# Derivation of deformation equation given thermal expansion

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.

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?
• I think it is a typo in printing. Does the solution of delta T make sense?
– r13
Mar 23, 2022 at 19:57

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 $$- \frac{PL}{E_b A_b}$$.