# Can there be strain without stress (e.g. Thermal Expansion)?

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?

PS: and also here they say strain is -3.6*10^-4?

• please don't add bit.ly link here. The site is blocked for some. Oct 29, 2021 at 10:21
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Oct 29, 2021 at 10:23

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, $$\sigma_1=E(\epsilon_1-\alpha \Delta T)$$
This may be to do with definitions. Stress is force per unit area and I think for this there is just this definition and it is unambiguous. Strain is extension divided by length, but now there is room for confusion because one has to specify the extension compared to what. That is, if we write $$s = \frac{L - L_0}{L}$$ where $$L$$ is the length of our system, then what is $$L_0$$? That is what your question is asking, I think. Ordinarily we would say that $$L_0$$ is whatever the length would be if there were no stress, so in this case when the stress is zero then the strain is zero as well. But someone might prefer to use as $$L_0$$ the length at some initial temperature, and then they might say that after thermal expansion you can have strain without stress. However I think that would be a non-standard way to define $$L_0$$ when discussing stress and strain.
The video in your first link incorrectly concludes the strain is zero because no external forces were put on the rod. However, strain is a transformation of an object that is defined without any need to mention stress. He drew the rod of length $$L$$ that now appeared elongated to $$L+\Delta L$$. Hence the rod is strained by $$\frac{\Delta L}{L}\quad radians$$. To emphasize the mathematical existence of strain without the need for stress, I stuck in radians because 6 strains and 3 rotations make up the Lie Group GL(3) where all 9 "angles" are measured in radians.