v2 = 0

Assemble appropriate parts of [k]’s 0 u3  0 1.3535 0.3535 0  0 1.3535 0  1 v3    9    = 2  10   1 0 u4   0  Symmetry  0 1 v4  

256

(a)

Similarly assembling [k(6)] through [k(10)] we obtain u3

0 0 0 0 0 0 1 1  [ K ee ]  0 1 0 0 0 0 0   [ K ei ] 1   0 1.3535 0.3535  0.3535  0.3535  1 0  0   0  1 0.3535 1.3535  0.3535  0.3535 0 0 9   2  10 [K ]   0.3535 0 0  1 0  0.3535  0.3535 1.3535   [ Kie ]   0 0  0.3535  0.3535 0.3535 1.3535 0  1  [ Kii ] 0 0 0 0 0 1 0 1    0 0 0 0 0 0 1  1 1 1 Now [Kii] – [Kie] [K ee ] [Kei] {ui} = {fi} – [Kie] [ K ee ] {fe}

1.3535 0.3535 0 0.3535 1.3535 0   0 1  0  0 0 1 0 1 0 0 1 0  0 0 1.3535  0  1 0.3535

0  1 0 0   =   –  0 0  0 0 0 0 0 1  2  109  0 0  0  1

0   1  1  0  0  0   1  0

0  0.3535  0.3535 0  0.3535  0.3535  0 0 1   0 0 0 

0  1   1 0 0   1 0 0 0   0.3535   0.3535  0.3535  1   1.3535   – 0.3535  0.3535 0

0   u3   0   v3     0   u4   0   v4 

0  0.3535  0.3535 1 0 0 0 1  0  0  0.3535  0.3535 0 1 0  1   20000      0 0  0 0 1.3535 0.3535   0  1    0 0 0  0  1 0.3535 1.3535  0  0 0 u3   20000 0  1 v3   20000   =   0 0 u4   20000   0 1 v4   0 

(b)

Adding two sections (a) and (b) 0  u3  1.3535 0.3535 0 0.3535 2.3535 0  2  v    3  = 2  109  0 1 0  u4   0   2 0 2  v4   0

 20000  20000     20000    0 

Solving u3 = 2.832  10–11 m, v3 = – 2.828  10–5 m 257

u4 = 1.0  10–5 m, v4 = – 2.828  10–5 m 5.17

[k]’s for each element are (1)

(2)

720 720   12  12  29  10  10  720 57600  720  28800  [k(1)] = [k(2)] = [k(3)] = [k(4)] = 12  720   (120)3  Symmetry  57600   3

Adding [k]’s of elements (1) and (2) for substructure (1) and apply boundary conditions

v1 = 1 = 0 0  12 720  v2    10 24  115200  720 28800 2   0   =  16.78    720  v3    10 12   Symmetry  57600  3   0  

Now rearrange the equations with interface displacement first  720  12  720  v3    10   12   720 57600 720 28800     0     3  =  16.78   720 24 0  v2    10    12   0 115200  2   0    720 28800 1 1 ] [Kei] {ui} = {fi} – [Kie] [ K ee ] {fe} Using Equation (5.6.6) [Kii] – [Kie] [ K ee

258

0  1   12 720    720   –12  720  24  12   16.78           720 57600  720 28800  0 115200  720 28800   0 1  10   10   12  720  24 v3  –    =        3   0   720 28800  0 115200  0   25.17  3020   v3    15   3020 483260   =  300     3  

(1)

Considering substructure (2) with boundary conditions v5 = 5 = 0  12 720   v3   0  12 720  57600  720 28800  3   0    =  16.78   24 0  v4    10    Symmetry  115200  4   0   Simplifying as per substructure (1)  25.17 3020  v3    5   3020 483264     =  – 300    3  

(2)

Adding (1) and (2) 0  v3    20  50.34    =   0 966528  3   0   Solving v3 = – 0.3973 in. 3 = 0 5.18

12 I

12  2  104 12

= 2.4  10–3

6I 6  2  104 = = 1.2  10–3 L 1

N E = 200  109 3 L m Considering substructure (1) and applying boundary conditions 259

u 1 = v1 = 0 C = 0, S = 1

[k(1)]

{d}

0 0.0004   1   0  0.0008 0.0012  0.0024 0 0.0012   u2   0    =  = 200  109   0.01 0   v2   0    Symmetry  0.0008 d 2  80  103   Rearranging equations (rows and columns) we place interface displacements first 0 0.0012 0.0012  u2   0  0.0024  0.01 0 0   v2   0  9   = 200  10 0.0008 0.0004  2  80  103      Symmetry  0.0008 1   0  

Using Equation (5.6.6)

0.0024 0 0.0012 0.0012  u2        0.01 0 0  [1250][0.0012 0 0.0004]v2   200  10      Symmetry    0.0008 0.0004   2  9

 0  0.0012     =  0  –  0  [1250] [0]    3 40  10  0.0004 Simplifying 0.12 0 0.12  u2   0      1  10  2.0 0   v2  =  0     0.12  2  80  103  9

(1)

Considering substructure (2) and applying boundary conditions C = 1, S = 0 0 0  0.01 u2  0  0.01  0 0.0024 0.0012 0   v2  0  = 200  109  0.0012 0.0008 0  2  0  0         0.01 0 0 0.01   0   3  



 Simplifying by applying Equation (5.6.6) 0 0   0.01 0.01  u2     1     0.0024 0.0012   0   200  10  0   0.01 0 0   v2     0.01        0.0008  0     2 9

= {0} – {0} 260



0 0  u2  0  2.0      0.12 0.12   v2  = 0 1  10       Symmetry 0.12  2  0 9

(2)

Adding (1) and (2) 0 0.12  u2   2.12    1  10 2.12 0.12   v2  =    Symmetry 0.24  2  9

 0     0   3 80  10 

Solving u2 = v2 = – 2.83 × 10−5 m

2 = 2.92  10–4 rad 5.19

NUMBER OF ELEMENTS = 3 NUMBER OF NODES = 4 NODE POINTS K 1 2 3 4

IFIX 000 000 111 111

XC(K) 0.000000 96.000000 0.000000 96.000000

YC(K) 120.000000 192.000000 0.000000 0.000000

FORCE(2, K) 0.000000 0.000000 0.000000 0.000000 ELEMENTS K NODE(I, K) 1 3 1 2 1 2 3 2 4

E(K) 3.0000000E+07 3.0000000E+07 3.0000000E+07

ZC(K) 0.000000 0.000000 0.000000 0.000000

FORCE(1, K) 10000.000000 15000.000000 0.000000 0.000000

FORCE (3, K) 0.000000 180000.000000 0.000000 0.000000 G(K) 1.0000000E+00 1.0000000E+00 1.0000000E+00

261

A(K) 1. 0000000E+01 1.0000000E+01 1.0000000E+01

XI(K) 2.0000000E+02 2.0000000E+02 2.0000000E+02

DISPLACEMENT X Y 0.70180E+00 0.79708E–02 0.72656E+00 – 0.12753E–01 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00

Z-ROTATION THETA – 0.44578E–02 – 0.49949E–03 0.00000E+00 0.00000E+00

ELEMENTS K NODE(I, K) X-FORCE 1 3 1 – 0.1993E+05 2 1 2 – 0.1843E+05 3 2 4 0.1993E+05

Y-FORCE Z-MOMENT X-FORCE Y-FORCE Z-MOMENT 0.1810E+05 0.1309E+07 0.1993E+05 – 0.1810E+05 0.8629E+06 – 0.1108E+05 – 0.8629E+06 0.1843E+05 0.1108E+05 – 0.4671E+06 0.6903E+04 0.6471E+06 – 0.1993E+05 – 0.6903E+04 0.6783E+06

5.20

PLANE FRAME PROBLEM 5.20 NUMBER OF ELEMENTS = 6 NUMBER OF NODES = 6 NODE POINTS K 1 2 3 4 5 6

1 0 0 0 0 1

IFIX 1 0 0 0 0 1

0 0 0 0 0 0

ELEMENTS K NODE(I, K)

XC(K) 0.000000 0.000000 0.000000 300.000000 300.000000 300.000000

YC(K) 0.000000 180.000000 360.000000 360.000000 180.000000 0.000000

FORCE(2, K) 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000

FORCE(3, K) 0.000000 0.000000 36000.000000 0.000000 0.000000 0.000000

E(K)

G(K)

262

ZC(K) 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000

A(K)

FORCE(1, K) 0.000000 2400.000000 1200.000000 0.000000 0.000000 0.000000

XI(K)

1 2 3 4 5 6

1 2 3 4 5 2

2 3 4 5 6 5

2.9000000E+07 2.9000000E+07 2.9000000E+07 2.9000000E+07 2.9000000E+07 2.9000000E+07

0.0000000E+00 0.0000000E+00 0.0000000E+00 0.0000000E+00 0.0000000E+00 0.0000000E+00

1.0000000E+01 1.0000000E+01 1.0000000E+01 1.0000000E+01 1.0000000E+01 1.0000000E+01

DISPLACEMENTS X Y 0.00000E+00 0.00000E+00 0.95468E+00 0.17130E–02 0.12408E+01 0.20315E–02 0.12403E+01 − 0.20315E–02 0.95333E+00 – 0.17130E–02 0.00000E+00 0.00000E+00 ELEMENTS K NODE(I, K) X-FORCE Y-FORCE 1 1 2 – 0.2760E+04 0.1787E+04 2 2 3 – 0.5131E+03 0.6953E+03 3 3 4 0.5045E+03 – 0.5131E+03 4 4 5 0.5131E+03 0.5046E+03 5 5 6 0.2760E+04 0.1813E+04 6 2 5 0.1308E+04 – 0.2247E+04 1 2 3 4 5 2

2 3 4 5 6 5

Z-MOMENT 0.7438E–01 0.1686E+05 – 0.7230E+05 0.8162E+05 0.3263E+06 – 0.3386E+06

2.0000000E+02 2.0000000E+02 2.0000000E+02 2.0000000E+02 2.0000000E+02 3.0000000E+02

Z-ROTATION THETA – 0.69680E–02 – 0.19754E–02 – 0.55648E–03 – 0.79750E–03 – 0.19213E–02 – 0.69838E–02 X-FORCE 0.2760E+04 0.5131E+03 – 0.5045E+03 – 0.5131E+03 – 0.2760E+04 – 0.1308E+04

Y-FORCE – 0.1787E+04 – 0.6953E+03 0.5131E+03 – 0.5046E+03 – 0.1813E+04 0.2247E+04

Z-MOMENT 0.3217E+06 0.1083E+06 – 0.8162E+05 0.9200E+04 0.3091E–01 – 0.3355E+06

5.21 For the slant-legged rigid frame shown in Figure P5–21, size the structure for minimum weight based on a maximum bending stress of 20 ksi in the horizontal beam elements and a maximum compressive stress (due to bending and direct axial load) of 15 ksi in the slantlegged elements. Use the same element size for the two slant-legged elements and the same element size for the two 10-foot sections of the horizontal element. Assume A36 steel is used.

263



I chose to use an ‘I’ beam for my cross sectional area because it is commonly used in bridges. My final design was chosen because the design meets the constraints of the problem while not being overly large.



The center part of the cross member (25 foot section) was taken to be W14  26. This has a thickness of 0.420 inches, a depth of 13.91 inches and a width of 5.025 inches.



The two angle members and the outside 10 foot members are also designed as ‘I’ beams. These members were taken to be W14  22. This size has a thickness of 0.335 inches, a depth of 13.74 inches and a width of 5.00 inches.



The maximum bending stress (about the local axis 3) in this member is 19.51891 ksi. This is under the 20 ksi constraint put on the design in the problem. It is near the center of the cross member.



The maximum stress in the angle member is – 11.51989 ksi and is well below the 15 ksi allowed in the problem.



The sizes were determined by taking some commonly used sizes from my Mechanics of Materials book and using trial and error. When I got the lower members too small the bending stress went too high in the cross member. The final design was chosen because it minimized the size of the cross section while also minimizing the size of the angle members. 5.22 For the rigid building frame shown in Figure P5–22, determine the forces in each element and calculate the bending stresses. Assume all the vertical elements have A = 10 in.2 and I = 100 in4 and all horizontal elements have A = 15 in.2 and I = 150 in.4. Let E = 29  106 psi for all elements. Let c = 5 in. for the vertical elements and c = 6 in. for the horizontal elements, where c denotes the distance from the neutral axis to the top or bottom of the beam cross section, as used in the bending stress formula  =  Mc . I 

Figure P5–22 1**** BEAM ELEMENT STRESSES ELEMENT CASE P/A NO. (MODE) 1 1 1.500E+02 1.500E+02 2 1 1.500E+02 1.500E+02 3 1 5.591E+01 5.591E+01

P/A+M2/S2

P/A–M2/S2 P/A+M3/S3 P/A–M3/S3 WORST SUM

1.500E+02 1.500E+02 1.500E+02 1.500E+02 5.591E+01 5.591E+01

264

1.500E+02 1.500E+02 4.676E+03 –4.376E+03 4.676E+03 –4.376E+03 9.203E+03 –8.903E+03 1.605E+02 –4.866E+01 2.054E+03 –1.942E+03

1.500E+02 4.676E+03 4.676E+03 9.203E+03 1.605E+02 2.0543+03

5.591E+01 5.591E+01 1.480E+01 1.480E+01 1.480E+01 1.480E+01 8.096E–02 8.096E–02 8.096E–02 8.096E–02 1.920E–01 1.920E–01 1.920E–01 1.920E–01 1.098E–01 1.098E–01 1.098E–01 1.098E–01 –1.500E+02 –1.500E+02 –1.500E+02 –1.500E+02 –5.610E+01 –5.610E+01 –5.610E+01 –5.610E+01 –1.491E+01 –1.491E+01 –1.491E+01 –1.491E+01

5.591E+01 2.054E+03 5.591E+01 3.947E+03 1.480E+01 1.329E+02 1.480E+01 7.855E+02 1.480E+01 7.855E+02 1.480E+01 1.438E+03 8.096E–02 8.096E–02 8.096E–02 5.962E+03 8.096E–02 5.962E+03 8.096E–02 1.192E+04 1.920E–01 –4.031E+03 1.920E–01 1.177E+03 1.920E–01 1.177E+03 1.920E–01 6.386E+03 1.098E–01 –8.724E+02 1.098E–01 8.076E+02 1.098E–01 8.076E+02 1.098E–01 2.488E+03 –1.500E+02 –1.500E+02 –1.500E+02 4.361E+03 –1.500E+02 4.361E+03 –1.500E+02 8.873E+03 –5.610E+01 3.192E+01 –5.610E+01 1.930E+03 –5.610E+01 1.930E+03 –5.610E+01 3.828E+03 –1.491E+01 8.670E+01 –1.491E+01 7.542E+02 –1.491E+01 7.542E+02 –1.491E+01 1.422E+03

–1.942E+03 –3.835E+03 –1.033E+02 –7.559E+02 –7.559E+02 –1.409E+03 8.096E–02 –5.962E+03 –5.962E+03 –1.192E+04 4.032E+03 –1.177E+03 –1.177E+03 –6.386E+03 8.726E+02 –8.074E+02 –8.074E+02 –2.487E+03 –1.500E+02 –4.661E+03 –4.661E+03 –9.173E+03 –1.441E+02 –2.042E+03 –2.042E+03 –3.940E+03 –1.165E+02 –7.840E+02 –7.840E+02 –1.451E+03

2.054E+03 3.947E+03 1.329E+02 7.855E+02 7.855E+02 1.438E+03 8.096E–02 5.962E+03 5.962E+03 1.192E+04 4.032E+03 1.177E+03 1.177E+03 6.386E+03 8.726E+02 8.076E+02 8.076E+02 2.488E+03 –1.500E+02 –4.661E+03 –4.661E+03 –9.173E+03 –1.441E+02 –2.042E+03 –2.042E+03 –3.940E+03 –1.165E+02 –7.840E+02 –7.840E+02 –1.451E+03

5.23

Problem 5-23 NUMBER OF ELEMENTS = 10 NUMBER OF NODES = 6 NODE POINTS K IFII 1 110 2 110

XC(K) 0.000000 300.000000

YC(K) 0.000000 0.000000

265

ZC(K) 0.000000 0.000000

FORCE(1, K) 0.000000 0.000000

3 000 4 000 5 000 6 000 ELEMENTS K NODE (I, K) 1 1 3 2 2 4 3 3 5 4 4 6 5 5 6 6 3 4 7 3 2 8 1 4 9 5 4 10 3 6

NODE 1 2 3 4 5 6

0.000000 300.000000 0.000000 300.000000

180.000000 180.000000 360.000000 360.000000

0.000000 0.000000 0.000000 0.000000

E(K) 3.0000000E+07 3.0000000E+07 3.0000000E+07 3.0000000E+07 3.0000000E+07 3.0000000E+07 3.0000000E+07 3.0000000E+07 3.0000000E+07 3.0000000E+07

G(K) 0.0000000E+00 0.0000000E+00 0.0000000E+00 0.0000000E+00 0.0000000E+00 0.0000000E+00 0.0000000E+00 0.0000000E+00 0.0000000E+00 0.0000000E+00

A(K) 1.0000000E+01 1.0000000E+01 1.0000000E+01 1.0000000E+01 1.2000000E+01 1.2000000E+01 2.0000000E+00 2.0000000E+00 2.0000000E+00 2.0000000E+00

FORCE(2, K) FORCE(3, K) 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 36000.000000 0.000000 0.000000 DISPLACEMENTS X Y 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 0.15236E–01 0.10352E–02 0.14269E–01 – 0.99244E–03 0.20440E–01 0.12193E–02 0.20082E–01 – 0.11183E–02

2400.000000 0.000000 1200.000000 0.000000 XI(K) 1.5000000E+02 1.5000000E+02 1.5000000E+02 1.5000000E+02 2.0000000E+02 3.0000000E+02 1.0000000E+00 1.0000000E+00 1.0000000E+00 1.0000000E+00

Z-ROTATION THETA – 0.90513E–04 – 0.11175E–03 – 0.72478E–04 – 0.13558E–04 0.20690E–03 – 0.74145E–04

ELEMENTS K NODE(I, K) X-FORCE 1 1 3 – 0.1725E+04 2 2 4 0.1654E+04 3 3 5 – 0.3069E+03 4 4 6 0.2097E+03 5 5 6 0.4288E+03 6 3 4 0.1160E+04 7 3 2 0.2149E+04 8 1 4 – 0.2011E+04 10 3 6 – 0.5227E+03

Y-FORCE 0.5246E+01 0.2770E+02 0.1602E+03 – 0.1926E+02 0.5934E+02 – 0.4351E+02 – 0.1976E+00 – 0.8417E–01 0.1791E+00

Z-MOMENT X-FORCE Y-FORCE Z-MOMENT 0.2132E+02 0.1725E+04 – 0.5246E+01 0.9230E+03 0.3793E+02 0.1654E+04 – 0.2770E+02 0.4948E+04 0.7434E+04 0.3069E+03 – 0.1602E+03 0.2140E+05 – 0.2187E+03 – 0.2097E +03 0.1926E+02 – 0.3248E+04 0.1452E+05 – 0.4288E+03 – 0.5934E+02 0.3279E+04 – 0.8294E+04 – 0.1160E+04 0.4351E+02 – 0.4759E+04 – 0.3119E+02 – 0.2149E+04 0.1976E+00 – 0.3793E+02 – 0.2132E+02 0.2011E+04 0.8417E–01 – 0.8126E+01 – 0.3119E+02 0.5227+03 0.1791E+00 – 0.3148E+02

Reactions

Node 2 From x and yforces in elements (2) and (7)

266

F2x = –27.7 – 1843 = 1871 lb  F2y = 1654 + 1106 = 2760 lb 

Similarly F1x = 1730 lb  F1y = 2760 lb  5.24

Problem 5.24 NUMBER OF ELEMENTS = 6 NUMBER OF NODES = 6 NODE POINTS K 1 2 3 4 5 6

IFIX 110 110 000 000 000 000

XC(K) YC(K) 0.000000 0.000000 300.000000 0.000000 0.000000 180.000000 300.000000 180.000000 0.000000 360.000000 300.000000 360.000000

ZC(K) 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000

FORCE(1, K) 0.000000 0.000000 4500.000000 0.000000 2250.000000 0.000000

FORCE(2, K) FORCE(3, K) 0.000000 –67500.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 67500.000000 0.000000 0.000000

ELEMENTS K 1 2 3 4

NODE(I, K) 1 3 3 5 2 4 4 6

E(K) 3.0000000E+07 3.0000000E+07 3.0000000E+07 3.0000000E+07

G(K) 0.0000000E+00 0.0000000E+00 0.0000000E+00 0.0000000E+00

267

A(K) 1.5000000E+01 1.5000000E+01 1.5000000E+01 1.5000000E+01

XI(K) 2.0000000E+02 2.0000000E+02 2.0000000E+02 2.0000000E+02

5 6

3 5

NODE 1 2 3 4 5 6

4 6

3.0000000E+07 3.0000000E+07

0.0000000E+00 0.0000000E+00

DISPLACEMENTS X(in.) Y(in.) 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 0.21212E+01 0.21601E–02 0.21193E+01 – 0.21601E–02 0.28221E+01 0.26614E–02 0.28215E+01 – 0.26614E–02

1.5000000E+01 1.5000000E+01

2.0000000E+02 2.0000000E+02

Z-ROTATION THETA – 0.15591E–01 – 0.15054E–01 – 0.51828E–02 – 0.52135E–02 – 0.13916E–02 – 0.17770E–02

ELEMENTS K NODE X-FORCE Y-FORCE Z-MOMENT X-FORCE Y-FORCE Z-MOMENT (I, K) (lb) (lb) (lb in.) (lb) (lb) (lb in.) 1 1 3 – 0.5400E+04 0.3105E+04 – 0.6750E+05 0.5400E+04 – 0.3105E+04 0.6264E+06 2 3 5 – 0.1253E+04 0.1349E+04 – 0.4968E+04 0.1253E+04 – 0.1349E+04 0.2478E+06 3 2 4 0.5400E+04 0.3645E+04 0.8592E–01 – 0.5400E+04 – 0.3645E+04 0.6561E+06 4 4 6 0.1253E+04 0.9017E+03 – 0.3340E+05 – 0.1253E+04 – 0.9017E+03 0.1957E+06 5 3 4 0.2744E+04 – 0.4147E+04 – 0.6214E+06 – 0.2744E+04 0.4147E+04 – 0.6227E+06 6 5 6 0.9014E+03 – 0.1253E+04 – 0.1803E+06 – 0.9014E+03 0.1253E+04 – 0.1957E+06

5.25 (a)

268

5.26

269

Solution using Autodesk Simulation

270

5.28

271

Figure P5-28

5.29

272

Figure P5–29

Displacements/Rotations (degrees) of nodes Node X– Y– number translation translation 1 0.0000E+00 0.0000E+00 2 0.0000E+00 0.0000E+00 3 0.0000E+00 –1.8330E+00 4 0.0000E+00 –1.2242E+00 5 0.0000E+00 0.0000E+00 6 0.0000E+00 0.0000E+00 7 0.0000E+00 0.0000E+00 8 0.0000E+00 0.0000E+00

Z– translation 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00

X– rotation 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00

Y– rotation 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00

Z– rotation 3.4030E–01 –6.8060E–01 –3.7774E–02 7.6168E–01 7.4186E–01 –3.7093E–01 0.0000E+00 0.0000E+00

5.30

NUMBER OF ELEMENTS = 5 NUMBER OF NODES = 6 NODE POINTS K IFIX XC(K) 1 111 0.000000 2 000 0.000000 3 0 0 0 180.000000 4 0 0 0 420.000000 5 0 0 0 600.000000 6 1 1 1 600.000000

YC(K) 0.000000 360.000000 480.000000 480.000000 360.000000 0.000000

ZC(K) FORCE(1, K) FORCE(2, K) FORCE(3, K) 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 –5000.000000 0.000000 0.000000 0.000000 –5000.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000

273

5.32

Problem 5.32 NUMBER OF ELEMENTS = 3 NUMBER OF NODES = 4 NODE POINTS K 1 2 3 4

IFIX 111 000 000 111

XC(K) 0.000000 0.000000 6.000000 6.000000

ELEMENTS K NODE(I,K)

YC(K) 0.000000 6.000000 6.000000 0.000000

ZC(K) 0.000000 0.000000 0.000000 0.000000

E(K)

FORCE(1,K) FORCE(2,K) 0.000000 0.000000 15000.000000 0.000000 0.000000 0.000000 0.000000 0.000000

G(K)

274

A(K)

FORCE(3,K) 0.000000 10000.000000 0.000000 0.000000

XI(K)

1 2 3

2 3 4

2.1000000E+11 2.1000000E+11 2.1000000E+11 NODE 1 2 3 4

1.0000000E+00 1.0000000E+00 1.0000000E+00

2.0000000E-02 2.0000000E-04 2.0000000E-02 2.0000000E-04 2.0000000E-02 2.0000000E-04

DISPLACEMENTS X Y 0.00000E+00 0.00000E+00 0.42966E–02 0.71361E–05 0.42871E–02 – 0.71361E–05 0.00000E+00 0.00000E+00

Z-ROTATION THETA 0.00000E+00 – 0.24093E–03 – 0.47744E–03 0.00000E+00

ELEMENTS K NODE X-FORCE Y-FORCE Z- MOMENT X-FORCE Y-FORCE Z-MOMENT (I,K) 1 1 2 – 0.4995E+04 0.8339E+04 0.2620E+05 0.4995E+04 – 0.8339E+04 0.2333E+05 2 2 3 0.6661E+04 – 0.4995E+04 – 0.1333E+05 – 0.6661E+04 0.4995E+04 – 0.1664E+05 3 3 4 0.4995E+04 0.6661E+04 0.1664E+05 – 0.4999E+04 – 0.6661E+04 0.2333E+05

Reactions

F1x = – 8339 N, F1y = – 4995 N, M1 = 26,700 N  m F4x = – 6661 N, F4y = 4995 N, M4 = 23,330 N  m 5.33

***** Frame Problem 5.33 ****** NUMBER OF ELEMENTS = 9 NUMBER OF NODES = 8 NODE POINTS K IFIX XC(K) 1 111 –5.000000 2 111 5.000000 3 000 –5.000000 4 000 5.000000

YC(K) 0.000000 0.000000 4.000000 4.000000

ZC(K) FORCE(1,K) FORCE(2,K) FORCE(3,K) 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 20.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000

275

5 6 7 8

000 000 000 000

–5.000000 7.000000 5.000000 7.000000 –5.000000 10.000000 5.000000 10.000000

ELEMENTS K NODE E(K) (I,K) 1 1 3 2.1000000E+06 2 2 4 2.1000000E+06 3 3 4 2.1000000E+06 4 3 5 2.1000000E+06 5 4 4 2.1000000E+06 6 5 6 2.1000000E+06 7 5 7 2.1000000E+06 8 6 8 2.1000000E+06 9 7 8 2.1000000E+06 NODE 1 2 3 4 5 6 7 8 ELEMENTS K NODE(I,K) 1 1 3 2 2 4 3 3 4 4 3 5 5 4 6 6 5 6 7 5 7 8 6 8 9 7 8

0.000000 0.000000 0.000000 0.000000

20.000000 0.000000 20.000000 0.0000000

0.000000 0.000000 0.000000 0.000000

A(K)

XI(K)

XJ(K)

9.9999998E–03 9. 9999997E–03 9.9999998E–03 9. 9999997E–03 2.0000000E–02 1. 9999999E–04 9.9999998E–03 9. 9999997E–03 9.9999998E–03 9. 9999997E–03 2.0000000E–02 1. 9999999E–04 4.9999999E–03 4. 9999999E–03 4.9999999E–03 4. 9999999E–03 2.0000000E–02 1. 9999999E–04

0.0000000E+00 0.0000000E+00 0.0000000E+00 0.0000000E+00 0.0000000E+00 0.0000000E+00 0.0000000E+00 0.0000000E+00 0.0000000E+00

G(K) 0.0000000E+00 0.0000000E+00 0.0000000E+00 0.0000000E+00 0.0000000E+00 0.0000000E+00 0.0000000E+00 0.0000000E+00 0.0000000E+00

DISPLACEMENTS X Y 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 0.13109E–01 0.40236E–04 0.13092E–01 –0.40236E–04 0.22043E–01 0.50737E–04 0.21994E–01 –0.50737E–04 0.26380E–01 0.46314E–04 0.26376E–01 –0.46314E–04

0.000000 0.000000 60.000000 0.000000

Z-ROTATION THETA 0.00000E+00 0.00000E+00 –0.26944E–02 –0.27869E–02 –0.19861E–02 –0.15658E–02 0.17098E–02 –0.11140E–02

X-FORCE –0.2112E+02 0.2112E+02 0.7451E+01 –0.7351E+01 0.7351E+01 0.2046E+02 0.1548E+01 –0.1548E+01 0.1691E+01

Y-FORCE 0.3040E+02 0.2960E+02 –0.1377E+02 0.1785E+02 0.2215E+02 –0.8899E+01 0.1831E+02 0.1693E+01 0.1548E+01

Z-MOMENT 0.7493E+02 0.7383E+02 –0.4847E+02 0.2182E+02 0.2468E+02 –0.4626E+02 0.1453E+02 0.9578E+00 0.1960E+02

X-FORCE 0.2112E+02 –0.2112E+02 –0.7451E+01 0.7351E+01 –0.7351E+01 –0.2046E+02 –0.1548E+01 0.1548E+01 –0.1691E+01

Y-FORCE 0.3040E+02 –0.2960E+02 0.1377E+02 –0.1785E+02 –0.2215E+02 0.8899E+01 –0.1831E+02 –0.1693E+01 –0.1548E+01

Z-MOMENT 0.4665E+02 0.4457E+02 –0.6925E+02 0.3173E+02 0.4177E+02 –0.4273E+02 0.4040E+02 0.4120E+01 –0.4120E+01

5.34

276

NUMBER OF ELEMENTS = 3 NUMBER OF NODES = 4 NODE POINTS K IFIX XC(K) 1 110 0.000000 2 110 7.000000 3 010 14.000000 4 010 21.000000 ELEMENTS K NODE(I,K) 1 1 2 2 2 3 3 3 4

E(K) 2.1000000E+11 2.1000000E+11 2.1000000E+11 NODE 1 2 3 4

ELEMENTS K NODE(I,K) X-FORCE 1 2 3

1 2 3

2 3 4

YC(K) 0.000000 0.000000 0.000000 0.000000

ZC(K) FORCE(1,K) FORCE(2,K) 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 G(K) A(K) 1.0000000E+00 1.0000000E–02 1.0000000E+00 1.0000000E–02 1.0000000E+00 1.0000000E–02

FORCE(3,K) 28583.000000 0.000000 0.000000 –28583.000000

XI(K) 1.0000000E–04 1.0000000E–04 1.0000000E–04

DISPLACEMENTS Z-ROTATION X Y THETA 0.00000E+00 0.00000E+00 0.28583E–02 0.00000E+00 0.00000E+00 – 0.95277E–03 0.00000E+00 0.00000E+00 0.95277E–03 0.00000E+00 0.00000E+00 – 0.28583E–02 Y-FORCE Z-MOMENT X-FORCE

Y-FORCE Z-MOMENT

0.0000E+00 0.4900E+04 0.2858E+05 0.0000E+00 – 0.4900E+04 0.5717E+04 0.0000E+00 0.1201E–03 –0.5717E+04 0.0000E+00 – 0.1201E–03 0.5717E+04 0.0000E+00 – 0.4900E+04 –0.5717E+04 0.0000E+00 0.4900E+04 0.2858E+05

Element (1)  f 1x(1)  0N   0   0           (1)  24500 29400 N 4900  f 1 y         m    28580 28580 57160 N m        –   =    1  =  0N  0   0     f 2 x   4900   24500   19600 N   f         2y   5717   28580   – 22860 N  m   m2 

277

Element (2)  f 2 x(2)  0N   0   0     (2)       f 2 y  24500 N  – 24500 0        (2)   5717   – 28580   22,860 N  m   m2  =    –   =   (2)  0  0   0     f 3 x   0   – 24500   24500 N   (2)         f 3 y   5717   28580    22860 N  m   m (2)   3 

Element (3)  f 3 x(3)  0N    0   0    (3)  19600 N    4900   24500  f 3 y         (3)   22860 N  m    5717   28580  m3  =   –   =    (3)  0N  0   0     f 4 x   4900   24500   29400 N  (3)       f 4 y     28580  28580   – 57160 N  m   m (3)   4 

Reactions Node 1 278

F1x = – f 1x (1) = 0 N F1y = – f 1 y (1) = – 29400 N M1 = – m1(1) = 0 N  m

Node 3 F3x = – f 3 x (2) = 0 N





F3y = – f 3 y (2)  f 3 y (1) = – 44100 N M3 = 0 N  m

Node 2 F2x = – f 2 x (2) = 0 N





F2y = – f 2 y (1)  f 2 y (2) = – 44100 N M2 = 0 N  m

Node 4 F4x = – f 3 x (3) = 0 N F4y = – f 4 y (3) = – 29400 N M4 = – m4(3) = 0 N  m

5.35

279

NUMBER OF ELEMENTS = 12 NUMBER OF NODES = 10 NODE POINTS K IFIX 1 111 2 111 3 000 4 000 5 000 6 000 7 000 8 000 9 000 10 000

ELEMENTS K NODE(1,K) 1 1 3 2 3 5 3 5 7 4 7 9 5 2 4 6 4 6

XC(K) 0.000000 10.000000 0.000000 10.000000 0.000000 10.000000 0.000000 10.000000 0.000000 10.000000

YC(K) 0.000000 0.000000 6.000000 6.000000 10.000000 10.000000 14.000000 14.000000 18.000000 18.000000

ZC(K) 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000

FORCE(1,K) 0.000000 0.000000 12000.000000 0.000000 8000.000000 0.000000 8000.000000 0.000000 4000.000000 0.000000

FORCE(2,K) 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000

FORCE(3,K) 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 2800.000000 0.000000

E(K) 2.1000000E+11 2.1000000E+11 2.1000000E+11 2.1000000E+11 2.1000000E+11 2.1000000E+11

G(K) 0.00000000E+00 0.00000000E+00 0.00000000E+00 0.00000000E+00 0.00000000E+00 0.00000000E+00

280

A(K) 2.0000000E–02 1.5000000E–02 1.0000000E–02 1.0000000E–02 2.0000000E–02 1.5000000E–02

XI(K) 2.0000000E–04 1.5000000E–04 1.0000000E–04 1.0000000E–04 2.0000000E–04 1.5000000E–04

7 8 9 10 11 12

6 8 3 5 7 9

8 10 4 6 8 10

2.1000000E+11 2.1000000E+11 2.1000000E+11 2.1000000E+11 2.1000000E+11 2.1000000E+11 NODE 1 2 3 4 5 6 7 8 9 10

ELEMENTS K NODE (I,K) 1 1 3 2 3 5 3 5 7 4 7 9 5 2 4 6 4 6 7 6 8 8 8 10 9 3 4 10 5 6 11 7 8 12 9 10

0.00000000E+00 0.00000000E+00 0.00000000E+00 0.00000000E+00 0.00000000E+00 0.00000000E+00

1.0000000E–02 1.0000000E–02 4.0000000E–02 4.0000000E–02 4.0000000E–02 4.0000000E–02

DISPLACEMENTS X Y 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 0.92499E–02 0.32295E–04 0.92428E–02 – 0.32295E–04 0.13440E–01 0.45836E–04 0.13435E–01 – 0.45836E–04 0.16322E–01 0.53376E–04 0.16317E–01 – 0.53376E–04 0.17395E–01 0.54706E–04 0.17393E–01 – 0.54706E–04

X-FORCE

Y-FORCE Z-MOMENT

X-FORCE

– 0.2261E+05 – 0.1066E+05 – 0.3959E+04 – 0.6981E+03 0.2261E+05 0.1066E+05 0.3959E+04 0.6981E+03 0.5993E+04 0.3968E+04 0.4237E+04 0.1799E+04

0.1601E+05 0.5360E+05 0.1000E+05 0.1728E+05 0.5969E+04 0.1078E+05 0.2206E+04 0.3331E+04 0.1599E+05 0.5355E+05 0.1000E+05 0.1730E+05 0.6032E+04 0.1085E+05 0.1796E+04 0.2896E+04 – 0.1194E+05 – 0.5973E+05 – 0.6704E+04 – 0.3350E+05 – 0.3261E+04 – 0.1643E+05 – 0.6981E+03 – 0.2692E+04

0.2261E+05 0.1066E+05 0.3959E+04 0.6981E+03 – 0.2261E+05 – 0.1066E+05 – 0.3959E+04 – 0.6981E+03 – 0.5993E+04 – 0.3968E+04 – 0.4237E+04 –0.1795E+04

1.0000000E–04 1.0000000E–04 6.0000000E–04 6.0000000E–04 6.0000000E–04 6.0000000E–04

Z-ROTATION THETA 0.00000E+00 0.00000E+00 – 0.79668E–03 – 0.79610E–03 – 0.45169E–03 – 0.45347E–03 – 0.23126E–03 – 0.22138E–03 – 0.25437E–04 – 0.88782E–04 Y-FORCE

Z-MOMENT

– 0.1601E+05 0.4244E+05 – 0.1000E+05 0.2272E+05 – 0.5969E+04 0.1310E+05 – 0.2206E+04 0.5492E+04 – 0.1599E+05 0.4241E+05 – 0.1000E+05 0.2270E+05 – 0.6032E+04 0.1328E+05 – 0.1796E+04 0.4288E+04 0.1194E+05 – 0.5971E+05 0.6704E+04 – 0.3354E+05 0.3261E+04 – 0.1618E+05 0.6981E+03 – 0.4288E+04

5.36

Figure P5–36

281

6 7

5 8

2 Displacement magnitude m

6.653052e – 005 5.987747e – 005 5.322442e – 005 4.657136e – 005 3.991831e – 005 3.326526e – 005 2.661221e – 005 1.995916e – 005 1.33061e – 005 6.653052e – 006 0

5.37

Figure P5–37

282

7 8

0.0000E+00 0.0000E+00

0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00

0.0000E+00 0.0000E+00

5.38

Figure P5–38 NODE Xnumber translation 1 0.0000E+00 2 1.4372E–01 3 1.4300E–01

Ytranslation 0.0000E+00 –1.1344E-04 –1.3696E–04

Ztranslation 0.0000E+00 0.0000E+00 0.0000E+00

283

Xrotation 0.0000E+00 0.0000E+00 0.0000E+00

YZrotation rotation (deg) 0.0000E+00 0.0000E+00 0.0000E+00 1.4397E–01 0.0000E+00 –4.1178E–01

4 5 6 7



1.4282E–01 1.4265E–01 0.0000E+00 0.0000E+00

–2.1948E–03 –4.7155E–04 0.0000E+00 0.0000E+00

0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00

0.0000E+00 2.2790E–01 0.0000E+00 –5.1623E–01 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00

BEAM ELEMENT FORCES AND MOMENTS ELEMENT CASE NO. (MODE) 1

AXIAL FORCE R1 –2.907E+04 –2.907E+04 –2.122E+05 –2.122E+05 –1.024E+05 –1.024E+05 –1.024E+05 –1.024E+05 –1.208E+05 –1.208E+05 –3.510E+04 –3.510E+04

SHEAR FORCE R2 –9.878E+05 2.122E+05 –2.907E+04 7.593E+04 4.084E+04 4.084E+04 1.208E+05 1.208E+05 –1.024E+05 –1.024E+05 –1.098E+05 –1.098E+05

SHEAR FORCE R3 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00

TORSION MOMENT M1 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00

BENDING MOMENT M2 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00

BENDING MOMENT M3 –1.538E+06 –3.605E+04 –3.605E+04 –2.001E+05 2.022E+05 5.932E+04 5.932E+04 –3.636E+05 –4.560E+05 3.636E+05 –4.760E+05 4.023E+05

5.39

(a) Truss model NUMBER OF ELEMENTS (NELE) = 15 NUMBER OF NODES (KNODE) = 8 NODE POINTS K 1 2 3

IFIX 111 111 001

XC(K) 0.000000E+00 4.000000E+00 0.000000E+00

YC(K) ZC(K) FORCE (1,K) FORCE (2,K) FORCE (3,K) 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 3.000000E+00 0.000000E+00 2.000000E+04 0.000000E+00 0.000000E+00

284

4 5 6 7 8

001 001 001 001 001

4.000000E+00 0.000000E+00 4.000000E+00 0.000000E+00 4.000000E+00

3.000000E+00 6.000000E+00 6.000000E+00 9.000000E+00 9.000000E+00

0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00

0.000000E+00 2.000000E+04 0.000000E+00 1.000000E+04 0.000000E+00

0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00

ELEMENTS K

NODE(I,K)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

1 1 2 2 3 3 3 4 4 5 5 5 6 6 7

3 4 3 4 4 5 6 5 6 6 7 8 7 8 8

E(K)

A(K)

2.0000E+11 2.0000E+11 2.0000E+11 2.0000E+11 2.0000E+11 2.0000E+11 2.0000E+11 2.0000E+11 2.0000E+11 2.0000E+11 2.0000E+11 2.0000E+11 2.0000E+11 2.0000E+11 2.0000E+11

2.0000E–04 2.0000E–04 2.0000E–04 2.0000E–04 2.0000E–04 2.0000E–04 2.0000E–04 2.0000E–04 2.0000E–04 2.0000E–04 2.0000E–04 2.0000E–04 2.0000E–04 2.0000E–04 2.0000E–04

NUMBER OF NONZERO UPPER CO-DIAGONALS (MUD)-11 DISPLACEMENTS X Y NODE NUMBER 1 NODE NUMBER 2 NODE NUMBER 3 NODE NUMBER 4 NODE NUMBER 5 NODE NUMBER 6 NODE NUMBER 7 NODE NUMBER 8

0.0000E+00 0.0000E+00 0.7935E–02 0.7315E–02 0.1738E–01 0.1681E–01 0.2603E–01 0.2572E–01

0.0000E+00 0.0000E+00 0.3730E–02 – 0.3583E–02 0.5276E–01 – 0.4849E–02 0.5662E–02 – 0.5026E–02

STRESSES IN ELEMENTS (IN CURRENT UNITS) ELEMENT NUMBER 1= 2= 3= 4= 5= 6= 7= 8= 9= 10 = 11 = 12 = 13 = 14 = 15 =

Z 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00

STRESS

0.24864E+09 0.14809E+09 – 0.16441E+09 – 0.23886E+09 – 0.30998E+08 0.10311E+09 0.78155E+08 – 0.10934E+09 – 0.84393E+08 – 0.28261E+08 0.25697E+08 0.19671E+08 – 0.42828E+08 – 0.11803E+08 – 0.15737E+08

(b) Rigid frame model NUMBER OF ELEMENTS = 15 NUMBER OF NODES = 8 285

NODE POINTS K 1 2 3 4 5 6 7 8

IFIK 110 110 000 000 000 000 000 000

XC(K) 0.000000 4.000000 0.000000 4.000000 0.000000 4.000000 0.000000 4.000000

YC(K) 0.000000 0.000000 3.000000 3.000000 6.000000 6.000000 9.000000 9.000000

ZC(K) 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000

FORCE(1,K) 0.000000 0.000000 20000.000000 0.000000 20000.000000 0.000000 10000.000000 0.000000

FORCE(2,K) FORCE (3,K) 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000

ELEMENTS K NODE(I, K) 1 1 3 2 1 4 3 2 3 4 2 4 5 3 4 6 3 5 7 3 6 8 4 5 9 4 6 10 5 6 11 5 7 12 5 8 13 6 7 14 6 8 15 7 8

E(K) 2.0000000E+11 2.0000000E+11 2.0000000E+11 2.0000000E+11 2.0000000E+11 2.0000000E+11 2.0000000E+11 2.0000000E+11 2.0000000E+11 2.0000000E+11 2.0000000E+11 2.0000000E+11 2.0000000E+11 2.0000000E+11 2.0000000E+11

G(K) 0.0000000E+00 0.0000000E+00 0.0000000E+00 0.0000000E+00 0.0000000E+00 0.0000000E+00 0.0000000E+00 0.0000000E+00 0.0000000E+00 0.0000000E+00 0.0000000E+00 0.0000000E+00 0.0000000E+00 0.0000000E+00 0.0000000E+00

NODE

DISPLACEMENTS

1 2 3 4 5 6 7 8

X 0.00000E+00 0.00000E+00 0.54772E–02 0.51685E–02 0.11556E–01 0.11199E–01 0.18021E–01 0.17667E–01

A(K) 2.0000000E–04 2.0000000E–04 2.0000000E–04 2.0000000E–04 2.0000000E–04 2.0000000E–04 2.0000000E–04 2.0000000E–04 2.0000000E–04 2.0000000E–04 2.0000000E–04 2.0000000E–04 2.0000000E–04 2.0000000E–04 2.0000000E–04

XI(K) 4.0000000E–04 4.0000000E–04 4.0000000E–04 4.0000000E–04 4.0000000E–04 4.0000000E–04 4.0000000E–04 4.0000000E–04 4.0000000E–04 4.0000000E–04 4.0000000E–04 4.0000000E–04 4.0000000E–04 4.0000000E–04 4.0000000E–04

Z-ROTATION

Y 0.00000E+00 0.00000E+00 0.32294E–02 – 0.32554E–02 0.40230E–02 – 0.40706E–02 0.42395E–02 – 0.42509E–02

THETA – 0.14936E–02 – 0.14200E–02 – 0.17329E–02 – 0.16980E–02 – 0.20908E–02 – 0.20924E–02 – 0.21639E–02 – 0.21228E–02

ELEMENTS K NODE (I,K) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

1 3 1 4 2 3 2 4 3 4 3 5 3 6 4 5 4 6 5 6 5 7 5 8 6 7 6 8 7 8

X-FORCE

Y-FORCE Z-MOMENT

X-FORCE

– 0.4306E+05 0.2267E+05 0.4038E+05 – 0.1745E+05 – 0.1746E+05 – 0.4038E+05 0.1955E+05 – 0.1545E+05 – 0.3363E+05 0.4340E+05 0.1748E+05 0.3363E+05 0.3087E+04 – 0.5654E+04 – 0.1201E+05 – 0.1058E+05 0.1220E+05 0.2784E+05 – 0.1582E+04 – 0.2226E+04 0.1879E+03 0.5942E+04 0.1406E+04 0.9799E+04 0.1087E+05 0.1228E+05 0.2893E+05 0.3563E+04 – 0.4091E+04 – 0.8150E+04 – 0.2887E+04 0.2978E+04 0.6416E+04 0.6004E+03 – 0. 1903E+04 – 0.4245E+04 0.3772E+04 0.7733E+03 0.3077E+04 0.2405E+04 0.5163E+04 0.8555E+04 0.3541E+04 – 0.1243E+04 – 0.3307E+04

0.4306E+05 0.1745E+05 – 0.1955E+05 – 0.4340E+05 – 0.3087E+04 0.1058E+05 0.1582E+04 – 0.5942E+04 – 0.1087E+05 – 0.3563E+04 0.2887E+04 – 0.6004E+03 – 0.3772E+04 – 0.2405E+04 – 0.3541E+04

286

Y-FORCE Z-MOMENT – 0.2267E+05 0.1746E+05 0.1545E+05 – 0.1748E+05 0.5654E+04 – 0.1220E+05 0.2226E+04 – 0.1406E+04 – 0.1228E+05 0.4091E+04 – 0.2978E+04 0.1903E+04 – 0.7733E+03 – 0.5163E+04 0.1243E+04

0.2762E+05 – 0.4692E+05 – 0.4364E+05 0.1860E+05 – 0.1061E+05 0.8749E+04 – 0.1132E+05 – 0.2770E+04 0.7899E+04 – 0.8214E+04 0.2517E+04 – 0.5270E+04 0.7894E+03 0.6933E+04 – 0.1664E+04

(c) Use program PFRAME to model a truss (Use PFRAME to model a Truss, i.e., MAKE I  0). NUMBER OF ELEMENTS = 15 NUMBER OF NODES = 8 NODE POINTS K IFIX 1 110 2 110 3 000 4 000 5 000 6 000 7 000 8 000

XC(K) 0.000000 4.000000 0.000000 4.000000 0.000000 4.000000 0.000000 4.000000

ELEMENTS K NODE(I,K) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

1 1 2 2 3 3 3 4 4 5 5 5 6 6 7

3 4 3 4 4 5 6 5 6 6 7 8 7 8 8

NODE 1 2 3 4 5 6 7 8

ELEMENTS K NODE X-FORCE (I,K) 1 2

YC(K) 0.000000 0.000000 3.000000 3.000000 6.000000 6.000000 9.000000 9.000000

ZC(K) 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000

FORCE(1,K) 0.000000 0.000000 20000.000000 0.000000 20000.000000 0.000000 10000.000000 0.000000

A(K)

XI(K)

2.0000000E–04 2.0000000E–04 2.0000000E–04 2.0000000E–04 2.0000000E–04 2.0000000E–04 2.0000000E–04 2.0000000E–04 2.0000000E–04 2.0000000E–04 2.0000000E–04 2.0000000E–04 2.0000000E–04 2.0000000E–04 2.0000000E–04

1.0000000E–06 1.0000000E–06 1.0000000E–06 1.0000000E–06 1.0000000E–06 1.0000000E–06 1.0000000E–06 1.0000000E–06 1.0000000E–06 1.0000000E–06 1.0000000E–06 1.0000000E–06 1.0000000E–06 1.0000000E–06 1.0000000E–06

DISPLACEMENTS

Z-ROTATION

X 0.00000E+00 0.00000E+00 0.79234E–02 0.73056E–02 0.17337E–01 0.16772E–01 0.25985E–01 0.25670E–01

Y 0.00000E+00 0.00000E+00 0.37278E–02 – 0.35821E–02 0.52713E–02 – 0.48468E–02 0.56570E–02 –0.50248E–02

THETA – 0.21033E–02 – 0.19714E–02 – 0.23957E–02 – 0.23447E–02 – 0.28898E–02 – 0.29231E–02 – 0.27580E–02 – 0.27544E–02

Y-FORCE Z-MOMENT at 1st node

X-FORCE

Y-FORCE Z-MOMENT at 2nd node

1 3 – 0.4970E+05 0.1044E+03 0.1762E+03 1 4 – 0.2956E+05 – 0.7432E+02 – 0.1762E+03

287

0.4970E+05 – 0.1044X+03 0.1372E+03 0.2956E+05 – 0.7432E+02 – 0.1955E+03

3 4 5 6 7 8 9 10 11 12 13 14 15

2 2 3 3 3 4 4 5 5 5 6 6 7

3 4 4 5 6 5 6 6 7 8 7 8 8

0.3282E+05 0.4776E+05 0.6178E+04 – 0.2058E+05 – 0.1548E+05 0.2170E+05 0.1686E+05 0.5641E+04 – 0.5143E+04 – 0.3912E+04 0.8543E+04 0.2373E+04 0.3153E+04

– 0.6108E+02 0.7391E+02 – 0.8141E+02 0.1320E+03 – 0.2166E+02 0.2894E+00 0.1391E+03 – 0.5654E+02 0.1572E+02 – 0.1677E+02 – 0.5227E+01 0.3387E+02 – 0.1286E+02

– 0.1357E+03 – 0.3282E+05 0.6108E+02 – 0.1697E+03 0.1357E+03 – 0.4776E+05 – 0.7391E+02 0.8597E+02 – 0.1654E+03 – 0.6178E+04 0.8141E+02 – 0.1603E+03 0.2309E+03 0.2058E+05 – 0.1320E+03 0.1650E+03 – 0.3305E+02 0.1548E+05 0.2166E+02 – 0.7524E+02 0.2253E+02 – 0.2170E+05 – 0.2894E+00 – 0.2108E+02 0.2472E+03 – 0.1686E+05 – 0.1391E+03 0.1701E+03 – 0.1114E+03 – 0.5641E+04 0.5654E+02 – 0.1148E+03 0.1480E+02 0.5143E+04 – 0.1572E+02 0.3237E+02 – 0.4735E+02 0.3912E+04 0.1677E+02 – 0.3652E+02 – 0.1967E+02 – 0.8543E+04 0.5227E+01 – 0.6461E+01 0.3956E+02 – 0.2373E+04 – 0.3387E+02 0.6206E+02 – 0.2591E+02 – 0.3153E+04 0.1286E+02 – 0.2554E+02

Comparison of TRUSS, PFRAME and modeling a truss using PFRAME DISPLACEMENTS u5 v5 0.01738 0.005276 0.011556 0.004023

TRUSS PFRAME Truss using PFRAME 0.017337 0.0052713 Note: Global displacements in meters

u7 0.02603 0.018021

v7 0.005662 0.0042395

0.025985

0.005657

FORCES ELEMENT 1 f1x f1y –49728 0 –43060 22670

Truss PFRAME Truss using PFRAME –49700

104.4

ELEMENT 2 f1x f1y –29618 0 –17450 –17460

ELEMENT 3 f2x f2y 32882 0 19550 –15450

–29560

32820

–74.32

–61.00

Note 1: From equilibrium, only forces for one node of an element are shown Note 2: All forces are in local element coordinates and in Newtons. 5.40

For the two-story, two-bay rigid frame shown, determine (1) the nodal displacement components and (2) the shear force and bending moments in each member. Let E = 200 GPa, I = 2 10–4 m4 for each horizontal member and I = 1.5  10–4 m4 for each vertical member.

288

Figure P5–40 Displacements/Rotations (degrees) of nodes NODE

X–

Y–

Z–

X–

Y–

Z–

number

translation

rotation

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

0.0000E+00 0.0000E+00 0.0000E+00 –7.2482E-04 0.0000E+00 7.2482E-04 –7.7467E–07 –3.8733E–07 0.0000E+00 3.8733E–07 7.7467E–07 –6.2096E–04 0.0000E+00 6.2096E–04 1.1921E–06 5.9603E–07 0.0000E+00 –5.9603E–07 –1.1921E–06

0.0000E+00 0.0000E+00 0.0000E+00 –1.4007E–06 –3.1986E–06 –1.4007E–06 –2.8014E–06 –9.2660E–03 –6.3972E–06 –9.2660E–03 –2.8014E–06 –3.4868E–06 –8.0263E–06 –3.4868E–06 –4.1723E–06 –1.0511E–02 –9.6555E–06 –1.0511E–02 –4.1723E–06

0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00

0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 1.6616E–02 0.0000E+00 0.0000E+00 0.0000E+00 –1.6616E–02 0.0000E+00 –6.6411E–02 0.0000E+00 1.6572E–02 0.0000E+00 0.0000E+00 0.0000E+00 –1.6572E–02 0.0000E+00 6.6411E–02 0.0000E+00 4.7408E–02 0.0000E+00 0.0000E+00 0.0000E+00 –4.7408E–02 0.0000E+00 –1.2336E–01 0.0000E+00 3.0792E–02 0.0000E+00 0.0000E+00 0.0000E+00 –3.0792E–02 0.0000E+00 1.2336E–01

1 **** BEAM ELEMENT FORCES AND MOMENTS ELEMENT CASE AXIAL NO. (MODE) FORCE R1 1

–1.121E+05 –1.121E+05 –2.559E+05 –2.559E+05 –1.121E+05 –1.121E+05 –1.121E+05 –1.121E+05 –2.559E+05 –2.559E+05 –1.121E+05 –1.121E+05 –5.484E+04 –5.484E+04 –1.303E+05 –1.303E+05 –5.484E+04 –5.484E+04 –5.484E+04 –5.484E+04 –1.303E+05 –1.303E+05 –5.484E+04 –5.484E+04

SHEAR FORCE R2

SHEAR FORCE R3

TORSION MOMENT M1

BENDING BENDING MOMENT MOMENT M2 M3

8.348E+03 8.348E+03 0.000E+00 0.000E+00 –8.348E+03 –8.348E+03 8.348E+03 8.348E+03 0.000E+00 0.000E+00 –8.348E+03 –8.348E+03 2.384E+04 2.384E+04 0.000E+00 0.000E+00 –2.384E+04 –2.384E+04 2.384E+04 2.384E+04 0.000E+00 0.000E+00 –2.384E+04 –2.384E+04

0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00

1.391E+04 –6.955E+03 0.000E+00 0.000E+00 –1.391E+04 6.955E+03 –6.955E+03 –2.782E+04 0.000E+00 0.000E+00 6.955E+03 2.782E+04 5.364E+04 –5.963E+03 0.000E+00 0.000E+00 –5.364E+04 5.963E+04 –5.963E+03 –6.557E+04 0.000E+00 0.000E+00 5.963E+03 6.557E+04

5.41 For the two-story, three-bay rigid frame shown, determine (1) the nodal displacements and (2) the member end shear forces and bending moments. (3) Draw the shear force and bending 289

moment diagrams for each member. Let E = 200 GPa, I = 1.29  10–4 m4 for the beams and I = 0.462  10–4 m4 for the columns. The properties for I correspond to a W 610  155 and a W 410  114 wide-flange section, respectively, in metric units.

Figure P5–41 1**** BEAM ELEMENT FORCES AND MOMENTS ELEMENT CASE NO. (MODE) 1

AXIAL SHEAR FORCE FORCE R2 R1 1.347E+04 –1.750E+04 1.347E+04 –1.750E+04 2.471E+03 –2.000E+04 2.471E+03 –2.000E+04 –2.470E+03 –2.000E+04 –2.470E+03 –2.000E+04 –1.347E+04 –1.750E+04 –1.347E+04 –1.750E+04 1.347E+04 –1.750E+04 1.347E+04 –1.750E+04 2.471E+03 –2.000E+04 2.471E+03 –2.000E+04 –2.470E+03 –2.000E+04 –2.470E+03 –2.000E+04 –1.347E+04 –1.750E+04 –1.347E+04 –1.750E+04 2.686E+03 –3.022E+03 2.686E+03 –3.022E+03 1.063E+03 –9.477E+03 1.063E+03 –9.477E+03 –1.063E+03 –9.478E+03 –1.063E+03 –9.478E+03 –2.686E+03 –3.024E+03 –2.686E+03 –3.024E+03 2.686E+03 –3.022E+03 2.686E+03 –3.022E+03 1.063E+03 –9.477E+03 1.063E+03 –9.477E+03 –1.063E+03 –9.478E+03 –1.063E+03 –9.478E+03 –2.686E+03 –3.024E+03

SHEAR FORCE R3 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00

290

TORSION MOMENT M1 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00

BENDING BENDING MOMENT MOMENT M2 M3 0.000E+00 –5.719E+04 0.000E+00 –4.685E+03 0.000E+00 –6.219E+04 0.000E+00 –2.184E+03 0.000E+00 –6.218E+04 0.000E+00 –2.184E+03 0.000E+00 –5.718E+04 0.000E+00 –4.685E+03 0.000E+00 –4.685E+03 0.000E+00 4.782E+04 0.000E+00 –2.184E+03 0.000E+00 5.782E+04 0.000E+00 –2.184E+03 0.000E+00 5.782E+04 0.000E+00 –4.685E+03 0.000E+00 4.781E+04 0.000E+00 –5.747E+02 0.000E+00 5.469E+03 0.000E+00 –1.668E+04 0.000E+00 2.269E+03 0.000E+00 –1.669E+04 0.000E+00 2.269E+03 0.000E+00 –5.786E+02 0.000E+00 5.469E+03 0.000E+00 5.469E+03 0.000E+00 1.151E+04 0.000E+00 2.269E+03 0.000E+00 2.122E+04 0.000E+00 2.269E+03 0.000E+00 2.122E+04 0.000E+00 5.469E+03

–2.686E+03 –3.024E+03

0.000E+00

1.152E+04

5.42 For the rigid frame shown, determine (1) the nodal displacements and rotations and (2) the member shear forces and bending moments. Let E = 200 GPa, I = 0.795  10–4 m4 for the horizontal members and I = 0.316  10–4 m4 for the vertical members. These I values correspond to a W 460  158 and a W 410  85 wide-flange section, respectively.

Figure P5–42 Displacements/Rotations (degrees) of nodes

291

5.43 For the rigid frame shown, determine (1) the nodal displacements and rotations and (2) the shear force and bending moments in each member. Let E = 29  106 psi, I = 3100 in.4 for the horizontal members and I = 1110 in.4 for the vertical members. The I values correspond to a W 24  104 and a W 16  77.

Figure P5–43 Displacements/Rotations (degrees) of nodes NODE number

X– translation

Y– translation

Z– translation

X– rotation

Y– rotation

Z– rotation

1 2 3 4 5 6 7 8

0.0000E+00 1.8762E–03 3.0993E-03 0.0000E+00 1.5614E–03 2.7811E–03 3.6829E–03 0.0000E+00

0.0000E+00 1.4645E–04 2.1717E–04 0.0000E+00 3.5423E–05 7.7093E–05 1.2865E–04 0.0000E+00

0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00

0.0000E+00 –1.5869E–03 –1.4409E–03 0.0000E+00 –1.3154E–03 –1.4087E–03 –1.5467E–03 0.0000E+00

292

9 10 11 12 13 14 15 16 17 18 19 20

1.4331E–03 2.6228E–03 3.5926E-03 0.0000E+00 1.3394E–03 2.4458E–03 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00

–4.4935E–05 –8.8701E–05 –1.4025E-04 0.0000E+00 –1.3694E–04 –2.0556E–04 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00

0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00

–1.2573E–03 –1.3909E–03 –1.5614E–03 0.0000E+00 –1.3297E–03 –1.3014E–03 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00

TORSION MOMENT M1

BENDING MOMENT M2

BENDING MOMENT M3

0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00

–1.444E+05 2.653E+04 –3.692E+04 4.777E+04 –1.200E+05 2.231E+04 –4.766E+04 4.074E+04 –3.150E+04 2.125E+04 –1.117E+05 1.828E+04 – 4.814E+04 3.822E+04 –3.620E+04 2.355E+04 –1.091E+05 1.031E+04 –3.797E+04 4.008E+04

1 **** BEAM ELEMENT FORCES AND MOMENTS ELEMENT CASE AXIAL NO. (MODE) FORCE R1

SHEAR FORCE R2

SHEAR FORCE R3

Columns 1

6.363E+03 6.363E+03 3.073E+03 3.073E+03 1.539E+03 1.539E+03 1.810E+03 1.810E+03 2.240E+03 2.240E+03 –1.952E+03 –1.952E+03 –1.902E+03 –1.902E+03 –2.240E+03 –2.240E+03 –5.950E+03 –5.950E+03 –2.982E+03 –2.982E+03

–1.139E+04 –1.139E+04 –5.646E+03 –5.646E+03 –9.487E+03 –9.487E+03 –5.894E+03 –5.894E+03 –3.517E+03 –3.517E+03 –8.662E+03 –8.662E+03 –5.757E+03 –5.757E+03 –3.983E+03 –3.983E+03 –7.958E+03 –7.958E+03 –5.203E+03 –5.203E+03

0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00

5.44 A structure is fabricated by welding together three lengths of I-shaped members as shown in Figure P5-44. The yield strength of the members is 36 ksi, E = 29e6 psi, and Poisson’s ratio is 0.3. The members all have cross-section properties corresponding to a W 18  76. That is, A = 22.3 in.2, depth of section is d = 18.21 in., Ix = 1330 in.4, Sx = 146 in.3, Iy = 152 in.4, and Sy = 27.6 in.3. Determine whether a load of Q = 10,000 lb downward is safe against general yielding of the material. The factor of safety against general yielding is to be 2.0. Also, determine the maximum vertical and horizontal deflections of the structure.

293

Figure P5-44

max = 12,329 psi <

36000 = 18,000 psi 2

 Safe against yielding 5.45

Tapered beam using 1 element

294

x x  I (x) = I0 1  n  = 100 1   = 100 + x    100  L L I   = 150 in.4  2

Tapered beam using 2 elements

L 3L I1–2   = 125 in.4 ; I2–3 =   = 175 in.4  4  4

Tapered beam using 4 elements

L 3L 5L I1–2   = 112.5 in.4 ; I2–3   = 137.5 in.4 ; I3–4   = 162.5 in.4  8  8  8 7L I4 –5 =   = 187.5 in.4  8

Tapered beam using 8 elements

L I1–2   = 106.25  16 

3L I2–3   = 118.75  16 

5L I3–4   = 131.25 in.4  16 

7L I4–5   = 143.75 in.4  16 

9L I5–6   = 156.25 in.4  16 

11L  I6–7  = 168.75 in.4  16  295

13L  I7–8  = 181.25 in.4  16 

15L  I8–9  = 193.75 in.4  16 

Analytical solution v=

Pl 2  n 2 l 1 n   n   x  2 x    1  x  ln 1  x   + Ax +B 2  n n l l  n EI 0  2l

Pl 2 [ln (1 + n) – (1 + n)] n 2 EI 0

Pl 3  1 n 1 ln (1 + n ) +  1   2  2 n n EI 0  n

x = 0, n = 1, P = 500, l= 100, E = 30  106, I0 = 100 A=

(500) (100 in.)2 (1)2 (30  106 ) (100 in.4 )

[ln(2) – 2]

= – 2.1781  10–3

500 (100 in.)3

ln (2) + 1  1  1  2 (1)2 (30  10 6 ) (100 in.4 ) 

= – 1.3448  10–1

500 (100 in.)2

  100 in.   –1   – 1.3448  10  (1) (30  10 ) (100 in.)   (1)   2

= 0.032187 in. PROBLEM 2 USING 1 ELEMENT-NUMBER OF ELEMENTS = 1 NUMBER OF NODES = 2 NODE POINTS K 1 2

IFIX 000 111

XC(K) YC(K) ZC(K) FORCE(1,K) FORCE(2,K) 0.000000 0.000000 0.000000 0.000000 500.000000 100.000000 0.000000 0.000000 0.000000 0.000000

FORCE(3,K) 0.000000 0.000000

ELEMENTS K 1

NODE(I,K) 1 2

E(K) 3,0000000E+07 NODE 1 2

G(K) 1.0000000E+00

A(K) 1.0000000E–01

DISPLACEMENTS X Y 0.00000E+00 0.37037E–01 0.00000E+00 0.00000E+00

XI(K) 1.5000000E+02

Z-ROTATION THETA – 0.55556E–03 0.00000E+00

ELEMENTS K 1

NODE(I,K) X-FORCE Y-FORCE Z-MOMENT X-FORCE Y-FORCE Z-MOMENT 1 2 0.0000E+00 0.5000E+03 0.2434E–03 0.0000E+00 – 0.5000E+03 0.5000E+05

PROBLEM 2 USING 2 ELEMENTS-296

NUMBER OF ELEMENTS = 2 NUMBER OF NODES = 3 NODE POINTS K IFIX XC(K) 1 000 0.000000 2 000 50.000000 3 1 1 1 100.000000

YC(K) ZC(K) FORCE(1,K) FORCE(2,K) 0.000000 0.000000 0.000000 500.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000

ELEMENTS K NODE(1,K) E(K) 1 1 2 3.0000000E+07 2 2 3 3.0000000E+07 NODE

A(K) 1.0000000E–01 1.0000000E–01

DISPLACEMENTS X Y 0.00000E+00 0.33333E–01 0.00000E+00 0.99206E–02 0.00000E+00 0.00000E+00

1 2 3 ELEMENTS K NODE X-FORCE (I,K) 1 1 2 0.0000E+00 2 2 3 0.0000E+00

G(K) 1.0000000E+00 1.0000000E+00

Y-FORCE Z-MOMENT

X-FORCE

FORCE(3,K) 0.000000 0.000000 0.000000

XI(K) 1.2500000E+02 1.7500000E+02

Z-ROTATION THETA – 0.52381E–03 – 0.35714E–03 0.000000E+00 Y-FORCE Z-MOMENT

0.5000E+03 0.6515E–02 0.0000E+00 – 0.5000E+03 0.2500E+05 0.5000E+03 – 0.2500E+05 0.0000E+00 – 0.5000E+03 0.5000E+05

For 1 element (I1 evaluated at x = L/2) 7 I1 = 100 1   = 450 in.4  2

For 2 elements 11 I1 = 100   = 275 in.4  4 25 I2 = 100   = 625 in.4  4

For 4 elements 15 I1 = 100   = 187.5 in.4  8 29 I2 = 100   = 362.5 in.4  8 297

43 I3 = 100   = 537.5 in.4  8 57 I4 = 100   = 712.5 in.4  8

For 8 elements 23 I1 = 100   = 143.75 in.4  16  37 I2 = 100   = 231.25 in.4  16  51 I3 = 100   = 318.75 in4  16  65 I4 = 100   = 406.25 in.4  16  79 I5 = 100   = 493.75 in.4  16  93 I6 = 100   = 581.25 in.4  16  107  I7 = 100  = 668.75 in.4  16  121 I8 = 100  = 756.25 in.4  16 

The analytical solution is

vmax = ymax =

1 (500) (100)3 = 0.0095 in. 17.55 (30  106 ) (100)

  FEM   

ymax 0.0095 in. 0.0123 in. 0.0103 in. 0.0097 in. 0.0096 in.

Analytical 1 element 2 elements 4 elements 8 elements

NUMBER OF ELEMENTS = 1 NUMBER OF NODES = 2 NODE POINTS K IFIK XC(K) YC(K) ZC(K) 1 000 0.000000 0.000000 0.000000 2 1 1 1 100.000000 0.000000 0.000000

FORCE(1,K) FORCE(2,K) FORCE(3,K) 0.000000 500.000000 0.000000 0.000000 0.000000 0.000000

ELEMENTS

298

K 1

NODE(I,K) 1 2

E(K) 3.0000000E+07 NODE

G(K) 0.0000000E+00

A(K) 1.0000000E–01

DISPLACEMENTS X Y 0.00000E+00 0.12346E-01 0.00000E+00 0.00000E+00

1 2

XI(K) 4.5000000E+02

Z-ROTATION THETA –0.18519E-03 0.00000E+00

ELEMENTS K NODE(I,K) X-FORCE Y-FORCE Z-MOMENT X-FORCE Y-FORCE Z-MOMENT 1 1 2 0.0000E+00 0.5000E+03 –0.7706E–02 0.0000E+00 –0.5000E+03 0.5000E+05 NUMBER OF ELEMENTS = 4 NUMBER OF NODES = 5 NODE POINTS K IFIX XC(K) 1 000 0.000000 2 000 25.000000 3 000 50.000000 4 000 75.000000 5 1 1 1 100.000000 ELEMENTS K NODE(I,K) 1 1 2 2 2 3 3 3 4 4 4 5

YC(K) 0.000000 0.000000 0.000000 0.000000 0.000000

E(K) 3.0000000E+07 3.0000000E+07 3.0000000E+07 3.0000000E+07 NODE 1 2 3 4 5

ELEMENTS K NODE X-FORCE (1,K) 1 1 2 0.0000E+00 2 2 3 0.0000E+00 3 3 4 0.0000E+00 4 4 5 0.0000E+00

ZC(K) 0.000000 0.000000 0.000000 0.000000 0.000000

FORCE(1,K) 0.000000 0.000000 0.000000 0.000000 0.000000

G(K) 0.0000000E+00 0.0000000E+00 0.0000000E+00 0.0000000E+00

FORCE(2,K) 500.000000 0.000000 0.000000 0.000000 0.000000

A(K) 1.0000000E–01 1.0000000E–01 1.0000000E–01 1.0000000E–01

DISPLACEMENTS X Y 0.00000E+00 0.97156E–02 0.00000E+00 0.56845E–02 0.00000E+00 0.25953E–02 0.00000E+00 0.67008E–03 0.00000E+00 0.00000E+00

Y-FORCE

Z-MOMENT X-FORCE

FORCE(3,K) 0.000000 0.000000 0.000000 0.000000 0.000000 XI(K) 1.8750000E+02 3.6250000E+02 5.3750000E+02 7.1250000E+02

Z-ROTATION THETA – 0.17050E–03 – 0.14272E–03 – 0.99620E–04 – 0.51170E–04 0.00000E+00 Y-FORCE Z-MOMENT

0.5000E+03 0.3812E –01 0.0000E+00 – 0.5000E+03 0.1250E+05 0.5000E+03 – 0.1250E+05 0.0000E+00 – 0.5000E+03 0.2500E+05 0.5000E+03 – 0.2500E+05 0.0000E+00 – 0.5000E+03 0.3750E+05 0.5000E+03 – 0.3750E+05 0.0000E+00 – 0.5000E+03 0.5000E+05

5.46

1   dV 2 V

= 299

Tr = G J

1  2 r2 dV 2 V G J 2 1 T2 2  GJ 2

= 

=

r  L

  r dA dx 2

T2 dx 2GJ ( x)

Now

=

r  2 x  1x  L

 1  r   L ||

[ B]

1  1x    L  2 x  ||

{}

max = Gmax = G [B] {} U=

1 r {}T [ B]T {G}[ B]{} r dx dA 2 A 

L 1 2 R 2 r r dr d  {}T [ B]T G [ B]{} dx 0 2 0 0

L 1 2 R 3 r dr d  {}T [ B]T G [ B]{} dx 0 2 0 0

(If J constant)

x r = r0 1    L

dr =

2 L U x 3 =    r0  r0  0 0  L 1x

r0 dx L r0 GL dx  2 1x  2 2 x  2 L

r4 L U x 3 GL = 2 0  1   dx 2  – 21x  22 x  L 0 L  2 x 2L

u = 1

Let 2

1

u3 L du =



x dx , du = L L

u4 2 L 4 1

2  r04 L 4 2 G U = u  21x – 22 x  4 4 1x 1 2L = J0(16 – 1)

G  2 x – 1x  L 300

U G = J0(16 – 1)  2 x – 1x  L  2 x 

 [K] =

15 G J 0  1  1  1 1 L 

5.47

1 T2 2  2 GJ 2

1 T2 dx 2  2 GJ

T= tx

  r dA dx 2

lb  in.  t   in. 

1 L   tx  2 dx 2 0 GJ

L 1   tx  2 dx  0 2 GJ

t 2 x3 2 GJ 3 0

t 2 L3 Total strain energy 6 GJ

5.48

301

NUMBER OF ELEMENTS = 2 NUMBER OF NODES = 3 NODE POINTS K IFIX XC(K) YC(K) ZC(K) FORCE(1,K) FORCE(2,K) 1 1 1 1 0.0000000 0.000000 0.000000 0.000000 0.000000 2 0 0 0 120.000000 0.000000 0.000000 –5.000000 0.000000 3 1 1 1 120.000000 0.000000 –120.000000 0.000000 0.000000 ELEMENTS K NODE(I,K) 1 1 2 2 2 3 NODE 1 2 3

E(K) 3.0000000E+04 3.0000000E+04

XI(K) 2.0000000E+02 2.0000000E+02

DISPLACEMENT 0.00000E+00 –0.21429E+00 0.00000E+00

XJ(K) 1.0000000E+02 1.0000000E+02

THETA-X 0.00000E+00 0.25714E–02 0.00000E+00

FORCE(3,K) 0.000000 120.000000 120.000000

G(K) 1.0000000E+04 1.0000000E+04

THETA-Z 0.00000E+00 –0.25714E–02 0.00000E+00

ELEMENTS K NODE Y-FORCE X-MOMENT Z-MOMENT Z-MOMENT Y-FORCE X-MOMENT (I,K) 1 1 2 0.2500E+01 –0.2143E+02 0.2786E+03 0.2143E+02 –0.2500E+01 0.2143E+02 2 2 3 –0.2500E+01 0.2143E+02 –0.2143E+02 –0.2786E+03 0.2500E+01 –0.2143E+02

5.51

302

5.52

Figure P5–52

5.58-5.59 Determine the displacements and reactions for the space frames shown in Figures P5–58 and P5–59. Let Ix = 100 in.4, Iy = 200 in.4, Iz = 1000 in.4, E = 30,000 ksi, G = 10,000 ksi, and A = 100 in.2 for both frames. 303

Figure P5–58 Displacements/Rotations (degrees) of nodes NODE X– Y– Z– number translation translation translation 1 0.0000E+00 0.0000E+00 0.0000E+00 2 6.8927E–01 –2.6320E–03 4.4137E–01 3 8.3682E–01 1.0091E+00 –5.8406E–02 4 0.0000E+00 0.0000E+00 0.0000E+00

X– rotation 0.0000E+00 3.1601E–01 –3.6119E–01 0.0000E+00

Y– rotation 0.0000E+00 –7.5875E–01 1.1418E–01 0.0000E+00

Z– rotation 0.0000E+00 –4.8913E–01 –4.8006E–01 0.0000E+00

Y– rotation

Z– rotation

5.59

Figure P5–59 Displacements/Rotations (degrees) of nodes NODE number

X– translation

Y– translation

Z– translation

X– rotation

1 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 2 3.1055E–04 2.5111E–01 1.0906E–05 –1.1112E–01 304

0.0000E+00 0.0000E+00 2.1045E–08 3.6115E–02

3 0.0000E+00 4 0.0000E+00 5 –3.1055E–04 6 0.0000E+00 7 0.0000E+00 8 0.0000E+00

0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 2.4992E–01 –2.5905E–05 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00

0.0000E+00 0.0000E+00 1.1109E–02 0.0000E+00 0.0000E+00 0.0000E+00

0.0000E+00 0.0000E+00 6.8756E–02 0.0000E+00 0.0000E+00 0.0000E+00

0.0000E+00 0.0000E+00 3.5829E–02 0.0000E+00 0.0000E+00 0.0000E+00

5.60 Design a jib crane as shown in Figure P5–60 that will support a downward load of 6000 lb. Choose a common structural steel shape for all members. Use allowable stresses of 0.66 Sy (Sy is the yield strength of the material) in bending, and 0.60 Sy in tension.

FigureP5–60 Horizontal load beam Vertical support beam Cross brace

S12  50 S12  50 S6  12.5

All members A36 structural steel 

The required maximum deflection governed the selection of the material section size, as smaller sizing would be lighter and more than adequate to support the load of 6000 lb, but provide too much deflection than the required 0.400 in.



The force in the cross brace (21066 lbf) does not yield to buckling as shown in the first set of calculations and with a S6  12.5 section the cross brace is designed above the imposed load of 6500 lb



The horizontal load beam is designed to withstand above the imposed bending moment of 516000 lbf  in. The minimum required section of S10  25.4 was exceeded, as shown in the second set of calculations, to accommodate the required deflection constraint. Also, the excessive section will allow additional safety against failure from overloading.

5.61 Design the support members, AB and CD, for the platform lift shown in Figure P5-61. Select a mild steel and choose suitable cross-sectional shapes with no more than a 4 : 1 ratio of moments of inertia between the two principal directions of the cross section. You may 305

choose two different cross sections to make up each arm to reduce weight. The actual structure has four support arms, but the loads shown are for one side of the platform with the two arms shown. The loads shown are under operating conditions. Use a factor of safety of 2 for human safety. In developing the finite element model, remove the platform and replace it with statically equivalent loads at the joints at B and D. Use truss elements or beam elements with low bending stiffness to model the arms from B to D, the intermediate connection, E to F, and the hydraulic actuator. The allowable stresses are 0.66Sy in bending and 0.60Sy in tension. Check buckling using either Euler’s method or Johnson’s method as appropriate. 1 of the length of member Also check maximum deflections. Any deflection greater than 360 AB is considered too large.

Figure P5-61 

Many viable solutions are possible.



This design recommends 1020 steel with cross sections of 5 in.  3 in.  83 in. rectangular tubing.



The maximum deflection is then 0.244 in. which is less than the maximum allowable of 0.25 in.



The bending stress is 11,418 psi which is less than the allowable of 31,600 psi.



The axial stress is 2520 psi which is less than the allowable of 28,700 psi.

5.62 A two-story building frame is to be designed as shown in Figure P5-62. The members are all to be I-beams with rigid connections. We would like the floor joists beams to have a 15-in. depth and the columns to have a 10 in. width. The material is to be A36 structural steel. Two horizontal loads and vertical loads are shown. Select members such that the allowable bending in the beams is 24,000 psi. Check buckling in the columns using Euler’s or Johnson’s method as appropriate. The allowable deflection in the beams should not exceed 1 of each beam span. The overall sway of the frame should not exceed 0.5 in. 360

306

Figure P5-62 

Many viable solutions are possible.



This design recommends A36 structural steel I-beams.



W 16  26 beams are recommended for the horizontal and diagonal members with largest bending stress of 7000 psi which is less than the allowable of 24,000 psi.



W 10 49 sections are recommended for the vertical members. Column buckling was verified to be satisfied.



Maximum sway in the horizontal direction is 0.246 in. which is less than the allowable of 0.50 in.



Another satisfactory solution is W 10  26 beams for horizontal and diagonal members and W 10  33 sections for the vertical members. The sway then becomes 0.417 in.

5.63 A pulpwood loader as shown in Figure P5–63 is to be designed to lift 2.5 kip. Select a steel and determine a suitable tubular cross section for the main upright member BF that has attachments for the hydraulic cylinder actuators AE and DG. Select a steel and determine a suitable box section for the horizontal load arm AC. The horizontal load arm may have two different cross sections AB and BC to reduce weight. The finite element model should use beam elements for all members except the hydraulic cylinders, which should be truss elements. The pinned joint at B between the upright and horizontal beam is best modeled with end release of the end node of the top element on the upright member. The allowable bending stress is 0.66 Sy in members AB and BC. Member BF should be checked for 1 of the length of BC. As a buckling. The allowable deflection at C should be less than 360

bonus, the client would like you to select the size of the hydraulic cylinders AE and DG.

307

Figure P5–63 

Many viable solutions are possible.



The design recommends AISI 1020 rolled steel.



The horizontal beam, AC, is recommended to be a rectangular tube 4 in. by 16 in. with 0.25 in. thickness. The maximum bending stress in member AC is 7736 psi less than the allowable of 3100 psi. The maximum deflection is 0.299 in. less than the allowable of 0.300 in.



The vertical member, BF, is recommended to be a square tube 10 in. by 10 in. with 0.5 in. thickness.

5.65 A small hydraulic floor crane as shown in Figure P5-65 carries a 5000 lb load. Determine the size of the beam and column needed. Select either a standard box section or a wide-flange section. Assume a rigid connection between the beam and column. The column is rigidly connected to the floor. The allowable bending stress in the beam is 0.60Sy. The allowable 1 of the beam length. Check the column for buckling. deflection is 360

Figure P5-65 

Many viable solutions are possible.



The design recommends A36 structural steel. 308



The horizontal and vertical members are recommended to be W 10  68.



The largest bending stress in the horizontal beam is 4756 psi less than the allowable of 21,600 psi. The maximum deflection is 0.215 in. less than the allowable of 0.222 in.



The column, ACD, has a bending stress of 5284 psi.



The column should be checked for buckling.

5.68 Design the gabled frame subjected to the external wind load shown (comparable to an 80 mph wind speed) for an industrial building. Assume this is one of a typical frame spaced every 20 feet. Select a wide flange section based on allowable bending stress of 20 ksi and an allowable compressive stress of 10 ksi in any member. Neglect the possibility of buckling in any members. Use ASTM A36 steel.

Figure P5-68    

Many viable solutions are possible. The design recommends A36 structural steel. The frame members are recommended to be W 10  12 wide flange shapes. The maximum worst stress (combined bending and compression) in any member is 19.5 ksi.  The maximum displacement is 0.719 in. 5.69 Design the gabled frame shown for a balanced snow load shown (typical of the Mid-west) for an apartment building. Select a wide flange section for the frame. Assume the allowable bending stress not to exceed 140 MPa. Use ASTM A36 steel.

Figure P5-69 309

   

Many viable solutions are possible. The table below lists some W sections that were considered. The recommended W 6  12 with bending stress of 120.4 MPa is less than the allowable value of 140 MPa. The maximum displacement is 0.0147 m. Table 1: Beam Trial Runs Beam Section

Bending Stress (Local 3)

Displacement

W30  173

1.6 MPa

6.26*10^–5

W12  45

15.2 MPa

0.0010 m

W8  13

88.7 MPa

0.0082 m

W6  12

120.4 MPa

0.0147 m

5.70 Design a gantry crane that must be able to lift 10 tons as it must lift compressors, motors, heat exchangers, and controls. This load should be placed at the center of one of the main 12-foot-long beams as shown in Figure P5-70 by the hoisting device location. Note that this beam is on one side of the crane. Assume you are using ASTM A36 structural steel.

    

Many viable solutions are possible. The design is based on the load being applied to the center span of a 12 ft long beam. The design recommends W 10  100 for the horizontal beams. The bracing members are recommended to be W 4  13. The vertical columns are recommended to be 4 in.  4 in.  14 in. thick hollow square

tubes. The table below is a summary of the final member sizes and deflection, stress, and buckling calculated and allowable results. Table 1: This table shows all the members with corresponding material and size. 

Member

Quantity

Material

Size (in. 

lb ) ft

loaded 12 ft beam

ASTM A36 St. Steel

W10  100

unloaded 12 ft beam

ASTM A36 St. Steel

W10  100

8 ft Beams

ASTM A36 St. Steel

W10  100

Corner Braces

ASTM A36 St. Steel

W4  13

Columns

ASTM A36 St. Steel

4  4 hollow  14 thick (in.)

Table 2: This table shows that the maximum deflections are less than the allowable deflection, and that the calculated bending stresses are less than the allowable stresses in the beams. Calculations Allowable Deflection (in.)

Calculated

Allowable

Bending Stress (psi)

Member

Maximum Deflection (in.) Using By Hand Autodesk

Loaded 12 ft Beam

0.0722

0.0847

0.2667

6405

7200

Unloaded 12 ft Beam

–

0.0141

0.2667

23.145

7200

8 ft Beams

–

0.01977

0.2667

1.286

7200

Corner Braces

–

0.02921

0.1333

256.654

7200

310

Columns

–

0.2863

0.4000

–

Table 3: This table shows that the corner braces and the columns have loads smaller than the load that would cause buckling.

Member

Buckling Strength (lb) Allowable load Calculated load

Loaded 12 ft Beam

–

Unloaded 12 ft Beam

–

8 ft Beams

–

Corner Braces

24000

330000

Columns

24572

60000

5.71 Design the rigid highway bridge frame structure shown in Figure P5-71 for a moving truck load (shown below) simulating a truck moving across the bridge. Use the load shown and place it along the top girder at various locations. Use the allowable stresses in bending and compression and allowable deflection given in the Standard Specification for Highway Bridges, American Association of State Highway and Transportation Officials (AASHTO), Washington, D.C. or use some other reasonable values.

Figure P5-71 

Many viable solutions are possible.



A36 structural steel is chosen in the design.



After some iteration, W 24  94 wide flange sections were selected for all members.



The largest bending stress of 12960 psi with the truck in the center span location is less than the allowable of 20,000 psi.

311



1 of the span The largest deflection of 0.731 in. is less than the allowable of 0.75 in. ( 800

length). 5.73 The curved semi-circular frame shown in Figure P 5-73 is supported by a pin on the left end and a roller on the right end and is subjected to a load P = 1000 lb at its apex. The frame has a radius to centerline of cross section of R = 120 in. Select a structural steel W shape from Appendix F such that the maximum stress does not exceed 20 ksi. Perform a finite element analysis using 4, 8, and then 16 elements in your finite element model. Also determine the maximum deflection for each model. It is suggested that the finite element answers for deflection be compared to the solution obtained by classical methods, such as using Castigliano’s theorem. The expression for deflection under the load is given by using Castigliano’s theorem as

0.178 PR 3 0.393 PR 0.393 PR   EI AE Av G where A is the cross sectional area of the W shape, Av is the shear area of the W shape (use depth of web times thickness of web for the shear area), E = 30  106 psi, and G = 11.5  106 psi. Now change the radius of the frame to 20 in. and repeat the problem. Run the finite element model with the shear area included in your computer program input and then without. Comment on the difference in results and compare to the predicted analytical deflection by using the equation above for y.

y =

For

R = 20 in. A = 8.79 in. tweb = 0.260 in. depth = 12.34 in. SA2 = 3.2084  100 in.

SA2 = tweb depth Ix = 238 in.4

E = 29  106 psi 3

G = 11.6  106 psi

m =

0.178 PR EI x

n =

0.393 PR AE

n = 3.08344  10–5 in.

v =

0.393 PR SA2G

v = 2.11191  10–4 in.

P = 1000 lb

m = 2.06317  10–4 in.

max = m + n + v max = 4.48343  10–4 in. maxnoshear = m + n maxnoshear = 2.37151  10–4 in. P

Figure P5-73 312



The table below shows the results for the 2 – 16 element models without shear area and with shear area included in some cases for both radii. Radius = 120 in. No. of Elements

Max Def. (in.)

Max. Stress (psi)

5.92E-02

1595

4.69E-02

1576

4.52E-02

1566

4.48E-02

1560

8 (SA2 incl.)

4.65E-02

1566

16 (SA2 incl.)

4.62E-02

1560

Longhand

4.48E-02

Longhand (SA2 incl.)

4.60E-02 Radius = 20 in.

No. of Elements

Max Def. (in.)

Max. Stress (psi)

3.01E-04

299

2.47E-04

281

2.39E-04

270

2.39E-04

265

8 (SA2 incl.)

4.55E-04

270

16 (SA2 incl.)

4.55E-04

265

Longhand

2.37E-04

Longhand (SA2 incl.)

4.48E-04

313

314

Chapter 6 6.1 For sketch of Ni see Figure 6.8. Others follow similarly By Equation (6.2.18) 1 Ni + Nj + Nm = [(i + j + m) + (i + j + m) x  [i + j + m)y] 2A By Equation (6.2.10)

(1)

i + j + m = xj ym – yj xm + yi xm – xi ym + xi yj – yi xj = 2A (by Equation (6.2.9))

(2)

i + j + m = yj – ym + ym – yi + yi – yj = 0

(3)

i + j + m = xm – xj + xi – xm + xj – xi = 0

(4)

By using (2)–(4) in (1), we obtain Ni + Nj + Nm = 1 identically 6.2 By Equation (6.2.47) 1 p = {d}T 2

1 [B] = 0 2A

1 v E v 1 [D] = 1 – v2 0 0

[B]T [D] [B] dV{d} – {d}T { f }

0 0

, {d} =

1 v 2

ui vi uj vj um vm i



p =

1 [ui vi uj vj um vm] 2

1 j 2A 0

E 1 v2

1 v v 1

0 0

1 v 2

1 0 2A

ui vi uj

dV 

vj um vm

– [ui vi uj vj um vm] { f } 1 1 = 2 ui 2 2A

E 1 1 v2 2 A

i 0

1 v v 1

0 0

1 v 2

= 2C

j 0

= 2C

0 0

1 v 2

0 0

1 v 2

1 v

v 1

0 0

1 v 2

1 v v 1

0 0

1 v 2

1 v v 1

0 0

1 v 2

dV vi – f1y

0 0

dV ui – f1x

1 v v 1



m 0

1 v v 1

dV uj – f2x j

0 i

dV vj – f2y

dV um – f3x m

0 m

dV vm – f3y

1 v v 1

0 0

1 v 2

f1x

ui vi uj

 dV

f1 y f2 x

–

f2 y f3 x

um vm

f3 y

From Equation (3.10.27) or Equation (6.2.48) d



= 0 =

B T [D] [B] dV{d} – { f } = 0

6.3 (a)

E  10 106 psi v  0.25 t  1 in.

[k] = tA[B]T [D] [B] xi = 0, yi = –1, xj = 2, yj = 0, xm = 0, ym = 1 A=

1 1 b h = (2)(2) = 2 in.2 2 2

i = yj – ym = 0 – 1 = – 1 j = ym – yi = 1 – (–1) = 2 m = yi – yj = – 1 – 0 = – 1 i = xm – xj = 0 – 2 = – 2 j = xi – xm = 0 – 0 = 0 m = xj – xi = 2 – 0 = 2 i

1 0 [B] = 2A

Since it is plane stress [D] =

E (1 v 2 )

1 v v 1

0 0

1 v 2

 1 0 2   0 2 1   1 0.25 0  10  106  2 0 0   T So [B] [D] = 1 0    0.25  4  0.9375   0 0 2   0 0.375  0   1 0 2     0 2 1  1  0.25  0.75    0.5 2  0.375   0.5 0  10  106  2 =   0 0.75  4  0.9375   0  –1  0.25 0.75    2  0.375  0.5

[k] = t A[B]T [D] [B]



 1  0.25  0.75    0.5 2  0.375   0.5 0  10  106  2 [k] = (1 in.)(2)   0 0.75  4  0.9375   0  1  0.25 0.75    2  0.375  0.5

1 4

1 0 2

0 2 0 2 0 0 1 0 2

1 0 2

0 2 1

i 1

[k] = 1.333  106

2.5 1.25 2 1.5 0.5 0.25

1.25 4.375 1 0.75 0.25 3.625

j 2 1 4 0 2 1

2 1.5 0.75 0 1.5 1.5 0.75

m 3 0.5 0.25 2 1.5 2.5 1.25

0.25 3.625 1 0.75 1.25 4.375

(b) xi = 1.2, yi = 0, xj = 2.4, yj = 0, xm = 1.2, ym = 1

i = yj – ym = 0 – 1 = – 1 j = ym – yj = 1 – 0 = 1 m = yi – yj = 0 – 0 = 0 i = xm – xj = 1.2 – 2.4 = – 1.2 j = xi – xm = 1.2 – 1.2 = 0 m = xj – xi = 2.4 – 1.2 = 1.2 A=

1 (1.2) (1) = 0.6 in.2 2

1.2

0 1

1.2 0

1 0

0 0

1 1.2

1.2

0.25

0.45

0.3 1

1.2 0.25

0.375 0

0 0

0.375 0.45

0.3

1.2

10  10 So [B] [D] = (1.2)  0.9375  T

8.33  10 0.9375

1 0.25 0.25 1 0

0 0 0.375

[k] = t A[B]T [D] [B]

[k] =

(1in.) (6in. )8.3310 2(0.6) (0.9375)

0.25

0.45

0.3 1

1.2 0.25

0.375 0

0 0

0.375 0.45

0.3

1.2

1 0

0 1 0 1.2 0 0

0 0

0 1.2

1.2

0 1 1.2

i 1 1.54 0.75 1 0.45 0.54 0.3

[k] = 4.44  106

j 0.75 1.815 0.3 0.375 0.45 1.44

1 0.3 1 0 0 0.3

m 3

0.45 0.375 0 0.375 0.45 0

0.54 0.45 0 0.45 0.54 0

0.3 1.44 0.3 0 0 1.44

(c) E = 10  106  = 0.25 t=1 Triangle coordinate definition This defines an array variable x coordinate is the top y coordinate is the bottom

0 0

x 0 y 1

2 0

Area of triangle = 12 base  height

0 1

1 (jx – ix) (my – iy) 2

i = jy – my

i = –1

i = mx – jx

i = –2

j = my – iy

j = 1

j = ix – mx

j = 0

m = iy – jy

m = 0

m = jx – ix

m = 2

A= 1 Develop stiffness matrix

1 0 [Bi] = 2A

1 [Bm] = 0 2A

1 [Bj] = 0 2A

i i

0 i

m m

Gradient matrix [B] = augment (Bi, Bj, Bm)

[B] =

0.5 0 1

0 0.5 0 0 0 1 0 0 0 1 0.5 0 0.5 1 0

Plane stress Constitutive matrix

[D] =

E 1 v2

1 v v 1

0 0

1 v 2

1.067  107 2.67  106  [D] =  2.67  106 1.067  107  0 0 

  0  7 0.4  10  0

[k] = t A [B]TD]B] Constant-strain triangular element stiffness matrix

[k] =

1 3

2 107

1 107

8 106

6 106

1.2 107

4 106

1 107

3.5 107

4 106

3 106

6 106

3.2 107

8 106

4 106

8 106

4 106

6 106

3 106

6 106

1.2 10

6 10

1.2 10

4 106

3.2 107

4 106

3.2 107

6.4 In general we know that

{} =

E 2

(1 v )

1 v v 1

0 0

1 v 2

1  0 2A

u1 v1 u2 v2 u3 v3

(a) For the first element we have

0.25 0   1  1 0 2 0 1 0 10  106  1   =  0.25 1 0 0 2 0 0 0 2   2(2)  (1  0.252 )  0 0.375  0  2 1 0 2 2 1

x y xy

0  0.0025     6400 psi  0.0012        1600 psi  0  5000 psi   0.0     0.0025  The principal stresses are given by the equations x

1 2

2 xy = 1,2

2 xy

1 2

and the plane that are acting upon is

p =

1 tan–1 2

xy x

6400  1600  6400  1600 2 2 2 1 =      5000   2 2   

= 9546 psi 1

6400  1600  6400  1600 2 2 2   2 =    –5000   2 2   

= – 1545 psi

p =

1 tan–1 2



5000 6400 – 1600 2



= – 32.2°

(b) For the second element we have x

10  106 (1  0.25)2

1 0.25 0 1 0.25 1 0  2(0.6) 0 0 0.375

1 0 1.2

0 1 0 0 0 1.2 0 0 0 1.2 1 0 1 1.2 0

0 0.0025 0.0012 0 0.0 0.0025

 x  10667 psi      y  2667 psi     xy  8333 psi  1

10667  2667  10667  2667 2 2 2 1 =      8333  2 2   



1 = 15910 psi 1

10667  2667  10667  2667 2 2 2 2 =      8333  2 2   



2 = – 2577 psi p =

1 tan–1 2



8333 10667  2667 2



p = –32.2° (c) For third element we have [B] =

0.5 0

0 1

0.5 0

0 0

0 0 0 1

0.5

0.5 1 0

1.067  107 2.67  106  [D] =  2.67  106 1.067  107  0 0 

1 in.

  lb 0  2 in. 7 0.4  10 

u1 = 0.0 in.

v1 = 0.0025 in.

u2 = 0.0012 in.

v2 = 0.0 in.

u3 = 0.0 in.

v3 = 0.0025 in.

u1 v1

{d} =

u2 v2

Displacement matrix

u3 v3

Stress evaluation x y

= [D]B]d}

x y xy

 6.4  103    lb =  1.6  103  2 in.  3    5  10 

Principal stresses

1 =

2 =

x –

2 xy

2 x –

2 xy

lb = 1 in.2 lb min = – 1.546  103 2 =  in. Principal angle

max = 0.955  104

p =

atan

2 xy x

2 p = – 32.179 deg.

6.5 Von Mises stress for biaxial stress state (a)

1 = 9546 e =

(b)

(c)

2 1

2 = – 1546 2

e = 1.041  104 psi

1 2

1 = 15910

2 = –2577

e =

2 1

e =

2 1

e = 1.041  104

e = 1.734  104 psi

1 2 1 2

lb in.2

6.6

i = yj – ym = 30 – 120 = – 90

i = xm – xi = 50 – 80 = – 30

j = ym – yi = 120 – 30 = 90

i = xi – xm = 20 – 50 = –30

m = yi – yj = 30 – 30 = 0

m = xj – xi = 80 – 20 = 60

2A = xi(yj – ym) + xj(ym – yi) + xm(yi – yj) = 20(– 90) + 80(90) + 50(0) = 5400 mm2 90 0 30

1 [B] = 5400

[D] =

0 30 90

90 0 30

0 0 0 30 0 60 90 60 0

0.25 0  1 0.25 0  1 105  109   = 1.12  1011 0.25 0.25 1 0 1 0  1  (0.25)2  0 0.375 0 0 0.375  0

[k] = t A [B]T [D] [B]

5.4 10 [k] = (0.01) 2

–3

[k] = 0.56  109

1 5.4 10 3

22.5

11.25

7.5 90

30 22.5

33.75 11.25

7.5 0

30 0

33.75 22.5

[k] = 1.037  105

1 0.25 0 0.25 1 0 [B] 0 0 0.375

(1.12  1011)

1 5.4 10 3

90 0 30

0 30 90

90 0 30

0 0 0 30 0 60 90 60 0

8437.5

1687.5

7762.5

337.5

675

1350

1687.5 7762.5

3937.5 337.5

337.5 8437.5

2137.5 1687.5

2025 675

1800 1350

337.5 675

2137.5 2025

1687.5 675

3937.5 2025

2025 1350

1800 0

1350

1800

1350

1800

3600

(b) Similarly

i = – 5

i = 0

j = 2.5

j = –5 325

m = 2.5

m = 5 25.0

12.5

6.25

12.5

6.25

9.375

4.6875

9.375

4.6875

15.625

7.8125

3.125

1.5625

27.343

1.5625

22.656 7.8125 27.343

[k] = 2.24  107

15.625 Symmetry

Now solve P6-6c for stiffness matrix t = 0.01 A =

tA 2

A = 5  10–5

[k] = t A [B]T [D] [B]

[k] = 0.5 

1.225 109

3.5 108

1.015 109

7 107

2.1 108

2.8 108

3.5 108

7 108

7 107

1.4 108

4.2 108

5.6 108

1.015 109

7 107

1.225 109

– 3.5 108

–2.1 108

2.8 108

7 107

1.4 108

–3.5 108

7 108

4.2 108

–5.6 108

2.1 108

4.2 108

–2.1 108

4.2 108

1.12 109

2.8 10

5.6 10

2.8 10

–5.6 10

6.7 (a) By Equation (6.2.36) {} = [D] [B] {d} Using results of Problem 6.5 (a) x y xy

0.25 0   1 1.12 1011  0.25 1 0  =  5400 103  0 0.375  0

90 0 30

0 30 90

90 0 30

0 0 0 30 0 60 90 60 0 0.002

10–3

m mm

0.001 0.0005 0 0.003 0.001

2.645 GPa    = 0.078 GPa   0.1165 GPa    326 © 2023 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1,2 =

2.645  ( 0.078) 2.645   0.078 2 2     0.1165 2 2  

= – 1.86

1.29

1 = – 0.07 GPa tan 2

p1 =

2 = – 2.65 GPa

2(0.1165) = – 0.091 2.645  0.078

p = – 2.59° (b) From Problem 6.5 (b) ’s and ’s given x

0.25 0   1 1.12 1011  = 0.25 1 0   25 106  0 0.375  0

5 0

0 0

2.5 0

0 2.5 5 0

5 2.5

0 5 2.5

 0.002   0.001     0  m 0.0005   –3  10  21.0GPa     mm  0   16.8GPa   0.003      0.001 

1,2 =

0  21.0 21 2 2     16.8 2  2 

= 10.5

19.8

1 = 30.3 GPa tan 2p =

2 = – 9.3 GPa

2(16.8) 0  21.0

p = – 29° (c)

i = – 10 mm

i = – 5 mm

j = 10 mm

j = – 5 mm

m = 0

m = 10 mm

A = 2.5  10 m –5

{} = [D] [B] {d}

[D] =

E 1 v2

1 v v 1

0 0

1 v 2

 1 0.25 (105 109 N2 )  m 1 [D] = 0.25 1  (0.25)2  0  0

[D] = 112  109

N m2

0   0  1  0.25   2 

1 0.25 0 0.25 1 0 0 0 0.375

u1 v1

{d} =

u2 v2 u3 v3

Determine the stresses in the element with nodal displacements listed u1 = 0.002

v1 = 0.001

u2 = 0.0005

v2 = 0

u3 = 0.003

v3 = 0.001

y2 = 0

x3 = 0.005

y3 = 0.01

Here are the coordinates for the element x1 = 0

y1 = 0

x2 = 0.01

E = 105 109 v = 0.25 u1 v2 u2 {d} = v2 u3 v3

1 v

[D] = v 1

0 0

1– v 2

[D] = [ D ]

E 1 – v2

1.12  1011 2.8  1010  0   10 11 [D] =  2.8  10 1.12  10 0    0 0 4.2  1010  

Equation (6.1.8) Equation (6.2.10)

1 = y2 – y3

1 = – 0.01

1 = x3 – x2

1 = – 5  10–3

2 = y3 – y1

2 = 0.01

2 = x1 – x3

2 = – 5  10–3

3 = y1 – y2

3 = 0

3 = x2 – x1

3 = 0.01

2 A = x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2) 2 A = 1  10–4 twice the area, Equation (6.2.9) Equation (6.2.32) combined 1

1 0 [B] = 2A

0 – 50

[B] =

–100 0

0 – 50

100 0

0 0

0 100

– 50

–100 – 50 100 100

In-plane stresses { = [D] [B] {d}  –1.54  1010    { =  1.4  109   9   3.15  10 

 x   0  x  1.54  1010  y  1

 y  4.2 109

 xy   2

 xy  3.15  109

Note: use the left bracket after the sigma then the 0, or 1 or 2 for the values in the sigma matrix Principal stresses

1 =

(

x –

2 xy

N 1 = 1.971  109 2 m

ave =

1 2

sqrt =

(

x –

2 y) xy

1 2

2 = ave – sqrt

p =

1 a tan 2 2

xy x –

2 = – 1.597  1010

N m2

p = – 0.179 p = – 10.278°

Principal angle

6.8 Von Mises stress (a) 1 = – 0.07

e =

2 = – 2.65

2 1

(b) 1 = 30.3

e =

e = 2.615 GPa

1 2

2 = – 9.3

2 1

e = 35.87 GPa

1 2

2 = – 15.95 GPa

P6.8c von Mises stress, Equation (6.5.37a)

vm =

( 1–

2 2)

vm = 1.705  1010

N m2

1 2 2 1

6.9 (a)

Plane strain

i = yj – ym = 1–5 = – 4

i = xm – xj = 2 – 4 = – 2

j = ym – yi = 5 – 1 = 4

j = xi – xm = 2 – 2 = 0

m = yi – yj = 1 – 1 = 0

m = xj – xi = 4 – 2 = 2

x y xy

0  0.75 0.25 10 106  0.25 0.75 0  =  (1  0.25) 1  2(0.25)  0 0.25  0

0.001



1 8

4 0

0 4 0 0 0 2 0 0 0 2

4 0 4 2 0

0.005 0.001 0.0025 0 0

5000    = 15000  psi 6000   

1,2 =

5000  (15000)  5000  (15000) 2 2    (6000) 2 2  

= –10000 ± 7813

1 = – 2187 psi tan 2p =

2 = –17810 psi

2(6000) 5000  (15000)

p = – 25.1° (b)

i = 2 – 4 = – 2 i = 4 – 4 = 0 j = 4 – 2 = 2

j = 2 – 4 = – 2

m = 2 – 2 = 0

m = 4 – 2 = 2

x y xy

= 16  106

0.75 0.25 0 0.25 0.75 0 0 0 0.25

1 4

2 0 0

0 0 2

2 0 2

0 0 0 2 0 2 2 2 0

0.001 0.005

5,000    = 15,000  psi 0.0025 7000    0 0.001

1,2 =

5000  15000 5000  (15000)  2    (7000)2  2 2  

1 = – 1398 psi tan 2p =

2 = –18600 psi

2(7000) 5000  (15000)

p = – 27.2° (c) Given displacements (in.) u1 = 0.001

v1 = 0.005

u2 = 0.001

v2 = 0.0025

u3 = 0

v3 = 0 u1 v1

{d} =

u2 v2 u3 v3

Material definition E = 10  106 psi

v = 0.25

Geometry description

i = 0 – 2

(yj – ym)

j = 2 – 0

(ym – yi)

m = 0 – 0

(yi – yj)



i = 0 – 2

(xm – xj)



j = 0 – 0

(xi – xm)



m = 2 – 0

(xj – xi)

1 2

2 2

A=2

1 [B] = 0 2A

Plane strain constitutive matrix 1 v v v 1 v

E [D] = (1 + v) (1 − 2v)

0 0 1 2v 2

Stress matrix { = [D]B]{d}  1  104   =  4  3  10   4   0.7  10 

 x  1  104 (psi)  y   3  104 (psi) 4

 xy   0.7  10 (psi)

1 =

2 xy

1 = – 0.7793  104 psi 2 =

2 xy

2 = – 3.221  104 psi p =

1 atan 2

2 xy x

p = – 0.305 (rad) 3 = 0 1

vm =

vm = 2.91  104 (psi) (d)

1 = – 4

1 = 0

2 = 2

2 = – 2

3 = 2

3 = 2

x y xy

0.75 0.25 0 0 = 16  106 0.25 0.75 0 0 0.25

1 8

4 0 0

0 0 4

2 0 2

0 2 0 2 0 2 2 2 2

0.001 0.005  5.5  0.001    =  8.5 ksi 0.0025  8.5    0 0 2

 5.5  8.5  5.5  8.5 2 1,2 =      8.5   2 2

1 = 1.63 ksi tan 2p =

2 = – 15.6 ksi

2 (8.5) 5.5  8.5

p = – 40.0° (e) 1 1 1 1 A = 2 3 2.25 2 3 1 3.25 334 © 2023 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

= 0.375 in.2

i = 1 – 3.25 = – 2.25

i = 2.25 – 3 = – 0.75

j = 3.25 – 3 = 0.25

j = 2 – 2.25 = – 0.25

m = 3 – 1 = 2

m = 3 – 2 = 1

{} = [D] [B] {d} 0.75 0.25 0 0.75 0.25 0 x = 1.6  0107 0.25 0.75 0.25y 0.75 0 0xy 0 0.25 0 0 0.25

0.001



1 0.75

2.25 0

0 0.75

0.25 0

0 2 0 0.25 0 1

0.75

2.25

0.25

1 2

0.005 0.001 0.0025 0 0

x y xy

55.3    = 80.67  ksi 62   

1,2 =

55.3  (80.67) 55.3  80.67 2 2      62  2 2  

1 = – 4.73 ksi p =

2 = – 131.3 ksi

 1  1 2(62)  tan  = – 39.2° 2 55.3  80.67 

(f) Material properties E = 10  106 psi

Modulus of elasticity.

v = 0.25

Poisson ratio

Nodal coordinates (coordinates defined CCW around element) x1 = 0 in. y1 = 0 in. x2 = 2 in. y2 = 0 in. x3 = 1 in. y3 = 2 in. 335 © 2023 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Nodal displacements u1 = 0.001 in. v1 = 0.005 in. u2 = 0.001 in. v2 = 0.0025 in. u3 = 0 in. v3 = 0 in. Set-up displacement vector {d} = (u1 v1 u2 v2 u3 v3)T {d}T = (0.001 0.005 0.0001 0.0025 0 0) in. Area of triangular element ( 12  base  height) 1 (x2 – x1) (y3 – y1) 2 A = 2 in.2 Calculate gradient matrix, B, as given in text Equation (6.2.32) Elements of B given by text Equation (6.2.10).

1 = y2 – y3

2 = y3 – y1

3 = y1 – y2

1 = x3 – x2

2 = x1 – x3

3 = x2 – x1

1 [B] = 0 2A

[B] =

0.5 0

0 0.25

0.5 0

0 –0.25

0.25

0.5

0.25

0.5

0 0

0 1 0.5 in. 0.5 0

Calculate constitutive matrix for plane strain E [D] = (1 + v) (1 − 2v)

1 v v v 1 v 0

 1.2  107 0.4  107  [D] =  0.4  107 1.2  107  0 0 

0 0 1 2v 2

  0  psi 7 0.4  10  0

= [D]B] {d}

y xy

 7500  =  22500  psi    7000   

x y xy

Principal stresses x

1 =

2 xy

1 = – 0.474  104 psi x

2 =

2 xy

2 = – 25.30 103 psi Angular location of principal stress plane

p =

2 xy

atan

p = – 21.513° 6.10 (a)

i = – 15

i = – 10

j = 15

j = 0

m = 0

m = 10



2A = 150 mm2 = 150  10–6 m2 x y xy

0.7 0.3 0  105 106  0.3 0.7 0  =  1  0.3 1  2(0.3)  0 0.2  0

5.0 1 150

15 10

2.0 0 0

 10–3

5.0 0

x y xy

78.75   = 49.2  MPa 8.07   

1,2 =

78.75  (49.2) 78.75  49.2 2 2     (8.07) 2 2  

= – 63.98 ± 16.83

1 = – 47.15 MPa 2 = – 80.81 MPa 2p = tan–1

2(8.07) = 28.64° 78.75  (49.2)

p = 14.32° (b)

x y xy

0  141.3 60.6 1 141.3 0  = 109    225 106 40.35  0.005



15 0

0 0

15 0

0 15

0 0

0 15

0.002 0 0

 10–3

0.005 0

x y xy

47.1    = 20.25 MPa  8.07   

1,2 =

47.1  20.5 47.1  20.25 2 2     (8.07) 2 2  

= – 33.66 ± 15.69

1 = – 17.97 MPa tan 2p =

2 = – 49.35 MPa

2(8.07)  47.1  20.25

p = – 15.5° (c)

x y xy

0.7 0.3 0  1  150   =  0.3 0.7 0     200  1.3 0.4  0 0.2  0 5



10 0

0 10

10 0

0 10

0 0

0 20

2 0 0

 10–3

5 0

x y xy

41.4    =  29.25 MPa 6.06   

1,2 =

41.4  29.25 41.4  29.25 2 2     6.06 2 2  

1 = – 26.85 MPa p =

2 = – 43.95 MPa

1 2  6.06  tan–1   2  41.4  29.25  339

p = – 22.5° (d)

E = 105  109

v = 0.3

t=1

y1 = 0.015

y2 = 0.005

y3 = 0.015

x1 = 0.005

x2 = 0.015

x3 = 0.025

1 = y2 – y3

2 = y3 – y1

3 = y1 – y2

1 = x3 – x2

2 = x1 – x3

3 = x2 – x1

1 [x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)] 2 1

1 0 [B] = 2A

E [D] = (1 v ) (1 – 2v )

1– v

1 v

1 – 2v 2

[k] = t A [B]T [D] [B] u1 = 0.000005

u2 = 0

u3 = 0.000005

v1 = 0.000002

v2 = 0

v3 = 0 u1

v1 u2 {d} = v2 u3 v3

{} = [D] [B] {d} x y xy

0.7 0.3 0   105    1 =   0.3 0.7 0  200 1.3  0.4   0 0.2  0



0 0

0 10 0 10 0 10

10 0

0 10

10 10

2 0 0

 10–3

5 0

x y xy

1 =

1.575   =  1.05  MPa  5.25    x –

2 xy

1,2 =

2 =

–

x –

2 xy

1.575  1.05 1.575  1.05 2 2     (5.25) 2 2  

1 = –5.145 MPa p =

2 = – 5.67 MPa

1  2(5.25)  tan–1   2  1.575  1.05 

p = 38° 6.11 (a)

Equation (6.3.7)  N1 0  t L  N2 {fs} =    0 0 0  N3  0

 N1 P0 y  0   L  N1   0   P y N P y 0   0L  L 2 0  dz dy  t   0  0L  dy N 2   0     N3 P0 y  0   L   N3  evaluated at x  a  0  yy 341

By Equations (6.3.12) – (6.3.17) L a x Lx ay ay , N2 = , N3 = 2A 2A 2A

N1 =

{fs} =

t P0 2 aL 2

f s1x

f s1 y

1 aL2 6

fs2x

1 P Lt 6 0

fs2 y

1 aL2 3

f s3 x

1 P Lt 3 0

f s3 y

(b)

By Equation (6.3.11) N1 Px

{fs} = t

0 N 2 Px

0 N3 Px

x y

Now

N1 = 0, N2 =

a y

ay Lx  ay , N3 = 2A 2A

0 0

{fs} = t

y 2

Lx ay 2A

P0 L

0 ay 2A

y 2 P0 L

x y

a y

Simplifying and integrating

0 0

{fs} =

Lay 3 3 L2

P0 t 2 ( 12 aL)

ay 4 4 L2

L 0

P0 tL 12

0 P0 tL 4

L ay 4 4 L2 0

or f2x =

P0 tL 12

f3x =

P0 tL 4

6.12 (a) Py(x) = ax2 + bx + c Given L a 2

L + p1 = p2 2

aL2 + bL + p1 = p3 c = p1 p3 – 2 p2 ) L2 (– 3 p1 – p3 4 p2 ) (a, b, c)  L p1 2

Find

a(p1, p2, p3) = 2 b(p1, p2, p3) =

( p1

(1)

p3 – 2 p2 ) L2

(– 3 p1 – p3 L

(2)

4 p2 )

(3)

c(p1, p2, p3) = p1

Forces in y direction at nodes 1 and 3 are N1 = Ni f s1y =

L 0

1–

x L

a ( p1 , p2 , p3 ) x2

b( p1 , p2 , p3 ) x

c ( p1 , p2 , p3 ) dx 

1 Lp1 6

1 Lp2 3

N1 Py ( x ) dx x

N3 = Nm f s3 y =

L 0

x a( p1, p2 , p3 ) x2 L

b( p1, p2 , p3 ) x

c ( p1 , p2 , p3 ) dx 

1 Lp3 6

1 Lp2 3

N m Py ( x ) dx 0

(Special case) If p1 = 0

p3 = p0

p(x) = ax2 + bx + c Want p(x) = p0 

 a =

x 2 = ax2 + bx + c L

p0 2 = (p0 – 2p2) using (1) L2 L2 344

2 p0 p 4p – 20 = 22 2 L L L p0

p2 =

4 p2 L2

p0 4

Then f s1 y =

1 p L 1 1 p pL L p10 + Lp2 = L 0 = 0  Like 6.11 b with f sx = 12 12 6 4 3 3 p

f s3 y =

p0 4

1 || 3 p L 1 || Lp0 = 0 Lp3 + Lp2 = 4 6 12 3

 f3x in 6.11b

(These answers match P6.11 for special case) (b)

{ fs} =

[Ts ] ds

[Ts ] = Surface tractions Ts x

Tsy

0 P0 sin

N L

[Ns] = Shape function matrix evaluated along edge 1-2 = Let

i= 1 j= 2 m= 3 Ni = N1 =

1 (i + i x + i y) 2A 345

N1(y = 0) =

1 (i + i x) 2A

i = xj ym – yj xm = 0(ym) – 0(xm) = 0

i = yj – ym = 0 – ym = – ym N1(y = 0) =

1 (0 – ym x) = 2A

1 (j + j x + j y) 2A

N j = N2 = N2( y = 0) =



ym x 2A

1 (i + i x) 2A

j = yj xm – xj ym = 0(xm) – Lym = – Lym



j = ym – yi = ym – 0 = ym N2 (y = 0) =

1 y [– Lym + ym x] = m [x – L] 2A 2A 1 [m + mx + my] 2A

Nm = N3 = N3(y = 0) =

1 [m + mx] 2A

m = xi yj – yi xj = L(0) – 0(0) = 0 m = yi – yj = 0  Nm(y = 0) = 0

{ fs} =

As expected

x 0

z 0

P0 sin Lx

dz dx

ym x P0 sin 2A

N1P0 sin Lx

= t

x L

dx = t

x L

N 2 P0 sin

x L

L ym 2A

L P0 sin Lx

(A)

N 3 P0 sin Lx

2nd term in (A) (ym = y3) t y3 P0 2A

fs1y =

 udv  uv –  vdu

x dx x sin L

ux

du  dx

dv  sin

t y3 P0 2A

fs1y =

t y3 P0 L2 2A

1 Ly3 2

cos

x L

cos

x L

L2 2

x

v  – cos L L

x dx L

cos

x L

sin

x

( 1) 0 0 0

t P0 L

4th term in (A) fs2y = t =

L 0

y3 x P0 x sin 2A L

t y3 P0 2A

L 0

x sin

y3 x P0 L sin 2A L

x dx L

t y3 P0 L 2A

L 0

dx sin

x dx L

DONE IN 2nd TERM

t y3P0 L2 2A

t y3P0 L L x cos 2A L

t y3 P0 L2 2A

t y3 P0 L L [–1 + 1] 2A

L 0

fs2y =

 fs =

t y3 P0 L2 = 2A

t y3 L2 P0 2

f s1x

f s1 y

t P0 L

fs2x fs2 y f s3 x f s3 y

1 L y3 2

t L P0

0 t P0 L

0 0

6.13

Refer to Section 6.5 for [K] Since

u1 = v1 = 0, u2 = v2 = 0

48 0 28  0  25,000  87 12 75000 5     = 48 0.91  0  Symmetry 25,000 

14 80 26 87

u3 v3 u4 v4

Solving u3 = 2.50  10–3 in. u4 = – 3.04  10–3 in.

v3 = – 1.376  10–2 in. v4 = – 1.466  10–2 in.

Using element (1) By Equation (6.2.36), {} = [D] [B] {d} 348 © 2023 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1 0.3 0 30 106 {} = 0.3 1 0 (0.91)(200) 0 0 0.35

0 0

0 10 20 0

0 0

10 0

0 20

0     0    2.5 103    2  1.376 10    0   0   x y xy

 4123    =  1237  psi 7938  

1,2 =

4123  1237 4123  1237 2 2     (7938) 2 2  

1 = 10748 psi p =

2 = – 5388 psi

1 2  7938  tan–1   2  4123  1237 

p = – 40° Using element (2) 1 0.3 0 30 106 0.3 1 0 {} = 0.91(200) 0 0 0.35

10 0 0

0 0 10

10 0 20

0 0 0 20 0 20 10 20 0

0     0     3.04  10 3     2   1.466  10    2.50  103     1.376  102  x y xy

  4123    =  1469  psi  2062   

1,2 =

 4123  1469  4123  1469 2 2     ( 2062) 2 2  

1 = 2147 psi

2 = – 4800 psi

p =

1  2  2062  tan–1   2   4123  1469 

p = 18.15° 6.14 (a)

INPUT TABLE 1. BASIC PARAMETERS NUMBER OF NODAL POINTS. . . . . . . . . . . . . . . NUMBER OF ELEMENTS. . . . . . . . . . . . . . . . . . . NUMBER OF DIFFERENT MATERIALS. . . . . . . NUMBER OF SURFACE LOAD CARDS. . . . . . . . . 1 = PLANE STRAIN, 2 = PLANE STRESS . . . . . . BODY FORCES (1 = IN – Y DIREC., 0 = NONE)

5 4 1 0 2 0

INPUT TABLE 2. MATERIAL PROPERTIES MATERIAL MODULUS OF NUMBER ELASTICITY

POISSON’S RATIO

1 0.105E+12 0.3000E+00 INPUT TABLE 3. NODAL POINT DATA NODAL POINT TYPE X 1 3 0.0000E+00 2 0 0.5000E+00 3 3 0.5000E+00 4 3 0.0000E+00 5 0 0.2500E+00 X-DISP. OR LOAD

MATERIAL DENSITY

MATERIAL THICKNESS

0.0000E+00

0.5000E–02

Y 0.0000E+00 0.0000E+00 0.2500E+00 0.2500E+00 0.1250E+00 Y-DISP. OR LOAD

0.0000E+00 0.2000E+05 0.0000E+00 0.0000E+00 0.0000E+00

0.0000E+00 – 0.3464E+05 0.0000E+00 0.0000E+00 0.0000E+00

INPUT TABLE 4. ELEMENT DATA GLOBAL INDICES OF ELEMENT NODES ELEMENT 1 2 3 4 1 1 5 4 4 2 1 2 5 5 3 5 2 3 3 4 4 5 3 3 OUTPUT TABLE 1.

MATERIAL 1 1 1 1

NODAL DISPLACEMENTS, m

NODE

U = X-DISP.

V = Y-DISP.

1 2 3 4 5

0.00000000E+00 0.5624948E–04 0.00000000E+00 0.00000000E+00 0.229959E–04

0.00000000E+00 – 0.6597062E–04 0.00000000E+00 0.00000000E+00 – 0.206952E–04

OUTPUT TABLE 2. STRESSES AT ELEMENT CENTROIDS ELEMENT 1 2 3 4

X 0.083 0.250 0.417 0.250

Y 0.125 0.042 0.125 0.208

SIGMA (X) 0.0613E+07 1.6384E+07 1.1502E+07 5.7310E+06

SIGMA (Y) 3.1840E+06 1.5239E+07 3.1158E+07 1.9103E+07

TAU (X, Y) 3.3431E+06 – 6.9854E+06 –1.1072E+07 –7.4294E+06

SIGMA (1) 1.1896E+07 2.2820E+07 3.6135E+07 2.2412E+07

SIGMA (2) 1.9012E+06 8.8026E+06 6.5251E+06 2.4221E+06

ANGLE 2.0993E+01 – 4.2657E+01 – 6.5798E+01 – 6.5993E+01

(c)

5 4 1 0 2 0

INPUT TABLE 2. MATERAL PROPERTIES MATERIAL MODULUS OF NUMBER ELASTICITY 1 0.105E+12

POISSON’S RATIO, 0.3000E+00

MATERIAL MATERIAL DENISTY THICKNESS 0.0000E+00 0.5000E–02

INPUT TABLE 3. NODAL POINT DATA POINT 1 2 3 4 5

TYPE 0 0 3 3 0

X 0.0000E+00 0.4000E+00 0.4000E+00 0.0000E+00 0.2000E+00

Y 0.0000E+00 0.0000E+00 0.4000E+00 0.4000E+00 0.2000E+00

OR LOAD 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00

OR LOAD 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 –0.3000E+05

GLOBAL INDICES OF ELEMENT NODES ELEMENT 1 2 3 4 1 1 2 5 5 2 2 3 5 5 3 3 4 5 5 4 4 1 5 5 OUTPUT TABLE 1. NODAL DISPLACEMENTS (m) NODE 1 2 3 4 5

U = X–DISP. – 0.3303082E–05 0.33030884E–05 0.00000000E+00 0.00000000E+00 0.27411561E–12

V = Y–DISP. – 0.25009076E–04 – 0.2500907E–04 0.00000000E+00 0.00000000E+00 – 0.32558982E–04

OUTPUT TABLE 2. STRESSES AT ELEMENT CENTROIDS

ELEMENT 1 2 3 4

X 0.20 0.33 0.20 0.07

Y 0.07 0.20 0.33 0.20

MATERIAL 1 .1 1 –.1

N m2

SIGMA(X) SIGMA(Y) TAU(X, Y) 5.9891E+05 –3.7840E+06 4.0454E–01 3.1171E+06 7.5000E+06 3.7160E+06 5.6352E+06 1.8784E+07 –1.1070E–01 3.1171E+06 7.5000E+06 –3.7160E+06 SIGMA(1) SIGMA(2) ANGLE 5.9891E+05 –3.7840E+06 5.2883E–06 9.6226E+06 9.9449E+05 6.0265E+01 1.8784E+07 5.6352E+06 –9.0000E+01 9.6226E+06 9.9449E+05 –6.0265E+01

6.15

6.16 (a)

 kN  F = (0.5 m ) (0.25 m ) (0.005 m ) 154.2  = 0.0964 kN  m3  

f1 = f2 = f3 = f4 =





N  (2) – 12 (0.25)(0.25)(0.005) 154.2 kN3  1000 1kN 

= – 16.06 N

f5 = (2)  f1 = 32.12 N  f5y = –32.12 N  (c)

Equation (6.3.6)

f B(1) 1x f B(1) 1y f B(1)2x f B(1)2y f B(1)5x f B(1)5y

 0   0  154.2  10.28      0  (0.4 m)(0.2 m)(0.005m)  0  –3 =  =    10 kN (2) (3) 154.2  10.28  0   0      154.2  10.28

All body force matrices for each element identical to above Adding the 4 element body force matrices f B1x f B1 y fB2x fB2 y

{FB} =

f B3x f B3 y fB4x fB4 y f B5 x f B5 y

 0   20.56    0     20.56  0  =  N  20.56  0     20.56  0     41.12

6.17 The triangular element is called a constant strain triangle (CST) because the strain is constant throughout the element. 6.18 The stresses are also constant as the strains are constant. 6.19 a. No, bending in the plane takes place b. Yes, a plane strain problem c. Yes, a plane stress problem d. Yes, loads in-plane of the rod, a plane stress problem e. Yes, a plane strain problem f. Yes, a plane stress problem g. No, loads out of the plane of the wrench h. Yes, as loads in the plane i. No, bending in the plane takes place 6.20 We must connect the beam element to two or more nodes of a plane stress element. The beam must be along the edge of the plane stress element. 6.21

nb = n0 (m + 1) (a) nb = 2(3 + 1) = 8 (b) nb = 2(5 + 1) = 12 for model (a) nb

model(b)

1 2 3 4 5 6 7 8 0

0 0

Symmetry

1 2 3 4 5 6 7 8 0 0 0 0 0 0 0 0 0 0

0 0

Symmetry

6.22

By (6.2.10)

i = xjym – yjxm =

b 2b 2 3

j =

h 3

(0) –

b 2

m =

b 2

h 3

h bh (0) = 3 3 2h bh = 3 3

2A 3

b 2

2A 3

bh 3

2A 3



{fB} =

Ni =

1 2A 1 1 = and Nj = Nm = 2A 3 3 3

or { fb i } =

Ni 0 1 3

0 Ni

Xb 1 3

tdA

Xb V Yb 3

{ fb i } = Similarly

{ fb j } = {fbm} =

Xb V Yb 3

(6.3.6)

6.25

N1 = N3 =

x h 4bh

, N2 = , N4 =

x h 4bh

(1)

at center (x = 0, y = 0) N1 =

1 1 1 1 , N2 = , N3 = , N4 = 4 4 4 4

N1 + N2 + N3 + N4 = 1 at point x

b ,y 2

h 2 b

N1 = N2 =

b 2

4bh

h 2

1 16

3 9 3 , N3 = , N4 = 16 16 16

In general add the functions in Equation (1) and you get for all x and y on the element N1 + N2 + N3 + N4 = 1 6.26

{} = [D] [B] {d} 1 [B] = 4bh

At center 1 [B] = 8

(h y ) 0

0 b x

h y 0

0 h y 0 b x 0 b x

h y 0

b x

h y

b x

h y

b x

h y

b x h y

(x = 0, y = 0)

1 0 2

0 2 1

1 0 2

0 1 0 2 0 2 1 2 1

1 0 2

0 2 1 0 0

{} =

1 8

1 0 2

0 2 1

1 0 2

0 1 0 2 0 2 1 2 1

1 0 2

0 2 1

0.005 0.0025 0.0025 0.0025 0 0

{} =

0.0009375 in. 0.00125 in. 0.000625

{} = [D] {} =

30 106 1 0.32

1 0.3 0 0.3 1 0 0 0 0.35

0.0009375 0.00125 0.000625

x y xy

18.54 31.94 ksi 7.21

Chapter 7 7.1 For the simple 4 noded elements it is a violation of displacement compatibility to have a mid-side node. Some of the elements have mid-side nodes in this model. Use ‘transition’ triangle to go from smaller to larger rectangular elements.

7.2 The mesh sizing is not fine enough in the reentrant corner region at C. We need smaller elements near point C and small radius at C. 7.3 Based on the formulation used here we cannot have ν = 0.5 for the plane strain case as the denominator in the material property matrices [D] (see Equation (6.1.10) and [K] (see Equation (6.4.3) becomes zero. A penalty formulation see Reference [7] can be used to avoid this problem. 7.4 The structure is plane strain if this section represents a cross section of a long structure in which the loads do not vary in the z direction. The structure is a plane stress problem if this section is a thin plate type structure with loads in the plane of the structure only. Also see Section 6.1 for descriptions of plane stress and plane strain and examples of each. 7.5 When abrupt changes in thickness or E’s occur from element to element. 7.6 Unit thickness 7.7 (a) Best aspect ratio. 7.8 See answer to Problem 6.20. 7.9 (a) No, as replacing a portion of the patch by a different material with different mechanical properties will in general produce non-uniform strain under constant state of applied stress. For rigid body mode tests, however, different mechanical property materials still result in rigid body displacement. (b) Yes, the patch can be arbitrary in shape. If we apply a test displacement field of ux = 1, uy = 0 at the external nodes of a patch of say 4 elements and set the internal nodal force to zero, then solve for the displacement components at internal node i, these displacement components should agree with the value of the displacement function at that node. Also the strain function or field should vanish identically at any point over each element. (c) Yes, we can mix triangular and quadrilateral elements in a 2-d patch test as long as the material properties are the same. (d) No. Mixing bars with plane elements would alter the constant strain states as the plane element and bar are of different structural types.

(e) The patch test should be applied when developing new finite elements, to determine if the element can represent rigid body motion as well as states of constant strain when these conditions occur. 7.10 Using Mathcad A = 1  10–4 1

E = 200  109 –1 0

[k1] = –1 0 0

L1 = 0.6

E [k1] = [k1] A L1

[k] = [k1] + [k2]

3.333 107

– 3.333 107

[k1] = – 3.333 10 0 0

E [k2] = [k2] A L2

[k2] = 0 1 –1 0 –1 1

L2 = 1.4

3.333 10 0

0 7

1.429 10

–1.429 107

0 –1.429 107

1.429 107

[k2] = 0

3.333 107

– 3.333 107

–1.429 107

1.429 107

[k] = – 3.333 10 0

4.762 10

Set these 3 values to defined quantities of u1 = u3 = 1 for the rigid body patch test u1 = 1

u3 = 1

F2 = 0

Guess at F1, F3, and u2 as shown below. F1 = 1

u2 = 0

Given

Use the given command to create a solve block.

F1 F2 F3

F3 = 1

u1 [ k ] u2 u3

Use control and equal sign here.

= Find (F1, u2, F3)

Use the ‘Find’ command to find F1, u2, and F3.

F1 u2 F3

F1 = 0

u2 = 1

F3 = 0

The rigid body motion patch test is satisfied as u2 = 1. Now check the constant strain test. Let u(x) = x for the nodes at the boundaries, i.e., u1 = 0 and u3 = 2, Verify that u2 (x = 0.6) = 0.6. u1 = 0

u3 = 2

F2 = 0

Initial these values

F1 = 1

F3 = 1

u2 = 0

Guesses for these values.

Given F1 F2 F3

u1 = [ k ] u2 u3

F1 u2 F

= Find (F1, u2, F3)

Use the ‘Find’ command to solve for F1, u2, and F3.

F1 = – 2  107

F3 = 2  107

u2 = 0.6

Now upon solving the system of equations u2 = 0.6 as it should to satisfy the patch test for constant strain. 7.12

INPUT TABLE 1.. BASIC PARAMETERS NUMBER OF NODAL POINTS. . . . . . . . . . . . . . . NUMBER OF ELEMENTS. . . . . . . . . . . . . . . . . . . NUMBER OF DIFFERENT MATERIALS. . . . . . . NUMBER OF SURFACE LOAD CARDS. . . . . . . . . 1 = PLANE STRAIN, 2 = PLANE STRESS . . . . . . BODY FORCES (1 = IN – Y DIREC., 0 = NONE)

5 4 1 0 2 0

INPUT TABLE 2.. MATERIAL PROPERTIES MATERIAL NUMBER 1

MODULUS OF ELASTICITY 0.3000E+08

POISSON’S RATIO, 0.3000E+00

MATERIAL DENSITY 0.0000E+00

MATERIAL THICKNESS 0.1000E+01

X-DISP. OR LOAD 0.0000E+00 0.5000E+04 0.5000E+04 0.0000E+00 0.0000E+00

Y-DISP. OR LOAD 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00

INPUT TABLE 3.. NODAL POINT DATA NODAL POINT

TYPE

1 2 3 4 5

3 0 0 3 0

0.0000E+00 0.2000E+02 0.2000E+02 0.0000E+00 0.1000E+02

Y 0.0000E+00 0.0000E+00 0.1000E+02 0.1000E+02 0.5000E+01

INPUT TABLE 4.. ELEMENT DATA ELEMENT 1 2 3 4

GLOBAL 1 1 2 5 1

INDICES OF 2 2 3 3 5

ELEMENT 3 5 5 4 4

NODES 4 MATERIAL 5 1 5 1 4 1 4 1

OUTPUT TABLE 1.. NODAL DISPLACEMENTS NODE 1 2 3 4 5

U = X-DISP. 0.00000000E+00 0.64664544E–03 0.61664509E–03 0.00000000E+00 0.30527671E–03

V = Y-DISP. 0.00000000E+00 0.66631597E–04 –0.66630528E–04 0.000000000E+00 0.24373945E–09

ELEMENT 1 2 3 4

7.13

7.14

X 10.00 16.67 10.00 3.33

Y 1.67 5.00 8.33 5.00

SIGMA(X) 1.0000E+03 9.9359E+02 1.0000E+03 1.0064E+03

SIGMA(Y) 1.0011E+02 –1.0171E+02 1.0011E+02 3.0192E+02

TAU(X, Y) –3.2032E+00 1.2599E–05 3.2035E+00 2.8124E–04

SIGMA(1) 1.0000E+03 9.9359E+02 1.0000E+03 1.0064E+03

SIGMA(2) 1.0010E+02 –1.0171E+02 1.0010E+02 3.0192E+02

ANGLE –2.0395E–01 6.5906E–07 2.0396E–01 2.2873E–05

STRESS IN PSI AT VARIOUS DISTANCES ALONG THE TENSILE PLATE WITH A 1000 # LOAD ELEMENT # 14 28 42 56 STRESS 26 –2026 –2026 26

(AT 1)

ELEMENT # STRESS

13 –563

27 –1437

41 –1437

55 –563

(AT 2)

ELEMENT # STRESS

11 –969

25 –1031

39 –1031

53 –969

(AT 4)

ELEMENT # STRESS

10 –1002

24 –998

38 –998

52 –1002

(AT 6)

ELEMENT # STRESS

9 –1002

23 –998

37 –998

51 –1002

(AT 8)

ELEMENT # STRESS

8 –1001

22 –1000

36 –1000

50 –1001

(AT 10)

ELEMENT # STRESS

6 –1000

20 –1000

34 –1000

48 –1000

(AT 15)

ELEMENT # STRESS

5 –1000

21 –1000

33 –1000

47 –1000

(AT 20)

ELEMENT # STRESS

3 –1002

17 –998

31 –998

45 –1002

(AT 30)

ELEMENT # STRESS

1 –1005

15 –995

29 –995

43 –1005

(AT 40)

7.15

The maximum von Mises stresses are at the top and bottom of the circular opening of the connecting rod. The maximum stresses can be seen in the Figure above.

7.16

7.17

7.19

Model Parameters Implant material modulus of elasticity 1.6  106 psi Bony material modulus of elasticity 1.0  106 psi Implant depth below bony material 0.100 in. 400 Mesh Density

7.20

From computer program output YmaxA = – 0.4993 in. Ymax exact = – 1.152 in. =

AR =

12 1 2

PL3 3 EI

= 24 (56% error due to large aspect ratio)

For other results see Example in Section 7.1, Table 7.1

7.21

7.22

The figure above is the maximum principal stress. The maximum is 3505 location of maximum stress occurs at the corners of the hole (with 1 mm radii) 7.23

N . The m^ 2

7.25

7.26

The largest von Mises stress of 6.17 MPa occurs at the top and bottom inside edges of the hole. The second largest principal stress of 5.78 MPa occurs at the elbow between the smallest cross section and where the taper begins. 7.28

7.31

Figure 1: Mesh with Boundary Conditions and Nodal Force Loads and Constraints: Fixed at bottom half of hole Concentrated load of 5000 lbf Thickness = 1 in.

Figure 2: von Mises stresses (psi)

Figure 3: Displacement (in.)

7.32 600 Mesh Density

Mesh Density: 25-600 Bracket without Fillet

25 Mesh Density 50 100 150 200 300 400 500 600 Bracket with Fillet 25 Mesh Density 50 100 150 200 300 400 500 600

Maximum Principal Stress lb in.2 22811.17 26114.15 27050.65 28179.32 28967.93 28800.52 35102.97 32852.23 33678.14

47481.11 47492.06 47502.16 47511.98 47521.59 47832.01 47688.08 47658.56 47674.97

In the FEA world re-entrant corners are a bad thing. These represent an infinite change in stiffness inside the part, which will result in an infinite stress concentration.

7.33

7.37

1080 as-rolled steel has a yield strength for 84800 psi. Dividing by a factor of safety of 1.5 results in an allowable yield strength of 56533 psi. A thickness of 0.3 inches causes a maximum von Mises stress of 53293.74 along the curve inside the wrench grip which is within the allowable yield strength value of 56533 psi. Note: The image reports a von Mises stress of 59215.26 psi, but that is an imaginary stress due to its location being at the node where a load of 1800 lbf is applied. 7.38 The model is shown first with the boundary and loading conditions then applied. The nodes of the far left hex were constrained from all movement. Next a material was chosen and an initial guess at the thickness t was made. ASTM A-36 was chosen. A thickness of 3.8 mm was chosen as this is an easy number to work with and it is compatible with the other wrench dimensions. A pressure of 2631.6 Pa was applied. The largest von Mises stress was 166.6 MPa as shown.

7.39 Zoomed in of the previous. To simulate a real cut, I inserted a very small radius at the point of concern.

Chapter 8 8.1 Triangular element From Section 8.2 3y b

N2 =

x b

2 x2

N4 =

4y 4xy , N5 = h bh

N6 =

4x b

(a) At

2 x2

3x b

N1 = 1–

x= 0

y h

2 y2

4 xy bh

4 y2

, N3 = 2

4 x2 b2

2 y2

4 xy bh

evaluate N ’s

y= 0 N1 = 1, N2 = 0 = N3 = N4 = N5 = N6

(b) At

b 2 y= 0

N1 = 1

and

3( b2 ) b

3(0) b

N2 = N3 = N4 = N5 = 0,

2 b2 b

N6 =

4 (0) (0) bh 4 ( b2 )

4 ( b2 )

2(0)2 h

3 2

1 2

8.2

Strains u1 x

1 = 2A

y xy

v1 u2 v2

(1)

Evaluate ’s and ’s at centroid

1 = – 3h +

4h b3 4hx + 4y = – 3h + b b

4h b3 4hx =–h+ b b

2 = – h +

3 = 0, 4 = 4y = 4 5 = – 4y = 6 = 4h –

h 3

1 h 3

4h 3

4h 3 b

8h 3 8hx – 4y = 4h – b b

1 = – 3b + 4x +

h =0 3

4b ( h3 ) b b 4by = – 3b + 4 + = 3 3 h h

 2 = 0, 3 = – b +

8 by 4b , 5 = 4b – 4x – h 3

 4 = 4x = 6 =

4b ( h3 ) b 4by =–b+ = 3 h h 4b

8b h3

b 3

4b 3

(2)

Performing the multiplications in (1) (After substituting ’s and ’s from (2))

x =

h u1 3

h u2 3

4h u4 3

4h 1 u5 3 bh

y =

b v1 3

b v3 3

4b v4 3

4b 1 v6 3 bh

xy =

b u1 3

h v1 3

h v2 3

b u3 3

4b u4 3

x =

1 h [– u1 + u2 + 4u4 – 4u5] bh 3

y =

1 b [– v1 + v3 + 4v4 – 4v6] bh 3

xy =

b ( u1 3

4u6)

4h v4 3

h ( v1 3

4u4

1 v v 1

0 0

1 v 2

h ( u1 3

4u4

4u5 )

h ( u1 3

4u4

b ( u1 3

4u4

4h v5 3

4v4

4b 1 u6 3 bh

1 4v5) bh

1 bh

Stresses {} = [D] {} x y

E 1 v2

x = y = xy =

E 1 v

E 2(1 v)

x y

b ( v1 3

4v4

4v6 )

1 bh

4u5)

b ( v1 3

4v4

4v6 )

1 bh

4u6)

h ( v1 3

4v4

4v5 )

1 bh

The equation is {fs} =

[ N s ]T {T } ds

{T } =

p …. is the surface traction 0

py N1 0

[N ] =

(1)

0 N1

N2 0

0 N2

N3 0

0 N3

N4 0

0 N4

(2) N5 0

0 N5

N6 0

0 N6

(3)

Substituting (2) and (3) in Equation (1), we have t

0 0

h 0

N1 0 N2 0 N3 0

0 N1 0 N2 0 N3

p dy dz 0

N6 0

0 N6

at x y

0 y

N1 p dy 0 N2 p 0 N3 p

(4)

From Section 8.2 for this particular element we have N1 = 1 –

3y 2 y2 + 2x2 + 4xy + 2 h h

3x b

2x2

N2 =

x b

N4 =

4y 4xy , N5 = h bh

N6 =

4x b

4 x2 b

y h

2 y2

4 xy bh

4 y2

, N3 = 2

(5)

4 xy bh

Substitute (5) into (4) and evaluating N’s at x = 0, y = y, we have

2 y2 h2

3y h

p dy

0 0 0 2 y2 h2

y h

0 0 0 4 y2 h2

4y h

0 0 0

fs1x = pt y

fs3x =

3y2 2h

– y2 2h

fs5x = pt

4 y2 2h

2 y3 3h

4 y3 3h 2

2 0

pt = h 0

pth 6

2 pth 3

Nodal equivalent forces 8.4

{fs} =

[Ns]T {T }ds

p0 y h

{T } =

py N1 0

[N ] =

(1) (2)

0 N1

N2 0

0 N2

N3 0

0 N3

N1 0 N2

0 N1 0

p0 y h

dy dz

0 N3 0

N2 0 N3

N6 0

0 N6

N4 0

0 N4

N5 0

0 N5

N6 0

0 N6

(3)

Substituting (2) and (3) in (1) fs

0 0

at x 0 y y

p0 y h

0 p0 y h

(4)

p0 y h

0 p0 y h

at x 0 y y

(Shape functions are same as in Problem 8.3). Upon substituting shape functions into (4) evaluating the N’s at x = 0, y = y, we have

2 y2 h2

3y h

p0 y h

0 0 0 2 y2 h2

y h

p0 y h

0 0 0 4 y2 h2

4y h

p0 y h

0 0 0

fs1x = fs3x = fs5x =

p0 t y 2 h 2 p0 t h

y3 3h

3 y3 3h

2 y4 4h 2

2 y4 4h

p0 t 4 y 3 4 y 4 – 2 h 3h 4h

0 h 0

=0 0

p0 th 6 p0 th 3

Nodal equivalent forces 8.5 (a)

{} = [B] {d} 1 [B] = 2A

Element is oriented as in Section 8.2  ’s and ’s as in Section 8.2, Eq. (8.2.8)

1 = – 3h +

4hx + 4y = 6x + 4y – 18 6

4hx = 6x – 6, 3 = 0 6

2 = – h +

 4 = 4y,  5 = – 4y 

6 = 4h –

8hx – 4y = – 12x – 4y + 24 b

1 = – 3b + 4x +

4by 8 = 4x + y – 12 h 3

2 = 0 3 = – b +

4by 8 = y–4 h 3

4 = 4x  5 = 4b – 4x –

16 8by = – 4x – y + 16 3 h

 6 = – 4x 

2A x =  2 u2 +  4 u4 +  6 u6 

= 0.001 (6x – 6) + 0.0002 (4y) + 0.0005 (– 12x – 4y + 24) 2A x = – 0.0012y + 0.006

x = – 5  10– 5 y + 2.5  10– 4



2A y =  3 v3 +  4 v4 +  5 v5 +  6 v6 



8 y 3

= 0.0002



4 + 0.0001 (4x) + 0.0001

16 y 16 + 0.001 (– 4x) 3

2A y = – 0.004x + 0.0008

y = – 1.67  10–4x + 3.33  10–5



2A xy = 0.002 (6x – 6) + 0.0005

8 y 3

4 + 0.0002 (4x) + 0.0001(4y)

+ 0.0001 (– 4y) + 0.0005 (– 4x) + 0.001 (– 12x – 4y + 24) 2A xy = – 0.0012x – 0.00267y + 0.01

xy = – 5  10–5 x – 1.11  10–4 y + 4.167 10–4



Evaluate stresses at centroid {} = [D] {} 0.00015

{ }

4 ,2 3

{} =

1.89 10 –4 0.000128

E 1 v

1 v v 1

0 0

1 v 2

1.5 10 4 1.89 10 4 1.28 10 4

3288 4848 psi 1536

(b)

Using expressions for ’s and ’s from part (a) with h = 4 in. and b = 6 in. now 391 © 2023 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in

1 =

8 8 x + 4y – 12, 2 = x – 4, 3 = 0 3 3

4 = 4y,  5 = – 4y,  6 = 

16 x – 4y + 16 3

1 = 4x + 6y – 18,  2 = 0,  3 = 6y – 6  4 = 4x, 5 = – 4x – 12y + 24,  6 = – 4x 8 x 4 + 0.0002 (4y) + 0.0005 3

2A x = 0.001

16 x 4 y 16 3

2A x = – 0.0012y + 0.004

x = – 5  10– 5 y + 1.67  10–4



2A y = 0.0002 (6y – 6) + 0.0001 (4x) + 0.0001 (– 4x – 12y + 24) + (0.001) (– 4x) 2A y = – 0.004 x + 0.0012

y = – 1.67  10– 4 x + 5  10– 5



8 x 4 + 0.0005 (6y – 6) 3

2A xy = 0.002

+ 0.0002 (4x) + 0.0001 (4y) + 0.0001 (– 4y) 16 x 4 y 16 3

+ 0.0005 (– 4x) + 0.001 2A xy = – 0.0012x – 0.001y + 0.005 

xy = – 5  10– 5 x – 4.167  10– 5 y – 2.083  10– 4 {} = [D] {} at centroid 2,

{ }

0.0001 0.000284

= (2, 43 )

{} =

4 3

0.0000527

E 1 v2

1 v v 1

0 0

0.0001 0.000284

0 0

1 v 2

0.0000527

928

{} =

8288 psi

– 632

8.6

Using Equation (8.1.14) in (8.1.13) x y xy

0 1 0 2x y 0 0 0 0 0 0 0 = 0 0 0 0 0 0 0 0 1 0 x 2 y [X ]–1 {d } 0 0 1 0 x 2 y 0 1 0 2x y 0

(1)

where by Equation (8.1.7) {a} = [X]–1 {d } and

{a} =

1 0

1 6

36 36 36

1 6

36 18

1 3

O6 6

– 1 u1 u2 O6 6

1 0

1 6

36 36 36

1 6

36 18

1 3

(2)

Using computer, we invert [X ] in Equation (2) and reorder {d } to normal form [u1 v1 u2 v2 …]T = {d }T





[X ]–1 {d } =

0.5 0

0 0

0.167 0.167

0 0

0 0.167

0 0

0 0.667

0 0

0.667 0.667

0 0

0.056

0.111

0.056

0.111

0 0

0.5 0

0 0

0.167 0.167

0 0

0 0.167

0 0

0 0.667

0 0

0.667 0.667

0.056

0.111

0.056

0.111





u1 v1 u2 v2

(3)

u6 v6

at centroid (x = 4, y = 2) x y xy

0 1 0 8 2 0 0 0 0 0 0 0 = 0 0 0 0 0 0 0 0 1 0 4 4 [X ]–1 {d } 0 0 1 0 4 4 0 1 0 8 2 0

(4)

Multiplying matrices in Equation (4) yields

x y

0.052 u1 0.059 u2 0.222 u4 0.222 u5 0.001 u6 0.053 v2 0.391 v3 0.223 v5 0.223 v6 ( 0.052 v1 0.059 v2 0.222 v4 0.222 v5 0 v6 0.053 u2

0.391 u3

0.223 u5

0.223 u6 )

Then x y xy

E 1 v2

1 –v 2

8.7

u1 = u(0, 0) = a1

(1)

u2 = u(60, 0) = a1 + 60 a2 + 3600 a4

(2)

u3 = u(60, 60) = a1 + 60 a2 + 60 a3 + 3600 a4 + 3600 a5 + 3600 a6

(3)

u4 = u(60, 30) = a1 + 60 a2 + 30 a3 + 3600 a4 + 1800 a5 + 800 a6

(4)

u5 = u(30, 30) = a1 + 30 a2 + 30 a3 + 900 a4 + 900 a5 + 900 a6

(5)

u6 = u(30, 0) = a1 + 30 a2 + 900 a4

(6)

By (1)  a1 = u1 By (2) – 2 (6)  a4 = By – (2) + 4 (6)  a2 = By 2(4) – (3)  a6 =

4u6  u2  3u1 60

u3 2u4 1800

(4) – (5)  a5 = (4)  a3 =

2u6 u1 1800

u4 u5 900 u3

4u5 60

u6 4u6

Can verify by substituting all a’s into Equation (3) 



u2  4u  u2  3u1  u = u1 +  6  x+ 60  

2u6 u1 1800

x2 +

u3 2u4 1800

4u5 60

u4 u5 900

4u6 u6

 Shape functions are N1 = 1 –

3x 60

x2 (From all u1 coefficient) 1800 395

y2 (From all u2 coefficients) 1800

x 60

y 60

N3 =

y 60

y2 (From all u3 coefficients) 1800

N4 =

xy 900

2 y2 (From all u4 coefficients) 1800

N5 =

4y 60

xy (From all u5 coefficients) 900

N6 =

4x 60

4y 60

2A = 2

1 2

 1 = 2A 

xy 900

2 x2 1800

xy (From all u6 coefficients) 900

(60) (60) = 3600 N1 x

= 3600

3 60

2x = – 180 + 4x 1800

1 60

2x 1800

y = – 60 + 4x – 4y 900

 3 = 0, 4 = 3600

y 900

= 4y

 2 = 3600

 5 = 3600

y = – 4y 900

 6 = 3600

4 60

 1 = 2A

N1 4 = 0,  5 = 3600 y 60

 2 = 3600

1 60



x2 1800

N2 =

4x 1800

y = 240 – 8x + 4y 900



x 900

2y 1800

x = 240 – 4x 900

= 60 – 4x + 4y

3 = 3600

1 60

2y = – 60 + 4y 1800

 4 = 3600

x 900

4y = 4x – 8y 1800

 6 = 3600

4 60

x 900

= – 240 + 4x

Chapter 9 9.1 (a)

[k] = 2  r A [ B ] T [D] [ B ] ri = 0, zi = 0, r j = 2, zj = 0, rm = 0, zm = 2

i = rj zm – zj rm = 2.2 – 0 = 4 j = rm zi – zm ri = 0.0 – 2.0 = 0 m = ri zj – zi rj = 0.0 – 0.2 = 0 i = zj – zm = 0 – 2 = – 2 j = zm – zi = 2 – 0 = 2 m = zi – zj = 0 – 0 = 0

r =

1 (2) 3

2 3

i = rm – rj = 0 – 2 = – 2

z =

1 (2) 3

2 3

j = ri – rm = 0 – 0 = 0

1 (2) (2) = 2 2

m = rj – ri = 2 – 0 = 2 2 1 0 [B] = 4 2 2

2 0 0 2 0 0 0 0 2 0 2 2 0 2 2

397

0 2 0 0

0.75 0.25 0.25 0    15  106 0  0.25 0.75 0.25 [D] = (1  0.25)(1  0.5) 0.25 0.25 0.75 0     0 0 0 0.25 0 1 – 0.5  –1  – 0.5 –1.5 – 0.5 – 0.5   6 2 1 2 0   15  10 [ B ]T [D] =   0 0 0.5  2.5  0  0.5 0.5 1.5 0.5    1.5 0.5 0   0.5 5

1 4

[k] = 12.556  106

1 2

0 3

4 1

2 0

1 0

4 1

1 3

1 1 0

2 1

2 8 0

1 0

2 3

4 2

1 0

(b)

Figure 9.1b



 

ri = 0, zi = 0, rj = 2, zj = 0, rm = 2, zm = 2 i = 4, j = 0, m = 0 i = – 2, j = 2, m = 0 i = 0, j = – 2, m = 2 r = 2

2 2 = 1.333, z = , A = 2 3 3

2 1 0 [B] = 4 1 0

0 0 0 2

2 0 1 2

0 0 2 0 2 0 1 0 2

398

2 0

0.75 0.25 0.25 0    15  106 0  0.25 0.75 0.25 [D] =  (1  0.25) 1  0.5 0  2  0.25 0.25 0.75    0 0 0 0.25 0   –1.25 – 0.25 0.25  0 0 0 – 0.5   0.75 1.25 – 0.5 15  106  1.75 T [ B ] [D] =   2.5  – 0.5 –1.5 – 0.5 0.5   0.25 0.25 0.75 0.5    1.5 0.5 0   0.5

[k] = 25.1325  106

2.75

– 2.25

0.5

0.25 –0.5

–1

– 2.25

5.75

– 2.5 0.25

1.5

0.5

–1

– 2.5

0.5

–3

0.25

–1

0.25

0.5

1.75

0.5

– 0.5

1.5

–3

0.5

lb v = 0.25 (Mathcad used here) in.2 Triangle coordinate definition

0 in.

z 1

This defines an array variable x coordinate is the top y coordinate is the bottom

2 in.

Area of triangle

0 in.

1 base  height 2

1 in.

2 in.

1 (jr – ir) (mz – iz) A = 2 in.2 2

Develop stiffness matrix

i = jr mz – jz mr j = mr iz – mz ir m = ir jz – iz jr

i = 4 in.2 j = 0 m = 0

i = jz – mz j = mz – iz m = iz – jz

i = – 2 in. i = mr – jr i = – 1 in. j = 2 in. j = ir – mr j = – 1 in. m = 0 in. m = jr – ir m = 2 in.

Evaluate [B] at centroid of element rbar =

jr mr 3

zbar =

jz mz 3

399

rbar = 1 in.

zbar = 0.667 in.

[Bi] =

1 2A

0 i

rbar

i i zbar

[Bj] =

rbar i

1 2A

0 j

rbar

0 j

0 m m zbar

rbar

rbar j

0 m

j j zbar

rbar

i m

[Bm] =

augment( Bi , B j , Bm )

Gradient matrix at centroid of element

[ B]

0.5

0.25

0.3333

0.25

0.5

0.25

0.5

1 v

1 2v 2

0.6  107

1.8  107

0.6  107

1.8  107

0.5 1 0.3333 0 in. 0

0.5

1 v E [D] = (1 v )(1 2v )

1.8  107  7 0.6  10 [D] =  7 0.6  10   0

Axisymmetric stress constitutive matrix

   lb 0  2 0  in. 0.6  107  0

[k] = 2rbar A [B] [D] [B] Axisymmetric element stiffness matrix T

 1.225  108  7  2.513  10  – 5.341  107  k   0.5 x  7  1.257  10  6.283  106   1.257  107

2.513  107

– 5.341  107

1.257  107

6.283  10 6

6.597  107

1.257  107

9.425  106

5.027  10 7

1.257  107

2.231  108

5.027  107

5.655  10 7

9.425  106

5.027  107

6.597  107

2.513  107

5.027  107

5.655  107

2.513  10 7

8.796  10 7

5.655  107

6.283  107

5.655  107

2.513  10 7

1.257  107   5.655  107   6.283  107  lb 5.655  107  in. 2.513  107   1.131  108 

9.2

400

 N1  0 p   {fs} =  [ N s ]T  r  ds = s  N 2 s  pz   0  N3  0

Now

0  N1   p z 0   h0     2 rdz N 2   0  0  Evaluated  @ r b N3  z  z

Ni =

1 1 (i + ir + iz), Nj = (j + jr + jz) 2A 2A

Nm =

1 (m + mr + mz) 2A

ri = 0, rj = b, rm = b, zi = 0, zj = 0, zm = h

i = rj zm – zj rm = bh – 0b = bh, j = rm zi – zm ri = 0 i = zj – zm = 0 – h = – h, j = zm – zi = h – 0 = h i = rm – rj = b – b = 0, j = ri – m = 0 – b = – b m = ri zj – zi rj = 0, m = zi – zj = 0 – 0 = 0 m = rj – ri = b – 0 = b So the shape functions evaluated at r = b and z = z Ni =

1 (bh + (– h) b + 0 z) = 0 bh

Nj =

1 1 (0 + h b + (– b)z) = (hb – bz) bh bh

Nm =

1 1 (0 + 0 b + b z) = (bz) bh bh

401

 0  0   1 (hb  bz ) h bh {fs} =   0  0  1  bh (bz )   0

     p0 z  0   h  2 rdz 1    ( hb  bz ) bh  0   0  1 (bz )  bh 0 0

0 0 p bz 2

0 2 b h p0bz – h = bh 0 0

p0bz 2 h

  0  h 0    p0bz 2  p0bz 3  2  2 3h  dz =   h  0    p0bz 3  0 3h     0

 0     f s1r  0    f s1z  2  p bh  0   2  6   f s 2 r  =    = 2 b h  0   fs2z   p0 bh2   f s 3r   3       f s3z   0 

0 0 p0 h 6

0 p0 h 3

9.3

RB ZB

Equation to be evaluated is {fB} =

2 rA RB 3 ZB RB ZB

r = 4+2

2 3

 r = 5.333 in. 402

ZB = 0.283

lb in.3

lb rev. rad 1 min 2 0.283 in.3 [5.333 in.] RB = ω  r = 20 2 min rev 60 s 32.2 12 lb3 2

in.

RB = 0.01712

lb in.3

2 2 rA =  (5.333 in.) (2 in.) = 22.34 in.3 3 3

So fB1r = (22.34) (0.01712) = 0.382 lb fB1z = (–22.34) (0.283) = – 6.32 lb fB2r = (22.34) (0.01712) = 0.382 lb fB3z = (– 22.34) (0.283) = – 6.32 lb fB3r = (22.34) (0.01712) = 0.382 lb fB3z = (– 22.34) (0.283) = – 6.32 lb 9.4 (a)

Element Figure 9.4 a The equation to be evaluated is {} = [D] [B} {d} ri = 0, zi = 0, rj = 2, zj = 0, rm = 1, zm = 3

i = 6, j = 0, m = 0, i = – 3, j = 3, m = 0, i = – 1, j = – 1, m = 2 r = 1, z = 1, A =

1 (3) (2) = 3 in.2 2

 3 0 3 0 1  0 1 0 1 B = 6 2 0 2 0   1 3 1 3

0 0  0 2 0 0  2 0

0.75 0.25 0.25 0    15  10 0  0.25 0.75 0.25 [D] = (1  0.25)(1  0.5) 0.25 0.25 0.75 0     0 0 0 0.25 6

r z

0.75 0.25 0.25 0   15  106 0.25 0.75 0.25 0  1 = (1.25)(0.5) 0.25 0.25 0.75 0  6    0 0 0 0.25



403

 3 0 3 0  0 1 0 1  2 0 2 0   1 3 1 3

0 0 0 2

 1   0  2  2   5  10–4   0   3  0  0    0

 2.58     0.54  4 =    10 psi  2.58  0.54 

(b) Element Figure 9.4b ri = 1, zi = 0, rj = 3, zj = 0, rm = 3, zm = 3

i = 9, j = – 3, m = 0, i = – 3, j = 3, m = 0 i = 0, j = – 2, m = 2 r = 2.333, zm = 1.00, A = 3 3 1 0 B = 6 0.858 0

0 3 0 0 0 1.14 3 2

0 0 0 2 2 0 0.512 0 3 0 2

 0.0001  0.0002 0.75 0.25 0.25 0      6 15  10 0  0.25 0.75 0.25  D = ,  d  =  0.0003 (1  0.25)(1  0.5) 0.25 0.25 0.75 0      0    0  0 0 0.25  0  {} = [D] B {d} r z

(c)

 3514    85.7  =   psi  2143   700 

u1 = 0.0001 in.

w1 = 0.0002 in.

u2 = 0.0005 in.

w2 = 0.0003 in.

u3 = 0.0 in.

w3 = 0.0 in.

404

 u1  w   4.8  103   1  r    3  u2    z  0.6  10  lb   d=  , = [D] [B] {d} =   2 4  w2      0.6  10  in.  u3    rz   3 0.45  10     w3 

9.5 By Equation (9.1.35) zm

{fsi} =

1 2A

j rj

0 j rj

pr pz

2rj dz

Now

j = rm zi – ri zm = rj zj – ri zm j = zm – zj



{fsj} =

Since rj = rm

j = ri – rj

zi = zm

1 (rj – ri) (zm – zj) 2

zm 2

rj pr [rj z j

ri zm ( zm

z j )r j

(ri

rj ) z

pz [ r j z j

ri zm ( zm

z j )r j

(ri

rj ) z

Integrating, we obtain

{fsj} =

2 rj 2A

Pr (rj z j

ri zm ) ( zm

zj)

r j ( zm

z j )2

(ri

rj )

Pz (rj z j

ri zm ) ( zm

zj)

r j ( zm

z j )2

(ri

rj )

( zm2

z 2j ) 2

( zm2

z 2j ) 2

Factoring out zm – zj and simplifying {fsj}=

{fsj} =

2 rj ( zm

z j ) pr

2 rj ( zm

zj)

rj z j ri zm 2 2 – rj z j r j zm – 2 2

ri zm 2 ri zm 2

pr [ 12 ( zm

z j )( rj

ri )

pz [ 12 ( zm

z j )( rj

ri )

2 r j ( zm

z j ) pr A

pz A

2 r j ( zm

rj z j 2 rj z j 2

zj)

9.6 (a)

405

    



 i = 50 (50) – (0) (0) = 2500 mm2 i = – 50 mm, i = – 50 mm j = 50 mm, j = 0 mm m = 0 mm, m = 50 mm 50 1 0 B = 2(1250) 50 50

0 50 0 0 0 50 0 0 0 50 1 0 50 0 50 0 mm 50 0 50 50 0

0.75 0.25 0.25 0    105  10 0.25 0.75 0.25 0  N [D] = 2 (1.25)(0.5) 0.25 0.25 0.75 0 m    0 0 0 0.25 9

 25 105  109  0 [D] B = 1250(1.25)  25   12.5

12.5 50 0 37.5 25 0 12.5 50 0 12.5 0 12.5

12.5 12.5   12.5 37.5 37.5 12.5   12.5 0 

[k] = 2 r A [ B ] T [D] [ B ] =

2 (0.01667 m)(1250)(105  109 )  106 [ B ] T [D] [ B ] (1250)(1.25)

Multiplying [ B ] T times [D] [ B ] , we obtain 3125

625

2500

1250 5000

625 0 625

1250 2500 625

1875 1250 N m 0

2500

625 1875

[k] = 3.5195  106 

Symmetry

9.6 (b)

406

A = 1800 mm2

i = – 60 mm, j = 60 mm, m = 0 i = 0, j = – 60 mm, m = 60 mm [k] = 2 r A B

[D] B

60 1 0 B = 2(1800) 30 0

0 0 0 60

60 0 30 60

0 0 0 60 0 60 1 0 30 0 mm 60 60 0

[D] as in 9.6 (a)  37.5 0 52.5 15 7.5 15   105  109  7.5 0 22.5 45 7.5 45 [D] [ B ] = 1.25(1800)  7.5 0 37.5 15 22.5 15     0 15 15 15 15 0

[k] =

2 (0.04 m)(1800 )(105  109 ) T [ B ] [ D][ B ] (1.25)(1800)(2)(1800 )

2475

0 900

[k] = 5.865  106

Symmetry

2025 900

450 900

225 900

450 0

5175

2250 3600

225 450

1350 2700

1575

450 2700

(c)

407

ri = 0

zi = 0

rj = 0.002

zj = 0

rm = 0.001

zm = 0.002

[B] =

1 2A

1 1 0.002 (rj – ri) (zm – zi), E = 105  109, z =  0.002, r = , v = 0.25 2 2 3

i = rj zm – zj rm

j = rm zi – zm ri

m = ri zj – zi rj

i = zj – zm

j = zm – zi

m = zi – zj

i = rm – rj

j = ri – rm

m = rj – ri

0 j

E (1 v) (1 2v)

m j

jz r

r i

[D] =

r j

m mz

1 v

v v

1 v

v v

1 v

0 0 0

1 2v 2

0 m

[k] = 2r A [B]T [D] [B]  8.577  108  8  1.759  10  3.738  108  k   0.5 x  7  8.796  10  4.398  107   8.796  107 

1.759  108

3.738  108

8.796  107

4.398  10 7

4.618  108

8.796  107

6.597  107

3.519  108

8.796  107

1.561  109

3.519  108

3.958  108

6.597  107

3.519  108

4.618  108

1.759  108

3.519  108

3.958  108

1.759  108

6.158  108

3.958  108

4.398  108

3.958  108

1.759  108

8.796  107   3.958  108   4.398  108   3.958  108  1.759  108   7.917  108 

9.7 (a)

408

From Problem 9.6 (a), we have [D] [ B ]  {} = [D] [ B ] {d}

 25  = 105  10 MPa  0 (1250)(1.25)  25   12.5 3

12.5 50

12.5

37.5 25 12.5 50

0 0

12.5 37.5

12.5

12.5 12.5

 75.6    =  58.8  MPa  92.4   58.8 

0.05   12.5  0.03  37.5  0.01   12.5   0.01  0  0     0 

(b)

From Problem 9.6 (b), we have [D] B



 37.5 0  0 = 105  10  7.5 (1800)(1.25)  7.5 0   0 15 3

52.5 22.5 37.5 15

0.05   15 7.5 15  0.03  72.8    45 7.5 45 0.01 = 50.4 MPa   15 22.5 15  0.01  39.2     15 15 0   0  39.2    0 

9.8 No, not in general, as the axisymmetric elements are rings, not plane quadrilaterals or triangles. So axisymmetric nodes are actually nodal circles whereas plane stress elements have node points. 409

9.9 No, the element circumferential strain is a function of r and z (see Equation (9.1.15). 9.10 Make ur = 0 for all nodes acting on the axis of symmetry. 9.11 How would you evaluate circumferential strain  at r = 0? From text Equation (9.1.15)

 =

a z a1 + a2 + 3 r r

(1)

r = a2

(2)

Also from text Equation (9.1.1e)

 =



(3) u r

(4)

u =  r

(5)

r = 

u r

Substituting (1) into (5) u =

a1 r

a3 z r = a1 + a2r + a3z r

(6)

Partial of (6) with reference to r  

u = a2 r

Compare to (2)

(7)

= r = a2 as stated in problem statement

9.12 What will be the stresses r and  at r = 0? From Equation (9.1.2)

r =

a2 E || (1 v )(1 2v ) r (1 v )

|| z (v)

(v)

E (a2  a2v  a6v  a2v ) (1 v)(1 2v) E (a2 + a6 v) (1 v)(1 2v) 410

  =



 =

E (1 v)(1 2v)

r (v )

z (v)

E ( a2v (1 v)(1 2v)

a6v

a2 (1 v )

a2 – a2v )

E (a2 + a6 v) (1 v)(1 2v)

 at r = 0,

 = r

9.13 Axisymmetric model pressure load of 13.26

lb in.2

Plane stress with a thickness of 18.85 inches

411

9.14 von Mises stresses (with filleted corners)

412

9.15

413

Note: Without the arc (inside radius), we have a 90° re-entrant corner where stress is approaching infinity. We have a singularity in the linear-elastic solution based on linear theory of elasticity. Therefore, we need the arc as in good practice or elastic-plastic model where an upper bound on the corner stress is the yield strength of the material.

9.18

414

9.19

Figure 1 Thick walled open-ended cylinder Theoretical Solution for hoop stress at inner radius q = 35  106 a = 1.5 b = 1.2

 =

r = 1.2

qb2 a 2 r 2 r 2 a 2 – b2

 = 1.594  108 Pa 415

Autodesk Results

Theoretical Results

Hoop Stress

159.5 MPa

159.4MPa

Maximum Principal Stress

159.5 MPa

Minimum Principal Stress

–35 MPa

Deflection in y-direction

0.93mm

The Autodesk results for hoop stress and the theoretical solution for hoop stress are very close which proves that the Algor model is correct. The pipe has a very minimum internal and external deflection, less than 1mm on the inner radius. The stresses are also manageable at 159 MPa. 9.20 A steel cylindrical pressure vessel with flat plate end caps is shown in the figure with vertical axis of symmetry. Addition of thickened sections helps to reduce stress concentrations in the corners. Analyze the design and identify the most critically stressed regions. Note that inside sharp re-entrant corners produce infinite stress concentration zones, so refining the mesh in these regions will not help you get a better answer unless you use an inelastic theory or place small fillet radii there. Recommend any design changes in your report. Let the pressure inside be 1000 kPa. 503 elements and 645 nodes. Stresses are highest at sharp corners and the middle of the top and bottom of the pressure vessel. The design is acceptable as the von Mises stresses do not reach the yield strength of the material.

9.22

416

The recommended head shape of the hemispherical ends versus the ellipsoidal ends would be the hemispherical ends due to a lower stress concentration at the transition between the head and the cylindrical body. 9.23 According to the von Mises stress analysis, the average stress through the glass is around 15 of the tensile strength of the glass. If the maximum force used with this syringe is 45 N, the design should be fine. However, if 45 N is the normal operating force which may increase, I 417

would recommend analyzing this again with a safety factor of 4 (180 N force) to make sure it will still be under 5 MPa. As for the maximum principal stresses, they are well below the tensile strength of the glass and do not appear to be an issue.

N Another analysis with a safety factor of 4 (28.64789 radian ) reveals that this syringe is still within the tensile strength of glass in all areas. With this information, I would conclude that the syringe design is indeed safe with this material specification.

9.25 Steel hole punch

418

9.27 a)

419

420

ρs = 7.36 X 10-4

a=1

b=4

r=1

υs = 0.29

υa = 0.33

ρaw = 0.0975

g = 386

ρa =

 aw

ρa = 2.526 X 10-4

b) What speed can the steel flywheel tolerate before yielding based on the maximum distortional energy theory (MDET)? Sys = 65300

Sya = 40000

Sys

s  s 

(3  s)  2 b 2 (1  3  s) 2   a  b 2  a 2  2  r  8 r 3  s  

ωs = 2.579 X 103 rad/s ωsrpm = ωs·

60 2

ωsrpm = 2.463 X 104 rpm Maximum speed of disk for steel to have hoop stress reach yield stress

c) Maximum rpm of aluminum alloy 6061-T6 to reach yield:

421

Sya

a  a 

(3  a)  2 b 2 (1  3  a) 2   a  b 2  a 2  2  r  8 r 3  a  

ωa = 3.427 X 103 ωarpm = ωa·

60 2

ωarpm = 3.272 X 104 rpm Max speed of disk for aluminum alloy 6061-T6 to reach yield

422

Chapter 10 10.1

dx ds

u = a1 + a2s + a3s2 x = a1 + a2s + a3s2

x1 = a1 + a2(– 1) + a3 (– 1)2

(1)

x2 = a1 + a2 (1) + a3 (1)2

(2)

(1) – (2) gives x1 – x2 = – 2a2, a2 =

(3)

x3 = a1 + a2(0) + a3(0)  a1 = x3

(4)

(3) and (4) into (1) a3 = x1 – x3 +

Now and

x2 2 x3 2 x2 2 x3 2 s 2

2 x x2  x1  x = x3 + s + 1 2 x2  x1 dx J= = + (x1 + x2 – 2x3)s ds 2 x x2 x3 = 1 for x3 at s = 0 2

(5) (6)

x2 – x1 = L  J= J=

L + [x1 + x2 – (x1 + x2)]s 2 L 2

10.2 Using Equation (10.1.1 b) For Figure (a) (a)

s = [x –

(x1 + x2 ) ] 2

2 x2

s = [15 –

2 (10 + 20) ] 20 10 2

s = (15 – 15)

2 =0 10

(b) By Equation (10.1.5) N1 =

1 0 1 0 = 0.5, N2 = = 0.5 2 2

u   0.005  N 2   1   0.5 0.5  0 0.005 u 2 

 1 1   0.005     0.0010  10 10  0.005

(d)  x  

By Equation (10.1.1 b) For Figure (b) (a) At A  XA = 7 in. s= 7–

5 10 2

s = [7 – 7.5]

2 10 5

2 5

s = – 0.2 (b) By Equation (10.1.5) N1 =

1 ( 0.2) = 0.6 2

N2 =

1 (0.2) = 0.4 2

 1  5

(d)  x   

 0.005  0.4    0.001 in. 0.005 1   0.005     0.002 5  0.005

10.3 Using Equation (10.1.1 b) Figure (a) (a) x = xA = 50 mm 20  60   2  s = 50        2   60  20 

s = [50 – 40]

2 40 424

s = 0.5 (b) N1 =

1  0.5 1 1  0.5 3 = , N2 = = 4 4 2 2

0.10 3 / 4    0.175 mm 0.20

1  0.10    0.0025 40  0.20

 1  40

(d)  x   

For Figure (b) (a)

10 30 2

2 30 10

2 20

s = [20 – 20] s= 0 (b)

N1 =

1 1 1 0 1 0 = , N2 = = 2 2 2 2

0.05   0.075 mm 0.10

 1  20

(d)  x   

1  0.05    0.0025 20  0.10

10.4

(1)

u = a1 + a2 s + a3 s2 + a4 s3

(A)

x = a1 + a2 s + a3 s2 + a4 s3

(B)

x1 = a1 + a2(–1) + a3(–1)2 + a4(–1)3

(1)

x2 = a1 + a2

1 + a3 2

1 2 + a4 2

x3 = a1 + a2

1 2

1 2 + a4 2

+ a3

x4 = a1 + a2(1) + a3(1)2 + a4(1)3

1 3 2

(2) (3) (4)

(1) + (4)  x1 + x4 = 2a1 + 2a3

(5)

a3 2

(6)

(2) + (3)  x2 + x3 = 2a1 + (5) – (6) gives

x1 + x4 – (x2 + x3) = 2a1 + 2a3 – 2a1 or

a3 =

a3 2

2 (x1 + x4 – x2 – x3) 3

(7)

(7) into (5) 2 (x1 + x4 – x2 – x3) 3

x1 + x4 = 2a1 + 2 or

1 3

a1 =

4 3

(8)

(1) – (4)  x1 – x4 = – 2a2 – 2a4 (2) – (3)  x2 – x3 = – a2 –

(9)

a4 4

(10)

(9) – 2 (10) gives 3a4 2

x1 – x4 – 2(x2 – x3) = 

a4 =

2 [2(x2 – x3) – (x1 – x4)] 3

(11)

(11) into (9) yields a2 =

1 3

8 3

(12)

Substituting (7), (8), (11) and (12) into (B) x=

4( x2

4 x1

x3 ) x1 6

8 x2 6

s2 +

8 x2

x2 6

x3 s x4 s3

4 x1 6

(13)

Combine like x1, x2, x3 and x4 coefficients x=

2 3 s 3

2 2 s 3

4 3 2 2 s – s 3 3

s 6

1 x1 6

4 s 3

2 x3 3

4 3 s 3 2 3 s 3

2 2 s 3 2 2 s 3

4 s 3

2 x2 3

s 6

1 x4 6

(14)

By (14) then

x1 x {x} = [N1 N2 N3 N4] 2 x3 x4 2 3 2 2 s 1  N1 = s s 3 3 6 6 4 3 2 2 4 2 N2 = s s s 3 3 3 3

N3 =

4 3 s 3

N4 =

2 3 s 3 2s

du = ds

(2)

2 2 s 3

4 s 3

2 2 s 3

s 6

4 s 3

1 6

2 3

1 6 4s

4 s 3

4 3

4 s 3

2s2

4 3 4 s 3

1 6

u2 u3 u4

Differentiating (13) dx = ds

2s 2

4 s 3

1 x1 + 4s 2 6

4 s 3

4 3

4s 2

4 s 3

4 3

4 s 3

1 6

x3 + 2s 2

Simplifying = 2s2 (x4 – x1) + – 4s2 (x3 – x2) –

4 1 s(x4 + x1) – (x4 – x1) 3 6

4 4 s (x3 + x2) + (x3 – x2) 3 3

= 2s2 L +

x x1 L 1 8 s 4 – L – 4s2 2 6 2 3

= 2s2 L +

2 8 L 8 s xc – – 2s2 L – s xc + L 3 3 6 3

–

8 s 3

4 L 3 2

dx L = ds 2

Now du = dx

du ds dx ds

and

du = x = [B] {d} dx



x =



[B] =

12s

8s 1 12s

4s 4

12s

3L 2

12s 2 8s 1 12s 2 4s 4 3L 3L 2

4 12s

8s 1 3L

4 12s 2

8s 1 3L

3L 2

12s 2 3L 2

u1 u2 u3 u4

10.5 For Figure (a) (a) Using Equation (10.5.6) 20 10 10 20 2 15 2 s + s 2 2

x = xA = 13 = 15 +  0s2 + 5s + 2 = 0

(b)

2 = – 0.4 5

N1 =

s( s – 1) = 2

0.4 ( 0.4 1) 2

= 0.28 N2 =

s ( s 1) = 2

0.4 ( 0.4 1) 2

= – 0.12 N3 = 1 – s2 = 1 – (– 0.4)2 = 0.84 N’s = 0.28 – 0.12 + 0.84 = 1 u = a1 + a2 s + a3 s2

(c)

u1 = 0.006 = a1 + a2(– 1) + a3(– 1)2 u3 = 0 = a1 + a2(0) + a3(0) u2 = – 0.006 = a1 + a2(1) + a3(1)2 

a3 = 0, a2 = – 0.006, a1 = 0



u = – 0.006s and s = – 0.4 at xA = 13



u = – 0.006 (– 0.4) = 0.0024 in. at A

(d)



x =

2s – 1 2s 1 4s u1 + u2 – u3 L L L

x =

2s – 1 2s 1 (0.006) + (– 0.006) = 0 L L

x =

– 0.012 (L = 10 in.) L

x = – 0.0012

in. in. 428

10.6 For Figure (a) (a) Using Equation (10.5.6) x = xA = 1.5 = 1 +

2 0 s 2

2 2 (1) 2 s 2

1.5 = 1 + s + 0s2 s – 0.5 = 0 s = 0.5 (b)

0.5 0.5 1 s s –1 s s –1 N1 = = 2 2 2

= – 0.125 N2 =

0.5 0.5 1 s s 1 = 2 2

= 0.375 N3 = 1 – s2 = 1 – 0.52 = 0.75 N ’s = – 0.125 + 0.375 + 0.75 = 1.0

u2 u (0.5)  1 (0.5) 2 2 u u  1 (0.5) 2  2 (0.5) 2 2 2 2u  3 (0.5) 2 2 0.002  0.001  (0.5)  0 2 0.002 0  (0.5) 2 2 2(0.001)  (0.5) 2 2 u A  0.0015 mm uA  u3 

(d) By Eq. (10.5.14)

2(0.5)  1 )(0) 2 2(0.5)  1 ( )(0.002) 2 4(0.5) ( )(0.001) 2  0.001

x  (

10.8

{F} = [K] {d}

(A)

where by Equation (10.5.22) 4.67 0.667 –5.33 4.67 –5.33 Symmetry 10.67

AE [K] = L

(1)

By Equation (10.5.9) for N ’s and using Equation (10.1.21) {fs} =

{fs} =

1 –1

[ N s ]T {T } dx , {T} = 2 s( s – 1) 2 s( s 1) 2 1 – s2

1 –1

kN m

kN L , dx = ds m 2

L ds 2

(2)

Upon integrating Equation (2) s3 s2 – 6 4 3

{fs} =

–1

2 1

s s – 6 4

(2) –1

4 2

(3)

s–

s3 3 –1

{fs} =

1 3 1 3 4 3

(4)

4 3 4 kN 3 16 3

(4)

Using Equations (4) and (1) in (A) and applying boundary condition u1 = 0, 4 3 16 3

2.393 – 2.732 – 2.732

104

5.468

(5)

Solving (5) for u2 and u3 u2 = 3.885  10–4 m, u3 = 2.916  10–4 m

(6)

Stress in bar E = 205  109 At s = – 1 or x = 0 2 (–1) – 1 2 (–1) 1 – 4 (–1) 1 = E 4 4 4

0 3.885 10–4 , 1  (3.987  107 ) 2.916 10–4

N m2

39.87 MPa

At s = 1 or x = L 2 1–1 2 1 –4 2 = E 4 4 4

0 3.885 10 –4 , 2 = (– 4.357  104) Pa 2.916 10–4

= – 0.04357 MPa Note small number compared to stress at fixed end. 10.9 x=

1 [(1 – s) (1 – t) x1 + (1 + s) (1 – t) x2 + (1 + s) (1 + t) x3 4 + (1 – s) (1 + t) x4] 1 [(1 – s) (1 – t) y1 + (1 + s) (1 – t) y2 + (1 + s) (1 + t) y3 + (1 – s) 4

(1 + t) y4] [J] =

x s x t

y s y t

Equation (10.2.10)

x 1 = [– x1 + t x1 + x2 – t x2 + x3 + t x3 – x4 – t x4] s 4 x 1 = [– x1 + s x1 – x2 – s x2 + x3 + s x3 + x4 – s x4] t 4 y 1 = [– y1 + t y1 + y2 – t y2 + y3 + t y3 – y4 – t y4] s 4 y 1 = [– y1 + s y1 – y2 – s y2 + y3 + s y3 + y4 – s y4] t 4

[J] =

J11 J 21

J12 J 22

J11 =

1 (– x1 + t x1 + x2 – t x2 + x3 + t x3 – x4 – t x4) 4

J12 =

1 (– y1 + t y1 + y2 – t y2 + y3 + t y3 – y4 – t y4) 4

J21 =

1 (– x1 + s x1 – x2 – s x2 + x3 + s x3 + x4 – s x4) 4

J22 =

1 (– y1 + s y1 – y2 – s y2 + y3 + s y3 + y4 – s y4) 4

where

Find determinate | J | | J | = J11 J22 – J21 J12 Multiplying and collecting terms |J|=

1 [2x1y2 – 2t x1y2 + 2t x1y3 – 2s x1y3 + 2s x1y4 – 2x1y4 – 2x2y1 + 2t x2y3 16

+ 2x2y3 + 2s x2y3 – 2s x2y4 – 2t x2y4 + 2s x3y1 – 2t x3y1 – 2x3y2 – 2s x3y2 + 2x3y4 + 2t x3y4 + 2x4y1 – 2s x4y1 + 2s x4y2 + 2x4y2 + 2t x4y2 – 2x4y3 – 2t x4y3] Factor out xi’s |J|=

1 [x1 (y2 – ty2 + ty3 – sy3 + sy4 – y4) 8

+ x2 (– y1 + t y1 + y3 + s y3 – s y4 – t y4) + x3 (s y1 – t y1 – y2 – s y2 + y4 + t y4) + x4 (y1 – s y1 + s y2 + t y2 – y3 – t y3)]

|J|=

1 [x1 x2 x3 x4] 8

y2 t y2 ty3 y1 t y1 y3

s y3 s y3

s y4 s y4

y4 t y4

s y1 t y1 y2 s y2 y1 s y1 s y2 t y2

y4 y3

t y4 t y3

y1 (0) y2 1 t y3 t s y4 (s 1) y1 1 t y2 (0) y3 (0) y4 1 t 1 |J|= [x1 x2 x3 x4] y1 s t y2 1 s y3 (0) y4 1 t 8 y1 1 s y2 s t y3 1 t y4 (0)

|J|=

1 [x1 x2 x3 x4] 8

0 1 t s t 1 s

1 t 0 1 s s t

t s s 1 0 1 t

s 1 s t 1 t 0

y1 y2 y3 y4

10.10 y s y t

x s x t

[J] =

x and y from Problem 10.9 x 1 1 1 1 = (– 1) (1 – t) x1 + (1 – t) x2 + (1 + t) x3 + (– 1) (1 + t) x4 s 4 4 4 4 = N1,s x1 + N2,s x2 + N3,s x3 + N4,s x4 1 1 1 1 x = (– 1) (1 – s) x1 + (– 1) (1 + s) x2 + (1 + s) x3 + (1 – s) x4 t 4 4 4 4 = N1,t x1 + N2,t x2 + N3,t x3 + N4,t x4 y 1 1 1 1 = (– 1) (1 – t)y1 + (1 – t)y2 + (1 + t)y3 + (– 1) (1 + t)y4 s 4 4 4 4 = N1,s y1 + N2,s y2 + N3,s y3 + N4,s y4 1 1 1 1 y = (– 1) (1 – s)y1 + (– 1) (1 + s)y2 + (1 + s)y3 + (1 – s)y4 t 4 4 4 4 = N1,t y1 + N2,t y2 + N3,t y3 + N4,t y4

[J] =

N1, s

N 2, s

N 3, s

N 4, s

N1, t

N 2, t

N 3, t

N 4, t

x1 x2 x3 x4

y1 y2 y3 y4

10.11 (a) [J] =

x s x t

y s y t

 x1  2, x2  2   x  2, x  2   3  4

1 1 1 1 x = (–1)(1 – t)(–2) + (1 – t)(2) + (1 + t)(2) + (–1)(1 + t)(–2) = 2 4 4 4 4 s 1 1 1 x = (– 1) (1 – s) (– 2) + (– 1) (1 + s) (2) + (1 + s) (2) t 4 4 4

1 (1 – s) (– 2) = 0 4

y1 = 1, y2 = –1 y3 = 1, y4 = 1 1 1 1 y = (–1) (1 – t) (–1) + (1 – t) (–1) + (1 + t) (1) 4 4 4 s

1 (–1) (1 + t) (1) = 0 4

1 1 1 y = (–1) (1 – s) (–1) + (–1) (1 + s) (–1) + (1 + s) (1) 4 4 4 t

+ |[J] | =

2 0 0 1

1 (1 – s) (1) = 1 4

= 2–0= 2

By Equation (10.2.22) 0

|[J] | =

t 1 1 [–2 2 2 –2]  s t 8 1 s

1 t 0 s 1 s t

t s s 1 0 t 1

s 1 s t t 1 0

1 1 1 1

(A)

Simplifying by multiplying the matrices in Equation (A) yields | [ J ] | = 2 also and

| [J] | =

A as 4

A = 4 × 2 = 8 (area of element) 

  

= |[J] | =

8 =2 4

10.12 By Equation (10.2.18) 1 [B1] [B2] [B3] [B4] [J ]

[B(s, t)] = By Equation (10.2.3) x=

1 [(1 – s) (1 – t) x1 + (1 + s) (1 – t) x2 + (1 + s) (1 + t) x3 + (1 – s) 4 434

(1 + t) x4] y=

(1)

1 [(1 – s) (1 – t) y1 + (1 + s) (1 – t) y2 + (1 + s) (1 + t) y3 + (1 – s) 4 (1 + t) y4]

By Equation (10.2.16) y t

1 [D’] = [J ]

() s

() t

y s

0 x s y t

0 x s

() t

() s

x t

() t () s

() s () t

x t y s

(2)

Let a=

y 1 = [y1(s – 1) + y2(– 1 – s) + y3(1 + s) + y4(1 – s)] t 4

y 1 = [y1(t – 1) + y2(1 – t) + y3(1 + t) + y4(– 1 – t)] s 4

1 x = [x1(t – 1) + x2(1 – t) + x3(1 + t) + x4(– 1 – t)] s 4

x 1 = [x1(s – 1) + x2(– 1 – s) + x3(1 + s) + x4(1 – s)] t 4

By Equation (10.3.5) N1 =

(1 - s) (1 - t) (1 + s) (1 - t) , N2 = 4 4

N3 =

(1 + s) (1 + t) (1 - s) (1 + t) , N4 = 4 4

[N] =

N1 0

Ni, s =

Ni , s

0 N1

N2 0

0 N2

Ni, t =

Ni t

N3 0

0 N3

N4 0

0 N4

Let

Now [B] = [D [N] Using [D from Equation (2) above, we obtain a

1 [B] = [J ]

0 c

c d

1 s 1 t 4

col (1)

[B] =

1 [J ]

a ( t 1) 4

d ( t 1) 4

c ( s 1) 4 a ( t 1) 4

0 c ( s 1) 4

cos(7)

col (2)

b ( s 1) 4

a (1 t ) 4 d ( t 1) 4 b ( s 1) 4

0 c (1 s ) 4

cos(8)

b (1 s ) 4

d ( 1 t) 4

0 c (1 s ) 4 a( 1 t) 4

d ( 1 t) 4 b (1 s ) 4

By Equation (10.2.19) 1 =[[B1] [B2] [B3] [B4]] [J ]

[B] =

where the submatrices are a Ni, s

[Bi] =

b N i ,t 0

c Ni, t

d Ni,s

0 c Ni, t a Ni,s

d Ni,s b N i ,t

where, for instance N1,s

t 1 N1 = s 4

N1,t

s 1 N1 = t 4

etc. 10.13

{fs} = At

[Ns]T {T} h

L ds 2

N3 1 0

N4 0

1 1

t= 1 f s3s f s 3t fs 4s f s 4t

0 N3

0 T N4

ps L h ds pt 2



{fs} =

1 1

0 N3 p 0 N4 p

L ds 2

{fs} =

s ) (1 t ) p 4

1 (1

s ) (1 t ) p 4

L ds 2

 f s 3s  0   0  1  pLh   ps ps 2   f s 3t     4  Lh  2  = 2    0  2  f s 4s  0      ps – ps 2  pLh  2 4  1  f s 4t    2 

10.14 (a)

{ fs} =

1 1

[Ns ]T {T}

L h dt 2

L = 5 in., pt = 4000 psi, ps = 0 1 s 1 s and N4 = for t = 1 2 2 N3 0 1 0 N 3  ps  0  h L dt =  1 N 0  pt  4000 2  4 0 N4

N3 =



f s3s f s 3t fs 4s f s 4t

0    1 1 s   0  4000   0.1  52  dt      2  1000  =   1   lb 0    0   1 1 s  1000  5   1 2  4000   0.1  2  dt 

   

(b)

f s1s f s1t

f s4s f s 4t

1 1

N1 0 N4 0

0 N1 0 N4

fs1s = fs4s = 0 N1 = fs1t =

1 1

250 t

250

L dt 2

1 t 1 t , N4 = for s = – 1 2 2

1 t 5 (– 250 t + 250) (0.1) 2 2

= 83.33 lb fs4t =

1 1

1 t 2

(– 250 t + 250) (0.1)

5 2

= 41.67 lb 10.15 (a)

1 1

cos

s ds 2

Use Table 10.1 (3 Gauss Points) 3

Wi cos i 1

si s s s = W1 cos 1 + W2 cos 2 + W3 cos 3 2 2 2 2

0.7746 5 8 5 cos + cos (0) + cos 2 9 9 9

I = 1.9176

0.7746 2

(Analytical I = 1.9176)

(2 Gauss Points)

0.57735 )) 2 0.57735 1.0(cos( )) 2  1.9172 I  (cos(

(2 Newton-Cotes Points)

1 1 I  2( yo  y1 ) 2 2 1 1  2( (0.87758)  (0.87758)) 2 2  1.7552 That is 1

cos

s 1 s 1 ds = 2 sin = 2 sin – 2 sin 2 2 2 1

1 2

1 = 4(0.47) 2

= 4 sin

= 1.9176 (b)

1 1

s 2 ds

(3 Gauss Points) 3

Wi si2 = W1 s12 + W2 s22 + W3 s32

i =1

5 8 5 (0.7746)2 + (0)2 + (– 0.7746)2 9 9 9

I = 0.667 (c)

1 1

s4 ds =

(Analytical I = 0.667)

5 8 5 (0.7746)4 + (0)4 + (– 0.7746)4 9 9 9

= 0.400

(3 Gauss Points)

(3 Newton-Cotes Points)

1 4 1 I  2( y o  y1  y 2 ) 6 6 6 4 yo  (1)  1 y1  0 y 2  (1) 4  1 1 1 I  2( (1)  (1)) 6 6  0.6667 cos s

(d)

ds =

5 9

cos 0.7746 1 0.77462

8 cos 0 9 1 02

5 cos 0.7746 = 2.873 9 1 0.7746 2

(Exact is 3.86) 1

(e) (f)

1 1 1

s3 ds =

s cos s ds =

5 8 5 (0.7746)3 + (0)3 + (– 0.7746)3 = 0 9 9 9 5 5 8 (0.7746) cos (0.7746) + (0) cos (0) + (– 0.7746) cos (– 0.7746) 9 9 9

= 0.30756 + 0 – 0.30756 = 0 (g) Two Gauss points 1

1 (4  2s)ds s

I  4( 0.57735)  2(0.57735) 4(0.57735)  2(0.57735)  2.6755 Exact is 2.7051

Two Newton-Cotes Sampling Points 1 1 I  2( y0  y1 ) 2 2 ( 1) y0  4  2(1)  2.25 y1  4(1)  2(1)  2 1 1 I  2( (2.25)  (2)) 2 2  4.25

Three Gauss Points x1  x3  0.77460, x2  0,W1  W3  0.5555,W2  0.8888 (40.7746  2(0.7746))0.5555 (40  2(0))0.8888  (40.7746  2(0.7746))0.5555  2.7045 Three Newton-Cotes Sampling Points 1 4 1 I  2( y0  y1  y2 ) 6 6 6 1 y0  4  2(1)  2.25 y1  4(0)  0  1 y2  41  2(1)  2 1 4 1 I  2( (2.25)  (1)  (2)) 6 6 6  2.75

Comparing Results: In all cases, the Gaussian Quadrature Method is more accurate than the Newton-Cotes Method. Using 2 Gauss Points was more accurate than using 3 Newton-Cotes Sampling Points in all cases. This is as expected. The text says that “Gaussian Quadrature is more accurate with fewer sampling points.” Also, for N sampling points, the highest order polynomial the Newton-Cotes Method can evaluate exactly is N - 1, whereas for the Gaussian Quadrature Method, it is 2N - 1.

10.16

[k] = [B]T (s1, t1) [D] [B] (s1, t1) | [J ] (s1, t1) | h W1 W1 + [B]T (s2, t1) [D] [B] (s2, t1) | [J ] (s2, t1) | h W2 W1 + [B]T (s1, t2) [D] [B] (s1, t2) | [J ] (s1, t2) | h W1 W2 + [B]T (s2, t2) [D] [B] (s2, t2) | [J ] (s2, t2) | h W2 W2 where s1 = – 0.5773, s2 = 0.5773 441 © 2023 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in

t1 = – 0.5773, t2 = 0.5773 Using computer program (a)

ENTER THE GAUSS POINTS S AND T FOR POINT 1 –0.57735, –0.57735 ENTER THE WEIGHT FOR POINT 1 1.0 ENTER THE NODAL VALUES X AND Y FOR POINT 1 3.0, 2.0 ENTER THE GAUSS POINTS S AND T FOR POINT 2 0.57735, –0.57735 ENTER THE WEIGHT FOR POINT 2 1.0 ENTER THE NODAL VALUES X AND Y FOR POINT 2 5.0, 2.0 ENTER THE GAUSS POINTS S AND T FOR POINT 3 0.57735, 0.57735 ENTER THE WEIGHT FOR POINT 3 1.0 ENTER THE NODAL VALUES X AND Y FOR POINT 3 5.0, 4.0 ENTER THE GAUSS POINTS S AND T FOR POINT 4 –0.57734, 0.57734 ENTER THE WEIGHT FOR POINT 4 1.0 ENTER THE NODAL VALUES X AND Y FOR POINT 4 3.0, 4.0 ENTER THE VALUE FOR YOUNGS MODULUS 30000000.0 ENTER THE VALUE FOR POISSONS RATIO 0.25 ENTER THE VALUE FOR THE THICKNESS, h 1.0 442 © 2023 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in

THE GAUSS VALUES S AND T AND WEIGHTS ARE POINT S T WEIGHT 1 –5.773500E–001 –5.773500E–001 1.0000000 2 5.773500E–001 –5.773500E–001 1.0000000 3 5.773500E–001 5.773500E–001 1.0000000 4 –5.773400E–001 5.773400E–001 1.0000000 THE NODAL COORDINATE VALUES ARE NODE X Y 1 3.0000000 2.0000000 2 5.0000000 2.0000000 3 5.0000000 4.0000000 4 3.0000000 4.0000000 THE ELEMENT PARAMETERS ARE YOUNGS POISSON’S THICKNESS MODULUS RATIO 30000000.0000000 2.500000E–001 1.0000000 DO YOU WISH TO VIEW THE VALUES OF J (Y/N)? THE VALUES OF I J I ARE THE VALUE OF J 1 1.0000000 THE VALUE OF J 2 1.0000000 THE VALUE OF J 3 1.0000000 THE VALUE OF J 4 1.0000000 DO YOU WITH TO VIEW THE B MATRIX (Y/N)? THE B MATRIX VALUES ARE ([B] 3  8) –1.0566 E–1

– 3.943 E–1

1.0566 E–1 –1.0566 E–1 3.943 E–1 1.0566 E–1 3.943 E–1

0 0

–3.943 E–1 –1.0566 E–1 –1.0566 E–1 1.0566 E–1 0 –3.943 E–1 0 0 3.943E–1 1.0566 E–1 3.943 E–1 –3.943 E–1

} = one row of the 3  8 [B] DO YOU WISH TO VIEW THE D MATRIX (Y/N)? THE VALUES OF THE D MATRIX ARE 32000000.0000000 8000000.0000000 0.0000000

8000000.0000000 32000000.0000000 0.0000000

0.0000000 0.0000000 12000000.0000000

DO YOU WISH TO VIEW THE K MATRIX (Y/N)? THE K MATRIX VALUES ARE 14666660.0000000

5000015.0000000 443

–8666672.0000000

–1000005.0000000

5000015.0000000 –8666672.0000000 –1000005.0000000 –7333369.0000000 –4999981.0000000 1333383.0000000 999970.6000000

14666610.0000000 1000001.0000000 1333353.0000000 –4999991.0000000 –7333369.0000000 –1000025.0000000 –8666592.0000000

1000001.0000000 14666690.0000000 –5000011.0000000 1333353.0000000 –1000005.0000000 –7333369.0000000 5000015.0000000

1333353.0000000 –5000011.0000000 14666690.0000000 1000001.0000000 –8666672.0000000 5000015.0000000 –7333369.0000000

–7333369.0000000 –4999991.0000000 1333353.0000000 1000001.0000000 14666610.0000000 5000015.0000000 –8666592.0000000 –1000025.0000000

–4999981.0000000 –7333369.0000000 –1000005.0000000 –8666672.0000000 5000015.0000000 14666660.0000000 999970.6000000 1333383.0000000

1333383.0000000 –1000025.0000000 –7333369.0000000 5000015.0000000 –8666592.0000000 999970.6000000 14666580.0000000 –4999961.0000000

999970.6000000 –8666592.0000000 5000015.0000000 –7333369.0000000 –1000025.0000000 1333383.0000000 –4999961.0000000 14666580.0000000

(b)

THE K MATRIX VALUES ARE 8-1 Column

8-2

14990860.0000000 3483641.0000000 –11385300.0000000 –1626729.0000000 –4661870.0000000 –4309764.0000000 1056312.0000000 2452853.0000000

3483641.0000000 13670480.0000000 370145.5000000 –565783.1000000 –4327808.0000000 –5451907.0000000 474022.0000000 –7652789.0000000

8-3

8-4

–11385300.0000000 370145.6000000 19631590.0000000 –6267461.0000000 4403077.0000000 222705.5000000 –12649370.0000000 5674610.0000000

–1626729.0000000 –565783.1000000 –6267461.0000000 14127760.0000000 2237624.0000000 –5114752.0000000 5656566.0000000 –8447218.0000000

8-5

8-6

–4661870.0000000 –4327808.0000000

–4309764.0000000 –5451907.0000000 444

4403076.0000000 2237624.0000000 11571920.0000000 3807131.0000000 –11313130.0000000 –1716948.0000000

222705.4000000 –5114752.0000000 3807131.0000000 11105380.0000000 279927.3000000 –538717.6000000

8-7

8-8

1056312.0000000 474021.9000000 –12649370.0000000 5656566.0000000 –11313130.0000000 279927.4000000 22906180.0000000 –6410515.0000000

2452853.0000000 –7652789.0000000 5674610.0000000 –8447218.0000000 –1716948.0000000 –538717.6000000 –6410515.0000000 16638720.0000000

10.18 [B (s, t)] =

1 [B1] [B2] [B3] [B4] [B5] [B6] [B7] [B8] [J ]

N1, s =

1 1 (1 – t) (s + t + 1) – (1 – s) (1 – t) 4 4

N2, s =

1 1 (1 – t) (s – t – 1) + (1 + s) (1 – t) 4 4

N3, s =

1 1 (1 + t) (s + t – 1) – (1 + s) (1 + t) 4 4

N4, s =

1 1 (1 + t) (s – t + 1) – (1 – s) (1 + t) 4 4

N5, s = (t – 1)s N6, s =

1 (1 – t2) 2

N7, s = – (1 + t)s N8, s =

1 2 (t – 1) 2

N 1, t =

1 1 (1 – s) (s + t + 1) + (1 – s) (t – 1) 4 4

N 2, t =

1 1 (1 + s) (t + 1 – s) – (1 + s) (1 – t) 4 4

N 3, t =

1 1 (1 + s) (s + t – 1) + (1 + s) (1 + t) 4 4

N 4, t =

1 1 (1 – s) (– s + t – 1) + (1 – s) (1 + t) 4 4 445

N 5, t =

1 (1 + s) (s – 1) 2

N6, t = – (1 + s) t N 7, t =

1 (1 – s2) 2

N8, t = (s – 1) t |[J] | =

x y s t

y x s t

= [N1, s x1 + N2, s x2 + … + N8, s x8]  [N1, t y1 + N2, t y2 + … + N8, t y8] – [N1, s y1 + N2, s y2 + … + N8, s y8]  [N1, t x1 + N2, t x2 + … + N8, t x8] a Ni , s

[Bi] =

b Ni ,t 0

c Ni ,t

d Ni , s

0 c Ni ,t a Ni , s

d Ni , s b Ni ,t

where a=

y = N1, t y1 + N2, t y2 + … + N8, t y8 t

y = N1, s y1 + N2, s y2 + … + N8, s y8 s

x = N1, s x1 + N2, s x2 + … + N8, s x8 s

x = N1, t x1 + N2, t x2 + … + N8, t x8 t

10.21 The 2-pt rule works as we have a 2nd order in s for the integrand see Equation (10.6.19) and for integrand of order 2n – 1 = 2  2 – 1 = 3 we get exact solution. 10.22 Compare the Q4 and Q6 elements. What property makes the Q6 element better in modeling beam bending? What is the weakness of the Q6 element that is not inherent in the Q4 element? The Q4 element does not have complete polynomial displacement functions. The Q4 does not include the x2 and y2 terms (or s2 and t2 terms) and does not include the bubble modes (g terms in Eq. (10.5.27)). The Q6 includes the quadratic terms as bubble modes that describe the state of constant curvature and hence yields better answers for modeling beam bending type problems. 446 © 2023 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in

The Q4 element insures compatibility as there are four a’s in each global coordinate relation defining the element geometric shape to the local coordinates as seen by Eq. (10.2.2) or analogously the displacement functions u and v. Having four generalized unknown degrees of freedom for u and for v allows enforcement of displacement compatibility at each of the four nodes. The Q6 element has compatibility issues because the g1-g4 degrees of freedom are internal and not nodal degrees of freedom and are then not connected to other elements. There is then the possibility that the edges of two adjacent elements may have different curvatures, and thus the displacement field along a common edge may be incompatible. That is why the element is called incompatible. See also Figure 10-17. This incompatibility is thought to be a weakness of the Q6 element that is not inherent in the Q4 element. However, for beam bending type problems, the Q6 element yields better results. 10.23 Compare the Q4 and Q8 elements. What makes the Q8 a better element to model beam bending? The Q8 element uses higher order displacement functions (incomplete cubic as the x3 and y3 terms are missing). These are shown by Eq. (10.5.28) for the element geometry shape, and hence for the higher order shape functions as well. Using the higher order displacement functions yields more accurate results with fewer elements than the Q4 element in many cases and definitely for models using material properties that are constant and that have simple loading. The Q4 element uses incomplete quadratic displacement functions (the x2 and y2 terms are missing), as seen by Eq. (10.2.2) for the element geometric shape equations that are analogous to the form of the displacement functions.

Chapter 11 11.1 (a)

[B] =

1 6V

0 0

By Equations (11.2.4) to (11.2.8) 1 2 0 1 0 2 1 0 2 1 = – 1 0 0 = 0, 2 = 1 0 0 = 0, 3 = – 1 2 0 = 4 1 0 0 1 0 0 1 0 0 1 0 2 1 0 0 4 = 1 2 0 = – 4, 1 = 1 2 0 = 0 1 0 0 1 0 0 1 0 2 1 0 2 1 0 2 2 = – 1 2 0 = 4, 3 = 1 0 0 = 0, 4 = – 1 0 0 = – 4 1 0 0 1 0 0 1 2 0

1 0 2 1 0 0 1 0 0 1 = – 1 2 0 = 4, 2 = 1 2 0 , 3 = – 1 0 2 = 0 1 0 0 1 0 0 1 0 0

449

1 0 0 4 = 1 0 2 = – 4 1 2 0

1 1 6V = 1 1

0 0 2 0 2 0 1 0 2 1 0 2 0 2 0 2 2 1)03 1 2 0 = 8 = (1) (– 1) 2 0 0 +12 (– 2 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 4 0 0 0 0 0 0 4 0 0 0 0

1 0 0 4 0 0 0 0 0 [B] = 8 0 0 0 4 0 0 0 4 0 4 0 0 0 4 0 0 4 0 0 0 0 0 0 0

4 0

0 0

4 0

4 4 0

4 0 4 4

Problem 11-1: ‘B’ matrix for tetrahedral solid element (b) x1 = 1

y1 = 0

z1 = 0

x2 = 0

y2 = 0

z2 = 2

x3 = 3

y3 = 0

z3 = 0

x4 = 0

y4 = 2

z4 = 0

Geometry description (m)

1 =

x2 x3

y2 y3

z2 z3

1 y2 1 = – 1 y3 1 y4

z2 z3 z4

450

1 x2 1 = 1 x3 1 x4

z2 z3 z4

1 x2 1 = – 1 x3 1 x4

y2 y3 y4

x1 x3

y1 z1 y3 2z3=

x1 x3

y1 1 z1 y1 z1 1 y1 y3 1 z3 y3 z23= 1 y3 y 4 1 z4 y 4 z 4 1 y 4

3 =

x1 x2 x4

y1 y2 y4

z1 z2 z4

x1 4 = – x2 x3

y1 y2 y3 1 1 1 1 |

[B1] =

1 6

1 6V

1 y1 3 = – 1 y2 1 y4

z1 z2 z4

z1 z2 z3

1 y1  4 = 1 y2 1 y3

z1 z2 z3

x1 x2 x3

y1 y2 y3

z1 z2 z3

| 1|

0 0

| 1|

1 6V |

0 0

1 x1 2 = 1 x3 1 x4

1 x1 3 = 1 x2 1 x4

z1 z2 z4

1 x1 3 = – 1 x2 1 x4

y1 y2 y4

1 x1 4 = – 1 x2 1 x3

z1 z2 z3

1 x1  4 = 1 x2 1 x3

y1 y2 y3

2 |

[B2] =

y1 y3

0 3|

4 |

0 |

[B4] =

0 0

2 |

| 2| 1 0 0  | 2| | 2| 6V 0 | | | 0 2 2| | 2| 0 | 2| |

0 | 3|

0 |

| 3| | 0 | 3| 0 |

z1 z3

V = 1.333

| 1| | 1| 0 0 | 1| | 1|

[B3] =

z11 x1 z1 1 x1 z13 x32 = z3 – 1 x3 z14 x4 z4 1 x4

0 0

4 |

| 4| 1 0 0 6V | 4 | | 4 | 0 | | | 0 4 4|

[B] = augment ([B1], [B2], [B3], [B4]) [B] = 0.5 0 0

0 0.75 0

0 0 0.75

0 0 0

0 0 0.5

0.5 0 0

0 0.25 0

0 0 0.25

0 0 0

0 0.5 0

0 0 0

0.75 0 0.75

0.5 0.75 0

0 0.75 0.5

0 0 0.5

0 0.5 0

0 0 0

0.25 0 0.25

0.5 0.25 0

0 0.25 0.5

0.5 0 0

0 0 0

0 0.5 0

451

11.2(a) Use Equation (11.2.18) and substitute [B] from 11.1 (a) and [D] from Equation (11.1.5) into Equation (11.2.18)  [k] = [B]T [D] [B] V 1

[B] =

0.5

–0.5

0.5

–0.5

0.5

–0.5

0.5

–0.5

0.5

–0.5

0.5 v=

0 0 0 E = 30  106

0.5

–0.5

1– v

0 1 – 2v 2

1 – 2v 2

0.3

[D] =

E (1 v ) (1 – 2v )

[k] = [B]T [D] [B] V 1

[k] =

1 x 3

3.846

2 0

3 0

4 0

5 0

6 0

7 0

8 0

3.846

–3.846

3.846

–3.846

13.462

3.846

– 3.846

5.769

13.462

–5.769

5.769

13.462

–5.769

3.846

–3.846

3.846

7.692

–3.846

–7.692

1 0

–3.846

–5.769

–3.846

–5.769

13.462

–3.846

21.154

9.615

1 1

–3.846

–5.769

–3.846

13.462

–3.846

3.846

21.154

9.615

1 2

–3.846

13.462

–7.692

3.846

21.154

 106 in.

(b) Evaluate the stiffness matrix for the element shown 452

[D] =

E (1 v ) (1 – 2v )

1– v

0 1 – 2v 2

1 – 2v 2

[k] = [B]T [D] [B] V

453

454

{} = [B] {d}

11.3 (a)

0 0 0 0 0 0 0 4 0 4 0 0 0 0 4 0 0 4 0 0 0

4 0 0 0 0 0 0 0 4

x y z xy



0.005 0.0 0.0 0.001 0.0 0.001 0.005 0.0 0.0 0.001 0.0 0.005

0 0 1 0 = 8 0

0 4 0 0 0 0 0 0 0 0 0 0 0 0 4 0 4 0 0 0

0 0 4

4 0 4 4

0 4

0 0 4 0 4 4

0.003 0.0 0.0025 in. 0.001 in. –0.002 0.0005

{} = [D] {} x y z xy yz zx

0.7 0.3 0.3 0 0 0   0.003   25.97       0 0   0.0   2.88   0.3 0.7 0.3 0  0.3 0.3 0.7 0 10  103 0 0  0.0025 =  16.33  ksi =      (1  0.3)(1  2(0.3))  0 0 0 0.2 0 0   0.001   3.83  0 0 0 0 0.2 0   –0.002    7.70       0 0 0 0 0 0.2   0.0005  1.92 

(b)

455

Nodal displacements (in.) u1 = 0.005

v1 = 0

w1 = 0

u2 = 0.001

v2 = 0

w2 = 0.001

u3 = 0.005

v3 = 0

w3 = 0

u4 = – 0.001

v4 = 0

w4 = 0.005

u1 v1 w1

u2 {d2} = v2 w2

{d1} =

u3 {d3} = v3 w3

u4 {d4} = v4 w4

{d} = stack ({d1}, {d2}, {d3}, {d4}) Material properties E = 10  106

v = 0.3

Element strain matrix ([B] from P11.b) {} = [B] {d} 0

=

0 5 10 4

in. 3 10 3 in.

2.5 10 3 2 10 3

Determine constitutive matrix

[D] =

E (1 v) (1 2v)

1 v v v v 1 v v v v 1 v 0

0 0 0

1 2v 2

Determine element stresses {} = [D]{}

456

 2.885  103    3  2.885  10   3  6.730  10  {} =   (psi) 4 –1.154  10   3  9.617  10   –7.693  103   

11.4 The strains and stress are constant in the 4-noded tetrahedral element. 11.5 Use Equation (11.2.10) for the shape functions N1 – N4. Substitute [N]3 12 from Equation (11.2.9) and {X} from Equation (11.2.20a) into Equation (11.2.19), {fb}12  1 = Xb Xb V T [ N ]12 3 Yb to show at node i, { fbi } Yb . V 4 Zb Zb 11.6 (a)

1 25 0 1 = – 1 0 25 = – 625, 1 0 0

1 0 0 2 = 1 0 25 = 0 1 0 0

1 0 0 3 = – 1 25 0 = 0, 1 0 0

1 0 0 4 = 1 25 0 = 625 1 0 25

1 25 0 1 = 1 25 25 = – 375, 1 40 0

1 10 0 2 = – 1 0 25 = 750 1 40 0

457

1 10 0 4 = – 1 25 0 = – 375 1 25 25

1 10 0 3 = 1 25 0 = 0, 1 40 0

1 25 25 1 10 0 0 = – 375, 2 = 1 25 0 = 0 1 40 0 1 40 0

1 = – 1 25

1 10 0 1 10 0 3 = – 1 25 25 = 750, 4 = 1 25 25 = – 375 1 40 0 1 25 0 1 1 6V = 1 1

10 0 0 25 25 0 25 25 0 2 25 0 25 = 1 (– 1) 25 0 25 40 0 0 40 0 0

1 25 0 1 25 0 0 1) 253 1 0 25 = 18750 1+ 10(– 1 0 0 1 0 0

[B] =

1 18750

625 0 0

0 375 0

0 0 375

0 0 0

0 750 0

0 0 0

0 0 750

625 0 0

0 375 0

0 0 375

375 0

625 375

0 750 375 0

0 0

0 750

0 0

0 750

0 0

375 0

625 375

0 375

375

625

750

375

625

(b)

458

x1 = 4

x2 = 10

x3 = 10

x4 = 12

y1 = 2

y2 = 7

y3 = 2

y4 = 2

z1 = 0

z2 = 0

z3 = 5

z4 = 0

1 1 1 v= 6 1 1

x1 x2 x3

y1 y2 y3

z1 z2 z3

1 y2 1 = – 1 y3 1 y4

z2 z3

z1 z3

1 y1 2 = 1 y3 1 y4

1 x1 2 = – 1 x3 1 x4

y1 y2 y4

z1 z2 z4

1 y1  3 = – 1 y2 1 y4

z1 z2 z4

1 x1 3 = 1 x2 1 x4

y1 y2 y3

z1 z2 z3

1 y1  4 = 1 y2 1 y3

z1 z2 z3

1 x1 4 = – 1 x2 1 x3

x2 x3

y2 y3

z2 z3

x1 2 = – x3

y1 y3

z1 z3

x1 x2 x4

x1 4 = – x2 x3

1 =

3 =

[B1] =

1 6V

0 0

1 x2 1 = 1 x3 1 x4

z2 z3 z4

1 x2 1 = – 1 x3 1 x4

y2 y3

1 x1 2 = 1 x3 1 x4

y1 y3 y4

1 x1 3 = – 1 x2 1 x4

y1 y2 y4

1 x1 4 = 1 x2 1 x3

y1 y2 y3

z1 z3 z4 z1 z2 z4

z1 z2 z3

0 0

1 1 [B2] = 6V 0

1 [B3] = 6V

[B4] =

1 6v

0 0

[B] = ([B1] [B2] [B3] [B4])

459

[B] =

0.125 0 0

0 0.05 0

0 0 0.05

0 0 0

0 0.2 0

0 0 0

0 0 0.2

0.125 0 0

0 0.15 0

0 0 0.15

0.05 0

0.125 0.05

0 0.05

0.2 0

0 0

0 0.2

0 0

0 0.2

0 0

0.15 0.125 0 0.15

0 0.15

0.05

0.125

0.2

0.15

0.125

11.7 (a)  563568.  200321.   200321.   96153.8  240385.  0. [k]  106  96153.8  0.   240385   371261   40064.1  40064.1

200321.

96153.8 240385.

349893.

120192.

160256. 336538. 96153.8

96153.8

240385. 371261.

40064.1

96153.8

144231

40064.1

82799.1

144231. 96153.8 160256. 96153.8

120192.

349893.

336538

40064.1

160256.

192308.

96153.8

160256.

336538

144231.

673077.

288462.

240385.

336538. 96153.8

96153.8 96153.8

192308.

160256.

192308.

96153.8

96153.8 96153.8

192308.

96153.8

144231. 336538.

288462.

673077.

240385.

144231.

40064.1 40064.1 96153.8

240385.

96153.8

240385.

563568.

200321.

82799.1

120192.

160256.

336538. 96153.8

96153.8 144231. 200321.

349393.

120192.

82799.1

144231. 96153.8

160256.

96153.8 336538. 200321.

120192.

11.8 (a) {} = [B] {d} (see [B] from Problem 11.6 above) u1 = 0

v1 = 0

w1 = 0

u2 = 0.01

v2 = 0.02

w2 = 0.01

u3 = 0.02

v3 = 0.01

w3 = 0.005

u4 = 0

v4 = 0.01

w4 = 0.01

ln[B1] = d = {{u1}, {v1}, {w1}, {u2}, {v2}, {w2}, {u3}, {v3}, {w3}, {u4}, {v4}, {w4}}; ln[B2] = ε = B.d; ln[B3] = MatrixForm [ε]

0.  x       0.0006   y     0. mm   z      0.000733333 mm  xy   yz   0.0004       zx   0.00113333  460

40064.1  120192.  82799.1   0.  144231.  96153.8  160256.   96153.8 336538.  200321.  120192.  349393. 

ln[B4] = σ =Dmatrix.ε; ln[B5] = MatrixForm [σ]

 x   34.6154    80.7692   y    34.6154    z    MPa     xy   28.2051  yz  15.3846       43.5397   zx  0.0 0.0 0.0 x

0.01

0.02

= [B]

0.01

0.00 0.0006 0.0

0.02

0.01

0.000733 0.0004

0.005

0.00113

0.0 0.01 0.01 x y z xy yz zx

0.7 0.3 0.3 0 0 0  0     0 0   0.0006   0.3 0.7 0.3 0  0.3 0.3 0.7 0 100  109 0 0  0  =    (1  0.3)(1  2(0.3))  0 0 0 0.2 0 0  0.00073  0 0 0 0 0.2 0   0.0004      0 0 0 0 0 0.2  0.00113 34.6    80.8  34.6  =   MPa  28.2  15.4     43.6 

11.8 (b) u1 = 0

v1 = 0

w1 = 0

u2 = 0.00001

v2 = 0.00002

w2 = 0.00001

461

u3 = 0.00002

v3 = 0.00001

w3 = 0.000005

u4 = 0

v4 = 0.00001

w4 = 0.00001

[D] =

E (1 + v) (1 − 2v)

1 v v v v 1 v v v v 1 v 0

[B] =

0 0 – 0.125 0 1 0 2 3 – 0.05 0 4 5 – 0.05

1 2 0 0 – 0.05 0 0 – 0.05 0.125 0 – 0.05 – 0.05 0 – 0.125

0 0

3 0 0 0 0.2 0 0

4 0 0.2 0 0 0 0

0 0 0

1 2v 2

5 0 0 0 0 0.2 0

6 0 0 0 0 0 0.2

7 0 0 0 0 0.2 0

8 0 0 0.2 0 0 0

9 0.125 0 0 0.15 0 0.15

10 0 – 0.15 0 0.125 0.15 0

u1 v1 w1

disp =

= [D] B] {disp}

u4 v4 w4 x y z xy yz zx

 115.4   307.7     76.9  =   MPa  125.0   38.5     201.9 

11.9 462

u = a1 + a2 x + a3 y + a4 z + a5 xy + a6 xz + a7 yz + a8 x2 + a9 y2 + a10 z2 Similar expressions for v and w.

Figure P11-9

11.10 Loads must be in the y-z plane (in the plane of the plane elements). 11.11

Using Equation (11.3.3) and Figure 11.5 Ni =

(1 ssi )(1 tti )(1 z zi ) 8

N1 =

(1 – s)(1 – t )(1 z ) , (s1 = – 1, t1 = – 1, z1 = 1) 8

N2 =

(1 – s)(1 – t )(1 – z ) , (s2 = – 1, t2 = – 1, z2 = – 1) 8

N3 =

(1 – s)(1 t )(1 – z ) , (s3 = – 1, t3 = 1, z3 = – 1) 8

N4 =

(1 – s)(1 t )(1 z ) , (s4 = – 1, t4 = 1, z4 = 1) 8

N5 =

(1 s)(1 – t )(1 z ) , (s5 = 1, t5 = – 1, z5 = 1) 8

N6 =

(1 s)(1 t )(1 – z ) , (s6 = 1, t6 = – 1, z6 = – 1) 8

N7 =

(1 + s)(1 − t )(1 – z ) , (s7 = 1, t7 = 1, z7 = – 1) 8

N8 =

(1 s)(1 t )(1 z ) , (s8 = 1, t8 = 1, z8 = 1) 8

11.12 Quadratic hexahedral element (see Figure 11.6) By Equation (11.3.11) 463

Ni =

(1 ssi )(1 tti )(1 z zi ) (ssi + tti + zzi – 2) 8

Node 1 s1 = – 1, t1 = – 1, z1 = 1 N1 =

(1 s )(1 t )(1 z ) (– s – t + z – 2) 8

Node 2 s2 = – 1, t2 = – 1, z2 = – 1 N2 =

(1 s)(1 t )(1 z ) (– s – t – z – 2) 8

Node 3 s3 = – 1, t3 = 1, z3 = – 1 N3 =

(1 s)(1 t )(1 z ) (– s + t – z – 2) 8

Node 4 s4 = – 1, t4 = 1, z4 = 1 N4 =

(1 s)(1 t )(1 z ) (– s + t + z – 2) 8

Node 5 s5 = 1, t5 = – 1, z5 = 1 N5 =

(1 s)(1 t )(1 z ) (s – t + z – 2) 8

Node 6 s6 = 1, t6 = – 1, z6 = – 1 N6 =

(1 s )(1 t )(1 z ) (s – t – z – 2) 8

Node 7 s7 = 1, t7 = 1, z7 = – 1 N7 =

(1 s)(1 t )(1 z ) (s + t – z – 2) 8

Node 8 = σ1 464

s8 = 1, t8 = 1, z8 = 1 N8 =

(1 s)(1 t )(1 z ) (s + t + z – 2) 8

11.14

Outer free face corner node deflections (z-direction) Pt A = – 0.01461 m, Pt B = – 0.01461 m, Pt C = – 0.01445 m, Pt D = – 0.01445 m Mechanics of materials,  =

PL3 (0.2) (0.06) 3 = – 0.0147 m, where I = = 3.6  10–6 m4 3 EI 12

N m2 The corner node answers from computer program Autodesk. Note the classical mechanics of materials solution gives the maximum deflection for a load applied through the centroid not offset.

L = 2.0 m, P = 4000 N, E = 200  109

465

11.15

466

11.16

Figure 1 von Mises Stress Using a factor of safety against yielding of 1.75 and a yield strength of Sy = 66,700 psi, the largest allowable von Mises stress is 38,114 psi. Figure 1 above shows the highest von Mises stress is 34,377 located at the inside of the curved part of the hook. Therefore the load of 22,500 lbs is acceptable. The largest displacement of 0.0965 inches is located at the free tip end of the hook.

11.17

467

Allowable stress is 50 ksi / (FOS of 1.5) = 33.33 ksi.. A thickness of 0.6 inches is needed due to the allowable stress and the displacement is minimal so the design is good. 11.18 Determine the thickness of the device such that the maximum deflection is 0.1 in. vertically. Thickness chosen was 0.625 inches, giving a maximum displacement of 0.03 inches and a maximum yield stress of 15,000 psi, less than the acceptable value of 18,000 psi (36,000/F.S.)

Figure 1 Displacement magnitude (0.625 in. thick) 468

Figure 2 von Mises Stress Distribution (0.625 in. thick) 11.19 Model Variables Variable

Value

Material

1035 quenched & tempered steel

Modulus of Elasticity

200 MPa

Force

150 N

Yield Strength

615 MPa

Maximum von Mises Stress

758 MPa

Maximum Displacement

4.13 mm

469

Autodesk Results

Figure 1 von Mises Stress (MPa)

Figure 2 Displacement (mm) 11.20

470

Maximum principal stress is 10080 psi. 11.21 von Mises Stress

The yield strength of AISI 4130 is approximately 360.69 MPa (from eFunda). Converting the maximum von Mises stress of 218.8 N 2 into MPa gives approximately 218.8 MPa. This is mm

over 100 MPa below the yield strength, so this will not fail due to static loading.

471

11.22

Figure 1 von Mises Stress of part. With yield strength of 6,000 psi and a max von Mises stress of 5250 psi, it is getting close to failing due to the total weight of the entire car placed on one wheel. Under normal operation conditions the actual weight placed on a front wheel is less than one quarter of the entire car weight. 11.23 With 6,282 elements Algor calculates higher stresses. Figure 1 shows the von Mises stress for analysis with 6282 elements.

472

Figure 1 The yield strength of the material is 53,700 psi and the maximum von Mises stress is 24,530 psi so the hitch has a factor of safety of 2.18 when analyzed with 6282 elements. 11.24

Maximum deflection = 0.01251 inches

473

Maximum von Mises stress = 5193.678 psi 11.25 The von Mises plot of the loader under loading is shown below in Figure 1. I am looking for the load that caused the loader to break. For this to happen, there is no factor of safety, However, based on the von Mises plot, you can see that much of the loader has a near infinite safety factor.

Figure 1 von Mises Plot I found that a load of 2200 pounds applied on three sides of the right lower arm (left in the above figure) in a counter clockwise direction caused the loader to fail. This loader put the loader under 44251 psi, which is just over the yield strength. It appears it failed roughly six to eight inches above the end of the arm. This is very near where the loader broke this spring. 11.26

474

Figure 1 von Mises plot due to 250 lb rider (maximum stress is 534.3 MPa)

11.28

(Largest von Mises stress is 186 N/mm2) located inside surface of hole.

475

476

Chapter 12 Solve these problems using the plate element from a computer program. 12.1 A square steel plate of dimensions 20 in. by 20 in. with thickness of 0.1 is clamped all around. The plate is subjected to a uniformly distributed loading of 1 lb2 . Using a 2 by 2 in.

mesh and then a 4 by 4 mesh, determine the maximum deflection and maximum stress in the plate. Compare the finite element solution to the classical one in [1].

Figure P12–1 AISI 4130 Steel

Plate bending 20  20  0.1 4130 steel plate Clamped all the way around with a 1 psi load. Mesh size

Minimum stress

Maximum stress

Maximum deflection

22

3440.01 psi

3573.76 psi

0.0785 in.

44 668.06 psi 7046.06 psi Analytical solution: (A.C. Ugural, Plates and Shells McGraw Hill) wmax =

0.06788 in.

Pa 4 (1 psi) (20 ) 4 (0.00126) = (0.00126) = 0.07339 in. D 2.7470 103

where D =

max =

Et 3 12 (1 – v 2 ) 6 M max = t2



(30  106 ) (0.1)3 12 (1 – 0.32 )



lb 0.03  106 = 2.747  103 in. (0.91) 12

(0.0513 p0 a 2 ) t2

= 12312 psi 12.2 An L-shaped plate with thickness 0.1 in. is made of ASTM A-36 steel. Determine the deflection under the load and the maximum principal stress and its location using the plate element. Then model the plate as a grid with two beam elements with each beam having the stiffness of each L-portion of the plate and compare your answer.

Figure P12–2

12.3 A square simply supported 20 in. by 20 in. steel plate with thickness 0.15 in. has a round hole of 4 in. diameter drilled through its center. The plate is uniformly loaded with a load of 5 lbf2 . Determine the maximum principal stress in the plate. in.

Figure P12–3

From the Autodesk analysis, the maximum principal stress was determined to be 24,728 psi. The precision of the von Mises stress was very close to 0.1, therefore, the results were deemed feasible. Stress (psi)

Maximum Value

Displacement (in.)

von Mises

Maximum Principal

21,978

24,728

Magnitude

–0.11699

12.4 A C-channel section structural steel beam of 2 in. wide flanges, 3 in. depth and thickness of both flanges and web of 0.25 in. is loaded as shown with 100 lb acting in the y direction on the free end. Determine the free end deflection and angle of twist. Now move the load in the z direction until the rotation (angle of twist) becomes zero. This distance is called the shear center (the location where the force can be placed so that the cross section will bend but not twist).

Figure P12–4

12.5 For the simply supported structural steel W 14  61 wide flange beam shown, compare the plate element model results with the classical beam bending results for deflection and bending stress. The beam is subjected to a central vertical load of 22 kip. The cross-sectional area is 17.9 in.2, depth is 13.89 in., flange width is 9.995 in., flange thickness of 0.645 in., web thickness of 0.375 in., and moment of inertia about the strong axis of 640 in.4

Figure P12–5

=

PL3 48EI

(22 K )(240 )3 48.29 103 ksi × 640 in.4

= – 0.34 in.

12.6 For the structural steel plate structure shown, determine the maximum principal stress and its location. If the stresses are unacceptably high, recommend any design changes. The initial thickness of each plate is 0.25 in. The left and right edges are simply supported. The load is a uniformly applied pressure of 15 lb2 over the top plate. in.

Figure: P12–6

Maximum principal stress of 56420 psi occurs towards the bottom of the right vertical plate on the underside. Maximum displacement is 0.2303 in. at center of top plate. ASTM A-36 steel has the following strength properties.  St = 58 to 80 ksi 484 © 2023 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in

 Sy = 36 ksi minimum Therefore, the stress levels in the structure would lead to failure with the 0.25 in. thick plate using the MDET (maximum distortional energy theory). Try 0.375 in. thick plate. 12.7 Design a steel box structure 4 ft wide by 8 ft long made of plates to be used to protect construction workers while working in a trench. That is, determine a recommended thickness of each plate. The depth of the structure must be 8 ft. Assume the loading is from a side load acting along the long sides due to a wet soil (density of 62.4 lb3 ) and varies linearly with the ft

depth. The allowable deflection of the plate type structure is 1 in. and the allowable stress is 20 ksi.

Figure P12–7 Solution method I first attempted to solve the problem by examining the long side and short side of the box individually, rather than as a welded assembly. It was felt that once believable results were obtained from modeling each side as an individual plate, the model could be expanded to include the entire box. However, Autodesk appears to have a glitch so that its ‘surface hydrostatic pressure’ feature does not work. Therefore, I was unable to easily apply a linearly varying pressure along the surface of a plate. To get around this problem, I tried to break the side plate of the box into 4 vertically stacked sections (different Autodesk parts) and then applying a different pressure to the surface of each section so that it approximated the linear pressure distribution of the soil. However, Autodesk failed to recognize the applied surface pressure on 3 of the 4 sections. So this model was deficient also. To solve the problem, I was forced to apply loads directly to the nodes. To facilitate the selection of the nodes and to make the nodal forces regular and repeatable, the plate was manually meshed into a convenient rectangular pattern. Four different nodal loads were used to approximate the linear pressure distribution. Results For the short side of the box (48 wide  96 deep), it was found that a 38 thick plate was as thin as could be used and still meet the given design criteria. Using simply supported boundary conditions, the maximum von Mises stress was 19.9 ksi and the maximum deflection was 0.76. The maximum stress was located at the bottom corners, which is typically an area of high stress for rectangular plates. These results were than compared to equations found in Roark’s ‘Formulas for Stress and Strain’. From Roark, the maximum stress was 18.2 ksi and the maximum deflection was 0.68 in. The same analysis was performed for the long side of the box (96 square). It was found that a 0.75 plate was right at the margins of acceptability for deflection. The Autodesk and hand calculations for both sides of the box are summarized below. 485 © 2023 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in

Long side of box ( 43

Hand Calc. von Mises (ksi) 11.36

Deflection (m) 0.93

19.9

18.18

0.68

thk  96 sq)

Short side of box ( 83

Autodesk model von Mises (ksi) Deflection (in.) 14.5 1.06 0.76

thk  48  96 dp)

Comments For both sides, the Autodesk model predicted higher stress and higher deflection than the hand calculations. I suspect the difference in results can be attributed to two sources. One, the stepwise variation of pressure that I was forced to use in the Autodesk model to approximate the hydrostatic pressure. The other is due to the coarseness of the mass. Nodal loads to approximate hyprostatic pressure variation as function of depth q0 =  a = Maximum pressure = 62.4

lb 1 ft 2 8ft 12 in. ft 3

= 3.467 psi soil pressure at 8 depth Assume linear pressure profile x = depth q(x) = q0 – q0 = q0 (1 –

x a x ) a

For Step 1 q (x =

7 1 a) = q0 = 3.04 psi 8 8

For Step 2 q (x =

3 5 a) = q0 = 2.17 psi 8 8

For Step 3 q (x =

5 3 a) = q0 = 1.30 psi 8 8

For Step 4 q (x =

7 1 a) = q0 = 0.43 psi 8 8

For a g  g mesh, the area supported by a interior node is, AI = g2 For a g  g mesh, the area supported by a perimeter node is AP =

g2 2

Let g = 4  nodal force for step 1 interior node = (3.04) (4)2 = Let g = 4  nodal force for step 1 perimeter node = (3.04)

48.64 lb node

lb 42 = 24.32 2 node

etc. …

Manual mesh Stepwise approximation of soil pressure Simple support boundary conditions Short side 7 48 wide  96 Deep  thick 8 4 square mesh

Manual mesh Stepwise approximation of soil pressure Simple support boundary conditions Short side of box 3 48 wide  96 deep  thick 8 4 square mesh

Manual 4 square mesh Stepwise approximation of soil pressure Simple boundary conditions Long side (96 square) 0.75 thick

12.8 Determine the maximum deflection and maximum principal stress of the circular plate shown in Figure P12–8. The plate is subjected to a uniform pressure p = 50 kPa and fixed along its outer edge. Let E = 200 GPa, v = 0.3, radius r = 500 mm, and thickness t = 20 mm. Analytical solution wmax =

Pr 4 (50,000) (0.5)4 = = 0.333 m 64 D 64 (0.1465  106 )

D = 0.0916 E t 3 = 0.0916 (200  109) (0.020)3 = 0.1465  106 N  m

Figure P12–8

wmax = 0.331 m Compares closely to analytical solution

12.9 Determine the maximum deflection and maximum stress for the plate shown. The plate is fixed along three sides. A uniform pressure of 100 kPa is applied to the surface. The plate is made of steel with E = 200 GPa, v = 0.3, and t = 10 mm and sides equal to a = 0.75 m and b = 1 m.

Figure P12–9

12.11 A square steel plate 2 m by 2 m and 10 mm thick at the bottom of a tank must support salt water at a height of 3 m, as shown in Figure P12–11. Assume the plate to be built in (fixed all around). The plate allowable stress is 100 MPa. Let E = 200 GPa, v = 0.3 for the steel properties. The weight density of salt water is 10.054 kN3 . Determine the maximum principal m

stress in the plate and compare to the yield strength.

Figure P12–11

Ftot = (V) () = (3 m ) (2 m ) (2 m ) (10.054 E 3 P=

N m3

) = 120,648 N

120, 648 N N F = = 30,162 2 (2m)(2m) Aplate m

Find Maximum principal stress = ? 347.43 MN2 m

Safe or not safe = ? Since stress allowable = 100 MPa < actual stress 347.43 MPa Tank plate is not safe. 12.12 A stockroom floor carries a uniform load of p = 80 lb2 over half the floor as shown in Figure ft

P12–12. The floor has opposite edges clamped and remaining edges and mid-span simply supported. The dimensions are 40 ft by 20 ft. The floor thickness is 4 in. The floor is made of reinforced concrete with E = 3  106 psi and v = 0.25. Determine the maximum deflection and maximum principal stress in the floor. 494 © 2023 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in

Material used: AISI 4130 Steel Uniform load of 0.1 psi applied to top surface.

12.15 Model variables Variable

Value

Material

1010 cold rolled

Modulus of Elasticity

29  106 psi

Maximum von Mises Stress

1351 psi

Maximum Displacement

0.00565 in.

Figure 6 Displacement (mm) 12.16

Figure 2 The Boundary Conditions on the bucket. The results of the analysis are shown below in Figure 3.

Figure 3 The von Mises stress in psi for the bucket plate analysis.

Chapter 13 13.1

Element [K] ’s W

(0.1m 2 )(200 m  °C )  1 1  10 10  W [k ] = 1 1   10 10  °C 2m     (1)

[k(2)] =

[k(3)] =

(0.1m2 )(100 mW 1  10 10   °C )  1    1m 1 1  10 10  (0.1m 2 )(50 mW  °C ) 0.5m

q* = 0 

 1 1  10 10  =    1 1  10 10 

Q=0

{f (1)} = {f (2)} = {f (3)} = 0

Assemble equations 0   t1   F1   10 10 0  10 20 10 0  t   0     2  =    0 10 20 10  t3   0    0 10 10  t4   F4   0 Boundary conditions t1 = 100°C, t2 = 300°C



0 1 0  0 20 10   0 10 20  0 0 0

0   t1   100  0  t2  1000    =   0  t3  3000  1  t4   300 

Solving t2 = 166.7°C t3 = 233.3°C

499

13.2

(1)

(2)

[k ] = [ k ] =

 (2)2 6 h Btu in.  F  1

 1 

3 in.

1  25.13 25.13 Btu  1   25.13 25.13  h  °F

Now there is convection through right end Btu  25.13 25.13    (2)2 0 0   1 [k(3)] =      2  25.13 25.13   h  in.  ° F  0 1 

 25.13 25.13 [k(3)] =    25.13 37.70 

Now q* = 0, Q = 0  { f (1) } = { f (2) } = 0 0   0  0 { f (3) } = h T A   = (1) (0°F)  22   =    1  1 0 

Assemble equations and boundary condition t1 = 200°F 0 0  t1  200F  F1x   25.13 25.13     0  t2 50.26 25.13 0       =    t3 50.26 25.13    0   Symmetry  t4 37.70    0    Solve Equations (2-4) of above 0  t2  5026  50.26 25.13       50.26 25.13 t3  =  0     0  Symmetry 37.70  t4   

t2 = 160°F, t3 = 120°F, t4 = 80°F 13.3

500

Btu

(2 in. ) (3 h  in. F ) AK xx 1 Btu = = 5 h  °F L 30 in.

hPL = 0, hA = 0, h T P L = 0 6

h= 0 

[k(1)] =

1  1 1 = [k(2)] 5  1 1  L

L  30

{ f q(1) } =  q* [N]T dx = 

1  X  ( 0 .1 x )  X L  dx  L 

 L  Btu 15  = 0.1  62  =   L 30   3  h 2

Then

15  45  60  { f q(2) } =   =   30  45 75 Solve {F} = [K] {t} Heat flow 1 0  t1  50F 15  F1  1 1      1 1  1 1  t2 30  60  =    5  75    1 1   t3    0 

(A)

Solve the 2nd and 3rd equations 90 + 10 =

1 (2 t2 – t3) 5

(1)

501

+ 75 =

1 (– t2 + t3) 5

(2)

--------------------------------------Adding (1) and (2) 175 =

2 1 t2 – t2 5 5

175 =

1 t2 5 1.5 m

t2 = 875°F Back-substitution into (1) yields t3 = 1750 – 500 t3 = 1250 °F Solve for F1 using Equation (A) above 15 + F1 =

1 (t1 – t2) 5 1 (50 – 875) 5

F1 = – 15 +

Btu = heat flow out left end. h

= – 180

13.4

T ∞ = 100°F

Q = 5000

Btu h  ft 3

[k(1)] = [k(2)] =

A K x x  1 1 L  1 1 

  121  12  1 



12 in. ft



 1 

1  1 1 Btu =    1 1 1  h  °F

0 0  [k(3)] = [k(1)] + hA   0 1  502

 

1 2 0 0   1 1 =   + 100   12 0 1   1 1 

1  1 [k(3)] =     1 1.694  QAL 1 (5000)  121  121 1    2 1 2 1 2

{ f (1) } =

1.447  =    1.447  { f (2) } = { f (1) }

0 1.447  { f (3) } = { f (1) } + { f hend } =    + h T A   1 1.447 

 

1 2 0 1.447  =     + (100) (100)  12 1 1.447 

1.447   0  { f (3) } =      1.447  69.4  Assemble global equations 0   t1  1 1 0  2 1 0  t2    =   4.55  2 1  t3     4.55 4.55       t 1.694    4   4.55 4.55   4.55  109.05 Solving simultaneously t1 = 75.5°F,

t2 = 74.1°F,

t3 = 69.7°F,

t4 = 62.5°F,

13.5

(0.1m )(6 m  C ) AK xx W = =6 °C L 0.1m

[k(1)] = [k(2)] = [k(3)] =

AK xx  1 1  1 1 W =6     L 1 1  1 1  °C

[k(4)] = [k(1)] also { f (1) } = { f (2) } = { f (3) } = 0 as Q = 0, q* = 0

503

W 0 0  { f (4) } = qA   =  500 2  (0.1m 2 )   m  1  1 

0 = 50   W 1  Assemble global equations  6 6 0 0 0  t1  50C   F1x    0   12 6 0 0   t2       12 6 0   t3   =  0      0  12 6   t4      6   t5  Symmetry   50 

(1)

Now t1 = 50°C into Equation (1) 1 0 0 0 0   t1  50o  F1x   0 12 6 0 0  t      2   300  0 6 12 6 0  t3  =  0       0 0 0 6 12 6  t4    0 0 0 6 6  t5   50 

(2)

Solving Equations (2-5) of Equation (2) t2 = 58.3°C, t3 = 66.7°C t4 = 75°C, t5 = 83.3°C 13.6

Area = A

(Can use unit A)

[k(1)] =

A(5)  1 1  50 50  W = (0.1m) 1 1  50 50  °C

[k(2)] =

A(0.8)  1 1  10 10  = (0.08 m)  1 1   10 10 

[k(3)] =

A(1.5)  1 1  15 15 = (0.10 m)  1 1   15 15  504

{f }’s = 0 Assemble global equations with t1 = 500°C and t4 = 100°C

50 0 0  t1  500 C  0 50   0 10 50  10 0  t2      = A     t 0   10 15 15   3     Symmetry    0 15 t  100  C   4 Solving the 2nd and 3rd equations above t2 = 457.1°C, t3 = 242.86°C q(3) = – Kxx q(3) = 2145

(100  242.86)C 0.10 m W m2

13.7

(1) AK xx (0.1m 2 )5 = =5 L 0.1m

[k(1)] =

(1) AK xx  1 1  1 1 =5     L 1 1  1 1 

 1 1  1 1 (3) [k(2)] = 10   , [k ] = 15 1 1   1 1     Assemble global equations 0   t1   F1x   5 5 0  15 10 0  t2   0     =   25 15 t3   0     15  t4   F4 x  Symmetry

(1)

Boundary conditions t1 = 100°C, t4 = 500°C  Equation (1) becomes 0 1 0  0 15 10  25   Symmetry 

0  t1   100  0 t2   5(100)    =   0 t3  15(500)   1  t4   500 

(2)

Solving the 2nd and 3rd of Equations (2) 505

t2 = 318.2°C, t3 = 427.3°C 13.8 A composite wall is shown below. For element 1, let Kxx = 5

W , for element 2 let m C

, for element 3 let Kxx = 15 mW . The left end has a heat source of 600 W Kxx = 10 mW C C applied to it. The right end is held at 10°C. Determine the left end temperature and the interface temperatures and the heat flux through element 3.

[k(1)] =

(0.1)(5)  1 –1  5 –5  0.1  –1 1  –5 5

[k(2)] =

(0.1)(10)  1 –1  10 –10  –1 1   –10 10 0.1    

[k(3)] =

(0.1)(15)  1 –1  15 –15  –1 1   –15 15 0.1     1  { f (2) }  { f (3) }  0 { f (1) }  600   0  0 0  t1  600  5 –5  0   –5 15 –10 0 t2       0   0 –10 25 –15 t3       0 –15 15 t4   F4 x   0

600 = 5t1 – 5t2

t1 = 230°C

0 = –5t1 + 15t2 – 10 t3

t2 = 110°C

0 = –10t2 + 25t3 – 15 (10)

t3 = 50°C

F4x = –15 (50) + 15 (10) = – 600 W W  1 1  50  6000 2 qx = –15  –  0.1 0.1  10  m 13.9

506

Find T1 T2, T3, T4, Q (heat transfer through the double pane)



(use A = 1 cm2)



2 W AK g  1 1 W 1 0  1 0 1m 0.80 m  K  1 1  10 2 [k ] = (1m 2 )  =  hi A        1 1   1 1 0 0 0.004 m L  m C   0 0    (1)

 210 200 W =   200 200  K W AK A  1 1 1m  0.025 m  K   1 1  2.5 2.5 W =  1 1  = 2.5 2.5  K L 1 1  0.01m     2

[k(2)] =





2 W AK g  1 1 W  0 0 0 0  1m 0.80 m  K  1 1  30 2 (1m2 )  = [k ] =  h0 A        1 1   1 1 0 1 0.004 m L  m C      0 1 (3)

 200 200  =   200 230  W 1  1 1  (693 K) (1m2)   = 2630 W   { f (1) } = hi T A   = 10 2 m K 0 0 0 0 { f (2) } =   0 W 0 0 0 { f (3) } = h0 T A   = 30 2 (263 K) (1m2)   = 7990 W   m K 1 1  1 {F} = [K] [T ] 6 0   T1  200  210  F1  2630 W  200 200  2.5 F  0  0  T2  2.5  2      =  2.5  200 200  T3  2.5  0  F3  0     F4  7990 W  0 230  T4  200  0

507

T1  287.5 K  14.5C T2  287.3 K  14.2C T3  265.1 K   7.88C T4  264.8 K   8.15C 13.10

W W K (2) xx  0.038 m C m C AK xx  1 1 hpL  2 1  K   K hend L  1 1  6 1 2 

K (1) xx  0.20

Assumptions A = 1 m2

K (3) xx  0.12

W m C

Determine temperatures at inner/outer surfaces and at interfaces. Also, determine heat transfer through wall.

1 0  1(0.20)  1 1  hA     0.025  1 1  0 0  8 8 1 0  18 8 W K (1)    10(1)     2  8 8   0 0   8 8  m 1(0.038)  1 1  0.42 0.42  W  K (2)   0.09  1 1   0.42 0.42  m 2 K (1) 

0 0   9.6 9.6  W 1(0.12)  1 1  20(1)      2 0.0125  1 1  0 1   9.6 29.6  m 0   0  {fh}rt end = hR T∞R A   =   1  400  1  200  {fh}lt end = hL T∞L A   =   0   0  K (3) 

F = Kt

508

8 0 0   t1  t1  18.54C  200  18  0   8 8.42 0.42 0   t 2  t 2  16.72C         0   0 0.42 10.02 9.6   t 3  t 3  17.76C  400   0 0 9.6 29.6   t 4  t 4  19.27C 1   t1  1 1  18.54  W  (1)  1    q (1) 0.20      14.56 2 x   K xx       L L  t 2  m  0.025 0.025  16.72  13.11

Unit definition Given in problem statement (Mathcad used to solve this one) Ltotal = 20 cm W Kxx = 15 m C h1 = 50 h2 = 80

W 2

m C W 2

m C

Tleft end = 100°C Tinf = 20°C dia = 0.5 cm dia radius = 2 P =  dia L=

A =  radius2

Ltotal 4

P = 0.016 m A = 0.00002 m2 radius = 0.0025 m M = 150

W  m C 3

y1(x) = M x + h1

 

0.05m 0.05m   x x2   x 2  P y ( x ) 1 dx P y ( x )  0 1 0 1  L  L2  dx  L   kh1 =  0.05m  0.05m  x x2   x2  y1 ( x)   2  dx P y1 ( x)  2  dx  P     0 L L  0  L

Develop stiffness matricies for each element x=

Ltotal 4

509

 0.0136 0.007  W [ kh1 ] =   0.007 0.0146 C [k1] =

A K xx  1 1 + [ kh1 ] L  1 1 

 0.0195 0.0011 W [k1] =   0.0011 0.0205 C y2(x) = M x + y1(x)

 

0.05m 0.05m   x x2   x 2  P y ( x ) 1 dx P y ( x )   2  dx   0 2 2 0 L L L   [kh 2 ] = 2  0.05m  0.05m x x   x2  y2 ( x)   2  dx P y2 ( x)  2  dx  P L L  L  0  0 

 0.0141 0.0075 W [kh 2 ] =   0.0075 0.016  C [k2] =

A K xx  1 1 + [kh 2 ] L  1 1 

 0.02 0.0016 W [k2] =   0.0016 0.0219 C y3(x) = Mx + y2(x)

 

0.05m 0.05m   x x2   x 2 y3 ( x) 1  dx P  y3 ( x)   2  dx   P 0 L L  0 L  [kh 3 ] =   0.05m  0.05m  x x2   x2  y3 ( x)   2  dx P y3 ( x)  2  dx  P L L  L  0  0 

 0.015 8.018  103  W [kh 3 ] =    8.018  103 0.018  C [k3] =

A K xx  1 1 + [kh 3 ] L  1 1 

 0.02 2.127  103  W [k3] =    2.127  103 0.023  C y4(x) = M x + y3(x)

 

0.05m 0.05m   x x2   x 2  P y ( x ) 1 dx P  y4 ( x)   2  dx   0 4 L L  0 L  [kh 4 ] =  2  0.05m  0.05m x x   x2  y4 ( x)   2  dx P y4 ( x)  2  dx  P L L  L  0  0 

 0.0151 0.0085 W [kh 4 ] =   0.0085 0.019  C [k4] =

A K xx  1 1 0 0  [kh 4 ]  h2 A     L  1 1  0 1 510

 0.0209 0.0026 W [k4] =   0.0026 0.0264 C  k10,0   k10,0  [ K] =  0  0   0

k10,1

k11,1  k2 0,0

k2 0,1

k2 1,0

k2 1,1  k3 0,0

k3 0,1

k31,0

k31,1  k4 0,0

k4 1,0

0   0  0   k4 0,1   k4 1,1 

Global [K] 0 0 0   0.0195 0.0011  0.0011 0.0404 0.0016 0 0   W 0.0016 0.0424 0.0021 0  [K] =  0 C  0 0 0.0021 0.0443 0.0026    0 0 0 0.0026 0.0264 Element force matrices for each element h  y1 ( x) W hlave = 1 y1(x) = 57.5 2 2 m C y2(x) = 65

W m C 2

y3(x) = 72.5 y4(x) = 80

W m C 2

h1ave = 53.75

W m C 2

h2ave =

y1 ( x)  y2 ( x) 2

h2ave = 61.25

W m C

h3ave =

y2 ( x)  y3 ( x) 2

h3ave = 68.75

W m C

h4ave =

y3 ( x)  y4 ( x) 2

h4ave = 76.25

W m C

{f1} =

h1ave Tinf P L 1 1 2

 6.188 f1 =  W  6.188

{f2} =

h2ave Tinf P L 1 1 2

 7.051 f2 =  W  7.051

{f3} =

h3ave Tinf P L 1 1 2

 7.914 f3 =  W  7.914

 h T P L 1    0   h2 Tinf A    {f4} =  4ave inf       1   1  2 

 8.778 f4 =  W  9.238

F1 = f10,0

F2 = f11,0  f 20,0 F3 = f 21,0  f30,0 F4 = f31,0  f 40,0 F5 = f 41,0 F1 = 6.188 W, F2 = 13.239 W, F3 = 14.966 W, F4 = 16.692 W, F5 = 9.238 W 511

Set up equations to solve for temperature

 k11,1  k20,0   k21,0 [Kmod] =  0   0 

k20,1

k21,1  k30,0

k30,1

k31,0

k31,1  k40,0

k41,0

0   0  k40,1   k41,1 

Guess

t2 = 75°C t3 = 60°C t4 = 50°C t5 = 25°C Given

 F2  ( K1,0 Tleft end )   t2    t  F3   = Kmod  3  F4    t4     t  F5   5  t2   t3    = Find (t2, t3, t4, t5)  t4   t  5  t2   30.717  t3   51.077   =   °C  t4   69.113  42.348  t  5 T  F1 = (K0, 0 K0, 1)  left end   t2  F1 = 7.614 W 13.12 A tapered aluminum fin (k =

200 W ) shown in Figure P13-12, has a circular cross-section m  °C

with base diameter of 1 cm and tip diameter of 0.5 cm. The base is maintained at 200°C and looses heat by convection to the surrounding at T = 10°C, h = 150 2W . The tip of the fin is m C

insulated. Assume one-dimensional heat flow and determine the temperatures at the quarter points along the fin. What is the rate of heat loss in Watts through each element? Use four elements with an average cross-sectional area for each element. (Autodesk results)

512

Temperature ˚C 199.9998 198.5246 197.0403 195.5741 194.009 192.6238 191.1487 180.6735 188.1084 186.7232 185.248

Heat Flow Rate J s

438.817 351.0536 263.2902 176.5268 87.76341 3.814697 e-008 –87.7634 –175.5268 –263.2902 –351.0538 –438.817

Heat Flux Magnitude J (m^2 s) 72679.05

72679.05

13.13

[k (1) ] 

1 0  AK xx  1 –1 2  hA     Use unit area, A = 1 ft L  –1 1  0 0  

Btu   (1 ft 2 )  0.10 h  ft  °F   1 –1  Btu.1ft 2 1 0 = + 1.5     0.50 in. h  ft 2  °F 0 0  –1 1 (1 ft) 12 in.

 3.9 –2.4  [k (1) ]     –2.4 2.4 

513

[k (2) ] 

[k (3) ] 

(1 ft 2 )(0.02)  1 –1  0.048 – 0.048  –1 1   – 0.048 5 in. 0.048    12 in. (1 ft) (1 ft 2 )(0.8)  1 –1 Btu 0 0 +4     0.5 in.  –1 1 h  ft 2  °F 0 1  12 in. (1 ft)

 19.2 –19.2  [k (3) ]     –19.2 23.2  1   1  105 Btu  (1 ft 2 )(70°F)      {f (1)} = h T A    1.5  2 h  ft  °F  0   0   0  0   0   0  Btu  (1ft 2 )(-10°F)      {f (3)} = h T A     4.0  2 h  ft  °F  1   1  -40  0  {f (2)} =   0  {F} = [K] {t} –2.4 0 0   t1   105  3.9  0   –2.4 2.448 –0.048 0  t2           0   0 –0.048 19.248 –19.2  t3   40   0 0 –19.2 23.2  t4 

Solving for t’s t1 = 67.6°F, t2 = 66.1°F, t3 = -8.91°F, t4 = -9.10°F 13.15 Base plate of an iron is 0.6 cm thick. The plate is subjected to 100 W of power over a base surface area of 250 cm2 resulting in a uniform flux generated on the inside surface. The thermal conductivity of the metal base plate is k = 20 mW °C . The outside ambient temperature

of plate is 20°C at steady state. Assume 1-D heat transfer through the plate thickness. Using 3 elements, model the plate to determine the temperatures at the inner surface and interior 13 points. h = 20 W/m2-oC

514

From Mathcad solution A = 0.025 m

Kxx = 20

100 A

q = 4  103

 1 –1 0 0  –1 1 0 0  [k1] =   0 0 0 0  0 0 0 0

L = 0.002 m

W/m-°C h = 20

k1 = A

Tinf = 20

qA = 100  250 –250 0 0  –250 250 0 0  [k1] =  0 0 0 0   0 0 0 0

K xx k L 1

 0 0 0 0 K  0 1 –1 0  [k2c] = A xx  L  0 –1 1 0  0 0 0 0

k2 = k2 c

0 0 0  0 250 –250 [k2] =   0 –250 250  0 0 0

0 K 0 [k3c] = A xx  L 0  0

0 0 [k3h] = h A  0  0

0 0 0 0 0 0  0 0 0 0 0 1

0 0  0 1 –1 0 –1 1 0 0

0 0

0 0 [k3] =  0  0

[k3] = [k3c] + [k3h]

 1  0 {fq} =   q A  0  0

100  0  {fq} =   0  0

h A = 0.5

0 0 0 0 0 0  0 250 –250  0 –250 250.5 

 0  0 {fh} = h Tinf A    0  1

515

0 0  0 0

 0  0 {fh} =    0 10

100  0  {f } =   0  10

{f } = {fq} + {fh}

[K] = [k1] + [k2] + [k3]

0 0  250 –250  –250 500 –250 0  [K] =  0 –250 500 –250    0 0 –250 250.5 

temps = 1solve (K, f)

q1 = – Kxx

 

 221.2    220.8  temps =   220.4     220 

–1 1  temps1   L L  temps2 

Remember to use the left bracket key to get the subscript temps 1 and temps 2. q1 = 4  103 W/m2 13.16 A hot surface of a plate is cooled by attaching fins (called pin fins) to it. The surface of the plate (left end of the fin) is at 90°C. The typical fin is 6 cm (0.06 m) long and has a crosssectional area of 5  10–6 m2 with a perimeter of 0.006 m. The fin is made of copper with k = . The temperature of the surrounding air is T = 20°C with heat transfer coefficient 400 mW C

on the surface (including the right end) estimated to be 10

W . Use three elements in your m 2 C

model to estimate the temperature distribution along the fin length.

[k(1)] =

A K xx  1 –1 hPL  2 1 5  10 –6  400  1 –1  =  –1 1 6  1 2  0.02 L  –1 1   

10  0.006  0.02  2 1  1 2 6  

 0.1004 – 0.0998 W = [k(2)] [k(1)] =  0.1004  C  – 0.0998 0 0  0.1004 – 0.0998  0 0 [k(3)] = [k(1)] + hA   + 10(5)  10–6     0.1004  0 1  – 0.0998 0 1 516

 0.1004 – 0.0998  W  [k(3)] =     – 0.0998 0.10045  C 



{f (1)} =

hT PL 1 10  20C  0.006 m  0.02 m 1 0.012      W 2 2 1 1 0.012 

{f (2)} = {f (1)} 0  0  {f (3)} = {f (1)} + h T A   = {f (1)} + 10.(20°C)  5  10 –6 m2   1  1  0.012  {f (3)} =  W 0.013 0 0  t1  90C 0.1004 – 0.0998   2(0.1004) – 0.0998 0   t2      t 2(0.1004) – 0.0998    3     0.10045   t4 Symmetry 0.012  F1   2(0.012)    =    2(0.012)   0.013  Solving Equations (2-4), t2 = 87.95°C, t3 = 86.72°C, t4 = 86.4°C 13.17

Fourier’s law q = – Kxx

dT Btu ( ) dx ft 2

(1)

Want to link thermal inputs F1 and F2 to nodal temperatures t1 and t2 x x   t1  [T] = [N] {t} = 1     L L  t2 

(2)

1   t1    = [B] {t} L  t2 

(3)

1 d [T ] =    L dx

Total heat input at node 1 is F1 = q A

(4) 517

and at node 2 F2 = – qA (negative sign accounts for the positive direction of F2 being an output at node 2) (5) Using Equations (1) and (3) in (4) and (5), we have F1 = – Kxx [B] {t} A

F2 = Kxx [B] {t} A

(6)

or K xx A  1 1  t1   F1    =   L 1 1  t2   F2   [k] =

K xx A  1 1 L  1 1 

(7)

element conductivity matrix

13.18

See Equation (13.4.28) Now convection from left end 1 0  [ K h left ] = hA   0 0  1  { f h left} = h T A   0 13.19

[k] = [B]T [D] [B] t A +  h[N]T [N] ds s

i = xj ym – xm yj = (10) (6) – (8) (1) = 52 j = xm yi – xi ym = (8) (2) – (6) (6) = – 20 m = xi yj – xj yi = (6) (1) – (10) (2) = – 14 i = yj – ym = 1 – 6 = –5 518

j = ym – yi = 6 – 2 = 4 m = yi – yj = 2 – 1 = 1 i = xm – xj = 8 – 10 = – 2  j = xi – xm = 6 – 8 = – 2 m = xj – xi = 10 – 6 = 4 1 xi

2A = 1 xi 1 xm

yi ym

1 6

= 1 10 1 = 18 ft2 1 8 6

Conduction part of [k] = [kc]

 i 1  i [kc] = 2A    m

i   K xx  j   0  m 

 1  i K yy  2 A  i

j

m 

j

m

 tA

 5 2  1  10 0  1  5 4 1  (1 ft) (9 ft2) = 4 2    0 10  18  2 2 4  18  4  1 8.056 4.444 3.611 [kc] =  5.556 1.111    Symmetry 4.772 

Convection part of [k] = [kh] Lij = 4.123 ft 2 1 0  2 1 0 hLij t  (20)(4.123)(1)   1 2 0 = 1 2 0 [kh] =    6 6  0 0 0  0 0 0  27.487 13.74 0  =  27.487 0    0   Symmetry

Total [k] = [kc] + [kh] 35.54 9.30 3.61 Btu [k] =  33.04 1.11   h  °F  Symmetry 4.72 

Force matrix {f } =  [N]T Q dV +  [N]T q ds +  [N]T h Tds V

q= 0 For point source Q* = Q

519

 Ni  { fQ } = Qt  N j  N  x  8  m

y3

Ni =

1 ( i  i x   i y ) x  8 2A

y3

Ni =

1 (52 + (– 5) (8) + (– 2) (3)) = 0.333 18

Nj =

1 (– 20 + (4) (8) + (– 2) (3)) = 0.333 18

Nm =

1 (– 14 + (1) (8) + (4) (3)) = 0.333 18

0.333 100      Btu  { fQ } = 300 (1) 0.333 = 100  0.333 100  h     1  1  hT Lij t   (20)(70)(4.123)1   { fh } = 1  = 1  2 2   0    0 100  2886  2986  2886      Btu   { f h } = 2886   { f } = 100  2886   2986   100  0   100  h  0       

13.20

2 0 1  h Lim  [k] = t A [B] [D] [B] + 0 0 0  6  1 0 2  T

(1)

1 2 2

2A = 1 1

0 = 44

6 520

Lim = 8.246 m

i = – 6, j = 8, m = – 2 i = – 4, j = – 2, m = 6 Ni =

1 [24 + (– 6) (0) + 0] = 0.545 44

Nj =

1 [12 + 8(0) + 0] = 0.278 44

Nm =

1 [8 + (– 2) (0) + 0] = 0.181 44

By (1)  6 4  1  10 0   6 8 2  8 2   [K] =   0 10   4 2 6  4(22)   2 6  2 0 1  20(8.246)  + 0 0 0   6 1 0 2  60.88 4.55 26.12  W [k] =  7.73 3.18   °C 59.52  Symmetry

{f } =  [ N ]T (0,0) Q* dV +  hT [ N ]TAlong S3 ds S3

 Ni  1  hT Lim   N   0  = Q*   j  2   N  1   m (0, 0) 0.545 1    (20)(15)(8.246)   = 200 0.273  0 2 0.183 1      1346    { f } =  54.6  W 1273  

13.21

521

Temperature deg F 100 97.77778 95.55558 93.33333 91.11111 88.88889 84.44445 82.22223 77.77779

100 °F

77.78 °F

Load Case: 1 of 1 Maximum Value: 100 deg F

0.000

0.620

1.258

1.885

Minimum Value: 77.7778 deg

13.22 For the square plate in figure P13-22, determine the temperature distribution. Let Kxx = Kyy = , and h = 20 2W . The temperature along the left side is maintained at 10 mW C m C

100 °C and that along the top side is maintained at 200 °C. Temperature deg C

200°C

200 186.9014 173.9027 160.7041 147.6054 134.5068 121.4082 108.3095 95.2109 82.11226 69.01362

T• = 50°C

100°C

Load Case: 1 of 1 Maximum Value: 200 deg C Minimum Value: 69.0136 deg C

522

The maximum temperature is along the top edge of the plate and is 200 °C. The smallest temperature is at the lower right edge of the plate and is 69.01 °C. 13.23

HEAT—Problem 13–23 KXX = 1.0 KYY = 1.0 CONVECTION COEFF = 0.0 FLUID TEMPERATURE = 0.0 SEMI–BANDWIDTH = 4 NEL 1 2 3 4 5 6 7 8

NODE 1 2 3 1 4 5 6 4 X(2) 0.0000 1.5000 1.5000 0.7500 1.5000 3.0000 3.0000 2.2500

NUMBER 2 3 5 3 5 4 3 4 5 6 e 6 8 7 6 7

X(1) 0.0000 0.0000 0.7500 0.0000 1.5000 1.5000 2.2500 1.5000

Y(1) 2.0000 0.0000 1.0000 2.0000 2.0000 0.0000 1.0000 2.0000

Y(2) 0.0000 0.0000 0.0000 1.0000 0.0000 0.0000 0.0000 1.0000

X(3) 0.7500 0.7500 1.5000 1.5000 2.2500 2.2500 3.0000 3.0000

Y(3) 1.0000 1.0000 2.0000 2.0000 1.0000 1.0000 2.0000 2.0000

*PRESCRIBED NODAL TEMPERATURE VALUES* 1 0.10000E+03 2 0.10000E+03 7 0.00000E+00 8 0.00000E+00 RESULTING NODAL TEMPERATURE VALUES 1 0.10000E+03 2 0.10000E+03 3 0.75000E+02 4 0.50000E+02 5 0.50000E+02 6 0.25000E+02 7 0.00000E+00 8 0.00000E+00 523

ELEMENT RESULTANTS ELEMENT 1 2 3 4 5 6 7 8

GRAD (X) –0.3333E+02 –0.3333E+02 –0.3333E+02 –0.3333E+02 –0.3333E+02 –0.3333E+02 –0.3333E+02 –0.3333E+02

GRAD (Y) 0.0000E+00 0.0000E+00 –0.1907E–05 –0.1907E–05 –0.1907E–05 0.0000E+00 0.0000E+00 –0.1907E–05

AVE TEMP 0.9167E+02 0.7500E+02 0.5833E+02 0.7500E+02 0.4167E+02 0.2500E+02 0.8333E+01 0.2500E+02

13.24

524

13.25

13.26

525

13.27

13.28

526

13.29 The temperature distribution of the earth is shown below with a 60 °F oil pipe 15 ft under the earth’s surface at 50 °F.

13.30

527

K = 0.50

q* = 50

Btu 778 lb  ft 1 h in.  lb = 0.108056 h  ft  °F 1 Btu 3600 s s  in. F

lbf  in. Btu  1 hr  1 min   778lbf ft  12in  = 129.66 s hr  60 min   60s   1 Btu   1 ft 

Heat source q* = 129.66

lbf  in. s

528

13.31

13.32

529

13.34

13.35

530

13.36

13.37 Determine the temperature distribution and rate of heat flow through the plain carbon steel ingot shown in Figure P13-37. Let k = 60 mW K for the steel. The top surface is held at 60oC,

while the underside surface is held at 0oC. Assume that no heat is lost from the sides.

531

13.38

532

13.39

13.40 For the basement wall, determine the temperature distribution and the heat transfer through the wall and soil.

533

13.41

534

13.42

13.44 The Allen Wrench, shown in Figure 1, is exposed at one end to at temperature of 300 K, W . Determine the temperature distribution while the other end has a heat flux of 10 m^2 535

throughout the wrench. It has a thermal conductivity of 43.6

W mK

and a specific heat

capacity of 0.000486 g JK . Part dimensions

Boundary condition

Figure 3: Temperature Distribution (K) 13.45 Temperature distribution

536

13.46 The thermal aspect of this component is that the base has an applied temperature of 100 °F. This is theoretically due to the warmed aluminum plate being heated by the electric motor. The lower surface of the lower finger has an applied surface temperature of 50 °F.

13.48  (1.5 )2

(0.017) 4  144 K xx A Btu = = 0.001 Small so neglect 10 / 12 L h  °F 4 1 

m c = 10 (0.24) = 2.4

Btu h  F

hPL 3   (1.5 )  104 Btu = = 0.04096 6  121  121 h  F 6

537

hPLT = (3) 

[k(1)] =

(1.5)  104  Btu   (200°F) = 49.087 12  12  h

2.4 1 1 2 1  + 0.04096     2 1 1 1 2 

 1.118 1.24091 [k(1)] =    1.159 1.2818  [k(2)] = [k(3)] = [k(4)] = [k(1)] also 1.2409 0 0  1.2818  1.118  1.159 1.2818  1.118 1.2409 0   [K] =  1.159 0 1.2818 – 1.118 1.2409   1.159 0 0 1.2818 0 0   0.164 1.241  1.159 0.164 1.241 0   [K] =  1.159 0.164 1.241  0  0 1.159 1.282 0

 49.087  1.159 (50)   49.087  {F} =  49.087     24.5435  64.719  t2   77.715   t3     [K –1]{F} =   89.747  t4  100.296 t   5

5 4 3 2 1

t1 = 50°F

 1  (10)(0.24)(50F)=120 Btu/h qin  mct  5 (10)(0.24)(100.296)  240.7Btu / h qout  mct

538

Chapter 14 14.1

Global [K] [K] = 4

{P} =

m s

1 –1 1 3 0 2 0 0

0 2 5 3

0 0 3 3

10 p2

p3 0 Accounting for the boundary conditions

p1 = 10, 12

p4 = 0, we get

40 0

Solving p2 = 4.545 m,

p3 = 1.818 m p1 1 1 (1) v x(1) = – k xx L L p2 40 m = – 2 [– 1 1] = 10.91 4.545 s m v x(2) = – 4 [– 4.545 + 1.818] = 10.91 s m v x(3) = – 6(– 1.818) = 10.91 s (1) Q (1) f = Av x = 21.82

Q (2) = 21.82 f

m3 s

Q (3) = 21.82 f

m3 s

14.2 539

[k(1)] = [k(2)] = [k(3)] = [f (3)] = q* A

= – 50

1 –1

1 2 1

–1

0 m  0  (2m2) =   1 s  100 

Global equations 2

p1 = 10

 0   0    =    0    100 

Rewriting 4 2

2 4

0 2

p2 p3

 20    =  0  100   

Solving p2 = – 40 m, p3 = – 90 m, v x(1) = – Kxx[B]

= 50

p1 p2

p4 = – 140 m

1 1   10  = –     (L = 1)  L L   40

m s

1 1  40 m v x(2) =  1     = 50 s  1 1  90  1 1  90  m v x(3) =  1     = 50  s  1 1  140

Volumetric flow rates Q1 = Q2 = Q3 = v x(1) A1 = v x(2) A2 = v x(3) A3 = 50 (2) = 100

m3 s

14.3

540

[k(1)] = 0.6 [k(2)] = 0.4 [k(3)] = 0.2

0.6 – 0.6 0 0 –0.6 1.0 0.4 0 0 0.4 0.6 0.2 0 0 0.2 0.2 Using the 2nd and 3rd equations above 1

0.4

0.6

6 0

p2 = 8.182 in., v x(1) = – Kxx [B]

p3 = 5.455 in. p1 p2

vx(1) = 0.182

in. s

vx(2) = – 1

1 1 10 10

vx(2) = 0.273

in. s

vx(3) = –

10 in. 0 p2 0 = p3 0 p4 = 0 0

1 1 10 10

= 0.545

1 1 10 10

=–1

10 8.182

8.182 5.455

5.455 0

in. s

(1) 2 Q (1) f = A v x = (6 in. ) (0.182

= 1.091

in.3 s

Q (2) = 1.091 f

in.3 s

Q (3) = 1.091 f

in.3 s

in. ) s

14.4

541

[k(1)] =

2 5 5

[k(2)] =

2 3 5

cm 3 , s

F3 = – 4  3 = – 12

6 5 6 5

Negative, as water flows out of right edge

Assemble global equations 2 2

2 3.2

0 1.2

1.2

p2 p3

3.2

1.2

 0    =  0    12   

Now assume p1 = 0  0  =    12 

Solving for p2 and p3 p2 = – 6

p3 = – 16

v x(1) = – Kxx

1 1 L L

= –2

1 1  0   5 5  6

v x(1) = 2.4

cm  (to right) s

v x(2) = – 2

1 1  6    5 5  16 

Velocities

v x(2) = 4

p1 p2

cm  (to right) s

Flow rates cm   (1) 2 Q (1)  (5 cm ) f = v x A =  2.4 s  

Q (1) f = 12

cm 3 s

542

4 cm  2 = v x(2) A =  Q (2)  (3 cm ) f  s 

= 12 Q (2) f

cm 3 s

14.5 [k] =

[B]T [D] [B] dV

For 1-D formulation



[B] =

1 1 LL

[k] =

1 L 1 L

[D] = [Kxx]

1 1 dV L L

[Kxx]

For element with constant cross sectional area A L

[k] =

Kxx

[k] =

K xx A L

Kxx =

1 in. 10 s

[k(1)] =

2 1 10

1 L2 1 L2

A dx

14.6

1 –1 –1

[k(2)] =

1 1 1 –1 10 – 1 1

‘Force’ at node 1 3

4 in.  in. 2 F1x =   (2 in. ) = 8 s  s 

Assume

p3 = 0

Assemble equations 0.2

0.2

0.2 0.2 0.1

p1 p2

8  =   0 

Solving p1 = 120 in.,

p2 = 80 in. 543

Velocities p1

1 1 L L

v x(1) = – Kxx

v x(1) = –

1 10

1 1  120  in.   = 4.0 1 1  80  s

v x(2) = –

1 10

1 1 80  in.   = 8.0 1 1  0 s

Volumetric flow rates 3

 in.  (2 in.2) = 8 in. (1) (1) Q (1)  f = vx A =  4 s  s 

3  8 in.  (1 in.2) = 8 in. (2) (2) Q (2) = A = v   x f s  s 

14.7 (a)

[k(1)] =

[k(3)] =

[k(5)] = Assemble [k]’s 1

1 1

1 1 1

4 1 1 , [k(2)] = 1 2

1 1 1

2 1 1

1 3

2 1 1

3 1 1 , [k(4)] = 1 4

2 1 1

3 1 1

3 1

4 1 1

1 5

544

1 2 1 2

1 2 1 3

1 2

0 1 4

1 3

1 4

1 2 1 3

1 4 1 4

1 5

m N s

p1 p2 p3 N m2

1 Q1 0 = 0 m3 s

Solve in Mathcad for p1, p2, p3 p1 = 0.897 q1 = q4 =

N , m2

p2 = 0.691

N , m2

P P P = 1 4 = 0.897, R1 R1

PPP P P P q2 = 1 =2 1 2 = 0.103, R2 RR R2 22

q5 =

P3 R4

= 0.044,

P3 0 = 0.103 R5

p3 = 0.515 q3 =

N m2

P3 R3

= 0.059

(all qs in

m3 ) s

q1 + q4 = 0.103 (check equals q2) q1 + q5 = 1 (check equals Q = 1)

(b)

R1 = 10

lb s in.5

R2 = 20 R3 = 30 R4 = 40 R5 = 50 545

[k(1)] =

[k(3)] =

[k(5)] =

1 10

1 1 1

4 1 1 , [k(2)] = 1 20

1 1 1

2 1 1

1 30

2 1 1

3 1 1 , [k(4)] = 1 40

2 1 1

3 1 1

1 50

3 1 1

4 1 1

1000 0 = 0

Assemble 1 1 10

1 20 1 20

1 – 20 1 20

1 30 1 30

1 40 1 40

1 30 1 30

1 40 1 1 40 50

in.5 lb s

p2 p3 lb in.2

in.3 s

Solve for p1, p2, p3 using Mathcad as p1 = 8971 psi, q1 =

q4 =

p2 = 6912 psi,

p3 = 5147 psi,

P4 = 0 = 897, R1

P1 – P2 P1 – P2 q2 = = 103, R2 R2

q3 =

q5 =

= 44,

P3 0 = 103 R5

P3 R3

= 59

(all qs in

in.3 ) s

q3 + q4 = 103 (check equals q2) q1 + q5 = 1000 (check equals Q) 14.8

546

 fQi  Ni    fQj  = Q* t N j   Nm  fQm 

Ni =

x y

4 2

1 (i + i x + iy) 2A

i = xi ym – yj xm = 61 j = yj xm – xi ym = – 12 m = xi yj – yi xj = – 7 i = yi – ym = – 6 i = xm – xj = – 7 j = ym – yi = 6

j = xi – xm = 0

m = yi – yj = 0

m = xj – xi = 7

Ni x = 4 =

23 1 (61 – 6(4) – 7(2)) = 42 42

Nj x=4 =

12 1 (–12 + 6(4)) = 42 42

Nm x = 4 =

7 1 (– 7 + 7(2)) = 42 42

y=2

m2  {fQ} = 100 s

23 54.76  1 m  12 = 28.57    42  16.67 7

m3 s

14.9

547

1 q * Lt {f } = 0 2 1

1  10  10  2  1     in.3 = 0  = 0  2 1  10  s    

f2 f3

14.10

From Equations (14.3.16) 25

0 25

[K] =

0 0

25 25

0 25

25  10–5 25 100

Symmetry  F1   25    m3 F =  25   10–5 s F   4  0 

Boundary conditions p1 = p4 = 10 m 25



0 25 25 25 Symmetry 100

p2 p3 p5

25 = 25 500

Solving p2 = 12 m, p3 = 12 m, p5 = 11 m 14.11 (33 nodes, 56 elements) FLUIDS PROBLEM 14–11 BOUNDARY VALUES NODAL FORCES LOADING CASE 1 548

0 0.00000E+00 LOADING CASE 2 1 0.50000E+04 *PRESCRIBED NODAL VALUES* 26 0.50000E+03 27 0.50000E+03 28 0.50000E+03 29 0.50000E+03 30 0.50000E+03 31 0.50000E+03 32 0.50000E+03 33 0.50000E+03 FLUIDS PROBLEM 14–11 NODAL VALUES, LOADING CASE 1 1 – 0.46404E+03 5 0.14359E+07 9 0.45441E+02 13 0.25077E+03 17 0.26088E+03 3 – 0.17302E+02 7 – 0.75913E+02 11 0.23328E+03 15 0.16262E+03 19 0.38851E+03

2 6 10 14 18 4 8 12 16 20

0.37597E+02 – 0.26094E+03 0.25835E+03 – 0.57844E+02 0.39845E+03 – 0.16157E+03 0.13183E+03 0.19417E+03 0.23693E+03 0.37488E+03

21 25 29 33

0.38550E+03 0.39925E+03 0.50000E+03 0.50000E+03

22 26 30

0.28400E+03 0.50000E+03 0.50000E+03

23 27 31

0.35570E+03 0.50000E+03 0.50000E+03

24 28 32

0.38834E+03 0.50000E+03 0.50000E+03

NODAL VALUES, LOADING CASE 2 1 – 0.49446E+03 5 0.14476E+07 9 0.30588E+02 13 0.23855E+03 17 0.25296E+03 21 0.38006E+03 25 0.39590E+03 29 0.50000E+03 33 0.50000E+03

2 6 10 14 18 22 26 30

0.23005E+02 – 0.28970E+03 0.25067E+03 – 0.84934E+02 0.39522E+03 0.27361E+03 0.50000E+03 0.50000E+03

4 8 12 16 20 24

– 0.18077E+03 – 0.34748E+01 0.18421E+03 0.22760E+03 0.37068E+03 0.38433E+03

FLUIDS PROBLEM 14–11

3 7 11 15 19 23

– 0.33072E+02 – 0.96780E+02 0.22496E+03 0.14912E+03 0.38500E+03 0.34982E+03

549

27 31

0.50000E+03 0.50000E+03

28 32

0.50000E+03 0.50000E+03

FLUIDS PROBLEM 14–11 ELEMENT VELOCITY COMPONENTS ELEMENT 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50

VEL(X) – 0.20066E+05 – 0.13173E+05 0.81240E+04 0.81240E+04 0.81240E+04 0.28676E+04 – 0.97320E+04 – 0.20066E+05 – 0.88300E+04 – 0.88300E+04 – 0.12222E+05 – 0.49369E+04 – 0.14055E+05 0.74983E+04 0.81240E+04 0.81240E+04 0.81240E+04 0.81240E+04 0.21425E+04 0.24686E+04 – 0.13329E+04 – 0.43866E+04 – 0.48610E+04 – 0.85681E+04 – 0.56043E+04 – 0.56043E+04 – 0.67102E+04 – 0.27549E+04 – 0.41023E+04 0.31961E+04 0.33656E+04 0.40359E+04 0.13674E+05 0.13674E+05 0.33137E+04 0.27655E+04 0.40783E+03 – 0.27156E+04 – 0.27727E+04 – 0.55241E+04 – 0.40618E+04 – 0.40618E+04 – 0.45923E+04 – 0.18482E+04 – 0.23311E+04 0.20741E+04 0.20397E+04 0.45798E+04 0.86399E+04 0.86399E+04

VEL(Y) – 0.52061E+04 – 0.12099E+05 – 0.12099E+05 – 0.12094E+05 0.13833E+05 0.19089E+05 0.19089E+05 0.87555E+04 0.41014E+04 – 0.29504E+04 – 0.19572E+04 – 0.92425E+04 – 0.14229E+05 – 0.14229E+05 – 0.12092E+05 – 0.12097E+05 – 0.12099E+05 0.96035E+04 0.11355E+05 0.11029E+05 0.89500E+04 0.89500E+04 0.73297E+04 0.36225E+04 0.23942E+04 – 0.21350E+04 – 0.20719E+04 – 0.60273E+04 – 0.72286E+04 – 0.72286E+04 – 0.42572E+04 – 0.35868E+04 – 0.14397E+05 0.70190E+04 0.76100E+04 0.81583E+04 0.60563E+04 0.60563E+04 0.50557E+04 0.23043E+04 0.16983E+04 – 0.16833E+04 – 0.17156E+04 – 0.44597E+04 – 0.50048E+04 – 0.50048E+04 – 0.44382E+04 – 0.18980E+04 – 0.54948E+04 0.35806E+04

550

51 52 53 54 55 56

0.48159E+04 0.23920E+04 0.12355E+04 – 0.18510E+04 – 0.18156E+04 – 0.40301E+04

0.33480E+04 0.57719E+04 0.44664E+04 0.44664E+04 0.38847E+04 0.16702E+04

RESULTANT NODAL VALUES 1 – 0.49446E+03 2 0.23005E+02 3 – 0.33072E+02 4 – 0.18077E+03 5 0.14476E+07 6 – 0.28970E+03 7 – 0.96780E+02 8 – 0.34748E+01 9 0.30588E+02 10 0.25067E+03 11 0.22496E+03 12 0.18421E+03 13 0.23855E+03 14 0.84934E+02 15 0.14912E+03 16 0.22760E+03 17 0.25296E+03 18 0.39522E+03 19 0.38500E+03 20 0.37068E+03 21 0.38006E+03 22 0.27361E+03 23 0.34982E+03 24 0.38433E+03 25 0.39590E+03 26 0.50000E+03 27 0.50000E+03 28 0.50000E+03 29 0.50000E+03 30 0.50000E+03 31 0.50000E+03 32 0.50000E+03 33 0.00000E+00

14.12 BOUNDARY VALUES

NODAL FORCES LOADING CASE 1 0

0.00000E+00

*PRESCRIBED NODAL VALUES* 1 0.60000E+01 2 0.60000E+01 3 0.60000E+01 6 0.30000E+01 7 0.30000E+01 8 0.30000E+01 NODAL VALUES, LOADING CASE 1

551

1 5 3 7

0.60000E+01 0.45000E+01 0.60000E+01 0.30000E+01

2 6 4 8

0.60000E+01 0.30000E+01 0.45000E+01 0.30000E+01

14.13 Use 14 of the whole system due to symmetry (25 nodes, 32 elements) NODAL FORCES LOADING CASE 1 0 LOADING CASE 2 25

0.00000E+00 0.12500E+03 (pumping rate)

*PRESCRIBED NODAL VALUES* 1 6 11 16 21

0.10000E+03 0.10000E+03 0.10000E+03 0.10000E+03 0.10000E+03

FLUID PROBLEM 14-12 NODAL VALUES, LOADING CASE 1 1 5 9 13 17 21 25 3 7 11 15 19 23

0.10000E+03 0.10000E+03 0.10000E+03 0.10000E+03 0.10000E+03 0.10000E+03 0.10000E+03 0.10000E+03 0.10000E+03 0.10000E+03 0.10000E+03 0.10000E+03 0.10000E+03

2 6 10 14 18 22

0.10000E+03 0.10000E+03 0.10000E+03 0.10000E+03 0.10000E+03 0.10000E+03

4 8 12 16 20 24

0.10000E+03 0.10000E+03 0.10000E+03 0.10000E+03 0.10000E+03 0.10000E+03

NODAL VALUES, LOADING CASE 2 1 5 9 13 17 21 25

0.10000E+03 0.10202E+03 0.10177E+03 0.10125E+03 0.10063E+03 0.10000E+03 0.10363E+03

2 6 10 14 18 22

0.10062E+03 0.10000E+03 0.10209E+03 0.10185E+03 0.10128E+03 0.10063E+03

3 7 11 15 19 23

0.10121E+03 0.10062E+03 0.10000E+03 0.10231E+03 0.10198E+03 0.10129E+03

4 8 12 16 20 24

0.10174E+03 0.10122E+03 0.10062E+03 0.10000E+03 0.10277E+03 0.10207E+03

RESULTANT NODAL VALUES 1

0.10000E+03

552

2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

0.10062E+03 0.10121E+03 0.10174E+03 0.10202E+03 0.10000E+03 0.10062E+03 0.10122E+03 0.10177E+03 0.10209E+03 0.10000E+03 0.10062E+03 0.10125E+03 0.10185E+03 0.10231E+03 0.10000E+03 0.10063E+03 0.10128E+03 0.10198E+03 0.10277E+03 0.10000E+03 0.10063E+03 0.10129E+03 0.10207E+03 0.00000E+00

ELEMENT VELOCITY COMPONENTS ELEMENT 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

VEL (X) 0.12207E–03 0.79155E–04 0.26226E–04 0.17643E–04 0.11921E–03 0.92506E–04 0.12875E–04 0.33379E–04 0.12207E–03 0.94414E–04 0.14305E–04 0.32902E–04 0.11921E–03 0.92506E–04 0.12875E–04 0.45300E–04 0.12207E–03 0.94414E–04 0.14305E–04 0.48161E–04 0.11921E–03 0.10443E–03 0.95367E–05 0.30041E–04 0.12207E–03 0.10967E–03 0.17643E–04 0.36240E–04 0.11921E–03 0.80585E–04

VEL (Y) 0.57220E–05 0.21935E–04 0.47684E–05 0.41962E–04 0.00000E+00 0.57220E–05 0.21935E–04 0.47684E–05 0.57220E–05 – 0.19073E–05 0.47684E–05 0.24796E–04 0.00000E+00 0.57220E–05 – 0.19073E–05 0.47684E–05 0.57220E–05 0.28610E–04 0.35286E–04 0.95367E–06 0.00000E+00 0.57220E–05 0.28610E–04 0.35266E–04 0.57220E–05 – 0.56267E–04 – 0.19073E–04 0.95367E–06 0.00000E+00 0.57220E–05

553

31 32

0.28133E–04 0.45300E–04

– 0.56267E–04 – 0.19073E–04

14.17

[k(1)] =

[k(3)] =

[k(5)] =

[k(7)] =

I1 80 80

0 80 , 80

I1 40 40

0 40 40

I2 10 10

I3 10 , 10

I3 5 5

0 5 , 5

[k(2)] =

, [k(4)] =

[k(6)] =

[k(8)] =

I1 5 5

I2 5 5

I2 10

0 10

I2 10 10

0 10 10

I3 5 5

0 5 5

I1 I2 I3 80 5 40 5 0 5 5 10 10 10 10 0 10 10 5 5

554

I1 I2 I3

125 5

5 35

0 I1 10I 2

20I 3

5 10

IAD = I1 – I2 IBC = I2 – I3 Using Mathcad 125 [K] = – 5

–5 35

0 – 10

– 10

[V] =

5 –10

0.161 AMPS = 0.027

AMPS = 1solve (K, V)

– 0.487

IAD = AMPS0 – AMPS1

IAD = 0.134

IBC = AMPS1 – AMPS2

IBC = 0.513

14.18

Resistor element stiffness matrices. I1 0 1 1 [k(1)] = 10 , [k(2)] = 5 1 1 I2 1 [k(3)] = 20 1

I3 1 , [k(4)] = 60 1

I1 1 1

I3 1 1

0 1 1

Assemble equations 555

I1 10 5 0 5

I2 I3 0 5 I1 20 20 I 2 20 5 20 60I 3

15 0

0 20

5 20

I1 I2

6 0

Solve in Mathcad IAB = I1 – I3 IBC = I2 – I3 15 [K1] = 0

0 20

–5 – 20

– 5 – 20

– 12

[V1] = – 6 0 – 0.853

AMPS1 = 1solve (K1, V1)

AMPS1 = – 0.458 – 0.158

I AB1 = AMPS10 – AMPS12

I AB1 = – 0.695

I BC1 = AMPS11 – AMPS12

I BC1 = – 0.3

14.19

[k(1)] =

I1 301 301

I2 301 , [k(2)] = 301

I1 301 301

I2 301 301 392

I1 I2

I2 392 392

0 392 392

9 0

And upon solving in Mathcad IAD = I1 – I2

if < 0.015 amp R1 556

acceptable

IBC = I2 [k] =

if < 0.015 amp R2 301

– 301

acceptable

[v] =

– 301 301 392

9 0

Ist iteration AMPS = 1solve (k, v)

AMPS =

0.053 0.023

IAD = AMPS0 – AMPS1

IAD = 0.03

IBC = AMPS1

IBC = 0.023

Note: amps through diodes must be < 0.015 AMPS, so try larger resistors. Try changing 301 to 392 as well. [k2] =

392

– 392

784

AMPS2 = 1solve (k2, v)

AMPS2 =

0.046 0.023

No good. Try larger ohm resistor (Try 523 ohms) [k3] =

523

– 523

– 523 1046

AMPS3 = 1solve (k3, v)

AMPS3 =

0.034 0.017

I AD3 = 0.03 > 0.015  no good

I AD3 = AMPS30 – AMPS3l

Try 549 ohms for both resistors. [k4] =

549

– 549

1098

AMPS4 = 1solve (k4, v)

AMPS4 =

0.033 0.016

Final iteration Try 715 for R1 and 806 for R2 [k5] =

715

– 715

– 1521 1521

AMPS5 = 1solve (k5, v)

AMPS5 =

I AD5 = AMPS50 – AMPS5l

I AD5 = 0.013

I BC5 = AMPS5l

I BC5 = 0.011



R2 = 806 

Let R1 = 715 ,

0.024 0.011

This works. Now amps < 0.015 amps

14.20 557

[k(1)] =

I1 221

I2 221

221

, [k(2)] =

221

221 383

IAD = I1 – I2

I2 383

0 383

383

12 0

if < 0.015 amp R1

If I2 < 0.015 amp R2

acceptable

IBD = I1 ICD = I2 [k20] =

221

– 221

– 221 221 383

AMPS20 = 1solve (k20, v20)

{v20} =

AMPS20 =

12 0 0.086 0.031

Must increase ohms size so amps < 0.015. Try previous final R’s of 715 and 806 ohms from P14.19. AMPS20 = 1solve (k5, v20)

AMPS20 =

0.032 0.015

Still amps too large in diodes under R1 (Try 2000 ohm for R1 and 1270 ohms for R2) k20 f =

AMPS20



2000

– 2000

3270

= 1solve (k20 f, v20)

AMPS20 =

Let R1 = 2000 ,

R2 = 1270 

0.015 9.449 10 –3

I  = 1   I2 

14.22

558

14.23

559

560

Chapter 15 15.1

[k(1)] =

AE 60

1 1 1

2 1 , 1

2 [k(2)] =

E TA , E TA

{f (1)} =

{f (2)} =

AE 60

3 1 1

1 1

E TA E TA

{F} = [K ] {d} becomes E TA AE = 0 60 E TA

Solving

1 1 0

0 u1 0 1 u2 1 u3

1 2 1

u2 =  T L = (7  10–6) (100°F) (60 in.) = 0.042 in. u3 = 2  T L = 0.084 in.

Reactions and actual nodal forces {F} = [K] {d} – {F0} F1x F3 x

1 1 0

F1x

F2 x

F2 x F3 x

1 2 1

0 1 1

AE L

0 TL 2 TL

E TA 0 E TA

0 0

(1) = (2) =

0 4 in.2

15.2

561

[k(1)] =

AE 1.5

1 1

1 , 1

[k(2)] =

AE 1.5

1 1

E TA E TA

{f (1)} = {f (2)} = Global equations AE 1.5

1 1 0

0 u1 0 = 1 u2 1 u3 0

1 2 1

E TA 0 E TA

Solving AE (2u2) = 0 1.5

u2 = 0 Forces in elements f1(1) x f 2(1) x

0 0

E TA = E TA

E TA E TA

E  T A = (200 GPa) (11.7  10–6/°C) (– 30°C)  (1  10–2 m2) = – 702 kN (1)  f1x(1) = – 702 kN, f 2x = 702 kN

FBD element 1

(1) =

702 kN

1  102 m 2 = 70.2 MPa

Similarly

= 70,200 KPa

(2) = 70.2 MPa F1x = – 702 kN, F3x = 702 kN

15.3

562

{f } = [T ]T {f}, f1x   ETA f 1x(2) f 3(2) x

25, 200  =    25, 200 

{f } = [T ]T {f} f1x f1 y f3 x f3 y

0 0   f 1x  25, 200  0.707  0.707 0.707 0.707  0 0   0  =    0 0.707  0.707   25, 200  0    0 0 0.707 0.707 0   

f1x = –17,816 lb, f1y = –17,816 lb f3x = 17,816 lb, f3y = 17,816 lb Boundary conditions u2 = v2 = u3 = v3 = u4 = v4 = 0 1.354 0.354 u1  F1x  17,816    = 500000 0.354 1.354 v1  F1 y   17,816 Solving u1 = v1 = – 0.0209 in. By Equation (14.1.57) u1

(1) =

E v [– C – S C S] 1 L u2

– {0}

v2   0.0209  6  0.0209  30 10  (1) = [0 – 1 0 1]   120 in. 0     0 (1)  = 5219 psi (T) u1

(2) =

E [– C – S L

v1 u3

– ET

v3 563

 0.0209   –0.0209    30 106   = [– 0.707 – 0.707 0.707 0.707]  0  120 2  0   10500

(2) = – 7381 psi (C) (3) = 5219 psi (T) 15.4

{f (1)} =

f 1x

f 2x

E TA  16800 =   lb E TA 16800

{f } = [T ]T {f} f1x f1 y f2 x f2 y

 1  2  3 =  2 0  0 

 3 2

1 2

1 2 3 2

0   f   16800   1x   0   f1y  0     3 f 2x  16800   2   f 0   1   2y  2

f1x = 8400 lb f1y = 14549 lb f2x = -8400 lb f2y = -14549 lb Boundary conditions u2 = v2 = u3 = v3 = u4 = v4 = 0 Global equations 2 in.2 30 106  F1x  8400   = 120 in.  F1 y  14549 

3 4

3 3 4

8400 = 216,506 u1

u1 = 0.0388 in.

14549 = 1,149,519 v1

v1 = 0.0127 in.

564

(1) =

v1 E [– C – S C S] u2 L

– ET

v2 6

30  10  1  3 1  2(120)   2 2 2 3

0.0388  3  0.0127    2  0   0 

– 30  106  7  10–6  -40°F

(1) = (216506) (-0.0304) + 8400 = 1827 psi (T)

(2) =

30 10 120

 0.0388  0.0127    [0 – 1 0 1]  –0  0   0 

(2) = -3164 psi (C)

(3) =

30  10  1 3 1   2(120)  2 2 2 3

0.0388  3  0.0127  – 0  2  0   0 

(3) = 1827 psi (T) 15.5

[k(1)] = [k(3)] =

AE L(1) AE (3)

1 1

1 , 1

1 1

[k(2)] =

565

AE L(2)

1 1

E TA E TA

{f (1)} = Global equations 1 AE 1 1 L 0 0

1 1 3

1 3 1 3

1 3

1 3 1 3

1 3

0 1 3

0 u2

E TA F1x E TA F3 x F4 x

0 0

Solving equation (2) above AE 5 L 3

u2 = ETA u2 =

3 TL 3(12  106 )(40)(1m) = 5 5

u2 = 2.88  10–4 m Global nodal forces F1x F2 x F3 x



2  10

2



200  10

F4 x



1  1  0  0

1

1 32  13  13

 13 1 3 0

0 0  1   3 2.88  104    0 0  1    0 3  

E TA E TA 0 0

ETA = (200  109) (12  10–6) (40°) (2  10–2) = 1,920,000 N = 1,920 kN F1x = – 1,152 + 1,920 = 768 kN F2x = 1,920 – 1,920 = 0 F3x = – 384 kN F4x = – 384 kN FBD

(1) = [C ] {d} – ET

566

u1  0  v  0  1  C S]   E T 4  u2  2.88  10  v  0   2 

E [– C – S L

E (2.88  10–4) – ET L

= 57.6 – 96

(1) = – 38.4 MPa (C)

(2) =

200  109 [– 1 3m

= – 192  105

2.88  104    0   0]  0 0     0

N m2

(2) = – 19.2 MPa (C) = (3) 15.6

1 4

[k(1)] =

AE 3.46m

3 4

1 4

3 4

1 4

3 4 3 4

Symmetry

0 AE 0 [k(2)] = 3m 0 0

0 1 0 1

0 0 0 0

0 1 0 1

567

3 4 3 4

1 4

[k(3)] =

AE 3.46 m

3 4 3 4 3 4 3 4

1 4 3 4 1 4

Symmetry

Boundary conditions u2 = v2 = u3 = v3 = u4 = v4 = 0 Thermal forces f 1x(2) = – ETA, f 3x(2) = ETA

Convert to global forces using {f } = [T ]T {f } f1(2) x f1(2) y

0 1 0 0

f3(2) x (2) f3 y

1 0 0 0

0 0 0 1

0 0 1 0

E TA 0 = E TA 0

0 E TA 0 E TA

 f1y(2) = –ETA = –(70  106) (23  10–6) (– 30°C)  (4  10–2 m2) = 1932 kN f3(2) y = – 1932 kN

Assemble equations {F} = [K ] {d} – {F0} u1 0 AE 0.5 3.46 0 2.65 v1

 0  =   1932 

Solving u1 = 0 v1 =



1932(3.46)

2.65 4  10

2

 70  10  6

= 9.0  10–4 m

Element forces {f } = [k ] {d } – { f 0 } = [k ] [T]* {d} – { f 0 } u1 (1)  f '1x   (1)  =  f '2 x 

AE 3.46

1 1

1 C 1 0

S 0 0 C

0 S

v1 u2 v2

568

AE  C 3.46  C

S S

u1  0   4   S  v1  9  10    S  u3  0  v  0   3 

C C

 4  102  70  106   23  9  104  3.46

 3  4  2  9  10   

 630  =   kN 630 

Stresses

 = [C {d} – E  T or (1) = f 2 x A

Element 1

(1) =

630 4  102

= – 15.75 MPa (C)

Element 2 u1  0   6 4  70 10 v  9  10  (2) = [0 –1 0 1]  1  3.0 m 0     0

– 70  106 (23  10–6) (– 30°C) = – 21000 + 48300 = 27300 kPa

(2) = 27.3 MPa (T) (3) = (1) = – 15.75 MPa (C) 15.7

ETA=

210 106

kN m2

10 6  30°C  (3  10–2) C

= 2.27  103 kN

569

0 AE 0 [k (1)] = L 0 0

0 1 0 1

0 0 0 0

0 1 0 1

0 0 0 0

1 0 1 0

0 0 0 0

E TA

f 0(1) =

E TA

[k(2)] =

AE L

f 0(2)

0 0

1 0 1 0

Transform initial forces to global forces f1(1) x f1(1) y f 2(1) x f 2(1)y

0 1 1 0 =  0 0  0 0

0 0   E TA  0      0 0  0    E TA   =   0 1  E TA   0   1 0   0    E TA 

Assemble global equations 3 10 2 m2 210 106 kN2 m

where Solving

1 0 0 1

0 2.27 103

u2 = v2 = u3 = v3 = 0 u1 = 0

v1 = – 3.6  10–4 m

Stresses

(1) =

E L

0 ||

1 ||

0 ||

1 ||

S C S

0 3.6 10 4 0

– ET

210  109 (3.6  10–4 m) – 210  109  12  10–6  30 1m

(1) = 0 210 10 9 (0) – 0 = 0 1m For statically determinate structure thermal stresses are zero.

(2) =

15.8 570

(1)

[k ] =

2 30 106

1 1

60 6

[k ] = 2 15 10 60 (2)

1 = [k(3)] 1

Global equations  2.5 2.5 u1  0  F1x  31200  2  24000  106    =  31200  2  24000  2.5 2.5 u2   where

(1)

f (1) = – ETA = – 30  106  6.5  10–6  80  2 = –31200 lb  –31200  (3) {f (1)} =   = {f } 31200  

f (2) = 15  106  10  10–6  80  2 = – 24000 lb 24000  {f (2)} =   lb  24000 

Solving Equation (1) above u2 = 0.0346 in. Stresses

st = =

AE u2 – ET L A 30 106 60

(0.0346) – 30  106 (6.5  10–6) 80

= 17280 – 15600

st = 1680 psi (T) sr =

E u2 – ET L

571

15 106 (0.0346) – 15  106 (10  10–6)  80 60

= 8640 – 12000 = – 3360 psi 15. 9 A uniform temperature increase of 10°C in each element yields zero stress for this special symmetric arrangement of the truss elements. See the table and figure 1 below showing the stresses from Autodesk to be zero in each element of the truss. Note that if the truss is not symmetric as shown in figure 2 and then is uniformly heated, the middle element has a stress of –3.46 MPa in it, while the top element has a stress of 2.83 MPa in it. **** 3-D truss elements Number of elements = 3 Number of materials = 1 **** Nodal stresses for 3-D truss elements El. # -------1 1 2 2 3 3

LC -----1 1 1 1 1 1

ND ------I J I J I J

Stress ---------------–1.118E-08 1.118E-08 –1.118E-08 1.118E-08 –1.118E-08 1.118E-08

Force --------------–1.341E-11 1.341E-11 –1.341E-11 1.341E-11 –1.341E-11 1.341E-11

When we uniformly heat the truss the stresses go to zero for this symmetric structure.

When we uniformly heat the truss the stresses are not zero for this unsymmetric statically indeterminate structure.

572

15.10 Bodies that are statically indeterminate will have stress due to uniform temperature change. (see figure 2 in solution to P 15.9) except in special symmetry cases (see figure 1 and table of results in P15.9). Also see, for instance, example 15.1, figure 15-5 and P 15.3, P 15.4 and P 15.6.

15.11

[k st] = A L

A Est L

1 1

1 A EAl , [k A1] = 1 L

1 1

F1x

Est

st TA

Est

EAl

Est

EAl

Est

EAl

Est

EAl

1 1 EA

EAl

Al TA

Boundary conditions u1 = 0 A (Est + EA1) u2 = (Est st + EA1 A1) T A L

u2 =

Est

EA1 A1 TL Est EA1

30 106

6.5 10 6 10 106

13 10 6 TL

(30 10) 106

u2 = 8.125  10–6 TL 573

= (8.125  10–6) (50°F) L u2 = 406.25  10–6 L

st = [C ] {d} – T

Est (– 1 L

0 1 0)

0 0 406.25 10 6 L

– Est st T

= 30  106  406.25  10–6 – 30  106  6.5  10–6  50 = 12187.5 – 9750

st = 2437.5 psi (T) A1 = 10  106  406.25  10–6 – 10  106  13  10–6  50 = 4062.5 – 6500

A1 = – 2437.5 psi (C) st = – A1 15.12 To close gap of 0.005 in. gap = br TLbr + m T Lm T = =

gap br Lbr

m Lm

0.005 in. 11.3 10

1 in. 14.5 10 6

1.5 in.

T = 151.3°F to close gap br close

= br T Lbr = 11.3  10–6  151.3°  1 in. = 0.0017097in.

m close

= m T Lm = 14.5  10–6  151.3  15 in. = 0.0032908 in.

(a)

[k br] =

Abr Ebr Lbr

1 1

1 A E , [k m] = m m 1 Lm 574

1 1

Abr Ebr Lbr

F1x

Ebr

Am Em Lm

u1 0 u2 u3 0

Ebr

br TAbr

Em m TAm

Am Em u2 = – Em m TAm + Ebr br TAbr Lm

u2 =

Am Em Lm

br TAbr

Em m TAm F3 x

Abr Ebr Lbr

Am Em Lm Am Em Lm

mTAm Abr Ebr Lbr

Ebr brTAbr Am Em Lm

4.5 106 14.5 10 6

0.15 in.2 15 106 11.3 10 6 6

0.10 15 10 1

0.15 4.5 10 1.5

0.1in.2

100

u2 = 3.673 10–4 in. 

br =

Ebr u2 – Ebr brT Lbr

Ebr Lbr

1.5 106 1

mTAm Abr Ebr Lbr

Ebr brTAbr Am Em Lm

– Ebr br T

4.5 106 14.5 10 6 0.15 in.2 15 106 11.3 10 6 0.1in.2 0.10 15 106 1

0.15 4.5 106 1.5

– 15  106  11.3  10–6  100 = 15  106

9.7875 16.95 1.5 0.45

106

107.4375 100 – 16950 1.95

10743.75 –16950 1.95

100 – 16950

= 5509.6 – 16950

br = – 11440 psi (C) Fbr = (– 11440 psi) (0.1 in.2) = – 1144 lb 575

100°

m = [C {d } – Em m T 3.673 10 4 in. u2

Em [– 1 Lm

0 1 0]

0 v2 0 u3

– Em m T

0 v3

4.5 106 (–3.673  10–4 in.) – 4.5  106  14.5  10–6  100°F 1.5 in.

= –1100 – 6525

m = – 7625 psi (C) Check equations Fm = m Am = (– 7625) (0.15 in.2) Fm = – 1144 lb Same as Fb = – 1144 lb. So equilibrium check satisfied.

 Reactions RL and Rm equal but opposite in direction. 15.13

i i

{fT} =

EtT 2(1 v)

j j m m

i = yj – ym = 0 – 1 = – 1

i = xm – xj = 2 – 4 = – 2

j = ym – yi = 1 – 0 = 1

j = xi – xm = 0 – 2 = – 2

m = yi – yj = 0 – 0 = 0

m = xj – xi = 4 – 0 = 4

576

 1 2    12.5 106 10 106 1 50   1 {fT} =   2(1  0.3) 2   0    4







 4464  8929    4464  {fT} =   lb 8929   0     17857 

15.14 E =70 GPa, v = 0.3 T = 50°C,  = 23  10–6/°C t = 5 mm

1 = – 250 mm,

2 = 250 mm, 3 = 0

1 = 0, 2 = – 500 mm, 3 = 500 mm

23106   {f }= T

  0.25m   0    9 70 10  0.005m  50C  0.25m    2(1  0.3)  0.50 m   0     0.50 m 



 71.9   0    71.9  {fT} =   kN 143.8  0    143.8

15.15

577

i = yj – ym = 0 – 4 = – 4,

i = xm – xj = 0 – 6 = – 6

j = ym – yi = 4 – 0 = 4,

j = xi – xm = 0 – 0 = 0

m = yi – yj = 0 – 0 = 0,

m = xj – xi = 6 – 0 = 6

4  6    6 6 7.0 10 30 10 1 50F   4  {fT} =   2(1  0.3)  0  0    6 30, 000   45, 000     30, 000  =   lb  0   0     45, 000 







15.16

i i

{fT} =

EtT 2(1 v)

j j m m

578

i = yj - ym j = ym - yi m = yi - yj

i = xm - xj j = xi - xm m = xj - xi

i = – 0.4 m j = 0.4 m m = 0

i = – 0.4 m j = 0 m = 0.4 m 0.4 0.4

{fT} =

12 10

210 10

0.01

2 1 0.25

20 C

0.4 0 0 0.4

134.4

{fT} =

134.4 134.4 0 0

134.4

15.17

Thermal force matrix Element 1 i = 1, j = 2, m = 5

i = yj – ym = 0 – 10 = – 10,

i = xm – xj = 20 – 40 = – 20

j = ym – yi = 10 – 0 = 10,

j = xi – xm = 0 – 20 = – 20

m = yi – yj = 0 – 0 = 0,

m = xj – xi = 40 – 0 = 40

579

{ fT(1) } =

12.5 10

10 10

1 50

20 10 20 0

2(1 0.3)

40 44643 89286

{ fT(1) } =

44643 – 89286 0 178572

Element 2 i = 2, j = 3, m = 5

i = 20 – 10 = 10, j = 10 – 0 = 10, m = 0 – 20 = – 20,

{ fT(2) } = 4464.3

i = 20 – 40 = – 20 j = 40 – 20 = 20 m = 40 – 40 = 0

44643

20 10

–89286

20 20 0

44643 89286 – 89286 0

Element 3 i = 3, j = 4, m = 5

i = 20 – 10 = 10 j = 10 – 20 = – 10 m = 20 – 20 = 0

{ fT(3) } = 4464.3

j = 20 – 0 = 20 j = 40 – 20 = 20 m = 0 – 40 = – 40

44643 10 89286 20 44643 10 = 89286 20 0 0 –178572 40

Element 4 i = 4, j = 1, m = 5

i = 0 – 10 = – 10

i = 20 – 0 = 20

j = 10 – 20 = – 10

j = 0 – 20 = – 20

m = 20 – 0 = 20

m = 0 – 0 = 0 580

{ fT(4) } = 4464.3

–44643

20 10

89286

20 20

44643

89286 89286

{F0} = [K] {d} By direct superposition, we have 3 2 89, 286 6 178,572 89, 286 178,572 89, 286 10 106 = 178,572 4.16 89, 286 178,572 0 0

0.1 0.2 0.2 2.6 3 2 6

Symmetry u1 v1

0 0 0.1 0.2 3

0 0 0.2 2.6 2 6

0.1 0.2 0 0 0.1 0.2 3

0.2 2.6 0 0 0.2 2.6 2 6

3 2 3 2 3 2 3 2 12

2 6 2 6 2 6 2 6 0 24

0 0

 u5 v5

Solving 0=

10 106 12 us  us = 0 4.16

10 106 24 vs  vs = 0 4.16

Stresses {} = { L} – { T} 



{ L} = [D] [B] {d } = 0 as {d } = 0 

 {} = – { T} = – [D] { T} 



Element 1

581

{} =

E 1 v2

1 v v 1

0 0

1 v 2

T T 0

 x  1 0.3 0   10 106 0.3 1 0  y  = 1 0.32   0 0 0.35  xy 

6.25 10 4 6.25 10 4 0

8929 8929 psi 0

Since [D] and { T} are same for all elements, all element stresses are equal. 

15.18

Based on use of symmetry us = vs = 0 (Also see solution to Problem 15.17)  {} = { L} – { T} = [D] [B] {d } – [D] {T} 



{} = – [D] { T} 

All stresses in elements are equal x y xy

E 1 v2

1 v v 1

0 0

1 v 2

T T 0

0.25  1 210 10  = 0.25 1 1  0.252   0 0 9

582

6 0  12 10 (20)      0  12 106 (20)   10.25   0 2    

67.2 67.2 MPa 0

x y

15.19 For bar with

 = 0 1  

x  L

T = constant By Equation (15.1.18) {fT} = A

L 0

[ B ]T [D] T dx

L, A, E, T constant  T =  T 

{fT} = A

1 L 1 L

L 0

AE 0T L

x E 0 1   T dx L  X L X L

1 1

L 0

 

 X2 AE 0T  x  2 L =  2 L  x  X2 L 

AE 0T L

L L

{fT} =

3 AE 0T 2

1 1

     L

0 L 0

L 2 L 2

15.20

T = t1 + t2 x By Equation (15.1.18) {fT} = A

L 0

[ B ]T [D] {T} dx

T = t1 + t2 x

[N] = 1

x x L L 583

1 L 1 L

{fT} = A

= AE 

AE L

{fT} =

E[N] {t} dx 1 L 1 L

x x L L

1 1

x L x L

L 0 L 0

1 Lx t1 1 Lx t1

L 0

AE L

x2 2L

x 2L

L L

L 2 L 2

L 0 L 0

t1 t2 x t L 2 x t L 2

t1 t2

L x2 t 2L 2 0

Lt 2 2 L t 2 2

t1 t1

For

t1 = t2 = T (constant temperature over element) TL AE T AE {fT} = = TL AE T L Equation (15.1.18)

15.21 {fT} =

[ B ]T [D] {T} d s

{T} = T

1 1 0

(1)

(2)

Using centroidal approximation {fT} =  [ B ]T [D] {T} d s s

= 2 r A  T B

[D]

1 1 0

(3)

where for axisymmetric case

584

[D] =

E (1 v)(1 2v)

1 v v v 1 v

v v

0 0

1 v

1 2v 2

(4)

substituting (4) into (3) and multiplying

{ fT} =

2 rA E T B 1 2v

1 1 1 0

15.22 Using modified CSFEP to account for thermal stress due to element temperature change. a) Pin-roller supports at bottom

b) Pin-pin supports at bottom

585

Fixing the bottom nodes creates much more stress. 15.23 Data File 0 Verification.5 4, 2, 2, 0, 2, 0 1, 1 0.30E+8, 0.33333, 0., 0.1, 1.E–5 .15E+8, 0.25, 0., 0.1, 5.E–5 1, 3, 0., 0., 0., 0. 2, 2, 1., 0., 0., 0. 3, 0, 0., 1., 0., 0. 4, 0, 1., 1., 0., 0. 1, 1, 2. 3, 3. 1. 80 2, 2, 4, 3, 3, 2. 50.

Run using CSFEP modified for temperature changes in elements verification.5 0 INPUT TABLE 1.. BASIC PARAMETERS NUMBER OF NODAL POINTS. . . . . . . . . . . . . NUMBER OF ELEMENTS. . . . . . . . . . . . . . . . . NUMBER OF DIFFERENT MATERIALS. . . . . NUMBER OF SURFACE LOAD CARDS . . . . . 1 = PLANE STRAIN, 2 = PLANE STRESS. . . . BODY FORCE (1 = IN – Y DIREC., 0 = NONE)

4 2 2 0 2 0

0 INPUT TABLE 2.. MATERIAL PROPERTIES MATERIAL MODULUS OF POISSON’S MATERIAL MATERIAL ALPHA NUMBER ELASTICITY RATIO DENSITY THICKNESS 1 0.3000E+08 0.3333E+00 0.0000E+00 0.1000E+00 0.1000E–04 2 0.1500E+08 0.2500E+00 0.0000E+00 0.1000E+00 0.5000E–04 INPUT TABLE 3.. NODAL POINT DATA NODAL POINT TYPE X Y 1 3 0.0000E+00 0.0000E+00 586

X-DISP. OR LOAD 0.0000E+00

Y-DISP. OR LOAD 0.0000E+00

2 3 4

2 0 0

0.1000E+01 0.0000E+00 0.1000E+01

0.0000E+00 0.1000E+01 0.1000E+01

0.0000E+00 0.0000E+00 0.0000E+00

0 INPUT TABLE 4.. ELEMENT DATA ELEMENT 1 2

GLOBAL INDICES OR ELEMENT NODES 1 2 3 4 MATERIAL TEMP 1 2 3 3 1 0.800E+02 2 4 3 3 2 0.500E+02

0 OUTPUT TABLE 1.. NODAL DISPLACEMENTS NODE 1 2 3 4

U = X-DISP. 0.00000000E+00 0.98888970E–03 –0.75555370E–03 0.13194460E–02

V = Y-DISP. 0.00000000E+00 0.00000000E+00 0.98888970E–03 0.20750000E–02

1OUTPUT TABLE 2.. STRESSES AT ELEMENT CENTROIDS ELEMENT X Y SIGMA(X) SIGMA(Y) 1 0.33 0.33 8.5000E+03 8.5000E+03 2 0.67 0.67 –8.5000E+03 –8.5000E+03 SIGMA(1) SIGMA(2) 1.7000E+04 –3.9063E–03 –1.9531E–03 –1.7000E+04

TAU(X, Y) –8.5000E+03 8.5000E+03 ANGLE 0.0000E+00 4.5000E+01

15.24 Solve Problem 15.3 using the Autodesk Program.

587

15.25 For the plane truss shown in Figure P15-6, bar element 2 is subjected to a uniform temperature drop of T = 30°C. Let E = 70 GPa, A = 4  10^–4m^2, and alpha = mm

23  10–6 mm . Determine the stresses in each bar and the displacement of node 1. °C

588

Figure 1 Axial Stress As shown in Figure 1, the stress in bar 1 and 3 is 15.76 MPa (C) and the stress in bar 2 is 27.3 MPa (T).

Figure 2 Displacement in Y Direction 589

Figure 2 shows the Y displacement at node 1 is 0.0009 m in the positive Y direction. There is no displacement at node 1 in the X direction. 15.27

590

Chapter 16 16.1

[m(1)] =

AL 6

[M] =

AL 6

1 2 2 1 , [m(2)] = 1 2

AL 6

2 3 2 1 1 2

2 1 0 1 4 1 0 1 2

16.2

(a) Lumped mass matrix

[m(1)] =

1 2 AL 1 0 , [m(2)] = 2 0 1

[m(3)] =

3 4 AL 1 0 2 0 1

1 0 [M] = 0 0

0 0 2 0 0 2 0 0

0 0 0

2 3 AL 1 0 2 0 1

AL 2

(b) Consistent mass matrix [m(1)] =

AL 6

[m(3)] =

AL 2 1 6 1 2

2 1 , 1 2

[m(2)] =

591

AL 2 1 6 1 2

[M] =

2 1 AL 1 4 6 0 1 0 0

0 0 1 0 4 1 1 2

16.3

 1 1 0  AE   [K] = 1 2 1  L  0 1 1   

[M] =

2 1 0 AL 1 4 1 6 0 1 2

([K] – 2 [M]) {X} = 0 with x1 = 0 AE L

2 1

divide by AL and let  =

1 1

AL 4 1 6 1 2

x2 x3

0 0

2 = 

Let



2 3

22 –

4 2  + 2 – 2 – 3 9 3

( –

) – (– –

)2 = 0 2

5 7 2  +  =0 3 36 60 36 2 2 –  +  =0 7 7

2 –

592

1,2 = =

60 7



4 (1) 36 7

2 8.571

7.273 2

1 = 0.649  

1 2

1 =

2 =

2 = 7.922 

= 0.806

1 2

= 2.815

16.4 (a) Two equal length elements

From Problem 16.3 results 1 2

1 = 0.806 =

E 2

, 2 = 2.815

1 2

30 106 0.00073 30

= 45.66 

106 s2

1 = 0.806

45.66 106 = 5.446  103

rad s

2 = 2.815

45.66 106 = 19.02  103

rad s

(b) 3 equal length elements

[K] =

1 1 0 0

1 2 1 0

0 1 2 1

0 0 AE 1 L 1

593

0 0 1 0

2 1 AL 1 4 6 0 1 0 0

[M] =

4 1 1 2

x1 = 0 ([K] – 2 [M]) {X} = 0

Now 

2 1

AE L

2 1

4 1 0 1 4 1 =0 6 0 1 2

0 1

2 1

2 3

6 8 3

4 2

3 3

6 6

4 1 0 AL 1 4 1 =0 6 0 1 2

2 1

0 1

4 9

6 2

3 2

= 0, 1 = 3

or 13 2 11 2 –  + = 0 3 36

2, 3 =

132

4 13 36 2

2 13

2 = 9.873 , 3 = 0.2805  =

30 106 = L2 0.00073 20

1 = 2 =

0.2805

= 1.0274  108

= 5.368  103

= 17.556  103

rad s

594

3 =

= 31.85  103

9.873

rad s

16.5

t0 = 0 d0 = d0dot = 0 d0dotdot =

ft 1 (50 – 0) = 25 2 2 s 0.03 2  25 = 0.01125 ft 2

d–1 = 0 – 0 + t1 = 0.03 s d1 =

1 {(0.03)2 (50) + (2  2 – 0.032 000) 0 – 2  0.01125} 2

= 0.01125 ft d2 =

1 {(0.03)2 (66.67) + (2  2 – 0.032  2000  0.01125} 2

d2 = 0.04238 ft d1dotdot =

1 {66.67 – 2000  0.01125} 2

d2dotdot = 22.09 d1dot =

ft s2

0.04238 0 ft = 0.71 2 0.03 s

t2 = 0.06 s 1 d3 = {(0.03)2 (83.33) + (2  2 – (0.03)2  2000)  (0.04238) – 2  0.01125} 2 d3 = 0.07287 ft 1 d2dotdot = {83.33 – 2000  0.04238} 2 d2dotdot = – 0.715 d2dot =

ft s2

0.07287 0.01125 ft = 1.03 2 0.03 s 595

t3 = 0.09 s 1 {0.032  100 + (2  2 – 0.032  2000)  (0.07287) – 2  0.04238} 2 d4 = 0.08278 ft

d4 =

d3dotdot =

1 (100 – 2000  0.07287) 2

d3dotdot = – 22.87 d3dot =

ft s2

0.08278 0.04238 ft = 0.67 2 0.03 s

t4 = 0.12 s 1 d5 = {(0.03)2 75 + (2  2 – 0.032  2000) (0.08278) – 2  0.07287} 2 d5 = 0.05194 ft 1 d4dotdot = (75 – 2000  0.08278) 2 d4dotdot = – 45.28 d4dot =

ft s2

0.05194 0.07287 ft = – 0.35 2 0.03 s

t5 = 0.15 s d6 =

1 {(0.03)2 50 + (2  2 – 0.032  2000) (0.5194) – 2  0.08278} 2

d6 = – 3.146  10–3 ft d5dotdot =

1 {50 – 2000 (0.05194)} 2

d5dotdot = – 26.94 d5dot =

ft s2

3.146 10 3 0.08278 ft = – 1.43 2 0.03 s

Summary t, s 0 0.03 0.06 0.09 0.12 0.15

F(t) lb 50 66.67 83.33 100 75 50

di, ft 0 0.01125 0.04238 0.07287 0.08278 0.05194

didotdot 25 22.09 –0.715 –22.87 –45.28 –26.94

ft s

didot

ft s

0 0.71 1.03 0.67 –0.35 –1.43

(b) By Newmark’s method 596

1 1 ,= M = 2 slugs 6 2 50 lb F= t + 50 K = 2000 0.09 ft F(0.03) = 555.6 (0.03) + 50 = 66.67 lb

=

F i 1 = Fi + 1 +

Fi = 66.67 +

1 6

0.03

d0dotdot = M –1 (F0 – K d0) =

and

1 6

0.03

t 2 didotdot 1 2

0.03 0

1 6

0.03 2 d0dotdot

ft 50 2000 (0) = 25 2 2 s

 Fi = 66.67 +

1 2

t didot

1 2

1 6

0.03 2 25

= 166.67 lb 

d1 =

166.67 F1 = K K

and 1

K = K +

t 2

= 2000 + 

d1 =

0.03 2

1 6

(2) = 15333

166.67 = 0.01087 ft 15333

d1dotdot = =

1 1 6

d1dotdot = 22.47

0.03

t d0dot

0.01087 0 0

t 2

1 2

0.03 2

d0dotdot 1 2

1 25 6

ft s2

d1dot = d0dot + t [(1 – ) d0dotdot +  d1dotdot] = 0 + (0.03) d1dot = 0.71205

1 25 2

1 22.47 2

ft s

Table below summarizes the results using Newmark’s method 597

Time 0 0.03 0.06 0.09 0.12 0.15 0.18

Displacement 0 0.01087 0.039319 0.069606 0.081832 0.059363 0.011632

Velocity 0 0.711957 1.084121 0.825234 –0.13384 –1.31425 –1.62917

Acceleration 25 22.46377 2.347196 –19.6063 –44.3317 –34.3627 13.3684

Force 50 66.66667 83.33333 100 75 50 50

Force prime 166.6667 602.8986 1067.297 1254.753 910.2281 178.3512

16.6

(a) Using central difference d0 = 0,

d0dot = 0

t = 0.02 s Step 1

t = 0.02s

F1 = 16 lb

[M] = m = 2 slugs, [M–1] = [K] = k = 1200 d0dotdot =

lb ft

ft 1 [20 – 1200 (0)] = 10 2 2 s

{d–1} = 0 – (0.02) (0) + {d1} =

1 2

0.022 (10) = 0.002 ft 2

1 [(0.02)2 (20) + {2(2) + (0.02)2 (1200)} (0) – 2 (0.002)] 2

= 0.002 ft {d2} =

1 [0.022 (16) + {2(2) – 0.022 (1200)} (0.002) – 2(0)] 2

= 0.00672 ft d1dotdot = d1dot = Step 2

ft 1 (16 – 1200 (0.002)) = 6.8 2 2 s 0.00672 0 ft = 0.168 2 0.02 s

t = 0.04 s

F2 = 12 lb 598

d3 =

1 [0.022 (12) + {2(2) – 0.022(1200)} (0.00672) – 2 (0.002)2] 2

d3 = 0.01223 ft d2dotdot = d2dot = Step 3

ft 1 (12 – 1200 (0.00672)) = 1.968 2 2 s

0.01223 – 0.002 ft = 0.2558 2 0.02 s

t = 0.06 s d4 =

F3 = 8 lb

1 [0.022 (8) + 3.52 (0.01223) – 2 (0.00672)] 2

= 0.0164 ft d3dotdot = d3dot = Step 4

ft 1 (8 – 1200 (0.01223)) = – 3.338 2 2 s 0.0164 – 0.00672 ft = 0.242 2 0.02 s

t = 0.08 s d5 =

F4 = 4 lb

1 [0.022 (4) + 3.52 (0.0164) – 2 (0.01223) 2

= 0.01743 ft d4dotdot = d4dot = Step 5

ft 1 (4 – 1200 (0.01640)) = – 7.84 2 2 s 0.01743 – 0.01223 ft = 0.13 2 0.02 s

t = 0.10 s d6 =

F5 = 0

1 [0.022(0) + 3.52 (0.01743) – 2 (0.01640)] 2

= 0.01428 ft

d5dotdot = d5dot =

ft 1 (0 – 1200 (0.01743)) = – 10.46 2 2 s 0.01428 0.01640 ft = – 0.053 2(0.02) s

Summary t, s

d, ft

ddot,

ft s

599

ddotdot, 2 s

0 0.02 0.04 0.06 0.08 0.10

0 0.002 0.00672 0.01223 0.01640 0.01743

0 0.168 0.2558 0.242 0.130 –0.053

10 6.8 1.968 –3.338 –7.89 –10.46

(b) Newmark’s time integration method (Mathcad solution) F0 = 20 lb

lb ft

K = 1200

1 6 At time t = 0

Assume  =

=

d0 = 0 ft

M = 2 slug

M=2

lb s 2 ft

t = 0.02s

1 for linear acceleration within each time step 2

d0dot = 0

ft s

Acceleration at t = 0 K d0 M Displacement at t = 0.02

d0dotdot =

Kprime = K + F1 =

d0dotdot = 10

ft s2

Kprime = 2.6  103

4 F0 5

lb in.

F1 = 16 lb M

F1prime = F1 +

t d 0do t

1 2

t 2 d0 dotdot

F1prime =

d0 dotdot

d1dotdot =

56 lb d1 =

F1prime

Acceleration at t = 0.02 1 d1dotdot = 6.923

d1 = 1.795  10–3 ft

K prime

t 2

t d0dot

1 2

s2 Velocity at t = 0.02

d1dot = d0dot + (t)[(1 – ) d0dotdot +  d1dotdot] d1dot = 0.169 Displacement at t = 0.04 F2prime = F2 +

F2 = M t

3 F0 5

F2 = 12 lb

t d1dot

1 2

ft s

t 2 d1dotdot

F2prime = 195.07 lb

600

d2 =

F2prime

d2 = 6.252  10–3 ft

K prime

Acceleration at t = 0.04 1

d2dotdot =

t d1dot

t 2

1 2

d1dotdot

d2dotdot = 2.249

s2 Velocity at t = 0.04

d2dot = d1dot + (t) [(1 – )d1dotdot +  d2dotdot] d2dot = 0.261

ft s

Displacement at t = 0.06 F3 =

2 F0 5

F3 = 8 lb M

F3prime = F3 +

t d2dot

1 2

t 2 d 2dotdot

F3prime =361.136 lb d3 =

F3prime

d3 = 0.012 ft

K prime

Acceleration at t = 0.06 d3dotdot = d3dotdot =– 2.945

t d 2dot

t 2

1 2

d 2dotdot

s2 Velocity at t = 0.06

d3dot = d2dot + (t) [(1 – ) d2dotdot +  d3dotdot] d3dot = 0.254

ft s

Displacement at t = 0.08 1 F4 = F0 5 F4prime = F4 +

F4 = 4 lb M t

t d3dot

1 2

t 2 d3dotdot

F4prime = 491.85 lb d4 =

F4prime K prime

d4 = 0.016 ft

Acceleration at t = 0.08

601

d4dotdot = d4dotdot = – 7.459

1 2

t 2

t d3dot

d3dotdot

ft s2

Velocity at t = 0.08 d4dot = d3dot + (t)[(1 – )d3dotdot +  d4dotdot]

d4dot = 0.15

ft s

Displacement at t = 0.10 F5 =

0 F0 5

F5 = 0 lb M

F5prime = F5 +

1 2

t d4dot

t 2 d 4dotdot

F5prime = 533.071 lb d5 =

F5prime

d5 = 0.017 ft

K prime

Acceleration at t = 0.10 1

d5dotdot = d5dotdot =– 10.251

t 2

t d 4dot

1 2

d 4dotdot

ft s2

Velocity at t = 0.10 d5dot = d4dot + (t) [(1 – ) d4dotdot +  d5dotdot]

d5dot = – 0.027

ft s

K = 1200

lb ft

M = 2 slug

M=2

lb s 2 t = 0.02s ft

= 1

At time t = 0 d0 = 0 ft

d0dot = 0

ft s

Acceleration at t = 0 d0dotdot =

K d0 M

d0dotdot = 10

ft s2

Displacement at t = 0.02 Kprime = K +

6 t

Kprime = 2.6  103

602

lb in.

F1 =

4 F0 5

F1 = 16 lb M

F1prime = F1 +

[6d0 + 6(t)d0dot + 2(t)2d0dotdot]

t 2

F1prime = 56 lb F1prime

d1 =

d1 = 1.795  10–3 ft

K prime

Acceleration at t = 0.02 6 d1dotdot = 2

(d1 – d0) –

d0dot – 2 d0dotdot

ft d1dotdot = 6.923 2 s

Velocity at t = 0.02 d1dot =

3 t

(d1 – d0) – 2 d0dot –

t 2

d0dotdot

d1dot = 0.169

ft s

Displacement at t = 0.04 F2 =

3 F0 5

F2 = 12 lb 3

F2prime = F2 +

t 2

[6d1 + 6(t)d1dot + 2(t)2d1dotdot]

F2prime = 195.077 lb F2prime

d2 =

d2 = 6.252  10–3 ft

K prime

Acceleration at t = 0.04 6

d2dotdot =

(d2 – d1) –

6 t

d1dot – 2 d1dotdot

d2dotdot = 2.249

ft s2

Velocity at t = 0.04 d2dot =

3 t

(d2 – d1) – 2 d1dot –

t 2

d1dotdot

d2dot = 0.261

ft s

Displacement at t = 0.06 F3 =

2 F0 5

F3prime = F3 +

F3 = 8 lb M 2

t 2

[6 d2 + 6  (t) d2dot + 2 (t)2 d2dotdot]

F3prime = 361.136 lb d3 =

F3prime K prime

d3 = 0.012 ft

Acceleration at t = 0.06 603

d3dotdot =

(d3 – d2) –

d2dot – 2 d2dotdot

ft d3dotdot = – 2.945 2 s

Velocity at t = 0.06 3

d3dot =

(d3 – d2) – 2d2dot –

d2dotdot

d3dot = 0.254

ft s

Displacement at t = 0.08 1 F0 5

F4 =

F4 = 4 lb M

F4prime = F4 +

[6 d3 + 6(t)d3dot + 2(t)2d3dotdot]

t 2

F4prime = 491.865 lb F4prime

d4 =

d4 = 0.016 ft

K prime

Acceleration at t = 0.08 6

d4dotdot =

d4dotdot = – 7.459

(d4 – d3) –

d3dot – 2 d3dotdot

ft s2

Velocity at t = 0.08 d4dot =

3 t

(d4 – d3) – 2 d3dot –

d3dotdot

d4dot = 0.15

ft s

Displacement at t = 0.10 F5 =

0 F0 5

F5 = 0 lb M

F5prime = F5 +

[6d4 + 6(t)d4dot + 2(t)2d4dotdot]

F5prime = 533.071 lb d5 =

F5prime

d5 = 0.017 ft

K prime

Acceleration at t = 0.10 d5dotdot =

6 2

d5dotdot = – 10.251

(d5 – d4) –

6 t

d4dot – 2d4dotdot

ft s2

Velocity at t = 0.10 604

d5dot =

3 t

(d5 – d4) – 2d4dot –

t 2

d4dotdot

d5dot = – 0.027

ft s

Newmark’s time integration method. Wilson’s method (Linear acceleration) Assume linear acceleration within each time step  = 1  = 0.167 K 31200  = 0.5 K31200 Summary table Newmark and Wilson same results.

time

F(t) (lb)

di (ft)

0 1 2 3 4 5

0 0.02 0.04 0.06 0.08 0.1

20 16 12 8 4 0

0 0.00179 0.00625 0.01157 0.01576 0.01709

di velocity ( 0 0.169 0.261 0.254 0.150 – 0.027

ft ) s

ft di accel ( 2 ) s 10 6.923 2.249 –2.945 –7.459 –10.251

F

56 195 361 492 533

16.9

Use time step Δt = 2.5 x 10-4 s

605

16.10

AE [K] = L

1 1 0

1 2 1

0 1 = 104 1

1 1 0

1 2 1

0 1 1

1 0 0 1 0 0 AL 0 2 0 = 50 0 2 0 2 0 0 1 0 0 1

[M] =

[K] {d} + [M] {ddotdot} = {F(t)} Find proper time step [M] {ddotdot} + [K] {d} = 0 ([K] – 2 [M]) {d = 0

104

= 0

2 0 0 1

u u

= 606

0 0

400

200

200 200

= 0

2 = 300 ± 100 13  = 25.7 t

3 4

max

rad s

3 2 = 0.058 s 4 25.7

 use t = 0.05 s

=

1 1 ,= 6 2

Displacement Node 3

Velocity Node 3

607

Acceleration Node 3

16.11

Global stiffness matrix

[k] =

EI L3



2

3

v1 12

v2 12

v3 0

6L 12

4 L2 6L

6L 24

2 L2 0

0 12

0 6L

6L 0

2 L2 0

0 12

8L2 6L

6L 12

2 L2 6L

2 L2

Lumped mass matrix v1  v2 2 v3 3

608

1 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0

AL 2

[m] =

0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0

Invoking boundary conditions v1 = 1 = 0 24

2 L2

4 L2

2 0 0 AL 0 0 0 2 0 0 1 0 0 0

–

0 0

Let 2 =  and divide AL and  =

EI A 24 L4

8 L2 6 L3 2 L2

(a)

12 L4 6 L3

12 L4 –6 L3

12 L4

2 6 L3

L2 6

L3 2

L3 4

0 6 L3

11 4

6 L2 2 L2 6 L3 4 L2

12 L4 6 L3

12 L4

– 12 11 3 4 L 6

L2 6

L3 2

L3 4

6 L3

8 L2 6 L3 2 L2

12 L4

2 6 L3

36 2 2 L6 L2

12 L4

32 2 12 L4 L4

36 2 2 L6 L2

4 2 2 L4

36 2 8 L6 L2

36 2 4 L6 L2

36 2 8

4 2

36 2 2

32 2

48 2

12 12 2

36 2

48 2 12

609

12 L4

28 2 L4

12 L4

Let

12 L4

288 3

120 3

12 L4

288 3

2448 4

L12

12 L4

864 3

2448 4

L12

56 2

28 2

48 2

648 3

288 3 L8

= x 56 2 2 864 3 x – x L4 L8

x1, 2 =

x1 = x2 = x1 = x2 =

56x2 –

864 x L4

864 L4

2448 4 =0 L12

2448 2 =0 L8

4(56)

2448 2 L8

2(56) 864 L4

445.1 L4

112 864 L4

445.1 L4

112

12 =

0.62

1 =

0.62EI

22 =

16.52EI

AL4 4

= =

11.69 4

3.74 4

864 L4

198144 L8

112 11.69



x1 =



x2 =



1 =

0.62



2 =

16.52



0.787 EI 2 1 = A L2



2 =



4.06 EI 2 2 = 2 A L

3.74 L4 L4 L4

0.62 EI AL4 1

16.52 EI A L4 1

The exact solution from simple beam theory yields

610

1 =

EI 2 1.875 2 0.879 EI 2 = A 2L A L2

2=

EI 2 4.694 2 5.5 EI 2 = 2 A 2L A L

(a)

Note: L = one-half actual length of beam in the analysis. Expressing answers in terms of full length l = 2L, we obtain 1

3.15 EI 2 0.787 EI 2 1 = = l 2 A A l2 2

and

16.24 EI 2 2 = A l2

(b)

[k(1)] = [k(2)] = [k(3)] =

2l 2

EI L3

4l 2

Symmetry

[m(1)] = [m(2)] = [m(3)] =

1 Al 0 2 0 0

0 0 0 0

0 0 1 0

0 0 0 0

Boundary conditions v1 = 1 = v4 = 4 = 0 | [K] – 2 [M] | = 0 24 EI l

0 8l

Symmetry

Let

=

6l 2

6l 2l 24 0 8l 2

1 0 0 0 0 0 0 1 0 Symmetry 0

EI Al 4 611

2l 2

0 8l 2

Symmetry

Rewrite 4  4 determinant as 2

24l

30l 2

2l 2 2





24l 6

8l 2

30l 2

24l

Evaluate 972 l2 – 192 L2 + 5 l4 = 0

2 =

192l 2 4 5l 2 2 10l

192 l

2 = 6   = 2.45

 = 5.69 Let

3l = L



972

192 l

132l 10l

2 = 32.4 

22.04

4 A 3l

51.23 l

EI A

L = whole length

 1 =

22.04

2 =

51.23

L 2 3

198.4 EI = A L2

EI A

461.07 EI = A L2

EI A

(c)

612

Boundary conditions v1 = v5 = 0 4l 2

[K] =

6l 24

2l 2 0

0 1 6l v2

8l 2

2l 2 4l

0 0 0 Al 0 2 0 0 0 0 0

[M] =

0 1 0 v2 0 2 0 3

l3 1 A 3 8

| [K] – 2 [M] | = 0 4l 2 AE 24

6l 2l 2 24 0

0 6l

0 2

4l 2

2l 2

3l 12 1

4l 2

2.45 l

0 0 0 0 0 0 0

2l 2

Symmetry

=

AE 24

0 24l E

EI A

(d)

Boundary conditions v1 = 1 = v3 = 0

613

[K] =

[M] =

EI l3

8l 2

2l 2

4l 2

2 0 0 0 0 0 0 0 0

AL 2

Divide by Al and let 2 =  and

AL4

2 0 0 AL 0 0 0 2 0 0 0

=

Determinant becomes 24

8l 2

2l 2 4l

Simplifying 672 – 288 l 4 3 = 28 l 4 2 

 = 13.71  =

= 3.703

, 2l = L

14.81 EI 2 = A L2

16.12

614

615

16.13

616

617

16.14 Note that even though damping data was not given in the problem. This solution includes damping.

Let the time step or increment = 0.002 s and the number of time steps = 200. Also use the following data in the Autodesk program

 = 7800

kg , E = 200 GPa, m3

= 0.3, ys = 250 MPa

A = 2  10–2 m2, J1 = 16  10–4 m4, I2 = I3 = 8  10–4 m4, S2 = S3 = 16  10–4 m3 Damping is to be included so use mass damping coefficient Cm =  = 3.00 and stiffness damping coefficient Ck =  = 0.001, Δt = 0.002 s.

16.16

618

619

16.17

For the first 3 natural frequencies found none of them are close to the range of the driving frequency, therefore the driving frequency should be acceptable.

16.18

620

Using Equation (16.8.16) 1 [M ] t

[ K ] {Ti }+ 1 =

1 M t

{Ti} + (1 – ) {Fi} +  {Fi}+ 1

) K

h = 0 assume unit area AK 1 [K] = L 1

= 500 M=

W C

1 1

kg (lm2 )(0.4 cm) m3 2 100 cm m

800 kg ·°C 2500

= 4000

Let [C] =

C AL 1 0 2 0 1 W ·s

(lm )2 (m· C) 1 1 = 0.004 m 1 1

1 0 0 1

1 [M ] t

1 0 0 1

Ws C 1

[K ]

1 [ M ] (1 t

)[ K ]

Then {Ti}+ 1 = C{Ti} as {Fi} = 0, {F i}+ 1 = 0 For [T1] (t = 8 s) and eliminating 1st row and column for boundary condition t1 = 0 we have t2 t3

{T1} =

t4 t5 t6

200 200 = C 200 200 200

where [C is [C with 1st row and column deleted with row and column one deleted

621

1000

500 1000

[K =

0 500 1000

0 0 500 1000

0 0 0 500 500

0 0 8000

0 0 0 8000

0 0 0 0 4000

0 333.3 1666.7

0 0 333.3 1666.7

Symmetry 8000

0 8000

[M = Symmetry 1666.7

M + [K = t

333.3 1666.7

Symmetry

0 0 0 333.3 833.3

666.7 166.7 0 0 0 666.7 166.7 0 0 M – (1 – ) [K = 666.7 166.7 0 t 666.7 166.7 Symmetry 333.3 0.5247

0.2139 0.4839

[C = Symmetry

1 2 3 4 5 6 7 8 9 10

Node t1 s 0 8 16 24 32 40 48 56 64 72 80

1 200 0 0 0 0 0 0 0 0 0 0

0.04472 0.00972 0.00194 0.2057 0.04472 0.00894 0.4839 0.2139 0.0428 0.5247 0.2049 0.4820

Temperatures, °C 2 3 4 200 200 200 159 191 198 135 178 193 120 165 187 109 155 180 101 146 173 94 138 167 88 131 160 84 125 154 79 119 148 76 114 142 622

5 200 199.6 198.2 195.5 191.7 187.1 182.0 176.5 170.8 165.1 159.3

6 200 199.8 199.1 197.5 194.8 191.1 186.7 181.6 176.3 170.7 165.0

16.19

Two element solutions

0.4 cm diameter [K =

AK 1 L 1

1 1

h L 2 1 W 6 1 2 °C

= (0.12567 cm2) 400 + 150

1m 100 cm

W (1.2567 cm)(1 cm) 2 1 2 m °C 1 2 6 100 cm m 2

0.5027 0.5027

W m °C

0.5027 0.5027

0.00628 0.00314 0.00314 0.00628

Using consistent mass [m =

c AL 2 1 1 2 6

W s kg °C

kg m3

375

0.1398 0.0699 W· s 0.0699 0.1398 °C

{f} =

hT PL 1 1 2

8900

0.12567

cm2 1002

1 cm 100

2 1

1 2

0.23561 W 0.23561

Global [K and [M 623

0.50893

0.49951 1.01786

[K = Symmetry

0 0.49951 0.50893

[M =

0.1398 0.06991 0 0.2796 0.06991 Symmetry 0.1398

{F} =

0.2356 0.4712 0.2356

Using Equation (16.8.16) with [T0 =

25°C 25°C (initial temperature of rod) 25°C

Suddenly end (left) temperature becomes 85°C ( = 32 ) For t = 0.1 s 1 [M ] t

[K ] {T1} =

1 [ M ] (1 t

)[ K ] {T0} + {F0}

As {Fi} = {Fi} + 1 for all time t 1.7374 0.3660 0 3.4747 0.3660 Symmetry 1.7374

1.228 0.865 0 85 C = 2.457 0.865 t2 1.228 t3

25 25 25

0.2356 0.4712 0.2356

Since t1 = 85° boundary condition adjust equations 1 0 0 0 3.4747 0.366 0 0.366 1.37

t1 t2 t3

85 = 105.17 85(0.3660) 52.585

Solving t1 = 85°,

t2 = 18.536°C,

t3 = 26.362°C

For t = 0.2 s 1.228 0.865 0 85 1.737 0.366 0 85 2.457 0.865 18.536 3.475 0.366 t2 = Symmetry 1.228 26.362 1.737 t3

0.2356 0.4712 0.2356

Solving 1

0 0 3.475 0.366 1.737

t1 t2 t3

85 = 141.93 0.865(85) 48.427 624

t2 = 29.613°C, t3 = 21.635°C For t = 0.3 s 1

0 0 3.475 0.366 1.737

85 t1 t2 = 134.0 52.687 t3

t2 = 36.404°C,

t3 = 22.662°C

For t = 0.4 s 1

0 0 3.475 0.366 1.737

t1 t2 t3

85 = 152.0 59.58

t2 = 41.03°C,

t3 = 25.655°C

For t = 0.5 s 1

0 0 3.475 0.366 1.737

t1 t2 t3

85 = 165.9 67.26

t2 = 44.665°C, t3 = 29.31°C For t = 0.6 s 1

0 0 3.475 0.366 1.737

t1 t2 t3

t2 = 47.75°C,

85 = 178.03 74.90

t3 = 33.06°C

For t = 0.7 s 1

0 0 3.475 0.366 1.737

t1 t2 = t3

t2 = 50.48°C,

85 188.85 82.17

t3 = 36.67°C

For t = 0.8 s 1

0 0 3.475 0.366 1.737

t1 t2 t3

85 198.7 88.97

t2 = 52.96°C, t3 = 40.06°C For t = 0.9 s

625

85 0 0 t1 3.475 0.366 t2 = 207.7 95.28 1.737 t3 t2 = 55.22°C, t3 = 43.22°C For t = 1.0 s 85 1 0 0 t1 3.475 0.366 t2 = 215.99 101.11 1.737 t3 1

t2 = 57.30°C, t3 = 46.14°C etc. (A computer solution follows)

Time (s) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 2.1 2.2 2.3 2.4 2.5

NODE 1 25 85 85 85 85 85 85 85 85 85 85 85 85 85 85 85 85 85 85 85 85 85 85 85 85 85

2 25 18.53611 29.61303 36.18435 40.72491 44.27834 47.29072 49.95809 52.37152 54.57756 56.60353 58.46814 60.1859 61.76908 63.22852 64.574 65.81448 66.95818 68.01265 68.98485 69.88121 70.70765 71.46961 72.17214 72.81986 73.41705 626

3 25 26.36189 21.63526 22.42717 25.30428 28.85201 32.49614 36.01157 39.31761 42.39278 45.23933 47.86852 50.29457 52.53218 54.59557 56.49814 58.25235 59.86974 61.36096 62.73586 64.0035 65.17226 66.24984 67.24336 68.15938 69.00393

2.6 2.7 2.8 2.9 3

85 85 85 85 85

73.96766 74.47531 74.94336 75.3749 75.77277

627

69.78261 70.50053 71.16246 71.77274 72.33542

Appendix A A.1  4 0   2 0  6 0  (a) [A] + [B] =      1 8  2 4  3 12 (b) [A] + [C], Nonsense, [A] and [C] not same order

(c) [A] [C]T, Nonsense, columns [A]  rows [C]T  5 2 1  3    (d) [D] [E] =  2 10 0  2    1 0 5  1  5(3)  2(2)  1(1)   20      = 2(3)  10(2)  0(1)  =  26  1(3)  0(2)  5(1)   8     

(e) [D] [C], Nonsense, columns [D]  rows [C] 5 2 1   3 2 0   (f) [C] [D] =   2 10 0   1 0 2   1 0 5  3(5)  (2)(2)  0 6  20  0 3  0  0  =  2  0  0 1  0  10  5  0  2

 19 26 3 =    3 2 9

A.2 [A] =

1 0 1 4

[A]–1 =

[C ]T |[ A] |

C11 = (–1)1 + 1 (8) = 8,

C12 = (–1)1 + 2 (1) = -1

C21 = (–1)2 + 1 (0) = 0,

C22 = (–1)2 + 2 (4) = 4

 8 1 [C] =   0 4 

| [A] | = A11 C11 + A12 C12 = (4) (8) + (0) (-1) = 32  8 0 [C]T =    1 4 

 8 0  1 4   1   –1  [A] = =  4 32   1

0  1  8 629

Verify by multiplying [A] [A]–1 = [I] A.3 [D]–1 =

[C ]T |[ D ]|

5 2 1  [D] =  2 10 0    1 0 5   50 10 10  Cofactor matrix, [C] =  10 24 2   2 46   10

| [D] | = 5(50) + (2)(-10) + (1)(-10) = 220  5  50 10 10   22 1  [D]–1 = 10 24 2    1   22 220  2 46    1  10  22

1 1   22 22  6 1  55 110  1 23  110 110 



A.4 Nonsense  2 0 A.5 [B] =    2 4

2 0 1 0 st (1)   divide 1 row by 2  2 4 0 1  1 0 12 0 (2)    2 4 0 1

st row by -2 and add to row 2

 1 0 12 0  nd (3)   divide 2 row by 4 0 4  1 1   1 0 1 0  12 0  2 –1 = (4)   [B]   1 1 1 1 0 1  4 4   4 4  –1 A.6 [D] by row reduction 5 2 1   5 2 1 1 0 0  [D]   2 10 0   2 10 0 0 1 0      1 0 5   1 0 5 0 0 1

a. Divide R1 by 5  1 2 1 1 0 0 5 5 5    2 10 0 0 1 0     1 0 5 0 0 1 b. Multiply R1 by -2 + Add to R2

630

 1 52  1 0 9 5   1 0

1 5 2  5

0 0  1 0  0 0 1

c. Multiply R1 by -1 + Add to R3 2 1 5  1 0 9 5 0  2 5

1 5  52 4 54

1 5  52  15

0 0  1 0 0 1

d. Multiply R2 by 1/23 + Add to R3 1 1 0 0  1 52 5 5   1 2 2 0 9 5  5  5 1 0  5 1 0 0 110  23 1 23 23 e. Multiply R3 by 23/275 + Add to R2 1 1 0 0  1 52 5 5  23 276 23  1 0  55 275 275  0 9 5 110 5 1 0 0 23  23 1 23 f. Multiply R3 by -23/550 + Add to R1

23 23 1 0 110  550  550  1 52   23 276 23  1 0  55 275 275  0 9 5 110 5 1 0 0 23  23 1 23 g. Multiply R2 by -1/23 + Add to R1 5 1 1 0  22  22 1 0  22  46 23 276 23  0  55 275 275  0 5 110 5 1 0 0  23 1 23 23 h. Divide R2 by 46/5 + R3 by 110/23 5 1 0 0 22  1 0 1 0  22 0 0 1  1 22

5  22  1 [D]–1 =   22   1 22

1  22

6 55 1 110

1  22

6 55 1 110

1  22  1  110  23  110 

A.7 Show that ([A] [B])T = [B]T [A]T by using [A] =

a11 a21

a12 a22

[B] =

631

b11 b12 b21 b22

b13 b23

([A] [B]) =

a11 (b11 ) a12 (b21 ) a11 (b12 ) a12 (b22 ) a11 (b13 ) a12 (b23 ) a21 (b11 ) a22 (b21 ) a21 (b12 ) a22 (b22 ) a21 (b13 ) a22 (b23 )

a11 (b11 ) a12 (b21 ) a21 (b11 ) a22 (b21 ) ([A] [B]) = a11 (b12 ) a12 (b22 ) a21 (b12 ) a22 (b22 ) a11 (b13 ) a12 (b23 ) a21 (b13 ) a22 (b23 ) T

b11 b21 [B] = b12 b22 b13 b23 T

[A]T =

b11 (a11 ) b21 (a12 ) T

a11 a12

a21 a22

b11 (a21 ) b21 (a22 )

[B] [A] = b12 (a11 ) b22 (a12 ) b12 (a21 ) b22 (a22 ) b13 (a11 ) b23 (a12 ) b13 (a21 ) b23 (a22 ) a11 (b11 ) a12 (b21 )

a21 (b11 ) a22 (b21 )

= a11 (b12 ) a12 (b22 ) a21 (b12 ) a22 (b22 ) a11 (b13 ) a12 (b23 ) a21 (b13 ) a22 (b23 ) Answer: ([A] [B])T = [B]T [A]T A.8 [T] =

C S S C

[C] =

C S S C

[C]T =

C S

S C

| [T] | = C2 + S2 = 1 [T]–1 =

C [C ]T = S |[T ] |

[T]T =

C S

S C

and



S C

[T]T = [T]–1 and T is an orthogonal matrix

A.9 Show {X}T [A] {X} is symmetric. Given {X} = {X}T = {X}T [A] {X} = =

, [A] =

a b b c

x 1 y x x 1 a b x y x b c 1

y x

ax b bx c x ay bx by cx 1 632

y x

ax 2

axy by bx 2

bx bx c

axy bx

by cx ay

bxy bxy cx 2

as the 1–2 term = 2–1 term {X}T [A] {X} is symmetric. A.10 Evaluate [K] =

L 0

1 L

[ B ]T E [B] dx, [B] =

[K] =

1 L 1 L

L 0 1 L2 1 L2

1 L2

1 L –1 L

1 L

L 0

[K] = E

1 L

1 dx L

E dx

1 L2

1 L

1 1

E L

1 1

(Should multiply by A to get actual [K] for a bar) A.11 The following integral represents the strain energy in a bar U=

where

{d} =

A L T {d } [B]T [D] [B] {d} dx 2 0 d1

[B] =

1 L

[D] = E

Show that ddU yields [k] {d}, where [k] is the bar stiffness matrix given by {d } [k] =

AE L

1 1

{d}T = [d1 d2]

[B]T =

1 L 1 L

A L T {d } [B]T [D] [B] {d} dx 2 0

AL AL {d}T [B]T [D]T [B] {d} = [d1 2 2

AL d 2 d1 1 [E ] 2 L L

AEL d 2 d1 2 L

1 L

633

1 L 1 L

1 L

d1 d2

1 L 1 L

d2]

d2 d1 d2

AEL d1 d 2 2 L2

d1 2

U= dU d d

Thus

dU d d

U d1

= dU d d

AEL d12 2

U d2

AE L

d1d 2

d1d2 2

AE 2L AE 2L

1 1

1 d1 1 d2

1 1

d22

2d1

2d 2

2d1

1 d1 1 d2

= [k]

d1 d2

AE [ d12 – 2d1d2 + d 22 ] 2L d1 d1

d2 d2

= [k] {d} knowing that [k] =

AE L

d1 d 2 d 2 d1

AE L

1 1

= [k] {d}

634

1 1

Appendix B B.1 Given:

2x 1  4x 2  20 4x 1  2x 2  10

Find: Determine the solution of the simultaneous eq. by Cramer’s Rule. Solution:

20   2 4   x1      10   4 2   x 2 

 d i  xi  a 

20 4 10 2 80 x1    4; 2 4 20 4 2 B.2 Given:

2 20 4 10 60 x2   3 2 4 20 4 2

x1  4 x2  3

2x 1  4x 2  20 4x 1  2x 2  10

Find: Determine the solution to the set of eq. by the inverse method. Solution:

20   2 4   x1      10   4 2   x 2 

x  A

1

C

 0.1 0.2   0.2 0.1

A   1

 x1   0.1 0.2  20       x 2  0.2 0.1 10  x1  (0.1)(20)  (0.2)(10)  4 x 2  (0.2)(20)  (0.1)(10)  3 x1  4 x2  3 B.3 Given: 2x 1  4x 2  5x 3  6

635

2x 2  4x 3  1 

 x 1   2 4 5   6        x 2   0 2 4   1  x  1 1 2   2   3   

1x1  1x 2  2x 3  2 Find: Solve the system of eq. by Gaussian elimination. Solution: Multiply R1 by -1/2 and add to R3

x1  2 4 5   6    x 2  0 2 4  1 x 3  0 1 9 / 2  1 Multiply R2 by -1/2 and add to R3

x1  2 4 5   6    x 2  0 2 4   1  x 3  0 0 5 / 2  1/ 2 

x 3  (1/ 2)(2 / 5)  1/ 5 x 2  1/ 2(4(1/ 5)  1)  1/10 x1  1/ 2(4(1/10)  5(1/ 5)  6)  23 /10 x1  23 /10 x 2  1/10 x 3  1/ 5 B.4 2x1  x 2  3x 3  11

4x1  2x 2  3x 3  8 2x1  2x 2  x 3  6 1 3 11  2  4 2 3 8 1)   1 6   2 2

 1 1 3 11   2 2 2   9 7  4) 0 1  4 2 0 0 1 2     

11   1 1 3  2 2 2   9 14  2) 0 4 0 3 4 5    

1 1 0 5  2 2   5) 0 1 0 1 0 0 1 2     

636

 1 1 3  2 2  9 3) 0 1  4  11 0 0 4 

11  2  7 2 11  2 

3 1 0 0  6) 0 1 0 1   0 0 1 2 

x1  3, x 2  1, x 3  2

B.5

 x1   3 2   y 1       x 2   2 1  y 2 

 z1   1 2   x1      z 2   4 2   x 2 

z1  3y1  2y 2  4y1  2y 2  7y1  4y 2 z 2  12y1  8y 2  4y1  2y 2  16y1  10y 2

 z1   7 4   y1      z 2  16 10   y 2 

 z1  1  10 4   y1       z 2  6  16 7   y 2 

2 5     y1  3 3  z1       y 2   8  7  z 2   3 6 

B.6

XT = (1, 1, 1, 1, 1) First iteration:

1 2x1  x 2  1  x1  (x 2  1) 2 1  (1  1)  0 2 1 6x 2  x1  x 3  4  x 2  (x1  x 3  4) 6 1  (0  1  4) 6  0.833 1 4x 3  2x 2  x 4  4  x 3  (2x 2  x 4  4) 4 637

1 x 3  (2(0.833)  1  4) 4  1.667 1 4x 4  x 3  x 5  6  x 4  (x 3  x 5  6) 4 1 x 4  (1.667  1  6) 4  2.167 1 2x 5  x 4  2  x 5  (x 4  2) 2 1  (2.167  2) 2  0.083 Second iteration: x1 =

1 1 (x2 – 1) = (0.833 – 1) = – 0.084 2 2

x2 =

1 1 (– 0.084 + 1.667 + 4) (x1 + x3 + 4) = 6 6

= 0.93 x3 =

1 1 (2x2 + x4 + 4) = (2(0.93) + 2.167 + 4) 4 4

= 2.007 x4 =

1 1 (x3 + x5 + 6) = (2.007 + 0.083 + 6) 4 4

= 2.023 x5 =

1 1 (x4 – 2) = (2.023 – 2) 2 2

= 0.011 Third iteration: x1 = – 0.035 x2 = 0.995 x3 = 2.003 x4 = 2.004 x5 = 0.002

Fourth iteration: x1 = – 0.003 x2 = 1.000 x3 = 2.001 x4 = 2.001 x5 = 0.000

Fifth iteration: x1 = 0 x2 = 1 x3 = 2 x4 = 2 x5 = 0

B.7 Given: Solve Problem B.1 by Gauss-Seidel iteration.

4x 2  2x1  20

 4 2   x 2   20        2x 2  4x1  10  2 4   x1  10  638

Find: Determine the solution of the following simultaneous equations. Solution: Initial Guesses: x i  Li a ii

x2 

20 5 4

x1 

10  2.5 4

Eq. (1) solving x2:

1 1 x 2  (20  2x1 )  (20  2(2.5))  3.75 4 4 Eq. (2) solving x1:

1 1 x1  (10  2x 2 )  (10  2(3.75))  4.375 4 4 Gauss-Seidel Method Table Iteration 0 1 2 3 4 5 6

x1 2.5 4.375 3.90625 4.02344 3.9941 4.0014 3.9996

x2 5 3.75 2.8125 3.046875 2.98828 3.0029 2.9992

x1 = 4 x2 = 3

B.8 Given: a)

2x1  6x 2  10 4x1  12x 2  20

6x1  3x 2  9 2x1  6x 2  12

8x1  4x 2  32 4x1  2x 2  8

Find: Classify the system of equations according to Section B.2 as unique, nonunique, or nonexistent. Solution: a)

2x1  6x 2  10 (4x1  12x 2  20) / 2  2x1  6x 2  10 Same as eq. 1 Nonunique

6x1  3x 2  9 2x1  6x 2  12 639

6 3  36  6  30  0 2 6 Unique c)

8x1  4x 2  32 4x1  2x 2  8 8 4  4 2   16  16  0  0   Nonexistent

1 1 1 d)

2 2 2  1(6  6)  1(6  6)  1(6  6)  0 3 3 3 Nonunique

B.9 First Figure: nd = 2, m = 3 nb = nd(m + 1) = 2(3 + 1) = 8

Second Figure: nd = 2, m = 5 nb = 2(5 + 1) = 12

640

Appendix D D.1 Using table D–1 (a) Load case 1 P = 10 kip L = 20 ft f1y = f2y =

– 10 = – 5 kip 2

m1 = – m2 =

10 (20) = – 100 kip  ft 2

(b) Load case 3 P = 5 kip L = 20 ft

=

1 4

f1y = f2y = – 5 kip 1 1 1 (5) (20) 4 4

m1 = – m2 =

= – 18.75 kip  ft (c) Load case 4 w = 1000

lb ft

f1y = f2y =

(1000) (30) = – 15000 lb 2

m1 = – m2 =

L = 30 ft

(1000) (30)2 = – 75000 lb  ft 12

(d) Load cases 1 and 7 P = 5 kip,

L = 20 ft

w=2

kip ft

f1y =

–5 2

13(20)(2) = – 18.75 kip 32

f2y =

5 2

3(20)(2) = – 6.25 kip 32

m1 =

5(20) 11(2)(20) 2 = – 12.5 – 45.83 8 192

= – 58.33 kip  ft 641

m2 =

5(2)(20)2 = 12.5 + 20.83 192

5(20) 8

= 33.33 kip  ft

(e) Load case 5 Note: Switch nodes 1 and 2 w = 2000

lb ft

L = 20 ft

f1y =

3(2000)20 = – 6000 lb 20

f2y =

7(2000)(20) = – 14,000 lb 20

m1 =

(2000)(20)2 = – 26,667 lb  ft 30

m2 =

(2000)(20)2 = 40,000 lb  ft 20

(f) Load case 2 P = 5 kN,

5(2)2 7 2(5)

f1y =

73 5(5)2 7 2(2)

f2y =

a = 5 m,

b=2m

= – 0.99 kN = – 4.01 kN

5(5) (2)2 = – 2.04 kN  m 72

m1 = m2 =

L = 7 m,

5(5)2 (2) 72

= 5.10 kN  m

(g) Load case 6 w= 4

kN ,L=6m m

f1y = f2 y = m1 = – m2 =

4(6) = – 6 kN 4

5(4)(6)2 = – 7.5 kN  m 96

(h) Load case 4 642

w= 5

kN ,L=4m m

f1y = f2y = m1 = – m2 =

5(4) = – 10 kN 2

5(4)2 = – 6.67 kN  m 12

643

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