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Bearing Strength Theory

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devoeger123

Mechanical
Aug 9, 2012
1
Can anyone explain why bearing allowables (pin/bolt in a hole) are greater than the yield strength? It seems to me that as the stresses increased around the hole that once the stress hit the compressive yield strength the edge of the hole would start to yield before reaching the bearing strength. I would appreciate help understanding the theory behind the bearing allowable.
 
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For one thing, the bearing area used with Fbru is d*t, whereas the contact area of bearing is greater, something less than half of the circumference.

d*t < A < Pi/2*d*T

Like most basic calcs, the standard bearing calculation and associated Fbry / Fbru value is a simplification of the actual complicated situation. The value has been calculated from test data and is an "effective" stress not an actual stress. MMPDS notes some issues with the methods used to calculated the bearing allowables for aluminum but states they are conservative.

Flabel has some nice pictures and further discussion.
 
There is no "theory" behind the bearing allowable, it is just the test specimen load divided by the area of the hole. As stated above, the actual stress state is NOT uniform, and further the fastener provides a thru-thickness constraint which further changes the stress state and the local yielding response.
 
'cause bearing is bearing and tension (yield) is tension ? ... different failure modes ...

you might as well ask why is fty greater the fcy.
 
devoeger123... Also... do not forget the essence of aero design...

We work hard to define LIMIT LOAD [LL] that will cover all reasonable/rational flight limits for stress on the structure.

For safety's sake we design to ULTIMATE LOAD [UL = LL X Safety Factor] accepting that damage [deformation] will probably occur... but that the structure will remain intact enough for safe recovery of the vehicle.

This concept allows us to design to the ultimate limits of the materials and fabrications prcess variables... yet have a light-enough, strong-enough and safe-enough vehicle that is economical to operate; but is NOT indestructable when abused.

In the case You cited, plastic strain [irreversible deformation] is occuring, and the material complexities [strain hardening, necking, etc] of what is happening at this extreme case were simplified by the graphs and equations developed by master stress theoreticians into good approximations that are [generally] conservative... and placed in texts/manuals for us dumb engineers to use safely.

Regards, Wil Taylor

Trust - But Verify!

We believe to be true what we prefer to be true.

For those who believe, no proof is required; for those who cannot believe, no proof is possible.
 
With regard to the bearing strength, consider a Hertzian contact problem. For example, a sphere on a surface can have very high contact stresses, yet may not yield per the values in MMPDS. Try out the following calculator:


If the sphere was really yielding at such low applied loads, ball bearings would not work very well. When faced with this problem, I have seen calculations based on the subsurface shear rather than the compression yield stress.

In any case, the local contact stress generated by bearing does not seem to correlate well to a compression coupon (which undergoes uniform stress through the length of the coupon, has a certain length, and given cross section). I think that a very short coupon would exhibit a higher apparent compression yield stress as well (i.e. bulk compression strength).

Brian
 
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