Basic Friction Joint 2

[illini8181]

I am trying to understand how a bolted connection I am designing behaves. The connection is attached (Figure 1). There are actually four steel “feet” which are bolted to a large steel plate (only one foot is shown) with four bolts per foot. Specifically, I am trying to understand at what point the friction at the joint will be overcome, depending on how much contact area there is at the joint. Both structures being bolted together are very large, and so it is very possible that each of the feet will be within tolerance, but that the angle of the foot plate will not match the angle of the base plate, and there will only be partial contact (the shaded region in the attached example) at each foot. Searching online, I have come up with the following equation:

Rs = u*P*(# shear planes)*(# bolts)

Where Rs = slip resistance, u = coefficient of friction, P = preload per bolt

As I understand it, as long as the shear force, Fs, is less than the resistance to slip calculated by this equation, the joint won’t slip. This makes sense to me. I understand this visually as shown in the attached Figure 2.

What I don’t understand is, how does this take into account the contact area at the joint? I would think that as the contact area decreases, the joint becomes more and more likely to slip until a certain threshold is reached and it slips. Is this the case? If so, how do I write this as an equation? My first thought would be to write:

Fshear = Rs (at slip threshold)
Fshear/Ashear = Rs/Acontact

However, this doesn’t make sense. If this is the equation, as the contact area decreases, the Rs/Acontact factor increases, meaning that more shear force is required for the joint to slip.

Does anyone have some guidance on where my understanding is flawed?

 

[LittleInch]

It's beacause Slip friction is independant of surface area for the same force - see

http://www.engin.brown.edu/courses/en3/Notes/Statics/friction/friction.htm

Of course as surface area decreases then pressure increases for identical force and if area gets very small then the foot would simply start to shear or bend the plate. In your case it also requires the foot to be perfectly rigid, which of course it isn't. If the foot wasn't flat against the plate then tightening the bolts will start to deform the plate and also introduce moments on the foot which might not be desirable.

Therefore the practical solution is to pack with steel shims either side of any bolts holes which have a gap such that prior to tightening, the foot is in direct contact with the plate at all the bolt holes. As the bolts are tightened, the foot is simply loaded normal to the plate and no other moments are applied to the foot. As friction is independant of area, then the friction force remains the same and so long as whatever your final surface pressure on the packing is, is lower than the yield sheer stress of the plate you won't overload the plate.

It is simple to prove - get any rectanular cuboid (e.g. a brick) put it on a flat surface and push with your finger. turn it onto a different side and repeat - friction force should be equal- I can remember doing this in a physics lesson in school and that was a loooooong time ago.

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

 

[illini8181]

Thanks LittleInch! Now that you say it, I also remember a similar physics experiment way back when. Thanks for the indepth response.