Answer
$$\tau_{FAIL} = 888.9 \space psi $$
Work Step by Step
Solution:
1) Area of Shear force Distribution:
From Diagram in textbook, area of shear force distribution is:
$A = 3\space in * 0.6\space in = 1.8 \space in^2$
2)Failure Stress:
$\tau_{FAIL} = \frac{F_{FAIL}}{A} = \frac{1600 \space lb}{1.8 \space in^2} = 888.9 \space psi$
$\boxed{\tau_{FAIL} = 888.9 \space psi} \leftarrow \space\space ANS$
