Slip Critical Joint

[TravisMcC]

Hi, I'm working on some joints, an example of which is attached. I am looking for some calculation or formula for calculating the resistance or ability of these bolts to resist slip from a moment and shear load.
The plain shear load is fairly easy to figure out, but throw in the moment and I cannot for the life of me figure out how to relate the developed friction to the moment.
In the example the moment is centred within the tube welded to the flange. The flange is bolted to another flange which is firmly secured. Bolts are all the same size. Holes are laser cut with generous (.030 clearance). Bolts are pretensioned using a torque wrench.

 

[desertfox]

hi TravisMcC

Having looked at your sketches I would ask if your twisting the square 85" tube then isn't that the twisting centre for the screws?
I have calculated the maximum load on the bolt assuming the above centre of rotation and I get the resultant load on the screw at 6.328" to be 980lbf this includes the 700lbf shear load.
Basically the two extreme righthand screws are doing all the work, the two inside the square tube do very little in terms of resisting the moment and the screw in the middle of the plate falls somewhere in between.
I think you need to get as many screws as possible, as far as possible away from the centre of the tube to try and even out the screw loads.
I also noticed that your torquing them to 200lbs-ft but that is usually a figure without screw lubrication and you mentioned wd-40, if your torquing to the above figure with wd-40 I would guess your over stressing the screw and tightening it to above yield, if that is the case you will lose preload on the joint and might also be the cause of the screw loosening.
Any chance of a picture of a broken plate?

desertfox

 

[TravisMcC]

DesertFox,
I get a value for an 3/4 UNC SAE grade 5 bolt, assuming a lubricated coefficient of friction at 0.18 of 223.5 ft-lbs, and using a preload of 0.7 proof(source Bowman Distribution, from Mechanical Engineering Design, 7th ed, by Joseph Shigley). From this i expect it is unlikely the bolts are being over-stressed.
I was a little perplexed at this exact situation, the moment is being applied to the plate at the centre of the tube, however the moment resisting that motion is being generated at the centroid of the bolt pattern. So long as the bolts dont slip. If they slip it becomes a bearing load on (murphy's law) the two farthest bolts, about the centre between them. This starts bending the plate, back and forth, and we have a fatigue failure at that nice sharp radius at the bottom.
At least that is what my best guess is

 

[TravisMcC]

Yeah, and there is no picture. I drove out and inspected the failure myself, the second was just described to me and it was exactly as the first. Unsuprisingly. They also were being operated in similar environments. We built a couple "one-off" parts with alot of "extra" steel to be sure these guys didn't have a repeat failure, but that is not a final solution.

 

[desertfox]

Hi TravisMcC

I would normally agree about the two plates one with an offset load and connected with bolts resisting movement about the bolt pattern centroid but I think in your case it is different, imagine gripping the square box with your hands and twisting it, it would want to rotate about the tubes centre line, however the bolts would prevent that.
I can't see if you twist the tube about its centre, why the plates would want to rotate about a theorectical bolt centre
which is a distance away from the tube centre.

desertfox

 

[paddingtongreen]

I still take the view that the two bolt group has to slip before the three bolt group becomes operative. It may be that those that work are working on just two bolts. Consider if the three bolt group was say two feet away, one foot away, at what point does the three bolt group work together with, and simultaneously with the two bolt group?

Michael.
Timing has a lot to do with the outcome of a rain dance.

 

[TravisMcC]

desert fox,
its the same as a bolt group on the end of a cantelever, why would they "rotate" about the bolt centroid, you would think they would pivot about the bolt closest the load, but that is not the case. Its an egg before the chicken scenario i think. Which is applying the load to which? It may seem that the load is obviously being applied by the tube, but an equal reaction is generated by an arm and a bolted plate.
paddingtongreen,
so long as the bolts dont slip, resultant forces should be about the bolt pattern centroid.

This is my understanding of how this system works, i'm definately open to references suggesting otherwise. In the meantime i'll wait patiently for my Steel Construction Manual.

 

[paddingtongreen]

@desertfox, think of torquing the wheel nuts on an auto, while the brake is off, the wheel tends to turn in the same direction as the applied torque.

@Travis, My worry is that the plate does not have sufficient stiffness. The torque is applied at the two bolt group and for them to work together, the plate must be stiff enough to carry the load to the three bolt group without bending. In structural work, we usually have the connection more central and with a much meatier plate.
Just check the deflection of the plate under the load from the three bolt group. I could be wrong in practice, it might be negligible, but it exists.

Michael.
Timing has a lot to do with the outcome of a rain dance.

 

[TravisMcC]

Ahh, i understand.
We have two plates bolted together,
one is 1/2" 50w and the other is 3/4" 44w. My thoughts being that if the friction generated is sufficient to prevent slip, that both plates can work together as one. When that doesnt happen, and slip occurs, obviously the 1/2" plate fails.
I expect the deflection is negligable, especially if the 2 plates are considered together. The geometry here lends itself to high bending stresses and very little deflection. I havnt the time today to work through that but it may be worth it when time permits.