Question

A cube of the isotropic, linear elastic epoxy resin E=2GPa, ν=0.3, α=50×10−6K−1 at 20∘C is unloaded. It is then heated to 60∘C.

What is the corresponding strain matrix?

The temperature is held constant at 60∘C. What stress is required to reduce all the components of the strain matrix to zero?

σ (in MPa):

Answer 1

e=a*ÄT=0.002
what about ó??

Answer 2

epsilon_x=sigma_x/E-nu*sigma_y/E-nu*sigma_z/E
sigma_x=sigma_y=sigma_z=sigma
epsilon_x=epsilon_y=epsilon_z=epsilon
so...
epsilon=sigma/E-nu*sigma/E-nu*sigma/E
epsilon=(sigma/E)*(1-2*nu)
=> sigma = (epsilon*E)/(1-2*nu)
You want the cube to shrink back to its original position, so plus -0.002 in to the formula. The stress should be negative (compressive).

Is this solution correct?

Answer 3

εx=

εy=

εz=

γyz=

γxz=

γxy

Answer 4

åx=åy=åz=o.oo2
ãx=ãy=ãz=0
I believe that your solution (anonymous) is correct so
ó=(epsilon*E)/(1-2*nu)= 10MPa
The stress is compressive so -10MPa
Do you agree?

Answer 5

The temperature is held constant at 60∘ C. What stress is required to reduce all the components of the strain matrix to zero?

σ (in MPa ): -10 MPa