Answer
${\text { (a) }} $
$$ {} {P=287 \mathrm{kN}} $$
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$ {\text { (b) }} $
$${} {\sigma_{b}=140.0 \mathrm{MPa}} $$
Work Step by Step
Let $P_a$ = Portion of axial force carried by shell
and $P_b$ = 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
$$A_{a}=\frac{\pi}{4}\left[(0.060)^{2}-(0.025)^{2}\right]=2.3366 \times 10^{-3} \mathrm{m}^{2}$$
$$ A_{b} =\frac{\pi}{4}(0.025)^{2}=0.49087 \times 10^{-3} \mathrm{m}^{2} $$
$$ P =\left[\left(70 \times 10^{9}\right)\left(2.3366 \times 10^{-3}\right)+\left(105 \times 10^{9}\right)\left(0.49087 \times 10^{-3}\right)\right] \frac{\delta}{L}=215.10 \times 10^{6} \frac{\delta}{L} $$
with
$$ \delta =0.40 \mathrm{mm}, L=300 \mathrm{mm} $$
${\text { (a) }} $
$$ {P=\left(215.10 \times 10^{6}\right) \frac{0.40}{300}=286.8 \times 10^{3} \mathrm{N}} $$
$$ {} {P=287 \mathrm{kN}} $$
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$ {\text { (b) }} $
$$ {\sigma_{b}=\frac{P_{b}}{A_{b}}=\frac{E_{b} \delta}{L}=\frac{\left(105 \times 10^{9}\right)\left(0.40 \times 10^{-3}\right)}{300 \times 10^{-3}}=140 \times 10^{6} \mathrm{Pa}} $$
$${} {\sigma_{b}=140.0 \mathrm{MPa}} $$
