Workshop_5_SOLUTION

Workshop_5_SOLUTION - Name:___$QL__QIiQM ______ _- CE 441...

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Unformatted text preview: Name:___$QL__QIiQM ______ _- CE 441 Reinforced Concrete Design — Workshop 5 Thursday November 52009 Due at the end of class For the following 15 by 20 inch tied column find: 1) maximum concentric factored load Pu that can be applied. What is the maximum spacing of #3 ties? 2) If the load may have an eccentricity e parallel to the 20-inch edge from the centroid, compute the balanced eccentricity eb, the corresponding axial compression Pnb, and the bending moment Mnb. The material properties are fc'=4 ksi,fy=fy,=60 ksi. The column has four # 11 bars that are arranged as shown. +3‘#——9"—»+ 3'14 D R :— ObeF: (Ag’AS‘Q + As; 3" T Z"Fl/03¢) @ riDo 7055514)st ZO~4054>>>+465454>0 20""? Wiflqiéfw 377.Lr;|373.2_a J ‘R 403% ;. menisci \1 3., @ w d) 20‘ Ply—Fl ¢Pn:O.GS(\OCl‘3-9 5M: 1For coma/Choc, \Ode H2 39%} \‘g “Alma tog Ac: 740.9 AccOrchwgt c: 7.:o,5.\ om‘xmvmuvm dict bars are required analogQ :tell. AC: 7:05.“; Tfe chgc/‘mg jug, $ {cos-l— CLiVWQJ/‘SlOl/l .’ L 'T‘Vz 5F8CXW§S g 4% :24.” 0% colon/vi m < \e» aw" \ g \5” wwmlg $01“ waximuw ._-———_../,: _He % 30‘?’ Even lg wQ were, abound ’X; afigfit “Hag “WWEMWUW Alan/maiom 0% cabin/m, 15”) EL, 9 $0.003 fl .3513: £2. =_ '7 A 9 I cl 5 0.003 o‘oosggo-ao a 2.1m! d‘l—i’ Cb :- [OaOG' [7ft .<_— GO ab: oasscb = 8,552." 6 = ——>‘ a 27000 As‘F‘a' ’Frvm forces &‘a%ram) 2¥X:O “m, z 0.35sggbawA/Sggflzsjcg),Asg aha: :10 A; $45 ‘3 0% es “@003 fig =0.00ZIOS 6, DR (0,064)" (0.09 > S > gyKSg‘eHg “mar—03%) 666.662} zQuEQCéO-‘gg@>—2<l5®éo Mnbzfib 85 //—::—/—/// 2MT=0 awasegz o.<55(4):g@.552_ ;7_§§,Z§z>+2459@_$5@m7_® 425,521, 25549.4, +24723 r. $021.? 30114 62,—. 175.35\ in 8,0: eg—(ég);1<s,<5st — Lg. ; Hrgst in Mnlo =4255®$5tj 3 5043 k—in :420‘24, Ht ...
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Workshop_5_SOLUTION - Name:___$QL__QIiQM ______ _- CE 441...

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