Elasticity Theory, Applications, and Numerics, Fourth Edition Martin H. Sadd Solution Manual

SOLUTION MANUAL

1.1.

(a) aii  a11  a 22  a33  1  4  1  6 (scalar) aij aij  a11a11  a12 a12  a13 a13  a 21a 21  a 22 a 22  a 23 a 23  a31a31  a32 a32  a33 a33  1  1  1  0  16  4  0  1  1  25 (scalar) 1 1 1 1 1 1  1 6 4  aij a jk  0 4 2 0 4 2  0 18 10 (matrix) 0 1 1 0 1 1  0 5 3   3 aij b j  ai1b1  ai 2 b2  ai 3 b3  4 (vector) 2 aij bi b j  a11b1b1  a12b1b2  a13b1b3  a 21b2 b1  a 22b2 b2  a 23b2 b3  a31b3b1  a32b3 b2  a33b3b3  1  0  2  0  0  0  0  0  4  7 (scalar) b1b3  1 0 2 b2 b3   0 0 0 (matrix) b3b3  2 0 4 bi bi  b1b1  b2 b2  b3b3  1  0  4  5 (scalar)

 b1b1 b1b2 bi b j  b2 b1 b2 b2 b3 b1 b3 b2

(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 1 2 0 1 6 2 aij a jk  0 2 1 0 2 1  0 8 4 (matrix) 0 4 2 0 4 2 0 16 8  4 aij b j  ai1b1  ai 2 b2  ai 3b3  3 (vector) 6 aij bi b j  a11b1b1  a12b1b2  a13b1b3  a21b2 b1  a22b2 b2  a23b2 b3  a31b3b1  a32b3b2  a33b3b3  4  4  0  0  2  1  0  4  2  17 (scalar)  b1b1 b1b2 bi b j  b2 b1 b2 b2 b3b1 b3b2

b1b3  4 2 2 b2 b3   2 1 1 (matrix) b3b3  2 1 1 bi bi  b1b1  b2 b2  b3b3  4  1  1  6 (scalar)

(c) aii  a11  a 22  a33  1  0  4  5 (scalar) aij aij  a11a11  a12 a12  a13 a13  a 21a 21  a 22 a 22  a 23a 23  a31a31  a32 a32  a33 a33  1  1  1  1  0  4  0  1  16  25 (scalar) 1 1 1  1 1 1  2 2 7  aij a jk  1 0 2 1 0 2  1 3 9  (matrix) 0 1 4 0 1 4 1 4 18  2 aij b j  ai1b1  ai 2 b2  ai 3b3  1  (vector) 1  aij bi b j  a11b1b1  a12b1b2  a13b1b3  a 21b2 b1  a 22b2 b2  a 23b2 b3  a31b3b1  a32b3b2  a33b3b3  1  1  0  1  0  0  0  0  0  3 (scalar)  b1b1 b1b2 bi b j  b2 b1 b2 b2 b3b1 b3b2

b1b3  1 1 0 b2 b3   1 1 0 (matrix) b3b3  0 0 0 bi bi  b1b1  b2 b2  b3b3  1  1  0  2 (scalar) (d) aii  a11  a 22  a33  1  2  0  3 (scalar) aij aij  a11a11  a12 a12  a13 a13  a 21a 21  a 22 a 22  a 23a 23  a31a31  a32 a32  a33 a33  1  0  0  0  4  1  0  9  0  15 (scalar) 1 0 0 1 0 0 1 0 0 aij a jk  0 2 1 0 2 1  0 7 3 (matrix) 0 3 1 0 3 1 0 9 4 1 aij b j  ai1b1  ai 2 b2  ai 3b3  0 (vector) 0 aij bi b j  a11b1b1  a12b1b2  a13b1b3  a 21b2 b1  a 22b2 b2  a 23b2 b3  a31b3b1  a32b3b2  a33b3b3  1  0  0  0  0  0  0  0  0  1 (scalar)  b1b1 b1b2 bi b j  b2 b1 b2 b2 b3b1 b3b2

b1b3  1 0 1 b2 b3   0 0 0 (matrix) b3b3  1 0 1 bi bi  b1b1  b2 b2  b3b3  1  0  1  2 (scalar)

1.2.

1 1 (a ) aij  (aij  a ji )  (aij  a 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 a(ij ) and a[ij ] satisfy th e appropriat e conditions 1 1 (b) aij  (aij  a ji )  (aij  a 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 a(ij ) and a[ij ] satisfy the appropriat e conditions 1 1 (c) aij  (aij  a ji )  (aij  a 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 a(ij ) and a[ij ] satisfy the appropriat e conditions 1 1 (aij  a ji )  (aij  a 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 a(ij ) and a[ij ] satisfy the appropriat e conditions

(d ) aij 

1.3. aij bij  a ji b ji  aij bij  2aij bij  0  aij bij  0  2 1   From Exercise 1 - 2(a) : a(ij ) a[ij ]  tr 1 4   1   2 1  From Exercise 1 - 2(b) : a(ij ) a[ij ]  tr  2 4   0   2 1   From Exercise 1 - 2(c) : a(ij ) a[ij ]  tr 2 4   1   2 1   From Exercise 1 - 2(d) : a(ij ) a[ij ]  tr 0 4   0 

T 1 1   0 1 1  8 3  1 0 1   0  3 2  1  1 0   T 2 0  0 2 0   4 5  2 0  3   0  5 4  0 3 0    T 2 1  0 0 1  0 3  0 0 1   0  3 8  1  1 0   T 0 0 0 0 0   4 4 0 0  2   0  4 0 0 2 0   

1.4.

 11a1  12 a2  13a3   a1   ij a j   i1a1   i 2 a2   i 3 a3   21a1   22 a2   23a3   a2   ai  31a1   32 a2   33a3   a3  11a11  12 a21  13a31 11a12  12 a22  13a32 11a13  12 a23  13a33    ij a jk             a11 a12  a21 a22 a31 a32

a13  a23   aij a33 

1.5. det( aij )  ijk a1i a2 j a3k  123a11a22a33   231a12a23a31   312a13a21a32

 321a13a22a31  132a11a23a32   213a12a21a33  a11a22a33  a12a23a31  a13a21a32  a13a22a31  a11a23a32  a12a21a33  a11 (a22a33  a23a32 )  a12 (a21a33  a23a31 )  a13 (a21a32  a22a31 ) a11 a12  a21 a22 a31 a32

a13 a23 a33

1.6.

0 1  45 rotation about x 1 - axis  Qij  0 2/2 0  2 / 2

0  2 / 2 2 / 2

0 1  From Exercise 1 - 1(a) : bi  Qij b j  0 2/2 0  2 / 2

0  1   1  2 / 2 0   2  2 / 2 2  2 

0 1  a ij  Qip Q jq a pq  0 2/2 0  2 / 2

T 1 0  2 0    2 / 2  0 4  1 0  2 1  2 / 2   0   2  2  2 / 2 1    2  2 / 2 1   0 

o

0  1 1 1  1 0     2 / 2  0 4 2  0 2/2     2 / 2  0 1 1  0  2 / 2

0 1  From Exercise 1 - 1(b) : bi  Qij b j  0 2/2 0  2 / 2 0 1  a ij  Qip Q jq a pq  0 2/2 0  2 / 2

0 1  From Exercise 1 - 1(c) : bi  Qij b j  0 2/2 0  2 / 2 0 1  a ij  Qip Q jq a pq  0 2/2 0  2 / 2

0  1 1 1  1 0     2 / 2 1 0 2 0 2/2 2 / 2 0 1 4 0  2 / 2

0 1  From Exercise 1 - 1(d) : bi  Qij b j  0 2/2 0  2 / 2 0 1  a ij  Qip Q jq a pq  0 2/2 0  2 / 2

T 1 0  2  2    2 / 2  0 4.5  1.5  0 1.5  0.5 2 / 2   0  1   1  2 / 2 1   2 / 2  2 / 2 0  2 / 2

0  1 2 0 1 0     2 / 2  0 2 1  0 2/2 2 / 2 0 4 2 0  2 / 2

0  1  2 / 2 0   2 / 2 1 

0  1 0 0 1 0     2 / 2  0 2 1   0 2/2 2 / 2 0 3 0 0  2 / 2

T  1 0  2 0   2 / 2   2 / 2 3.5 2.5  2 / 2 1.5 0.5 2 / 2   1  2 / 2 2 / 2 T

0  1 0 0   2 / 2  0 3  2 0 0  1  2 / 2

1.7.

 cos( x1 , x1 ) cos( x1 , x 2 )   cos  cos(90 o  )  cos  sin   Qij        o cos    sin  cos  cos( x 2 , x1 ) cos( x 2 , x 2 ) cos(90  )  cos  sin    b1   b1 cos   b2 sin   bi  Qij b j         sin  cos  b2   b1 sin   b2 cos   cos  sin    a11 a12   cos  sin   aij  Qip Q jq a pq       sin  cos  a 21 a 22   sin  cos  a cos 2   (a12  a 21 ) sin  cos   a 22 sin 2  a12 cos 2   (a11  a 22 ) sin  cos   a 21 sin 2    11 2 2 2 2   a 21 cos   (a11  a 22 ) sin  cos   a12 sin  a11 sin   (a12  a 21 ) sin  cos   a 22 cos  T

1.8. a ' ij  QipQ jq a pq  aQipQ jp  aij

1.9.

' ijkl  ' ik jl  ' il jk  QimQ jnQkpQlq (mn pq  mpnq  mqnp )  QimQ jmQkpQlp  QimQ jnQkmQln  QimQ jnQknQlm  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 If a   0  2  0 0

0 0   3 

I a  aii  1   2   3 II a 

1 0  2  0 0 2

1 0 III a  0  2 0 0

0 1 0   1 2   2 3  1 3 3 0 3

0 0  1 2 3 3

1.12.

  1 1 0 (a) aij   1  1 0  I a  1, II a  2 , III a  0  0 0 1  Characteri stic Eqn is  3  2  2  0  (2    2)  0  (  2)(  1)  0 Roots  1  2 ,  2  0 ,  3  1 1  2 Case : (1) n1(1)  n2(1)  0 1 1 0 n1  1 1 0 n (1)   0  n3(1)  0  n1(1)  n2(1)   2 / 2 , n (1)  ( 2 / 2)( 1,1, 0 )   2  2 2 2 0 0 3 n3(1)  n1(1)  n2(1)  n3(1)  1  2  0 Case :

 n1( 2 )  n2( 2 )  0  1 1 0  n1   1  1 0  n   0   n1  n2   2 / 2  n ( 2 )  ( 2 / 2)(1,1,0) n3( 2 )  0   2 2 2 2  0 0 1  n3  n1( 2 )  n2( 2 )  n3( 2 )  1  3  1 Case :  2n1( 3)  n2(3)  0  2 1 0  n1   1  2 0  n   0  ( 3) ( 3) n1( 3)  2n2( 3)  0  n1  n2  0, n3  1  n  (0,0,1)   2 2 2 2  0 0 0  n3  n1(3)  n2(3)  n3(3)  1 0  1  1  The rotation matrix is given by Qij  2 / 21 1 0  and 0 0 2 / 2  0   1 1 0 1  1 0  1  1   2 0 0 1 aij  Qip Q jp a pq  1 1 0   1  1 0 1 1 0    0 0 0 2 0 0 2 / 2   0 0 1 0 0 2 / 2   0 0 1 T

1.12.

  2 1 0 (b) aij   1  2 0  I a  4 , II a  3 , III a  0  0 0 0  Characteri stic Eqn is  3  42  3  0  (2  4  3)  0  (  3)(  1)  0 Roots  1  3 ,  2  1,  3  0 1  3 Case : (1) n1(1)  n2(1)  0 1 1 0 n1  1 1 0 n (1)   0  n3(1)  0  n1(1)  n2(1)   2 / 2 , n (1)  ( 2 / 2)( 1,1, 0 )   2  2 2 2 0 0 3 n3(1)  n1(1)  n2(1)  n3(1)  1  2  1 Case :

 n1( 2 )  n2( 2 )  0  1 1 0  n1   1  1 0  n   0   n1  n2   2 / 2  n ( 2 )  ( 2 / 2)(1,1,0) n3( 2 )  0   2 2 2 2  0 0 1  n3  n1( 2 )  n2( 2 )  n3( 2 )  1  3  0 Case :  2n1( 3)  n2(3)  0  2 1 0  n1   1  2 0  n   0  ( 3) ( 3) n1( 3)  2n2( 3)  0  n1  n2  0, n3  1  n  (0,0,1)   2 2 2 2  0 0 0  n3  n1(3)  n2(3)  n3(3)  1 0  1  1  The rotation matrix is given by Qij  2 / 21 1 0  and 0 0 2 / 2  0   2 1 0 1  1 0  1  1   3 0 0 1      aij  Qip Q jp a pq  1 1 0   1  2 0 1 1 0    0  1 0 2 0 0 2 / 2   0  0 0 0 0 0 2 / 2  0 0 T

1.12.

  1 1 0 (c) aij   1  1 0  I a  2 , II a  0 , III a  0  0 0 0  Characteri stic Eqn is  3  22  0 or 2 (  2)  0 Roots  1  2 ,  2   3  0 1  2 Case : (1) n1(1)  n2(1)  0 1 1 0 n1  1 1 0 n (1)   0  n3(1)  0  n1(1)  n2(1)   2 / 2 , n (1)   2 / 2(1,1, 0 )   2  2 2 2 0 0 2 n3(1)  n1(1)  n2(1)  n3(1)  1  2   3  0 Case :

 1 1 0  n1   1  1 0 n   0   n1  n2  0  n  n , n 2  1  2n 2  n  (k,k, 1-2k 2 ) 2 2 2 1 2 3 1   2 n1  n2  n3  1  0 0 0  n3  for arbitrary k , and thus directions are not uniquely determined . For convenienc e we may choose k  2 / 2 and 0 to get n ( 2 )   2 / 2(1,1, 0 ) and n (3)  (0 , 0 ,1) 0  1  1  The rotation matrix is given by Qij  2 / 21 1 0  and 0 0 2 / 2  0   1 1 0 1  1 0  1  1   2 0 0 1      aij  Qip Q jp a pq  1 1 0   1  1 0 1 1 0    0 0 0 2 0 0 2 / 2   0 0 0 0 0 2 / 2   0 0 0 T

1.12.  6  3 0 (d) a ij   3 6 0  I a  18 , II a  99 , III a  162  0 0 6  Characteristic Eqn is  3  182  99  162  0 or (   9)(  6)(  3)  0 Roots   1  3 ,  2  6 ,  3  9  1  3 Case : (1) n1(1)  n 2(1)  0  3  3 0 n1   3 3 0 n (1)   0  3n3(1)  0  n1(1)  n 2(1)   2 / 2 , n3(1)  0 , n (1)   2 / 2(1 , 1 , 0 )   2  2 2 2  0 0 3 n3(1)  n1(1)  n 2(1)  n3(1)  1  2  6 Case :

n1( 2)  n 2( 2)  0  0  3 0  n1    3 0 0  n   0   n ( 2 )   (0 , 0 , 1 )   2 2 2 2  0 0 0  n3  n1( 2)  n 2( 2)  n3( 2)  1  3  9 Case : (1) n1(3)  n 2(3)  0  3  3 0  n1   3  3 0  n (1)   0   3n3(3)  0  n1(3)  n 2(3)   2 / 2 , n3(3)  0 , n (3)   2 / 2(1 ,  1 , 0 )   2  2 2 2  0 0  3 n3(1)  n1(3)  n 2(3)  n3(3)  1

0  1 1  The rotation matrix is given by Qij  2 / 20 0 2 / 2  and 1  1 0  T

0   6  3 0 1 1 0  1 1 3 0 0 1      a ij  Qip Q jp a pq  0 0 2 / 2   3 6 0 0 0 2 / 2   0 6 0 2 1  1 0 0 9 0   0 0 6 1  1 0 

1.13*.

1.14. (a) u  2 x1e1  x1 x2 e2  2 x1 x3e3

  u  u1,1  u2,2  u3,3  2  x1  2 x1 e1 e2   u   / x1  / x2 2 x1 x1 x2

e3  / x3  0e1  2 x3e2  x2 e3 2 x1 x3

 2 u  0e 1  0e 2  0e 3  0  2 u   x2  2 x3

0 x1 0

0  0  , tr (u)  2  x1  2 x1 2 x1 

(b) u  x12 e1  2 x1 x2 x3e2  2 x32 e3   u  u1,1  u2,2  u3,3  2 x1  2 x1 x3  4 x3 e1   u   / x1 x12

e2  / x2 2 x1 x2 x3

e3  / x3  2 x1 x2 e1  0e2  2 x2 x3e3 2 x32

 2 u  2e 1  0e 2  4e 3  0  2 x1 u   2 x2 x3  0

0  2 x1 x2  , tr (u)  2 x1  2 x1 x3  4 x3 4 x3 

0 2 x1 x3 0

(c) u  x2 e1  2 x2 x3e2  4 x12 x3e3   u  u1,1  u2,2  u3,3  0  2 x3  4 x12 e1 e2   u   / x1  / x2 x2 2 x2 x3

e3  / x3  2 x2 e1  8 x1 x3e2  e3 4 x12 x3

 2 u  0e1  0e2  8e3  0 u   0 8 x1 x3

1 2 x3 0

0  2 x2  , tr (u)  2 x3  4 x12 4 x12 

1.15.

1 ai   ijk a jk 2 ii 1 1 imnai   ijk imna jk    ji 2 2  ki

im  jm  km

in 1  jn a jk   ( jm kn   jn km )a jk 2  kn

1 1   (amn  anm )   (amn  amn )  amn 2 2  a jk  ijk ai

1.16.

(a) ()  () ,k  ,k  ,k       2 ()  () ,kk  (,k  ,k ) ,k  ,kk  ,k  ,k  ,k  ,k  ,kk   ,kk   ,kk  2,k  ,k  ( 2 )  ( 2 )  2     (u)  (u k ) ,k  u k ,k  ,k u k    u  (  u) (b)   (u)   ijk  u k ) , j   ijk  u k , j   , j u k )   ijk  , j u k  ijk u k    u  (  u)    u  v )  ( ijk u j vk ) ,i   ijk (u j vk ,i  u j ,i vk )  vk  ijk u j ,i  u j  ijk vk ,i  v  (  u)  u  (  v )      ijk ( ,k ) , j   ijk  ,kj  0 because of symmetry and antisymmet ry in jk     ( ,k ) ,k   ,kk   2 

(c)   (  u)  ( ijk u k , j ) ,i   ijk u k , ji  0 , because of symmetry and antisymmet ry in ij   (  u)   mni ( ijk u k , j ) ,n   imn ijk u k , jn  ( mj  nk   mk  nj )u k , jn  u n ,nm  u m,nn   (  u)   2 u u  (  u)   ijk u j ( kmnu n ,m )   kij  kmnu j u n ,m  ( im  jn   in  jm )u j u n ,m  u n u n,i  u m ui ,m 1   ( u  u )  u  u 2

1.17.

Cylindrica l coordinate s : 1  r ,  2   , 3  z ( ds ) 2  ( dr ) 2  ( rd) 2  ( dz ) 2  h1  1 , h2  r , h3  1 eˆr  cos e 1  sin e 2 , eˆθ   sin e 1  cos e 2 , eˆ z  e 3 eˆ eˆ eˆ eˆ eˆ eˆr eˆ  eˆθ , θ   eˆr , r  θ  z  z  z  0   r r r  z  1     eˆr  eˆθ  eˆ z r r  z f 1 f f f  eˆr  eˆθ  eˆ z r r  z 1  1 u u z u  ( ru r )   r r r  z 2 1   f  1  f  2 f 2 f   r   r r  r  r 2  2 z 2 u   1 u z u  ˆ  u r u z  ˆ 1   u     e r   eθ   ( ru  )  r eˆ z z  r  r  r    z  r 

1.18.

Spherical coordinate s : 1  R ,  2   , 3   x1  1 sin  2 cos 3 , x 2  1 sin  2 sin 3 , x 3  1 cos  2 Scale factors : ( h1 ) 2 

x k x k  (sin  cos ) 2  (sin  sin ) 2  cos 2   1  h1  1 1 1

( h2 ) 2 

x k x k  R 2  h2  R  2  2

x k x k  R 2 sin 2   h3  R sin  3 3 Unit vectors : eˆ R  cos  sin e 1  sin  sin e 2  cos e 3 eˆ  cos  cos e 1  sin  cos e 2  sin e 3

( h3 ) 2 

eˆθ   sin e 1  cos e 2 eˆ R eˆ eˆ  0 , R  eˆ , R  sin eˆθ R   eˆ eˆ eˆ  0 ,    eˆr ,   cos eˆθ R   eˆ eˆ eˆ  0 ,   0 ,    cos eˆ R   Using (1.9.12) - (1.9.16)   1  1    eˆR  eˆ  eˆ R R  R sin   f 1 f 1 f f  eˆR  eˆ  eˆz R R  R sin   1  1  1  u  2 ( R 2 sin uR )  2 ( R sin u )  2 ( Ru  ) R sin  R R sin   R sin   1  1  1   2 ( R 2uR )  (sin u )  ( u ) R R R sin   R sin   2 f  

1   2 f  1  f 1  1 f (sin  )  2 ( )  R sin    2 R sin  R  R  R sin    R sin   sin   2

1   2 f  1  f 1 2 f R  (sin  )    R 2 R  R  R 2 sin    R 2 sin 2  2

1-18. Continued

 1   1        u   2 [ ( R sin u )  ( Ru  )] eˆR   [ (u R )  ( R sin u )] eˆ  R  R sin     R sin    1      [( Ru  )  (u R ) eˆ   R R  u   1   1 u R 1    (sin u )    eˆ R     ( Ru  ) eˆφ    R sin   R R   R sin    1   u     ( Ru  )  R eˆθ    R  R

2.1. (a) u  Axy , v  Byz 2 , w  C ( x 2  z 2 )

1    Ay 2 Ax Cx    1 eij  (ui , j  u j ,i )    Bz 2 Byz  2    2Cz      1   Ax Cx   0 2   1 1  ij  (ui , j  u j ,i )   Ax 0 Byz   2  2  Cx  Byz 0     (b) u  Ay 2 , v  Bx 2 y , w  Cyz 0  1 eij  (ui , j  u j ,i )    2   

Ay  Bxy Bx 2 

  0  1 ij  (ui , j  u j ,i )    Ay  Bxy  2   0 

0  1  Cz 2  Cy  Ay  Bxy 0 1 Cz 2

 0   1  Cz  2   0  

(c) u  Axz 3 , v  Bzy 2 , w  C ( x 2  y 2 )  3  Az  1 eij  (ui , j  u j ,i )     2    

0 2 Bzy 

1  (3 Axz 2  2Cx)  2  1 2 ( By  2Cy )   2  2Cz  

 0 0   1 ij  (ui , j  u j ,i )   0 0  2  1   (3 Ayz 2  2Cx)  1 ( By 2  2Cy )  2 2

1  (3 Axz 2  2Cx)  2  1 2 ( By  2Cy )   2   0 

2.2. u  k ( x 2  y 2 ) , v  k (2 x  y ) , w  0 u v u v ex   2kx , e y   k ,  xy  2e xy    2ky  2k  2k (1  y ) x y y x z 

1  v u  1     (2k  2ky)  k (1  y ) 2  x y  2

y

2k(dy)2

-kdy dy 2kdx x

dx 2k(dx)2

2.3.

(a) u  A  u  Ax  f ( y ) x v ey    B  v   By  g ( x ) y

ex 

1  u v  df dg df dg exy      0   0   constant   2  y x  dy dx dy dx  f  y  uo and g  x  vo  u  Ax  y  uo , v   By  x  vo  , uo and vo are associated with rigid body motion terms as per (2.2.9) (b) u A  Ax  u  x 2  f ( y ) x 2 v B ey    By  v   y 2  g ( x) y 2

ex 

1  u v  df dg df dg exy      0   0   constant   2  y x  dy dx dy dx A B  f  y  uo and g  x  vo  u  x 2  y  uo , v   y 2  x  vo 2 2  , uo and vo are associated with rigid body motion terms as per (2.2.9) (c) u  Ay  u  Axy  f ( y ) x v ey    Bx  v   Bxy  g ( x) y

ex 

1  u v  df dg df dg exy      0  Ax   By   0   By    Ax   constant   2  y x  dy dx dy dx B A  f  y 2  y  uo and g  x 2  x  vo  2 2 B A u  Axy  y 2  y  uo , v   Bxy  x 2  x  vo 2 2  , uo and vo are associated with rigid body motion terms as per (2.2.9)

2.4.

u  Az  u  Azx  f ( y, z ) x v ey   Az  v  Azy  g ( x, z ) y w 1 ez   Bz  w  Bz 2  h( x, y ) z 2 1  u v  f g f g e xy      0   0   L( z ) 2  y x  y x y x

ex 

e yz 

1  v w  g h g h     0  Ay      Ay   M ( x) 2  z y  z y z y

1  w u  h f h f   0    Ax     Ax   N ( y)  2  x z  x z x z From the previous three relations, L ( z )   N ( y ) , L ( z )   M ( x) , M ( x)   N ( y )  L ( z )  M ( x)  N ( y )  0  L( z )  a , M ( x)  b , N ( y )  c f   L( z )  a  f  ay  f1 ( x, z ) and y f   Ax  N ( y )   Ax  c  f   Axz  cz  f 2 ( x, y ) z To satisfy each form f   Axz  ay  cz  d1 . Forms for g and h following in similar fashion giving e zx 

1 2 Ay  by  cx  d 3 .

2 Combining the individual results and redefining the constants of integratio n gives u  Azx   z y   y z  u o g  Azy  ax  bz  d 2 and h  

v  Azy   x z   z x  vo w

1 ( Bz 2  Ay 2 )   y x   x y  wo 2

2.5.

u*  u o   z y   y z , v*  vo   x z   z x , w*  wo   y x   x y ex 

u * v * w *  0, ey   0 , ez  0 x y z

e xy 

1  u * v *  1    (  z   z )  0  2  y x  2

e yz 

1  v * w *  1    (  x   x )  0  2  z y  2

1  w * u *  1     (  y   y )  0 2  x z  2 u ex   0  u  f ( y, z ) x v ey   0  v  g ( z , x) y w ez   0  w  h ( x, y ) z 1  u v  f g f g e xy      0   0   L( z ) 2  y x  y x y x

e zx 

e yz 

1  v w  g h g h     0   0   M ( x) 2  z y  z y z y

1  w u  h f h f   0   0   N ( y)  2  x z  x z x z From the previous three relations, L ( z )   N ( y ) , L ( z )   M ( x) , M ( x)   N ( y )  L ( z )  M ( x)  N ( y )  0  L( z )  a , M ( x)  b , N ( y )  c f f   L( z )  a  f  ay  f 1 ( x, z ) and   N ( y )  c  f  cz  f 2 ( x, y ) y z To satisfy each form f  ay  cz  d1 . Forms for g and h following in similar fashion giving e zx 

g  ax  bz  d 2 and h  by  cx  d 3 . Combining the individual results and redefining the constants of integratio n gives the rigid body motion form u  uo   z y   y z v  vo   x z   z x w  wo   y x   x y

2.6.

 cos  sin   Qij     sin  cos  Use two - dimensiona l transform ation theo ry  e e r   cos  sin    e x e xy  cos   sin  eij   r  Qip Q jq eij        sin  cos  e xy e y   sin  cos   er e  e y  ex  ex  e y ex  e y   cos 2  e xy sin 2 sin 2  e xy cos 2   2 2  2  e y  ex ex  e y ex  e y  sin 2  e xy cos 2  cos 2  e xy sin 2   2 2 2

2.7.

ea  e x cos 2 30 o  e y sin 2 30 o  2e xy sin 30 o cos 30 o  3 1 3  e y  e xy 4 4 2 2 o 2 eb  e x cos 90  e y sin 90 o  2e xy sin 90 o cos 90 o 

0.001  e x

0.002  e y ec  e x cos 2 150 o  e y sin 2 150 o  2e xy sin 150 o cos 150 o  3 1 3  e y  e xy 4 4 2 Solving for the strains : e x  0.0027 , e y  0.002 , e xy  0.0017 0.004  e x

2.8*.

ex  0.001  0.003 cos 2  0.001sin 2 ey  0.001  0.003 cos 2  0.001sin 2 exy  0.003 sin 2  0.001cos 2 The maximum values are found to be (ex ) max  0.0022 @   9.2o ,189.2o (ey ) max  0.0022 @   99.2o , 279.2o (exy ) max  0.0032 @   144.2o , 324.2o The MATLAB Plot is given by

2.9*.

Define directions n and t in the shaded plane as shown.

1 1 For direction n : l1  cos  , m1  sin  , n1  cos  2 2 1 1 For direction t : l2   sin  , m2  cos  , n2   sin  2 2 Relations (2.3.3)  enn  exl12  e y m12  ez n12  2(exy l1m1  e yz m1n1  ezxn1l1 )

y

t n B A 45o

z  (3 cos 2   4 sin 2   2 sin 2)  10 3 ent  exl1l2  e y m1m2  ez n1n2  exy (l1m2  m1l2 )  e yz (m1n2  n1m2 )  ezx (n1l2  l1n2 )  (3.5 sin 2  2 cos 2)  10 3 The MATLAB plot of these relations are



x

2.10*.

MATLAB CODE % Principal Value Problem % Enter strain matrix e=[2,-2,0;-2,-4,1;0,1,6]*0.001; % Calculate principal values L and directions N [N,L]=eig(e); fprintf('Principal Values') disp(diag(L)') fprintf('Principal Directions') N1=N(:,1)' N2=N(:,2)' N3=N(:,3)' SCREEN OUTPUT >> Principal Values -0.0047 0.0026 Principal Directions N1 = -0.2852 -0.9543 0.0893 N2 = 0.9570 -0.2784 0.0815 N3 = -0.0529 0.1087 0.9927

0.0061

2.11.

Volume of Undeforme d Element  Vo  dxdydz Volume of Deformed Element  V f  (1  e1 )dx(1  e2 )dy (1  e3 )dz  dxdydz  (e1  e2  e3 )dxdydz  terms of O(e 2 , e 3 )  (1  )dxdydz  (1  )Vo  V  V f  Vo  (1  )Vo  Vo  Vo   

V Vo

2.12.

 2  2 0 eij   2  4 1  10 3  0 1 6 0  4 / 3 0 1  ~ Spherical Strain : eij  ekk  ij   0 4 / 3 0   10 3 3  0 0 4 / 3 2 0  2 / 3  Deviatoric Strain : eˆij  eij  e~ij    2  16 / 3 1   10 3  0 1 14 / 3 eˆkk  2 / 3  16 / 3  14 / 3  0  No volumetri c changes associated with eˆij

2.13.

ex 

 2 ex u  3u   x y 2 xy 2

 2ey v  3v ey    y x 2 yx 2  2 e xy 1  u v   3u  3v e xy      2   2  y x  xy yxy xxy If the displaceme nts are single - valued and continuous , we can interchang e the order of differenti ation 

2  2 e xy  2 ex  e y   2 xy y 2 x 2

2.14.

 ij ij   ikl  jmpelp,km   kj  lj



 im  km  lm

 ip  kp elp,km  lp



  ij ( km  lp   kp  lm )   im ( kj  lp   kp  lj )   ip  kj  lm   km  lj )  elp,km   ij (ell ,kk  ekl ,kl )  ell ,ij  e jk ,ki  eli, jl  e ji,kk

If indices k and l are taken to be the same, the first term will drop, and we get ij  (eij,kk  ekk ,ij  eik , jk  e jk ,ik )  0 , which is the same as (2.6.1) with k  l   e        e   e   imne jn,m      e   pqj  imne jn,mq   imn pqj enj ,mq  ip

2.15. ij   ikl  jmpelp,km 11  1kl 1mp elp,km  e22,33  e33, 22  2e23, 23  22   2 kl  2 mp elp,km  e33,11  e11,33  2e31,31 33   3kl  3mp elp,km  e11, 22  e22,11  2e12,12 12  1kl  2 mp elp,km  e33,12  (e12,3  e23,1  e31, 2 ) ,3  23   2 kl  3mp elp,km  e11, 23  (e23,1  e31, 2  e12,3 ) ,1 31   3kl 1mp elp,km  e22,31  (e31, 2  e12,3  e23,1 ) , 2 ij, j  i1,1  i 2, 2  i 3,3 

i  1 : 1 j , j  11,1  12, 2  13,3  e22,331  e33, 221  2e23, 231    e33,122  (e12,3  e23,1  e31, 2 ) ,32    e22,313  (e31, 2  e12,3  e23,1 ) , 23   0

Likewise for the i  2 and i  3 cases

2.16.

Starting with the first thre e compatibil ity equations, 2  2 e xy  4ey  4 e xy  2 ex  e y  4 ex  2  2 2  2 2 2 (a) xy y 2 x 2 y z x z xyz 2

 2ey

 2 e yz  4ey  4 e yz  2 ez  4 ez  2  2 2  2 2 2 (b) yz z 2 y 2 z x y x yzx 2

 2 e zx  4 ex  4 e zx  2 ez  2 ex  4 ez  2 2  2 2  2 2 2 (c ) zx x 2 z x y z y zxy 2 (a )  (c)  (b)  (b)  (a )  (c)  (c)  (b)  (a ) 

 4 ex  3  e yz e zx e xy       y z  y 2 z 2 xyz  x  4ey z 2 x

 2

 3  e zx e xy e yz      xyz  y z x 

 4 ez  3  e xy e yz e zx       x y  x 2 y 2 xyz  z

Using the last three compatibil ity equations (2.6.2) 4,5,6 , 2 (2.6.2) 4 gives the first fourth order equation yz 2 (2.6.2) 5 gives the second fourth order equation zx 2 (2.6.2) 6 gives the third fourth order equation xy

2.17.

(a) ex  Ay , e y  ez  0, exy  ( Ax  Bz 2 ) / 2, e yz  Bxz  Cy , ezx  Cx (2.6.1)1  0  0  0 (2.6.1) 2  0  0  0 (2.6.1)3  0  0  0 (2.6.1) 4  0  0 (2.6.1)5  0  0 (2.6.1)6  0  0 satisfies compatibility equations (b) ex  2 Ax , ey  Bx , ez  Cxy , exy  By / 2, eyz  Cxz / 2, ezx  Cyz / 2 (2.6.1)1  0  0  0 (2.6.1) 2  0  0  0 (2.6.1)3  0  0  0 (2.6.1) 4  0  0 (2.6.1)5  0  0 (2.6.1)6  0  0 satisfies compatibility equations (c) ex  2 Ax , ey  By 2 , ez  Cxz , exy  By, e yz  Cx 3 , ezx  Cyz 2 (2.6.1)1  0  0  0 (2.6.1) 2  0  0  0 (2.6.1)3  0  0  0  ( 3Cx 2  Cz 2  0)  6Cx ! x does not satisfy compatibility equations (2.6.1) 4  0 

2.18.

The given strains must satisfy th e compatibil ity equations (2.6.2) 2  2 e xy  2 ex  e y 2   2  6 Ay  6 Ax  2 B(2 x  2 y )  6 A  4 B  A  B 2 2 xy 3 y x (2.6.2) 2,3, 4,5,6  0  0

(2.6.2)1 

2.19.

b y tan 1 2 x u v w 1  u v  ex   0 , ey   0 , ez   0 , e xy      0 x y z 2  y x 

u  v  0, w 

1  v w  b  y b x e yz      (tan 1 )  2 2  z y  4 y x 4 x  y 2 1  w u  b  y b y ezx     (tan 1 )   2 2  x z  4 x x 4 x  y 2 The six compatibil ity equations (2.6.2) yield 00 00 00 0

  e yz e zx        ( 0)  0  0  0  x  x y  x

0

  ezx e yz        (0)  0  0  0  y  y x  y

00 Thus all compatibil ity equations are satisfied even thoug h the displaceme nts are multi - valued.

2.20. (

1 ur u  )rd r  r

D'

D

C'

B'  C

u u    d 

*





C rd d



A' *

A 

dr

A(r,)

From the figure geometry, u u [( dr  r dr ) 2  (  dr ) 2 ]1 / 2 AB  AB u r r er    r AB dr r * * * * A C  AC AC   A C e   AC A*C * u (u    rd  u )  rd (r  ur )d  rd    rd rd 1 u ur   r  r 1 1  u 1 ur u  er  (CAB  C AB)       2 2  r r  r 

 B

u  dr r

2.21.

u  u r 1 1  1 u r u  u   , e   u r    , er      r r   2  r  r r 

er 

A 1 A B  , e    B sin   , er   cos  2 r r 2r r  1 1 (b) er  2 Ar , e  Ar 2  Br cos  , er  B sin   B sin   0 r 2 1 (c) er  0 , e   A sin   B cos   A sin   B cos   0 r 11 1  er   ( A cos   B sin )  C  ( A cos   B sin   Cr )   0 2r r 

(a) er  

2.22.

From relation (A.2)  er 

u  ur u 1 , e   ur    , ez  z r r   z

1  1 ur u u  1  u 1 u z  1  ur u z  er      , ez        , ezr   2  r  r r  2  z r   2  z r  with no  - dependence  u u u er  r , e  r , ez  z r r z 1  u u  1 u 1  u u  er       , ez  , ezr   r  z  2  r r  2 z 2  z r  with no u  er 

ur u u 1  u u  , e  r , ez  z , er  0  ez  0, ezr   r  z  r r z 2  z r 

2.23.

From relation (A.3)  eR 

u  uR 1 1  u  , e   u R    , e   sin u R  cos u   R R   R sin    

1  1 uR u u  eR      2  R  R R   1  1 u u e    cot u   2 R  sin     1  1 uR u u  eR      2  R sin   R R  with all fields depending only on R  u u 1 eR  R , e  R , e   sin uR  cos u  R R R sin  1  u u  1 1 u u eR       , e    cot u  , eR       2  R R  2R 2  R R  with u  u  0  eR 

uR u , e  e  R , eR  e  eR  0 R R

2.24*.

For the case u  Ax , v  Bx u  u   v  ex  1  2        1  1  2 A  4 A2  B 2  1 x  x   x  2

MATLAB Plots:

2

3.1. y

X  ij   0  0

0 0 Y 0 0 0 Ti n   ij n j , n  ( sin , cos )

S

S

S

n 

Txn   x n x   X sin 

x

T yn   y n y  Y cos  Stresses on oblique plane S :  S  y   x sin 2    y cos 2   2 xy sin  cos   X sin 2   Y cos 2   S  xy   x sin  cos    y sin  cos    xy (cos 2   sin 2 )   X sin  cos   Y sin  cos   (Y  X ) sin  cos  3.2*.

 2 1  4 (a)  ij   1 4 0  , n  (cos , sin ,0)  4 0 1  Txn   x n x   xy n y   xz n z  2 cos   sin 

4 1 0 (b)  ij  1  6 2 , n  (cos , sin ,0) 0 2 1 Txn   x n x   xy n y   xz n z  4 cos   sin 

T yn   yx n x   y n y   yz n z  cos   4 sin 

Tyn   yx n x   y n y   yz n z  cos   6 sin 

Tzn   zx n x   zy n y   z n z  4 cos 

Tzn   zx n x   zy n y   z n z  2 sin 

2

2

2

T n  Txn  T yn  Tzn  19  2 cos 2  6 sin 2

2

MATLAB Plots : 6.5 part(a) part(b) 6

|Tn|

5.5

5

4.5

4

3.5

0

0.5

1

1.5 2 Angle, 

2.5

2

2

T n  Txn  Tyn  Tzn  29  12 cos 2  sin 2

3

3.5



3.3.

 r

r  r

y y’ x’  x

 r  x   x cos 2    y sin 2   2 xy sin  cos  

x  y



x  y

cos 2   xy sin 2 2 2    y   x sin 2    y cos 2   2 xy sin  cos  

x  y



x  y

cos 2   xy sin 2 2 2  r  xy   x sin  cos    y sin  cos    xy (cos 2   sin 2 ) 

y  x 2

sin 2   xy cos 2

3.4.

cos   sin  2 - D Rotation Matrix : Q     sin  cos   cos   sin    r  r   cos  sin     QQ T       sin  cos    r     sin  cos   x   r cos 2     sin 2   2 r sin  cos   y   r sin 2     cos 2   2 r sin  cos   xy   r sin  cos     sin  cos    r (cos 2   sin 2 )

3.5.

 x Plane stress  ij   xy  0 det(  ij  ij )  0 

 xy y 0

0 0 0

x    xy 0  xy y  

  2  ( x   y )   x  y   2xy  0 1 1, 2  [( x   y )  ( x   y ) 2  4( x  y   2xy ) 2 

x  y 2

 x  y     2xy )   2  

  2  max  1  2

2

 x  y      2xy ) 2   2

3.6.

 1 1 1  Using principal coordinate s with n   , ,   3 3 3 1 1 1  oct   ij ni n j  (1   2   3 )   kk  I1 3 3 3 2 n 2 2 2  oct | T |  N   ij n j  ik nk  ( ij ni n j ) 1 1  (12   22   32 )  (1   2   3 ) 2 3 9 1  2(12   22   32 )  2(1 2   2  3   3 1 ) 9 1 1  2( I12  2 I 2 )  2 I 2  2 I12  6 I 2 9 9 1  ( 1   2 ) 2  (  2   3 ) 2  (  3   1 ) 2 9 Next using given x,y,z axes 1 1 1  oct   kk  I1  ( x   y   z ) 3 3 3 1 2  oct   ij n j  ik nk  ( ij ni n j ) 2  2 I12  6 I 2 9 1  ( x   y ) 2  ( y   z ) 2  ( z   x ) 2  6 2xy  6 2yz  6 2zx 9







 















3.7.

1 1, 2  [( x   y )  ( x   y ) 2  4( x  y   2xy ) 2 1 1 2 1 1 2  I1  I1  4I 2  I1  I 1  I 2 (or could start with 2 - D Characteri stic Equation) 2 4 2 4





1   2 1 2  I1  I 2 2 4 the principal and max shear stresses are functions of the invariants and thus must be independen t of the coordinate system, so  max 

1, 2 

x  y 2

 x  y             2xy  r     r    2r 2 2 2     2

2

 x  y          2xy   r  max      2r 2 2     2

2

3.8.

 x  ij    xy

 xy  3P  2 xy   y  4c 3 c 2  y 2

x  y

c2  y2   0 



 x  y  3P    2xy  3 xy  (c 2  y 2 ) 2  x 2 y 2 1, 2    2 4c  2  Principal Directions : 2

3P 2 xy  xy  (c 2  y 2 ) 2  x 2 y 2  4c 3  c2  y2

 n1(1, 2)    (1, 2)   0  0  xy  (c 2  y 2 ) 2  x 2 y 2  n2  c2  y2

xy  (c 2  y 2 ) 2  x 2 y 2 (1, 2) n2 c2  y2 Thus the principal directions can be expressed by n1(1, 2 ) 









 

n (1, 2)  xy  (c 2  y 2 ) 2  x 2 y 2 e1  c 2  y 2 e 2  constant

3.9*.

Using MATLAB the first principal stress contours are:

3.10.

Setting up the principal value problem with  ij  Pf ij ( xk ) 

   ( ij  ij )n j  0  Pf ij ( xk )  ij n j  0   f ij ( xk )   ij n j  0 P   and thus the principal values of this problem would be of the form 1, 2,3  g1, 2,3 ( xk )  1, 2,3  Pg1, 2,3 ( xk ) P The principal directions follow from

Pf ( x )   ij

k

 n (j1, 2,3)  0  Pf ij ( xk )  Pg1, 2,3  ij n (j1, 2,3)  0   f ij ( xk )  g1, 2,3  ij n (j1, 2,3)  0

1, 2 , 3 ij

and thus the principal directions will not depend on P

3.11*.

x  

2 Px 2 y 2 Py 3 2 Pxy2 ,    ,    y xy ( x 2  y 2 ) 2 ( x 2  y 2 ) 2 ( x 2  y 2 ) 2

 x   y     2xy )   max   2   MATLAB Code and Plot : 2

2

2

  Py( x 2  y 2 )    2 Pxy2 )  Py       2 2 2  2 2 2  2 ( x  y 2 )  ( x  y )   ( x  y ) 

% Exercise 3-6 % Maximum Shear Stress Contours for Flamant Problem clear; % Generation of variable space [x,y]=meshgrid(-3:0.05:3,0:0.05:3); % Calculation of Nondimensional Maximum Shear Stress sxy=y./(x.^2+y.^2); hold on axis equal axis off % Plot contours with reversed y-axis contour(x,-y,sxy,20); 3.12.

| T n |2  12 n12   22 n22  32 n32 N  1n12   2 n22  3n32 2

S 2  T n  N 2  12 n12   22 n22  32 n32  (1n12  2 n22  3n32 ) 2  n12 n22 (1  2 ) 2  n22 n32 ( 2  3 ) 2  n32 n12 (3  1 ) 2

3.13.

1 Using S max  (1   3 ) in the equation of circle : S 2  ( N   3 )( N  1 )  0  2 1 1 (1   3 ) 2  ( N   3 )( N  1 )  0  N  (1   3 ) 4 2 Using (3.4.9)  1 1  1  (1   3 ) 2   (1   3 )   2  (1   3 )   3  S  ( N   2 )( N   3 ) 4 2  2 1 n12   (1   2 )(1   3 ) (1   2 )(1   3 ) 2 2

n22 

S 2  ( N   3 )( N  1 ) 0  0 ( 2   3 )( 2  1 ) ( 2   3 )( 2  1 )

1 1  1  (1   3 ) 2   (1   3 )  1  (1   3 )   2  S  ( N  1 )( N   2 ) 4 2  2 1 n32   ( 3  1 )( 3   2 ) ( 3  1 )( 3   2 ) 2 2

3.14.

3 1 1   ij  1 0 2 is the given stress matrix and using the previously determined principal directions , 1 2 0  2/ 6 1/ 6 1/ 6    Qij   1 / 3 1 / 3 1 / 3  would be the proper tra nsformatio n matrix  0  1 / 2 1 / 2   Using standard tensor tr ansformati on theory  2/ 6 1 / 6 1 / 6  3 1 1   2 / 6 1/ 6 1/ 6      ij  QimQ jn  mn   1 / 3 1 / 3 1 / 3  1 0 2  1 / 3 1 / 3 1 / 3   0  1 / 2 1 / 2  1 2 0  0  1 / 2 1 / 2   4 0 0   0 1 0  0 0  2

T

3.15.

The characteri stic equation for the deviatoric stress tensor is given by det[ ˆ ij   d  ij ]  0.

1 Since ˆ ij   ij   kk  ij , this equation can be written as 3 1 1 det[  ij   kk  ij   d  ij ]  det[  ij  (  kk   d ) ij ]  0 .

3 3 However, the characteri stic equation for the stress tensor  ij is det[  ij  ij ]  0, and thus 1 1 it follows that    kk   d or  d     kk .

3 3 If n is the principal direction of the stress tensor  ij , then ( ij  ij )n j  0, and this can 1   be written as   ij  (  kk   d ) ij n j  0 or (ˆ ij   d  ij )n j  0. Therefore n also the principal 3   direction of the deviatoric stress ˆ ij .

3.16.

 2 1  4 (a)  ij   1 4 0   4 0 1  0  7 / 3 0 1  ~  ij   kk  ij   0 7 / 3 0  3  0 0 7 / 3 4   1 / 3 1 1  ˆ ij   ij   kk  ij   1 5/3 0  3   4 0  4 / 3

4 1 0 (b)  ij  1  6 2 0 2 1 0 0   1 / 3 1  ~  ij   kk  ij   0  1/ 3 0  3  0 0  1 / 3 1 0  13 / 3 1  ˆ ij   ij   kk  ij   1  17 / 3 2  3  0 2 4 / 3

3.17.

3 1 1  With  ij  1 0 2  1  4 ,  2  1,  3  2 1 2 0 1 1  vonMises  [(1   2 ) 2  ( 2   3 ) 2  ( 3  1 ) 2 ]1 / 2  [(3) 2  (3) 2  (6) 2 ]1 / 2  27  5.196 2 2 1  oct  (1   2   3 )  1 3 1 1 1  oct  [(1   2 ) 2  ( 2   3 ) 2  ( 3  1 ) 2 ]1 / 2  [(3) 2  (3) 2  (6) 2 ]1 / 2  54  6  2.449 3 3 3

3.18.

0  0   ij    0 0 0 0 0   0 det[  ij  ij ]     0  ( 2   2 )  0 0 0   1   ,  2  0 ,  3    1   Case : ( ij  1 ij )n (j1)  0  n (1)  1 / 2 (1,1, 0)  2  0 Case : ( ij   2  ij )n (j2 )  0  n ( 2)  (0 , 0 ,1)  3   Case : ( ij   3  ij )n (j3)  0  n ( 3)  1 / 2 (1,  1, 0) For the octahedral plane, n1  n2  n3  1 / 3 2 1 1 2 T n  12 n12   22 n22   32 n32  (12   22   32 )  ( 2   2 )   2 3 3 3 1   N  1n12   2 n22   3 n32  (1   2   3 )    0 3 3 3 2 2 S  Tn N2    6 / 3 3

3.19.

 x Plane stress ij    xy  0 x   y

 xy y 0

0 0   0 

   y  I1 2 1,2    x    xy )   max , 3  0 2 2  2  From (3.5.5)  2

2

2

I  I  I  I        1 2   1  max    1   max    1   max  1   max  2  2  2  2  2 e

2 1

2 2

2

I    1   32max 2 I2 1  2max  (e2  1 ) 3 4

3.20*.

 x From Exercise 3 - 7,  ij    xy 1, 2 

x  y 2

 xy  3P  2 xy   y  4c 3 c 2  y 2

von Mises Stress Contours 2



c2  y2   0 



 x  y  3P    2xy  3 xy  (c 2  y 2 ) 2  x 2 y 2 ,  3  0   4c  2 

1 [(1   2 ) 2  ( 2   3 ) 2  ( 3  1 ) 2 ]1 / 2  [12  1 2   22 ]1 / 2 2 MATLAB stress contours :

 vonMises 

3.21.

 dx d  x  y x  y    cos 2   xy sin 2   0  d d  2 2  

x  y 2

sin 2   xy cos 2  0  tan 2 p 

2 xy x  y

dy 2  x  y  x  y dy dx  tan 2 p      1  2  2  dx 2 xy 1  tan 2  p  dy  xy   1   dx   2 tan  p

2

3.22.

F  0 

y 

x

 ( x  x dx)dy   x dy x  yx  ( yx  dy )dx   yx dx  Fx dxdy  0 y Divide through by dxdy and let dx, dy  0  x  yx    Fx  0 x y

F  0   y

dy )dx   y dx y  xy  ( xy  dx)dy   xy dy  Fy dxdy  0 x Divide through by dxdy and let dx, dy  0  xy  y    Fy  0 x y

 M  0 (@ center)  ( yx 

 yx

dy )dx

y

dy  yx  Fy

x

Fx

 xy  yx

y

y

( y 

 y

dy dy   yx dx 2 2

y  xy dx dx  ( xy  dx)dy   xy dy 0 x 2 2 Divide through by dxdy and let dx, dy  0   xy   yx

 yx y

dy

 xy 

 xy x x 

dx  x dx x

3.23.  x  yx    2     2  0   0   2     x y x  y  y  xy 

 xy



x

 y y

0

   2     2     0 x  xy  y  x 2 

 both equilibriu m equations are identicall y satisfied  r  3.24.

 

 r d 

  d 

r 

F  r 

Fr rd

F  0 

d

r

 r

 r dr r

 r dr r

dr

 r



r

  r dr )( r  dr )d   r rd  ( r  r d)dr cos( d / 2) r     r dr cos( d / 2)  (    d)dr sin( d / 2)    dr sin( d / 2)  Fr rdrd  0  Divide through by rdrd and let dr , d  0  r 1  r  r        Fr  0 r r  r  F  0  ( r 

 r  dr )( r  dr )d   r rd  ( r  r d)dr sin( d / 2)   r dr sin( d / 2) r   (    d)dr cos( d / 2)    dr cos( d / 2)  F rdrd  0  Divide through by rdrd and let dr , d  0  r 1       2 r  F  0 r r  r  M  0 (@ center)  ( r 

 r  dr dr rd rd dr )( r  dr )d   r rd  ( r  r d)dr   r dr 0 r 2 2  2 2 Divide through by rdrd and let dr , d  0   r   r

( r 

3.25.

My V (R2  y2 ) x   ,  xy  ,  y   z   xz   yz  0 I 3I  x  yx  zx y dM 2 y    Fx  0    V 0 x y z I dx 3I  xy  y  zy ( R 2  y 2 ) dV    Fy  0  0 x y z 3I dx  xz  yz  z    Fz  0  0  0 x y z  The approximat e mechanics of materials stresses do not satisfy th e equilibriu m equations.

3.26.

 x   x ( x) ,  y   z   xy   yz   xz  0 , Fx  g , Fy  Fz  0 x  x  yx  zx     Fx  0  x  g  0   x    gdx  g (l  x) l x y z x  xy  y  zy    Fy  0  0  0 x y z  xz  yz  z    Fz  0  0  0 x y z

3.27.

0   p 0  ij   p ij   0  p 0    0  p   0  ij, j  Fi  0   p, j  ij  Fi  0  p,i  Fi or p  F Using a scalar approach  x  yx  zx p    Fx  0   Fx x y z x  xy  y  zy p    Fy  0   Fy x y z y  xz  yz  z p    Fz  0   Fz x y z z  p  F

3.28.

  e rTr  eθTθ  e zTz Tr  σ r e r  τ rθ eθ  τ rz e z Tθ  τ rθ e r  σ θ eθ  τ θz e z Tz  τ rz e r  τ θz eθ  σ z e z From (1.9.14) or exercise 1 - 16,   u 

1  1 u u z ( ru r )   r r r  z

Tr 1 1 Tθ Tz  Tr   r r r  z  1  (σ r e r  τ rθ eθ  τ rz e z )  (σ r e r  τ rθ eθ  τ rz e z ) r r 1    ( τ rθ e r  σ θ eθ  τ θz e z )  τ rz e r  τ θz eθ  σ z e z  r  z  r  r  rz 1  er  eθ  e z  (  r e r   r eθ   rz e z ) r r r r   1      r e r   r eθ   eθ    e r  z e z  r          rz e r  z e θ  z e z z z z Collecting terms in each coordinate direction and adding the appropriat ed body force gives  r 1  r  rz 1    (  r    )  Fr  0 r r  z r  r 1    z 2     r  F  0 r r  z r  rz 1  z  z 1     rz  Fz  0 r r  z r  σ 

4.1.

From relation (4.2.1)  x  C11ex  C12e y  C13ez  2C14exy  2C15e yz  2C16ezx  y  C21ex  C22e y  C23ez  2C24exy  2C25e yz  2C26ezx  z  C31ex  C32e y  C33ez  2C34exy  2C35e yz  2C36ezx  xy  C41ex  C42e y  C43ez  2C44exy  2C45e yz  2C46ezx  yz  C51ex  C52e y  C53ez  2C54exy  2C55e yz  2C56ezx  zx  C61ex  C62e y  C63ez  2C64exy  2C65e yz  2C66ezx While from (4.2.3)  x  C1111ex  C1122e y  C1133ez  2C1112exy  2C1123e yz  2C1131ezx  y  C2211ex  C2222e y  C2233ez  2C2212exy  2C2223e yz  2C2231ezx  z  C3311ex  C3322e y  C3333ez  2C3312exy  2C3323e yz  2C3331ezx  xy  C1211ex  C1222e y  C1233ez  2C1212exy  2C1223e yz  2C1231ezx  yz  C2311ex  C2322e y  C2333ez  2C2312exy  2C2323e yz  2C2331ezx  zx  C3111ex  C3122e y  C3133ez  2C3112exy  2C3123e yz  2C3131ezx Comparing these two relations implies the result C1111 C  2211 C Cij   3311 C1211 C2311  C3111

C1122 C2222 C3322 C1222 C2322 C3122

C1133 C2233 C3333 C1233 C2333 C3133

C1112 C2212 C3312 C1212 C2312 C3112

C1123 C2223 C3323 C1223 C2323 C3123

C1131  C2231 C3331   C 1231 C2331  C3131 

4.2.

ij  Cijkl ekl ,  ji  C jikl ekl ij   ji  Cijkl ekl  C jikl ekl  Cijkl  C jikl which is (4.2.4)1 1 1 1 (ekl  elk )  Cijkl ekl  Cijkl elk 2 2 2 1 1 1 *  Cijkl ekl  Cijlk ekl  (Cijkl  Cijlk )ekl  Cijkl ekl 2 2 2 * but from the definition of Cijkl it is symmetric in k and l , and thus Cijkl  Cijlk which is (4.2.4)2

ij  Cijkl ekl  Cijkl

4.3. ij  Cijklekl  (ij kl  ik  jl  il  jk )ekl  ekk ij  eij  e ji  ekk ij  (   )eij letting    and (   )  2, ij  ekk ij  2eij

4.4.

From (4.2.6) 

ij  Cijklekl  (ij  kl  ik  jl  il  jk )ekl  ekk ij  eij  e ji  ekk ij  eij  eij Comparing this with relation (4.2.7)     and    ,     Cijkl  ij  kl  (il  jk  ik  jl ) 2 2 Since   k    Cijkl  (il  jk  ik  jl )  (k  )ij  kl 3 3 E E Since   and    2(1  ) (1  )(1  2) E E Cijkl  ij  kl  (il  jk  ik  jl ) (1  )(1  2) 2(1  ) 4.5.

 kk (3  2) 1  1  (4.2.7)  eij  ij  ekk ij  ij   kk ij 2 2 2 2(3  2) 1 1    This will equal form (4.2.10)providing  and  2 E 2(3  2) E (3  2)  and this will be true if E  and   (   ) 2(  )

ij  ekk ij  2eij   kk  (3  2)ekk  ekk 

4.6.

From exercise 4 - 3,

E

 (3  2)   E   2    (2  2)    (   )  2(1  ) 

 1 (   )  E   1 1 2(  ) 2   2(1  ) E 1 1  2 E   1   2(1  ) 2 2 (1  )(1  2) 

4.7.

If i and ni are the principal values and directions for the stress, then (ij  i ij )ni  0 For isotropic materials, ij  ekk ij  2eij and thus   1 (ij  i ij )ni  (ekk ij  2eij  i ij )ni  0   eij  (i  ekk )ij ni  0 2   1 which can be written as (eij  ei ij )ni  0 where ei  (i  ekk ) 2 However, the principal value problem for the strain can be expressed as (eij  ei ij )nie  0 where ei and nie are the principal values and directions for the strain.

These results for isotropic materials then imply that ni  nie and i  2ei  ekk

4.8.

ea  ex cos 2 60o  e y sin 2 60o  2exy sin 60o cos 60o  1 3 3  e y  exy 4 4 2 2 o 2 o eb  ex cos 90  e y sin 90  2exy sin 90o cos 90o 

300  10 6  ex

400  10 6  e y ea  ex cos 2 120o  e y sin 2 120o  2exy sin 120o cos120o  100  10 6  ex

1 3 3  e y  exy 4 4 2

Solving for the strains : ex  400  10 6 , e y  400  10 6 , exy 

200  10 6 3

All other strain components are assumed to vanish.

Using Hooke' s law with elastic moduli for steel (  111GPa,   80.2GPa)  x  (ex  e y )  2e x  - 64.16MPa  y  (ex  e y )  2e y  64.16MPa  xy  2e xy  18.52MPa

4.9.

u M (1   2 ) v M (1  ) ex   y , ey   y x EI y EI 1  u v  M (1   2 ) M (1   2 ) exy       x x  0 , ez  exz  e yz  0 2  y x  EI EI Hooke' s law gives the following in - plane stress components My  x  (ex  e y )  2e x  I  y  (ex  e y )  2e y  0  xy  2e xy  0 Thus this state of stress correspond s to pure bending of a rectangula r beam in the x,y  plane

4.10.

FromTable 4 - 1, E  3k 1  2  and so if E  0 and k  0, then  

1 2 E  21    and so if E  0 and   0, then   1 1  1    2 E Noting that   , and if we choose the more restrictiv e inequality 1   1  2  1 0 0 2

4.11.

 0 , 0 E    2 E / 5 ,   1 / 4 1   1  2     ,   1/ 2  E/2 ,  0 E    2 E / 5 ,   1 / 4 21      E / 3 ,   1/ 2  E /3 ,   0 E  k  2 E / 3 ,   1 / 4 31  2     ,   1/ 2 For   1/2,  and k become unbounded.

4.12.

4.13.

3(1  2) p  0    1 / 2 (for general E and p) E E E k   and    1   1  2  3(1  2) From  ij  ekk  ij  2eij , the term ekk  0   and thus becomes indetermin ate

ekk  

 ij   p ij  2eij   kk  3 p  2ekk  3 p  p   kk / 3

4.14.

(a) Aluminum : E  68.9GPa ,   0.34 ,   25.7GPa , k  71.8GPa

  0 0  E 2.17 0 0      Simple Tension : eij   0   0    0  0.74 0   10 -3 E    0 0  0.74  0 0     E   / 2 0  0 1.46 0  0 Pure Shear : eij   / 2 0 0  1.46 0 0  10 -3  0 0 0  0 0 0  p  0 0   3k  2.32 0 0     p Hydrostati c Compressio n : eij   0  0    0  2.32 0   10 -3 3 k   0  2.32 p   0  0 0    3k  (b) Steel : E  207GPa ,   0.29 ,   80.2GPa , k  164GPa   0 0  E 1.45 0 0       Simple Tension : eij   0   0  0  0.42 0   10 -3 E    0 0  0.42  0 0    E   / 2 0  0 0.935 0  0    Pure Shear : eij   / 2 0 0  0.935 0 0  10 -3  0 0 0  0 0 0  p  0 0   3k  1.02 0 0     p  Hydrostati c Compressio n : eij   0  0  0  1.02 0   10 -3 3k   0  1.02 p   0  0 0   3k  (c) Rubber : E  0.0019GPa ,   0.499 ,   0.000654GPa , k  0.326GPa   0 0  E 7894 0 0       Simple Tension : eij   0   0  0  3939 0   10 -3 E    0 0  3939  0 0    E   / 2 0  0 5351 0  0    Pure Shear : eij   / 2 0 0  5351 0 0  10 -3  0 0 0  0 0 0  p  3k  Hydrostati c Compressio n : eij   0   0 

0 

p 3k 0

 0   511 0 0     0  0  511 0   10 -3   0 0  511 p   3k 

4.15. ~  1   , ˆ    1   , e~  1 e  , eˆ  e  1 e   ij kk ij ij ij kk ij ij kk ij ij ij kk ij 3 3 3 3 ~  3ke~ ij  ekk ij  2eij  kk  (3  2)ekk   ij ij

1 1 ˆ ij  ekk ij  2eij  (3  2)ekk ij  2(eij  ekk ij )  2eˆij 3 3 4.16.

With  x  2 y ,  z   xz   yz  0 , Hooke' s law gives





1 1 1  2  x  ( y   z )   x  x   x E E 2 2E  2E 2E  x  ex and the slope of the uniaxial stress - strain curve is 2 2 ex 

4.17.*

Unconfined :  r  (er  e  e z )  2er  0  (  2)er  e  e z    (er  e  e z )  2e  0  (  2)e  er  e z  z  (er  e  e z )  2e z  e r  e 

 ez  

   (3  2) e z  (  2)e z   z  (er  e )  (  2)e z   e z  Ee z   Confined : 1 er   r  (    z )  0   r      z E 1 e     ( z   r )  0      r   z E 1 e z   z  ( r    ) E 2  r    z  1  1 1 2  1 (1  )(1  2) e z   z  ( r    )  e z   z   z   z  E E 1   E 1  E (1  ) z  ez  E *ez (1  )(1  2)

4.17.* 18 16 14 12

E/E*

10 8 6 4 2 0

0

0.05

0.1

0.15

0.2 0.25 0.3 Poisson's Ratio, 

0.35

0.4

4.18.

For steel, E  207GPa ,   0.29 1 1 20  0.29(30) 10 3  5.4  105 e x   x   ( y   z )  E 207 1 1 30  0.29(20) 103  11.7  10 5 e y   y   ( z   x )  E 207 1 1  0.29(50) 103  7.0  105 e z   z   ( x   y )  E 207 x  l x ex  (300)(5.4  10 5 )  0.0164mm













y  l y e y  (200)(11.7  10 5 )  0.0234mm z  l z ez  (4)( 7.0  10 5 )  0.00028mm

0.45

0.5

4.19.

For steel, E  207GPa ,   0.29 1 1  50  0.29(50) 10 3  3.1  10 4 e x   x   ( y   z )  E 207 1 1 50  0.29(50) 10 3  3.1  10 4 e y   y   ( z   x )  E 207 1 1  0.29(50  50) 10 3  0 e z   z   ( x   y )  E 207 x  l x ex  (300)( 3.1  10 4 )  0.093mm













y  l y e y  (200)(3.1  10 4 )  0.062mm z  l z ez  (4)(0)  0 4.20.

No axial strain  ex  0 and no transvers e stresses   y   z  0

Hooke' s law (4.4.4) then gives 1 ex   x  ( y   z )  T  0   x  - ET E 1 e y   y  ( z   x )  T  ET  T  (1  )T E 1 ez   z  ( x   y )  T  (1  )T E













4.21. ij  ekk ij  2eij  (3  2)(T  To )ij 

 x  (ex  ey  ez )  2ex  (3  2)(T  To )  (  2)ex  (e y  ez )  (3  2)(T  To ) Using Table 4 - 1  E E x  [(1  )ex  (e y  ez )]  (T  To ) (1  )(1  2) (1  2) Similarly E E y  [(1  )e y  (ez  ex )]  (T  To ) (1  )(1  2) (1  2) E E z  [(1  )ez  (ex  e y )]  (T  To ) (1  )(1  2) (1  2) Shear stresses follow from the original equation E E E τ xy  2exy  exy , τ yz  e yz , τ zx  ezx 1  1  1 

11.1.

From Hooke' s, 

ij  Cijkl ekl

ij  kl   Cijkl  Cklij  Cij  C ji ekl eij

and thus the general 6  6 anisotropi c stiffness matrix is symmetric and thus implies that only 21 independen t elastic moduli exist

11.2.

It is equivalent to either do three rotations about the coordinate axes or three reflection s about each of the coordinate planes. Choose the three reflection s. 1 0 0  Reflection about the x1 , x 2  plane : Q ij  0 1 0  0 0  1 Transforma tion law for the elasticity tensor, C ijkl  Qim Q jn Qkp Qlq C mnpq  C14  C1123  Q1m Q1n Q2 p Q3q C mnpq  C1123  C14  0 C15  C1131  Q1m Q1n Q3 p Q1q C mnpq  C1131  C15  0 C 24  C 2223  Q2 m Q2 n Q2 p Q3q C mnpq  C 2223  C 24  0 C 25  C 2231  Q2 m Q2 n Q3 p Q1q C mnpq  C 2231  C 25  0 C 34  C 3323  Q3m Q3n Q2 p Q3q C mnpq  C 3323  C 34  0 C 46  C 2312  Q2 m Q3n Q1 p Q2 q C mnpq  C 2312  C 46  0 C 56  C 3112  Q3m Q1n Q1 p Q2 q C mnpq  C 3112  C 56  0 C 35  C 3331  Q3m Q3n Q3 p Q1q C mnpq  C 3331  C 35  0   1 0 0 Reflection about the x 2 , x3  plane : Q ij   0 1 0  0 0 1 Transforma tion law for the elasticity tensor, C ijkl  Qim Q jn Qkp Qlq C mnpq  C16  C1112  Q1m Q1n Q1 p Q2 q C mnpq  C1112  C16  0 C 26  C 2212  Q2 m Q2 n Q1 p Q2 q C mnpq  C 2212  C 26  0 C 36  C 3312  Q3m Q3n Q1 p Q2 q C mnpq  C 3312  C 36  0 C 45  C 2231  Q2 m Q3n Q3 p Q1q C mnpq  C 2331  C 45  0 C11 C12   C 22     These two transform ations  C ij            

0  0  0   0  0   C 66  giving the desired form for orthotropi c materials with nine independen t elastic constants.

Employing the final (third) reflection about the x3 , x1  plane would not produce any new relations.

C13 C 23 C 33   

0 0 0 C 44  

0 0 0 0 C 55 

11.3.

Material Symmetry : Cijkl  QimQ jnQkp QlqCmnpq For this case, the following moduli are zero : C14  C1123  C15  C1131  C16  C1112  C24  C2223  0 C25  C2231  C26  C2212  C34  C3323  C35  C3331  0 C36  C3312  C45  C2331  C46  C2312  C56  C3112  0 and : C44  C55 , C66  (C11  C22 ) / 2 The particular rotation is given by :  cos  sin  0 Qij   sin  cos  0  0 0 1 C55 Check :   Q3mQ1nQ3 p Q1q Cmnpq  Q33Q1nQ33Q1q C3n 3q  Q1nQ1q C3n 3q C55  C3131  Q11Q11C3131  Q11Q12C3132  Q12Q11C3231  Q12Q12C3232  cos 2 C55  sin 2 C44  (cos 2   sin 2 )C55  C55 C22 Check :   Q2 mQ2 nQ2 p Q2 q Cmnpq C22  C2222 4 2 2 2 2 2 2  Q21 C1111  Q21 Q22 C1122  Q21 Q22 C1212  Q21 Q22 C1221 2 2 2 2 2 2 4  Q21 Q22 C2112  Q21 Q22 C2121  Q22 Q21 C2211  Q22 C2222

 sin 4 C11  2 sin 2  cos 2 C12  4 sin 2  cos 2 C66  cos 4 C22  C11 sin 4   2C12 sin 2  cos 2   2(C11  C12 ) sin 2  cos 2   C22 cos 4   C22 (sin 4   2 sin 2  cos 2   cos 4 )  C22 ( sin 2   cos 2 ) 2  C22

11.4.

Cij must be positive definite  all principal minors pi and diagonal elements are  0 Orthotropic Case: p1  0  C11  0 p2  0 

C11 C12

C11 p3  0  C12 C13

C12  0  C11C22  C122  0 C22 C12 C22 C23

C13 2 C23  0  C11C22C33  C11C23  C22C132  C33C122  2C12C23C31  0 C33

p4  0  C44  0 , p5  0  C55  0 , p6  0  C66  0 Transversely Isotropic Case: p1  0  C11  0 p2  0 

C11 C12

C11 p3  0  C12 C13

C12  0  C112  C122  0 C11 C12 C11 C13

C13 2 C13  0  C11C22C33  C11C23  C22C132  C33C122  2C12C23C31  0 , with C22  C11 , C C33

p4  0 and p5  0  C44  0 , p6  0  C11  C12  0 Cubic Case: p1  0  C11  0 p2  0 

C11 C12

C11 p3  0  C12 C12

C12  0  C112  C122  0 C11 C12 C11 C12

C12 C12  0  C113  3C11C122  2C123  0 C11

p4,5,6  0  C44  0 Isotropic Case:   0 p1  0    2  0 p2  0 

  2 

  2 p3  0    p4,5,6  0    0

  0  4(  )  0      0   2    2 

   0  4 2 (3  2)  0  3  2  0   2

11.5.

 S11 S12   S 22     For orthotropi c materials , Sij            

0 0  0  0 0  S66  Positive definite strain energy function  that each principal minor of Sij be positive  S13 S 23 S33   

0 0 0 S 44  

0 0 0 0 S55 

2

S11 S12

S12 1 1   21  E    221  2  0  S11S 22  S122     S 22 E1 E2  E2  E1

S 22 S 23

S 23 1 1   32  E 2 2    32  0  S 22 S33  S 23      3 S33 E2 E3  E3  E2

S33 S13

S13 1 1   31  E 2    31  0  S33S11  S132      3 S11 E3 E1  E3  E1

2

  ij2 

2

Ei Ej

Using moduli Ei and E j from Table 11 - 1  2 S - Glass/Epox y : 12 

2 Boron/Epox y : 12 

E1 50   2.94 E2 17

E1 205   10.25 E2 20

2 Carbon/Epo xy : 12 

E1 205   20.5 E2 10

2 Kevlar49/E poxy : 12 

E1 76   13.82 E2 5.5

Thus the inequaliti es  12 could be greater th an one.

11.6.

Hydrostatic Stress: ij   pij  ex   S11 e    y     ez    Cubic Material      2eyz     2ezx        2exy   

S12 S11    

S12 S12 S11   

0 0 0 S 44  

0 0 0 0 S 44 

0   p  0    p    0   p    0  0  0  0    S44   0 

 ex  ey  ez  e  ( S11  2S12 ) p , exy  eyz  ezx  0  x   y   z   p  C11ex  C12 ey  C12 ez  (C11  2C12 )e  e   Dilatation    3e   p

3p 1  p   (C11  2C12 ) C11  2C12 3

1 108  2(61)   2 105   153.3 105 GPa  1.53MPa 3

p C11  2C12

11.7.

For material with a plane of symmetry and with   x   y   z   xy  0  0 S16   0   e x   S11 S12 C13 0  e    S C23 0 0 S 26   0  22  y      ez     S 33 0 0 S 36   0          S 44 S 45 0    yz  2e yz     2ezx       S55 0    zx           S 66   0  2e xy    e x  e y  ez  e xy  0 , 2e yz  S 44  yz  S 45 zx , 2ezx  S 45 yz  S55 zx Compatibil ity equations : 2  2 e xy  2 ex  e y  2  00 y 2 x 2 xy

 2ey z

 2

 2 e yz  2 ez  2  00 y 2 yz

 2 ezx  2 ez  2 e x   2  00 x 2 z 2 zx e e   2 ez   e    xy  yz  zx   0  0 xy z  z x y   2 ex   e yz ezx e xy    e yz ezx       0      yz x  x y z  x  x y        ( S 44  yz  S 45 xz )  ( S54  yz  S55 xz )  0   x  x y  

  ( S 44  yz  S 45 xz )  ( S54  yz  S55 xz )  f ( y ) x y

 2ey zx



  ezx e xy e yz    ezx e yz       0     y  y z x  y  y x 

      ( S54  yz  S55 xz )  ( S 44  yz  S 45 xz )  0   y  y x    ( S54  yz  S55 xz )  ( S 44  yz  S 45 xz )  g ( x ) y x Adding the previous results  f ( y )  g ( x )  0  f ( y )   g ( x )  constant  C    ( S 44  yz  S 45 xz )  ( S54  yz  S55 xz )  C x y



11.8.

Homogeneou s equation : S 44 xx  2 S 45 xy  S55 yy  0 Look for solutions of the form ( x, y )  f ( x  y ) , where  is a parameter  ( S 44  2 S 45  S55 2 ) f   0  S55 2  2 S 45  S 44  0  characteri stic equation 2 S 45  S 45  S 44 S55 Solving the quadratic for the roots  i  1, 2  S55

1   2  2

S 45 S , 1 2  44 S 55 S55

The original differenti al equation can be written as S 45  2   2  S 44  2   2  0  S55 x 2 S 55 xy y 2 1 2

 2  2  2  (    )  0  1 2 x 2 xy y 2

         1   2   0 x  y x   y

11.9.

Relations (11.5.3) B11 

2 S11S 33  S132 S S  S13S 23 S S  S 23 , B12  12 33 , B22  22 33 S 33 S 33 S 33

B16 

2 S16 S 33  S13S36 S S  S36 S S  S 23S 36 , B66  66 33 , B26  26 33 S 33 S 33 S 33

Using these relations into Hooke' s law for plane strain  S11S 33  S132 S S  S13S 23 S S  S13S 36 e x  B11 x  B12 y  B16  xy   x  12 33  y  16 33  xy S 33 S 33 S 33 Now we wish to develop plane stress relations with  z  0  S13  S 23  0  e x  S11 x  S12 y  S16  xy e y  B12 x  B22 y  B26  xy 

2 S12 S 33  S13S 23 S S  S 23 S S  S 23S 36  x  22 33  y  26 33  xy S 33 S 33 S 33

 S12 x  S 22 y  S 26  xy 2e xy  B16 x  B26 y  B66  xy 

2 S16 S 33  S13S 36 S S  S 23S 36 S S  S 36  x  26 33  y  66 33  xy S 33 S 33 S 33

 S16 x  S 26 y  S 66  xy 11.10.

Case 2 : 1   2      i ,  3   4   Governing Equation : D1 D2 D3 D4   0      , D3  D4   y x y x Integratin g as before yields the form   2 Re[ z1 F1 ( z1 )  F2 ( z1 )] , where z1  x  y with D1  D2 

In terms of the variable z1 , this formulatio n is similar to the isotropic case given by (10.2.7) Another scheme of demonstrat ing this result is to expand the governing equation as 2

2

                  0  x   y x   y  2  2 , xxxx  2 (   ), xxxy  ( 2  4   2 ), xxyy  2(   ), xyyy  , yyyy  0    2   2 ,     2 ,  2  4   2  (   ) 2  2  4 2  2(  2   2 ) Through a change in variables , this equation can be transform ed into the standard biharmonic form for isotropic theory.

11.11.

S11 4  2 S16  3  ( 2 S12  S 66 ) 2  2 S 26  S 22  0 For orthotropi c materials S16  S 26  0  S11 4  ( 2 S12  S 66 ) 2  S 22  0 Solving as a quadratic equation for  2 : 2    2 S12  S 66  1   2 S12  S 66  S      4 22      2  S11 S11 S11       For S - Glass/Epox y material : 2 S  S 66 S 22 E1 50 E 50    2.94 , 12  2 12  1  2(0.27)   6.6 S11 E 2 17 S11 12 7 2 1, 2

1 2  6.6  6.6  4( 2.94)   0.48 ,  6.12     2 1, 2  0.69i ,  2.47i

 12, 2 

11.12. ________

________

  F1 ( z1 )  F1 ( z1 )  F2 ( z2 )  F2 ( z2 ) x 

 2  12 F1( z1 )  12 F1( z1 )   22 F2( z2 )  22 F2( z2 ) 2 y  12 F1( z1 )  12 F1( z1 )   22 F2( z2 )   22 F2( z2 )  2 Re[12 F1( z1 )   22 F2( z2 )]

 2  y  2  F1( z1 )  F1( z1 )  F2( z2 )  F2( z2 ) x  F1( z1 )  F1( z1 )  F2( z2 )  F2( z2 )  2 Re[ F1( z1 )  F2( z2 )]  xy  

 2  1 F1( z1 )  1 F1( z1 )   2 F2( z2 )  2 F2( z 2 ) xy  1 F1( z1 )  1 F1( z1 )   2 F2( z2 )   2 F2( z2 )  2 Re[1 F1( z1 )   2 F2( z2 )]

11.13.  x  2 Re[12  1 ( z1 )   22  2 ( z 2 )]  y  2 Re[ 1 ( z1 )   2 ( z 2 )]  xy  2 Re[1 1 ( z1 )   2  2 ( z 2 )] For plane stress, e x  S11 x  S12  y  S16  xy  S11Re[12  1 ( z1 )   22  2 ( z 2 )]  2 S12 Re[ 1 ( z1 )   2 ( z 2 )]  2 S16 Re[1 1 ( z1 )   2  2 ( z 2 )]  2 Re[ p1 1 ( z1 )  p2  2 ( z 2 )] 

u x

e y  S12  x  S 22  y  S 26  xy  2 Re[ q11 1 ( z1 )  q2 2  2 ( z 2 )] 

v y

2e xy  S16 x  S 26 y  S 66  xy  2 Re[( p11  q1 ) 1 ( z1 )  ( p2 2  q2 ) 2 ( z 2 )] 

u v  y x

where pi  S11 i2  S16 i  S12 and qi  S12 i  S 26  ( S 22 ) /  i Integratin g the stain - displaceme nt relations  u  2 Re[ p1 1 ( z1 )  p2  2 ( z 2 )]  f ( y ) v  2 Re[ q1 1 ( z1 )  q2  2 ( z 2 )]  g ( x ) Using these results in the expression for the shear strain  u v   2 Re[ p11 1 ( z1 )  p2 2  2 ( z 2 )]  f ( y )  2 Re[ q11 1 ( z1 )  q2 2  2 ( z 2 )]  g ( x ) y x  2 Re[( p11  q1 ) 1 ( z1 )  ( p2 2  q2 ) 2 ( z 2 )]  f ( y )  g ( x )  0  f ( y )   g ( x )  constant  C  f ( y )  Cy  C1 , g ( x )  Cx  C2 Note that f and g are rigid - body motion terms and may be dropped

11.14.

Stresses :  x  2 Re[12  1 ( z1 )   22  2 ( z 2 )]  y  2 Re[ 1 ( z1 )   2 ( z 2 )]  xy  2 Re[1 1 ( z1 )   2  2 ( z 2 )] From stress transform ation theory,  r   x cos 2    y sin 2   2 xy sin  cos   2 Re[(12 cos 2   sin 2   21 sin  cos ) 1 ( z1 )  ( 22 cos 2   sin 2   2 2 sin  cos ) 2 ( z 2 )]  2 Re[(sin   1 cos ) 2  1 ( z1 )  (sin    2 cos ) 2  2 ( z 2 )]     x sin 2    y cos 2   2 xy sin  cos   2 Re[(12 sin 2   cos 2   21 sin  cos ) 1 ( z1 )  ( 22 sin 2   cos 2   2 2 sin  cos ) 2 ( z 2 )]  2 Re[(cos   1 sin ) 2  1 ( z1 )  (cos    2 sin ) 2  2 ( z 2 )]  r   x sin  cos    y sin  cos    xy (cos 2   sin 2 )  2 Re[( 12 sin  cos   sin  cos   1 cos 2   1 sin 2 ) 1 ( z1 )  (  22 sin  cos   sin  cos    2 cos 2    2 sin 2 ) 2 ( z 2 )]  2 Re[(sin   1 cos )(cos   1 sin ) 1 ( z1 )  (sin    2 cos )(cos    2 sin ) 2 ( z 2 )] Displaceme nts : u  2 Re[ p1 1 ( z1 )  p2  2 ( z 2 )] v  2 Re[q1 1 ( z1 )  q2  2 ( z 2 )] From displaceme nt transfo rmation theory, ur  u cos   v sin   2 Re[( p1 cos   q1 sin ) 1 ( z1 )  ( p2 cos   q2 sin ) 2 ( z 2 )] u  u sin   v cos   2 Re[( q1 cos   p1 sin ) 1 ( z1 )  ( q2 cos   p2 sin ) 2 ( z 2 )]

11.15*.

General characteri stic equation : S11 4  2 S16  3  ( 2 S12  S 66 ) 2  2 S 26  S 22  0 For orthotropi c materials S16  S 26  0  S11 4  ( 2 S12  S 66 ) 2  S 22  0 Solving as a quadratic equation for  2 : 2    2 S12  S 66  S 22  1   2 S12  S 66       4    2   S11 S11 S11       2 2 Defining 1,2  i1, 2  1, 2  1, 2 2 1, 2

For the isotropic case,





2 S  S 66 S 22 1  1 , 12  2  12, 2    2  2 2  4  1 S11 S11 2

Using MATLAB : S - Glass/Epox y : 1, 2  0.693 , 2.474 Boron/Epox y :

1, 2  0.577 , 5.545

Carbon/Epo xy :

1, 2  0.788 , 5.747

Kevlar49/E poxy : 1, 2  0.643 , 5.784

11.16.

With u  v  0 , w  w( x, y )  e x  e y  e z  e xy  0 , and therefore Hooke' s law for monclinic materials becomes   x  C11     y   z       yz      zx        xy   

C12 C 22    

C13 C 23 C33   

0 0 0 C 44  

0 0 0 C 45 C55 

C16   0  C 26   0  C36   0     0  2e yz  0   2e zx    C66   0 

 x   y   z   xy  0 w w  C55 y x w w  yz  2C 44e yz  2C 45e xz  C 44  C 45 y x  yz  2w 2w 2w Equilbrium Eqns.  xz   0  C55 2  2C 45  C 44 2  0 x y xy x y

 xz  2C 45e yz  2C55e xz  C 45

Looking for solutions w  F ( x  y ) gives (C55  2C 45  C 44 2 ) F   0 Cancelling the common F  term produces the quadratic characteri stic equation C 44 2  2C 45  C55  0 Solving for the two roots gives :  

2 - C 45  C 45  C 44C55

C 44

2 As previously shown C 44C55  C 45  roots are complex conjugate pairs (,  )

and so general solution t o governing equation can be written as w  F1 ( x  y )  F2 ( x  y )  F1 ( z*)  F2 ( z *) , where z*  x  y Since w must be real  w  F1 ( z*)  F1 ( z *)  2 Re{F1 ( z*)}  2 Re{F ( z*)} and the stresses then can be written as  xz  2 Re{( C 45  C55 ) F ( z*)}  yz  2 Re{( C 44  C 45 ) F ( z*)}

11.17*.

For the case : X  0 ,  i  i i A1 

( X   2Y )  2Y ( X  1Y ) 1Y  , A2   2i( 2  1 ) 2i( 2  1 ) 2i(1   2 ) 2i(1   2 )

 1 ( z1 ) 

A1  1 2Yy  i 2Yx A  1 2Yy  i 2Yx  ,  2 ( z 2 )  2  2 2 2 z1 2( 2  1 )( x  1 y ) z 2 2(1   2 )( x 2   22 y 2 )

 r  2 Re[(sin   1 cos ) 2  1 ( z1 )  (sin    2 cos ) 2  2 ( z 2 )]  

Y1 2 (1   2 ) sin  r (cos   12 sin 2 )(cos 2    22 sin 2 ) 2

1 2Y sin   1  cos 2   12 1  cos 2    22   r (1   2 )  cos 2   12 sin 2  cos 2    22 sin 2  

   2 Re[(cos   1 sin ) 2  1 ( z1 )  (cos    2 sin ) 2  2 ( z 2 )] 

1 2Y sin   cos 2   12 sin 2  cos 2    22 sin 2    0 r (1   2 )  cos 2   12 sin 2  cos 2    22 sin 2  

 r  2 Re[(sin   1 cos )(cos   1 sin ) 1 ( z1 )  (sin    2 cos )(cos    2 sin ) 2 ( z 2 )] 

1 2Y cos  1 2Y cos   0 r (1   2 ) r (1   2 )

2Y sin  ,     r  0 r MATLAB Plots for orthotropi c and isotropic cases :

For the isotropic case, 1   2  1   r  

r/Y

Orthotropic Case Carbon/Epoxy

Isotropic Case

11.18*.

The hoop stress solution for the pressurize d circular hole problem was given by (11.5.47)  ie i    p Re   (sin   1 cos )(sin    2 cos )  [(1 2  i1  i 2 ) sin 3   i (1 2  2) sin 2  cos    ( 21 2  1) sin  cos 2   (1   2  i ) cos 3 ]   For the orthotropi c case with 1,2  i1, 2   ie i    p Re   (sin   i1 cos )(sin   i 2 cos )  [( 1 2  1   2 ) sin 3   i (1 2  2) sin 2  cos    ( 21 2  1) sin  cos 2   i (1   2  1) cos 3 ]   MATLAB Plot : (Note higher stress regions for orthotropi c materials)

 /p Orthotropic Case Carbon/Epoxy

Isotropic Case

11.19*.

From solution (11.5.51) for the circular hole case    S 12  1 y  22  2 y   1   1  (1   2 )  a 2   2 ( y 2  a 2 )  (1   2 )  (1  1 )  a 2  12 ( y 2  a 2 ) 2      Solution for the isotropic case comes from (8.4.15)  x (0, y )  S 

 x (0, y )    (  / 2, y ) 

S a2  S  3a 4  S  a 2 3a 4  1  2   1  4   S   2  4  2 y  2 y  2y y 

MATLAB Plots :

x(0,y)/S

Orthotropic Case Carbon/Epoxy

Isotropic Case

11.20.

From solution (11.5.50) with a  0 ,      iSb 12 z1 iSb 22 z2    x  S  Re   1   1  (1   2 )(i 2b)  z 2   2b 2   (1   2 )(i1b)  z12  12b 2 2   2     S1  z1 S 2  z2    S  Re  1   1  (1   2 )  z 2   2b 2   (1   2 )  z12  12b 2 2   2  Now for x  0 , and y  b , z1  1 y and z2   2 y        S1  y S 2  y y   x  S  Re   1   1  S  Re  S  S  2 2 2 2 2 2       (    ) (    )   y b y b 1 2  1 2     y  b    y   2S  S   y 2  b2    and likewise for the other stress components ,        S z1 S z2 S y     y  Re   1   1   1   (1   2 ) 2  z 2   2b 2  Re (1 2 )  y 2  b 2  (1   2 )1  z12  12b 2 2     2       S z1 S z2    xy   Re   1   1  0  (1   2 )  z 2   2b 2   (1   2 )  z12  12b 2 2   2  Note the singular nature of the stresses at the crack tip of order O(1/ r )

11.21.

1 ( z1 )  A1 z1 





1 Sa 2 2 Sa 2 2 z1  z12  a 2  1 ( z1 )  A1  2(1   2 ) 2(1   2 )

 2 ( z2 )  A2 z2 







1

z  a z1  z12  a 2

1 Sa 21 Sa 21 z2  z22  a 2  2 ( z1 )  A2  2(1   2 ) 2(1   2 )

2 1

2



1



z22  a 2 z2  z22  a 2



 x  2 Re[121 ( z1 )   222 ( z2 )]  Sa 2 212  2 Re  A112  2(1   2 )   y  2 Re[1 ( z1 )  2 ( z2 )]



1

z12  a 2 z1  z12  a 2



 A2 22 

Sa 21 22 2(1   2 )

 Sa 2 2 1 Sa 21  2 Re  A1   A2  2(1   2 ) z12  a 2 z1  z12  a 2 2(1   2 )   xy  2 Re[11 ( z1 )   22 ( z2 )]











  z22  a 2 z2  z22  a 2 

 Sa 21 2 1 Sa 21 2  2 Re  A11   A2 2  2(1   2 ) z12  a 2 z1  z12  a 2 2(1   2 )  Far - Field Behavior : | z1 |  , | z2 | 



  z22  a 2 z2  z22  a 2  1



1



  z22  a 2 z2  z22  a 2 



1



 x  2 Re[ A112  A2 22 ]  ( 22  22 )(12  12 ) S (12  12  21 2 )( 22  22 ) S 2 22 [ 2 (12  12 )  1 ( 22  22 )]S   2    2 2 2[( 2  1 ) 2  (2  1 ) 2 ] 22 [( 2  1 ) 2  (22  12 )]  2[( 2  1 )  (2  1 ) ]  0 [( 2  22 )  (12  12  21 2 )]S  [( 2  1 ) 2  (2  1 ) 2 ]  y  2 Re[ A1  A2 ]  2  2   [(   ) 2  (   ) 2 ] S  S 2[( 2  1 ) 2  (2  1 ) 2 ] 2 1 2 1    xy  2 Re[ A11  A2 2 ]  1 ( 22  22 ) S  2 (12  12  21 2 ) S 2 [ 2 (12  12 )  1 ( 22  22 )]S   2   2 2 2 2 22 [( 2  1 ) 2  (22  12 )]   2[( 2  1 )  (2  1 ) ] 2[( 2  1 )  (2  1 ) ] 0

11.21. Continued

At Crack Tip : z1  a  zˆ1  r cos   1r sin  , z1  a , z1  a  2a 



1

z  a z1  z  a 2 1

2

2 1

2



1 1  2azˆ1 a  2azˆ1 a 2azˆ1







z2  a  zˆ2  r cos    2 r sin  , z2  a , z2  a  2a 



1

z  a z2  z  a 2 2

2

2 2

2





1 a 2azˆ2

 S a     S a      2 1 1 2  2 1 2    x  Re   1   Re       ˆ ˆ 2 (    ) z z 2 r (    ) cos    sin  cos    sin    1 2  2 1  1 2  2 1    S a  1 2  2 1   Re        2r cos    sin  cos    sin   1 2  2 1  Likewise at the crack tip, the other stress components reduce to y 

 S a  1  1 2  Re   2r cos   1 sin    1   2  cos    2 sin 

 xy 

 S a  1 2  1 1  Re   2r cos    2 sin    1   2  cos   1 sin 

11.22.* Anisotropic Crack Tip  y /KI Contours (1=0.753i , 2=2.870i)

0.8 0.6 0.4 0.2 0 -0.2 -0.4 -0.6 -0.8 -1 -1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

11.23.

From Exercise 11-16: w  2 Re{F ( z*)} ,  xz  2 Re{(C45  C55 ) F ( z*)} ,  yz  2 Re{(C44  C45 ) F ( z*)} From characteristic equation : C45  C55  (C45  C44 ) Choosing F ( z*)  A z * , where A   2 K 3 / (C55  C45 )  F ( z*)  A / 2 z * w  2 Re{F ( z*)}  2 Re{ A r cos    sin }   cos    sin    cos    sin     K3 2r Re    K 3 2r Re    C55  C45   C45  C44     C45  C55  K3 1   xz  2 Re{(C45  C55 ) F ( z*)}  Re  A Re    2 r cos    sin   2r   cos    sin      C44  C45  K 3 1 1  yz  2 Re{(C44  C45 ) F ( z*)}  Re  A Re    2 r cos    sin   2r   cos    sin   K  1  K3 Note  yz (r , )  3 Re  Re i  0  2r 2r  1 

11.24.

r 

E Er u du   r  r  du u     r  ,       r  , 1   r  r  dr r 1   r  r  r dr  E  Er

dσ r σ r  σ θ  0 dr r  d2 u Er 1 du u  1 du u   2  νθr  νθr 2   ( E r  E   r)   E r r  E   2   0  1  νθr ν rθ  dr r dr dr r  1   r  r  r 

Equilibriu m Eqn :

d 2u 1 du u d 2 u 1 du u  E  E  0    n 2 2  0 , where n 2  E  / E r r  2 2 2 r dr r dr dr r dr r m Looking for solutions of the form : u  Cr  Er

m(m  1)  m  n 2  0  m 2  n 2  m   n , Solution is u  Ar n  Br  n r 

Er Er Er u  du   r   A (n   r )r n 1  B (n   r )r  n 1  1   r  r  dr r 1   r  r 1   r  r

 

E E rn E rn du  u (n   r )r n 1  B (n   r )r  n 1    r  A 1   r  r  r dr  1   r  r 1   r  r

Rewriting the radial stress form as :  r  C1 r n 1  C 2 r  n 1 and applying the Boundary Conditions : pb n 1 pb n 1 a 2 n  r (a )  0 ,  r (b)   p  C1   2 n , C2  2n b  a 2n b  a 2n pb n 1 pb n 1 n n 1 n 1 2 n  n 1   r   2n r  a r ,    r  a 2 n r  n 1  b  a 2n b 2n  a 2n









11.25.

eR 

 1  R  R (     ) . . . . (1) ER ER

e 

  1    R    R  R . . . . (2) E E ER

e 

  1    R  R  R   . . . . (3) E ER E

e    / 2  

1   R   , e R   R / 2 R , eR   R / 2 R . . . . (4) E

(1)   R  E R e R   R (     )  1  R   2 E  2 (     )  R  R         e  e  R  R  (2)  (3)  e  e    ER 1   R  ER   E E  Combine previous two lines   R  C11e R  C12 (e  e ) Likewise following similar steps     C12 e R  C 22 e  C 23e ,    C12 e R  C 23e  C 22 e Solving relations (4) for the shear stresses gives  R  C 44 e R ,    (C 22  C 23 )e ,  R  C 44 eR where E R (1   R ) E  , C12   R m m  E E  1   2R   C 22  (1   R )m  ER 

C11 

C 23 

 E E    R   2R   (1   R )m  ER 

C 44  2 R , C 22  C 23  2   m  1   R  2 2R

E ER

E (1   R )

11.26.

General Equilib. Eqns  R 1  R 1  R 1    (2 R         R cot )  0 R R  R sin   R  R 1   1   1    [(     ) cot   3 R ]  0 R R  R sin   R  R 1   1   1    (2  cot   3 R )  0 R R  R sin   R d R 2 With  R   R ( R) ,         ( R) ,  R      R  0   ( R    )  0 dR R du u du u Using  R  C11  2C12 ,       C12  (C 22  C 23 )  dR R dR R d  du u 2 du u du u  2C12    C11  2C12  C12  (C 22  C 23 )    C11 dR  dR R R dR R dR R 2 C  C 23  C12 d u 2 du u   k 2 2  0 , where k 2  2 22 2 R dR C11 dR R

This is Cauchy - Euler ODE, so look for solutions u  CR m  1 1 1 m 2  m  k 2  0  m  [1  1  4k 2 ]     k2 2 2 4 So solution is given by u  AR n (1 / 2 )  BR  n (1 / 2 ) C  C 23  C12 1 2 E  (1   R ) 1 2 R (1   R ) 1 1 where n 2   k 2   2 22     4 4 C11 4 E R (1   R ) 4  R (1   R ) du u  2C12   R  C n* R n 3 / 2  C *n R  n 3 / 2 dR R * where C n  (n  1 / 2)C11  2C12 A , C *n  (n  1 / 2)C11  2C12 B

 R  C11

Boundary Conditions :  R (a )  0   R  C n* a n 3 / 2  C *n a  n 3 / 2  0  R (b)   p   R  C n*b n 3 / 2  C *n b  n 3 / 2   p Solving two eqns. for constants  C n*  



pb n 3 / 2 pb n 3 / 2 a 2 n * , C  n b 2n  a 2n b 2n  a 2n



pb n  ( 3 / 2 ) n (3 / 2 ) R  a 2 n R ( n  (3 / 2 )) , 2n 2n b a du u pb n  (3 / 2 )       C12  (C 22  C 33 )   2 n C n R n ( 3 / 2)  C  n a 2 n R ( n  (3 / 2 )) dR R b  a 2n 1  [n  (1 / 2)] R 1  [n  (1 / 2)] R Cn  , C n   R [2  (1   R )( n  (1 / 2)) /  R ]  R [2  (1   R )( n  (1 / 2)) /  R ]

 R  





11.27.

(11.7.3)   r 

E Er u du   du u   r  ,        r  1   r  r  dr r 1   r  r  r dr 

d r  r      2 r  0  dr r Er d  du u  E   r E  1 du  r E r  E  u   r   r   2 r   2 1   r  r dr  dr r  (1   r  r ) r dr (1   r  r ) r (8.4.74) 

(1   r  r ) 2 3 d 2u du E  r r  u  r . . . . Cauchy - Euler non - homogeneou s ODE type.

2 dr E r Er dr E Look for homogeneou s solution of the form u h  r n , n 2    u h  C1 r n  C 2 r  n Er 2

Particular solution, try u p  Kr 3  K   u  u h  u p  C1 r n  C 2 r  n 

(1   r  r ) 2 E r [9  ( E  / E r )]

(1   r  r ) 2 r 3 E r [9  ( E  / E r )]

11.28.

For solid disk C 2  0  u  C1 r n  Kr 3 , K   r  

(1   r  r ) 2 E r [9  ( E  / E r )]



Er Er u  du   r   nC1 r n 1  3Kr 2  C1 r r n 1   r Kr 2  1   r  r  dr r  1   r  r



Er C1 (n   r )r n 1  K (3   r )r 2 1   r  r





Boundary Condition :  r (a)  0  C1 (n   r )a n 1  K (3   r )a 2  0  C1    r   

3   r Er  K (3   r )a 3 n r n 1  K (3   r )r 2  2 (a 3 n r n 1  r 2 ) 2 1   r  r 9n







E E du  u C1 (1   r n)r n 1  K (1  3 r )r 2    r  1   r  r  r dr  1   r  r

  

K (3   r ) 3 n a (n   r )

2 [n(3   r )a 3 n r n 1  (n 2  3 r )r 2 ] 2 9n



11.29.

 3P  2 xy3 S16   3 c xy   (2c 2 y 2  y 4 ) 3 6S11 4c    xx  0   xxxx  0 ;  yy 



 S 3P  3P  4 S   2 xy  16 (4c 2  12 y 2 )   yyyy   3  16  3  6 S11 4c  4c  S11  



3P 2 c  y 2   xxyy  0 3 4c For plane stress, (11.5.6) 

 xy 

 4U  4U  4U  4U  4U S 22 4  2 S 26 3  (2S12  S 66 ) 2 2  2S16  S11 4  0  x x y x y xy 3 y  2S16

  3P  2  S11 3P3  4S16   S16 3P3 (4  4)  0  satisfies governing equation 3 4c 4c  S11  4c

 x   yy 

 S16 3P  3P 3P S16  c 2 2 2    y 2   2 xy  ( 4 c  12 y )   xy   3  3 3 6 S11 4c  2c 2c S11  3  

 y   xx  0 3P 2 3P  y2  2    xy   xy   3 c  y   xy   1 4c  c 2  4c





Check Boundary Conditions :  y ( x,  c )  0  xy ( x,c)  0 c

  x (0, y)dy  -c

3P S16 c c 2 (  y 2 )dy 0 3  c 3 2c S11

3P c y2 -c  xy (0, y)dy   4c -c (1  c 2 )dy   P c 3P S16 c c 2 y  ( 0 , y ) ydy  (  y 3 )dy 0 3 - c x  c 3 2c S11 c

c

  x ( L, y)dy   -c

3PL c 3P S16 c c 2 ydy  (  y 2 )dy 0 3 - c 3  c 3 2c 2c S11

3P c y2 -c  xy ( L, y)dy   4c -c (1  c 2 )dy   P c 3PL c 2 3P S16 c c 2 y  ( L , y ) ydy   y dy  (  y 3 )dy   PL 3 - c 3 - c x  c 3 2c 2c S11 Orthotropi c Case : S16  S 26  0  same stresses as isotropic case c

12.1.

Thermal Strains : eij(T )  (T  To ) ij 1    ij   kk  ij E E 1   eij  eij( M )  eij(T )   ij   kk  ij  (T  To ) ij  E E 1  2 E ekk  3(T  To )  ekk   kk  3(T  To )   kk  E 1  2 E   ekk  3(T  To ) ij  (T  To )ij    ij  eij   1   1  2   ekk  2eij  (3  2)(T  To ) ij Mechanical Strains : eij( M ) 

12.2.

Strain Compatibil ity : eij,kl  ekl ,ij  eik , jl  e jl ,ik  0 1    ij   kk ij  (T  To )ij E E Using Hooke' s law in the strain compatibil ity relations and setting k  l   ij,kk   kk ,ij   ik , jk   jk ,ik  Hooke' s Law : eij 

 E (  mm,kk ij   mm,ij  kk   mm, jk ik   mm,ik  jk )  (T,kk ij  T,ij  kk  T, jk ik  T,ik  jk ) 1  1  Note that with zero body forces,  ij,j  0  1  E  kk ,ij   mm,kk ij  (T,kk ij  T,ij ) 1  1  1  2 E Setting i  j   mm,kk   T,kk 1  Combining these results gives 1 E 1  (T,ij  ij,kk  kk,,ij   ij T,kk ) (1  ) 1  1   ij,kk 

12.3.

Unrestrict ed thermal expansion : eij  T ( x, y, z )ij  ex  e y  ez  T , exy  e yz  ezx  0 Using strain compatibil ity relations  2 2  2exy  2 ex  e y  2 ex  e y  2T  2T   2    0   0 y 2 x 2 xy y 2 x 2 y 2 x 2

 2 e y  2 ez  2eyz  2 e y  2 ez  2T  2T  2 2  2  2 0 2  2 0 z 2 y yz z y z y  2 e z  2 ex  2ezx  2 ez  2 ex  2T  2T  2 2  2  2 0 2  2 0 x 2 z zx x z x z 2 2 2 T T T The above three relations  2  2  2  0 x y z e   2 ex   e e  2 ex  2T    yz  zx  xy   0 0 yz x  x y z  yz yz  2e y  2ey   ezx exy e yz   2T        0 0 zx y  y z x  zx zx e  2 ez   e e   2 ez  2T    xy  yz  zx   0 0 xy z  z x y  xy xy  2T T  2T T  2T T 0  f ( y, z ) , 0  g ( x, z ) , 0  h ( x, y ) 2 x x xy x zx x T T T   constant  a , and following similar steps it can be shown that b, c x y z T  a  T  ax  F ( y, z ) x T F b  b  F  by  G ( z ) y y T dG c  c  G  cz  d z dz  T  ax  by  cz  d

12.4.

For Plane Stress : 1 1 1  e x  ( x   y )  (T  To ) , e y  ( y   x )  (T  To ) , e xy   xy  E E E E E x  ( e x  e y )  (T  To ) 2 1  1  E E y  ( e y  e x )  (T  To ) 2 1  1  E  xy  e xy 1  Txn   x n x   xy n y  (Txn ) s   E  u v   E u  v E  (  ) (T  To ) n x     n y  (Txn ) s  2  x  y 1   2 ( 1   )  y x  1       T yn   xy n x   y n y  (T yn ) s   E  u v   E  v u E    n x   (  ) (T  To ) n y  (T yn ) s  2 x 1   1   y   2(1  )  y x 

12.5.

Plane Strain :  x  (ex  e y )  2ex  (3  2)(T  To ) ,  y  (ex  e y )  2e y  (3  2)(T  To ) ,  xy  2exy Solving for the strains  1  1  1  ex  [(1  ) x   y  E(T  To )] , e y  [(1  ) y  x  E(T  To )] , exy   xy E E E 2  2exy  2 ex  e y Using these strains in the compatibil ity relation  2 2  y 2 x xy  2  xy 2 2 [(1  ) x   y  E(T  To )]  x 2 [(1  ) y  x  E(T  To )]  2 xy  y 2  2  xy  2 x  2 y   xy x 2 y 2 But from equilibriu m equations with zero body forces,

(1  ) 2 ( x   y )  E 2T  2

2  xy  y  2 y  2  xy  x  xy  2 x   xy  0   0 ,   0   0 x y x 2 xy x y y 2 yx

 2  xy  2 x  2 y 2    0 and thus the previous compatibil ity statement reduces to xy x 2 y 2 E 2  2 ( x   y )   T 0 1  To determine the correspond ing plane stress result, simply use the transform ation Table 12 - 1  E (1  2)  1  E E , ,   E , 2 (1  ) 1  1  2 1  and thus the compatibil ity relation for plane stress becomes  2 ( x   y )  E  2T  0

12.6.

From Example 12 - 1, the stresses were given by  x   2 C2 cosh x  C3 x sinh x sin y  ETo sin y   2  y   2 C2 cosh x  C3 ( x sinh x  cosh x ) sin y      1  xy   2 C2 sinh x  C3 ( x cosh x  sinh x ) cos y    Stress free boundary conditions :  x (  a, y )   xy (  a, y )  0  C2 cosh a  C3a sinh a   ETo /  2 1 sinh a )  0  Solving for the two constants  ET a cosh a  sinh a  ETo sinh a C2   2 o , C3   a   sinh a cosh a  a   sinh a cosh a  MATLAB Plots ( a  1 ,   2) : C2 sinh a  C3 ( a cosh a 

y/ETo siny = 1

siny = -1

12.7.

For the radially symmetric case, 2 

d2 1 d 1 d  d  1 d  d  1 d  d  4 r    r  ,    r  2 dr r dr r dr  dr  r dr  dr  r dr  dr   

 the plane stress governing equation  4   E 2T  0  1 d  d  1 d  d    1 d  dT  1 d  r    r     E r   0 , and since  r  r dr  dr  r dr  dr    r dr  dr  r dr 1 d  d  1 d 2  1 d  dT   r  r     E r  r  r dr  dr  r dr r dr  dr   Integratin g this result thr ee times  d 1 d 2  dT  r  r    Er  C1   dr  r dr dr  1 d 2 r r    ET  C1 log r  C2  r dr

r

r 2  r   E Tdr  C1  r log rdr  C2 r 2  C3  r  

C E C Tdr  1 ( 2 log r  1)  C2  23 2  r 4 r

12.8.

Using axisymmetr ic stress forms : E du u r  [    (1  )T ] 2 r 1   dr E u du   [   (1  )T ] 2 dr 1  r d r  r    Into Equilibriu m Eqn. :  0 dr r d 2 u 1 du u dT   2  (1  ) , which can be written as 2 r dr r dr dr d 1 d dT  ( ru )  ( 1   )   dr  r dr dr Integratin g the governing equation  1 d (ru )  (1  )T  constant , intergrati ng again  r dr A (1  ) u  A1 r  2   Trdr r r

12.9.

 2T 1 T 1 d  dT    ho  0  r   ho 2 r r r r dr  dr  h Integratin g  T  C1 log r  C2  o r 2 4 Boundary Conditions : T (0) must remain bounded  C1  0 T (a)  To  C2  To   T  To 

ho 2 a 4

ho 2 (a  r 2 ) 4

Governing equation for plane strain,  4 

E 2 Eho  T  0   4   1  1 

1 d  d  1 d 2   Eho r  r    r  r dr  dr  r dr  1 





Integratin g the governing equation t hree times  C C Eho 2  r  1 (2 log r  1)  C2  23  r 4 r 16(1  ) d 2 d C C 3Eho 2  (r r )  1 (2 log r  1)  C2  23  r ,  r  0 2 dr dr 4 r 16(1  ) h  z  ( r   )  E(T  To )  ( r   )  E o (a 2  r 2 ) 4 Since the cylinder is solid, the stresses must be bounded at r  0  C1  C3  0  

Boundary condition  r (a)  0  C2    r 

Eho 2 a 16(1  )

Eho Eho r 2  a 2 ,   3r 2  a 2 , r  0 16(1  ) 16(1  )







 Eho 2 Eho 2  h  z   r  a   E o (a 2  r 2 ) 8(1  )  4  4(1  )



12.10.

From the general solution given by (12.7.6), A (1  ) r u  A1r  2   Td r r Bounded displaceme nts at r  0  A2  0 (1  ) a Td a2  (1  ) r (1  )r a u  T  d     Td r a2 For the case where T  To : u (a)  0  A1  

u

 1 r2 (1  )To r (1  )To r a r a2     0  d    d   ( 1   )  T  o 2   r a2 r 2 a 2 

r 

E E  u u  [er  e  (1  )(T  To )]     (1  )(T  To )  0 2 2  1  1    r r 

 

d 2 d  (r r )  0 dr 2 dr

12.11.

The plane stress governing equation  4   E 2T  0 with axisymmetr y reduces to 1 d  d  1 d  d    1 d  dT   r   r     E r 0 r dr  dr  r dr  dr    r dr  dr  This result can be integrated to give r 1   r r r r   A1 log  A2   log  A3    A4  E    Tdd a  a a a a a   2

2

and the resulting stress field then becomes 2 A E r 1 d A1 A2 r r   2  2 ( 2 log  1)  23  2  Trdr r dr r a a a r a 2A A A d 2 d r 1 r      2  ( r r )   21  22 ( 2 log  3)  23  E T  2  Trdr  dr dr r a a a r a   With zero tractions on r  a and r  b , general boundary conditions (12.5.9)  ( a )  0  A3  A4  0 b 1   b r b b  A2   log  A3    A4  E    Tdd a a a a a a   2

(b)  0  A1 log

2

d( a )  0  A1  A2  2 A3  0 dr 2 A E b A A d(b) b  0  21  22 ( 2 log  1)  23  2  Trdr  0 dr b a a a b a Solving for the constants gives A1 

2 2 2  b E    b  b b b b r  b b 2 log ( 2 log  1 )   1 Trdr  4 log Tr log dr       a    N    a  a a a a a  a a 

A2 

2 2  b  b  2  b E    b  b b r  2 log   1 Trdr  2  1 Tr log dr     a    a    N    a  a a a    a  

A3   A4 

2 2 2   b 2  b E   b   b b b b r   2 log Trdr  2 log   a    1 a Tr log dr         N   a   a a a a    a  

2 2  b   b  b where N  4 log      1 a   a  a 

2

12.12.

Spherical Coordinate s with Spherical Symmetry : u u eR  R , e  e  R , eR  e  eR  0 R R  R  ( eR  e  e )  2eR  (3  2)(T  To ) u R u  2 R  (3  2)(T  To ) R R u u  E E   (1  ) R  2 R   (T  To )  (1  )(1  2)  R R  (1  2) Likewise,       ( eR  e  e )  2e  (3  2)(T  To )  (  2)

uR u  2 R  (3  2)(T  To ) R R u  E E  uR   R (T  To )  (1  )(1  2)  R R  (1  2)  R      R  0  (  2)

Using these results into the equilibriu m equations  d R 1 d  1 d  1   dT  ( 2 R       )  0  ( R 2uR )     2 dR R dR  R dR  1   dR Integratin g this differenti al equation  C 1  1 R 2 uR   2  T d  C1 R  22 1  R R R u 2C 1   2  Note : R   T  3  T 2 d   C1  32 R 1    R R  Using these results back into Hooke' s Law gives the stresses EC1 2 EC2 1 2E 1 R 2 R   T d   3  1  R 1  2 1   R 3 EC1 EC2 1 ET E 1 R 2     T d    3  1  R 1  2 1   R 3 1  

12.13.

Spherical Symmetry P roblem :  2T  0 

C d  2 dT  R   0  T   1  C2 dR  dR  R

Ti ab Ta , C2   i ba ba R R Ta  b Ta Ta 1 1    T  i   1   T 2 d  i  (b   2 )d  i  bR 2  R 3  0 baR  ba ba 2 3  EC1 2 EC2 1 2E Ti a 1  R (b)  0     0 1   b  a 6 1  2 1   b 3 2 EC2 1 2E Ti a  1 b 1  EC1  R (a )  0      0  1   b  a  2 a 3  1  2 1   a 3 Temperatur e Boundary Conditions : T ( a )  Ti , T (b)  0  C1  

(1  2)Ti a (1  )Ti a 3b 3 2 2 Solving for the constants, C1  (b  ab  2a ) , C2   3(1  )( b 3  a 3 ) 2(1  )( b 3  a 3 ) Inserting these expression s into the general solution from Exercise 12 - 11 gives the stress field R 

ETi ab  1 2 a 2b2  2 a  b  ( b  ab  a )  1   b 3  a 3  R R 3 

   

ETi ab  1 2 a 2b2  2 a  b  ( b  ab  a )  1   b 3  a 3  2R R 3 

On the inner surface R  a with b  a (1  ) ETi ETi b 1      ( a 2  ab  2b 2 )  [1  (1  )  2(1  ) 2 ] 3 3 3 1  b  a 2(1  ) [(1  )  1] ETi 2  ( 1   ) 2(1  ) 3 On the outer surface R  b with b  a (1  ) ETi ETi a 1      (b 2  ab  2a 2 )  [(1  ) 2  (1  )  2] 3 3 2(1  ) b  a 2(1  ) [(1  ) 3  1] ETi 2  (1  ) 2(1  ) 3

12.14.

Using definition s of the complex and integrated temperatu re function it was determined that t t T  R  I , and this suggested the decomposit ion of the displaceme nt field x y u  u  t R , v  v  t I , where  is a constant t o be determined Using this decomposit ion definition in Hooke' s Law for plane strain  u v  u  x      2  (3  2)T x  x y   u v   t t u t    2     (3  2)T  ( R  I )  2 R  x x   x y   x y  u v  u t   t   2     (3  2)T  2 R  2 R  x x   x  x y   u v  u t   2     (3  2)T  2(  ) R x x  x y   u v  u   2     [(3  2)  2(  )]T x  x y  Now if the factor  (3  2)  2(  )  0 , the temperatu re terms will vanish. (3  2) This will occur if    (1  ) , and thus for this case  x  x ,  y  y and  xy  xy 2(  ) For the plane stress case, simply use the interchang e of elastic constants in Table 12 - 1 to show that 

1     1   1  2  1   

12.15.

From relations (12.8.15)    z2  ri 2 ro2      ( z )  A log z  2 , ( z )  A log z  2 2 2  2  r  r z ( r  r ) i o  i o    Stress free boundary conditions : ______   ( r  ir ) r  ri , ro   ( z )  ( z ) e 2i [ z ( z )  ( z )]  0   r  ri , ro ______ 1 2z 1 2z 2z 1 2ri 2 ro2  2 i 2 i  z         ( z )   ( z ) e [ z  ( z )   ( z )]    2   2 e  2  2   2 2 2 2 2 3  z r  r z r  r z r  r z ( r  r ) z i o i o i o i o    i At z  ri e  3i  e  i 2r ei ei 2ri e  i 2ri ei e  i 2ri 2 ro2e  3i  2 i  e   ( r  ir ) r  ri    2i 2  2  e    2 3   r ri  ro ri ri  ro2 ri 2  ro2 ri (ri  ro2 )ri   ri  i 2   i 2r 2ro2 e  0    2 i 2  2 2   ri ri  ro ri (ri  ro ) 

Likewise at z  ro ei  e  i 2ro ei ei 2ro e  i   3i 2ro ei e  i 2ri 2 ro2e  3i  2 i e   ( r  ir ) r  ro    2    e  r  r 2  r 2  r  (r 2  r 2 )r 3  ri  ro2 ro ri 2  ro2  ro i o o i o o   o  2 2   i 2r 2ri e  0    2 o 2  2 2   ro ri  ro ro (ri  ro )   stress free boundary conditions are satisfied

12.16.

Conduction Equation :  2T 1 T 1  2T   0 r 2 r r r 2 2 Using separation of variables , let T (r , )  f (r ) g () 

 2T  0 

1   r 2  f (r )  f (r ) 1 1 r    g ()  constant  c 2  f  g  f  g  2 fg   0   r r f (r ) g () 1 c2  f (r )  2 f (r )  0 and g ()  c 2 g ()  0 r r Solution t o  - equation is g ()  A sin c  B cos c , but temper ature field should be odd function of  with c  1  g ()  A sin  C Solution t o r - equation is f (r )  C1r  2 r C    Temperatur e solution is T   C1r  2  sin  , where we have absorbed the constant A in C1 and C2 r   T (a, ) C Insulated boundary condition on hole :  0  C1  22  0  C2  a 2C1 r a q q Conditions at Infinity : T (, )  y  C1  k k 2 q a   T   r   sin  k r  f (r ) 

12.17.

The potentials from Example 12 - 4 where given by  a2  2iqa 2   ( z )  A log z , ( z )   A 2  log z  1 , where A   (1   )k z  ______ 8qa 2   A    r     2  ( z )   ( z )   4 Re  ( z )   4 Re     sin  , (1   )kr   z 3  4 Eqa 2  1    r     sin  1  1  kr  z a2 1      r  2i r  2e 2i z  ( z )  ( z )   2 Ae 2i   2  2 3   z z  z For plane stress,  

1  2a 2 1  2 Ae 2i  e 3i  3 e 3i  e i  r r r  

 8qa 2 i  a2 a2  cos   2 cos   i 2 sin   (1   )kr  r r 

Separating real and imaginary parts    r  

Eqa 3 1 Eqa  a a 3     cos  sin  and   r kr 3 2 k  r r 3 

Solving for the individual normal stresses  r  

1 Eqa  a a 3  1 Eqa  a a 3    3  sin  ,        sin  2 k r r  2 k  r r 3 

12.18*.

Temperatur e Field for Exampl e 12 - 4 : T 

q a2   r   sin  k r 

Temperatur e Field for Exampl e 12 - 5 : T 

q a2      sin  , k  

where ,  are the coordinate s in the mapped region with     m m m z  R e i  e i   x  R    cos  and y  R    sin         MATLAB Plots of isotherms for each problem :

Example 12-4: Circular Hole Case

Example 12-5: Elliptical Hole Case

12.19.

To extract th e circular hole case from Example 12 - 5, let a  b  m  0 and R  a  r z  a  re i  ae i    a Eqa    ( 2  m)[ 4   2 (1  m 2 )  m 2 ] sin  2kh( ) Eqa 2 4   (   2 ) sin  8 2 k Eqa  a a 3     sin   2k  r r 3  Eqa ( 2  m){[  4   2 (1  m) 2  m 2 ] sin   2 2 m sin 3} 2kh( ) Eqa 2 4   (   2 ) sin  8 2 k

  



Eqa  a a 3     sin  2k  r r 3 

Eqa ( 2  m)[ 4   2 (1  m 2 )  m 2 ] cos  2kh( ) Eqa 2 4   (   2 ) cos  2 k 8

 



Eqa  a a 3     cos  2k  r r 3 

12.20*.

From Example 12 - 5, the non - dimensiona l hoop stress around hole boundary is gvien by equation (12.8.32) 2 with   1  (1  m){[(1  (1  m) 2  m 2 ] sin   2m sin 3}    Eqa / k 2(1  2m cos 2  m 2 ) 2 (1  m)[(1  m  m 2 ) sin   m sin 3] (1  2m cos 2  m 2 ) 2 Note :   ( / 2)  1 /(1  m) 

MATLAB Plots for m  0 ,  1 / 2 ,  1: m = -1

m=0

m = 1/2

m = -1/2 m=1

12.21.

Uncoupled energy equation for steady state case : qi ,i  0 Anisotropi c heat conduction law : qi  k ijT, j Combining these two relations  k ijT,ij  0 , which for two - dimensions gives k xx

 2T  2T  2T  2 k  k 0 xy yy xy x 2 y 2

Looking for solutions : T  T ( x  y )  (k xx  2k xy   k yy 2 )T   0 we get the characteri stic equation : k xx  2k xy   k yy 2  0 with solution :  



1  2k xy  4k xy2  4k xx k yy 2k yy



with k xx k yy  k xy2 the roots will be complex conjugate pairs : 

k xy k yy

i

k xx k yy  k xy2 k yy

 , 

So the general solution becomes : T ( x, y )  F1 ( x  y )  F2 ( x   y ) but since the temperatu re must be real  T ( x, y )  F ( x  y )  F ( x  y )  2 Re{F ( z*)} , where z*  x  y

13.1.

  x 2  4 y 2 ,   R 2 e 3  ( x 2  y 2  z 2 )e 3   z   2 x  2 y  2( x  y ) x y   z v   8 y  2 x  2( x  4 y ) y x  w 0 z Since  2 u  0 and (  u)  0  Navier' s equations will be satisfied. u

13.2.

u       ekk    u    (          , kk





1 1 1 1 1     u    (    )         (  )   2    2 2 2 2 2 2 1 or i   i , kk 2 13.3.

Navier' s Equation :  2 u  (  ) (  u)  0 Using the vector identity (1.8.5)9   (  u)   (  u)   2 u   (  u)    (  u)   2 u and using this result in Navier' s equation 





 2 u  (  )   (  u)   2 u  0  (  2) 2 u  (  )  (  u)   0 Taking the divergence of this result gives





  (  2) 2 u  (  )  (  u)   0  (  2) 2 (  u)  0 Taking the Laplacian of the original form of Navier' s equations    2  2 u  (  ) (  u)  0   4 u    2 (  u)  0  Thus the displaceme nt vector is biharmonic , and the stress and strain wil l also be biharmonic .





13.4.

Lame' s Potential :  2  0 ,   0  2ui  ,i 1 1 1 eij  (ui , j  u j ,i )  (,ij  , ji )  ,ij 2 4 2  ij  ekk ij  2eij  , kk ij  ,ij , but , kk  0 2  ij  ,ij 13.5.

Galerkin Vector Representa tion : 2u  2(1  ) 2V  ( V )  u  2u 

1  2 1 V ( V )  2

1  4 1 2 V  [( V )]  2

1   2  1  2 1 2 1  2 2  (  u)     ( V )   ( V )   2 ( V )]   ( V )] 2 2 2    Navier' s Equation :  2 u  (  ) (  u)  F  0  1   4  1  2 2  1 2  V  [( V )]  (  )   ( V )]  F  0  2     2   1 1  2  2 (1  ) 4V     (  )  [( V )]  F  0  2   2 (1  ) 4V  F  0 or F  4V   1 

13.6.

Helmholtz Representa tion : u       From (13.1.3)    u   2  and   u   2  Galerkin Vector Representa tion : 1  2 1 2u  2(1  ) 2V   ( V )  u  V  ( V )  2 u 

1  2  1  1 2    2V   ( V )   2  ( V )  2  2 

u 

1    1  1    2V     ( V )   2  (  V )  2   

Comparing results for   u and   u    Note that    





1  2 1  ( V ) ,    (  V ) 2 

1  2  V   ( V ) which provides proper match for Galerkin form. 

13.7.

Galerkin Vector Representa tion : 1 1 2(1  )Vi, kk  Vk , ki   u 2(1  ) 2V  ( V )  ui  2 2 1 1  2(1  )V j , kki  Vk ,kji  , ui , j  2(1  )Vi , kkj  Vk , kij  , u j ,i  2 2 1 2(1  )Vk , mmk  Vm,mkk   1 2(1  )Vm, mkk  Vm, mkk   1  2 Vm, mkk uk , k  2 2 2 ij  uk , k ij  (ui , j  u j ,i )





1  2 1 Vm , mkk ij  2(1  )Vi , kkj  2(1  )V j , kki  2Vk , kij  2 2  Vm , mkk ij  (1  )Vi , kkj  (1  )V j , kki  Vk , kij 

Specific components : 11   x  Vm , mkk  (1  )V1, kk 1  (1  )V1, kk 1  Vk , k11   2 ( V )  2(1  )

  2 2  2   Vx  2 ( V )  2(1  )  2Vx    2  2 ( V ) x x x x  

The other normal stress expression s follow in similar fashion 12   xy  (1  )V1, kk 2  (1  )V2, kk 1  Vk , k12  2  2 2  (1  )  Vx  (1  )  Vy  ( V ) y x xy    2   (1  )  2Vx   2Vy   ( V ) x  y  xy The other shear stress expression s follow in similar fashion

13.8.

Cylindrica l Coordinate Case : V  Vr er  Ve  Vz ez  1    eθ  ez r r  z 1    Vr er   ez  Vr er   2 Vr er      Vr er     er Vr er   eθ r  z  r  From Section 1 - 9 or Exercise 1 - 16 :   er

 V 1 V  V     er r er  eθ Vr eθ  r er   e z r er  r   z   r 1     V 1 V  V      er  eθ  e z   er r er  eθ Vr eθ  r er   e z r er  r  z   r r   z   r   2V 1 Vr 1  2Vr  2Vr 1  2 V 1  2 V    2r   2  2  2 Vr er  2 r eθ    2Vr  2 Vr er  2 r eθ 2 r r r  z r r  r r     r  In similar fashion V  2 V   2 (Veθ )    2V  2 eθ  2  er ,  2 (Vz e z )   2Vz e z r  r   To determine biharmonic forms, reapply th e previous Laplacian operators   V  2 V  V    2 V   4 Vr er    2 2 Vr er    2   2Vr  2r er  2 r eθ    2   2Vr  2r er    2  2 r eθ  r  r   r    r       V  1 V  2   2 V    2   2Vr  2r   2   2Vr  2r  er  2   Vr  2r eθ r  r  r  r   r      2 V  1  2 V  2   2 Vr    2  2 r   2  2 r  eθ  2  e r r   r 2     r   r  r    2 1  2 4 4  2Vr  1  V      2  Vr  4 e   2   2  2  r  eθ 2  r r  r   r    r   The other two biharmonic forms follow in similar fashion

13.9.

Proposed solution t o Boussinesq ' s Problem is combinatio n of Galerkin v ector and Lame potential (1  2) P P ,B Vx  Vy  0 , Vz  AR ,   B log( R  z ) with A  2 2 4 2 These forms satisfy th e governing equations :    0 and  Vz  0 Displaceme nts : Px  z 1  2  1  1  2Vz    u   2 x 2 xz 4R  R 2 R  z  Py  z 1  2  1  1  2Vz    v   2 y 2 yz 4R  R 2 R  z  w

z2  P   2V  1  1   2(1  )  2    2(1  ) 2Vz  2z   R  z  4R  2 z 2 

Stresses : x  

 2   2  2      2 Vz x  x 2 z  1 P(1  2) x 2 1 P(1  2) 1 P (1  2) x 2 Pz 3 Px 2 z 1 Pz      2 R 3 ( R  z ) 2 R( R  z ) 2 R 2 ( R  z ) 2 R 3 2 R 5 2 R 3

z x 2 (2 R  z )  R P  3x 2 z      2  3  (1  2)  2  ) z  R ( R z  R R R  R   y 

z y 2 (2 R  z )  R P 3y2z  2   2  2       )  2  1 (    V      z  R R  z R( R  z ) 2  2R 2  R 3 y 2  y 2 z   

3Pz 3 2   2   2  z  2   (2  )  2 Vz   2R 5 z  z  z  xy 

P  3 xyz (1  2)( 2 R  z ) xy   3Vz  2     R( R  z ) 2 2R 2  R 3 xy xyz 

 yz 

3Pyz 2 2     2   (1  ) 2  2 Vz   2R 5 z  yz y 

3Pxz2 2   2   2   (1  )  2 Vz    zx  2R 5 z  zx x  Boundary Conditions Check :  yz ( x, y,0)   zx ( x, y,0)   z ( x, y,0)  0 (except at R  0, stresses are singular at origin) 

Finally it can be shown that on any plane z  constant,   z (r , z )2rdr   P 0

13.10.

Proposed solution t o Cerruti' s Problem is combinatio n of Galerkin v ector and Lame potential

Vx  AR , Vy  0 , Vz  Bx log( R  z ) ,  

Cx P (1  2) P (1  2) P , where A  ,B ,C Rz 4(1  ) 4(1  ) 2

These forms satisfy th e governing equations :  2  0 and  4Vz  0 Displaceme nts : u

1  1    V V     2(1  ) 2Vx   x  z   2 x 2  x  x z  

 R 1  1 P(1  2) 1 P (1  2) x 2 P Px 2  P  x2 x 2         (1  2)  1  2  2  2 R  z 2 ( R  z ) 2 R 2R 2R 3  4R  R 2  R  z ( R  z )  v

1  1    Vx Vz   Pxy  1 1  2     2        2 y 2  y  x z   4R  R ( R  z ) 2 

w

1  1    V V   Px  z 1  2     2(1  ) 2Vz   x  z      2 z 2  z  x z   4R  R 2 R  z 

Stresses : x 

 2  2  Vx Vz   2  2    2   2 ( 1   )  V    x x 2 x x  x z  



3P(1  2) x P(1  2) x 3 P(1  2) x 3 Px Px 3Px 3Px3       2( R  z ) 2 R ( R  z )3 R 2 2( R  z ) 2 R 2 R 3 R 3 2R 3 2R 5



Px  3x 2 (1  2)  2 2 Ry 2  2     R  y   2R 3  R 2 ( R  z ) 2  R  z 

y 

 2  2  2  Vx Vz  Px  3 y 2 (1  2)  2 2 Rx 2  2             3 R  x     y 2  y 2  x z  2R 3  R 2 ( R  z ) 2  R  z 

z 

 2  2  Vx Vz   2  2 3Pxz2    2 ( 1   )  V          z  z 2 z z 2  x z  2R 5 

 xy 

 2   2  Vx Vz  Py  3x 2 (1  2)  2 2 Rx 2  2    (1  )  2Vx     R  x     xy y xy  x z  2R 3  R 2 ( R  z ) 2  R  z 

 yz 

 2   2  Vx Vz  3Pxyz  (1  )  2Vz     yz y yz  x z  2R 5

2  2  3Px 2 z     Vx Vz   (1  )  2Vx   2Vz      zx x z  2R 5  z  zx  x Boundary Conditions Check :  yz ( x, y,0)   zx ( x, y,0)   z ( x, y,0)  0 (except at R  0 , stresses are singular at origin)

 zx 

Finally it can be shown that on any plane x  constant, 

 -



  ( x, y, z )dydz   P 

x

13.11.

Stresses from Galerkin vector representation V are given by (13.3.4):  x  2(1  )

  2 2   Vx    2  2   V x x  

 2 2   2  y  2(1  )  Vy     2   V y y    z  2(1  )

  2 2   Vz    2  2   V z z  

   2   xy  (1  )   2Vx   2Vy    V x  y  xy   2   yz  (1  )   2Vy   2Vz    V y  z  yz  2  2   2  zx  (1  )   Vz   Vx    V z  x  zx Clearly with  2Vx   2Vy   2Vz  0 all terms with Poisson's ratio will drop out Starting with relations (13.3.7), similar arguements can be made for the Love strain potential Vz 13.12

For the case  =1/2, Kelvin solution (13.310) reduces to 2ur 

P rz P  1 z2  , u  0 , 2  u      z 4 R 3 4  R R 3 

P 3r 2 z P 3z 3 P 3rz 2 ,       0 ,    ,     z r z zr 4 R 5 4 R 5 4 R 5 For the case  =1/2, Boussinesq solution given in Exercise 13.9 reduces to

r  

u

Px z Py z P z2 , v  , w  4R R 2 4R R 2 4R R 2

P 3x 2 z P 3y2 z 3Pz 3 ,    ,    y z 2R 2 R 3 2R 2 R 3 2R 5 P 3 xyz 3Pyz 2 3Pxz 2  xy   ,    ,    yz xz 2R 2 R 3 2R 5 2R 5

x  

13.13.

 A R  Papkovich Representa tion : 2u  A    B  4(1  )   1   From formulatio n result (13.4.3), 2u  2 h      1  2  2 A R 2h  A ,   B 1  2 4(1  ) F Now since  2 h     2 A  2 F  2 A R B   1  2 4(1  )  2 A R  2 1    2 B   2    2  2(  A)  ( 2 A)  R 4(1  )  1  2 4(1  )  1  2 Using formulatio n result (13.4.6) 



 2  



 1  2  R  F 1  1  2  R  F 1 2   ( R  h)   2(  h)  R  ( 2 h)    2(1  )  2 2 2  2(1  )  2 



1  2  R  F (  A) R  F  1  2   A      2(1  )  2 2 2  4(1  )

Combining these results  2 1  2   A  1 2(  A)  2( R  F )  2B  1  2 4(1  ) 4(1  ) 1   A    A  ( R  F )  R  F  2(1  ) 2(1  )



13.14.

 A R  General Papkovich Represent ation : 2u  A    B  4(1  )   Axisymmetr ic Form - Boussines q Potential s : Ar  A  0 , Az  Az ( r, z ) , B  B( r, z ) , with  2 B  0 and  2 Az  0  ur  

Az z  Az z  1   1     B   , u  0 , u z    Az   B  2 r  4(1  )  2  z  4(1  ) 

Az z  Az z  Az z  1 2  1   1     er   B   , e    B   , ez    Az   B  2  2 r  4(1  )  2r r  4(1  )  2 z  z  4(1  )  Az z  1   Az z  1 2    erz    B      Az   B  4 rz  4(1  )  4 r  z  4(1  )  

Az z  1 Az 1 2    B   4 r 2 rz  4(1  ) 

 r  ( er  e  ez )  2er 

Az z   Az  2  Az z    2 1   2   2   2  B   2  B     2  r r r z  4(1  )  2 z r  4(1  ) 



 Az z  Az z    Az  2   2   2  B     1  2  4(1  )  1  2 z r  4(1  ) 

Likewise ,   

 Az z  Az z    Az 1    2      B   1  2  4(1  )  1  2 z r r  4(1  ) 

z  

 Az z  Az z    Az Az  2   2    2  B     1  2  4(1  )  1  2 z z z  4(1  ) 

Az z  1 Az 2   rz  2erz    B   2 r rz  4(1  ) 

13.15.

Boussinesq Problem - Papkovich Solution 2(1  ) P (1  2) P Az  , B log( R  z ) R 2 Using results from Exercise 13 - 12, Az z  1    B   2 r  4(1  ) 

ur    uz 

1   (1  2) P Pz  P  rz (1  2) r  log( R  z )     2 r  2 2R  4R  R 2 R  z 

Az z  1    P  z2  A  B   2 ( 1   )  , u  0    z  2  z  4(1  )  4R  R 2 

r  

 Az z  Az z    Az  2   2    2  B    1  2  4(1  )  1  2 z r  4(1  ) 

    2(1  ) P   2  (1  2) P Pz  2  Pz     log( R  z )     2   1  2  2R  1  2 z  R  r  2 2R  

P  3r 2 z (1  2) R    2R 2  R 3 R  z 

   

    2(1  ) P  1   (1  2) P Pz   Pz  2  log( R  z )       1  2  2R  1  2 z  R  r r  2 2 R 

(1  2) P  z R   2  2 R  R R  z 

    2(1  ) P    2(1  ) P   2  (1  2) P Pz  2  Pz  z     log( R  z )       2   1  2  2R  1  2 z  R  z  R  z  2 2R  

3Pz 3 2 R 5

1   2(1  ) P   2  (1  2) P Pz  3P rz 2 log( R  z )      2 r  R  rz  2 2 R  2R 5 rz (1  2) r z (1  2) R z Note ur  0  2    , and defining cos    R Rz R Rz R (1  2) cos    cos 2   cos   (1  2) 1  cos  u r  0 for   o where o is determined from  rz  2erz 

cos 2 o  cos o  (1  2)  0

o z

13.16*.

Displaceme nt Field - Boussines q Problem : ur 

P  rz (1  2)r  P  z2   , u  2 ( 1   )  , u  0 z 4R  R 2 R  z  4R  R 2 

Note surface displaceme nts : ur ( r,0)   MATLAB Vector Dis tribution Plot :

P(1  2) P(1  )  0 , u z ( r,0)  0 4r 2r

13.17.

 A R  General Papkovich Represent ation : 2u  A    B  4(1  )   Axisymmetr ic Form - Boussines q Potential s : Ar  A  0 , Az  Az ( r, z ) , B  B( r, z ) , with  2 B  0 and  2 Az  0  2u r  

Az z  Az   B z   B     r  4(1  )  r 4(1  ) r

2u z  Az 

Az z  Az Az   B z    B    Az  z  4(1  )  z 4(1  ) 4(1  ) z

Az 3  4 B z Az   4(1  ) z 4(1  ) z From Exercise 13 - 12  z  ( er  e  ez )  2e z 



 Az z  Az z    Az Az  2   2    2  B     1  2  4(1  )  1  2 z z z  4(1  ) 

 z ( r,0)  0   

Az Az   Az Az  2 B 1    2  0 4(1  )(1  2) z 1  2 z z z 2(1  ) z

1 Az  2 B  2 2 z z

Integratin g  Az  2

B , where we have dropped the arbitrary function of integratio n z

13.18.

 A R  Two - Dimensiona l Papkovich Represent ation : 2u  A    B  4(1  )   where A  A1 ( x, y )e 1  A2 ( x, y )e 2 and B  B( x, y ) 2u  A1 

( xA1  yA2 )  A    B 1  A1 B  A1   x  A1  y 2     x  4(1  )  x 4(1  )  x x 

A A  B 1  (3  4) A1  x 1  y 2    4(1  )  x x  x Likewise 

2v 

 A A  B 1 (3  4) A2  x 1  y 2    4(1  )  y y  y

Forming the complex displaceme nt 2(u  iv ) 

  A1 A   A A   B 1 B    y 2   i  x 1  y 2    i  (3  4)( A1  iA2 )   x 4(1  )  x   y y   x y    x

Define γ(z) 

A1  iA2 B B and ψ(z)  i with   3  4  4(1  ) x y

2(u  iv )  γ(z)  z γ (z)  ψ(z)

13.19.

Trial Papkovich Functions for Kelvin Problem : B  0 , Az 

P 2R

From Exercise 13 - 12 2ur  

   Az z    Pz Prz        r  4(1  )  r  8R(1  )  8(1  ν ) R 3

    P   Az z   P   Pz Pr 2 2u z   Az            3 z  4(1  )   2R z  8R(1  )   2R 8R (1  )   

 4(1  ) r 2   2(1  2) 1 z 2  P P     3 8(1  ν )  R R 3  8(1  ν )  R R R 

2u  0 This displaceme nt field matches that given in relation (13.3.10)1 for the Kelvin Problem 13.20.

For the Doublet Force Problem, use superposit ion of two Kelvin Problems and take limit d  0    (r , z  d )   r (r , z )   rD  lim  r (r , z )   r (r , z  d )  d lim  r  d r  d 0 d 0 d z   D   (1  2) z (r 2  z 2 ) 3 / 2  3r 2 z (r 2  z 2 ) 5 / 2 8(1  ) z Likewise for the other nonzero stress components D   Dz  (1  2) z (r 2  z 2 ) 3 / 2  3 z 3 (r 2  z 2 ) 5 / 2 8(1  ) z D   D   (1  2) z (r 2  z 2 ) 3 / 2 8(1  ) z D   rzD  (1  2)r (r 2  z 2 ) 3 / 2  3rz 2 (r 2  z 2 ) 5 / 2 8(1  ) z Using the transform ation relations (B.6) to go from cylindrica l to spherical coordinate s r z  R   r sin 2    z cos 2   2 rz sin  cos  , with sin   r (r 2  z 2 ) 1 / 2  , cos   z (r 2  z 2 ) 1 / 2   R R 2(1  ) D  2( 2   )  R    sin 2   cos 2  3  1  8(1  ) R  















 R  ( r   z ) sin  cos    rz (sin 2   cos 2 )   R  

2(1  ) D sin  cos  8(1  ) R 3



13.21.

Since it is expected that the Center of Dilatatio n will produce a sphericall y symmetric stress state, we consider a special direction in the x, z - plane as shown to apply the superposit ion principle.

For a Doublet acting along the z - axis, the solution was given in Exercise 13 - 18 as 2(1  ) D  2( 2   ) 2(1  ) D   sin 2   cos 2  ,  Rz    sin  cos  3  8(1  ) R  1  8(1  ) R 3   For a Doublet acting along the x - axis, replace  by   in the previous expression s  2 2(1  ) D  2( 2   ) 2  x 2(1  ) D  Rx    cos 2   sin  ,  R  sin  cos  3  8(1  ) R  1  8(1  ) R 3   For a Doublet acting along the y - axis, replace  by in the previous z - axis expression s  2 2 ( 1   ) D  Ry  ,  Ry   0 3 8(1  ) R Applying superposit ion 

 Rz  

σ R   Rx   Ry   Rz  

2(1  ) D  2( 2   )  (1  2) D 1  1   3  8(1  ) R  1  2(1  ) R 3 

 R   Rx    Ry    Rz   0 These results should then be valid for all directions

R

z 

  2

 2

y x

13.22.

For the sphericall y symmetric case : u R  u ( R), u   u   0 , and relation (A.12) 1 reduces to 2u d 2 u 2 du 2  u 2 0   2u0 2 R dR R R dR Look for solutions of the form : u  AR m and substitue into governing equ.  2

m(m  1)  2m  2  0  m 2  m  2  0  (m  1)( m  2)  0  roots m  1,  2 C  u  C1 R  22 R Using relations (A.2) into (A.9) gives the stress relations u du du u  du  R    2   2  (  2)  2 R dR dR R  dR u u u du  du         2   2  2(  )   R R R dR  dR which allowes the stresses to be written as :  R  K 1  Note that C1 

2K 2 K ,       K 1  32 3 R R

K 1  2 K1 , C 2  2 2(1  ) 2

13.23.

2K2 K ,     K1  32 3 R R Boundary Conditions on problem (raduis of cavity taken as a ): 2K 2K  R (a)  0  K1  3 2  0  K1  3 2 a a  R ()   ()   ()  S  K1  S  K 2  Sa 3 / 2 From Exercise 13-22   R  K1 

 a3   a3   the stresses are given by:  R  S 1  3  ,     S 1  3   R   2R  3 3  max   (a)   (a)  S  Stress Concentration Factor  K  2 2 From section 8.4.2, the corresponding two-dimensional result was K 2 D  2 So K 2 D  K 3 D which is to be expected since a three-dimensional domain has an additional dimension to relieve or reduce the stress concentration.

13.24.

2K2 K ,     K1  32 3 R R Boundary Conditions on spherical shell problem:  R ( R1 )   p1 ,  R ( R2 )   p2  From Exercise 13-22   R  K1 

p1 R13  p2 R23 ( p1  p2 ) R13 R23 K1  , K2  and the stresses become R23  R13 2( R23  R13 ) R 

p1 R13  p2 R23 ( p1  p2 ) R13 R23 1  R23  R13 ( R23  R13 ) R 3

   

p1 R13  p2 R23 ( p1  p2 ) R13 R23 1  R23  R13 2( R23  R13 ) R 3

For the special case: p1  p , p2  0 R 

pR13  R23  pR13  R23  1  ,     1    R23  R13  R 3  R23  R13  2 R 3 

13.25*.

From Exercise 13-22: C 2K K u  C1 R  22 ,  R  K1  32 ,     K1  32 R R R K 1  2 with C1  K1 , C2  2 2(1  ) 2 Inclusion Problem Boundary Conditions: 1  2  R ()  S  K1  S  C1  S 2(1  ) C 1  2 3 u (a)  0  C1a  22  0  C2  C1a 3  K 2   Sa a 1    1  2 a 3  1  2 a 3    R  S 1  2 ,     S   1  1   R 3  1   R 3     For the case with   1/ 2 :  R      S On the boundary of the inclusion R  a : 1  2  3(1  )   1  2  3  R  S 1  2  S ,     S 1   S  1   1    1    1   MATLAB Plots of boundary stresses:

3

 /S R

Dimensionless Stress

2.5

2

1.5

1

 /S ,  /S

0.5

0



0

0.05



0.1

0.15

0.2

0.25

0.3

Poisson's Ratio

0.35

0.4

0.45

0.5

13.26.

Given stress relations for the spherical cavity problem with tens ion S along z direction : 3  15   (a,   0)   S z 2(7  5) 27  15  z (a, z  0)  S 2(7  5) y x

  (a,    / 2) 

15  3 S 2(7  5)

(a) Tension S along x and z directions :  max   z (a, z  0)    (a,   0) 27  15 3  15 24  30 S S S 2(7  5) 2(7  5) 2(7  5) (b) Tension S along z and compressio n x directions :  max   z (a, z  0)    (a,   0) 

27  15 3  15 15 S S S 2(7  5) 2(7  5) 7  5 which also correspond es to a pure shear far - field loading (c) Tension S along x, y and z directions :  max   z (a, z  0)    (a,   0)    (a,    / 2) 

27  15 3  15 15  3 21  15 3 S S S S S 2(7  5) 2(7  5) 2(7  5) 2(7  5) 2 which matches with results of Exercise 13 - 21. 

13.27*.

Stress Concentration Factor,  max/S

3.5

Case (b) has highest stress 3

(a) (b) (c)

2.5

2

1.5

1

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

Poisson's Ratio

13.28*.

2.25

Stress Concentration Factor

2.2 2.15 2.1 2.05 2

0.42

1.95 1.9 1.85 1.8 1.75 -1

-0.5

0

Poisson's Ratio

0.5

13.29.

1 1  Use the Morera stress function w ith 13   z,1 ,  23   z, 2 , 12,12    2 ,   ( x, y ) 2 2 2 11  - 2  23, 23  z, 2 , 23  , 22  22  - 2  31,31  z,1 ,31  ,11

33  - 2 12,12   2  (11   22 ) 1 1  1   1  12  - 12,33 +  23,13 + 13, 23  0    z, 2     z,1    , 21   ,12  ,12 2 2  2 ,13  2 , 23  1   1   23  -  23,11 + 13, 21 + 12,31    z, 2     z,1   0  0  2 ,11  2 , 21  1   1  31  -  31, 22 + 12,32 +  23,12    z,1   0    z, 2   0  2 , 22  2 ,12  stresses are given by the usual form of the Airy stress function ( x, y )

13.30. ij  ,kk ij  ,ij ij , j  ,kkj ij  ,ijj  ,kki  , jji  0 For this case: kk  3,kk  ,kk  2,kk ij ,kk  ,kkll ij  ,ijkk General compatibility equations (5.3.3) 1 1 ij ,kk  kk ,ij  0  ,kkll ij  ,ijkk  2,kkij  0  1  1  1  ,kkll ij  ,kkij  0 1  For 2-D plane stress case in x1 , x2 plane,   ( x1 , x2 ) and ij  , kk ij  ,ij  11  ,kk  ,11  ,22 22  ,kk  ,22  ,11 12   ,12 which is usual Airy stress function case

14.1.

 2ey

 2 e xy  2 ex The governing compatibil ity equation is  2 xy x 2 y 2 and for plane stress using relation (7.2.2) this would give 2  1   2  1  2 1            2  xy   0      y x x y 2 2 E  y  E E  xy  E x  E  Introducin g the usual Airy stress function gives  2  1  2   2   2  1  2   2   2  1    2     2  0   xy  E xy  x 2  E x 2 E y 2  y 2  E y 2 E x 2  We could also simply start with plane strain relation (14.1.5) and use conversion Table 7 - 1.

For the special case of E  E ( x) and   constant :  1  4  2  1  2   2  1  4   4   1   3           2 1     E x 2 y 2 x  E  xy 2   0  x 2  E x 2 E y 2  E y 4 E x 2 y 2     1   4  4    1   3  2  1   2    4   1   3  4  4   2   3  2   2  2  2    E  x x  E  x E x 2 y 2 x  E  xy 2 y  x  E  x  1  4  2  1   2   1   3   0   2   2  21       2 2 2  E  x E x  E  y  x  y  x  y     1   4  4  4    1   3  2  1   2   2  1   2   1   3  4  2 2 2  4   2   3  2   2   2   2  2   0 E  x x  E  x x  E  xy 2 x y y  x  E  x x  E  y 1   4  4  4    1     2   2    2  1   2   2   4  2 2 2  4   2    2  2   2   2   2   0 E  x x  E  x  x x y y  y  x  E  x y 

14.2.

u v u v    (k  2)  k x y x y v u v u  y  (  2)    (k  2)  k y x y x  u v   xy     , ( x)   o (1  ax) , ( x)  k( x)  y x   x  (  2)

Using the new form of Hooke' s in equilibriu m equations :  x  xy  0  x y   u v     u v    (k  2)  k         0 x  x y  y   y x      2u  2u  2 v  u      o a k  2   0  o (1  ax) k 2 2  2  x y xy  x    x  xy  y  0  x y    u v   2 v  2 v  2u     o a    0  o (1  ax) k 2 2  2  y x xy   y x   y

14.3.

  constant, E ( x)  E o e ax 

E 1 E 1 E a  aE o e ax  aE  a 2  x E x E x E

1 1 ( x   y )  ( , yy  , xx ) E E 1 1 e y  ( y   x )  ( , xx  , yy ) E E 1  1  e xy   xy    , xy E E 2  2 e xy  2 ex  e y Compatibil ity Relation :  2  xy y 2 x 2

ex 

2 2  1 2  1    1   (    )  (    )  2  , xy   , yy , xx  , xx , yy  2  2  xy  E y  E  x  E  

1  1 1 E 1 E  1  ( , yyyy  , xxyy )   ( , xxx  , yyx )  2 ( , xx  , yy )   2(1  )  , xxyy  2  , xyy  E x  E E x E x  E  1 1 a ( , yyyy  , xxyy )   , xxxx  , yyxx  a( , xxx  , yyx )    , xxx  , yyx  a( , xx  , yy  E E E 2(1  )  ( , xxyy  a , xyy )  E   2  4   2a ( 2 )  a 2  2   a 2 (1  ) 2  0 x y

14.4. u T T T T  x2     f ( y) (14.1.10)1    u   dx  f ( y )  (1  Kx)dx  f ( y )  xK x E E Eo  E o  2  (14.1.10) 2 

v T T T T    v    dy  g ( x)   y  g ( x)   (1  Kx) y  g ( x) y E E E Eo

 xy u v T T   2e xy   0  f ( y )  Ky  g ( x)  0  f ( y )  Ky   g ( x)  constant  o  y x  Eo Eo g ( x)  o x  vo ; f ( y ) 

T y 2 K  o y  u o Eo 2

Fixity Conditions : u (0,0)  0  u o  0 ; v(0,0)  0  vo  0 ;  z (0,0)  0  o  0 u 

 x2 T  y 2   T   ; v   x  K   (1  Kx) y E o  2   Eo  2

14.5.* v/(T/Eo)-displacement (K = -0.5)

1

1

0.5

0.5

y

y

u/(T/Eo)-displacement (K = -0.5)

0 0

0

0.5 1 x v/(T/E o)-displacement (K = 5)

1

1

0.5

0.5

0 0

y

y

0.5 1 x u/(T/Eo)-displacement (K = 5)

0.5 x

1

0

0

0

0.5 x

1

14.6.

For plane stress, (14.1.6)

 2  1  2   2   2  1  2   2   2  1    2       0     2 xy  E xy  x 2  E x 2 E y 2  y 2  E y 2 E x 2 

For the case E  E ( y ) ,   constant , and   Ty 2 / 2 , (14.1.6) 

d2  1   0 dy 2  E 

1  Ay  B or E  E o /(1  Ky) , where E o and K are constants E The stresses are the same as Example 14 - 1, given by  x  T ,  y   xy  0 Integratin g twice 

u T T T (1  Ky)   u   dx  f ( y )  x  f ( y) x E E Eo v T T T T  y2     g ( x)    v    dy  g ( x)   ( 1  Ky ) dy  g ( x )   y  K y E E Eo  E o  2   xy u v TK TK   2e xy  0 x  f ( y )  g ( x)  0  x  g ( x)   f ( y )  constant  o  y x  Eo Eo TK x 2 f ( y )   o y  u o ; g ( x )    o x  vo Eo 2 Fixity Conditions : u (0,0)  0  u o  0 ; v(0,0)  0  vo  0 ;  z (0,0)  0  o  0 u 

 y 2 x2  T T   y  K  (1  Ky) x , v    2  2   Eo E o   

14.7.

For plane stress, (14.1.6)

 2  1  2   2   2  1  2   2   2  1    2       0     2 xy  E xy  x 2  E x 2 E y 2  y 2  E y 2 E x 2 

For the case with E  E ( y )  1 /( Ay  B) ,   constant , and   ( y )  d2   2  d 4 d 3   ( Ay  B) 2   0  ( Ay  B) 4  2 A 3  0 dy 2  y  dy dy Letting Y  

C1 d 3 dY 2A  previous ODE becomes  Y  0 with solution Y  3 dy ( Ay  B) dy ( y  B / A) 2

C1 C1 d 3 d 2     C 2 , where C1 and C 2 are arbitrary constants of integratio n 3 2 2 ( y  B / A) dy ( y  B / A) dy

and so the stresses become  x 

C1 d 2   C 2 ,  y   xy  0 2 y  B/ A dy

14.8.

Equation (14.1.6) for plane stress:  2  1  2   2   2  1  2   2   2  1    2      2      0 x 2  E x 2 E y 2  y 2  E y 2 E x 2  xy  E xy  Using   Txy  2  1  2  1   (  T )  0       0 . . . for constant Poisson's ratio xy  E xy  E   1 Integrating this result   f ( x)  g ( y) , where f and g are arbitrary functions E

14.9.

Hooke' s Law plane stress  er 

1 1  1 d d 2  ( r  (r )  )     2  E (r ) E  r dr dr 

e 

1 1  d 2 1 d   (   (r ) r )   2   E (r ) E  dr r dr 

d (re )  er  0  dr d  1  d 2 d  1  1 d d 2     0 r   ( r )    ( r )   dr  E (r )  dr 2 dr  E (r )  r dr dr 2 

(8.4.76) 

14.10.

For the hollow cylinder problem the displaceme nts were given by : u  Ar  ( n k ) / 2  Br (  nk ) / 2 Using the relations for the stresses gives r 

 E  du u E   (n  k )   (n  k )    A     r ( n  k  2 ) / 2  B    r (  n k 2 ) / 2  2  2   1    dr r  1    2 2    

 

E u du  E   ( n  k )  ( n  k  2 ) / 2  (  n  k )  (  n k 2 ) / 2     A 1  r  B 1   r 2  2    1   r dr  1     2 2    

Boundary Conditions : 2  n  k  2  ( nk 2 ) / 2   n  k  2  (  nk 2 ) / 2 1    r (a)  pi  A   B  pi a a 2 2 E     2  n  k  2  ( n k 2 ) / 2   n  k  2  (  nk 2 ) / 2 1    r (b)  po  A   B  po b b 2 2 E     Solving the two equations for A and B 

  n  k  2  (  n k 2 ) / 2  po a (  n  k  2 ) / 2   pi b 1  2   A E  n  k  2   n  k  2  ( nk 2 ) / 2 (  n k 2 ) / 2 b  a (  n  k  2 ) / 2 b ( n  k  2 ) / 2   a 2 2    (  n k 2 ) / 2 2 2 pi b  po a (  n  k  2 ) / 2 1   E (n  k  2) a ( nk 2 ) / 2b (  nk 2 ) / 2  a (  n k 2 ) / 2b ( nk 2 ) / 2



2







 2 p a 1  B E (n  k  2)a 2







( n  k  2 ) / 2 ( n  k  2 ) / 2 o i ( n  k  2 ) / 2 (  n  k  2 ) / 2 (  n  k  2 ) / 2 ( n  k  2 ) / 2

 pb

b a Back substituti ng into the stress realtions gives



b





a n / 2b n / 2 r ( 2k n ) / 2 r    a k n / 2b ( 2 k ) / 2 po  a n / 2b ( 2k ) / 2 po r k  a ( 2k ) / 2b n / 2 pi (b k  r k k k b a  n / 2  n / 2 ( 2 k  n ) / 2  (a ( 2k ) / 2b n / 2 pi  a n / 2b ( 2k ) / 2 po )r k (2  k  n) a b r    bk  a k k  n  2  

a k / 2b k / 2 (ab ( k n ) / 2 pi  a ( k n ) / 2bpo )( 2  k  n)   k  n  2 



14.11.

From the general solution :



a n / 2b n / 2 r ( 2k n ) / 2  a k n / 2b ( 2k ) / 2 po  a n / 2b ( 2k ) / 2 po r k  a ( 2k ) / 2b n / 2 pi (b k  r k k k b a n / 2 n / 2 ( 2 k  n ) / 2  (a ( 2k ) / 2b n / 2 pi  a n / 2b ( 2k ) / 2 po )r k (2  k  n) a b r    bk  a k k  n  2 

r  



a k / 2b k / 2 (ab ( k n ) / 2 pi  a ( k n ) / 2bpo )( 2  k  n)    k  n  2  For the special case of internal pressure only : po  0 









pi a ( 2k n ) / 2 ( 2k n ) / 2 a n / 2b n / 2 r ( 2k n ) / 2 ( 2k ) / 2 n / 2 k k a b p ( b  r   (r  b k r ( 2k n ) / 2 i k k k k b a b a ( 2 k ) / 2 n / 2 k n / 2 n / 2 ( 2 k  n ) / 2  (a b pi )r (2  k  n) a k / 2b k / 2 (ab ( k n ) / 2 pi )( 2  k  n)  a b r      bk  a k k  n  2 k  n  2  

r  

pi a ( 2k n ) / 2  2  k  n ( 2k n ) / 2 2  k  n k ( 2k n ) / 2   r  b r  b k  a k  k  n  2 k  n  2

14.12.

With k  n 2  4  4n  k 2  n 2  4(1  n) For the case with 0    1/2 , n  0  n 2  4  k  n 2  4  2n n 2  4  2  n  2  k  n  n 2  2n  4  2  n   2  k  n  0 (equality w ith n  0)  n 2  4  2  n  2  k  n   n 2  2n  4  2  n   2  k  n  0 (equality w ith n  0) 2 2  k  n 2  (k  n) / 2 2  k  n     1 (equality w ith n  0) k  n k  n  2 k  n 1 k  n  2 2 2  k  n 2  (k  n) / 2 2  k  n     1 (equality w ith n  0) k  n k  n  2 k  n 1 k  n  2 Using this results in relations (14.2.7) it follows that  r  0 ,    0

14.13.*

Use the special solution for the case of internal pressure only and let b   r    





pi a ( 2k n ) / 2 ( 2k n ) / 2 a (r  b k r ( 2k n ) / 2   pi a ( 2k n ) / 2 r ( 2k n ) / 2   pi   k k b a r

pi a ( 2k n ) / 2  2  k  n ( 2k n ) / 2 2  k  n k ( 2k n ) / 2  r  b r  b k  a k  k  n  2 k  n  2

 pi a

( 2 k  n ) / 2

2  k  n ( 2k n ) / 2 2  k  n  a  r  pi   k  n  2 k  n  2  r 

0 -0.1 -0.2 -0.3

n=2 n=1

 r/pi

-0.4

n = 1/2 n=0

-0.5 -0.6 -0.7 -0.8

 = 0.25

-0.9 -1

1

2

3

4

5

6

7

8

9

10

r/a

1 0.9

 = 0.25

0.8 0.7 n=0



 /pi

0.6 0.5

n = 1/2

0.4

n=1

0.3 n=2 0.2 0.1 0

1

2

3

4

5

6

r/a

7

8

9

10

( 2 k  n ) / 2

( 2 k  n ) / 2

14.14.

For the case   0 and n  3  k  13  n Solution is given by relations (14.3.11)  r  0 ,    2 r 2 Displaceme nt solution follows from (14.3.6) : u (r )  Ar u ( 0) 

( n  k ) / 2

 Br

( 133) / 2

2 a 3   Eo

2 a 3 0 Eo

14.15*.

MATLAB plot of hoop stress in rotating disk problem

1



Dimensionless Stress,  /2a2

0.9 0.8 0.7 0.6

n=-0.5

0.5 0.4

n=0

0.3

n=0.5 n=1

0.2

n=2 0.1 0

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Dimensionless Distance, r/a

Similar to the radial stress behavior, the hoop stress becomes unbounded at the origin for the case with n < 0, while for n > 0, the hoop stress goes to zero at the origin.

14.16.

du u  dr r Intergrati ng gives the displaceme nt soluton u  C1r

For the case with  r    , relation (14.3.2) 

Using Hooke' s law   r 

E (r )  du u  C E (r )  (r )   1 2  1   (r )  dr r  1  (r )

Using equilibriu m equations :

d r  r    d  C1 E (r )     2 r   2 r  0  dr r dr  1  (r ) 

Intergrati ng gives the soluton :  r 

C1 E (r )  2 r 2  C2 1  (r )

Using the boundary conditon  r (a )  0  C2  and 

E (r ) 1  2 (a 2  r 2 ) 1  (r ) 2C1

2 a 2 2

14.17.

 x  (ex  ez )  2ex  2

u u u w  u w        2   ,   x x x z  x z 

2    u   u     u   u   2     2  1  2    2     2( z )   *   2  x   x 1  2   x   x 2       where *  1  2  w  Likewise for z - component :  z  (ez  ex )  2ez  2( z )   *   z   u w   xz  2exz  ( z )     ( z )  z x       u     u w  Equilibriu m equation : x  xz  0  2( z )   *    ( z )    0   x z x   x  z   z x    2u   2u  2 w  d( z )  u w        2( z ) 2   *   ( z ) 2    0   x  x  z  x  z dz  z x       1 d( z )  u w   2u  (1  2*)    0 x ( z ) dz  z x  Likewise :

 xz  z    u w     w   0  ( z )    *    0    2( z )  x z x   z x  z   z 

 2 d( z )  w     *   0  z ( z ) dz  z  d( z ) 1 d( z ) Special case : ( z )   o e z   o e z   dz ( z ) dz Thus the equilibriu m equations reduce to :  2 w  (1  2*)

 2u  (1  2*)

  u w     0 x  z x 

 2 w  (1  2*)

  w   2   *   0 z  z 

14.18.

u u u u u  u u u   r   (er  e  ez )  2er    r  r  z   2 r  2 r   ,   r  r  z r z  r r r r z  r 2( z )     u   u u 1  2( z )    ( z )  u   2  r     2  r     2  r     2  r   *( z )  2  r   r   r 2   r 1  2( z )      u u  u u u   z  (er  e  ez )  2ez    r  r  z   2 z  2 z   r z  z z  r  u    u   2  z     2  z   *( z )   z   z 2  u u  u u u    (er  e  ez )  2e    r  r  z   2 r  2 r   r z  r r  r u   u   2  r     2  r   *( z )   r   r 2   u u   zr  2ezr    r  z  , r  z  0 r   z General Equilibrium Equations (A.5), no body forces: r 1 r rz 1    ( r    )  0  r r  z r   u   u u  1  u u   2  r   *( z )     r  z   2  r  r   0 r  r z  z r  r  r r    2ur 1 ur  2ur ur  1  2ur 1  2u z      *( z )   0 r 2 r r z 2 r 2 r 2 z 2 2 r z  1  2ur 1  2u z  2 1   0    2  ur   *( z ) r  r 2 z 2 2 r z   1   ur 1    2 1     2   2ur   0    2  2  ur   1  2 *( z )     0    2  ur   *( z ) r  r 2  r r r  r    r 1  z 2    r  0  0  0 r r  z r rz 1 z  z 1    rz  0  r r  z r   u u    u 1  u u     r  z   2  z   *( z )     r  z   0 r  z r  z  z r  z r   2 2 2 2  u z 1 u z  u z  ur  u z  1 ur    2   2   2 *( z )    0   2u z   1  2 *( z )     0 2 r r r z r z z z r z z

14.19.

Using transform ation (14.5.5) into governing equation (14.5.4)    w    w      W       W     0     0      x  x  y  y  x  x     y  y        1 W W       1 W W         0 x    x 2  x   y    y 2  y     W W     W W        0 x  x 2  x  y  y 2  y     2W  2W  1  1 1 1   2  2    ( 2x   2y )  ( xx   yy )   0  y  2  2    x   1 1  2W   ( 2x   2y )   xx   yy W  0 2  2  Separation of Variables : W  X ( x)Y ( y ) ,    o p ( x)q( y )   1 1  o pq X Y  XY    ( o2 p2 q 2   o2 p 2 q2 )   o pq   o pq W  0 2  2 o pq  Divide through by  o pqXY  X  Y  1 p2 1 q2 1 p 1 q      0  X Y 4 p2 4 q2 2 p 2 q  Y  1 q2 1 q  X  1 p2 1 p     constant  n 2       2 2 X 4 p 2 p 2 q Y 4 q  2 1  p  2 1  p   2 1  q  2 1  q yy  x xx Y  0  X  0 , Yxx   n   x    X xx  n      4 p 2 p 4  q  2  q        

14.20.

Using the functions , p( x)  e

x

, q( y )  e

y

2

1 d 2 p 1  1 dp  2 1 2 2   ao         ao 2 p dx 2 4  p dx  2 4 4 2

1 d 2 q 1  1 dq  2 1 2 2     b      bo o 2q dy 2 4  q dy  2 4 4

14.21.

For the circular section case, we have axisymmetr y, no warping displaceme nt and the section rotates as a rigid body  ur  u z  0 , u  rz r 2 The stress can then be calculated using (A.8)  r      z   rz   r  0 , z  r

The strains then follow from relation (A.2) as er  e  ez  erz  er  0 , ez  For the axisymmetr ic case :  d r x d  d r y d   ,   x dr x r dr y dr y r dr   1     1   x d  1 x d  y d  1 y d    2           2  x   x  y   y  r dr   r dr  r dr   r dr  x 2  y 2 d  1 1 d  1 x d dx 1 y d dy   2    r dr   r dr   r dr dr  r dr dr d  1  d 1 1 d 1 d 2  1 d  r d    2        2 2 dr    dr  r dr  dr r dr   dr  d        x d   y d               dr  x   x   r dr   r dr  2

z      2 xz

2 yz

2

2

2

14.22.

From relation (14.6.9) 

d  r d     2r dr   dr 

r d d C  r 2  C1   r  1   dr dr r ( r )     (r )rdr  C1  dr  C2 r Integratin g 

As discussed in text, C1  0 , and boundary condition (a)  0  C2    (r )rdr a

a

solution is (r )    ()d r

d  z  r(r ) dr a 2 a 2 J T /   r dr  2  r 3(r )dr  z   0 0 

z  

14.23. m

 n  Using the gradation model : (r )   o 1  r  , solutuon (14.6.11) gives  a  m

 n  (r )    ()d  o  1    d r r  a  For the m  1 case, this easily integrates out to a

a

o 2  n r3 (a  r 2 )  o (a 2  ) 2 3 a while for the m  1 case, gives (r ) 

a

2 a  a  n  (r )   o   d   o       log 1    r n n  a  r  n 1  a 2 a  a 2  a 2  n  a   o   r    log 1  r    o      log 1  n  a  n  n  n  n   a

 n  Using relation (14.6.12)1  z  r(r )   o r 1  r   a 

m

14.24.

Using the gradation model : (r )   o e

n  r a

, solutuon (14.6.11) gives a

a

a

r

r

(r )    ()d  o  e   o e

n  r a

n   a

n     a  n  e  d  o    1 2 2   a  (n / a )    r

 ar  a  2         o a 2  1  1 e n 2  n n  n n    n  r

Using relation (14.6.12)1  z  r(r )   o re a

14.25.

Relation (14.6.12)1 : z  r(r )  dz (ro )  r(r )  (r )  0  ro   , where ro is location of max stress point dr (ro ) Since r  a  (r )  

( r ) a m

 n   n  n  Using the gradation model : (r )   o 1  r   (r )   o m 1  r   a   a  a  (ro ) a  n  a  ro    m1  ro    ro   (ro ) n(1  m)  a  n  Using the gradation model : (r )   o e ro  

(ro ) a  (ro ) n

n  r a

n  r  (r )   o  e a  a n

m1



14.26.

Using the gradation model (14.6.13) 2 : (r )   o e

n  r a

, the torsional rigidity is

 r a a 2 a J T /  z r 2 dr 2 r 3(r )dr  2o  r 3e a dr  0 0  0 n

a

n  r 3 3r 2 6r 6  ar    2o     e  2 (n / a) 3 (n / a) 4   (n / a) (n / a) 0

 6 1 3 6 6   2o a 4  4    2  3  4 e n  n n   n  n n For small n : e

n

n2 n 2 n3 n  1 n    1 e  n    2!

2! 3!

  6 1 3  n 2 n3 6  n2    J  2o a  n     4    2  3 1  n    2! 3! n  2! n n n   4

4 1 1  a  o  4n   2o a 4   n  (n 2 )  1  ( n 2 )   2  5 4 5   4 a  o For the homogeneou s case, n  0, and thus J  which checks with previous study 2

15.1.

Using complex va riable methods, choose the potentials ( z) 

ib ib log z ,  ( z )   log z 4(1  ) 4(1  ) ______

______

2(u  iv )   ( z )  z  ( z )   ( z )  (for plane strain case) ib ib 1 ib log z  z  log z 8(1  ) 8(1  ) z 8(1  ) i (bx  ib y ) i (bx  ib y ) i (bx  ib y )  (3  4) (log r  i)  (cos 2  i sin 2)  (log r  i) 8(1  ) 8(1  ) 8(1  ) Check cyclic property : ib ib b [u  iv ]C  (3  4) (2i )  (2i )   [(3  4)  1]  b 8(1  ) 8(1  ) 4(1  ) Separating real and imaginary parts  u  iv  (3  4)

u

by  bx  1 y 1 xy  x2  y2  2 2 tan   ( 2   1 ) log( x  y )      2  x 2(1  ) x 2  y 2  8(1  )  x2  y2 

b y  1 y  bx 1 xy  x2  y2  2 2 v  (1  2) log( x  y )  2   tan  2  x 2(1  ) x 2  y 2  8(1  )  x  y2  Stress Field : ______  ib 1      x   y  2  ( z )   ( z )   4 Re   (bx sin   b y cos )     4(1  ) z  r (1  ) i  y   x  2i xy  2 z  ( z )   ( z )    be 3i  b e i 2r (1  ) i  bx cos 3  b y sin 3  bx cos   b y sin  2r (1  )   b y cos   bx sin   b y cos 3  bx sin 3 2r (1  )  y  x   b y cos   bx sin   b y cos 3  bx sin 3 2r (1  )   xy   bx cos 3  b y sin 3  bx cos   b y sin  4r (1  )  x  3bx sin   b y cos   b y cos 3  bx sin 3 4r (1  )  y  bx sin   3b y cos   b y cos 3  bx sin 3 4r (1  )





























15.1. Continued

Expressing the stresses in terms of Cartesian coordinate s  x 

 y (3x 2  y 2 )  x( x 2  y 2 )   bx   b y 2(1  )  ( x 2  y 2 ) 2 ( x 2  y 2 ) 2 

y 

 y( y 2  x 2 )  x( x 2  3 y 2 )   bx 2   b y 2(1  )  ( x  y 2 ) 2 ( x 2  y 2 ) 2 

 xy  

 x( x 2  y 2 )  y( x 2  y 2 )   bx 2   b y 2(1  )  ( x  y 2 ) 2 ( x 2  y 2 ) 2 

15.2.

Edge Dislocati on : u

b  1 y 1 xy  b  1  2 1 y2  2 2 tan  , v   log( x  y )  2  x 2(1  ) x 2  y 2  2  4(1  ) 2(1  ) x 2  y 2 

Cyclic Behavior : [u ]C 

b  1 y  tan  b , [v ]C  0 2  x  C

Strain Field : u b  y 1 (x2  y2 ) y  2x2 y  b  y 1 y( y 2  x 2 )         2  x 2  y 2 2(1  ) ( x 2  y 2 ) 2  x 2  x 2  y 2 2(1  ) ( x 2  y 2 )2   

ex  

b  y( y 2  x 2 )  b 2 2  y ( x  y )   [ y ( y 2  x 2 )  2(1  ) y ( x 2  y 2 )]  4  4 2r  2(1  )  4r (1  )

ey 

v b  1  2 2y 1 ( x 2  y 2 )2 y  2 y 3     y 2  4(1  ) x 2  y 2 2(1  ) ( x 2  y 2 )2 

 e xy  

b  1  2 yx 2  b 2 2 y ( x  y )   [(1  2) y ( x 2  y 2 )  2 yx 2 ]  4  4 2r  2(1  ) (1  )  4r (1  )

1  u v  b  x 1 x ( x 2  y 2 )  b  1  2 2x 1 2 xy 2         2  y x  4  x 2  y 2 2(1  ) ( x 2  y 2 ) 2  4  4(1  ) x 2  y 2 2(1  ) ( x 2  y 2 ) 2  b  x( x 2  y 2 )  b  1  2 xy 2  2 2 2 2 x ( x  y )   x ( x  y )  4r 4  2(1  )  4r 4  2(1  ) (1  ) 

b x( x 2  y 2 ) 4r (1  ) Plane Strain Stress Field : 

4

x 

2 b b y (3x 2  y 2 ) 2 2 ( e x  e y )  2e x   y ( y  3 x )   1  2 2r 4 (1  ) 2(1  ) ( x 2  y 2 ) 2

y 

2 b b y( x 2  y 2 ) 2 2 ( e x  e y )  2e y  y ( x  y )  1  2 2r 4 (1  ) 2(1  ) ( x 2  y 2 ) 2

b b x( x 2  y 2 ) 2 2  xy  2e xy  x( x  y )  2r 4 (1  ) 2(1  ) ( x 2  y 2 ) 2 In Cylindrica l Coordinate s : b  r   x cos 2    y sin 2   2 xy sin  cos    sin  2(1  ) r b     x sin 2    y cos 2   2 xy sin  cos    sin  2(1  ) r  r   x sin  cos    y sin  cos    xy (cos 2   sin 2 ) 

b cos  2(1  ) r

15.3.

Screw Disl ocation : b y uv0, w tan 1 2 x b  1 y  Cyclic Behavior : [ w]C  tan b 2  x  C Strain Field : e x  e y  e z  e xy  0 , e yz 

1 w b x 1 w b y  , e xz   2 2 2 2 y 4 x  y 2 x 4 x  y 2

Stress Field :  x   y   z   xy  0 b x b y ,  xz  2e xz   2 2 2 2 x  y 2 x  y 2 In Cylindrica l Coordinate s :

 yz  2e yz 

 r   x cos 2    y sin 2   2 xy sin  cos   0     x sin 2    y cos 2   2 xy sin  cos   0 z  0  r   x sin  cos    y sin  cos    xy (cos 2   sin 2 )  0  z   yz cos    zx sin  

b  r cos  r sin   b cos   sin   2 2  2  r r  2 r

 zr   yz sin    zx cos  

b  r cos  r sin   sin   cos   0 2 2  2  r r 

15.4.

Stress Solution for Edge Dislocati on : bB bB r     sin  ,  r  cos  r r Sum Forces on Cylinder of radius a , 2

 F   (a cos   a sin )d x

r

r

0

y

rad

rad

2

 2bB  sin  cos  d 0

2

 bB  sin 2 d  0 0

2

 F   (a sin   a cos )d y

r

r

0

2

 bB  (  sin 2   cos 2 )d 0

2

 bB  cos 2 d  0 0

15.5.

Stress Field for Screw Disl ocation : b 2r On the ends of a cylinder of finite length,  z will not vanish ,  r      z   r   rz  0 ,  z 

and the resultant twisting moment is given by a 2

a 2

0 0

0 0

T     z rrddr   

a b ba 2 rddr  b  rdr  0 2 2

15.6.

Strain Energy De nsity for Screw Disl ocation : 1 1 b 2  z  z  2 2 4 2 r 2 Strain Energy for Screw Disl ocation in Region Ro  r  Rc :

U

Wscrew  

2

0



Ro

 U rdrd  Rc

R b 2 log o 4 Rc

2 Ro b 1 2  dr Rc 4  2 r 2

a

x

15.7.

bB bB sin  ,  r  cos  r r Strain Field for Edge Dislocati on (Plane Strain) : 1  1  1  2 1  2 bB er  [(1  ) r   ]  (1  2) r  r   sin  E E 2 2 r 1  1  1  2 1  2 bB e  [(1  )   r ]  (1  2)      sin  E E 2 2 r 1 bB  r   r  cos   r Strain Energy De nsity for Edge Dislocati on :

Stress Field for Edge Dislocati on :  r     

U

1 1  2 b 2 B 2 b2 B 2 2 (  r er    e   r  r )  sin   cos 2  2 2 2 2 r 2r

b2 B 2  b2 B 2 b2 B 2 2 2 2 (sin   cos  )  sin   (1  2 sin 2 ) 2 2 2 2r  r 2r Strain Energy for Edge Dislocati on in Region Ro  r  Rc : 

Wscrew  

2

0



Ro

 U rdrd  Rc

b 2 B 2 2  Ro 1 (1  2 sin 2 )drd 2 0 Rc r

R R b2 B 2 b 2 2(1  ) log o  log o 2 Rc 4(1  ) Rc

15.8.

1 Kelvin State with Unit Load s :   0 , i  C iα , where C  R 8(1 - ) x x x  2ui  (   x k k ) ,i  4(1  ) i  C (  ) ,i  4C (1  ) iα  C   3 i  (3  4) iα   R R R  R ui 

C  x  xi   2  (3  4) iα  2R  R 

 ij  ,ij  2 ij  k ,k  (1  2)(  i, j   j ,i )  x k  k ,ij x iα x j 3C C  jα xi kα x k  C 3 ,  i, j  C ,  j ,i  C ,  k ,ij   5 kα xi x j  3 kα ij  3 3 3 R R R R R R  C  3x x x  ij   3   2i j  (1  2)( iα x j   jα xi  ij x  )  R  R  C  For the specail case a   [0,0,1] : z  0 , iz  0 , 0,    R 

 k ,k  C

uiz 

C  zxi   2  (3  4) i 3   2R  R 

u xz 

 C zx Czx Czy C  z2 z z  2  (3  4)   , u  , u  y z 2 3 3 2R R 2R 2R 2R  R 

 ijz  

 C  3zxi x j   (1  2)( i 3 x j   j 3 xi  ij z )   3  2 R  R 

 zx  

   C  3zx 2 C  3zy 2 C  3z 3 z z        ( 1  2  ) z ,     ( 1  2  ) z ,     ( 1  2  ) z y z 3  2 3 2 3 2   R  R R  R R R     

 zxy  

 z  C 3zxy 3Czxy z C  3z 2 y C  3z 2 x    2  (1  2) x    ,     ( 1  2  ) y ,    yz zx 3 2 5 3  2 3   R R R R  R R  R  

Using the displaceme nt vector transforma tion relations (B.8), 2C (1  ) cos  u R  u xz sin  cos   u yz sin  sin   u zz cos    R C (3  4) sin  u  u xz cos  cos   u yz cos  sin   u zz sin    2 R u  u xz sin   u yz cos   0

15.8. Continued

Using the stress transform ation relations (B.9),  R   zx sin 2  cos 2    zy sin 2  sin 2    zz cos 2   2 zxy sin 2  sin  cos   2 zyz sin  cos  sin   2 zzx sin  cos  cos   2C ( 2  )

cos  2

R z 2 2     x cos  cos    zy cos 2  sin 2    zz sin 2   2 zxy cos 2  sin  cos   2 zyz sin  cos  sin   2 zzx sin  cos  cos   C (1  2)

cos  R

2

    zx sin 2    zy cos 2   2 zxy sin  cos   C (1  2)

cos  2

R z 2 z 2 z  R   x sin  cos  cos    y sin  cos  sin    z sin  cos   2 zxy sin  cos  sin  cos    zyz (sin 2   cos 2 ) sin    zzx (sin 2   cos 2 ) cos   C (1  2)

sin  2

R z     x cos  sin  cos    zy cos  sin  cos    zxy cos (cos 2   sin 2 )   zyz sin  cos    zzx sin  sin  0  R   zx sin  sin  cos    zy sin  sin  cos    zxy sin (cos 2   sin 2 )   zyz cos  cos    zzx cos  sin  0

15.9.

Center of Compressio n :  =  o

Noting that R 2  x k x k , R,i 

3 1 1 xi , oi = 2(1  2) R 2(1  2) R 3

xi  1  x ,     i3  R  R  ,i R

 3 1 1 1 2(1  ) xi 2uio  ( o  xk ok ) ,i  4(1  ) oi       3  2(1  2) R 2(1  2) R  ,i (1  2) R xi 2(1  ) xi x 1     i3  3 3 (1  2) R (1  2) R R x uio   i 3 2R ekko  

1  xk  1 o 1  3xi x j ij  o o   3   3   0 , eij  ui , j  u j ,i    2  R  ,k 2 2  R5 R 

 oij  ekko ij  2eijo 

3 xi x j R5



 ij R

 3

 1  3 xi x j  2   ij  3  R  R 

For spherical displaceme nt components , use transform ation relations (B.8) 1 u Ro  u xo sin  cos   u oy sin  sin   u zo cos    2R 2 uo  u xo cos  cos   u oy cos  sin   u zo sin   0 uo  u xo sin   u oy cos   0 For spherical stress components , use transfor mation relations (B.9)  oR   ox sin 2  cos 2    oy sin 2  sin 2    oz cos 2   2 oxy sin 2  sin  cos   2 oyz sin  cos  sin   2 ozx sin  cos  cos  

2 3

R o     ox cos 2  cos 2    oy cos 2  sin 2    oz sin 2   2 oxy cos 2  sin  cos   2 oyz sin  cos  sin   2 ozx sin  cos  cos  

1 R

3

 o   ox sin 2    oy cos 2   2 oxy sin  cos   

1 R

3

15.9. Continued

 oR   ox sin  cos  cos 2    oy sin  cos  sin 2    oz sin  cos   2 oxy sin  cos  sin  cos    oyz (sin 2   cos 2 ) sin    ozx (sin 2   cos 2 ) cos  0 

o 

  ox cos  sin  cos    oy cos  sin  cos    oxy cos (cos 2   sin 2 )   oyz sin  cos    ozx sin  sin  0

 oR   ox sin  sin  cos    oy sin  sin  cos    oxy sin (cos 2   sin 2 )   oyz cos  cos    ozx cos  sin  0

15.10.

The state for a line of Centers of Dilatation along x1 - axis from 0 to a is given by, x

a

S o ( x )    S o ( x; )d , where S o ( x; )  S o ( x1  a, x 2 , x3 )  0

x

u1o 

( x1  ) x1 a d ( x1  ) 2 1 a d   2 0 [( x1  ) 2  x 22  x33 ]3 / 2 2 0 [( x1  ) 2  x 22  x33 ]3 / 2

1 1 1 2 2 3 2 2 3    , where Rˆ  ( x1  a)  x 2  x3 and R  x1  x 2  x3 ˆ 2  R R  x a x a d ( x1  ) d x o u2  2   2  2 2 3 3 / 2 2 0 [( x1  )  x 2  x3 ] 2 0 [( x1  ) 2  x 22  x33 ]3 / 2 

x



x 2  x1 ( x1  a)  1  1 x1 x 2 1 ( x1  a) x 2  1        2 3 2 3 2 2  R x 2  x3 Rˆ x 2  x3  2 x 2  x33  R Rˆ 

u 3o 

x3 a x a d ( x1  ) d  3  2 2 3 3 / 2  2 0 [( x1  )  x 2  x3 ] 2 0 [( x1  ) 2  x 22  x33 ]3 / 2



x3  x1 ( x1  a)  1    2 2 x 2  x33  R Rˆ 

15.11.*

Isotropic Self - Consistent Crack Dist ribution Case with   0.5 : 45(    )( 2   ) 45(0.5   )( 2   )   2 16(1   )(10  3   ) 16(1   2 )( 5  1.5   ) 45(0.5)(1  2  )( 2   ) 9 (1  2  )    16(1   2 )( 2.5)( 2   ) 16 (1   2 ) 16 2  18  (9  16)  0 , solving by quadratic formula   

9  9 2  16(9  16) 16

When   9 / 16  0 E 16(1   2 )(10  3 ) 16(10)9 1 1  11  0 E 45( 2   ) 45( 2)16  32(1   )( 5   ) 32(5)9 1 1  11  0  45( 2   ) 45( 2)16 Dilute Case with   0.5 : 45(    )( 2  ) 45(0.5   )( 2  0.5) 45(1  2  )     2 16(1   )(10   3  ) 16(1  0.25)(10   1.5  0.5) 8(17   1) 45  8  90  136 When   9 / 16  11   0.297 37 E 45( 2  ) 45(1.5)   2 E 45( 2  )  16(1   )(10  3) 45(1.5)  16(0.75)(8.5) 1 1 1     0.54 68 68 9 1.85 1  1 45 45 16  45( 2  ) 45(1.5)    45( 2  )  32(1  )( 5  ) 45(1.5)  32(0.5)( 4.5) 1 1 1     0.625 48 48 9 1.6 1  1 45 45 16

15.11.* Continued MATLAB Plots of effective moduli ratios and comparison with isotropic dilute case : 1 Self-Consistent Isotropic Dilute

0.9

Effective Poissons Ratio, / 

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0

0

0.1

0.2

0.3 0.4 0.5 Crack Density Parameter, 

0.6

0.7

1 Self-Consistent Isotropic Dilute

Effective Shear Modulus Ratio, / 

0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0

0

0.1

0.2

0.3 0.4 0.5 Crack Density Parameter, 

0.6

0.7

1 Self-Consistent Isotropic Dilute

Effective Youngs Modulus Ratio, E/E

0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0

0

0.1

0.2

0.3 0.4 0.5 Crack Density Parameter, 

0.6

0.7

15.12.

For Two - Dimensiona l Couple Stress Theory : ex 

u v v u 1  v u  , ey  , e xy    z , e yx    z , with  z     x y x y 2  x y 

 2ey  2 ex  3u  3v  ,  , y 2 y 2 x x 2 x 2 y  2 e xy xy



 2  v  2  v 1  v u   1  2  v u  1   3v  3u                      z 2      2 xy  x  xy  x 2  x y   2 xy  x y  2  x y y x 

2  2 e xy  2 ex  e y   2 y 2 x 2 xy

Standard principles of calculus 

 2 z  2 z  xy yx

2 2  z 1   2 v  2 u  e xy   v    v 1  v u   1   v  u   ,    2      z           2  x 2  x xy  x x  x xy   x  x 2  x y   2  x

 z e xy e x   x x y 2 2  z 1   2 v  2 u  e xy   v    v 1  v u   1   v  u     2  ,     z            2   y 2  yx y  y y  x  y  x 2  x y   2  yx y 

 z e y e xy   y x y Stresses :  x  ( e x  e y )  ( 2   )e x ,  y  ( e x  e y )  ( 2   )e y ,  xy  ( 2   )e xy  1 1 1 [ x  (  x   y )] , e y  [ y  (  x   y )] , e xy   xy ( 2   ) ( 2   ) ( 2   ) where    /( 2  2   )

ex 

2  2 e xy  2  xy  2 ex  e y 2 2   2  [    (    )]  [    (    )]  2  x x y y x y y 2 x 2 xy y 2 x 2 xy 2  2 x   y 2 2    (  x   y )  (  xy   yx ) y 2 x 2 xy m xz m yz   m xz   z , m yz   z   x y y x

Now m xz  

 e xy e x   z 1 1  and e xy      (  xy   yx ) , e x  [ x  (  x   y )]  x y  2( 2   ) ( 2   )  x

  2    (  xy   yx )  [ x  (  x   y )]  l 2 (  xy   yx )  2l 2 [ x  (  x   y )] 2( 2   ) x 2( 2   ) y x y    where l 2  , Likewise for the relation m yz  2l 2 [ y  (  x   y )]  l 2 (  xy   yx ) 2( 2   ) x y

m xz 

15.13.

Couple Stress Function Formulati on :  2  2   2  2   2  2   2  2   ,    ,     ,     y xy yx y 2 xy x 2 xy xy y 2 xy x 2   m xz  , m yz  x y Using these forms in the equlibrium relations (14.4.7)   x  yx  3  3  3  3  0      0 , checks x y xy 2 x 2 y xy 2 yx 2  xy  y  3  3  3  3  0   2     0 , checks x y x y xy 2 yx 2 xy 2 m xz m yz  2   2   2  2   2  2     xy   yx  0  2  2      0 , checks x y x y xy y 2 xy x 2

x 

15.14.

Couple Stress Function Formulati on :  2  2   2  2   2  2   2  2   ,    ,     ,     y xy yx y 2 xy x 2 xy xy y 2 xy x 2   m xz  , m yz  x y Using these forms in the compatibil ity relations (14.4.13) :

x 

2  2 x   y 2 2    (  x   y )  (  xy   yx )  y 2 x 2 xy 2  4  4   4  4   2   4 4  4 4 2                   x 2 y 4 xy 3 x 4 x 3y y 2  x 2 y 2 xy 3 x 2 y 2 yx 3  2  4  4  4  2  2     2 2 2  4    2  2   0  (1  ) 4   0   4   0 4 x x y y y   x

m xz m yz 2 2    , satisfied identicall y y x yx xy   m xz  l 2 (  xy   yx )  2l 2 [ x  (  x   y )]  x y   2   2      2 2  2     2  2   2  2  2   2l 2  2  2    l2     x x  x y xy  y  y 2 xy y   x   2       l 2  2    2(1  )l 2 x y   Likewise using m yz  2l 2 [ y  (  x   y )]  l 2 (  xy   yx )  x y   2       l 2  2    2(1  )l 2 y x Following the suggested differenti ations on the top of page 393 in text  2   l 24   0

15.15.

Stress Concentrat ion Around Circular Hole - Micropolar Case Hoop Stress Solution :  

  6l 2  2 A  3l A  6A 6A  T (1  cos 2)  21   42  44  cos 2  5  1 K o (r / l1 )  1  21  K 1 (r / l1 ) cos 2 2 l1 r  r r r  r   r  

4T (1  )a 2 l 22 Tal1 F T 2 Ta 4 (1  F ) Ta 2 where A1   a , A2   , A3  , A4  , A5   2 4(1  F ) 2(1  F ) 1 F (1  F ) K 1 (a / l1 ) l 22  a 2 2a K o (a / l1 )  F  8(1  ) 2 4  2   l 1 K 1 (a / l1 )  l1  l1

1

On Hole Boundary : r  a   (a, )  

T T  6T (1  F ) 24T (1  )l 22   cos 2 (1  cos 2)      2 2 2  4(1  F ) a (1  F )   3l1   6l12  2TF 1  2  K 1 (a / l1 ) cos 2 K ( a / l )   o 1 (1  F ) K 1 (a / l1 )  a a   

 1 6(1  F ) 24(1  )l 22 6 Fl1 K o (a / l1 ) 12 Fl12  2F  T  T    2    cos 2 2   2 4(1  F ) a (1  F ) (1  F )a K 1 (a / l1 ) (1  F ) (1  F )a  Fl12  T  1 3(1  F ) 24(1  )l 22 6 Fl1 K o (a / l1 ) T   ( 1  F )     2 F  12   cos 2 1 F  2 2 a K 1 (a / l1 ) a2 a2  K o (a / l1 ) l1  8(1  ) l 22 a2  But from definition of F    4    and thus K 1 (a / l1 ) 2a  F l12 l12    (a, )  T 

Fl12  T  1 3(1  F ) 24(1  )l 22 3Fl12  8(1  ) l 22 a2       4   2 F  12  (1  F )   cos 2 1  F  2 2 a2 a 2  F l12 l12  a 2 

T  1 3(1  F )   2 cos 2   (1  F )   3 F  2 F   T 1    1 F  2 2 1 F    Clearly th is expression takes on a maximum value when cos 2  1      / 2 T 

2   3 F   (  ) max    (a,  / 2)  T 1    T   1 F   1 F  When l1  l 2  l  0  l2  a 2 2a K o (a / l1 )  F  8(1  ) 22 4  2   l 1 K 1 (a / l1 )  l1  l1 (  ) max  3T

1

 8(1  )

l 22 K (a / l1 ) 4l  a  2al1 o K 1 (a / l1 ) 2 1

2

0

15.16.

Constituti ve Equations for Linear Isotropic Elastic Materials with Voids   e   2e   , h   , g      e ij

kk

ij

ij

ij

i

,i

kk

Under Plan e Stress Condtions :  x  ( e x  e y  ez )  2e x    y  ( e x  e y  ez )  2e y    z  ( e x  e y  ez )  2ez    0  (  2)ez    ( e x  e y )  xy  2e xy ,  xz   yz  0 Using the relation for ez     ( e x  e y )   2 2   2e x      x   e x  e y  (ex  e y )    2e x   2   2   2   2 2 y   (ex  e y )    2e y   2   2 2 Combining these results gives :  ij  (ekk  ) ij  2eij , where ekk  e x  e y   2    ( e x  e y )    g      ekk      ( e x  e y  ez )       e x  e y    2   2 2 (ex  e y )        2                 e x  e y  (ex  e y )   2   2    2    2  

15.17.

For elasticity with void s, the plane stress compatibil ity relation is the same as that originally developed in Chapter 7, given by (7.2.6) 2  2exy  2 ex  e y  2 2 y 2 x xy From the stress - strain relations developed in Exercise 14 - 16 2 2 4(  ) 2 2 x   ( ex  e y )    2ex  ex  ey    2   2   2   2   2



 2 4(  )    ex  e y     2  2(  )    2

y 

 2 4(  )    e y  ex     2  2(  )    2

These relations can be cast in the identical form as the original classical elasticity results given in Exercise 7 - 6, Eˆ Eˆ ˆ ˆ ˆ x  ( e   e ) ,   (e y  ˆ ex ) , with x y y 1  ˆ 2 1  ˆ 2 2 2 ˆ (3  2)  ˆ x   x  , ˆ y   y  ,E , ˆ    2   2 (   ) 2(  ) Thus the solution for the strains follows as 1 1 1 1  ˆ ex  (ˆ x  ˆ ˆ y ) , e y  (ˆ y  ˆ ˆ x ) and also exy   xy   xy 2 Eˆ Eˆ Eˆ Substituti ng these into the strain compatibil ity relation and following similar steps as in Exercise 7 - 7   2  xy  2ˆ x  2ˆ y 1 2 2  2 (ˆ x  ˆ y )  2      2 2 (1  ˆ ) xy x y   2 3  2 2 2  2 (ˆ x  ˆ y )   0 2(  )   2 4 2 3  2   2 ( x   y )     2  0    2 (  )   2  2  2 ( x   y )   0  Introducin g the usual Airy stress function  x   4 

 2  0 

 2  2 ,    y x 2 y 2

15.18.*

Hoop stress solution for elasticity with voids  

 2 F ( r )  T  a2  a 4        1   cos 2  a  1  3    2  r 2  r 4    F (a ) 

F (r)  1 

h 2  1 2hK 2 ( r / h )  h2  2 2 K 2 ( r / h )  h2  6 2 K ( r / h )       F ( r )     F ( r )   3 2  2      3 3 3 4 N r a K 2 ( a / h )  N  r a K 2 ( a / h )  N r a hK 2 ( a / h ) 

1 a a When N  , L   2  h   2 h 2 2 a 1 K 2 ( 2)  a 2  6 4 K 2( 2r / a )  F (a )  1     , F ( r )    4   2  a 2 a 2 K 2 ( 2)  2  r4 a K 2 ( 2)  From the properties of Bessel Functions : 2 K 2 ( x )   K1 ( x )  K 2 ( x ) x 1 2 2  2 K 2( x )  K 0 ( x )  K1 ( x )   K1 ( x )  K 2 ( x )   2 K 2 ( x ) x x x  x    ( r ,  / 2) / T 

 1  a2  a 4  2 F ( r )  1  3 4   1  2   a 2  r  F (a )  r 

From classical elasticity , the correspond ing result is 1 a 2 3a 4    ( r,  / 2) / T   2  2  4  2 r r  MATLAB Plots :

Voids Theory

Classical Elasticity

15.19.* MATLAB Contour Plots of Doublet Mechanics Solution – p1 and p2 Microstresses for Flamant Problem

+

-

p2 – Contours

+

-

p1 – Contours

15.20.

One dimensional displacement field: u  u ( x ) , v  w  0  du Only one non-zero strain component: ex  dx Equation (15.7.3): ij  ekk ij  2eij  c[ekk ij  2eij ],ll  du d 3u du d 3u  c(  2) 3 ,  y   z    c 3 ,  xy   yz   zx  0 dx dx dx dx d x Equilibrium equations ij , j  0 dx d 2u d 4u d 4u 1 d 2u (  2) 2  c(  2) 4  0  4  0 dx dx dx c dx 2 Using standard ODEsolution techniques, look for solution form e mx 

 x  (  2)

1  characterist equation: m 2  m 2 +   0  c  1 four roots: m  0 (double root) , m   i , and thus solution is given by c  x   x  u  c1  c2 x  c3 sinh    c4 cosh    c  c

15.21. u  v  0, w  w( x, y ) 1 w 1 w , eyz  2 x 2 y Constitutive law: ij  ekk ij  2eij  c[ekk ij  2eij ],ll

Strain-displacement relations  exz 

gives only two non-zero expressions for  xz and  yz  xz  

 2   2 w  2 w  w  2  w    c  2  2    w  c  2  2  x x  y    x y  x  x

 yz  

 2   2 w  2 w  w  2  w    c  2  2    w  c  2  2  y y  y    x y  y  x

With only two non-zero stresses, equilibrium equations reduce to single z-relation   2 w  2 w   2    2 w  2 w   xz  yz 2    0  2 w  c  2  2   2  w  c  2  2   0  x y x  y   y  y    x  x  4w 2w 2w 4w 4w    c  2   4 0 x 2 y 2 x 2 y 2 y 4   x

15.22.

Use typical table results:  sin mx    dx  0 cos mx  to confirm results: 

 e

 ax

  sin mx  ,  xe ax   dx  0 cos mx 

x  

P  2 Px 2 y y (1   y ) e cos(  x ) d     0 ( x 2  y 2 ) 2

y  

P  2 Py 3 y (1   y ) e cos(  x ) d     0 ( x 2  y 2 ) 2

 xy  

P  2 Pxy 2 y  ye sin(  x ) d     0 ( x 2  y 2 ) 2

16.1.

Linear Approximat ion : u( x, y )  c1  c2 x  c3 y Evaluating at each node  u( x1 , y1 )  u1  c1  c2 x1  c3 y1 u( x2 , y 2 )  u2  c1  c2 x2  c3 y 2 u( x3 , y 3 )  u3  c1  c2 x3  c3 y 3 Solving the three equations for the constants  1 1 1 c1  ( 1u1   2 u2   3u3 ) , c2  (1u1   2 u2   3u3 ) , c3  (  1u1   2 u2   3u3 ) 2 Ae 2 Ae 2 Ae where where Ae is the area of the element, and i  x j y k  xk y j ,  i  y j  y k ,  i  xk  x j Back substituti ng these results into the original form for u( x, y )  1 u( x, y )  [( 1u1   2 u2   3u3 )  (1u1   2 u2   3u3 ) x  (  1u1   2 u2   3u3 ) y ] 2 Ae 

1 1 1 ( 1  1 x   1 y )u1  (  2   2 x   2 y )u 2  (  3   3 x   3 y )u 3 2 Ae 2 Ae 2 Ae 3

  ui  i ( x , y ) i 1

where  i ( x, y ) 

1 ( i   i x   i y ) are the interpolat ion or approximat ion functions 2 Ae

16.2.

For the constant strain triangula r element, [ K ]  he Ae [ B ]T [C ][ B ] , with 1 0 1  [ B]  0 1 2 Ae    1 1  1 0 0  1  0 h  [K ]  e  2 4 Ae  0  2  3 0   0 3  1 0  he  2   4 Ae  0  3  0

0 1 0 2 0 3

2 0 2

0 2 2

3 0  C11 C12  0  3 , [C ]  C12 C22    3  3  0  0

0  0   C66 

1  1   C11 C12 0  1 0  2 0  3 0   2     C12 C22 0   0  1 0  2 0  3  2  0 C66    1 1  2  2  3  3   0  3   3  1  1   1C11  1C12  2 C11  2 C12  3C11  3C12   2    1C12  1C22  2 C12  2 C22  3C12  3C22  2    1C66 1C66  2C66  2C66  3C66  3C66   3   3 

12 C11   12 C66 1  1C12  1  1C66 1 2C11   1  2 C66         h   e  4 Ae             

     

     

        

16.3.

 Fx  In general, the body force vector is given by {F }  he  []T  dxdy e  Fy  0  2 0  3 0   Fx   Fx   [ ]     1     Fy   0  1 0  2 0  3   Fy   Fx 1  F    y 1  Fx  2     Fy  2   Fx  3     Fy  3  For the case of element - wise constant body forces, 1 e Fx 1dxdy  Fx e 1dxdy  3Fx Ae , and likewise for all other components hA T  {F }  e e  Fx Fy Fx Fy Fx Fy  3 T

T

16.4.

For the linear tri angular element, with constant boundary t ractions  1Txn  Txn    n n  1T y  T y  n n  T  T  h L T n  he  []T  xn ds  he   2 xn ds  e 12  xn  12 12  T 2 T y  T y   2 y n  3Tx  0    n  0 12  3T y  hL 1 Now he  1Txn ds  heTxn  1ds  heTxn  ( 1  1 x   1 y )ds  e 12 12 12 12 2 A 2 e Integrals of the other components involving 1 and  2 follow identical patterns, while integrals containing the  3 terms will vanish since  3  0 on 12 .

Collecting all of the integral evaluation s  Txn   n T y  n T  h L T n  he  []T  xn ds  e 12  xn  12 2 T y  T y  0    0 12

16.5.

For the isotropic plane stress case : E E E C11  C22  , C12  , C66    2 2 1  1  2(1  ) For element 1 in Example 15 - 1, 1  1 ,  2  1 ,  3  0 ,  1  0 ,  2  1 ,  3  1 , A1  1 / 2 , h  1 E E , 1  1C12  1  1C66  0 , 1 2C11   1  2 C66   , ...

2 1  1  2 and similarly for all other components in the stiffness matrix 

 12 C11   12 C66  

0 1  1    2    E (1)  [K ]  2(1   2 )          

1 1  2 3  2

 1   2 1   2 3  2

 







0 1   2 1   2 1  2 1  2 

  0       1   0  1 

For element 2 , 1  0 ,  2  1 ,  3  1 ,  1  1 ,  2  0 ,  3  1 , A1  1 / 2 , h  1 E , 1  1C12  1  1C66  0 , 1 2 C11   1  2C66  0 , . . .

2(1  ) and similarly for all other components in the stiffness matrix 

 12 C11   12 C66 

1    2      E ( 2)   [K ]  2(1   2 )        

0

0

1   1

1  2 0 0 1  2



















1  2  1 1  2 3  2





1   2  1     1    2  1     2  3   2 

16.6.

From Example 15 - 1, the reduced global system matrix was given by (1)  K 33        

(1) K 34 (1) K 44  

(1) K 35 (1) K 45 (1) ( 2) K 55  K 33 

(1)  U 2  T / 2 K 36     (1) K 46  V2    0   (1) ( 2)  K 56  K 34 U 3  T / 2  (1) ( 2)  K 66  K 44   V3   0 

1  1  3         2 2 2   U 2  T / 2 3 1   1  V   0  E    2 2   2     3 U T / 2 2(1  )    3   0        2  V 0  3    3        2   151.8  69  41.4 27.6  U 2  T / 2   151.8 41.4  110.4 V2   0  10 9        151.8 0  U 3  T / 2     151.8   V3   0    U 2   0.492  V   0.081      11 Solving the system   2    T  10 m U 3   0.441   V3   0.030

16.7.* MATLAB PDE Toolbox: FEA Solution to Biaxial Loading Problem in Exercise 8-25

y

x

1000

1000 FEA Model: (w/d ≈ 7, 2792 Nodes, 5376 Elements)

B

B 



A

A

x - Contours

y - Contours

Comparison of x Results: FEA Solution: (x)A  200 , (x)B  1900 Analytical Solution: (x)A  0 , (x)B  2000

Comparison of y Results: FEA Solution: (y)A  1900 , (y)B  300 Analytical Solution: (y)A  2000 , (y)B  0

16.8.* MATLAB PDE Toolbox: FEA Solution to Biaxial-Shear Loading Problem in Exercise 8-15 (Figure 8-15)

y

x

1000

1000 FEA Model: (w/d ≈ 7, 2792 Nodes, 5376 Elements)

B

B  A

 A

x - Contours

y - Contours

Comparison of x Results: FEA Solution: (x)A  -270 , (x)B  3300 Analytical Solution: (x)A  0 , (x)B  4000

Comparison of y Results: FEA Solution: (y)A  -3500 , (y)B  280 Analytical Solution: (y)A  4000 , (y)B  0

16.9.* MATLAB PDE Toolbox: FEA Solution to Curved Beam Problem shown in Figure 8-32.

Choose case b/a = 4, to provide simple match with analytical results given in Figure 8-33 1000

FEA Model: b/a = 4 , 2009 Nodes , 3872 Elements

A

B

x - Contours

Comparison of x Results: FEA Solution: (x)A  -8973 , (x)B  3875 Analytical Solution: (x)A  -10,000 , (x)B  2571

16.10.* MATLAB PDE Toolbox: FEA Solution to Torsion Problem in Exercise 9-17/18

FEA Model: (b = 0.267, a = 0.8, 2129 Nodes, 4096 Elements)

A



B

FEA |  | - Contours Resultant shear stress :     

2 yz

2

             () 2    x   y  2

2 xz

Analytical Results :  A  2 B ; FEA Results :

 A  1.2 ,  B  0.75

16.11.

For the two - dimensiona l, plane strain case : Gij 

1   1 (3  4) ln   ij  r,i r, j   8(1  )  r 

Using the fact that r,i  Gij,k 

xi  r

r,k   1  ( 3  4  )  ij  ( r,i r, j ) ,k   8(1  )  r 

1  (3  4)r,k ij  r,i  jk  r, j ik  2r,i r, j r,k  8(1  ) r The stress form is given by Tikj  Glj,l  ik  (Gij,k  Gkj ,i ) 

  (3  4)r,l lj  r,l  jl  r, j ll  2r,l r, j r,l  8(1  ) r 1  (3  4)r,k ij  r,i  jk  r, j ik  2r,i r, j r,k   8(1  ) r 1  (3  4)r,i kj  r,k  ji  r, j ik  2r,i r, j r,k   8(1  ) r 2  2(1  2)r, j ik   8(1  2)(1  ) r 1  2(1  2)( r,k ij  r,i  jk )  2r, j ik  4r,i r, j r,k   8(1  ) r 1 (1  2)r, j ik  (1  2)( r,k ij  r,i  jk )  2r,i r, j r,k   4(1  ) r



pij  Tikj nk  

1 r    r  (1  2) r, j ni  (1  2)  ij  r,i n j   2r,i r, j   4(1  ) r  n   n 

1 r    r  (1  2)  ij  r,i n j  r, j ni   2r,i r, j   4(1  ) r  n   n 

where we have used r,k nk 

r n