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Question On Slip/Bolting Joint

Question On Slip/Bolting Joint

Question On Slip/Bolting Joint

Just a quick question, and grateful for any guidance.

I have an simple application that has 3 Bolts all have the same diameter, d, and has a slight clearance with the hole, D. They are responsible for clamping 2 plates together, or it could be 3 dependent on design.
They are tensioned, and a clamping force is exterted on the 2/3 plate surfaces, F.
A axial shear force, P is then applied to the joint.

The Plates are made from Steel.

The Shear Force Per Bolt = P / 3 (No. Bolts)

The Total Clamping Force on joint is = F x 3

If the joint was to move and slip and enter a bearing situation, then the Axial Shear Force has to be greater than the Total Clamping Force.

However, as there will be a co-efficient of friction between the plates i.e. 0.33 (Steel/Steel), then I presume that the total force to stop slippage would be Shear Load, P * Coefficient of Friction

Therefore if the Force to move the Joint that is under the Clamping Load must, then P x Co Eff Friction >= F.

Am I correct in my summation?

RE: Question On Slip/Bolting Joint


Wouldn't your total clamping force be F, not 3 x F?

I dont think adding more bolts increases the clamping force if they are all torqued the same.

RE: Question On Slip/Bolting Joint


Your assumption is correct. Your sliding force P=3Fμ. This is Newtonian friction. I am assuming your bolts are close together such that there are no weird moments developing.

Assume your coefficient of friction is not very accurate. Assume your bolt tension due to torque is not very accurate. Assign factors of safety accordingly.


RE: Question On Slip/Bolting Joint

I do not know what an axial shear force is and your drawing does not show it correctly. Your drawing shows some assembly that will be accelerating off to the left. In Statics a free-body diagram should be used, which will show all of the forces and they will add up to zero.

The bolts should not carry any shear load unless they are in contact with both plates at the interface and are being sheared, as in a a pair of scissors. The bolts are in tension and the shear loads are transferred through friction between the plates.

There is no way to calculate the actual shear on these bolts because it depends entirely on the fit and tolerances of the whole assembly. For calculation purposes it is assumed the shear is zero, and that the shear is transferred through friction.

RE: Question On Slip/Bolting Joint

Because one cannot guarantee contact between the bolts and the hole edge the shear force is resisted by the friction between the clamped plates. So force to overcome friction is the total clamping force provided by the bolts multiplied by the friction coefficient.

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

RE: Question On Slip/Bolting Joint

Cheers guys.
Desertfox, if the Shear Force was 500 kN and the total Clamping Load from the Bolts was 300 kN i.e 100 kN per Bolt.
Take CoF 0.5

I am confused, please bear with me. The way I saw it...
The shear force on the plates with friction from the plates would be 500 kN × 0.5 = 250 kN
As the clamping force is greater than the force trying to move it, the joint wouldnt move?

But as you put it, 300 kN x 0.5 = 150kN across the plate which is less than the original clamping force? How can the joint lose clamping force ?

The way I tried to visualize what was happening. If I held a piece of plate between my fingers, and exerted little force on the surface, then it would take little force to overcome the friction to cause it to slip through my fingers. If I then put more pressure on the plates with my fingers, it would take more force to move the plates. Thats why I am finding it hard to understand that, say 300 kN clamping load x 0.5 CoF is less clamping pressure on the plates. Taking 0.5 as roughish surface CoF If the CoF was 1 as Glass/Glass it would take more force to move the plates.

Thanks for you understandmg

RE: Question On Slip/Bolting Joint


I'm basically restating what others have already said, but the following must be true to prevent slip:

(shear force) = (faying surface friction force) <= ( (clamp force) * (coefficient of friction) )

You had it backwards in your original post. Any coefficient of friction less than one will mean the shear force required to cause slip will be less than the clamp force. Look up "slip-critical joint" or "slip-critical connection" for more information on the specific topic, but you may want to read more about friction in general first.


RE: Question On Slip/Bolting Joint


The joint hasn't lost clamping force but if you have a total bolt clamping force of 300KN then the shear force acting at right angles to overcome the the bolt clamping is resisted only by friction, therefore in accordance with the sliding friction law

mu = F/R

F is the shear force in your case.

R = the normal reaction ( this is the bolt clamping force)

Imagine having to push a large heavy crate say 100kg, to move it one only as to overcome sliding friction, so if mu is 0.25 then exerting 25kg will slide the crate but to lift it you would need at least 100kg force

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

RE: Question On Slip/Bolting Joint

Cheers guys, and the help is much appreciated.

I can fulled understand that say a 100Kg load with a mu=0.25 would take 25 Kg of force to move it, but if mu=1 then it would take 100Kg of force to move something on a more rougher surface - makes sense.

Its been a confusion that an application from 30 + years ago that I am looking and have found that the clamping force on the 3 off bolts under friction is alot less than the shear force that can be acting on it - therefore it will move. However, this has been deemed accpetable years ago!

RE: Question On Slip/Bolting Joint

Hi daparojo

you're very welcome.

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

RE: Question On Slip/Bolting Joint

Over that long a service history, are you sure the original assumed torques are correct? Has the joint ever actually been twisted that much?
Or has it twisted under use EVERY time and the twist torque forces are actually being resisted by the 3x bolts rotating in their bolt holes to act as shear pins?

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