Title: Solution Manual for Elasticity: Theory, Applications, and Numerics – 5th Edition by Martin H. Sadd | Complete Step-by-Step Solutions for Continuum Mechanics, Stress-Strain Relationships, Elastic Deformation, Boundary Value Problems, Energy Methods,

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SOLUTIONS MANUAL

,Table of contents
Part 1: Foundations and elementary applications

1. Mathematical Preliminaries

2. Deformation: Displacements and Strains

3. Stress and Equilibrium

4. Material Behavior – Linear Elastic Solids

5. Formulation and Solution Strategies

6. Strain Energy and Related Principles

7. Two-Dimensional Formulation

8. Two-Dimensional Problem Solution

9. Extension, Torsion, and Flexure of Elastic Cylinders

Part 2: Advanced applications

10. Complex Variable Methods

11. Anisotropic Elasticity

12. Thermoelasticity

13. Displacement Potentials and Stress Functions: Applications to Three-Dimensional Problems

14. Nonhomogeneous Elasticity

15. Micromechanics Applications

16. Numerical Finite and Boundary Element Methods

,1


1-1.

(a) aii a11 a22 a33 1 4 1 6 (scalar)
aij aij a11a11 a12 a12 a13 a13 a21a21 a22 a22 a23 a23 a31a31 a32 a32 a33 a33
1 1 1 0 16 4 0 1 1 25 (scalar)
1 1 1 11 1 1 6 4
a a 04 2 04 2 0 18 10 (matrix)
ij jk

0 1 1 05 3
1 0 1

3
ab b a b a a b 4 (vector)
ij j i1 1 i2 2 i3 3
2
aij bib j a11b1b1 a12b1b2 a13b1b3 a21b2b1 a22b2b2 a23b2b3 a31b3b1 a32b3b2 a33b3b3
1 0 2 0 0 0 0 0 4 7 (scalar)
b1b1 b1b2 b1b3 1 0 2
bb bb bb bb 0 0 0 (matrix)
i j 2 1 2 2 2 3

b3b1 b3b2 b3b3 2 0 4
bibi b1b1 b2b2 b3b3 1 0 4 5 (scalar)

(b) aii a11 a22 a33 1 2 2 5 (scalar)
aij aij a11a11 a12 a12 a13a13 a21a21 a22 a22 a23a23 a31a31 a32 a32 a33a33
1 4 0 0 4 1 0 16 4 30 (scalar)
1 2 0 12 0 1 6 2
a a 02 1 02 1 0 8 4 (matrix)
ij jk
4 4 0 8
2 2 16
0 0
4

a b a b a b a b 3 (vector)
ij j i1 1 i2 2 i3 3
6
aijbib j a11b1b1 a12b1b2 a13b1b3 a21b2b1 a22b2b2 a23b2b3 a31b3b1 a32b3b2 a33b3b3
4 4 0 0 2 1 0 4 2 17 (scalar)
b1b1 b1b2 b1b3 4 2 2
bb bb bb bb 2 1 1 (matrix)
i j 2 1 2 2 2 3

b3b1 b3b2 b3b3 2 1 1
bibi b1b1 b2b2 b3b3 4 1 1 6 (scalar)



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,(c) aii a11 a22 a33 1 0 4 5 (scalar)
aij aij a11a11 a12 a12 a13a13 a21a21 a22 a22 a23a23 a31a31 a32 a32 a33a33
1 1 1 1 0 4 0 1 16 25 (scalar)
1 1 1 11 1 2 2 7
a a 10 2 10 2 1 3 9 (matrix)
ij jk

0 4 4 4 18
1 0 1
21

a b a b a b a b 1 (vector)
ij j i1 1 i2 2 i3 3
1
aijbib j a11b1b1 a12b1b2 a13b1b3 a21b2b1 a22b2b2 a23b2b3 a31b3b1 a32b3b2 a33b3b3
1 1 0 1 0 0 0 0 0 3 (scalar)
b1b1 b1b2 b1b3 1 1 0
bb bb bb bb 1 1 0 (matrix)
i j 2 1 2 2 2 3

b3b1 b3b2 b3b3 0 0 0
bibi b1b1 b2b2 b3b3 1 1 0 2 (scalar)



1-2.
1 1
(a) aij (aij a ji ) (aij a ji )
2 2
1 2 1 1 1 0 1 1
18 3 10 1
2 2
3 1 0
2 1
1
clearlya(ij ) and a[ij ] satisfy the appropriate conditions

1 1
(a  a ) (a  a )
(b) aij ij ji
2 ji
2 ij
1 2 2 0 1 0 2 0
24 5 2 0 3
2 2
0 4 3 0
5 0

clearlya(ij ) and a[ij ] satisfy the appropriate conditions

,3


1 1
(a  a ) (a  a )
(c) aij ij ji
2 ji
2 ij
1 2 2 1 1 0 0 1
20 3 0 0 1
2 2
3 1 0
8 1
1
clearlya(ij ) and a[ij ] satisfy the appropriate conditions


1-3.
aijbij a jibji aijbij 2aijbij 0 aijbij 0
21 1 0 1 1 T
1
From Exercise1- 2(a) : a(ij ) a[ij ] tr 8 0 0
4  3 1
1 1 1
3 1
2 0
1
T
22 0 0 2 0
1
From Exercise1- 2(b) : a(ij ) a[ij ] tr 4 5 2 0
4  3
2 0 0
5 4 0 0
3
T
2 2 1 0 0 1
1
From Exercise1- 2(c) : a(ij ) a[ij ] tr 0 3 0 0
4  1
2 1 0
3 1
8 0
1



1-4.
11a1 12a2 13a3 a1
a a aa a a a a
a
ij j i1 1 i2 2 i3 3 21 22 2 23 3 2 i

1
a
31 1 32 2 a a
33 3 a3
a
11 11 a
12 21 11a 12 12 a 22 a
11 13 a
12 23 a
13 33
a
13 31 13a32
 a11
ij a
jk

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,a12 

a13
 a
 a 21 a22 a23 ij

a31 a32 a33

,5



1-5.
det(aij ) ijk a1ia2 ja3k a a a
123 11 22 33 a a a
231 12 23 31 a a a
312 13 21 32

a a a
321 13 22 31132a11a 23a 32 a a a
213 12 21 33

a11a22a33 a12a23a31 a13a21a32 a13a22a31 a11a23a32 a12a21a33
a11 (a22a33 a23a32 ) a12 (a21a33 a23a31 ) a13 (a21a32 a22a31 )
a11 a12 a13
a21 a22 a23
a31 a32 a33


1-6.
1 0 0
45o rotation about x 1 - axis Qij 2 /2 2 /2

0
 2 /2 2 /2

0
1 0 0 1 1
From Exercise1 -1(a) : b Qb 0 2/2 2 /2 0 2
i ij j
0 2 /2 2/2 2 2
T
1 0 0 11 1 1 0 0 2 0
1
 0 
ai Qip jq a pq 2 /2  0 4 20 2 /2  4 1
Q
0
j 0 1 
0 2/ 1 0 2/2 2
2 /2 2 0 2/2
1
0 2 2
1 0
From Exercise1 -1(b) : b Qb 0 2/2 2/2 1 2
i ij j
0 2 /2 2/2 1 0
T
1 0 0 2 0 1 0 0 2 2
1
1
ai Qip a pq 2 /2 2 /2 02 1 0 2 /2 2/2 4.5 1.5
jq  
Q 0 2 /2 2/ 2 2 /2 2/ 0 1.5 0.5
j

2 0 0 2 0
1
0 4 1
0
1 0
From Exercise1 -1(c) : b Q b 2 /2 2 /2 1 2/2
0
i ij j



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, 0 2 /2 2/2 0 2/2
T
1 0 0 11 1 1 0 0 1 0
2
2/
ai Qip jq a pq 2 /2  1 0 2 /2   3.5 2.5
0 2 0 2
j
Q  2 /2 2/ 2 /2   1.5 0.5
2 0 2/
0 1 4 2
0

,7


1-7.
cos(x1 , cos(x1 , x2 ) cos cos(90o ) cos sin
x1 )
Qij cos(x , cos(x , x ) cos(90o cos sin cos
x ) )
2 1 2 2



bb Q cos sin b1 b1 cos b2 sin
i ij j sin cos b b sin b cos
2 1 2
T
ai cos sin a12 cos sin
Qip jq a pq cos
Q a11 a
j
 sin cos
a 21 sin
22
2
a cos a ) sin cos sin 2 a cos2 a ) sin cos sin 2
(a a (a a
11 a a
cos2 12
(a a21 ) sin cos 22
sin
a
2
a12 sin 2 (a11 22
) sin
a cos
21
cos2 
21 11 22 12 11 12 21 22




1-8.

a' ij QipQjq a pq aQipQjp a ij




1-9.

' ij kl ' ik jl ' il jk QimQjnQkpQlq ( mn pq mp nq mq np )
QimQjmQkpQlp QimQjnQkmQln QimQjnQknQlm ij kl ik jl il jk



1-10.

Cijkl ij kl ik jl il jk ij kl ( ik jl il jk )
kl ij ( ki lj kj li ) Cklij

1-11.
1 0 0
If a 0 2 0
 
0 0 3

Ia aii 1 2 3

0 0 0
II a 1 3

1
0 2 1
2 3
2 0 3 0 3
1 2


1 0 0

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, III a 2 0 1 2 3
0 0
0
3