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Clamping force required

Clamping force required

(OP)
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.

RE: Clamping force required

If your design must resist a turning moment as shown, it sucks. Use a four bolt pattern.

You can make a crude estimate of your force through the Newtonian physics you learned either in high school, or by second year college at the latest. For anything more accurate, I would suggest building something and testing it.

--
JHG

RE: Clamping force required

I second @drawoh.

If you're worried about the bolts/washers not keeping it stationary under duress, bolting in addition to dowels may be necessary.

RE: Clamping force required

Can you add a boss to the plate so it binds on the beam when the force is applied?
Can you bend the plate in a 90 toward the beam and then a 90 down so the bend binds on the beam?

I used to count sand. Now I don't count at all.

RE: Clamping force required

I smell homework.

RE: Clamping force required

Surely homework is not that bad poor these days...

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

RE: Clamping force required

(OP)
Let me elaborate a little on the intent here. The 1/4 plate acts as a "stop" for a sliding tube assembly (almost like a forklift arm, if you want to think of it like that) which moves along the top HSS. That's why its bolted on and not welded, so the fork arm can be removed if need be. There is another HSS member along the bottom, which the fork arm butts up against. So the two horizontal HSS members are capped at the ends with these bolted end plates. They are connected in the middle to a carriage assembly but that is irrelevant.

Obviously, when a load is placed on the end of the fork arm a force couple is created on the two horizontal HSS members. If the fork arms are extended all the way to the ends, then you get more deflection/twisting. Yes, I know a multiple bolt pattern would more effectively resist the tendency of the beams to "twist." Any small step to reduce cost is important, though, so if it was theoretically possible to stop any movement with a single bolt, then that removes extra cost when fabricating this. Or maybe not, maybe the clamping force would necessitate a very large bolt. Hence why I'm posing the question.

Now, maybe I'm overthinking this but the clamping force needed would be a function of the resisting moment (needs to be equal to or greater than applied moment), the coefficient of static friction between steel, and the distance away this friction force acts from the pivot point, correct? So:

Clamping force = Resisting moment/µ*d
Clamping force = 17100 in-lbs/(0.5*1.375) = 24,872 pounds.

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. Seems quite large to me, but maybe this is too simplistic? Does the contact area between the rotating 1/4 plate and the tube wall have an affect?

RE: Clamping force required

Sounds like those old friction shock absorbers (even Jimmie can't help me here). There is data on friction shocks but you would need a consistent bearing surface. You can read up on AISC structural non slip connections and their requirements for faying surfaces but that might not be relevant.
On some cars, they peen faying surfaces to prevent slip.

RE: Clamping force required

You may think you're saving money but not really. In the real world you have to depend on the ability and consistency of those that have to work with your designs. If you somehow arrive at a torque number that theoretically meets your needs, and even go so far as to confirm it in tests, from then on you are completely dependent on the consistency of the workers that have to assemble this thing. When one of them "under torques" it, which WILL happen, and there is a failure, guess what - it won't be his fault. ALL eyes will look at you. You are making a very basic mistake here in asking ANY single point connection to resist ANY moment load. And that's not even considering the possible effects of vibration, material creep over time, stripped threads, and any number of other sources of failure. Don't expose yourself and your reputation to cruel ridicule.

A very wise man decades ago told me about designing equipment,"Five years from now they won't know if it was a few dollars over budget or two weeks late. But they will by God know if it works." Those words have served me well.

RE: Clamping force required

mill it from a single piece of stock, I also say for get that design. very poor, and it will fail.

RE: Clamping force required

If you insist on a single bolt, and can not weld the arm, then "keying" the arm to the tube by welding stops on the arm might work. Or a slice of larger rectangular tubing could be used instead.

A short chunk of smaller rectangular or a square of 3/8 inch plate with nipped corners could be welded to the arm, engaging the tube inside walls.

Similar to this -
https://encrypted-tbn0.gstatic.com/images?q=tbn:AN...

The feature that the bolt screws into is missing.

RE: Clamping force required

Your design could fail if somebody spills oil on it, or if the roof leaks on it, or if condensation forms on it, or...lots of things.

Your line of thinking isn't far off, though. Consider the above advice regarding welded/bent stops or dowels and the costs associated with these, then the cost of more bolts, drilling more holes, and so on...

900 pounds acting at 19 inches isn't a tremendous force to resist; you'll be able to come up with something clever and economical. Report back to the thread and show us when you do!

RE: Clamping force required

I have to agree with others that relying on a single bolt is not good design and I would suggest a minimum of two bolts because then you are not relying solely on friction to hold the plate in position.

Now regarding the question of the clamping force, I calculate that it would need to be a minimum of 3600lbf to resist the 900lbf you intend to impose on it. This figure however is purely theoretical because I have assumed that the coefficient of friction is 0.25 and area of the clamped parts is independent of friction (in practice it isn't).Now if the actual coefficient of friction was only 0.2 then the clamping then the force would need to increase to 4500lbf as a minimum to prevent movement of the stop plate. From my example above you can see a variation in friction coefficient significantly alters the minimum clamping force and I wonder how you are going to control that friction coefficient consistently so that a known clamping force guarantee's the connection won't fail? In addition to this you will have to ensure a consistent bolt load in the fixing which will probably mean pre-tensioning the bolt as torque cannot be relied on.

If you use two bolts or more then the friction problem completely goes away because now for the plate to move the bolts have to fail and that is much easier to calculate or predict than relying on friction. Now you mention saving money, so consider the cost of:- buying bolt tensioning gear, spending hours testing the clamping joint in conjunction with variable friction and then compare that with the cost of drilling another hole and adding a bolt.

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

RE: Clamping force required

(OP)
"Now regarding the question of the clamping force, I calculate that it would need to be a minimum of 3600lbf to resist the 900lbf you intend to impose on it." And what variables, equation did you use to calculate this?

RE: Clamping force required

All I used was the theory of sliding friction , F= mu * R where F = 900lbf and mu= 0.2

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

RE: Clamping force required

One requires that you likely need to use a higher grade/strength of bolt, control the quality of the two mating surfaces, the cleanliness of the surfaces during assembly and control the torque value of the single bolt being installed, and maybe also use a visual indicator to show when the bolt has begun backing off.

Two bolts means you can use two barely strong-enough bolts that can keep the plate in place, and not have to worry about the surface conditions, and the only indicator that the bolts are backing out can be the fact that the plate is getting wobbly.

I don't follow the same logic you do in identifying the single bolt as 'cheaper'.

RE: Clamping force required

(OP)
Single bolt is cheaper (assuming it would not have to be of much greater size) because its one less tapped hole to drill, one less fastener to be tightened during assembly, etc. I know, it seems like splitting hairs. Not my idea.

desertfox, your calculation assumes that the plate only has to resist 900 lbf of lateral movement, but this is a torque we are talking about, so I don't believe a simple F = mu*R is appropriate for this example.

RE: Clamping force required

...but this is a torque we are talking about...

Yeah. Regardless of whether or not you want to factor surface area into your analysis, the location (radially) of that frictional force will be critical to know.

For a given load, a simple thrust bearing washer arrangement will take more torque to turn as the washer grows in diameter.

RE: Clamping force required

Yes I forgot the moment arm however how are you going to ensure that 24,872lbf in your calculation is going to occur at the edge of the box section, when a bolt is tightened its pressure from under the head spreads out in a 60 degree cone (30 degrees either side of vertical).
My guess is if you draw the cone from under the bolt head you won't even reach the edge of the box section and even if you do, you're back to controlling the bolt preload accurately, so you're calculation is really also inappropriate if you think about it.

To me it's a no brainier putting at least two fixings in the joint, far cheaper, more reliable and a damn sight safer.

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

RE: Clamping force required

Looks to like it would be simple to mock-up and test. Then you'd know.

Regards,

Mike

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

RE: Clamping force required

...

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

RE: Clamping force required

(OP)
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?

RE: Clamping force required

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.

RE: Clamping force required

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

RE: Clamping force required

(OP)
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...

RE: Clamping force required

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.

RE: Clamping force required

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

RE: Clamping force required

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/Bolt...

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

RE: Clamping force required

(OP)
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.

RE: Clamping force required

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.

RE: Clamping force required

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

RE: Clamping force required

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?

RE: Clamping force required

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

RE: Clamping force required

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)

RE: Clamping force required

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%.

RE: Clamping force required

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

RE: Clamping force required

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.

RE: Clamping force required

Quote (BrianE22)

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

RE: Clamping force required

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?

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