slab, 36m box girder and 10m 10m T-girder all simply supported. Only the slab bridge will be designed.
Roadway Grade
=
1660.00m asl
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: concrete: Cylinder strength
Steel: f y = 400Mpa
f’c = 28Mpa 28Mpa
[A5.4.2.1] [A5.4.2.4]
Es = 200Gpa
Design method is Load and Resistance Factor Design (LRDF)
Reference: AASHTO LRFD Bridge Design Specifications, SI units, 2 nd 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
1
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=540mm, d= 540- ø/2-25 = 499mm S=10.415m≤Clear span + d = 10000 + 499 = 10.499m Ok! 2.
(cover) [Art.4.6.2.3]
Live Load Strip Width
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. W 1 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 2
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 p
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) Lane: MLn = 2.856*(1/2)*2.595*10.415
=156.2 kNm this
→governs
= 38.59kNm
MLL+Im = 1.33*156.2 + 38.59 = 246.34kNm
b)
Edge Stri p
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]
3
Strength Limit States (RC)
5.
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
Select Load Modifiers, η1
Strength
service
fatigue
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
1.0
1.0
[Art. 1.3.5]
i)
Ductility η0
1.05
η0 * ηR * ηI = 1.0 6.
7.
Select Applicable Load Combinations
[Table 3.4.1-1]
Strength I
U= η(1.25DC + 1.50DW + 1.75(LL+IM)+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)
Dead Load Force Effects
a) I nteri or Stri p :- We take 1m Strip, ρcm=2400 kg/m3
[Table 3.5.1-1]
WDC = (2400*9.81)* 10 -3KN/m3 * 0.54 m = 12.71kN/m 2 WDW = (2250*9.81)* 10 -3KN/m3 * 0.075m = 1.66kN/m 2 75mm bituminous wearing surface, ρ bit = 2250kg/m3 VDC = ½ * 12.71*10.415 = 66.21kN/m
[Table 3.5.1-1]
V DW = ½ * 1.66*10.415 = 8.64kN/m
b) Edge Str i p:
( )
4
VDC = ½ * 16.06*10.415 = 83.63kN/m 8.
Investigate Service Limit State i) Durability: Cover for main reinforcement steel for
deck surface subjected to tire wear = 60mm
[Art. 5.12]
bottom of cast in-place slab = 25 mm
ηD = ηR = ηI = 1.0η = 1.0 a) M oment – I nteri or Stri p
∑
M =1.0(172.34 + 22.51 + 246.34) = 441.2 kNm
Reinforcement:
∑ Assume j=0.875 and f s = 0.6 f y = 0.6*400 = 240
b) M oment – Edge str ip:
M =1.0(217.76 + 0 + 534.81) = 752.5 kNm
ii)
Control of Cracking
[Art.5.7.3.4]
a) I nteri or strip
5
190 <394.6 Okay!
b) Edge Stri p
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) I nteri or Stri p – check concr ete tensil e 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 = 29630.16mm 2
Equivalent concrete area
6
Determine x from ½*1000*x 2 = 29630.16(499-x)
x=144.87mm
Now Icr = 1/3*1000*144.87 3 + 29630.1(499-144.86) 2 = 4.729*10 9 mm4/m. Steel stress over n, f s/n = M(d-x)/Icr =(440.61*10 6*354.13)/(4.729*109) = 32.99MPa f s=7*32.99=230.93MPa≤0.6fy
Now, f sa can be computed:
f s = 230.93≤f sa = 240Mpa OK!
b)
Edge Stri p
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 9mm4 f s/n = M(d-x)/Icr = 751.32x10 6*(749-219.655)/12.227x109) = 32.53Mpa f s = 7*32.53Mpa=227.71Mpa 7
f s
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 inert ia as either the effective moment of inertia, Ie or the gross moment of inertia, I g 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. 4 a)
if immediate deflection is calculated using Ig.
Dead L oad 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
WDW=1.66*7.32
Ma – actual maximum moment (Nmm) f r – modulus of rupture
yt – distance from N.A to extreme tension fiber (mm)
√ √ f r = 0.63
= 0.63
= 3.33Mpa,
Location of N.A,
8
Since the section does not crack under DL, Ig should be used
Camber 5*4.53=22.65mm upward
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>M cr Multiple presence factor
() () Design Truck Load
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=1.804+4.06+0.054=5.92<<13mm Ok!
9
Design Lane Load +25% of design Truck Load: W=9.3*2*1=18.6kN/m
ΔLL+IM=1.33+1.48=2.79mm<<13.0mm Ok! Tandem Load Single concentrated tandem load at mid-span (spaced at zero meter) P=1.33*220*2*1 = 585.2KN With average Ie over 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
f s max = 7*5.58 = 39.06 Mpa b) Rein for cing Bars:
The stress range in straight reinforcement bars resulting from fatigue load combination shall not exceed. f f=145-0.33f min+55(r/h) f f-is stress range f min-minimum LL stress, where there is stress reversal=0 for our case r/h=0.3 10
f f=145-0.33(0)+55(0.3)=161.5Mpa f max
F lexur e: Equi valent Rectangular str ess Di str ibuti on
a)
I nteri or stri p
[A r t. 5.7.2.2.2]
Mu=ηΣγiQI=1.05[1.25MDC + 1.5MDW + 1.75M LL +IM+γTGMTG] For simple span bridges, temperature gradient effect reduces 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(246.34)] = 714.3kNm/m Mu = φAsf yd(1-0.588 ρfy/f’c) D=540-32/2-25 = 499mm
=
0.0086 > ρmin = 0.03*f’c/fy = 0.03*28/400=0.0021
As = 0.0086*1000*499=4291.4mm 2
Use
b) Edge Str i p
Mu=ηΣγiQI=1.05[1.25(217.76) +0+ 1.75(534.81)+0) = 1268.52KN/m D=540 + 250 – 32/2-25 = 749mm
* + ρ=0.00906 > ρmin=0.0021
As =ρ bd = 0.00906*750*749=5089.46m 2
11
ii ) Shear
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) Di stri bution Rein for cement: 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 stri p:
Transverse reinforcement = 0.1715*4347.34mm2 = 745.6mm2
m
b)
Edge str ip:
Transverse reinforcement = 0.1715 * 5061.93 mm 2 = 868.44mm2
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
12
a)
I nteri or Strip:
m, transverse.
13
