Answer
$$P = 9.817 \space kN$$
$$\sigma = 500 \space MPa$$
Work Step by Step
1. Lateral Elongation formula:
$\delta = \epsilon * L = \frac{\sigma}{E} * L$
Solving the above formula for sigma:
$\delta = \frac{\sigma}{E} * L \Rightarrow \sigma = \frac {\delta * E}{L} = =\frac{45*10^{-3}\space m * 200 * 10^9\space N/m^2}{18\space m} = \boxed{500\space MPa} $ $\leftarrow ANS (b)$
2. Axial Stress Formula
$\sigma = \frac{P}{A}$
for a constant circular section of cable:
$A = \frac{\pi * D^2}{4} = 1.963 * 10^{-5}\space m^2$
$500*10^6 \space N/m^2 = \frac {P}{1.963 * 10^{-5}\space m^2} \Rightarrow P = 9817\space N = \boxed{9.82 \space kN}$ $\leftarrow ANS (a)$
