Page 252 - Intro to Tensor Calculus

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We conclude that if the material undergoes a strain, with the x 1 -x 2 plane as a plane of symmetry then

e 5 and e 6 change sign upon reversal of the x 3 axis and e 1 ,e 2 ,e 3 ,e 4 remain unchanged. Similarly, we find σ 5
and σ 6 change sign while σ 1 ,σ 2 ,σ 3 ,σ 4 remain unchanged. The equation (2.4.6) then becomes

     
e 1 s 11 s 12 s 13 s 14 s 15 s 16 σ 1
 s 21 s 22 s 23 s 24 s 25
 e 2  s 26   σ 2 
     
 s 31 s 32 s 33 s 34 s 35
 . (2.4.12)
 e 3  s 36   σ 3 
  =   
 s 41 s 42 s 43 s 44 s 45
 e 4  s 46   σ 4 
     
−e 5 s 51 s 52 s 53 s 54 s 55 s 56 −σ 5
−e 6 s 61 s 62 s 63 s 64 s 65 s 66 −σ 6
If the stress strain relation for the new orientation of the x 3 axis is to have the same form as the
old orientation, then the equations (2.4.6) and (2.4.12) must give the same results. Comparison of these
equations we find that
s 15 = s 16 =0
s 25 = s 26 =0

s 35 = s 36 =0
(2.4.13)
s 45 = s 46 =0
s 51 = s 52 = s 53 = s 54 =0

s 61 = s 62 = s 63 = s 64 =0.
In summary, from an examination of the equations (2.4.6) and (2.4.12) we find that for an aelotropic
material (crystal), with one plane of symmetry, the 36 constants s ij reduce to 20 constants and the generalized
Hooke’s law (constitutive equation) has the form

     
e 1 s 11 s 12 s 13 s 14 0 0 σ 1
 s 21 s 22 s 23 s 24 0
 e 2  0   σ 2 
   0   
 s 31 s 32 s 33 s 34
 . (2.4.14)
 e 3  0   σ 3 
  =  0  
 s 41 s 42 s 43 s 44
 e 4  0   σ 4 
   0 0 0 0   
e 5 s 55 s 56 σ 5
0 0 0 0
e 6 s 65 s 66 σ 6
Alternatively, the Hooke’s law can be represented in the form
     
σ 1 c 11 c 12 c 13 c 14 0 0 e 1
 c 21 c 22 c 23 c 24 0
 σ 2  0   e 2 
 
   
 c 31 c 32 c 33 c 34 0
 .
 σ 3  0   e 3 
  =   
 c 41 c 42 c 43 c 44 0
 σ 4  0   e 4 
     
σ 5 0 0 0 0 c 55 c 56 e 5
σ 6 0 0 0 0 c 65 c 66 e 6