# Clamping force needed?

1. Jan 9, 2017

### gomerpyle

• 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?

2. Jan 9, 2017

### haruspex

What is an HSS? I have trouble envisioning a tube with four walls. Can you supply a diagram?

3. Jan 10, 2017

### gomerpyle

Hollow Structural Steel. I'm mostly trying to figure out how you calculate the resisting frictional torque? I can't find any information about this, only stuff about sliding friction or rotational friction (I.E., stopping a spinning disk).

Where is the frictional force acting here? Its probably not along the tube wall because the pressure of a bolt is not going to distribute out that far to make a difference, so I'm guessing its actually acting right around the surface that it bolts into, but how to calculate it? What is the moment arm?

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4. Jan 10, 2017

### Nidum

Any calculation would be so complex and in the end so meaningless that you may as well just guess an answer . Reality is that you will always have a potential pivot .

Last edited: Jan 10, 2017
5. Jan 10, 2017

### malemdk

The frictional force that resists the turning moment is due to tightening of nut ,

6. Jan 10, 2017

### Staff: Mentor

That doesn't seem like a very good design. Why aren't there two bolts spaced a bit apart to resist the torque?

Last edited: Jan 10, 2017
7. Jan 10, 2017

### haruspex

It's not clear in the diagram what the rectangular box looks like with the bar removed. If it were hollow (i.e. open ended) then it wouldn't be too bad. But from your remarks above I'm assuming it ends in a flat plate.
That being so, it is extremely difficult to say what the distribution of the normal force across it will be. I suggest most of it would be concentrated in the area around the head of the bolt.

8. Jan 10, 2017

### Nidum

Indeed - almost all under the bolt head and a very small distance around . Situation can be improved a little by modifying the geometry of the mating surfaces .

Better to use more bolts as Berkeman suggests (set at max practical radius from c/l) and/or use positive keying .

Better still though would be to do a basic rethink on the joint design and eliminate most of the the problem .

9. Jan 10, 2017

### malemdk

Putting two bolts apart or keyed joint is the best solution, in these cases the moment is resisted by shear force in the bolt or key

10. Jan 10, 2017

### haruspex

Sure, but it is not clear from the original question whether those are options. Maybe the physical arrangement is a given and the requirement is to calculate on that basis.