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HW 8 - 6.6-1 Determine the axial compressive design...

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Unformatted text preview: 6.6-1 Determine the axial compressive design strength. Use K; for the unbraced condition. KXL _ 1.7!14! _ _ rx/ry — 2.44 —9.754fi<KyL—14fi From the column load tables with KL = 14 ft, (#an = 774 kips i — M = 0.5168 > 0.2 use Eq.6.3(AISCEq.H1-1a). 4:510" ‘ 774 Check the braced condition first. For the axis of bending, cm = 0.6—0.4(A—A2—j = 0.6—0.4(%%) = 0.3867 2 2 P9] ”2151 _ 7:131. _ 7: (29,000)g8812 =8934kips =(K,L)2 (1<,,L)2 (1.0x14x12)2 Bu _ Cm Cm _ 0.3867 = 0.405 < 1.0 Sway condition: use [6-27] 1—(aP,/Pe1,.) = 1-(1.00P,./P.1,.) ” 1—(400/8934) u5631 = 1.0 32 = __1 = ____1 = .___—1 = 1.176 aZPm _ 1.002191, ._ l.00(6000] 213:2 1 ZPeZ 1 40000 The total amplified moment at the top is M = 31M," + BzMg, = 1.0(45) + 1.176(40) = 92.04 fi-kips 14 mp 1- The total amplified moment at the bottom is M 1,0, = 1.0(24) + 1.176(95) = 135. 7 fi-kips u . Use M“ = 135. 7 fi—kips. Compute the moment strength. From the beam design charts with Lb = 14 ft, (15an = 479 fi-kips for C1, = 1.0 and ¢bMp = 521 ft-kips Using the total amplified moment, compute C1,: = 12.51171max 2. 5Mm5x + 3MA + 4MB + 3M5 = 12.5 135.7) 2.50357) + 3(35.11) + 4(21. 33) + 3(78.77) For C), = 2.208, $5M” = 2.208(479) = 1058 fi-kips Since 1058 fi-kips > ¢bMp, use (15an = ¢bMp = 521 fi-kips Eq. 6.3 (AISC Eq. Hl-la): Cb = 2. 208 L i & fl} 1 m4 = ¢an + 9&me + mm, —0.5168+ 9( 521 +0) 0.74s <1.o (OK) Member is satisfactory. E (a) LRFD Solution Pu = 1.2(135)+1.6(415) = 826.0 kips The factored moments are Mmx (top) = 1.2(90) + 1.6(270) = 540. 0 fi-kips Mm (bot) = 1.2(30) + 1.6(90) = 180.0 ft—kips The amplification factor 31 can be estimated as 1.0 for purposes of making a trial selection. Mu, = 5'1me = 1.0(540) = 540 fi-kips Try a W14 x 159. From Table 6-1, with KL = L5 = 20 13, [6-42] p = 0.619 x10‘3, b, = 0863 x10"3 pPu = (0.619 x 10‘3)(826) = 0.5113 > 0.2 Equation 6.8 controls. As a preliminary check, pP, + b,M,, + 12va = pPu + bew‘ + b,Muy = (0.619 x 10‘3)(826) + (0.863 x10‘3)(540)+ 0 = 0.977 < 1.0 (OK) Calculate 31 for the axis of bending: Cm = 06—04%) = 06—04—397) = 0.4667 251 7:209 000)§1900) - PH= 7! x = , =1.475x104k1 »‘ (IgL)2 (0.8 x 20 ><12)2 ps 31,. = C—m = ——°—-i5—57—— = 0.4944<1.o Bl, = 1.0as assumed 1_ P. 1_ 826 Pm 14750 (Since this shape is adequate for C1, = 1.0, the steps shown below, computation of the actual value of C1, and an adjustment of b,, are not necessary.) Compute C1, and modify bx to account for C1,. = 12.5Mmax 2. 5Mme + 3MA + 4M3 + 3MC = 12.5940) =2 143 2.5(540)+3(0)+4(180)+3(360) ' = i _1_= £1 ____1 = -' Cbx¢anx Cbx9xbx 2.143x9x0.863th3 2207ftk1ps From the Zx table, (13me = 1080 ft-kips < 2207 fi-kips 45me = 1080 fi-kips = ___8 = _8 = —4 bx 9(¢me) 90080) 8. 23X 10 pp“ + beu, + QM", = (0.619 x 10‘3)(826) + (s. 23 x 10-4)(540) + o = 0.956 <1.o (OK) UseaW14 x 159 Cb 6.7-7 (a) LRFD Solution Pu = 1.2(0.25 x 80) + 1.6(0.75 x 80) = 120.0 kips [6-48] The factored moments at the top are Mm = 1.2(0.25 x 133) + 1.6(O.75 x133) = 199.5 fi—kips Mmy = 1.2(0.25 x 43) + l.6(0.75 x 43) = 64.5 ft—kips The factored moments at the bottom are Mm = l.2(0.25 x 27) + 1.6(0.75 x 27) = 40.5 ft—kips Mmy =1.2(0.25 x 9) + 1.6(0.75 x 9) = 13. 5 fi-kips The amplification factor B 1 can be estimated as 1.0 for purposes of making a trial selection. For the two axes, M,“ = leMm = 1.0(199.5) = 199.5 ft-kips MW = Brme = 1.0(64.5) = 64.5 ft-kips Check the W103. Try a W10 x77. From Table 6-1, withKL = [.17 = 16 fl, p = 1.46 x10‘3, bx = 2.62 x10‘3, by = 5.16 ><10‘3 pPu = (1.46 x 10‘3)(120) = 0.1752 < 0.2 Equation 6.9 controls. As a preliminary check, 0. 5pP, + %(b,.M,,. + byMU) = 0.5pP. + %(b,Mw. + byM...) = O.5(1.46 x10‘3)(120)+ %((2.62 x 10-3)(199.5) + (5. 16 x10'3)(64.5)) = 1.05 > 1.0 (but close; retain this as a possibility.) Check the W125. Try 3. W12 x 72. From Table 6-1, p = 1.41 >< 10'3, bx = 2.37 x10‘3, by = 4.82 x10'3 pPu = (1.41 x 10'3)(120) = 0.1692 < 0.2 Equation 6.9 controls. As a preliminary check, 0.51519. + %(b,Mw+byMuy) = 0.5(1.41x10‘3)(120)+ %((2.37 x 10-3)(199.5) + (4. 82 x10‘3)(64.5)) = 0.966 < 1.0 (OK) Check the W145. Try a W14 x 68. From Table 6-1, p = 1.74 X 10'3, l7Jr = 2.38 X 10—3, by = 6.42 x10"3 pPu = (1.74 x 10‘3)(120) = 0.2088 > 0.2 Equation 68 controls. As a preliminary check, pPu + bew, + byMuy [6-49] = (1.74 x10‘3)(120)+(2.38 x10‘3)(199.5)+(6.42 x 10-3)(64.5) = 1. 093 = 1.10 > 1.0 (N.G.) TryanZ x 72, p = 1.41 x 10-3, bx = 2.37 x10‘3, by = 4.82 x 10-3 Calculate B1 for each axis: Cmx = 0.6—0.4(&) = 0.6—0.4(—&) = 0.6812 M2 199.5 — _ ML) = _ _& = Cm, 4 0.6 0.4(M2 0.6 0.4( 64.5) 0.6837 71'2EI, n2 29 000) 597 . x = = ’ = 4635 k1 s “1 (K,L)1 (16x 12)2 p » B], = —§mx—_ = M = 0.6993<1.0 Bl, =1.0asassumed 1_ P. 1_ 120 Pm 4635 ”211:1 712(29 000) 195) . P, = y = —J—’ = 15141q 1’ (K,L)2 (16x 12)2 '35 31y = J11— : M = 0.7426<1.0 1_ Pu 1__fl pm 1514 31y = 1.0 as assumed Since the amplification factors are as assumed, and this shape is adequate for C1, = 1 0 ' a computation of the actual value of Cb and an adjustment of bx are not necessary. The preliminary evaluation is sufficient. \‘ \., mw‘ UseaW12x72 \ \ ...
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