Answer
$(a)\space 4.9\times10^{6}\space N/m^{2}$
$(a)\space 6\times10^{-6}\space m$
Work Step by Step
(a) We know that the shear stress = $\frac{F}{A}=\frac{mg}{A}=\frac{(160\space kg)(9.80\space m/s^{2})}{3.2\times10^{-4}m^{2}}=4.9\times10^{6}N/m^{2}$
(b) Let's apply equation 10.18 $F=S(\frac{\Delta Y}{L_{0}})A$ to the bar.
$\Delta Y=(\frac{F}{A})\frac{L_{0}}{S}=4.9\times10^{6}N/m^{2}\times\frac{0.1\space m}{8.1\times10^{10}N/m^{2}}\approx6\times10^{-6}m$
The vertical deflection of the right end of the bar = $6\times10^{-6}m$
Here we get the shear modulus S of steel from table 10.2
