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bolt torque linear force

Hi Guys: I think I have a simple question.

200ft/lbs on a bolt. 3/4" diameter bolt, about 16 turns/inch (pitch) 200*12inches=2400in/lbs, right? 3/4" bolt radius=.375", right? that's 6400 pounds at the thread, pushing up the inclined plane, right? The inclined plane is 2.356" long in X, and travels 1/16" in y, right? Is that a ratio of about 37.7, right? Would that make 241,275 pounds of linear force? Excluding all friction.. Even with the friction of the bolt, isn't this alot of force being exerted on a 3/4" bolt?

Thanks! Harry

Reply to
harryhydro
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Yes. That is the torque required to further tighten the bolt.

No.

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David A. Smith

Reply to
dlzc

I though of this shortly afterward. Someone mentioned a LOT of loss to friction. I think there MUST be. If it were a perfect 'gear', You could HOLD it at 200ft/lbs with a torque wrench, but as you release the wrench, it would wind back out, at the same force, like a spring. So, how about this thought? Tighten the bolt to 200 ft/lbs, then measure how much torque to Loosen the bolt. Subtract that from 200, then divide by two. The remainder is the actual torque going to the stretch, maybe. If the bolt was just 50% loss torque, after tightening to 200 ft/lbs, you could wind it out with almost no effort, since

100ft/lbs would cover the friction, Hence the /2) Harry. PS sound awefully lossy. . .
Reply to
harryhydro

Right. Torque wrench right in the mouth. kind of thing.

No there is a lot of distortion / plastic yield and such on the mating surfaces, that will not be symmetric on an initial "torque reversal".

If you've tightened to 200 ft/pounds, then it'll be 175 to 200 ft/ pounds on reversal.

Thank God it is, or we'd have to loc-tite or stake *everything*.

David A. Smith

Reply to
dlzc

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