# Stress on a statically indeterminate beam with nonuniform temperature distribution

The bar below has length $$a$$, an uniform cross-section and has both ends fixed to walls. The temperature at the left end is raised by $$\Delta T_1$$ and that of the right by $$\Delta T_2$$, where $$\Delta T_2 > \Delta T_1$$. The temperature change $$\Delta T$$ within the bar is linear from one end to the other. Take $$E$$, $$\alpha$$ and $$A$$ as constants. Determine the stress in the bar and disregard buckling.

I am assuming that the variation of temperature at any given point is given by

$$\Delta T = \Delta T_1 + \Delta T_x$$

Where

$$\Delta T_x = \frac{x}{a} (\Delta T_2 - \Delta T_1)$$

Knowing that this is a statically indeterminate beam, we have that the deflection given by the thermal increase $$(\delta_T)$$ is "compensated" by the load-displacement given by the reaction of the wall $$(\delta_P)$$. The compatibility equation is therefore

$$\delta_{T} - \delta_{P} = 0$$

For a small $$dx$$, $$\delta_T$$ is

$$d\delta_T = \alpha \Delta T dx$$

and $$\delta_P$$ is

$$d\delta_P = \frac{\sigma}{E}~dx$$

After replacing the differentials in the compatibility equation and integrating from $$0$$ to $$a$$, I have found the following result

$$\sigma_T = E\alpha \frac{\Delta T_1 + \Delta T_2}{2}$$

I was wondering if my approach to this problem is correct. If not, what did I do wrong?

your approach is correct. Because the function $$f(T)$$ is linear, we could break the beam into an infinitesimal number of small dx lengths.