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    Introduction to

    Bridge Engineering

    Lecture 4 (II)

    CONCRETE BRIDGES

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    Presented By:

    YASIR IRFAN BADRASHI&

    QAISER HAYAT

    Presented To: PROF. DR. AKHTAR NAEEM KHAN

    &

    CLASSMATES

    CONCRETE BRIDGES

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    Topics to be Presented:

    Example Problem on:

    (i). Concrete Deck Design(ii). Solid Slab Bridge Design

    (iii). T-Beam Bridge Design

    CONCRETE BRIDGES

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    7.10.1

    CONCRETE DECK DESIGN

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    CONCRETE DECK DESIGN

    Use the approximate method ofanalysis [4.6.2] to design the deckof the reinforced concrete T-Beambridge section of Fig.E-7.1-1 for aHL-93 live load and a PL-2performance level concrete barrier

    (Fig.7.45).The T-Beams supporting the deckare 2440 mm on the centers andhave a stem width of 350 mm. Thedeck overhangs the exterior T-Beam approximately 0.4 of the

    distance between T-Beams. Allowfor sacrificial wear of 15mm ofconcrete surface and for a futurewearing surface of 75mm thickbituminous overlay. Use fc=30MPa, fy=400Mpa, and compare theselected reinforcement with that

    obtained by the empirical method[A9.7.2]

    Problem Statement:

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    The minimum thickness for concrete deck slabs is 175 mm [A9.7.1.1].

    Traditional minimum depths of slabs are based on the deck span length Sto control deflection to give [ Table A2.5.2.6.3-1]

    A. DECK THICKNESS

    Use hs = 190 mm for the structural thickness of the deck. By adding the15 mm allowance for the sacrificial surface, the dead weight of the deckslab is based on h= 205mm. Because the portion of the deck that

    overhangs the exterior girder must be designed for a collision loadon the barrier, its thickness has been increased by 25mm to ho=230mm

    mmmm

    S

    h 17518130

    30002440

    30

    3000

    min

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    B. WEIGHTS OF THE COMPONENTS[ TABLE A3.5.1-1 ]

    For a 1mm width of a transverse strip.

    Barrier

    Pb = 2400 x 10-9 Kg/mm3 x 9.81 N/Kg x 197325 mm2

    = 4.65 N/mm

    Future Wearing Surface

    WDW = 2250 x 10-9 x 9.81 x 75 = 1.66 x 10-3 N/mm

    Slab 205mm thick

    Ws = 2400 x 10-9 x 9.81 x 205 = 4.83 x 10-3 N/mm

    Cantilever Overhanging

    Wo = 2400 x 10-9 x 9.81 x 230 = 5.42 x 10-3 N/mm

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    C. BENDING MOMENTFORCE EFFECTS GENERAL

    An approximate analysis of strips perpendicularto girders is considered acceptable [A9.6.1]. Theextreme positive moment in any deck panelbetween girders shall be taken to apply to allpositive moment regions. Similarly, the extremenegative moment over any girder shall be takento apply to all negative moment regions[A4.6.2.1.1]. The strips shall be treated ascontinuous beams with span lengths equal to thecenter-to-centre distance between girders. Thegirders are assumed to be rigid [A4.6.2.1.6]

    For ease in applying the load factors, thebending moments will separately be determinedfor the deck slab, overhang, barrier, future

    wearing surface, and vehicle live load.

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    1. DECK SLAB

    h = 205 mm,

    Ws = 4.83 x 103 N/mm,S = 2440 mm

    Placement of the deck slabdead load and results of amoment distribution analysis fornegative and positive momentsin a 1-mm wide strip is given infigure E7.1-2

    A deck analysis design aidbased on influence lines is givenin Table A.1 of Appendix A. Fora uniform load, the tabulatedareas are multiplied by S forShears and S2 for moments.

    mmNmmWsS

    FEM /239612

    )2440)(1083.4(

    12

    232

    Fig.E7.1-2: Moment

    distribution for deck slabdead load.

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    R200 = Ws (Net area w/o cantilever) S= 4.83 x 10-3 (0.3928) 2440 = 4.63 N/mm

    M204 = Ws (Net area w/o cantilever) S2

    = 4.83 x 10-3 (0.0772) 24402

    = 2220 N mm/mm M300 = Ws (Net area w/o cantilever) S2

    = 4.83 x 10-3 (-0.1071) 24402= - 3080 N mm/mm

    Comparing the results from the design aid with thosefrom moment distribution shows good agreement. Indetermining the remainder of the bending momentforce effects, the design aid ofTable A.1 will beused.

    1. DECK SLAB

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    2. OVERHANG

    The parameters are

    ho = 230 mm,Wo = 5.42 x 10

    -3 N/mm2

    L = 990 mm

    Placement of the overhang dead load is shown in the figure E7.1-3. Byusing the design aid Table A.1, the reaction on the exterior T-Beamand the bending moments are:

    Fig.E7.1-3

    Overhangdead loadplacement

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    2. OVERHANG

    R200 = Wo (Net area cantilever) L= 5.42 x 10-3 (1+ 0.635 x 990/2440) 990 = 6.75 N/mm

    M200 = Wo (Net area cantilever) L2

    = 5.42 x 10-3 (-0.5000) 9902 = -2656 N mm/mm

    M204 = Wo (Net area cantilever) L2

    = 5.42 x 10-3 (-0.2460) 9902 = -1307 N mm/mm

    M300 = Wo (Net area cantilever) L2

    = 5.42 x 10-3 (0.1350) 9902 = 717 N mm/mm

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    3. BARRIER The parameters are

    Pb = 4.65 N/mmL = 990 127 = 863 mm

    Placement of the center of gravity of the barrier dead load isshown in figure E7.1-4. By using the design aid Table A.1 for theconcentrated barrier load, the intensity of the load is multiplied

    by the influence line ordinate for shears and reactions. Forbending moments, the influence line ordinate is multiplied by thecantilever length L.

    Fig.E7.1-4

    Barrierdead loadplacement

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    3. BARRIER

    R200 = Pb (Influence line ordinate)= 4.65(1.0+1.27 x 863/2440) = 6.74 N/mm

    M200 = Pb (Influence line ordinate) L

    = 4.65(-1.0000) (863) = -4013 N mm/mm

    M204 = Pb (Influence line ordinate) L

    = 4.65 (-0.4920) (863) = -1974 N mm/mm

    M300 = Pb (Influence line ordinate) L

    = 4.65 (0.2700) (863) = 1083 N mm/mm

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    4. FUTURE WEARING SURFACE

    FWS = WDW = 1.66 x 10-3 N/mm2

    The 75mm bituminous overlay is placed curb to curb asshown in figure E7.1-5. The length of the loaded cantilever isreduced by the base width of the barrier to giveL = 990 380 = 610 mm.

    Fig. E7.1-5: Future wearing surface dead load placement

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    If we use the design aid Table A.1, we have

    R200 = WDW [(Net area cantilever) L + (Net area w/o cantilever) S]= 1.66 x 10-3 [(1.0 + 0.635 x 610/2440) x 610 + (0.3928) x 2440)]= 2.76 N/mm

    M200 = WDW (Net area cantilever) L2

    = 1.66 x 10-3 (-0.5000)(610)2 = -309 N mm/mm

    M204 = WDW [(Net area cantilever) L2 + (Net area w/o cantilever) S2 ]

    = 1.66 x 10-3 [(-0.2460)(610)2 + (0.0772)24402 ] = 611 N mm/mm

    M300 = WDW [(Net area cantilever) L2 + (Net area w/o cantilever) S2 ]

    = 1.66 x 10-3 [(0.1350)(610)2 + (-0.1071)24402 ] = -975 N mm/mm

    4. FUTURE WEARING SURFACE

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    D. VEHICULAR LIVE LOAD

    Where decks are designed using the

    approximate strip method [A4.6.2.1], and thestrips are transverse, they shall be designedfor the 145 KN axle of the design truck[A3.6.1.3.3]. Wheel loads on an axle areassumed to be equal and spaced 1800 mmapart [Fig.A3.6.1.2.2-1]. The design truckshould be positioned transversely to producemaximum force effects such that the centerof any wheel load is not closer than 300mm

    from the face of the curb for the design ofthe deck overhang and 600mm from theedge of the 3600 mm wide design lane forthe design of all other components[A3.6.1.3.1]

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    D. VEHICULAR LIVE LOAD

    The width of equivalent interior transverse strips

    (mm) over which the wheel loads can be considereddistributed longitudinally in CIP concrete decks isgiven as[Table A4.6.2.1.3-1]

    Overhang, 1140+0.883 X Positive moment, 660+0.55 S Negative moment, 1220+0.25 S

    Where X is the distance from the wheel load tocenterline of support and S is the spacing of the T-Beams. Here X=310 mm and S=2440 mm(Fig.E7.1-6)

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    D. VEHICULAR LIVE LOAD

    Figure E 7.1-6 : Distribution of Wheel load

    on Overhang

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    D. VEHICULAR LIVE LOAD

    Tire contact area [A3.6.1.2.5] shall beassumed as a rectangle with width of 510mm and length given by

    P

    IM

    l

    100128.2

    Where is the load factor, IM is the dynamicload allowance and P is the Wheel load.

    Here = 1.75, IM = 33% , P = 72.5 KN.

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    D. VEHICULAR LIVE LOAD

    Thus the tire contact area is510 x 385mm

    with the 510mm in thetransverse direction asshown in Figure.E7.1-6

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    D. VEHICULAR LIVE LOAD

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    D. VEHICULAR LIVE LOAD

    Figure E 7.1-6 : Distribution of Wheel loadon Overhang

    Back

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    D. VEHICULAR LIVE LOAD

    3

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    D. VEHICULAR LIVE LOAD

    mm

    m

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    D. VEHICULAR LIVE LOAD

    Fig.E7.1-7: Live load placement for maximum positive moment

    (a) One loaded lane, m = 1.2

    (b) Two loaded lanes, m = 1.0

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    D. VEHICULAR LIVE LOAD

    If we use the influence line ordinates fromTable A-1, the exterior girder reaction andpositive bending moment with one loaded lane(m=1.2) are

    200

    204

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    D. VEHICULAR LIVE LOAD

    For two loaded lanes(m=1.0)

    Thus, the one loaded lane case governs.

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    D. VEHICULAR LIVE LOAD

    3. MAXIMUM INTERIOR NEGATIVE LIVE LOAD MOMENT.

    the critical placement of live load for maximum negativemoment is at the first interior deck support with oneloaded lane (m=1.2) as shown in Fig.E7.1-8.

    The equivalent transverse strip width is

    1220+0.25S = 1220+0.25(2440) = 1830 mm Using Table A-1, the bending moment at location 300 is

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    D. VEHICULAR LIVE LOAD

    4. MAXIMUM LIVE LOAD REACTION ON EXTERIOR GIRDER

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    E. STRENGTH LIMIT STATE

    The gravity load combination can be statedas [Table A.3.4.1-1]

    P P

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    E. STRENGTH LIMIT STATE

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    The T-Beam stem width is 350mm, so the design

    sections will be 175mm on either side of the supportcenterline used in the analysis. The critical negativemoment section is at the interior face of the exteriorsupport as shown in the free body diagram

    [Fig. E7.1-10]

    E. STRENGTH LIMIT STATE

    Back

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    The values of the loads in Fig E7.1-10 are for a 1-

    mathematical model strip. The concentrated wheelload is for one loaded lane, that is,

    W = 1.2(72500)1400 = 62.14 N/mm

    1. Deck Slab:

    E. STRENGTH LIMIT STATE

    s

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    2. Overhang

    3. Barrier

    E. STRENGTH LIMIT STATE

    o

    200

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    4. Future Wearing Surface

    5. Live Load

    E. STRENGTH LIMIT STATE

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    6. Strength-I Limit State

    E. STRENGTH LIMIT STATE

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    F. Selection Of Reinforcement

    The effective concrete

    depths for positive andnegative bending will bedifferent because of thedifferent coverrequirements as indicated

    in this Fig shown.

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    F. Selection Of Reinforcement

    u

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    F. Selection Of Reinforcement

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    F. Selection Of Reinforcement

    Maximum reinforcement keeping in view theductility requirements is limited by [A5.7.3.3.1]

    Minimum reinforcement [5.7.3.3.2] forcomponents containing no prestressing steel issatisfied if

    da 35.0

    y

    cs

    f

    f

    bd

    A '03.0

    )(

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    F. Selection Of Reinforcement

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    F. Selection Of Reinforcement

    1. POSITIVE MOMENT REINFORCEMENT :

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    F. Selection Of Reinforcement

    Check Ductility

    Check Moment Strength

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    F. Selection Of Reinforcement

    2. Negative Moment Reinforcement

    Back

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    F. Selection Of Reinforcement

    Check Moment Strength

    For transverse top bars,

    Use No. 15 @225 mm.

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    F. Selection Of Reinforcement

    3. DISTRIBUTION REINFORCEMENT:

    Secondary reinforcement is placed in the bottom of the slab todistribute the wheel loads in the longitudinal direction of the bridgeto the primary reinforcement in the transverse direction. Therequired area is a percentage of the primary positive momentreinforcement. For primary reinforcement perpendicular to traffic

    [A9.7.3.2]

    Where Se is the effective span length [A9.7.2.3]. Se is the distanceface to face of stems, that is,

    Se=2440-350= 2090mm

    %673840

    eS

    Percentage

    %67%,842090

    3840UsePercentage

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    So

    Dist.As = 0.67(Pos.As)=0.67(0.889)

    = 0.60 mm2/mm

    For longitudinal bottom bars,Use No.10 @ 150 mm,

    As = 0.667 mm2/mm

    F. Selection Of Reinforcement

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    F. Selection Of Reinforcement4. SHRINKAGE AND TEMPRATURE REINFORCEMENT.

    The minimum amount of reinforcement in each direction shall be[A5.10.8.2]

    Where Ag is the gross area of the section for the full 205 mm thickness.

    For members greater than 150 mm in thickness, the shrinkage andtemperature reinforcement is to be distributed equally on both faces.

    Use No.10 @ 450 mm, Provided As = 0.222 mm2/mm

    y

    g

    sf

    AATemp 75.0.

    mmmmATemp s /38.0200

    )1205(75.0. 2

    mmmmATemp s /19.0).(2

    12

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    G. CONTROL OF CRACKING-GENERAL

    Cracking is controlled by limiting the tensile stress in

    the reinforcement under service loads fs to an allowabletensile stress fsa [A5.7.3.4]

    WhereZ = 23000 N/mm for severe exposure conditions.dc =Depth of concrete from extreme tension fiber to

    center of closest bar 50 mmA = Effective concrete tensile area per bar having thesame centroid as the reinforcement.

    y

    c

    sas f

    Ad

    Zff 6.0

    )(

    3/1

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    G. CONTROL OF CRACKING-GENERAL

    M = MDC + MDW + 1.33 MLL

    c

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    G. CONTROL OF CRACKING-GENERAL

    c

    .2770030)2400(043.0 5.1 MPaEc

    Where

    = density of concrete = 2400 Kg/m3.

    fc = 30 MPa.

    So that

    Use n = 7

    ,2.7

    27700

    200000n

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    G. CONTROL OF CRACKING-GENERAL

    1. CHECK OF POSITIVE MOMENT REINFORCEMENT.

    The service I positive moment at Location 204 is

    The calculation of the transformed section properties is based on a 1-mm widedoubly reinforced section shown in the Figure E7.1-12

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    G. CONTROL OF CRACKING-GENERAL

    Sum of statical moments about the neutral axis yields

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    G. CONTROL OF CRACKING-GENERAL

    The positive moment tensile reinforcement of No.15 bars at 25mm

    on centers is located 33 mm from the extreme tension fiber.Therefore,

    c

    sa y

    sa y s

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    G. CONTROL OF CRACKING-GENERAL

    2. CHECK OF NEGATIVE REINFORCEMENT:

    The service I negative moment at location 200.72 is

    The cross section for the negative moment is shown in Fig.E7.1-13.

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    G. CONTROL OF CRACKING-GENERAL

    Balancing the statical moments about the

    neutral axis gives

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    G. CONTROL OF CRACKING-GENERAL

    The negative moment tensile reinforcement of

    No.15 bars at 225 mm on centers is located 53mm from the tension face. Therefore dc is themaximum value of 50mm, and

    sa

    sa

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    H. FATIGUE LIMIT STATE

    The investigation for fatigue is notrequired in concrete decks formultigirder applications [A9.5.3]

    I TRADITIONAL DESIGN FOR INTERIOR

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    I. TRADITIONAL DESIGN FOR INTERIORSPANS

    The design sketch in Fig.E7.1-14 summerizes the

    arrangement of the transverse and longitudinalreinforcement in four layers for the interior spans of thedeck. The exterior span and deck overhang have specialrequirements that must be dealt with separately.

    J EMPERICAL DESIGN OF CONCRETE DECK

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    J. EMPERICAL DESIGN OF CONCRETE DECKSLABS

    Research has shown that theprimary structural action of theconcrete deck is not flexure, but

    internal arching. The archingcreates an internal compressiondome. Only a minimum amount of

    isotropic reinforcement is requiredfor local flexural resistance.

    J EMPERICAL DESIGN OF CONCRETE DECK

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    J. EMPERICAL DESIGN OF CONCRETE DECKSLABS

    1. DESIGN CONDITIONS [A9.7.2.4]

    Design depth excludes the loss due to wear,h=190mm. The following conditions must be satisfied:

    J EMPERICAL DESIGN OF CONCRETE DECK

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    J. EMPERICAL DESIGN OF CONCRETE DECKSLABS

    2. REINFORCEMENT REQUIREMENTS [A9.7.2.5]

    J. EMPERICAL DESIGN OF CONCRETE DECK

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    J. EMPERICAL DESIGN OF CONCRETE DECKSLABS

    3. EMPERICAL DESIGN SUMMARY

    while using the empirical design approach there is no need of usingany analysis. When the design conditions have been met, theminimum reinforcement in all four layers is predetermined. Thedesign sketch in the Fig.E7.1-15 summarizes the reinforcementarrangement for the interior deck spans.

    K. COMPARISON OF REINFORCEMENT

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    K. COMPARISON OF REINFORCEMENTQUANTITIES

    The weight of reinforcement for the traditional and

    empirical design methods are compared in Table.E7.1-1for a 1-m wide transverse strip. Significant saving, inthis case 74% of the traditionally designedreinforcement is required, can be made by adopting theempirical design method.

    (Area = 1m x 14.18m)

    L DECK OVERHANG DESIGN

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    L. DECK OVERHANG DESIGN

    The traditional and the empirical methods

    does not include the design of the deckoverhang.

    The design loads for the deck overhang areapplied to a free body diagram of a

    cantilever that is independent of the deckspans.

    The resulting overhang design can then beincorporated into either the traditional or the

    empirical design by anchoring the overhangreinforcement into the first deck span.

    L DECK OVERHANG DESIGN

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    L. DECK OVERHANG DESIGN

    Two limit states must be investigated.

    Strength I [A13.6.1] and ExtremeEvent II [A13.6.2]

    The strength limit state considersvertical gravity forces and it seldomgoverns, unless the cantilever span isvery long.

    L DECK OVERHANG DESIGN

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    L. DECK OVERHANG DESIGN

    The extreme event limit stateconsiders horizontal forces causedby the collision of a vehicle with

    the barrier.

    The extreme limit state usually

    governs the design of the deckoverhang.

    L DECK OVERHANG DESIGN

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    L. DECK OVERHANG DESIGN

    1. STRENGTH I LIMIT STATE:

    The design negative moment is taken at theexterior face of the support as shown in theFig.E7.1-6 for the loads given in Fig.E7.1-10.

    Because the overhang has a single load pathand is, therefore, a nonredundant member,then 05.1R

    L DECK OVERHANG DESIGN

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    L. DECK OVERHANG DESIGN

    L DECK OVERHANG DESIGN

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    L. DECK OVERHANG DESIGN

    L DECK OVERHANG DESIGN

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    L. DECK OVERHANG DESIGN

    2. EXTREME EVENT II LIMIT STATE

    the forces to be transmitted to the deck overhanddue to a vehicular collision with the concrete barrierare determined from a strength analysis of thebarrier.

    In this design problem, the barriers are to bedesigned for a performance level PL-2, which issuitable for

    High-speed main line structures on freeways,

    expressways, highways and areas with a mixture ofheavy vehicles and maximum tolerable speeds

    L DECK OVERHANG DESIGN

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    L. DECK OVERHANG DESIGN

    The maximum edge thickness of the deck overhand is

    200mm[A13.7.3.1.2] and the minimum height of barrierfor a PL-2 is 810mm.

    The transverse and longitudinal forces are distributedover a length of barrier of 1070mm. This length

    represents the approximate diameter of a truck tire,which is in contact with the wall at the time of impact.

    The design philosophy is that if any failures are to occurthey should be in the barrier, which can readily be

    repaired, rather than in the deck overhang. The resistance factors are taken as 1.0 and the

    vehicle collision load factor is 1.0

    M CONCRETE BARRIER STRENGTH

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    M. CONCRETE BARRIER STRENGTH

    All traffic railing systems shall be proven

    satisfactory through crash testing for adesired performance level [A13.7.3.1]. If apreviously tested system is used with onlyminor modification that do not change itsperformance, then additional crash testing is

    not required [A13.7.3.1.1] The concrete barrier shown in the

    Fig.E7.1-17 (Next Slide) is similar to theprofile and reinforcement arrangement to

    traffic barrier type T5 analyzed byHirsh(1978) and tested by Buth et al (1990)

    M CONCRETE BARRIER STRENGTH

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    M. CONCRETE BARRIER STRENGTH

    c t

    Fig. W7.1-17 (Concrete Barrier and connection to deckoverhang.)

    M CONCRETE BARRIER STRENGTH

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    M. CONCRETE BARRIER STRENGTH

    H

    LMHMMLLR

    ccwb

    tc

    w

    2

    882

    2..(E7.1-8)

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    M. CONCRETE BARRIER STRENGTH

    t

    t

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    M. CONCRETE BARRIER STRENGTH

    1. MOMENT STRENGTH OF WALL ABOUT

    VERTICAL AXIS,MWH.

    The moment strength about the verticalaxis is based on the horizontal

    reinforcement in the wall. The thickness ofthe barrier wall varies and it is convenientto divide it for calculation purposes intothree segments as shown in Fig. E7.1-18

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    M. CONCRETE BARRIER STRENGTH

    M CONCRETE BARRIER STRENGTH

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    M. CONCRETE BARRIER STRENGTH

    Neglecting the contribution of compressive

    reinforcement, the positive and negative bendingstrengths of segment I are approximately equal andcalculated as

    nI

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    M. CONCRETE BARRIER STRENGTH

    For segment II, the moment strengths are slightly

    different. Considering the moment positive if it producestension on the straight face, we have

    n neg

    n pos

    n II

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    M. CONCRETE BARRIER STRENGTH

    For segment III, the positive and negative

    bending strengths are equal and

    nIII

    nIInI nIII

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    M. CONCRETE BARRIER STRENGTH

    Now considering the wall to have uniform

    thickness and same area as the actual wall andcomparing it with the value of MwH.

    This value is close to the one previously calculated and is

    easier to find

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    M. CONCRETE BARRIER STRENGTH

    2. MOMENT STRENGTH OF WALL ABOUT HORIZONTAL

    AXISThe moment strength about the horizontal axis isdetermined from the vertical reinforcement in thewall.

    The yield lines that cross the vertical reinforcement(Fig.E7.16-16) produce only tension in the slopingwall, so that the only negative bending strengthneed to be calculated.

    Matching the spacing of the vertical bars in thebarrier with the spacing of the bottom bars in thedeck, the vertical bars become No.15 at 225mm

    (As = 0.889 mm2/mm) for the traditional design

    (Fig.E7.1-14).

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    M. CONCRETE BARRIER STRENGTH

    For segment I, the average wall thickness is 175mm

    and the moment strength about the horizontal axisbecomes

    At the bottom of the wall the vertical reinforcement atthe wider spread is not anchored into the deckoverhang. Only the hairpin dowel at a narrower spreadis anchored. the effective depth of the hairpin dowel is[Fig.E7.1-17]

    d=50+16+150+8 = 224 mm

    M. CONCRETE BARRIER STRENGTH

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    M. CONCRETE BARRIER STRENGTH

    II+III

    M. CONCRETE BARRIER STRENGTH

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    M. CONCRETE BARRIER STRENGTH

    3. CRITICAL LENTH OF YIELD LINE PATTERN,LC

    Now with moment strengths and Lt=1070mm known,Eq.E7.1-9 yields

    c

    t t b w

    c

    M. CONCRETE BARRIER STRENGTH

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    M. CONCRETE BARRIER STRENGTH

    4. NOMINAL RESISTANCE TO TRANVERSE

    LOAD,RWFrom Eq.E7.1-8, We have

    w c tb w

    cc

    M. CONCRETE BARRIER STRENGTH

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    M. CONCRETE BARRIER STRENGTH

    5. SHEAR TRANSFER BETWEEN BARRIER AND DECK

    The nominal resistance Rw must be transferred acroass a cold jointby shear friction. Free body diagrams of the forces transferred fromthe barrier to the deck overhang are shown in the Fig.E7.1-19

    c

    M. CONCRETE BARRIER STRENGTH

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    M. CONCRETE BARRIER STRENGTH

    The nominal shear resistance Vn of the

    interface plane is given by [A5.8.4.1]

    cvn vf c

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    M. CONCRETE BARRIER STRENGTH

    The last two factors are for concrete placed

    against hardened concrete clean and free oflaitance, but not intentionally roughened.Therefore for a 1-mm wide design strip

    n cv

    vf fy

    M. CONCRETE BARRIER STRENGTH

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    M. CONCRETE BARRIER STRENGTH

    The minimum cross-sectional area of dowels

    across the shear plane is [A5.8.4.1]

    vf

    y

    v

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    M. CONCRETE BARRIER STRENGTH

    The basic development length lhb for a hooked bar with

    fy = 400 MPa. Is given by [A5.11.2.4.1]

    and shall not be less than 8db or 150mm. For a No.15bar, db=16mm and

    which is greater than 8(16) = 128mm and 150mm. Themodifications factors of 0.7 for adequate cover and 1.2

    for epoxy coated bars [A5.11.2.4.2] apply, so that thedevelopment length lhb is changed to

    lhb=0.7(1.2)lhb = 0.74(292) = 245mm

    '

    100

    fc

    dl bhb

    mmlhb 292

    30

    )16(100

    M. CONCRETE BARRIER STRENGTH

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    M. CONCRETE BARRIER STRENGTH

    c c w

    M. CONCRETE BARRIER STRENGTH

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    M. CONCRETE BARRIER STRENGTH

    The standard 90o hook with an extension of 12db=12(16)=192mm atthe free end of the bar is adequate [A5.10.2.1]

    M. CONCRETE BARRIER STRENGTH

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    CO C S G

    6. TOP REINFORCEMENT IN DECK OVERHANG

    The top reinforcement must resist the negative bendingmoment over the exterior beam due to the collision andthe dead load of the overhang. Based on the strength ofthe 90o hooks, the collision moment MCT (Fig.E7.1-19)

    distributed over a wall length of (Lc+2H) is

    M. CONCRETE BARRIER STRENGTH

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    The dead load moments were calculated

    previously for strength I so that for the ExtremeEvent II limit state, we have

    u

    M. CONCRETE BARRIER STRENGTH

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    Bundling a No.10 bar with No.15 bar at 225mm

    on centers, the negative moment strengthbecomes

    s

    n

    M. CONCRETE BARRIER STRENGTH

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    this moment strength will be reduced because

    of the axial tension forceT = Rw/(Lc+2H)

    By assuming the moment interaction curve

    between moment and axial tension as a straightline (Fig.E7.1-20]

    M. CONCRETE BARRIER STRENGTH

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    u

    st

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    M. CONCRETE BARRIER STRENGTH

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    The development length available for the hook in the overhang before reachingthe vertical leg of the hairpin dowel is

    available ldh=16+150+8=174mm>155mm

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    db

    M. CONCRETE BARRIER STRENGTH

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    7.10.2: SOLID SLAB BRIDGE DESIGN

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    PROBLEM STATEMENT:

    Design the simply supported solid slab bridgeof Fig.7.2-1 with a span length of 10670mmcenter to center of bearing for a HL-93 liveload. The roadway width is 13400mm curb to

    curb. Allow for a future wearing surface of75mm thick bituminous overlay. Usefc=30MPa and fy=400 MPa. Follow the slabbridge outline in Appendix A5.4 and the

    beam and girder bridge outline in section 5-Appendix A5.3 of the AASHTO (1994) LRFDbridge specifications.

    7.10.2: SOLID SLAB BRIDGE DESIGN

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    A. CHECK MINIMUM RECOMMENDEDDEPTH [TABLE A2.5.2.6.3-1]

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    DEPTH [TABLE A2.5.2.6.3 1]

    B. DETERMINE LIVE LOAD STRIPWIDTH [A4.6.2.3]

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    WIDTH [A4.6.2.3]

    1. One-Lane loaded:

    Multiple presence factor included [C4.6.2.3}

    1 1

    B. DETERMINE LIVE LOAD STRIPWIDTH [A4.6.2.3]

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    WIDTH [A4.6.2.3]

    C. APPLICABILITY OF LIVE LOADS FOR DECKSAND DECK SYSTEMS

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    1. MAXIMUM SHEAR FORCE AXLE LOADS [FIG.E7.2-2]

    C. APPLICABILITY OF LIVE LOADS FOR DECKSAND DECK SYSTEMS

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    C. APPLICABILITY OF LIVE LOADS FOR DECKSAND DECK SYSTEMS

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    1. MAXIMUM BENDING MOMENT AT MIDSPAN-

    AXLE LOADS [FIG.E7.2-3]

    D. SELECTION OF RESISTANCEFACTORS (Table 7.10 [A5.5.4.2.1]

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    FACTORS (Table 7.10 [A5.5.4.2.1]

    E. Select load modifiers [A1.3.2.1]

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    F. SELECT APPLICABLE LOAD COMBINATION(TABLE 3.1 [TABLE A3.4.1-1])

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    1. STRENGTH I LIMIT STATE

    2. SERVICE I LIMIT STATE

    3. FATIGUE LIMIT STATE

    G. CALCULATE LIVE LOAD FORCEEFFECTS

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    1. INTERIOR STRIP.

    G. CALCULATE LIVE LOAD FORCEEFFECTS

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    2. EDGE STRIP [A4.6.2.1.4]

    G. CALCULATE LIVE LOAD FORCEEFFECTS

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    H. CALCULATE FORCE EFFECTS FROM

    OTHERloads

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    1. INTERIOR STRIP, 1-mm WIDE

    H. CALCULATE FORCE EFFECTS FROM

    OTHERloads

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    2. EDGE STRIP, 1-MM WIDE

    I. INVESTIGATE SERVICE LIMIT STATE

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    1. DURIBILITY

    I. INVESTIGATE SERVICE LIMIT STATE

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    a. MOMENT- INTERIOR STRIP

    s y

    I. INVESTIGATE SERVICE LIMIT STATE

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    b. MOMENT-EDGE STRIP

    I. INVESTIGATE SERVICE LIMIT STATE

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    2. CONTROL OF CRACKING

    a. INTERIOR STRIP

    s sa

    r

    c r

    c

    s

    I. INVESTIGATE SERVICE LIMIT STATE

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    Location of neutral axis

    cr

    I. INVESTIGATE SERVICE LIMIT STATE

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    STEEL STRESS

    s

    s y

    c

    y

    sa

    I. INVESTIGATE SERVICE LIMIT STATE

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    b. EDGE STRIP

    (103)(x2) = (35 x 103)(510-x)

    cr

    I. INVESTIGATE SERVICE LIMIT STATE

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    STEEL STRESS

    s

    I. INVESTIGATE SERVICE LIMIT STATE

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    3. DEFORMATIONS [A5.7.3.6]

    e

    c e

    cr

    a

    cr

    crae

    I. INVESTIGATE SERVICE LIMIT STATE

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    t

    g

    cr

    e

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    I. INVESTIGATE SERVICE LIMIT STATE

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    4607mm

    I. INVESTIGATE SERVICE LIMIT STATE

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    Back

    I. INVESTIGATE SERVICE LIMIT STATE

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    I. INVESTIGATE SERVICE LIMIT STATE

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    DESIGN LANE LOAD

    Lane

    I. INVESTIGATE SERVICE LIMIT STATE

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    The live load deflection estimate of 17mm is

    conservative because Ie was based on themaximum moment at midspan rather than anaverage Ie over the entire span.

    Also, the additional stiffness provided by the

    concrete barriers has been neglected, as wellas the compression reinforcement in the topof the slab.

    Bridges typically deflect less than the

    calculations predict and as a result thedeflection check has been made optional.

    I. INVESTIGATE SERVICE LIMIT STATE

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    5. Concrete stresses [A5.9.4.3].

    As there is no prestressing thereforeconcrete stresses does not apply.

    I. INVESTIGATE SERVICE LIMIT STATE

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    5. FATIGUE [A5.5.3]

    Fatigue load should be one truck with 9000-mm axle

    spacing [A3.6.1.1.2]. As the rear axle spacing is large,therefore the maximum moment results when the twofront axles are on the bridge. as shown in Fig.E7.2-8,the two axle loads are placed on the bridge.

    No multiple presence factor is applied (m=1). FromFig.E7.2-8

    I. INVESTIGATE SERVICE LIMIT STATE

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    I. INVESTIGATE SERVICE LIMIT STATE

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    a. TENSILE LIVE LOAD STRESSES:

    One loaded lane, E=4370mm

    s

    I. INVESTIGATE SERVICE LIMIT STATE

    b [ ]

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    b. REINFORCING BARS:[A5.5.3.2]

    min

    J. INVESTIGATE STRENGTH LIMIT STATE

    [ ]

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    1. FLEXURE [A5.7.3.2]

    RECTANGULAR STRESS DISTRIBUTION [A5.7.2.2]

    a. INTERIOR STRIP:

    (2/7)

    J. INVESTIGATE STRENGTH LIMIT STATE

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    J. INVESTIGATE STRENGTH LIMIT STATE

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    J. INVESTIGATE STRENGTH LIMIT STATE

    For simple span bridges temperature gradient effect

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    For simple span bridges, temperature gradient effectreduces gravity load effects. Because temperature gradientmay not always be there, so assume = 0TG

    J. INVESTIGATE STRENGTH LIMIT STATE

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    So the strength limit state governs.

    Use No.30 @ 150 mm for interior strip.

    J. INVESTIGATE STRENGTH LIMIT STATE

    b

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    b. EDGE STRIP

    J. INVESTIGATE STRENGTH LIMIT STATE

    STRENGTH I

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    STRENGTH I:

    Use No. 30 @ 140mm for edge strip.

    J. INVESTIGATE STRENGTH LIMIT STATE

    2 SHEAR

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    2. SHEAR

    Slab bridges designed for moment inconformance with AASHTO[A4.6.2.3]

    maybe considered satisfactory forshear.

    K. DISTRIBUTION REINFORCEMENT[A5.14.4.1]

    Th t f b tt t

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    The amount of bottom transverse

    reinforcement maybe taken as a percentageof the main reinforcement required forpositive moment as.

    K. DISTRIBUTION REINFORCEMENT[A5.14.4.1]

    a INTERIOR SPAN:

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    a. INTERIOR SPAN:

    K. DISTRIBUTION REINFORCEMENT[A5.14.4.1]

    b EDGE STRIP

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    b. EDGE STRIP:

    L. SHRINKAGE AND TEMPRATURE REINFORCEMENT

    Transverse reinforcement in the top of the slab

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    Transverse reinforcement in the top of the slab

    [A5.10.8]

    M. DESIGN SKETCH

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    TABLE A-1

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    BACK

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