Answer
$\text{(a)}$
$$\sigma_{a}=65.1 \mathrm{MPa}$$
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$\text{(b)}$
$$\delta=0.279 \mathrm{mm}$$
Work Step by Step
Let $\quad P_a$ = Portion of axial force carried by shell.
$\quad \quad P_b$ = Portion of axial force carried by core.
$$\delta=\frac{P_{a} L}{E_{a} A_{a}}, \quad \ or\ \quad P_{a}=\frac{E_{a} A_{a}}{L} \delta$$
$$\delta=\frac{P_{b} L}{E_{b} A_{b}}, \quad \ or \ \quad P_{b}=\frac{E_{b} A_{b}}{L} \delta$$
Thus
$$P=P_{a}+P_{b}=\left(E_{a} A_{a}+E_{b} A_{b}\right) \frac{\delta}{L}$$
with
$$A_{a}=\frac{\pi}{4}\left[(0.060)^{2}-(0.025)^{2}\right]=2.3366 \times 10^{-3} \mathrm{m}^{2}$$
$$A_{b}=\frac{\pi}{4}(0.025)^{2}=0.49087 \times 10^{-3} \mathrm{m}^{2}$$
$$P=\left[\left(70 \times 10^{9}\right)\left(2.3366 \times 10^{-3}\right)+\left(105 \times 10^{9}\right)\left(0.49087 \times 10^{-3}\right)\right] \frac{\delta}{L}$$
$$P=215.10 \times 10^{6} \frac{\delta}{L}$$
Strain:
$$\varepsilon=\frac{\delta}{L}=\frac{P}{215.10 \times 10^{6}}=\frac{200 \times 10^{3}}{215.10 \times 10^{6}}=0.92980 \times 10^{-3}$$
$\text{(a)}$
$$\sigma_{a}=E_{a} \varepsilon=\left(70 \times 10^{9}\right)\left(0.92980 \times 10^{-3}\right)=65.1 \times 10^{\circ} \mathrm{Pa}$$
$$\sigma_{a}=65.1 \mathrm{MPa}$$
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$\text{(b)}$
$$\delta=\varepsilon L=\left(0.92980 \times 10^{-3}\right)(300 \mathrm{mm})$$
$$\delta=0.279 \mathrm{mm}$$
