Solution to Problem 207 Axial Deformation

Based on maximum allowable stress:
$\sigma = \dfrac{P}{A}$

$20\,000 = \dfrac{500}{\frac{1}{4} \pi d^2}$

$d = 0.1784 \, \text{in}$
 

Based on maximum allowable deformation:
$\delta = \dfrac{PL}{AE}$

$0.20 = \dfrac{500(30 \times 12)}{\frac{1}{4} \pi d^2 (29 \times 10^6)}$

$d = 0.1988 \, \text{in}$
 

Use the bigger diameter, d = 0.1988 inch.           answer