Using the appendix, the moment of inertia, web thickness, and flange thickness are given as,
I = 926 in^{4}
t_{w} = 0.711 in
t_{f} = 0.691 in
All that is missing is Q. Since the maximum shear is needed, the largest Q is required. The largest Q of most symmetrical cross sections occurs at the middle. Thus, determining Q for top half of cross section is
Q_{max} = A_{1} y_{1} +A_{2} y_{2}
= 0.691 (6.251) [(9 - 0.691 / 2)]
+ (9 - 0.691) (0.711) [(9 - 0.691)/2]
= 37.38 + 24.54 = 61.92 in^{2}
This gives the final shear stress as
τ = VQ / It
= (1,090)(61.93) / [(926)(0.711)]
τ = 102.5 psi |