I'm working on a thermal expansion problem out of Beer & Johnston's Mechanics of Materials and I'm a bit confused by the underlying rationale of one of the steps in the solution process.

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In the exercise text, $\alpha_s$ and $\alpha_b$ are given along with a $\sigma_{max}$ of the inner steel core that cannot be exceeded. The core is considered fully bonded to the shell and the goal is to derive the max $\Delta T$ that can be observed while remaining under $\sigma_{max}$. I understand the general solution process, but a sign change in the deformation equation has me confused, as the book does not give a specific rationale for its change. The equation the text gives is:

$$\delta = \alpha (\Delta T) L + \frac{PL}{EA}$$

I have verified the final solution with a solutions manual, and the solution process for this problem proceeds:

$$\begin{align} P &= A_s \sigma_{max} & (1) \\ \delta_b &= \alpha_b (\Delta T) L - \frac{PL}{E_b A_b} & (2) \\ \delta_s &= \alpha_s (\Delta T) L + \frac{PL}{E_s A_s} & (3) \\ & \text{solve for} \,\,\Delta T \,\, \text{below} \end{align}$$

The specific element in question is the sign change of the $\delta_P$ term in Equation (2). In particular,

  1. Why does this sign change occur?
  2. For what reason is it assigned to the brass shell and not the inner steel core?
  • $\begingroup$ I think it is a typo in printing. Does the solution of delta T make sense? $\endgroup$
    – r13
    Mar 23, 2022 at 19:57

1 Answer 1


The solution depends on the coefficients of thermal expansion (CTE) for brass ($\alpha_b$) and steel ($\alpha_s$) and whether the temperature is increasing ($\Delta T>0$) or decreasing ($\Delta T<0$).

  • If $\Delta T$ is positive, both will expand due to the temperature. This is the term $+\alpha (\Delta T) L$ in both equations.
  • If brass has a larger CTE than steel, the larger expansion of the brass will put the steel in tension. The resulting force P will cause the steel to get longer than the thermal expansion alone. This is the term $+ \frac{PL}{E_s A_s}$.
  • At the same time, the steel will put the brass in compression. The resulting force P will cause the brass to get shorter than due to thermal expansion alone. This is the term $- \frac{PL}{E_b A_b}$.

In summary, the sign change is because one part is in compression due to the force P and the other part is in tension.


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