Bridge Ch 5 Example on Slab Bridge

8/13/2019 Bridge Ch 5 Example on Slab Bridge

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Chapter 5- Superstructures

Design Example on Slab Bridge Fundamentals of Bridge Structures

AAiT, Department of Civil & Environmental Engineering Page 1

Chapter 5

SUPERSTRUCTURES

Example on Design of Slab Bridge

Design Data and Specifications

Superstructure consists of 10m slab, 36m box girder and 10m T-girder all simply supported.

Only the design of Slab Bridge will be used for illustration.

Roadway Grade = 1660.00 m, amsl

HWM = 1643.56 - Roadway grade dictates elevation of

superstructure and not minimum free board requirement.

I. Slab II. T-Girder III.Box-Girder

Clear span = 10m Clear span 10m Clear span = 36m

Road way width = 7.32m Road way width = 7.32m Road way width = 7.32m

Curb width = 0.8m Curb width = 0.80m Curb width = 0.80m

-Materials

Concrete: Class A concrete: Cylinder strength fc = 28MPa [A5.4.2.1]

[A5.4.2.4]

Steel: fy= 400MPa

Es= 200GPa

Design method is Load and Resistance Factor Design (LRDF)


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Reference: AASHTO LRFD Bridge Design Specifications, SI units, 2ndEdition, 2005.

Slab Bridge Design

1. Depth Determination [A2.5.2.6.3]

Minimum recommended depth for slabs with main reinforcement parallel to traffic is

Where S is the span, S=c/c of supports clear span + d, S=10+0.4/2+0.43/2=10.415m

Use D = 540 mm, d= 540- F/2-25 = 499mm S=10.415mClear span + d = 10000 + 499 =

10.499m Ok! (Cover)

2. Live Load Strip Width [Art.4.6.2.3]

a) I nteri or Strip

i) One lane loaded: multiple presence factor included [C.4.6.2.3]

L1 is smaller of 10415 or 18000. W1is the smaller of 8920 or 9000

L1 = 10415 W1= 8920

ii) Multiple lanes loaded


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W=Actual edge to edge width = 8920mm

NL= Int(clear roadway width/3600)

Use E=3256.63mm

Equivalent concentrated and distributed loads

Truck: P1=35/3.2566=10.75; P2 = 145/3.2566 = 44.52Tandem: P3=110/3.2566 = 33.78

Lane: w = 9.3/3.2566 = 2.856

b) Edge Strip Longitudinal edge strip width for a line of wheels [Art.4.6.2.1.4]

E= distance from edge to face of barrier + 300+1/4* strip width

E= 800 + 300+3256.63/4 = 1914.08mm > 1800mm

E=1800mm

3. Influence Lines for Shear Force and Bending Moment

Slab bridges shall be designed for all vehicular live loads specified in AASHTO

Art 3.6.1.2, including the lane load [Art.3.6.1.3.3]

a) I nter Stri pi) Maximum Shear Force

This governs


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Impact factor = 1+IM/100 = 1+33/100 = 1.33, not applied to lane load [Art.3.6.2.1]

VLL+IM=1.33*72.52+14.87 = 111.32

ii) Maximum bending Moment

Truck: MTr

= 44.52(0.703+2.553) + 10.75(0.103) = 146.06 kNm

Tandom: MTa= 33.78(2.304*2) =155.66 kNm this governs

Lane: MLn = 2.856*(1/2)*2.604*10.415 =38.73kNm

MLL+Im= 1.33*155.66+38.73 = 245.76kNm

b) Edge Strip

Because E= 1800mm, one lane loaded with a multiple presence factor of 1.2 will be

critical

4. Select resistance factor, [Art. 5.5.4.2.1]

Strength Limit States (RC)

Flexure & Tension 0.90Shear & Torsion 0.90

Axial Compression 0.75

Bearing On concrete 0.70

Compression in strut and tie model 0.70

5. Select Load Modifiers, 1

Strength service fatigue

i) Ductility 0 0.95 1.0 1.0 [Art. 1.3.3]ii) Redundancy R 1.05 1.0 1.0 [Art. 1.3.4]iii) Importance I 1.05 1.0 1.0 [Art. 1.3.5]


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0= R= I= 1.0

6. Select Applicable Load Combinations [Table 3.4.1-1]

Strength I U= (1.25DC + 1.50DW + 1.75(LL+1M)+1.0FR+TGTGService I U=1.0(DC+DW) +1.0(LL+IM) + 0.3(WS+WL+1.0FR

Fatigue U=0.75*(LL+IM)

7. Dead Load Force Effects

a) I nteri or Strip:- Consider a 1m Strip, con=2400 kg/m3 [Table 3.5.1-1]WDC= (2400*9.81)* 10

-3 kN/m3* 0.54 m = 12.71kN/m2

WDW= (2250*9.81)* 10-3

kN/m3* 0.075m = 1.66kN/m

2

75mm bituminous wearing surface, bit= 2250kg/m3 [Table 3.5.1-1]VDC= * 12.71*10.415 = 66.21kN/m VDW= * 1.66*10.415 = 8.64kN/m

b) Edge Str ip:

VDC= * 16.06*10.415 = 83.63kN/m

8. Investigate Service Limit State


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i) Durability: Cover for main reinforcement steel for [Art. 5.12]

deck surface subjected to tire wear = 60mm bottom of cast in-place slab = 25 mm

D= R= I= 1.0 = 1.0

a) Moment I nteri or Strip

M=1.0(172.34 + 22.51 + 245.76) = 440.61 kNm

Reinforcement:Assume j=0.875 and fs= 0.6 fy= 0.6*400 = 240

b) Moment Edge str ip:

M=1.0(217.76 + 0 + 533.56) = 751.32kNm

ii) Control of Cracking [Art.5.7.3.4]

Components shall be so proportioned that the tensile stress in the mild steel

Reinforcement at the service limit state, fs, does not exceed fsa


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Zcrack width parameter (N/mm) = 23000N/mm for severe exposure

dcdepth of concrete measured from extreme tension fiber to center of bar located closest

there to. Clear cover used to compute dc50mm

a) I nteri or strip

190


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Now, steel stress should be calculated for elastic cracked section. The moment of inertia

of the composite transformed section should be used for the stress calculation

N=7, nAsprove= 7*4232.88 = 29630.16mm2 Equivalent concrete area

Determine x from *1000*x2= 29630.16(499-x) x=144.87mm

Now Icr= 1/3*1000*144.873+ 29630.1(499-144.86)2= 4.729*109mm4/m.

Steel stress over n, fs/n = M(d-x)/Icr=(440.61*106*354.13)/(4.729*10

9) = 32.99MPa

fs=7*32.99=230.93MPa0.6fy

Now, fsacan be computed:

fs= 230.93fsa= 240Mpa OK!

b) Edge Strip

Medge= 751.32KNm/m


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*750*X2 = 7*4882.93(749-x) x = 219.655mm


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a) Dead Load Camber:

Total dead load of the bridge and the whole bridge cross-section is consideredWDC= 12.71*8.62+(2.53+0.59+0.23)*1.8*2=121.62KN/m

Maactual maximum moment (Nmm)

frmodulus of rupture

ytdistance from N.A to extreme tension fiber (mm)

fr= 0.63 = 0.63 = 3.33Mpa,

Location of N.A,

Since the section does not crack under DL, Ig should be used

Chamber 4*4.53=18.12mm upward

WDW=1.66*7.32


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b) Live Load Deflection (Optional) [Art.

2.5.2.6.2]

Use design truck alone or design lane plus 25% of truck load. [Art. 3.6.1.3.2]

When design truck is used alone, it should be placed so that the distance between its resultant

and the nearest wheels is bisected by span centerline. All design lanes should be loaded.

MDC+DW+LL+IM = 1813.79+1.33*146.06*3.2566*2*1.0 = 3079.04KNm>Mcr

Multiple presence factor

Design TruckLoad

First load, P=385.7KN,a=8.78,b=1.635m,X=4.48m

Second load, P=385.7,a = x = 4.48m, b = 5.935m

Third load, P=93.1kKN,a=10.235,b=0.18m,X=5.935

(LL+IM)1=1.75+3.83+0.003=5.583


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LL+IM=1.33+1.48+2.79mm


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ff-is stress range fmin-minimum LL stress, where there is stress reversal=0 for our case

r/h=0.3

ff=145-0.33(0)+55(0.3)=161.5Mpafmaxmin

As =bd = 0.0086*750*749=5082.19mm2


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ii ) Sheaf

Slab bridges designed in conformance with AASHO, Art 4.6.2.3 may be considered

satisfactory for shear. Art. 4.6.2.3 deals with approximate method of analysis of slab bridges

using equivalent strip method.

But if longitudinal tubes are placed in the slab as in pre stressed concrete, and create voids

and reduce the cross section, the shear resistance must be checked.

iii) Distri bution Rein forcement:The amount of bottom transverse reinforcement may be

taken as a percentage of the main reinforcement required for positive moment as

a) I nteri or strip:

Transverse reinforcement = 0.175*4347.34mm2= 745.6mm2

m

b) Edge str ip:

Transverse reinforcement = 0.1715 * 5063.8 mm2= 868.44mm2

m


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iv) Shrinkage& Temperature Reinforcement: Reinforcement for shrinkage & temperature

stresses shall be provided near surfaces of concrete exposed to daily temperature changes.The steel shall be distributed equally on both sides

a) I nteri or Strip:

m, transverse.

2.1 LIMIT STATESGENERAL

Bridges shall be designed for specified limit states to achieve the objectives of

constructibility, safety, and serviceability, with due regard to issues of inspectibility,

economy, and aesthetics, as specified in Chapters 311.

Regardless of the type of analysis used, Equation 2.1 shall be satisfied for all specified force

effects and combinations thereof. Equation 2.1 below is the basis of the LRFD methodology.

Each component and connection shall satisfy Equation 2.1 for each limit state, unless

otherwise specified. For service and extreme event limit states, resistance factors shall be

taken as 1.0, except for bolts, for which the provisions of Chapter 8: Bridge Detailsapply.

All limit states shall be considered of equal importance.

iiQi Rn = Rf (2.1)


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Where:

for loads for which a maximum value of iis appropriate:

i = DRI 0.95 (2.2)

for loads for which a minimum value of iis appropriate:

i = 1 1.0 (2.3)

DRI

Where: i = load modifier: a factor relating to ductility, redundancy, and operational

importance

i = load factor: a statistically based multiplier applied to force effects

Qi = force effect

= resistance factor: a statistically based multiplier applied to nominal resistance

(see chapters 5,6,7 8, 10 and 12).

Rn = nominal resistance

D = a factor relating to ductility, as specified below

R= a factor relating to redundancy as specified below

I= a factor relating to operational importance as specified below

Rf = factored resistance: Rn

Ductility, redundancy, and operational importance are significant aspects affecting the margin

of safety of bridges. Whereas the first two directly relate to physical strength, the last

concerns the consequences of the bridge being out of service. The grouping of these aspects

on the load side of Equation 2.1 is, therefore, arbitrary. However, it constitutes a first effort at

codification. In the absence of more precise information, each effect, except that for fatigue

and fracture, is estimated as 5 percent, accumulated geometrically, a clearly subjective

approach. With time, improved quantification of ductility, redundancy, and operationalimportance, and their interaction and system synergy, shall be attained, possibly leading to a

rearrangement of Equation 2.1, in which these effects may appear on either side of the

equation or on both sides.

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Bridge Ch 5 Example on Slab Bridge

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