Chapter 9 - CHAPTER 9 — COMPOSXTE CONSTRUC'HON 9.1-«1(a...

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Unformatted text preview: CHAPTER 9 — COMPOSXTE CONSTRUC'HON 9.1-«1 (a) E. zwg-S f.’ m (145)1-5J4 4 3492 ksi, n: E: — 295000 EC —-3492 2-8.3 uses %2%=10.13m. ForanS >440, AS 211.8111}, dz 17.9 111,1); a 612 ill.“ Component A y Ay f 61‘ 1+ A612 Slab 40.52 2 81.04 54.03 2.469 301.0 w 11.8 12.95 152.8 612 8.481 1461 Sum 52.32 233.8 '77 1762 — _ EA)” _. 233.8 _, - y — WZA — #5232 .- 4. 46916., (13) Top of 81661: f * _Ml : (290x12)(4.469m4) = s m 1.4 1762 Bottom of steel: f3 : My 2 (290 x 12 4 + I7'9 "4469) = 34. 4 1:51 (tension) In, 2 1762 in.4 0.926 ksi (compression) Irr 1762 Top ofslab: m i119" : (290x12 4.469) 2 . . f; HI” 80762) 1.10 ks1(compresszon) 9.1-3 (:0 E. =wé-5 f; 2(14518-57‘ =34928si, n: g: = 232.230 98.3 uses %m%§m9.75in. ForaW24 X 55, A; 516.211122, (2' = 23.6111. L = 1350111.“ Component A y Ay f a? 7+ Adz Slab 43.33 2.25 93.73 74.04 3.789 704.0 W 16.2 16.3 264.1 1350 10.26 3055 Sum 60.08 362.8 3759 ZAy m 256.2132 = 603916., 3”» = if ” 60.08 16 z 3759 111.4 [9-11 (b) Top ofsteel: f _ M m (450x12)(6.039m4.5) m t” I” ‘ 3759 ‘ Bottom of steel: f; : :ng = (450 x12)(4.5+23.6m6.039“= 31‘7ksi (tension) H” mm 2. 21 ksi (compression) 3759 Top ofsiab: a 542 = W : . . f5 “Itr 3(3759) WEE!) 9.1—5 Determine location of plastic neutral axis: ASP}, w 16.7(50) a 835.0 kips, 0.85fc'bt = 0.85(4)(75)(5) 2 1275 kips Since 835 kips < 1275 kips, PNA is in the slab and C2 835 kips. From C m T, 0.8512321!) = 55F}, 0.85(4)a(75) = 835, Solution is: {a = 3. 275} in. _i _g_= 21.1 _ 3.275 2 - —2+t 2 -————~2 +5 ————2 13.9Im. Mn 2 Ty 2 83503.91) 2 11,6101n.»kips=968 ft—kips Mn = 968 ft-kips 9.2—1 Loads applied before the concrete cures: slab weight: (%)(150) = 50 psf, 5005) = 3751515: 3423 a 375 + 22 = 397 ib/ft, w; a 200.5) a 150 lb/ft Loads applied after the concrete cures: wD 2 375 + 5 I 3801b/ft, WL 2 (60 +10)(7.5) 2 525 Ila/ft Strength of the composite section: AsFy : 6.149(50) m 324. 5 kips, 0.85]; c = 0.85(4)(4 x 90) = 1224 kips Use C m 324. 5 kips. e Hagan = wsgmm fl . a {185be 0.85(4)(9O) 1'0601n, [9“2] y=%+rm%xfifi+4wia92§fl=10.3zin. Mn = Cy m 324500.32) : 33491n—kip5 m 279.1 ftwkips (a) LRFD solution Before the concrete cures: From the Z table, gben = 955M? = 125 finkips wu =1.2wD +1.6er, =1.2(397)~+«1.6(150)= 716.41b/ft M = gequ = %(0.7164)(30)2 = 80.6 ftukips< 125 ft—kips I (0K) After the concrete cures: (fian = 0.90(279.1) m 251 ft-kips wu = 1.2m; +1.6w; = 1.2(380)+1.6(525) m 12961b/fc Ma : %(1.296)(30)2 a 146 ft-kips < 251 ft-kips (OK) Flexural strength is adequate. (b) ASD solution Before the concrete cures, Mn 2 3.12. 95 95 wa == wD+wL = 397+ 150 = 547 lb/fi From the Z; table, a: 82. 8 ft—kips M, = —§—waL2 m §(0.547)(30)2 m 61.5 ft—kips < 82.8 ft-kips (OK) After the concrete cures, Mn _ 279.1 m _ - Qb — 1'67 167ftkips w = W); +wL = 330 + 525 m 905 we Mg m %(0.905)(30)2 a 102 ft—kips <16? ft-kips (OK) Fiexurai strength is adequate. 9.3-1 Loads appiied before the concrete cures: slab weight w (% ) (150) x 50 psf, 50(7) 2 3501bfft 14213 = 350+ 16 = 366 lb/ft, wL x 20(7) = I401b/ft [9-3] Loads applied after the concrete cures: wD 2 35011301, wL : (I25 +15)(7) : 9801b/ft Strength of the composite section: Effective flange Width = {25 x 12):“4 : 75 in. or 7(12) 2: 84- in, use b 2 75 in. ASP}, w 4.7160) 2 235.5 kips, 0.851%C =~—- O.85(4)(4 x 75) = 1020 kios Use C = 235.5 kips. C 235.5 ~ = mm z W m o. 2 . a 085be 0.85(4)(75) 9 35 m _mci _£m 12.0 _ 0.9235 3 - y—2+t 2 —2 +4 2 9.5381n. Mn = Cy = 235.509.5315) 2 2246 in.«kips = 187.2 ftwkips (a) LRFD solutiort Before the concrete cures: From the 2,; tabIe= tban = WM}, = 75.4 ft-kips w“ -~= lawn—t 1.61421; =1.2(366)+1.6(140)= 663. 2 1b/ft M“ = é-WMLZ a —§~(0.6632)(25)2 a 51.5 fiz—kips < 75.4 fimkips (OK) After the concrete cures: £5an : 0.900872) = 168 ft—kips wt, $1.2wD+1.6wL =1.2(350)+1.6(980)x19881b/fi Mu a %(1.988)(25)2 = 155 ft—kips < 168 ft-kips (OK) Shear: gbv V" m 79.1 kips V1, 3 W513 2 198g 25 =24.9kips<79.1kips (OK) Beam is satisfactory (1)) ASD solution Before the concrete cures: From the 2;. tab1e. m Mp __ . . -— h—b— — 50.1 11-1331“; 142., 2 my +wL m 366 +140 = 506 lb/ft M“ = «mg—way = ~§~(0.506)(25)2 = 39.5 fiwkips < 50.1ftwkips (OK) After the concrete cures, [9-4] MML _ 187.2 m _‘ Qb m 1.67 112ftk1ps we 2 1420 +14»); 2: 350 + 980 = 1330 Eb/ft M, = —§~(1.33o)(25)2 m 104 fiwlcips <112 ft-kips (OK) Shear: V” = 52.8kips Qt Va : W313 2 = 16.6kips< 52.8 kips (OK) Beam iswgmatisfactory " 9.4-1 Loads applied before the concrete cures: slab weight 2 (163)050) = 75 psf, 75(9) = 675 we m; = 675 + 57 2 7321b/ft, WI, 2 20(9) = 1801b/ft Loads applied after the concrete cures: m; = 7321b/ft, w; :2 250(9) = 22501b/ft Strength of the composite section: Effective flange width = (40 x 123/4 2 120 in. or 9(12) m 108 in, use b a 108 in. ASFy m 16. 7(50) 2 835.0 kips, 0.85chC : 0.85(4)(6 x 108) m 2203 kips Use C =~ 835111135. 2 C : AL” = ' _“g,lm _g=21.1 _2.274= - y—2+t 2 2 +6 2 15.41111. Mn = Cy 3 83505.41) = 12,870 in.-kips 2 1073 ftukips (a) Before the concrete cures: From the 2;, table, ()5an = (fibMp 2 484 ft—kips wu : 1.2wD+ 1.614); 21.2(732)+1.6(180)=11661b/ft Mu : é'wuLz = —§~(1.166)(40)2 2 233 fiwkips <484 ft-kips (OK) After the concrete cures: (fian = 090(1073) = 966 ftwkips wu =1.2wD +1.6w; $1.2(732)+1.6(2250)= 44781b/ft [95>] M“ = %(4.478)(40)2 m 896 661665 < 966 ft—kips (OK) Shear: qfiv V}, m 256 kips V“ z W“: 5: 4-478 4‘91 = 896 kips < 256 kips (OK) 2 2 Beam is satisfactory (1)) Before the concrete cures: “My” _ % W _ - From the 2;; table, gb — Qb —— 322 ft klps w, = wD+wL m 732 +180 = 9121b/ft Ma 2 é—waLl : %(0.912)(40)2 = 182 ftmkips <322 ft-kips (OK) After the concrete cures, Ml m 1073 = _ - 96 m 1.67 643ftk1ps w, = m; + m = 732+ 2250 m 29821b/ft Ma 2 ga. 982)(4o)2 m 596 6-16135 < 643 fi-kips (OK) Shear: V” = 171 kips 96 ya = W315 = = 59.6 kips <171kips (OK)- Beam is satisfactory 2 (c) A“ = flggfl— = 0.4418 16.2, EC = wgs f; : (145)1-5flf = 3492 ksi Qn = 0514561046136 5 RgRPASCFu == 0.5(0.4418)J4(3492 = 26. 11 kips RngAscFu m 1.0(0.6)(0.4418)(65) = 17.23 kips < 26.1} kips use Q, = 17.23 kips 2 V’ = 835 : N1 Q” 1723 48.5, round. up to 49. total number = 2(49) = 98 USe 98 studs 9.4-3 AsFy = 11.8(50) = 590.0 kips, 0.85fc/ic = 0.85(4)(4 x 81) m 1102 kips Use C =~ V 1-: 590 kips. [9-61 2 A“ z ’“3/4 = o_4413in_2, EC ___ wg-S fc’ = (145)194’1 = 34.92 ksi 4 911 z S RngAscFu m 0.5(0.4418)1/4(3492) = 26. 11 kips RngAscFu = 1.0(0.6)(0.4418)(65) 2 17.23 kips < 26.11 kips use Q” m 17.23 kips - 15:. = M” 5 N1 —~ Q" £7.23 34.2, round up to 35. total number 3 2(35) = 70 Use ’70 studs 9.6-1 (3) Before concrete cures: Slab: wig-(150x15) = 375.0 ib/ft 993 = 375 +22 = 3971003, WW = 2007.5) = 150.0 13/0, Is 9 199 01.4 4 4 AD“ 5WDL __ 5 0.397/12i30X12) =1.254in. ‘ 384515 “ 384(29,000)(199) ___ 5me4 ____ 5(0.150/12)(30><12)4 m . Acamt* AzAD+Acansr= A: in. (b) After concrete has cured: Transformed moment of inertia: 3:c = was fi' 2 (145)1-5JZ = 3492 ksi, n m = 3%? 2 3.3 13568 b _ 90 _. - ‘ “if” “'"Sww- ForaWMx 22, As = 6.49 ml, d = 13.7 £11., Ix = 199 in.4 Slab 4.500E+01 2.0003900 9.000E+01 6.000E+01 1.11513+00 10159E+02 -49OE+-+01 739+“ 1E+02 7-35 8733*” 5-; = 4%? a = 3115111., I“ z 703.201.‘3 Use 135 = 0.751;, 2 0.750012) = 527.4i1’1.4 [9—7] wL = 60(7.5) = 450.01bfft WW : 100.5) = 75.01001 W...” a 5025) = 37.51010 5091. ‘1‘ Wparr '1“ Wain-£4 384E131? $0.450 + 0. 0330 0. 03131412130 x 12)4 384(29,000)(527.4) 1.254+0.6703 = 1. 924» in. A = 1.92111. A r” AD +Apm~f +Acgj] 3 AD ‘1' m 1254+ 9.66 (a) From Problem 9.3-1, 0 W12 x 16 is used, with r = 4111., s = 7 ft, L a 25 ft, qcm, "2 20 psi Q'part = 15 psf, q; = 125. psf, A992 steel and 4 ksi concrete. Before concrete cures: Slab: {534150)(7) n 3501100 wD = 350 + 16 z 366 113/112, wean“ x 20(7) = 140 lb/ft, Is =2 103 111.4 w 51001.4 a 5g0.366/12)(25x12)4 m . A” 384E]. 384(29,000){103) “mm 4 _ swcomlL4 .. 5(0-140/12 (25x12 =0.41]9in. Am“ _ 384E115 w 384(29,000)(103) AwAD+Amm=1.077+0.4119=1.489111. 152149111. After concrete has cured: Wparr = 15(7) m 105 Ib/fi, w; = 125(7) x 875 ib/ft Transformed moment of inertia: 2 g 2 29,000 E. 3492 Effective flange width = (25 x 12)/4 = 75 in. or 7(12) 2 84 in., use b m 75 in. EC = 025%]? = (Mala/Z m 3492 ksi, n = 8.3 use 8 ForanZ x16, A; = 4.71109, dr- 12.0 111,11 = 1031:].4 Slab ' .501 2.0003000 .50E+1 5.000£+01 8.30E017.990E+01 W 4.5400 1.00013+01 1.03013+02 7.100 .409+02 J.) m a? 33 £221 2 2.393111, jar = 420.8 01.4 Use leg = 0.751”, = 0.7501238) w 315. 6 inf” _ swam m 5(0.105/12)(25x1224 _ . AW” “ 334230? ‘ 384(29,000)(315.6) ‘O'mogm' 4 4 AL _ SWLL a W 2 0.3403111. ‘ 384E131? 384(29,000)(315.6) A m AD Mpg... +AL =1.077 + 0.1008 + 0.8403 n 2. 01s in. A m 2.02 in. - - r M L = 25 x 12 = - - (b) Mammum pertmssxble AL — 360 W360 O. 833 3 110. < 0.8403 In. (N.G.) Use LRFD and seiect another shape. . " 502w z 5g0.875/12g(25x12)4 m . 4 Requlre‘ueff “ 384mm,. - 384(29,000)(0.8333) 31mm“ Required 1” = Jag/0.75 m 318.2/0.75 m 424.3104 Try 21 W12 x 19. Loads applied before the concrete cures: slab weight: i (150) z 50 psi 50(7) = 350 113/0 12 my = 350 + 19 z 3691b/fi, Wcons! = 20(7) = 140 {in/ft Loads applied after the concrete cures: m; = 3691b/ft, w = (125 +15)(7) = 9801b/ft Check strength before the concrete cures: From the 2;. table, 42an = quMP = 92.6 ft—kips wfl =1.2wD + 1.614»; £1.2(369)+1.6(140)= 666. 8 1113/13 Mu = é—wuLZ = l§(0.6668)(25)2 m 52.1fi-kips < 92.6 ftwkips (OK) After the concrete cures: Compute the strength of the composite section. AsFy = 5.5760) 2 278.5 kips, 0. SSfiAc = 0.85(4)(4 x 75) = 1020 kips Use C = 278.5 kips. = C 2 273.5 :10- a 0.85111: 0.85(4)(75) '92“ M4 “91:12.2 _1.092 = ‘ y'“ 2 2 2 2 9. M,1 fl Cy = 278.5(9.554) = 2661 in.-kips = 221.8 ft—kips (12an z 0.90(221. 8) = 200 ft—kips [9—91 w“ a 1.2wD+1.6wL 21.2{369)+1.6(980)= 2011mm M. : J§(2.011)(25)2 m 157 fi-kips < 200 ft—kips (OK) Shear: qvan = 85.7 kips Wui = 301125) =25.1kips<85.7kip3 (OK) 2 2 Compute deflections. Transformed moment of inertia: Slab mm 5.000+01 W 5.570E+00 1.010E+01 5.626E+01 4.0694‘02‘ a“ 214.1) m. 131.3 _ - y~:j—Ww3.049m., 1,. = 498.2 in.4 > 424.3 111.4 required (OK) 1.5 = 0.751,, m 0.75(498.2) m 373. 7 m.“ 5wa4 m 5(0.369/12)(25x12)4 2 38451.. 384(29,000)(130) Snagmfi = 5 0.105/12x25 x12“ 2 0.03516 in AW” 2 3343145 384(29,000)(373.7) 55ml.4 3 5(0.875/12)(25x12)4 : 384mg? 384(29,000)(373.7) A 2 AD-i—Apm +13; = 0.8603 +0.08516+0.7096 w L 66 in. AD = 0. 8603 in. AL = 0.7096 in. Use 21 W12 x19 9.6-5 (a) From Probiem 9.4—2, 21 W14 x 22 is used, with r = 4111., s 2 8 ft, L = 27 ft, gem; m 20 psf, gm” m 20 psf, 4L 2 120 psf, A992 steel and 4 ksi concrete. Before concrete cures: Slab: “ffasoxm a 400 Ib/ft W}; m 400 + 22 = 422 lb/ft, wow,“ = 20(8) : 1601b/ft, I; = 199 in.”1 Z ngL“ m 5g0.422/12)(27x12)4 : . A” 384% 384(29,000)(199) 087441” A = #Sanva“ 2 W “m” 384% 384(29,000)(199) : 0.3315in. [940] A 1“ AD+Aconsf : = in. A m in. After concrete has cured: Wm” = 20(8) :- 1601b/ft, w; = 120(8) 2 9601b/ft Transformed moment of inertia: _. 1.5 r , 1.5 H, - x £3“ _ 29,000 Ec—wc f. 4145) 71 449216;, in EC _m_3492 Effective flange Width 2 (27 x 12.)/4 = 81 in. or 8(12) w 963111., use b w 81 in. m 8.3 use 8 b_81__ - fi—TW10.1311’1. ForaW14 x 22, AS a 6.49 111.2, d = 13.7 in.,1,, m 199 in.4 Sla .05+01 2.000E+08.100E+01 5.400E+02E+00 1.145E+02 w 6.00 10353101 199051432 7.6280 5.766E+02 -“ZAyd151.4_ - - °4 ymW—W—3.2221nq Irrm691-1in' Use Iefi" m 0.7512, = 0.713(691. I) z 518. 3 111.4 A m Swami.4 = 5(0.160/12)(27><12)4 Pa" 384mg 384(29,000)(518.3) = SwLL4 _ 5(0.960f122(27><12)4 = . AL 384E131?” 384(29,000){518.3) 0'7637m' A = AD+Apm+AL = 0.8744+0.1273+0.7637 = 1. 765 in. A = 1.77 in. x 0.127 3 in. (13) Maximum permissible AL 2 3—‘é0— = Liz—012w = 0.9000 in. > 0.7637 in. (OK) No redesign necessary 9.7-1 (a) Transformed moment of inertia: __ 1.5 r __ 1.5 ... - L" Q, m 29,000 _. EC —wc fc “(145) J? ~3492 1(51, kn EC — 3492 m 8.31 use8 Effective flange width = (30 x 12)/4 = 90 in. or 6(12) == 72 in. (controls). mg— : % = 9 in. For area and moment of inertia computations for the siah, use a depth of 4.5 — 2 a 2. 5 in. ForaW18 x 35, AS = 10.3 in}, d 217.7in.,Ix = 510 1111.4 [9-111 mWflmWWW Slab 2.250E+01 1.2SOE+00 2.813E+01 1.172E+01 3.7993800 3.366E+02 2’” — M = 5.04916. 1.. = 15561114 «‘7’: 2A “ 32.80 Use L; m 0.751} x 0.75(1556) 1“ 1167111.4 =gflgm : 5(1.0/12)g30><m1gm)i= . 2 . AL 38456. 384(29,000)(1167) “385‘” A__..._L 053.9813; (b) ASF). m 10.3(50) a 515.0 kips, 0.85fc’bt : 0.85(4)(72)(4.5 ~2) 2 612.0 kips Use C: 515 kips. 0. SSfC’ab = AsFy 0. 85(4)a(72) m 515, Solution is: {a = 2. 104} in. mi __c_z_=17.7” W2.104: v y~2+t 2 2 44.5 2 12.30111. Mn = Ty = 51502.30) 2 6335 in.-kips = 528 ftwkips Mn = 528 ft—kips Shear connectors: Maximum diameter x 2. 55» : 2.5(0. 420) = 1. 05 in. > 3/4 in. (OK) Maximum diameter = 3/4 in. with formed steel deck. (OK) 2 “(344) : 0.4418162, EC :wg-S f; = (145)1~5J_ : 3492831 Q1 2 0.514541%; :3 RngAxFy m 0.5(0.4418)W9“72”5‘ : 26.11 kips RngASCFu a 1.0(0.6)(0.4418)(65) = 17.23 kips <26.11 kips use Q, 2 17.23 kips N1=totain0.0fstuds+2 = (30 x 12/9 x 6 m 15 239,, = Q” xN1:17.23(15)—- 258. Skips ASF}. z 769(50) = 384. 5 kips 0.85fi’b1‘ = 0. 85(4)(66)(4-.5 “1.5) a 673. 2 kips Am a [942] Since 2: Qn is the smallest of the three possibiiities, C = 258. 5 kips, there is partial composite action, and the PNA is in the steel section. If the PNA were at the bottom of the compression flange, ny = bfthy 2 5.03(0.420)(50) = 105. 6 kips The net force to be transferred between steel and concrete would be Tm C; m T—— ny : (A3133, «w ny) — ny : ASF}, —« 2ny T: 384.5 m— 2(lOS.6) = 173. 3 kips < 258.5 kips PNA is within the steel flange. V r» T— Cs =2 (A311?) — bft’Fy) ~- bfr’Fy = ASP), ~— beflFy or 258.5 = 384.5 -— 2[5.03t’(50)], Solution is: {t’ 2 0.2505} G. = bfr’Fy : 5.03(0.2505)(50) 2 63.0 kips Compute j), the distance from the top of the steel to the centroid of the area below the PNA. EA); m. 5339 8. 288 in. l i! Location of concrete compressive force: C 2 258.5 = 035be 0. 85(4)(66) 1‘ 152 Moment arm for concrete compressive force is a: jawwg— = 8.288+4.5—WL1§5—2— =12.211n. Moment arm for compressive force in the steel is 5,... 12; m 3233- 0-22505 = 8. 163 in. Mn : EMT = C(12.21) + 048.163) = 258. 5(12. 21) + 63.0(8.163) m 3671 in.»kips : 305.9 ft-kips Mn = 306 fi—kips NWWNMWWm~Mw [9-13] 9.8% From the solution to problem 9.76, for 3Awin. studs andfc' 2 4 ksi, Q” = 17.23 kips : 30 x12 : N; 12(2) 15.0 239,. : NlQn m15(17.23): 253. 5 kips Aug, n 7.69(50) = 384. s kips 0.85;?!” n 0. 85(4)(66)(4.5 “1.5) a 673. 2 kips The smallest of these three controls; C x 258. 5 kips (this is denoted as 2Q" in the tables) W C = 258.5 0.85127; 0.85(4)(66) _ “.61: _.1_._.1.§._2.= - 12—: 2 4.5 2 3.924111. =-« 1.152in. Since we are seeking a nominal strength, we can use either the LRFD value or the A313 value from the table. We will use the LRFD value. Interpolate for 96M”. First, interpolate vertically (create an intermediate row), then horizontally. —50E0 —3924E+{}D 31601302 i 571610 .00+02 « ctMn _ 276.0 ... - Mn w W — 0.90 — 306. 7ftwl<1ps m. = 307 ft-kips The following dimensions and properties from Part 1 of the Manual will be needed: AS = 10.4 i119, design wall thickness 2 0.349 in., and [y ~—" 80.4 in.4 EC m w]-5 ff, 2 (145)1'5fl = 3492 ksi The area of concrete As can be estimated as bd — AS, or more accurately as foliows: [9~ l 4} 2r The corner radius is given in Part 1 of the Manual as 21‘. The area of an quarter-circle spandrel with a radius r is r2 —— 211—an = (2:)? — 7%(202 "11:2(4 -— fl”) = (0.349)2(4 — 71') a 0.1046 in? and since there are four of these segments to be deducted, Ac = bdes ~— 4(0.1046) = 7(9) m 10.4 — 4(0.1046) = 52.18in,2 For computing the moment of inertia of the concrete, the moment of inertia of the spandrel about an axis parallel to the 9-inch side through the point of tangency is e~ = e—siw = (ii—w = —n')(0.349)4 = 3. 252 x10“2in,4 The distance to the centroid of the spandrel frOm this axis is -2 2r = 2(2x0.349! = . x ————-3(4WTC) 3(4_fl) 0.54211n. From the parailel axis theorem, the moment of inertia of the spandrei about a centroidai axis paraliei to the 9~inch side is f: I—«AJT:2 m 0.03252—0.1046(05421)2 :1.781x10“3 Use the paraliebaxis theorem and the following table to obtain the moment of inertia of the concrete: {9-15] Segment A [bar d Ibar +Ad2 Outorrectangle 63003-01 2.57SE+02 0.000E+00 2.573E+02 Steelshape -1.040E+01 —8.040E+01 DODGE-+00 -8.040E+01 spandrei 4.046501 —1.781E—03 3.344E+00 —1.171E+00 spandrel “1.046301 -1.781E—03 3.344E+00 ~1.171E+00 spandret "1.046501 -1.781E—03 3.344E+00 “1.171E+00 spandrel 4.046301 «1.781E-03 3.344E+00 -1.171E4~00 Sum 1.722E+02 16 m 172.2 in.4 Note that a slightly unoonservative approximate solution can be: obtained by using . 3 1.: a 8321—1. 2 31329—404 2 I77in.4 From AISC Equation 12—13, Pa n AsFy +A5rFyr + (72.4ch m 10.4(46) + 0 + 0.85(52.18)(4) = 655. 8 kips From AISC Equation 12-15, z... A: ) C3 0.6+2[AC+AS 50.9 W 10.4 m . m _ 0.6+2(-————-——-—-52'18+10.4) 0.9324> 0.9 .. useC3 0.9 From AISC Equation 12-14, (EDW : 181.15 + E515.» + CgEcIC : 29,000(80.4) + 0 + 0.9(3492)(172.2) = 2. 873 x 106 kipdn.2 From AISC Equation 12-5, 2 2 6 Pg 3 75 gEflgg m It (2.873 x10 3 = 5k. (KL)2 (0.65 x20 x .12)2 “6 “35 Then 0.441% = 0.446558) 3 288. 6kips Pa = 1165 kips > 288.6 kips use AISC Equation 12-2: (£32.) (655.8 ) anPO 0.658 P2 2655.8 0.658 1165 :518.1kips Pn = 518 k__iP_§ From the tables, with KL = 0.65 x 20 2 13.0 f2, 95‘; n = 388 kips. P" m qban/qb = 388/075 2 SI? kips ...
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This note was uploaded on 10/06/2010 for the course CE 406 taught by Professor Drake during the Spring '10 term at Cal Poly Pomona.

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Chapter 9 - CHAPTER 9 — COMPOSXTE CONSTRUC'HON 9.1-«1(a...

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