Span Lb bMn bMn bMp (ft.) Cb for Cb = 1.0 for Cb from ______ col. 3 x col. 4 limit AB 12 549 1.67 916.8 664 kip-ft BC 8 633 1.0 633 CD 10 591 1.67 986.97 664 kip-ft. Thus, for span AB, = 664 kip-ft. > Mu - OK! bMn for span BC, = 633.0 kip-ft. > Mu -OK! bMn For span CD, = 664 kip-ft. > Mu -OK! bMn Step IV. Check for local buckling = bf / 2tf = 7.67; Corresponding = 0.38 (E/Fy)0.5 = 9.152  p Therefore, < - compact flange  p = h/tw = 50; Corresponding = 3.76 (E/Fy)0.5 = 90.55  p Therefore, < - compact web  p Compact section. - OK! This example demonstrates the method for designing beams with several laterally unsupported spans with different Lb and Cb values.
Span Lb (ft.) bMn for Cb = 1.0 from ______ Cb bMn for Cb col. 3 x col. 4 bMp limit AB 12 549 1.67 916.8 664 kip-ft BC 8 633 1.0 633 CD 10 591 1.67 986.97 664 kip-ft. Thus, for span AB, bMn = 664 kip-ft. > Mu - OK! for span BC, bMn = 633.0 kip-ft. > Mu -OK! For span CD, bMn = 664 kip-ft. > Mu -OK! Step IV. Check for local buckling  = bf / 2tf = 7.67; Corresponding p = 0.38 (E/Fy)0.5 = 9.152 Therefore,  < p - compact flange  = h/tw = 50; Corresponding p = 3.76 (E/Fy)0.5 = 90.55 Therefore,  < p - compact web Compact section. - OK! This example demonstrates the method for designing beams with several laterally unsupported spans with different Lb and Cb values.
CE470-Ch-5-Beam-Design.docx