Solutions Manual For
A First Course in the Finite Element Method 6th EDITION DARYL L. LOGAN
(Additional Files Download Link Provided at the End of this Document)
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
2
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
{F} = [K] {d} [K] –1 {F} = [K ]–1 [K] {d} 3 © 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] –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] =
k2
k3
k2
k2
k1
and CT =
k2
k2
k3
k2
k2
k1
k3
k2
k2
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
(k1
k2 ) (k2
k2
k3
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
k2
0 k1 k2 P k1 k3 k2 k3
(k1
2
k2 k3
k2 ) P k1 k3 k2 k3
(c) In order to find the reaction forces we go back to the global matrix F = [K]{d} F1x F2 x F3 x
=
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
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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 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
k
u2
0
2k u2
k
k
k
F3x
k u2
k
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)
=
k
k
u1
0
k
k
u2
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.
Element (2) 5 © 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 k
uf 2 x (2)0.5 = uf 33x (2)1
k k
k k
k k
u2
0.5
u3
1
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
0
F3 x
P =
F4 x
0
F5 x
?
0
2k
k
0
P =
k 0
2k k
k 2k
0
u2 =
k k 0 0 0
k 2k k 0 0
0 k 2k k 0
u1
0
0 0 k 2k k
0 0 0 k k
u2
0
2ku2
u3
P 0
ku2 2ku3 ku4 ku3 2ku4
u4
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 = ku3
u3 = u2 =
u3 2
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 = – ku2 = – k
P F1x = 2k
P 2
F5x = – ku4 = – k
P F5x = 2k
P 2
Check Fx = 0 F1x + F5x + P = 0 6 © 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.
P 2
P + 2
+P=0
0=0 2.4
(a) [k(1)] = [k(2)] = [k(3)] = [k(4)] =
k k
k k
By the method of superposition the global [K] is constructed. {F} = [K] {d} and u1 = 0 and u5 =
Also F1x
?
F2 x
0
F3 x
0 =
F4 x
0
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
u1
0
u2
?
u3
?
u4
?
u5
(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 (2u2) – k 2 – u2 + 4 u2 – u2 – u3 = 2 u4 =
4
u3 = 2 4 2
2
= 0 u2 =
4
2
u4 =
3 4
(c) Going back to the global equation 7 © 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.
{F} = [K]{d} F1x = – k u2 = k
4
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
u1
0 0
5 5
5 14
u3
0 9
u2 u4
0
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
9
u2
9
14
u4
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
u1
=
u2
2.8
k1 = 1000
1 1
1 ; k2 = 1000 1
1 1
1 1
So
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[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
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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
?
u1
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)
f 3 x (3)
(3)
(3)
f4 x
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
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[k(1)] =
1000 1000
1000 1000
[k(2)] =
500 500
500 500
[k(3)] =
500 500
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
u1
0
u2
?
u3
0
u4
0
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
Element (3) 12 © 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.
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
0
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
1 1
3 3 [k(2)] = 10000 3 3
{F} = [K] {d} 13 © 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.
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 =
u1
0
u2
?
u3
?
u4
0
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
1
1
1
1
1
1
1
1
1
1
1
1
0 f1x (1) 257 N 2 f 2 x (1) 257 N 2.57 10
Element (2) f2 x = 30000 f3 x
2.57 102 f 2 x (2) 193 N (2) f3 x 193 N 1.93 102
Element (3) f3 x f3 x = 10000 f4 x f4 x Reactions
f3 x (3) 193 N 1.93 102 0 f 4 x (3) 193 N
{F1x} = (10000
0 N ) [1 – 1] F1x = – 257 N 2 m 2.57 10
{F4x} = (10000
1.93 102 N ) [–1 1] m 0
F4x = – 193 N
2.13
[k(1)] = [k(2)] = [k(3)] = [k(4)] = 60
1 1
1 1
{F} = [K]{d}
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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
u1
0
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
1
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
1 1
{F} = [K]d} 15 © 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.
F1x ? = 4000 F2 x 100 F3 x 200
1 1 0
1 2 1
0 1 1
u1
0
u2
?
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)
1
1
1
1
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
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
500
500
500
f1x 1.0 kN 0 = 0.002 f3 x 1.0 kN 16
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Element (2)
500 f 2 x 500 = 500 f3 x 500 Element (3)
500 500
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
0
0
–300 – 300
–400
0
–300 – 300
(300 300 400)
–400
0
–400
– 400
400 400
u1
0
u2 u3 u4
0
0 = 1500 u2 – 600 u3 1000 = – 600 u2 + 1000 u3
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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, lbin.
– 3.0 – 2.0 – 1.0 0.0 0.5 1.0
6000 3000 1000 0 – 125 0 18
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2.0 p
1000
= 2000 x – 1000 = 0 x = 0.5 in. yields minimum p as table verifies.
x
(b)
p =
p
x
1 2 kx – Fx = 250 x2 – 1000 x 2
x, in.
p, lbin.
– 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)
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p = p
x
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 Nmm
p =
(d)
p
x
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 Nmm 2.19
p =
1 2 kx – Fx 2
p =
1 (500) x2 – 1000 x 2
p = 250 x2 – 1000 x 20 © 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.
p
x
= 500 x – 1000 = 0
x = 2.0 in. 2.20
F = k2
(x = )
dU = F dx U=
U=
x 0
(kx2) dx
k x3 3
= – Fx
p = p
x
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 lbin. 2.21 Solve Problem 2.10 using P.E. approach
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p =
3
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
p
u3 p
u4
= 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
0
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
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p =
3
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)
u4
In matrix form k1 0 k1 0
0 k2 k2 0
k1
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 =
I2
I1 L
I = I1 +
I2
I1 L
x
Now V = IR V = – V1 = R (I2 – I1) V = V2 = R (I2 – I1) V1 V2
=R
1 1
1 1
I1 I2
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24 © 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.
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
0
A1E1 AE L2 2 L1 2
– A2 E2 L2
– A2 E2 L2
AE A2 E2 L3 3 L2 3
0
– 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
0=
2AE AE u2 – u3 u3 = 2 u2 L L
P=
AE 2AE u2 + u3 L L
P=
AE 2AE u2 + (2 u2) L L 25
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u2 =
1 PL 3 AE
u3 = 2
1 PL 3 AE
u3 =
2 PL 3 AE
(c) A = 1 in.2; E = 10 106 psi; L = 10 in. P = 1000 lbs (i) u2 =
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 6.66 104 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
26
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(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 13.34 104 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)
6
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
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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 1.19 104 f2 x x 10000 N 6 = 84 10 1 1 2.38 104 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 3.81 103 f2 x x 4532 lb 6 = 1.2 10 (2) 0 1 1 f3x f3 x 4532 lb 3.4
28
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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 1.25 103 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
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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
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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 6.32 103 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 6.32 103 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 8.70 105 in. 1505 5 u2 0 3 = 10 5 1505 u3 10000 u3 6.58 103 in. Reactions F1x = (– 1500 103) (u2) F1x = – 130 lb F4x = (– 1500 103) (u3) F4x = – 9870 lb 31
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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 6.58 103 f3(2) x 130 lb
f3(3) 1 1 6.58 103 f3x x 9870 lb 6 = 1.5 10 (3) 0 f4 x 1 1 f 4 x 9870 lb 3.8
[k(1)] =
(4 104 m 2 )(200 106 kN2 ) 1 1 m 1 1 1m
1 1 kN [k(1)] = 800 102 1 1 m (2 104 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
32
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(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 0.50 103 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
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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 1.870 103 f 2 x 2.90 f2 x 3 = 7 10 kN 1 1 1.454 103 f3 x 2.90 f3x Element (3) 1 1 1.454 103 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 1.454 103 F4x = 103 [–2 2] F4 x 2.90 kN 0
3.11
34
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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)
35
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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
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 43
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
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(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
L
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
3
l 2
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)
u(
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)
37
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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
l2
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
b=
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
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C=
1 2
,S=
C 2 EA [K] = L
1 2 CS
C 2
2
CS
S
C2
–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
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(c)
3 , 2
C=
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
3 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
2
S = 0.117
0.883 0.321 0.883 0.321 (70 104 )(2 104 ) 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
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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.
u1 = 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)
C=
3 , 2
1 2
S=–
u1 = u1 C + v1 S =
1 3 + (0) 1 2 2 2
u1 = 0.433 in.
41
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u 2 = u2 C + v2 S =
1 3 + 3 4 2 4
1 2
u2 = – 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
S
0
C
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
42
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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) =
C=
=
2 , 2
S=
u1 v C S] 1 , 45 u2 v2 2 , E = 30 106 psi, L = 60 in. 2
6
30 10 2 60 2
2 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
kN
3 2
0.25 210 106 –3 1 0 10 2 3 1.00 0
m2
= 45.47 MPa
3.19
43
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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]
44
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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
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
C=
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
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2 2 ; S=– ; L2–3 = 2 2
C=
12 1 A2 E2 2 [k2–3] = L2 3 12 1 2
12
12
1 2 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 1x = L 1 1 0 f 2 x
3
=
10 10 C 2 100 C
S 0 0 C
S S
C
C
u1 0 v1 S u2 v2 0 S 0 S 0 0.283
f 1x =
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
46
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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
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
3
34
3 2
43 0 23 43
3 3 ( ) 1 23 ( 43 ) 2 4
0
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
47
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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
CS
C 2
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
3 4
3 4
1 4
43
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
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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
3 4
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
43
3 4
14
3 4
49
43 422 L AE 14 1570 L AE 3 0 4 1 0 4
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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
50
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(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
x
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
0
[k3–4] =
AE 1 20 0 1 0
y
x
(3) y
x
x
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
0
x
y
51
x
(4) y
y
y
0 (2) 0 1 0 0 (3) 0 1
(1)
(4)
(3) x
(2)
(1) y
y
0 (1) 0 1 0 0 (4) 0 1
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(2)
(4)
x [k2–4] =
y
x
y
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] =
y
(3) y
x
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 = [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
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(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
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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
54
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0. 0256 0.0192 0 0 0.0756 0.0192 0.05 0
(3)
(4)
0.0192
0
0.0144
0
0
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 = [005 u2 + 0 v2 + 0 u3 + 0 v3] AE u2 = 0 (1)
1000 = [0 u2 + 00667 v2 + 0 u3 – 00667 v3] AE
(2)
0 = [0 u3 + 0 v2 + 00756 u3 + 00192 v3] AE
(3)
u3 = – 0254 v3
(3)
1000 = [0 u2 – 0.0667 v2 + 00192 u3 + 00811 v3] AE
(4)
Adding (2) and (4) 2000 = [00192 u3 + 00144 v3] AE Substituting (3) in (5) 2000 [00192 (– 0254 v3) + 00144 v3] AE
v3 =
210000 AE
u3 = (– 0254)
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
55
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(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
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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 = 005 u2
u2 = 0
1000 = 00667 v2 – 00667 v3 0 = 005 u3
(1)
u3 = 0
1000 = – 00067 v2 + 00667 v3
(2)
Adding (1) to (2) 2000 = 0v2 + 0v3 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
57
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L
(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
(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
0
u3 v3
1.28 1 20 50
58
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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 (– 00352) 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
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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 [TT ] = 0 0
But
SC SC 2
S C
0
2
0
0
C2 S 2
0
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 (4 104 ) (210 106 ) kN = = 280 102 L 3 m
Element (1) C = 0, S = 1
60
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(1)
(2)
0 0 1 [k(1)] = 280 102 0 0 0 0 1
0 0 0 1 0 0 0 1
Element (2)
C=
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
1.76 104 8.93 104 0 1] 0 0
(1) = 62.5 MPa (T)
(2) = 70 106 22
22
2 2
1.76 104 4 2 8.93 10 2 0 0
(2) = 52.9 MPa (T)
61
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(3) = 70 106 [– 1 0
1.76 104 8.93 104 0] 0 0
1
(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
0
0
0
1 3
0 0 0 0
Element 1–4
C = – 05, 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
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