MECH 4301 Fall 2022  
Intermediate Mechanics of Materials  

Homework Assignment #5  
(Solution)  

1. (Anisotropic stiffness) In general, the constitutive relation for linear elastic solids is expressed  
in the following form:  
1111 1122 1133 1123 1113 111211 
2211 2222 2233 2223 2213 221222 
3311 3322 3333 3323 3313 331233 
2311 2322 2333 2323 2313 231223 
1311 1322 1333 1323 1313 131213 
1211 122212 
C C C C C C 
C C C C C C 
C C C C C C 
C C C C C C 
C C C C C C 
C C C 
σ 
σ 
σ 
σ 
σ 
σ 
  
  
  
  
=  
  
  
   
  
11 
22 
33 
23 
13 
1233 1223 1213 1212 12 
2 
2 
2C C C 
ε 
ε 
ε 
ε 
ε 
ε 
   
   
   
   
   
   
   
     
   
.  
Prove the following relation in the stiffness of isotropic linear elastic solids.  
( )1212 1111 1122 
1 
2 
C C C= − .  
[Hint] Considering the symmetry of stiffness tensor of hexagonal crystal, you can start from the  
stiffness of tetragonal crystals, as shown  
1111 1122 1133 
1111 1133 
3333 
2323 
2323 
1212 
0 0 0 
0 0 0 
0 0 0 
0 0 
0 
C C C 
C C 
C 
C 
sym C 
C 
  
  
  
  
=   
  
  
  
   
C  
, where there are 6 independent constants. The hexagonal system exhibits a sixfold rotation  
around x3 axis, so the structure superimposes upon itself after rotation of 60º along x3 axis. To  
derive the relation, perform the coordinate transformation of 1122C with respect to the rotation  
of 60º along x3 axis.  

2  
(Solution)  
Following the above figure, we could compute the transformation matrix due to the rotation.  
Only nonzero components in the transformation matrix are  
11 1 1 12 1 2 
21 2 1 22 2 2 
1 3,  
2 2 
3 1,  
2 2 
Q e e Q e e 
Q e e Q e e 
′ ′= ⋅ = = ⋅ = 
′ ′= ⋅ = − = ⋅ = 
    
    
.  
From the crystal symmetry in cubic, we know that stiffness tensor is invariant with respect to the  
rotation.  
1122 1122C C′ =  
Now, performing coordinate transformation of rotation gives us the following relation.  
1122 1 1 2 2p q r s pqrsC Q Q Q Q C′ =  
Following summation convention, free indices p, q, r, and s can be expanded as  
1122 1 1 2 2 
11 11 21 21 1111 11 11 21 22 1112 11 11 22 21 1121 11 11 22 22 1122 
11 12 21 21 1211 11 12 21 22 1212 11 12 22 21 1221 11 12 22 22 1222 
12 11 21 



p q r s pqrsC Q Q Q Q C 
Q Q Q Q C Q Q Q Q C Q Q Q Q C Q Q Q Q C 
Q Q Q Q C Q Q Q Q C Q Q Q Q C Q Q Q Q C 
Q Q Q 
′ = 
= + + + 
+ + + + 
+ 21 2111 12 11 21 22 2112 12 11 22 21 2121 12 11 22 22 2122 
12 12 21 21 2211 12 12 21 22 2212 12 12 22 21 2221 12 12 22 22 2222  
Q C Q Q Q Q C Q Q Q Q C Q Q Q Q C 
Q Q Q Q C Q Q Q Q C Q Q Q Q C Q Q Q Q C 
+ + + 
+ + + + 

,where only nonzero components of transformation matrix( ijQ ) are considered.  

From the symmetry in the stiffessness tensor for tetragonal structures, we can also simplify the  
stiffness tensor, as follows.  
( ) ( ) 
( ) ( ) ( ) 
( ) ( ) ( ) 
11 11 21 21 1111 11 11 21 22 11 11 22 21 11 11 22 22 1122 
11 12 21 21 11 12 21 22 1212 11 12 22 21 1212 11 12 22 22 
12 11 21 21 12 11 21 22 1212 12 11 22 21 1212 
0 0  
0 0  
0  
Q Q Q Q C Q Q Q Q Q Q Q Q Q Q Q Q C 
Q Q Q Q Q Q Q Q C Q Q Q Q C Q Q Q Q 
Q Q Q Q Q Q Q Q C Q Q Q Q C Q 
= + + + 
+ + + + 
+ + + + ( ) 
( ) ( ) ( ) ( ) 
12 11 22 22 
12 12 21 21 1122 12 12 21 22 12 12 22 21 12 12 22 22 1111 
11 11 21 21 1111 11 11 22 22 1122 11 12 21 22 1212 11 12 22 21 1212 
12 11 21 22 1212 12 11 22 21 1212 12 
0  
0 0  
Q Q Q 
Q Q Q Q C Q Q Q Q Q Q Q Q Q Q Q Q C 
Q Q Q Q C Q Q Q Q C Q Q Q Q C Q Q Q Q C 
Q Q Q Q C Q Q Q Q C Q 
+ + + + 
= + + + 
+ + + 12 21 21 1122 12 12 22 22 1111Q Q Q C Q Q Q Q C+ 


Plugging nonzero component of transformation matrix( ijQ ) gives  
3  
1111 1122 
1212 1212 
1 1 3 3 1 1 1 1 
2 2 2 2 2 2 2 2 
1 3 3 1 1 3 1 3 
2 2 2 2 2 2 2 2 
3 1 3 1 
2 2 2 2 
C C 
C C 
         = − − +                     
             + − + −                                 
    + −            
1212 1212 
1122 1111 
3 1 1 3 
2 2 2 2 
3 3 3 3 3 3 1 1 
2 2 2 2 2 2 2 2 
C C 
C C 
       + −                  
         + − − +                          

After simplifying all terms, this equation ends up with  
1122 1111 1122 1212 1212 1212 1212 1122 1111 
1122 1111 1122 1212 
3 1 3 3 3 3 9 3 
16 16 16 16 16 16 16 16 
6 10 12 
16 16 16 
C C C C C C C C C 
C C C C 
= + − − − − + + 
= + − 

( )1212 1111 1122 
1 
2 
C C C∴ = −  

2. (Compatibility) Prove the following compatibility equation,  
, , , , 0ij k k ij ik j j ikε ε ε ε+ − − =     
using the kinematic relation:  
1 
2 
ji 
ij 
j i 
uu 
x x 
ε 
 ∂∂ 
= +  ∂ ∂  
.  
(Solution)  
( ) ( ) ( ) ( ), , , , , , , , , , , ,, ,, , 
, , , , , , , , 
1 1 1 1 
2 2 2 2 
1 
2 
ij k k ij ik j j ik i j j i k k i k k i j jij jk ik 
i jk j ik k ij kij i kj k ij j ij jik 
u u u u u u u u 
u u u u u u u u 
ε ε ε ε+ − − = + + + − + − + 
 = + + + − − − +  
        
        

Since the order of partial differentiation does not matter, it goes to zero.  

3. (Hydrostatic stress) The stress field in a stationary fluid is hydrostatic,  
0 0 
0 0 
0 0 
ij i j ij i j 
p 
e e p e e p 
p 
σ δ 
−  
 = ⊗ = − ⊗ = −  
 −  
σ 
    

, where ( ), ,p p x y z= is the pressure field. Show that the equilibrium equations ,ij j iF oσ + =  
imply that the pressure must satisfy the following relation p∇ = F .  
(Solution)  
Plugging the hydrostatic stress state into the equilibrium equation gives  
4  
( ) 
, 
, ,, 
, 
0 
0 
ij j i 
ij i j ij i i ij 
i i 
F 
p F p F p F 
p F p 
σ 
δ δ 
+ = 
− + = − + = − + = 
∴ = ⇔∇ = F 



4. (Equilibrium and compatibility) Show that the following stress component satisfy the  
equations of equilibrium with zero body force, but it does not satisfy compatibility  
( ) 
( ) 
( ) 
2 2 2 
2 2 2 
2 2 
2 
0 
0 
xx 
yy 
zz 
xy 
yz 
zx 
c y x y 
c x x y 
c x y 
c xy 
σ ν 
σ ν 
σ ν 
σ ν 
σ 
σ 
 = + −  
 = − −  
= + 
= − 
= 
= 

, where c is a nonzero constant. Note that there are only six independent equations in the  
compatibility equation. You don’t need to check all equations from  
, , , , 0ij k k ij ik j j ikε ε ε ε+ − − =    which gives rise to 81 equations.  
2 2 2 2 22 2 2 
2 2 
2 2 22 
2 
2 2 22 22 2 
2 2 2 2 
0, 0 
0 
2 0, 2 0, 2 
yz xy yy xy yzxx zx zx 
xy yz zxzz 
yz yy xyzx xxzz zz 
y z x x y x z z x y y z y x 
x y z z x z y 
y z z y z x x z x y 
ε ε ε ε εε ε ε 
ε ε εε 
ε ε εε εε ε 
∂ ∂ ∂ ∂ ∂∂ ∂ ∂ 
+ − − = + − − = 
∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ 
∂ ∂ ∂∂ 
+ − − = 
∂ ∂ ∂ ∂ ∂ ∂ ∂ 
∂ ∂ ∂∂ ∂∂ ∂ 
− − = − − = − 
∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ 
22 
2 2 0 
yyxx 
y x 
εε ∂∂ 
− = 
∂ ∂ 

(Solution)  
First of all, the equilibrium condition with zero body force can be expressed as  
0 
0 
0 
xyxx xz 
yx yy yz 
zyzx zz 
x y z 
x y z 
x y z 
σσ σ 
σ σ σ 
σσ σ 
∂∂ ∂ 
+ + = 
∂ ∂ ∂ 
∂ ∂ ∂ 
+ + = 
∂ ∂ ∂ 
∂∂ ∂ 
+ + = 
∂ ∂ ∂ 

To check if the stress state satisfies the equilibrium condition, plugging all stress components  
into the equilibrium equation shows that  
5  
( )( ) ( ) ( ) 
( ) ( )( ) ( ) 
( ) ( ) ( )( ) 
2 2 2 
2 2 2 
2 2 
2 0 
2 2 0 
2 0 
2 2 0 
0 0 
0 
c y x y c xy 
c x c x 
x y z 
c x x yc xy 
c y c y 
x y z 
c x y 
x y z 
ν ν 
ν ν 
νν 
ν ν 
ν 
 ∂ + − ∂ − ∂  + + = − = 
∂ ∂ ∂ 
 ∂ − −∂ − ∂ + + = − + = 
∂ ∂ ∂ 
∂ +∂ ∂ 
+ + = 
∂ ∂ ∂ 
.  
Therefore, the stress state satisfies the equilibrium condition.  
Next, to check the compatibility conditions, the stress components needs to be expressed in  
terms of strains. Using the generalized Hooke’s law, the strains are computed as  

( ) ( )( ) ( ) ( )( ) 
( ) 
( ) ( )( ) ( ) ( )( ) 
2 2 2 2 2 2 2 2 
2 2 
2 2 2 2 2 2 2 2 
1 
1 2 
1 

xx xx yy zz 
yy yy xx zz 
c y x y x x y x y 
E E 
c y 
E 
c x x y y x y x y 
E E 
ε σ ν σ σ ν ν ν ν 
ν ν 
ε σ ν σ σ ν ν ν ν 
  = − + = + − − − − + +    
= − − 
  = − + = − − − + − + +    
( ) 
( ) ( ) ( ) ( )( ) 
( ) ( ) 
2 2 
2 2 2 2 2 2 2 2 
1 2 
1 
0 
1 1 12 2 
2 
1 0 
2 
1 0 
2 
xx zz xx yy 
xy xy 
yz yz 
zx zx 
c x 
E 
c x y y x y x x y 
E E 
c xy c xy 
E E 
ν ν 
ε σ ν σ σ ν ν ν ν 
ν νε σ ν ν 
µ 
ε σ 
µ 
ε σ 
µ 
= − − 
  = − + = + − + − + − −    
= 
+ + 
= = − = − 
= = 
= = .  
Since this is the plane strain condition, the only equation for compatibility is  
, , ,2 0xx yy yy xx xy xyε ε ε+ − = .  
To check if it holds with strain components,  
6  
( ) ( ) ( ) 
( ) ( ) ( ) 
( ) ( ) 
2 2 2 2 
, , , 
2 2 
2 2 2 2 
12 1 2 1 2 2 2 
2 1 2 2 1 2 2 2 1 
2 1 2 1 2 2 2 2 1 
xx yy yy xx xy xy 
yy xx xy 
c cy x c xy 
E E E 
c c c 
E E E 
c c 
E E 
νε ε ε ν ν ν ν ν 
ν ν ν ν ν ν 
ν ν ν ν ν ν ν 
 +   + − = − − + − − − −     
      
= − − + − − + + 
= − − + − − + + = − 

For isotropic materials, the Poisson’s ratio needs to be 11 2ν− < ≤ , so that the compatibility  
equation cannot be zero.