10/18/2014 1 ตัวอย่าง การจําลอง โครงสร้างสะพานส่วนบน โดย นายเกรียงไกร คําพา ตุลาคม 2557 DESIGNING OF REINFORCE CONCRETE DECK SLAB BRIDGE WITH AASHTO (LRFD) DESIGN SPECIFICATION Abutment A Bent 1 Bent 2 Abutment B Abutment A Bent 1 Bent 2 Abutment B Pier Cab Wingwall Pile Approach Slab Deck Slab
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DESIGNING OF REINFORCE CONCRETE DECK SLAB BRIDGE WITH AASHTO (LRFD) DESIGN SPECIFICATION
Abutment A Bent 1 Bent 2 Abutment BAbutment A Bent 1 Bent 2 Abutment B
Pier Cab
Wingwall
PileApproach Slab
Deck Slab
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SuperstructureDesign Example
Reinforce Concrete Deck Slab Bridge
Bridge Geometry
10m. 10m.10m.
30m.
b bAbutment A Bent 1 Bent 2 Abutment B
5m.
E.J. E.J. E.J. E.J.
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Bridge Deck Geometric Data
Side Railing CL of Bridge
Roadway width = 9 m.
Overall width = 10 m.
T
‐Overall bridge width = 10 m.‐Roadway width = 9 m.
‐Asphalt wearing surface = 60 mm. thickness.‐Bridge skew angle = 0 Degree‐Concrete Parapet Railing = 2 side (Section Area = 0.6m2 /side)
Section View
T
Wearing Surface
p g ( / )‐No. of Lane = 2 Lane
Elevation View
Bridge Span Length= 10 m.
Load Criteria
Dynamic Load Allowance (IM) Dynamic Impact Factor
‐Dynamic load factor for Truck & Tandem load (Service & Ultimate) = IM = 1.33‐Dynamic load factor at fatigue limit state = IM =1.15‐ (AASHTO, Table 3.6.2.1‐1)
Braking Force (BR) and Wind Load (W)
In this case, For superstructure design, braking forces and wind on live load are not applicable
Dead Load Components (DC) Structural and nonstructural attachment
‐For Truck Weight limit is 51 Ton ‐Thai Truck Factor for simple span 5 to 20 m.Increasing Factored = 1.0 For Bending MomentIncreasing Factored = 1.05 For Shear Force & Reaction
3600 mm
Lane Load TransverseDistribution
Load CriteriaLive Load (LLFATIGUE)Design Vehicular Live Load for Fatigue Check (AASHTO, 3.6.1.4)
9.0
Configuration for Fatigue Load
IMPACT LOAD FOR FATIGUE CHECK = 1.15 (AASHTO 3.6.2)
Load Distribution Factored for Fatigue
Load Distribution for Fatigue, where the bridge is analyzed by approximate load distribution, as specified in Article 4.6.2, the distribution factor for one traffic lane shall be used.
Live Load Analysis (Approximate Methods)STEP‐3‐Calculate Forces on Bridge Deck per Strip Width.
Live Load for Interior slab design LL = Transient Loads, HL93
Consider moment at 5.3 m from supportMoment with impact factorMHL93+IM = 762.18 kN.m/laneMLL_INTERIOR = 762.18xr/[Em] = 762.18x1.0/[3.3] =230.97 kN.m/mConsider shear force at supportVHL93+IM = 355.8 kN/laneThai Truck Increase Factored for Shear = 1.05VLL_INTERIOR = 1.05x355.80xr/[Em] = 1.05x355.80x1.0/[3.3]
= 113.21 kN/m
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Live Load Analysis (Approximate Methods)STEP‐3‐Calculate Forces on Bridge Deck per Strip Width.
Live Load for edge slab design
Edge beam is assumed to support one line of truck axle
Ee
Live Load Analysis (Approximate Methods)STEP‐3‐Calculate Forces on Bridge Deck per Strip Width.
Live Load for edge slab design LL = Transient Loads, HL93
Edge beam is assumed to support one line of truck axle. Lane load is considered by tributary portion by= [Ee‐Wedge]/3 = [1.8‐0.5]/3 = 0.43 RatioMoment with impact factorMONE LINE TRUCK AXLE+IM = 485.98x1.33/2 = 323.18 kN.m/laneMLANE LOAD = 116.25 kN.m/laneMLL_EDGE SLAB = 323.18+0.43x116.25 = 373.37 kN.m
= 373.37xr/[Ee] = 373.37x1.0/[1.8] =207.43 kN.m/m
Consider shear force at supportThai Truck Increase Factored for Shear 1 05Thai Truck Increase Factored for Shear = 1.05VONE LINE TRUCK AXLE+IM = 232.55x1.33/2= 154.65 kN/laneVLANE LOAD = 46.50 kN/laneVLL_EDGE SLAB = 154.65+0.43x46.5 = 174.80 kN/lane
Live Load Analysis (Approximate Methods)STEP‐3‐Calculate Forces on Bridge Deck per Strip Width.
Live Load for Fatigue Check
LLFATIGUE = Transient Loads, Fatigue
Consider moment at 5.3 m from supportMoment with impact factor (IM factor =1.15, Distribution Factor =1.2)MFATIGUE_EDGE = 1.2x[1.15x377.89/2] x r/[Es]
= 1.2x217.3x1.0/[1.80] = 144.9 kN.m/mMFATIGUE_INTERIOR = 1.2x[1.15x377.89] x r/[Em]
= 1.2x435.6x1.0/[3.3] = 158.1 kN.m/m
9.0
Live Load Analysis (Approximate Methods)STEP‐3‐Calculate Forces on Bridge Deck per Strip Width.
DC = Dead load structural components and nonstructural attachments
DCDECK = 0.6x24.03 =14.42 kN/m2
Consider at X = 5.3 m. from support and r = 1.0MDECK = 14.42x5.3x(10.0‐5.3) x1.0 /2 x1.0 =179.61 kN.m/mConsider at support and r = 1.0VDECK = 14.42x10.0 x1.0 /2 x1.0 =72.1 kN/m
DCBARRIER = 0.5x24.03/10 =1.2 kN/m2
Consider at X = 5.3 m. from support and r = 1.0MBARRIER = 1.2x5.3x(10.0‐5.3) x1.0 /2 x1.0 =14.97 kN.m/mConsider at support and r = 1.0V = 1 2x10 0 x1 0 /2 x1 0 =6 01 kN/m
NOTE: Assume traffic barriers load are distribute equally over the full bridge cross‐section.
VBARRIER = 1.2x10.0 x1.0 /2 x1.0 =6.01 kN/m
Consider at X = 5.3 m. from support and r = 1.0MDC= 179.61 + 14.97 =194.58 kN.m/mConsider at support and r = 1.0VDC = 72.1 + 6.01 =78.11 kN/m
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Live Load Analysis (Approximate Methods)STEP‐3‐Calculate Forces on Bridge Deck per Strip Width.
DW = Dead load of wearing surfaces and utilities
DWWEARING SURFACE = 0.06x22.07x9.0/10 =1.19 kN/m2
Consider at X = 5.3 m. from support and r = 1.0MWEARING SURFACE = 1.19x5.3x(10.0‐5.3) x1.0 /2 x1.0 =14.85 kN.m/mConsider at support and r = 1.0VWEARING SURFACE = 1.19x10.0x1.0/2 x1.0 = 5.96 kN/m
FORCES DC DW LL+IM LLFATIGUE
Moment (kN.m/m)
Summary Design Forces
MINTERIOR STRIP 194.58 14.85 230.97 158.1
MEDGE STRIP 194.58 14.85 207.43 144.9
SHEAR (kN/m)
VINTERIOR STRIP 78.11 5.96 113.21 ‐
VEDGE STRIP 78.11 5.96 101.97 ‐
Load Combination
Service I = 1.0DC+1.0DW+1.0[LL+IM]Strength I = 1.25DC+1.5DW+1.75[LL+IM]Fatigue I = 1.5[LLFATIGUE]
Interior StripMomentMoment
Service I = 1.0x194.58+1.0x14.85+1.0x230.97 = 440.39 kN.m/m
Strength I = 1.25x194.58+1.5x14.85+1.75x230.97= 669.68 kN.m/m
Fatigue I = 1.5x158.1= 237.1 kN.m/m
Shear
Service I = 1.0x78.11+1.0x5.96+1.0x113.21 = 197.28 kN.m/m
Strength I = 1.25x78.11+1.5x5.96+1.75x113.21= 304.7 kN.m/m
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Load Combination
Service I = 1.0DC+1.0DW+1.0[LL+IM]Strength I = 1.25DC+1.5DW+1.75[LL+IM]Fatigue I = 1.5[LLFATIGUE]
Edge StripMomentMoment
Service I = 1.0x194.58+1.0x14.85+1.0x207.43 = 416.85 kN.m/m
Strength I = 1.25x194.58+1.5x14.85+1.75x207.43= 628.5 kN.m/m
Fatigue I = 1.5x144.9= 217.4 kN.m/m
Shear
Service I = 1.0x78.11+1.0x5.96+1.0x101.97= 186.04 kN.m/m
Strength I = 1.25x78.11+1.5x5.96+1.75x101.97= 285.1 kN.m/m
Bridge Deck Reinforcement Design
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Reinforce Concrete Design
Concrete‐Compressive strength at 28 days (Cylinder) fc
’ = 25 Mpa.‐Modulus of elasticity, Ec = Wc
1.5 .0.0428.fc1/2
= 2,3201.5 x0.0428x251/2
E 23 914 M
Material Properties
‐‐5.4.2.4
Ec = 23,914 Mpa.‐Thermal expansion coefficient t = 0.0000108 m/m/ ˚C‐Relative humidity of the ambient temperature = 80 % ==> Use for Design Expansion Joint (Creep & Shrinkage)‐Minimum Concrete Covering = 50 mm.SteelGrade SD40Yield strength fy = 390 Mpa.Tensile Strength = 560 Mpa
‐‐5.4.2.2
‐‐TIS & ASTM
‐‐Table 5.12.3.1
Tensile Strength = 560 Mpa.Modulus of elasticity, Es = 200,000 Mpa.Temp.(˚C)High = 50 ˚C , Low = 15 ˚C , Mean = 30 ˚C
Reinforce Concrete Design
Resistance Factor
Figure C5.5.4.2.1‐1 Variation of with net tensile strain
‐‐5.5.4.2.1‐2
c = distance from the extreme compression fiber to neutral axis (mm)
dt = distance from the extreme compression fiber to centroid of the extreme tension steel element (mm)
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Reinforce Concrete Design
Resistance Factor
For Non‐ Prestress Concrete ReinforcementUsed Resistance Factored = 0.9 for Tension Controlled
Nominal Flexural Resistance (Need to design with tension controlled)
C = 0.85.fc’.a.bfc’
ᵋc = 0.003 compression
c
de
b
N.A. N.A.
a
a = 1.c 0.85.fc’
1 = 0.85 > 0.85‐(fc'‐28)/140 > 0.65 ‐‐5.7.2.2
de – a/2
fs = fy
Section Strain
ᵋs = 0.005 TensionAs
Section Dimension
Force Equilibrium
T = As .fy
Actual Stress
Reinforce Concrete Design
Nominal Flexural Resistance
Force Equilibrium: Compression = TensionC = T
0.85.fc’.a.b = As.fya = As.fy / 0.85.fc’.b A f / 0 85 f ’ b1.c = As.fy / 0.85.fc’.bc = As.fy / 0.85.fc’.b .1
From strain compatibility condition for tension controlledc/0.003 = de /[0.003+0.005]
For Non‐ Prestress Concrete ReinforcementUsed Resistance Factored = 0.9
c/de = 0.375
‐‐Eq 5.7.3.1.2‐4
Used Resistance Factored 0.9
If the ratio of c/de ≤ 0.375 ==> Tension Controlled
Flexural capacity .Mn = .As.fy.[de‐a/2] ‐‐5.7.3.2
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Reinforce Concrete Design
Positive Moment DesignStrength Limit State: Strength I
Mu‐max = Max(669.7, 628.5) = 669.7 kN.m/mTry to provide steel bar = [email protected] { d2/4} 1 000/100 4 909 2 Eq 5 7 3 1 2 4
Use the bottom bar parallel to traffic are [email protected] = 848.5 kN.m/m > 349.2 kN.m/m O.K.
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Reinforce Concrete Design
LRFD states that for cast‐in‐place slab superstructures designed for moment in conformance with LRFD 4.6.2.3,
Shear Capacity Check
Positive Moment DesignStrength Limit State: Strength I
‐‐ 5.14.4.1
may be considered satisfactory for shear.Vu = Max(304.7, 285.1) = 304.7 kN.v = 0.9.Vc = .0.083∙β∙f c’1/2∙b∙dβ = 2.0b = 1,000 mm., d = 537.5 mm., f c‘= 25 Mpa..Vc = 0.90x0.083x2.0x251/2x1000x537.5/1,000
= 401.5 kN. > 304.7 kN. O.K.
‐‐ Eq. 5.8.3.3‐3
‐‐5.8.3.4.1
‐‐5.5.4.2
Reinforce Concrete Design
Positive Moment DesignService Limit State: Service ICrack Control by Distribution Reinforcement
‐‐ Eq.5.7.3.4‐1Spacing of reinforcement in the layer closest to tension face S ≤ {123,000.e/[s.fss]}‐2.dcs = 1+dc/{0.7.[h‐dc]}h = overall thicknesse = exposure factor
= 1.00 for Class 1 exposure condition (crack width 0.430mm)(Upper bound in regards to crack width for appearance and corrosion.)
= 0.75 for Class 2 exposure condition (crack width 0.325mm)(Include decks and substructures exposed to water.)d = thickness of concrete cover measured from extremedc = thickness of concrete cover measured from extreme tension fiber to center of the flexural reinforcement located closest thereto (mm)fss = tensile stress in steel reinforcement at the service limit stateIn the application of these provisions to meet the needs of the Authority having jurisdiction. The crack width is directly proportional to the e exposure factor by Authority.e = 0.5/0.21x (crack width‐0.22) +0.5
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Reinforce Concrete Design
Positive Moment DesignService Limit State: Service ICrack Control by Distribution Reinforcement
M+ = Max(440.39, 416.85 ) = 440.4 kN.m/mCheck concrete cracked at bottom fiber (Concrete tension)C ec co c ete c ac ed at botto be (Co c ete te s o )Modulus of rupture = fr = 0.63.fc’
Positive Moment DesignService Limit State: Service ICrack Control by Distribution Reinforcement
Elastic Cracked Section Property at Service State (Cracked, Linear Stage , approximately fc<0.45fc’)f M/ A j d 440 4/ [49 09 0 89 0 54] 187 M
C = 0.5.k.d.fc
fck.d /3
h
ᵋc
k.d
d
b
N A N A
fss = M/ As.j.d = 440.4/ [49.09x0.89x0.54] = 187 Mpa.dc = 50covering+[25/2] = 62.5 mm., h = 600 mm.In this design example used e = 0.50 (crack width 0.22mm)
fss
Section Dimension
Force Equilibrium
Section Strain
ᵋs
N.A.
As
N.A.
T = As .fss
M j.d
Actual Stress
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Reinforce Concrete Design
Positive Moment DesignService Limit State: Service ICrack Control by Distribution Reinforcement
s = 1+dc/{0.7.[h‐dc]} = 1+62.5/{0.7x[600‐62.5]}/{ [ ]}= 1.17
Minimum Spacing of reinforcement in the layer closest to tension face to control concrete cracked (0.22mm.).
S = {123,000.e/[s.fss]}‐2.dc = [123,000x0.5/(1.17x187)]‐2x62.5
S = 157 mm.Use the bottom bar parallel to traffic are DB25@100Sprovide = 100 mm. ≤ 157mm. O.K.
Reinforce Concrete Design
Distribution Reinforcement
Reinforcement shall be placed in the secondary direction in the bottom of slabs as a percentage of the primary reinforcement for positive moment
‐‐9.7.3.2 & 5.14.4
Bottom reinforcement parallel to trafficBottom reinforcement parallel to traffic = DB25@100 = 4,909 mm2 , L = 10,000 mm.%.As = 1,750/ L
1/2 ≤ 50%= 1,750/ 10,0001/2
= 17.5%As, For bottom bar perpendicular to traffic
= 17.5% x 4,909 = 860 mm2
Use the bottom bar perpendicular to traffic are DB12@125(As = 905 mm2)
‐‐ Eq 5.14.4.1‐1
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Reinforce Concrete Design
Shrinkage & Temperature Reinforcement
Reinforcement for shrinkage and temperature stresses shall be provided near to daily temperature changes and in structural mass concrete in each direction and each face.
Use the Top bar perpendicular & parallel to traffic are DB12@250 (As = 452 mm2)
Eq. 5.10.8 1
‐‐ Eq. 5.10.8‐2
Reinforce Concrete Design
Fatigue Limit State Check
Maximum Moment due to Permanent Load (MPE) = DC+DW =194.58+14.85 = 209.42 kN.m/mMaximum Moment due to Fatigue I (Mtruck)=Max(237.1, 217.4) = 237.1 kN.m/m
Elastic Cracked Section Property at Fatigue Limit State (Cracked, Linear Stage, approximately fc<0.45fc’)For permanent load ,at the service and fatigue limit states ,use an effective modular ratio of 2n.n = 2.Es/Ec =2x8.36 =16.7 = As/ b.d = 4,909/[1,000x537.5] = 0.0087
‐‐5.7.1
Load Combination= 209.42+237.1 = 446.5 kN.m/m
s
k = [2n.+(n.)2]1/2‐n. = 0.421J = 1‐k/3 =0.860, Ig =0.018 m4/m.Check property of sectionlimit for crack section fc ≥ 0.25.f c’1/2 = 1.25 Mpa.fc‐tension = M.(h/2)/Ig = 446.5x(0.3)/0.018x1000
= 7.42Mpa. > 1.25 Mpa.Section is cracked
‐‐5.5.3.1
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Reinforce Concrete Design
Fatigue Limit State Check
Elastic Cracked Section Property at Fatigue Limit State (Cracked, Linear Stage, approximately fc<0.45fc’)fc,comp =2.M /(k∙j∙b∙d2)
Fatigue Limit Stress in Steel Bar Checkedγ(Δf) ≤ (ΔF)TH Δf = force effect, live load stress range due to the passage of the fatigue load as specified in Article 3.6.1.4 (Mpa.)γ = load factor specified in Table 3.4.1‐1 for the (ΔF)TH = 166 ‐ 0.33 fMIN = constant‐amplitude fatigue thresholdfMIN = minimum live‐load stress resulting from the Fatigue I load
bi i bi d i h h f i h
‐‐Eq.5.5.3.2‐2
‐‐Eq.5.5.3.2‐1
combination, combined with the more severe stress from either the permanent loads or the permanent loads, shrinkage, and creep‐induced external loads;
Reinforce Concrete Design
Fatigue Limit State Check
Fatigue Limit Stress in Steel Bar Checkedγ(Δf) = Mtruck /As.j.d = 237.1x1,000/ [4,909x0.86x0.54]
= 104 Mpa.fMIN = MPE /As.j.d = 209.42x1,000/ [4,909x0.86x0.54] MIN PE s
= 92.39 Mpa.(ΔF)TH = 166 ‐ 0.33 fMIN
= 166 ‐ 0.33x92.39= 135.54 Mpa.
γ(Δf) = 104 Mpa. ≤ (ΔF)TH = 135.54 Mpa. O.K.
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Expansion Joint DesignTemperature Movement
High = 50 ˚C , Low = 15 ˚C , Mean = 30 ˚CTemp. Rise = TT‐Rise = 50‐30 = 20 ˚CTemp. Fall = TT‐Fall =15‐30 = ‐15 ˚CThermal expansion of concrete = t = 1.08x10
‐5 m/m/˚CSlab span length =10 mSlab span length =10 m.Displacement XT‐Rise = t.TT‐Rise.L = 1.08x10‐5 x 20x10,000 = 2.16 mm.XT‐Fall = t.TT‐Fall.L = 1.08x10‐5 x ‐15x10,000 = ‐1.62 mm.
Creep and Shrinkage Movement
Creep of the concrete for expansion joint design is ignored.
Shrinkage Strain = ᵋSH = Kvs.Khs.Kf.Ktd.0.48x10‐3
In the absence of more accurate data the shrinkage
‐‐Eq. 5.4.2.3.3‐1
5 4 2 3 1In the absence of more accurate data, the shrinkage coefficients may be assumed to be 0.0002 after 28 days and0.0005 after one year of drying. So, Consider the effect of shrinkage after one year of drying
ᵋSH = 0.0005 (Approximately)
Xsrinkage = ᵋSH.L = 0.0005x10,000 = ‐5 mm.
‐‐ 5.4.2.3.1
Expansion Joint DesignFor conventional concrete structures,the movement is based on the greater of two cases:Movement from Shrinkage and Temperature
X = XT‐Fall+Xsrinkage = ‐1.62 – 5 = ‐6.62 mm.
Movement from factored effects of Temperature
Factored effects of temperature =1.15X = 1.15xXT‐Rise = 1.15x2.16 = 2.48 mm.
‐‐ FDOT criteria (SDG 6.4.2 )
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Camber DiagramEffective Moment of Inertia of Slab
Time to remove of formwork = 14 days.fc‘ at 14 days = 20 Mpa. Ec at 14 days = = 2,320
1.5 x0.0428x201/2 = 21,389 Mpa. fr at 14 days = 0.623.fc’
Pre‐Camber of formwork to compensate immediately sag of the deck profile at time
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Live Load Deflection Check
The deflection should be taken as the larger of:‐That resulting from the design truck alone, or‐That resulting from 25 percent of design truck + lane load
Limit of deflection for simple span = L/80010 000/800 12 5 mm
‐‐2.5.2.6.2
‐‐3.6.1.3.2
= 10,000/800 = 12.5 mm.Design equivalent strip widthEm = 3.3 m.Ie = 0.0142 m4/m x 3.3 m. = 0.04686 m4
Ec = 23,914 Mpa.
Maximum Deflection due to Lane Load at Mid Span = 1.0896 mm.Maximum Deflection due toMaximum Deflection due to Truck Load at Mid Span = 4.1419 mmMaximum Deflection due to Tandem Load at Mid Span = 4.0406 mm
Maximum Deflection = Max{4.1419, 1.0896+[4.1419x25/100]} = 4.1419 mm. < 12.5 mm. O.K.
Summary of Steel Reinforcement
Steel Reinforcement No. of Steel Bar Bars Sign
Bottom bar parallel to traffic
DB25@100 AS4
Bottom bar perpendicular DB12@125 AS2p pto traffic
@
Top bar parallel to traffic
DB12@250 AS3
Top bar perpendicular to traffic
DB12@250 AS1
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Steel Reinforcement DetailsExample Slab Bridge Details from DOT Standard Drawing
Steel Reinforcement DetailsExample Slab Bridge Details from DOT Standard Drawing
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Steel Reinforcement DetailsExample Slab Bridge Details from DOT Standard Drawing
Steel Reinforcement DetailsExample Slab Bridge Details from DOT Standard Drawing
Bridge Approach Slab Analysis & Design Criteria 1. The bridge approach slab is designed as a slab in accordance with Section 4.6 of the AASHTO LRFD Bridge Design Specifications (Strength I loading, IM = 1.33). 2. The end support is assumed to be a uniform soil reaction with a bearing pp glength that is approximately 1/3 the length of the approach slab. 3.The Effective Span Length (Seff) is assumed to be: Seff = L‐ (L/3)/2 ‐ a/2 ; L= Approach Slab Length, a = Support Seat Width
4. Longitudinal steel reinforcing bars do not require modification for skewed applications. 5. The maximum skew accommodated by the standard design is 30 degrees.
/a/2L
L/3
SeffL/6a/2
Seff
Simple Support
Wing Wall Design
Wingwalls may either be designed as monolithic with the abutments, or be separated from the abutment wall with an expansion joint and designed to be free standing.
The wingwall lengths shall be computed using the required roadway slopes. Wingwalls shall be of sufficient length to retain the roadway p g g yembankment and to furnish protection against erosion.
Ref. 11.6.1.4
Lateral Earth Pressure &
Wingwall Length
Lateral Pressure due to Top Surcharge Load
Edge of Abutment
Mu
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SubstructurePier Cab Design
Load Transfer to Pier CabLive Load Analysis Case 1
E.J.
9.3kN/m
Pattern Live Load due to Maximum Moment on Pier Cab and Maximum Axial Force on Pile
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Load Transfer to Pier CabLive Load Analysis Case 1
Dynamic load allowance need not be applied to foundation components that are
Estimate No. of Pile
Dynamic load allowance need not be applied to foundation components that are entirely below ground level. Maximum Load = {[156.3+5.96]x1.1+101.52}x10.2 = 2,923 kN/BentUsed Pile Diameter 0.4x0.4 m Safe Load 700 kN/PileNo. of Pile = 2,923/700 = 4.2 So, Used 5‐Pile 0.4x0.4 Safe Load 700kN/Pile
Pile Bent Dimension & Geometry
0.6m.
0.7m.
10m.
Cap Beam Section
2.25m.0.6m. 0.6m.
10.2m.2.25m.2.25m.2.25m.
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Pier Cap Design Force
DCSelf Weight = 10.1 kN/m
SFD
0.6m. 2.25 m. 2.25 m. 2.25 m. 0.6m.2.25 m.
BMD
Pier Cap Design Force
DCsuper= 156.3 kN/m
SFD
0.6m. 2.25 m. 2.25 m. 2.25 m. 0.6m.2.25 m.
BMD
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Pier Cap Design Force
DW= 11.92 kN/m
SFD
0.6m. 2.25 m. 2.25 m. 2.25 m. 0.6m.2.25 m.
BMD
Pier Cap Design ForceLive Load Case 1 LL1+IM = 125.76 kN/m Positive Max
SFD
0.6m. 2.25 m. 2.25 m. 2.25 m. 0.6m.2.25 m.
BMD
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Pier Cap Design ForceLive Load Case 2 LL1+IM = 125.76 kN/m Negative Max
SFD
0.6m. 2.25 m. 2.25 m. 2.25 m. 0.6m.2.25 m.
BMD
Pier Cap Design ForceSummary Design Force
FORCES SW DW DC LL1+IM LL2+IM
Moment (kN.m)
M 3 2 48 8 3 7 64 1M+ 3.2 48.8 3.7 64.1 ‐
M‐ ‐4.8 ‐74.8 ‐5.7 ‐ ‐72.6
Shear (kN)
V 12.7 196.5 15 156.1 173.8
Moment M+ (Service I) = 1.00x[3.2+48.8]+1.0x[3.7]+1.00x[64.1] = 119.8 kN.m
Design Force Moment & Shear
Moment M+ (Strength I) = 1.25x[3.2+48.8]+1.5x[3.7]+1.75x[64.1] = 182.8 kN.m
1.33Mu = 1.33x182.8 = 243.2 kN.mMr = 243.2 kN.mUse the bottom bar are 4 DB20
‐‐ 5.4.2.6
Use the bottom bar are 4‐DB20.Mn = 272.8 kN.m > 243.2 kN.m O.K.
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Reinforce Concrete Design
Positive Moment DesignService Limit State: Service ICrack Control by Distribution Reinforcement
M+ = 119.8 kN.mCheck concrete cracked at bottom fiber (Concrete tension)C ec co c ete c ac ed at botto be (Co c ete te s o )Modulus of rupture = fr = 0.63.fc’
Positive Moment DesignService Limit State: Service ICrack Control by Distribution Reinforcement
Elastic Cracked Section Property at Service State (Cracked, Linear Stage , approximately fc<0.45fc’)A t l St i th t lActual Stress in the steelfss = M/ As.j.d = 119.8/ [12.57x0.93x0.6255] = 164 Mpa.dc = 50covering+12+[20/2] = 72 mm., h = 700 mm.In this design example used e = 0.50 (crack width 0.22mm)
s = 1+dc/{0.7.[h‐dc]} = 1+72/{0.7x[700‐72]}= 1.16
Minimum Spacing of reinforcement in the layer closest to i f l k d (0 22 )tension face to control concrete cracked (0.22mm.).
S = {123,000.e/[s.fss]}‐2.dc = [123,000x0.5/(1.16x164)]‐2x72
S = 179.3 mm.Use the bottom bar 4‐DB20 , Sprovide = 600/4 = 150 mm. ≤ 179.5 mm. O.K.
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Reinforce Concrete Design
Negative Moment DesignStrength Limit State: Strength I
Mu‐max = ‐235.1 kN.mTry to provide steel bar = 5‐DB20.A { d2/4} 5 1 571 2 Eq 5 7 3 1 2 4
1.33Mu = 1.33x235.1 = 312.7 kN.mMr = 285.2 kN.mUse the bottom bar are 5 DB20
‐‐ 5.4.2.6
Use the bottom bar are 5‐DB20.Mn = 331.7 kN.m > 285.2 kN.m O.K.
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Reinforce Concrete Design
Negative Moment DesignService Limit State: Service ICrack Control by Distribution Reinforcement
M‐ = ‐157.9 kN.mCheck concrete cracked at bottom fiber (Concrete tension)C ec co c ete c ac ed at botto be (Co c ete te s o )Modulus of rupture = fr = 0.63.fc’
Negative Moment DesignService Limit State: Service ICrack Control by Distribution Reinforcement
Elastic Cracked Section Property at Service State (Cracked, Linear Stage , approximately fc<0.45fc’)A t l St i th t lActual Stress in the steelfss = M/ As.j.d = 157.9/ [15.71x0.92x0.6255] = 175 Mpa.dc = 50covering+12+[20/2] = 72 mm., h = 700 mm.In this design example used e = 0.50 (crack width 0.22mm)
s = 1+dc/{0.7.[h‐dc]} = 1+72/{0.7x[700‐72]}= 1.16
Minimum Spacing of reinforcement in the layer closest to i f l k d (0 22 )tension face to control concrete cracked (0.22mm.).
S = {123,000.e/[s.fss]}‐2.dc = [123,000x0.5/(1.16x175)]‐2x72
S = 159 mm.Use the bottom bar 5‐DB20 , Sprovide = 600/5 = 120 mm. ≤ 159 mm. O.K.
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Reinforce Concrete Design
Shrinkage & Temperature Reinforcement
Reinforcement for shrinkage and temperature stresses shall be provided near to daily temperature changes and in structural mass concrete in each direction and each face.
‐‐5.10.8
Use DB16 A = 201 mm2Use DB16, As 201 mmAg = 600x700 = 420,000 mm2
As‐required = 0.0015.Ag = 630 mm2 < 1,257 mm2 O.K.233 ≤ As ≤ 1,270 mm2 /m. O.K.At near face of side pier capSrequired = Min[700/(630/201),300] = 223 mm
Slenderness and Second ‐Order EffectMoment magnification (Approximate method)
1.Check Column SlendernessEffective length factor, Kx = 1.0 (Assume piles are braced with top slab)Lx = 6.30 m, Ag = 0.4x0.4 = 0.16m
2, Ix = 0.4x0.43/12 = 0.0021333 m4
rx =(Ix/Ag)1/2 = 0.115 m, Kx.Lx/rx =1.0x6.3/0.115= 54.8
For members braced against side sway the effects of slenderness may be neglected where KL/r is less than 34‐12(M1/M2) = 34‐12(65.7/199) = 3030 < Kx.Lx/ rx =54.8 < 100 Second‐order effect should be considered ‐‐ 5.7.4.3
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Slenderness and Second ‐Order EffectMoment magnification (Approximate method)
2. Determine EI for use in determining Pe are the greater ofEI = {[Ec.Ig/5]+[Es.Is]}/[1+d]EI = [Ec.Ig/2.5]/[1+d]More conservative solution being
‐‐ Eq 5.7.4.3‐1
‐‐ Eq 5.7.4.3‐2
EI = [Ec.Ig/2.5]/[1+d]Ec = 23,914 Mpa., Ig = 0.4x0.4
3/12 = 0.0021333 m4
d = ratio of maximum factored permanent load moments to maximum factored total load momentMDL= 0Mtotal‐Strength I = M2s = 1.75x[113.7] = 199 kN.md = MDL/Mtotal = 0/199 = 0EI = [23,914x106x 0.0021333/2.5]/1.0 = 20,406 kN.m2
= 87.2 kN. > 42.35 kN. O.K.Transverse shear reinforcement is not required
‐‐ Eq. 5.8.3.3‐3
‐‐5.8.3.4.1
‐‐5.5.4.2
Transverse shear reinforcement is not required.Used Minimum transverse shear reinforcementDB12@300 at Typical sectionThe spacing of ties along the longitudinal axis ofthe compression member shall not exceed the leastdimension of the compression member or 300 mm.
‐‐5.10.6.3
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The transverse reinforcement for confinement at the plastic hinges shall be determined as followsThe total cross‐sectional area (Ash) of stirrup reinforcement shall be eitherA 0 30 S h [f ’/f ] {[A /A ] 1}
Ductility of ColumnTransverse Shear Reinforcement for confinement
E 5 10 11 4 ld 2Ash = 0.30.S.hc.[fc’/fyh].{[Ag/Ae]‐1}or,Ash = 0.12.S.hc.[fc’/fy ].{0.5+[1.25Pu/Ag.fc’]}where,S = vertical spacing (not exceed 100 mm.)Ae = area of column core measured to the outside of the transverse reinforcement (mm2)fc’ = compressive strength of concrete (Pa)f = yield strength of stirrup or spiral reinforcement (Pa)
Eq‐‐5.10.11.4.ld‐2
Eq‐‐5.10.11.4.ld‐3
fy = yield strength of stirrup or spiral reinforcement (Pa)Ag = gross area of column (mm2)Ash = total cross‐sectional area of stirrup reinforcement (mm2)hc =core dimension of tied column in the direction under consideration (mm.)Pu = factored axial load (MN)
Use 12 mm bar diameter (DB12, Ash = 113.14 mm2)Ag = 400x400 = 160,000 mm2
= 2x113.14/{0.3x300x[25/400]x{[160,000/122,500]‐1}= 131 mm.
DB12@125 at plastic hinge region of column
Eq‐‐5.10.11.4.ld‐2
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• Provided at the top and bottom of the column over a length not less than the greatest of the maximum cross‐sectional column dimensions, one‐sixth of the clear height of the column, or 450mmP id d h f il i il b h
Ductility of ColumnTransverse Shear Reinforcement for confinement
5.10.11.4.1
• Provided at the top of piles in pile bents over the same length as specified for columns• Provided within piles in pile bents over a length extending from 3.0 times the maximum cross sectional dimension below the calculated point of moment fixity to a distance not less than the maximum cross‐sectional dimension or450 mm above the mud line