Continuum Mechanics Modeling of Material Behavior Martin H. Sadd 1st edition solution manual

SOLUTION MANUAL

2.1 1  1 0 1   ai  2 , Aij  1 2 0 1  1 2 1 ai ai  6 (scalar)

(a)

1 2 1  ai a j  2 4 2 (matrix) 1 2 1  Aii  A11  A22  A33  4 (scalar)  2 Aij a j  5 (vector) 6

0  1 2 1   ai  2 , Aij  1 0 0 1  1 2 1 ai ai  5 (scalar) 0 0 0  (b) ai a j  0 4 2 (matrix)   0 2 1  Aii  A11  A22  A33  2 (scalar) 5  Aij a j  0 (vector) 5  2 0 2 1    ai  1  , Aij  3 2 0 1  1 0 4 ai ai  6 (scalar)  4 2 2 (c) ai a j  2 1 1  (matrix)   2 1 1  Aii  A11  A22  A33  6 (scalar)  3 Aij a j  8 (vector) 6

2.2 1 0 1 Aij  1 2 0 1 2 1  1 1 / 2 1 1 (a) A(ij )  Aij  A ji  1 / 2 2 1 , six independent components 2  1 1 1  0 1/ 2 0  1 A[ij ]  Aij  A ji  1 / 2 0  1 , three independent components 2  0 1 0 









1 2 1 Aij  1 0 0 1 2 1  1 3 / 2 1 1 (b) A(ij )  Aij  A ji  3 / 2 0 1 , six independent components 2  1 1 1 1/ 2 0   0 1  A[ij ]  Aij  A ji   1 / 2 0  1 , three independent components 2 1 0   0









0 2 1  Aij  3 2 0 1 0 4  0 5 / 2 1 1 (c) A(ij )  Aij  A ji  5 / 2 2 0 , six independent components 2  1 0 4  0  1 / 2 0 1 A[ij ]  Aij  A ji  1 / 2 0 0 , three independent components 2  0 0 0









2.3.

Aij Bij   A ji B ji   Aij Bij  2 Aij Bij  0  Aij Bij  0

2.4 Aij Cij  A jiCij  Aij C ji





 Aij C(ij )  C[ij ]  Aij C(ij ) Bij Cij   B jiCij   Bij C ji





 Bij C(ij )  C[ij ]  Bij C[ij ] 0

B12

B13

det Bij   B12

0

B23   B12 B13 B23  B13 B12 B23  0

 

 B13

 B23

0

2.5 11a11  12 a21  13 a31 11a12  12 a22  13 a32 11a13  12 a23  13 a33   ij a jk             a11 a12 a13   a21 a22 a23   aik  a31 a32 a33  11a11  12 a12  13 a13 11a21  12 a22  13 a23 11a31  12 a32  13 a33   ij akj             a11  a12  a13

a21 a22 a23

a31  a32   aki a33 

2.6 det( Aij )  ijk A1i A2 j A3k  123 A11 A22 A33   231 A12 A23 A31  312 A13 A21 A32  321 A13 A22 A31  132 A11 A23 A32   213 A12 A21 A33  A11 A22 A33  A12 A23 A31  A13 A21 A32  A13 A22 A31  A11 A23 A32  A12 A21 A33  A11 ( A22 A33  A23 A32 )  A12 ( A21 A33  A23 A31 )  A13 ( A21 A32  A22 A31 ) A11

A12

A13

 A21

A22

A23 (using the usual first row expansion)

A31

A32

A33

Likewise the same for the form det[ Aij ]  ijk Ai1 A j 2 Ak 3

2.7 det[ Aij ] 

A11

A12

A21

A22

 A11 A22  A12 A21 , det[ Bij ] 

B11

B12

B21

B22

 B11 B22  B12 B21

 det[ Aij ] det[ Bij ]  ( A11 A22  A12 A21 )( B11 B22  B12 B21 )  A11 A22 B11 B22  A12 A21 B12 B21  A11 A22 B12 B21  A12 A21 B11 B22 det[ AB ] 

A11 B11  A12 B21

A11 B12  A12 B22

A21 B11  A22 B21

A21 B12  A22 B22

 ( A11 B11  A12 B21 )( A21 B12  A22 B22 )  ( A11 B12  A12 B22 )( A21 B11  A22 B21 )  A11 A22 B11 B22  A12 A21 B12 B21  A11 A22 B12 B21  A12 A21 B11 B22  det[ Aij ] det[ Bij ] Orthogonality Properties :





Given Aik A jk  ij  det Aik A jk  det ij  det Adet A  1  det A  1

2.8 Aip

Aiq

Air

Property (2.5.4) ijk  pqr det[ Amn ]  A jp

A jq

A jr

Akp

Akq

Akr

ip

iq

ir

For the case Aij  ij  ijk  pqr   jp

 jq

 jr

 kp

 kq

 kr

ii

iq

ir

From (2.5.5) with p  i  ijk iqr   ji

 jq

 jr   jq kr   jr  kq (2.5.6)

 ki

 kq

 kr

(2.5.5)

From (2.5.6) with q  j  ijk ijr   jj  kr   jr  kj  3 kr   kr  2 kr (2.5.7) From (2.5.7) with r  k  ijk ijk  2 kk  6 (2.5.8)

2.9 e1  e1  1 , e2  e2  1 , e3  e3  1 e1  e2  e2  e1  0 , e1  e3  e3  e1  0 , e2  e2  1 , e3  e3  1  ei  e j  ij From (2.6.4) e1  e2  e3  e2  e1 , e2  e3  e1  e3  e2 , e3  e1  e2  e3  e1  ei  e j  ijk ek

2.10

cos 90 cos 90 cos180  0 0  1 For 90 rotation about x2  axis : Qij  cos 90 cos 0 cos 90   0 1 0   cos 0 cos 90 cos 90  1 0 0  Cases from Exercise 2.1 : o

0 0  1 1   1 1/ 2 1/ 2 (a) ai  Qij a j  0 1 0  2   2  , a  ai ai   6 , a  aiai   6 1 0 0  1   1  T

0 0  1 1 0 1 0 0  1  1  2  1       Aij  Qip Q jq Apq  0 1 0  1 2 0 0 1 0    0 2 1  , Aii  6 , Aii  6 1 0 0  1 2 1 1 0 0   1 0 1  0 0  1 0  1 1/ 2 1/ 2 (b) ai  Qij a j  0 1 0  2   2  , a  ai ai   6 , a  aiai   6 1 0 0  1   1 T

0 0  1 1 2 1 0 0  1  1  2  1       Aij  Qip Q jq Apq  0 1 0  1 0 0 0 1 0    0 0 1  , Aii  2 , Aii  2 1 0 0  1 2 1 1 0 0   1 2 1  0 0  1 2  1 1/ 2 1/ 2 (c) ai  Qij a j  0 1 0  1    1  , a  ai ai   6 , a  aiai   6 1 0 0  1   2  T

0 0  1 0 2 1  0 0  1  4 0  1 Aij  Qip Q jq Apq  0 1 0  3 2 0 0 1 0    0 2 3  , Aii  6 , Aii  6 1 0 0  1 0 4 1 0 0   1 2 0 

2.11





d QQT  I  Q Q T  QQ T  0  dt T Q Q T  QQ T   Q Q T





2.12  cos( x1 , x1 ) cos( x1 , x2 )   cos  cos(90 o  )  cos  sin   Qij        o cos    sin  cos  cos( x2 , x1 ) cos( x2 , x2 ) cos(90  )  cos  sin    b1   b1 cos   b2 sin   bi  Qij b j         sin  cos  b2   b1 sin   b2 cos   cos  sin    A11 Aij  Qip Q jq Apq     sin  cos   A21

A12   cos  sin   A22   sin  cos 

 A cos2   ( A12  A21 ) sin  cos   A22 sin 2    11 2 2  A21 cos   ( A11  A22 ) sin  cos   A12 sin 

2.13



T

A12 cos2   ( A11  A22 ) sin  cos   A21 sin 2   A11 sin 2   ( A12  A21 ) sin  cos   A22 cos 2 



1 1 1 ai   ijk A jk   ijk A( jk )  A[ jk ]   ijk A[ jk ]  2 2 2 1 1 1 ai imn   ijk imn A[ jk ]   ( jm kn   jn km ) A[ jk ]   A[ mn ]  A[ nm]    A[ mn ] 2 2 2  A[ mn ]  ai imn

2.14 Property 7 : Given Aij Bij  c  scalar and Bij is a second order tensor  Aij Bij  c  Aij Qim Q jn Bmn  c  Amn Bmn  Amn  Qim Q jn Aij  A is a second order tensor Property 8 : Given Aij b j  ci with b j and ci being first order tensors  Aij bj  ci  Aij Q jmbm  Qin cn  Qin Anmbm  Aij Q jm  Qin Anm 





Qkm Aij Q jm  Qin Anm  Aij Q jmQkm  Qin AnmQkm  Aij  jk  Aik  Qin Qkm Anm  A is a second order tensor

2.15 1 1 ikl jqr Aqk Arl A jm  ikl det[ A] mkl , where we have used (2.5.10) 2 det[ A] 2 det[ A] 1 1 1  ikl mkl   kli klm  2im  im , where we have used (2.5.7) 2 2 2

Aij1 A jm 

2.16 I1  trA  I A I 2  trA 2  Aij A ji From (2.11.3) II A 

1 ( Aii A jj  Aij Aij )  Aij Aij  Aij A ji  Aii A jj  2 II A 2

 I 2  I A2  2 II A I 3  trA 3 , Using CT Theorem (2.13.1)  A3  I A A 2  II A A  III A I Taking trace operation gives trA 3  I A ( I A2  2 II A )  II A I A  3III A  I 3  I A3  3II A I A  3III A

2.17 1 0 0  Given Aij   0  2 0   0 0  3  I A  Aii  A11  A22  A33  1   2   3 II A 

A 1 ( Aii A jj  Aij Aij )  11 A21 2

 1 2   2 3   31 III A  det[ Aij ]  1 2 3

A12 A22



A22

A23

A32

A33



A11

A13

A31

A33

2.18  2 1 0 (a) Aij   1  2 0  0 0 0 A I A  Aii  4 , II A  11 A21

A12 A22



A22

A23

A32

A33



A11

A13

A31

A33

 3 , III A  det[ Aij ]  0

Characteristic Eqn.  3  42  3  0  (  1)(  3)  0 , 1  0 ,  2  1,  3  3 1  0 , ( Aij  ij ) n j  0   2n1(1)  n2(1)  0 n1(1)  2n2(1)  0  n1(1)  n2(1)  0 and thus n3(1)  1  n (1)  e3  2  1, ( Aij  ij ) n j  0   n1( 2)  n2( 2)  0 n1( 2)  n2( 2)  0 n3( 2)  0  n1(1)  n2(1) and thus  n ( 2)  (e1  e2 ) / 2  3  3 , ( Aij  ij )n j  0  n1(3)  n2(3)  0 n1(3)  n2(3)  0 3n3(3)  0  n1(3)  n2(3) and thus  n (3)  (e1  e2 ) / 2  0 Rotation Matrix to Principal Axes : Qij  1 / 2 1 / 2 0 0 0  A  QAQ  0  1 0  0 0  3 T

0 1/ 2 1/ 2

1 0  0

2.18  1 1 0 (b) Aij   1  1 0  0 0 2 A I A  Aii  0 , II A  11 A21

A12 A22



A22

A23

A32

A33



A11

A13

A31

A33

 4 , III A  det[ Aij ]  0

Characteristic Eqn.  3  4  0   (2  4)  0 , 1  0 ,  2  2,  3  2 1  0 , ( Aij  ij )n j  0   n1(1)  n2(1)  0 n1(1)  n2(1)  0 2n3(1)  0  n3(1)  0  n1(1)  n2(1) and normalizing n1(1)  n2(1)  1 / 2  n(1)  (e1  e2 ) / 2  2  2 , ( Aij  ij )n j  0  n1( 2)  n2( 2)  0 n1( 2)  n2( 2)  0 4n3( 2)  0  n3( 2)  0  n1( 2)  n2( 2) and thus  n (3)  (e1  e2 ) / 2  3  2 , ( Aij  ij )n j  0   3n1(3)  n2(3)  0 n1(3)  3n2(3)  0 00  n1(3)  n2(3)  0 and thus n3(3)  1  n(3)  e3 1 / 2  Rotation Matrix to Principal Axes : Qij  1 / 2  0  0 0 0  T A  QAQ  0  2 0 0 0 2

1/ 2 1/ 2 0

0  0  1

2.18  6  3 0 (c) Aij   3 6 0  0 0 6 A I A  Aii  18 , II A  11 A21

A12 A22



A22

A23

A32

A33



A11

A13

A31

A33

 99 , III A  det[ Aij ]  162

Characteristic Eqn.  3  182  99  162  0  (  9)(  6)(  3)  0  1  9 ,  2  6,  3  3 1  9 , ( Aij  ij )n j  0   3n1(1)  3n2(1)  0  3n1(1)  3n2(1)  0  3n3(1)  0  n3(1)  0  n1(1)   n2(1) and normalizing n(1)  (e1  e2 ) / 2  2  6 , ( Aij  ij )n j  0   3n2( 2)  0  n2( 2)  0  3n1( 2)  0  n1( 2)  0 00  n3( 2)  1 and thus  n(3)  e3  3  3 , ( Aij  ij )n j  0  3n1(3)  3n2(3)  0  3n1(3)  3n2(3)  0 3n3(3)  0  n3(3)  0  n1(3)  n2(3) and normalizing n1(1)  n2(1)  1 / 2  n (3)  (e1  e2 ) / 2 1 / 2 1 / 2  Rotation Matrix to Principal Axes : Qij   0 0 1 / 2 1 / 2  9 0 0  A  QAQ  0 6 0 0 0 3 T

0  1  0

2.19 Mulitply basic equation An  n by A1 to get

1 n   1 /  are the principal values of A1 , and n are the principal directions A1 An  A1n  n  A1n  A1n 

Using principal coordinates, I A  1   2   3 , II A  1 2   2 3   31 , III A  1 2 3  II A / III A 

1 2   2 3   31 1 1 1     I A 1 1 2 3 1  2  3

2.20

Using principal axes Aij bi b j  1b12   2b22   3b32 If  i  0 , then clearly Aij bi b j  0 If Aij bi b j  0  1b12   2b22   3b32  0 Since vector b is arbitrary, choose b  (1,0,0)  1  0 Likewise for other choices b  (0,1,0)   2  0 , b  (0,0,1)   3  0

2.21 3 0 1  (a) Aij  1 2 0  Amm  9 1 2 4 3 0 0 ~ 1 Aij  Amm ij  3ij  0 3 0 3 0 0 3 0 0 1  ~ Aˆij  Aij  Aij  1  1 0 1 2 1 0 0 1  (b) Aij  1 2 0  Amm  3 1 2 1 1 0 0 ~ 1 Aij  Amm ij  ij  0 1 0 3 0 0 1  1 0 1 ~ Aˆij  Aij  Aij   1 1 0  1 2 0

0 2 1  (c) Aij  3 2 0  Amm  6 1 0 4 2 0 0 ~ 1 Aij  Amm ij  2ij  0 2 0 3 0 0 2  2 2 1  ~ ˆ Aij  Aij  Aij   3 0 0  1 0 2

2.22

From (2.13.2)  A3  I A A2  II A A  III A I Multiply by A  A4  I A A3  II A A2  III A A ; substituting in for A3 from (2.13.2)  A4  I A ( I A A2  II A A  III A I )  II A A2  III A A  ( I A2  II A ) A2  ( III A  I A II A ) A  I A III A I Next multiplying by A again  A5  ( I A2  II A ) A3  ( III A  I A II A ) A2  I A III A A ; substituting in for A3 from (2.13.2)  A5  ( I A2  II A )( I A A2  II A A  III A I )  ( III A  I A II A ) A2  I A III A A A5  ( I A3  2 I A II A  III A ) A2  ( I A III A  I A2 II A  II A2 ) A  ( I A2 III A  II A III A ) I

2.23

A  RU  VR  AT A  U T RT RU  U T IU  U T U  U T U  U 2 AAT  VRR T V T  VI V T  V V T  V V V 2

2.24 (a) u  x1e1  x1 x2e 2  2 x1 x2 x3e 3   u  u1,1  u2, 2  u3,3  1  x1  2 x1 x2 e1

e2

  u   / x1  / x2 x1

x1 x2

e3  / x3  2 x1 x3e1  2 x2 x3e 2  x2e 3 2 x1 x2 x3

 2 u   2u1e1   2u2e 2   2u3e 3  0e1  0e 2  0e 3  0  1 u   x2 2 x2 x3

0 x1 2 x1 x3

0  0  , tr (u)  1  x1  2 x1 x2 2 x1 x2 

(b) u  x12e1  2 x1 x2e 2  x33e 3   u  u1,1  u2, 2  u3,3  2 x1  2 x1  3 x32 e1

e2

  u   / x1  / x2 x12

2 x1 x2

e3  / x3  0e1  0e 2  2 x2e 3 x33

 2 u   2u1e1   2u2e 2   2u3e 3  2e1  0e 2  6 x3e 3  0  2 x1 u  2 x2  0

0 2 x1 0

0  0  , tr (u)  4 x1  3 x32 3 x32 

(c) u  x22e1  2 x2 x3e 2  4 x12e 3   u  u1,1  u2, 2  u3,3  0  2 x3  0 e1

e2

  u   / x1  / x2 x22

2 x2 x3

e3  / x3  2 x2e1  8 x1e 2  2 x2e 3 4 x12

 2 u   2u1e1   2u2e 2   2u3e 3  2e1  0e 2  8e 3  0 2 x2 u   0 2 x3 8 x1 0

0  2 x2  , tr (u)  2 x3 0 

2.25 (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 v k ) ,i   ijk (u j vk ,i  u j ,i v k )  v k  ijk u j ,i  u j  ijk vk ,i  v  (  u)  u  (  v )      ijk ( ,k ) , j   ijk  ,kj  0 because of symmetry and antisymmetry in jk     ( ,k ) ,k   ,kk   2 

(c)   (  u)  ( ijk u k , j ) ,i   ijk u k , ji  0 , because of symmetry and antisymmetry 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

2.26 Cylindrical coordinates : 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   e θ   ( ru )  r eˆ z z  r  r  r    z  r 

2.27 Spherical coordinates : 1  R ,  2   , 3   x 1  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

( h3 ) 2 

eˆ  cos  cos e 1  sin  cos e 2  sin e 3 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   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     eˆR

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

2.27 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

2.28

 1  R ,  2   ,  3   . . . Orthogonal Coordinate s x1  R sin  cos  , x 2  R sin  sin  , x 3  R cos  x m x n x m x m g ij  gi  g j   mn  i   i  j   j x m x m x1 x1 x 2 x 2 x 3 x 3 g11   1 1  1 1  1 1  sin 2  cos 2   sin 2  sin 2   cos 2   1 1 1         x m x m g 22   R 2 cos 2  cos 2   cos 2  sin 2   sin 2    R 2 2 2   x m x m g 33   R 2 sin 2  sin 2   sin 2  cos 2   0   R 2 sin 2  3 3   g12  g 21  g 23  g 32  g13  g 31  0 1  gij  0  0

0 0  1     2 ij 2 R 0  g  0 1/ R 0    0 0 R 2 sin 2   0 1 / R 2 sin 2    2 x m  i 1 kji   ijk   g km, j  g jm, k  g kj, m  g mi k j m   x 2 1 1 111  g1m ,1  g1m ,1  g11, m  g m1  g11,1  g11,1  g11,1  g 11  0 2 2 1 1 1 1 1 133  g 3m ,1  g1m ,3  g31, m  g m 3  g33,1  g13,3  g 31,3  g 33  2r sin 2   2  2 2 2 2 r sin  R 1 1 332  g 3m ,3  g3m ,3  g33, m  g m 2  g 32 ,3  g 32 ,3  g 33, 2  g 22 2 2 1 1   2 R 2 sin  cos   2   sin  cos  2 R 1 1 1 1 233  g 3m , 2  g 2 m ,3  g 23, m  g m 3  g33, 2  g 23,3  g 23,3  g 33  2 R 2 sin  cos   2  2 2 2 R sin 2  0

0

3.1 (a)

3

3

2.5

2.5

2

2

1.5

1.5

1

1

0.5

0.5

0

0

-0.5 -0.5

(b)

0

0.5

1

1.5

2

2.5

-0.5 -0.5

3

4

4

3.5

3.5

3

3

2.5

2.5

2

2

1.5

1.5

1

1

0.5

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0

0

-0.5 -0.5

0

0.5

1

1.5

2

2.5

3

3.5

4

-0.5 -0.5

(c) 6

6

5

5

4

4

3

3

2

2

1

1

0

0 0

1

2

3

4

5

6

0

0

0

0.5

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1.5

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1.5

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2

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2.5

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4

3

3.5

5

4

6

3.2

  4 x3 t

  x1  2 x2  4tx3    {1, 2, 4t } , x1  2tX1  3tX 2  v1  2 X 1  3 X 2 (a) x2  X 1  2tX 2  v2  2 X 2

x3  X 3  v1  0 D    v    4 x3  2 X 1  3 X 2  4 X 2  2 X 1  7 X 2  4 X 3 Dt t

  2tx1  4 x2  x3    { 2t ,4,1} ,

  2 x1 t

x1  X 1  3tX 2  v1  3 X 2 (b) x2  2tX1  3t 2 X 2  v2  2 X 1  6tX 2

x3  X 3  v3  0 D    v    2 x1  (3 X 2 ) (2t )  (2 X 1  6tX 2 )(4)  10 X 1  30tX 2 Dt t

  x1  4t 2 x2  x3    {1,4t 2 ,1} ,

  8tx2 t

x1  X 1  2 X 2  v1  0 (c) x2  2 X 1  3tX 2  v2  3 X 2

x3  2 X 3  v3  0 D    v    8tx2  (3 X 2 ) (4t 2 )  16tX1  36t 2 X 2 Dt t

3.3

x1  X 1  3t 2 X 2  v1  6tX 2  a1  6 X 2 x2  X 1  2tX 2  v2  2 X 2  a2  0 x3  X 3  v3  0  a3  0 (a) To express in terms of spatial coordinates, invert the motion relations

X 1  x1  3t 2 ( x1  x2 ) /(3t 2  2t ) , X 2  ( x1  x2 ) /(3t 2  2t ) , X 3  x3  v1  6( x1  x2 ) /(3t  2) , v2  2( x1  x2 ) /(3t 2  2t ) , v3  0 a1  6( x1  x2 ) /(3t 2  2t ) , a2  0 , a3  0

x1  2tX1  4 X 2  v1  2 X 1  a1  0 x2  X 1  2t 3 X 2  v2  6t 2 X 2  a2  12tX 2 x3  X 3  v3  0  a3  0 (b) To express in terms of spatial coordinates, invert the motion relations

t3  x2 ) /(t 4  1) X 2  ( x1  2tx2 ) / 4(1  t 4 ) , X 3  x3  2 t3 3 v1  2( x1  x2 ) /(t 4  1) , v2  t 2 ( x1  2tx2 ) /(1  t 4 ) , v3  0 2 2 4 a1  0 , a2  3t ( x1  2tx2 ) /(1  t ) , a3  0 X 1  ( x1

x1  tX1  3 X 2  v1  X 1  a1  0 x2  2t 2 X 1  X 2  v2  4tX1  a2  4 X 1 x3  X 3  v3  0  a3  0 (c) To express in terms of spatial coordinates, invert the motion relations

X 1  ( x1  3x2 ) /(t  6t 2 )  v1  ( x1  3 x2 ) /(t  6t 2 ) , v2  4t ( x1  3 x2 ) /(1  6t ) , v3  0 a1  0 , a2  4( x1  3 x2 ) /(t  6t 2 ) , a3  0

3.4 2 x2 2 x1 0 v  2 x1 x2 e1  3 x2 e 2  e3  v   0 3 0  0 0 0 v From (3.4.4) ai  i  v j vi , j  t (a) a1  0  2 x2 (2 x1 x2 )  2 x1 (3 x2 )  0(1)  4 x1 x22  6 x1 x2 a2  0  0(2 x1 x2 )  3(3 x2 )  0(1)  9

a3  0  0(2 x1 x2 )  0(3 x2 )  0(1)  0

4t 0 0 v  4 x1te1  2 x2 e 2  2t e3  v   0 2 0  0 0 0 v From (3.4.4) ai  i  v j vi , j  t (b) a1  4 x1  4t (4 x1t )  0(2 x2 )  0(2t 2 )  4 x1  16 x1t 2 2





  a  4t  0(4 x t )  0(2 x )  0(2t )  4t

a2  0  0(4 x1t )  2(2 x2 )  0(2t )  4 x2 2

2

3

1

2

2t 0 0 v  2 x1te1  4 x2t e2  3e3  v   0 4t 2 0  0 0 0 v From (3.4.4) ai  i  v j vi , j  t (c) a1  2 x1  2t (2 x1t )  0(4 x2t 2 )  0(3)  2 x1  4 x1t 2 2









a2  8 x2t  0(2 x1t )  4t 2 (4 x2t 2 )  0(3)  8 x2t  16 x2t 4





a3  0  0(2 x1t )  0(4 x2t 2 )  0(3)  0

3.5 x1  X 1  3 X 2  u1  x1  X 1  3 X 2 x2  2 X 1  4 X 2  u 2  x2  X 2  2 X 1  3 X 2

(a)

x3  4 X 1 X 3  u3  x3  X 3  4 X 1 X 3  X 3  0 u  u   2 X  4 X 3

  1 x   3 0 ,F  2 X  0 4 X 1  1 4 X 3 3

0

0  4 0  0 4 X 1  3

x1  X 1  3 X 2 X 3  u1  x1  X 1  3 X 2 X 3 x2  2 X 1  4 X 2  X 3  u 2  x2  X 2  2 X 1  3 X 2  X 3

(b)

x3  6 X 1 X 2  u3  x3  X 3  6 X 1 X 2  X 3  0 u  u   2 X  6 X 2

3X 3 3 6 X1

3X 2   1  x 1  , F   2 X  6 X 2  1 

3X 3 4 6 X1

3X 2  1  0 

x1  3 X 1 X 3  2 X 2  u1  x1  X 1  3 X 1 X 3  2 X 2  X 1 x2  2 X 1  4 X 2 X 3  u 2  x2  X 2  2 X 1  4 X 2 X 3  X 2

(c)

x3  4 X 1  3 X 3  u3  x3  X 3  4 X 1  2 X 3 2 3 X1  3 X 3  1 3 X 3 u  x   u   2 4X3 1 4X 2  , F   2 X  X   4  4 0 2 

2 4X3 0

3X1  4 X 2  3 

3.6

1  0  1 0 0    ( 2) F  0 1 0 , F   0  2 0 0 0 1  0 0 1 1  0 1 0 (1) ( 2 ) dx  F F dX  0 1 0  0  2 0 0 1  0 0 1 0 0 1  ( 2 ) (1) dx  F F dX   0  2 0 0 1  0 0 1 0 0 (1)

0 1  0 dX   0  0 1 0 1  0 dX   0  0 1

 2 2 0 1 2 0

0  dX 1  1dX 1   2 dX 2   0 dX 2     2 dX 2   1  dX 3   dX 3 0  dX 1  1dX 1  1dX 2   0 dX 2     2 dX 2   1  dX 3   dX 3

3.7 x1  X 1  2 X 2

1 2 0  x  x2  2 X 1  X 2  F   2 1 0 X 0 0 1 x3  X 3 Inverse Motion

(a)

X 1  (2 x2  x1 ) / 3

 1 / 3 2 / 3 0 X  X 2  (2 x1  x2 ) / 3  F   2 / 3  1 / 3 0 x   0 X 3  x3 0 1 1 2 0  1 / 3 2 / 3 0 1 0 0 1 Check : FF  2 1 0  2 / 3  1 / 3 0  0 1 0 0 0 1  0 0 1 0 0 1 1

x1  4 X 1  X 2

4 1 0 x  x2  X 1  4 X 2  X 3  F   1 4 1 X 0 0 1 x3  X 3 Inverse Motion

(b)

X 1  (4 x1  x2  x3 ) / 15

 4 / 15  1 / 15 1 / 15  X  X 2  ( x1  4 x2  4 x3 ) / 15  F    1 / 15 4 / 15  4 / 15 x  0 X 3  x3 0 1  4 1 0  4 / 15  1 / 15 1 / 15  1 0 0 1 Check : FF  1 4 1  1 / 15 4 / 15  4 / 15  0 1 0 0 0 1  0 0 1  0 0 1 1

x1  X 1  X 2

1 1 0  x  x2  2 X 1  4 X 2  F   2 4 0 X  4 0 3 x3  4 X 1  3 X 3 Inverse Motion

(c)

X 1  (4 x1  x2 ) / 2

1/ 2  2 X  X 2  ( x2  2 x1 ) / 2 F   1 1/ 2 x   8 / 3 2 / 3 X 3  (8 x1  2 x2  x3 ) / 3  1 / 2 0  1 0 1 1 0   2    1 Check : FF  2 4 0   1 1/ 2 0   0 1 4 0 3  8 / 3 2 / 3 1 / 3 0 0 1

0  0  1 / 3 0 0 1

3.8 2



2



RBM  dx  dX  dX F T F  I dX  0  F T F  I  0 and so F T F  I , taking the transpose gives FF T  I Using (2.5.3)  detF  1 2

2

Likewise if F T F  I  dx  dX  0 ,  RBM Now if F T F  I 

xi xi   mn X m X n

  2 xi xi x  2 xi   xi xi    mn    i  0 (1) X p  X m X n X m X p X n X m X n X p  Interchange indices m  p, n  m, p  n  (1) - (2) 

 2 xi xi x  2 xi  i  0 (2) X p X n X m X p X m X n

 2 xi xi x  2 xi  i 0 X m X p X n X p X m X n

Interchange indices m  p, p  m  (1)  (3)  2

 2 xi xi x  2 xi  i  0 (3) X p X m X n X m X p X n

 2 xi xi  2 xi xi X n  2 xi 0 0 il  0  X p X m X n X p X m X n xl X p X m

 2 xl xl 0  Rip  xl  Rip X p  cl  x  Q (t ) X  c X p X m X p

3.9

F  Fv Fd  det F  J  det Fv det Fd  det Fv





det Fd  1  det J 1 / 3 F  1  J 1 det F  1  1  1

3.10 (a) x1  X 1  3 X 2

1 3 0    2 3 / 2 0 x   1 x2  2 X 1  4 X 2  F   2 4 0 , F   1  1 / 2 0 X  0 0 1  0 x3  X 3 0 1 1 2 0 1 3 0  5 11 0 T C  F F  3 4 0 2 4 0  11 25 0 0 0 1 0 0 1  0 0 1 1 3 0 1 2 0 10 14 0 T B  FF  2 4 0 3 4 0  14 20 0 0 0 1 0 0 1  0 0 1  5 11 0 1 0 0   4 11 0 1 T 1 1 1   E  F F  I  11 25 0  0 1 0  11 24 0 2 2 2 2  0 0 1 0 0 1  0 0 0 1 0   2 3 / 2 0 1 0 0  2 1 1 1  1 T 1  e  I  F F   0 1 0  3 / 2  1 / 2 0  1  1 / 2 0  2 2 2 0 0 1  0 0 1  0 0 1  7 / 2 0 1 0 0   5   4 7 / 2 0 1 1 1    0 1 0   7 / 2 5 / 2 0  7 / 2  3 / 2 0 2 2 2 0 0 1  0  0 0 1 0 1





 

3.10 (b) x1  X 1  3 X 2

1 3 0   0 x   1 x2  2 X 1  4 X 3  F   2 0 4 , F  1 / 3 X  0 0 1   0 x3  X 3 1 2 0 1 3 0 5 3 8  T C  F F  3 0 0 2 0 4  3 9 0  0 4 1 0 0 1  8 0 17  1 3 0 1 2 0 10 2 0 T B  FF  2 0 4 3 0 0   2 20 4 0 0 1  0 4 1  0 4 1  5 3 8  1 0 0  4 1 T 1 1 1   E  F F  I  3 9 0    0 1 0    3 2 2 2 2 8 0 17  0 0 1 8

2  1 / 6 2 / 3 0 1  1/ 2

8 8 0  0 16  1 / 3 0  0 1/ 2  2  1 0 0   0 1 1 1    1 T 1  e  I  F F   0 1 0  1 / 2  1 / 6 0 1 / 3  1 / 6 2 / 3  2 2 2 0 0 1   2 2 / 3 1  0 0 1   1 / 18 4/9  1 / 18  4/9  1 0 0   1/ 9  8/9 1 1 1    0 1 0   1 / 18 10 / 36  10 / 9   1 / 18 26 / 36 10 / 9  2 2 2 0 0 1  4 / 9  10 / 9 49 / 9   4 / 9 10 / 9  40 / 9





3

 

3.11

Since invariants I , II , III are related to I1 , I 2 , I 3 , we choose to work with the later set I1B  trB  tr ( FF T )  tr ( F T F )  trC  I1C I 2 B  trB 2  tr ( BB )  tr ( FF T FF T )  tr ( FCF T )  tr (CC )  I 2C I 3 B  trB 3  tr ( BBB )  tr ( FF T FF T FF T )  tr ( FCCF T )  tr (CCC )  I 3C

3.12 F  RU  VR  F T F  ( RU )T ( RU )  U T RT RU  U T U  U 2 FF T  (VR )(VR )T  VRR T V T  VV T  V 2 FU 1  RUU 1  R , V 1F  V 1VR  R ( RU  VR ) RT  V  RURT Consider the invariants of V and U I1V  tr (V )  tr ( RURT )  tr (U )  I1U I 2V  tr (V 2 )  tr ( RURT RURT )  tr ( RUURT )  tr (UU )  I 2U I 3V  tr (V 3 )  tr ( RURT RURT RURT )  tr ( RUUURT )  tr (UUU )  I 3U Therefore the invariants of V and U are the same and thus the principal values must be identical.

3.13 (a)

2 0 0 x1  2 X 1 , x2  4 X 3 , x3  4 X 2  F  0 0 4 0 4 0 2 0 0 2 0 0  4 0 0  2       2 T U  F F  0 0 4 0 0 4  0 16 0   U  0 0 4 0 0 4 0 0 0 16  0 2 0 0  2 0 0  4 0 0  2       2 T V  FF  0 0 4 0 0 4  0 16 0   V  0 0 4 0 0 4 0 0 0 16  0 0  1 0 0 2 0 0 1 / 2 0    1 R  FU  0 0 4  0 1 / 4 0   0 0 1 0 4 0  0 0 1 / 4 0 1 0 1 0 0 1 0 0 1 0 0 T RR  0 0 1 0 0 1  0 1 0 0 1 0 0 1 0 0 0 1

0 0 4 0 0 4 0 0 4 0 0 4

(b)

 2 0 0 x1  2 X 1 , x2  4 X 1 , x3  4 X 2  F  4 0 0 0 4 0 2 4 0 2 0 0 16 0 0 4       2 T U  F F  0 0 0 4 0 0   0 16 0  U  0 0 4 0 0 4 0  0 0 4 0 2 0 0 2 4 0 4 0 0  2       2 T V  FF  4 0 0 0 0 0  0 16 0   V  0 0 4 0 0 4 0 0 0 16  0 0  0 0 1  2 0 0 1 / 4 0    1 R  FU  4 0 0  0 1 / 4 0   1 0 0 0 4 0  0 0 1 / 2 0 1 0 0 0 1 0 1 0 1 0 0 RR T  1 0 0 0 0 1  0 1 0 0 1 0 1 0 0 0 0 1

0 0 4 0 0 2 0 0 4 0 0 4

(c)

4 0 0 x1  4 X 1 , x2  4 X 2 , x3  2 X 3  F  0 4 0 0 0 2 4 0 0 4 0 0 16 0 0 4       2 T U  F F  0 4 0 0 4 0   0 16 0  U  0 0 0 2 0 0 2  0 0 4 0 4 0 0 4 0 0 16 0 0 4       2 T V  FF  0 4 0 0 4 0   0 16 0  V  0 0 0 2 0 0 2  0 0 4 0 0  1 0 0 4 0 0 1 / 4 0    1 R  FU  0 4 0  0 1 / 4 0   0 1 0 0 0 2  0 0 1 / 2 0 0 1 1 0 0 1 0 0 1 0 0 RR T  0 1 0 0 1 0  0 1 0 0 0 1 0 0 1 0 0 1

0 0 4 0 0 2 0 0 4 0 0 2

3.14 x1  X 1  3 X 2

 4 11 0 1 T 1 (a) From Exercise (3.10) x2  2 X 1  4 X 2  E  F F  I  11 24 0 2 2  0 0 0 x3  X 3





 5  1     1  2 Ei i  1   5  1   0    2 Eij sin  (ij )   1  2 Ei i 1  2 E j j (i )

sin  (12 ) 

2 E12 22  1  2 E11 1  2 E22 5 5

sin  ( 23 ) 

2 E23 2 E31  0 , sin  (31)  0 1  2 E22 1  2 E33 1  2 E33 1  2 E11

x1  X 1  3 X 2

4 3 8  1 T 1 (b) From Exercise (3.10) x2  2 X 1  4 X 3  E  F F  I  3 8 0  2 2 8 0 16  x3  X 3





 5 1     (i )  1  2 Ei i  1   2   17  1   2 Eij sin  (ij )   1  2 Ei i 1  2 E j j sin  (12 ) 

2 E12 3 1   1  2 E11 1  2 E22 3 5 5

sin  ( 23 ) 

2 E23 2 E31 8  0 , sin  (31)   1  2 E22 1  2 E33 1  2 E33 1  2 E11 17 5

3.15

Example 3.5.1 Simple Shear Deformation : 1  0 1 / 2  1      1  1  , dx  1 2  2    2 dx  FdX  0 1 0 1 / 2    2 2 0 0 1  0   0    Example 3.5.2 Extensiona l Deformation : 0  1 / 2   1    1   , dx  1 2  2 0  1 / 2   2 1 2 2  2      3   0  0

1 0 dx  FdX   0  2  0 0

3.16





dx ( i )   (i )  1 dX ( i ) 2

2

dx ( i )  dX ( i )  2ei i dx ( i )   (i ) 

2

1 1 1  2ei i

sin  (ij )  cos (ij ) 

dx ( i )  dx ( j ) dx ( i ) dx ( j )

,i j

dX ( i ) F 1 F 1dX ( j ) T

 sin  (ij ) 

  1dX  (i )

(i )

( j)

2eij 1  2ei i 1  2e j j



 1 dX ( j ) (i  j )

,i j

3.17 (a) x1  X 1  3 X 2

1 3 0    2 3 / 2 0  5 11 0 x     1 x2  2 X 1  4 X 2  F   2 4 0 , F   1  1 / 2 0 , C  11 25 0 X 0 0 1  0  0 0 1 x3  X 3 0 1 5 (i ) (i ) dx  Ci i dX  25 dX ( i )  1  1 0  N1   2 N1  N 2 2        1 T nda  J ( F ) NdA  2 3 / 2  1 / 2 0  N 2  dA  2 (3 / 2) N1  (1 / 2) N 2  dA  0   0 1  N 3  N3 dv  JdV  2dV

(b) x1  X 1  3 X 2

1/ 2  2  1 3 0   0 5 3 8  x     1 x2  2 X 1  4 X 3  F   2 0 4 , F  1 / 3  1 / 6 2 / 3 , C  3 9 0  X 0 0 1   0 8 0 17  x3  X 3 0 1  5 (i ) (i ) dx  Ci i dX   9  dX ( i ) 17  1 / 3 0   N1  (1 / 3) N 2  0        1 T nda  J ( F ) NdA  6 1 / 2  1 / 6 0  N 2  dA  6  (1 / 2) N1  (1 / 6) N 2  dA   2 2 / 3 1  N 3   2 N1  (2 / 3) N 2  N 3  dv  JdV  6dV

3.18 (a) u  x1 x2 e1  3 x2 e 2  x32 e3 x1 / 2 0   x2 1   ij  ui , j  u j , i   x1 / 2 3 0  2  0 0 2 x3  x1 / 2 0  0 1  ij  ui , j  u j , i   x1 / 2 0 0 2  0 0 0









(b) u  2 x1 x32 e1  2 x1 x2 e2  2 x33e3  2 x32  1  ij  ui , j  u j , i   x2 2 2 x x  1 3





 0  1 ij  ui , j  u j , i   x2 2  2 x x 1 3 



(c) u  2





 0  1 ij  ui , j  u j , i   x / x22 2  3/ 2 



2 x1 0

 x2 0 0

2 x1 x3   0  0 

x1 e1  4 x2 x3e 2  3 x1e3 x2

 2 / x2  1  ij  ui , j  u j , i   x1 / x22 2  3/ 2 



2 x1 x3   0  6 x32 

x2



 x1 / x22 4 x3 2 x2  x / x22 0  2 x2

3 / 2  2 x2  0 

 3 / 2  2 x2  0 

3.19

Using Principal Coordinates Volume of Undeformed Element  dV  dx1dx2 dx3 Volume of Deformed Element  dv  (1  e1 )dx1 (1  e2 )dx2 (1  e3 )dx3  dx1dx2 dx3  (e1  e2  e3 )dx1dx2 dx3  terms of O(e 2 , e3 )  (1  )dx1dx2 dx3  (1  )dV  dv  dV  (1  )dV  dV  dV   

dv  dV   kk  I  dV

3.20

1 ˆ ij   ij   kk ij 3 1 I ˆ  ˆ ii   ii   kk ii   ii   ii  0 3 Vanishing of dilatation  ˆ ij   ij 3.21

Principal value problem for strain tensor : ( ij  ij )n j  0 , where  are the principal values 1 1 1   Using ˆ ij   ij   kk ij  (ˆ ij   kk ij  ij )n j  0   ˆ ij  (   kk )ij n j  0 3 3 3   Thus principal value problem for the strain deviator has identical principal directions as 1 the strain tensor itself but with principal values given by    kk 3 3.22 ij ij  ikl  jmp lp , km   kj



lj

im

ip

 km  kp lp , km lm

lp





 ij ( kmlp   kplm )  im ( kjlp   kplj )  ip  kjlm   kmlj ) lp , km  ij (ll , kk   kl, kl )  ll , ij   jk , ki  li , jl   ji, kk If indices k and l are taken to be the same, the first term will drop, and we get ij  (ij , kk   kk, ij  ik , jk   jk , ik )  0 , which is the same as (3.11.2) with k  l   ε        ε   ε  imn  jn, m      ε   pqj imn  jn, mq  imn  pqj  nj , mq  ip

3.23 Starting with the first three compatibility equations, and using x, y, z notation for coordinates 2  2  xy  4 y  4  xy  2 x   y  4 x  2  2 2  2 2 2 (a) xy y 2 x 2 y z x z xyz 2

 2 y z

 2

 2  yz  4 y  4  yz  2 z  4 z  2    2 (b) yz y 2 z 2 x 2 y 2 x 2 yzx 2

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

 2

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

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

Using the last three compatibility 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

3.24

(a) 11  Ax 2 x3 ,  22   Ax1 x3 , 33  Bx3 , 12  13   23  0  211    23 31 12       A  0  0  0 . . . not satisfied    x2x3 x1  x1 x2 x3  x1  strain field is not compatible

(b) 11  Ax 23 ,  22  Ax13 , 12 

3 Ax1 x2 ( x1  x2 ) ,  33  13   23  0 2

 2 11  2  22  2 12  2  6 Ax 2  6 Ax1  3(2 Ax1  2 Ax 2 ) . . . satisfied x1x2 x22 x12 All other comp. equations are satisfied with 0  0.  strain field is compatible

(c) 12  A

x1 2 x1  x22

, 13   A

x2 2 x1  x22

, 11   22   33   23  0

  2 11  2  22  2 12 2  x  A 2 1 2  . . . not satisfied   2  0  0  2 2 2  x1x2 x1x2  x1  x2  x2 x1  strain field is not compatible

3.25

11  Ax1 ,  22  Ax 2 , 12  Bx1 11 

u1 A  Ax1  u1  x12  f ( x2 ) , arbitrary function x1 2

 22 

u 2 A  Ax 2  u 2  x22  f ( x2 ) , arbitrary function x2 2

12 

1  u1 u 2  df ( x2 ) dg ( x1 ) df ( x2 ) dg ( x1 )    Bx1     Bx1    Bx1  constant 2  x2 x1  dx2 dx1 dx2 dx1

 f ( x2 )  ax2  b , g ( x1 )  Bx12  ax1  c , with a,b,c all arbitrary constants  u1 

A 2 A x1  ax2  b , u 2  x22  Bx12  ax1  c 2 2

Note that portions from f ( x2 ) and g ( x1 ) are rigid body motion terms.

3.26

u1  o x2  uo , u 2  o x1  vo 1 (ui , j  u j , i )  2 u u u 11  1  0 ,  22  2  0 ,  33  3  0 x1 x2 x3  ij 

12 

u  1  u1 u 2  1  u 1  u u     0 ,  23   2  3   0 ,  31   3  1   0  2  x2 x1  2  x3 x2  2  x1 x3 

Thus all strains vanish  rigid - body motion

3.27

(a) x1  2t 2 X 1  4tX 2 , x2  4 X 2 , x3  2tX 3 Inverse Motion : X 1  ( x1  x2 t ) / 2t 2 , X 2  x2 / 4 , X 3  x3 / 2t 2t 2  F  0  0 

4t 4 0

0  0  , J  det F  16t 3 2t 

4tX 1  4 X 2  2( x1  x2 t ) / t  x2  D  , v ( x, t )    Velocity : v ( X , t )  x ( X , t )   0 0    Dt  2 X 3    x3 / t 2 / t  2 0  2 / t  1 0  0  1 0  v  1 1    T T L  0 0 0  , D  L  L   1 0 0  , W  L  L  1 0 0 x 2 2  0  0 0 0 0 0 1 / t  0 1 / t  0  1 1 1 wi    ijkW jk   ( i12W12   i 21W21 )   ( i12   i 21 )  0  e3 2 2 2 1



2t 2  E  F T DF   4t  0 

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

4t  2t 2    8  4t 0 0 

0 2t 2  0  0 2  0

J  Jvk ,k  JLkk  16t 3 (3 / t )  48t 2 3.27



4t 4 0

0   8t 3   0   16t 2 2t   0



4t 4 0

0  0 2t   8t 2 16t 0

0  0 4t 



(b) x1  2tX 1  4t 2 X 2 , x2  4 X 2  2tX1 , x3  X 3 Inverse Motion : X 1   2t 4t 2  F   2t 4  0 0 

x1 t ( x1  x2 )  , X 2  ( x1  x2 ) / 4(t 2  1) , X 3  x3 2 2t 2(t  1)

0  0 , J  det F  8t  8t 3  8t (1  t 2 ) 1

 x1 t ( x1  x2 )   t  (t 2  1)  2 X  8 tX  1 2   D x1 t ( x1  x2 )     Velocity : v ( X , t )  x ( X , t )    2 X 1  , v ( x, t )    2  t Dt (t  1)    0   0     t t t 1 t 1   1    2 0  t  t 2  1 t 2  1 0  t  t2 1 2t t  1   1  v  1 t t 1 t t T   , L    2 0 , D  L  L    0 2 x  t t  1 t 2  1  2 t2 1  2t t  1  0 0 0 0 0 0       1 / t 0  0 1  T W  L  L   1 / t 0 0 2  0 0 0  0  1 1 1 wi    ijkW jk   ( i12W12   i 21W21 )   ( i12 / t   i 21 / t )   0   (1 / t )e3 2 2 2  1 / t  t 1 t  1   2   2 0   2t 4t 2 0 2t t  1  2t  2t 0  t t  1    1 t t E  F T DF  4t 2 4 0   2 0  2t 4 0 2 t 1  2t t  1   0 0 1  0 1 0 0 0  0  







 8t 16t 2  4 0    12t 3  4 32t 3  8 0  0 0 0  1 2t J  Jvk , k  JLkk  8t (1  t 2 )(  2 )  8(1  3t 2 ) t t 1

3.27



(c) x1  2tX 1  4 X 2 , x 2  4t 2 X 2  2tX 1 , x3  6t 2 X 3 Inverse Motion : X 1   2t F   2t  0

4 4t

2

0

x1 ( x1  x 2 )  , X 2  ( x1  x 2 ) / 4(t 2  1) , X 3  x3 / 6t 2 2 2t 2t (t  1)

0  0  , J  det F  48t 3 (1  t 2 ) 6t 2 

  x1t 2  x 2 )   2 t ( t  1 )   2X1   2 2   x t  2 t x  x D 2 2 Velocity : v ( X , t )  x ( X , t )   2 X 1  8tX 2  , v ( x , t )   1  2 Dt t (t  1)    12tX 3  2 x3     t    0 2t (t  1)   1 / 2t 0  0 2  2t  1  0  , W  1 / 2t 0 0 2 t (t  1)   0 0 0 2 0 t   0  1 1 1 wi    ijkW jk   ( i12W12   i 21W21 )   ( i12 / 2t   i 21 / 2t )   0   (1 / 2t )e 3 2 2 2 1 / 2t  1  t  t 2  1  t (t 2  1)  v  t 2t 2  1 L  2 x t  1 t (t 2  1)   0 0 

2t T E  F DF   4  0

 t  0  2  t 1   1 t 2  0 , D   2  2t (t  1)  2  0   t 

 t  2  2t 0   t  12  1 t 4t 2 0   2t (t 2  1) 0 6t 2    0  

  8t    4  12t 2   0 

4t 

1 t 2 2t (t  1) 2t 2  1 t (t 2  1)

16t 4  12t 2  4

t 1 4t (8t 4  6t 2  2) 8 t2 1 0 2

2

0

1 t 2 2

 0   2t  0   2t  0 2  t 

4 4t

2

0

0  0  6t 2 

 0    0   72t 3  



 t 2t 2  1 2  J  Jv k ,k  JLkk  48t 3 (1  t 2 ) 2  2    48t 2 t 2  4t 2  3  t  1 t (t  1) t 

3.28



For objective second order tensors A*  QAQ T or A  Q T A * Q Taking the time deriative  A *  Q AQ T  QA Q T  QAQ T Note the previous relations for the spin tensor W *  QWQ T  Q Q T , W  W T  A *  Q Q T A * QA Q T  A * QQ T  (W * QWQ T ) A * QA Q T  A * (W * QWQ T )  W * A * QWAQ T  QA Q T  A * W * QAWQ T  A * -W * A *  A * W *  Q ( A  WA  AW )Q T o

 A  A  WA  AW is objective

3.29

C t ()  Ft ()T Ft () and Ft ()  F () F 1 (t )  1

Ft* ()  F * () F * (t )  QF () F 1 (t )Q T  QFt ()Q T  C t* ()  Ft* ()T Ft* ()  QFt ()T Q T QFt ()Q T  QFt ()T Ft ()Q T  QCt ()Q T Bt* ()  Ft* () Ft* ()T  QFt ()Q T QFt ()T Q T  QFt () Ft ()T Q T  QBt ()Q T

3.30 (a) x1  2 X 1  4tX 2 , x2  4 X 2 , x3  2tX 3 0 0  1 / 2  t / 2   1 0  F   0 1/ 4 0   0 2t  0 1 / 2t  0  1  t   0  2 4 0  1 / 2  t / 2    1 Ft ()  F () F (t )  0 4 0   0 1/ 4 0   0 1 0  0 0 2  0 0 1 / 2t  0 0  / t  2 4t F  0 4 0 0

0 0  1  t   0   1 t  0   1      2 C t ()  Ft () Ft ()   t   1 0  0 1 0     t   (  t  )  1 0   0 0  / t  0 0  / t   0 0  2 / t 2  T

0 0   ( t   ) 2  1  t   0  1  t   0   1       T Bt ()  Ft () Ft ()  0 1 0   t   1 0     t   1 0  0 0  / t   0 0  / t   0 0  2 / t 2 

3.30

(b) x1  2tX 1  X 2 , x2  4 X 2  2tX 1 , x3  X 3  2t 1 0 2 / 5t   1 F   2t 4 0  F   1 / 5  0 0 1  0

 1 / 10t 0 1/ 5 0 0 1

 1  4 1     0  5t 5  2 1 0 2 / 5t  1 / 10t 0  5t 5  4  4  4 Ft ()  F () F 1 (t )   2 4 0  1 / 5 1/ 5 0     0  5t 5 5t 5   0 0 1  0 0 1  0 0 1   4 4  1  4 1   4 1   5t  5  5t  5 0  5t  5  5t  5 0   1   4 4   4  4 C t ()  Ft ()T Ft ()     0     0 5t 5  5t 5   5t 5 5t 5  0 1  0 0 1  0     2 2   4 1   4 4   4 1   1   4 4   4           0               5 t 5 5 t 5 5 t 5 5 t 5 5 t 5  5t 5             2 2  4 1   1   4 4   4     1   4               0         5t 5   5t 5   5t 5  5t 5   5t 5  5t 5   0 0 1       1 4 4  4 1   4 1   5t  5  5t  5 0  5t  5  5t  5 0  4 4   1   4  4 Bt ()  Ft () Ft ()T     0     0 5t 5  5t 5 5t 5   5t 5  0 0 1 0 0 1       2 2   4 1    1   4 1  4 4    1   4            0              5t 5   5t 5   5t 5  5t 5   5t 5  5t 5    2 2  4 1  4 4    1   4    4 4    4                    0      5t 5   5t 5   5t 5  5t 5   5t 5  5t 5   0 0 1     

3.30

(c) x1  2tX 1  4 X 2 , x2  4 X 2  2tX 1 , x3  6t 2 X 3 0  0  2t 4 1 / 4t 1 / 4t    1 F  2t  4 0   F   1 / 8  1 / 8 0   0 0 6t 2   0 0 1 / 6t 2   1  1   0  1 / 4t 1 / 4t 0   2t 2 2t 2  2 4  1  1 Ft ()  F () F 1 (t )  2  4 0   1 / 8  1 / 8 0       2t 2 2t 2 2 2  0 0 6   0 0 1 / 6t   0 0   1  1   1  1  0    0   2t  2 2t  2 2t 2 2t 2  1  1   1  1  C t ()  Ft ()T Ft ()     0    0   2t 2 2t 2   2t 2 2t 2   0 0 2 / t 2   0 0 2 / t 2        1  2   1  2         2t 2   2t 2     1   1    2       2t 2  2t 2   0    1  2t  2  1 Bt ()  Ft () Ft ()T     2t 2  0    1  2   1  2         2t 2   2t 2     1   1    2       2t 2  2t 2   0  

  1   1  2      2t 2  2t 2  2 2   1   1        2t 2   2t 2  0

 0    0   4 / t 4   

 1  2t 2  1  2t 2 0

 1  2t 2  1  2t 2 0

  1 0   2t 2   1 0     2t 2 2 / t 2   0  

  1   1  2      2t 2  2t 2  2 2   1   1        2t 2   2t 2  0

 0   0   2 / t 2  

 0   0   2 / t 2  

 0    0   C t () , . . . since Ft () Ft ()T  4 / t 4   

3.31

Z , AB For reference rectangular domain : X 1  X  X 2 , Y1  Y  Y2  Using deformation : r 2  2 AX ,   BY , z 

2 AX 1  r  2 AX 2 , BY1    BY2 (circular ring sector in current deformed configuration) Using relation (3.17.11)   r  X   F  r  X  z  X det F 

r Y  r Y z Y

r   A Z      2 AX  r 0 Z   z   0  Z  

A 1   rB  1 r  AB 

0 rB 0

 0  A r  0   0  1  0 AB  

0 rB 0

 0  0  1   AB 

4.1

n  (e1  2e2  e3 ) / 6  2 1 1  1  5 1  1      1 0 2 2  3 (a) t  Tn  6 6  1 2 0 1  5

n  (2e1  e 2  e3 ) / 6 (b)

1 0 1   2  3 1  1      t  Tn  0 2 2  1   4 6 6  1 2 0 1  4 n  (e1  2e2 ) / 3

(c)

 2 1 0  1   4 1  1      t  Tn  1 4 2  2  9 3 3  0 2 0 0 4

4.2

 cos 0 cos 90 cos 90  1 0 0 Q  cos( xi , x j )  cos 90 cos 90 cos 0   0 0 1 cos 90 cos180 cos 90  0  1 0 1  1 1 0 0   2 1 1  1 0 0   2        T (a) T   QTQ  0 0 1 1 0 2 0 0  1   1 0  2 0  1 0 1 2 0 0 1 0   1  2 0  0 1 0 0 1 0 1  1 0 0  1 1        T (b) T   QTQ  0 0 1 0 2 2 0 0  1  1 0  2 0  1 0 1 2 0 0 1 0  0  2 2  0  1 1 0 0   2 1 0  1 0 0   2        T (c) T   QTQ  0 0 1 1 4 2 0 0  1   0 0  2 0  1 0 0 2 0 0 1 0   1  2 4 

4.3  cos  sin   2 - D Rotation Matrix : Q     sin  cos   cos  sin   T11 T12  cos   sin  T   QTQ T       sin  cos  T12 T22   sin  cos   T11  T11 cos 2   T22 sin 2   2T12 sin  cos  T22  T11 sin 2   T22 cos 2   2T12 sin  cos  T12  T11 sin  cos   T22 sin  cos   T12 (cos2   sin 2 ) Using trig identites : sin  cos  

1 1 1 sin 2 , sin 2   (1  cos 2) , cos 2   (1  cos 2)  2 2 2

T11  T22 T11  T22  cos 2  T12 sin 2 2 2 T  T22 T11  T22 T22  11  cos 2  T12 sin 2 2 2 T  T11 T12  22 sin 2  T12 cos 2 2 T11 

4.4 T

Tr

Trr

Tr

x2  r  x1 Since r corresponds to x1 and  corresponds to x2 T11  Trr  T11 cos 2   T22 sin 2   2T12 sin  cos  T22  T  T11 sin 2   T22 cos 2   2T12 sin  cos  T12  Tr  T11 sin  cos   T22 sin  cos   T12 (cos2   sin 2 )

4.5 cos   sin  2 - D Rotation Matrix : Q     sin  cos   cos   sin  Trr Tr   cos  sin   T   QTQT       sin  cos   Tr T   sin  cos  T11  Trr cos 2   T sin 2   2Tr sin  cos  T22  Trr sin 2   T cos 2   2Tr sin  cos  T12  Trr sin  cos   T sin  cos   Tr (cos2   sin 2 )

4.6 T11 T12 Plane stress Tij  T12 T22  0 0 det(Tij  Tij )  0 

0 0 0

T11  T

T12

T12

T22  T

0

 T 2  (T11  T22 )T  T11T22  T122  0 T1, 2 



1 (T11  T22 )  (T11  T22 ) 2  4(T11T22  T122 ) 2 2

T  T22 T T   11   11 22   Txy2 ) 2 2   2

 max 

T1  T2 T T    11 22   Txy2 ) 2 2  



4.7





1 (T11  T22 )  (T11  T22 ) 2  4(T11T22  T122 ) 2 1 1 2 1 1 2  I1  I1  4 I 2  I1  I1  I 2 (or could start with 2 - D Characteristic Equation) 2 4 2 4

T1, 2 





T1  T2 1 2  I1  I 2 2 4  the principal and max shear stresses are functions of the invariants S max 

and thus must be independent of the coordinate system

4.8 1 (T1  T3 ) in the equation of circle : S 2  ( N  T3 )( N  T1 )  0  2 1 1 (T1  T3 ) 2  ( N  T3 )( N  T1 )  0  N  (T1  T3 ) 4 2 Using (4.4.10)  Using S max 

1 1  1  (T1  T3 ) 2   (T1  T3 )  T2  (T1  T3 )  T3  S  ( N  T2 )( N  T3 ) 4 2  2 1 n12   (T1  T2 )(T1  T3 ) (T1  T2 )(T1  T3 ) 2 2

n22 

S 2  ( N  T3 )( N  T1 ) 0  0 (T2  T3 )(T2  T1 ) (T2  T3 )(T2  T1 )

1 1  1  (T1  T3 ) 2   (T1  T3 )  T1  (T1  T3 )  T2  S  ( N  T1 )( N  T2 ) 4 2  2 1  n32   (T3  T1 )(T3  T2 ) (T3  T1 )(T3  T2 ) 2 2

n

1 (e1  e3 ) 2

4.9 3 1 1  Tij  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 transformation matrix  0  1 / 2 1 / 2   Using standard tensor transformation theory  2/ 6 1 / 6 1 / 6  3 1 1   2 / 6 1/ 6 1/ 6     T   QTQ T   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

4.10 0  0  Tij    0 0 0 0 0 T



0

det[Tij  T ij ]  

T

0  T ( 2  T 2 )  0

0

0

T

 T1   , T2  0 , T3   T1   Case : (Tij  T1ij )n (j1)  0  n (1)  1 / 2 (1,1, 0) T2  0 Case : (Tij  T2  ij )n (j2)  0  n ( 2)   (0 , 0 ,1) T3    Case : (Tij  T3 ij ) n (j3)  0  n (3)  1 / 2 (1,  1, 0) For the octahedral plane, n1  n2  n3  1 / 3 1 1 2 2 t  T12 n12  T22 n22  T32 n32  (T12  T22  T32 )  ( 2   2 )   2 3 3 3 1   N  T1n12  T2 n22  T3 n32  (T1  T2  T3 )    0 3 3 3 2 2 S  t  N2    6 / 3 3

4.11

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





T   (Tij  Tij )n j  0  Pf ij ( xk )  T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 T1, 2,3

 g1, 2,3 ( xk )  T1, 2,3  Pg1, 2,3 ( xk ) P The principal directions follow from

Pfij ( xk )  T1,2,3ij n (j1,2,3)  0  Pf ij ( xk )  Pg1,2,3ij n(j1,2,3)  0   fij ( xk )  g1,2,3ij n(j1,2,3)  0

and thus the principal directions will not depend on P

4.12 (a)

1 1  2 1 1  1 0 0 4 / 3 2 2 ~ 1 ~     Tij  1 0 2  Tij  Tkk ij  ij  0 1 0 , Tˆij  Tij  Tij   1  2 / 3 2  3 3 3 1 2 0 0 0 1  1 2  2 / 3

1 0 1 1 0 0 0 0 1  1 ~ ~       (b) Tij  0 2 2  Tij  Tkk ij  ij  0 1 0 , Tˆij  Tij  Tij  0 1 2  3 1 2 0 0 0 1 1 2  1 2 1 0 1 0 0 0 1 0  ~ 1 ~       ˆ (c) Tij  1 4 2  Tij  Tkk ij  2ij  20 1 0 , Tij  Tij  Tij  1 2 2  3 0 2 0 0 0 1 0 2  2

4.13 The characteristic equation for the deviatoric stress tensor is given by det[Tˆij  Tˆij ]  0.

1 Since Tˆij  Tij  Tkk  ij , this equation can be written as 3 1 1 det[Tij  Tkk ij  Tˆij ]  det[Tij  ( Tkk  Tˆ )ij ]  0 .

3 3 However, the characteristic equation for the stress tensor Tij is det[Tij  Tij ]  0, and thus 1 1 it follows that Ti  Tkk  Tˆi or Tˆi  Ti  Tkk .

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

4.14





1 ˆ ˆ 1 1 2 2 2 2 TiiT jj  Tˆij Tˆij   Tˆij Tˆij  - (Tˆ 11 + Tˆ 222 + Tˆ 33 ) - T 12 - T 223 - T 31 2 2 2 1 2 2 2 2 2 2 Tˆ 11 + Tˆ 22 + Tˆ 33  [(Tˆ 11 - Tˆ 22 ) +(Tˆ 22 - Tˆ 33 ) +(Tˆ 33 - Tˆ 11 ) ]  [Tˆ 11 Tˆ 22  Tˆ 22 Tˆ 33  Tˆ 33 Tˆ 11] 2 1 2 2 Tˆii  0  Tˆ 11 Tˆ 22  Tˆ 22 Tˆ 33  Tˆ 33 Tˆ 11   Tˆ 11 + Tˆ 222 + Tˆ 33  2 1 2 2 2 2 2 2 Tˆ 11 + Tˆ 22 + Tˆ 33  [(Tˆ 11 - Tˆ 22 ) +(Tˆ 22 - Tˆ 33 ) +(Tˆ 33 - Tˆ 11 ) ] 3 1 2 2  II Tˆ = - [(Tˆ 11 - Tˆ 22 ) 2 +(Tˆ 22 - Tˆ 33 ) 2 +(Tˆ 33 - Tˆ 11 ) 2] - T 12 - T 223 - T 31 6 1 2 2 = - [(T 11 - T 22 ) 2 +(T 22 - T 33) 2 +(T 33 - T 11 ) 2] - T 12 - T 223 - T 31 6 II Tˆ 





4.15 To construct solution use previously developed relations : Tii  I T ; Tij Tij  I T2  2 IIT ; Tij T jkTki  trT 3  3IIIT  3IIT I T  I T3 1 1 1 1   IITˆ   Tˆij Tˆij    Tij  ij I T  Tij   ij I T  2 2 3 3   1 2 1  1    Tij Tij  I T2  I T2   IIT  I T2 2 3 3  3 Using the Cayley - Hamilton Theorem for the deviatoric stress tensor  Tˆ 3  IITˆ Tˆ  IIITˆ 1  0 Taking the trace of this relation  1 1 1 1 1    IIITˆ  trTˆ 3   Tij  ij I T  T jk   jk I T  Tki   ki I T  3 3 3 3 3    1 1 3 3   Tij T jkTki  Tij Tij I T  I T3  IT  3 3 27  1 1 3 3   3IIIT  3IIT I T  I T3  ( I T2  2 IIT ) I T  I T3  IT  3 3 27  2 3 1  I T  I T IIT  IIIT 27 3

4.16  1 1 1  Using principal coordinates with n   , ,   3 3 3 1 1 1  oct  Tij ni n j  (T1  T2  T3 )  Tkk  I1 3 3 3 2 2 2 2  oct | t |  N  Tij n jTik nk  (Tij ni n j ) 1 1  (T12  T22  T32 )  (T1  T2  T3 ) 2 3 9 1  2(T12  T22  T32 )  2(T1T2  T2T3  T3T1 ) 9 1 1  2( I12  2 I 2 )  2 I 2  2 I12  6 I 2 9 9 1  (T1  T2 ) 2  (T2  T3 ) 2  (T3  T1 ) 2 9 Next using given x1,x2 ,x3 axes

  

 







1 1 1  oct  Tkk  I1  (T11  T22  T33 ) 3 3 3 1 2  oct  Tij n jTik nk  (Tij ni n j ) 2  2 I12  6 I 2 9 1  (T11  T22 ) 2  (T22  T33 ) 2  (T33  T11 ) 2  6T122  6T232  6T312 9









4.17

 e   vonMises 

1 [(T11  T22 ) 2  (T22  T33 ) 2  (T33  T11 ) 2  6(T122  T232  T312 )]1 / 2 2

1  oct  Tkk ,  oct  ( 2 / 3) e 3 2 1 1  (a) T  1 0 2  1 2 0 1 2 e  [(2  0) 2  (0  0) 2  (0  2) 2  6(12  2 2  1)]1 / 2  22 ,  oct  ,  oct  44 / 3 3 2 1 0 1  (b) T  0 2 2  1 2 0 1 e  [(1  0) 2  (2  0) 2  (0  1) 2  6(0 2  2 2  12 )]1 / 2  3 2 ,  oct  1 ,  oct  2 2 2 1 0 (a) T  1 4 2  0 2 0 1 e  [(2  4) 2  (4  0) 2  (0  2) 2  6(12  2 2  0 2 )]1 / 2  3 3 ,  oct  2 ,  oct  6 2

4.18 T11  

2 Px12 x2 2 Px23 2 Px1 x22 , T   , T   22 12 ( x12  x22 ) 2 ( x12  x22 ) 2 ( x12  x22 ) 2 2

T T   max   11 22   T122 )  2  

2

MATLAB Code and Plot :

% Exercise 4.18 % 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);

2

  Px2 ( x12  x22 )    2 Px1 x22 )  Px2        ( x 2  x 2 ) 2   ( x 2  x 2 ) 2  ( x12  x22 ) 1 2 1 2    

4.19

3Px1 x2 N x22  3P   T11    , T22  0 , T12   1  2c 4c  c 2  2c 3 Top Boundary ( x2  c)  T22  T12  0 Bottom Boundary ( x2  c)  T22  T12  0 Left Boundary ( x1  L)  T11  

3PLx2 N x22  3P     , T   1  12 2c 4c  c 2  2c 3

c

Resultant Normal Force :  T11 dx2  N -c

c

Resultant Vertical Force :  T12 dx2   P -c

c

Resultant Moment :  T11 x2 dx2   PL , -c

Right Boundary ( x1  0)  T11 

x2  N 3P  1  2  , T12   2c 4c  c 2 

c

Resultant Normal Force :  T11 dx2  N -c

c

Resultant Vertical Force :  T12 dx2   P -c

c

Resultant Moment :  T11 x2 dx2  0 -c

These boundary tractions indicate that the problem is a cantilever beam with right end shear loading P and normal loading N

4.20

Extensiona l deformation x1  1 X 1 , x2   2 X 2 , x3   3 X 3  1 0 x  F  0 2 X   0 0

0 0 0  1 / 1 X   1 0,F   0 1/ 2 0  , J  1 2  3 x  0  3  0 1 /  3  0 0  2 1 1  1 / 1  2 / 1 1 /  2     o 1 T (a) PK1 Tensor : T  JT ( F )  1 2  3 1 0 2  0 1/ 2 0   1 2  3 1 / 1 0 1 2 0  0 1 / 1 2 /  2 0 1 /  3  0 0  2 / 1 1 /  2 1 /  3  1 / 1  1 o PK2 Tensor : S  F T  1 2  3  0 1/ 2 0  1 / 1 0 2 /  3   0 0 1 /  3  1 / 1 2 /  2 0   2 / 21   1 2  3 1 / 1 2 1 / 1 3 

1 / 1 2 0 2 /  2 3

1 / 1 3   2 /  2 3  0 

0 0  0 1 0 1  1 / 1 1 / 1      (b) PK1 Tensor : T  JT ( F )  1 2  3 0 2 2  0 1/ 2 0   1 2  3  0 2 / 2 0 1 /  3  1 2 0  0 1 / 1 2 /  2 0 0  1 / 1 0 1 / 3  1 / 1    1 o PK2 Tensor : S  F T  1 2  3  0 1/ 2 0  0 2 /  2 2 /  3   0 0 1 /  3  1 / 1 2 /  2 0  o

1 T

 1 / 21   1 2  3  0 1 / 1 3 

0 2 / 22 2 /  2 3

1 / 1 2 4 / 22 2 /  2 3

1 / 3  2 /  3  0 

1 / 1 3   2 /  2 3  0 

0 0  2 1 0 1 / 1  2 / 1 1 /  2     (c) PK1 Tensor : T o  JT ( F 1 )T  1 2  3 1 4 2  0 1/ 2 0   1 2  3 1 / 1 4 /  2 0 2 0  0  0 0 1 /  3  2 / 2 0 0  2 / 1 1 /  2 0  1 / 1    1 o PK2 Tensor : S  F T  1 2  3  0 1/ 2 0  1 / 1 4 /  2 2 /  3   0 0 1 /  3   0 2 / 2 0   2 / 21   1 2  3 1 / 1 2  0 

1 / 3  2 /  3  0 

  2 /  2 3  0  0

0  2 /  3  0 

4.21

Deformation x1  2 X 1 , x2  X 2 / 2 , x3  X 3 / 2  0 0  2 1 / 2 F  0 1 / 2 0  , F 1   0 0  0 0 1 / 2 

0 2 0

0  0  , det F  J  1 2 

From Nanson' s Formula (3.7.3) nda  J ( F 1 )T NdA , with n  N  e1  1 / 2 da   0  0

0  1 1 1 2 0  0 dA  da  dA  2 2 0 2  0 T11 0 0 Given Cauchy Stress T   0 0 0  0 0 0 0  T11 / 2 0 0 T11 0 0 1 / 2 0    o 1 T PK1 Stress T  JT ( F )   0 0 0  0 2 0    0 0 0  0 0 0  0 0 2   0 0 0 0  T11 / 2 0 0 T11 / 4 0 0 1 / 2 0  1 o PK2 Stress S  F T   0 2 0   0 0 0   0 0 0  0 0 2   0 0 0  0 0 0 0

4.22

Deformation x1  X 1 , x2  X 2 , x3  X 3   0 0  1 /  0 0   1 F   0 1 0 , F   0 1 0 , det F  J    0 0 1  0 0 1 T11 0 0 Given Cauchy Stress T   0 0 0  0 0 0 T11 0 0 1 /  0 0 T11 0 0 o 1 T PK1 Stress T  JT ( F )    0 0 0  0 1 0   0 0 0  0 0 0  0 0 1  0 0 0 1 /  0 0 T11 0 0 T11 /  0 0 1 o PK2 Stress S  F T   0 1 0  0 0 0   0 0 0  0 0 1  0 0 0  0 0 0

5.1 From (5.1.2)

D D G dv   (G ) dv   G vk ,k dv  Dt R Dt R R m

  [ R

  R

DG DG   G  Gvk ,k ] dv   [  (  vk ,k )G ] dv Dt Dt R

DG dv Dt

5.2

    (v )  0 t Using (2.18.20) 3 and including the z - component  1  1   (rvr )  (v )  (v z ) r r r  z   1          (v )    (rvr )  (v )  r (v z )   0 t t r  r  z    (v ) 

5.3

Given v      v      0 . . . from identity (2.17.11) 6     (v )  0 and with constant mass density    v  0 t Combining these results gives   v       2   0 Conservation of mass

5.4 (a)

v1  a(t ) x1 , v2  b(t ) x2 , v3  c(t ) x3   v  a(t )  b(t )  c(t )  0 does satsify

k 2 x2 , v3  0 (b) 2   v  kx2  kx2  0 does satsify v1  kx1 x2 , v2  

(c)

v1  2k ( x12  x22 ) , v2  2kx1 x2 , v3  kx1 x2   v  4kx1  2kx1  0 does not satsify

5.5 Conservation of mass with constant mass density    v  0 For the case with v1 

             0 , v2   , v  x2 x1 x1  x2  x2  x1 

and so conservation of mass is identically satisfied for all continuous  functions

5.6

 v  T  b  a    v  v   t   v  With T   pI   p  b  a    v  v  or  t  1 v  p  b  a   v  v  t

5.7

 a 2  2 x1  (a) T   ax1 x2   0 

 0  (b) T   x3  x2 ax2  (c) T  bx3  0

 ax1 x2 a 2 x2 2 0

 0 T1 j , j  ax1  ax1  0  0  0  T2 j , j   ax2  ax2  0  0  T3 j , j  0  0  0  0 0 

 x3 x2  T1 j , j  0  0  0  0  0 0   T2 j , j  0  0  0  0 0 0  T3 j , j  0  0  0  0 bx3 ax1 0

0 T1 j , j  0  0  0  0  0  T2 j , j  0  0  0  0 0 T3 j , j  0  0  0  0

5.8

Plane stress : T11  T11 ( x1 , x2 ) , T22  T22 ( x1 , x2 ) , T12  T12 ( x1 , x2 ) , T13  T23  T33  0 For this case the equilbrium equations with no body forces reduces to T11 T21 T T   0 , 12  22  0 x1 x2 x1 x2  2  2  2 Now with T11  2 , T22  2 , T12    x1x2 x2 x1 T11 T21    2     2     0   x1 x2 x1  x22  x2  x1x2  T12 T22    2     2   0     x1 x2 x1  x1x2  x2  x12  Thus equilibrium equations are satisfied identically and the stress form is self - equilibriated

5.9 For the case with v  0 and no  dependence, (5.5.3)  Trr Trz 1 v v   v   (Trr  T )  br   r  vr r  v z r  r z r r z   t Tr Tz 2   Tr  b  0 r z r Trz Tzz 1 v v   v   Trz  bz   z  vr z  v z z  r z r r z   t

5.10

For the case with v  v z  0 and no  and z dependence, (5.5.3)  Trr 1 v   v  (Trr  T )  br   r  vr r  r r r   t Tr 2  Tr  b  0 r r Trz 1  Trz  bz  0 r r

5.11 Relation (4.7.6)  Tij  J 1 F jkTiko Assuming symmetry of the Cauchy stress (5.3.5)  Tij , j  bi  ai  ( J 1 F jkTiko ) , j  bi  ai (J

1

 F jkTiko ) , j  

F jk  Tiko   F jk  o   Tik  det F  x  x det F j    j 

 F jk  Tiko  F jk  Tiko X m  F jk  Tiko 1 1 Tiko    First term :      Fmj  det F  x det F  X  x det F  X J X k j m j m         F jk  1   x j  1        F jk det F  Second term : 2 x j  det F  det F x j  X k  (det F ) x j 1   x j  X m 1    det F  X m   F jk 2 det F X m  X k  x j x j (det F ) X m 1   x j  X m 1    det F Fmj1   F jk 2 det F X m  X k  x j  X (det F ) m 1   x j  X m 1    det F    2 det F X m  X k  x j (det F ) X k 1   x j  X m 1      (det F ) Fmj1 F jm . . . using (2.5.11) 2 det F X m  X k  x j (det F ) X k 1   x j  X m 1 X m   x j     0   det F X m  X k  x j det F x j X k  X m 

 

1

 Tij , j  bi  ai  ( J Tijo X j

F jkTiko ) , j  bi  ai

Tiko 1 Tiko   bi  ai   Jbi  Jai  J X k X k

  o Bi   o ai

5.12 Sress Power   Tij Dij dv R

 Tij Dij dv   Tij Lij dv   ( J F jkTik )( Fim Fmj ) JdV    mk Tik Fim dV   Tik Fik dV 1

R

R

1

o

o

Ro

Ro

o

Ro

 Tij Dij dv   ( J Fik F jl S kl )( Fmi E mn Fnj ) JdV    mk  nl S kl E mn dV   S mn E mn dV 1

R

1

Ro

Ro

  Tij Dij dv   R

1

Ro

Tijo Fij dV 

 Sij E ij dV Ro

Ro

5.13 x1  2tX 1  4t 2 X 2

(a) x2  4tX 2 x3  X 3

T11 T12 Tij  T12 T22  0 0

0 0 0

Velocity : v1  2 X 1  8tX 2 , v2  4 X 2 , v3  0 x1 x2 x  , X 2  2 , X 3  x3 2t 2 4t x x Spatial Velocity : v1  1  x2 , v2  2 , v3  0 t t 1 / t 1 0 1 / t 1 / 2 0    L   0 1 / t 0  D  1 / 2 1 / t 0  0  0 0 0 0 0 Inverse Motion : X 1 

 Tij Dij 

T11  T22  T12 t

x1  2tX 1

(b) x2  tX 1  2tX 2 x3  X 3

T11 T12 Tij  T12 T22  0 0

0 0 0

Velocity : v1  2 X 1 , v2  X 1  2 X 2 , v3  0 x1 x x , X 2  2  1 , X 3  x3 2t 2t 4t x x x Spatial Velocity : v1  1 , v2  2  1 , v3  0 t t 2t 0 0  1/ t  1 / t 1 / 4t 0   L  1 / 2t 1 / t 0  D  1 / 4t 1 / t 0  0  0 0 0 0 0 Inverse Motion : X 1 

 Tij Dij 

T11  T22 T12  t 2t

5.14

Energy equation no sources :  - Tij Dij  qi ,i  0 Ideal non - viscous fluid : Tij   p ij and Fourier law qi  k,i    pij Dij  k,kk  0    pDkk  k,kk  0    k 2   p (  v )

5.15 s 

h 1 q h 1 q h 11 1     ( )  s   ( i ) ,i  s    qi ,i  qi ( ) ,i            

s 

h 11 1 h 1 1    qi ,i  qi ( 2 ),i   s   (  q)  2 (q  )       

1 (q  )  Using energy equation :  - tr(TD)    q - ρh  0  ρh    q   - tr(TD)

 s  h  (  q ) 

 s   - tr(TD) 

1 1 (q  ) or ( s   )  Tij Dij  (q  )  0  

5.16

Equation (5.7.14) ( s   )  Tij Dij  0 Energy Equation with no sources  - Tij Dij  qi ,i  0    Tij Dij  qi ,i Combining  s  Tij Dij  qi ,i  Tij Dij  0   s  qi ,i  0   s  

5.17

 T D  0 For isothermal process (5.7.11)    ij ij   

1 Tij Dij 

qi ,i 

6.1 Tij  Cijkl kl   (6.2.7)

Tij  kl

Tij  mn 



  (Cijkl kl )  Cijkl kl  Cijkl km jn  Cijmn  mn  mn

Tkl  Cijkl  Cklij  ij

6.2 From (6.2.17)  Tij  Cijkl kl  (ij  kl   ik  jl   il  jk ) kl   kk  ij   ij   ji   kk  ij   ij   ij let :    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  )

6.3 From basic relations : 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

Noting that   0

E , and if we choose the more restrictiv e inequality 1   1  2 

1  0 2

6.4 1 1 1 ~ 1 Tij  Tkk ij , Tˆij  Tij  Tkk ij , ~ij   kk ij , ˆ ij   ij   kk ij 3 3 3 3 ~ ~ Tij  kk ij  2ij  Tkk  (3  2) kk  Tij  3k ij 1 1 Tˆij  kk ij  2ij  (3  2) kk ij  2( ij   kk ij )  2ˆ ij 3 3

6.5 From hydrostatic pressure loading case :  kk  

3(1  2) p  0    1 / 2 (for general E and p) E

E E   and    1   1  2  3(1  2) From Tij  kk  ij  2ij , the term kk    0 and thus becomes indeterminate k

Thus we write : Tij   p ij  2ij  Tkk  3 p  2kk  3 p  p   kk / 3

6.6

The compatibility relations (6.2.29) with k  l are given by  ij ,kk   kk,ij   ik , jk   jk ,ik  0 Using Hooke' s law (6.2.31) 2 into these relations   (Tmm ,kk  ij  Tmm ,ij  kk  Tmm , jk  ik  Tmm ,ik  jk 1     (Tmm ,kk  ij  3Tmm ,ij  Tmm , ji  Tmm ,ij )  (Tmm ,kk  ij  Tmm ,ij ) 1  1  From the equilibrium equations Tij , j   Fi and using this result in the above relation gives

Tij ,kk  Tkk,ij  Tik , jk  T jk ,ik 

Tij ,kk 

1  Tkk,ij  Tmm ,kk  ij  Fi , j  F j ,i 1  1 

For the case i  j , the above relation gives Tii ,kk   compatibility statement yields the desired result Tij ,kk 

1  Tkk,ij    ij Fk ,k  Fi , j  F j ,i 1  1 

1  Fi ,i , and using this result back into the 1 

6.7

Equaton (6.2.36) : Tij  (  )u k ,k  ij  (ui , j  u j ,i ) Using in equilibrium equations : Tij , j  Fi  0  (  )u k ,kj ij  (ui , jj  u j ,ij )  Fi  0 (  )u k ,ki  (ui , jj  u j ,ij )  Fi  0 ui ,kk  (  )u k ,ki  Fi  0

6.8 1 1 1 Tij  ij  (kk  ij  2ij ) ij   jj  kk  ij  ij 2 2 2 1  ( xx   yy   zz ) 2  ( 2xx   2yy   2zz  2 2xy  2 2yz  2 2zx ) 2 1 1 1      1  U  Tij  ij  U  Tij  Tij  Tkk  ij   Tij Tij  T jjTkk 2 2  E E 2E  2E 1  2   (Txx  Tyy2  Tzz2  2Txy2  2Tyz2  2Tzx2 )  (Txx  Tyy  Tzz ) 2 2E 2E U

6.9 U (ε ) 

1  jj  kk  ij  ij 2

1 ( ii  jj   ij  ij )   ij  ij  I 2  2 II  2 1 2 1 U (ε )  I   ( I 2  2 II  )  (   ) I 2  2II  2 2 1   Also U (T )  Tij Tij  T jjTkk 2E 2E Tkk  I T and Tij Tij  I T2  2 IIT Now I    jj   kk , II  

 U (T ) 



1  2  2 1 2 ( I T  2 IIT )  IT  I T  2(1  ) IIT 2E 2E 2E



6.10

Using Txx 

 2  2  2 , T  , T   yy xy xy y 2 x 2

Txx Txy    2     2      0  0  x y x  y 2  y  xy  Txy x



Tyy y

0 

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

1 1 1  (Txx  Tyy ) ,  y  (Tyy  Txx ) ,  xy  Txy E E E 2  2  xy  2 x   y Compatibility Relation   2  xy y 2 x 2 From Hookes (Plane Stress)   x 

2 2  1 2 1     1   ( T   T )  ( T   T )  2 Txy        xx yy yy xx 2 2 xy  E y  E  x  E  

 2Txy  2Txx  2Tyy 1 2  (Txx  Tyy )  2   (1  ) xy x 2 y 2 But from equilibrium equations (zero body forces),  2Txy Txx Txy  2Txx  0    x y xy x 2 Txy x



Tyy y

0 

 2Txy

 2Tyy y 2



 2Txy yx

2  2Txx  Tyy 2    0 , and using this result into compatibility relation gives xy x 2 y 2

 2 (Txx  Tyy )  0

6.11

s ly 2 ly 3 xy 2 xy 3      xy   2   2  4 c c c c  s y 2 y3     , xx  , xxx  , xxxx  0 , x   y   4 c c 2  , xy 

s  2 y 3y2  1   2   , xxyy  0 4  c c 

, yy 

s  2l 6ly 2 x 6 xy   2   , yyyy  0   2  4 c c c c 

  4  0 Txx  , yy 

s  2l 6ly 2 x 6 xy  s  3 y   2    2  1  (l  x) 4 c c c  c c  2c 

s  2 y 3y2  s y  3 y  Tyy  , xx  0 , Txy  , xy   1   2    1  1   4 c 4 c  c  c  Boundary Conditions : Tyy ( x, c)  0 , satisfied Txy ( x, c)  s , satisfied Txy ( x,c)  0 , satisfied Conditions at fixed end x  0 : c 

sl

c

3y 

  c Txx (0, y )dy  2c   c 1  c dy  sl , satisfied s c  2 y 3y2    c Txy (0, y )dy   4   c 1  c  c 2 dy  0 , satisfied c

sl

c

c 

3y 

  c Txx (0, y ) ydy  2c   c 1  c  ydy  slc , satisfied Conditions at free end x  l : Note that Txx (l , y )  0 c

c

c

c

  Txx (l , y )dy  0 &  Txx (l , y ) ydy  0 , satisfied s c  2 y 3y2    c Txy (l , y )dy   4   c 1  c  c 2 dy  0 , satisfied c

6.12   a4   Trr 

1  1  2   2   1   a4  2 2  0 , T  2  0 , Tr     r r r  r  r   r 2 r

r2 Boundary Condition : Tr (r2 )    a4  r22  Tr   22 r du Displacements :  rr  r  0  u r  constant , but u r (r1 )  0,  u r  0 dr 1  1   r22 1  1 u r u u   r  Tr        E E r 2 2  r  r r  du u 2(1  )r22 1 d  u  2(1  )r22 1      dr r E dr  r  E r2 r3 (1  )r22 1 (1  )r22 1 Solution : u    Cr , but u (r1 )  0  C  E r E r12  u 

(1  )r22  r 1   2  r  E  1 r

6.13 u x  yz , u y  xz , u z  u z ( x, y ) Strain - Displacements Relations   xx   yy   zz   xy  0 ,  xz  Hooke' s law  Txx  Tyy  Tzz  Txy  0  w  Txz    y   Txz ( x, y )  x   w  Tyz    x   Tyz ( x, y )  y 

 1  w 1  w   y  ,  yz    x   2  x 2  y  

6.14 e1

e2

e3

Along streamline : v  dx  v x

vy

0  0  v x dy  v y dx  0

dx dy

0

  dy  dx  0  d  0 ,    constant y x

6.15   dx  dy  v x dx  v y dy  0 x y   d  dx  dy  v y dx  v x dy  0 x y vy v  dy   dy   dy   dy     x ,         1 ,  orthogonal v y  dx   v x  dx    dx    dx   d 

6.16 v x  Ax , v y  By Incompressible :   v  0  A  B  0  A   B e1

e2

  v   / x  / y Ax

By

e3 0  0  irrotational 0

 1  Ax    Ax 2  f ( y ) x 2  1 vy   By  f ( y )  By  f ( y )  By 2 (dropping the constant) y 2 1    ( Ax 2  By 2 ) 2 vx 

6.17

  K 1  K   , v   0 r  r r Potential Function :   K log r

vr 

Streamlines

Flow is radial sink flow Potential lines

6.18

  K log r 1   K  0 , v    r  r r Potential Function :    K

vr 

Potential lines

Flow is tangentia l inversely proportional to radial coordinate

6.19 Tij   pij  Dkk ij  2Dij  Tij , j  bi  ai   v   p, i  vk , ki  (vi , jj  v j , ij )  bi   i  v j vi , j    t   v   p, i  (  )vk , ki  vi , kk  bi   i  v j vi , j   t   v   p  (  )(  v )   2 v  b    v  v   t 

6.20 Energy Equation :   Tij Dij  qi , i  h  0 2   ( pij  Dkk ij  2Dij ) Dij  qi , i  h  0 3 2   pDkk  ( Dkk ) 2  2Dij Dij  qi , i  h  0 3

Streamlines

6.21

NS Equations :  p 

  v  (  v )   2 v   b    v  v  3  t 

For incompressible slow flows with no body forces   p   2 v  0 Take   of reduced equation   2 p   2 (  v )  0 e1

e2

e3

v    ψ   / x  / y  / z  0

0



  e1  e2 y x

This form satisfies :   v  0 and thus p   2 v ,  2 p  0   p      ( 2 ) y  x y  2 p   v   0   2 2    4   0    p      ( 2 ) x  y x  6.22 v 1  v  v   p    2 v t   v  1     v  v   p   2 v     t   (  v )    (v  v )   2 ω . . . using indentity   (  )  0 t Next using the following vector identities 1 1  v  v   v    v  (v  )v   v  v   v  ω 2 2    (v  ω)  ω(  v )  (ω  )v  (v  )ω  v  v  ( v   ) v 

ω  (ω   )v  (v   )ω   2 ω t Dω ω    v  ω  (ω   )v    2 ω Dt t

6.23 NS equation for incompressible, steady state case with no pressure gradient   2 v  0 For the given velocity field vr  0 , v  v (r ) , v z  0 , this reduces to v d 2 v 1 dv v  0     0 , which is a Cauchy - Eluer ODE with solution r2 dr 2 r dr r 2 C v  C1r  2 , C1 and C2 being arbitrary constants r Boundary Conditons : v (ri )  0 , v (ro )  ro determine the two constants and give

 2 v 

ri2 ro2   r 1     v  2 ro  ri2  ri2 r 

6.24

Tr (ro )  2Dr (ro )  2





1 d v   dv v  v  (v )T r        ro    r  ro 2 r  r r dr  r  r  r  dr o

ri2 ro2   r 1  2ri2 ro2  1 2ri2 d  v    v  2   Tr (ro )  ro    ro 2  2 2 3 dr  r  r  r ro  ri2  ri2 r  r  r r ro  ri2 o i o o Torque  T  2ro2Tr (ro ) 

4ri2 ro2 ro2  ri2

6.25   Tij Dij  ( pij  Dkk ij  2Dij ) Dij   pDkk  ( Dkk ) 2  2Dij Dij For incompressible case Dkk  0    Tij Dij  2Dij Dij

o

6.26 Viscoelast ic Maxwell Model : E T12  T12  E 12  E o  Solution subject to zero intial condition 

T12 /  o 1.0

E   t  T12 (t )   o 1  e       Linear Viscous Model : T12  2 12   12   o

6.27

Ts  TK  T T T E    K    E*   T T E T T E T   (   s )      (  )    E*   E*   E* E * E   T T    E  T  p1T  qo   q1 E* E*  EE * E * where p1  , qo  , q1  E  E* E  E* E  E*

 s   K     s   K   

6.28

Td  TK  T T T E    K    *   T T E T T E T   (   d )      (  )    *   *   *   *   *   T  p1T  q1  q2 T T   *   E E   *  * where p1  , q1   * , q2  E E  d   K     d   K   

Viscous Fluid Maxwell Model

t

6.29 Three - parameter solid model Stress relaxation : (t )   o H (t ) T  p1T  qo   q1  T  p1T  qo  o , t  0 General Solution T  Th  T p T p  qo  o , Th  Ke t / p1 where K  constant q  Initial Condition : T (0)  E *  o  K  qo  o  E *  o  K   o E * qo    o  1  qo   p1  q   q T  q1  T   o  1  qo e t / p1  qo  o  G (t )     qo e t / p1  qo  1 e t / p1  qo (1  e t / p1 )  o  p1 p1  p1   Creep Function : T(t )  To H (t ) T  p T  q   q   T  q   q  , t  0 1

o

1

o

o

1

General Solution    h   p p 

To ,  h  Ke tqo / q1 where K  constant qo

Initial Condition : T (0)  To  (0) 

 1 p To T T 1  1   K  o  o  K  To     To  1   E* qo E *  E * qo   q1 qo 

p T p 1    p1 1  tqo / q1 1 1    To  1  e tqo / q1  o  J (t )     e   1 e qot / q1  (1  e qot / q1 ) q q q T q q q q q o  o o o  o 1 o  1  1

6.30 Three - parameter fluid model Stress relaxation : (t )   o H (t ) T  p T  q   q   T  p T  0 , t  0 1

1

2

1

General Solution T  Ke

t / p1

where K  constant

Initial Condition : T (0)  E o  T 

q12 q12 o  K  o p1q1  q2 p1q1  q2

 t / p1 q12 q12 T  e  o e t / p1  G (t )    p1q1  q2  o  p1q1  q2 

Creep Function : T(t )  To H (t ) T  p T  q   q   T  q   q  , t  0 1

1

2

o

1

2

General Solution    h   p , p 

To q t , q1 h  q2h  0   h  K1e q1t / q2 ,  h   2 K1e q1t / q2  K 2 where K1, 2  constants q1 q1

Initial Conditions : Zero Strain and Strain Rate only from dashpots   ( 0)  0  K 2 

T T q2 p p q q K1 ,  (0)  o  o  To 1  K 2  To 1 1 2 2 q1 *  q2 q1

 t  p q q p q q (t ) t  (t )  To   1 1 2 2 (1  e q1t / q2 )   J (t )    1 1 2 2 (1  e q1t / q2 ) To q1 q1 q1  q1 

6.31 2

1.8

1.8

1.6

1.6

Relaxation and Creep Functions

Relaxation and Creep Functions

Three Parameter Fluid Model 2

1.4 1.2 1 0.8 0.6

Relaxation Function Creep Function

1.2 1 0.8 0.6 0.4

0.4 Relaxation Function Creep Function

0.2 0

1.4

0

2

4

6

8

10 Time, t

12

14

16

18

0.2 0

20

0

2

4

Three Parameter Solid Model 2

1.8

1.8

Relaxation and Creep Functions

Relaxation and Creep Functions

1 0.8 0.6

8

10 Time, t

12

14

20

0.8 0.6

0.2

6

18

1

0.4

4

16

16

18

20

Three Parameter Solid (p1=0.01, qo=0.5, q1=10) Results look similar to Kelvin Model

Relaxation Function Creep Function

1.2

0.2

2

14

1.4

0.4

0

12

1.6

Relaxation Function Creep Function

1.2

0

10 Time, t

Three Parameter Fluid Model

2

1.4

8

Three Parameter Fluid (p1=5, q1=10, q2=10)

Three Parameter Solid (p1=5, qo=0.5, q1=10)

1.6

6

0

0

2

4

6

8

10 Time, t

12

14

16

18

20

Three Parameter Fluid (p1=5, q1=10, q2=0.1) Results look similar to Maxwell Model

6.32 Stress - Strain Response for Three Parameter Solid 70

60

Stess, T

50

40

30

20

10

0

0

0.05

0.1

0.15

0.2

0.25 0.3 Strain, 

0.35

0.4

0.45

0.5

Parameters: p1=0.3, qo=200, q1=300

6.33 Maxwell Model for  = 1

Kelvin Model

1

10

G1/E

0.8

8

0.7

7

0.6 0.5 0.4

5 4 3

0.2

2

0.1

1

0

1

2

3

4 5 6 Frequency ()

7

8

9

10

G2/ 

6

0.3

0

G1/E

9

G2/ 

Complex Modulus

Complex Modulus

0.9

0

0

1

2

3

4 5 6 Frequency ()

7

8

9

10

6.34

Three - Parameter Solid : T  p1T  qo   q1 Using (t )   o e it gives T (t )  To e it  To  p1iTo  qo  o  q1io  To  

qo  q1i q  q1i  1  p1i    o o  o 1  p1i 1  p1i  1  p1i  qo  q1i  qo p1i  q1 p12 qo  q1 p12  i (q1  qo p1)   o o 1  p12 2 1  p12 2

qo  p1q12 (q  qo p1 )  G1  , G2  1 2 2 1  p1  1  p12 2 6.35 t d(t )   L  T (t )   G (t  t ) dt    T ( s )  G ( s )  ( s )  G ( s ) s  ( s )  0 dt    T (s) G (s)  s (s) t dT (t )   L  (t )   J (t  t ) d t     ( s )  J ( s ) sT ( s )  0 dt     (s) J (s)  sT ( s ) 1  G (s) J (s)  2 s Taking the inverse transform using (6.5.43) 3 

0 G (t ) J (t  t )dt   L G (s) J (s)  L 1 / s   t t

-1

-1

2

6.36

For Example 6.5.9 use  ( s )   o  s ,  ( s )   o  s       ot   1 -1   o /   1  1  1  -1  -1   L  2 L  2 L  t [1  e              2 o o o o  s  (s)   s ( o  s )     s (   s)          1 1 1   L-1     L-1  2 L-1  2 2      s [  ( s )   ( s )] s [    s     s ] s [(    )  2  s ]   o o o o       (  )     o o t 1 1 2 - 1  ( o   o ) / 2    L  t [1  e 2      (    ) o  o o  o o  o 2 o o  s]    s [ 2   Solution follows as done in the Example : r12 p  r22  r12 p  r22    1   Trr  2 1 , T  2 r2  r12  r 2  r2  r12  r 2  ur 

  r12 r22 p r12 rp 1  1 -1     L  L-1  2 2 2 2 2 2    2(r2  r1 )r  s  ( s )  2(r2  r1 )  s [  ( s )   ( s )] 

 (  )   ot   o o t r12 r22 p r12 rp 1   1 2     t [1  e  t [1  e 2  2 2 2 2     o o  o 2(r2  r1 )r  o 2(r2  r1 )  o   o    

6.37 From Example 6.5.10 Trr 

P  3r 2 z 1  2 ( s) 3 ( s )  1  2 ( s)  R   3  L-1     , with 2  s s s[3k   ( s )] 2R  R   R  z

o s  1/  3 o         3 o 1  2  ( s ) 3  ( s )   -1 s  1/    L-1     L-1  L-1    L   s s[3k ( s  1 / )   o ]   s[3k   o ]     s[3k   ( s )]     s  1/    For this problem use (t )   o e t /    ( s ) 

 

 (  3 k /  )     o t o / k 3 o 3k    L  1 e   s[ s  ( o  3k / ) ]   o  3k /      3k   -1 

6.38 2 1.8

m=2 Dimensionless Stress, T/T

o

1.6 1.4 1.2

 A

1

m=50

0.8 0.6 0.4 0.2 0

0

1

2

3 4 5 6 7 Dimensionless Strain, E /To

8

9

10

For the case m  , the response is elastic perfectly plastic. The parameter  is related to the location of point A.

6.39 1 ~~ 1 1  2 Tij ij  T jj  kk  T jjTkk 2 6 6E 1  2 2 1  2 Since Tkk  I T  U v  IT  (T1  T2  T3 ) 6E 6E 1 2 From Exercise 6.9  U  I T  2(1  ) IIT 2E 1 2 1  2 2 Ud  U  Uv  I T  2(1  ) IIT  IT 2E 6E 2(1  ) 2 2(1  ) 1 2 Ud  IT  IIT  I T  3IIT 6E 2E 6 1 It can be shown that [(T1  T2 ) 2  (T2  T3 ) 2  (T3  T1 ) 2 ]  I T2  3IIT 2 1 2 1  Ud  I T  3IIT  [(T1  T2 ) 2  (T2  T3 ) 2  (T3  T1 ) 2 ] 6 12 Exercise 4.14  Uv 

















1 1 1 [(Tˆ1 - Tˆ 2 )2 +(Tˆ 2 - Tˆ 3 )2 +(Tˆ 3 - Tˆ1 )2]  - [(T 1 - T 2 )2 +(T 2 - T 3 )2 +(T 3 - T 1 )2]   Tˆij Tˆij 6 6 2 1 1  U d  Tˆij Tˆij  J2 4 2 II Tˆ = -

6.40 f  [(T1  T2 ) 2  4k 2 ] [(T2  T3 ) 2  4k 2 ][(T3  T1 ) 2  4k 2 ]  [(Tˆ  Tˆ ) 2  4k 2 ] [(Tˆ  Tˆ ) 2  4k 2 ][(Tˆ  Tˆ ) 2  4k 2 ] 1

2

2

3

3

Expanding  f  A  4 Bk  16Ck  64 k 2

4

1

4

where A  (Tˆ1 - Tˆ 2 )2 (Tˆ 2 - Tˆ 3 )2 (Tˆ 3 - Tˆ1 )2 B  (Tˆ1 - Tˆ 2 )2 (Tˆ 2 - Tˆ 3 )2 +(Tˆ 2 - Tˆ 3 )2 (Tˆ 3 - Tˆ1 )2 +(Tˆ 3 - Tˆ1 )2 (Tˆ1 - Tˆ 2 )2 C  (Tˆ1 - Tˆ 2 )2  (Tˆ 2 - Tˆ 3 )2  (Tˆ 3 - Tˆ1 )2 Note that I Tˆ = Tˆ1  Tˆ 2  Tˆ 3  0 II Tˆ = Tˆ1 Tˆ 2  Tˆ 2 Tˆ 3  Tˆ 3 Tˆ1  

1 2 1 (Tˆ1 + Tˆ 22 + Tˆ 32)  - [(Tˆ1 - Tˆ 2 )2 +(Tˆ 2 - Tˆ 3 )2 +(Tˆ 3 - Tˆ1 )2] 2 6

1 2 (Tˆ + Tˆ 22 + Tˆ 32) 3 1 These results  (Tˆ1 - Tˆ 2 )2  ( II Tˆ  3 Tˆ1 Tˆ 2) ; (Tˆ 2 - Tˆ 3 )2  ( II Tˆ  3 Tˆ 2 Tˆ 3) ; (Tˆ 3 - Tˆ1 )2  ( II Tˆ  3 Tˆ 3 Tˆ1) II Tˆ = Tˆ1 Tˆ 2 Tˆ 3 

A  4 IIT3ˆ  27 IIIT2ˆ ; B  9 IIT2ˆ ; C  6 IITˆ  f  4 IIT3ˆ  27 IIIT2ˆ  36 IIT2ˆ k 2  96 IITˆ k 4  64 k 4 ; or f  4 J 23  27 J 32  36 k 2 J 22  96 k 4 J 2  64 k 6

6.41 Along line OH : T1  T2  T3  Tˆij  0  J2  J3  0 both Mises and Tresca predict no yielding

6.42

(4.5.6)  3 ˆ ˆ 1 Tij Tij  [(T11  T22 ) 2  (T22  T33 ) 2  (T33  T11 ) 2  6(T122  T232  T312 )]1/ 2 2 2 1  [(T1  T2 ) 2  (T2  T3 ) 2  (T3  T1 ) 2 ]1/ 2 2 1  6k . . . from Mises condition (6.6.13) 2

e 

  e2  3k 2

6.43

T11 T12 Plane Strain  T  T12 T22  0 0

0 0  with all fields depending only on x1 , x2 T33  and with  33  0 , incompressibility  11   22  0  we have a state of pure shear in the x1 , x2 plane  that the in - plane stresses are also pure shear  the stress component T33 must equal the in - plane hydrostatic pressure  T33 

1 (T11  T22 ) 2

Mises Yield Condition :  e2  3k 2  [(T11  T22 ) 2  (T22  T33 ) 2  (T33  T11 ) 2  6(T122  T232  T312 )]  6k 2  1 1 (T22  T11 ) 2  (T22  T11 ) 2  6T122  6k 2  (T11  T22 ) 2  4T122  4k 2 4 4 Tresca Yield Condition :  max  k  (T11  T22 ) 2 

2

2

 T11  T22  T T  2 2 2 2 2 2    T12  k   11 22   T12  k  (T11  T22 )  4T12  4k 2 2    

6.44 Trr 

r12 p  r22  r2 p  r2  1   , T  1 1  2  , Tzz  (Trr  T ) , Tr  Tz  Trz  0  r22  r12  r 2  r22  r12  r 2 

Trr r22 T r22 Tzz r12 p 1 2 , 1 2 ,  2 , with K  2 , and note that 0    1/2 K K K r r r2  r12  T  Tzz  Trr Tresca Condition : T  Trr  2k  T  Trr 

r12 p  r2 r2  2r12 p r22 1  2  1  2   r22  r12  r2 r 2  r22  r12 r 2

Mises Condition : (Trr  T ) 2  (T  Tzz ) 2  (Tzz  Trr ) 2  6k 2  2

2

2

2

   r4 1  r2  r2 r2  K   2 22   K 2 1  22  2   K 2  2  1  22   6k 2  K 2  24  (1  2) 2   k 2 3 r  r r      r 

6.45

 rp2  rp2 p (6.6.34)   1  2  log 2  , r1  rp  r2  r1  k r2    rp2 / r22   1  2  log 2 2  , r1 / r2  rp / r2  1  r1 / r2  r2   MATLAB Plot rp2

5 4.5

r1/r2 = 0.1

Dimensionless Pressure, p/k

4 3.5

0.2 3

0.3

2.5 2

0.4

1.5

0.5

1 0.5 0.1

0.2

0.3

0.4 0.5 0.6 0.7 0.8 Dimensionless Plastic Radius, rp/r2

0.9

1

Plot shows much higher pressures are needed to cause plastic deformation for smaller values of r1/r2 . This is expected since there is more material to withstand the pressure loadings.

6.46 Governing equation :  2  

1 d  d   r   2 r dr  dr 

d  d  d  2  r  C1     r  C1r  C2  r   2r  dr  dr  dr 2 d(0)  2 Boundary conditions :  0  C1  0 , (a )  0  C2  a dr 2  2 d  (a  r 2 )  Tz      r 2 dr a 1 2T 2Tr T  2  dxdy  2 (a 2  r 2 )rdr  a 4    4    4 R 0 2 a a Integrating 

 T  T   k , k = Y/ 2 Mises Yield Condition :   3k  6(T  T )  6k   T  T   k , k = Y/ 3

Tresca Yield Condition :  max  k   max  Tz max  2 e

2

2 xz

2 yz

2 xz

2

2 yz max 2 xz

2 yz max

6.47 2T x 2 y 2 2T 2a 2b      max  (0,b)   ab a 4 b 4 ab 2 a 2  b 2 2T 2a 2b At yield both Tresca and Mises gives same result   max   k ab 2 a 2  b 2 kab 2 k (a 2  b 2 )  Tp  and  p  2 2a 2b

7.1 (7.2.6) 3  qi ,i  ρ(h  s) (7.2.8) 2  s  ij  ij 

cv   o

Fourier's law (7.2.9)  qi  kij , j   k ij , ji  h   ij  ij  

cv   o

Linearizin g by letting    o       h  0 k ,  c  ij

ij

v

ij o ij

7.2 Tij  kk  ij  2ij  (3  2) T  ij   Tkk  (3  2) kk  3(3  2) T    kk   ij  

Tkk  3 T  (3  2)

  Tkk  1 1   Tij     Tij  kk  ij  (3  2) T  ij    3     ( 3   2  )    T ij T ij   2 2  ( 3   2  )   





 1  Tkk 1    Tij   ij  2T  ij    Tij  Tkk  ij   T  ij 2  (3  2) E E 

7.3

Tij  (  ) kk  ij  2ij  (3  2)T  ij  (  )u k ,k  ij  (ui , j  u j ,i )  (3  2) T  ij Tij , j  Fi  0  (  )u k ,ki  (ui ,ij  u j , ji )  (3  2) T  ,i  ui ,kk  (  )u k ,ki  (3  2)T  ,i  Fi  0

7.4 Uncoupled energy equation for steady state case : qi ,i  0 Anisotropic heat conduction law : qi  kijT, j Combining these two relations  kij T,ij  0 , which for two - dimensions gives k xx

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

7.5

For plane stess problem : Txx  Txx ( x, y ) , T yy  T yy ( x, y ) , Txy  Txy ( x, y ) , Tzz  Txz  T yz  0 Hooke' s law reduces to E T E E u v E T (    )    (   )  x y 1  y 1   1 2 1   2 x E E E v u E T yy  ( y   x )  T   (  ) T  2 2 1  x 1   1  1   y E E u v Txy   xy  (  ) 1  2(1  ) y x Using equilibrium equations with no body forces Txx 

E   u v  E T  Txx Txy     2u  0  0 2(1  ) x  x y  1   x 2 2 x y  , where  2  2  2 Txy T yy E   u v  E T  x y  0      2 v  0 x y 2(1  ) y  x y  1   y 2  2  xy  2 x   y Compatibility relation  2 2  xy y 2 x 2 2  1 2 1     1  Txy    (Txx  T yy )   T    2  (T yy  Txx )   T    2  xy  E y 2  E  x  E  

 2 (Txx  T yy )  E T  2   0

7.6 1 1 1  (Txx  T yy )   T  ,  y  (T yy  Txx )   T  ,  xy  Txy E E E Inverting as in exercise 7.5 

x 

E T E T E E u v ( x   y )   (  )  2 2 1  y 1   1  1   x E T E T E E v u T yy  ( y   x )   (  )  2 2 1  x 1  1  1   y E E u v Txy   xy  (  ) 1  2(1  ) y x Txx 

 E  u v   E E T  u v  t xn  Txx n x  Txy n y   (  )  n x     n y 2 y 1    1   x  2(1  )  y x   E  u v   E E T  v u  t yn  Txy n x  T yy n y     n x   (  )  ny 2 x 1    1   y  2(1  )  y x 

7.7

  f ( x) sin y 

E T To sin  y , f ( x)  C 2 cosh x  C3 x sinh  x 2

Txx 

 2   2 C 2 cosh x  C3 x sinh x sin  y  E T To sin  y y 2

T yy 

  2 2 2   C cosh  x  C ( x sinh  x  cosh  x ) 2 3   sin  y  x 2      2 1   2 C 2 sinh  x  C3 ( x cosh  x  sinh x) cos y xy   

Txy  

On lateral stress free boundaries x   a , n x  1 and n y  0  t xn  Txx x a  0  C 2 cosh  a  C3 a sinh  a   E T To /  2 1  0  C 2 sinh  a  C3 (a cosh a  sinh  a )  0  Solve two equations for two unknowns C 2 and C3  t yn  Txy

C2  

x a

ETo a cosh  a  sinh  a  ETo sinh a , C3  2  a  sinh  a cosh  a   a  sinh a cosh  a 

7.8 From the general solution given by (7.2.51), A2 (1  ) T r   d r r Bounded displacements at r  0  A2  0

u  A1r 

(1  ) T a  d a2 (1  ) T r (1  )T r a u  d    d r a2 For the case where    o  constant : u (a )  0  A1  

u

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

Trr 

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

T 

d 2 d  (rTrr )  0 dr 2 dr

7.9  2  1  1 d  d    ho  0  r    ho 2 r r r dr  dr  r h Integrating    C1 log r  C 2  o r 2 4 Boundary Conditions : (0) must remain bounded  C1  0 ( a )   o  C 2   o 

ho 2 a 4

ho 2 (a  r 2 ) 4 Governing equation for plane stress,  4   E T  2   0   4   E T ho 

   o 





1 d  d 1 d 2  r Trr    E T ho  r  r dr  dr  r dr  Integrating the governing equation three times  Trr 

C E T ho 2 C1 (2 log r  1)  C 2  23  r 4 16 r

C 3E T ho 2 C d 2 d T  2  (rTrr )  1 (2 log r  1)  C 2  23  r ,  r  0 dr 4 16 dr r Since the plate is solid, the stresses must be bounded at r  0  C1  C3  0 Boundary condition Trr (a )  0  C 2    Trr 







E T ho 2 a 16

E T ho 2 E T ho r  a 2 , T  3r 2  a 2 16 16



7.10  ij 

 1 1 1  1 Tij    Tkk  ij  p ij  2 3H  6 9k 

 kk 

 1 1 1  1 1 p p  Tkk    Tkk  p  Tkk   Tkk  3k   kk   2 H 3k H H   2 3k 

  ij 

 k 1  1 p 1 Tij      kk   ij  p ij  2 H 3H  2 3 

   k 1  p 1 Tij  2  ij      kk   ij  p ij   H 3H  2 3    Tij  p ij  (  ) kk  ij  2ij

7.11

Undrained conditions   0, and thus (7.3.6)   H 2  Rk  PR P PR  p and  kk    2   P 2  H k H kH   kH 2 kR    k 1  2  Ku 2 H  Rk  H  kR  P   kk     Ku but

P kH 2 H 2  Rk

7.12

The compatibility relations (6.2.29) with k  l are given by  ij ,kk   kk,ij   ik , jk   jk ,ik  0 Hooke' s law (7.3.4)1 can be re - written as 1   1 Tij  Tkk  ij  p ij and substituting this form into the compatibility relations  E E 3H Tij ,kk  Tkk,ij  Tik , jk  T jk ,ik  ij 

 E (Tmm ,kk  ij  Tmm ,ij  kk  Tmm , jk  ik  Tmm ,ik  jk )  ( p,kk  ij  3 p,ij  p, ji  p,ij ) 1  3(1  ) H  E  (Tmm ,kk  ij  3Tmm ,ij  Tmm , ji  Tmm ,ij )  ( p,kk  ij  p,ij ) 1  3(1  ) H  E  (Tmm ,kk  ij  Tmm ,ij )  ( p,kk  ij  p,ij ) 1  3(1  ) H From the equilibrium equations Tij , j   Fi and using this result in the above relation gives 

1  E Tkk,ij  Tmm ,kk  ij  Fi , j  F j ,i  ( p,kk  ij  p,ij ) 1  1  3(1  ) H 1  2E For the case i  j , the above relation gives Tii ,kk   Fi ,i  p,kk , and using this result back into the 1  3(1  ) H compatibility statement yields

Tij ,kk 

Tij ,kk 

1   2E E Tkk,ij    ij Fk ,k  Fi , j  F j ,i  p,kk  ij  ( p,kk  ij  p,ij )  1  1  1   3(1  ) H 3(1  ) H

Tij ,kk 

1 (1  2)  1    Tkk,ij  pij     ij Fk ,k  Fi , j  F j ,i   ij p,kk  1  1   1   1 

7.13 (7.3.12) Tij ,kk 

1 (1  2)  1    Tkk,ij  p,ij     ij Fk ,k  Fi , j  F j ,i    ij p,kk  1  1   1  1  

1 (1  2)  1    Tkk,ii  p,ii    3Fk ,k  2 Fi ,i   3 p,kk  1  1   1  1   2 3(1  2)(1  )  (1  2)(1  ) 2 Tkk,ij  p,ii   Fk ,k  1  (1  )(1  ) 1  1   2 (Tkk  4* p )   Fk ,k 1  3(1  2)(1  )  (1  2)(1  ) where *  4(1  )(2  )

Tii ,kk 

7.14

(7.3.5)  Tij  p ij  (  ) kk  ij  2ij  p ij  (  )u k ,k  ij  (ui , j  u j ,i ) Tij , j  Fi  0   p,i  (  )u k ,ki  (ui ,ij  u j , ji )  ui ,kk  (  )u k ,ki  p,i  Fi  0

7.15 For plane strain : 1  (1  ) p 1  (1  ) p 1  [(1  )Txx  T yy ]  , y  [(1  )T yy  Txx ]  ,  xy  Txy E 3H E 3H E 2  2  xy  2 x   y Compatibility relation for plane strain :  2  xy y 2 x 2 x 

2  E  2  E  2 ( 1   ) T   T  p  ( 1   ) T   T  p  2 Txy      xx yy yy xx 3H  x 2  3H  xy y 2 

 

 2Txy  2Txx  2T yy E 2 (1  ) (Txx  T yy )   p2   3H xy x 2 y 2 2

But from equilibrium equations with zero body forces, 2 Txy T yy  2T yy  2Txy Txx Txy  2Txx  Txy  0   0,  0   0 x y xy x y yx x 2 y 2

 2Txy

2  2Txx  T yy 2    0 and thus the previous compatibility statement reduces to xy x 2 y 2 E  2 (Txx  T yy )  2 p  0 3(1  ) H 1  2 E  2 Txx  T yy  2p  0 , with    2(1  ) 6(1  ) H





7.16 T11  C11 T  C  22   12 T33  C13    T23   0 T31    0    T12   0 D   0  1   D2   0 D  e  3   31

C12

C13

0

0

0

0

0

C11

C13

0

0

0

0

0

C13

C33

0

0

0

0

0

0

0

C44

0

0

0

 e24

0

0

0

C44

0

 e15

0

0

0

0

0

C66

0

0

0

0

e15

0

0  11

0

0

e24

0

0

0

0   22

e32

e33

0

0

0

0

0

 e31   11   e32    22   e33    33    0  2 23  0   2 31    0   212  0   E1    0   E2   33   E3 

7.17 For plane strain in x2 , x3  plane 11  13  12  0 , and E1  0 (7.4.8)  T12  T13  0 , and out of plane normal stress is T11  C12  22  C13  33  e31 E3 Thus we can re - write (7.4.8) in reduced form as 0    22   0 e32   E2  C33 0    33    0 e33    E 0 C 44  2 32  e24 0   3  Likewise (7.4.9) can be re - written as (removing D1 e15  23 ) T22  C11 T   C  33   13 T23   0

C13

  22  e24    33  e33 0   2 32  Combining in a similar fashion as done in Exercise 7.16 gives the 5x5 system   D2   22 D    0  3 

T22  C11 T  C  33   13 T23    0     D2   0  D3   e32

0   E2   0       33   E3  e32

0

C13

0

0

C33

0

0

0

C 44

0

e24

e24   22

e33

0

0

e32    22  e33    33  0  2 32    0   E2    33   E3 

8.1 T  f F   T *  f * F * Frame Indifference  QTQ T  f (QF ) Use Q  R T  R 1





 R T TR  f ( R T F )  T  Rf R T F R T but from P.D. theorem F  RU  U  R T F  T  Rf U R T Now PK2 is defined as S  JF 1T ( F 1 )T F 1TF 1  F 1 Rf U R T ( F 1 )T  U 1 R 1 Rf U R T RU 1  U 1 f U U 1 T

 S  JU 1 f U U 1  h(U )

8.2 For incompressible case : U  U ( I B , II B )  U U I B U II B   Bij I B Bij II B Bij I B  Bkk   ik  jk  ij  Bij Bij II B 1  Bkk Bll  Bkl Bkl   Bkkij  Bij  Bij 2 Bij 

U U U  ij  I B ij  Bij Bij I B II B

Tij  pij  Bik





U U  Bkj  Bkj Bik

 U   U  U U I B ik  Bik  Bkj Tij  pij  Bik   kj  I B  kj  Bkj    ik  II B II B  I B   I B 





 U U  U  Bij  2 Tij  pij  2  IB Bik Bkj II B  II B  I B

8.3 Example 8.3.1 : Cauchy Stresses were given by 1     T11   p  12   2  2   2   1  2       1 T22  T33   p  1   2   0  T12  T23  T31  0 Neo - Hookean Model : U  C10 ( I B  3)  1  2C10 ,  2  0 1  T11  2C10  2     T22  T33  T12  T23  T31  0 1 1  T11  2C10    2      Mooney - Rivlin Model : U  C10 ( I B  3)  C01 ( II B  3)  1  2C10 ,  2  2C01 PK1 : T o  T ( F 1 )T  T110 

C  1  T11  2 C10  01  2       T22  T33  T12  T23  T31  0 PK1 : T o  T ( F 1 )T  T110 

C  1 1  T11  2 C10  01    2       

8.4

S  F 1T o 0   1 / 2 F  0   0 0

0  1 /  0  1 0   F   0 1/ 2  0 1/ 2  0

1 /  0 1 o S  F T   0 1/ 2  0 0

0  0  1/ 2 

0  T11 /  0 0 T11 / 2  0   0 0 0   0 1/ 2   0 0 0  0

0 0  0 0 0 0

8.5 Plots with particular choices of 1 and 2 25 Mooney:  1 = 3.9  2 = 0.2 Neo-Hookean:  1 = 3.9  2 = 0 20

Cauchy Axial Stress, T

11

Material X (Linearized Data Fit)

15

10

5

0

1

1.5

2 Stretch Ratio, 

8.6 From Exercise : t 

(T11  T12 ) 1 

2

e1 

(T12  T22 ) 1 2

e2

 (T  T )  (T  T22 )   1 12 N  t  n   11 e1  12 e2  (e1  e 2 )   1 2   1 2  1 2     (T  T ) (T  T22 ) T  11 2 12  12   T22  122 , 2 1  1  1   (T  T )  (T  T22 )   1 12 S  t  s   11 e1  12 e2  ( e1  e 2 )   1 2   1 2  1 2     (T  T ) (T  T22 ) T  11 2 12   12  12 2 2 1  1  1 

2.5

8.7 From Exercise : T11  1 2 , T22   2  2 , T12  (1   2 )  , T33  T32  T31  0 T11 T12 PK1 : T  T ( F )  T12 T22  0 0 1 T

o

0  1 0 0 T11  T12 0   1 0  T12  T22 0  0 0 1  0

1   0 T11  T12 1 o PK2 : S  F T  0 1 0 T12  T22 0 0 1  0

T12 T22 0

0 T11  2 T12   2T22  0   T12  T22 0  0

T12 T22 0

8.8

1   2  Simple shear : B     0 

 0  1 0 0 1

U  C10 ( I B  3)  C 20 ( I B  3) 2 U U  C10  2C 20 ( I B  3) , 0 I B II B From (8.3.9) Tij   p ij  2

U U 1 Bij  2 Bij  I B II B





Tij   p ij  2C10  2C 20 ( I B  3) Bij   p ij  2 C10  2C 20  2 Bij 





T11   p  2 C10  2C 20  2 (1   2 )



T22  T33   p  2 C10  2C 20  2







T12  2 C10  2C 20  2   2C10   2C 20  3 , T23  T31  0 For neo - Hookean model :

0 0 0

U U  C10 ,  0  T12  2C10  I B II B

For C10  C 20  1, MATLAB plots are shown below

T12  T22 T22 0

0  0 0

1.5

Cauchy Shear Stress, T

12

Original Model neoHookean Model

1

0.5

0

0

0.05

0.1

0.15

0.2

0.25 0.3 Shear, 

0.35

0.4

0.45

0.5

8.9

For Neo - Hookean Model : 1  2C10 ,  2  0 From Exercise : a

a

Trr    1 2 r dr   2C10  2  r dr   C10  2 (r 2  a 2 ) r

r

a

a

r

r

T  1 2 r 2   1 2 r dr   2C10  2 r 2  2C10  2  r dr   C10  2 (3r 2  a 2 ) a

Tzz   2  2 r 2   1 2 r dr   C10  2 (r 2  a 2 ) r

Tz  (1   2 ) r  2C10 r a

a

M  2 (1   2 )r 3dr  4C10  r 3 dr  C10 a 4 0

0

a

a

a

0

0

0

N  2  2  2 r 3 dr    1 2 r 3dr  2C10 2  r 3 dr  C10 2 a 4 / 2 From linear elasticity : Trr  T  Tzz  0 , Tz  r , (  shear modulus ,   angle of twist)  a 4   ,N 0 M    2   So shear stress and twisting moment match, but other expressions are different

8.10

r   r R ,     Z , z   z Z  r (3.17.10)  F   0  0

r

 r B  FF T   0  0

0   r r   0  z   0

r2  B 1   0  0 

0 0

0 r 0 0

0 r  , det F  2r  z  1  z  0  2r   r 0    0 r  z   0 0

0

0 2r  r 2  2  z r

0    z r  2z 



  r / 2r  z  (2r  r 2  2 ) / 2r 2z 

r2  r / 2r  z

T   p1  1 B   2 B 1  Trr   p  12r   2 r2 T   p  1 (2r  r 2  2 )   2 r2 Tzz   p  12z   2 (2r  r 2  2 ) / 2r 2z Tz  1 z r   2 r / 2r  z , Tr  Trz  0 a

a

0

0

M   Tz 2r 2 dr  2 [1 z  a

a

0

0

2 3 ]r dr 2r  z

N   Tzz 2rdr  2  [12z   2  2 (2r  r 2  2 ) / 2r 2z ]rdr Expression s for M and N can be further reduced following procedures in Example 8.3.3

8.11

x1  1 X 1 , x2   2 X 2 , x3  X 3  1 0 F   0  2  0 0

0 0  , det F  1  1 2  1    1 / 1 2  

21 0  B   0 22 0 0 

12 0   0  , B 1   0  0  2  

0 22 0

0   0   2 

T   p 1  1 B   2 B 1  T11   p  121   2 12 T22   p  122   2 22 T33   p  1 / 2122   2 2122 T12  T23  T31  0 Now since T33  0  p  1 / 2122   2 2122 and so



 1  T11  (1 / 2122   2 2122 )  121   2 12   21  2 2  1  22  2 1 2  





 1  T22  (1 / 2122   2 2122 )  122   2 22   22  2 2  1  21 2 1 2   Next apply boundary conditions :



 1  F1  T11 2hb  2hb 21  2 2  1  22  2 1 2  





 

 1  F2  T22 2ha  2ha 22  2 2  1  21 2 1 2   Solving the system of two equations for the two unknows 1 and  2          2 2 F11 / b F2  2 / a  F1 / b F2 / a  1 1   1   , 2   2 2  1  1 1  2 2 2 2h(21  22 )  2  1 2 h (    ) 1 2 1  2 2   1  1  2 2 1  2 2  2 2 1 2 1 2  1 2 1 2   

8.12

v1  kx1 , v2  kx2 , v3  0    v  k  k  0  motion is isochoric k 2  k 0 0  D  0  k 0  D 2   0 0 0 0 0 

0 k2 0

0  0 0

Reiner - Rivlin Model : T  pI  1 D   2 D 2  T11   p  (1   2 k )k T22   p  (1   2 k )k T33   p , T12  T23  T31  0

8.13 2 2   , T12  1  4 2 2 1 / 2   o  1 ,  2 / 4  o  1 2 

T11  p  T22  p 

T11  p  T22  p  T12 

   2 2   (o  1 2 )  2  o 1  1  2   2 4  o 

   1   ( o  1 2 )    o 1  1  2  2  o 

MATLAB Plots : 0

10

-0.01

10

-1

10

-0.03

o

logT / 

-2

12

11

log(T +p)/ 

o

10

10

-0.05

10

-0.07

10

-3

10

-0.09

10 -4

10 -2 10

-1

10 log

0

10

-2

10

-1

10 log

0

10

8.14

Second Order Fluid Model under simple shearing flow : T   pI  1 A 1  2 A 12   3 A 2 ,  i   i ( )  T11   p   2  2 T22   p   2  2  2 3  2 T33   p T12  1 , T23  T31  0 Basic definitions : ( )  T12  1 ( )  1 ( )  T11  T33   2 ( )  2  2 ( )  T22  T33   2 ( )  2  2 3 ( )  2

8.15 v1  kx1 , v2  kx2 , v3  0  k 0 D  0  k 0 0

0 0  0

k 0 T A1  L  L  2 D  2 0  k 0 0

k 2 0  0 , A 2  A(1) L  LT A(1)  2  0 0 0 

Second Order Fluid Model : T   pI  1 A 1  2 A12   3 A 2  T11   p  21k  4 2 k 2  2 3 k 2 T22   p  21k  4 2 k 2  2 3 k 2 T33   p T12  T23  T31  0

0 k

2

0

0  0 0

8.16

T   pI  1 A 1  2 A 12   3 A 2   4 A 22   5 ( A 1 A 2  A 2 A 1)   6 ( A 1 A 22  A 22 A 1)   7 ( A 12 A 2  A 2 A 12 )   8 ( A 12 A 22  A 22 A 12 ) A 1  ( N  N T ) , A 2  2  2 N T N , 0 0 0  1 0 0 0 0 0      T T with N  1 0 0  N N  0 0 0 , NN  0 1 0  0 0 0 0 0 0 0 0 0 A 12   2 ( NN T  N T N ) , A 22  4  4 ( N T N )( N T N )  4  4 N T N A 1 A 2  2  3 ( N  N T ) N T N  2  3 N , A 2 A 1 2  3 N T N ( N  N T )  2  3 N T A 1 A 22  4  5 ( N  N T ) N T N  4  5 N , A 22 A 1 4  5 N T A 12 A 2  2  4 ( NN T  N T N ) N T N  2  4 N T N , A 2 A 12  2  4 N T N A 12 A 22  4  6 ( NN T  N T N ) N T N  4  6 N T N , A 22 A 12  4  6 N T N  T   pI  ( )( N  N T )  1 ( ) N T N   2 ( ) NN T , with ( )  1  2 5  3  4 6  5 1 ( )   2  2  2 3  2  4 4  4  4 7  4  88  6  2 ( )   2  2

8.17

0   2k 0  v1  2kx1 , v2   kx2 , v3  kx3  L   0  k 0    0 0  k  0 0   4k  4 0 0 A 1 2 D   0  2k 0  , A 2  A(1) L  LT A(1)  4k 2 0 1 0  0 0 0 1 0  2k   4 0 0 2 2 A1  4k 0 1 0  trA12  24 k 2 0 0 1 T   pI  1 A 1  2 A12   3 A 2  3 (trA12 ) A1  T11   p  4k1  4k 2 ( 2   3 )  4k3 (k 2 ) T22   p  2k1  4k 2 ( 2   3 )  2k3 (k 2 ) T33   p  2k1  4k 2 ( 2   3 )  2k3 (k 2 ) T12  T23  T31  0

8.18  ~ T  pI   ( I i , t , s ) Et (t  s )ds 0

 ~   ( s ) , Et (t  s )  Gt (t  s )  C t (t  s )  I  T  pI   ( s )C t (t  s )  I ds 0

 s  v1  0 , v2  x1 , v3  0  C t (t  s )  I    s  0 

2 2



 s 0  0 0 0 0 

T11   p   2  ( s ) s 2 ds , T22  T33  p , T12     ( s ) sds , T23  T31  0 0

0



( )  T12     s( s )ds 0



1 ( )  T11  T33   2  s 2( s )ds 0

 2 ( )  T22  T33  0

8.19 From (8.5.3) 

T11  T33   p   ( s )ds 0



T22   p   (1   2 s 2 )( s )ds 0



T12    s( s )ds 0

T23  T31  0  o  s / 1 e  12

With ( s )  

 (s)ds  0

  o  ,  s( s )ds   o ,  (1   2 s 2 )( s )ds  o (1  2  2 12 ) 0 0 1 1

T11  T33   p  T22   p 

o 1

o (1  2  2 12 ) 1

T12   o  T23  T31  0

8.20   U U  From the basic solution, T12  2   s    ds 0  I1 I 2 

The potential function has a functional form U  U ( s, I1 , I 2 ) , but I1  I 2  3   2 s 2  U  U ( s ), and we can use the chain rule U U U  ( s ) U  ( s )    I1 I 2  ( s ) I1  ( s ) I 2 

U 1 U 1 1 U    ( s ) 2 s  ( s ) 2 s s  ( s )

  U  U U   ( )  T12  2   s   ds   ds  0 0  ( s )  I1 I 2 

8.21

Shear flow : v1  0 , v2  x1 , v3  0  I1  I 2  3   2 s 2 U 

 I  15  (t ) 9  I  I 3   c(t )( I1  3) ( I1  3) 2  (t ) log 1 2   24(t )  c(t ) log 1 2 2 9    I 2  15 



 9  2 2 s 2  (t ) 4 4 9   24(t )  c(t ) log(1)  c(t )  2 s 2  s  (t ) log  2 2 9  



(t ) 4 4 9  2   s  (t ) log1   2 s 2   c(t )  2 s 2 2 2  9 

8.22   U   (t ) 4 4 9  2  ds     s  (t ) log1   2 s 2   c(t )  2 s 2  ds 0  ( s ) 0  ( s )  2 2  9   

( )  T12  





2  2 s  9    2( s )  3 s 3  (t ) 9  2c( s ) s  ds 0 2 2   1 2s 2 9    9( s )   2   ( s )  2 s 2   c( s ) sds 2 2 0  9  2 s  

8.23  9( s )  ( )  T12 /   2  ( s )  2 s 2   c( s ) sds 2 2 0  9  2 s 

Choosing (t )  Ao e aot , (t )  Bo e bot , c(t )  Co e cot   9  ( )  2  ao Ao e ao s  2 s 2  b B e bo s  co Co e co s  sds 2 2 o o 0  9  2 s   e bo s sds  2co Co  e co s sds 2 2 0 0 9  2 s 0  6   1   1  3b    2ao Ao  2  4   18bo Bo  2 g  o    2co Co  2   2   2    ao   co  

 2ao Ao  2  e ao s s 3 ds  18bo Bo 



 zt 6A  te 9 B b  3b  C   2  3o  2  o 2 o g  o   o  , with g ( z )   dt 0 1 t 2 c a 2   2   o  o 

8.24

For Uniaxi al Extensiona l Isochoric Deformation :   ( )  ( ) x1 (t )   (t )   (t ) x1   (t ) X 1  (t ) ξ ()  x2  1/ 2 (t ) X 2   2 ()  x2 (t )  Ft ()   0  (  )  x ( t )  x3  1/ 2 (t ) X 3   (t )  3 ( )  x3 (t )  0  ()  1 () 

0  (t )  ( ) 0

    0    (t )   ()  0

  ( )    ( )    ()  2 0 0  0 0       (t )    (t )    (t )       (t )  (t ) C t ()  Ft ()T Ft ()   0 0  0 0  0  ( )  ( )       (t )    (t )   0 0 0 0   0  ()    ()    Since C t () is diagonal its inverse is easily found by inverting the diagonal terms   (t )  2     ()   C t1 ()   0   0  

0  ( )  (t ) 0

 0    0    ( )   (t )  

0  (t )  ( ) 0

 0    0    (t )   ()  

8.25 The stress difference follows from (8.5.17) 2 t  U  (t ) t  U  ( ) U 2 ()  U  (t )    T11  T22  T11  T33  2    d  2      d 2 2  I  ( )  I  (t ) I 2  (t )  I 2 ()   1  1 2 2 t   (t ) t   (t )  ()   U () U   ()   (t )  2  2   d    h  , t    d       2   ( )   ( ) (t )   I1  (t ) I 2  (t )    ()   

 U () U   (t )   (t ) where h  , t     2    , since I1 and I 2 are functions of  ()  ()   I1 (t ) I 2  1, t  0 Now for single step stress relaxation (t )    , t  0  1 t 1  1       2    h , t    d   2    h ,   d   2   H (, t ) , H (, t )   h(, ) d t     t    

8.26 1 () 

 () 1 ()  () x1 (t ) ,  2 ()  2 x2 (t ) ,  3 ()  3 x3 (t ) 1 (t )  2 (t )  3 (t )

 1 ()   1 (t ) ξ ()  (a) Ft ()   0 x (t )    0   () 1 (t )  2 (t )  3   3 (t ) 1 ()  2 ()

0  2 ()  2 (t ) 0

 1 ()   1 (t ) (b) C t ()  Ft ()T Ft ()   0    0   21 ()  2  1 (t )   0    0   21 (t )  2  1 ()  C t1 ()   0    0 

0 22 (t ) 22 () 0

 0    ()  2 ()  3 () 0  , det F  1  1 1  1 (t )  2 (t )  3 (t )  3 ()    3 (t ) 

0  2 ()  2 (t ) 0

0 22 () 22 (t ) 0

  2 (t ) 0   21   1 ()   0  0   2  3 (t )     0 23 ()   

   () 0  1   1 (t ) 0  0   3 ()    0  3 (t )     21 () 0   2   1 (t )   0  0   2  3 ()     0 23 (t )    0 22 (t ) 22 () 0

2 21 (t ) 22 (t )  1 ()  2 ()  1  I1  trC t ()  2   1 () 22 ()  1 (t )  2 (t ) 

2 () 2 ()  1 (t )  2 (t )   I 2  trC t ()  12  22  1 (t )  2 (t )  1 ()  2 () 

2

0  2 ()  2 (t ) 0

0 22 () 22 (t ) 0

 0   0    3 ()    3 (t )       0  2  1 (t )  2 (t )      (  )  (  ) 2  1  

     0  2  1 ()  2 ()      1 (t )  2 (t )   0

0

8.26 Continued   U  U (c) Stresses : T  pI  2   C t1 (t  s )  C t (t  s ) ds  0 I I 2  1  2   U  (t ) U 21 ()  1 T11   p  2     ds 2 0 I 2 ( )  I  ( t ) 2 1  1 1  2   U  (t ) U 22 ()  2 T22   p  2     ds 2 0 I 2 ( )  I  ( t ) 2 2  1 2    U   ( )  ( )  2  1  T33   p  2  0  I   (t )  (t )  1  1 2 



 T12  T23  T31  0

2

2 U  1 (t )  2 (t )      ds  I 2  1 ()  2 ()   

9.1 Two dimensional theory (plane strain) u  u ( x, y ), v  v( x, y ), w  0 1  v u   x   y  0 ,  z   z ( x, y )     2  x y   ij  u j ,i   ijl l  u v v u , y  ,  xy    z   yx    z ,  z   zx   zy  0 x y x y General Constituti ve Equations (9.2.4) x 

Tij  kk  ij  (  ) ij   ji mij   k ,k  ij   i , j   j ,i  Txx   ( x   y )  (2  ) x T yy   ( x   y )  (2  ) y Txy  (2  ) xy  T yx m xz  

 z  , m yz   z x y

9.2

Using (9.2.8) and (9.2.9) x 

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

2  2  xy  2 x   y  2  u   2  v   2  v u   2     2   xy  x y  xy y 2  x  x 2  y  y 2 x 2

 2 z  2 z  follows directly from math continutiy xy yx  xy  x    u   z    u     z      z   x x  y  y  x  x  x y     y  xy  z    v    v         z   z   y x  y  y  x y x y 

9.3

Txx T yx    2  2     2  2      0  0   x y x  y 2 xy  y  xy x 2  Txy x



T yy y

0

   2  2     2  2     0   x  xy y 2  y  x 2 xy 

m xz m yz           Txy  T yx  0   Txy  T yx  0     x y x  x  y  y  all forms satisfy all equations

9.4 Couple Stress Function Formulation :  2  2   2  2   2  2   2  2   , T   , T    , T    yy xy yx xy y 2 xy x 2 y 2 xy x 2 xy   m xz  , m yz  x y Using these forms in the compatibility relations (9.2.13)

Txx 

2  2Txx  T yy 2 2     ( T  T )  (Txy  T yx )  xx yy xy y 2 x 2 2  4  4   4  4   2   4  4  4  4 2           2  2    2 2     y 4 xy 3 x 4 x 3y y  x y xy 3 x 2 y 2 yx 3  x 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 identically 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     2l 2    l   2     x 2  y 2   x x  x 2 xy  y  y 2 xy y 2  













  2   l 2  2   2(1  )l 2   . . . (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   . . . (3) y x Differentiating equation (2) by x and equation (3) by y and adding   2   l 2 4   0





9.5 Stress Concentration Around Circular Hole - Micropolar Case Hoop Stress Solution : T 

  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  

where A1  

4T (1  )a 2 l 22 Tal1 F T 2 Ta 4 (1  F ) Ta 2 a , A2   , A3  , A4  , A5   2 4(1  F ) 2(1  F ) 1 F (1  F ) K 1 (a / l1 )

l2  a 2 2a K o (a / l1 )  F  8(1  ) 22 4  2   l 1 K 1 (a / l1 )  l1  l1 On Hole Boundary : r  a T (a, )  

1

T T  6T (1  F ) 24T (1  )l 22  (1  cos 2)      cos 2 2 2  4(1  F ) a 2 (1  F )   3l1   6l12  2TF 1   K 1 (a / l1 ) cos 2 K ( a / l )   o 1 2  (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 K1 (a / l1 ) (1  F ) (1  F )a  T 

Fl12  T  1 3(1  F ) 24(1  )l 22 6 Fl1 K o (a / l1 )    2 F  12  (1  F )   cos 2 1 F  2 2 a K 1 (a / l1 ) a2 a2 

But from definition of F  T (a, )  T 

K o (a / l1 ) l1  8(1  ) l 22 a2    4    and thus K 1 (a / l1 ) 2a  F l12 l12 

Fl12  T  1 3(1  F ) 24(1  )l 22 3Fl12  8(1  ) l 22 a 2   ( 1  F )     4   2 F  12   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 this expression takes on a maximum value when cos 2  1      / 2 T 

2   3 F   (T ) 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 (T ) max  3T

1

 8(1  )

l 22 K (a / l1 ) 4l12  a 2  2al1 o K1 (a / l1 )

0

9.6

1 U  kk  mm  ( km km  l 2  km  km ) 2 1 1 1  ij  (ui , j  u j ,i ) ,  ij  (i , j   j ,i ) , i   ijk u k , j 2 2 2 U  1  2 Tij    kk  mm  ( km km  l  km  km )   ij  ij  2         1     kk mm   mm kk     km km   km km   kk  ij  2ij 2   ij  ij    ij  ij  mij  

U  1  2   kk  mm  ( km km  l  km  km )   ij  ij  2 





 l 2  km  km  2l 2  ij  ij

9.7

u1  0 , u2   x1 x3 , u3  x1 x2 (torsional deformation)  x3 / 2 x2 / 2  0 1   ij  (ui , j  u j , i )   x3 / 2 0 0  2  x2 / 2 0 0  0 0   x1   1 1    i   ijk uk , j   x2 / 2 , ij  (i , j   j , i )   0   / 2 0  2 2   x3 / 2   0 0   / 2  x3 x2   0  Tij  kk ij  2ij   x3 0 0   x2 0 0  0 0  2 l 2   mij  2l 2  ij  2l 2  0   / 2 0    0  0 0   / 2  0

0  l 0

  0   l 2  0

2

9.8

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

kk ij

ij

ij

i

,i

kk

Under Plane Stress Condtions : Txx  ( x   y   z )  2 x   T yy  ( x   y   z )  2 y   Tzz  ( x   y   z )  2 z    0  (  2) z    ( x   y ) Txy  2 xy , Txz  T yz  0 Using the relation for e z     ( x   y )   2 2   2 x     Txx    x   y  ( x   y )    2 x   2   2   2   2 2 T yy   ( x   y )    2 y   2   2 2 Combining these results gives : Tij  (kk  ) ij  2ij , where  kk   x   y   2    ( x   y )    g       kk      ( x   y   z )        x   y    2    ( x   y )    2 2  2                  x   y      2 ( x   y )   2    2    2   

9.9 For elasticity with voids, the plane stress compatibility relation is the same as that originally developed in Chapter 6, given by (6.2.45)  2 x

 2 y

 2  xy

Txx  

2 2 4(  ) 2 2 ( x   y )    2 x  x  y    2   2   2   2   2

 2 xy y 2 x 2 From the stress - strain relations developed in Exercise 14 - 16



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

T yy 

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

These relations can be re - cast in the form ˆ Eˆ ˆ  E (  ˆ  ) , with ˆ Tˆxx  (     ) , T x y yy y x 1  ˆ 2 1  ˆ 2 2 ˆ 2 ˆ (3  2)  Tˆxx  Txx  , T yy  T yy  ,E , ˆ    2   2 (   ) 2(    ) Thus the solution for the strains follows as 1 1 1 1  ˆ  x  (Tˆxx  ˆ Tˆyy ) ,  y  (Tˆyy  ˆ Tˆxx ) and also  xy  Txy  Txy 2 Eˆ Eˆ Eˆ Substituti ng these into the strain compatibility relation  2Txy  2Tˆxx  2Tˆyy 1 2 2 2 ˆ ˆ  (Txx  T yy )  2      2 2 (1  ˆ ) xy   2 x y 3  2 2 2  2 (Tˆxx  Tˆyy )   0 2(  )   2 4 2 3  2   2 (Txx  T yy )     2  0    2 (  )   2  2  2 (Txx  T yy )   0  Introducing the usual Airy stress function Txx   4 

 2  0 

 2  2 , T   yy x 2 y 2

9.10

  0 ,   0 ,   0 ,   0 , 3  2  0 , M  (9.3.13)  h 2 

3  2 3 2

 /  2  2  1  (   )

3  2 3  3   2   3    3   3    1  1 (   ) 2 2 2 2 and with   0 ,   0 we conclude that h 2  0

9.11 Hoop stress solution for elasticity with voids

T 

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

F (r )  1 

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

1 a a When N  , L   2  h   2 h 2 2  K ( 2)  a 1 a 2  6 4 K 2 (2r / a )    F (a)  1   2  2 2 , F ( r )     4  2 a 2 r4 a K 2 (2)  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 1  a 2  2 F (r )  a 4      T (r ,  / 2) / T  1  2   a  1  3 4  2  r  F (a )  r  From classical elasticity , the corresponding result is 1 a 2 3a 4   T (r ,  / 2) / T   2  2  4  2 r r  MATLAB Plots :

9.12

 (q11 ) 2  Q  (q12 ) 2  q11 q12 

(q31 ) 2   (q32 ) 2  q31 q32 

(q21 ) 2 (q22 ) 2 q21 q22

 cos 2   cos 2  cos 2 0   2 2 2   sin  sin  cos ( / 2)   cos  sin  cos  sin  cos 0 cos( / 2)  

P2

P1





P3

 cos 2  cos 2  1     sin 2  sin 2  0  cos  sin  cos  sin  0  

(Microstresses)

9.13 MATLAB Contour Plots Doublet Mechanics Solution – p1 and p2 Microstresses for Flamant Problem

+

 = 60

+

-

-

p2 – Contours

p1 – Contours

9.14 u  u ( x) , v  w  0   du  dx 1  ij  (ui , j  u j ,i )   0  2 0 

 0 0 0 0  0 0 

Tij  kk  ij  2ij  c[kk  ij  2ij ],ll  Tij  

  du  du du  du    ij  2  i1 j1  c    ij  2  i1 j1   dx dx  dx  ,kk    dx  ,kk

 du  du d 3u  d 3u  Txx  (  2)  c 3  , T22  T33    c 3  , T12  T23  T31  0 dx  dx   dx  dx Equilibriu m Equaitons : Tij , j  0 

dTxx d 2u d 4u  0  2 c 4  0  dx dx dx

d 4u 1 d 2u d 2  d 2u 1  d 2u 1  0   0   u  u  C1  C 2 x  dx 4 c dx 2 dx 2  dx 2 c  dx 2 c  x   x  u  c1  c2 x  c3 sinh    c4 cosh   c  c

9.15   0 1 1 u  v  0 , w  w( x, y )   ij  (ui , j  u j ,i )  0 2 2  w   x

 0 0 0  w 0 y  Tij  kk  ij  2ij  c[kk  ij  2ij ],ll  2ij  2cij ,kk  0

Txx  T yy  Tzz  Txy  0 Txz  

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

Equilibriu m equaitons : Tij , j  0  T31,1  T32 , 2  0 

Txz T yz  0 x y

  2 w  2 w         2 w  2 w         w  c 2  2       w  c 2  2    0  x  x  y   y  y  y    x  x 2w 2w  4w 4w 4w   0   c  2  2 2 4   x 4 x 2 y 2  x  y  y  

9.16  P  Py  y Py  2 xy 2 Pxy 2 y Txy    ye sin(x)d   e sin(x)d      0  0   (x2  y 2 )2  ( x 2  y 2 ) 2  P  P  (1  y )e y cos(x)d     e y cos(x)d  y  e y cos(x)d    0 0 0   P y y2  x2  2 Py 3   2  y      x  y2 (x2  y 2 )2  ( x 2  y 2 ) 2  P  P  Txx    (1  y )e y cos(x)d     e y cos(x)d  y  e y cos(x)d  0  0   0

T yy  



P y y2  x2  2 Px 2 y  y       x2  y2 (x2  y 2 )2  ( x 2  y 2 ) 2

9.17

Using standard translation from direct to index notation : I   ij , M  M ij , M 2  M ij M jk , ε   ij , Mε  M ij  jk εM   ij M jk , M 2 ε  M ij M jk kl , εM 2   ij M jk M kl trε   kk , tr ( Mε )  M ij  ji , tr ( M 2 ε )  M ij M jk  ki T  f1 I  f 2 M  f 3 M 2  f 4 ε  f 6 ( Mε  εM )  f 7 ( M 2 ε  εM 2 )  Tij  f1 ij  f 2 M ij  f 3 M ik M kj  f 4 ij  f 6 ( M ik  kj   ik M kj )  f 7 ( M ik M kl lj   ik M kl M lj ) For linear in strain, (9.6.5)  f1  a1 kk  a2 M rp pr  a3 M rq M qp pr f 2  d1 kk  b1M rp pr  b2 M rq M qp pr f 3  d 2 kk  d 3 M rp pr  b3 M rq M qp pr f 4  2c1 , f 6  2c2 , f 7  2c3 Tij   ij (a1 kk  a2 M rp pr  a3 M rq M qp pr )  M ij (d1 kk  b1M rp pr  b2 M rq M qp pr )  M is M sj (d 2 kk  d 3 M rp pr  b3 M rq M qp pr )  2c1 ij  2c2 ( M ir  rj   ir M rj )  2c3 ( M ip M pr rj   ir M rp M pj )

9.18 Cijkl  a1 ij  kl  a2 ( M ij  kl   ij M kl )  a3 ( ij M kq M ql   kl M iq M qj )  b1M ij M kl  b2 ( M ij M kq M ql  M is M sj M kl )  b3 M is M sj M kq M ql  c1 ( ki lj   li  kj )  c2 ( M ik  lj  M kj li  M il  kj  M lj  ki )  c3 ( M ir M rk  lj  M kr M rj  li  M ir M rl  kj  M lr M rj  ik ) For the case M ij  M o  ij Cijkl  a1 ij  kl  a2 M o ( ij  kl   ij  kl )  a3 M o2 ( ij  kq ql   kl  iq  qj )  b1M o2  ij  kl  b2 M o3 ( ij  kq ql   is  sj  kl )  b3 M o4  is  sj  kq ql  c1 ( ki lj   li  kj )  c2 M o ( ik  lj   kj li   il  kj   lj  ki )  c3 M o2 ( ir  rk  lj   kr  rj  li   ir  rl  kj   lr  rj  ik )  a1 ij  kl  2a2 M o  ij  kl  2a3 M o2  ij  kl  b1M o2  ij  kl  2b2 M o3 ij  kl  b3 M o4  ij  kl  c1 ( ki lj   li  kj )  2c2 M o ( ik  lj   kj li )  2c3 M o2 ( ik  lj   kj li )  C1111  C 2222  C3333  a1  2a2 M o  2a3 M o2  b1M o2  2b2 M o3  b3 M o4  2c1  4c2 M o  4c3 M o2  a1  2c1  (2a2  4c2 ) M o  (2a3  b1  4c3 ) M o2  2b2 M o3  b3 M o4 C1122  C1133  C 2233  a1  2a2 M o  (2a3  b1 ) M o2  2b2 M o3  b3 M o4 C1212  C1313  C 2323  c1  2c2 M o  2c3 M o2 From classical linear elasticity - see (6.2.19) C1111  C 2222  C3333    2 , C1122  C1133  C 2233   , C1212  C1313  C 2323      a1  2a2 M o  (2a3  b1 ) M o2  2b2 M o3  b3 M o4   c1  2c2 M o  2c3 M o2

9.19 0  M 1 0  For the case with M ij   0 M 2 0   M 1 i1 j1  M 2  i 2  j 2  M 3 i 3 j 3 , (9.6.8)   0 0 M 3  Cijkl  a1 ij  kl  a2 ( M 1 i1 j1  M 2  i 2  j 2  M 3 i 3 j 3 ) kl   ij ( M 1 k1 l1  M 2  k 2  l 2  M 3 k 3 l 3 )





  ij ( M 1 k1 q1  M 2  k 2  q 2  M 3 k 3 q 3 )(M 1 q1 l1  M 2  q 2  l 2  M 3 q 3 l 3 )    a3     kl ( M 1 i1 q1  M 2  i 2  q 2  M 3 i 3 q 3 )(M 1 q1 j1  M 2  q 2  j 2  M 3 q 3 j 3 )     b1 ( M 1 i1 j1  M 2  i 2  j 2  M 3 i 3 j 3 )(M 1 k1 l1  M 2  k 2  l 2  M 3 k 3 l 3 )  ( M 1 i1 j1  M 2  i 2  j 2  M 3 i 3 j 3 )(M 1 k1 q1  M 2  k 2  q 2  M 3 k 3 q 3 )(M 1 q1 l1  M 2  q 2  l 2  M 3 q 3 l 3 )    b2    ( M 1 i1 s1  M 2  i 2  s 2  M 3 i 3 s 3 )(M 1 s1 j1  M 2  s 2  j 2  M 3 s 3 j 3 )(M 1 k1 l1  M 2  k 2  l 2  M 3 k 3 l 3 )     ( M 1 i1 s1  M 2  i 2  s 2  M 3 i 3 s 3 )(M 1 s1 j1  M 2  s 2  j 2  M 3 s 3 j 3 )(M 1 k1 q1  M 2  k 2  q 2  M 3 k 3 q 3 )    b3   ( M 1 q1 l1  M 2  q 2  l 2  M 3 q 3 l 3 )     c1 ( ki lj   li  kj )  ( M 1 i1 k1  M 2  i 2  k 2  M 3 i 3 k 3 ) lj  ( M 1 k1 j1  M 2  k 2  j 2  M 3 k 3 j 3 ) li    c2    ( M 1 i1 l1  M 2  i 2  l 2  M 3 i 3 l 3 ) kj  ( M 1 l1 j1  M 2  l 2  j 2  M 3 l 3 j 3 ) ki     ( M 1 i1 r1  M 2  i 2  r 2  M 3 i 3 r 3 )(M 1 r1 k1  M 2  r 2  k 2  M 3 r 3 k 3 ) lj      ( M 1 k1 r1  M 2  k 2  r 2  M 3 k 3 r 3 )(M 1 r1 j1  M 2  r 2  j 2  M 3 r 3 j 3 ) li   c3     ( M 1 i1 r1  M 2  i 2  r 2  M 3 i 3 r 3 )(M 1 r1 l1  M 2  r 2  l 2  M 3 r 3 l 3 ) kj    ( M    M    M   )(M    M    M   )  1 l1 r1 2 l2 r2 3 l3 r3 1 r1 j1 2 r2 j2 3 r 3 j 3 ik   For orthotropic materials, elasticity matrix must be of the form C1111      Cij         

C1122

C1133

0

0

C 2222

C 2233

0

0



C3333

0

0





C 2323

0







C3131









0  0  0   , with the indicated nine non - zero components 0  0   C1212 

Explore some specific components : C1111  a1  2a2 M 1  2a3 M 12  b1M 12  2b2 M 13  b3 M 14  2c1  4c2 M 1  4c3 M 12 .. . (note similarity with previous exercise) Other components follow a similar pattern but with C1111  C 2222  C3333

9.20

Two - dimensional case : (0    2) ()  2

0

1 1 (1  a cos 2)  (1  a cos 2   a sin 2 )  2 2 2

()d 

1 2 1  a  (1  a cos 2)d  2   sin 2  1   0 2 2  2 0

n  (cos , sin ) 





1 1 D11n1n1  2 D12 n1n2  D22 n2 n2   1 D11 cos 2   2 D12 sin  cos   D22 sin 2  Dij ni n j  2 2 2 1 1 1 ( n)  Dij ni n j  (sin 2   cos 2   a cos 2   a sin 2 )  D11 cos 2   2 D12 sin  cos   D22 sin 2   2 2 2 0  1  a D11  1  a , D12  D21  0 , D22  1  a  D     0 1  a



2

1 2 (1  a cos 2)ni n j d  0 2 0 1 2 1 a M 11  (1  a cos 2) cos 2 d  1    2 0 2 2 1 2 M 12  (1  a cos 2) sin  cos d  0 2 0 1 2 1 a M 22  (1  a cos 2) sin 2 d  1    0 2 2 2 0  1 1  a / 2  M  1  a / 2 2 0 M ij   ()ni n j d 

9.21

For a given bar  

Pi l  Pi  ki  where k  ki  Ai Ei / l Ai Ei N

N

i 1

i 1

The total load capacity is P   Pi   ki   Nk . . . for no breakage n  Now if n - bars fail P  ( N  n)k  N 1  k  K (1  D)  N where K  Nk and the damage paramerter is given by D  n / N Note the similarity with relation (9.7.5)



9.22  1 D 1 2  2 2 U1 D       E  (1  D) 2 E 2 2 2 E 2 E 2 E (1  D) 

9.23 0.8

1.6

0.7

1.4

Damage Parameter, D

Nondimensional Stress,  /E

m=5

0.6 0.5 0.4

m=10

0.3 0.2

m=20

0.1 0

1.2

m=10 1 0.8

m=5 0.6 0.4 0.2 0

0

0.2

0.4

0.6

0.8

1 1.2 Strain, 

1.4

1.6

1.8

m=20

2

0

0.2

0.4

0.6

0.8

1 1.2 Strain, 

1.4

1.6

9.24 Using (6.611) and (6.612) decompose the strain energy into volumetric and distortional parts Y 

U 1 U v  U d   1  1  2 (Tkk ) 2  1 TˆijTˆij   1 D 1 D 1  D  6E 4     2   e2 1  1  2 2 1    2 2    3(1  2)  H   2 (1  )  9 H     e    2    1  D  6E 2 E  3   2 E (1  D)   e  3 

2 2  H    e2  (1  )  3(1  2)   , where we have used E  (1  D) E   e   2 E (1  D) 2  3   

1.8

2