Answer
$$ (a)\ \ \ 0.0358 \mathrm{m}\mathrm{m}$$
$$ (b)\ \ \ -0.00258 \mathrm{m}\mathrm{m}$$
$$ (c)\ \ \ -0.000344 \mathrm{m}\mathrm{m}$$
$$ (a)\ \ \ -0.00825 \mathrm{m}\mathrm{m}^2$$
Work Step by Step
$\begin{aligned} A &=(1.6)(12)=19.20 \mathrm{mm}^{2} \\ &=19.20 \times 10^{-6} \mathrm{m}^{2} \\ P &=2.75 \times 10^{3} \mathrm{N} \\ \sigma_{x} &=\frac{P}{A}=\frac{2.75 \times 10^{3}}{19.20 \times 10^{-6}} \\ &=143.229 \times 10^{6} \mathrm{Pa} \\ \varepsilon_{x} &=\frac{\sigma_{x}}{E}=\frac{143.229 \times 10^{6}}{200 \times 10^{9}}=716.15 \times 10^{-6} \\ \varepsilon_{y} &=\varepsilon_{z}=-v \varepsilon_{x}=-(0.30)\left(716.15 \times 10^{-6}\right)=-214.84 \times 10^{-6} \end{aligned}$
$\begin{array}{lll}(a) & L=0.050 \mathrm{m} & \delta_{x}=L \varepsilon_{x}=(0.50)\left(716.15 \times 10^{-6}\right)=35.808 \times 10^{-6} \mathrm{m} \\ (b) & w=0.012 \mathrm{m} & \delta_{y}=w \varepsilon_{y}=(0.012)\left(-214.84 \times 10^{-6}\right)=-2.5781 \times 10^{-6} \mathrm{m} \\ (c) & t=0.0016 \mathrm{m} & \delta_{z}=t \varepsilon_{z}=(0.0016)\left(-214.84 \times 10^{-6}\right)=-343.74 \times 10^{-9} \mathrm{m}\end{array}$
$\begin{aligned}(d) \quad A &=w_{0}\left(1+\varepsilon_{y}\right) t_{0}\left(1+\varepsilon_{z}\right)=w_{0} t_{0}\left(1+\varepsilon_{y}+\varepsilon_{z}+\varepsilon_{y} \varepsilon_{z}\right) \quad A_{0}=w_{0} t_{0} \\ \Delta A &=A-A_{0}=w_{0} t_{0}\left(\varepsilon_{y}+\varepsilon_{z}+\text { negligible term }\right)=2 w_{0} t_{0} \varepsilon_{y} \\ &=(2)(0.012)(0.0016)\left(-214.84 \times 10^{-6}\right)=-8.25 \times 10^{-9} \mathrm{m}^{2} \end{aligned}$
