DESIGN OF HORIZONTALLY CURVED GIRDERS

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DESIGN OF HORIZONTALLY DESIGN OF HORIZONTALLY CURVED GIRDERS CURVED GIRDERS

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CURVED BRIDGES ARE PROVIDEDCURVED BRIDGES ARE PROVIDED

BECAUSE OF LIMITED RIGHTS OF WAYBECAUSE OF LIMITED RIGHTS OF WAY TO MITIGATED TRAFFIC PROBLEMTO MITIGATED TRAFFIC PROBLEM TO SIMPLIFY COMPLICATED GEOMETRICESTO SIMPLIFY COMPLICATED GEOMETRICES

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OBJECTIVEOBJECTIVE

The objective of this presentation is to furnish The objective of this presentation is to furnish design of curved bridges with a basic design of curved bridges with a basic

understanding of the configuration, geometric understanding of the configuration, geometric approximations and structural behavior and their approximations and structural behavior and their

interrelationship in simple structural loading interrelationship in simple structural loading system.system.

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CURVED BRIDGE CURVED BRIDGE CONFIGURATIONSCONFIGURATIONS

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USE OF CHORDSUSE OF CHORDS

Curved bridge beams are made as a Curved bridge beams are made as a series of short segments/chords to series of short segments/chords to approximate the theoretical arc. approximate the theoretical arc.

Max offset b/w Arc and its Chord is Max offset b/w Arc and its Chord is s = Lcs = Lc22/8R/8RWhere, Where, Lc = Chord Length/Arc Length Lc = Chord Length/Arc Length R = Radius of CurvatureR = Radius of Curvature

Arc-to-Chord offset be limited to 1.5 ft.Arc-to-Chord offset be limited to 1.5 ft.

CURVED BRIDGE CONFIGURATIONS

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BEAM CROSS-SECTIONBEAM CROSS-SECTION BOX BEAMS VERSUS I-BEAMSBOX BEAMS VERSUS I-BEAMS

For full-span-length curved beams made in plant (using post tensioning), box section is For full-span-length curved beams made in plant (using post tensioning), box section is preferred because of high torsional rigidity and handling considerations.preferred because of high torsional rigidity and handling considerations.

For segmental construction I-beams may be used. Straight segments are supported on For segmental construction I-beams may be used. Straight segments are supported on temporary shores and post-tensioned in the field after constructing diaphragm at the segment temporary shores and post-tensioned in the field after constructing diaphragm at the segment joints.joints.

CONTINUITYCONTINUITYContinuity is very desirable in curved bridges. It reduces the effects of torsion.Continuity is very desirable in curved bridges. It reduces the effects of torsion.

CROSSBEAMSCROSSBEAMSCrossbeams/Intermediated diaphragm are required to counteractCrossbeams/Intermediated diaphragm are required to counteract Effects of torsionEffects of torsion Lateral forces resulting from curvatureLateral forces resulting from curvature

CURVED BRIDGE CONFIGURATIONS

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PRELIMINARY DESIGN PRELIMINARY DESIGN

USEFUL GEOMETRIC APPROXIMATIONSUSEFUL GEOMETRIC APPROXIMATIONS USEFUL STRUCTURAL APPROXIMATIONSUSEFUL STRUCTURAL APPROXIMATIONS

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USEFUL GEOMETRIC APPROXIMATIONSUSEFUL GEOMETRIC APPROXIMATIONS

Arc to chord offset is Arc to chord offset is s = Ls = Lcc

22/8R/8R

Excess of arc length over chord length Excess of arc length over chord length Arc length = (8sArc length = (8s22/3L/3Lcc) x Chord length) x Chord length

Center of gravity of an arcCenter of gravity of an arcC.G. of an arc is offset from the chord by 2s/3, or C.G. of an arc is offset from the chord by 2s/3, or LLcc

22/12R/12R

Curved surfacesCurved surfacesC.G. of the curved surface lies outside the center of C.G. of the curved surface lies outside the center of gravity of the centerline arc by an amount “e”.gravity of the centerline arc by an amount “e”.e = Be = B22/12R/12R

PRELIMINARY DESIGN

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USEFUL STRUCTURAL APPROXIMATIONSUSEFUL STRUCTURAL APPROXIMATIONSARE DISCUSSED IN THE CONTEXT ARE DISCUSSED IN THE CONTEXT

OFOF STRUCTURAL BEHAVIOUR OF CURVED-STRUCTURAL BEHAVIOUR OF CURVED-

BEAM BRIDGESBEAM BRIDGES

PRELIMINARY DESIGN

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STRUCTURAL BEHAVIOUR OF CURVED-BEAM BRIDGESSTRUCTURAL BEHAVIOUR OF CURVED-BEAM BRIDGES

Longitudinal FlexureLongitudinal Flexure TorsionTorsion CrossbeamsCrossbeams

PRELIMINARY DESIGN

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Longitudinal FlexureLongitudinal Flexure

Analysis as a straight beamAnalysis as a straight beam

For vertical load the beam is analyzed as a straight beam with span length equal to the arc length.For vertical load the beam is analyzed as a straight beam with span length equal to the arc length.

Loads on outside beamLoads on outside beam

The shears and moments in the outside-exterior beam are larger than other beams. This is caused The shears and moments in the outside-exterior beam are larger than other beams. This is caused

by the following factors:by the following factors:

-Arc length on the outside of the curve is longer than the nominal length at the centerline of the -Arc length on the outside of the curve is longer than the nominal length at the centerline of the bridge.bridge.

-Other beams will shed some of the their torsional moment by shifting load toward the next beam -Other beams will shed some of the their torsional moment by shifting load toward the next beam to the outside. The outermost beam is the final resting place for this shifted load.to the outside. The outermost beam is the final resting place for this shifted load.

PRELIMINARY DESIGN

STRUCTURAL BEHAVIOUR OF CURVED-BEAM BRIDGES

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Torsion Torsion It will be seen that torsional moment are related to the flexural moment M divided by the radius of It will be seen that torsional moment are related to the flexural moment M divided by the radius of curvature R.curvature R.

TorsionTorsion in a Simple-Span Curved Beamin a Simple-Span Curved Beam Consider a short segment near mid-spanConsider a short segment near mid-span

PRELIMINARY DESIGN


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Torsion in a Fixed-Ended BeamTorsion in a Fixed-Ended Beam

Continuous span are intermediated between simple-span and fixed-ended beams.

Interior spans resemble the fixed case more closely, and the free ends of the exterior spans may be closer to the simple-span case.

PRELIMINARY DESIGN


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CrossbeamsCrossbeams

Crossbeams must be designed for the shears and moments resulting from the change in Crossbeams must be designed for the shears and moments resulting from the change in direction of the primary bending moment at the location of the crossbeams.direction of the primary bending moment at the location of the crossbeams.

The longitudinal forces in the bottom flange have a transverse component at the location of The longitudinal forces in the bottom flange have a transverse component at the location of the crossbeam. the crossbeam.

The crossbeam must be deep enough to brace the bottom flange to resist this component.The crossbeam must be deep enough to brace the bottom flange to resist this component.

PRELIMINARY DESIGN


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DETAILED DESIGNDETAILED DESIGN

Detailed design is done by using a beam gridwork computer model.Detailed design is done by using a beam gridwork computer model. Computer model may be created in a horizontal plane, ignoring grade and super Computer model may be created in a horizontal plane, ignoring grade and super

elevation.elevation. The extra weight caused by super elevation should be taken into account.The extra weight caused by super elevation should be taken into account.

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Loading Stages-I-beamsLoading Stages-I-beams

1.1. Individual SegmentsIndividual SegmentsThe segments are pre-tensioned in the plant to compensate for self-weight bending of the The segments are pre-tensioned in the plant to compensate for self-weight bending of the individual segment.individual segment.

2.2. Shoring LoadsShoring LoadsThe individual segments are erected in the field, supported by final bearings and by shores The individual segments are erected in the field, supported by final bearings and by shores at intermediate locations. Post-tensioning ducts are spliced and crossbeams are cast.at intermediate locations. Post-tensioning ducts are spliced and crossbeams are cast.During this loading stage, stresses in the beams do not change. Loads are added to the During this loading stage, stresses in the beams do not change. Loads are added to the shoring.shoring.

3.3. Non-composite GridworkNon-composite GridworkPost-tensioning is applied to the non-composite gridwork after the crossbeams have cured Post-tensioning is applied to the non-composite gridwork after the crossbeams have cured sufficiently. This lifts the beams from the shores. The load that was present in the shores sufficiently. This lifts the beams from the shores. The load that was present in the shores becomes a load applied to the non-composite beam gridwork.becomes a load applied to the non-composite beam gridwork.The weight of the deck and haunch is applied to the non-composite gridwork.The weight of the deck and haunch is applied to the non-composite gridwork.

DETAILED DESIGN

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4.4. Composite GridworkComposite GridworkThe weights of future wearing surface, barriers, live load plus impact, and centrifugal force The weights of future wearing surface, barriers, live load plus impact, and centrifugal force are applied to the composite gridwork. The simplifying assumptions for distribution of are applied to the composite gridwork. The simplifying assumptions for distribution of these loads in straight bridges cannot be used for curved bridges.these loads in straight bridges cannot be used for curved bridges.

5.5. Other Design ChecksOther Design Checks-- Allowable stressesAllowable stresses

-- Deflection and camberDeflection and camber

-- Prestress lossesPrestress losses

-- Ultimate strengthUltimate strength

-- Torsion (additional consideration)Torsion (additional consideration)

DETAILED DESIGN

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DESIGN PROBLEMDESIGN PROBLEM

Data Data -- radius 600ftradius 600ft-- span length 120ft(measured along the arc span length 120ft(measured along the arc

at the centerline of the bridge)at the centerline of the bridge)-- superelevation 6%superelevation 6%-- design speed 40mphdesign speed 40mph-- girder cross section girder cross section → bulb-tee-beam → → bulb-tee-beam →

6 in number spaced at 9ft6 in number spaced at 9ft-- deck slab 8”deck slab 8”-- wearing surface ½”wearing surface ½”-- design live load is HL-93design live load is HL-93

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Plan geometryPlan geometry- Arc-to-Chord offset LArc-to-Chord offset Laa

22/8R = (120/8R = (12022) / (8x600) = 3ft >1.5ft (min. recommended offset)) / (8x600) = 3ft >1.5ft (min. recommended offset)- Subdividing the beam into three segmentsSubdividing the beam into three segments

max offset = (120/3)max offset = (120/3)22 / (8x600) / (8x600) = .3333ft or 4” → barely detectable visually and acceptable.= .3333ft or 4” → barely detectable visually and acceptable.

- Over hangs: from beam centerline on the outside will be 2’-8”and 3’-4” on the inside.Over hangs: from beam centerline on the outside will be 2’-8”and 3’-4” on the inside.

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MaterialsMaterials• Cast-in-place slab: Actual thickness, ts =8.0 in.Cast-in-place slab: Actual thickness, ts =8.0 in.

Structural thickness =7.5 in.Structural thickness =7.5 in.Note that a 1/2-in. wearing surface is considered an integralpart of the 8-in. deck.Note that a 1/2-in. wearing surface is considered an integralpart of the 8-in. deck.Concrete strength at 28 days, f ´ c =4.0 ksiConcrete strength at 28 days, f ´ c =4.0 ksi

• Precast beams: AASHTO-PCI Bulb-tee with 2-in.-added width as shown in Figure 12.9.1.2-1Precast beams: AASHTO-PCI Bulb-tee with 2-in.-added width as shown in Figure 12.9.1.2-1Concrete strength of beam at post-tensioning, f ´ ci =6.5 ksiConcrete strength of beam at post-tensioning, f ´ ci =6.5 ksiConcrete strength at 28 days, f ´ c =6.5 ksiConcrete strength at 28 days, f ´ c =6.5 ksiConcrete unit weight, wc =0.150 kcfConcrete unit weight, wc =0.150 kcfDesign span =120.0 ft (Arc length at centerline of bridge)Design span =120.0 ft (Arc length at centerline of bridge)

• Post-tensioning strands: 0.6-in. dia, seven-wire, low-relaxationPost-tensioning strands: 0.6-in. dia, seven-wire, low-relaxationArea of one strand =0.217 in.2Area of one strand =0.217 in.2Ultimate strength, fpu =270.0 ksiUltimate strength, fpu =270.0 ksiYield strength, fpy =0.9fpu =243.0 ksi [LRFD Table 5.4.4.1-1]Yield strength, fpy =0.9fpu =243.0 ksi [LRFD Table 5.4.4.1-1]Stress limits for post-tensioning strands: [LRFD Table 5.9.3-1]Stress limits for post-tensioning strands: [LRFD Table 5.9.3-1]at jacking: fpj =0.80fpu =216.0 ksiat jacking: fpj =0.80fpu =216.0 ksiat service limit state (after all losses):at service limit state (after all losses):fpe <0.80fpy =194.4 ksifpe <0.80fpy =194.4 ksiModulus of elasticity, Ep =28,500 ksi [LRFD Art. 5.4.4.2]Modulus of elasticity, Ep =28,500 ksi [LRFD Art. 5.4.4.2]

• Reinforcing bars: Yield strength, fy =60 ksiReinforcing bars: Yield strength, fy =60 ksiModulus of elasticity, Es =29,000 ksi [LRFD Art. 5.4.3.2]Modulus of elasticity, Es =29,000 ksi [LRFD Art. 5.4.3.2]

• Future wearing surface: 2 in. additional concrete, unit weight =0.150 kcfFuture wearing surface: 2 in. additional concrete, unit weight =0.150 kcf• New Jersey-type barrier: Unit weight =0.300 kip/ft/sideNew Jersey-type barrier: Unit weight =0.300 kip/ft/side

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Cross-Section PropertiesCross-Section Properties• Non-Composite SectionsNon-Composite Sections

A =area of cross-section of precast beam =911 in.2A =area of cross-section of precast beam =911 in.2

h =overall depth of beam =72 in.h =overall depth of beam =72 in.

I =moment of inertia about the centroid of the precast beam =608,109 in.4I =moment of inertia about the centroid of the precast beam =608,109 in.4

yb =distance from centroid to extreme bottom fiber of the precast beam =36.51 in.yb =distance from centroid to extreme bottom fiber of the precast beam =36.51 in.

yt =distance from centroid to extreme top fiber of the precast beam =35.49 in.yt =distance from centroid to extreme top fiber of the precast beam =35.49 in.

Sb =section modulus for the extreme bottom fiber of the precast beam =16,657 in.3Sb =section modulus for the extreme bottom fiber of the precast beam =16,657 in.3

St =section modulus for the extreme top fiber of the precast beam =17,134 in.3St =section modulus for the extreme top fiber of the precast beam =17,134 in.3

Ilat =lateral moment of inertia of precast beam =46,014 in.4Ilat =lateral moment of inertia of precast beam =46,014 in.4

wg =beam self-weight per unit length =0.949 kip/ftwg =beam self-weight per unit length =0.949 kip/ft

Ec =modulus of elasticity, ksi =33,000(wc)Ec =modulus of elasticity, ksi =33,000(wc)1.5 1.5 √√fc [LRFD Eq. 5.4.2.4-1]fc [LRFD Eq. 5.4.2.4-1]

wherewhere

wc =unit weight of concrete =0.150 kcfwc =unit weight of concrete =0.150 kcf

The The LRFD SpecificationsLRFD Specifications, Commentary C5.4.2.4, indicates that the unit weight of, Commentary C5.4.2.4, indicates that the unit weight of

normal weight concrete is 0.145 kcf. However, precast concrete mixes typically have a relatively low normal weight concrete is 0.145 kcf. However, precast concrete mixes typically have a relatively low water/cementitious materials ratio and high density. Therefore, a unit weight of 0.150 kcf is used in this water/cementitious materials ratio and high density. Therefore, a unit weight of 0.150 kcf is used in this example. For high strength concrete, this value may need to be increased based on test results.example. For high strength concrete, this value may need to be increased based on test results.

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Fc’ =specified strength of concrete, ksiFc’ =specified strength of concrete, ksi

Therefore, the modulus of elasticity for:Therefore, the modulus of elasticity for:

cast-in-place slab, Ecs =33,000(0.150)1.5 =3,834 ksicast-in-place slab, Ecs =33,000(0.150)1.5 =3,834 ksi

precast beam at transfer of post-tensioning (at 28 days minimum)precast beam at transfer of post-tensioning (at 28 days minimum)

Eci =33,000(0.150)1.5 =4,888 ksiEci =33,000(0.150)1.5 =4,888 ksi

precast beam at service loads, Ec =33,000(0.150)1.5 =4,888 ksiprecast beam at service loads, Ec =33,000(0.150)1.5 =4,888 ksi

The torsional constant, J, is estimated in accordance with LRFD Specifications, and Section 7.6.5.The torsional constant, J, is estimated in accordance with LRFD Specifications, and Section 7.6.5.

J = AJ = A44 / (40.0 I / (40.0 Ipp))

The polar moment of inertia Ip is equal to the sum of I and Ilat. Ip =654,123 in.4The polar moment of inertia Ip is equal to the sum of I and Ilat. Ip =654,123 in.4

J = (911^4) / (40.0 x 654,123) = 26,324 inJ = (911^4) / (40.0 x 654,123) = 26,324 in44

Properties of the 12- by 66-in. crossbeam:Properties of the 12- by 66-in. crossbeam:

A =792 in.2A =792 in.2

I =287,496 in.4I =287,496 in.4

Ilat =9,504 in.4 (for lateral bending)Ilat =9,504 in.4 (for lateral bending)

J =33,120 in.4J =33,120 in.4

wg =0.825 kip/ftwg =0.825 kip/ft

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Composite SectionsComposite Sections Effective Flange WidthEffective Flange Width

Effective flange width for interior beams shall be the lesser of: [LRFD Art. 4.6.2.6.1]Effective flange width for interior beams shall be the lesser of: [LRFD Art. 4.6.2.6.1]

-- (1/4) span =(120 x 12)/4 =360 in(1/4) span =(120 x 12)/4 =360 in

-- 12ts plus greater of web thickness or 1/2 beam top flange width12ts plus greater of web thickness or 1/2 beam top flange width

= (12 x 7.5) +(0.5 x 44) =112 in.= (12 x 7.5) +(0.5 x 44) =112 in.

-- average spacing between beams =(9 x 12) =108 in.average spacing between beams =(9 x 12) =108 in.

Therefore, the effective flange width is 108 in. for the beam.Therefore, the effective flange width is 108 in. for the beam.

For the interior crossbeams, the effective flange width is (12 x 7.5) +12 =102 in.For the interior crossbeams, the effective flange width is (12 x 7.5) +12 =102 in.

Modular RatioModular RatioModular ratio between slab and beam materials, n=Ecs/Ec=3,834/4,888=0.7845Modular ratio between slab and beam materials, n=Ecs/Ec=3,834/4,888=0.7845

Transformed Section PropertiesTransformed Section PropertiesTransformed flange width for interior beams =n (effective flange width)Transformed flange width for interior beams =n (effective flange width)

=(0.7845)(108) =84.73 in.=(0.7845)(108) =84.73 in.

Transformed flange area for interior beams =n(effective flange width)(structural thickness)Transformed flange area for interior beams =n(effective flange width)(structural thickness)

=(0.7845)(108)(7.5) =635.45 in.2=(0.7845)(108)(7.5) =635.45 in.2 Ac =total area of the composite section =1,564 in.2Ac =total area of the composite section =1,564 in.2 hc =overall depth of the composite section =80 in.hc =overall depth of the composite section =80 in. Ic =moment of inertia of the composite section =1,208,734 in.4Ic =moment of inertia of the composite section =1,208,734 in.4 ybc =distance from the centroid of the composite section to the extreme bottom fiber of the precast beam ybc =distance from the centroid of the composite section to the extreme bottom fiber of the precast beam

=53.05 in.=53.05 in. ytg =distance from the centroid of the composite section to the extreme top fiber of the precast beam ytg =distance from the centroid of the composite section to the extreme top fiber of the precast beam

=18.95 in.=18.95 in.

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ytc =distance from the centroid of the composite section to the extreme top fiber of the deck =26.95 in.ytc =distance from the centroid of the composite section to the extreme top fiber of the deck =26.95 in. Sbc =composite section modulus for the extreme bottom fiber of the precast beam =22,784 in.3Sbc =composite section modulus for the extreme bottom fiber of the precast beam =22,784 in.3 Stg =composite section modulus for the top fiber of the precast beam =63,792 in.3Stg =composite section modulus for the top fiber of the precast beam =63,792 in.3 Stc =composite section modulus for the extreme top fiber of the deck slab =57,176 in.3Stc =composite section modulus for the extreme top fiber of the deck slab =57,176 in.3 Iclat =moment of inertia of composite section for lateral bending =666,423 in.4Iclat =moment of inertia of composite section for lateral bending =666,423 in.4Composite properties of interior crossbeams:Composite properties of interior crossbeams: Ac =1,397 in.2Ac =1,397 in.2 Ic =765,432 in.4Ic =765,432 in.4 Iclat =529,860 in.4 for lateral bendingIclat =529,860 in.4 for lateral bending Jc =54,204 in.4Jc =54,204 in.4

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LoadsLoads

For a first approximation, all loads except the truck load will be assumed to be distributedFor a first approximation, all loads except the truck load will be assumed to be distributed

Dead LoadsDead Loads Dead Loads Acting on the Non-Composite StructureDead Loads Acting on the Non-Composite Structure1.1. Beam and crossbeam weight:Beam and crossbeam weight:

Beams =(6)(120 ft)(0.949 kip/ft) =683 kipsBeams =(6)(120 ft)(0.949 kip/ft) =683 kipsCrossbeams =(4)(45 ft)(0.825 kip/ft) =149 kipsCrossbeams =(4)(45 ft)(0.825 kip/ft) =149 kipsTotal weight of beams and crossbeams =683 +149 =832 kipsTotal weight of beams and crossbeams =683 +149 =832 kips

2.2. Deck weight:Deck weight:Gross area of deck =(120 ft)(51 ft) =6,120 ft 2Gross area of deck =(120 ft)(51 ft) =6,120 ft 2Actual thickness =8 in.Actual thickness =8 in.Deck weight =[8 in./(12 in./ft)](0.150 kcf )(6,120 ft 2 ) =612 kipsDeck weight =[8 in./(12 in./ft)](0.150 kcf )(6,120 ft 2 ) =612 kipsFor a minimum haunch thickness of 0.5 in., the superelevation of 0.06 will cause theFor a minimum haunch thickness of 0.5 in., the superelevation of 0.06 will cause theaverage haunch thickness to be 0.5 in. +0.06(22 in.) =1.82 in., say 2 in. The haunchaverage haunch thickness to be 0.5 in. +0.06(22 in.) =1.82 in., say 2 in. The haunchweight is 0.150 kcf (2 in.)(44 in.)/(144 in.2 /ft 2 ) =0.092 kip/ft/beamweight is 0.150 kcf (2 in.)(44 in.)/(144 in.2 /ft 2 ) =0.092 kip/ft/beamHaunch weight =(6)(120 ft)(0.092 kip/ft) =66 kipsHaunch weight =(6)(120 ft)(0.092 kip/ft) =66 kipsWeight of deck, including haunch =678 kipsWeight of deck, including haunch =678 kips

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Dead loads (cont’d)Dead loads (cont’d)

Dead Loads Acting on the Composite StructureDead Loads Acting on the Composite StructureBarrier weight is given as 0.3 kip/ft/sideBarrier weight is given as 0.3 kip/ft/sideBarrier weight =(2)(120 ft)(0.3 kip/ft) =72 kipsBarrier weight =(2)(120 ft)(0.3 kip/ft) =72 kipsFuture wearing surface is 0.025 ksfFuture wearing surface is 0.025 ksf(0.025 ksf )(120 ft)(48 ft) =144 kips(0.025 ksf )(120 ft)(48 ft) =144 kipsDead load on composite structure =72 +144 =216 kipsDead load on composite structure =72 +144 =216 kipsTotal dead load =832 +678 +216 =1,726 kipsTotal dead load =832 +678 +216 =1,726 kips

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Live LoadsLive Loads

Design live loading is HL-93 which consists of a combination of:Design live loading is HL-93 which consists of a combination of:1.1. Design truck or design tandem with dynamic allowance.Design truck or design tandem with dynamic allowance.2.2. Design lane load of 0.64 kip/ft without dynamic allowance.Design lane load of 0.64 kip/ft without dynamic allowance.

IM =33%IM =33%where IM =dynamic load allowance, applied to design truck or design tandem onlywhere IM =dynamic load allowance, applied to design truck or design tandem only

The number of design lanes is computed as:The number of design lanes is computed as:Number of design lanes =the integer part of the ratio of w/12, where w is the clear roadway width, ft, Number of design lanes =the integer part of the ratio of w/12, where w is the clear roadway width, ft, between the curbs:between the curbs:w =48 ftw =48 ftNumber of design lanes =integer part of (48/12) =4 lanesNumber of design lanes =integer part of (48/12) =4 lanes

Multiple presence factor, m:Multiple presence factor, m:For 4 lanes, m =0.65.For 4 lanes, m =0.65.Stresses from truck and lane loads obtained from refined analysis will be multiplied by 0.65.Stresses from truck and lane loads obtained from refined analysis will be multiplied by 0.65.

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Live Loads (cont’d)Live Loads (cont’d) Lane loadingLane loading

To maximize the effect of the live load, the 10-ft loaded width is shifted to the left within each design lane. To maximize the effect of the live load, the 10-ft loaded width is shifted to the left within each design lane. This causes the lane load to have an eccentricity of 1ft relative to the lane centerline, and the four lane This causes the lane load to have an eccentricity of 1ft relative to the lane centerline, and the four lane loads have an eccentricity of 1ft relative to the bridge centerline. The average arc length increases by the loads have an eccentricity of 1ft relative to the bridge centerline. The average arc length increases by the ratio of 601-ft radius/600-ft radius, to 120.2 ft.ratio of 601-ft radius/600-ft radius, to 120.2 ft.

The total lane loading for the four design lanes is (4)(120.2 ft)(0.64 kip/ft)(0.65)=200.0 kips.The total lane loading for the four design lanes is (4)(120.2 ft)(0.64 kip/ft)(0.65)=200.0 kips.

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Live Loads (cont’d)Live Loads (cont’d) Truck LoadingTruck Loading

The total weight of the design truck is 8 +32 +32 =72 kipsThe total weight of the design truck is 8 +32 +32 =72 kipsIncluding 33% impact, 1.33 x 72 =95.76 kips.Including 33% impact, 1.33 x 72 =95.76 kips.For 4 trucks, including the multiple presence factor, m:For 4 trucks, including the multiple presence factor, m:4(95.76)(0.65) =249.0 kips4(95.76)(0.65) =249.0 kips

Total Live LoadTotal Live LoadTotal live load =200.0 +249.0 =449.0 kipTotal live load =200.0 +249.0 =449.0 kip

Centrifugal ForceCentrifugal ForceThe design speed is 40 mph. The centrifugal force coefficient is The design speed is 40 mph. The centrifugal force coefficient is given by:given by:C = [4/3] vC = [4/3] v22/gR/gRwhere,where,C =coefficient to compute centrifugal forceC =coefficient to compute centrifugal forcev =highway design speed, ft/secv =highway design speed, ft/secg =gravitational acceleration, 32.2 ft/secg =gravitational acceleration, 32.2 ft/sec22

R =radius of curvature of traffic lane, ftR =radius of curvature of traffic lane, ftThe design speed in ft/sec =40 mph/0.682 =58.65 ft/secThe design speed in ft/sec =40 mph/0.682 =58.65 ft/secC=0.2374C=0.2374This is applied to the truck axle loads only, without the dynamic This is applied to the truck axle loads only, without the dynamic load allowance, and with the factor, m. The centrifugal force for load allowance, and with the factor, m. The centrifugal force for four trucks is 4(72 kip)(0.2374)(0.65) =44.4 kips.four trucks is 4(72 kip)(0.2374)(0.65) =44.4 kips.

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Correction FactorsCorrection FactorsThe bending moments in the exterior beam on the outside of the curve will be greater than in a The bending moments in the exterior beam on the outside of the curve will be greater than in a straight bridge for two reasons:straight bridge for two reasons:

Additional Span Length FactorAdditional Span Length FactorThe outside beam is on a radius of 622.5 ft. This increases the span length by a factor of 622.5/600 The outside beam is on a radius of 622.5 ft. This increases the span length by a factor of 622.5/600 =1.0375.=1.0375.

Shift in Center of GravityShift in Center of GravityCG of centerline arc is offset from a line through the centerline chord by an amount equal to 2/3 of CG of centerline arc is offset from a line through the centerline chord by an amount equal to 2/3 of the arc-to-chord offset. the arc-to-chord offset. 2/3 x 3 = 2ft2/3 x 3 = 2ftThe additional eccentricity caused by the extra area outside the centerline is equal to The additional eccentricity caused by the extra area outside the centerline is equal to BB22 /12R=(51 ft) /12R=(51 ft)22 / (12)(600) =0.36 ft, / (12)(600) =0.36 ft,Hence the eccentricity due to dead load is 2.36 ft.Hence the eccentricity due to dead load is 2.36 ft.

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The next step is to find how much the load on the outside beam is increased because of this The next step is to find how much the load on the outside beam is increased because of this eccentricity.eccentricity.

For six unit areas at 9-ft spacing,For six unit areas at 9-ft spacing,The moment of inertia,I = 1,417.5 ft 4 The moment of inertia,I = 1,417.5 ft 4 The section modulus, S= 63 ft 3The section modulus, S= 63 ft 3For an arbitrary load of 1 kip per bearing, For an arbitrary load of 1 kip per bearing, or 6 kips, at 2.36 ft eccentricity,or 6 kips, at 2.36 ft eccentricity,

P/A +Pe/S =1 +6(2.36)/63 =1.2248.P/A +Pe/S =1 +6(2.36)/63 =1.2248.This is the increase in load on the outside This is the increase in load on the outside exterior beam caused by the eccentricity of exterior beam caused by the eccentricity of the load. The total correction factor for the load. The total correction factor for bending moment due to dead load is bending moment due to dead load is (1.0375)(1.2248) =1.271.(1.0375)(1.2248) =1.271.

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For the lane loading, the LRFD requirement to place the load off-center of the lane adds 1 ft For the lane loading, the LRFD requirement to place the load off-center of the lane adds 1 ft to the eccentricity.to the eccentricity.

For a 6-kip load at 3.36-ft eccentricity, the load on the outside beam isFor a 6-kip load at 3.36-ft eccentricity, the load on the outside beam is

1 +6(3.36)/63 =1.32. 1 +6(3.36)/63 =1.32.

The total correction factor for lane loading is (1.0375)(1.32) =1.370.The total correction factor for lane loading is (1.0375)(1.32) =1.370.

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Correction factor for the truck loading.Correction factor for the truck loading.

For the truck loading, LRFD Article 3.6.1.3.1 specifies that the center of the wheel load be For the truck loading, LRFD Article 3.6.1.3.1 specifies that the center of the wheel load be placed 2 ft from the curb. This causes the center of the vehicle to be 5 ft from the curb (also placed 2 ft from the curb. This causes the center of the vehicle to be 5 ft from the curb (also the lane edge), so the eccentricity from the centerline of the lane is 1ft. The trucks are in the the lane edge), so the eccentricity from the centerline of the lane is 1ft. The trucks are in the center of the bridge, which has a 3-ft eccentricity with respect to the supports. Thus, the center of the bridge, which has a 3-ft eccentricity with respect to the supports. Thus, the vertical truck loading has an eccentricity of 4 ftvertical truck loading has an eccentricity of 4 ft

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Effects of Centrifugal ForceEffects of Centrifugal Force

The total centrifugal force of 44.4 kips acts at a height of 6 ft. The vertical truck loading is 249 The total centrifugal force of 44.4 kips acts at a height of 6 ft. The vertical truck loading is 249 kips. The horizontal force acting at 6 ft increases the eccentricity of the vertical load by kips. The horizontal force acting at 6 ft increases the eccentricity of the vertical load by (44.4/249)(6 ft) =1.07 ft. (44.4/249)(6 ft) =1.07 ft. The total eccentricity of the vertical truck load is 5.07 ft, and the correction is The total eccentricity of the vertical truck load is 5.07 ft, and the correction is 1 +6(5.07)/63 =1.4831 +6(5.07)/63 =1.483The total correction factor due to centrifugal force and truck loading is The total correction factor due to centrifugal force and truck loading is (1.0375)(1.483) =1.538.(1.0375)(1.483) =1.538.

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Bending Moments – Outside Exterior BeamBending Moments – Outside Exterior Beam

For all loads, the bending moment may be estimated as that for a 120-ft straight beam multipliedFor all loads, the bending moment may be estimated as that for a 120-ft straight beam multipliedby the correction factor.by the correction factor.

For all loads except the truck loadings, the 120-ft straight beam bending moment is WL/8 divided by six For all loads except the truck loadings, the 120-ft straight beam bending moment is WL/8 divided by six beams in the bridge.beams in the bridge.

For the truck loading, the bending moment is scaled from that for a standard truck on a 120ft straight span.For the truck loading, the bending moment is scaled from that for a standard truck on a 120ft straight span. Table shown below comprises summary of the estimated midspan bending moments for the outside Table shown below comprises summary of the estimated midspan bending moments for the outside

exterior beam.exterior beam.

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The next step is to check stresses which is not cover here.The next step is to check stresses which is not cover here.

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Beam GridworkBeam GridworkComputer ModelsComputer Models

Model 1 – Beam Segments on ShoresModel 1 – Beam Segments on Shores Model 2 – Shore LoadsModel 2 – Shore Loads Model 2 is the non-composite beam gridwork on Model 2 is the non-composite beam gridwork on

the nominal 120-ft span.the nominal 120-ft span. In model 2 shores are removed.In model 2 shores are removed. The loads previously existing in the shores are The loads previously existing in the shores are

applied to Model 2.applied to Model 2. The self-weight loads applied directly to Model The self-weight loads applied directly to Model

2.2. The difference in total self-weight bending The difference in total self-weight bending

moment in the out-side exterior beam is less than moment in the out-side exterior beam is less than 0.1 percent.0.1 percent.

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Beam GridworkBeam GridworkComputer Models (cont’d)Computer Models (cont’d)

Model 3 – Weight of Deck and Model 3 – Weight of Deck and HaunchesHaunches

The total weight of the deck and the The total weight of the deck and the haunches = 678 kipshaunches = 678 kips

This load is assumed to be applied as a This load is assumed to be applied as a uniform load of 110.8 psf over the 6,120 sf uniform load of 110.8 psf over the 6,120 sf gross area of the deck.gross area of the deck.

The finite elements are only used as a means of The finite elements are only used as a means of applying a uniform load.applying a uniform load.

The structural properties of the deck are The structural properties of the deck are zeroed out, because this is a non-zeroed out, because this is a non-composite model.composite model.

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Model 4 –Weight of Barriers and Future Wearing SurfaceModel 4 –Weight of Barriers and Future Wearing Surface Model 4 represents the composite structure. The general appearance of the model is the Model 4 represents the composite structure. The general appearance of the model is the

same as Model 3same as Model 3 Composite section properties are used in the beam gridwork.Composite section properties are used in the beam gridwork. The 0.025 ksf uniform load is applied over the entire 51-ft width of the deck.The 0.025 ksf uniform load is applied over the entire 51-ft width of the deck. A net barrier load of 0.263 kip/ft (0.3 kip/ft less the 0.025 ksf acting over the 1.5-ft widthA net barrier load of 0.263 kip/ft (0.3 kip/ft less the 0.025 ksf acting over the 1.5-ft width occupied by the barrier) is applied as a line load along the longitudinal edges of the model.occupied by the barrier) is applied as a line load along the longitudinal edges of the model.

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Model 5 – Lane LoadingModel 5 – Lane Loading The upper part of the below mention figureThe upper part of the below mention figure shows the specified location of the lane loads in a cross-section shows the specified location of the lane loads in a cross-section

through the bridge.through the bridge. The lower part of The lower part of Figure Figure shows the actual loads applied to the model.shows the actual loads applied to the model. The loads were chosen so that deck elements would be loaded uniformly and the total load would have the The loads were chosen so that deck elements would be loaded uniformly and the total load would have the

correct location of the resultant load.correct location of the resultant load.

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Model 6 – Truck Loading with Centrifugal ForceModel 6 – Truck Loading with Centrifugal Force

For maximum positive moment, the minimum rear axle spacing of 14 ft controls. The maximum bending For maximum positive moment, the minimum rear axle spacing of 14 ft controls. The maximum bending moment occurs with the middle axle load placed 2.33 ft from midspan.moment occurs with the middle axle load placed 2.33 ft from midspan.

The main axle wheel loads are 16 kips each, plus a 33 percent dynamic allowance, or 21.28 kips. For the The main axle wheel loads are 16 kips each, plus a 33 percent dynamic allowance, or 21.28 kips. For the design speed of 40 mph, the centrifugal force is 0.2374 of the truck weight (without dynamic allowance). design speed of 40 mph, the centrifugal force is 0.2374 of the truck weight (without dynamic allowance). This force acts 6 ft above the roadway. The overturning moment per main axle is 0.2374 times 32 kips This force acts 6 ft above the roadway. The overturning moment per main axle is 0.2374 times 32 kips times 6 ft, or 45.58 ft-kips. Dividing by the 6-ft wheel spacing, the wheel loads due to centrifugal force are times 6 ft, or 45.58 ft-kips. Dividing by the 6-ft wheel spacing, the wheel loads due to centrifugal force are ±7.6 kip. The total main axle wheel loads, including the 0.65 factor, m, are 0.65 (21.28 ±7.6) =18.77 kips ±7.6 kip. The total main axle wheel loads, including the 0.65 factor, m, are 0.65 (21.28 ±7.6) =18.77 kips and 8.89 kips. The front axle wheel loads are 1/4 of this, or 4.69 kips and 2.22 kips.and 8.89 kips. The front axle wheel loads are 1/4 of this, or 4.69 kips and 2.22 kips.

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The wheel loads are placed on fictitious, pin-The wheel loads are placed on fictitious, pin-ended members in order to transfer the loads to ended members in order to transfer the loads to the main beams, as shown for the heavier axles.the main beams, as shown for the heavier axles.

The added pin-ended members and loads that The added pin-ended members and loads that represent the truck loading for the condition represent the truck loading for the condition producing maximum moment are shown in producing maximum moment are shown in Figure.Figure.

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Summary of Bending MomentsSummary of Bending Moments

The bending moments for each of the six beams from the six loading models are summarized in The bending moments for each of the six beams from the six loading models are summarized in Table Table shown belowshown below. .

Pretensioning counteracts the moments from Model 1while post-tensioning is used to counteract the Pretensioning counteracts the moments from Model 1while post-tensioning is used to counteract the moments from Models 2 through 6. moments from Models 2 through 6.

The The LRFD Specifications LRFD Specifications [Article 3.4.1 and Table 3.4.1.1], states that for checking tension in prestressed [Article 3.4.1 and Table 3.4.1.1], states that for checking tension in prestressed members at service load, the Service III load combination may be used. This combination is 1.00 (DC members at service load, the Service III load combination may be used. This combination is 1.00 (DC +DW) +0.8(L +IM).+DW) +0.8(L +IM).

DESIGN OF HORIZONTALLY CURVED GIRDERS

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