Answer
$\begin{aligned}(a) & \delta=4.0041 \times 10^{-4} \end{aligned}$
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(b)
$$\Delta d_{o}=9.6098 \times 10^{-2} \mathrm{mm}$$
$$
\Delta t=4.0041 \times 10^{-3} \mathrm{mm}
$$
Work Step by Step
$\begin{aligned} d_{o} &=0.240 \quad t=0.010 \quad L=2.0 \\ d_{i} &=d_{o}-2 t=0.240-2(0.010)=0.220 \mathrm{m} \quad P=640 \times 10^{3} \mathrm{N} \\ A &=\frac{\pi}{4}\left(d_{o}^{2}-d_{i}^{2}\right)=\frac{\pi}{4}(0.240-0.220)=7.2257 \times 10^{-3} \mathrm{m}^{2} \end{aligned}$
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$\begin{aligned}(a) & \delta=-\frac{P L}{E A}=-\frac{\left(640 \times 10^{3}\right)(2.0)}{\left(73 \times 10^{9}\right)\left(7.2257 \times 10^{-3}\right)} \\ &=-2.4267 \times 10^{-3} \mathrm{m} \\ \varepsilon &=\frac{\delta}{L}=\frac{-2.4267}{2.0}=-1.21335 \times 10^{-3} \\ \varepsilon_{L A T} &=-v \varepsilon=-(0.33)\left(-1.21335 \times 10^{-3}\right) \\ &=4.0041 \times 10^{-4} \end{aligned}$
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(b) $\Delta d_{o}=d_{o} \varepsilon_{L A T}=(240 \mathrm{mm})\left(4.0041 \times 10^{-4}\right)=9.6098 \times 10^{-2} \mathrm{mm}$
$$
\Delta t=t \varepsilon_{L A T}=(10 \mathrm{mm})\left(4.0041 \times 10^{-4}\right)=4.0041 \times 10^{-3} \mathrm{mm}
$$
