Derivation of deformation equation given thermal expansion

Question

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?

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.