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Unformatted text preview: Print name i have abided by the Virginia Tech Honor Code. Sign name CEE 3424 — Reinforced Concrete Structures I Fall Semester 2007 - September 25, 2007 Exam #1.. . Problems 1a)-h) (3 points each) Provide concise answers to the following. 1a) Which type of test for tensile strength will give a higher apparent strength: split cylinder or beam? F Faunth ' beams 8 h ‘1 {L Beam ’ Cylinder P +9 8 1b) Why is the mean strength of concrete received from a ready-mix plant greater than the design strength, f’c? 5m“.— Conccdcc, is \rxi h\ 7 var- \‘ adolL J how‘mf) "t‘ke. Mm ‘S'H‘ev‘cjt‘k we“ ovt MM. desian Jd‘ shims-Hm fenSur‘es “Wok \IU’7 gnaw \OOGVCMQS wk\\ “NW!” {\N’ . - . . . . _ \ess We 1c) Which of the followrng cross-sections, which have equal area, wrll shrink more? Why? 25 15 3 5 G 7 2 B l me A ~ \"r Mo‘s wft- SW‘C’OKQ 0~C£0\ Lot Hui soda—awe v 1d) Give three of the basic assumptions that are made in the design or analysis of a reinforced concrete beam in flexure. T; C - Mung SzcfiHons remain \OML ’S‘Fce‘ {JAB ' CDMr‘c’fg Ms M tensfle S‘brev‘osik 7 ‘- P1F£e¢+ \Oovxd bzkween CDAUL+Q. CK'A S+2¢( " mm '9me tn Comm—eke is 0.003 ' '\dao~\k2¢d 6-5 be“0~\"‘°“ hr Sfe‘d {—C Predictions 1e) What are the two limit states that must be considered in the design of a reinforced concrete structure? . 6w¢n05+k Sew \RCQOsb \ *1 1f) For ultimate strength design: 4) * Resistance 2 Load Effects * Factor of Safety Name three variabilities or uncertainties accounted for with the ¢ (strength reduction) faCtor' -—- MoijrioA \l0~r{o~\oi “4V1 ’ d‘x mast onad \iox‘xodo‘l \i 'H —' (dear paczwxzmc' vartmloi my —- unccfiaflwfl 1A Cod cu\a‘Nor\S -—- + pa 04 40‘} lurt 19) Describe a ba anced condition failure in flexure. Is this a desirable or undesirable type 0f mime? Why? do make ecu shes so 5 ‘l’ 0L5 ~s Te e.\ \I i‘e,\c$.S , UndefikfoAda . No Qora wocrwh/xg 0'? ¥a\\v({. 1h) Given the following load-deflection plot for a simply supported beam: steel yields ' Name (with either a point or a line segment) the following: - Ultimate Strength D - Cracked—elastic stage A '6 —Cracked-inelastic stage 6 ._ D 2. (30 pts) Given the following floor fr‘aming plan: ’ 15ll| SECTION A-A PLAN VIEW M = 50 psf duct allowance = 4 psf tile flooring = 12 psf ceiling = 8 psf concrete unit weight = 150 pot a) What is the uniformly distributed factored load on joist A-A? kl \\ | b .. \ ’- ,/ 56:; W? a? woo ., as x 6054:? v '75 5‘ up J3 ' (pa -, l2 r75 r 4 + 8 = 951 4F “w/ - . — , 8 s F '\ gulls, w“ 1: SObe 4* flo’z— ’ ‘78 “7/ “38.3 x ‘55 = “‘90 h \ It)“ on bnowv‘ "— w/ Q Stem fiaX‘C VA" : Z42U5x 57/44 +o+al ‘b f ‘ Wm: ZlAD 4% if} m span of the joist, and assuming that the beam behaves as simply—supported, what is the factored mid-span moment, M? R: we xlooolb: lLoollp g 7’ (.00 $0 M“: wag—9— i. ‘ 4" = 262,750 «Cl-~00 “<7 A x / ' 7"" x e. ,w , l 3. Given the following beam (50 pts): f'c = 5500 psi r—e l0"——l fy = 60,000 psi concrete unit weight = l50 pcf /WD =8OO lbm wL =1500 lb/ft lllllllllllIIIIIIIIIIIHIIIIIIIllIIIIMIIIIIIIIIIIIIIIIIIIII‘ lIIIllIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIlIIIIIIIIIIIIIIIIIl l 9‘—O" for 1 #9 bar As = 1.0 in2 (distributed dead load does not include self weight of beam) a) Using gross cross-sectional properties, will the section be cracked under service { loads? 1 ‘8'533 : 1“‘@ ,7, = .Qr : 7.6 5500 = 555 P" (“$12) 4': 55¢. lszvla ' W : 49,4 ,9, ——-°> : “7.2:” - 317/253 ‘L 2 7 , a b ' ‘7 56;, “A: = '° “3‘s it l$z> = [92.7 ‘4? (27’ (44» ' 09 g“ “"9 ~L,_, ws -= aoo+lsoo+l¢lzfl = Z‘HB @rL ” ) M _ 2,4039“ x l‘i: _ - M5 ' 8 ' “2'6 K, Pr > M“ (Would: uv-K’“ ’5 b) Using the appropriate section properties based on your answer to part a), what is the average stress in the reinforcing steel at service? 5; = 57 ’55“) : 422‘] k5? i *3 59‘“ for a «8 mecca d A f \ : ' ’1' b , < a. I IIII y; 4 7,23] fl ,0 .(j' ' \\__ ,7?» .. Needle-cl app/'6 / r / \____// /, C 2 in ..-_, 7/ l M __, : wig; : \LSl 4 = A (cl-9) “6’6" 5 s 5 3 c) Assuming that the steel yields, what is the nominal moment capacity of this beam, M"? A x 4,0 u Ox" '8$.55,'o 1 C/ Mn ‘ 4%005- = 26mm“ 1’ Zfifl E44: d) What is the strain in the bottom layer of reinforcing steel at ultimate (at)? 6600 » g. = LOS' 0.06 (—a-gfl = 0,775 am‘ “7/ c= a??? W022“? "‘wkk o “00% a O ‘5 " Us.“ ’7/ ft -, m[1%.I’E,LZ7 ' O, 1 (KL-9 M— A e) What is the stress in the bottom layer of reinforcing steel at ultimate? “7 :5 ., oo ks: ...
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