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ESSENTIALS OF LRFD
EXAMPLE H-4
Given:
Check the adequacy of a W14×176 beam-column (Fy = 50 ksi) in an unbraced symmetrical frame subjected to the following factored forces: Pu Mux My KxLx
= 1,400 kips (due to gravity plus wind) = 300 kip-ft (due to wind only) =0 = Ky Ly = 14.0 ft
Drift index, ∆oh / L ≤ 0.0025 (or 1⁄400) ΣPu = 24,000 kips ΣH = 800 kips Solution:
As in Example H-3, for a W14×176 with KL = 14.0 ft, φcPn = 1,940 kips. Pu 1,400 kips = = 0.72 > 0.2, Interaction Equation H1-1a φcPn 1,940 kips governs.
Since
Because Mntx = Mnty = Mlty = 0 and only Mltx ≠ 0, Mux = B2Mltx and Muy = 0. Mltx = 300 kip-ft According to Equation C1-4, B2 =
1 1 = = 1.08 ΣPu ∆oh 1 − 24,000 kips (0.0025) 1− 800 kips ΣH L
Mux = 1.08 × 300 kip-ft = 324 kip-ft Because Lb < Lp = 14.2 ft, Mnx = Mpx = ZxFy; φb Mnx = 1,200 kip-ft as in Example H-3. By Interaction Equation H1-1a: 8 1,400 kips 8 324 kip−ft + = 0.72 + 0.27 = 0.96 < 1.0 9 1,940 kips 9 1,200 kip−ft W14×176 is o.k. Torsion and Combined Torsion, Flexure, and/or Axial Force
Criteria for members subjected to torsion and torsion combined with other forces are given in Section H2 of the LRFD Specification. They require the calculation of normal and shear stresses by elastic analysis of the member under the factored loads. The AISC book Torsional Analysis of Steel Members (American Institute of Steel Construction, 1983) provides design aids and examples for the determination of torsional stresses. Extensive coverage is given there to wide-flange shapes (W, S, and HP), channels (C and MC) and Z shapes. For these members, the charts and formulas simplify considerably AMERICAN INSTITUTE OF
STEEL CONSTRUCTION