Test 3 - Print name 60L U T1 0 M l have abided by the...

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Unformatted text preview: Print name 60L U T1 0 M l have abided by the Virginia Tech Honor Code. Sign name CEE 3424 — Reinforced Concrete Structures I Fall Semester 2005 — November 27, 2007 Exam #3 1. Name the three factors with the greatest influence on crack widths [3]. Swain in fine rebar Coggp over We. (doctr- 59cmth 0 ‘9 Felon“? 2. Give two reasons why the 4) factor for shear (0.75) is smaller than the 4) factor for tension controlled flexure (0.90) [2]. ' Less ("taunt a‘e Calculation EFKPHQ {OCAUVQ 3. Name 4 of the 6 requirements that a one-way slab system must meet in order to use the ACI approximate method to determine slab moments [2] +wo or- more, spans unit-ormw dust“ butefi LDMS New“: is pm‘ smmfig spams more or“ less ecbwti LL. 6351... braced (:r‘mwc 4. For the indicated T-beam, what is the effective flange width? [3] IOft—Oin ~~~~~~~~ \ determine . H '2 effective _ '2'” m flange F4— I81n spon length — 20ft width for ~ this beam ~~~~ 5. Is the following T-beam adequate to support a factored moment, Mu= 500 k-ft? [20] #— I8"——1 f’c = 3500 psi 26" fy = 60,000 psi “.990. No.8 bars 1 As’q 5.7911} 3 Zlhn I e0. No. 8 — As = 0.79 in2 4.5.. | in. clear between layers {5 a ‘m £\om<3e:) ° 45: 0L6 + 333600332002) 7.1lvbo _ 797“.- £1,893,645- ’ no 65; 0,76 Dawda amd Cahabver'. u?) x 79—. 4E4VH CF: bx(t8-103',%6'3I5 .- 142.3: CPMn ‘ 9 (men:- - . 7- r . LSOO‘V" 'A‘SF= bolts: - 2'38"“ ASN : -2~38 : 4‘7ginz £92 a: 4:134“) : 9.34m #833,540 qés M“ = 142.5(215-3)+ 4.7sxeo(2t.15‘ 2 Mn : 73‘90 ‘m'L ’ 9‘3 k'g c: ———-= 11.22pm 4,6 = 2m + a" +921 235;“ .007: : gt - (Tamas—H.227 .0037: 'h'ams PH 0 n 6. For the following slab layout [20}: . . . _ slob depth = 6.5" distributed live load _ 75 psf Fe : 4000 psi additional distributed dead Iood = l0 psf fy = 60,000 psi Fm? " Elm Wand—J L—l3‘-O"——| L—l4'—0"—~l L—t2'—O"——l Uzi/0* a) What is the maximum moment (absolute value) in the slab, and where does it occur? . D W§¢\(’ = (726 M50 : ‘4; u‘/ LU“: I.L,(-7s)+t.z (towns) = 224.5 5? m 7." + S ( zzctsY'B’Q‘) — 3.30%”, larges L) @E MK: 4 / 2H Z ’ C? C M“, f/O(ZZQ.6)(L€3) = 3,69 l: «Cl- b) At the location of maximum moment, what As is required for strength? cl = (as- ,‘76- I25 : 6.6" Mu ‘2. gr”: T0r3e+ Mn; {)5 : 3'8“ : 50.7 io'k ‘ 5 l A - As.“ «(3.9295 ' bot,92$-55 - O'Hob «Cf M h 5'0. .7 c) What size bars and bar spacing would you select for this location? (use #3 = 0.11in2, #4 = 0.20ln2, or #5 = 0.31 inz). ( Y b 5> #5 cm“: * ‘y : 7‘qsan ASMM:.5ol8lpln: ,00l8 tz , ' L‘V.’ .(r' 2 4401/3164» 7:5 ‘ . l \\ O ( do 89“.“ #4 [156 1 l2. " (A'S‘V‘ 1g3C? 7’7— ’ .. r’ \ n “ ’4 H o 2 K //, 5m: Sk=5o .r ta 4@ l4 (" > x/ d) What additional checks would you make? ~ctdvafl Mn \. gt J gt ) 9b '_ CfiMfiad— > M“ 7. For the given beam, if the service load moment is 90 k-ft , is the spacing of the flexural reinforcement adequate to control crack widths? (1 ea. #8’bar = 0.79 inz) (neutral axis depth, c = 3.71in.) [10] n—zw—fi T #4 stirrup l8" 2" clear cover over stirrup 3 ea #8 f'c= 5000 psi L fy = 60,000 psi note: inside bend diameter for #4 bar is 2" Cc : 2* pg : 20% in O~$$um¢ L, : 40 task :9 3: Eco _ : 8-7SU‘ “J 4 -AOL 4.0 > na+ 0k 0A 5 ' 5 5M: 26 - zaps“) : (as M » _ , : S" i"\ Chad; w/ MM; (5 a. I8-2 6 .s I “M flqo“; - Bai ~ 2 ‘ ' I ' o S. a : Ami— Z3 2.3106 - k 8. Given the following beam cross-section and loading diagram: [20] f'C = 6000 psi V A _ 5w24 fy = 60,000 psi m _ 384E] wD =I500 Ib/ft (including selfth /wL =I500 Ib/ft IIIIIIIIIIIIIIIIIIIIIIIIIlllllllllllllllllllllllIlllfiilllll IIIIIIIIIIIIIIIIIIllIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII 6 ed. #9 bars _ 2 25—0" As = 6.0 In a) What is the cracked transformed moment of inertia, la? 1 3 7&9) EC —— 574F760; = 4416 u: 1—“, 51mm”) J, (icy-rad 1 2 h = 222:0 : (057 + 5.9.97 [205 -7,88) 4 .. . , “x 1%; = aésfizos c3 I“ ; 33% m C : 7o Rh b) What is the cracking moment, Mcr? « 13 : ‘(ZUchqP : (8)432 in £(r : 7,5 Jéooox 1 {’5‘ “\WM/ . ‘2. . : : : 8%2“h-|"; 6" l1 _. 7 c) For the given loading conditions, what is le? , + . 7' I ‘ Ma= Us lag-5 : 234.4 MC" V * 5 m4 4 _ 74.3 _ 7:32. ) 890 = m ;” IC ' (254A) .8432 *0 (234.4) 8 . m d) What is the instantaneous mid-span deflection caused by dead plus live load? 5( aZlesqu 123 0 kg. ..\ : . W\ ” 2; 384'446 753‘ °ll‘l4‘m"t ,/ w: 9. For the following v.,/¢ diagram (for half’of a beam) and beam cross—section: [20] #— 36“ —~—~1 l00 centerline of span I V 80 l95" '2” I2" TU 60 60k ' kips L 40 20 f'c = 4000 psi fy : 60,000 psi 2 4 6 8 l0 I2 lec1.#3:0.l|in2 \ Distance from support, ft I 90. #4 z 0.20 in? a) What is the maximum Vul¢ for which the beam must be designed? Slope: gig-“O : 431$ “4+ p 4 , \%= fis-E‘altm = élik ‘ b) At the critical section, what is the required steel contribution to shear strength (VsVrequ)? V¢;Z \l4ooo ' l‘i'$'lZ 7 Z‘lo‘ck (22 V 0 Vv/gvb: 87.9—29.4, = 683k c) What is the reguired spacing of #3 bars to provide this strength? \I ,_ AV 4‘3 a 5 : M45 4 .. L fl/ O,\IK3X(DO‘lQ0$ _ w 5 = 6 8 .3 ’ é' E ‘7‘ d) What is the required, spacing of #4 bars to provide this-strength? 1 0,7.x3 : IZ’O“ 5 53.5 e) Whatisthe maXimum sPacin allowed at this location? t, W J g ' d’z “ Z4 L3 4 4000 ' H.341 = 64.2.!4 >585L -- C\/ 7s“ I‘-"’ ice/Vial} f) What bar size and spacing would you choose? 2 ; a. #5@ 0%" or “46 qfiuboverm 50.12! '93 : 0,0563: 4 a3? ...
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This test prep was uploaded on 04/01/2008 for the course CEE 3424 taught by Professor Clroberts-wollmann during the Fall '07 term at Virginia Tech.

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Test 3 - Print name 60L U T1 0 M l have abided by the...

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