Temperature Stresses on Unrestrained Cube

[Vatsy]

Hii all..

I have a question.Suppose there is a cube of side 1m .It is heated by 1°C. The cube is not restrained in any direction. We are required to find out the volumetric strain.

I am getting it as 3α.

But my question is that since the cube is not restrained , ∴ there should be no stresses in any direction.
∴ ε_v=(σ_x +σ_y+σ_z)(1-2μ)/E
which gives ε_v=0.

please help..

 

[SteamKing]

Hii all..

I have a question.Suppose there is a cube of side 1m .It is heated by 1°C. The cube is not restrained in any direction. We are required to find out the volumetric strain.

I am getting it as 3α.

But my question is that since the cube is not restrained , ∴ there should be no stresses in any direction.
∴ ε_v=(σ_x +σ_y+σ_z)(1-2μ)/E
which gives ε_v=0.

please help..

The equation you quote for the strain pre-supposes that there are applied stresses on the object.

 

[Chestermiller]

Hii all..

I have a question.Suppose there is a cube of side 1m .It is heated by 1°C. The cube is not restrained in any direction. We are required to find out the volumetric strain.

I am getting it as 3α.

But my question is that since the cube is not restrained , ∴ there should be no stresses in any direction.
∴ ε_v=(σ_x +σ_y+σ_z)(1-2μ)/E
which gives ε_v=0.

please help..

Hi Vasty. Welcome to Physics Forums.
Your equation needs to be modified when thermal expansion and contraction effects are involved, as follows:
$$(ε_v-3αΔT)=\frac{(σ_x+σ_y+σ_z)(1-2μ)}{E}$$
Now, can you figure out how the 6 Hooke's law strain equations have to be modified when thermal expansion effects are included?

 

[Vatsy]

Hi Vasty. Welcome to Physics Forums.
Your equation needs to be modified when thermal expansion and contraction effects are involved, as follows:
$$(ε_v-3αΔT)=\frac{(σ_x+σ_y+σ_z)(1-2μ)}{E}$$
Now, can you figure out how the 6 Hooke's law strain equations have to be modified when thermal expansion effects are included?

Thanks a lot...