Clamping force required 1

[310toumad]

Lets say you have a piece of 1/4 plate bolted to the end of a 3 x 3 x .25 HSS. A force of 900 pounds is applied at the end of the plate 19" away from the center of the bolt. What would be the clamping force required by the bolt to resist the plate moving? Picture attached.

So the moment acting on the pivot would be 900*19 = 17100 in-lbs, which I'm assuming is what the resisting frictional torque has to be equal to or greater than. I can't really find an equation related to this problem. The static coefficient of friction between steel on steel, clamping force, and maybe contact area all are variables? Any input would be appreciated, thanks.

 

[drawoh]

...

This is assuming a static coefficient of friction of about 0.5, and a radial distance to the tube wall of about 1.375 from the center of the pivot.

For your effective clamped radius, I would use the radius of the head of your screw. There some interesting Belleville washers in the McMaster Carr catalogue that could increase your clamping radius to that of the washer. Half the width of your arm is very optimistic.

--
JHG

 

[310toumad]

Even with two bolts, the holes in the end plate are still slightly over-sized, so slipping can still occur. I'm not necessarily worried about the bolted end connection failing, just reducing the deflection of the horizontal HSS members due to 'twisting' from the force couple acting on them.

The bolt threads into a piece of angle-iron welded to the inside of the HSS, so I suppose the resisting frictional torque will be more-so acting on that surface rather than the tube walls, because as you said the pressure distribution from a bolt will not spread out nearly that far to make enough difference.

So the question remains, when two flat surfaces are contacting, how do you calculate the clamping force needed to induce a resisting frictional torque on one surface to counteract the applied moment on the other?

 

[BrianE22]

For only the torque part, if your tube was circular and had a radius of r then the torque you would develop by friction would be:

Torque = Fbolt x mu x r.

For a rectangular tube I'd play it safe and assume r was about 1/2 the tube width. Also I think mu = .5 is a bit optimistic. I'd use mu = .1.

 

[3DDave]

Assume the radius is the radius of the clearance hole and use the formula already mentioned several times.

 

[310toumad]

So clearance hole is 9/32", that would yield:

Torque = Fbolt*mu*r
Fbolt = Torque/mu*r
Fbolt = 17100 in-lbs / (0.1*.140) = 1,221,429 pounds...

 

[rb1957]

as so many others have said already, using one bolt like this is bad design. You're relying on friction. If you insist on the single bolt, I'd add tabs so the moment has a better loadpath.

another day in paradise, or is paradise one day closer ?

 

[dhengr]

310toumad:
You don’t really seem interested in good design info. and advice, and you have been given plenty of it above, if you’d just read carefully and think a little. Maybe it can be summed up in saying that it is not wise engineering to count on friction to take a significant load or impulse. You be the judge. You’re not really hearing what’s being said, you haven’t explained the details of your problem and design very well, and we certainly can’t see it from here. So do it your way. Make one, and tighten the bolt until you get enough bolt tension to hold your load in whatever way is satisfactory for your needs. Call this the min. torque req’rd., and put some factor of safety on that and call it good. Then start worrying about some of the advice you’ve been given above. One failure, may well, cost many times what a dozen of these things done with good design would cost.

 

[SnTMan]

So clearance hole is 9/32", ...

1/4" bolt no less, I'd expect you'll break it before you get the clamp force you need.

Regards,

Mike



The problem with sloppy work is that the supply FAR EXCEEDS the demand

 

[desertfox]

Hi 310toumad

You are correct even with two bolts the joint can slip however it won't slip if the are tightened correctly. Look at this website below and see the calculation for bolts resisting shear generated by a torque.
With a single bolt the end stop can rotate 360?degrees if it comes loose but with two bolts it can only move till the clearence is taken up between the bolt and hole.

http://www.roymech.co.uk/Useful_Tables/Screws/Bolted_Joint.html

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[310toumad]

I apologize if it appears that I'm disregarding advice given on the function of the design itself. I understand functionality is of course one of the first things on some people's minds when evaluating these topics, but also understand my original question had nothing to do with seeking input related to that.

Is a single bolt a bad design to retain this assembly as a rigid frame because it relies upon friction? I think that's universally agreed upon, and obviously a multi-bolt pattern is much more effective at handling eccentric loading, however:

1) I'm much more interested in the physics behind the problem I stated.
2) The single bolt idea is not mine. Again, the value of design critique is not lost on me, but its not really the kind of information I'm looking for.

 

[JNieman]

I think you're finding out that when you run the calculations on this situation, you're going to end up with a much more expensive assembly when trying to satisfy requirements with a single fastener, if one can even be physically produced. So the originally stated goal (to be cheaper to make) is no longer within the realm of possibility.

 

[SnTMan]

310toumad, put that way I think BrianE22's equation is about as good as it gets.

Regards,

Mike

The problem with sloppy work is that the supply FAR EXCEEDS the demand

 

[rb1957]

I think brian's equation is very conservative. The bolt is clamping the plate against some other plate (or maybe it is just the head of the bolt ?); either way, I think using "r", the hole radius as the torque arm is conservative. Of course there are many imponderables with this set-up ... how does the bolt clamp up pressure distribute away from the bolt ?

Could you work the preload equation in reverse ? If you preload the bolt with an applied torque, isn't it reasonable to say that the bolt can resist that torque; ie tighten the bolt with the torque expected (obviously factoring this by 1.5 would give you some reassurance inservice) ? Use T = Pd/5.

wotcha think?

another day in paradise, or is paradise one day closer ?

 

[drawoh]

310toumad,

There is nothing wrong with relying on friction to hold something in place. There is something wrong with relying on friction that requires a 1/4" bolt to exert 1.2M[ ]pounds.

--
JHG

 

[rb1957]

I think OP was posting that to show the suggested equation produced an unreasonable number.

using T = Pd/5, 17100*5/0.25 = 342000 lbs ... equally unreasonable ... suggests the design is unreasonable (by my calcs a 1" bolt is down, good for 80,000 lbs ULT)

another day in paradise, or is paradise one day closer ?

 

[BrianE22]

Why all the concern about the bolt hole radius? Assuming the sketch is somewhat to scale and assuming the bolt length is at least as long as the tube then the torsional stiffness of the tube is probably 1000's x as stiff as the torsional stiffness of the bolt. Very little of the torque is reacted by the bolt, probably less than .1%.

 

[desertfox]

Hi Brian

We were not concerned with the torsional stiffness of the tube, what the OP was asking was if he fastens a flat plate to the end of a tube with a single bolt and then applies torque to the flat plate what clamping force does he need to generate between the flat plate and the tube end to prevent the flat plate rotating.

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[Tmoose]

About all the clamp force one 1/4 inch fastener could exert would be 3000 lbs.
3000 lbs X 0.1 for friction = 300 lbs sliding force resistance.
17,100 in-lbs of torque would require 17,100/300 = 57 " effective radius where the friction/contact must be applied.



I'd cut 3 pieces of 2.5 " long angle instead of 1.
Then I'd Weld one angle down inside the tube 2 inches or so(to provide plenty of fastener length for elongation and reliable preload retention when things shift/embed a little bit, with a cage nut or ??? to provide threads for the bolt.
Finally, I'd Weld two of the angles to the face of the lever, positioned to engage the square tube snugly, but inset to miss the fillet welds securing the angles to the lever.

 

[drawoh]

Why all the concern about the bolt hole radius?

The OP is applying a couple of 900lb[×]19"[ ]radius. He wants to resist this with a single, 1/4"[ ]bolt. We need to know what radius the bolt acts at. The radius of the clearance hole is a good, conservative estimate. The stiffness of the tube is not important. The stiffness of that plate/arm is what matters here.

Here is another way to visualize the problem. I install the bolt and nut with a wrench. How long is the wrench handle? How much force can I exert on the end of the wrench handle? Could these numbers be less than 19" and 900lb?

--
JHG

 

[SwinnyGG]

You may have a bigger problem on your hands. By my calculation 900lbf is nearly enough to cause the bolted plate to fail on the tension side, and that's without considering that there may be an impact component to the force in the real world. The tensile stress is in the neighborhood of 46,000 psi without considering buckling. Assuming we're talking mild steel that leaves you with a safety factor around 1.15. Think your estimate of the forces involved are accurate enough for that to be safe?