A W12x58 member, which comprises a vertical member in a truss span, is subjected to a factored axial compressive force of 163 kips. The W12x58 section is oriented such that the strong axis is subjected to wind loading. For a grade 50 specimen, what is the maximum permitted uniformly distributed wind loading on truss vertical member, assuming the distance between top and bottom chord of the truss is 12-ft.?
The section is subjected to a combined loading of axial force and strong axis bending. Because the member is part of a truss, it is assumed to be pinned at both ends.
The first thing to determine is what the ratio of applied axial force to allowable is, to determine which interaction equation to use.
We can determine the allowable axial force for a given K and L values. The problem doesn't explicitly use the term unbraced length. However, for a vertical member in a truss, it is sufficient to use the vertical distance between the top and bottom chords. Thus, L = 12 ft. The k value is set to 1.0. Therefore, the K*L value is then 12-ft. *1.0 = 12-ft. From Table 4-1, this results in an allowable axial capacity of 603 kips
φPn = 603 kips
Pr = 163 kips
Pr / φPn = 163 kips / 603 kips = 0.27 which is greater than 0.2. Therefore the first equation governs the capacity.
Next, the maximum permitted moment and the capacity is estimated at 306 kip*ft.
φMux = 306 kip*ft
Now, the interaction equation can be solved.
163 kip / (603 kip) + 8*Mr / (9*306 kip*ft) = 1.0
Solving for Mr = 251.2 kip*ft. From this, the required distributed loading can be determined from the equation:
Mmax = wL2 / 8. Solving for w, we obtain:
W = Mmax * 8 / L2 = 251.2 kip*ft * 8 / (12-ft.)^2 = 13.9 kip / ft.