Thermal stress in a restrained bar

When a bar is prevented from expanding freely, a compressive thermal stress develops.

Data

$L = 2500\ \text{mm}$, $A = \frac{\pi}{4}(25)^2 = 490.9\ \text{mm}^2$, $\Delta T = 70 - 20 = 50\,^\circ\text{C}$.

(a) Rigid walls

Free thermal expansion:

$$\delta_T = \alpha L \Delta T = 12\times10^{-6}\times 2500\times 50 = 1.5\ \text{mm}$$

The walls allow no expansion, so the wall reactions must push the bar back by the full 1.5 mm. The thermal strain $\alpha\Delta T$ is cancelled by an equal elastic strain:

$$\sigma = E\alpha\Delta T = 200000\times 12\times10^{-6}\times 50 = 120\ \text{MPa (compressive)}$$ $$P = \sigma A = 120\times 490.9 = 58.9\times10^3\ \text{N}$$

(The stress does not depend on the length.)

(b) Walls yield by 0.6 mm

Only the difference between the free expansion and the wall movement has to be suppressed:

$$\delta_{\text{restrained}} = 1.5 - 0.6 = 0.9\ \text{mm}$$ $$\sigma = E\,\frac{\delta_{\text{restrained}}}{L} = 200000\times\frac{0.9}{2500} = 72\ \text{MPa (compressive)}$$ $$P = 72\times 490.9 = 35.3\ \text{kN}$$

Answer: (a) $\sigma = 120$ MPa (compression), $P = 58.9$ kN; (b) $\sigma = 72$ MPa (compression), $P = 35.3$ kN.