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(Number_bolts)*(Diameter_bolt) = (2498.35lbf)/(15000 lbf/(in^2) * .9 09 in) = 1.1851 851 in
Ideally, we want the Number_bolts to be 6 so that even if half of them were loaded at any time (and there have to be at least 3 bolts fully loaded at any time because 3 points define a plane), they would be able to withstand the force since the safety factor was 2. Therefore, we have
Diameter_bolt = 1.1851851/6 = .031 308 in
All this means is that the bolt will be able to withstand the shear stresses on it, and it won't snap under those forces. Now, we need to calculate the yield strength of the bolts given that they're loaded in tension.
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