Question

Steel Structures: Design and Behavior (4th Edition) 4th Edition - chapter 8, problem-8.13 ( for the first load case)

Select the lightest W14 section, using the ??modified flexure analogy approach, to carry a concentrated load W at mid-span of a 24 ft beam, in addition to the weight of the beam. The ends of the simply supported span are assumed to have torsional simple support. The dead load, WD = 7 kips, the live load, WL = 22 kips, Fy = 50 ksi, and the concentrated load has an eccentricity of e = 2 in from the plane of the web.

Answer 1

Given data: Span length of the beam, l = 24ft Dead load, DL = 7 kips Live load, WL = 22 kips Yield strength, Fy = 50 ksi eccentricity, e = 2 in Calculate the ultimate load using LRFD load combination. Pu = 1.2DL + 1.6Wl = 1.2x7 + 1.6x22 = 43.6 kips Calculate the maximum bending moment. M_max = Pul/4 = 43.6x24/4 = 261.6 kips-ft \r\n Try W14 times 90 section Write the properties of the W14 times 90 section Area, A = 26.5 in^2 Depth, d = 14.02 in Thickness of web, t_w = 0.44 in Thickness of the flang, t_f = 0.71 in Width of the flange b_f = 14.52 in Moment of inertia about x-axis, I_x = 999 in^4 \r\n Calculate the design bending moment F_y = M' times y/I rightwardsdoublearrow M' = F_y times I/y = 50 times 10^3 times 999/(14.02) M' = 7125534.95 lb-in = 7125534.95/(1000 times 12) kips-ft \r\n M' = 593.79 kips-ft Since, M_max < M' Hence: the section W14 times 90 is Safe in flexure.