FIGURE 3. Gusset plateshear, and disregard interaction of direct stress and shearing stress in computing the ulti-
mate-load and ultimate-moment capacity.
Calculation Procedure:- Resolve the diagonal forces into their horizontal
and vertical components
Let Hu and V 11 denote the ultimate shearing force on a horizontal and vertical plane, re-
spectively. Resolving the diagonal forces into their horizontal and vertical components
gives (4^2 + 52 )^05 = 6.40. Horizontal components: 150(4/6.40) = 93.7 kips (416.8 kN);
110(4/6.40) = 68.7 kips (305.6 kN). Vertical components: 150(5/6.40) = 117.1 kips
(520.9 kN); 110(5/6.40) = 85.9 kips (382.1 kN). - Check the force system for equilibrium
Thus, ^Fn= 206.0 - 43.6 - 93.7 - 68.7 = O; this is satisfactory, as is 2F> = 117.1 - 85.9
-31.2 = 0. - Compare the ultimate shear at section a-a with the
allowable value
Thus, Hn = 206.0 - 43.6 = 162.4 kips (722.4 kN). To compute Hn^ assume that the
shearing stress is equal to the yield-point stress across the entire section. Then Hu>allow =
24(0.5)(18) = 216 kips (960.8 kN). This is satisfactory. - Compare the ultimate shear at section b-b with the
allowable value
Thus, Vn = 117.1 kips (520.9 kN); Fw?allow = 18(0.5)(18) = 162 kips (720.6 kN). This is sat-
isfactory. - Compare the ultimate moment at section a-a with the
plastic moment
Thus, cd = 4(6)/5 = 4.8 in (122 mm); Mn = 4.8(117.1 + 85.9) = 974 in-kips (110.1 kN-m).
Or, Mn = 6(206 - 43.6) = 974 in-kips (110.1 kN-m). To find the plastic moment Mp, use