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

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


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

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?


1 Answer 1


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

Each pair of dx slices symmetrically from extreme left and right have their sum of stresses equal to the next pair (with a bit less difference between their Ts but same average). until the two center dx slices are of the same T with the 2times stress.

So once you add say n pairs of slices the total stress is the average stress.

  • $\begingroup$ Thank you for checking my approach! Just so I can know, did you try to get the same answer? $\endgroup$
    – Iuri
    Mar 9, 2022 at 13:02

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