Answer
(a) Brass-core:
$$\sigma_{B} =140.6 \mathrm{MPa} $$
---
(b) Aluminum:
$$ \sigma_{A} =93.8 \mathrm{MPa} $$
Work Step by Step
$$ \delta_{A} =\delta_{B}=\delta ; \quad P=P_{A}+P_{B} $$
$$\delta=\frac{P_{A} L}{E_{A} A_{A}} \text { and } \quad \delta=\frac{P_{B} L}{E_{B} A_{B}}
$$
Therefore,
$$ P_{A}=\left(E_{A} A_{A}\right)\left(\frac{\delta}{L}\right) ; P_{B}=\left(E_{B} A_{B}\right)\left(\frac{\delta}{L}\right) $$
Substituting,
$$ P_{A}=\left(E_{A} A_{A}+E_{B} A_{B}\right)\left(\frac{\delta}{L}\right) $$
$$ \quad \in=\frac{\delta}{L}=\frac{P}{\left(E_{A} A_{A}+E_{B} A_{B}\right)} $$
$$\in=\frac{\delta}{L}=\frac{P}{\left(E_{A} A_{A}+E_{B} A_{B}\right)}$$
$$\in=\frac{ }{\left(70 \times 10^{9} \mathrm{Pa}\right)(0.06 \mathrm{m})(0.01 \mathrm{m})+\left(105 \times 10^{9} \mathrm{Pa}\right)(0.06 \mathrm{m})(0.04 \mathrm{m})}$$
$$\quad=1.33929 \times 10^{-3}$$
Now,$$\sigma=E \in$$
(a) Brass-core:
$$\sigma_{B}=\left(105 \times 10^{9} \mathrm{Pa}\right)\left(1.33929 \times 10^{-3}\right)$$
$$\quad=1.40625 \times 10^{8} \mathrm{Pa}$$
$$\therefore\sigma_{B} =140.6 \mathrm{MPa} $$
---
(b) Aluminum:
$$ \sigma_{A} =\left(70 \times 10^{9} \mathrm{Pa}\right)\left(1.33929 \times 10^{-3}\right) $$
$$ =9.3750 \times 10^{7} \mathrm{Pa} $$
$$\therefore \sigma_{A} =93.8 \mathrm{MPa} $$
