Example on Design of Slab Bridge - …docshare01.docshare.tips/files/19325/193256896.pdf · Only...

Chapter 5- Superstructures

Design Example on Slab Bridge Fundamentals of Bridge Structures

AAiT, Department of Civil & Environmental Engineering

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 f c’ = 28MPa [A5.4.2.1]

[A5.4.2.4]

Steel: fy = 400MPa

Es = 200GPa

Design method is Load and Resistance Factor Design (LRDF)




Reference: AASHTO LRFD Bridge Design Specifications, SI units, 2nd

Edition, 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=10.415m≤Clear span + d = 10000 + 499 =

10.499m Ok! (Cover)

2. Live Load Strip Width [Art.4.6.2.3]

a) Interior Strip

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

L1 is smaller of 10415 or 18000. W1 is the smaller of 8920 or 9000

L1 = 10415 W1 = 8920

ii) Multiple lanes loaded




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 = 44.52

Tandem: 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) Inter Strip

i) Maximum Shear Force

This governs




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, φ [Art. 5.5.4.2.1]

Strength Limit States (RC) φ

Flexure & Tension 0.90

Shear & Torsion 0.90

Axial Compression 0.75

Bearing On concrete 0.70

Compression in strut and tie model 0.70

5. Select Load Modifiers, η1

Strength service fatigue

i) Ductility η0 0.95 1.0 1.0 [Art. 1.3.3]

ii) Redundancy ηR 1.05 1.0 1.0 [Art. 1.3.4]

iii) Importance ηI 1.05 1.0 1.0 [Art. 1.3.5]




η0 = ηR = ηI = 1.0

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

Strength I U=η (1.25DC + 1.50DW + 1.75(LL+1M)+1.0FR+γTG TG

Service 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) Interior Strip:- Consider a 1m Strip, ρcon=2400 kg/m3 [Table 3.5.1-1]

WDC= (2400*9.81)* 10-3

kN/m3 * 0.54 m = 12.71kN/m

2

WDW = (2250*9.81)* 10-3

kN/m3 * 0.075m = 1.66kN/m

2

75mm bituminous wearing surface, ρbit = 2250kg/m3 [Table 3.5.1-1]

VDC = ½ * 12.71*10.415 = 66.21kN/m VDW = ½ * 1.66*10.415 = 8.64kN/m

b) Edge Strip:

VDC = ½* 16.06*10.415 = 83.63kN/m

8. Investigate Service Limit State




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

ηD = ηR = ηI = 1.0 η = 1.0

a) Moment –Interior 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 strip:

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




Z – crack width parameter (N/mm) = 23000N/mm for severe exposure

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

there to. Clear cover used to compute dc≤50mm

a) Interior strip

190 <394.6 Okay!

b) Edge Strip

140<418.98 Okay

A- Area of concrete having the same controid as the principal tensile reinforcement

and bounded by the surfaces of the cross-section and a line parallel to the neutral axis

divided by the number of bars (mm2), clear cover here also ≤ 50mm

The concrete is considered cracked if tensile stress in concrete ≥ 80% of the modulus of

rupture, [Art. 5.7.3.4&5.4.2.6]

a) Interior Strip – check concrete tensile stress against 0.8fr

Mint = 440.61kNm/m




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 ½*1000*x2 = 29630.16(499-x) x=144.87mm

Now Icr = 1/3*1000*144.873 + 29630.1(499-144.86)

2 = 4.729*10

9 mm

4/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=230.93MPa≤0.6fy

Now, fsa can be computed:

fs = 230.93≤fsa = 240Mpa OK!

b) Edge Strip

Medge = 751.32KNm/m




½*750*X2 = 7*4882.93(749-x) x = 219.655mm<250mm Curb height

Icr = 1/3*750*(219.655)3 + 7*4882.93(749-219.655)

2 = 12.227*10

9mm

4

fs/n = M(d-x)/Icr = 751.32x106*(749-219.655)/12.227x10

9) = 32.53Mpa

fs = 7*32.53Mpa=227.71Mpa

fs<fsa Ok!

iii) Deformations

Deflection and camber calculations shall consider dead load, live load, erection loads,

concrete creep and shrinkage.

[Art. 5.7.3.6.2]

Immediate (instantaneous) deflections may be computed taking the moment of inertia as

either the effective moment of inertia, Ie or the gross moment of inertia, Ig

The long-term deformation (due to creep and shrinkage) may be taken as the immediate

deflection multiplied by the following factor

3.0-1.2(A’s/AS)≥1.6 if immediate deflection is calculated using Ie.

if immediate deflection is calculated using Ig.




a) Dead Load Camber:

Total dead load of the bridge and the whole bridge cross-section is considered

WDC = 12.71*8.62+(2.53+0.59+0.23)*1.8*2=121.62KN/m

Ma – actual maximum moment (Nmm)

fr – modulus of rupture

yt – distance 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




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

(ΔLL+IM)1=1.75+3.83+0.003=5.583<<13mm Ok!

Design Lane Load +25% of design Truck Load:

W=9.3*2*1=18.6kN/m




ΔLL+IM=1.33+1.48+2.79mm<<13.02mm Ok!

TandemLoad

Single concentrated tandem load at mid-span (spaced at zero meter)

P=1.33*220*2*1 = 585.2KN

With average Ieover the entire span used instead of Ie at section of maximum moment as

done here, smaller deflection would result. The contribution of compression steel is also

neglected. For these reasons, live load deflections are made optional in AASHTO.

9. Investigate Fatigue Limit State.

U=0.75(LL+IM), IM=15%

Fatigue load shall be one design truck with 9m axle spacing.

Maximum moment results when the two from axles are on the span and the rear axle is out of

span.

a) Tensile Live Load stress:

One lane loaded E=4298.2mm

fs max = 7*5.58 = 39.05 Mpa

b) Reinforcing Bars:

The stress range in straight reinforcement bars resulting from fatigue load

combination shall not exceed.

ff=145-0.33fmin+55(r/h)




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)=161.5Mpa

fmax<ffok!

40.66<161.5 ok!

10. Investigate Strength Limit State

i) Flexure: Equivalent Rectangular stress Distribution [Art. 5.7.2.2.2]

a) Interior strip

Mu=ηΣγiQI=1.05[1.25MDC + 1.5MDW + 1.75MLL +IM+γTGMTG]

For simple span bridges, temperature gradient effect reduce gravity load effects.

Because temperature gradient may not always be there, assume γTG=0

Mu=1.05 [1.25(172.34) + 1.5(22.51) + 1.75(245.76)] = 713.23kNm/m

Mu = φAsfyd(1-0.588 ρfy/f’c)

D=540-32/2-25 = 499mm

Ρ = 0.0086 >ρmin = 0.03*f’c/fy = 0.03*28/400=0.0021

As = 0.0086*1000*499=4291.4mm2

Use

b) Edge Strip

Mu=ηΣγiQI=1.05[1.25(217.76) +0+ 1.75(533.56)+0) = 1266.22KN/m

D=540 + 250 – 32/2-25 = 749mm

ρ=0.00904>ρmin

As =ρbd = 0.0086*750*749=5082.19mm2




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) Distribution Reinforcement: The amount of bottom transverse reinforcement may be

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

a) Interior strip:

Transverse reinforcement = 0.175*4347.34mm2 = 745.6mm

2

m

b) Edge strip:

Transverse reinforcement = 0.1715 * 5063.8 mm2 = 868.44mm

2

m




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) Interior Strip:

m, transverse.

2.1 LIMIT STATES

GENERAL

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 3 – 11.

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 Details apply.

All limit states shall be considered of equal importance.

i i Qi Rn = Rf (2.1)




Where:

for loads for which a maximum value of i is appropriate:

i = D R I 0.95 (2.2)

for loads for which a minimum value of i is appropriate:

i = 1 1.0 (2.3)

D R I

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 operational

importance, 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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