Consider the one-dimensional thermoelastic problem of a uniform bar constrained in the axial $x$ direction but allowed to expand freely in the $y$ and $z$ directions, as shown in the following figure. Taking the reference temperature to be zero, show that the only nonzero stress and strain components are given by
$\sigma_x = -E \alpha T$
$\epsilon_y = \epsilon_z = (1+\nu) \alpha T$
Hints: Use $\epsilon_{ij} = \frac{1+\nu}{E} \sigma_{ij} - \frac{\nu}{E} \sigma_{kk} \delta_{ij} + \alpha (T-T_0) \delta_{ij}$ (p90)
Use $\epsilon_x = 0$, $\sigma_y = \sigma_z = 0$ (No transverse stresses)
Verify that Hooke's law for isotropic thermoelastic materials can be expressed in the form
$\sigma_x = \frac{E}{(1+\nu)(1-2\nu)} \left[ (1-\nu) \epsilon_x + \nu (\epsilon_y + \epsilon_z) \right] - \frac{E}{1-2\nu} \alpha (T-T_0)$
$\sigma_y = \frac{E}{(1+\nu)(1-2\nu)} \left[ (1-\nu) \epsilon_y + \nu (\epsilon_x + \epsilon_z) \right] - \frac{E}{1-2\nu} \alpha (T-T_0)$
$\sigma_z = \frac{E}{(1+\nu)(1-2\nu)} \left[ (1-\nu) \epsilon_z + \nu (\epsilon_x + \epsilon_y) \right] - \frac{E}{1-2\nu} \alpha (T-T_0)$
$\tau_{xy} = \frac{E}{1+\nu} \epsilon_{xy}$, $\tau_{yz} = \frac{E}{1+\nu} \epsilon_{yz}$, $\tau_{zx} = \frac{E}{1+\nu} \epsilon_{zx}$
Hints: Read P.90, Use $\sigma_{ij} = 2\mu \epsilon_{ij} + \lambda \epsilon_{kk} \delta_{ij} - \alpha (T-T_0) \delta_{ij} (3\lambda + 2\mu)$
Use the table attached for $\lambda$, $\mu$, $E$, $\nu$, $K$
Conversion
