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ESSENTIALS OF LRFD
where = equivalent axial load to be checked against the column load table, kips Pu eq Pu, Mux, Muy are defined in the Interaction Equations for compression and bending m, u are factors tabulated in this LRFD Manual, Part 3 As soon as a satisfactory trial section has been found (i.e., one for which Pu eq ≤ tabulated φc Pn), a final verification should be made with the appropriate Interaction Equation, H1-1a or H1-1b
EXAMPLE H-3
Given:
Check the adequacy of a W14×176 beam-column, 14.0 ft in height floor-to-floor, in a braced symmetrical frame in 50 ksi steel. The member is subjected to the following factored forces due to symmetrical gravity loads: Pu = 1,400 kips; Mx = 200 kip-ft, My = 70 kip-ft (reverse curvature bending with equal end moments about both axes); and no loads along the member.
Solution:
For a braced frame, K = 1.0 KxLx = KyLy = 14.0 ft For a W14×176: A Zx Zy rx ry Kl / rx Kl / ry
= 51.8 in.2 = 320 in.3 = 163 in.3 = 6.43 in. = 4.02 in. = (14.0 ft × 12 in./ft) / 6.43 in. = 26.1 = (14.0 ft × 12 in./ft) / 4.02 in. = 41.8
From Table E-1, above, φcFcr = 37.4 ksi for Kl / r = 41.8 in 50 ksi steel. φcPn = (φc Fcr) A = 37.4 ksi × 51.8 in.2 = 1,940 kips Pu 1,400 kips = = 0.72 > 0.2, Interaction Equation H1-1a φc Pn 1,940 kips governs.
Since
For a braced frame, Mlt = 0. From Equation C1-1: Mux = B1x Mntx , where Mntx = 200 kip-ft; and Muy = B1y Mnty , where Mnty = 70 kip-ft From Equations C1-2 and C1-3: B1 =
Cm > 1.0 (1 − Pu / Pe1 )
where in this case (a braced frame with no transverse loading), Cm = 0.6 − 0.4(M1 / M2) AMERICAN INSTITUTE OF
STEEL CONSTRUCTION