Answer
$$(a) =-0.0724 \mathrm{mm}$$
$$(b) -0.01531 \mathrm{mm}$$
Work Step by Step
$\begin{aligned} \sigma_{x} &=\sigma_{z}=-p=-48 \times 10^{6} \mathrm{Pa}, \quad \sigma_{y}=0 \\ \varepsilon_{x} &=\frac{1}{E}\left(\sigma_{x}-v \sigma_{y}-v \sigma_{z}\right) \\ &=\frac{1}{105 \times 10^{9}}\left[-48 \times 10^{6}-(0.33)(0)-(0.33)\left(-48 \times 10^{6}\right)\right] \\ &=306.29 \times 10^{-6} \\ \varepsilon_{y} &=\frac{1}{E}\left(-v \sigma_{x}-\sigma_{y}-v \sigma_{z}\right) \\ &=\frac{1}{105 \times 10^{9}}\left[-(0.33)\left(-48 \times 10^{6}\right)+0-(0.33)\left(-48 \times 10^{6}\right)\right] \\ &=301.71 \times 10^{-6} \end{aligned}$
(a) Change in length: only portion $B C$ is strained. $L=240 \mathrm{mm}$
$$
\delta_{y}=L \varepsilon_{y}=(240)\left(-301.71 \times 10^{-6}\right)=-0.0724 \mathrm{mm}
$$
(b) Change in diameter: $d=50 \mathrm{mm}$
$$
\delta_{x}=\delta_{z}=d \varepsilon_{x}=(50)\left(-306.29 \times 10^{-6}\right)=-0.01531 \mathrm{mm}
$$
