slab design

Introduction to Introduction to

Bridge EngineeringBridge Engineering

Lecture 4 Lecture 4 (II)(II)

CONCRETE BRIDGESCONCRETE BRIDGES

22

Presented By:Presented By: YASIR IRFAN BADRASHIYASIR IRFAN BADRASHI

&&

QAISER HAYATQAISER HAYAT

Presented To:Presented To: PROF. DR. AKHTAR NAEEM PROF. DR. AKHTAR NAEEM KHANKHAN

&&

CLASSMATESCLASSMATES


33

Topics to be Topics to be Presented:Presented:

Example Problem on:Example Problem on:

(i).(i). Concrete Deck Concrete Deck DesignDesign

(ii).(ii). Solid Slab Bridge Solid Slab Bridge DesignDesign

(iii).(iii). T-Beam Bridge T-Beam Bridge DesignDesign


44

7.10.17.10.1

CONCRETE DECK CONCRETE DECK DESIGNDESIGN

55

CONCRETE DECK DESIGNCONCRETE DECK DESIGN

Use the approximate method of Use the approximate method of analysis [4.6.2] to design the deck analysis [4.6.2] to design the deck of the reinforced concrete T-Beam of the reinforced concrete T-Beam bridge section of Fig.E-7.1-1 for a bridge section of Fig.E-7.1-1 for a HL-93 live load and a PL-2 HL-93 live load and a PL-2 performance level concrete barrier performance level concrete barrier (Fig.7.45). (Fig.7.45).

The T-Beams supporting the deck The T-Beams supporting the deck are 2440 mm on the centers and are 2440 mm on the centers and have a stem width of 350 mm. The have a stem width of 350 mm. The deck overhangs the exterior T-deck overhangs the exterior T-Beam approximately 0.4 of the Beam approximately 0.4 of the distance between T-Beams. Allow distance between T-Beams. Allow for sacrificial wear of 15mm of for sacrificial wear of 15mm of concrete surface and for a future concrete surface and for a future wearing surface of 75mm thick wearing surface of 75mm thick bituminous overlay. Use fc’=30 bituminous overlay. Use fc’=30 MPa, fMPa, fyy=400Mpa, and compare the =400Mpa, and compare the selected reinforcement with that selected reinforcement with that obtained by the empirical method obtained by the empirical method [A9.7.2][A9.7.2]

Problem Statement:

66

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

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

A. DECK THICKNESSA. DECK THICKNESS

Use hUse hss = 190 mm for the structural thickness of the deck. By = 190 mm for the structural thickness of the deck. By adding the adding the 15 mm allowance for the sacrificial surface, the dead weight of the 15 mm allowance for the sacrificial surface, the dead weight of the deck deck slab is based on h= 205mm. Because the portion of the deck that slab is based on h= 205mm. Because the portion of the deck that overhangs the exterior girder must be designed for a collision load overhangs the exterior girder must be designed for a collision load on the barrier, its thickness has been increased by 25mm to on the barrier, its thickness has been increased by 25mm to hhoo=230mm=230mm

mmmmS

h 17518130

30002440

30

3000min

77

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

For a 1mm width of a transverse strip.For a 1mm width of a transverse strip.

BarrierBarrier

PPbb = 2400 x 10 = 2400 x 10-9-9 Kg/mm Kg/mm33 x 9.81 N/Kg x 197325 mm x 9.81 N/Kg x 197325 mm2 2

= 4.65 N/mm= 4.65 N/mm

Future Wearing SurfaceFuture Wearing SurfaceWDW = 2250 x 10WDW = 2250 x 10-9-9 x 9.81 x 75 = 1.66 x 10 x 9.81 x 75 = 1.66 x 10-3 -3

N/mmN/mm

Slab 205mm thickSlab 205mm thickWs = 2400 x 10Ws = 2400 x 10-9-9 x 9.81 x 205 = 4.83 x 10 x 9.81 x 205 = 4.83 x 10-3 -3

N/mmN/mm

Cantilever OverhangingCantilever Overhanging

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

88

C. BENDING MOMENT C. BENDING MOMENT FORCE EFFECTS – GENERALFORCE EFFECTS – GENERAL

An approximate analysis of strips perpendicular An approximate analysis of strips perpendicular to girders is considered acceptable [A9.6.1]. to girders is considered acceptable [A9.6.1]. The extreme positive moment in any deck The extreme positive moment in any deck panel between girders shall be taken to apply panel between girders shall be taken to apply to all positive moment regions. Similarly, the to all positive moment regions. Similarly, the extreme negative moment over any girder shall extreme negative moment over any girder shall be taken to apply to all negative moment be taken to apply to all negative moment regions [A4.6.2.1.1]. The strips shall be treated regions [A4.6.2.1.1]. The strips shall be treated as continuous beams with span lengths equal to as continuous beams with span lengths equal to the center-to-centre distance between girders. the center-to-centre distance between girders. The girders are assumed to be rigid [A4.6.2.1.6]The girders are assumed to be rigid [A4.6.2.1.6]

For ease in applying the load factors, the For ease in applying the load factors, the bending moments will separately be bending moments will separately be determined for the deck slab, overhang, barrier, determined for the deck slab, overhang, barrier, future wearing surface, and vehicle live load.future wearing surface, and vehicle live load.

99

1. DECK SLAB1. DECK SLAB h = 205 mm, h = 205 mm,

Ws = 4.83 x 10Ws = 4.83 x 10–3–3 N/mm, N/mm, S = 2440 mmS = 2440 mm

Placement of the deck slab Placement of the deck slab dead load and results of a dead load and results of a moment distribution analysis moment distribution analysis for negative and positive for negative and positive moments in a 1-mm wide strip moments in a 1-mm wide strip is given in figure E7.1-2is given in figure E7.1-2

A deck analysis design aid A deck analysis design aid based on influence lines is based on influence lines is given in given in Table A.1 of Appendix A. For a . For a uniform load, the tabulated uniform load, the tabulated areas are multiplied by S for areas are multiplied by S for Shears and SShears and S22 for moments. for moments.

mmNmmWsS

FEM /239612

)2440)(1083.4(

12

232

Fig.E7.1-2: Moment distribution for deck slab dead load.

1010

R200 = Ws (Net area w/o cantilever) SR200 = Ws (Net area w/o cantilever) S = 4.83 x 10= 4.83 x 10-3-3 ( (0.3928) 2440 = 4.63 ) 2440 = 4.63

N/mmN/mm M204 = Ws (Net area w/o cantilever) SM204 = Ws (Net area w/o cantilever) S22

= 4.83 x 10= 4.83 x 10-3-3 ( (0.0772) 2440) 244022 = 2220 N mm/mm= 2220 N mm/mm

M300 = Ws (Net area w/o cantilever) SM300 = Ws (Net area w/o cantilever) S22

= 4.83 x 10= 4.83 x 10-3-3 (- (-0.1071) 2440) 244022 = - 3080 N mm/mm= - 3080 N mm/mm

Comparing the results from the design aid Comparing the results from the design aid with those from with those from moment distribution shows shows good agreement. In determining the good agreement. In determining the remainder of the bending moment force remainder of the bending moment force effects, the design aid of effects, the design aid of Table A.1 will be will be used.used.

1. DECK SLAB1. DECK SLAB

1111

2. OVERHANG2. OVERHANG The parameters are The parameters are

hhoo = 230 mm, = 230 mm,

WWoo = 5.42 x 10 = 5.42 x 10-3-3 N/mm N/mm22

L = 990 mmL = 990 mm Placement of the overhang dead load is shown in the figure Placement of the overhang dead load is shown in the figure

E7.1-3. By using the design aid Table A.1, the reaction on the E7.1-3. By using the design aid Table A.1, the reaction on the exterior T-Beam and the bending moments are:exterior T-Beam and the bending moments are:

Fig.E7.1-3

Overhang dead load placement

1212

2. OVERHANG2. OVERHANG

R200 = WR200 = Woo (Net area cantilever) L (Net area cantilever) L = 5.42 x 10= 5.42 x 10-3-3 (1+ 0.635 x 990/2440) 990 = (1+ 0.635 x 990/2440) 990 =

6.75 N/mm6.75 N/mm

M200 = WM200 = Woo (Net area cantilever) L (Net area cantilever) L22

= 5.42 x 10= 5.42 x 10-3-3 (-0.5000) 990 (-0.5000) 99022 = -2656 N = -2656 N mm/mmmm/mm


= 5.42 x 10= 5.42 x 10-3-3 (-0.2460) 990 (-0.2460) 99022 = -1307 N = -1307 N mm/mmmm/mm


= 5.42 x 10= 5.42 x 10-3-3 (0.1350) 990 (0.1350) 99022 = 717 N = 717 N mm/mmmm/mm

1313

3. BARRIER3. BARRIER The parameters are The parameters are

PPbb = 4.65 N/mm = 4.65 N/mm L = 990 – 127 = 863 mmL = 990 – 127 = 863 mm

Placement of the center of gravity of the barrier dead Placement of the center of gravity of the barrier dead load is shown in figure E7.1-4. By using the design aid load is shown in figure E7.1-4. By using the design aid Table A.1 for the concentrated barrier load, the intensity Table A.1 for the concentrated barrier load, the intensity of the load is multiplied by the influence line ordinate for of the load is multiplied by the influence line ordinate for shears and reactions. For bending moments, the shears and reactions. For bending moments, the influence line ordinate is multiplied by the cantilever influence line ordinate is multiplied by the cantilever length L.length L.

Fig.E7.1-4

Barrier dead load placement

1414

3. BARRIER3. BARRIER

R200 = PR200 = Pbb (Influence line ordinate) (Influence line ordinate) = 4.65(1.0+1.27 x 863/2440) = 6.74 N/mm= 4.65(1.0+1.27 x 863/2440) = 6.74 N/mm

M200 = PM200 = Pbb (Influence line ordinate) L (Influence line ordinate) L = 4.65(-1.0000) (863)= 4.65(-1.0000) (863) = -4013 N = -4013 N

mm/mmmm/mm

M204 = PM204 = Pbb (Influence line ordinate) L (Influence line ordinate) L = 4.65 (-0.4920) (863) = -1974 N mm/mm= 4.65 (-0.4920) (863) = -1974 N mm/mm

M300 = PM300 = Pbb (Influence line ordinate) L (Influence line ordinate) L = 4.65 (0.2700) (863) = 1083 N mm/mm= 4.65 (0.2700) (863) = 1083 N mm/mm

1515

4. FUTURE WEARING SURFACE4. FUTURE WEARING SURFACE

FFWSWS = W = WDWDW = 1.66 x 10-3 N/mm = 1.66 x 10-3 N/mm22

The 75mm bituminous overlay is placed curb to curb The 75mm bituminous overlay is placed curb to curb as shown in figure E7.1-5. The length of the loaded as shown in figure E7.1-5. The length of the loaded cantilever is reduced by the base width of the barrier cantilever is reduced by the base width of the barrier to give to give L = 990 – 380 = 610 mm. L = 990 – 380 = 610 mm.

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

1616

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

R200 = WR200 = WDWDW [(Net area cantilever) L + (Net area w/o cantilever) [(Net area cantilever) L + (Net area w/o cantilever) S]S]= 1.66 x 10= 1.66 x 10-3-3 [(1.0 + 0.635 x 610/2440) x 610 + (0.3928) x [(1.0 + 0.635 x 610/2440) x 610 + (0.3928) x 2440)]2440)]= 2.76 N/mm= 2.76 N/mm

M200 = WM200 = WDWDW (Net area cantilever) L (Net area cantilever) L22

= 1.66 x 10= 1.66 x 10-3-3 (-0.5000)(610) (-0.5000)(610)22 = -309 N mm/mm = -309 N mm/mm

M204 = WM204 = WDWDW [(Net area cantilever) L [(Net area cantilever) L22 + (Net area w/o cantilever) + (Net area w/o cantilever) SS22 ] ]

= 1.66 x 10= 1.66 x 10-3-3 [(-0.2460)(610) [(-0.2460)(610)2 2 + (0.0772)2440+ (0.0772)244022 ] = 611 N mm/mm ] = 611 N mm/mm

M300 = WM300 = WDWDW [(Net area cantilever) L [(Net area cantilever) L22 + (Net area w/o cantilever) + (Net area w/o cantilever) SS22 ] ]

= 1.66 x 10= 1.66 x 10-3-3 [(0.1350)(610) [(0.1350)(610)22 + (-0.1071)2440 + (-0.1071)244022 ] = -975 N ] = -975 N mm/mmmm/mm

4. FUTURE WEARING SURFACE4. FUTURE WEARING SURFACE

1717

D. VEHICULAR LIVE LOADD. VEHICULAR LIVE LOAD Where decks are designed using the Where decks are designed using the

approximate strip method [A4.6.2.1], and approximate strip method [A4.6.2.1], and the strips are transverse, they shall be the strips are transverse, they shall be designed for the 145 KN axle of the designed for the 145 KN axle of the design truck [A3.6.1.3.3]. Wheel loads on design truck [A3.6.1.3.3]. Wheel loads on an axle are assumed to be equal and an axle are assumed to be equal and spaced 1800 mm apart [Fig.A3.6.1.2.2-1]. spaced 1800 mm apart [Fig.A3.6.1.2.2-1]. The design truck should be positioned The design truck should be positioned transversely to produce maximum force transversely to produce maximum force effects such that the center of any wheel effects such that the center of any wheel load is not closer than 300mm from the load is not closer than 300mm from the face of the curb for the design of the deck face of the curb for the design of the deck overhang and 600mm from the edge of overhang and 600mm from the edge of the 3600 mm wide design lane for the the 3600 mm wide design lane for the design of all other components design of all other components [A3.6.1.3.1][A3.6.1.3.1]

1818

D. VEHICULAR LIVE LOADD. VEHICULAR LIVE LOAD The width of equivalent interior transverse The width of equivalent interior transverse

strips (mm) over which the wheel loads can be strips (mm) over which the wheel loads can be considered distributed longitudinally in CIP considered distributed longitudinally in CIP concrete decks is given asconcrete decks is given as[Table A4.6.2.1.3-1][Table A4.6.2.1.3-1]

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

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

1919

D. VEHICULAR LIVE LOADD. VEHICULAR LIVE LOAD

Figure E 7.1-6 : Distribution of Wheel load on Overhang

2020

D. VEHICULAR LIVE LOADD. VEHICULAR LIVE LOAD Tire contact area [A3.6.1.2.5] shall be

assumed as a rectangle with width of 510 mm and length given by

PIM

l

100128.2

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

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

2121


Thus the tire contact area is Thus the tire contact area is

510 x 385mm 510 x 385mm

with the 510mm in the with the 510mm in the transverse direction as transverse direction as shown in Figure.E7.1-6shown in Figure.E7.1-6

2222


2323


Figure E 7.1-6 : Distribution of Wheel load on Overhang

Back

2424


3

2525


mm

m

2626


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

2727

D. VEHICULAR LIVE LOADD. VEHICULAR LIVE LOAD If we use the influence line ordinates from If we use the influence line ordinates from

Table A-1, the exterior girder reaction and Table A-1, the exterior girder reaction and positive bending moment with one loaded positive bending moment with one loaded lane (m=1.2) arelane (m=1.2) are

200

204

2828


For two loaded For two loaded lanes (m=1.0)lanes (m=1.0)

Thus, the one loaded lane case governs.

2929

D. VEHICULAR LIVE LOADD. VEHICULAR LIVE LOAD3. MAXIMUM INTERIOR NEGATIVE LIVE LOAD MOMENT.3. MAXIMUM INTERIOR NEGATIVE LIVE LOAD MOMENT.

the critical placement of live load for maximum the critical placement of live load for maximum negative moment is at the first interior deck support negative moment is at the first interior deck support with one loaded lane (m=1.2) as shown in Fig.E7.1-with one loaded lane (m=1.2) as shown in Fig.E7.1-8.8.

The equivalent transverse strip width is The equivalent transverse strip width is 1220+0.25S = 1220+0.25(2440) = 1830 mm 1220+0.25S = 1220+0.25(2440) = 1830 mm

Using Table A-1, the bending moment at location Using Table A-1, the bending moment at location 300 is 300 is

3030


4. 4. MAXIMUM LIVE LOAD REACTION ON EXTERIOR GIRDERMAXIMUM LIVE LOAD REACTION ON EXTERIOR GIRDER

3131

E. STRENGTH LIMIT STATEE. STRENGTH LIMIT STATE The gravity load combination can be The gravity load combination can be

stated as [Table A.3.4.1-1]stated as [Table A.3.4.1-1]

P P

3232

E. STRENGTH LIMIT STATEE. STRENGTH LIMIT STATE

3333

The T-Beam stem width is 350mm, so the design The T-Beam stem width is 350mm, so the design sections will be 175mm on either side of the sections will be 175mm on either side of the support centerline used in the analysis. The support centerline used in the analysis. The critical negative moment section is at the interior critical negative moment section is at the interior face of the exterior support as shown in the free face of the exterior support as shown in the free body diagrambody diagram[Fig. E7.1-10][Fig. E7.1-10]


Back

3434

The values of the loads in Fig E7.1-10 are for a The values of the loads in Fig E7.1-10 are for a 1-mathematical model strip. The concentrated 1-mathematical model strip. The concentrated wheel load is for one loaded lane, that is,wheel load is for one loaded lane, that is,

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

1.1. Deck Slab:Deck Slab:


s

3535

2. Overhang2. Overhang

3. Barrier3. Barrier


o

200

3636

4. Future Wearing Surface4. Future Wearing Surface

5. Live Load5. Live Load


3737

6. Strength-I Limit State6. Strength-I Limit State


3838

F. Selection Of F. Selection Of ReinforcementReinforcement

The effective concrete The effective concrete depths for positive and depths for positive and negative bending will negative bending will be different because of be different because of the different cover the different cover requirements as requirements as indicated in this Fig indicated in this Fig shown.shown.

3939


u

4040


4141


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

Minimum reinforcement [5.7.3.3.2] for Minimum reinforcement [5.7.3.3.2] for components containing no prestressing components containing no prestressing steel is satisfied if steel is satisfied if

da 35.0

y

cs

f

f

bd

A '03.0

)(

4242


4343


1. POSITIVE MOMENT REINFORCEMENT :

4444


Check DuctilityCheck Ductility

Check Moment StrengthCheck Moment Strength

4545

F. Selection Of F. Selection Of ReinforcementReinforcement2. Negative Moment Reinforcement2. Negative Moment Reinforcement

Back

4646


Check Moment StrengthCheck Moment Strength

For transverse top bars,For transverse top bars,

Use No. 15 @225 mm.Use No. 15 @225 mm.

4747


3. DISTRIBUTION REINFORCEMENT:3. DISTRIBUTION REINFORCEMENT:

Secondary reinforcement is placed in the bottom of the Secondary reinforcement is placed in the bottom of the slab to distribute the wheel loads in the longitudinal slab to distribute the wheel loads in the longitudinal direction of the bridge to the primary reinforcement in the direction of the bridge to the primary reinforcement in the transverse direction. The required area is a percentage of transverse direction. The required area is a percentage of the primary positive moment reinforcement. For primary the primary positive moment reinforcement. For primary reinforcement perpendicular to traffic [A9.7.3.2]reinforcement perpendicular to traffic [A9.7.3.2]

Where SWhere Se e is the effective span length [A9.7.2.3]. Sis the effective span length [A9.7.2.3]. Se e is the is the distance face to face of stems, that is,distance face to face of stems, that is,

SSee=2440-350= 2090mm=2440-350= 2090mm

%673840

eS

Percentage

%67%,842090

3840UsePercentage

4848

SoSo

Dist.ADist.Ass = 0.67(= 0.67(Pos.As

)=0.67(0.889))=0.67(0.889)

= 0.60 mm= 0.60 mm22/mm/mm

For longitudinal bottom bars,For longitudinal bottom bars,

Use No.10 @ 150 mm,Use No.10 @ 150 mm,

AAss = 0.667 mm = 0.667 mm22/mm/mm


4949


4. SHRINKAGE AND TEMPRATURE REINFORCEMENT.4. SHRINKAGE AND TEMPRATURE REINFORCEMENT.The minimum amount of reinforcement in each direction shall be The minimum amount of reinforcement in each direction shall be [A5.10.8.2][A5.10.8.2]

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

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

Use No.10 @ 450 mm, Provided AUse No.10 @ 450 mm, Provided Ass = 0.222 mm = 0.222 mm22/mm/mm

y

gs f

AATemp 75.0.

mmmmATemp s /38.0200

)1205(75.0. 2

mmmmATemp s /19.0).(2

12

5050

G. CONTROL OF CRACKING-G. CONTROL OF CRACKING-GENERALGENERAL

Cracking is controlled by limiting the tensile Cracking is controlled by limiting the tensile stress in the reinforcement under service loads fstress in the reinforcement under service loads fss to an allowable tensile stress fto an allowable tensile stress fsasa [A5.7.3.4] [A5.7.3.4]

Where Where Z = 23000 N/mm for severe exposure conditions.Z = 23000 N/mm for severe exposure conditions.ddc c == Depth of concrete from extreme tension fiber Depth of concrete from extreme tension fiber to to center of closest bar 50 mmcenter of closest bar 50 mmA = Effective concrete tensile area per bar A = Effective concrete tensile area per bar having the having the same centroid as the reinforcement.same centroid as the reinforcement.

yc

sas fAd

Zff 6.0

)( 3/1

5151


M = MDC + MDW + 1.33 MLL

c

5252


c

.2770030)2400(043.0 5.1 MPaEc

Where

= density of concrete = 2400 Kg/m3.

f’c = 30 MPa.

So that

Use n = 7

,2.727700

200000n

5353


1. 1. CHECK OF POSITIVE MOMENT REINFORCEMENT.CHECK OF POSITIVE MOMENT REINFORCEMENT.

The service I positive moment at Location 204 isThe service I positive moment at Location 204 is

The calculation of the transformed section properties is based on a The calculation of the transformed section properties is based on a 1-mm wide doubly reinforced section shown in the Figure E7.1-1-mm wide doubly reinforced section shown in the Figure E7.1-1212

5454


Sum of statical moments about the neutral axis yieldsSum of statical moments about the neutral axis yields

5555


The positive moment tensile reinforcement of No.15 bars The positive moment tensile reinforcement of No.15 bars at 25mm on centers is located 33 mm from the extreme at 25mm on centers is located 33 mm from the extreme tension fiber. Therefore,tension fiber. Therefore,

c

sa y

sa y s

5656


2.2. CHECK OF NEGATIVE REINFORCEMENT:CHECK OF NEGATIVE REINFORCEMENT:

The service I negative moment at location 200.72 isThe service I negative moment at location 200.72 is

The cross section for the negative moment is shown in Fig.E7.1-The cross section for the negative moment is shown in Fig.E7.1-13.13.

5757


Balancing the statical moments about the Balancing the statical moments about the neutral axis givesneutral axis gives

5858


The negative moment tensile reinforcement The negative moment tensile reinforcement of No.15 bars at 225 mm on centers is located of No.15 bars at 225 mm on centers is located 53 mm from the tension face. Therefore dc is 53 mm from the tension face. Therefore dc is the maximum value of 50mm, andthe maximum value of 50mm, and

sa

sa

5959

H. FATIGUE LIMIT STATEH. FATIGUE LIMIT STATE

The investigation for fatigue is not The investigation for fatigue is not required in concrete decks for required in concrete decks for multigirder applications [A9.5.3]multigirder applications [A9.5.3]

6060

I.I. TRADITIONAL DESIGN FOR INTERIOR TRADITIONAL DESIGN FOR INTERIOR SPANSSPANS

The design sketch in Fig.E7.1-14 summerizes the The design sketch in Fig.E7.1-14 summerizes the arrangement of the transverse and longitudinal arrangement of the transverse and longitudinal reinforcement in four layers for the interior spans reinforcement in four layers for the interior spans of the deck. The exterior span and deck of the deck. The exterior span and deck overhang have special requirements that must overhang have special requirements that must be dealt with separately.be dealt with separately.

6161

J. EMPERICAL DESIGN OF CONCRETE J. EMPERICAL DESIGN OF CONCRETE DECK DECK SLABS SLABS Research has shown that the Research has shown that the

primary structural action of the primary structural action of the concrete deck is not flexure, but concrete deck is not flexure, but internal arching. The arching internal arching. The arching creates an internal compression creates an internal compression dome. Only a minimum amount dome. Only a minimum amount of isotropic reinforcement is of isotropic reinforcement is required for local flexural required for local flexural resistance.resistance.

6262

J. EMPERICAL DESIGN OF CONCRETE J. EMPERICAL DESIGN OF CONCRETE DECK DECK SLABS SLABS1. DESIGN CONDITIONS [A9.7.2.4]1. DESIGN CONDITIONS [A9.7.2.4]

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

6363

J. EMPERICAL DESIGN OF CONCRETE J. EMPERICAL DESIGN OF CONCRETE DECK DECK SLABS SLABS2. REINFORCEMENT REQUIREMENTS [A9.7.2.5]2. REINFORCEMENT REQUIREMENTS [A9.7.2.5]

6464

J. EMPERICAL DESIGN OF CONCRETE J. EMPERICAL DESIGN OF CONCRETE DECK DECK SLABS SLABS

3. EMPERICAL DESIGN SUMMARY3. EMPERICAL DESIGN SUMMARY

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

6565

K. COMPARISON OF REINFORCEMENT K. COMPARISON OF REINFORCEMENT QUANTITIES QUANTITIES

The weight of reinforcement for the traditional The weight of reinforcement for the traditional and empirical design methods are compared in and empirical design methods are compared in Table.E7.1-1 for a 1-m wide transverse strip. Table.E7.1-1 for a 1-m wide transverse strip. Significant saving, in this case 74% of the Significant saving, in this case 74% of the traditionally designed reinforcement is required, traditionally designed reinforcement is required, can be made by adopting the empirical design can be made by adopting the empirical design method.method.

(Area = 1m x 14.18m)(Area = 1m x 14.18m)

6666

L. DECK OVERHANG DESIGNL. DECK OVERHANG DESIGN

The traditional and the empirical The traditional and the empirical methods does not include the design of methods does not include the design of the deck overhang. the deck overhang.

The design loads for the deck overhang The design loads for the deck overhang are applied to a free body diagram of a are applied to a free body diagram of a cantilever that is independent of the cantilever that is independent of the deck spans.deck spans.

The resulting overhang design can then The resulting overhang design can then be incorporated into either the be incorporated into either the traditional or the empirical design by traditional or the empirical design by anchoring the overhang reinforcement anchoring the overhang reinforcement into the first deck span.into the first deck span.

6767


Two limit states must be Two limit states must be investigated. investigated.

Strength I [A13.6.1] and Extreme Strength I [A13.6.1] and Extreme Event II [A13.6.2]Event II [A13.6.2]

The strength limit state considers The strength limit state considers vertical gravity forces and it vertical gravity forces and it seldom governs, unless the seldom governs, unless the cantilever span is very long.cantilever span is very long.

6868


The extreme event limit state The extreme event limit state considers horizontal forces considers horizontal forces caused by the collision of a caused by the collision of a vehicle with the barrier. vehicle with the barrier.

The extreme limit state usually The extreme limit state usually governs the design of the deck governs the design of the deck overhang.overhang.

6969


1. STRENGTH I LIMIT STATE:1. STRENGTH I LIMIT STATE:

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

Because the overhang has a single load Because the overhang has a single load path and is, therefore, a nonredundant path and is, therefore, a nonredundant member, then member, then

05.1R

7070


7171


7272


2. EXTREME EVENT II LIMIT STATE2. EXTREME EVENT II LIMIT STATE

the forces to be transmitted to the deck the forces to be transmitted to the deck overhand due to a vehicular collision with the overhand due to a vehicular collision with the concrete barrier are determined from a concrete barrier are determined from a strength analysis of the barrier.strength analysis of the barrier.

In this design problem, the barriers are to be In this design problem, the barriers are to be designed for a performance level PL-2, which designed for a performance level PL-2, which is suitable foris suitable for

““High-speed main line structures on freeways, High-speed main line structures on freeways, expressways, highways and areas with a expressways, highways and areas with a mixture of heavy vehicles and maximum mixture of heavy vehicles and maximum tolerable speeds”tolerable speeds”

7373


The maximum edge thickness of the deck The maximum edge thickness of the deck overhand is 200mm[A13.7.3.1.2] and the overhand is 200mm[A13.7.3.1.2] and the minimum height of barrier for a PL-2 is 810mm. minimum height of barrier for a PL-2 is 810mm.

The transverse and longitudinal forces are The transverse and longitudinal forces are distributed over a length of barrier of 1070mm. distributed over a length of barrier of 1070mm. This length represents the approximate diameter This length represents the approximate diameter of a truck tire, which is in contact with the wall at of a truck tire, which is in contact with the wall at the time of impact. the time of impact.

The design philosophy is that if any failures are The design philosophy is that if any failures are to occur they should be in the barrier, which can to occur they should be in the barrier, which can readily be repaired, rather than in the deck readily be repaired, rather than in the deck overhang.overhang.

The resistance factors are taken as 1.0 and The resistance factors are taken as 1.0 and the vehicle collision load factor is 1.0the vehicle collision load factor is 1.0

7474

M. CONCRETE BARRIER STRENGTHM. CONCRETE BARRIER STRENGTH

All traffic railing systems shall be All traffic railing systems shall be proven satisfactory through crash proven satisfactory through crash testing for a desired performance level testing for a desired performance level [A13.7.3.1]. If a previously tested [A13.7.3.1]. If a previously tested system is used with only minor system is used with only minor modification that do not change its modification that do not change its performance, then additional crash performance, then additional crash testing is not required [A13.7.3.1.1]testing is not required [A13.7.3.1.1]

The concrete barrier shown in the The concrete barrier shown in the Fig.E7.1-17 (Next Slide) is similar to the Fig.E7.1-17 (Next Slide) is similar to the profile and reinforcement arrangement profile and reinforcement arrangement to traffic barrier type T5 analyzed by to traffic barrier type T5 analyzed by Hirsh(1978) and tested by Buth et al Hirsh(1978) and tested by Buth et al (1990)(1990)

7575

M. CONCRETE BARRIER M. CONCRETE BARRIER STRENGTHSTRENGTH

c t

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

7676


H

LMHMM

LLR cc

wbtc

w

2

882

2…..(E7.1-8)

fm

7777


tt

7878


1. MOMENT STRENGTH OF WALL ABOUT 1. MOMENT STRENGTH OF WALL ABOUT

VERTICAL AXIS,MVERTICAL AXIS,MWWH.H.

The moment strength about the The moment strength about the vertical axis is based on the horizontal vertical axis is based on the horizontal reinforcement in the wall. The reinforcement in the wall. The thickness of the barrier wall varies thickness of the barrier wall varies and it is convenient to divide it for and it is convenient to divide it for calculation purposes into three calculation purposes into three segments as shown in Fig. E7.1-18segments as shown in Fig. E7.1-18

7979


8080


Neglecting the contribution of compressive Neglecting the contribution of compressive reinforcement, the positive and negative bending reinforcement, the positive and negative bending strengths of segment I are approximately equal strengths of segment I are approximately equal and calculated asand calculated as

nI

8181


For segment II, the moment strengths are slightly For segment II, the moment strengths are slightly different. Considering the moment positive if it different. Considering the moment positive if it produces tension on the straight face, we haveproduces tension on the straight face, we have

n neg

n pos

n II

8282


For segment III, the positive and negative For segment III, the positive and negative bending strengths are equal and bending strengths are equal and

nIII

nIInI nIII

8383


Now considering the wall to have uniform Now considering the wall to have uniform thickness and same area as the actual wall thickness and same area as the actual wall and comparing it with the value of Mand comparing it with the value of MwwH.H.

This value is close to the one previously calculated and is easier to find

8484


2. MOMENT STRENGTH OF WALL ABOUT HORIZONTAL 2. MOMENT STRENGTH OF WALL ABOUT HORIZONTAL AXISAXIS

The moment strength about the horizontal axis is The moment strength about the horizontal axis is determined from the vertical reinforcement in the determined from the vertical reinforcement in the wall. wall.

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

Matching the spacing of the vertical bars in the Matching the spacing of the vertical bars in the barrier with the spacing of the bottom bars in the barrier with the spacing of the bottom bars in the deck, the vertical bars become No.15 at 225mmdeck, the vertical bars become No.15 at 225mm

(As = 0.889 mm(As = 0.889 mm22/mm) for the traditional design /mm) for the traditional design (Fig.E7.1-14).(Fig.E7.1-14).

8585


For segment I, the average wall thickness is For segment I, the average wall thickness is 175mm and the moment strength about the 175mm and the moment strength about the horizontal axis becomeshorizontal axis becomes

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

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

8686


II+III

8787


3. 3. CRITICAL LENTH OF YIELD LINE PATTERN,LCRITICAL LENTH OF YIELD LINE PATTERN,LCC

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

ct t b w

c

8888


4. NOMINAL RESISTANCE TO TRANVERSE 4. NOMINAL RESISTANCE TO TRANVERSE

LOAD,RLOAD,RWW

From Eq.E7.1-8, We haveFrom Eq.E7.1-8, We have

wc t

b wcc

8989


5. SHEAR TRANSFER BETWEEN BARRIER AND DECK5. SHEAR TRANSFER BETWEEN BARRIER AND DECK

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

c

9090


The nominal shear resistance Vn of the The nominal shear resistance Vn of the interface plane is given by [A5.8.4.1]interface plane is given by [A5.8.4.1]

cvn vf c

9191


The last two factors are for concrete placed The last two factors are for concrete placed against hardened concrete clean and free of against hardened concrete clean and free of laitance, but not intentionally roughened. laitance, but not intentionally roughened. Therefore for a 1-mm wide design strip Therefore for a 1-mm wide design strip

n cv

vf fy

9292


The minimum cross-sectional area of The minimum cross-sectional area of dowels across the shear plane is [A5.8.4.1]dowels across the shear plane is [A5.8.4.1]

vfy

v

9393


The basic development length lThe basic development length lhbhb for a hooked bar for a hooked bar with fwith fyy = 400 MPa. Is given by [A5.11.2.4.1] = 400 MPa. Is given by [A5.11.2.4.1]

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

which is greater than 8(16) = 128mm and 150mm. which is greater than 8(16) = 128mm and 150mm. The modifications factors of 0.7 for adequate cover The modifications factors of 0.7 for adequate cover and 1.2 for epoxy coated bars [A5.11.2.4.2] apply, and 1.2 for epoxy coated bars [A5.11.2.4.2] apply, so that the development length lso that the development length lhbhb is changed to is changed to

llhbhb=0.7(1.2)l=0.7(1.2)lhbhb = 0.74(292) = 245mm = 0.74(292) = 245mm

'

100

fc

dl bhb

mmlhb 29230

)16(100

9494


c c w

9595


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

9696

M. CONCRETE BARRIER M. CONCRETE BARRIER STRENGTHSTRENGTH6. TOP REINFORCEMENT IN DECK OVERHANG6. TOP REINFORCEMENT IN DECK OVERHANG

The top reinforcement must resist the negative The top reinforcement must resist the negative bending moment over the exterior beam due to bending moment over the exterior beam due to the collision and the dead load of the overhang. the collision and the dead load of the overhang. Based on the strength of the 90o hooks, the Based on the strength of the 90o hooks, the collision moment Mcollision moment MCTCT (Fig.E7.1-19) distributed (Fig.E7.1-19) distributed over a wall length of (Lover a wall length of (Lcc+2H) is +2H) is

9797


The dead load moments were calculated The dead load moments were calculated previously for strength I so that for the previously for strength I so that for the Extreme Event II limit state, we haveExtreme Event II limit state, we have

u

9898


Bundling a No.10 bar with No.15 bar at Bundling a No.10 bar with No.15 bar at 225mm on centers, the negative moment 225mm on centers, the negative moment strength becomes strength becomes

s

n

9999


this moment strength will be reduced this moment strength will be reduced because of the axial tension force because of the axial tension force

T = RT = Rww/(L/(Lcc+2H)+2H)

By assuming the moment interaction By assuming the moment interaction curve between moment and axial tension curve between moment and axial tension as a straight line (Fig.E7.1-20]as a straight line (Fig.E7.1-20]

100100


u

st

101101


102102


103103


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

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

104104


105105


106106


107107


db

108108


109109

7.10.27.10.2

SOLID SLAB BRIDGE SOLID SLAB BRIDGE DESIGNDESIGN

110110

7.10.2: SOLID SLAB BRIDGE DESIGN7.10.2: SOLID SLAB BRIDGE DESIGN

PROBLEM STATEMENT: PROBLEM STATEMENT: Design the simply supported solid slab Design the simply supported solid slab bridge of Fig.7.2-1 with a span length of bridge of Fig.7.2-1 with a span length of 10670mm center to center of bearing for a 10670mm center to center of bearing for a HL-93 live load. The roadway width is HL-93 live load. The roadway width is 13400mm curb to curb. Allow for a future 13400mm curb to curb. Allow for a future wearing surface of 75mm thick bituminous wearing surface of 75mm thick bituminous overlay. Use foverlay. Use fcc’=30MPa and f’=30MPa and fyy=400 MPa. =400 MPa. Follow the slab bridge outline in Appendix Follow the slab bridge outline in Appendix A5.4 and the beam and girder bridge A5.4 and the beam and girder bridge outline in section 5-Appendix A5.3 of the outline in section 5-Appendix A5.3 of the AASHTO (1994) LRFD bridge AASHTO (1994) LRFD bridge specifications.specifications.

111111

7.10.2: SOLID SLAB BRIDGE DESIGN7.10.2: SOLID SLAB BRIDGE DESIGN

112112

A.A. CHECK MINIMUM CHECK MINIMUM RECOMMENDED RECOMMENDED DEPTH [TABLE A2.5.2.6.3-1]DEPTH [TABLE A2.5.2.6.3-1]

113113

B. DETERMINE LIVE LOAD STRIP B. DETERMINE LIVE LOAD STRIP WIDTH [A4.6.2.3] WIDTH [A4.6.2.3]

1. One-Lane loaded:

Multiple presence factor included [C4.6.2.3}

1 1

114114

B. DETERMINE LIVE LOAD STRIP B. DETERMINE LIVE LOAD STRIP WIDTH [A4.6.2.3] WIDTH [A4.6.2.3]

115115

C. APPLICABILITY OF LIVE LOADS FOR C. APPLICABILITY OF LIVE LOADS FOR DECKS DECKS AND DECK SYSTEMS AND DECK SYSTEMS

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

116116

C. APPLICABILITY OF LIVE LOADS FOR C. APPLICABILITY OF LIVE LOADS FOR DECKS DECKS AND DECK SYSTEMS AND DECK SYSTEMS

117117

C. APPLICABILITY OF LIVE LOADS FOR C. APPLICABILITY OF LIVE LOADS FOR DECKS DECKS AND DECK SYSTEMS AND DECK SYSTEMS1. MAXIMUM BENDING MOMENT AT MIDSPAN- 1. MAXIMUM BENDING MOMENT AT MIDSPAN-

AXLE LOADS [FIG.E7.2-3]AXLE LOADS [FIG.E7.2-3]

118118

D. SELECTION OF RESISTANCE D. SELECTION OF RESISTANCE FACTORS (Table 7.10 FACTORS (Table 7.10 [A5.5.4.2.1][A5.5.4.2.1]

119119

E. Select load modifiers E. Select load modifiers [A1.3.2.1][A1.3.2.1]

120120

F. SELECT APPLICABLE LOAD F. SELECT APPLICABLE LOAD COMBINATION COMBINATION (TABLE 3.1 [TABLE A3.4.1-1]) (TABLE 3.1 [TABLE A3.4.1-1])1. STRENGTH I LIMIT STATE1. STRENGTH I LIMIT STATE

2. SERVICE I LIMIT STATE2. SERVICE I LIMIT STATE

3. FATIGUE LIMIT STATE3. FATIGUE LIMIT STATE

121121

G. CALCULATE LIVE LOAD FORCE G. CALCULATE LIVE LOAD FORCE EFFECTS EFFECTS1. INTERIOR STRIP.1. INTERIOR STRIP.

122122

G. CALCULATE LIVE LOAD FORCE G. CALCULATE LIVE LOAD FORCE EFFECTS EFFECTS2. EDGE STRIP [A4.6.2.1.4]2. EDGE STRIP [A4.6.2.1.4]

123123

G. CALCULATE LIVE LOAD FORCE G. CALCULATE LIVE LOAD FORCE EFFECTS EFFECTS

124124

H. CALCULATE FORCE EFFECTS FROM H. CALCULATE FORCE EFFECTS FROM OTHER OTHER loads loads1. INTERIOR STRIP, 1-mm WIDE1. INTERIOR STRIP, 1-mm WIDE

125125

H. CALCULATE FORCE EFFECTS FROM H. CALCULATE FORCE EFFECTS FROM OTHER OTHER loads loads2. EDGE STRIP, 1-MM WIDE2. EDGE STRIP, 1-MM WIDE

126126

I. INVESTIGATE SERVICE LIMIT I. INVESTIGATE SERVICE LIMIT STATESTATE1. DURIBILITY1. DURIBILITY

127127

I. INVESTIGATE SERVICE LIMIT I. INVESTIGATE SERVICE LIMIT STATESTATEa. MOMENT- INTERIOR STRIPa. MOMENT- INTERIOR STRIP

s y

128128

I. INVESTIGATE SERVICE LIMIT I. INVESTIGATE SERVICE LIMIT STATESTATEb. MOMENT-EDGE b. MOMENT-EDGE

STRIPSTRIP

129129

I. INVESTIGATE SERVICE LIMIT I. INVESTIGATE SERVICE LIMIT STATESTATE2. CONTROL OF CRACKING2. CONTROL OF CRACKING

a.a. INTERIOR STRIPINTERIOR STRIP

s sa

r

c r

c

s

130130

I. INVESTIGATE SERVICE LIMIT I. INVESTIGATE SERVICE LIMIT STATESTATE

Location of neutral axis

cr

131131


STEEL STRESSSTEEL STRESSs

s y

c

y

sa

132132

I. INVESTIGATE SERVICE LIMIT I. INVESTIGATE SERVICE LIMIT STATESTATEb. EDGE STRIPb. EDGE STRIP

½(103)(x2) = (35 x 103)(510-x)

cr

133133


STEEL STRESSSTEEL STRESS

s

134134

I. INVESTIGATE SERVICE LIMIT I. INVESTIGATE SERVICE LIMIT STATESTATE3. DEFORMATIONS [A5.7.3.6]3. DEFORMATIONS [A5.7.3.6]

e

c e

cr

a

crcra

e

135135


t

gcr

e

136136

I. INVESTIGATE SERVICE LIMIT I. INVESTIGATE SERVICE LIMIT STATESTATEBy using Ig: [A5.7.3.6.2]

137137


b. LIVE LOAD DEFLECTION: (OPTIONAL)[A2.5.2.6.2]b. LIVE LOAD DEFLECTION: (OPTIONAL)[A2.5.2.6.2]

allowIMLL

138138


4607mm

139139


Back

140140


141141


DESIGN LANE LOADDESIGN LANE LOAD

Lane

142142


The live load deflection estimate of The live load deflection estimate of 17mm is conservative because I17mm is conservative because Iee was was based on the maximum moment at based on the maximum moment at midspan rather than an average Imidspan rather than an average Iee over over the entire span. the entire span.

Also, the additional stiffness provided by Also, the additional stiffness provided by the concrete barriers has been the concrete barriers has been neglected, as well as the compression neglected, as well as the compression reinforcement in the top of the slab.reinforcement in the top of the slab.

Bridges typically deflect less than the Bridges typically deflect less than the calculations predict and as a result the calculations predict and as a result the deflection check has been made optional.deflection check has been made optional.

143143


5. Concrete stresses [A5.9.4.3].5. Concrete stresses [A5.9.4.3].

As there is no prestressing As there is no prestressing therefore concrete stresses does therefore concrete stresses does not apply.not apply.

144144

I. INVESTIGATE SERVICE LIMIT I. INVESTIGATE SERVICE LIMIT STATESTATE5. FATIGUE [A5.5.3]5. FATIGUE [A5.5.3]

Fatigue load should be one truck with 9000-mm Fatigue load should be one truck with 9000-mm axle spacing [A3.6.1.1.2]. As the rear axle axle spacing [A3.6.1.1.2]. As the rear axle spacing is large, therefore the maximum moment spacing is large, therefore the maximum moment results when the two front axles are on the results when the two front axles are on the bridge. as shown in Fig.E7.2-8, the two axle loads bridge. as shown in Fig.E7.2-8, the two axle loads are placed on the bridge.are placed on the bridge.

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

145145


146146


a. TENSILE LIVE LOAD STRESSES:

One loaded lane, E=4370mm

s

147147


b. REINFORCING BARS:[A5.5.3.2]b. REINFORCING BARS:[A5.5.3.2]

min

148148

J. INVESTIGATE STRENGTH LIMIT STATEJ. INVESTIGATE STRENGTH LIMIT STATE

1. FLEXURE [A5.7.3.2] 1. FLEXURE [A5.7.3.2]

RECTANGULAR STRESS DISTRIBUTION RECTANGULAR STRESS DISTRIBUTION [A5.7.2.2][A5.7.2.2]

a. INTERIOR STRIP:a. INTERIOR STRIP:

(2/7)

149149


150150


151151


For simple span bridges, temperature gradient effect reduces gravity load effects. Because temperature gradient may not always be there, so assume = 0

TG

152152


So the strength limit state governs.

Use No.30 @ 150 mm for interior strip.

153153


b.b. EDGE STRIP EDGE STRIP

154154


STRENGTH I:

Use No. 30 @ 140mm for edge strip.

155155


2. SHEAR2. SHEAR

Slab bridges designed for moment Slab bridges designed for moment in conformance with in conformance with AASHTO[A4.6.2.3] maybe AASHTO[A4.6.2.3] maybe considered satisfactory for shear.considered satisfactory for shear.

156156

K. DISTRIBUTION REINFORCEMENT K. DISTRIBUTION REINFORCEMENT [A5.14.4.1] [A5.14.4.1]

The amount of bottom transverse The amount of bottom transverse reinforcement maybe taken as a reinforcement maybe taken as a percentage of the main reinforcement percentage of the main reinforcement required for positive moment as.required for positive moment as.

157157

K. DISTRIBUTION REINFORCEMENT K. DISTRIBUTION REINFORCEMENT [A5.14.4.1] [A5.14.4.1]a. INTERIOR SPAN:a. INTERIOR SPAN:

158158

K. DISTRIBUTION REINFORCEMENT K. DISTRIBUTION REINFORCEMENT [A5.14.4.1] [A5.14.4.1]

b.b. EDGE STRIP: EDGE STRIP:

159159

L. SHRINKAGE AND TEMPRATURE L. SHRINKAGE AND TEMPRATURE REINFORCEMENTREINFORCEMENT

Transverse reinforcement in the top of the Transverse reinforcement in the top of the slab [A5.10.8]slab [A5.10.8]

160160

M. DESIGN SKETCHM. DESIGN SKETCH

161161

TABLE A-1TABLE A-1

BACK

162162BACK

163163

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