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Unformatted text preview: 5.83
(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 ﬁkips > 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.) 5.84 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,, lateraltorsional 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 ftkips
(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 ftkip < 261ftkips (OK)
W16 x 45 is adequate. 5.101 (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 ftklps 54.4 k Compute Cb : Mmax = M3 = 509.3 ftkips
MA = MC = 40.7(7.5) — 0.9(7.5)2/2 = 279. 9 ftkips = 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 ftkips Cb Try a W16 x 89
42an = 410 ftkips for C1, = 1.0. For Cb = 1.276,
(l)an = 1.276(410) = 523 ftkips > 509.3 ﬁkips (OK)
¢bMp = 656 ﬁkips > 523 ﬁkips (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 ﬁkips < 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 5.103 (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 ZOﬁ 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 ﬁkips
MA = 253(15)—14.91(15)2/2 — 44(5) = 1898 ft—kips
M3 = 253(20) — 14.91(20)2/2 — 44(10) = 1638 ftkips MA = 253(25)—14.91(25)2/2 — 44(15) = 1006 ﬁkips = 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 ftkips Try a W36 x 160
45an = 1740 ftkips for Cb = 1.0. For Cb = 1.189,
([5an = 1.189(1740) = 2069 ﬁkips > 1905 ftkips (OK)
qbep = 2340 ﬁkips > 2069 ftkips (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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 Fall '07
 Kargahi
 Shear, Shear strength, kips, Beam weight

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