the first part is a very good point, and the last paragraph made me think here... if a bolt has (for arguments sake, lets use a round number) 100lbs tensile strength Tensile strength is the stress required to fail the material. It's unit for stress is essentially pressure, not force. psi not lbs. For this discussion we're talking about yield strength anyways, not ultimate tensile strength., it would have 60% of that as sheer Loading directions are independent of each other. You can load an isotropic material (steel) to the yield strength in any direction without it affecting the other directions (to keep it simple I won't get into the effect of Poisson's ratio), ok, so 60lbs sheer strength, but they don't really compound... so that being said... lets say for instance, that the 100/60lb limits were actual breaking limits... if you were to torque the bolt so that there were 99lbs of tensile pressure on the bolt, you cannot put much of any sheer force on the same bolt. basic math would imply that if you were to put 0.6lbs of sheer force on it, it would fail, however, to me, i would think it would take more than that... that being said, the 60lb normal sheer limit is not accurate with compounded stress... What I said earlier in the paragraph makes all of this void.
in other words, if you want to do it by the book, You obviously didn't read the book., for every lb of tensile pressure on the bolt, the sheer limit on that bolt should be lowered by 0.6 of a lb (60% threshold)
ie, the torque specs for a ford 8.8 are around 33ft/lbs? i'm going to guess with the ram mounted to the cover, it's going to be closer to the 50ft/lbs... with that, that gives us a clamping force, or tensile load, of about 660lbs (based off of 3/8" bolt) now subtract that from the 9300lbs of tensile strength that Kage calculated, that gives us 8640, of remaining tensile strength, multiply that by .6 (60%) gives us 5184lbs of available sheer pressure before the failure point I didn't read any of this.
here's where i lose it a little, maybe, and start guessing: lol
Kage, i assume you mean that by ksi, you mean kilos per inch^2? No. He's not mixing the Imperial and SI system. ksi = kips per square inch = kiloPOUNDS per square inch. and by that the 5184 is psi, based on the fact that the cross section of the 3/8 bolt is just that, 3/8, (measuring from the outside of the threads, which do not help the strength at all) we should be able to multiply the 5184 by .375, to give the actual sheer point of of 1944 axial lbs... You just took pounds per square inch * inch and got lbs. What happened to the other inch? You have to multiply by the shear area - exactly like Kage did. now that being said, the initial sheer won't be the perfect fit (hole for bolt) there will be a % of loss here, but lets ignore that...
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