Problem 125 from NCEES practice exam

Hello,

Can someone please explain to me how the formula of secondary shear force was obtained in this problem? Where did the denominator (2)(36) come from? I understand the way primary force was calculated but I am confused about the secondary shear force formula. Also, what will be this secondary shear force formula if there were 2, 4 or 5 bolts in this problem. I have attached the pictures for problem and solution from NCEES exam. I really appreciate your help.

Mod Note - Edited to remove copyrighted material.  

Looks like it's a polar moment of inertia calc. So for three bolts your polar moment of inertia for the bolt group is 2 x s^2= 2 x (6^2).  If you had 4 bolts it would be 2 x ((0.5s)^2+(1.5s)^2) and for 5 bolts its 2 x (s^2+(2s)^2), with "s" being the spacing of the bolts.  If the bolts spacing varies you need to sum the "d^2" terms, with "d" being the distance to the group centroid.

There are 2 loads acting on the highest loaded bolt: Direct and Torsional. Here's how I worked the problem:

View attachment Scanned (002).pdf

Also notice that the area, A, of the bolt will cancel out of the shear stress due to torsion equation and not affect the shear force (which makes sense).  That looks to be the way the NCEES solution did it.  Since the bolts all have the same area and same distance from centroid, J = Nr^2A (Where N is the number of bolts).

When calculating J and using the parallel axis theorem, the very small Jc value of pi*D^4/32 is ~ zero.  So you are only left with r^2*A as the meaningful value.  And this means that N should be 2 in this case, because the center bolt J is ~ 0 (As its axis lies on the bolt group centroidal axis).

Hope I didn't confuse the hell out you.  Check out the MERM equation 53.64, but especially read the paragraph above the equation.