Isotropic thermal expansion

Question

I frequently see the equation $$ \sigma_t = E\alpha \Delta T $$

as the equation for thermal stress.

Where $E$ is Young's modulus, $\alpha$ is the CTE, and $\Delta T$ is the change in temperature.

If we were to write out, for an isotropic material, the stiffness tensor we get the expression below

For thermal stress, the shear components of thermal strain would be zero, so we are left with the normal strains $\varepsilon_{ii}$ as the only potentially non-zero strains. These strains are identical for $i=1,2,3$ since the material is isotropic, and each is equal to $\alpha \Delta T$.

This is where I am confused. If we use the above relationship, one component of the normal stress would be $\sigma_{ii} = (2\mu+3\lambda) \alpha \Delta T$, but this is clearly not equal to $E\alpha \Delta T$, since $2\mu + 3\lambda \neq E$.

I must be misunderstanding something fundamental. Could someone point out what this might be?

Answer 1

You are comparing two different material models: the one dimensional Hooke's Law ($\sigma = E \varepsilon$) and a linear three dimensional one where all the quantities in $\sigma = E\varepsilon$ become tensors.

In three dimensions, a strain in one direction causes the material to contract in the other directions (Poisson's Ratio $\nu$). So, a thermal strain in one direction causes the material to contract in the others and thereby causing stresses in these directions. Your result is the the superposition of stresses due to thermal expansion and contraction. If my calculations are correct you get an isotropic stress of $\sigma = \frac{E}{1 - 2\nu} \alpha \Delta T$ which is bigger than $E\alpha\Delta T$ because the transverse contraction counteracts the thermal expansion. This effect is not taken into account by the one dimensional model.