Bearing load augmented due to a shim

[Benbarca7]

Hi all,

I am doing a repair engineering training for civil aircraft repairs and I am not sure to understand what is exactly done here.

It is an example to show the influence of a shim between two joined members.
The example assumes a P=2000lbf load transfered from one plate to the other by two fasteners.
there is a shim between the two plates. (all the thicknesses are the same)

In the example, we assume each fastener takes 2000/2=1000lbf.
then rivet shear margin is calculated (there is a positive margin).

Then we assume that the rivets are acting as little beams built in at each end, so that half the moment is taken out at each
end.
we recalculate the margin to take into account the bending (1/ sq root(Rb^2 + Rs^2)-1. we find a negative margin.

I am fine until this.

the problem comes now, please find attached the paper sheet that follows in the training.

http://files.engineering.com/getfil...-4448-8db4-153da10fa14d&file=Bearing_load.jpg

What is done is that the bearing load is increased because of the offset of the shim.

I dont understand the 2.33 which is taken to increase the bearing load, in the equation:
1000x2xt = Paug x2.33xt

I suppose 2.33 is 2+(1/3). What is "2" and what "1/3" ?

What I understand is because of the offset due to the shim, the bending increase and so the bearing load, I just dont get how the increase is calculated.
it looks very easy, like a factor due to geometry of something, but I just dont get it, sorry.

Can you please help me?

Thank you,

Ben

Ben
Nacelle Stress Engineer (repair on Civil Aircraft)

 

[rb1957]

they're replacing the bending with a couple ...

i think they're saying assume the couple is 2/3 of the thickness of the outer layers so 2.33t = (1+2*2/3)t

however ...
the height of the section is 3t, yes?
the height of the two 1/2s of the bending stress (triangles) is 1.5t
the centroid of the triangle is 2/3*1.5t = t (or 2t between the 2 forces, not 2.33t

 

[SWComposites]

just where did this "analysis method" come from??

it makes no sense at all, though it appears to be very conservative.

the moment due to the shim may or may not be reacted by the fastener, depending on the joint configuration. And the shim is unlikely to carry any bearing load, so the assumed bearing load distribution is not accurate.

 

[rb1957]

if conservative is good, then hyper-conservatism is better ?

 

[rb1957]

looking at the calc again, he's comparing a 1,000 lbs shear load to 1857 lbs "augmented" load, by adding the bending to the shear. but isn't there bending in the original config'n, with the 1,000 lbs the skin being sheared into the dblr.

 

[dktoao]

It looks like he is assuming that the pin is press fit. Otherwise I think that the deflection of the pin and the slop in the holes would cause the inner lip of the loaded flanges to carry most of the load. A press fit would make the load distributions make sense. It could also be the reason why friction is neglected because although friction will be carrying some of the load, it will be shared by the bearing loads in an indeterminate manner, justifying the conservatism. However, if it is press fit into that middle shim I would think that the moment load would transfer into the middle plate and would be carried at the interface between the plate and the shims. So I am guessing that this is accurate for two press fit plates and a non-press fit shim in the middle. That would explain why 2.33 is used instead of 2.0 as well. So, maybe correct solution, poor presentation?

 

[rb1957]

another "2/3" factor i noticed in the outer plies is that it's the average of the linear bending distribution (1 on the outside surface, 1/3 on the inner surface of the outer ply).

he's got the moment as P*2t, which is assuming P is mid-thickness (of the outer plies).

the peak of the bending stress dist'n is 6M/(d*(3t)^2) ... assuming a rectangular section d*3t. the average stress on the outer ply is 4M/(9dt^2), and the bearing load on the fastener Paug) is 4M/(9dt^2)*(dt) = 4(P*2t)/(9t) = 8/9*P (compares with 2/2.33 = 6/7).

but still some offset moment should be included in the baseline, which by the same logic would be 4(P*t)/(4t) = P ! ... oops, that's not the point he's trying to make !

 

[rb1957]

a couple more thoughts ...

maybe he's saying that the baseline has the fastener load at the interface between the sheets, so no bending. but then the bending with the shim would be P*t (not 2t).

also i think this bearing calc is appliciable to the sheet surrounding the fastener, in that it combines the fastener shear bearing with the off-set moment. the fastener should be stressed for the shear load and the bending, as you've outlined, and not include this augmented load.

 

[Benbarca7]

Hi guys, thanks for the replies.

Before responding, I will provide the previous page, I think it can help.

http://files.engineering.com/getfil...5-bb24-2ecfffcd3852&file=bearing_load_pg1.jpg

Then, I will add that this is a training, so it is not exactly very precise method, and there might be errors inside.
It is probably conservative and very direct to the solution.






Ben
Nacelle Stress Engineer (repair on Civil Aircraft)

 

[Benbarca7]

Looking at your first post rb1957, I think this is what has been done.
2.33 is 1t+2*2/3 t. o I think this is done because the shim is not structural and the stress distribution should be only in the external plies, and nothing in the shim.
It is a triangular distribution with an offset due to the shim.
On the drawing we can see a trapeze distribution so it makes me confused.

I am confused also with the baseline calculation. It looks like the baseline calculation is a bending calculation, but without the shim (L= t)

So what is done is first a calculation as if the shim is not structural.
Then it shows that the offset of the shim involves bending effects (and negative impact on the margin), the bending effects are calculated with an approach (offset of triangular distribution) but not a real bending calculation of the actual structure.
(is it conservative, I don't know).

Does it make sense?

Ben
Nacelle Stress Engineer (repair on Civil Aircraft)

 

[rb1957]

offset moment = Pt looks alittle light, the shear is close to the interface; a typical assumption is triangular, peaking on the interface.