At 02:51 PM 8/30/2015, you wrote: >Ignoring clamping friction in a bolted shear joint calculation seems >to be a worst case analysis in that the assumption is: the bolts are >torqued just enough to keep from rattling around. The total load >then would be applied in shear to the bolt. That assumption does >not seem realistic for practical applications. >Sid Wood +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
I'm not an engineer but the following from the web would indicate that the 3/8" bolts (4 / each main spar attach) are a great deal more than adequate for our application. Clamping friction would only increase the safety factor. Our 3/8" inch bolts yield considerable more shear strength than the 1/4" bolts in the example below. A 1200 pound KR pulling 6 G's is 7200 pounds of force spread across 8 each 3/8" wing attach bolts. That's possibly the reason I've never seen a failed attach fitting in the several crashed KR's I've witnessed. Like I said, I'm not an engineer so take with a grain of salt, whatever that means........... Larry Flesner P.S. Against my better judgement I'll even confess I never torqued my bolts but used a semi-educated "that feels about right". I'm not recommending that approach for other builder. Do it right. (grade 8 as I recall) A = Cross-sectional area of the fastener size (since bolt bodies/shanks have circular cross-sections, use area of a circle) = Pi x r2 where R (radius) = .250/2 = .125, therefore A = Pi x (.125)2 = .0491 square inches (in2) Capability in shear = 91,000 lbs / in2 x .0491 in2 = 4468 lbs Using the same .250-inch diameter grade 5 fastener results in the following: Capability in shear = 75,000 lbs / in2 x .0491 in2 = 3683 lbs
