Dissipating load 3

[DownHillHero]

I am trying to figure out how to mount a bar that will dissipate the forces from a torque.

The bar will be slipped into a hollow square beam that is located 6" from the point where the torque is applied and it extends 8". There will be 3 bolts placed along the hollow square beam at 2", 4" and 6"

The calculation to find the forces at bolts A, B and C would be T = A*8 + B*10 + C*12
Since I only have one equation and three unknowns how can I calculate the force at each bolt?

Attached is an image of the situation.

 

[BrianE22]

It's statically indeterminant. You'd have to look at the relative stiffnesses of each of the load paths to determine how much load goes where.

 

[Engdoitbetter]

First of all, in case of motion in a plane you have three degrees of freedom, and consequently three equations to use. These are:
a) equilibrium along horizontal axis: 0=0, an identity equation, so unuseful;
b) equilibrium along vertical axis: Ay+By+Cy=0;
c) rotation around axis perpendicular to the plane, the same equation you wrote above.

So you have 2 equations and 3 unknowns. The problem is one time statically indeterminate or hyperstatic.
This means that you have to introduce one further equation which should account for the way the beam buckles under the load, like Virtual Work principle or elastic line equation.

The problem here is that since the active force is only a pure moment, surely one of the costraint forces is upward and not downward as you are showing in your attachment. Furthermore, you need to take into account the stiffness of the bolts (but at first you can model them as simple rigid supports).

It goes without saying that using only 2 bolts would make your life easier.

Hope it helps.

Stefano

 

[MiketheEngineer]

Or go buy RISA and evaluate it

 

[desertfox]

Hi go to this site and scroll down till you hit this heading:-
"Strength of bolts withstanding torsion generated shear loading"

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

 

[Kwan]

Some problems can be easily solved by making assumptions. Conservatively ignore the middle fastener. Now you have two equations and two unknowns.

How are you getting a pure moment on the bar? Pure moments are not often seen. Are you sure there is no x or y loading?

 

[1gibson]

I have this terrible gut feeling that differential equations might be useful here.

 

[Engdoitbetter]

Kwan, in fact desert fox link reports a method which would make us unload ideally the middle fastenere.

As of differential equation, equation of elastic line is clearly useful here to find the value of the hyperstatic unkown.

 

[TheTick]

Dissipate the force? Energy is dissipated. Force is distributed.

 

[dwallace1971]

How about the method for analyzing a continuous beam in Roarke's?

"On the human scale, the laws of Newtonian Physics are non-negotiable"

 

[zekeman]

"Some problems can be easily solved by making assumptions. Conservatively ignore the middle fastener. Now you have two equations and two unknowns. "

If you really want to be conservative, you take the pair with minimum distance, which is the smaller of x2-x1 or x3-x2 so that
F1=-F2=T/(x2-x1)
or
F3=-F2=T/(X3-x2)

and take the maximum F

 

[Cockroach]

I think BrianE22 hit the nail on the head with his approach, consideration of relative stiffness. If I understand the problem correctly, torque is to be controlled by three bolts laterally pinning the piece in a rectangular tube. But your equations govern bending in the plane of the paper, not torque. So do you want bending or torque resistance?

The other thing to point out, do the bolts shear due to failure by torsion similtaneously? As pointed out by TheTick, forces are distributed, yet you infer dissipation as in some sort of work-energy.

I see the problem of three distibuted bolts holding input torque of a six inch bar, bolts are 2, 4 and 6 inches off the LHS face of that rectangular sleeve holding the bar. Is this correct? And do the bolts shear by torque similtaneously?

Regards,
Cockroach

 

[DownHillHero]

Thanks for the help everyone, yes the load will be distributed sorry about the small mistake on wording...

Yes shearing of the bolts will be something to consider but that was going to be taken into consideration after I figured out the forces acting at the bolts (need the force to determine if the bolt will fail). Also the bolt would not have to fail simultaneously.

I have the intention of analyzing two parts of this set up. First would be to design the bar with the torque on it so that limited bending will occur. The second part is to figure out what the forces on the bolts would be, and where there locations should be to help distributed the load from the torque.

The hollow square beam that it is mounted in then should distribute the load from the torque over a larger area?

 

[hydtools]

Can you fit the bar to the square tube close enough that the load is taken by the square tube and screws will only be needed to keep the bar in the tube? Then you have two points of contact. One where the bar enters the tube and the other where the end of the bar inside the tube contacts the tube.
Not that the the more complicated three-screw attachment loading can not be calculated. Just might simplify the problem.

Ted

 

[dhengr]

Seems you have a bar with a moment on it, inserted in a tube, which is then welded to the foundation structure. Maybe the moment is due to a vert. load out at the left end of the bar, that means bending moments, shears, associated stresses, etc. in the bar. Any other loads on the bar? Note that torque and moment have the same units of measure, but fairly different meaning in many cases. Why not use only the middle bolt to hold the bar in place, and have the tube and its welds provide the reactions for the bar. The welds are much more rigid than any bolted connection, with any bolt hole tolerance or yielding, so they will take the reaction loads and not the bolts. Put a .25" high by 2" long bar on top of your cantilevered bar on its right end, and a .25" x 2" bar on its bottom edge where it bears on the tube. This pretty well defines your back span reaction points. You might put stiffener plates from the tube t&b flanges to the foundation structure to help get the reactions out of the tube and into the foundation structure.

Otherwise, Desertfox give you a very good link to one method of analysis of the bolts.

 

[Cockroach]

HyTools most likely picked up the sent. Exactly, putting a bolt THROUGH the piece act.y circumvents what is required, hold the pice in torque.

So I give HyTools the star. You need to C,BORE a series of set screws, brass of course, to resist torque. The rod is necessarily circular, the housing may be whatever, square being the worst choice. The rod carries torque equally distributed over the span of shear screws.

Regards,
Cockroach

 

[DownHillHero]

As suggested by Hydtools if the bar fits close enough into the square tube that the screws are only used to hold the bar in place, as suggested there would only be two point of contact, but if the bar were fitted into the tube would not more of the bar would in contact with the tube than just the end points? This would then better distribute the load across the length of the tube.

Also if more of the bar is in contact then is the determining the load along the length of the bar/tube becomes more a problem of determining the relative stiffness of the bar within the tube?

 

[hydtools]

Except for a very precise fit or a press fit there will be some clearance which will allow the bar to rotate, even slightly, so that effective contact will be the bar end inside the tube and the entrance of the tube to bar. More detailed analysis into the deflection and deformation will show some contact areas as compared to initial edge contacts. It is resonable to start with the assumption of edge contacts to determine magnitude of the reaction forces.

Ted

 

[DownHillHero]

Ok so we are simplifying things.

Im having a little bit of trouble figuring out how to model the elastic line equation. My inclination is to model the edge contacts as a simple support such as a roller, however that would imply that the deflection at the edge contact points should be zero however the contact form the edge point is created by the deflection at those points. Or perhaps working out the elastic line equations and then specifying the deflection at the edge contacts points?

 

[dhengr]

Look back at my 2AUG12, 15:35 post, and reread it for its full meaning. You are making this problem way too complicated unless you really have a reason to do that. Don’t model it, just do a simple cantilever beam calc. You should have learned that in your first Statics and Strength of Materials courses. You need a small bolt where the middle bolt was, to hold your bar inside the outer tube. Eliminate the other two bolts because they just make for a highly indeterminate system. Make the outer tube .5" higher inside, than your basic cantilevered bar is high. Referring to your original picture, put a .25x.25" bar on the top edge of your cantilevered bar near the right end (back end) of the outer tube; put a .25x.25" bar on the bottom edge of your cantilevered bar near the left end of the outer tube (front end). This .25x.25" bar stock is as long as your cantilever bar is wide. Now you have a simply defined cantilever beam, whatever its actual dimensions. Put two .25" stiffeners outside the outer tube and back to the foundation structure, to match the reaction point inside the tube. The reaction loads are taken right into the stiffeners and back to the foundation structure.