ME Nebraska
NCEES #523 MD
I arrived at the solution a little different than they did. I used MERM (pages 51-16 to 51-17) as the template. Here is what I did...
1. Calculate the torsional shear stress on each fastener using Tau_t = Fer/J where F = 500 lbf, e = 19 in, r = 6 in, and J = approx. 10.22 (J is calculated using the parallel axis theorem for the two offset fasteners and then the basic pie*r^4/2 for the fastener located at the centroid. The fastener at the centroid hardly contributes to the total). (See example 51.6 in MERM).
-This comes out to 5,577 psi, which is per fastener. (This is directed completely vertically. See Figure 51.13 in MERM for an explanation on the direction)
2. Calculate the "direct vertical downward shear" using Tau_v = F/nA where F = 500 lbf , n = 3 , and A = .1419 in^2 (tensile stress area for a 1/2-13 UNC fastener - this can be found in table 51.5 in MERM)
- This comes out to 1,174 psi, which is also per fastener.
3. Sum the two downward shear stresses, 5,577 psi + 1,174 psi = 6,751 psi. Multiply 6,751 psi * .1419 in^2. This equates to 958 lb, thus option (D) 960 is correct.
NCEES # 537 MD - I missed this problem also. If I cannot find anything in MERM pertaining to it, I am not going to worry about it, as it seems like a special case that would probably not appear on the exam (hope i'm right about this!)
Hope this helps some.