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ตุลาคม 2557

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

‐Concrete BarrierC t D k‐Concrete Deck

‐Overlay and Wearing surface

Dead Load of Wearing Surface and Utilities (DW)

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Load CriteriaLive Load (LL)Design Vehicular Live Load HL93‐Design Vehicles Load (AASHTO, 3.6)

‐HS20‐44 Truck + Lane Load

4.3‐9.0

‐Tandem + Lane Load

Axle Configuration and Design Lane Width.

3000 mm

3600 mm

Design Thai Truck Load Thai Truck Factor x HL93 (AASHTO LRFD)

‐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.

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Unit Weight of Material

Concrete = 22.76 kN/m3 (2320 kg/m3)Reinforced Concrete = 24.03 kN/m3 (2450 kg/m3)Asphalt Wearing Surfaces = 22.07 kN/m3 (2250 kg/m3)Reinforcing Steel = 77.01 kN/m3 (7850 kg/m3)

Deck Thickness, T

For Simple Span T =1.2(S+3,000)/30

Determine Deck Thickness

=1.2(10,000+3000)/30=520mm > 175 mm O.K

Assume deck thickness = T = 600 mm. ‐‐Table 2.5.2.6.3‐1

Live Load Analysis byEquivalent Strip Width Method

(Approximate Methods)

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Live Load Analysis STEP‐1‐Determine Forces/Lane for HL93‐Design Vehicles Load

Maximum Moment Due to HS20‐44 Truck Loading

Direction

Span Length = 10 m.

M+max = 446.75 kN.m/Lane

NOTE: CALCULATE CHECK FOR DESIGN TRUCK LOAD

SIMPLE SPAN (L>10m.) PATTERN LOAD DUE TO MAXIMUM MOMENT

Lof BridgeC

2.85m 0.70m0.75m

145kN 35kN

RESULTANT OF FORCE =325kN

L/2‐0.70L/2+0.70

145kN

THE POSITION OF MAXIMUM MOMENT

4.30m

M+max = R1x{L/2+0.7} ‐ {145x4.3}= 187.05x5.7‐623.5= 442.685 kN.m/Lane

R1 = {162.5L+245.5}/L

= 187.05 kN

O.K.

R2 = {162.5L‐245.5}/L

= 137.95 kN

L = 10m

Difference =1%

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Live Load AnalysisSTEP‐1‐Determine Forces/Lane for HL93‐Design Vehicles Load

Maximum Shear Due to HS20‐44 Truck Loading

Direction

Span Length = 10 m.

Vmax = 232.55 kN/Lane


Maximum Moment Due to Tandem Loading

Direction

Span Length = 10 mSpan Length = 10 m.


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NOTE: CALCULATE CHECK FOR DESIGN TENDEM LOAD

SIMPLE SPAN (L>1.2m.) PATTERN LOAD DUE TO MAXIMUM MOMENT

Lof BridgeC

0.6m 0.3m0.3mTHE POSITION OF MAXIMUM

MOMENT

110kN 110kN

RESULTANT OF FORCE =220kN

L/2‐0.3L/2+0.3

M+max = R2x{L/2‐0.3} = 103.4x{5.0‐0.3}= 485.98 kN.m/Lane

R1 = {110.L+66}/L

= 116.6 kN

O.K.

R2 = {110.L‐66}/L

= 103.4 kN

L = 10m


Maximum Shear Due to Tandem Loading

Direction

Span Length = 10 m.


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Maximum Moment Due to Uniform Lane Load

Span Length = 10 m.

Uniform Lane Load 9.3kN/m



Maximum Shear Due to Uniform Lane Load

Span Length = 10 m.


Vmax = 46.5 kN/Lane

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[Dynamic Factor x HS20‐44] + Lane Load

Maximum Moment for LLCASE‐1‐Truck+Lane

Direction

Span Length = 10 m


[1.33]

Span Length = 10 m.



[Dynamic Factor x HS20‐44] + Lane Load

Maximum Shear for LLCASE‐1‐Truck+Lane

Direction

Span Length = 10 m.

Uniform Live Load 9.3kN/m

[1.33]


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[Dynamic Factor x Tandem] + Lane Load

Maximum Moment for LLCASE‐2‐Tandem+Lane

Direction

S h 0

[1.33]

Span Length = 10 m.



[Dynamic Factor x Tandem] + Lane Load

Maximum Shear for LLCASE‐2‐Tandem+Lane

Direction

Span Length = 10 m.

[1.33]


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Live Load AnalysisSTEP‐1‐Determine Forces/Lane for HL93‐Design Vehicles LoadResult Summary

Envelop Maximum Bending Moment of LLCASE‐1‐Truck+Lane& LLCASE‐2‐Tandem+Lane

Mmax At X = 5.3 m

1.2 m

110 kN 110 kN


Tandam+Lane Govern to Design Moment

Live Load AnalysisSTEP‐1‐Determine Forces/Lane for HL93‐Design Vehicles LoadResult Summary

Envelop Maximum Shear Force of LLCASE‐1‐Truck+Lane& LLCASE‐2‐Tandem+Lane

Truck+Lane Govern to Design Shear

4.3 m

145 kN 145 kN 35 kN

4.3 m


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FORCES VS SPAN LENGTH HL‐93 LIVE LOAD

TANDEM + LANE LOAD WILL BE GOVERN TO DESIGN SHEAR FORCE WHEN SPAN LENGTH

> 7.6 m.

NOTE:


TRUCK + LANE LOAD WILL BE GOVERN TO DESIGN MOMENT

WHEN SPAN LENGTH> 12.2 m.

NOTE:

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Reference : LRFD Bridge Design Manual: Minnesota Department of Transportation

Live Load Analysis (Approximate Methods)STEP‐2‐Determine the Equivalent Strip Width

Equivalent Strip Width for Interior SlabFor One Lane

Es = 0.25+0.42(L1xW1)1/2 < W/NL

L = Min(18 Or 10 m ) = 10m

For Multilane

Em = 2.10+0.12(L1xW1)1/2 < W/NL

L1 = Min(18 Or 10 m.) = 10m.W1 = Min(18 Or 10m.) = 10m.

L1 = Min(18 Or 10 m.) = 10m.W1 = Min(9 Or 10m.) = 9m.

Es = 0.25+0.42(10x9)1/2 = 4.23 m. < W/NL =10/2 = 5m. ‐‐Eq‐4.6.2.3‐1

1

Em = 2.10+0.12(10x9)1/2 = 3.3 m. < W/NL =10/2 = 5m. ‐‐Eq‐4.6.2.3‐2

Em

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Live Load Analysis (Approximate Methods)STEP‐2‐Determine the Equivalent Strip Width

Equivalent Strip Width for Edge Slab

Es = 4.23 m, Wedge = 0.5 m.

Ee = Wedge+0.30+Es/4

Ee = 1.8 m.

NL = 10/3.6 = 2.77 = 2 Design Lane

e edge s

= 0.5+0.3+[4.23/4]= 1.86 m.but not exceed ether 0.5xEs= 2.12m. or 1.8m.

So, Use the design equivalent strip widthEe = 1.8 m. and Em = 3.3 m. ‐‐4.6.2.1.4be m

r =1.05‐0.25(tan θ˚) = 1.05‐0.25(tan 0˚) = 1.05 ≤ 1.0 =1.0 ‐‐Eq‐4.6.2.3‐1

Ee

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 for edge slab design

Edge beam is assumed to support one line of truck axle

Ee


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

= 1.05x174.80xr/[Em] = 1.05x174.80x1.0/[1.8] =101.97 kN/m

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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


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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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



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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



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


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


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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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


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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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

Flexural Reinforcement

As‐provide = {.d2/4}x1,000/100 = 4,909 mm2

1 = 0.85, b = 1,000 mm.c = As.fy / 0.85.fc’.b .1

= 4,909x390 / {0.85x25x1,000x0.85}= 106 mm.

a = 1.c = 0.85x106 = 90.1 mm.de = 600‐(50)covering ‐(25/2) = 537.5 mm.Check c/de =106/537.5 = 0.197 ≤ 0.375 ==> Tension Controlled = 0 9

‐‐5.7.2.2

‐‐Eq 5.7.3.1.2‐4

‐‐5.5.4.2.1‐2 0.9.Mn = 0.9x4,909x390[537.5‐(90.1/2)] /1,000,000

= 848.5 kN.m/m > 669.7 kN.m/m O.K.Capacity Ratio = 669.7/848.5= 0.79Use the bottom bar parallel to traffic are DB25@100

NOTE: DB25@125==>(.Mn = 622.3 kN.m)

5.5.4.2.1 2


Mr is the lesser of 1.2Mcr and 1.33Mu

Modulus of rupture for cracking moment calculation

Minimum Reinforcement (Cracking Moment)


fr = 0.97.fc’1/2 = 0.97x251/2

= 4.85 Mpa.Sc = b.h

2/6 = 1.0x0.62/6 = 0.06 m3/m.

1.2Mcr = 1.2.Sc.fr = 1.2x0.06x4.85x1000 = 349.2 kN.m

1.33Mu = 1.33x628.5 = 835.9 kN.mMr = 349.2 kN.m

‐‐ 5.4.2.6

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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LRFD states that for cast‐in‐place slab superstructures designed for moment in conformance with LRFD 4.6.2.3,

Shear Capacity Check


‐‐ 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


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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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’

1/2 = 3.15 Mpa. h = 0.6 m., Ig = b.h

3/12 = 0.018 m4/1m. Widthfc‐tension = M.(h/2)/Ig = 7.34 Mpa. > 0.8fr = 2.5 Mpa.

Section is cracked

Elastic Cracked Section Property at Service State (Cracked, Linear Stage, approximately fc<0.45fc’)

/ b d /[ ]

‐‐ 5.7.3.4 &5.4.2.6

= As/ b.d = 4,909/[1,000x537.5] = 0.0087Modular Ratio, n = Es/Ec = 200,000/23,914 = 8.36k = [2n.+(n.)2]1/2‐n. = 0.3218J = 1‐k/3 =0.893fc,comp =2.M /(k∙j∙b∙d2) = 10.61 Mpa. < 0.45fc’=11.25 Mpa.

‐‐ 5.7.1.4



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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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.


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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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

A ≥ 0 75∙b∙h / { 2∙[b+h]∙fy } ‐‐ Eq 5 10 8‐1As ≥ 0.75 b h / { 2 [b+h] fy }As = 0.75x1.0x0.6/{2x[1.0+0.6]x390}= 3.6x10‐4 m2 /m.= 360.6 mm2 /m.

233 ≤ As ≤ 1,270 mm2 /m. O.K.

Use the Top bar perpendicular & parallel to traffic are DB12@250 (As = 452 mm2)

Eq. 5.10.8 1

‐‐ Eq. 5.10.8‐2


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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Elastic Cracked Section Property at Fatigue Limit State (Cracked, Linear Stage, approximately fc<0.45fc’)fc,comp =2.M /(k∙j∙b∙d2)

= 2x446.5/[0.421x0.86x1.0x0.53752]x1000= 8.54 Mpa. < 0.45fc’=11.25 Mpa.

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;



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’

1/2 = 2.79 Mpa. A 4 909 mm2 A / b d 0 0087

‐‐5.7.3.6.2

‐‐5.4.2.4

As = 4,909 mm2, = As/ b.d = 0.0087n = Es/Ec = 8.36k = 0.3218J =0.893Icr =b.(k.d)

3/3+n.As.(d‐k.d)2 = 0.007178 m4/m

Ig = 0.018000 m4/m

Mcr = fr .Ig/yt = 2.79x0.018/ (0.60/2) = 167.17 kN.m/mDCDECK+BARRIER = 14.42+1.2 =15.62 kN/m

2 /mMa = 15.62x10

2/8 = 195.28 kN.m/mMa 15.62x10 /8 195.28 kN.m/m

Ie =(167.17/195.28)3x0.018+[1‐(167.17/195.28)3]x0.007178

= 0.0142 m4/m ≤ Ig = 0.018000 m4/m

‐‐Eq. 5.7.3.6.2‐1

Camber Diagram

W0 = 15.62 kN/m2/m

Ec = 21,389 Mpa, Ie = 0.0142 m4/m

= W0 .X.(L3‐2.L.X2+X3)/(24Ec.Ie)

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


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ตัวอย่าง การจําลอง โครงสร้างสะพานส่วนล่างโดยนายเกรียงไกร คําพา

ตุลาคม 2557

SubstructureDesign Example AASHTO (LRFD)

Design Specification

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Bridge Geometry

10m. 10m.10m.

30m.

b bAbutment A Bent 1 Bent 2 Abutment B

5m.

E.J. E.J. E.J. E.J.

Out Line Design Member

SubstructureAbutment A ===> Wingwall, Pier Cab, PileAbutment B ===> Wingwall, Pier Cab, PileBent 1 ===> Pier Cab, PileBent 2 ===> Pier Cab, PileMiscellaneous StructureMiscellaneous Structure Approach SlabConcrete Railing

Abutment A Bent 1 Bent 2 Abutment B

Pier Cab Deck Slab

Wingwall

PileApproach Slab

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Approach Slab Design

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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9.3kN/m

9.70m.0.30m.

9.70m.

R1 R2 R3 R4

[R2+R3][R2+R3]max

Maximum Reaction Force = [Truck+Impact] + Lane LoadInterior Strip [Em = 3.3 m.]Interior Strip [R2+R3]max = {[1.33x242]+[9.3x10]}/3.3 = 125.76 kN/mEdge Strip [Ee = 1.8 m.] Edge Strip [R2+R3]max = {[1.33x242/2]+[0.43x9.3x10]}/1.8 = 111.62 kN/m


E.J.

9.3kN/m

Pattern Live Load due to Maximum Torsion on Pier Cab and Maximum Bending Moment on Pile

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Pier CabLoad Transfer to Pier CabSummary Load from Super Structure Transfer to Pier Cab

FORCES DC DW LL1(Truck+Lane)

Case‐1

LL2(Truck+Lane)

Case‐2

LL1+IM Case‐1

LL2+IM Case‐2

Reaction Force (kN/m)

INTERIOR STRIP 156.3 11.92 101.52 84.56 125.76 107.82

EDGE STRIP 156.3 11.92 89.44 75.71 111.62 97.02

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


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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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

Moment M‐ (Service I) = 1.00x[4.8+74.8]+1.0x[5.7]+1.00x[72.6] = ‐157.9 kN.mMoment M‐ (Strength I) = 1.25x[4.8+74.8]+1.5x[5.7]+1.75x[72.6] = ‐235.1 kN.m

Shear (Strength I) = 1.25x[12.7+196.5]+1.5x[15]+1.75x[173.8] = 588.15 kN

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Mu+max = +182.8 kN.mTry to provide steel bar = 4‐DB20.A { d2/4} 4 1 257 2 Eq 5 7 3 1 2 4


As‐provide = {.d2/4}x4= 1,257 mm2

1 = 0.85, b = 600 mm.c = As.fy / 0.85.fc’.b .1

= 1,257x390 / {0.85x25x600x0.85}= 45.23 mm.

a = 1.c = 0.85x45.23 = 38.45 mm.de = 700‐(50)covering ‐12‐(25/2) = 625.5 mm.Check c/de =45.23/625.5 = 0.07 ≤ 0.375 ==> Tension Controlled = 0 9

‐‐5.7.2.2

‐‐Eq 5.7.3.1.2‐4

‐‐5.5.4.2.1‐2 0.9.Mn = 0.9x1,257x390[625.5‐(38.45/2)] /1,000,000

= 267.5 kN.m > 182.8 kN.m O.K.Capacity Ratio = 182.8/267.5= 0.68

Use the bottom bar are 4‐DB20

5.5.4.2.1 2






fr = 0.97.fc’1/2 = 0.97x251/2

= 4.85 Mpa.Sc = b.h

2/6 = 0.6x0.72/6 = 0.049 m3/m.

1.2Mcr = 1.2.Sc.fr = 1.2x0.049x4.85x1000 = 285.2 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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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’

1/2 = 3.15 Mpa. b=0.6 m., h = 0.7 m., Ig = b.h

3/12 = 0.01715 m4

fc‐tension = M.(h/2)/Ig = 2.5 Mpa. >= 0.8fr = 2.5 Mpa.Section is cracked


/ b d /[ ]

‐‐ 5.7.3.4 &5.4.2.6

= As/ b.d = 1,257/[600x625.5] = 0.0033Modular Ratio, n = Es/Ec = 200,000/23,914 = 8.36k = [2n.+(n.)2]1/2‐n. = 0.2089J = 1‐k/3 =0.93fc,comp =2.M /(k∙j∙b∙d2) = 5.25 Mpa. < 0.45fc’=11.25 Mpa.

‐‐ 5.7.1.4



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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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


As‐provide = {.d2/4}x5= 1,571 mm2

1 = 0.85, b = 600 mm.c = As.fy / 0.85.fc’.b .1

= 1,571x390 / {0.85x25x600x0.85}= 56.55 mm.

a = 1.c = 0.85x67.86 = 48.1 mm.de = 700‐(50)covering ‐12‐(25/2) = 625.5 mm.Check c/de =56.55/625.5 = 0.09 ≤ 0.375 ==> Tension Controlled = 0 9

‐‐5.7.2.2

‐‐Eq 5.7.3.1.2‐4

‐‐5.5.4.2.1‐2 0.9.Mn = 0.9x1,571x390[625.5‐(48.1/2)] /1,000,000

= 331.7 kN.m > 235.1 kN.m O.K.Capacity Ratio = 235.1/331.7= 0.71

Use the Top bar are 5‐DB20

5.5.4.2.1 2





Negative Moment DesignStrength Limit State: Strength I

fr = 0.97.fc’1/2 = 0.97x251/2

= 4.85 Mpa.Sc = b.h

2/6 = 0.6x0.72/6 = 0.049 m3/m.

1.2Mcr = 1.2.Sc.fr = 1.2x0.049x4.85x1000 = 285.2 kN.m

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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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’

1/2 = 3.15 Mpa. b=0.6 m., h = 0.7 m., Ig = b.h

3/12 = 0.01715 m4

fc‐tension = M.(h/2)/Ig = 3.22 Mpa. >= 0.8fr = 2.5 Mpa.Section is cracked


/ b d /[ ]

‐‐ 5.7.3.4 &5.4.2.6

= As/ b.d = 1,571/[600x625.5] = 0.0042Modular Ratio, n = Es/Ec = 200,000/23,914 = 8.36k = [2n.+(n.)2]1/2‐n. = 0.23J = 1‐k/3 =0.92fc,comp =2.M /(k∙j∙b∙d2) = 6.4 Mpa. < 0.45fc’=11.25 Mpa.

‐‐ 5.7.1.4


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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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

Use 2‐DB16

‐‐ 5.10.8‐2


Vu = 588.15 kN.dv = Max.[de‐a/2, 0.9de ,0.72h] = Max.[606.3, 563, 504]

Shear Capacity CheckStrength Limit State: Strength INominal shear resistance

‐‐ 5.8.2.9

= 606.3 mm.bv = 600 mm.v = 0.9.Vc = .0.083∙β∙f c’1/2∙bv∙dvβ = 2.0, θ = 45˚f c‘= 25 Mpa..Vc = 0.90x0.083x2.0x251/2x600x606.3/1,000

= 271.7 kN. < 588.15 kN.Transverse reinforcement is required

‐‐ Eq. 5.8.3.3‐3

‐‐5.8.3.4.1

‐‐5.5.4.2

Transverse reinforcement is required.

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Nominal shear strength provided by shear reinforcementVn = Vc + Vs + Vp

Shear Capacity CheckStrength Limit State: Strength ITransverse reinforcement

‐‐ Eq. 5.8.3.3‐1

Vn = Min.[Vu/v ,0.25fc’.bv.dv] = Min.[653.5, 2274] = 653.5 kN.

Vp = 0 kN. (for non‐prestress concrete)Vs = 653.5‐ 301.9 = 352 kN.β = 2.0 ,θ = 45˚Size of stirrup bar = DB12 (2‐leg, Av = 2x113.1 =226.2 mm2)Angle of inclination of transverse reinforcement = =90˚Spacing of shear reinforcementS = A f d Cotθ/V = 226 2x390x606 3xCot 45˚/352x1 000

‐‐5.8.3.4.1

‐‐ Eq. 5.8.3.3‐2

Sreq = Av.fy.dv.Cotθ/Vs = 226.2x390x606.3xCot 45 /352x1,000Sreq = 152 mm.Used DB12@150

Pile Cab Design

Summary of Reinforcement

5‐DB20 Main Flexural Bar

0.6m.

(Negative)

4‐DB20 Main Flexural Bar(Positive)

2‐DB16 Each Face(Shrinkage &

Temperature Steel)

0.7m.

(Positive)

DB12@150(Shear Reinforcement)

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SubstructurePile Bent Design

Load on PileBraking Force

BR1‐1=25%x[242]=60.5 kN./Lane, BR1‐2=5%x[242+9.3x10]=16.75 kN./Lane= 60.5 kN/Lane x 2 Lane. = 121 kN. [Maximum]Apply braking force 121 kN at 1.8 m. above road surface level

PLL+PDC+PDW PLL+PDC+PDW

1.80m.

BR = 121/2 = 60.5 kN

1.3 m.

4.3 m.

Mt = 60.5x3.1= 187.6 kN.m

BR = 121/2 = 60.5 kN

9 4

3.1 m.

2.0 m.

Ground level

Pile fixed point level (Assume)

Mb = 60.5x9.4= 568.7 kN.m

Braking load apply

Pile bent structural model & support configuration

Bending Moment Diagram

9.4 m.

6.3 m.

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Load Transfer to PilePile Loads ‐ Service

FORCES DC DW LL1 (Truck+Lane)Case‐1

BR WA

Axial Force (kN/Pile) 435 27 228.5 ‐ ‐

M t (L it di l)Moment (Longitudinal)

MxTop Pile (kN.m/Pile) ‐ ‐ ‐ 37.5 ‐

MxBottom Pile (kN.m/Pile) ‐ ‐ ‐ 113.7 ‐

Pile Capacity Resistance Checked

Note: In this design example not consider water load on longitudinal direction (WA)

Maximum Load = [435+27+228.5]= 690.5 kN/Pile < 700 kN ==> O.K.

Load CombinationTable 3.4.1‐1 Load Combinations and Load Factors

In this design example consider only load combination group Strength I

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Pile Loads – Strength‐I

Maximum Axial LoadPu (Strength I) = 1.25x[352+83]+1.5x[27]+1.75x[228.5] = 984.2 kNMaximum MomentMut (Strength I) = 1.75x[37.5] = 65.7 kN.mMub (Strength I) = 1.75x[113.7] = 199 kN.mMaximum ShearMaximum ShearVu (Strength I) = 1.75x[60.5/5] = 21.2 kN

Vu = 21.2 kN/Pile

Mt = 65.7 kN.m/Pile

Pu = 984.2 kN/Pile

Vu = 21.2 kN/Pile

Mb = 199 kN.m/Pile

Pu = 984.2 kN/Pile

6.3 m.

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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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

3 Determine Euler buckling load P3. Determine Euler buckling load PexPex = 2.EI/(KL)2 = 2.20,406/[1.0x6.32] = 5,078 kN4.Determine the factor momentCm = 1.0 for braced against side swayPu = 1.25x435+1.5x27+1.75x228.5 = 984.2 kNk. = 0.75 ,for concretes = 1/{1‐[Pu/k.Pe]} =1/{1‐[984.2/0.75x5,078]} = 1.35

‐‐ Eq 4.5.3.2.2b‐5

‐‐ Eq 4.5.3.2.2b‐4


5. 4.Determine the magnified factor momentMc= s.M2s= 1.35x199 = 268.7 kN.m ‐‐ Eq 4.5.3.2.2b‐1

Summary Design Force

Pu = 984.2 kNVu = 21.2 kNMu = 268.7 kN.m

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Pile Bent Axial Compression & Flexural Combnation Design Checked

Section Dimension400x400 mm.Concretefc’ = 25 Mpa.Reinforcement Steelfy = 400 Mpa.CoveringS = 50 mmNo of BarUsed 12‐DB28 (12‐#9)Rebar Ratio = 4.8%

Shear Capacity CheckStrength Limit State: Strength INominal shear resistance

Vu = 42.35 kN.a = As.fy / 0.85.fc’.b = 4x616x400/0.85x25x400 = 116 mm.dv = Max.[de‐a/2, 0.9de ,0.72h] = Max.[266, 291.6, 288]

‐‐ 5.8.2.9

= 291.6 mm.bv = 400 mm.v = 0.9.Vc = .0.083∙β∙f c’1/2∙bv∙dvβ = 2.0, θ = 45˚f c‘= 25 Mpa..Vc = 0.90x0.083x2.0x251/2x400x291.6/1,000

= 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

hc = 400‐2x(50)covering = 300 mmAe = 300x300 = 122,500 mm2

f ’ 25 MP


fc’ = 25 MPa.fy = 400 MPa.Ash = 0.30.S.hc.[fc’/fyh].{[Ag/Ae]‐1}Srequired = Ash /{0.30.hc.[fc’/fyh].{[Ag/Ae]‐1}}

= 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


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

10/18/2014 - engfanatic.tumcivil.com · wingwall pile approach slab deck slab. 10/18/2014 2...

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