Solution Manual of Elasticity theory applications and numerics by Sadd 2nd & 4th

1. https://gioumeh.com/product/elasticity-theory-applications-numerics-solution/https://gioumeh.com/product/elasticity-theory-applications-numerics-solution/

2. https://gioumeh.com/product/elasticity-theory-applications-numerics-solution/https://gioumeh.com/product/elasticity-theory-applications-numerics-solution/ 1.1. (a) a ij = + + = + a + = a 1 4 1 a 6 a (scalar) + a a a a a 11 22 a 33 ii = + + + + + + + a a a a a a a a a a a a a 11 11 12 12 13 + 13 + 21 21 22 22 23 23 31 31 32 32 33 33 ij = + + + + + + = 1 1 1  0 16 4 0 1 1 25 (scalar)      1 1 1 1 1 1 1 6 4       = = 0 4 2 0 4 2 0 18 10 (matrix) a a       ij jk             0 1 1 0 1 1 0 5 3   3   = + + = 4 (vector) a b a b a b a b   1 1 2 2 3 3 ij j i i i     b 2 b = + + + + + + + + a b b a b b a b b a a b b a b b a b b a b b a b b a b b 11 + b 1 1 12 1 2 13 + 1 3 21 2 1 22 2 2 23 2 3 31 3 1 32 3 2 33 3 3 ij i j = + + + b + + b + = 1 0 2 0 0 b 0 0 b 0 4 7 (scalar)     1 0 2 b 1 1 1 2 1 3     = = 0 0 0 (matrix) b b b b b b b b     2 1 2 2 2 3 i j   b       2 0 4 b b b b b b 3 1 + 3 2 + 3 3 = = + + = 1 0 4 5 (scalar) b (b) a b b b b b b 1 1 2 2 3 3 i i = + + = + a + = 1 2 2 a 5 (scalar) + a a a a a 11 a 22 a 33 ii = + + + + + + + a a a a a a a a a a a a a a 11 11 12 12 13 + 13 21 = 21 22 22 23 23 31 31 32 32 33 33 ij ij = + + + + + + + 1 4 0  0 4 1  0 16 4 30 (scalar)     1 2 0 1 2 0 1 6 2       = = 0 2 1 0 2 1 0 8 4 (matrix) a a       ij jk             0 4 2 0 4 2 0 16 8   4   = + + = 3 (vector) a b a b a b a b   1 1 2 2 3 3 ij j i i i     b 6 b = + + + + + + + + a b b a b b a b b a a b b a b b a b b a b b a b b a b b 11 + 1 1 + b 12 1 2 13 + 1 3 21 2 1 22 2 2 23 2 3 31 3 1 32 3 2 33 3 3 ij i j = + + b + + b + = 4 4 0 b 0 2 b 1 0 b 4 2 17 (scalar)     4 2 2 1 1 1 2 1 3     = = 2 1 1 (matrix) b b b b b b b b     2 1 2 2 2 3 i j   b       2 + 1 = 1 b b b b b b 3 1 + 3 2 + 3 3 = = + 4 1 1 6 (scalar) b b b b b b b 1 1 2 2 3 3 i i

3. https://gioumeh.com/product/elasticity-theory-applications-numerics-solution/https://gioumeh.com/product/elasticity-theory-applications-numerics-solution/ = + + = + + = (c) a ij a 1 a 0 4 5 a (scalar) + a a a a a 11 22 a 33 ii = + + + + + + + a a a a a a a a a a a a a a 11 + 11 12 12 + 13 13 21 21 22 22 23 23 31 31 32 32 33 33 ij = + + + + + + = 1 1 1  1 0 4  0 1 16  25 (scalar)    1 1 1 1 1 1 2 2 7       = = 1 0 2 1 0 2 1 3 9 (matrix) a a       ij jk             0 1 4 0 1 4 1 4 18   2   = + + = 1 (vector) a b a b a b a b   1 1 2 2 3 3 ij j i i i     1 b = + + + + + + + + a b b a b b a b b a b a b b a b b a b b a b b a b b a b b 11 + b 1 + 1 12 + 1 2 13 1 3 21 2 1 22 2 2 23 2 3 31 3 1 32 3 2 33 3 3 ij i j = + + + b + + = 1 1 0 b 1 0 b 0 0 0  0  3 (scalar)   1 1 0 b b 1 1 1 2 1 3     = = 1 1 0 (matrix) b b b b b b b b     2 1 2 2 2 3 i j         0 + 0 = 0 b b b b b b 3 1 + 3 2 + 3 3 = = + 1 1 0 2 (scalar) b b b b b b b b 1 1 2 2 3 3 i i = + + = + a + = (d) a ij 1 2 0 a 3 a (scalar) + a a a a a 11 a 22 a 33 ii = + + + + + + + a a a a a a a a a a a a a 11 11 12 + 12 + 13 13 + 21 21 22 22 23 23 31 31 32 32 33 33 ij = + + + + + = 1 0 0  0 4 1  0 9 0 15 (scalar)     1 0 0 1 0 0 1 0 0       = = 0 2 1 0 2 1 0 7 3 (matrix) a a       ij jk             0 3 1 0 3 1 0 9 4   1   = + + = 0 (vector) a b a b a b a b   1 1 2 2 3 3 ij j i i i     0 b = + + + + + + + + a b b a b b a b b a b a b b a b b a b b a b b a b b a b b 11 + b 1 1 12 1 2 13 + 1 = 3 21 2 1 22 2 2 23 2 3 31 3 1 32 3 2 33 3 3 ij i j = + + + b + + b + 1 0 0 0 0 b 0 0 b 0 0 1 (scalar)     1 0 1 b 1 1 1 2 1 3     = = 0 0 0 (matrix) b b b b b b b b     2 1 2 2 2 3 i j         1 0 = 1 b b b b b b 3 1 + 3 2 + 3 3 = = + + 1 0 1 2 (scalar) b b b b b b b b 1 1 2 2 3 3 i i

4. https://gioumeh.com/product/elasticity-theory-applications-numerics-solution/https://gioumeh.com/product/elasticity-theory-applications-numerics-solution/ 1.2. 1 1 = + + − ) a ( ( ) ( ) a a a a a ij ij ji ij ji 2 2     2 1 1 0 1 1 1 1     = + − 1 8 3 1 0 1     2 2         − − 1 3 2 1 1 0 clearly and satisfy th appropriat e conditions e a a ( ) [ ] ij ij 1 1 = + + − ) b ( ( ) ( ) a a a a a ij ij ji ij ji 2 2     2 2 0 0 2 0 1 1     = + − − 2 4 5 2 0 3     2 2         0 5 4 0 3 0 clearly and satisfy th appropriat e conditions e a a ( ) [ ] ij ij 1 1 = + + − ) c ( ( ) ( ) a a a a a ij ij ji ij ji 2 2     2 2 1 0 0 1 1 1     = + 2 0 3 0 0 1     2 2         − − 1 3 8 1 1 0 clearly and satisfy th appropriat e conditions e a a ( ) [ ] ij ij 1 1 = + + − ) d ( ( ) ( ) a a a a a ij ij ji ij ji 2 2     2 0 0 0 0 0 1 1     = + − 0 4 4 0 0 2     2 2         0 4 0 0 2 0 clearly and satisfy th appropriat e conditions e a a ( ) [ ] ij ij

5. https://gioumeh.com/product/elasticity-theory-applications-numerics-solution/https://gioumeh.com/product/elasticity-theory-applications-numerics-solution/ 1.3. a = − = −  =  = 2 0 0 b a b a b a b a b ij ij ji ji ij ij ij ij ij ij   T     2 1 1 0 1 1   1       = − = From Exercise 1 - 2(a) : 1 8 3 1 0 1 0 a a tr     ( ) [ ] ij ij     4         − − 1 3 2 1 1 0   T     2 2 0 0 2 0   1       = − − = From Exercise 1 - 2(b) : 2 4 5 2 0 3 0 a a tr     ( ) [ ] ij ij     4         0 5 4 0 3 0   T     2 2 1 0 0 1   1       = = From Exercise 1 - 2(c) : 2 0 3 0 0 1 0 a a tr     ( ) [ ] ij ij     4         − − 1 3 8 1 1 0   T     2 0 0 0 0 0   1       = − = From Exercise 1 - 2(d) : 0 4 4 0 0 2 0 a a tr     ( ) [ ] ij ij     4         0 4 0 0 2 0 1.4.  +  +      a a a a 11 1 12 2 13 3 1      =  +  +  =  +  +  = = a a a a a a a a a     1 1 2 2 3 3 21 1 22 2 23 3 2 ij j i i i i          +  +  a  a a + a 31 1 32 2 33 3 3  +  +  +    +  +    a a  a a a  a a a  a 11 11 12 21 13 31 11 12 12 22 13 32 11 13 12 23 13 33    = a   ij jk          a a a 11 12 13   = = a a a a   21 22 23 ij     a a a 31 32 33 1.5. det( =  = 123  +  +  ) a a a a a a a a a a a a a 1 a 2 a 3 a + 11 a 22 33 231  12 a 23 a 31 a 312 13 21 32 ij ijk  i j k + + 132  + a + a 321 a 13 22 31 a − 11 23 a 32 a 213 − 12 a 21 a 33 a = − − a a a a a a a a a a a 11 22 a 33 a 12 a 23 a 31 ) 13 21 32 13 22 31 ) 11 ( 23 32 12 21 33 = − − + − ( ( ) a a a a a a a a a a a 11 a 22 a 33 23 32 12 21 33 23 31 13 21 32 22 31 a 11 12 13 = a a a 21 22 23 a a a 31 32 33

6. https://gioumeh.com/product/elasticity-theory-applications-numerics-solution/https://gioumeh.com/product/elasticity-theory-applications-numerics-solution/ 1.6.   1 0 0   o  = 45 rotation about x - axis 0 2 / 2 2 / 2 Q   1 ij   1   − 0 2 / 2 2 / 2       0 0 1 1        i = = = From Exercise 1 - 1(a) : 0 2 / 2 2 / 2 0 2 b Q b       ij j             − 0 2 / 2 2 / 2 2 2 T         1 0 0 1 1 1 1 0 0 1 2 0          ij = = = − 0 2 / 2 2 / 2 0 4 2 0 2 / 2 2 / 2 0 4 1 a Q Q a         ip jq pq                 − − − 0 2 / 2 2 / 2 0 1 1 0 2 / 2 2 / 2 0 2 1       1 0 0 2 2        i = = = From Exercise 1 - 1(b) : 0 2 / 2 2 / 2 1 2 b Q b       ij j             − 0 2 / 2 2 / 2 1 0 T   −       1 0 0 1 2 0 1 0 0 1 2 2          ij = = = − 0 2 / 2 2 / 2 0 2 1 0 2 / 2 2 / 2 0 5 . 4 5 . 1 a Q Q a         ip jq pq                 − − − 0 2 / 2 2 / 2 0 4 2 0 2 / 2 2 / 2 0 5 . 1 5 . 0       1 0 0 1 1        i = = = From Exercise 1 - 1(c) : 0 2 / 2 2 / 2 1 2 / 2 b Q b       ij j             T − − 0 2 / 2 2 / 2 0 2 / 2         1 0 0 1 1 1 1 0 0 1 2 0          ij = = = 0 2 / 2 2 / 2 1 0 2 0 2 / 2 2 / 2 2 / 2 5 . 3 5 . 2 a Q Q a         ip jq pq                 − − − 0 2 / 2 2 / 2 0 1 4 0 2 / 2 2 / 2 2 / 2 5 . 1 5 . 0       1 0 0 1 1        i = = = From Exercise 1 - 1(d) : 0 2 / 2 2 / 2 0 2 / 2 b Q b       ij j             − 0 2 / 2 2 / 2 1 2 / 2 T         1 0 0 1 0 0 1 0 0 1 0 0          ij = = = − 0 2 / 2 2 / 2 0 2 1 0 2 / 2 2 / 2 0 3 2 a Q Q a         ip jq pq                 − − − 0 2 / 2 2 / 2 0 3 0 0 2 / 2 2 / 2 0 0 1

7. https://gioumeh.com/product/elasticity-theory-applications-numerics-solution/https://gioumeh.com/product/elasticity-theory-applications-numerics-solution/ 1.7.  1  1        −    cos( , ) cos( , ) cos sin x x x x o cos cos( 90 ) 1 2 = = = Q          2  2 ij −   +   cos( , ) cos( , ) sin cos x x x x o cos( 90 ) cos 1 2  +          cos sin cos sin b b b  i 1 1 b 2 = = = b Q b       ij j −  +  −   sin cos sin cos b b 2 1 2 T           cos sin cos sin a a  ij 11 12 = = a Q Q a  cos   (   sin  ip jq pq −   −   sin cos sin cos a a 21 22    + +   +   − −   −  2 2 2 2 cos ) sin cos ( a ) sin cos sin a a a a a a a a = 11 12 21 22 12 11 22 21      − −   −   − +   +  2 2 2 2 cos ( ) sin cos sin sin ( ) sin cos cos a a a a a a a 21 11 22 12 11 12 21 22 1.8. a ' 1.9.  '  =  = =  Q Q a aQ Q a ij ip jq pq ip jp ij       +  +  =   +   +   ' ' ( ) Q Q Q  Q ij Q kl ik jl il jk im Q jn + kp Q lq Q mn pq mp  nq + mq  np + = 1.10. C  +  =     Q Q Q Q Q Q Q Q ln im jm kp lp im jn km im jn kn lm ij kl ik jl il jk =   +   +   =   +    +   ( ) ijkl ij kl ik  jl il jk ij kl ik jl il jk =   +   +   = ( ) C kl ij ki lj kj li klij 1.11.    0 0 1   =  If 0 0  a   2     0 = 0 + 3 =   +  I a 1 0  2  3 a ii   0  0  2 1 1 = + + =   +   +   II 1 2 2 3 1 3 a 0 0 0  3 3 2 0 0 1 =  =    0 0  III 2 1 2 3 a 0 0 3

8. https://gioumeh.com/product/elasticity-theory-applications-numerics-solution/https://gioumeh.com/product/elasticity-theory-applications-numerics-solution/ 1.12. − 1   1 1 0   = − 0  = − = − = (a) 1 0 , 1 , 2 0 a I II III   ij a a a     is 0 1  Roots − , 0  −  + 1  =    +  − =    +  − = 3 2 2 Characteri stic Eqn 2 0 ( ) 2 0 ( 2 )( ) 1 0  2  = −  =  = 2 , 1 2 3  = − Case : 1   + = ) 1 ( 1 ) 1 ( 1 ) 1 ( 2   1 1 0 0 n n n     =  =  = − =  =  − ) 1 ( 2 ) 1 ( 3 ) 1 ( 1 ) 1 ( 2 ) 1 ( n 1 1 0 0 0 n 2 , 2 / ( 2 / 2 )( 0 , 1 , 1 ) n n n n         2 2 2     + + = ) 1 ( 3 ) 1 ( 1 ) 1 ( 2 ) 1 ( 3 0 0 3 1 n n n  = 0 Case : 2 − + = − 1 ) 2 ( 1 n ) 2 ( 2     0 1 1 0 n n n 1     − 0 =   = =   =  ) 2 ( n 1 0 0 2 / 2 ( 2 / 2 )( 1 ) 0 1 , , n n n = 2 ) 2 ( 3 0     2 1 2     :     2 2 0 1 n + + = ) 2 ( 1 ) 2 ( 2 ) 2 ( 3 1 n n n 3  = 1 Case 3 − + = − ) 3 ( 1 n ) 3 ( 2 n     2 0 2 1 0 n n n 1     − =   = = =  =  ) 3 ( 3 ) 3 ( n 1 2 0 0 , 0 1 ) 1 , 0 , 0 ( n n n n − = 2 ) 3 ( 1 n ) 3 ( 2 n 2 0     2 1 2         2 2 0 0 0 n + + = ) 3 ( 1 ) 3 ( 2 ) 3 ( 3 1 n 3 − 1   1 1 0   = rotation The matrix given is by 2 / 2 1 0 and Q   ij     0 0 2 / 2 T − 1 − 1 − 1 −        1 1 0 1 1 0 1 1 0 2 0 0 1          ij = = − 0 = 1 0 1 0 1 0 0 0 0 a Q Q a         ip jp pq 2                 0 0 2 / 2 0 1 0 0 2 / 2 0 0 1

9. https://gioumeh.com/product/elasticity-theory-applications-numerics-solution/https://gioumeh.com/product/elasticity-theory-applications-numerics-solution/ 1.12. −   2 1 0   = −  = − = = (b) 1 2 0 , 4 , 3 0 a I II III   ij a a a     0 0 0  Roots −  , 1 −  −  =    +  + =    +  + = 3 2 2 Characteri stic Eqn is 4 3 0 ( 4 ) 3 0 ( 3 )( ) 1 0  3  = −  = −  = 3 , 0 1 2 3  = − Case : 1   + = ) 1 ( 1 ) 1 ( 1 ) 1 ( 2   1 1 0 0 n n n     =  =  = − =  =  − ) 1 ( 2 ) 1 ( 3 ) 1 ( 1 ) 1 ( 2 ) 1 ( n 1 1 0 0 0 n 2 , 2 / ( 2 / 2 )( 0 , 1 , 1 ) n n n n         2 2 2     + + = ) 1 ( 3 ) 1 ( 1 ) 1 ( 2 ) 1 ( 3 0 0 3 1 n n n  = − 1 Case : 2 − + = − 1 ) 2 ( 1 n ) 2 ( 2     0 1 1 0 n n n 1     − 0 =   = =   =  ) 2 ( n 1 0 0 2 / 2 ( 2 / 2 )( 1 ) 0 1 , , n n n = 2 ) 2 ( 3 0     2 1 2     :     2 2 0 1 n + + = ) 2 ( 1 ) 2 ( 2 ) 2 ( 3 1 n n n 3  = 0 Case 3 − + = − ) 3 ( 1 n ) 3 ( 2 n     2 0 2 1 0 n n n 1     − =   = = =  =  ) 3 ( 3 ) 3 ( n 1 2 0 0 , 0 1 ) 1 , 0 , 0 ( n n n n − = 2 ) 3 ( 1 n ) 3 ( 2 n 2 0     2 1 2         2 2 0 0 0 n + + = ) 3 ( 1 ) 3 ( 2 ) 3 ( 3 1 n 3 − 1   1 1 0   = rotation The matrix given is by 2 / 2 1 0 and Q   ij   1   0 0 2 / 2 T − 1 − − 1 −         1 1 0 2 0 1 1 0 3 0 0 1          ij = = − = − 0 1 0 1 2 0 1 0 0 1 0 a Q Q a         ip jp pq 2                 0 0 2 / 2 0 0 0 0 0 2 / 2 0 0

10. https://gioumeh.com/product/elasticity-theory-applications-numerics-solution/https://gioumeh.com/product/elasticity-theory-applications-numerics-solution/ 1.12. − 1   1 1 0   = − 0  = − = = (c) 1 0 , 2 , 0 0 a I II III   ij a a a     is 0 0  Roots −  −  =   + = 3 2 2 Characteri stic Eqn 2 0 or ( ) 2 0  2  = −  =  = 2 , 0 1 2 3  = − Case : 1   + = ) 1 ( 1 ) 1 ( 1 ) 1 ( 2   1 1 0 0 n n n     =  =  = − =  =  − ) 1 ( 2 ) 1 ( 3 ) 1 ( 1 ) 1 ( 2 ) 1 ( n 1 1 0 0 0 n 2 , 2 / 2 ( 2 / 0 , 1 , 1 ) n n n n         2 2 2     + + = ) 1 ( 3 ) 1 ( 1 ) 1 ( 2 ) 1 ( 3 0 0 2 1 n n n  = 1  = 0 Case : 2 3 − 1     1 0 n 1 − 2 + 2 2 = 0 2 n n     2 2 1 n 2 − 0 =   = = −  =  2 n 1 0 0 , 1 2 ( 1 2 ) n n n n n k,k, - k     2 1 2 3 1 + + = 1 n n   for   k   ,   1 3 0 0 n 3 arbitrary and thus directions uniquely not are determined convenienc For . we e may choose = =  =  ) 2 ( ) 3 ( n n 2 / 2 and 0 to get 2 0 , 1 , 1 ( 2 / ) and ) 1 , 0 , 0 ( k − 1   1 1 0   = rotation The matrix given is by 2 / 2 1 0 and Q   ij     0 0 / 2 2 T − 1 − 1 − 1 −        1 1 0 1 1 0 1 1 0 2 0 0 1          ij = = − 0 = 1 0 1 0 1 0 0 0 0 a Q Q a         ip jp pq 2                 0 0 2 / 2 0 0 0 0 2 / 2 0 0 0

11. https://gioumeh.com/product/elasticity-theory-applications-numerics-solution/https://gioumeh.com/product/elasticity-theory-applications-numerics-solution/ 1.12. −   6 3 0   = −  = = = (d) 3 6 0 18 , 99 , 162 a I II III   ij a a a     0 0 6 3 2  Roots −  +  −  + =  −  −  − = Characteri stic Eqn is 18 99 162 0 or ( 9 )( 6 )( ) 3 0  Case  = :  =  = , 3 , 6 9 1 2 3  = 3 1   ) 1 ( 1 ) 1 ( 1 ) 1 ( 2 − − ) 1 ( 3 n = 0 n   3 3 0 0 n n n     ) 1 ( 2 ) 1 ( 1 ) 1 ( 2 ) 1 ( 3 ) 1 ( − =  =  = =  = =  n 3 3 0 0 3 n 2 , 2 / , 0 2 0 , 1 , 1 ( 2 / ) n n n n         2 2 2     ) 1 ( 3 ) 1 ( 1 ) 1 ( 2 ) 1 ( 3 0 0 3 + + = n 1 n  = 6 Case : 2 ) 2 ( 1 ) 2 ( 2 − = =     0 3 0 0 n n n 1     ) 2 ( − =   =  n 3 0 0 0 ) 1 , 0 , 0 ( n     2 2 2 2         ) 2 ( 1 ) 2 ( 2 ) 2 ( 3 + + = 0 0 0 n 1 n n n 3  = 9 Case : 3   ) 1 ( 1 ) 3 ( 1 ) 3 ( 2 − − + n = 0 n   3 3 0 0 n n n     ) 1 ( 2 ) 3 ( 3 ) 3 ( 1 ) 3 ( 2 ) 3 ( 3 ) 3 ( − − =  − =  = − =  = =  − n 3 3 0 0 3 n 2 , 2 / , 0 2 , 1 ( 2 / 0 , 1 ) n n n n         2 2 2     ) 1 ( 3 − ) 3 ( 1 ) 3 ( 2 ) 3 ( 3 0 0 3 + + = n 1 n   1 1 0   = rotation The matrix given is by 2 / 2 0 0 2 / 2 and Q   ij     − 1 1 0 T −         1 1 0 6 3 0 1 1 0 3 0 0 1          ij = = − = 0 0 2 / 2 3 6 0 0 0 2 / 2 0 6 0 a Q Q a         ip jp pq 2                 − − 1 1 0 0 0 6 1 1 0 0 0 9 1.13*.