{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

old Exam 2 solution

old Exam 2 solution - AMERICAN UNIVERSITY OF SHARJAH Civil...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: AMERICAN UNIVERSITY OF SHARJAH Civil Engineering Department Prof. S.W. Tabsh CVE 313-01: Reinforced Concrete Design Fall 2009 Solution of Exam # 2 (Time = 90 minutes) Problem 1: The 6 m-long, simply-supported beam shown is subjected to uniformly distributed factored load wu. It is adequately reinforced for flexure. Determine the maximum w based on shear if the No. 12 stirrups are spaced at 150 mm within 1 m from the supports and 250 mm in the remaining middle region. Given. f”c —— 25 MPa (normal1wt. ) and fyt= 250 MPa. Ignore the beam’s weight. (a) ’cDol 0wJ07 Jim suppaflr V“ :00“ (3’0 S17—): 2 “rfiSw SleL 4. \ic: Orv—MW; LWJ 1 oflvdxfzsxfiwfst: 35,9, ktx/ IOOO V5:- bfi: ZK\\%KZSDK53’2_ _L : 200 Le [UV 5 |5o IO‘ N) CPVDZ CFQQJM/s) : 03:1551} +200,L\> : 252.‘ k “V (- ’- uc: Z____.—-52‘\ 1’02 M him Vw=¢Vn > 0" 2 442 00) ED 1— ‘M away gm 3%)?le \/ :Wu(3—\): 2w“ ) Vegaao? W \15— W: |20.2 16rd 250 660 kid/M Eflmfi cwn— ”V a; WW_.3£{L5£?_+_QLZ,_ 46.0 kfiovm: Problem 2: The roof structure shown below is composed of 150 mm-thick, 1-way slab supported on beams and girders. It is subjected to 12 kPa factored load (that includes the self weight). The slab is reinforced for flexure with one layer of No. 12 bars at its mid-depth, as shown below. Determine the re uired s acin of the No. 12 bars for flexure. Given: f’c = 25 MPa (normal weight) and fy = 420 MPa. Use the ACI coefficient method to determine the critical moment and shear in the slab. Do not check the slab thickness for deflection control and shear, and do not design the temperature and shrinkage reinforcement in the other direction. Centerline dimensions are shown Beam & girder width = 300 mm 3-D View of Roof Flexural Reinforcement in Slab Solution: MIMEMM in 1..me SLQI'O KW (QM AC1 (Giff. Nimcb MW: (MAL: : l2 k‘J/wi (3-5-03) ;— i345 kM‘Mfléi’ii‘O ‘i 0| dz—iyvvwi' (alum) { A: .- Mq y M“ l3.é§xl0 NAMM it Mygi—g) 2W7 (0.01301) ofingofiasns). L“? A: —; 50? Wi‘ML/W—S’i'l‘lf) ‘50?- MMLHM’OMM w‘ 2 M: 22$NM }6P .1 507 sq71220wvs M3427; H3 wwzfieysz? C\\LCiL Spacing ; 220 mm <ét7 £65va ; LiYonvx K lfb .. GM“: [DS‘iT : O‘OO\8< (a: i: M:O\Oog42 \ot looox ISO (1: M3 i :Q‘gx‘o‘m/Zzo #20 03599) 0.89m ZEX [Ooo ¢Mn=¢Aail7 (d ~ %>:0‘[email protected]/ZZO>XL¥QOGS ' L975! “of 1?).ékN'5nK : iO-i MM) $20,? Problem 3: |.35_0. 200 350 The T-beam shown below has a 200mmx400mm hollow ”4% _. $222925: :-::.-.=:-.=2:::::: part within the flange. It is reinforced with 5N0. 32 bars and subjected to positive bending. Determine the flexural cagacig of the section (thn) if PC = 20 MPa (normal wt.), fy = 460 MPa. 800 All dimensions Solution: r are in mm W Suiw Lathe vz—Arawn as: $4 C : a/E“:\Z658/O‘%5 '2. [Mi WW" (5?: ciao"; (A 43/6 ’- 0‘00$(806~l¢‘1r2>flqfi,z R >0ms>fiufi 9';th : <\>[(A3—A71e)fif @. am Hi, Q- :9] i z 93 [Que—731) 4 so (goo - L311» ”mi Ago (gab—Lgflzlzso 10” -0013) kV-M Problem 4: The beam cross-section shown below is subjected to a | 400 mm factored moment Mu = 200 kN-m. Note that the cross- section cannot carry the applied moment without d’=57mm com ression s eel. Design the doubly-reinforced sec 'on = figmusting No.14 bars in compression and Not.l32 d 334mm bars in tension. Given: R; = 20 MPa (normal wt.), fy = 250MPa,d=334mmandd’ =57mm. AS____olution: - _0 385+: LE1 Q gal-5‘1): 0 QSKQOX 40010 H8510 SHX'SSLI: 2836'“: As 97 250 :JL- 252% 2:0 2 ms W 03510? b 0359 Zoiqoo 2 Sam.» { .. D . mm. ., A» Q, g 27;) 28%,:250 (334-;— ’° i2- M A Mu Mn _ 200 _203;M.2 kN—M Y\7_ ~ — l “ -—'—" cp o.°\ Assent Comp. Stir-2A \ll\a\c14:é L AIS” \V‘WL 1 W‘- 2?8 wwm 197(0) J) 250 (334, 53‘) MAS-:ASHA/S 1 23361—278”; 3'?4MM1 L; 3/ 253.144....— 0.6214 > ,_:Ec/' .25 ":20 s; 60o 10'0”)? (0 (25 i] J>EA 666060437) A 6 [>50 $31Hj(€00—ZS> / a: COM? shd jvdéé (70/: ’ #4 No. 14 ms: 2%,?— 7%“ 4% if N). 31km: & 2 r513”; 13% \J% LtN°43Z 4* A’ 9, 8m 80” MM ' m 0\: A3 Y— s s : (Hxfioqazusg 250 2 10 Cl‘fitlzé 0 851000 0 $5x’20 x460 7 0.8: i=0 W: Motown/00735 :=.4) 0‘10 me WA <\>Mn =0 9%va ~368)(25o)(334— Lg)+3ogx250(334-51})]l—;—‘: 20; ...
View Full Document

{[ snackBarMessage ]}