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HW 7 - CHAPTER 6 BEAM-COLUMNS 6.2-1(a LRFD solution From...

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Unformatted text preview: CHAPTER 6 - BEAM-COLUMNS 6.2-1 (a) LRFD solution: From the column load tables, the compressive design strength of 21 W12 x 106 with F = 50 ksiandeL = 1.0x 14 = 14feet is 45an = 1130 kips ‘ From the design charts in Part 3 ofthe Manual, for L1, = 14 ft andCb = 1.0, 45an = 597 ft-kips (Since the bending moment is uniform, C5 = 1.0.) The factored axial compressive load is Pu =1.2PD+1.6PL =1.2(0.25 x 250) + 1.6(0.75 x 250) = 375.0 kips The factored bending moment is M" =1.2MD +1.6ML =1.2(0.25 x 240) + 1.6(0.75 x 240) = 360.0 ft-kips Determine which interaction equation controls: Pu _ 375 __ - ' ' - (1’an — —1130 —— 0.3319 > 0.2 .. use Equation 6.3 (AISC Equation H1 121) Pu 8 Mu: My ) _ 375 8 360 m n +9(¢M’ + «MSW 1130+9(597+0) —0.868<1.0 (OK) , , , , W, W, , Ibis—rumba szmsfiesjheAlSQSnecificatim_\ 0.3-5 KXL = 0.9(14) = 12. 6 fi, KyL = 1.0(14) = 14 fit. (a) LRFD solution: From the column load tables, for KL = 14 ft, (11an = 1130 kips From the design charts in Part 3 of the Manual, for L1, = 14 Pt and C b = 1.0, 41an = 642 ft—kips and ¢bMp = 646 fi-kips. For C5 = 1.6, .1:an = 1.6(642) = 1027 fi-kips > ¢bMp use ¢an = ¢bMp = 646 ft-kips P. =1.2PD +1.6PL =1.2(0.33 x 342) + 1.6(0.67 x 342) = 502.1kips Mm =1.2MD +1.6ML =1.2(0.33 x 246)+1.6(0.67 x 246) = 361.1ft-kips For the axis of bending, C m = 1.0 and ‘ 2 2 _ anI _ 1: E1, _ #129,000)11110) =1.390x104kips _ (K1L)2 (K.,L)2 (12.5x12)2 el _ Cm = Cm = 1.0 = 31‘ l—(aP,/P¢1) 1—(1.00P,,/P21) 1—(502.1/13900) 1'037 Mu = 31M,” +B2Mq, = 1.037(361. 1) +0 = 374.5 fi-kips Determine which interaction equation controls: P“ = flz—J— = 0.4443 > 0.2 use Equation 6.3 (AISC Equation Hl—la) (15.10,. 1130 A. i AIL A) = _3_ 374.5 _ <1:an + 9 (mm... + ¢any o.4443+ 9 646 +0) —0.960 <1.o (OK) This memberjatisfies the AISC Specification 6.5-5 (a) LRFD solution: Pu = 1.2(16) + 1.6(16) = 44. 8 kips, For the axis of bending, _ _ fl _ _ _& = C,,._0.6 0.4(M2)-—0.6 0.4( 84) 1.0 P1= ”251 _ n2E1,'_ ”2:29,000)g171) = 3399 k' (K102 (1902 (10 ><12)2 lps _ C». = Cm = 1.0 = Bl_1—(aP,/P,1) 1—(1.00P,./P,,1) 1—(44.8/3399) 1'013 Mu = 31M", = 1.013(84) = 85.09 fi-kips Mm = 1.2(30)+1.6(30) = 84.0 ft-kips Compute the moment strength. For this loading, CI, = 1.0. From the beam design charts with L1, =10 fland C1. = 1.0, ¢bM,. = 134 fl-ki ps. From the column load tables with KL = 10 ft, 45an = 330 kips. Pu _ 44.8 _ . - _ . ‘1,an — 330 — 0.1358<0.2 .. useEquation 6.4 (AISC Eq.H1 lb). P. M: h) = fl M 291;an + (45me + ¢any 2(330) + 134 +0) = 0.703 <1.0 (OK) £52 (a) LRFD Solution: The service load moments at each end are MD = 0.33(150) = 49.50 fi-kips and ML = 0.67(150) = 100.5 ft—kips The factored-load moment at each end is 1.2MD +1.6ML =1.2(49.50)+1.6(100.5)= 220. 2 fi-kips For the axis of bending, 2 2 _ n2E_I_ = 7r_E1x _ ” (29,000XS33 = 7359 kips _ (KlL)2 (KXL)2 (1.0x15x12)2 Cm = 0.6—0.4(%) = 06—04%) = 0.2 Bl=._.CL"__=_Cm_.___=__ 0_-2 1 —(aP,/P.1,) 1 -(1.0P../P,1) 1 — (Pu/7359) Assume B1 = 1.0 and check it later. ' Mm, = 31M,“ = 1.0(220.2) = 220. 2 fi—kips Compute the moment strength. From the beam design charts with L1, = 15 ft, 4’1an = 528 fi-kips for C1, = 1.0, and WM}, = 551 ft-kips From Figure 5.15e in the textbook, C5 = 2.27. For Cb = 2.27, 45an = 2.27(528) = 1199 fi—kips 21 [6-17] Since 1199 fi-kips > (fibMp, use (13an = MM}, = 551 ft—kips Determine the axial compressive design strength. From the column load tables with __ ’ KL = 15 ft, H «15an = 990 kips f]: > 0.2 and use Equation 5.3 (AISC Eq.H1-1a): C n Pu i( Max M197 ) < 0 ¢an + 9 (15me + 45me _ I. L i __220-2 = ' '- = Let 990+9 551 +0) 1.0,Solutlonls.{P., 638.3} Assume that Check assumptions. PU __ 683-3 > 0.2 4:an ‘ 990 (OK) ' 31 C. 0.2 =141.019.0431) = W = 0.221 <1.o 31-15631 = 1.0 (as assumed; OK) Let1.2D+1.6L=P,, 1.2(0.33P)+1.6(0.67P) = 638.3, Solution is: {P = 434. 8} P = 435 kips a'\ ...
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