Answer
$\text { (a) }$
$$ h=15.00 \mathrm{\ mm}$$
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$\text { (b) }$
$$P=288 \mathrm{\ kN}$$
Work Step by Step
$$\delta=\delta_{a}=\delta_{b} ; \quad P=P_{a}+P_{b}$$
$$\delta=\frac{P_{a} L}{E_{a} A_{a}} \quad$ and $\quad \delta=\frac{P_{b} L}{E_{b} A_{b}}$$
Therefore,
$$P_{a}=\left(E_{a} A_{a}\right) \frac{\delta}{L} ; \quad P_{b}=\left(E_{b} A_{b}\right)\left(\frac{\delta}{L}\right)$$
$\text { (a) }$
$$P_{a}=\frac{1}{2} P_{b}$$
$$\left(E_{a} A_{a}\right)\left(\frac{\delta}{L}\right) =\frac{1}{2}\left(E_{b} A_{b}\right)\left(\frac{\delta}{L}\right) $$
$$A_{a} =\frac{1}{2}\left(\frac{E_{b}}{E_{a}}\right) A_{b} $$
$$ A_{a} =\frac{1}{2}\left(\frac{105 \mathrm{GPa}}{70 \mathrm{GPa}}\right)(40 \mathrm{mm})(60 \mathrm{mm}) $$
$$ A_{a} =1800 \mathrm{mm}^{2} $$
$$1800 \mathrm{mm}^{2} =2(60 \mathrm{mm})(h) $$
$$\therefore h=15.00 \mathrm{\ mm}$$
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$\text { (b) }$
$$\quad \sigma_{b} {=\frac{P_{b}}{A_{b}} \Rightarrow P_{b}=\sigma_{b} A_{b} \text { and } P_{a}=\frac{1}{2} P_{b}} $$
$$\therefore {} {P=P_{a}+P_{b}}$$
$$P=\frac{1}{2}\left(\sigma_{b} A_{b}\right)+\sigma_{b} A_{b}$$
$$P=\left(\sigma_{b} A_{b}\right) 1.5$$
$$P=\left(80 \times 10^{6} \mathrm{Pa}\right)(0.04 \mathrm{m})(0.06 \mathrm{m})(1.5)$$
$$P=2.880 \times 10^{5} \mathrm{N}$$
$$P=288 \mathrm{\ kN}$$
