Answer
2.25 kips
Work Step by Step
The area which is used to bear the shear force is
Area (A) = $\frac{5}{8} \times \frac{1}{2} = \frac{5}{16} in^{2}$
Now the shear force is given as 1.2 ksi.
$τ= 1.2 ksi$
$\frac{F'}{A}=1.2$
$F'=1.2 \times A$
$F'= 1.2 \times \frac{5}{16}$
$F'= 0.375 kips$
So, The Total Forces P will be equal to
$P=6 \times F $
$P=6 \times 0.375 $
$P= 2.25 kips$
Hence, the force P required to fail the joint will be 2.25 kips.
