1 Having a minimum durometer hardness of 50 and utilizing ...



CIE 536Design of University Ave. Parking Garage ByKalpesh ParikhAcknowledgment Our deepest gratitude goes to Prof Riyad S. Aboutah for his continuous and constructive advice and follow up. His successive advisories and comments were the pillars in our every step during the Project. We are thankful to him for the fact that he has inspired and helped us know about the Prestressed Concrete Structural System.IntroductionLocationThe University Ave. Parking Garage building is located in Syracuse University Campus and it is surrounded by Harrison St., University ave , Walnut Place and East Adams St. It is a Ideal place because in its vicinity University offices and College is located. From planning point of view it is a best place for a Parking.AdvantagesIn recent decades, precast prestressed concrete (PC) structures have become more and more prevalent in the construction industries. A precast concrete structure is composed of Individual prefabricated members with different types of connections. Prestressed concrete structures have better corrosion resistance than reinforced concrete structures. Structural AspectsGenerally for the parking garagesection which are ideally used are Double Tee Section and Tee Girder below is the Snap shot for how it looks.Figure Showing 3-D View of Double tee and Inverted Tee SectionStructural System and MaterialsDouble Tee Section Figure Showing 15LDT26The double tee girder flange is designed to carry the load of the cast-in-place slab, the flange weight and a 50 lb./sq. ft. construction dead load. (or a 3000 lb. concentrated construction load per IBC). When determining the section properties of the double tee girder the center of gravity is based upon the total area (excluding corner fillets) of the entire exterior or interior girder and not on the individual stems. Moment of inertia and dead loads to each stem are figured to the center of the flange between the stems. The non-composite dead load of the external double tee girder is divided according to the following sketch. Figure showing different member stem and flangeInverted Tee GirderFigure Showing 28IT44The Section was decided on the basis of the total depth of the 26” (double tee ) and bearing pad which is should be l/180 or 3 inch whichever is maximum so in our case it is coming out to be 3 inch so on that basis section was examine for various check. The members are like flange and web. The main purpose of providing the inverted tee that Double tee section easily sits on to it. Loads coming into the tee girder as shown below. MaterialsLight Weight ConcreteLight weight concrete is preferred as a concrete materials for our design purpose.The majority of regular concrete produced is in the density range of 150 pounds per cubic foot (pcf). The last decade has seen great strides in the realm of dense concrete and fantastic compressive strengths (up to 20,000 psi) which mix designers have achieved. Yet regular concrete has some drawbacks. It is heavy, hard to work with, and after it sets, one cannot cut or nail into it without some difficulty or use of special tools. Some complaints about it include the perception that it is cold and damp. Still, it is a remarkable building material - fluid, strong, relatively cheap, and environmentally innocuous. And, it is available in almost every part of the world.Regular concrete with microscopic air bubbles added up to 7% is called air entrained concrete. It is generally used for increasing the workability of wet concrete and reducing the freeze-thaw damage by making it less permeable to water absorption. Conventional air entrainment admixtures, while providing relatively stable air in small quantities, have a limited range of application and aren't well suited for specialty lightweight mix designs.Lightweight concrete begins in the density range of less than 120 pcf. It has traditionally been made using such aggregates as expanded shale, clay, vermiculite, pumice, and scoria among others. Each have their peculiarities in handling, especially the volcanic aggregates which need careful moisture monitoring and are difficult to pump. Decreasing the weight and density produces significant changes which improves many properties of concrete, both in placement and application. Although this has been accomplished primarily through the use of lightweight aggregates, since 1928 various preformed foams have been added to mixes, further reducing weight. The very lightest mixes (from 20 to 60 pcf) are often made using only foam as the aggregate, and are referred to as cellular concrete. The entrapped air takes the form of small, macroscopic, spherically shaped bubbles uniformly dispersed in the concrete mix. Today foams are available which have a high degree of compatibility with many of the admixtures currently used in modern concrete mix designs. Gecko Stone of Hawaii is currently experimenting with one such foam. Foam used with either lightweight aggregates and/or admixtures such as fly ash, silica fume, synthetic fiber reinforcement, and high range water reducers (aka superplasticizers), has produced a new hybrid of concrete called lightweight composite concrete, or LWC.. for PrestressingHere we have chosen high strength steel that is s Strands (group of twisted wires). Low relaxation Strands are chosen because it has good stress- strain response of seven wire strands .Structural welded wire reinforcement (WWR) is selected for the top flange of the double tee .It has good ultimate strength (250 ksi.) Wires are used at the top of the flange for double tee section.Bearing Pads Bearing pad selected chloroprene (Neoprene) purpose is to absorb the energy produced by moving vehicle . Neoprene allows beam rotation at the bearing point because of deflection. Neoprene Bearing Pads, It is manufacture them in our purpose built plant using high-grade virgin Neoprene Polychloroprene Grade- 2 elastomer- mixed with carbon black and rubber chemicals to form into a rubber compound to match AASHTO M 251 Specification under controlled heat and pressure.Properties1 Having a minimum durometer hardness of 50 and utilizing 100% chloroprene as the elastomer. 2 Short and Long-Duration Compression Test up to 4200 kN with horizontal shear up to 400kN can be carried out.Bearing size is decided on the basis of the minimum in place distance from the face of the support to the end of the double tee in the direction of the span should be (length of Girder)/180 where l is the clear span , butit should be greater than 3” of the Structural System 1 Girder (Length = 30 ft) selected Inverted tee section 28IT44 2 Double Tee Section (Length =60 ft) selected Section 15LDT263 Bearing Pad (Neoprene) Selected =3 inch Durability and Green engineering aspect in the designIn the Design we can considered green engineering from the point of view of Materials used for designing that is Steel and Concrete. So main focus on Concrete In today world sustainability is in talk and lot of structures is now being used as green engineering .From the Point of view Green Engineering in the materials now Prestressed Structure are concentrated on the Concrete Mixed which are used.As we know, concrete which is used in order to achieve green engineering the cementitious materials such as use of Fly ash or Natural Pozzalano use of Ground Granulated Blast furnace and use of Silica Fume. Blast Furnace and Fly ash which are waste can be used as cementitious materials achieve the strength required. The mixing of cementious materials depends on the concrete strength required. Selection of use of cementitious materials in the concrete mix, use of fly ash or silica fumes would may affect the color of the finished concrete but from the stand point of the strength and durability and optimized use of waste can be made and some saving in the cost can be added. Apart from that Carbon dioxide can be reduce which is helpful from the enviormental point of view.Apart from that we are using Light weight concrete which gives more prestress losses but are beneficial as a concrete materials. The Precast also contains rebars , the rebars can be made from recycle steel which we can use as a Green materials.Design a Double-Tee Precast –PretensionedAnalysis and Design of Double Tee SectionA 60 feet Span Simply Supported Light Weight concrete double tee beam is subjected to un factored live load of 50 psf as it is pre-topped there is 2 inch topping is already considered into the flange. In order to design the double tee we have selected the section and procedure are carried and check are made for that it satisfies the condition.Double Tee SectionLoad Analysis:Live Load = 50 psfSpan Length = 60 ftLoad CombinationTotal Load = 1.2 Dead Load + 1.6 Live LoadAssuming a trial Section 15 LDT 26Material PropertiesConcrete: Light weight Concretef’ci = Compressive Strength of the concrete at the initial prestressf’ci = 4500 psif’c = Specified Strength of the concrete.f’c = 8000 psiw= Density of the concrete =115 lb/ft3Prestressing Steel?” φ seven wire strandsLow Relaxation Grade 270fpu = Specified Tensile strength of prestressing tendons = 270 ksiSection PropertiesSection 15 LDT26Ac = 7.486 ft2I = 53280 in4Yb = 19.82 inYt = 6.18 inSb = 2688 in3St = 8618 in3Self Wt = 861 plfV/S =2.38 inNumber of Strands Provided = 20All strands are straightFigure Showing Pretopped Double Tee Cross SectionDead Load = 0.861 k/ftLive Load = 50x 15/1000 =0.75 kip/ft Total Load = 1.2 Dead Load + 1.6 Live LoadTotal Load = 2.2332 kip/ftCalculating Stresses and Losses as Stage ProgressStages1. At Transfer2. After TransferType of PrestressingBonded Member Pretensioned.Losses StagesType of PrestressingBefore TransferAt TransferAfter TransferPre tensionedRelaxtion lossesElastic ShorteningRelaxation LossesCreep LossesShrinkage LossesPermissible Stresses for the Prestressed concrete flextural memberDescription Stresses At TransferStresses Under Service LoadExtreme Fibre in Compression-0.7 f’ci-0.45f’cExtreme Fibre in Tension6√f’c12√fcCalculation of Stresses At TransferFigure Showing Beam with its own self weightM@B =(25.83 x 2.5) –(0.861x 0.5 x 2.52) =61.88 kip/ftM@centre = wl2/8 = 387.45 kip/ftAt Transfer we know tension is at the top and compression is at the bottom. Since we do not know the eccentricity at this time but we know the Pre-stressed force and allowable stress at extreme fibers, we can easily calculate that.ftop= 6√f’c = 402.5 psifbottom= -0.7 f’ci = -3150 psiFi = Aps*fps = 619.65kftop= - FiA + Fi*e*YtI- M*YtIftop= - 619.651078+ 619.65*e8618- 61.88*128618 e = 14.75”fbottom= - FiA - Fi*e*YtI+ M*YtIe=12.4”Minimum is taken for calculation ahead for losses.Calculate Losses:Elastic Shortening: E.S = Kes*Es(fcirEci)Where: Kes=1 (Pre tensioned) Calculation of young modulus Ec and EciEc = (40000√f’c +106) x(wc145)1.5 =3.23 x 106 psiEci = (40000√f’ci +106) x(wc145)1.5 =2.601 x 106 psifcir= Kcir(FiA+Fi*e2I)+Mg*eIfcir= 0.9(619.651078+619.65*12.4253280)+61.88*12*12.453280 = 1.96 ksiThe elastic Shortening losses will be: E.S = 1*28*103*1.962.61*10^3 = 21.6 ksiStress Level at After transferSteel initial Stresses = 202.5 ksiElastic Shortening and Relaxation = 21.6 ksiNet Stress after transfer = 202.5 – 21.6 = 180.9 ksifps = 180.9 ksiFi = fps * Aps = 555.19 kipsSince we revised the Fi value by accounted stress losses so we need to calculate new “e” value with this force by keeping the same max allowable stress at extreme ends.So the new “e” is calculated below:Now calculating Stresses at top and bottomftop= - FiA + Fi*e*YtI- M*YtI0.4025= - 555.191078+ 555.19*e8618- 61.88*128618 e = 15.75”fbottom= - FiA - Fi*e*YtI+ M*YtIe=14.13”Minimum is taken for calculation ahead for losses.I have assumed “e= 14.13” but the value of Fi reduced to Fi=555.19 k so, the max stresses at extreme fiber ends are under allowable limit state as per PCI code so STRESSES are………O.KStage 2 : Under Service Load (Due to Double Tee beam)Fi =555.19 kipse = 14.13”Calculate losses:1. Creep loss C.R=KcrEsEc(fcir – fcds) Where: fcir = Kcir(FiA+Fi*e2I)-Mg*eI fcir =(555.191078+555.19*14.13253280)-61.88*14.13*1253280 = 2.398 ksi fcds= 0 The creep loss will be C.R=1.6*280003230(2.398) = 33.26 ksi2. Shrinkage loss S.H = 8.2x106Ksh*Es (1-0.06VS)(100-R.H) Where: Ksh = 1 V/S=2.38 R.H = 75 The shrinkage loss will be = 5.9 Ksi3. Steel relaxationR.E = [Kre-J(S.H+C.R+E.S)]cWhere: Kre=5000psiJ= 0.04………….Table4.5.1 (PCI)C= 1…………….Table4.5.2 (PCI)The relaxation loss will be = [5-0.04(5.9+33.26+21.6)]*1 = 4.756 ksifT = Total loss = C.R+R.E+S.H = 33.26+4.756+5.9= 43.916 ksiStress left in strands after losses = fpi – fT =180.9-43.916= 137.47 ksiRevise Fi with new f = 3.06*137.47 = 419.17 ksiTotal Loss at all Stage= 65.516 ksi% of Total Loss = 65.516/202.5 = 32.306 %Check stresses under Service LoadAllowable stress at super imposed loads isCompression side =- 0.45fc’ = -0.45*8000 = -3.6 ksiTension side = 12fc' = 12'8000 = 1.073 ksiftop= - FiA + Fi*e*YtI- (Mg+Msd)*YtIftop= - 419.171078 + 419.17*14.138618- 387.45*128618 = -0.241 ksi < -3.6 ksifbot= - FiA + Fi*e*YbI- (Mg+Msd)*YbIfbottom= - 419.71078- 419.17*14.132688+ 387.45*122688 = -2.502 ksi< 1.073 ksiStresses at the top and bottom are coming safe when check with the allowable stresses.Summary of Stresses Stresses Level at Various StagesSteel Stress, ksiPercentageAfter Tensioning (0.75 fpu)202.5100Elastic Shortening-21.6-10.67Creep Loss-33.26-16.42Shrinkage Loss-5.9-2.91Relaxation Losses-4.756-2.34Final net Stress137.47 67.66Checking for ultimate Flexural capacity?Mn ≥Mu?MnMcr ≥ 1.2Check stress at Mu.Ultimate moment calculation – Bonded tendons considering no reinforcementAssume a< hf Check fse >0.5fpu satisfies the check we can proceed by these below equationfps = fpu 1-?pβ1ρp*fpu/f'cFor Low Relaxation we have ?p= 0.28f’c > 4000 psi where f’c = 8000 psiβ1 = 0.85-0.05f'c-40001000= 0.65 which is okρp = Apsbdp = (3.06)/(15*12*20.31) = 8.37 x 10-4fps= 269.99 ksiTps = Aps Fps = 3.06*269.99 =826.16 KipsTps = 826.16 KipsC= Tpsa= Tps0.85*f'c*b=0.67a< hf hence assumption is correct Mn = Tps dp-0.5a= 16502.54Mn = 1375.21 Kips-ft?Mn=0.9*1375.21=1237.69 kip-ft Figure Showing Total factored load on the beamMmax= wl2/8 = 1004.4 kips-ft?Mn >Mu…………………….. Hence okCracking MomentFr = 7.5√f’c = 670.82 psiMcr = fe+r2cfr*IcF = 419.17 Kipse=14.13 inchMcr =730.96 kips-ftMuMcr=1237.69730.96= 1.69>1.2 ………………………………………………….okNow Checking Flexural Strength for Flange partAssume Ast WWR in Precast = W6 @ 6 inch = 0.12 in2/ft…………..From Design Aid 11.21 The cantilever flange controls the design since the negative moment over the stem reduces the moment between the stems. Construct the strain diagram.Try C= 0.25 inch?s+0.0030.003 =dc?s(d-c)=0.003c?s=0.003dc-0.003?y=fyEs=6028000=2.14 *10^-3?s=0.003*30.25-0.003 =0.033Check, ?y < ?s........... Hence oka=β1*c = 0.1625C= 0.85 f’cb a=13.26 kips/ftT=AstFy=0.12 *60 =7.2 kips/ft <13.26 kip/ftTherefore try,a=T0.85*f'c*b=0.088”C=aβ1=0.135Check, ?s=0.003*30.135-0.003 = 0.064 >0.00214…… Hence okTherefore reinforcement yield and analysis is validCdt = 0.135 < 0.375 from figure 4.2.1.3 PCI handbookTherefore tension controls, and ?=0.9?Mn=?T(d-0.5a)?Mn=9.43kip-inft=785.83lb-ft/ftCheck minimum and maximum reinforcementd=3 inchb=12 inchAsmin=3f'cfy*bw*d=0.16 inch2Provide minimum steel reinforcement As provided W8 0.319 in2 @6 inch SpacingCalculation of development lengthLd =(fse/3)db + (fps-fse)db =17.5”Calculate allowable load,flange self weight = 4*115/12 =38.33 wd (flange section) = 38.33 *12*3.3752*0.5 =261.96 lb-ft/ftMl = 785.83 -261.96 =523.873 lb-ft/ft.Calculation of Live load and checking with the actual live loadWl = 523.873*123.3752*1.6 =344.93 psf < 50 psf ………… Hence safeShear Design The shear design of precast concrete members is designed according to new provision of ACI 318-08.The shear resistance of precast concrete members must meet the requirement: ?Vn≥VuWhere: Vn = Vc+VsVc = nominal shear strength of concreteVs = nominal shear strength of shear reinforcement? = 0.75Since fse(137.47ksi) ≥ 40% fpu(108ksi),used below formula to calculate shear at each point along longitudinal length.Vc = [0.6λfc' +700(Vu*dpMu)]bwdp…………11.9 ACI CodeWhere: Vu*dpMu ≤ 1; 2λfc' bwdp ≤Vc≤ 5λfc' bwdpbw = 16.25 inchdp >=0.8h so use dp=0.8*h=20.3 inchλ=0.75(Lighweight concrete)Figure Showing Total factored loadLengthVuMuVcCheck For (Vu*dp/Mu)<1Vc minVc MaxVu/φ066.9600045.34746113.368689.28360.2642290.0322.2900320.54736842145.34746113.368680.352653.5684339.0084.3390080.25679012345.34746113.368671.424946.8726146.9286.1469280.15860566445.34746113.368662.4961240.1767713.7927.7137920.10833333345.34746113.368653.5681533.489039.69.03960.07703703745.34746113.368644.641826.78410124.3510.1243520.05502645545.34746113.368635.7122120.08810968.0510.9680480.03809523845.34746113.368626.7842413.39211570.6911.5706880.02407407445.34746113.368617.856276.69611932.2711.9322720.01167227845.34746113.36868.92830012052.812.0528045.34746113.36860Check for Shear ReinforcementLengthVuφVc/2Check066.9693.75Shear Reinforcement is required as Vu>φVc/2 as Per PCI Code 360.26422.599653.56820.088946.87217.5771240.17615.0661533.4812.5551826.78410.0442120.0887.5332413.3925.022276.6962.5113000Table Showing Area Check which is required to decide on the stirrupLengthVsSmaxAvAv min=50bw S/Fy A 1min=A2 minComment0-160.7219.5-1.08646720.26406250.0001775040.35427702Av min=50bw S/Fy for reinforced3-62.75960619.5-0.4242549370.26406250.0001775040.354277026-2.93678078519.5-0.0198526380.26406250.0001775040.35427702911.3656622119.50.0768318770.26406250.0001775040.354277021214.3320957619.50.0968849670.26406250.0001775040.354277021512.8087994619.50.0865874840.26406250.0001775040.35427702189.08850316619.50.0614382810.26406250.0001775040.35427702214.16642909219.50.0281650610.26406250.0001775040.3542770224-1.44416350119.5-0.0097625450.26406250.0001775040.3542770227-7.4378986319.5-0.0502801950.26406250.0001775040.3542770230-13.6042375819.5-0.0919646460.26406250.0001775040.35427702* A2 min= (0.75√fc'bw.S/Fy)* A1 min= (Aps fpu S/80 Fy d)*√(d/bw)Table showing at each Station Stirrups Provided and actual area of the stirrupsLengthAv ProvidedStirrups DesignationSpacing in inch00.395061728#4-U 2 legged19.530.395061728#4-U 2 legged19.560.395061728#4-U 2 legged19.590.395061728#4-U 2 legged19.5120.395061728#4-U 2 legged19.5150.395061728#4-U 2 legged19.5180.395061728#4-U 2 legged19.5210.395061728#4-U 2 legged19.5240.395061728#4-U 2 legged19.5270.395061728#4-U 2 legged19.5300.395061728#4-U 2 legged19.5Deflection CheckMost precast, pre stressed concrete flexural members will have a upward camber at the of pre-stress, caused by the eccentricity of the pre-stressing force. This camber may increase of decrease with time, depending upon the stress distribution across the member under sustained load, superimposed dead load even during the time of erection. So must need to calculate the possible deflection would at these stages and all should be under the maximum permissible deflection according to the PCI handbook.At TransferInitial , Eci = 2.601 x 106 psiBefore 28 days Ec = 3.23 x 106 psiDue to initial Prestress only,?=(P e l^28 Eci Ig) + (Pi ee-ecl^224EciIg)ee =14.75” , I=53280 in4Pi = 619.65 Kips?I =-619.65*14.5*60*1228*2.601*106*53280 =-4.2 “Self weight intensity = w = 0.861 k/ft = 861/12 lb/inSelf weight ?d =5wl4384 Eci Ig …………..For Un cracked Section.?d = 5*71.75*60*124384*2.601*106*53280 =1.81”Thus the net camber at transfer =-4.201+ 1.81 = -2.39 inch (↑)Immediate Service Load deflection fr = 7.5*√f'c = 670.82 psifbottom under service load = -2.052 ksi dp = 20.31”ρp = Apsbdp = (3.06)/(15*12*20.31) = 8.37 x 10-4n=EsEc = 8.67Icr = n Aps d2p(1-1.6√nρp) = 9452 in4Moment, Mcr due to that portion of live load that causes cracking are Mcr = Sb (7.5f'c+fce-fd) =2842.82 kips-ftMa= Unfactored maximum live load momentMmax = 337.5 k-ftMcrMmax = 2842.8337.5=8.42 hence we can say use Ig for check of the deflection ?l = 5wl4384 Ec I Wul= 50*15 =0.75 k/ft =750/12 =6.25?l = 5wl4384 Ec I = 1.27” ↓Long Term Deflection (camber) by PCI multipliersLoadTransfer ?p inchPCI Multiplier? (inch) finalPrestress Camber at Transfer-4.21.8-7.56Deflection due to own weight1.811.853.3485Net-2.39↑-4.2115↑Deflection due to live load1.27↓Final ?-2.39↑-2.941↑The Result are good because less loading condition there is not much deflection due to live load hence check for allowable deflection.Maximum permissible computed deflection PCI handbook pg 4-88 Table 4.8.1l/480 = (60*12/480) = 1.5”.Design of Inverted Tee GirderLoads Analysis:Loads acting over inverted tee beams are:-Concentrated loads acting due to the cumulative effect of Self weight of Double tee beam and uniformly concentrated live load are:Uniformly distributed live load = 50psf = 0.05k/ftWidth of slab = 15ftTotal live load = 15*0.05 = 0.75k/ftFactored Live load = 0.05*15*1.6 = 1.2k/ftDensity of TT = 115 lb/ft3Area of TT = 7.86ft2Self weight = 0.115*7.86 = 0.86k/ftFactored self weight = 1.2*0.86 = 1.032k/ftTherefore, the total load coming from Double tee = 1.2 + 1.032 = 2.232k/ftReaction = 2.232*604 = 33.48 k Figure Showing Load coming from Double tee beamSelf weight of Inverted Tee Beam Area = 5.44ft2Density = 115 lb/ft3Self weight = 0.115*5.44 = 0.626k/ftFactored self weight = 0.75k/ftFigure Showing Load on Inverted tee beam Section Selected: - 28IT44 (PCI handbook Pg 2-45) Section Properties:Designationh (in)h1/h2A(in)I(in)YbSbStNo.Strands28IT444428/1678412443717.437139468320-StraightMaterials Properties: Pre-stressing steel Concrete (Light weight concrete)fpu=270ksi fc’= 8000psi?” diameter f’ci= 4500psiLow-relaxation strands w = 115lb/ft3fpi= 0.75fpu= 202.5ksi Eci’= 377280ksf Es = 28000ksi Ec’ = 465120ksfAps= 20*0.153=3.06 in2 Check Stresses At transfer: Since we do not know the eccentricity at this time but we know the Pre-stressed force and allowable stress at extreme fibers, we can easily calculate that.Allowable stress at super imposed loads areCompression side = 0.7fci’ = 0.7*4500 = -- 453.6ksfTension side = 6fci' = 6'4500 = +57.95ksfFi = Aps*fps = 619.65kftop= - FiA + Fi*e*YtI- M*YtI (Under Tension) ftop= - 619.655.44 + 619.65*e*2.216- 25.78*2.216 e = 9.5”With fbot (Under compression)I was getting e = 27.27” but we supposed to considered the small value.Calculate Losses Elastic Shortening E.S = Kes*Es(fcirEci)Where: Kes=1 (Pre tensioned)fcir= Kcir(FiA+Fi*e2I)+Mg*eI fcir= 0.9(619.655.44+619.65*0.79226)+84.37*0.7926 = 165.15 ksfThe elastic Shortening losses will be: E.S = 1*403200*165.15377280 = 1764.96 ksf = 12.25 ksi Steel relaxation R.E = [Kre-J(S.H+C.R+E.S)]cWhere: Kre=5000psiJ= 0.04………….Table4.5.1 (PCI)C= 1…………….Table4.5.2 (PCI)The relaxation loss will be = [5000-0.04(0+0+12256.6)]*1 = 4509.73psi = 4.51 ksifT = Total loss = E.S+R.E = 12.25+4.51= 16.76ksiStress left in strands after losses = fpi – fT =202.5-16.76 = 185.74ksiRevise Fi with new f = 3.06*185.74= 568.36k Since we revised the Fi value by accounted stress losses so we need to calculate new “e” value with this force by keeping the same max allowable stress at extreme ends.So the new “e” is calculated below:Fi = Aps*fps = 568.36kftop= - FiA + Fi*e*YtI- M*YtIftop= - 568.365.44 + 568.36*e*2.216- 25.78*2.216 e = 9.85” > 9.5 (old value)But I in further calculation I have assumed “e= 9.5” but keep Fi=568.36kSince I have assumed “e= 9.5” but the value of Fi reduced to Fi=568.36k so, the max stresses at extreme fiber ends are under allowable limit state as per PCI code so STRESSES are………O.KAt Superimposed dead load (Due to Double Tee beam)Fi =568.36k e = 9.5inCalculate losses: Creep loss C.R=KcrEsEc(fcir – fcds)Where: fcir = Kcir(FiA+Fi*e2I)-Mg*eI fcir =0.9(568.365.44+568.36*0.79226)-84.37*0.7926 = 152.76 ksf fcds= MsdI*e = 400.3656*0.792 = 52.85ksf = 0.367ksiMsd is only due to the self wt of Double tee beam, there will be no live load consideredThe creep loss will be C.R=1.6*280003230(1.107 – 0.367) = 9.02 ksi Shrinkage loss S.H = 8.2x106Ksh*Es (1-0.06VS)(100-R.H)Where: Ksh = 1V/S=5.44R.H = 75The shrinkage loss will be = 8.2x106*1*28000 (1-0.06*5.44) (100-75) = 3.866 ksiSteel relaxation R.E = [Kre-J(S.H+C.R+E.S)]cWhere: Kre=5000psiJ= 0.04………….Table4.5.1 (PCI)C= 1…………….Table4.5.2 (PCI)The relaxation loss will be = [5000-0.04(9.01+3.866+0)*1000]*1 = 4484.96psi = 4.5 ksifT = Total loss = C.R+R.E+S.H = 9.02+3.866+4.5= 17.386ksiStress left in strands after losses = fpi – fT =185.74-17.386 = 168.35ksiRevise Fi with new f = 3.06*168.35.74= 515.16kCheck stressesAllowable stress at super imposed loads isCompression side = 0.45fc’ = 0.45*8000 = -518.4ksfTension side = 12fc' = 12'8000 = +154.56ksfftop= - FiA + Fi*e*YtI- (Mg+Msd)*YtI (Under compression)ftop= - 515.165.44 + 515.16*0.792*2.216- 484.74*2.216 = -94.7ksf < -518.4ksffbot= - FiA + Fi*e*YbI- (Mg+Msd)*YbI (Under compression)ftop= - 515.165.44 - 515.16*0.792*1.456+ 484.74*1.456 = -76.21ksf < -518.4ksfStresses at the top and bottom are coming safe……………O.K Check ultimate flexural strength In this section our two main objectives are:?Mn ≥Mu?MnMcr ≥ 1.2Ultimate flexural strengthEstimate the steel stress at ultimate by fps equation which is valid to use since fse=168.35ksi > 0.5fpu=135 ksi.ρp= Apsbw.d = 3.0612*36.07 = 0.00707fps= 270000 (1-0.5ρpfpufc') = 270000 (1-0.5*0.00707*270008000) = 266.778ksi T’=Apsfps= 3.06*266.778= 816.34 kC=0.85*fc’b*aBut T’=C therefore a= 816.340.85x8x12 = 10in Mn= T’(d-a/2) = 816.4(36.07-10/2) = 25363.684 k-in = 2113.64k-ft ?Mn = 0.9x2113.64=1902.27 k-ft Mu= 147.41x15’-66.96x11’-66.96x3.5’-0.9x15’x7.5’=1139.13k-ft Hence ?Mn(1902.27 k-ft) > Mu(1139.13k-ft) ………..condition satisfied.Cracking MomentCracking has to be check at the tension side which is basically at the bottom side of the girder and it is calculated by following equation:Mcr = Fi*e +FiA*Yb+fr*IYb=515.16*0.792 +515.166*1.45+96.6*61.45 = 1174.3k-ftTherefore; ?MnMcr=1902.271174.3 = 1.62 > 1.2……………….condition satisfied.Check stress at Mu:Allowable stresses at Mu are:Compression side = 0.45fc’ = 0.45*8000 = -518.4ksfTension side = 12fc' = 12'8000 = +154.56ksfftop= - FiA + Fi*e*YtI- (Mu)*YtI (Under compression)ftop= - 515.165.44 + 515.16*0.792*2.216- 1139.13*2.216 = -364 ksf < -518.4ksffbot= - FiA + Fi*e*YbI- (Mu)*YbI (Under Tension)fbot= - 515.165.44 - 515.16*0.792*1.456+ 1139.13*1.456 = 82 ksf < 154.56 ksfStresses at the top and bottom are coming safe……………O.KStresses at each stageAt Transfer(ksf)Stress at service load(ksf) At Compression0.7fci’ = 453.60.45fc’ = 518.36At Tension 6fci' = 57.9512fc' = 145.26 Shear Design The shear design of precast concrete members is designed according to new provision of ACI 318-08.The shear resistance of precast concrete members must meet the requirement: ?Vn≥VuWhere: Vn = Vc+VsVc = nominal shear strength of concreteVs = nominal shear strength of shear reinforcement? = 0.75Since fse(168.35ksi) ≥ 40% fpu(108ksi),used below formula to calculate shear at each point along longitudinal length. Vc = [0.6λfc' +700(Vu*dpMu)]bwd…………11.9 ACI CodeWhere: Vu*dpMu ≤ 1; 2λfc' bwd ≤Vc≤ 5λfc' bwdFigure Showing Shear and bending Moment calculated From Staad ProTable Showing Calculation of Shear StrengthDistance(ft)Vu(k)Vu/ФVc(k)Vc-minVc-maxVs=Vc-Vu/Ф (k)0146.187194.916320.4158.07145.17-49.772.5144.142192.19320.4158.07145.17-47.02575.138100.184124.8458.07145.1724.6567.573.09397.45798.3758.07145.170.9131071.04994.73281.9758.07145.17-12.7612.52.0442.72519.96858.07145.1755.345150017.42158.07145.1758.07The above table shows the comparison between Vc, Vu, Vs. The negative values in the last column showing that we need shear reinforcement Vs at respective stations along the length of girder.Calculate Av at 0 and 2.5ftAv = Vs*SFy*dWhere: Smax= ?*h= ?*44=33”<24”Spacing = 24”Av = 49.77*2460*36.07 = 0.55in2Provide #5@24” (Av=0.6in2)Av = Vs*SFy*dAv = 47.02*2460*36.07 = 0.55in2Provide #5@24” (Av=0.6in2)The next shear reinforcement needed at 10’ from support so it’s better to provide the same shear reinforcing till 12’from support and since we do not need any reinforcing at mid span (as you can see in below diagram of SFD and BMD) but I have provided minimum reinforcing (#3@24”) at the portion 3’ from centre of beam on each side.Shear strength of ledge portion: The design shear strength of beam ledge ,supporting concentrated loads from stem of double tee beam can be determined by the lesser of two equation below:The above figure shows detailing of ledge part of inverted beam and it also shows the all parameters which has been used in equations used below.For s (24”) > bt+hl (16+7.25=23.25”) ?Vn = 3? λfc' hl[2(bl-b)+ bt+ hl]……………..4.5.1.1PCI Hand book = 3x0.75x0.758000x16[2(20-12)+7.25+16] = 94.78 Kips ……………governs ?Vn = ? λfc' hl[2(bl-b)+ bt+ hl+2de]………….4.5.1.21PCI Hand book= 0.75x0.758000x16[2(20-12)+7.25+16+2x45] = 101.04 kipsSo the the ?Vn = 94.78k and Vu(reaction from each stem) = 66.96kTherefore, ?Vn≥Vu is satisfied.Design of Transverse (Cantilever) Bending of ledge: Transverse bending of the ledge requires flexural reinforcement ,As, which is computed by following equation .As = 1?Fy[Vuad] ……………..4.5.2.1 = 10.75*60[66.965.514.5] = 0.564 in2 Provide #5@16” (Av=0.6in2)Note: This reinforcement may be uniformly spaced over the width of 96in on either side of the bearing. Spacing should not exceed hl(16”) or 18”.Design of Longitudinal Bending of Ledge: Longitudinal reinforcing should be placed in both top and bottom of the ledge portion of beam.Al= 200(bl-b)dl/fy = 200(20-12)14.19/60000 = 0.38 in2Provide 1- #4 top and 1-#4 bottom” (Av=0.4in2)NOTE: I have considered the one side as ledge for analysis and design of shear reinforcing. Since we have ledge at both sides because it is inverted tee beam, i have provided the same reinforcing for that.Deflection and Camber CheckMost precast, pre stressed concrete flexural members will have a upward camber at the of pre-stress, caused by the eccentricity of the pre-stressing force. This camber may increase of decrease with time, depending upon the stress distribution across the member under sustained load, superimposed dead load even during the time of erection. So must need to calculate the possible deflection would at these stages and all should be under the maximum permissible deflection according to the PCI handbook. Deflection at Transfer of Pre-stressing force:?(Prestress) = L2*Fi*e8Eci*I = 302*568.36*0.7928*377280*6 = 0.2683in (↑)Where Fi = Pre stressing at transfer ?(Due to self wt.) = 5*w*L4384*Eci*I = 5*0.75*304384*377280*6 = 0.042in (↓)Estimate Deflection at Transfer = ?(Prestress)- ?(Due to self wt.) = 0.2683-0.042 = 0.2263inAt Erection: Using multipliers at erection from table……………?(Prestress) = 0.2683x1.85 = 0.496in (↑)?(Due to self wt.) = 0.042x1.8= 0.0756in (↓)Estimate Deflection at Erection= 0.496-0.0756= 0.4204inAt Final stage: ?Prestress = 0.2683x2.45 = 0.657in (↑) ?Superimposed load = 0.425 (↓) ?(Due to self wt.) = 0.042x2.7 = 0.1134 (↓)Estimate Deflection at Erection= 0.657-0.425-0.1134= 0.1186inFigure showing Deflection due to only superimposed load is calculated in STAAD.Pro.StagesDeflectionPermissibleConditionAt Transfer0.2263L/480= 0.75inGoodAt Erection0.4204GoodAt Final0.1186Good DrawingsReferencesBooksACI CodeInternational building Code –IBC Load calculation & SpecificationJames G Macgregor ,“Reinforced Concrete Mechanics and Design”, (2nd edition) by Prentice hall, A Simon and Schuster company, Upper saddle river ,NJ – 07458PCI design handbook -6th EditionDr. Edward G Nawy, Prestressed Concrete a Fundamental approach-2nd EditionT.Y.Lin, Design of Prestressed Concrete Structures. ................
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