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So if the load on the u-bolt clamp was eccentric (as in the image attached) how would I calculate the force required for slip?http://files.engineering.com/getfile.aspx?folder=b032d896-cf26-48a3-91be-a2a8fc9e38db&file=slip_question.jpg

Thanks Brian E22. That is a good point.

Thanks all for clearing up my doubts. Sliding force = 2x6000xFrictCoeff. This is precisely what I presented to a customer who laughed and told me I was wrong. I showed on a fbd how there must be an equal reaction force on the opposite side of the U-bolt, but he maintained that if the tension...

Thanks Cory!

No I don't think that's right rb1957. Coefficient of friction is determined by the materials only. Friction depends only on the normal forces and the friction coefficient. It is not dependant on the contact area.

Thanks for the input LittleInch. So the force to slide the bar out of the U-bolt is 12,000*friction coeff? That's what I was thinking but I'm not yet convinced. Only 6000 N is applied to the assembly after all.

In the picture attached, can somebody please explain if the axial force in each side of the U bolt creates 6000 N on each side of the bar , or is the 6000 N divided between each side? http://files.engineering.com/getfile.aspx?folder=aa1a0ed6-cc9d-4237-8430-c8b89c23161c&file=Clamp_Friction_force.JPG

Thanks for the advice. I should say this is not lifting/ hoist equipment in the traditional sense. The linkage passes through a limited space so the eccentricity and size reduction is necessary. The example I gave is very much a simplification though, (the rod is actually part of a telescopic...

Great. thanks

I'm designing a eccentrically loaded lifting linkage and I'm wondering the best way to calculate stress as there is a combination of tension and bending. The linkage is made up of 1m long 25mm diameter round bar with a rod end on one end. The other end has a plate welded along the side of the...