Strain energy density and thermal stress

(a) Strain energy density

$$U_0 = \frac{1}{2E}\left[\sigma_1^2+\sigma_2^2+\sigma_3^2 - 2\nu(\sigma_1\sigma_2+\sigma_2\sigma_3+\sigma_3\sigma_1)\right]$$

Terms: $\sigma_1^2+\sigma_2^2+\sigma_3^2 = 10000+2500+900 = 13400$; $\sigma_1\sigma_2+\sigma_2\sigma_3+\sigma_3\sigma_1 = 5000 - 1500 - 3000 = 500$.

$$U_0 = \frac{13400 - 2(0.3)(500)}{2\times200\times10^3} = \frac{13100}{4\times10^5} = 0.03275\ \text{N\,mm/mm}^3$$

Volumetric part (change of volume only):

$$U_v = \frac{1-2\nu}{6E}(\sigma_1+\sigma_2+\sigma_3)^2 = \frac{0.4}{1.2\times10^6}(120)^2 = 0.0048\ \text{N\,mm/mm}^3$$

Distortion part:

$$U_d = \frac{1+\nu}{6E}\left[(\sigma_1-\sigma_2)^2+(\sigma_2-\sigma_3)^2+(\sigma_3-\sigma_1)^2\right] = \frac{1.3}{1.2\times10^6}(25800) = 0.02795\ \text{N\,mm/mm}^3$$

Check: $U_v + U_d = 0.0048+0.02795 = 0.03275 = U_0$. About 85 percent of the energy is distortion energy (used in the von Mises theory).

(b) Fully constrained heated cube

Thermo-elastic equations for isotropic material:

$$\varepsilon_x = \frac{1}{E}\left[\sigma_x - \nu(\sigma_y+\sigma_z)\right] + \alpha\Delta T$$

Total strain is zero in all directions and the stress is the same in all three by symmetry, $\sigma_x=\sigma_y=\sigma_z=\sigma$:

$$\begin{aligned} 0 &= \frac{\sigma(1-2\nu)}{E} + \alpha\Delta T \\ \sigma &= -\frac{E\alpha\Delta T}{1-2\nu} = -\frac{200\times10^3 \times 12\times10^{-6}\times 50}{1-0.6} = -300\ \text{MPa} \end{aligned}$$

For comparison, a bar restrained in one direction only would have $E\alpha\Delta T = 120$ MPa; the triaxial restraint makes the stress 2.5 times larger. The stress is hydrostatic, so it causes no distortion and the cube does not yield by the shear-based theories.

Answer: (a) $U_0 = 0.03275$ N mm/mm$^3$ (= 32.75 kJ/m$^3$), $U_v = 0.0048$, $U_d = 0.02795$ N mm/mm$^3$; (b) $\sigma = -300$ MPa (compressive, equal in all three directions).