Answer
$$D_{PIN} = 1.030\space in$$
$$\sigma_{bearing}=38.83\space ksi$$
Work Step by Step
1) Internal Force in Link:
Average Axial stress in link (calculated across the entire cross sectional area of the link, so $A_{link} = 2 * 0.25\space in^2$)
$\sigma_{avg} = \frac{F}{A_{link}} = \frac{F}{0.5\space in^2} = -20 \space ksi$ (negative value for stress indicates compression)
Solving for F:
$F = |-20| \space ksi* 0.5\space in^2 = 10\space kip$ (directed in compression)
2) Diameter of Pin:
Shear stress in pin:
$\tau_{pin} = \frac{F}{A_{pin}} = \frac{10\space kip}{0.25 * \pi * D_{pin}^2} = 12.0 \space ksi$
Solving for unknown $D_{pin}$ :
$D_{pin} = \sqrt{\frac{10}{0.25\pi * 12}} = \boxed{1.030\space in}\space\space\leftarrow ANS1$
3)Bearing Stress (calculated across the width of the opening)
$\sigma_{bearing} = \frac{F}{t * D_{pin}} = \frac{10\space kip}{0.25 * 1.030} = \boxed{38.83\space ksi}\space\space\leftarrow ANS2$
