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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 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)
Chapter 5- Superstructures
Design Example on Slab Bridge Fundamentals of Bridge Structures
AAiT, Department of Civil & Environmental Engineering Page 2
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
Chapter 5- Superstructures
Design Example on Slab Bridge Fundamentals of Bridge Structures
AAiT, Department of Civil & Environmental Engineering Page 3
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
Chapter 5- Superstructures
Design Example on Slab Bridge Fundamentals of Bridge Structures
AAiT, Department of Civil & Environmental Engineering Page 4
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]
Chapter 5- Superstructures
Design Example on Slab Bridge Fundamentals of Bridge Structures
AAiT, Department of Civil & Environmental Engineering Page 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
Chapter 5- Superstructures
Design Example on Slab Bridge Fundamentals of Bridge Structures
AAiT, Department of Civil & Environmental Engineering Page 6
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
Chapter 5- Superstructures
Design Example on Slab Bridge Fundamentals of Bridge Structures
AAiT, Department of Civil & Environmental Engineering Page 7
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
Chapter 5- Superstructures
Design Example on Slab Bridge Fundamentals of Bridge Structures
AAiT, Department of Civil & Environmental Engineering Page 8
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
Chapter 5- Superstructures
Design Example on Slab Bridge Fundamentals of Bridge Structures
AAiT, Department of Civil & Environmental Engineering Page 9
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.
Chapter 5- Superstructures
Design Example on Slab Bridge Fundamentals of Bridge Structures
AAiT, Department of Civil & Environmental Engineering Page 10
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
Chapter 5- Superstructures
Design Example on Slab Bridge Fundamentals of Bridge Structures
AAiT, Department of Civil & Environmental Engineering Page 11
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
Chapter 5- Superstructures
Design Example on Slab Bridge Fundamentals of Bridge Structures
AAiT, Department of Civil & Environmental Engineering Page 12
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)
Chapter 5- Superstructures
Design Example on Slab Bridge Fundamentals of Bridge Structures
AAiT, Department of Civil & Environmental Engineering Page 13
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
Chapter 5- Superstructures
Design Example on Slab Bridge Fundamentals of Bridge Structures
AAiT, Department of Civil & Environmental Engineering Page 14
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
Chapter 5- Superstructures
Design Example on Slab Bridge Fundamentals of Bridge Structures
AAiT, Department of Civil & Environmental Engineering Page 15
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)
Chapter 5- Superstructures
Design Example on Slab Bridge Fundamentals of Bridge Structures
AAiT, Department of Civil & Environmental Engineering Page 16
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.