The bolt is used with a lifting device that does fall under the ASME B30.26 standard, but the standard does not call for a specific torque requirement or test for the bolt. I am doing this not out of a specification or requirement, but I want to give the customer a torque value to work with in the field. The bolt and lifting device will be attached and removed many times in its life, which also complicates things and makes it all the more important to get a value right.
I oversimplified my explanation earlier. The bolt is a 5/8" grade 5 bolt that threads into a 304 stainless block. The listed yield strength I have for a grade 5 bolt is 92,000 psi. 92,000 psi*.25*pi*(.625^2) = 28,225 lbs in tension theoretically for the bolt to yield. The lab reported yield beginning at 16,000 lbs of tension at 280 ft lbs. The tension value is lower likely because the 304 stainless will begin to yield before the grade 5 bolt. I probably need to calculate the thread pull out strength of a theoretical 304 stainless nut to see if that tension number makes sense. I know that 304 stainless nut has a yield strength of about 30,000 psi, which is only 30% of grade 5, but you have to factor in all of the engaged threads.
The torque value still puzzles me a little bit since the torque value to acheive the 16,000 lbs of tension on the connection was 280 ft lbs. That number is what is way over the calculated value (Torque = Friction Factor x Tension Force x bolt major diameter) for a simple grade 5 unlubricated zinc plated connection. However, I did find in another engineering forum that the friction of a stainless steel fastener may be 2 to 2.5 times that of a plain steel fastener. If you solve for a friction factor with the numbers from the lab test, Friction Factor = (280 Ft. lbs x 12 in/ft)/(16,000 lbs * .625) = .336. That may explain why it took so much torque to reach yield. Please let me know if you think that might be the explanation.