Question

A 3/8-16 UNC nut and bolt are used to assemble steel plates. The clearance holes in the plates are 7/16 in. The plates are clamped together by the nut and bolt with a force of 500 lb. If the torque coefficient is 0.22, determine the (a) torque that is applied and (b) resulting stress on the bolt

Answer 1

Answer:

a) T=41.25 in-lb

b) Resulting stress=6493.5 lb/in^2

Explanation:

a) The torque is:

$T=C*D*F$

where C is the torque coefficient, F the force and D is 3/8. Replacing:

$T=0.22*3/8*500=41.25 in-lb$

b) The area is:

$A=0.25\pi (D-\frac{0.9743}{n})^{2}=0.25\pi (3/8-\frac{0.9743}{16})^{2}=0.077 in^{2}$

The resulting stress is:

$Resulting stress=\frac{F}{A}=\frac{500}{0.077} =6493.5 lb/in^{2}$