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120 Finite Element Modeling and Simulation with ANSYS Workbench

We can also express stresses in terms of strains by solving the above equation


 σ  1 ν 0   ε x   ε x0 
x
  E      

 σ  = 2  ν 1 0   ε y  −  ε y0   (4.5)
y
  1 −ν 0 0 1 ( −ν 2      

)/


 τ xy     γ xy   γ xy0 
or,
σ= E
ε+ σ 0
where σ= −ε 0 is the initial stress.
E
0
For plane strain case, we need to replace the material constants in the above equations in
the following fashion:
E ν
E → ; ν → ; G → G (4.6)
1 − ν 2 1 − ν

For example, the stress is related to strain by


 σ  1 − ν ν 0   ε   ε x0 
x
x
  E      

 σ  =  ν 1 − ν 0    ε  −  ε y0  
y
y

  (1 +ν)(1 −ν)2  0 0 (1 −ν) 2/  γ    
2

 τ xy     xy   γ xy0 
in the plane strain case.
Initial strain due to a temperature change (thermal loading) is given by the following for
the plane stress case
∆
 ε x0   α T 
   
 ε y0  = α T∆  (4.7)

   0 
 γ xy0   
where α is the coefficient of thermal expansion, ΔT the change of temperature. For the
plane strain case, α should be replaced by (1 + ν)α in Equation 4.7. Note that if the struc-
ture is free to deform under thermal loading, there will be no (elastic) stresses in the
structure.

4.2.4 Strain and Displacement Relations

For small strains and small rotations, we have,

∂u ∂v ∂u ∂v
ε= , ε= , γ xy = +
x
y
∂x ∂y ∂y ∂x