307-2005 - CAE 307 Midterm Exam Spring 2005 H 10 ft 25 ft...

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Unformatted text preview: CAE 307 Midterm Exam Spring 2005 H 10 ft 25 ft The simply supported T-beam with overhangs shown above supports a uniform distributed service dead load min = 500 lbs/ft (does not include the self weight) and a uniform dis- tributed service live load w“, = 1.5 kip/ft. Assuming f; =4000 psi, fy = 60,000 psi, and normal weight concrete do the following: 20" i 4% l 18" l— éoa 3—#9 i: 14“ l+ cross section for positive moment 20” 2.5" 2.5" 18H F 14" H. cross section for negative moment a. Calculate the maximum positive and nggative servicefilgad bending moments the beam develops. b. Check to see if the beam is safe according to the Alternate Design Method ofthe ACI assuming the allowable stress in concrete is given by (fc).uow=0.45fé and the allowable stress in steel is given by (f,)a“ow‘=24,000 p51. c. Check to see if the deflections at the free ends of the overhangs satisfy the deflection criteria in Table 9.5b of the ACI code (treat the overhang as a cantilever beam). Assume the T-beam is part of a floor construction that will» support nonstructural elements that are likely to be damaged by large deflections. Also assume the live loads are sustained for at least 5 years and that the nonstructural elements are attached before the live loads occurahd before anxgrfiékdfiflfictions occur. “——“~._ Note: The deflection at the free end of the overhang A is given by w 0L3 (1sz A=23! 12 2 a3L +——+ 71:14 12 where a = 10 ft is the length of the overhang, L = 45 ft is the total length of the beam, in is the magnitude of the uniform distributed applied throughout the length of the beam, and EI‘ is the effective flexural stiffness of the beam. ' Problem 2[20pts] 17.5” 20" Assuming fé=5000 psi, fy = 40 ksi, and normal weight concrete calculate 9.5% the nominal moment resistance (Mn) of the cross section shown. Make ‘ sure to check the maximum and A minimum steel ratio requirements. fl-M— J l v i GAE 3(2) 7: PPOKIDIQM 0C) WMW :[ £:§.>(’jl> 4. erm. 3 $1995 11/0”- 7L I. S lax/+15 :- 8,2.‘38/1/1‘4 $09 1. =(JZ‘5-‘5I‘153 = 1'03 15,94 BARFS L St. MOKYltmur‘fl aojq+i\/L M°fl0u+ OCCu/Lj 61.7“ MIN = i (ask;ch (10 H): a Vl/ony'mmam ch',-)~;vL magma/F OCLU/és af£ mic/Spa,” + l 25H Mmm= *ns/z-H+ /;C:23.e/.;¢,J§( ‘2 )=‘63.%£.4+! 6) ’F’oa’i M55; Saw“ Q0 )posi-lfiu/IL Mom0A3+§ EC; S¥,ooo}&c' = 360$k5; Q CI 000 36.05 A5 ‘3 30.903: 3’95392 (30300CQ) 5 Y) A1 (31,5) [00 ‘3‘ (3183.015): Jag MA, .5 m um. 075 T—b Gk: QG L," . E; 3 I * (aoxoxvx-an (IV><’X~‘7’>Z/a :nA;(Is.s~x)' H" 159% :‘1:‘.€fi Y) A5 7x2~+He>< “#20 =—¢ X " 5.C15‘—I fix) (CA/so fiat) = ¢IZ%}52/_H_ " NW (m 8,53 4, _ SPR w 9m? f". NEW :ww %O'xvtbo +1 (xa-8x4le)=QL/(IS.S*X); E I Li .S ICC: I/3 (gamfi '— '/3(g>(7r-L/)3 = '851m” I+s -— n Aims—x32: QL/(IS.5-S,¢'-/)a‘ 9252mm Tea: Ewan" Crmlc‘Aiafle 4,624“;di 57’725565 [$4 + mama~+ 4” a $3 = r} meossrx) __ 8(93.8lz~u>Om~/m(/s.5-5.9w”) I‘m. W $5 : 18H» ’153 .4, 83" L9; OK 5;: Mil”. 7c : {68,g‘)(/2)(S.@V3 : JAM/9;? (£Ln...=(b,vs)v) —._.————-~ It»; EV??? :: 1,94,; OK Serf-1 ‘F'Ml 4- Mome+ CALck flaqm7LivL MomQM+t WW I l . As 1 3'¢ 7N 7' ;.$” "N AS, x Qpaé (H3(><)(*/2)+QnAS’Kx—a.s§ =nA5 (15.5- x) 1x1 + ()é3(3.o3(>r~a.s) = 8(V.€E)(r$.5vx3 7—x2' Ms.wa *7<2S¢n$2:<2’> =r> x: 5.41 1-,; ICC = 7.5 (HDCSAHS ‘1 8ng m” 13,5: nAg(.g,s-s.g;>z= 3M;sz 3:“: “195mm Ics= SMA; (x—g.5)z‘- NH "WV saw: Heer 5mm“ ‘Es ’ h Mr?” Crag-x): (g)(ns)(l;>)(;s.s- 5.6016934th Eta “195° (QVAJ; I OK r .— 335 1 2n Mm»: (“x—42,5) —. (10(1157027C5,61.,2,5)= J33; 4), (A J:M 9/950 SJVA/i OK .— ¥c = MMM (x) : (1:5)lu)(§,rs1>: [,56 1m 4 1,9 4;; or: I‘& f saw a 541% occQ/z.d/f~j 7L. ADM 07C AC3: C ) Ca J cw im+m X ‘ 5 edfir‘brmo» )2 {bf-(r 7qu furl 3f 04; axe?" __ 2kg; 2 (903(H)(i6>+ ammo» : 2A1 (9a3(v3+(I~/)(w) 51 57,61 7/": | f :1: - 1/;1(;203m)3+ (Evan—9,6131 ' 4 '/I:L(IL/3(w)3+ (lb/)(lv3C‘7-m—‘wz52 : 79710314” . £39 .. figm>miov : LH-mo = Emirfl MM = 3+ (13’191) I, 90/2. DEM) L091) ONLyg I 1 Ma : Lug—Db C": “)1 :C®.}.‘5453/fl3 Cwsszfifiz) r. 3915'“ L l 2 I Ca I Cu Int. TC”: 0’ (16 I: c‘lbu Ca /g M /Q+,‘o¢.§1 445-4 w l x, E; 5% {NBA} (WYXZ) + (n —J)A5' (242.3) = HA3 (ms—xi) ,5_5,X 7x1 +~ ¥(3.03Cx "53.5): 8(H.é%)(15.5~;f<) WhAs ¥>fl~ + 58.'-/L/>< *65233. = {6 "X:— (23.21)” th ’- 1/3Cl‘4)>(3 = [ubf 2 Z 1+5: 10A: C15$¢x5= 8(Ll'683(13»5~té.90 a3;.234. 7~7 l 1‘ Ic.$=(n—13Asl(><—Q.S)V_7 (7')(:§)(6.-1)~;.S)1; 399;," g | l ’- _.LU{= Mama” Cale“ (0,40. It 1&4! (3)2513 loaJ 0,9137 " I : (M >3; .. M a 31'1- '— 3, 3 '3 c #55 c +— 1 .9) am ?.3 ¥¢710+[ a 31.3 M“ J [ M“ “@0555 I Limos) L, 35 I¢= 725mm“ (ADI—51, 1’ C¢c1835§0 2).: (’°)(q_5)3~ (1032(95)?’ 5030/5) + 7—0037 ] 9(36053C72ar ) i; “W; M42 I; (ADJ; = 65.215183 m ‘PDK DEAD Lox—H: & Lave Loam Ma :(21355 HH§CI00Hf> 1m 194+ [19 ; L12 2 33m, 5‘ MW» 313 as: I: =-I(« (ADL.\TL_L3" : (I 223 302’] Z ¢, )9” 3373603) (L163?) Ca{wrafl¢ Joy? 1“an JQK/fic’L/‘MAT pl: git; (3.03 : ¢,qbr38 3:442 5 ow m p. 59L 0-0053) \7 mare >: i“ T. A“.-. '3 j+5¢/>' 1+5¢(<p,¢r353 ANCADHLJ; 9x: (fisswaAM: 95.ti (ALL-31 4' Am i 9/7190 (1,54%)(rsm3 : (filo—95M, C1135? - 95407—83 1L wan ‘3 ‘7’90 €0.23” NO 6001‘ AcchzJ.:/r +0 (9-SL) prololem fillqbfgfi) A? ‘CC/CS-léJ’. 95,5579] A5: 30am): 2.3mm Q We OK: As :~-‘(;.3¥)(‘7’°) : 1.59:”. 'gMHJE (¢.953(s)0~r) HAL) f‘ Mn = Asrrj (A 'q/2> =(9'3¥3(‘7’03[11§—/_§fl z I a ,l‘. Mn: /5?L/ k-;~=113Qflv}+l M- E+ Lat. FS/nnxi (AWE 1 9/ Jon Clo‘ = ,S (6.80: . {g C //”" 8+ : dwc. C «c 8+ : qs’wzozmv/mj: gum; ><15.cz:v5‘-i 3+ ['99 OK M m~ /) 3J4; LWA : 3J50~3Q (“130372571 ),3;,." Agni»: Z? me '“VOQQO 3°°'°~*:"—1 : $2ng a .33 1-2323‘ "iwoaa ...
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This note was uploaded on 04/14/2008 for the course CAE 307 taught by Professor Desantiago during the Fall '04 term at Illinois Tech.

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307-2005 - CAE 307 Midterm Exam Spring 2005 H 10 ft 25 ft...

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