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HW%209%20Solution - CE440l Fall 2009 Homework#9 Due...

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Unformatted text preview: CE440l Fall 2009 Homework #9 Due WEDENSDAY Nov. 18 For all problems (for this homework and future homeworks) assume a #3 tie and max agg size of 3/4” unless otherwise stated. You may make any other assumptions necessafl, but be sure to state them 1n your solutions 1. Make a list of all the references you have seen in class or doing homework to the ACI code this list should include the classes through Monday Nov. 16 and this homework assignment. Include the code section, code page number, the topic, and a brief summary of what the code says in that section if possible ( 1. e. if you can summarize it in a few words otherwise leave this column blank). Do not copy word for word out of the code. Please hand 1n your ENTIRE list (i e back from HW8) and indicate what 15 new on the list. 2. The beam shown carries its own dead load plus an additional uniform service dead load of 0.5 k/ft and a uniform service live load of l Sk/ft The dead load acts on the entire beam, but the live load can act on parts of the span Three ossible loading cases are shown. Service 11119 toad = 1.5 has“: stadiums: mice deed ma 2: 11.511111. [at 24 ft, 911 Liveéoad -fl_--I-- . x’fieadm" lb} (6) a. Draw factored bending—moment diagrams for the three loading cases shown and superimpose them to draw a bending ~moment envelope b. Design the beam, selecting b, h, and the reinforcing bars. Use fc’ =3750psi and fy= ~60 ,000psi and interior exposure. Assume a No.3 tie will be used c. Draw an elevation of the beam showing the reinforcement. Estimate the lengths of the top and bottom bars from the bending moment envelope d. Draw cross sections at the points of maximum positive and negative moment. 3. A footbridge is to be built, consisting of a one-way solid slab spanning 25 ft between masonry abutments, as shown. A service live load of 120 psf must be carried. In addition, a 1000 lb concentrated load, assumed to be uniformly distributed across the bridge width (NOT LENGTH), may act at any location on the span. A 3 inch asphalt wearing surface will be used, weighing 20 psf. Prepare a design for the slab, using material strengths fy = 60 ksi and fc’ = 4.0 ksi, and summarize your results in the form of a sketch showing all concrete dimensions and reinforcement. 3n 4‘“’“—6 4. A rectangular concrete beam measures 16 in. wide and has an effective depth 27.3 in. Compression steel consisting of three No. 11 bars is located 2.5 in. from the compression face of the beam. The tie is a NO. 4 bar. The tensile reinforcement consists of 4 No. ll bars. lffy = 60 ksi and fC’ = 4 ksi, what is the design moment capacity of the beam, according to the ACI code. Repeat the problem if the compressive reinforcement is composed of only 2 No. 11 bars. i (:5 ; Eiléaé g 02> bfethf ¥€E€IEQ"€§« QEMCEEMEE Moflwsfl‘lg (Efmngm {:9r 3f” ‘E‘E/‘Lr€flw E9555: §§$a$ 31%; “E émggfiEfifi'gfijgfi aniwg ngfafifi? «gaiyagigrgé asp-5‘25 géffi c \E w, \ M: , - A i 9-; a? 5 v ‘ ‘ WV? 3’ ' ”“3 ‘7“ ”#5:? Qafléf’i, fimfi 53; W43, ”'3: ”f‘r’ii (A Skew Egan,“ (“a ”igrgamgpji’ Elm eEfiVIQ‘E‘WV» JEflW K 6%”E’Cmm4g kmffliifi an? 'E°f 5 EzmiEam 5,5333% We? 2? C5443? \i’ m _ , E . 3 . , ,, a h - , N , ‘ 59) ‘u’ffi‘w? 5:, I" 5% {eagfz 9 w £35,”? 5px; 2%“? a a“: mwxfiggwfi 3,;W; $3“? 23%: 23’ 2:: Owfii: xi: figaEW 9, V91 1: mi. m3?” = i ”w SOLM‘T’E'OM i i E 5 i ER ‘ ‘Efl, i \ i7 7/ \) {A 1);? 569$ « I; a; .h E E9“) baagaefi EQAW mi‘fla \C : g7ggfé* L‘wa’“ 3;» g 735% (gégfiéd E i E i E % $5. ga ”1‘5 3‘3 fi (5?. § § @ 52a a: E) \Z/ fig E x, QEE mm E“? km iGng‘naEWm‘u 6\ VF; “E”: “9 ¥’% W359 E06 Eh 343;); $ New Airs wiEa Eg‘f MA E31" “Er ) 333% 9323533333535 Wamfi 3 We gag, 4: ”W“? a} 53qu r» m 3 M wwlgwgngkwafi W %’M. V 4:2 3 My?” 3, 3 x if,“ “A“ 3 W3 W3 ““ ‘W’i323%gg‘a WW ($333325 5,133.3. » . . . f E AC yvxougt MM,” Pygg-iqug V‘nf/WV‘A’ GCQUMF‘V 3% €36,733?» fizfi 8 Z 3 ‘3 “2» wuk/m zwuk/m Dead (k/ft) = 30.5 SeW(k/fi)=§03 Uve(k/fi)=§115 Envelope (kip—ft) Maximum Minimum 0 89 E 101 § 13 101 112 E 128 i 16 128 131 § 152 § 17 152 147 i 172 i 18 172 160 2 189 E 17 189 170 g 202 g 16 202 176 5 212 g 14 212 179 E 219 E 11 219 178 E 222 3 6 222 174 5 222 g f 1 222 E 167 i 219 é ‘ -5 219 i —5 156 i 213 E -12 213 § —12 M2 § KB 5 20 KB E 20 124 g 189 3 -29 189 § -29 104 E 172 E -39 172 § —39 5 5 152 5 -50 129 E -62 102 g -75 kDOO\lO‘1U'I-i>UJNl-I|O NNNHHI—‘l—‘Hi—‘HHHH NHOKDOONOWVU'l-DUJNHO "NU‘I\I SUSHNLD HHI—A UJ\l ONU‘I 0°“wa [LAIIII OOONONU'I “poo-1N0 UJ\J DON .111 E8 oooooooooooo N U) I LD N H l 1.: H L0 = = 1 24 -136 § -39 § —136 0 25 -108 § -31 g -108 0 26 -82 g -24 g -82 0 27 —6O 2 -17 E —60 0 5 28 —42 E -12 E -42 0 E —42 29 -27 i -8 3 —27 0 i 30 -15 § -4 § -15 0 31 -7 g —2 g -7 0 32 -2 g 0 3 -2 0 33 0 § 0 i 0 0 2,7/ mao_w>cm_ ES. wao_m>:m_ xm_>_ Ill 38 two; Pitt 7 $8 .83 vacumm $8 «53 $5 .I I a: x £325 :9. .3838 .5”. mao_m>:m tam EEmmE .EwEoS. A. com... omH, 03- ,, om ,, OOH ,, omH CON omN (u-dpl) wawow ‘ SUUAHES ‘ ~ ; \ k . V , 1 a‘f’we Mfivwsz‘ flatware; «as? flak “fiwFfo/‘é’ w: [051d :61“; 5 (Mad ’3 f? wag? igéii—figg "g3 § 3Q 5a m g “E £3365 % fig “Ff“flvw LXRS<2 wa‘i’g’a 3/1 1-; 5432,“ ”’vwa’w > 8m “/4, :D panacea? / ¥im§ Jzohg $894“ 3% A! , 1; \A: 3+ CW“ ’* é ‘ Jr ébfiz ‘2 2h“ I.C‘m+ ”g; at i, w .35: xx x 1A a w x 1. [we a y m 5%. H W. 1%. aw, Juli. hwy £52. all fi/M/ «kw! 1W M A??? lbw a W [a . s! «2 2 r x , w , 9 er kW 2 : 3 2 f 5 PT iv 353:; $3, 1 o E : S : t V W \W, q 2 g M: E? H, .,. ; 1%; fa \M/wi/ fiwx L MW 59 \ . v! a ? u; 3 m % My % a, / ‘ a 2,. if .w my 4/ m iffy,” w. «Mu * \xii Al? 5’ Gr! la M: g Eb! 77 .2 7.“. a I V!!! nah x w, x/liL m5, he by , a 6 Z 1%; «a \J r: /!ix Andi/n} Q“ 75 flxmw X @531 / mum. x % g \r O \ w 7 a... \x a?» 9 T AV .8 a 5;? rue m !M x, , 7y . .i; a \\\./ Hf: ._ 03d» : Ma \Mil 2 Z ,M, .m 3, ,5, i jimmy MW m z” aim L T ”x ; a! i a, i - a ‘ u! 2 1, R w anW ”w w?! 21/ NW» X23“; 0.. 5%; E? E WM. KW“: _ . U! o ’ w m We is.“ SEEM» w my WNW any $233925! 3 m. FA. ,, ”a. i i NW A?“ m w, i “x A a, ,6 1 a); ‘2... a i“ p )0 ; WW “H, C / {\w 7.! )0 a1; 00 W a 2W9 in 2 ”w x}; a,“ V km; A iii ya fia/ a! f!..\\ P? w; x X u } Q E 1.! is. a mi 6 3%“ {We 7 my... . a \, ”M {iii Va u z) A V MM? 2 7.. .Bmw MW #1, V i ii w. gm 5! _ aw i g z, x», 3 xx ,V PW : i w a; m y 2 fl WW . 3 $ is a!” a t. J. 2!: leMV m , // y \W Z _ M :m Map! a Aw”. 7 5 w a a % W ,m , w a x mm w 4% m a“, g. z , o a), w 3, ix :v x L: ? xx!» ... /x n 1 W m, 5 MW in & aw. H 7. ?; nu Km 2. H. s éé fly {by mu 9\ iii; x Q,“ k _ {a Ne) ‘ . = m” 9» MW .11.! » \AU 2? mam“ < Av M a. 3 5 x K a» z w an A” S a w w P» WW Q : \xaial .\ 1 1 v 1/21. Nb AM em“. a) a. -1 a a :0 V a 0». gym/M M. /Nw : CH :1 16 Q Q “lot? a, , {@s W & 5/“ M L h” a . g a! ML A. MW 2mm Dead (k/ft) = §.05 Self (k/ft) = $0.35 Live (k/ft) = 31.5 0 0 1 0 2 3 a 0 3 90 103 15 103 0 4 114 130 18 130 0 5 134 154 20 154 0 6 150 174 20 174 0 7 163 191 20 191 0 8 173 205 19 205 0 9 179 215 17 215 0 10 182 222 14 222 0 11 181 226 9 226 0 12 177 226 4 226 a 0 13 170 222 —2 222 -2 14 159 215 -9 215 -9 15 144 205 —18 205 -18 16 127 1‘91 -27 191 -27 17 105 174 -37 174 -37 18 81 5 154 -49 154 -49 19 53 130 —61 130 -61 20 21 102 -75 102 -75 21 -13 72 -89 72 -89 22 -52 37 -105 37 -105 23 -93 0 —121 0 -121 24 -139 -41 -139 0 -139 25 -109 -33 -109 0 -109 26 -84 -25 —84 0 -84 27 —62 -18 -62 0 -62 28 —43 -13 -43 0 -43 29 -27 -8 -27 0 -27 30 -15 -5 -15 o -15 31 -7 -2 -7 0 -7 32 -2 -1 -2 0 -2 33 0 0 0 0 0 25/ on_m>cm_ :__>_ on_m>cm xm_>_ III $8 .83 PEP 33 can: 958m wmmu Umog “mt”— nlu I a: x 2325 :mm .mBum .8 mno_m>:m ucm :5.me EwEoS. (H'dpl) luaLuow 4 42.. “2%“ fim‘imgflflmfi 5 1 l ' I i I 5%” MWMM” “£3995 W”? #in i flfifj :9 $®§J€ {M @w w ‘ “r W 3' afigfifi lb ”2% fias‘igmiwgmm .. if flewm‘mf 19%??? m ii“? I k, / V ‘ (f: w“: {m Kg emit) j i f L iyfvgm ‘2 59"; m: i955” %\ : cfi/fias’g) W [(3/3] ~ 2,6? 2:. 273%” MA as yix 2”; L , 1 -— —-——-—————. CE 4401 Problem 3: 11/18/2009 HW #9 KEY 3.1 A footbridge is to be built, consisting of a one way solid slab spanning 25 feet between masonry abutments. A service live load of 120 psf must be carried. in addition, a 1000 lb load uniformly distributed across the width of the bridge may act at any location on the span. A 3 inch asphalt wearing surface will be used weighing 20 psf. Design the slab using fy = 60 ksi and f0 = 4 ksi. Summarize your results in the form of a sketch showing all concrete dimensions and reinforcement. General Live load Asphalt dead load Concrete unit weight Bridge length Bridge width Concentrated load Approximate height from Table 9.5(a) and round to even number Design one foot wide strip Self weight Clear cover (bridge = exterior exposure) Material strengths Steel strain yield Concrete factor Determine reguired moment Total dead load Total factored ditributed load obvious controlling load case Total factored point load obvious controlling load case Maximum moment with concentrated load at midspan Assume tension control Required moment capacity LL :: 120-psf DLa := 20-psf "lc := 150~pcf L :2 ZS-fi W := 6'ft = 1000-lbf 20.167-fl W ft L . __ . happrox :2 ‘26 = 15-111 h .— 15111 := 12111 kip DLS :2 “(C-'hb = O.‘187f— t cc = 2m fy 2: 6O-ks1 fc z: 4-ksi E := 29000ksi fy s = — = 0.002069 y E 61 := 0.85 ki DL :2 DLaxb + DLS = 0207-—B ft kip Wu :2 1.2-DL + 1.6~LL-b = 0.441”;- t Pu :2 1.6.P-b = 0.267-kjp wu~L2 Pu-L Mu := 8 + 4 = 36.1-kip-ft cl) := 0.9 Mu Mnreq := —¢— : 40.1-kip-ft CE 4401 11/18/2009 HW #9 KEY 3.2 Problem 3: (Continued) Determine steel re uired in both directions based on assum tions Assume effective depth dousss ;= h _ 3.111 = 12.111 M Determine Rn Rn := ——n—r€—q— = 0.279-ksi 2 b'dgucss f Determine m m2: y , = 17.6 0.85-fC 1 ZmRn Aprroximate rebar ratio p 2: —- 1 — 1 — = 0.004853 m f Y Shrinkage and in2 temperature steel Asmin :: 0'0018'h = 0324'? i112 Flexure steel area AS :2 p-dguess = 0.699? . 2 Controlling steel area in in span direction Asrfiq '= maX(AS’Asmin) = 0699.? Design steel in span direction Span direction steel db :2 0,875-in s :2 10-in Ab z: 0.60-in size, spacing, and area Steel area provided A ‘_ Ab _ 0 720 £2 in one foot of width bprov“ 'S— * ' ' ft Check spacing limits SpaceCheck :2 if(s > min(3-h,18-in),"N0tOK" ,"OK") SpaceCheck = "OK" Calculate actual effective depth d ;= h _ cc _ 0-5'db = 12.563-in Depth of compression Abprov'fy _ block assuming f = f a I: : 1059-10 s y 0.85-fc Ultimate concrete strain ecu :2 0.003 Depth to neutral axis c ;= i = 1.246-in 51 _ Ecuid — 0) Steel strain a := ——-—— = 0.027 C Check assumptions of steel AssumpCheck := if(€ < 5y v s < 0.005, "Not OK" ,"OK") yielding and tension controlled section AssumpCheck = "OK" CE 4401 11/18/2009 HW #9 KEY ‘ 3.3 Problem 3: (Continued) . . a . Nominal moment capacuty Mn ;: b'Abprov'fy‘(d _ E) : 43_3-k1p-ft . Mnreq Check moment capaCIty MomentCheCk :: if M > 1,"N0t OK" , "OK" 11 MomentCheCk = "OK" Design steel in width direction Width direCtion $1961 db2 := 0.625'in 52 := 10in Ab2 := 0.31-in size, spacing, and area 2 Steei area provided A ._ Ab2 _ 0 372 _i_n_2_ in one foot of width bprovz ‘“ S ‘ ' ‘ ft 2 Check spacing iimits SpaceCheckZ :2 if(s > min(5~h,18-in),"N0t OK" ,"OK") SpaceCheckZ 2 "OK" A . Check 8 & T steel area STCheck := if [—33:12- Apr‘OVZ STCheck = ”OK" > 1,"N0t OK" , "OK’J Sketch $5191 , is"? g; f? CE 4401 11/18/2009 HW #9 KEY 41 Problem 4: A rectangular concrete beam measures 16 in wide and has an effective depth of 27.3 in. Compression steel consisting of three No. 11 bars is located 2.5 in from the compression face of the beam. The tie is a No. 4 bar. Tensile reinforcement consists of 4 No. 11 bars. lf fy = 60 ksi and fc = 4 ksi, what is the design moment capacity of the beam according to ACl code. Repeat the problem if compression reinforcement consisted of only 2 No. 11 bars. General Effective depths and width d ;: 273.11} b := 16-in dpfirfie :2 2.5-in Material strengths fyzz 60-ksi fC :2 4-ksi E := 29000-ksi Tension steel area AS :2 4~1.56-in2 : 6.24~in2 fy Steel strain yield 5y 2: — = 0.002069 E Concrete factor 01 2: 0.85 Ultimate concrete strain ecu ;= 0.003 Compression reinforcement 3 No. 11 bars Initial assumption that tension steel yields compression steel does not . . . 2 . 2 Compressron reinforcement Asprime := 3-1.56~1n = 4.68-1n sprimc :: —a__ [El Strain in compression steel 5 . . 0.006375-in Esprime sunphfy ——> 0.003 — ——;—-—— Force in compressron steel fsprime ;= Aspfime.gs§§i}_fic.E Force in tensmn steel fs := Asefy = 374-kip Force in concrete c ;= 0,35.fc.(b.a _ Asprimc) Solve force equilibrium for a using TI 89 see next page for hand calcs Depth of compression block a ;= 3336.111 Strain in compression ecu- i — dpfime steel. We see the ._ B1 ~0001338 asumption of not yielding ESINl‘ime " a T ‘ lS correct (esprime < sy) 6—1 . . . -= . . . . z 9. - Force in compressron steel fSpl‘lme ' ASprIme ESpl‘lme E 18‘ hp AH mwi‘g’s :22 i,0%rw,é?%MQs/‘§”Eam$ bi: £229.23 ”i 2229;242:232 3%?222 “125355? $2? 21222 NE RES SQUAHES < c- . \ g 1 20%: :2 2% 23% g: : 3 (252:) (20.90; 22 2.222222222/2“) 222w ; 932/2— 32252229 ' o2 C: 92%”; (‘9) (3262‘23622‘525; 5%.‘292’ 552%51 2% % m ' .g , ‘2 5% d? g‘:( 5:35: imifigz‘wgwq j: 1 a: if (2.: :22 2:: <2 2 g . 2%; Ewazafi'mgv 322222222 25222222222322 522:2; 2%: W? 5032/: 22922123 £¢2§Lmiq4§¥or 3291222“ 2% £32,: §ag§fc21fly 2 i p > f (x A \3 £9! 2i {.fib‘flfi‘f’gyfiéfiw PQanforcngh§ kh’fl MN :2, : 5’22?!sz 122.2228? C2¥2 H 2:? :29? g%g22§§;krimm f 2 372:4: 2722M~572$§ 2 5222221 2» 222222222 ”b ,m 3. 22:92:» E, 22 2% 222.222 2.2223 2’2 w ” 2 a 2 FWEéJEUiz/ efififiéajgf $ 2’5“: : Ll'egégfivfifa CE 4401 11/18/2009 HW #9 KEY 4.2 Problem 4: (Continued) Depth to neutral axis (3 := 3— : 4.513~m 31 _ ‘ . ecu-(d — c) TenSIon steel straln a := —— = 0.0151 C Check assumptions of AssumpCheck := if(e < ey v e < 0.005,"Not OK" ,"OK") tension steel yielding and tension controlled section AssumpCheck : "OK" Sum moments about a a . centroid of concrete force Mn ': fS'(d _ —) _ fSpfimfi'(dpfimfi _ E) = 783 hp'fi Based on strain in ._ O 9 tesnsion steel > 0.005 d) ’_ ' Design moment capacity DesignMomentCapacity3N011 := ¢~Mn : 705 kip-ft DcsignMomentCapacity3N011 = 705 kip-fr émj Compression reinforcement 2 No. 11 bars Initial assumption that tension steel yields compression steel does not - - . 2 . 2 Compressmn reinforcement ASprime2 :2 2-1.56-1n = 3.12-1n a2 .Ecu' — " dprime . . . 51 Strain in compressron steel 5 ————————— sprime2 : a2 0.0063753 2 simplify —> 0.003 — —13 Esprime a 2 Force in compression steel fsprimeZ :: AsprimeZ'EsprichE Force in tension steel fS = 374 kip Force in concrete C2 2: 0.85-fC-(b-32 — Aspfimez) freshen; Solve force equilibrium for a using TI 89 see next page for hand ca/cs Depth of compression block a2 ;= 4453411 . . . a2 Strain in compression Ecu' — — dpfime steel. We seethe ‘_ B1 _ 0001572 asumption of not yielding E5101011162 ‘_ 32 ‘ ' is correct (Esprimea < 2y) [6—) 1 f Force in compression steel SprimCZ '= A sprimeZ' EsprimeZ' E = 142' kip CE 4401 11/18/2009 HW #9 KEY 4.3 Problem 4: (Continued) . a2 Depth to neutral ax:s (:2 := — = 5.251‘in 51 6 ~ (1 —- C Tension steel strain 52 ;: 3M 2 0.0126 ‘32 Check assumptions of AssumCheckZ := if (52 < 8y v 52 < 0.005, "Not OK" , ""OK) tension steel yielding and tension controlled section AssumCheCkZ = "OK" Sum moments about 32 a2 . centroid of concrete force Mn2 ': fs’[ _ g] _ fsprim62(dprime ‘ ‘2‘] = 779101"ft Based on strain in tesnsion steel > 0.005 (1):: 0'9 Design moment capacity DesignMomentCapacityZNol1 := (13-an = 701kip~ft lDesignMomentCapacityZNo11 = 701 kip-f9 H ...
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