In here The solution says that strain is 0 in thermal expansion?
Doesn't this sound weird and contradictiory?
by taking two infinitesimally closer points the stress could be zero but in a article of quora everyone is saying stress is not zero in thermal expansion?
( https://www.quora.com/Can-there-be-strain-without-stress )
PS:
and also here they say strain is -3.6*10^-4?
Best Answer
If an isotropic material is not under stress and its temperature is changed, then the linear strain within the material is the same in all arbitrary directions, and equal to $\alpha \Delta T$, where $\alpha$ is the linear coefficient of thermal expansion. So, if we have any three perpendicular directions, say 1, 2, and 3, the strains in these directions are $$\epsilon_1=\alpha \Delta T$$$$\epsilon_2=\alpha \Delta T$$$$\epsilon_3=\alpha \Delta T$$Basically, what has happened here is that, by changing the temperature of the material, we have changed the state of zero stress from the original state to the new "thermally strained" state. Now if stresses are superimposed on this, the changes in the strains are measured relative to this new deformed state. So, for a material that obeys Hooke's law and that is subjected to a combined situation of changed temperature and imposed stresses, the principle stresses and strains in the material are related by: $$\epsilon_1=\alpha \Delta T+\frac{[\sigma_1-\nu(\sigma_2+\sigma_3)]}{E}$$$$\epsilon_2=\alpha \Delta T+\frac{[\sigma_2-\nu(\sigma_1+\sigma_3)]}{E}$$$$\epsilon_3=\alpha \Delta T+\frac{[\sigma_3-\nu(\sigma_1+\sigma_2)]}{E}$$where $\nu$ is the Poisson's ratio and E is the Young's modulus. If the state of stress is uniaxial, so that the stresses in the 2 and 3 directions are zero, the equation for the 1-direction reduces to $$\epsilon_1=\alpha \Delta T+\frac{\sigma_1}{E}$$or, equivalently, $$\sigma_1=E(\epsilon_1-\alpha \Delta T)$$