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]

310toumad,

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

 

[JNieman]

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.

 

[SandCounter]

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.

 

[Jboggs]

I smell homework.

 

[SnTMan]

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

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

 

[310toumad]

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?

 

[BUGGAR]

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.

 

[Jboggs]

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.

 

[mfgenggear]

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

 

[Tmoose]

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/...X84PFPD1FAjGaO6WBR6BM4OOwMzPUjokd3eKB2GlTcaAA

The feature that the bolt screws into is missing.

 

[Nescius]

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!

 

[desertfox]

Hi 310toumad

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

 

[310toumad]

"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?

 

[desertfox]

Hi 310toumad

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

 

[JNieman]

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'.

 

[310toumad]

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.

 

[Nescius]

...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.

 

[desertfox]

Hi 310toumad

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

 

[SnTMan]

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