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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.
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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.
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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:
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J. EMPERICAL DESIGN OF CONCRETE DECKSLABS
2. REINFORCEMENT REQUIREMENTS [A9.7.2.5]
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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)
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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.
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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.
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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.
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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
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L. DECK OVERHANG DESIGN
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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
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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
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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)
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M. CONCRETE BARRIER STRENGTH
c t
Fig. W7.1-17 (Concrete Barrier and connection to deckoverhang.)
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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
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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
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M. CONCRETE BARRIER STRENGTH
II+III
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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
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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
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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
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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
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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
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M. CONCRETE BARRIER STRENGTH
c c w
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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]
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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
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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
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Bundling a No.10 bar with No.15 bar at 225mm
on centers, the negative moment strengthbecomes
s
n
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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]
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u
st
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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
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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]
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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
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a. MOMENT- INTERIOR STRIP
s y
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b. MOMENT-EDGE STRIP
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2. CONTROL OF CRACKING
a. INTERIOR STRIP
s sa
r
c r
c
s
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Location of neutral axis
cr
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STEEL STRESS
s
s y
c
y
sa
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b. EDGE STRIP
(103)(x2) = (35 x 103)(510-x)
cr
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STEEL STRESS
s
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3. DEFORMATIONS [A5.7.3.6]
e
c e
cr
a
cr
crae
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t
g
cr
e
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I. INVESTIGATE SERVICE LIMIT STATE
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4607mm
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Back
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I. INVESTIGATE SERVICE LIMIT STATE
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DESIGN LANE LOAD
Lane
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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.
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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
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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)
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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
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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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