Answer
(a) $\delta_{x}=0.1973 \mathrm{mm}$
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(b) $\delta_{y}=-0.00651 \mathrm{mm}$
Work Step by Step
$P=75 \mathrm{kN}=75 \times 10^{3} \mathrm{N}$
$A=\frac{\pi}{4} d^{2}=\frac{\pi}{4}(0.022)^{2}=380.13 \times 10^{-6} \mathrm{m}^{2}$
$\sigma=\frac{P}{A}=\frac{75 \times 10^{3}}{380.13 \times 10^{-6}}=197.301 \times 10^{6} \mathrm{Pa}$
$\varepsilon_{x}=\frac{\sigma}{E}=\frac{197.301 \times 10^{6}}{200 \times 10^{9}}=986.51 \times 10^{-6}$
$\delta_{x}=L \varepsilon_{x}=(200 \mathrm{mm})\left(986.51 \times 10^{-6}\right)$
(a) $\delta_{x}=0.1973 \mathrm{mm}$
$\varepsilon_{y}=-v \varepsilon_{x}=-(0.3)\left(986.51 \times 10^{-6}\right)=-295.95 \times 10^{-6}$
$\delta_{y}=d \varepsilon_{y}=(22 \mathrm{mm})\left(-295.95 \times 10^{-6}\right)$
(b) $\delta_{y}=-0.00651 \mathrm{mm}$
