- #1

gomerpyle

- 46

- 0

Thread moved from the technical Engineering forums, so no Homework Template is shown

Say you have a piece of flat bar stock bolted to the end of a 3 x 3 x .25 HSS. A force of 900 pounds is exerted on the flat bar approx. 19" away from the center of where it is bolted. What clamping force would be needed to prevent the bar from rotating?

My attempt at this was to first calculate the applied moment around the bolt, so 19*900 = 17,100 in-lbs. Secondly, to assume that a force acts along each side of the HSS walls resisting the moment, and that the coefficient of static friction for steel-steel contact is 0.5.

The overall resisting frictional torque would be mu*N*d. If d is the same, and there are four walls of the tube, then it would be 4*mu*N*d.

Thus: N = 17,100/(4*0.5*1.38) = 6,195 pounds clamping. Is this methodology correct? Then I began thinking, what happens if instead of a hollow piece of structural tubing, the flat plate is bolted to another flat plate? Technically if you pick any point of contact between them, there are an infinitesimal amount of frictional forces with varying moment arms, so how would you approximate that?

My attempt at this was to first calculate the applied moment around the bolt, so 19*900 = 17,100 in-lbs. Secondly, to assume that a force acts along each side of the HSS walls resisting the moment, and that the coefficient of static friction for steel-steel contact is 0.5.

The overall resisting frictional torque would be mu*N*d. If d is the same, and there are four walls of the tube, then it would be 4*mu*N*d.

Thus: N = 17,100/(4*0.5*1.38) = 6,195 pounds clamping. Is this methodology correct? Then I began thinking, what happens if instead of a hollow piece of structural tubing, the flat plate is bolted to another flat plate? Technically if you pick any point of contact between them, there are an infinitesimal amount of frictional forces with varying moment arms, so how would you approximate that?