CE456 HW 6 Solution

# CE456 HW 6 Solution - 5.8-3(a Pu = 1.6131 = 1.6(60 = 96...

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Unformatted text preview: 5.8-3 (a) Pu = 1.6131; = 1.6(60) = 96 kips 96k 96k 48k _48k _ Mu = 144(2) —— 96(1) = 192 ﬁ—kip From the Z; table, g!)an = ¢bMp = 203 ﬁ-kips > 192 ft—kips (OK) Check shear: _h__ L = 29,000 _ tW—51.6<2.24 Fy 2.24/ 50 -5195 V” = 0.6FyAw = O.6(50)(15.9 x 0.275) = 131. 2kips ¢vVn = 1.0(131.2) = 131 ksi Vu = 144 kips > 131 kips (N .G.) Beam not adequate (05v Vn can also be found in the 2,, table.) 5.8-4 Nominal shear strength: 1'— = 41.1 < 2.24 i = 2.24 129,000 = 53. 95 tW Fy V" = 0.6FyAw = 0.6(50)(l6.1 x 0.345) = 166. 6 kips (15v = 1.00, Qv = 1.50 Nominal ﬂexural strength: The unbraced length is L1, = 10 ft. From the Zx table, a W16 x 45 is compact, LP = 5.55 ft, and L, = 16.5 ft. Since LP < L1, < L,, lateral-torsional buckling must be checked. Lr_Lp M,1 = C1, [Mp — (Mp — 0.7Fny)( )] 5 MP (inelastic LTB) For cantilevers, use C1, = 1.0 (AISC F1). Mp = Fny = 50(82.3) = 4115 in.-kips Mn =1.o[4115 —(4115 —o.7 x 50 x 72.7)(%)5—‘%)] = 3477 in.-kips = 289. 8 ft-kips (a) Pu = 1.6PL = 1.6(85)_ = 136 kips, r wu = 1.214313 = 1.2(0.1QO) = 0.12 kips/ft ¢vVn = 1.00(166.6) = 167 kips Vu = 136 + 0.12(10) = 137 kips <167 kips (OK) ¢an = O.90(289.8) = 261 ft—kips Mu = 136(1) + 0.12(10)(10/2) = 142 ft-kip < 261ft-kips (OK) W16 x 45 is adequate. 5.10-1 (a) Neglect beam weight and check it later. wu =1.2wD +1.6wL =1.2(O.75)= 0.9 kips/ft Pu =1.2PD +1.6PL =1.6(34)= 54.4kips M. = iwuLZ + Li = %(0.9)(30)2 + —————54":(3°) 8 4 = 509. 3 ft-klps 54.4 k Compute Cb : Mmax = M3 = 509.3 ft-kips MA = MC = 40.7(7.5) — 0.9(7.5)2/2 = 279. 9 ft-kips = 12.5Mmax 2. 5Mmax + 3MA + 4MB '1' 3MC = 12.5 509.3) = 1 276 2. 5(509. 3) + 3(279. 9) + 4(509. 3) + 3(279.9) ' Enter the beam design charts with Lb = 30 ft and AC? = = 399 ft-kips Cb Try a W16 x 89 42an = 410 ft-kips for C1, = 1.0. For Cb = 1.276, (l)an = 1.276(410) = 523 ft-kips > 509.3 ﬁ-kips (OK) ¢bMp = 656 ﬁ-kips > 523 ﬁ-kips (OK) Check beam weight. wD = 0.75 + 0.089 = 0.839 kips/ft wu = 1.2(0.839) = 1.007 kips/ft _ __1_ 2 Pull = .1. 2 Mu— 8wuL +-—-—4 8(1.007)(30) + 54.4(30) 4 = 521 ﬁ-kips < 523 ft—kips (OK) The inclusion of the beam weight changes the total load only slightly, and C), does not need to be recomputed. Check shear: From the Zx table, (in V" = 264 kips > V” z 40.7 kips (OK) Use a W16 x 89 5.10-3 (a) Assume a beam weight of 200 lb/ft wu =1.2wD +1.6wL =1.2(3.33 + 0.2) + 1.6(6.67) = 14. 91 kips/ft Pu =1.2PD +1.6PL = 1.2(10) + 1.6(20) = 44.0 kips The ZO-ﬁ unbraced length will control. Compute Cb : 44k 14.91 k/ft A B C Shear is zero when 253.0 —- 44 — 14.91x == 0, x = 14. 02 ft Mmax = 253(14.02) — 44(14.02 - 10) — 14.91(14.02)2/2 = 1905 ﬁ-kips MA = 253(15)—14.91(15)2/2 — 44(5) = 1898 ft—kips M3 = 253(20) — 14.91(20)2/2 — 44(10) = 1638 ft-kips MA = 253(25)—14.91(25)2/2 — 44(15) = 1006 ﬁ-kips = 12.5Mmax 2. 5Mmax + 3MA + 4M); + 3MC = 12.5 1905) 2.5(1905) + 3(1898) + 4(1638) + 3(1006) Cb = 1.189 Enter the beam design charts with L}, = 20 ft and 1g: = T19T%—5§ = 1602 ft-kips Try a W36 x 160 45an = 1740 ft-kips for Cb = 1.0. For Cb = 1.189, ([5an = 1.189(1740) = 2069 ﬁ-kips > 1905 ft-kips (OK) qbep = 2340 ﬁ-kips > 2069 ft-kips (OK) Beam weight = 160 lb/ft < assumed value of 200 lb/ft (OK) Check shear: From Z; table, 915W" = 702 kips > 253 kips (OK) Use a W36 x 160 ...
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