I agree with the 9324 lb/in^2 but my understanding was that I would then multiply by the distance from the center of the shaft which is .1771 Ft to reach 1651.1 ft/lbs of torque required. My initial post has an excel attachment where I show my work.
I really appreciate the help.
I have a scenario I would like to get double checked. I have a shaft spinning at 1800 rpm driven by a 500 hp electric motor. I am bolting another drive shaft to it with (qty 8) 3\8 dia shoulder bolts on a 4-1/4" bolt circle. The bolts have a shear strength of 10.5 ksi. The stresses on the bolts...
