(001) Bridges Book

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Toma, S.; Duan, L. and Chen, W.F. “Bridge Structures ”Structural Engineering Handbook Ed. Chen Wai-Fah

Boca Raton: CRC Press LLC, 199 9

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Bridge St ruct ures

Shouji TomaDepartm ent of Civ il Engineering,

Hok kai -Gakuen U niversity, Sapporo, Japan

Lian D uanDivision of Structures, California

Departm ent of Transportati on, Sacramento,

C A

Wai-Fah C henSchool of C ivi l Engineering,

Purdue U niversity,

West Lafayette, IN

10.1 General10.2 Steel Bridges10.3 ConcreteBridges

10.4 ConcreteSubstructures10.5 Floor System10.6 Bearings, Expansion Joints, and Railings10.7 Girder Bridges10.8 TrussBridges10.9 Rigid FrameBridges(Rahmen Bridges)10.10Arch Bridges10.11Cable-Stayed Bridges10.12Suspension Bridges10.13Dening Term sAcknowledgmentReferencesFurther ReadingAppendix: Design Examples

10.1 General

10.1.1 Introduction

A bridge isastructurethat crossesover ariver, bay, or other obstruction, permitting thesmooth andsafe passageof vehicles, trains, and pedestr ians. An elevation view of a typical bridge is shown in

Figure 10.1 . A bridgestructurei sdivided into an upper part (the superstructure ), which consistsof theslab, the oor system , andthemain trussor girders , andalower part (the substructure ), whicharecolumns, piers, towers, footings, piles, and abutments . Thesuperstructureprovideshorizontal spanssuch asdeck and girders and carriestrafc loadsdirectly. Thesubstructuresupports the horizontalspans, elevating above the ground surface. In this chapter, main structural features of commontypes of steel and concrete bridges are discussed. Two design examples, a two-span continuous,cast-in-place, prestressed concrete box girder bridge and a three-span continuous, composite plategirder bridge, aregiven in theAppendix.

c 1999by CRC PressLLC

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FIGURE10.1: Elevation view of a typical bridge.

c 1 9 9 9 b y C R C P r e s s L L C

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10.1.2 Classication

1. Classication by Materials

Steel bridges: A steel bridge may usea wide variety of structural steel componentsand systems: girders, frames, trusses, arches, and suspension cables.Concrete bri dges: There are two primary typesof concrete bridges: reinforced andprestressed.

Timber bridges: Wooden bridgesareused when the span is relatively short.

Metal alloy bridges: Metal alloyssuch asaluminum alloy and stainlesssteel are alsoused in bridgeconstruction.

2. Classication by Objectives

Highway br idges: bridgeson highways.

Railway br idges: bridgeson railroads.

Combined bridges: bridgescarrying vehiclesand trains.Pedestrian bridges: bridgescarrying pedestr ian trafc.

Aqueduct bridges: bridgessupportingpipeswith channeledwaterow.

Bridgescan alternatively beclassied into movable (for ships to pass the river) or xedand permanent or temporary categories.

3. Classicati on by Structural System (Superstructures)

Plate girder bridges: The main girders consist of a plate assemblage of upper andlower angesand aweb. H- or I -cross-sectionseffectively resist bendingand shear.

Box girder bri dges: The single (or multiple) main girder consists of a box beamfabricated from steel platesor formed from concrete, whichresistsnot only bendingand shear but also torsion effectively.

T-beam bridges: A number of reinforced concrete T-beamsare placed side by sideto support the liveload.

Composite girder br idges: Theconcrete deck slab worksin conjunctionwith thesteelgirders to support loadsas a united beam. The steel girder takes mainly tension,while theconcrete slab takesthe compression component of thebending moment.

Gri llagegirder bri dges: Themain girdersareconnected transversely by oor beamsto form agrid pattern which sharestheloadswith themain girders.

Trussbri dges: Trussbar membersare theoretically considered to beconnected withpins at their ends to form tr iangles. Each member resists an axial force, eitherin compression or tension. Figure 10.1 showsa Warren truss bridge with vert icalmembers, which is a “trough bridge”, i .e., the deck slab passes through the lowerpart of the bridge. Figure 10.2 showsa compari son of the four design alternativesevaluated for Minato Oh-Hasshi in Osaka, Japan. The truss frame design wasselected.

Arch bridges: Thearch is a structure that resists load mainly in axial compression.In ancient times stone was the most common material used to construct magnif-icent arch bridges . There is a wide variety of arch bridges as will bediscussed inSection 10.10


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FIGURE10.2: Design comparison for Minato Oh-Hashi, Japan. (From Hanshin Expressway PublicCorporation, Construction Recordsof Minato Oh-Hashi, Japan Society of Civil Engineers, Tokyo [inJapanese] , 1975. With permission.)

Cable-stayed bridges: Thegirdersaresupported byhighlystrengthened cables(oftencomposed of tightlyboundsteel strands) whichstem directly from thetower. Thesearemost suited to bridgelong distances.

Suspension br idges: The girders are suspended by hangers tied to the main cableswhich hang from the towers. The load is transmit ted mainly by tension in cable.


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This design is suitablefor very long span bridges.

Table 10.1 showsthe span lengthsappropriateto each typeof bridge.4. Classication by Support Condition

Figure 10.3 showsthreedifferent support conditionsfor girder bridges.Simply supported bridges: Themain girders or trussesare supported by a movablehingeat oneend and axed hingeat the other (simple support); thus they can beanalyzed using only theconditionsof equilibrium.

Continuously supported bridges: Girders or trusses are supported continuously bymore than three supports, resulting in a structurally indeterminate system. Thesetend to bemore economical since fewer expansion joints, which havea commoncause of servi ce and maintenanceproblems, are needed. Sinkage at the supportsmust beavoided.

Gerber bri dges (canti lever bridge): A continuous bridge is rendered determinateby placing intermediatehinges between the supports. Minato Oh-Hashi’s bridge,

shown in Figure 10.2 a, is an exampleof aGerber trussbridge.

10.1.3 Plan

Before the structural design of a bridge is considered, a bridge project will start with planning thefundamental design conditi ons. A bridgeplan must consider thefollowing factors:

1. Passing Line and LocationA bridge, beingacontinuation of a road, doesbest to follow thelineof theroad. A rightanglebridgeiseasy to design and construct but often forcesthel ine to bebent. A skewedbridgeor a curved bridge is commonly required for expresswaysor rail roadswhere theroad linemust bekept straight or curved, even at thecost of a moredifcult design (seeFigure 10.4 ).

2. WidthThewidth of ahighway bridgeisusually dened asthewidth of theroadway plusthat of the sidewalk, and often the samedimension asthat of the approaching road.

3. Typeof Structureand Span LengthThe types of substructures and superstructures are determined by factors such as thesurroundinggeographical features, thesoil foundation, thepassing lineandit swidth, thelength and span of thebridge, aesthetics, therequirement for clearancebelow thebridge,transportation of theconstruction materials and erection procedures, construction cost,period, and so forth.

4. AestheticsA bridge is required not only to fulll its function asa thoroughfare, but also to use itsstructureand form to blend, harmonize, and enhanceitssurroundings.

10.1.4 Design

The bridge design includes selection of a bridge type, structural analysis and member design, andpreparation of detailed plans and drawings. The size of members that sati sfy the requirementsof design codes are chosen [ 1, 17]. They must sustain prescribed loads. Structural analyses areperformed on a model of thebridge to ensure safety aswell as to judge the economy of thedesign.Thenal design iscommitted to drawingsand given to contractors.


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TABLE10.1 Typesof Bridgesand Applicable Span Lengths

From JASBC, Manual Design Data Book, Japan Association of Steel BridgeConstruction, Tokyo (in Japanese), 1981. With permission.


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FIGURE10.3: Supportingconditions.

FIGURE10.4: Bridgelines.

10.1.5 Loads

Designersshould consider thefollowing loadsin bridgedesign:

1. Primary loadsexert constantly or continuously on the bridge.

Dead load: weight of thebridge.Liveload: vehicles, trains, or pedestr ians, including theeffect of impact. A vehicularload is classied into three parts by AASHTO [ 1]: the truck axle load, a tandemload, and auniformly distributed laneload.

Other primary loadsmay begenerated by prestressing forces, thecreep of concrete, theshrinkage of concrete, soil pressure, water pressure, buoyancy, snow, and centrifugalactionsor waves.


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2. Secondary loadsoccur at i nfrequent intervals.

Wind load: a typhoon or hurricane.Earthquakeload: especially critical in itseffect on the substructure.

Other secondary loadscomeabout with changesin temperature, accelerati on, or tempo-rary loadsduringerection, collision forces, and so forth.

10.1.6 InuenceLines

Sincethel iveloadsby denition move, theworst casescenario along thebridgemust bedetermined.The maximum live load bending moment and shear envelopesare calculated conveniently usinginuencelines. Theinuencel inegraphically il lustratesthemaximum forces(bending moment andshear), reactions, and deectionsover a section of girder asa load travels along it slength. Inuencelinesfor thebending moment and shear forceof asimply supported beam areshown in Figure 10.5 .For aconcentrated load, thebending moment or shear at section A can becalculated bymultiplyingthe load and the inuencelinescalar. For auniformly distributed load, it is theproduct of the loadintensity and thenet areaof thecorresponding inuencelinediagram.

10.2 Steel Bridges

10.2.1 Introduction

The main part of a steel bridge is made up of steel plates which compose main girders or framesto support a concrete deck. Gas ame cutting is generally used to cut steel plates to designateddimensions. Fabrication by welding is conducted in the shop where the bridge components arepreparedbefore beingassembled (usually bolted) on the construction site. Several members for twotypical steel bridges, plate girder and trussbridges, are given in Figure 10.6 . The composite plategirder bridgein Figure 10.6 a is adeck typewhile thetrussbridge in Figure 10.6b is a through-decktype.

Steel has higher strength, ductili ty, and toughness than many other structural materials such asconcrete or wood, and thus makesan economical design. However, steel must bepainted to preventrustingand also sti ffened to prevent a local bucklingof thin membersand plates.

10.2.2 Welding

Welding is the most effective meansof connecting steel plates. Thepropert iesof steel changewhenheated and thischangeisusually for theworse. Molten steel must beshielded from theair to preventoxidization. Welding can be categorized by the method of heating and the shielding procedure.Shielded metal arc welding (SMAW), submerged arc welding (SAW), CO 2 gas metal arc welding(GMAW), tungsten arc inert gas welding (TIG), metal arc inert gas welding (MIG), electric beamwelding, laser beam welding, and friction welding arecommon methods.

Therst two weldingproceduresmentionedabove, SMAWand SAW, areused extensivelyin bridge

construction dueto their high efciency. Both usean electric arc, which is generally considered themost efcient methodofapplyingheat. SMAWisdonebyhandandissuitablefor weldingcomplicated jointsbut islessefcient than SAW. SAW isgenerally automated and can bevery effectivefor weldingsimplepartssuch astheconnection between theangeand web of plategirders. A typical placementof thesewelding methodsis shown in Figure 10.7 . TIG and MIG usean electric arc for heat sourceand inert gasfor shielding.

An electricbeam weld must not beexposed to air, and thereforemust belaid in avacuum chamber.A laser beam weld can beplaced in air but is lessversati le than other typesof welding. I t cannot be


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FIGURE10.5: Inuencelines.

used on thick platesbut i sideal for minuteor arti stic work. Sincethe welding equipment necessaryfor heatingand shielding isnot easy to handleon aconstruction site, all weldsareusually laid in thefabrication shop.

Theheating and cooling processesduringwelding induceresidual stressesto the connected part s.Thesteel surfacesor partsof thecrosssection at somedistancefrom the hot weld, cool rst. Whenthe area close to theweld then cools, i t tries to shrink but i s restrained by the moresolidied and


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FIGURE10.6: Member namesof steel bridges. (FromTachibana,Y.andNakai, H., BridgeEngineering,Kyoritsu Publishing Co., Tokyo, Japan [ in Japanese] , 1996. With permission.)

cooler parts. Thus, tensile residual stresses are trapped in the vicinity of the weld while the outerpartsareput into compression.

Therearetwo typesof welded joints: grooveandllet welds(Figure 10.8 ). Thellet weld isplacedat the junction of two plates, often between a web and ange. It is a relatively simple procedurewith no machining required. Thegrooveweld, also calledabutt weld, issuitablefor jointsrequiringgreater strength. Dependingon thethicknessof adjoiningplates, theedgesarebeveledin preparationfor theweld to allow themetal to ll the joint. Variousgrooveweld geometries for full penetrationwelding areshown in Figure 10.8b.

Inspection of welding is an important task since an imperfect weld may well havecatastrophicconsequences. It is difcult to nd faults such asan i nterior crack or ablowholeby observing onlythesurfaceof aweld. Manynondestructivetestingproceduresareavailablewhichusevariousdevices,suchasx-ray,ultrasonicwaves, color paint, or magneti cpart icles. Theseall havetheir ownadvantagesand disadvantages. For example, thex-rayand theultrasonic tests aresuitablefor interior faultsbut


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FIGURE 10.7: Welding methods. (From Nagai, N., Bridge Engineering, Kyoritsu Publishing Co.,Tokyo, Japan [in Japanese] , 1994. With permission.)

require expensiveequipment. Useof color paint or magneticpart icles, on theother hand, is acheapalternativebut only detectssurfaceaws. Thex-rayand ultrasonic tests areused in common bridgeconstruction, but ul trasonic testing isbecomingincreasingly popular for both its“ high tech” and itseconomical features.

10.2.3 Bolting

Bolt ingdoesnot requiretheskil ledworkmanship neededfor welding,andisthusasimpler alternative.

It is applied to the connections worked on construction site. Some disadvantages, however, areincurred: (1) spliceplatesare needed and the forcetransfer i s indirect; (2) screwing-in of theboltscreatesnoise; and (3) aesthetically boltsare lessappeali ng. In special cases that need to avoid thesedisadvantages, the welding may beused even for site connections.

Therearethreetypesof high-tensilestrength-bolted connections: theslip-crit ical connection, thebearing-type connection (Figure 10.9 ), and the tensile connection (Figure 10.10 ). Theslip-critical(friction) bolt ismost commonlyused in bridgeconstruction aswell asother steel structuresbecauseit issimpler than abearing-typebolt and morereliablethan atension bolt. Theforceistransferred by


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FIGURE 10.8: Types of welding joints. (From Tachibana, Y. and Nakai, H., Bridge Engineering,Kyoritsu Publishing Co., Tokyo, Japan [ in Japanese] , 1996. With permission.)

thefriction generated between thebaseplatesandthespliceplates. Thefriction resistanceisinducedby theaxial compression forcein thebolts.

Thebearing-typebolt transfers the forceby bearing against the plate aswell asmaking someuseof friction. Thebearing-typebolt can transfer larger forcethan thefriction bolts but islessforgivingwith respect to the clearancespaceoften existing between thebolt and the plate. Theserequire thatpreciseholesbedri lled and at exact spacings. Theforcetransfer mechanism for theseconnectionsisshown in Figure 10.9 . In thebeam-to-column connection shown in Figure 10.10 , theboltsattachedto thecolumn aretension boltswhile theboltson thebeam areslip-crit ical bolts.

Thetension bolt transfersforcein thedirectionof thebolt axis. Thetension typeof bolt connectionis easy to connect on site, but difcultiesarise in distr ibuting forces equally to each bolt, resultingin reduced reliabil ity. Tension bolts may also be used to connect box members of the towers of suspension bridges where compression forces are larger than the tension forces. In this case, thecompression is shared with butting surfacesof theplatesand thetension is carri ed by thebolts.

10.2.4 Fabrication in Shop

Steel bridgesarefabricated into members in theshopyard and then transported to theconstructionsite for assembly. Ideally all constructional work would becompleted in theshop to get thehighestquali ty in theminimum construction time. The larger and longer themembers can be, the better,within the restr ictions set by transportation l imi ts and erection tolerances. When crane ships forerection and bargesfor transportation can beused, oneblock can weigh asmuch asathousand tons


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FIGURE 10.9: Slip-crit ical and bearing-type connections. (From Nagai, N., Bridge Engineering,Kyoritsu Publishing Co., Tokyo, Japan [ in Japanese] , 1994. With permission.)

and beerected asawholeon thequay. In thesecasesthebridgeismadeof asinglecontinuousblock

and much of the hassle usually associated with assembly and erection is avoided.

10.2.5 Constructionon Site

Thedesigner must consider theloadsthat occur during construction, generally different from thoseoccurring after completion. Steel bridges are parti cularly proneto buckling during construction.Theerection planmust bemadeprior to themain design and must bechecked for every possibleloadcasethat may ariseduringerection, not only for strength but also for stabil ity. Truck craneand benterection (or staging erection); launching erection; cableerection; canti lever erection; and largeblockerection (or oating crane erecti on) are several techniques (see Figure 10.11 ). An example of thelargeblock erection isshown in Figure 10.43 , in whicha186-m, 4500-ton center block istransportedbybargeand lifted.

10.2.6 Painting

Steel must bepainted to protect it from rusting. There is a widevariety of paints, and the life of asteel structure is largely inuenced by its quali ty. In areas near the sea, the salty air is particularlyharmful to exposed steel. Thecost of painti ngishighbut isessential to thecontinued good conditionof thebridge. Thecolor of thepaint is also an important consideration in termsof it spublic appealor aesthetic quali ty.


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FIGURE10.10: Tension-typeconnection.

10.3 ConcreteBridges

10.3.1 Introduction

For modern bridges, both structural concrete and steel give sati sfactory performance. The choicebetween thetwo materialsdependsmainlyupon thecost of constructionandmaintenance. Generally,concrete structuresrequirelessmaintenancethan steel structures, but sincethe relativecost of steeland concrete is different from country to country, and may even vary throughout different partsof thesamecountry, it is impossible to put onedenitively abovethe other in termsof “economy”.

In thissection, the main featuresof common typesof concrete bridgesuperstructuresare brieydiscussed. Concrete bridge substructureswill be discussed in Section 10.4 . A design example of atwo-span continuous, cast-in-place, prestressed concretebox girder bridgeisgiven in theAppendix.For a moredetailed look at design proceduresfor concrete bridges, referenceshould bemadeto therecent booksof Gerwick [ 7], Troitsky [ 24], Xanthakos[ 26, 27], and Tonias[ 23] .

10.3.2 ReinforcedConcreteBridges

Figure 10.12 showsthe typical reinforced concrete sections commonly used in highway bridgesu-perstructures.

1. SlabA reinforced concrete slab (Figure 10.12 a) isthemost economical bridgesuperstructurefor spansof up to approximately 40ft (12.2m). The slabhassimpledetailsand standardformwork and isneat, simple, and pleasing in appearance. Common spans range from16 to 44 ft (4.9 to 13.4 m) with structural depth-to-span ratiosof 0.06 for simple spansand 0.045for continuousspans.

2. T-Beam (Deck Girder)TheT-beams(Figure 10.12 b) aregenerally economicfor spansof 40to 60ft (12.2 to 18.3m), but do require complicated formwork, part icularly for skewed bridges. Structural


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FIGURE 10.11: Erecti ons methods. (From Japan Construction Mechanizati on Association, Cost Estimation of BridgeErection, Tokyo, Japan [in Japanese] , 1991. With permission.)

depth-to-span ratios are 0.07 for simple spans and 0.065 for continuous spans. Thespacing of girders in a T-beam bridge dependson the overall width of the bridge, theslab thickness, and thecost of theformwork and may betaken as1.5 timesthestructuraldepth. Themost commonly used spacingsarebetween 6 and 10 ft (1.8 to 3.1m).

3. Cast- in-PlaceBox GirderBox girders liketheoneshown in Figure 10.12 c, are often used for spansof 50 to 120 ft


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FIGURE10.12: Typical reinforced concrete sectionsin bridgesuperstructures.

(15.2to 36.6m). Itsformwork for skewed structuresissimpler than that required for theT-beam. Dueto excessivedead loaddeections, theuseof reinforcedconcreteboxgirdersover simplespansof 100 ft (30.5m) or moremay not beeconomical. Thedepth-to-spanratiosaretypically 0.06for simplespansand 0.055 for continuousspanswi th thegirdersspaced at 1.5 times the structural depth. Thehigh torsional resistanceof the box girdermakesi t part icularly suitablefor curved alignments, such astherampsonto freeways. It ssmooth owing linesareappealing in metropolitan cities.

4. Design ConsiderationA reinforced concrete highway bridge should bedesigned to sati sfy the specication orcoderequirements, suchastheAASHTO- LRFD [1] requirements (American Associationof StateHighway and Transportation Ofcials—Load and ResistanceFactor Design) forall appropriateservice, fatigue, strength, and extremeevent limit states. In theAASHTO-LRFD [1], service limit states include cracking and deformation effects, and strengthlimit statesconsider thestrength and stabil ity of astructure. Abridgestructureisusuallydesigned for the strength limit statesand is then checked against the appropriateserviceand extremeevent limit states.


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10.3.3 PrestressedConcreteBridges

Prestressed concrete, using high-strength materials, makes an att ractive alternative for long-spanbridges. It hasbeen widely used in bridgestructuressincethe 1950s.

1. SlabFigure 10.13 showsFederal Highway Administration (FHWA) [ 6] standard typesof pre-cast, prestressed, voided slabsand their sectional propert ies. While cast- in-place, pre-stressed slab ismoreexpensivethan reinforced concrete slab, precast, prestressed slab iseconomical when manyspansareinvolved. Common spansrangefrom 20to 50ft (6.1to15.2 m). Structural depth-to-span ratiosare0.03for both simpleand continuousspans.

FIGURE10.13: Federal Highway Administration (FHWA) precast, prestressed, voided slab sections.(From Federal Highway Administration, Standard Plansfor Highway Bridges, Vol. 1, ConcreteSuper- structures, U.S. Department of Transportation, Washington, D.C., 1990. With permission.)

2. Precast I GirderFigure 10.14 showsAASHTO [ 6] standard typesof I-beams. These compete with steelgirders and generally cost more than reinforced concrete with the same depth-to-spanratios. Theformwork is complicated, particularly for skewed structures. Thesesectionsare applicable to spans30 to 120 ft (9.1 to 36.6 m). Structural depth-to-span ratiosare0.055 for simplespansand 0.05for continuousspans.


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FIGURE 10.14: Precast, prestressed AASHTO (American Association of State Highway and Trans-portation Ofcials) I-beam sections. (From Federal Highway Administration, Standard Plans for Highway Bridges, Vol. 1, Concrete Superstructures, U.S. Department of Transportation, Washington,D.C., 1990. With permission.)

3. BoxGirderFigure 10.15 showsFHWA [ 6] standard types of precast box sections. The shape of acast-in-place, prestressed concrete box girder is similar to the conventional reinforcedconcrete box girder (Figure 10.12 c). Thespacing of thegirderscan betaken astwicethestructural depth. It isused mostly for spansof 100to 600 ft (30.5to 182.9m). Structuraldepth-to-span ratios are 0.045 for simple spans and 0.04 for continuousspans. These


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sectionsareused frequently for simplespansof over 100 ft (30.5 m) and areparticularlysuitable for widening in order to control deections. About 70 to 80% of California’shighway bridgesystem is composed of prestressed concrete box girder bridges.

FIGURE10.15: Federal HighwayAdmini strati on (FHWA) precast, pretensionedboxsections. (FromFederalHighwayAdministration, Standard Plansfor HighwayBri dges, Vol. 1, ConcreteSuperstr uctures,U.S. Department of Transportation, Washington, D.C., 1990. With permission.)

4. Segmental BridgeThesegmentally constructed bridgeshavebeen successfully developedby combining theconcepts of prestressing, box girder, and the cantilever construction [ 2, 20]. The rstprestressed segmental boxgirder bridgewasbuilt in Western Europein 1950. Cali fornia’sPine Valley Bridge, as shown in Figure 10.16 (composed of three spans of 340 ft [103.6m], 450 ft [137.2m], and 380 ft [115.8ft] with thepier height of 340 ft [103.6m]), wastherst cast-in-placesegmental bridgebuilt in theU.S., in 1974.

Theprestressedsegmental bridgeswith precast or cast-in-placesegmental can beclassiedby the construction methods: (1) balanced cantilever, (2) span-by-span, (3) incremen-tal launching, and (4) progressive placement. The selection between cast- in-place andprecast segmental, and among various construction methods, is dependent on projectfeatures, siteconditions, environmental and public constraints, construction timefor theproject, and equipment available. Table 10.2 lists the range of application of segmentalbridgesbyspan lengths[ 20].


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FIGURE10.16:a PineValley Bridge, California. Construction state. (From CaliforniaDepartment of Transportation. With permission.)

FIGURE 10.16:b Pine Valley Bridge, California. Construction completed. (From CaliforniaDepart-ment of Transportation. With permission.)


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FIGURE10.17: A anged section at nominal moment capacity state.

TABLE10.2 Rangeof Application of Segmental BridgeTypeby Span LengthSpanft (m) Bridgetypes

0–150 (0–45.7) I-typepretensioned girder100–300 (30.5–91.4) Cast-in-placepost-tensioned box girder100–300 (30.5–91.4) Precast-balanced canti lever segmental, constant depth200–600 (61.0–182.9) Precast-balanced canti lever segmental, vari abledepth200–1000 (61.0–304.8) Cast-i n-placecanti lever segmental800–1500 (243.8–457.2) Cable-staywith balanced canti lever segmental

5. Design ConsiderationComparedtoreinforcedconcrete,themain design featuresof prestressedconcretearethatstressesfor concreteand prestressingsteel and deformationof structuresat eachstage(i.e.,during construction, stressing, handling, transportation, and erection aswell asduringtheservicelife)andstressconcentrationsneedto beinvesti gated. In thefollowing,weshallbriey discuss the AASHTO-LRFD [ 1] requirements for stress limits, nominal exuralresistance, and shear resistancein designing aprestressed member.

a) StressLimits Calculationsof stressesfor concrete and prestressing steel are based mainly on theelasti c theory.

Tables 10.3 to 10.5 li st the AASHTO-LRFD [ 1] stresslimits for concrete and prestressing tendons.b) Nominal Flexural Resistance, M n

Flexural strength isbasedontheassumptionsthat (1) thestrain isl inearlydistr ibutedacrossacross-section (except for deep exural member); (2) the maximum usable strain at extreme compressiveber is equal to 0.003; (3) the tensile strength of concrete is neglected; and (4) a concrete stress of 0.85 f c is uniformly distr ibuted over an equivalent compression zone. For a member with aangedsection (Figure 10.17 ) subjected to uniaxial bending, theequationsof equilibrium areused to giveanominal moment resistanceof:

M n = Aps f ps d p −a2 +As f y (d s −

a2

)

−As f y d s −a2 +0.85f c (b −bw )β 1h f

a2 −

h f

2(10.1)


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TABLE10.3 StressLimit sfor Prestressing TendonsPrestressing tendon type

Stress-relievedstrand and plain Deformed

Stress Prestressing high-strength Low Relaxation high-strengthtype method bars strand bars

At jacking Pretensioning 0.72 f pu 0.78 f pu —(f pj ) Post-tensioning 0.76 f pu 0.80 f pu 0.75 f puAf ter Pr et ensi oni ng 0.70 f pu 0.74 f pu —

transfer Post-tensioning(f pt ) At anchorages

andcoupler s 0.70 f pu 0.70 f pu 0.66 f puimmediatelyafter anchor set

General 0.70 f pu 0.74 f pu 0.66 f pu

At service After all losseslimit 0.80f py 0.80 f py 0.80 f py

state (f pe )

FromAmerican Associationof StateHighwayandTransportation Ofcials, AASHTO LRFD Bridge Design Specicati ons, First Edition, Washington, D.C.,1994. With permission.

a = βc (10.2)

c =Aps f pu +A s f y −A s f y −0.85β 1f c (b −bw )h f

0.85β 1f c bw +kA psf pud p

(10.3)

f ps = f pu 1 −kc

d p(10.4)

k = 2 1.04 −f py

f pu(10.5)

where A represents area; f is stress; b is the width of the compression face of member; bw isthe web width of a section; h f is the compression ange depth of a cross-section; d p and d s aredistancesfrom extremecompression ber to the centroid of prestressing tendonsand to centroid of tension reinforcement, respectively; subscripts c and y indicate specied strength for concrete andsteel, respectively; subscripts p and s signify prestressing steel and reinforcement steel, respectively;subscripts ps,py , and pu correspond to states of nominal moment capacity, yield, and speciedtensile strength of prestressing steel, respectively; superscript prime ( ) representscompression; andβ 1 is the concrete stress block factor, equal to 0.85 f c ≤ 4000 psi and 0.05 less for each 1000 psiof f c in excess of 4000 psi, and minimum β 1 =0.65. The aboveequations also can be used for arectangular section in which bw =b istaken.

Maximum reinforcement limit:

cd e ≤ 0.42 (10.6)

d e

=

Aps f ps d p +A s f y d s

Aps f ps +A s f y(10.7)

Minimum reinforcement limit:

φM n ≥1.2M cr (10.8)

in which φ istheexural resistancefactor 1.0for prestressed concreteand 0.9for reinforced concrete,and M cr is thecracking moment strength given bytheelastic stressdistri bution and themodulusof ruptureof concrete.


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TABLE10.4 Temporary Concrete StressLimits at Jacking StateBefore LossesDueto Creep andShrinkage—Fully PrestressedComponents

Stress Stresstype Areaand condition ksi (MPa)

Compressive Pretensioned 0.60 f ci

Post-tensioned 0.55 f ci

Precompressed tensilezonewithout bonded reinforcement N/AArea other than the precompressed tensil e zones and withoutbonded auxil iary reinforcement

0.0948 f ci ≤0.2

0.25 f ci ≤1.38Tensile

Nonsegmentalbridges

Areawith bonded reinforcement whichissufcient to resist 120%of the tension forcei n the cracked concretecomputed

0.22 f ci

on the basis of uncracked section 0.58 f ci

Handling stressesin prestressedpil es 0.158 f ci

0.415 f ci

TypeAjointswit hminimum bondedaux-ili ary reinforcement through the

0.0948 f ci max. tension

Longitudinal stressthrough joint inprecompressed

joints which is sufcient to carry thecal-culated tensil e forceat a stressof 0.5 f ywith internal tendons

(0.25

f

cimax. tension)

tensilezone Type A joints without the minimumbonded auxil iary reinforcement throughthej ointswi th internal tendons

No tension

TypeB with external tendons 0.2min. compression(1.38mi n. compression)

Segment al Tr ansverse st ress For any t ype of j oi nt 0.0948 f c max. tensionbridges through joints (0.25 f c max. tension)

Without bonded non-prestressed rein-forcement

No tension

Other area Bonded reinforcement is sufcient tocarry thecalculated tensile forcein the

0.19 f ci

concrete on the assumption of an un-cracked secti on at astressof 0.5 f sy

(0.50 f ci )

Note: TypeA joints are cast-in-place joints of wet concrete and/or epoxy between precast units. TypeB joints are dry j ointsbetween precast units.

From American Association of StateHighway and Transport ation Of cials, AASHT O LRFD Br idgeD esign Specications, FirstEdition, Washington, D.C.,1994. With permission.

c) Nominal Shear Resistance, V nThenominal shear resistanceshall bedetermined by the following formulas:

V n = the lesser of V c +V s +V p0.25 f c bν d ν +V p

(10.9)

where

V c =0.0316 β f c bν d ν (ksi)0.083 β

f c bν d ν (MPa)

(10.10)

V s = Aν f y d ν (cos θ +cos α) sin αs

(10.11)

where bν is the effective web width determined by subtracting the diameters of ungrouted ductsor one-half the diameters of grouted ducts; d ν is the effective depth between the resultants of thetensile and compressive forces due to exure, but not less than the greater of 0.9 d e or 0.72 h ; Aνis the area of t ransverse reinforcement within distance s ; s is the spacing of the stirrups; α istheangleof inclination of transversereinforcement to the longitudinal axis; β is a factor indicating the


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TABLE10.5 ConcreteStressLimitsat Servi ceLimit State After All Losses—Fully PrestressedComponents

Stress Stresstype Areaand condition ksi (MPa)

Nonsegmental bridge at servicestate 0.45 f c

Compressive Nonsegmental br idgeduringshippingandhandling 0.60 f c

Segment al br idge dur ing shi ppi ng and handl ing 0.45 f c

Withbondedprestressingtendons 0.19 f cother than piles (0.50 f c )

Precompressed Subjected to severecorrosiveTensile tensilezoneassuming

uncracked secti onsconditions 0.0948 f c

Nonsegmental 0.25 f cbridges With unbonded prestressing tendon No tension

TypeAjointswit hminimum bondedaux-ili ary reinforcement through the joint swhich is sufcient to carry thecalculatedtensile forceat a stressof 0.5 f y with in-ternal tendons

0.0948 f c(0.25 f c )

Longitudinal stressi nprecompressed tensilezone

Type A joints without the minimumbonded auxil iary reinforcement throughthejoints

No tension

TypeB with external tendons 0.2min. compression(1.38mi n. compression)

Segmentalbridges

Transversestressinprecompressed tensilezone

For any typeof joint 0.0948 f c

0.25 f cTypeA joint without minimum bondedauxiliary reinforcement through joints

No tension

Other area(without bondedreinforcement)

Bonded reinforcement is sufcient tocarry the calculated t ensile force in theconcrete on the assumption of an un-crackedsecti on at a stressof 0.5 f sy

0.19 f c0.50 f c

Note: TypeA joints are cast- in-placejoints of wet concrete and/or epoxybetween precast units. TypeB joints are dry jointsbetween precast units.

FromAmerican Association of StateHighway and Transportation Ofcials, AASHT O LRFD Bri dgeDesign Specications, FirstEdition, Washington, D.C.,1994. With permi ssion.

abilityof diagonallycracked concreteto transmit tension; and θ istheangleof inclinationof diagonalcompressivestresses(Figure 10.18 ). Thevaluesof β and θ for sectionswith transversereinforcementare given in Table 10.6 . In this table, the shear stress, ν , and strain, εx , in thereinforcement on theexural tension side of the member aredetermined by:

ν =V u −φV p

φb ν d ν(10.12)

εx =M ud ν +0.5N u +0.5V u cot θ −Aps f po

E s A s +E p Aps ≤ 0.002 (10.13)

where M u and N u are the factored moment and axial force (taken as positive if compressive),respectively, associated with V u , and f po is the stress in prestressing steel when the stress in thesurrounding concrete is zero and can beconservatively taken as the effective stressafter losses, f pe .When thevalueof εx calculated fromtheaboveequation isnegative, it sabsolutevalueshall bereduced


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FIGURE10.18: I llustrationof Ac for shear strengthcalculation. (FromAmeri can Associationof StateHighway and Transportation Ofcials, AASHTO LRFD Bri dge Design Specicati ons, First Edition,Washington, D.C., 1994. With permission.)

TABLE10.6 Valuesof θ and β for Sectionswith TransverseReinforcement

Angle εx ×1000ν

f c(degree) −0.2 −0.15 −0.1 0 0.125 0.25 0.50 0.75 1.00 1.50 2.00

≤0.05 θ 27.0 27.0 27.0 27.0 27.0 28.5 29.0 33.0 36.0 41.0 43.0β 6.78 6.17 5.63 4.88 3.99 3.49 2.51 2.37 2.23 1.95 1.72

0.075 θ 27.0 27.0 27.0 27.0 27.0 27.5 30.0 33.5 36.0 40.0 42.0β 6.78 6.17 5.63 4.88 3.65 3.01 2.47 2.33 2.16 1.90 1.65

0.100 θ 23.5 23.5 23.5 23.5 24.0 26.5 30.5 34.0 36.0 38.0 39.0β 6.50 5.87 5.31 3.26 2.61 2.54 2.41 2.28 2.09 1.72 1.45

0.125 θ 20.0 21.0 22.0 23.5 26.0 28.0 31.5 34.0 36.0 37.0 38.0β 2.71 2.71 2.71 2.60 2.57 2.50 2.37 2.18 2.01 1.60 1.35

0.150 θ 22.0 22.5 23.5 25.0 27.0 29.0 32.0 34.0 36.0 36.5 37.0β 2.66 2.61 2.61 2.55 2.50 2.45 2.28 2.06 1.93 1.50 1.24

0.175 θ 23.5 24.0 25.0 26.5 28.0 30.0 32.5 34.0 35.0 35.5 36.0β 2.59 2.58 2.54 2.50 2.41 2.39 2.20 1.95 1.74 1.35 1.11

0.200 θ 25.0 25.5 26.5 27.5 29.0 31.0 33.0 34.0 34.5 35.0 36.0

β 2.55 2.49 2.48 2.45 2.37 2.33 2.10 1.82 1.58 1.21 1.000.225 θ 26.5 27.0 27.5 29.0 30.5 32.0 33.0 34.0 34.5 36.5 39.0β 2.45 2.38 2.43 2.37 2.33 2.27 1.92 1.67 1.43 1.18 1.14

0.250 θ 28.0 28.5 29.0 30.0 31.0 32.0 33.0 34.0 35.5 38.5 41.5β 2.36 2.32 2.36 2.30 2.28 2.01 1.64 1.52 1.40 1.30 1.25

From American Association of State Highway and Transport ation Ofcials, AASHT O LRFD Br idge Design Specications,First Edition, Washington, D.C.,1994. With permi ssion.

bymultiplyingby thefactor F ε , taken as:

F ε =E s As +E p Aps

E c Ac +E s As +E p Aps(10.14)

where E s , E p , and E c are modulesof elasticity for reinforcement, prestressing steel, and concrete,respectively, and Ac is the area of concrete on the exural tension side of themember, asshown inFigure 10.18 .

Minimum transversereinforcement:

Aν min =0.0316 f c

bν S f y

(ksi)0.083 f c

bν S f y

(MPa)(10.15)


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Maximum spacing of t ransversereinforcement:

For V u < 0.1f c bν d ν smax = thesmaller of 0.8d ν24 in. (600 mm )

(10.16)

For V u ≥ 0.1f c bν d ν smax = thesmaller of 0.4d ν12 in. (300 mm )

(10.17)

10.4 ConcreteSubstructures

10.4.1 Introduction

Bridgesubstructurestransfer trafc loadsfrom the superstructureto the footingsand foundations.Vert ical intermediatesupports (piersor bents) and end supports (abutments) are included.

10.4.2 BentsandPiers

1. PileBentsPileextension, asshownin Figure 10.19 a, isused for slabandT-beam bridges. It isusuallyused to crossstreamswhen debrisis not aproblem.

FIGURE10.19: Bridgesubstructures—piersand bents. (From Cali forniaDepartment of Transporta-tion, Bridge Design Aids Manual, Sacramento, CA, 1990. With permission.)

2. Solid PiersFigure 10.19 b showsatypical solid pier, used mostly when stream debrisor fast currentsare present. These are used for long spans and can be supported by spread footingsorpilefoundations.

3. Column Bents


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Column bents(Figure 10.19 c) aregenerally usedondry landstructuresandaresupportedby spread footingsor pile foundations. Mult i-column bents are desirable for bridges inseismiczones. Thesingle-column bent, suchasaT bent (Figure 10.19 d),modied T bent(Cbent) (Figure 10.19 e), oroutrigger bent (Figure 10.19 f ), maybeusedwhen thelocation

of the columns is restricted and changes of the alignment are impossible. To achieve apleasing appearance at the minimum cost using standard column shapes, Calt rans [ 3]developed“standard architectural columns” (Figure 10.20 ). Prismaticsectionsof columntypes1 and 1W, with one-way aresof column types2 and 2W, and with two-way aresof column types3 and 3W may beused for varioushighway bridges.

10.4.3 Abutments

Abutments are the end supports of a bridge. Figure 10.21 shows the typical abutments used forhighway bridges. Theseven typesof abutmentscan bedivided into two categories: open and closedends. Selection of an abutment typedependson therequirementsfor structural support, movement,drainage, road approach, and earthquakes.

1. Open-End AbutmentsOpen-endabutmentsincludediaphragm abutmentsandshort-seat abutments. Thesearethemost frequentlyused abutmentsand areusually themost economical, adaptable, andattractive. The basic structural differencebetween the two types is that seat abutmentspermit thesuperstructureto moveindependentlyfrom theabutment whilethediaphragmabutment does not. Since open-end abutments have lower abutment walls, there islesssett lement in theroad approachesthan that experienced by higher backlled closedabutments. They also providefor moreeconomical widening than closed abutments.

2. Closed-End AbutmentsClosed-end abutments include canti lever, strutted, rigid frame, bin, and closure abut-ments. Thesearelesscommonlyused, but for bridgewideningsof thesamekind, unusualsites, or in tightly constrainedurban locations. Rigid frameabutmentsaregenerally usedwith tunnel-typesingle-span connectors and overhead structureswhich permit passagethrough a roadway embankment. Becausethe structural supports areadjacent to trafcthesehavea high ini tial cost and present aclosed appearanceto approaching trafc.

10.4.4 Design Consideration

After the recent 1989 Loma Prieta and the 1994 Northridge Earthquakes in the U.S. and the 1995Kobeearthquakein Japan, major damageswerefound in substructures. Special attention, therefore,must bepaid to seismic effects and thedetailing of theducti le structures. Boundary condit ionsandsoil–foundation–structureinteraction in seismic analysesshould also becarefully considered.

10.5 Floor System

10.5.1 Introduction

Theoor system of abridgeusually consistsof adeck, oor beams, and str ingers. Thedeck directlysupports the liveload. Floor beamsaswell asstr ingers, shown in Figure 10.22 , form a gri llageandtransmit the load from the deck to the main girders. The oor beams and stringers are used forframed bridges, i.e., truss, rahmen, and arch bridges(seeFigures 10.40 , 10.45 , and 10.47 ), in whichthe spacing of the main girders or trusses is large. In an upper deck type of plate girder bridge the


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FIGURE10.20: Caltrans(CaliforniaDepartment of Transportation) standard architectural columns.(FromCaliforniaDepartment of Transportation, BridgeD esign AidsManual, Sacramento, CA, 1990.With permission.)

deck is directly supported by the main girders, and often there is no oor system because the maingirders run in parallel and closetogether.

Theoor systemisclassiedassuitablefor either highwayor railroadbridges. Thedeckof ahighwaybridgei sdesigned for thewheel loadsof trucksusingplatebending theory in two dimensions. Oftenin design practice, however, thisplate theory is reduced to equivalent one-dimensional beam theory.Thematerials used arealso classied into concrete, steel, or wood.

The recent inux of trafc ow has severely fatigued existing oor systems. Cracksin concrete


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FIGURE10.21: Typical typesof abutments. (FromCaliforniaDepartment of Transportation, Bridge Design Ai dsM anual, Sacramento, CA, 1990. With permission.)

decks and connections of oor system are often found in old bridges that havebeen in service formany years.

10.5.2 Decks

1. ConcreteDeckA reinforced concrete deck slab is most commonly used in highway bridges. It is the

deck that is most susceptible to damagecaused by the ow of trafc,which continuestoincrease. Urban highwaysare exposed to heavy trafc and must berepaired frequently.Recently, a composite deck slab was developed to increase the strength, ducti li ty, anddurability of deckswithout increasing their weight or affecting the cost and duration of construction. In a composite slab, the bottom steel plate serves both as a part of theslab and the formwork for pouring the concrete. There are many waysof combiningthe steel plate and thereinforcement. A typical example is shown in Figure 10.23 . Thisslab i s prefabricated in the yard and then the concrete is poured on site after girders


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FIGURE10.22: Floor system. (From Nagai, N., Bri dge Engineeri ng, Kyoritsu Publishing Co., Tokyo,Japan [in Japanese] , 1994. With permission.)

havebeen placed. A precast, prestressed deck may reducethe timerequired to completeconstruction.

FIGURE10.23: Compositedeck. (From Japan Association of Steel BridgeConstruction, Planningof Steel Br idges, Tokyo [in Japanese] , 1988. With permission.)

2. Steel DeckFor longspans, the steel deck isused to minimizethe weight of the deck. Thesteel deckplate is stiffened with longitudinal and transverse ribsas shown in Figure 10.24 . Thesteel deck also works as the upper ange of the supporting girders. The pavement on


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thesteel deckshould becarefully nished to prevent water from penetrating through thepavement and causing the steel deck to rust.

FIGURE10.24: Steel plate deck. (From Japan Association of Steel Bridge Construction, Outlineof Steel Br idges, Tokyo [in Japanese] , 1985. With permission.)

10.5.3 Pavement

Thepavement on thedeck providesa smooth driving surfaceand prevents rain water from seepinginto thereinforcingbarsand steel deckbelow. A layer of waterproong may beinserted between the

pavement and the deck. Asphalt is most commonly used to pave highway bridges. Its thickness isusually 5 to 10 cm on highwaysand 2 to 3 cm on pedestrian bridges.

10.5.4 Stringers

The stringers support the deck directly and transmit the loads to oor beams, as can be seen inFigure 10.22 . They are placed in the longitudinal direction just l ike the main girders are in a plategirder bridgeand thusprovidemuch thesamekind of support.

Thestringersmust besufciently stiff in bendingto prevent cracksfrom formingin thedeck or onthe pavement surface. Thedesign codesusually limit thevert ical displacement caused by the weightof atruck.

10.5.5 Floor BeamsThe oor beams are placed in the transverse direction and connected by high-tension bolts to thetruss frameor arch, as shown in Figure 10.22 . Theoor beamssupport the str ingers and transmitthe loadsto main girders, trusses, or arches. In other words, themain trussor arch receivestheloadsindirectly via the oor beams. Theoor beamsalso providetransversestiffnessto bridgesand thusimprovethe overall torsional resistance.


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10.6 Bearings, Expansion Joints, andRailings

10.6.1 Introduction

Aside from the main components, such as the girders or the oor structure, other parts such asbearings(shoes), expansion joints, guardrailings, drainagepaths, lighting,and sound-proongwallsalso makeup the structureof a bridge. Each playsa minor part but providesan essential function.Drainsush rain water off and wash away dust. Guardrailingsand lightsadd to theaesthetic quali tyof thedesign aswell asproviding their obviousoriginal functions. A sound-proong wall may takeawayfrom thebeauty of thestructurebut might berequired bylawin urban areasto isolatethesoundof trafc from thesurrounding residents. In the following section, bearings, expansion joints, andguardrailingsare discussed.

10.6.2 Bearings (Shoes)

Bearingssupport the superstructure(themain girders, trusses, or arches) and transmit the loadstothe substructure (abutments or piers). The bearings connect the upper and lower structures andcarry thewholeweight of thesuperstructure. Thebearingsaredesigned to resist thesereaction forcesby providing support conditions that are xed or hinged. The hinged bearings may bemovableorimmovable;horizontal movement isrestrained or unrestrained, i.e., horizontal reaction isproducedor not. Theamount of thehorizontal movement isdetermined bycalculating theelongation duetoa temperature change.

Many bearingswere found to havesustained extensivedamageduring the 1995KobeEarthquakein Japan, dueto stressconcentrations, which are the weak spotsalong the bridge. Thebearingsmayplay the roleof a fuse to keep damage from occurring at vital sectionsof the bridge, but therisk of the superstructure falling down goes up. The girder-to-girder or girder-to-abutment connectionsprevent the girders from collapsing during strongearthquakes.

Many typesof bearingsareavailable. Someareshown in Figure 10.25 and briey explained in thefollowing:

Line bearings: The contacting l ine between the upper plate and the bottom round surfaceprovidesrotational capability aswell assliding. Theseareused in small bridges.

Platebearings: Thebearingplatehasa planesurfaceon thetop sidewhich allowsslidingand aspherical surfaceon thebottom allowingrotation. Theplateisplaced between theupperand lower shoes.

Hinged bearings(pin bearings): A pin is inserted between theupper and lower shoesallowingrotation but no translation in longitudinal direction.

Roller bearings: Lateral translation isunrestrainedbyusingsingleor multiplerollersfor hingedbearingsor spherical bearings.

Spherical bearings(pivot bearings): Convex and concave spherical surfaces allow rotation inall directions and no lateral movement. The two types are: a point contact for largedifferencesin the radii of each sphereand asurfacecontact for small differencesin theirradii.

Pendel bearings: An eyebar connectsthe superstructureand the substructureby apin at eachend. Longitudinal movement ispermitted byinclining theeyebar; therefore, thedistanceof the pins at endsshould beproperly determined. Theseareused to provideanegativereaction in cable-stayed bridges. There is no resistancein the transversedirection.

Wind bearings: This type of bearing providestransverseresistancefor wind and is often usedwith pendel bearings.


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FIGURE 10.25: Typesof bearings. (From Japan Association of Steel BridgeConstruction, A Guide Book of Bearing Design for Steel Bridges, Tokyo [in Japanese] , 1984. With permission.)

Elastomeric bearings: Theexibil ityof elastomericor lead rubber bearingsallowsboth rotationand horizontal movement. Figure 10.26 explains a principle of rubber-layered bearingsby comparing with a unit rubber. A layered rubber is stiff, unlike a unit rubber, forvertical compressionbecausethesteel platesplacedbetweentherubber restrain theverticaldeformation of therubber, but exible for horizontal shear forcelike aunit rubber. Theexibili ty absorbs horizontal seismic energy and is ideally suited to resist earthquakeactions. Since the disaster of the 1995 KobeEarthquake in Japan, elastomeric rubberbearingshavebecomemoreandmorepopular,but whether theyeffectively sustain severevert ical actionswi thout damageis not cert ied.

Oil damper bearings: Theoil damper bearingsmoveunder slow actions(such astemperaturechanges) but do not move under quick movements (such as those of an earthquake).They areused in continuousspan bridgesto distr ibute seismic forces.


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as they pass over the junction. To solve this problem, rubber joints are used on the road surfaceto provide a smooth transit ion for modern bridge construction (see Figure 10.27 e), or continuousgirdersaremorecommonly adopted than simplegirders.

FIGURE10.27: Typesof expansion joints. (From Japan Associati on of Steel BridgeConstruction, AGuideBook of Expansion Joint Design for Steel Bridges, Tokyo[in Japanese] , 1984. With permission.)

10.6.4 Railings

Guardrailingsareprovided to ensurevehiclesandpedestr iansdo not fall off thebridge. They may beahandrail for pedestr ians, aheavier guard for vehicles, or acommon railingfor both. Thesearemadefrom materials such asconcrete, steel, or aluminum. Theguardrailingsare located prominently andare thusopen to thecrit ical eyeof thepublic. It is important that they not only keep trafc withinboundariesbut also add to the aesthetic appeal of thewholebridge(Figure 10.28 ).


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FIGURE10.28: Pedestr ian railing. (From Japan Association of Steel BridgeConstruction, Outlineof Steel Br idges, Tokyo [in Japanese] , 1985. With permission.)

10.7 Girder Bridges

10.7.1 Structural Features

Girder bridges are structurally the simplest and the most common. They consist of a oor slab,girders, and thebearingswhich support and transmit gravity loadsto thesubstructure. Girdersresistbending moments and shear forces and are used to span short distances. Girders are classied bymaterial into steel plate and box girders, reinforced or prestressed concrete T-beams, and compositegirders. The box girder is also used often for prestressed concrete continuous bridges. The steelgirder bridgesareexplained in thissection; theconcrete bridgesweredescribed in Section 10.3 .

Figure 10.29 showsthestructural composit ion of plateand box girder bridgesandtheload transferpath. In plate girder bridges, the live load i s directly supported by the slab and then by the maingirders. In box girder bridges the forcesare taken rst by the slab, then supported by the str ingers


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and oor beamsin conjunction with themain box girders, and nally taken to thesubstructureandfoundation through the bearings.

FIGURE10.29: Steel girder bridges. (From Nagai, N., Bridge Engineeri ng, Kyoritsu Publishing Co.,Tokyo, Japan [in Japanese] , 1994. With permission.)

Girdersareclassied asnoncompositeor composite, that is,whether thesteel girdersact in tandemwith the concrete slab (using shear connectors) or not. Sincecompositegirdersmakeuseof thebestpropert iesofboth steel and concrete, theyareoften therational andeconomicchoice. LessfrequentlyH or I shapesareused for themain girders in short-span noncompositebridges.

10.7.2 PlateGirder (Noncomposite)

Theplate girder is the most economical shapedesigned to resist bending and shear; the moment of inertia is greatest for a relatively low weight per unit length. Figure 10.30 showsa plan of atypicalplategirder bridgewith four main girdersspanning30m and awidth of 8.5 m.

Thegravity loadsaresupported byseveral main plategirders, each manufactured byweldingthree


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FIGURE10.30: General plans of a typical plate girder bridge. (From Tachibana, Y. and Nakai, H.,Bri dgeEngineeri ng, Kyoritsu Publishing Co., Tokyo, Japan [ in Japanese] , 1996. With permission.)

plates: an upper and lower ange and a web. Figure 10.31 shows a block of plate girder and it sfabrication process. The web and the anges are cut from steel plate and welded. The block isfabricated in theshop and transported to theconstruction sitefor erection.

Thedesign procedurefor plategirders, primarily thesizingof the threeplates, is asfollows:

1. Web height: The web height i s the fundamental design factor affecting the weight andcost of the bridge. If the height is too small, the anges need to be largeand the deadweight increases. The height (h) is determined empirically by dividing the span length(L) bya“reasonable” factor. Common ratiosare h/L =1/18to 1/20for highway bridgesand ali tt lesmaller for railway bridges. Theweb height also inuencesthestiffnessof the


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FIGURE10.31: Fabrication of plate girder block.

bridge. Greater heightsgenerally producegreater sti ffness. However, if theheight is toogreat, the web becomesunstable and must havei ts thicknesssupplemented or sti ffenersadded. Thesemeasuresincrease the weight and the cost. In additi on, plategirders withexcessively deep web and small angesare liable to bucklelaterally.

2. Web thickness: Theweb primari ly resistsshear forces, which are not usually signicantwhen the web height is properly designed. The shear force is generally assumed to bedistr ibuteduniformly acrosstheweb instead of using theexact equation of beam theory.The web thickness (t) is determined such that thinner is better as long as buckling i sprevented. Sincetheweb doesnot contribute much to thebending resistance, thin websare most economical but the possibi li ty of buckling increases. Therefore, the web i susually stiffened by hori zontal and vert ical sti ffeners, which will bediscussed later (seeFigure 10.34 ). It is not primari ly strength but rather sti ffnessthat controlsthedesign of webs.

3. Areaofanges: After thesizesof webaredetermined, theangesaredesigned. Theangeswork mostly in bendingand therequired areais calculated using equilibrium condit ionsimposed on the internal and external bending moment. A selecti on of strength for thesteel material is principally madeat thisstagein thedesign process.

4. Width and thi ckness of anges: The width and thickness can be determined by ensur-ing that the area of the anges falls under the limiting width-to-thickness ratio, b/t (Figure 10.32 ), asspecied in design codes. If theangesaretoo thin (i.e., thewidth-to-thicknessratio istoo large), thecompression angemaybuckleor the tension angemaybedistorted bytheheat of welding. Thus, the thicknessof both angesmust bechecked.Sinceplate girdershavelit tletorsional resistance, special attention should bepaid to lat-eral torsional buckling. Toprevent this phenomenon, the compression angemust havesufcient width to resist “out-of-plane” bending. Figure 10.33 showsthelateral torsional

buckling that may occur bybending with respect to strongaxis.

After determining the member sizes, calculationsof the resisting moment capacity are made toensure code requirements are sati sed. If these fail, the above steps must be repeated until thespecicationsaremet.

A few other important factorsin thedesign of girder bridgeswill beexplained in thefollowing:

Design of web stiffeners: The horizontal and vert ical sti ffeners should be attached to theweb


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FIGURE10.32: Local buckling of compression ange.

FIGURE10.33: Lateral torsional buckling.

(Figure 10.34 ) when i t is relatively thin. Bending moment produces compression andtension in theweb,separated byaneutral axis. Thehorizontal sti ffener preventsbucklingdue to bending and is therefore attached to the compression side (the top half for asimply supported girder). Since the bending moment is largest near the midspan of asimply supported girder, the horizontal sti ffeners are usually located there. If the webis not too deep nor its thickness too small, no sti ffeners are necessary and fabricationcosts are reduced. Vert ical sti ffeners,on the other hand, prevent shear buckling,which isproduced bythetension and compression eldsin diagonal directions. Thecompressioneld causes shear buckling. Sincethe shear forceis largest near the supports, the mostvert ical sti ffenersareneeded there. Bearing sti ffeners,which aredesigned independently just as any other compression member would be, are also required at the supports tocombat largereaction forces. Buckling patterns of aweb are shown in Figure 10.34 .

Variable sections: Thevariable cross-sectionsmay beused to savematerial and cost where the

bending moment issmaller, that is, near theend of thespan (seeFigure 10.31 ). However,thisreduction increasesthe labor required for weldingand fabrication. Thecost of laborand material must bebalanced and traded off. In today’sindustrial climate, labor ismoreimportant and costly than thematerial. Therefore, thechangeof girder sectionisavoided.Likewise, thick plates are often specied to eliminate the number of sti ffeners needed,thus to reducethenecessary labor.


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FIGURE10.34: Buckling and sti ffeners of web.

10.7.3 Composite Girder

If two beamsaresimply laid oneupon theother, asshown in Figure 10.35 a, they act separately andonly share the load depending on their relativeexural stiffness. In thiscase, slip occurs along theboundary between thebeams. However, if the two beamsareconnected and slip prevented asshownin Figure 10.35 b, they act as a unit, i.e., a composite girder. For composite plate girder bridges,the steel girder and the concrete slab are joined by shear connectors. In thi sway, the concrete slabbecomesintegralwith thegirderandusuallytakesmost of thecompression component of thebendingmoment while the steel plategirder takes the tension. Composite girders are much more effectivethan thesimply tiered girder.

FIGURE 10.35: Principle of tiered beam and compositebeam. (From Tachibana, Y. and Nakai, H.,Bri dgeEngineeri ng, Kyoritsu Publishing Co., Tokyo, Japan [ in Japanese] , 1996. With permission.)

Let usconsider the two casesshown in Figure 10.35 and notethedifferencebetween tiered beamsand compositebeams. Both havethesamecross-sectionsand aresubjected to aconcentrated load atmidspan. Themoment of inert ia for thecompositebeam is four timesthat of thetiered beams, thus


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the resulting vert ical deection is one-fourth. Themaximum bending stressin theextreme(top orbottom) ber is half that of the tiered beam conguration.

The corresponding stress distr ibutions are shown in Figure 10.36 . Points “S” and “V” are thecenter of areaof thesteel section and the composite section, respectively. According to beam theory,the strain distr ibuti on is linear but the stressdistr ibuti on hasastep changeat theboundary betweenthe steel and concrete.

FIGURE10.36: Section of compositegirder. (From Tachibana, Y. and Nakai, H., BridgeEngineering,Kyoritsu Publishing Co., Tokyo, Japan [ in Japanese] , 1996. With permission.)

Threetypesof shear connectors—studs,horseshoes, and steel blocks—areshown in Figure 10.37 .Studsaremost commonly used sincethey areeasily welded to thecompression angeby theelectric

FIGURE10.37: Typesof shear connectors. (From Nagai, N., BridgeEngineering, Kyoritsu PublishingCo., Tokyo, Japan [in Japanese] , 1994. With permission.)

resistance welding, but the weld inspection is a cumbersometask. If the weld on a certain stud ispoor, thestud mayshear off and tr igger atotallyunforeseen failuremode. Other typesareconsideredto maintain morereliability.

Shear connectorsareneeded most near theendsof thespan, where the shear forcei slargest. Thisregion is illustrated in Figure 10.35 a, which showsthemaximum shift due to slip occursat the endsof tiered beams. It is thisslip that is restrained by theshear connectors.


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10.7.4 GrillageGirder

When girders are placed in a row and connected transversely by oor beams, the truck loads aredistributed by theoor beamsto the girders. This system iscalled agri llageof girders. If themain

girders are plate girders, no sti ffness in torsion is considered. On the other hand, box girders andconcretegirderscan beanalyzed assumingsti ffnessi savailableto resist torsion. Floor beamsincreasethe torsional resistanceof thewholestructural system of the bridge.

Let usconsider the structural system shown in Figure 10.38 ato observetheload distribution in agrillage system. This gri llage has threegirders with oneoor beam at midspan. In this case, there

FIGURE10.38: Grillagegirders. (From Tachibana, Y. and Nakai, H., Bridge Engineeri ng, KyoritsuPublishing Co., Tokyo, Japan [ in Japanese] , 1996. With permission.)

arethreenodal forcesat theintersectionsof thegirdersand theoor beam but only two equilibriumequations (V = 0 and M =0) . Thus, it becomesonedegreestaticallyindeterminate. If wedisconnectthe intersection between main girder B and theoor beam and apply apair of indeterminateforces,X , at point b, asshownin Figure 10.38 b, X can beobtained usingthecompatibility condition at point


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b . Oncetheforce, X , is found, the sectional forces in thegirders can becalculated. This structuralsystem is commonly applied to the practical design of plategirder bridges.

10.7.5 Box GirderStructural conguration of box girders is il lustrated in Figure 10.39 . Sincethe box girder is aclosedsection, its resistance to torsion is high with no loss of strength in bending and shear. On theother hand, plate girdersare open sections generally only considered effective in resisting bendingand shear. Steel plates with longitudinal and transversesti ffeners are often used for deckson boxgirder or thin-walled structuresinstead of a concreteslab(Figure 10.39 b) although aconcreteslabispermissible.

FIGURE10.39: Box girders. (From Nagai, N., Bri dge Engineering, Kyori tsu Publi shing Co., Tokyo,Japan [in Japanese] , 1994. With permission.)

Torsion isresisted in two parts: pure torsion (St. Venant torsion) and warping torsion. Thepure

torsional resistanceof I-plategirders is negligible. However, for closed sectionssuch asabox girder,the pure torsional resistance is considerable, making them part icularly suited for curved bridgesorlong-span bridges. On the other hand, the warping torsion for box sections is negligible. The I-section girder hassomewarping resistancebut it is not largecompared to thepuretorsion of closedsections.

10.8 TrussBridges


Thestructural layout of a trussbridge is shown in Figure 10.40 for a through bridge with thedecklocatedat thelevel of lower chords. Theoor slab, whichcarriestheliveload, issupportedbytheoorsystem of str ingersandcrossbeams. Theload istransmitted to themain trussesat nodal connections,oneon each side of thebridge, through the oor system and nally to the bearings. Lateral braces,which also are a truss frame, are attached to the upper and lower chords to resist horizontal forcessuch aswind and earthquakeloadsaswell as torsional moments. Theportal frameat the entranceprovidestransit ion of horizontal forcesfrom the upper chordsto thesubstructure.

Truss bridges can take the form of a deck bridgeas well as a through bridge. In this case, theconcrete slab is mounted on the upper chordsand the sway bracing is placed between the vert icalmembersof two main trussesto providelateral stabil ity.


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FIGURE10.40: Trussbridge. (From Nagai, N., Bri dge Engineering, Kyoritsu Publi shing Co., Tokyo,Japan [in Japanese] , 1994. With permission.)

A trussis composed of upper and lower chords, joined by diagonal and vert ical members (web

members). Thisframeaction correspondstobeam action in that theupper and lower chordsperformlikeangesand thediagonal bracesbehavein much thesameway as the web plate. Thechordsaremainly in chargeof bending moment while the web members take the shear force. Trusses are anassembly of bars, not plates, and thusarecomparatively easier to erect on siteandareoftenthechoicefor long bridges.

10.8.2 Types of Trusses

Figure 10.41 showssometypical trusses. AWarren trussisthemost commonand isaframecomposedof isoscelestriangles,wherethewebmembersareeither in compressionor tension. Thewebmembersof a Pratt truss are vert ical and diagonal members where the diagonals are inclined toward thecenter and resist only tension. The Pratt t russ is suitable for steel bridges since it is tension that ismost effectively resisted. It should be noted, however, that vert ical members of Pratt truss are incompression. A Howetruss is simi lar to the Pratt except that thediagonals are inclined toward theends, leading to axial compression forces, and the vertical members resist tension. Wooden bridgesoften makeuseof theHowetrusssincethe connectionsof thediagonals in wood tend to compress.A K-truss, so named sincetheweb membersform a“K”, ismost economical in largebridgesbecausethe short member lengths reducethe risk of buckling.

10.8.3 Structural Analysis andSecondary Stress

The truss is a framed structureof bars, theoretically connected by hinges, forming stable tr iangles.Trusses contain tr iangle framed units to keep it stable. Its members are assumed to resist onlytensile or compressiveaxial forces. A statically determinate trusscan beanalyzed using equil ibriumconditionsonly. If more than theleast number of members required for stability are provided, thetrussbecomesindeterminate and can no longer besolved using only the conditionsof equili brium.Thedisplacement compatibilityshouldbeadded. An internallyand/or externally indeterminatet russisbest solved using computer software.

In practi ce, trussmembersareconnected to gusset plateswith high-tensionbolts(seeFigure 10.42 ),not rotation-freehinges, simply becausethesearemucheasier to fabricate. The“ pinned” condit ionof theory isnot reectedin theeld. Thisdiscrepancyresults in “secondary stresses” (bendingstresses)in themembers. Secondary stressesaregiven byacomputer analysisof arigid frameand areusuallyfound to belessthan 20% of theprimary (axial) stresses. I f thetrussmembersareproperly designed,


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FIGURE10.41: Typesof trusses.

FIGURE10.42: Nodal jointsof at russbridge. (FromJapan Association of Steel BridgeConstruction,Out li ne of Steel Br idges, Tokyo [in Japanese] , 1985. With permission.)

that is, theslendernessratiosof the trussbarsaresufciently largewith no buckling, then secondarystressescan conveniently and reliably bedisregarded.

10.8.4 Gerber Truss BridgeFigure 10.43 is aphoto of aGerber trussbridgeduring theerection of thecentral part, which is theMinato Oh-Hashi in Japan. Itsplan view is shown in Figure 10.2 . A Gerber trusshas intermediatehingesbetween thesupportsto createastaticallydeterminatestructural system. In thecaseof MinatoOh-Hashi, thesoil condition at thebottom of theharbor wasfoundto benot stiff and solid; thustheGerber trussproved the wisest choice.


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FIGURE 10.43: Lifting erection of the Minato Oh-Hashi, Japan. (Gerber bridge, 1974). (FromHanshin Expressway Public Corporation, Techno Gallery, Osaka, Japan, 1994. With permission.)

10.9 RigidFrameBridges (RahmenBridges)


Themembers are rigidly connected in “rahmen” structuresor “r igid frames”. Unlike the trussandthe arch bridges, which will bediscussed in the following subsection, all themembersare subjectedto both an axial forceand bendingmoments. Figure 10.44 showsvarioustypesof rahmen bridges.

Themembersof arigid framebridgearemuchlarger thanthosein atypical building. Consequently,stressconcentrationsoccur at the junctionsof beamsand columnswhich must becarefully designedusing ni teelement analysesor experimental verication. Thesupportsof rahmen bridgesareeitherhinged or xed, makingit an externally indeterminatestructure, and it isthereforenot suitablewhenthefoundation is likely to sink. Thereactionsat supportsarehorizontal andvert ical forcesat hinges,with theaddition of abendingmoment at axed base.

10.9.2 Portal Frame

A portal frame is the simplest design (Figure 10.44 a) and is widely used for the piers of elevatedhighway bridgesbecausethespaceunderneath can beeffectively used for other roadsor parkinglots.Thesepierswereproved, in the1995KobeEarthquakein Japan, to bemoreresil ient, that is, to retainmorestrength and absorb moreenergy than single-legged piers.

10.9.3 π -Rahmen (StruttedBeam Bridge)

The π-rahmen design is usually used for bridges in mountainous regionswhere the foundation isrm, passing over deep valleyswith a relatively long span, or for bridgescrossing over expressways(Figure 10.44 b). Asshown in the structural layout of a π -rahmen bridge in Figure 10.45 , thetwolegssupport themain girders, inducing axial compression in thecenter span of thegirder. Liveloadon thedeck is transmitted to themain girdersthrough theoor system. Intermediatehingesmay be


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FIGURE10.44: Typesof rahmen bridges.

insertedin thegirdersto makeGerber girders. A V-legrahmen bridgeissimilar to a π -rahmen bridgebut can span longer distanceswith no axial forcein the center span of thegirder (Figure 10.44 c).

10.9.4 Vierendeel Bridge

The Vierendeel bridge is a rigid framewhose upper and lower chordsare connected rigidly to thevertical members (Figure 10.44 d). All the members are subjected to axial and shear forces as wellas bending moments. This is internally a highly indeterminate system. Analysis of the Vierendeelframe must consider secondary stresses (see Section 10.8.3 ). It is morestiff than Langer or Lohsearch bridgesin which somemembers takeonly axial forces.

10.10 ArchBridges10.10.1 Structural Features

An arch rib actslikea circular beam restrained not only vert ically but also horizontally at both ends,and thus results in vert ical and horizontal reactionsat the supports. Thehorizontal reaction causesaxial compression in additi on to bending moments in thearch rib. Thebending momentscaused bythehorizontal forcebalancesthosedueto gravity loads. Compared with theaxial force, theeffect of


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FIGURE10.45: π -rahmen bridge. (From Japan Association of Steel BridgeConstruction, Outlineof Steel Br idges, Tokyo [in Japanese] , 1985. With permission.)

thebendingmoment isusually small. That iswhy thearch isoften madeof materials that havehighcompressivestrength, such asconcrete, stone, or brick.

10.10.2 Types of Arches

An archbridgeincludestheroad deckand thesupportingarch. Varioustypesof archesareshown inFigure 10.46 . In the gure, the thick linerepresents the members carrying bending moment, shear,and axial forces. Thethin linerepresentsmemberstaking axial forcesonly. Archbridgesareclassiedinto the deck and the through-deck types according to the location of the road surface, asshownin Figure 10.46 . Since the deck in both types of bridges is sustained by either vert ical columns orhangersto thearch,structurally thesameaxial forceaction, either compression or tension, is in effectin themembers. Thedifferenceis that thevert ical membersof deck bridgestakecompressiveforcesand thehangersof through-deck bridgestaketension. Theliveload actson thearch only indirectly.

A basic structural typefor an arch isatwo-hingearch (seeFigure 10.46 a). Thetwo-hingearchhasonedegreeof indeterminacyexternally becausethereare four end reactions. If onehingeisadded atthecrown of thearch,creatingathree-hingearch, it isrendered determinate. If theendsareclamped,turningit into axed arch, it becomesindeterminateto thethird degree. Thetied arch is subtendedbytwo hingesby atieand simply supported (Figure 10.46 b). Thetied arch isexternally determinatebut internally has one degree of indeterminacy. The oor structures hang from the arch and areisolated from the tie. Other typesof arch bridgeswill bediscussed later in moredetail.

10.10.3 Structural Analysis

Almost all bridgedesignanalyses, in thi sageof super computingpower, useniteelement methods.Theanalysis of an arch i sbasically the sameas that for a frame. Theweb members are analyzed astruss bars which take only axial forces. The arch rib and the girders are analyzed aseither trussesor beam-columns depending on the type of arch considered. Beam-columns takeaxial and shearforcesandbending moments. An archrib isusually madeup of straight piece-wisecomponents, notcurved segments, and it isso analyzed.


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FIGURE 10.46: Typesof arch bridges. (From Shimada, S., Journal of Bridge and Foundati on Engi- neering, 25(8), 1991[in Japanese] . With permission.)

10.10.4 Langer Bridge

TheLanger archisanalyzedbyassumingthat thearchribtakesonlyaxial compression(Figure 10.46 c).The arch rib is thin, but the girders are deep and resist moment and shear as well as axial tension.The girders of the Langer bridge are regarded as being strengthened by the arch rib. Figure 10.47showsthestructural componentsof aLanger bridge.

If diagonals are used in the web, it is called a trussed Langer. The difference between a trussedLanger andastandard trussi sthat thelower chord isagirder instead of just a bar. TheLanger bridgeis also determinate externally and indeterminate internally. The deck-type bridge of the Langer isoften called areversed Langer.

10.10.5 LohseBridge

TheLohsebridgeis very simi lar to theLanger bridgeexcept theLohsebridgecarriesit sresistancetobendingin thearch rib aswell asthegirder (Figure 10.46 d). By this assumption, theLohsebridgeisstiffer than theLanger. Thedistribution of bendingmomentsin thearch rib and thegirder dependson the stiffnessratio of the two members, which is the designer’s decision. TheLohsearch bridgesmay bethought of as tiered beams(see Figure 10.35 ) connected by vert ical members. The vert ical


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FIGURE 10.47: Langer arch bridge. (From Japan Association of Steel BridgeConstruction, Outline of Steel Bridges, Tokyo [in Japanese] , 1985. With permission.)

members are assumed to takeonly axial forces. Aesthetically, the Lohse is more imposing than theLanger, and is thereforesuited to urban areaswhiletheLanger tsinto mountain areas.

10.10.6 Trussed Arch andNielsen Arch Bridges

Generally diagonal members are not used in arch bridges, thus avoiding difculty in structuralanalysis. However, recent advancements in computer technology havechanged this outlook. Newtypesof archbridges, such asthetrussed archin whichdiagonal trussbarsareused instead of verticalmembers or the Nielsen Lohsedesign in which tension rodsare used for diagonals, havenow beenint roduced (seeFigure 10.46 e, f). Diagonal web members increase the stiffnessof abridgemore sothan vert ical members.

All the members of the truss bridge takeonly axial forces. On theother hand, the trussed archbridge may resist bending in either the arch r ib or the girder, or both. Since the diagonals of theNielsen Lohse bridge carry only axial tension, they are prestressed by the dead load to compensatefor thecompression forcedueto theliveload.

10.11 Cable-StayedBridges


A cable-stayed bridgehangsthe girders from diagonal cablesthat are tensioned from the tower, asshown in Figure 10.48 . The cablesof cable-stayed bridgesare anchored in thegirders. The girdersare most often supported by movable or xed hinges. Due to the diagonally tensioned cables, axial

forcesand bendingmomentsareimposed on thegirder and thetower. Thebendingmoment in thegirder i s reduced when supported by the cables, and spans can be longer than conventional girderbridges(as long as300 to 500m). Themaximum span length is the 890 m of theTatara Bridge inJapan (see Table 10.1). Because of the wonder and beauty of this type, its design has been copiedin even relatively small bridges including ones carrying only pedestr ians. For long-span bridges,stabil ity under strong wind currents should be carefully considered in the design. The dynamiceffectsof wind and earthquakesmust be studied analytically and experimentally. Wind tunnel testsmay benecessary to ensure excessiveoscil lation doesnot occur along the length of the bridgeor in


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FIGURE10.48: Cable-stayed bridge. (From Japan Association of Steel BridgeConstruction,Outline of Steel Bridges, Tokyo [in Japanese] , 1985. With permission.)

thetower. Thecablesalso may resonate in thewind if they arethin and exible. In this case, devicesare necessary to curb the vibration. The stabil it y of bridges under wind loadswill bediscussed inmoredetail in Section 10.12 (seeFigure 10.61 ).

10.11.2 Types of Cable-StayedBridges

Cable-stayed bridges may be classied by the hanging formation of the cable and the shapeof thetower. Figure 10.49 il lustrates three typical cable formations. Structurally, the radial cable mosteffectively decreasestheaxial forcein thetower and girders; however, difculty in construction arises

dueto thestructural complexity at thetop of the tower. Thefan type is morecommon because thecableconnectionsat the tower aredistr ibuted. Theharp type is aesthetically themost pleasing.

FIGURE10.49: Typesof cableformation. (From Nagai, N., Bri dgeEngineering, Kyoritsu PublishingCo., Tokyo, Japan [in Japanese] , 1994. With permission.)

Figure 10.50 showsvarioustower designs. Asthespan length becomeslarge, columnssuch astheA, the H, or the upside-down Y shapeare selected; thesehavesignicant torsional resistance.


Thecable-stayed bridgeisusually analyzed using linear elastic frameanalysis. Thecable is modeledasabar element with hinged ends. Figure 10.51 showstheow of gravity loads. Most of the load is


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FIGURE 10.50: Types of towers. (From Nagai, N., Bridge Engineering, Kyoritsu Publishing Co.,Tokyo, Japan [in Japanese] , 1994. With permission.)

transmit ted to thesubstructurethrough thecablesandthetower, but somegoesto thegirder directly.Thesmaller the bending stiffnessof thegirder, the lessthe load is taken by the girder. Asthetowerbecomeshigher, the tension forceof thecablecan bereduced.

Because of the sag in thecable dueto itsown weight, a reduced elastic modulusmay beused inanalysis. Thisreduced modulusisslightly lower thantheactual elasticmodulusof thecablematerial.Thegirder and the tower aredesigned to takeaxial compression, bending, and shear. Sincethelargeforcein the cable is concentrated on thegirder and tower, stressconcentration at thoseconnectionsshould becarefully checked using ni teelement analysis. Taking into theconsideration the fact thatthe supports are subjected to largenegative reactions (uplif t) , Pendel bearingsare used. These, asmentioned previously (seeFigure 10.25 ), arecomposed of an eye-bar and two end hinges, whichmaymovehorizontally and rotatefreely.

In thepreliminary design, thebridgeis modeled asaplaneframe. For thedetails, however, morepreciseanalyses such as three-dimensional stress analyses may be used. Nonlinear effects may betaken into consideration for exible long-span bridges.


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FIGURE10.51: Forceow in cable-stayed bridges. (From Nagai, N., Bridge Engineeri ng, KyoritsuPublishing Co., Tokyo, Japan [ in Japanese] , 1994. With permission.)

10.11.4 Tension in Cable

Oneof theimportant aspectsin thedesign of acable-stayed bridgeisthedeterminationof thetensionforcein thecable, which isdirectly related to forcesin thetower and thegirder. Control onthetensionforce in the cables is critical. The pre-tension of the cables must be known because it changes thestresses in the girder and the tower. Figure 10.52 shows the bending moment distribution underdead loadsalong the bridgebefore and after theprestressing forceisapplied. It can beseen that theproper prestress reducesbending moments in the girder signicantly. If the vert ical component of the tension is selected to beequal to thereaction of thecontinuousgirder (supported at the junctionof the cable and girder), the bending moment in the girder can be reduced to match that of thecontinuousgirder.

FIGURE10.52: Bendingmoment distr ibution. (FromJapan Societyof Civil Engineers, Cable-Stayed Bridges—Technology and its Change, Tokyo, Japan [in Japanese] , 1990. With permission.)

Thefollowingthreegeneral principlesare to beconsidered in determiningcabletension [ 19]:

1. Avoid having any bending moments (generated by dead loads) in the tower. This is


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accomplished bybalancing thehorizontal componentsof thecabletension in theleft andright endsof thetower.

2. Keepthebendingmomentsin thegirdersmall. It dependsonthelocationandthedistancebetween joints to the cable. Small distances (such asa multi-cable) will result in smallbendingmoments in thegirders.

3. Closethegirder byconnectingthecenter block lastlywithout usinganycompellingforces.Thecabletension isselected such that zero sectional forceexistsat thecenter of thegirder.

10.12 Suspension Bridges


Suspension bridgesuse two main cablessuspended between two towers and anchored to blocksatthe ends. Figure 10.53 shows the structural components of a suspension bridge. Stiffening girders

FIGURE 10.53: Suspension bridge. (From Japan Association of Steel BridgeConstruction, Outline of Steel Bridges, Tokyo [in Japanese] , 1985. With permission.)

are either truss or box type (see Figure 10.54 ) and hung from the main cables using hangers. Thesuspension bridgeismost suitablefor longspans. Table 10.7 isa list of theworld’sten longest bridges,all of which aresuspension bridges. Thelongest is theAkashi Kaikyo Bridge, which hasamain spanof 1990.8m, in Japan. It wasoriginally designed with amain span of 1990 m (Figure 10.55 ), but wasextended by0.8m when theKobeEarthquakecamecloseto this mark in 1995.

The ow of forces in a suspension bridge is shown in Figure 10.56 . The load on the girder istransmitted to thetowersthrough thehangersandthemain cables, and then to theanchor blocks. It


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FIGURE10.54: Typesof stiffening girders. (From Japan Association of Steel Bridge Construction,Out li ne of Steel Br idges, Tokyo [in Japanese] , 1985. With permission.)

TABLE10.7 TheWorld’s 10Longest BridgesCenter span

Rank Name (m) Country Year completed

1 Akashi Kai kyo Br idge 1990 Japan 1998 (est.)2 Great Bel t East Br idge 1624 Denmark 1997(est .)3 Humber Bridge 1410 England 19814 TsingMaBridge 1377 China 1997 (est.)5 Verrazano GateBridge 1298 U.S. 19646 Golden GateBridge 1280 U.S. 19377 Mackinac StraitsBridge 1158 U.S. 19578 M inami Bisan-Seto Br idge 1100 Japan 19889 Fai th Sul ton M ehmet Br idge 1090 Tur key 198810 BosporusBridge 1074 Turkey 1973

From Honshu Shikoku BridgeAuthority, Booklet and Brochures, Japan. With permission.

can beseen that anchor blocksareessential to takethehorizontal reaction forcefrom thecables. Thegravity of theanchor blocksresiststhe upward component of thecable tension force, and the shearforcebetween theanchor blocksandthefoundation resists thehorizontal component. Constructiondifculty may arisewhere soil conditi onsare poor. Different from the cable-stayed bridge, no axialforceis induced i n the girders of a suspension bridgeunless it is a self-anchored suspension bridge

(seeFigure 10.57 d).Thesagin themain cableaffects the structural behavior of the suspension bridge: the smaller the

sag, the larger the stiffness of the bridge and thereby large horizontal forces are applied to anchorblocks. In general theratio of thesag to themain span is selected to beabout 1:10.

10.12.2 Types of Suspension Bridges

Suspension bridgescan beclassied bythesupport conditi on of their sti ffeninggirdersand themaincable (Figure 10.57 ). The three-span, two-hingetypeis most commonly used for highway bridges.Thecontinuousgirder isoften adopted for railroadbridgesto avoid “knucklepoints”,whichadverselyaffect the trains. When the side span is short, the single-span typeis selected. The main cablesof self-anchored bridgesarexed to thegirdersinstead of to theanchor blocks, making theconstructionof anchor blocksunnecessary; instead the axial compression is carr ied in the girdersasin the cable-stayed bridge. Therearespecial cases(suchastheSevern Bridgein England) wherediagonal hangershavebeen used.


If thedead load of thecableand thestiffening girders is assumed to beuniformly distributed alongthe bridge length, the deection of the cable is parabolic and all dead loadsare supported by the


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FIGURE10.55: Side view of Akashi Kaikyo Bridge, Japan (1998 expected). (From Honshu Shikoku Bridge Authority, TechnolBridge, Japan [in Japanese] . With permission.)


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FIGURE10.56: Forceow in a suspension bridge. (From Nagai, N., Bridge Engineeri ng, KyoritsuPublishing Co., Tokyo, Japan [ in Japanese] , 1994. With permission.)

FIGURE10.57: Typesof suspension bridges. (From Nagai, N., BridgeEngineering, Kyori tsuPublish-ing Co., Tokyo, Japan [in Japanese] , 1994. With permission.)

cable. In this case, only liveloadsact on thegirder.There are two analyt ical procedures: elastic theory, in which linear elastic material and small

displacement areassumed, and thedeection theory,whichconsidersthedeection of thecabledueto l ive loads. When the span becomes large, elastic theory is too conservative in its estimation of bending moments.

10.12.4 Cable Design

For thecable, thehigh-strength steel wire, i.e., usually 5mm in diameter with astrength of 160to 180kg/mm 2 (1760N/mm 2) and zinc-galvanized, isused. Thereareseveral typesof cables(Figure 10.58 ):strand rope, spiral rope, locked coil rope (LCR), and parallel wi re strand (PWS). The PWSisusedmost commonlyfor suspension bridges; thousandsof parallel wireelementsarebundled into acircleby asqueezing machine, then wrapped with steel wireand painted.

Thewireistreatedbyanair spinning(AS) methodor theprefabricatedparallel wirestrandmethod(PWSS). In the ASmethod, the 5-mm wire is elected by rounding between anchor blocksone byoneunti l theprescribed number of wires is obtained. In thePWSSmethod, a strand that bundles


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FIGURE10.58: Typesof cables. (FromNagai, N., BridgeEngineering, KyoritsuPublishingCo.,Tokyo,Japan [in Japanese] , 1994. With permission.)

100 to 200 wire elements is suspended between the anchor blocksby xing with a socket. In thismethod, theconstruction period can beshort becausemorewiresareelected at onetimethan in theASmethod. Thethick strand ismorestable to wind but harder to handleduring construction. Afoothold (or catwalk) must beprovided under thecablefor theworkers to attach the cableband tothe main cable.

10.12.5 StiffeningGirder

Trussor boxtypegirdersareused to sti ffen suspension bridges. Thegirdermust becarefullydesignedto havesufcientsti ffnessfor wind stabil ity. Forvery longspanstrussesaremost effectivein improvingthe stiffnessand stabil ity (see Figure 10.54 ). Thebox girder i salso often adopted dueto its easeof fabrication.

10.12.6 Tower

Thetower isdesigned to besubjected to largeaxial compression and bending moment. It isdesignedto have smaller bending sti ffness in the longitudinal direction since the horizontal forces comingfrom both sides of the tower keep i t balanced. Figure 10.59 shows a compari son of several towersused for variousstructures. TheSearsTower in Chicago, known asthe tallest building, hasaheightof 443 m.

A bridgetower usually consistsof morethan threecellsinside, each having adequate resistancetotorsion and local buckling under large axial forces. Mechanical dampers such as the TMD (tunedmassdamper) or theTLD (tuned liquid damper) areoften used duringconstruction to control toweroscillationscaused bywind forces. Figure 10.60 showsatypical construction procedureadopted fortheAkashi Kaikyo Bridge, in which a climbing tower craneis used. An alternativemethod is to useacreeper crane, which clambersup along the tower.

10.12.7 Stability for Wind

Suspension bridgesaresoexiblethat thedynamic stabil ityunder wind effectsshould beinvestigatedusing awind tunnel. Thedynamic responsesmay becategorized into threetypes, of which responsebehaviors are shown in Figure 10.61 : vortex-induced oscillations, buffeting, and torsional utters.The utter, also called “galloping”, is torsional oscillation and is especially dangerous since it is aself-diverging resonance and may incite failure quickly and easily. The ow of air increases theamplitude of oscil lations under certain combinationsof wind speed and structural characteristics(natural frequency), asillustrated in Figure 10.61 . Flexiblebridges, suchassuspension or cable-stayedbridges, must becarefully designed if thewind speedsareli kely to inciteutter.


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FIGURE 10.59: Compari son of towers. (From Honshu Shikoku Bridge Authorit y, Technology of Akashi Kaikyo Bridge, Japan [in Japanese] . With permission.)

Vortex-induced oscillations were once thought to be caused by the Karman vortex. Now it isunderstood to be the air ow coming from thesurface or edge of thegirders that yields vibrationswhichresonatewith thenatural frequencyof thestructure. Thisvibrationoccursat alowandrelativelynarrow range of wind speedsand does not develop dangerous degrees of amplitude amplicati on.Buffetingi sarandomvibration causedbyturbulencein theair owor spontaneousgusts. Horizontalmovementsaredominant and theamplitudesincreaseproport ionally with thesquareof wind speed.

10.13 DeningTerms

Abutment: An end support for a bridgestructure.Arch bridge: A bridgethat includestheroad deck and thesupportingarch.Bridge: A structure that crossesover ariver, bay, or other obstruction, permitting the smooth

and safe passageof vehicles, trains, and pedestr ians.Cable-stayed bridge: A bridgein whichthesuperstructureishungfrom thediagonal cablesthat

are tensioned from the tower.Cast-in-placeconcrete: Concrete placed in i ts nal position in the structure while still in a

plastic state.Compositegirder: A stell girder connected to a concrete deck so that they respond to force

effectsasaunit.Deck (slab): A component, with or without wearing surface, directly supportingwheel loads.Floor system: A superstructure in which the deck is integral with itssupporting components,

such asoor beamsand str ingers.Girder: A structural component whoseprimary function is to resist loadsin exureand shear.


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FIGURE 10.60: Construction of a tower. (From Honshu Shikoku BridgeAuthority, Technology of Akashi Kaikyo Bridge, Japan [in Japanese] . With permission.)

Generally, thisterm is used for fabricated sections.Girder bridge: A bridgesuperstructurethat consists of a oor slab, girders, and bearings.Inuenceline: A continuousor discretized function over a section of girder whose value at a

point, multiplied bya load actingnormal to thegirder at that point, yieldstheforceeffectbeing sought.

Lever rule: The static summation of moments about onepoint to calculate the reaction at asecond point .

LRFD (Load and ResistanceFactor Design): A method of proportioning structural compo-nents (members, connectors, connecting elements, and assemblages) such that no ap-plicable limit state is exceeded when the structure is subjected to all appropriate loadcombinations.


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FIGURE 10.61: Dynamic response of a tower against wind. (From Nagai, N., Bri dge Engineering,Kyoritsu Publishing Co., Tokyo, Japan [ in Japanese] , 1994. With permission.)

Precast member: Concrete element cast in a location other than itsnal posit ion.Prestressed concrete: Concretecomponents in which thestressesand deformationsare intro-

duced by application of prestressing forces.Rigid framebridge: A bridgein whichthesuperstructureand substructuremembersarerigidly

connected.Segmental bridge: A bridge in which primary load-supporting members are composed of in-

dividual members called segments post-tensioned together to act as a monolithic unitunder loads.

Substructure: Structural partsof thebridgewhich providethe horizontal span.Superstructure: Structural parts of thebridgewhich support thehorizontal span.Suspension bridge: A bridgein which thesuperstructureis suspended by two main cablesand

anchored to end blocks.Trussbridge: A bridgesuperstructure which consists of a oor system and main trusses.

Acknowledgment

Manyof theguresin thischapter arecopiedfrom other booksandjournals. Theauthorswould liketoexpresssinceregrati tude to theoriginal authors. Special thanksgo to Prof. N. Nagai of theNagaokaInsti tute of Science and Technology, Profs. Y. Tachibana and H. Nakai of Osaka City University,the Japan Association of Steel Bridge Construction, American Association of State Highway andTransportation Ofcials, and CaliforniaDepartment of Transportation for their generosity.

References

[1] Ameri can Association of State Highway and Transportati on Ofcials. 1994. AASHTO LRFD BridgeDesign Specications, 1st ed., AASHTO, Washington, D.C.

[2] Ameri canAssociationof StateHighwayandTransportationOfcials.1989. GuideSpecications for Design and Construction of Segmental Concrete Bri dges, AASHTO, Washington, D.C.

[3] Caltrans. 1990. Bridge Design Detail s M anual. California Department of Transportation,Sacramento, CA.


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[4] Calt rans.1993.BridgeDesignPracticeManual, vol.2, Cali forni aDepartment ofTransportation,Sacramento, CA.

[5] Calt rans. 1990. BridgeDesign AidsM anual. Cali forniaDepartment of Transportation, Sacra-mento, CA.

[6] Federal Highway Administration. 1990. Standard Plansfor Highway Br idges, Vol. I , Concrete Superstructures, U.S. Department of Transportation, FHWA, Washington, D.C.

[7] Gerwick, B.C., Jr. 1993. Construction of Prestressed Concrete Structures, 2nd ed., John Wiley& Sons, NewYork.

[8] Hanshin Expressway Public Corporation. 1975. Constr uction Records of Mi nato Oh-H ashi,HEPC, Japan Society of Civil Engineers, Tokyo, Japan (i n Japanese).

[9] Hanshin Expressway Public Corporation. 1994. Techno Gallery, HEPC, Osaka, Japan.[10] Honshu Shikoku Bridge Authorit y. Technology of Akashi Kaikyo Bridge, HSBA, Japan ( in

Japanese).[11] Japan Association of Steel Bridge Construction. 1981. Manual Design Data Book, JASBC,

Tokyo, Japan (in Japanese).[12] Japan Association of Steel BridgeConstruction. 1984. A GuideBook of BearingDesign for Steel

Bridges, JASBC, Tokyo, Japan (i n Japanese).[13] Japan Association of Steel BridgeConstruction.1984. A GuideBook of Expansion Joint Design

for Steel Bridges, JASBC, Tokyo, Japan (in Japanese).[14] Japan Association of Steel Bri dgeConstruction. 1985. Out li ne of Steel Bridges, JASBC, Tokyo,

Japan (in Japanese).[15] Japan Association of Steel BridgeConstruction. 1988. Planni ng of Steel Bridges, JASBC, Tokyo,

Japan (in Japanese).[16] Japan Construction Mechanization Association. 1991. Cost Esti mati on of Bri dge Erecti on,

JCMA, Tokyo, Japan (in Japanese).[17] Japan Road Association. 1993. Specicati onsfor Hi ghway Bri dges, Part I Common Provisions,

Part II Steel Br idges, and Part I II Concrete Bridges, JRA, Tokyo, Japan (in Japanese).[18] Japan Society of Civil Engineers. 1990. Cable-Stayed Br idges—Technology and its Change,

JSCE, Tokyo, Japan (in Japanese).[19] Nagai, N. 1994. BridgeEngineeri ng, Kyori tsu Publi shing Co., Tokyo, Japan (i n Japanese).[20] Podolny, W. and Muller, J.M. 1982. Construction and Designof Prestressed ConcreteSegmental

Bridges, John Wiley & Sons, New York.[21] Shimada, S. 1991. Basic theory of arch structures, Journal of Bri dgeand Foundati on Engineer-

ing, Kensetsu-Tosho, 25(8), 48-52, Tokyo, Japan (i n Japanese).[22] Tachibana, Y. and Nakai, H. 1996. Bri dge Engineering, Kyori tsu Publi shing Co., Tokyo, Japan

(i n Japanese).[23] Tonias, D.E. 1995. Bri dgeEngineeri ng, McGraw-Hill , New York.[24] Troit sky, M.S. 1994. Planni ng and Design of Bridges, John Wiley & Sons, New York.[25] VSL. 1994. VSL Post-Tensioning System, VSL Corporation, Campbell, CA.[26] Xanthakos, P.P. 1994. Theory and Design of Bridges, John Wiley & Sons, New York.[27] Xanthakos,P.P.1995. BridgeSubstr uctureand Foundati onDesign, Prentice-Hall ,Upper Saddle

River, NJ.

Further Reading

[1] Billington, D.P. 1983. TheTower and theBridge, Basic Books, Inc., New York.[2] Leonhardt, F. 1984. Bri dges, Aestheticsand Design, MIT Press, Cambridge, MA.[3] Chen, W. F. and Duan, L. 1998. Handbook of Bri dgeEngineeri ng, CRC Press, BocaRaton, FL.


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Appendix: Design Examples

10.A.1 Two-Span, Continuous, Cast-in-Place, PrestressedConcreteBoxGirderBridge

Given: A two-span, continuous, cast-in-place, prestressed concrete box girder bridgehas two equalspansof length 157ft (47.9 m) with acolumn bent. Thesuperstructure is 34 ft (10.4 m) wide. Theelevation of thebridgeis shown in Figure 10.62 a.

Material:

Ini tial concrete f ci = 3500 psi (24.13MPa), E ci =3372 ksi (23,250MPa)Final concrete f c = 4000 psi (27.58MPa), E c =3600ksi (24,860 MPa)Prestressing steel f pu = 270 ksi (1860 MPa) low relaxation strand, E p = 28,500 ksi(197,000 MPa)Mild steel f y =60ksi (414 MPa), E s =29,000ksi (200,000 MPa)

Prestressing:

Anchorageset thickness =0.375 in. (9.5 mm)Prestressing stressat jacking f pj =0.8f pu =216 ksi (1489 MPa)The secondary moments due to prestressing at the bent are M DA =1.118 P j (kips-ft)M DG =1.107 P j (kips-ft)

Loads:

Dead load =self-weight +barrier rail +futurewearing 3 in ACoverlayLiveload =AASHTO HS20-44 +dynamic load allowance

Specicati on:

AASHTO-LRFD [ 1] (referred to asAASHTO in this example)

Requirements:

1. Determinecross-section geometry2. Determinelongitudinal section and cablepath3. Calculate loads4. Calculate liveload distribution factors5. Calculateunfactored momentsand shear demandsfor interior girder6. Determine load factorsfor strength limit state I and servicel imit state I7. Calculate section propert iesfor interior girder8. Calculateprestresslosses9. Determineprestressing force, P j , for interior girder

10. Check concrete strength for interior girder, servicelimit state I11. Flexural strength design for interior girder, strength limit state I12. Shear strength design for interior girder, strength limit stateI


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FIGURE10.62: A two-span, continuous, prestressed concrete box girder bridge.

Solution

1. DetermineCross-Section Geometry1.1) Structural Depth, d For prestressed continuous spans, the structural depth, d , can be determined using a


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depth-to-span ratio (d/L) of 0.04.

d

=0.04L

=0.04 (157 )

=6.28 ft (1.91 m)

Use d =6.25 ft (1.91 m)

1.2) Girder Spacing, S To provide effective torsional resistance and a sufcient number of girdersfor prestresspaths, the spacingof girders should not belarger than twicetheir depth.

S max < 2d =2(6.25) =12 .5 ft (3.81 m)

Using an overhangof 4 ft (1.22 m), the center-to-center distance between two exteriorgirdersis26 ft (7.92 m).

Try threegirdersand two bays, S =26 / 2 =13ft > 12.5 ft N.G.

Try four girdersand threebays, S

=26 / 3

=8.67 ft < 12.5 ft O.K.

Useagirder spacing, S =8.67 ft (2.64 m)

1.3) Typical Section From past experi enceand design practice, weselect a thi cknessof 7 in. (178mm) at theedge and 12 in. (305 mm) at the face of exterior girder for theoverhang, the width of 12 in. (305 mm) for girders with the exterior girder aring to 18 in. (457 mm) at theanchorageend. Thelength of thisareisusuallytaken asone-tenthof thespan length 15.7ft (4.79m). Thedeck and soft thicknessesdepend on theclear distancebetween adjacentgirders. Wechoose7.875 in. (200 mm) and 5.875in. (149 mm) for thedeck and softthicknesses, respectively. A typical section for this example is shown in Figure 10.62 b.Thesection propertiesof thebox girder are :

Properties Midspan Bent (faceof support)

A ft2 (m2 ) 57.25(5.32) 68.98(6.41)I ft4 (m4 ) 325.45(2.81) 403.56(3.48)yb f t (m) 3.57(1.09) 3.09(0.94)

2. DetermineLongitudinal Section and CablePathTo lower the center of gravit y of the superstructure at the face of a bent cap in a cast-in-placepost-tensioned box girder, the thicknessof soft is ared to 12 in., asshown inFigure 10.62 c. A cablepath isgenerally controlled bythemaximum dead load momentsand the posit ion of thejack at theend section. Maximum eccentr icitiesshould occur atpoints of maximum dead load moment and almost no eccentr icity should bepresent atthe jacked end section. For thisexample, themaximum dead load momentsoccur at thebent cap, close to 0.4 L for span 1 and 0.6 L for span 2. A parabolic cable path i schosenasshown in Figure 10.62 c.

3. Calculate Loads3.1) Component Dead Load, DCThecomponent dead load, DC , includesall structural dead loadswith the exception of the future wearing surfaceand specied uti li ty loads. For design purposes, two partsof the DC are dened as:

DC 1: girder self-weight (150 lb/ft 3) acti ng at theprestressing state

DC 2: barr ier rail weight (784 kips/f t) acting at servicestateafter all losses


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3.2) Wearing Surface Load, DW Thefuturewearing surfaceof 3 in. (76 mm) with aunit weight of 140 lb/ft 3 isdesignedfor thisbridge.

DW = (deck width–barrier width) (thicknessof wearing surface) (unit weight)

= [34 −2(1.75)](0.25)( 140 ) = 1067 .5 lb/ft

3.3) LiveLoad, LL , and Dynami c Load Allowance, IM The design live load, LL , is the AASHTO HS20-44 vehicular live load. To considerthe wheel load impact from moving vehicles, the dynamic load allowance, IM =33%(AASHTO-LRFD Table 3.6.2.1-1), isused.

4. CalculateLiveLoad Distribution FactorsAASHTO-LRFD [ 1] recommends that approximate methods be used to distribute liveload to individual girders. The dimensions relevant to this prestressed box girder are:depth, d =6.25ft (1.91m); number of cells, N c =3;spacingof girders, S =8.67ft (2.64

m); span length, L =157 ft (47.9 m); half of thegirder spacing plusthe total overhang,W e =8.334 ft (2.54m); and thedistancebetween thecenter of an exterior girder and theinterior edgeof a barrier, d e =4–1.75 =2.25 ft (0.69 m). Thisbox girder i swithin therangeof applicability of theAASHTO approximate formulas. Thel iveload distributionfactorsare calculatedasfollows.

4.1) Live Load Di str ibuti on Factor for Bending Moments

(a) Interior girder (AASHTO Table4.6.2.2.2b-1)

Onelaneloaded:

LD M

=1.75

+

S

3.6

1

L

0.35 1

N c

0.45

= 1.75 +8.673.6

1157

0.35 13

0.45

= 0.432 lanes

Two or morelanesloaded:

LD M =13N c

0.3 S 5.8

1L

0.25

=133

0.3 8.675.8

1157

0.25

= 0.656 lanes (controls)

(b) Exterior girder (AASHTO Table 4.6.2.2.2d-1)

LD M =W e14 =

8.33414 =0.595 lanes (controls)

4.2) Live Load Distri buti on Factor for Shear

(a) Interior girder (AASHTO Table4.62.2.3a-1)


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Onelaneloaded:

LD V

=S

9.5

0.6 d

12L

0.1

=8.679.5

0.6 6.2512(157 )

0.1

= 0.535 lanes

Two or morelanesloaded:

LD V =S

7.3

0.9 d 12L

0.1

=8.677.3

0.9 6.2512 (157 )

0.1

= 0.660 lanes (controls)

(b) Exterior girder (AASHTO Table 4.62.2.3b-1)

Onelaneloaded: Lever rule

Thelever ruleassumesthat the deck in itstransversedirection is simply supportedbythegirdersand usesstaticsto determine theliveload distribution to thegirders.AASHTO-LRFD also requiresthat when thelever ruleisused, themultiplepresencefactor, m , should apply. For a oneloaded lane, m =1.2. Thelever rulemodel fortheexterior girder i sshown in Figure 10.63 . From static equilibrium:

R =5.928.67 =0.683

LD ν = mR = 1.2(0.683 ) = 0.820 (controls)

FIGURE10.63: Liveload distr ibuti on for exterior girder—lever rule.


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Two or morelanesloaded: Modify interior girder factor by e

LD V

=e(LD v ) interior girder

=0.64

+d e

12 .5(LD ν ) interior girder

= 0.64 +2.2512 .5

(0.66) =0.541

Theliveload distribution factorsat the strength limit state:

Strength limit stateI Interior girder Exterior girder

Bending moment 0.656lanes 0.595lanes

Shear 0.660lanes 0.820lanes

5. Calculate Unfactored Momentsand Shear Demandsfor Interior Girder

It is practically assumed that all dead loads are carr ied by the box girder and equallydistributed to each girder. Theliveloadstakeforcesto thegirdersaccording to liveloaddistr ibution factors(AASHTO Article4.6.2.2.2). Unfactoredmoment andshear demandsfor an interior girder areshownin Figures 10.64 and 10.65 . Detailsareli sted in Tables 10.8and 10.9 . Only theresultsfor span 1areshownin thesetablesandguressincethebridgeissymmetrical about thebent.

TABLE10.8 Moment and Shear Dueto Unfactored DeadLoad for theInterior Girder (Span 1)

Unfactored dead load

DC 1a DC 2b DW c

Location M DC 1 V DC 1 M DC 2 V DC 2 M DW V DW (x/L) (k-ft ) (kips) (k-ft ) (kips) (k-ft ) (kips)

0.0 0 125.2 0 11.4 0 15.60.1 1700 91.5 155 8.4 212 11.40.2 2871 57.7 262 5.3 357 7.20.3 3513 24.0 321 2.2 437 3.00.4 3626 −9.7 331 −0.9 451 −1.20.5 3210 −43.4 293 −4.0 399 −5.40.6 2264 −77.1 207 −7.1 282 −9.60.7 789 −111 72 −10.1 98 −13.80.8 −1215 −145 −111 −13.2 −151 −18.00.9 −3748 −178 −342 −16.3 −466 −22.21.0 −6833 −216 −622 −19.4 −847 −26.4

(−6292) ( −573) ( −781)

Note: Momentsin bracketsarefor faceof support at thebent. Moments inspan 2 are symmetrical about the bent. Shear in span are anti sym-metrical about thebent.

a DC 1, interior girder self-weight.b DC 2, barr ier self-weight.c DW , weari ngsurfaceload.

6. DetermineLoad Factorsfor Strength Limit StateI and ServiceLimit StateI6.1) General Design Equation (AASHTO Article1.3.2)

η γ i Q i ≤φR n (10.18)


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FIGURE10.64: Moment envelopesfor span 1.

FIGURE10.65: Shear envelopesfor span 1.


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TABLE10.9 Moment and Shear Envelopesand Associated Forcesfor theInterior Girder Dueto AASHTO HS20-44LiveLoad (Span 1)

Posit ive momentandassociated Negat ivemoment Shear and

shear andassociated shear associated moment

Location M LL +IM

V LL +IM

M LL +IM

V LL +IM

V LL +IM

M LL +IM

(x/L) (k-ft) (kips) (k-ft) (kips) (kips) (k-ft)

0.0 0 0 0 0 60.0 00.1 782 49.8 −85 −5.4 50.1 7870.2 1312 41.8 −169 −5.4 42.0 13200.3 1612 29.3 −253 −5.4 34.3 16140.4 1715 21.8 −337 −5.4 −27.7 16500.5 1650 −30.0 −422 −5.4 −35.1 16280.6 1431 −36.7 −506 −5.4 −42.0 14240.7 1081 −42.6 −590 −5.4 −49.9 8520.8 647 −47.8 −748 −8.3 −59.2 2160.9 196 −32.9 −1339 −50.1 −68.8 −6671.0 0 0 −2266 −67.8 −78.5 1788

−(2104)

Note: LL + I M = AASHTO H S20-44 live load plus dynamic load allowance. Moments inbrackets are for f aceof support at t hebent. Momentsi n span 2 aresymmetr ical aboutthebent. Shear in span 2i santisymmetr icalabout thebent. Liveloaddi str ibuti onfactorsareconsidered.

where γ i are load factors, φ is a resistance factor, Q i represents forceeffects, R n isthenominal resistance, and η is a factor related to ducti li ty, redundancy, and operationalimportanceof that being designed. η is dened as:

η =ηD ηR ηI ≥0.95 (10.19)

where

ηD =1.05 for nonducti le componentsand connections0.95 for ducti le componentsand connections (10.20)

ηR =1.05 for nonredundant members0.95 for redundant members (10.21)

ηI =1.05 operationally important bridge0.95 general bridge

only apply to strength and extremeevent limit states

(10.22)

For thisbridge, the following valuesareassumed:

Ductilit y Redundancy ImportanceLimi t states ηD ηR ηI η

Strength limit state 0.95 0.95 1.05 0.95Servicelimit state 1.0 1.0 1.0 1.0

6.2) Load Factorsand Load Combinations Theload factorsand combinationsarespecied as(AASHTO Table3.4.1-1):

Strength limi t stateI : 1.25(DC 1 +DC 2) +1.5(DW) +1.75(LL +M)

Serviceli mit stateI: DC 1 +DC 2 +DW +(LL +I M)


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7. Calculate Section Propert iesfor Interior GirderFor an interior girder asshown in Figure 10.66 , theeffectiveangewidth, beff , isdeter-mined (AASHTO Article 4.6.2.6) by

beff = thelesser of

Leff 4

12h f +bwS

(10.23)

where L eff is the effectivespan length and may be taken as the actual span length forsimply supported spans and the distance between points of permanent load inectionfor continuousspans; h f is thecompression angedepth; and bw is theweb width. Theeffectiveangewidth and thesection propert iesareshownin Table 10.10 for theinteriorgirder.

FIGURE10.66: Effectiveangewidth of interior girder.

TABLE 10.10 EffectiveFlangeWidth and Section Propert iesforInterior Girder

BentLocat ion Di mensi on M idspan ( face of suppor t)

h f in. (mm) 7.875(200) 7.875(200)L eff / 4 in. (mm) 353(8966) 235.5(11963)

Top 12 h f +bw in. (mm) 106.5(2705) 106.5(2705)ange S i n. ( mm) 104 ( 2642) 104 ( 2642)

beff i n. (mm) 104(2642) 104(2642)

h f i n. ( mm) 5.875 ( 149) 12 ( 305)L eff / 4 in. (mm) 353(8966) 235.5(11963)

Bottom 12 h f

+bw in. (mm) 82.5(2096) 156(2096)

ange S i n. ( mm) 104 ( 2642) 104 ( 2642)beff in. (mm) 82.5(2096) 104(2096)

Area A ft2 (m2 ) 14.38 (1.336) 19.17 (1.781)Moment of inerti a I ft4 (m4 ) 81.85 (0.706) 112.21(0.968)

C.G. yb f t (m) 3.55(1.082) 2.82(0.860)

Note: L eff = 117.8ft (35.9m) for mi dspan,L eff = 78.5f t (23.9 m) for thebent,bw = 12in. ( 305 mm).


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8. CalculatePrestressLossesFor a cast-in-placepost- tensioned box girder, two types of losses, instantaneous losses(friction, anchorageset, and elastic shortening) and time-dependent losses (creep andshrinkageof concrete and relaxation of prestressing steel) are signicant. Since the pre-

stresslossesarenot symmetr ical about thebent for thisbridge, thecalculationisperformedfor both spans.8.1) Frictional Loss, f pF

f pF = f pj 1 −e−(Kx +µα) (10.24)

where K is the wobble friction coefcient = 0.0002 1/ft (6.6 × 10−7 1/mm); µ isthecoefcient of friction =0.25(AASHTO Article5.9.5.2.2a); x isthelength of aprestressingtendon from the jacking end to the point considered; and α is the sum of theabsolutevaluesof angle changein theprestressing steel path from the jacking end.For aparaboliccablepath(Figure 10.67 ), theanglechangeis α = 2ep /L p , where ep isthevert ical di stance between two control points and L p is the horizontal distance betweentwo control points. Thedetailsaregiven in Table 10.11 .

FIGURE10.67: Parabolic cable path.

TABLE10.11 PrestressFrictional Loss

Segment ep (in.) L p (ft) α (rad) α (rad) L p (ft ) Point f pF (ksi)

A 31.84 0 0 0 0 A 0.0AB 31.84 62.8 0.0845 0.0845 62.8 B 7.13BC 42.50 78.5 0.0902 0.1747 141.3 C 14.90

CD 8.50 15.7 0.0902 0.2649 157.0 D 20.09DE 8.50 15.7 0.0902 0.3551 172.7 E 25.06EF 42.50 78.5 0.0902 0.4453 251.2 F 32.18FG 31.84 62.8 0.0845 0.5298 314.0 G 38.23

8.2) AnchorageSet Loss, f pAThe effect of anchorage set on the cable stress can be approximated by the Calt rans


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procedure[ 4], asshown in Figure 10.68 . It isassumed that theanchorageset losschangeslinearly within thelength, L pA .

FIGURE10.68: Anchorageset lossmodel. (From Cali forniaDepartment of Transportation, Bridge

Design Practice, Copyright 1983 (Figure 3–10, pages3–46, updated March, 1993), Sacramento, CA,1993. With permission.)

L pA = E( L)L pF

f pF (10.25)

f =2 f pF x

L pF (10.26)

f pA = f 1 −x

L pA(10.27)

where L is the thickness of the anchorage set; E is the modulus of elasticity of theanchorageset; f isthechangein stressduetotheanchorset; L pA isthelength inuencedbytheanchor set; L pF isthelength to apoint wherelossisknown;and x isthehorizontaldistancefrom thejacking end to thepoint considered.For an anchor set thicknessof L = 0.375 in. and E =29,000ksi, consider the point Bwhere L pF = 141 .3 ft and f pF = 14 .9 ksi:

L pA = E( L)L pF

f pF = 29 ,000 (3/ 8)( 141 .3)12 (14 .90 ) = 92 .71 ft < 141 .3 ft O.K

f

=2 f pF x

L pF =2(14 .90 )( 92 .71)

141 .3 =19 .55 ksi

f pA = f 1 −x

L pA = 19 .55 1 −x

92 .71

8.3) Elastic Short ening Loss, f pES The loss due to elastic shortening in post-tensioned members is calculated using thefollowing formula (AASHTO Article5.9.5.2.3b):


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f pES =N −1

2N E p

E cif cgp (10.28)

where N isthenumber of identical prestressingtendonsand f cgp isthesumof theconcretestressat thecenter of gravityof theprestressing tendonsdueto theprestressing forceafter jacking and the self-weight of member at the section with the maximum moment. Forpost-tensioned structures with bonded tendons, f cgp may be calculated at the centersection of thespan for simply supported structuresand at thesection with themaximummoment for continuousstructures. To calculate the elastic shortening loss, weassumethat theprestressingjack forcefor an interior girder P j =1800kipsand thetotal numberof prestressing tendons N =4. f cgp is calculated for the mid-support section:

f cgp =p j

A +P j e2

I x +M DC 1e

I x

=

1800

19 .17(12)2

+

1800 (28 .164 )2

112 .21(12 )4

+

(−6292 )( 12 )( 28 .164 )

112 .21 (12 )4

= 0.652 +0.614 −0.914 =0.352 ksi (2448 MPa)

f pES =N −1

2N E p

E cif cgp =

4 −12(4)

28 ,5003370

(0.352 ) = 1.12 ksi (7.7 MPa)

8.4) Time-Dependent Losses, f pT M AASHTO providesa table to estimate the accumulated effect of time-dependent lossesresulting from thecreep and shrinkageof concreteand therelaxation of thesteel tendons.From AASHTO Table 5.9.5.3-1:

f pT M =21 ksi (145 MPa) (upper bound)

8.5) Total Losses, f pT

f pT = f pF + f pA + f pES + f pT M

Detailsaregiven in Table 10.12 .9. DeterminePrestressing Force, P j , for Interior Girder

Since the live load is not in general equally distr ibuted to the girders, the prestressingforce, P j , required for each girder may differ. To calculate prestress jacking force, P j ,theini tial prestressforcecoefcient, F pCI , and nal prestressforcecoefcient, F pCF , aredenedas:

F pCI = 1 −f pF + f pA + f pES

f pj (10.29)

F pCF

=1

−f pT

f pj (10.30)

Thesecondary moment coefcientsare dened as:

M sC =xL

M DAP j

for span 1(1 − x

L ) M DGP j

for span 2(10.31)


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TABLE 10.12 Cable Path and PrestressLossesLocation Prestress losses (ksi) Force coeff.

Span (x /L) f pF f pA f pES f pT M f pT F pCI F pCF

0.0 0.00 19.55 41.67 0.904 0.8070.1 1.78 16.24 40.14 0.911 0.814

0.2 3.56 12.93 38.61 0.918 0.8210.3 5.35 9.93 37.40 0.924 0.8270.4 7.13 6.31 35.56 0.933 0.835

1 0.5 8.68 3.00 1.12 21 33.79 0.941 0.8440.6 10.24 32.36 0.947 0.8500.7 11.79 33.91 0.940 0.8430.8 13.35 0.00 35.47 0.933 0.8360.9 14.90 37.02 0.926 0.8291.0 20.09 42.21 0.902 0.805

0.0 20.09 42.21 0.902 0.8050.1 25.06 47.18 0.879 0.7820.2 26.49 48.61 0.872 0.7750.3 27.91 50.03 0.866 0.7680.4 29.34 51.46 0.859 0.7620.5 30.76 0.00 1.12 21 52.88 0.852 0.755

2 0.6 32.18 54.30 0.846 0.7490.7 33.69 55.81 0.839 0.7420.8 35.21 57.33 0.832 0.7350.9 36.72 58.84 0.825 0.7281.0 38.23 60.35 0.818 0.721

Note:

F pCI = 1 −f pF + f pA + f pES

f pj

F pCF = 1 −f pT f pj

where x is the distance from the left end for each span. The combined prestressingmoment coefcientsaredened as:

M psCI = F pCI (e) +M sC (10.32)

M psCF = F pCF (e) +M sC (10.33)

where e is the distance between the cable and the center of gravity of a cross-section;positivevaluesof e indicatethat thecableisabovethecenter of gravity, and negativeonesindicate thecableis below it.

Theprestressforcecoefcientsand thecombined moment coefcientsarecalculatedandlisted in Table 10.13 . Accordingto AASHTO,theprestressingforce, P j , can bedeterminedusing theconcrete tensilestressl imit in theprecompression tensilezone(seeTable 10.5):

f DC 1 +f DC 2 +f DW +f LL +I M +f psF ≥ −0.19 f c (10.34)

in which


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TABLE 10.13 PrestressForceand Moment CoefcientsLocat ion Cablepath Forcecoeff. Moment coefcients(f t)

Span (x/L) e (in.) F pCI F pCF F pCI e F pCF e M sC M psCI M psCF

0.0 0.240 0.904 0.807 0.018 0.016 0.000 0.018 0.0160.1

−13.692 0.911 0.814

−1.040

−0.929 0.112

−0.928

−0.817

0.2 −23.640 0.918 0.821 −1.809 −1.618 0.224 −1.586 −1.3940.3 −29.136 0.924 0.827 −2.244 −2.008 0.335 −1.908 −1.6720.4 −31.596 0.933 0.835 −2.456 −2.200 0.447 −2.008 −1.752

1 0.5 −29.892 0.941 0.844 −2.344 −2.101 0.559 −1.785 −1.5420.6 −24.804 0.947 0.850 −1.958 −1.757 0.671 −1.287 −1.0870.7 −13.608 0.940 0.843 −1.278 −1.146 0.783 −0.495 −0.3630.8 −4.404 0.933 0.836 −0.342 −0.307 0.894 0.552 −0.5880.9 10.884 0.926 0.829 0.840 0.752 1.006 1.846 1.7581.0 28.164 0.902 0.805 2.117 1.888 1.118 3.235 3.006

0.0 −28.164 0.902 0.805 2.117 1.888 1.107 3.224 2.9950.1 10.884 0.879 0.782 0.797 0.709 0.996 1.793 1.7050.2 −4.404 0.872 0.775 −0.320 −0.284 0.886 0.566 0.6010.3 −16.308 0.866 0.768 −1.176 −1.044 0.775 −0.401 −0.2690.4 −24.804 0.859 0.762 −1.776 −1.575 0.664 −1.111 −0.9100.5 −29.892 0.852 0.755 −2.123 −1.881 0.554 −1.570 −1.328

2 0.6 −31.596 0.846 0.749 −2.227 −1.971 0.443 −1.784 −1.5280.7 −29.136 0.839 0.742 −2.037 −1.801 0.332 −1.705 −1.4690.8 −23.640 0.832 0.735 −1.639 −1.447 0.221 −1.417 −1.2260.9 −13.692 0.825 0.728 −0.941 −0.830 0.111 −0.830 −0.7191.0 0.240 0.818 0.721 0.016 0.014 0.000 0.016 0.014

Note: e is the distancebetween the cable path and central gravity of the interior girder cross-section; positivemeanscable is abovethe central gravity and negativeindicatescable is below the central gravity.

f DC 1 =M DC 1 C

I x(10.35)

f DC 2 =M DC 2 C

I x(10.36)

f DW =M DW C

I x(10.37)

f LL +I M =M LL

+IM C

I x (10.38)f psF =

P peA +

(P pe e)CI x +

M s CI x =

F pCF P j A +

M psCF P j CI x

(10.39)

where C( = yb or yt ) is thedistancefrom theextremeber to thecenter of gravity of thecross-section; f c isin ksi; and P pe is the effective prestressing forceafter all losseshavebeen incurred. From Equations 10.34 and 10.39 , wehave:

P j = −f DC 1 −f DC 2 −f DW −f LL +IM −0.19 f cF pCF

A +M psCF C

I x

(10.40)

Detailedcalculationsaregiven in Table 10.14 . Most critical pointscoincidewith locationsof maximum eccentr icity: 0.4L in span 1, 0.6L in span 2, and at thebent. For thisbridge,thecontrollingsection isthrough theright faceof thebent. Herein, P j =1823kips(8109kN). Rounding P j up to 1830kips (8140kN) givesa required area of prestressing steelof Aps =P j /f pj = 1830 / 216 =8.47 in. 2 (5465mm 2).

10. Check Concrete Strength for Interior Girder, ServiceLimit StateITwo criteria are imposed on the level of concrete stresses when calculating requiredconcrete strength (AASHTO Article5.9.4.2):


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TABLE 10.14 Determination of Prestressing Jacking Forcefor an Interior GirderTop ber Bottom ber

Jacking JackingLocation Stress(psi) force Stress(psi) force

Span (x/L) f DC 1 f DC 2 f DW f LL +IM P j (kips) f DC 1 f DC 2 f DW f LL +IM P j (kips)

0.0 0 0 0 0 — 0 0 0 0 —0.1 389 36 48 179 — −512 −47 −64 −236 7490.2 658 60 82 301 — −865 −79 −108 −395 13070.3 805 73 100 369 — −1058 −97 −132 −485 15420.4 831 76 103 393 — −1092 −100 −136 −517 1573

1 0.5 735 67 91 378 — −967 −88 −120 −497 14820.6 519 47 64 328 — −682 −62 −85 −431 11930.7 181 16 22 248 — −238 −22 −30 −326 4550.8 −278 −25 −35 −171 242 366 33 46 225 —0.9 −859 −78 −107 −307 1210 1129 103 140 403 —1.0 −1336 −122 −166 −447 1818 1098 100 136 367 —

0.0 −1336 −122 −166 −447 1823 1098 100 136 367 —0.1 −859 −78 −107 −307 1264 1129 103 140 403 —0.2 −278 −25 −35 −171 254 366 33 46 225 —0.3 181 16 22 248 — −238 −22 −30 −326 5200.4 519 47 64 328 — −682 −62 −85 −431 1371

2 0.5 735 67 91 378 — −967 −88 −120 −497 16910.6 831 76 103 393 — −1092 −100 −136 −517 17820.7 805 73 100 369 — −1058 −97 −132 −485 17390.8 658 60 82 301 —

−865

−79

−108

−395 1474

0.9 389 36 48 179 — −512 −47 −64 −236 8431.0 0 0 0 0 — 0 0 0 0 —

Note: Posit ive stressindicatescompression andnegative stressindicatestension. P j areobtained byEquation 10.40.

f DC 1 +f psI ≤0.55f ci at prestressing statef DC 1 +f DC 2 +f DW +f LL +I M +f psF ≤0.45f c at servicestate (10.41)

f psI =P j I

A +P j I e C

I x +M sI C

I x =F pCI P j

A +M psCI P j C

I x(10.42)

Theconcretestressesin theextremebers(after instantaneouslossesand nal losses) are

given in Tables 10.15 and 10.16 . For the ini tial concretestrength in theprestressing state,thecontrolling locationisthebottom ber at 0.9L sectionin span 1. FromEquation 10.41wehave:

f ci, req ≥f DC 1+f psI

0.55 = 9300.55 =1691 psi < 3500 psi

... choose f ci =3500 psi (24 .13 MPa) O.K.

For the nal concrete strength at theservice limit state, the controlling location isagainin thebottom ber at 0.9L section in span 1. From Equation 10.41 wehave:

f c, req ≥ f DC 1+f DC 2+f DW +f LL +IM +f psF 0.45

= 15390.45 =3420 psi < 4000 psi

... choose f c =4000 psi (27 .58 MPa) O.K.

11. Flexural Strength Design for Interior Girder, Strength Limit StateIAASHTO requiresthat for thestrength limit state I


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TABLE 10.15 Concrete Stressesafter InstantaneousLossesfor the Interior GirderTop ber st ress (psi ) Bot tom ber st ress (psi )

Total Total

Location F ∗pCI M ∗psCI initial F ∗pCI M ∗psCI initialSpan (x/L) f DC 1 P j /A P ∗j Y t /I f psI stress f DC 1 P j /A P ∗j yh /I f psI stress

0.0 0 799 8 807 807 0 799 −10 789 7890.1 389 805 −389 416 806 −512 805 512 1317 8050.2 658 812 −665 147 805 −865 812 874 1686 8210.3 805 817 −800 17 821 −1058 817 1052 1868 8100.4 831 824 −842 −18 813 −1092 824 1107 1931 839

1 0.5 735 831 −748 83 819 −967 831 984 1815 8480.6 519 837 −540 298 816 −682 837 710 1547 8650.7 181 831 −208 623 804 −238 831 723 1104 8660.8 −278 825 231 1056 778 366 825 −304 520 8660.9 −859 818 774 1592 733 1129 818 −1017 −199 9301.0 −1336 598 1257 1854 519 1098 598 −1033 −435 663

0.0 −1336 598 1252 1850 514 1098 598 −1030 −432 6660.1 −859 777 752 1528 670 1129 777 −988 −212 9170.2 −278 771 237 1008 730 366 771 −312 −459 8250.3 181 765 −168 597 777 −238 765 221 986 7490.4 519 759 −466 293 812 −682 759 613 1372 690

2 0.5 735 753 −658 95 830 −967 753 865 1619 6520.6 831 748 −748 0 830 −1092 748 983 1731 6390.7 805 741 −715 27 831 −1058 741 940 1681 6230.8 658 735

−594 141 799

−865 735 781 1516 651

0.9 389 729 −348 381 770 −512 729 458 1187 6751.0 0 723 7 730 730 0 723 −9 714 714

Note: Posit ive stressindicatescompression andnegative stressindicatestension.

TABLE 10.16 ConcreteStressesafter Total Lossesfor theInterior GirderTop ber st ress (psi ) Bot tom ber st ress (psi )

Total TotalLocation F ∗pCF M ∗psCF nal F ∗pCF M ∗psCF nal

Span (x/L) f LOAD P j /A P ∗j Y t /I f psF stress f LOAD P j /A P ∗j yb /I f psF stress

0.0 0 713 7 720 720 0 713 −9 704 7040.1 653 720 −343 377 1030 −858 720 450 1170 3120.2 1100 726 −584 141 1241 −1466 726 768 1494 480.3 1348 731 −701 30 1377 −1772 731 922 1652 −1190.4 1403 738 −735 4 1407 −1844 738 966 1704 −140

1 0.5 1272 746 −647 99 1371 −1672 746 850 1596 −760.6 958 751

−455 296 1254

−1260 751 599 1350 90

0.7 467 745 −152 593 1060 −614 745 200 945 3310.8 −510 739 −246 985 475 670 739 −324 415 10850.9 −1351 732 737 1469 119 1776 732 −969 −237 15391.0 −2070 533 1168 1701 −368 1702 533 −960 −427 1275

0.0 −2070 533 1164 1697 −373 1702 533 −957 −423 12780.1 −1351 691 715 1406 55 1776 691 −940 −249 15270.2 −510 685 252 937 427 670 685 −331 353 10240.3 467 679 −113 566 1033 −614 679 148 828 2130.4 958 673 −382 292 1250 −1260 673 502 1175 −85

2 0.5 1272 667 −557 111 1383 −1672 667 732 1399 −2730.6 1403 662 −641 21 1424 −1844 662 842 1504 −3400.7 1348 655 −616 40 1387 −1772 655 809 1465 −3070.8 1100 649 −514 135 1235 −1466 649 676 1325 −1220.9 653 643 −302 341 994 −858 643 397 1040 1811.0 0 637 6 643 643 0 637 −8 629 629

Note: f LOAD = f DC 1 +f DC 2 +F DW +f LL +IM . Positivestressindicatescompression andnegative stressindicatestension.

M u ≤ φM n

M u = η γ i M i = 0.95[1 .25 (M DC 1 +M DC 2)

+ 1.5M DW +1.75M LLH ] +M ps

where φ is the exural resistance factor 1.0 and M ps is the secondary moment due to


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prestress. Factored moment demands, M u , for theinterior girder in span 1arecalculatedin Table 10.17 . Although the moment demands are not symmetrical about the bent(due to different secondary prestress moments), the results for span 2 are similar andthe differenceswill not be considered in this example. Thedetailed calculationsfor the

exural resistance, φM n , areshown in Table 10.18 . It isclear that noadditional mild steelisrequired.

TABLE 10.17 Factored Moments for an Interior Girder (Span 1)M DC 1 M DC 2 M DW

(kips-ft) (kips-ft) (kips-ft) M LL +IM M ps M uLocation Dead Dead Wearing (kips-ft) (kips-ft) (kips-ft)

(x/L) load-1 load-2 surface Positive Negative P /S Positi ve Negative

0.0 0 0 0 0 0 0 0 00.1 1700 155 212 782 −85 205 4009 25690.2 2871 262 357 1312 −169 409 6820 43580.3 3513 321 437 1612 −253 614 8469 53680.4 3626 331 451 1715 −337 818 9012 55990.5 3210 293 399 1650 −422 1023 8494 50500.6 2264 207 282 1431 −506 1228 6942 37210.7 789 72 98 1081

−590 1432 4392 1613

0.8 −1215 −111 −151 647 −748 1637 922 −13970.9 −3748 −342 −466 196 −1339 1841 −3355 −59061.0 −6292 −573 −781 0 −2104 2046 −7219 −10716

Note: M u =0.95[1.25(M DC 1 +M MDC 2 ) +1.5M DW +1.75M LL +IM ]+M ps

TABLE 10.18 Flexural Strength Design for I nterior Girder, Strength Limit StateI (Span 1)Location Aps d p As d s b c f ps d e a φM n M u

(x/L) (in.2 ) (in.) (in.2 ) (in.) (in.) (in.) (ksi) (in.) (in.) (k-ft) (k-ft)

0.0 32.16 0 72.06 104 7.14 253.2 32.16 6.07 5206 00.1 46.09 0 72.06 104 7.27 258.1 46.09 6.18 7833 40090.2 56.04 0 72.06 104 7.33 260.1 56.04 6.23 9717 68200.3 61.54 0 72.06 104 7.35 261.0 61.54 6.25 10759 84690.4 64.00 0 72.06 104 7.36 261.3 64.00 6.26 11226 90120.5 8.47 62.29 0 72.06 104 7.36 261.1 62.29 6.25 10903 84940.6 57.20 0 72.06 104 7.34 260.3 57.20 6.24 9937 69420.7 48.71 0 72.06 104 7.29 258.7 48.71 6.20 8328 43920.8 38.20 0 71.06 82.5 21.19 228.1 38.20 18.01 −4965 −13970.9 53.48 0 71.06 82.5 23.36 237.0 53.48 19.86 −7822 −59061.0 62.00 0 71.06 104 8.13 261.0 62.00 6.25 −10848 −10716

Note: 1. Prestr essing steel, f ps = f pu (1 −k c

d p), k =2(1.04 −

f pyf pu

)

2. For anged section, c/d e ≤ 0.42 ,

M n =Aps f ps (d p − a2 ) +As f y (d s − a

2 )

−A f y (d s − a2 ) +0.85f c (b −bw )β 1 h f ( a

2 −hf 2 )

a =β1 C

c =Aps f pu

+As f y

−As f y

−0.85β

1f c (b

−bw )h

f 0.85β1 f c bw +kA ps

f pud p

3. For anged section, c/d e > 0.42—over-reinforced,

M n = (0.36β1 −0.08β 21 )f c bw d 2e +0.85β1 f c (b −bw )h f (d e −0.5h f )

d e =Aps f ps d p +As f y d s

Aps f ps +As f y

4. For rectangular secti on, i.e., when c < h f take b = bw in theaboveformulas.


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12. Shear Strength Design for Interior Girder, Strength Limit StateIAASHTO requiresthat for thestrength limit state I

V u ≤

φV n

V u = η γ i V i

= 0.95 1.25(V DC 1 +V DC 2) +1.5V DW +1.75V LL +I M +V ps

where φ is shear resistance factor 0.9 and V ps is the secondary shear due to prestress.Factored shear demands, V u , for the interior girder are calculated i n Table 10.19 . Todeterminetheeffectiveweb width, assumethat theVSL post-tensioningsystem of 5 to 12tendon units [ 25] will beused with agrouted duct diameter of 2.88 in. In this example,bν = 12 −2.88 / 2 =10.56in. (268mm). Detailed calculationsof the shear resistance,φV n (using two-leg #5 stirrups, Aν = 0.62 in. 2 [419 mm 2]) for span 1, are shown inTable 10.20 . The results for span 2 are similar to span 1 and the calculationsare notrepeated for thi sexample.

TABLE10.19 Factored Shear for an Interior Girder (Span 1)V DC 1 V DC 2 V DW (kips) (kips) (kips) V LL +IM M LL +IM V ps M u

Locat ion Dead Dead Wear ing ( ki ps) ( k- ft ) ( ki ps) V u (k-ft)(x/L) load-1 load-2 surface Envelopes Associated P /S (kips) Associated

0.0 125.2 11.4 15.6 60.0 0 13.03 297.1 00.1 91.5 8.4 11.4 50.1 787 13.03 231.0 40170.2 57.7 5.3 7.2 42.0 1320 13.03 168.0 68830.3 24.0 2.2 3.0 34.3 1614 13.03 105.4 84720.4 −9.7 −0.9 −1.2 −27.7 1650 130.3 −47.3 89030.5 −43.4 −4.0 −5.4 −35.1 1628 13.03 −109.2 84570.6 −77.1 −7.1 −9.6 −42.0 1424 13.03 −170.3 69290.7 −111 −10.1 −13.8 −49.9 852 13.03 −233.1 40110.8 −145 −13.2 −18.0 −59.2 216 13.03 −298.3 205.40.9 −178 −16.3 −22.2 −68.8 −667 13.03 −364 −47901.0

−216

−19.4

−26.4

−78.5

−1788 13.03

−434.3

−10191

Note: V u =0.95[1.25 V DC 1 +V DC 2 +1.5V DW +1.75V LL +IM ] +V ps

10.A.2 Three-Span, Continuous, Composite PlateGirder Bridge

Given: A three-span, continuous, compositeplategirder bridgehastwo equal spansof length 160 ft(48.8m) and onemidspan of 210 ft (64m). Thesuperstructureis44ft (13.4m) wide. Theelevation,plan, and typical cross-section are shown in Figure 10.69 .Structural steel: A709Grade50for web and anges,

F yw =F yt =F yc =F y = 50 ksi (345 MPa)

A709 Grade 36 for stiffeners, etc.,F ys = 36 ksi (248 MPa)

Concrete:

f c = 3250 psi (22 .4 MPa),

E c = 3250 ksi (22 , 400 MPa),

modular ratio, n =9


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TABLE 10.20 Shear Strength Design for Interior Girder, Strength Limit StateI (Span 1)Location d ν y V p εx θ V c S φV n |V u |(x/L) ( in.) ( rad) (kips) ν/f c (1000) (degree) β (kips) ( in.) (kips) (kips)

0.0 54.00 0.084 124.1 0.090 −0.028 23.5 6.50 234.4 12 460.7 297.10.1 54.00 0.063 93.9 0.071 −0.093 27 5.60 201.8 12 400.4 231.00.2 52.90 0.042 63.1 0.055 0.733 33 2.37 83.7 24 194.0 168.00.3 58.87 0.021 31.8 0.034 1.167 38 2.10 82.5 24 167.6 105.40.4 60.84 0.000 0.0 0.020 1.078 36 2.23 90.6 24 150.2 47.30.5 59.14 0.018 27.8 0.037 1.026 36 2.23 88.0 24 171.0 109.20.6 54.06 0.036 56.2 0.058 0.539 30 2.48 89.5 24 196.4 170.30.7 54.00 0.054 83.5 0.077 −0.106 27 5.63 202.9 12 392.0 233.10.8 54.00 0.072 110.4 0.097 −0.287 23.5 6.50 234.3 12 448.4 298.30.9 54.00 0.090 136.8 0.117 −0.137 23.5 3.49 125.8 9 420.5 364.01.0 57.42 0.000 0.0 0.199 2.677 36 1.0 38.3 3.5 478.9 434.3

Note: 1. bν =10.56i n. and y is slopeof the prestressing cable.2. Aν =0.62in. 2 (2#5)

V n = the lesser of V c +V s +V p0.25f c bν d ν +V p

V c =0.0316 β f c bν d ν , V s =Aν f y d ν cos θ

s

ν =V u −φV p

φb ν d ν, ε x

M ud ν +0.5N u +0.5V u cot θ −Aps f po

E s As +E p Aps ≤ 0.002

F ε

=

E s As +E p ApsE c Ac

+E s As

+E p Aps

(when εx is negative, multiply by F ε )

Aν min =0.0316 f cbν S f y

For V u < 0.1f c bν d ν , S max = smaller of 0.8d ν24 in.

For V u ≥ 0.1f c bν d ν , S max = smaller of 0.4d ν12 in.

Loads:

Dead load = self-weight + barrier rail + futurewearing 3 in ACoverlayLiveload = AASHTO HS20-44 + dynamic load allowanceSingle-laneaveragedaily truck trafc (ADTT)

=3600 (oneway)

Deck: Concrete slabdeck with thicknessof 10.875in. (276 mm) hasbeen designedConstruction: Unshored; unbraced length for compression ange, L b = 20 ft (6.1 m)Specicati on: AASHTO-LRFD [ 1] (referred to asAASHTO in this example)Requirements: Design the following portions of an interior girder for maximum positive exureregion at span 1:

1. Calculate loads2. Calculate liveload distribution factors3. Calculate unfactored moments and shear demands4. Determine load factorsfor strength limit state I and fatiguel imit state5. Calculatecompositesection propert iesfor posit iveexureregion6. Flexural strength design, strength limit stateI7. Shear strength design, strength limit stateI8. Fatiguedesign, fatigueand fracturelimit state9. Intermediate transversesti ffener design

10. Shear connector design11. Constructabil ity check


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FIGURE10.69: A three-span, continuousplate girder bridge.

Solution

1. Calculate Loads1.1) Component Dead Load, DC for an Interior Girder

Thecomponent dead load, DC , includesall structural dead loadswith the exception of thefurther wearingsurfaceandspecieduti li ty loads. For design purposes, thetwo partsof DC are dened as:

DC 1: Deck concrete (self-weight, 150 lb/ft 3) and steel girder including bracingsystem and details(estimated weight, 300lb/ft for each girder). Assumethat DC 1 isactingon thenoncompositesectionand isdistributed to each girder bythetri butaryarea. Thetributary width for theinterior girder is16 ft (4.9 m).


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DC 1 = [(10 .875 / 12 )( 16) +(1.5)( 15 .25 −10 .975 )/ 12 (1.5)] (0.15) +0.3

= 2.557 kips/ft (37 .314 kN/m)

DC 2: Barrier rail weight (784kips/ft). Assumethat DC 2 isactingon thelong-termcompositesection and is equally distr ibuted to each girder.

DC 2 = 0.784 / 3 = 0.261 kips/ft (3.809 kN/m)

1.2) Wearing SurfaceLoad, DW A futurewearing surfaceof 3 in. (76mm) with aunit weight of 140 lb/ft 3 is assumed tobecarried bythe long-term compositesection and equally distributed to each girder.

DW = (deck width–barrier width) (thicknessof wearing surface)(unit weight) / 3

= [44 −2(1.75)] (0.25 )( 0.14)/ 3 = 0.473 kips/ft (6.903 kN/m)

1.3) LiveLoad, LL , and Dynami c Load Allowance, IM Thedesign live load, LL , is the AASHTO HS20-44 vehicular liveload. To consider thewheel load impact from movingvehicles, thedynamic load allowance, IM =33%for thestrength limit stateand 15%for thefatiguelimit stateareused [AASHTO Table3.6.2.1-1].

2. CalculateLiveLoad Distribution Factors2.1) RangeApplicabil it y of AASHTO Approximate Formulas AASHTO-LRFD [ 1] recommends that approximate methods be used to distribute liveload to individual girders. For concretedeckonsteel girders, liveload distribution factorsaredependent onthegirderspacing, S , spanlength, L , concreteslabdepth, t s ,longitudinalstiffnessparameter, K g , andnumber ofgirders, N b . Therangeofapplicabili ty ofAASHTOapproximate formulas are 3.5 ft

≤S

≤16 ft; 4.5 in.

≤t s

≤12 in.; 20 ft

≤L

≤240

ft; and N b ≥ 4. For this design example, S = 16 ft, L 1 = L 3 = 160 ft, L 2 = 210 ft,t s = 10.875 in., and N b = 3 < 4. It is obvious that thisbridge is out of therange of applicability of AASHTO formulas. Theconventional level rule is used to determineliveload distribution factors.2.2) Level Rule Thelevel ruleassumesthat thedeck in itst ransversedirection is simply supported bythegirders and usesstaticsto determine the liveload distr ibution to the girders. AASHTOalso requiresthat when the level rule is used, themultiplepresencefactor, m (1.2 for oneloaded lane, 1.0 for two loaded lanes, 0.85for threeloaded lanes, and 0.65for morethanthreeloaded lanes), should apply.2.3) Live Load Distr ibuti on Factors for Strength Limi t State Figure 10.70 showslocationsof trafc lanes for the interior girder. For a 12-ft (3.6-m)trafc lanewidth, thenumber of trafc lanesfor thisbridgeis three.

(a) Onelaneloaded (Figure 10.70 a)

R =1316 = 0.8125 lanes

LD =mR = 1.2(0.8125 ) = 0.975 lanes


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FIGURE10.70: Liveload distribution—lever rule.

(b) Two lanesloaded (Figure 10.70 b)

R =1316 +

916 =1.375 lanes

LD =mR = 1.0(1.375 ) = 1.375 lanes (controls)

(c) Threelanesloaded (Figure 10.70 c)

R =(13 +3)

16 +7

16 =1.4375 lanes

LD =mR = 0.85(1.4375 ) = 1.222 lanes


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2.4) LiveLoad Distribution Factorsfor FatigueLimit State AASHTO requiresthat onetrafc laneload beused and multiplepresencefactorsnot beapplied to the fatigue limit state. The live load distribution factor for the fatigue limitstate, therefore, is obtained byonelaneloaded without amultiplepresencefactor of 1.2.

LD =0.813

3. Calculate Unfactored Momentsand Shear DemandsFor an interior girder, unfactored moment and shear demandsareshownin Figures 10.71and 10.72 for the strength l imit state and Figures 10.73 and 10.74 for thefatigue limitstate. Thedetails are li sted in Tables 10.21 to 10.23 . Only the results for span 1 and onehalf of span 2 areshown in thesetablesand guressincethebridgeis symmetrical aboutthecenterline of span 2.

4. DetermineLoad Factorsfor Strength Limit StateI and FatigueLimit State4.1) General Design Equation (AASHTO Article1.3.2)

η γ i Q i

≤φR n (10.43)

where γ i areload factorsand φ resistancefactors; Q i representsforceeffectsor demands;and R n is the nominal resistance. η is a factor related to ducti li ty, redundancy, andoperational importanceof that being designed and is dened as:

η =ηD ηR ηI ≥0.95 (10.44)

where

ηD =1.05 for nonducti le componentsand connections0.95 for ducti le componentsand connections (10.45)

ηR =1.05 for nonredundant members0.95 for redundant members (10.46)

ηI =1.05 operationally important bridge0.95 general bridge

only apply to strengthand extremeevent l imit states

(10.47)

For thisbridge, the following valuesareassumed:

Ductilit y Redundancy ImportanceLimi t states ηD ηR ηI η

Strength limit state 0.95 0.95 1.05 0.95

Fatigueli mit state 1.0 1.0 1.0 1.0

4.2) Load Factorsand Load Combinations Theload factorsand combinationsarespecied as(AASHTO Table3.4.1-1):

Strength limi t stateI: 1.25(DC 1 +DC 2) +1.5(DW) +1.75(LL +I M)

Serviceli mit state: 0.75(LL +I M)

5. Calculate Composite Section Propert iesfor PositiveFlexure RegionTry steel section (Figure 10.75 ) as:


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FIGURE10.71: Moment envelopesdue to unfactored loads.

FIGURE10.72: Shear envelopesdue to unfactored loads.


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FIGURE10.73: Unfactored moment dueto fatigueloads.

Topange: bf c =18in. t f c =1in.

Web: D =96in. t w =0.625i n.

Bottom ange: bf t =18in. t f t =1.75in.

5.1) EffectiveFlangeWidth (AASHTO Article4.6.2.6)

For an interior girder, theeffectiveangewidth

beff = the lesser of

L eff 4 =

115 (12 )4 = 345 in.

12 t s +bf 2 = (12 )( 10 .875 ) +18 / 2 =140 in. (controls)

S = (16)( 12 ) =192 in.

where L eff is the effectivespan length and may be taken as the actual span length forsimplysupported spansandthedistancebetween thepointsof permanent load inectionfor continuousspans; bf is the top angewidth of thesteel girder.5.2) Elastic Composit eSection Properti es For the typical section (Figure 10.75 ) in the posit iveexure region of span 1, the elasticsection propert ies for the noncomposite, the short- term composite ( n = 9), and thelong-term composite( 3n

=27), respectively, are calculated in Tables 10.24 to 10.26 .

5.3) Plasti c Moment Capacit y, M pTheplasti cmoment capacity, M p , isdetermined using equilibrium equations. Therein-forcement in the concrete slab is neglected in thisexample.

(a) Determine thelocation of theplastic neutral axis(PNA)Assuming that the PNA is within the top ange of the steel girder (Figure 10.76 )and that yP NA is thedistance from the top of thecompression ange to thePNA,weobtain:


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TABLE 10.21 Moment Envelopesfor Strength Limit StateIM DC 1 M DC 2 M DW

(kips-ft) (kips-ft) (kips-ft) M LL +IM M uLocation Dead Dead Wearing (kips-ft) (kips-ft)

Span (x/L) load-1 load-2 surface Positive Negative Positive Negative

0.0 0 0 0 0 0 0 00.1 2047 209 379 1702 −348 6049 26410.2 3439 351 636 2949 −696 10310 42500.3 4177 426 773 3784 −1042 12858 48350.4 4260 435 788 4202 −1390 13684 4387

1 0.5 3688 376 682 4212 −1738 12800 29080.6 2462 251 455 3829 −2086 10236 4020.7 582 59 108 3069 −2434 6017 −31310.8 −1954 −199 −361 1951 −2782 173 −76960.9 −5143 −525 −951 941 −3736 −6522 −142971.0 −8988 −917 −1663 637 −5720 −13074 −23641

0.0 −8988 −917 −1663 637 −5720 −13074 −236410.1 −3913 −399 −724 924 −2998 −4616 −111360.2 33 3 6 2230 −1695 3759 −2767

2 0.3 2852 291 528 3499 −1607 10302 18120.4 4544 464 841 4448 −1607 14540 44730.5 5108 521 945 4766 −1607 15954 5359

Note: Live load distri bution factor, LD = 1.375 . Dynamic load allowance, IM = 33%. M u =0.95 1.25 M DC 1 +M DC 2 +1.5M DW +1.75M LL +IM

TABLE 10.22 Shear Envelopesfor Strength Limit State IV DC 1 V DC 2 V DW

(kips-ft) (kips-ft) (kips-ft) V LL +IM V uLocation Dead Dead Wearing (kips-ft) (kips-ft)

Span (x/L) load-1 load-2 surface Positive Negative Positive Negative

0.0 148.4 15.1 27.4 133.7 −23.4 455.4 194.30.1 107.5 11.0 19.9 110.1 −24.9 352.2 127.70.2 66.6 6.8 12.3 90.6 −34.2 255.3 47.80.3 25.6 2.6 4.7 75.2 −45.7 165.2 −35.70.4 −15.3 −1.6 −2.8 60.5 −59.0 76.5 −122.1

1 0.5 −56.2 −5.7 −10.4 46.6 −74.3 −10.8 −211.80.6 −97.1 −9.9 −18.0 33.7 −91.2 −96.7 −304.30.7 −138.0 −14.1 −25.5 22.1 −109.5 −180.2 −398.90.8 −178.9 −18.3 −33.1 12.0 −129.0 −261.5 −495.80.9 −219.8 −22.4 −40.7 6.5 −149.5 −334.9 −594.11.0 −260.7 −26.6 −48.2 4.4 −170.5 −402.5 −693.3

0.0 268.5 27.4 49.7 181.6 −15.0 724.2 397.30.1 214.7 21.9 39.7 154.4 −15.8 594.2 311.20.2 161.1 16.4 29.8 128.3 −22.1 466.5 216.4

2 0.3 107.4 11.0 19.9 104.0 −32.5 341.8 115.00.4 53.7 5.5 9.9 81.8 −45.9 220.4 8.10.5 0 0 0 62.4 −62.4 103.8 −103.8

Note: Live load distri bution factor, LD = 1.375 . Dynamic load allowance, IM = 33%. V u =0.95 1.25 V DC 1 +V DC 2 +1.5V DW +1.75V LL +IM

P s +P c1 =P c2 +P w +P t (10.48)

whereP s = 0.85 f c beff t s = 0.85 (3.25)( 140 )( 10 .875 ) = 4206 kips(18,708kN)

P c1 = yP NA bf c F ycP c2 = A f c F yc −P c1 = (t f c −yP NA )b f c F yc

P c = P c1 +P c2 =A f c F yc = (18)( 1)( 50) = 900 kips(4,003 kN)P w = Aw F yw = (96 )( 0.625 )( 50 ) =3,000 kips (13,344kN)P t = A f t F yt = (18 )( 1.75 )( 50) =1,575 kips(7,006kN)Substi tuting theaboveexpressionsinto Equation ( 10.48 ) and solvingfor yP NA , weobtain


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FIGURE10.74: Unfactored shear dueto fatigueloads.

FIGURE10.75: Typical cross-section in posit iveexureregion.

yP NA =t f c

2P w +P t −P s

P c +1 (10.49)

yP NA =12

3000 +1575 −4206900 +1 = 0.705 in. < t cf =1.0 in. O.K.

(b) Calculate M p


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TABLE 10.23 Moment and Shear Envelopesfor FatigueLimit StateM LL +IM V LL +IM (M LL +IM )u (V LL +IM )u

Location (kips-ft) (kips) (kips-ft) (kips)

Span (x/L) Positi ve Negative Positi ve Negative Positi ve Negative Positi ve Negative

0.0 0 0 68.2

−11.9 0 0 51.1

−8.9

0.1 868 −177 56.2 −12.7 651 −133 42.1 −9.50.2 1504 −355 46.2 −17.5 1128 −266 34.7 −13.10.3 1930 −532 38.4 −23.3 1447 −399 28.8 −17.50.4 2143 −709 30.9 −30.1 1607 −532 23.1 −22.6

1 0.5 2148 −886 23.8 −37.9 1611 −665 17.8 −28.40.6 1953 −1064 17.2 −46.5 1465 −798 12.9 −34.90.7 1565 −1241 11.3 −55.8 1174 −931 8.5 −41.90.8 995 −1419 6.1 −65.8 746 −1064 4.6 −49.30.9 480 −1905 3.3 −76.2 360 −1429 2.5 −57.21.0 325 −2917 2.2 −87.0 243 −2188 1.7 −65.2

0.0 325 −2917 92.6 −7.6 243 −2188 69.5 −5.70.1 471 −1529 78.7 −8.1 353 −1146 59.1 −6.00.2 1137 −865 65.4 −11.3 853 −648 49.1 −8.5

2 0.3 1785 −820 53.0 −16.5 1338 −615 39.8 −12.40.4 2268 −820 41.7 −23.4 1701 −615 31.3 −17.60.5 2430 −820 31.8 −31.8 1823 −615 23.9 −23.9

Note: Live load distribution factor, LD = 0.813. Dynamic load allowance, I M = 15%. M LL +IM u =0.75 M LL +IM u and V LL +IM u =0.75 V LL +IM u

TABLE10.24 NoncompositeSection Propertiesfor Posit ive Flexure RegionA y i A i yi yi −ysb A i yi −ysb

2 I oComponent ( in. 2 ) (in.) (in.3 ) (in.) (in.4) (in.4)

Top ange, 18 x 1 18 98.25 1,768.5 54.587 53,636 1.5Web, 96 x 0.625 60 49.75 2,985.0 6.087 2,223 46,080

Bottom ange, 18x 1.75 31.5 0.875 27.6 −42.788 57,670 8.04

109.5 — 4,781.1 113,529 46,090

ysb =Ai yi

Ai =4,781 .1109 .5 =43 .663 in.

yst = (1.75 +96 +1) −43 .663 =55 .087 in.

I girder I o + A i yi −ysb2

= 46 ,090 +113 ,529 =159 ,619 in.4

S sb =I girder

ysb =159 ,61943.663 =3, 656 in.3

S st =I girder

yst =159 ,61955.087 =2,898 in.3

Summing all forcesabout thePNA, weobtain:

M p = M P NA =P s d s +P c1yP NA

2 +P c2t f c −yP NA

2 +P w d w +P t d t

(10.50)

whered s = 10 .875

2 +4.375 −1 +0.705 = 9.518 in. (242 mm)d w = 96

2 +1 −0.705 =48 .295 in. (1,227 mm)d t = 1.75

2 +96 +1 −0.705 = 97 .17 in. (2,468 mm)

M p = (4,206 )( 9.518 ) +(18)( 50 ) (0.705 )2

2 +(18)( 50 ) (1−0.705 )2

2

+(3,000 )( 48 .295 ) +(1,575 )( 97 .17 )


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TABLE 10.25 Short-Term CompositeSection Propert ies( n =9)A y i A i yi yi −ysb −n A i yi −ysb−n

2 I oComponent (in. 2 ) (in.) (in.3 ) (in.) (in.4 ) (in.4 )

Steel sect ion 109.5 43.663 4,781.1 –38.791 164,768 159,619Concreteslab140/9x10.875 169.17 107.563 18,196 25.109 106,653 1,667

278.67 — 22,977 — 271,421 161,286

ysb −nAi yi

Ai =22,977278 .67 =82 .454 in.

yst −n = (1.75 +96 +1) −82 .454 =16 .296 in.

I com −n = I o + A i yi −ysb−n2

= 161 ,286 +271 , 421 =432 , 707 in.4

S sb −n =I con −nysb −n =

432 ,70782.454 =5,248 in.3

S st −n =I com −n

yst −n =432 ,70716.296 =26 , 553 in.3

TABLE 10.26 Long-Term CompositeSection Propert ies( 3n =27)A y i A i yi yi −ysb −3n A i yi −ysb −3n

2 I oComponent (in. 2 ) (in.) (in.3 ) (in.) (in.4) (in.4 )

Steel section 109.5 43.663 4,781.1 −21.72 51,661 159,619Concrete slab140/ 9 x 10.875 56.39 107.563 6,065.4 42.18 100,320 556

165.89 — 10,846.4 — 151,981 160,174

ysb−3n =Ai yi

Ai =10, 846 .4165 .89 =65.383 in.

yst −3n = (1.75 +96 +1) −65.383 =33 .367 in.

I com−3n =

I o +

Ai

yi −

ysb −3n

2

= 160 ,174 +151 , 981 =312 , 155 in.4

S sb−3n =I con −3nysb−3n =

312 ,15565 .383 =4,774 in.3

S st −3n =I com −3n

yst −3n =312 ,15533.367 =9,355 in.3

= 338 ,223 kips-in. =28 ,185 kips-ft (38 ,212 kN-m)

5.4) Yield Moment, M y (AASHTO Article6.10.5.1.2)

The yield moment, M y , corresponds to the rst yielding of either steel ange. It isobtained by thefollowing formula:

M y =M D 1 +M D 2 +M AD (10.51)

where M D 1 , M D 2 , and M AD aremomentsdueto thefactored loadsapplied to thesteel,the long-term and theshort- term compositesection, respectively. M AD can beobtainedby solving equation:


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FIGURE10.76: Plastic moment capacity state.

F y =M D 1

S s +M D 2

S 3n +M AD

S n(10.52)

M AD = S n F y −M D 1

S s −M D 2

S 3n(10.53)

(10.54)

where S s , S n , and S 3n (seeTables 10.24 to 10.26 ) aresection moduli for thenoncomposite

steel, theshort-term and thelong-term compositesection, respectively. From Table 10.21 ,maximumfactoredposit ivemoments M D 1 and M D 2 in span 1areobtainedat thelocationof 0.4L 1 .

M D 1 = (0.95)( 1.25)(M DC 1) = (0.95)( 1.25)( 4260 ) = 5,059 kips-ftM D 2 = (0.95) (1.25M DC 2 +1.5M DW )

= (. 095 ) [1.25(435 ) +1.5(788 )] = 1,640 kips-ft

For the top ange:

M AD = (26 ,553 ) 50 −5,059 (12)

2, 898 −1,640 (12 )

9, 355

= 715 , 552 kips-in. = 59 ,629 kips-ft (80 , 842 kN-m)

For thebottom ange:


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M AD = (5,248 ) 50 −5,059 (12 )

3, 656 −1,640 (12 )

4,774

= 153 ,623 kips-in. = 12 , 802 kips-ft (17 ,356 kN-m) (controls)

... M y = 5,059 +1,640 +12 , 802 = 19 ,501 kips-ft (26 ,438 kN-m)

6. Flexural Strength Design, Strength Limit StateI6.1) Compactnessof Steel Gi rder Section Thesteel section isrst checked to meet therequirementsof acompact section (AASHTOArticle6.10.5.2.2).

(a) Ductility requirement:

D p ≤d + t s + t h

7.5

where D p isthedepth from thetop of theconcretedeck to thePNA, d isthedepth of

thesteel girder, and t h isthethicknessof theconcretehaunch abovethetopangeof the steel girder. Thepurposeof thisrequirement isto prevent permanent crashingof the concrete slab when the composite section approaches it s plastic momentcapacit y. For this example, referring to Figure 10.75 and 10.76 , weobtain:

D p = 10 .875 +4.375 −1 +0.705 =14 .955 in. (381 mm )

D p = 14 .955 in. <d + t s + t h

7.5 =98 .75 +10 .875 +3.375

7.5

= 15 .067 in. O.K.

(b) Web slendernessrequirement, 2D cp

t w ≤3.76

E

F yc

where D cp is the depth of the web in compression at the plastic moment state.SincethePNA is within the top ange, D cp is equal to zero. The web slendernessrequirement issatised.

(c) Compression angeslendernessand compression angebracing requirement

It isusuallyassumedthat thetopangeisadequatelybracedbythehardenedconcretedeck; thereare, therefore, no requirements for the compression angeslendernessand bracing for compact compositesectionsat the strength limit state.... the section is acompact compositesection.

6.2) Moment of Inerti a Ratio Limit (AASHTO Article6.10.1.1)Theexural members shall meet the following requirement:

0.1 ≤I yc

I y ≤0.9

where I yc and I y are the moments of i nert ia of thecompression ange and steel girderabout thevert ical axisin theplaneof web, respectively. This limit ensuresthat the lateraltorsional bucking formulasarevalid.


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I yc =(1)( 18 )3

12 =486 in.4

I y

=486

+(96)( 0.625 )3

12 +(1.75)( 18 )3

12 +1338 in.4

0.1 <I yc

I y =486

1338 = 0.36 < 0.9 O.K.

6.3) Nomi nal FlexureResistance, M n (AASHTO Article10.5.2.2a)It isassumed that theadjacent interior-pier section isnoncompact. For continuousspanswith thenoncompactinterior support section, thenominal exureresistanceof acompactcompositesection is taken as:

M n = 1.3R h M y ≤M p (10.55)

where Rh is aangestressreduction factor taken as1.0 for thishomogeneousgirder.

M n =1.3(1.0)( 19 ,501 ) = 25 ,351 kips-ft < M p = 28 ,185 kips-ft

6.4) Strength Limi t State I AASHTO-LRFD [ 1] requiresthat for strength limit stateI

M u ≤φf M n (10.56)

where φf istheexural resistancefactor =1.0. For thecompositesection in theposit iveexureregion in span 1, themaximum moment occursat 0.4L 1 (seeTable 10.21 ).

M u = 13 , 684 kips-ft < φ f M n = (1.0)( 25 ,351 ) =25 ,351 kips-ft O.K.

7. Shear Strength Design, Strength Limit State I7.1) Nominal Shear Resistance, V n

(a) V n for an unstiffened web (AASHTO Article 6.10.7.2)

V n =V p = 0.58F yw Dt w For D

t w ≤ 2.46 EF yw

1.48 t 2w EF yw For 2.46 EF yw

< Dt w ≤ 3.07 E

F yw

4.55t 3w ED For D

t w> 3.07 E

F yw

(10.57)

where D isdepth of web and t w isthicknessof web.

... Dt w =

960.625 =153 .6 > 3.07 EF yw

3.07 29 , 00050 =73 .9

... V n =4.55 t 3w E

D =4.55 (0.625 )3(29 , 000 )

96 =335 .6 kips (1,493 kN)

(b) V n for an end-stiffened web panel (AASHTO Article 6.10.7.3.3c)


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V n = CV p (10.58)

C =

1.0 For Dt w < 1.10

Ek

F yw

1.10(D/t w )

Ek

F yw For 1.10 Ek

F yw ≤Dt w ≤ 1.38

EkF yw

1.52(D/t w )2 Ek

F ywFor D

t w > 1.38 EkF yw

(10.59)

k = 5 +5

(d o /D) 2(10.60)

in which d o is the spacingof transversesti ffeners (Figure 10.77 ).

FIGURE10.77: Typical steel girder dimensions.

For d o =240 in. and k = 5 + 5(240 / 96 )2 =5.80

...

Dt w = 153 .6 > 1.38

EkF yw = 1.38

29 ,000 (5.8)50 = 80

... C =152

(153 .6)2 = 29 ,000 (5.80 )50 = 0.374

V p = 0.58F yw Dt w =0.58(50)( 96 )( 0.625 ) =1,740 kips (7,740 kN)

V n = CV p = 0.374 (1740 ) = 650 .8 kips (2, 895 kN)


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(c) V n for interior-sti ffened web panel (AASHTO Article 6.10.7.3a)

V n =

V p C +0.87 (1−C)√ 1+(d o /D) 2

For M u ≤ 0.5φf M p

RV p C +0.87 (1−C)√ 1+(d o /D) 2 ≥CV p For M u > 0.5φf M p

(10.61)

where

R = 0.6 +0.4φf M n −M u

φf M n −0.75φf M y ≤ 1.0 (10.62)

7.2) Strength Limi t State I AASHTO-LRFD [ 1] requiresthat for strength limit stateI

V u ≤φν V n (10.63)

where φν istheshear resistancefactor =1.0.

(a) Left end of span 1:

... V u = 445 .4 kips > φ ν V n (for unstiffened web) =335 .6 kips... Stiffenersare needed to increaseshear capacity.

In order to facili tatehandlingof web panel sections, the spacing of transversestiff-eners shall meet (AASHTO Article6.10.7.3.2) the following requirement:

d o ≤D

260(D/t w )

2

(10.64)

Try d o =240 in. for end-sti ffened web panel

d o =240 in. < D260

(D/t w )

2

= 96260

96 / 0.625

2

=275 in. O.K.

and then

φν V n = (1.0)650 .8 = 650 .8 kips > V u = 445 .4 kips O.K.

(b) Location of therst intermediatestiffeners, 20ft (6.1m) from theleft end in span 1:Factored shear for this location can be obtained using l inear interpolation fromTable 10.22 . Since V u = 328.0 kips (1459 kN) is less than the shear capacity of the unsti ffened web, φν V n = 335.5 kips (1492 kN), the intermediate transversesti ffenersmay beomit ted after therst intermediatestiffeners. Similar calculationscan beused to determine theremainingstiffenersalongthegirder.

8. FatigueDesign, Fatigueand Fracture Limit StateThebasemetal at theconnectionplateweldstoanges, and webslocatedat 96ft (29.26m)(0.6L 1) from theleft end of span 1 will bechecked for thefatigueload combination.


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8.1) Load-Induced Fatigue (AASHTO Article6.6.1.2)Thedesign requirements for load-induced fatigueapply only to (1) detailssubjected to anet applied tensile stressand (2) regionswhere the unfactored permanent loadsproducecompression, and only if the compressive stress is less than twice the maximum tensile

stress resulting from thefatigue load combination. In the fatiguelimit state, all stressesarecalculated using theelastic section propert ies(Tables 10.24 to 10.26 ).

(a) Top-angeweldThecompressivestressat the top-angeweld dueto unfactored permanent loadsisobtained:

f DC =M DC 1(y st −t f c )

I girder +(M DC 2 +M DW )(y st −t f c )

I com −3n

=2462 (12 )( 55 .087 −1.0)

159 ,619 +(251 +455 )( 12 )( 33 .367 −1.0)

312 ,155

= 10 .89 ksi (75 .09 MPa)

Assume that the negativefatiguemoments are carr ied by the steel section only inthe positive exureregion. Themaximum tensile stress at the top-ange weld atthis location dueto factored fatiguemoment is

f LL +I M = −(M LL +I M )u yst −t f c

I girder =798 (12 )( 54 .087 )

159 , 619

= 3.25 ksi (22 .41 MPa)... f DC =10 .89 ksi > 2f LL +I M =6.49 ksi... no need to check fatiguefor the top-angeweld

(b) Bottom-angeweld

• Factored fatiguestressrange, ( f ) uFor the posit ive exure region, weassume that posit ive fatiguemoments areapplied to theshort- term compositesection and negativefatiguemomentsareapplied to thenoncompositesteel section only.

( f ) u =(M LL +I M )u ysb −n − t f t

I com−n + −(M LL +I M )u ysb − t f t

I girder

=1465 (12 )( 82 .454

−1.75)

432 ,707 +798 (12 )( 43 .663

−1.75 )

159 , 619

= 5.79 ksi (39 .92 MPa)

• Nominal fatigueresistancerange, ( F) nFor let-welded connectionswith weld linesnormal to the direction of stress,the basemetal at transversestiffeners to angeweldsisfatiguedetail categoryC (AASHTO Table 6.6.1.2.3.-1).


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( F) n =AN

1/ 3

≥12

( F) T H (10.65)

where A is aconstant dependent on detail category =44(10)8

for category Cand

N = (365 )( 75)n(ADT T ) ST (10.66)

ADTT ST = p(ADT T ) (10.67)

where p is afraction of atruck in asinglelane(AASHTO Table3.6.1.4.2-1) =0.8 for three-lane trafc, and n is the number of stress-rangecyclesper truckpassage(AASHTO Table 6.6.1.2.5-2) =1.0 for thepositiveexureregion.

N = (365 )( 75)( 1.0)( 0.8)( 3600 ) =7.844 (10 )7

For category C detail, ( F) T H

=12ksi (AASHTO Table 6.6.1.2.5-3).

... AN

1/ 3

=44 (10 )8

7.844 (10 )7

1/ 3

= 3.83 ksi <12

( F )T H = 6 ksi

... ( F) n =12

( F) T H = 6 ksi (41 .37 MPa)

• Fatiguelimit stateAASHTO requiresthat each detail shall sati sfy:

( f ) u ≤ ( F) n (10.68)

For top-angeweld( f ) u = 5.79 ksi < ( F) n = 6 ksi O.K.

8.2) Fati gue Requirements for Web (AASHTO Article6.10.4)Thepurpose of these requirements is to control out-of-plane exing of theweb duetoexure and shear under repeated liveloadings. The repeated live load is taken as twicethe factored fatigueload.

(a) Flexure requirement

f cf ≤

Rh F yc For 2D ct w ≤ 5.76 EF yc

Rh F yc 3.58 −0448 2D ct w F yc

E For 5.76 EF yc ≤2D ct w ≤6.43 EF yc

28 .9Rh E t w2D

c

2For 2D c

t w

> 6.43

E

F yc

(10.69)

where f cf is the maximum elastic exural stress in the compression angeduetotheunfactored permanent loadsand repeatedliveloadings; F yc istheyield strengthof thecompression ange; and D c is thedepth of theweb in compression.

• Depth of web in compression, DcConsidering the algebraic sum of stressesactingon different sectionsbased onelasti c section propert ies, D c can beobtained by thefollowing formula:


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D c =f DC 1 +f DC 2 +f DW +f LL +IM f DC 1

yst +f DC 2+f DW

yst −3n +f LL +IM

yst −n− t f c

=M DC 1

S st +M DC 2

+M DW

S st −3n +2(M LL

+IM )u

S st −nM DC 1I girder +

M DC 2+M DW I com −3n +

2(M LL +IM )uI com−n

− t f c (10.70)

Substi tut ing moments (Tables 10.21 and 10.23 ) and section propert ies (Ta-bles 10.24 and 10.26 ) into Equation 10.70 , weobtain:

D c =4260 (12 )

2, 898 +(435+788 )( 12 )

9,355 +2(1607 )( 12 )

26 ,5534260 (12 )159 , 629 +

(435+788 )( 12 )312 ,155 +

2(1607 )( 12 )432 ,707

−1

=17 .640 +1.569 +1.4520.320 +0.047 +0.089 −1 = 44 .29 in. (1,125 mm )

2D c

t w = 2(44 .29)0.625 = 141 .7 < 5.76 E

F yc = 183 .7

• Maximum compressivestressin ange, f cf (at location 0.4L 1)

f cf = f DC 1 +f DC 2 +f DW +f LL +I M

=M DC 1

S st +M DC 2 +M DW

S st −3n +2(M LL +I M )u

S st −n

= 17 .64 +1.57 +1.45 = 20 .66 < R h F yc = 50 ksi

(b) Shear (AASHTO Art icle10.6.10.4.4)Theleft end of span 1 is checked asfollows:

• Fatigueload

V u = V DC 1 +V DC 2 +V DW +2(V LL +I M )u

= 148 .4 +15 .1 +27 .4 +2(51 .1) =293 .1 kips (1304 kN)

• Fatigueshear stress

νcf =V u

Dt w =293 .1

96 (0.625 ) = 4.89 ksi (33 .72 MPa)

• Fatigueshear resistance

C = 0.374 (seeStep 7)νn = 0.58CF yw = 0.58 (0.374 )( 50 )

= 10 .85 ksi > ν cf = 4.89 ksi O.K.


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8.3) Distortion-Induced Fatigue (AASHTO Article6.6.1.3)All transverse connection plates wil l be welded to both the tension and compressionanges to provide rigid load paths so distorti on-induced fatigue (the development of signicant secondary stresses) can beprevented.

8.4) FractureLimit State (AASHTO Article6.6.2)Materials for main load-carrying components subjected to tensile stresseswill meet theCharpy V-notch fracture toughnessrequirement (AASHTO Table 6.6.2-2) for tempera-ture zone2 (AASHTO Table6.6.2-1).

9. Intermediate TransverseStiffener DesignThe intermediate transverse stiffener consists of two plates welded to both sides of theweb. Thedesign of therst intermediatetransversesti ffener isdiscussed in thefollowing.9.1) Projecti ng Width, bt , Requirements (AASHTO Article6.10.8.1.2)Toprevent local bucklingof thetransversestiffeners, thewidth of each projectingstiffenershall satisfy theserequirements:

2.0 + d 30

0.25

bf ≤

bt

≤

0.48 t p

E

F ys

16 t p(10.71)

where bf isthefull widthof thesteel angeand F ys isthespeciedminimumyield strengthof the stiffener. To allow adequate spacefor cross-frameconnections, try stiffener widthbt =6 in. (152 mm):

bt =6 in. > 2.0 + d 30 =2.0 + 98 .75

30 = 5.3 in.0.25bf = 0.25(18) = 4.5 in. O.K.

Try t p =0.5 in. (13mm) and obtain:

bt = 6 in. < 0.48 t p EF ys = 0.48 (0.5) 29 ,000

36 =6.8 in.16 t p = 16(0.5) = 8 in. O.K.

Usetwo 6 in. x 0.5 in. (152 mm x 13 mm) transversestiffener plates.9.2) Moment of Inertia Requirement (AASHTO Article6.10.8.1.3)Thepurposeof thisrequirement isto ensure sufcient rigidity of transversestiffeners toadequately develop atension eld in theweb.

I t ≥ d o t 2w J (10.72)

J = 2.5D p

d o

2

−2.0 ≥0.5 (10.73)

where I t isthemoment of inertiafor thetransversestiffener takenabouttheedgein contactwith thewebfor singlestiffenersandabout themid-thicknessof thewebfor stiffener pairs(Figure 10.78 ); D p is the web depth for webswithout longitudinal sti ffeners.


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FIGURE10.78: Cross-section of web and transversestiffener.

... J = 2.596

240

2

−2.0 = −1.6 < 0.5 ... Use J = 0.5

I t = 263(0.5)

12 +(6)( 0.5)( 3.313 )2 =83 .86 in.4

> d o t 2w J = (240 )( 0.625 )2(0.5) = 46 .88 in.4 O.K.

9.3) Area Requirement (AASHTO Article6.10.8.1.4)Thisrequirement ensuresthat transversesti ffenershavesufcient areato resist thevert icalcomponent of the tension eld, and is only applied to t ransverse sti ffeners required tocarry the forcesimposed by tension-eld action.

A s ≥A s min = 0.15BDt w (1 −C)V u

φν V n −18 t 2wF yw

F ys(10.74)

where B =1.0 for stiffener pairs. From thepreviouscalculation:

C = 0.374 , F yw = 50 ksi, F ys = 36 ksi

V u =328 .0 kips, φf V n =335 .5 kips, t w =0.625 in.

As = 2(6)( 0.5) = 6 in.2 > A s min

= 0.15(1.0)( 96 )( 0.625 )( 1 −0374 )328 .0335 .5 −18(0.625 )2 50

36

= −0.635 in.2


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Thenegativevalueof A s min indicatesthat theweb hassufcient areato resist thevert icalcomponent of thetension eld.

10. Shear Connector DesignIn acompositegirder, stud or channel shear connectorsmust beprovided at theinterfacebetween the concrete deck slab and the steel section to resist the interface shear. For acompositebridgegirder, the shear connectors should benormally provided throughoutthe length of the bridge(AASHTO Article 6.10.7.4.1). Stud shear connectorsarechosenin thisexample and will bedesigned for the fatigue limit stateand then checked againstthe strength l imit state. The detailed calculations of the shear stud connectors for theposit iveexure region of span 1 are given in the following. A simi lar procedure can beused to design theshear studsfor other portionsof thebridge.10.1) Stud Size (AASHTO Article6.10.7.4.1a)To meet the limits for cover and penetration for shear connectors specied in AASHTOArticle6.10.7.4.1d, try:

Stud height, H stud = 7 in. > t h +2 = 3.375 +2 = 5.375 in. O.K.Stud diameter, d stud =0.875 in. < H stud / 4 = 1.75 in. O.K.

10.2) Pitch of Shear Stud, p , for FatigueLimit State

(a) Basic requirements (AASHTO Article 6.10.7.4.1b)

6d stud ≤ p =n stud Z r I com −n

V sr Q ≤ 24 in. (10.75)

where n stud is the number of shear connectors in a cross-section; Q is the rstmoment of transformedsection (concretedeck) about theneutral axisof theshort-term composite section; V sr is the shear force range in the fatigue limit state; andZ r isthe shear fatigueresistanceof an individual shear connector.

(b) Fati gueresistance, Z r (AASHTO Article6.10.7.4.2)

Z r = αd 2stud ≥ 5.5d 2stud (10.76)

α = 34 .5 −4.28 log N (10.77)

where N is the number of cycles specied in AASHTO Art icle 6.6.1.2.5, N =7.844(10) 7 cycle(seeStep 8).

α = 34 .5 −4.28 log (7.844 ×107) =0.711 < 5.5

Z r = 5.5d 2stud = 5.5(0.875 )2 =4.211 ksi

(c) First moment, Q , and moment of initi al, I com−n (seeTable 10.25 )


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Q =beff t s

9yst −n −t h +

t s2

=140 (10 .875 )

916 .296

+3.375

+10 .875

2 =4247 .52 in.3

I com −n = 432 ,707 in.4

(d) Required pitch for thefatiguelimit stateAssume that shear studsare spaced at 6 in. transversely across the top ange of asteel section (Figure 10.75 ) and, using n stud =3 for this example, obtain

P required =3(4.211 )( 432 ,707 )

V sr (4,247 .52 ) =1,286 .96

V sr

The detailed calculations for the posit ive exure region of span 1 are shown inTable 10.27 .

TABLE 10.27 Shear Connector Design forthe Posit iveFlexureRegion in Span 1

Location V sr P required P nal(x/L) (kips) (in.) (i n.) n total-stud

0.0 60.1 21.4 12 30.1 51.6 24.9 12 510.2 47.8 26.9 18 990.3 46.2 27.9 18 1320.4 45.7 28.2 18 1650.4 45.7 28.2 12 1620.5 46.2 27.9 12 1140.6 47.8 26.9 12 660.7 50.3 25.6 9 3

Note:

V sr = + V LL +IM u + − V LL +IM uP required =

nstud Z r I com−nV sr Q = 1286 .96

V sr

n total-stud is the summation of number of shear studs be-tween the locationsof thezero moment and that location.

10.3) Str ength Limi t StateCheck

(a) Basic requirement (AASHTO Art icle6.10.7.4.4a)The result ing number of shear connectors provided between the section of max-imum positivemoment and each adjacent point of zero moment shall sati sfy thefollowing requirement:

n total-stud ≥V h

φsc Q n(10.78)

where φsc is the resistance factor for shear connectors, 0.85; V h is the nominalhorizontal shear force; and Q n is the nominal shear resistanceof one stud shearconnector.

(b) Nominal horizontal shear force(AASHTO Article6.10.7.4.4b)


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V h = the lesser of 0.85f c beff t sF yw Dt w +f yt bf t t f t +F yc bf c t f c

(10.79)

V h−concrete = 0.5f c beff t s =0.85(3.25 )( 140 )( 10 .875 ) = 4, 206 kipsV h−steel = F yw Dt w +F yt bf t t f t +F yc bf c t f c

= 50 [ (18)( 1.0) +(96)( 0.625 ) +(18 )( 1.75 )] = 5, 475 kips... V h = 4,206 kips (18 ,708 kN)

(c) Nominal shear resistance(AASHTO Art icle 6.10.7.4.4c)

Q n = 0.5Asc f c E c ≤Asc F u (10.80)

where A sc isacross-sectional areaof astud shear connector and F u isthespeciedminimum tensilestrength of a stud shear connector =60ksi (420 MPa).

... 0.5 f c E c =0.5 3.25(3,250 ) = 51 .4 kips < F u = 60 kips

... Q n =0.5A sc f c E c = 51 .4π( 0.875 )2

4 =30 .9 kips

(d) Check result ing number of shear stud connectors (seeTable 10.27 )

n total-stud =165 from left end 0.4L 1162 from 0.4L 1 to 0.7L 1

>V h

φsc Q n =4206

0.85(30 .9) =160 O.K.

11. Constructabil ity CheckFor unshored construction, AASHTO requires that all I-section bending members beinvestigated for strength and stabil it y during construction stagesusing appropriateloadcombinationsgiven in AASHTO Table 3.4.1-1. The following checksare made for thesteel girder section only under factored dead load, DC 1. It isassumed that thenal totaldead load, DC 1, producesthecontrollingmaximum moments.11.1) Web SlendernessRequi rement (AASHTO 6.10.10.2.2)

2D c

t w ≤ 6.77 Ef c

(10.81)

where f c is the stressin compression angedueto the factored dead load, DC 1, and D cis thedepth of theweb in compression in theelastic range.


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D c = yst −t f c = 55 .087 −1 = 54 .087 in. (1,374 mm )

f c =(0.95)( 1.25)M DC 1

S st =0.95(1.25 )( 4260 )( 12)

2,898 = 20 .95 ksi (145 MPa)

2D c

t w =2(54 .087 )

0.625 = 173 .1 ≤ 6.77 Ef c = 6.77 29 ,000

20 .95 = 251 .9 O.K.

... no longitudinal sti ffener is required

11.2) Compression FlangeSlendernessRequi rement (AASHTO Art icle6.10.10.2.3)This requirement prevents the local buckling of the top ange before the concrete deckhardens.

bf

2t f ≤ 1.38E

f c 2D ct w

(10.82)

bf

2t f =18

2(1.0) = 9 ≤ 1.38E

f c 2D ct w

= 1.38 29 , 000

20 .95√ 173 .1 =14 .2 O.K.

11.3) Compression FlangeBracing Requirement (AASHTO Art icle 6.10.10.2.4)

(a) Flexure(AASHTO Article6.10.6.4.1)To ensure that a noncomposite steel girder hassufcient exural resistancedur ingconstruction, themoment capacityshould becalculatedconsideringlateral torsionalbuckling with an unbraced length, L b (Figure 10.77 ).For a steel girder without longitudinal sti ffeners and (2D c /t w ) > λ b E/F yc , thenominal exural resistanceis

M n =1.3Rh M y ≤M p For L b ≤L p

Cb Rb Rh M y 1 −0.5L b−L pL p −L r ≤Rb Rh M y For L p < L b ≤L r

Cb Rb RhM y

2L rL b

2

≤Rb Rh M y For L b > L r

(10.83)

L p ≤ 1.76 r t EF yc

(10.84)

L r = 19 .71I yc d S xc

EF yc

(10.85)

where λ b equals 4.64 for a member with a compression ange area less than thetension angeareaand 5.76 for members with a compression angeareaequal toor greater than the tension angearea; r t is theminimum radiusof gyration of thecompression angeof the steel section about the vert ical axis; S xc is the sectionmodulusabout thehorizontal axis of thesection to the compression ange(equalto S st in Table 10.24 ); Cb is the moment gradient correction factor; and R b is aangestressreduction factor consideringlocal bucklingof aslender web (AASHTOArticle6.10.5.4.2).


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Cb =1.75 −1.95P lP h +0.3

P lP h

2

≤ 2.3 (10.86)

where P l is the forcein thecompression angeat the braced point with the lowerforcedueto the factored loading, and P h is the forcein the compression angeatthebraced point with higher forcedueto thefactored loading. Cb isconservativelytaken as1.0 in this example.

r t = I yf

A f = (18)3(1.0)/ 12(18)( 1.0) = 5.20 in. (132 mm )

L p = 1.76r t EF yc = 1.76(5.2) 29 ,000

50 = 220 in. < L b = 240 in.

L r = 19 .71 (486 )( 98 .75)

2,898

29 ,000

50 =435 in. (11 ,049 mm )

2D c

t w = 173 .1 < λ b Ef c =4.64 29 ,000

20 .95 = 172 .6

Sincethesetwo valuesarevery close, take Rb =1.0 (AASHTO Article 6.10.5.4.2).

M y = S st F y =2, 898 (50 ) = 144 ,900 kips-in. = 12 ,075 kips-ft... L p = 220 in. < L b =240 in. < L r =435 in.

... M n

=(1.0)( 1.0)( 1.0)( 12 ,075 ) 1

−0.5

240 −220

435 −220= 11 ,513 kips-ft < R n Rh M y = 12 , 075 kips-ft

M u = 0.95(1.25 )( 4,260 ) = 5,059 kips-ft (6,859 kN-m)

< φ f M n +(1.0)( 11 ,513 )

= 11 ,513 kips-ft (15 , 609 kN-m) O.K.

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