Answer:
Option B
1025 psi
Explanation:
In a single shear, the shear area is [tex]\frac {\pi d^{2}}{4}=\frac {\pi 0.25^{2}}{4}[/tex]
The shear strength=[tex]0.58\sigma_y[/tex] and in this case [tex]\sigma_y=36 000 psi[/tex]
Shear strength=[tex]\frac {Load}{Shear area}[/tex] hence making load the subject then
Load=Shear area X Shear strength
Load=[tex]\frac {\pi 0.25^{2}}{4} \times 0.58\times 36000\approx 1025 psi[/tex]
