Chap-5-T-Girder Example-1.pdf

Chapter 5- 18 m span Girder Bridge Fundamentals of Bridge Structures
AAiT, Department of Civil & Environmental Engineering Page 1
Chapter 5: T-Girder Bridge Example
SIMPLE SPAN R.C. T- GIRDER BRIDGE
PART- I : SUPERSTRUCTURE DESIGN
Consultant: XXXXX
Project: YYYYY
Name of Bridge: RRRRR
Prepared by: Designer
Checked by: Client
Case: Statical calculation
Date: Date-Month-Year
1. DESIGN DATA AND SPECIFICATIONS
1.1 Subject Information:
Superstructure type:- Simple span R.C.T Girder
Clear span : 18 m
Clear Roadway 7.3 m
concrete ( section
9.3)
Curb 0.85 m
1.2 Material properties:
Concrete :- Grade C - 30 concrete ( section 9.3)
fc'= = 24 MPa (fc' cylinder )
fc=0.4*fc' = 10 MPa (For serviceability)
Ec=.043γC
1.5
sqrt(fc') = 24,768 MPa

=
1.3 Reinforcement steel:
Grade 400 steel: For rebars diam. above 16mm
fy = 400 MPa
fs = 200 MPa
Es = 200,000 MPa S 5.4.3.2
Grade 300 steel: For rebars less than diam. 16
fy = 300 MPa
fs = 150 MPa
Es = 200,000 MPa S 5.4.3.2
Modular ratio n = Ec / Es = 8.07 Use n = 8
Live Loading: (1) Design Truck : HL-93 live load
(2) Design Tandem
Ptruck = 72.5 kN
Lane load = 9.3 kN/m
Ptandem = 55 kN

P/4 P P
(4.3m - 9.0m)
(4.3m)
P=wheel load =72.5KN
LONGITUDNAL ARRANGEMENT
P P
(1.8m)
TRANSVERSE ARRANGEMENT
P P
(1.8m)
P P
(1.2m)
P=wheel load =1/2*110KN=55KN
DESIGN TANDEM
Design method: Load and Resistance Factor Design (LRFD) Method
Reference: ERA's Bridge Design Manual 2002
AASHTO Standard Specifications for Highway Bridges, 1998, and latest version
2. BRIDGE CROSS SECTION

3.1 Given Data
Clear Road way width, W1= 7.30 m
Clear Road way width, W2= 0 m
Sidewalk width, SW= 0.85 m
Total bridge width, WT= 9.00 m
No. of lane= 2
Multiple presence factor for double lane
loading= 1.00
Multiple presence factor for single lane
loading= 1.20
3.2 Preliminary Dimensions
No. of girders= 4.00
width of support= 0.50 m
expansion length left at both sides= 0.05 m
c/c of support= 18.50 m
Clear span plus depth of member = 19.40 m
Design span length, S= 18.50 m
Top slab thickness, T = 0.20 m
Recommended min. girder depth=0.07*S= 1.30 m
Use Girder depth D= 1.40 m
c/c girder spacing, a = 2.20 m
End of slab to center of exterior girder, c = 1.20 m
Minimum Web width recommended, bw = 0.20 M

Use web width, bw = 0.40 m
Clear span b/n girders = 1.80 m
Wall slenderness ratio = 3.50 m
Top slab thickness = 0.20 m
Overhang slab thickness near ext.
girder= 0.30 m
Overhang slab length= 1.000 m
fillet= 0.10 m
Exterior diaphragm depth= 1.20 m
Width of diaphragm= 0.25 m
Number of exterior diaphragm= 2.00
Number of interior diaphragm= 2.00
Total number of diaphrams= 4
c/c spacing of diaphragms= 6.22 m
Assumed wearing surface thickness= 0.10 m
Crown slope in the transverse
direction= 2.50 %
depth of post= 0.30 m
width of post= 0.25 m
Average spacing between posts= 1.50 m
Height of post 0.85 m
depth of rail= 0.40 m

Width of rail= 0.25 m
Face of rail from end of curb= 0.10 m
Edge of post from exterior edge of
curb= 0.05 m
curb Top width= 0.85 m
curb Bottom width= 1.00 m
curb Top Height = 0.30 m
curb Bottom Height= 0.25 m
unit weight of rein. concrete= 25.00 kN/m3
unit weight of Asphalt= 22.50 kN/m3
ф for moment= 0.90
ф for shear= 0.90

4. DESIGN OF OVER HANG SLAB
4.1 Loads
4.1.1DeadLoad&Moments
Dead loads
(kN/m)
Moment arm about face
of exterior girder
(m)
MDL
(kNm/m)
W1= 2.50 0.775 1.938
W2= 1.06 0.825 0.877
W3= 6.38 0.575 3.666
W4= 3.13 0.333 1.042
Sum WDL = 13.06 MDL = 7.521

4.1.2 Live Loads
i ) Design Truck Load & Moments
The application of live load for the design of deck overhang is done according to AASHTO 1998, Art. 3.6.1.3
In designing sidewalks, slabs and supporting members, wheel load located on the sidewalk shall be
300mm from the face of the rail 0.3 m
Distribution width for wheel loads (Art. 3.24.5.1.1) (AASHTO 4.6.2.1.3-Table 1)
x = 0.350 m
kN Assuming the worst case
where the vehicle climb on the curb P = 72.5

E=0.833x+1140, where x=the distance in mm from load to point of support
E = 1431.55 mm 1.43 m
Live load moment per linear meter width of slab
MLL= (P/E)*x= 17.73 kNm/m
Dynamic Allowance Factor
IM = 33%
Live Load plus impact:
MLL+IM = 23.57 kNm/m
Railing Load
E=0.833x+1140, where x=the distance in mm from load to point of support
E = 1681.45 mm 1.681 m x =0.650
Live load moment per linear meter width of slab Ph=44.51
MRL= (Ph/E)*h= 34.41 kNm/m h=1.3
ii ) Design
Tandem
The design tandem wheel load, P = 55kN, is smaller than the design truck load, P=72.5kN, and hence not considered for
for overhang design
iii)Lane Load
Longutidinal lane load distrbution=9.3kN/M 9.30 kN/M
Transversal lane load distrbution=(9.3/3)kN/M= 3.10 kN/M

Mlane load = 1.55 kNm/m
iv ) Pedestrian Live Load
Pedestrian live load = 4.00 kN/m2 (AASHTO 2004 Art.3.6.1
Moment due to pedestrian load = 2.00 kNm/m
Total Design Moment
MTOT = 1.25*MDL + 1.75*MLL+IM = 75.84 kNm/m
4.2 Reinforcement
4.2.1 Reinforcement for Mtot
As= Mu / ( Ø fy (d - a/2 ) )
a = As*fy / ( 0.85 * fc' b ) Mu = 75.84 kN-m/m
Ø = 0.90
b = 1000 mm
fy = 300 N/mm2
fc' = 24 N/mm2
D = 300 mm
diam = 16 mm
cover = 50 mm
d = 242 mm
Goal Seek 0.00
Assume a = 17.72 mm
As = Mu / ( Ø fy (d - a/2 ) ) = 1,205 mm2/m

a = As*fy / ( 0.85 * fc' b ) = 17.72 mm
Required As = 1205 mm2/m
Required
spacing = 160 mm
Minimum Reinforcement (Art. 8.17.1)
Fr = 0.63 * SQRT(fc') = 3.09 N/mm2
Icr = bh3
/12 = 2.250E+09 mm4
Mcr = fr * Icr / (yt) = 46.30 kNm/m Yt = d/2
1.2 * Mcr = 55.55 kNm/m
Mdesign = 75.84 kNm/m OK!!
Mu = 55.55 KNm/m
Ø = 0.90
b = 1000 mm
fy = 300 N/mm2
fc' = 24 N/mm2
D = 300 mm
diam = 16 mm
cover = 50 mm
d = 242 mm

Assume a = 12.84 mm Goal Seek 0.00
As = Mu / ( Ø fy (d - a/2 ) ) 873 mm2
a = As*fy / ( 0.85 * fc' b ) = 12.84 mm
OK !!
Required As = 873 mm2/m
Required spacing for minimum steel area = 230 mm
S Provided= 160 mm
Use diameter 16
mm bars
c/c 160 mm
As provided
= 1,257 mm2
5. DESIGN OF DECK SLAB
5.1 Interior span slab
5.1.1 Loadings
a) Dead Loads
Dead loads computation
Slab = 5.00 kN/m2
Asphalt 10cm thickness
= 2.25 kN/m2
wDL = 7.25 kN/m2
MDL= (WDL*S2
*0.80)/8 where 0.80 is a continuity factor

Span length S = Clear span
= 1.80 m
MDL = 1/8( wDL*S2
*0.80) = 2.35 kNm/m
b) Live Loads
Live Load moment for continuous slab ( Art. 3.24.3.1)
MLL = 1/32(S+2)P20*0.80 where S= span length in feet (Art. 3.24.1.2), S= 1.80 m
MLL = = 3221.39 lb-ft/ft = P= 72.50 kN
MLL = = 14.33 kNm/m
Impact Factor (Art. 3.8.2.1)
IM = 33%
Live load plus impact
MLL+IM = = 19.06 kNm/m
Lane load P=9.3/3 = 3.10 kN/m
Mlane = 1.00 kNm/m
c) Factored Design moment
Total Design moment
MTOTAL = 1.25*MDL + 1.75*(MLL+IM +Mlane) = 38.04 kNm/m
5.1.2Reinforcement
Goal Seek -0.01 Mu =38.04
Assume a = 15.44 mm Ø =0.90

As = Mu / ( Ø fy (d - a/2 ) ) 1,049 mm2 b =1000
a = As*fy / ( 0.85 * fc' b ) = 15.43 mm fy =300
OK !! fc' =24
D =200
diam =16
Required As = 1049 mm2/m cover =50
Spacing, s = 192 mm d =242
Use diameter 16 mm bars c/c 190 mm ( top and bottom reinf.-transverse)
As provided = 1,058 mm2; S1,S2
For main reinforcement perpendicular to traffic, the distribution reinf. is given as percentage of the main slab reinf. as given
below:
As(distr.) % = 3840/sqrt(S) <= 67%
Therefore, % As dist. = 91 % ( S is in feet) where; S= 1.8 m
% As (distr.) = 67% 721.7847769
As dist. = 0.67 * As provided = 709 mm2/m
spacing = 217.12 mm diam. of bar= 14
Use diam. 14 mm bars c/c 210 mm (bottom reinf. - Longtiudinal)
As provided = 733 mm2 S3
For components less than 1200mm thick, the area of reinforcement in each direction shall not be less than:
As (temp. & shrinkage) > 0.11*(Ag/fy) = 73.33 mm2/m

As (temp and shrink.)= 73.33 mm2/m
Using diam. 12 bar
spacing = 1542 mm diam. of bar =
Use diam. 12 c/c 300 mm (top. reinf. ) S4
6. DESIGN OF LONGITUDINAL GIRDERS
6.1. Loads
6.1.1 DeadLoads
a)ExteriorGirder

C= 1.20 m, cant. Span to center of support
a= 2.20 m
bw= 0.40 m
Uniform dead loads per linear meter span (kN/m)
W1= 2.50
W2= 1.06
W3 (EXTERIROR
CURB)= 6.38
W4 3.13
W5=girder 14.00
W6=top slab 4.50
W8=Wearing surface 3.26
W9=Fillet 0.25
sum 35.08
Weight of Diaphragms:
Conc. Load P1 = 5.63 kN

L
WDL
P1 ( KN)
P1
WDL= 35.075 kN/m
P1= 5.63 kN
P2= 11.25 kN
L= 18.50 m
Shear Forces and Bending Moments due to Dead Loads on Exterior Girder
VDL(x)= P1/2+W(L/2-X)
MDL(x)= P2X/2+WX/2(L-X)
x
(m)
VDL
(kN)
MDL
(kNm)
0.00 330.07 0.00

0.925 297.62 295.51
1.850 265.18 561.01
2.775 232.74 796.50
3.700 200.29 1001.98
4.625 167.85 1177.45
5.550 135.40 1322.90
6.475 102.96 1438.35
7.400 70.51 1523.78
8.325 38.07 1579.20
9.250 5.63 1604.61
b) Interior Girder

Uniform dead loads per linear meter span
(kN/m)
Top slab = 11.00 tts=thickness of
girder = bw*(D-tts)*25 = 12.00 top slab
Wearing surface = 4.95
sum WDL = 27.95 kN/m
Diaphragms:
L
WDL
P1 ( KN)
P1
WDL= 27.95 kN/m
P1= 11.25 kN
P2= 11.25 kN
L= 18.50 m

Shear Forces and Bending Moments due to Dead Loads on
Interior Girder
VDL(x)= P1/2+W(L/2-X)
MDL(x)= P2X/2+WX/2(L-X)
x
(m)
VDL
(kN)
MDL
(kNm)
0.00 330.07 0.00
0.925 297.62 295.51
1.850 265.18 561.01
2.775 232.74 796.50
3.700 200.29 1001.98
4.625 167.85 1177.45
5.550 135.40 1322.90
6.475 102.96 1438.35
7.400 70.51 1523.78
8.325 38.07 1579.20
9.250 5.63 1604.61

6.1.2 Live Loads
P/4 P P
(4.3m - 9.0m)
(4.3m)
P=wheel load =72.5KN
P P
(1.8m)
P P
(1.8m)
P P
(1.2m)
P=wheel load =1/2*110KN=55KN
a) Design Truck Load : HL-93
b) Design Tandem
The design lane load shall consist of a load of 9.3kN/m, uniformly distributed in the longitudinal direction.
Transversely, the design lane load shall be assumed to be uniformly distributed over 3m width.
Lane load for exterior girder 4.495 kN/m
Load width=1.45
Ra=Rb= 41.58 kN
Lane load for interior girder 6.82 kN/m Load width=2.2
Ra=Rb= 63.09 kN

6.1.2.1 Dynamic Load Allowance
Section 3.13, the vehicular dynamic load allowance IM
IM = 33%
The live loads shall be factored by 1+IM/100 = 1.33
6.1.2.2 Transverse Load Distribution
In designing sidewalks, slabs and supporting members, a wheel load should be located one foot from the face of the barrier
For the design of components other than the deck overhang, the design truck or tandem shall be positioned transversely such tha
load
is not closer than 600mm from the edge of the design lane. (Art 3.6.1.3.1 - AASHTO 2004 )
Distribution Factor for Shear (Sec. 13.4: Table 13-7 & 13-8)
Exterior Girder:
Case-1: One Design lane loaded
The lever rule is applied assuming that the slab is simply supported over the longitudinal beams (Table 13-8)

The distribution coefficient to the exterior girder for shear
REX1 (shear)= if (C<(0.8+0.6), P/a*[(a -
d1)+(d2)],P/a*[(1.8-(C-(0.8+0.6))-(C-(0.8+0.6))]) = 0.95 P

Case-2: Two or more design lanes loaded
The distribution of live load per lane for shear in exterior girder is determined according to the formulas given in Table 13-8.
REX2 (shear) = (0.6+de/3000)*Rin shear = 0.692 per lane de= 0.900
This factor is for one wheel load, for axle load which is equivalent to two lines of wheels should be multiplied by 2 Ok
REX2 (shear) = (0.64+de/3800)*Rin shear = 1.384 P Ok
Therefore, REX (shear) in exterior girder is maximum of the above two values, REX1 or REX2
REX (shear) = 1.384 P
Interior Girder:
The distribution of live load per lane for shear in interior girder is determined according to the formulas given in Table 13-7.
RINT 1 (shear) = (0.36+S/7600) = 0.649 per lane where 1100<=S<=4900
6000<=L<=73000
890<=d<=2800
For two lines of wheels (axle load) Nb>=4 ok
RINT 1 (shear) = 1.299 P S=
de= distance from outside face of exterior girder to interior edge of curb or traffic barrier. It is positive if the exterior
web is inboard of the curb and negative if it is outboard de=C-bw/2-0.4

L=SPAN LENGTH
d=over all depth of a girder
S is the girder spacing
Nc is the number of cells
Case-2: Two or more design lanes are loaded
The distribution of live load per lane for shear in interior girder
RINT 1 (shear) = 0.2+S/3600-[S/10700]^2 =
For two lines of wheels (axle load)
RINT 2 (shear) =
Therefore, RINT (shear), in interior girder is maximum of the above two values, RINT 1 or RINT 2
RINT (shear) =
Distribution Factor for moment (Sec.13.4: Table 13-3 and 13-4)
Exterior Girder
The distribution of live load per lane for moment in exterior girder (Use Lever
Rule)
(Table 4.6.2.2.2d-1 Distribution of Live
Loads Per Lane for Moment in Exterior
Longitudinal Beams.)
REXT1 (moment)= if (C<(0.8+0.6), P/a*[(a-d1)+(d2
)],P/a*[(1.8-(C-(0.8+0.6))-(C-(0.8+0.6))]) = 1.045 P S=2.2
Case-2: Two or more design lanes loaded de= 0.9

REXT2 (moment) = (0.77+de/2800) = 1.091 P
The distribution factor for wheel load (i.e. two lines of wheels) is 2 times the maximum of the above two values
Rext (moment)= 2.183 P
Interior Girder
The distribution factor of live load per lane for moment in interior girder:
RINT1(moment) =0.06+(S/4300)^0.4*(S/L)^0.3*(Kg/Lts^3)^0.1
Case-2: Two or more design lanes loaded
RINT2(moment) =0.075+(S/4300)^0.64*(S/L)^0.2*(Kg/Lts^3)^0.1 0.512
The distribution factor for wheel load (i.e. two lines of wheels) is 2 times the maximum of the above two values
RINT (moment) = 1.024 P
6.1.2.3 Shear Forces and Bending Moments due to Live Loads
a) Influence Lines for Shear Forces and Bending Moments
a-1) Design Truck: HL-93

Influence Lines for Shear Force at "x" distance from end support
Influence Lines for Bending Moment at "x" distance from end support

a-2) Design Tandem
Influence Lines for Shear Force at 'x' distance from end support

Influence Lines for Bending Moments at 'X' distance from end support
b) Shear forces due to Live load plus impact
b-1) Exterior Girder
Design Truck: VLL + IM = (1.00+IM )*(Dist. Factor)*(Ptruck)* (A+B+C/4) IM = 33%
Design Tandem: VLL + IM = (1.00+IM )*(Dist. Factor)*(Ptandem)*(A+B+C) L = 18.50
Dist. Factor = 1.38
P truck= 72.50
P tandem= 55.00

x
(m)
Live Load 1:
Design Truck
Coefficients VLL+IM
A B C D E F
0 1.000 0.768 0.535 - - -
253.72
0.925 0.950 0.718 0.485 - - - 238.71
1.85 0.900 0.668 0.435 - - - 223.70
2.775 0.850 0.618 0.385 - - - 208.69
3.7 0.800 0.568 0.335 - - - 193.67
4.625 0.750 0.518 0.285 - - - 178.66
5.55 0.700 0.468 0.235 - - - 163.65
6.475 0.650 0.418 0.185 - - - 148.64
7.4 0.600 0.368 0.135 - - - 133.62
8.325 0.550 0.318 0.085 - - - 118.61
9.25 0.500 0.268 0.035 - - - 103.60

Live Load 2:
Design Tandem
VLL+I
(MAX)
Coefficients VLL+IM
A B C D (kN)
1.000 0.935 0.124 0.059 214.50 253.72
0.950 0.885 0.074 0.009 194.26 238.71
0.900 0.835 0.024 - 178.11 223.70
0.850 0.785 - - 165.53 208.69
0.800 0.735 - - 155.41 193.67
0.750 0.685 - - 145.28 178.66
0.700 0.635 - - 135.16 163.65
0.650 0.585 - - 125.04 148.64
0.600 0.535 - - 114.91 133.62
0.550 0.485 - - 104.79 118.61
0.500 0.435 - - 94.67 103.60

b-2) Interior Girder
Design Truck: VLL + IM = (1.00+IM )*(Dist. Factor)*(Ptruck)* (A+B+C/4) IM = 33%
Design Tandem: VLL + IM = (1.00+IM )*(Dist. Factor)*(Ptandem)*(A+B) L = 18.50m
Dist. Factor = 1.538
P truck= 72.50kN
P tandem= 55.00 kN
x
(m)
Live Load 1:
Design Truck
Coefficients VLL+IM
A B C D E F (kN)
0 1.000 0.768 0.535 - - - 281.91
0.925 0.950 0.718 0.485 - - - 265.23
1.85 0.900 0.668 0.435 - - - 248.55
2.775 0.850 0.618 0.385 - - - 231.87
3.7 0.800 0.568 0.335 - - - 215.19
4.625 0.750 0.518 0.285 - - - 198.51
5.55 0.700 0.468 0.235 - - - 181.83
6.475 0.650 0.418 0.185 - - - 165.15
7.4 0.600 0.368 0.135 - - - 148.47
8.325 0.550 0.318 0.085 - - - 131.79
9.25 0.500 0.268 0.035 - - - 115.11

Live Load 2:
Design Tandem
VLL+I
(MAX)
Coefficients VLL+IM
A B C D (kN) (kN)
1.000 0.935 0.124 0.059 238.34 281.91
0.950 0.885 0.074 0.009 215.84 265.23
0.900 0.835 0.024 - 197.91 248.55
0.850 0.785 - - 183.92 231.87
0.800 0.735 - - 172.67 215.19
0.750 0.685 - - 161.43 198.51
0.700 0.635 - - 150.18 181.83
0.650 0.585 - - 138.93 165.15
0.600 0.535 - - 127.68 148.47
0.550 0.485 - - 116.43 131.79
0.500 0.435 - - 105.18 115.11

c) Bending Moments due to Live load plus impact
c-1) Exterior Girder
Design Truck: VLL + IM = (1.00+IM )*(Dist. Factor)*(Ptruck)* Max(Sum Coeff's)
IM = 33%
Design Tandem: VLL + IM = (1.00+IM )*(Dist. Factor)*(Ptandem)*(A+B) L = 18.5 m
Dist. Factor =2.18
P truck=72.5 kN
P tandem=55.0 kN
Design Truck Load Case 1
x Coef. ( Loading 1 ) sum Coef.
(m) A B C/4 D E F/4 A+B+C/4+D+E+F/4
0 - - - - - - -
0.925 0.88 0.66 0.11 - - - 1.65
1.85 1.67 1.24 0.20 - - - 3.10
2.775 2.36 1.71 0.27 - - - 4.34
3.7 2.96 2.10 0.31 - - - 5.37
4.625 3.47 2.39 0.33 - - - 6.19
5.55 3.89 2.60 0.33 - - - 6.81
6.475 4.21 2.70 0.30 - - - 7.21
7.4 4.44 2.72 0.25 - - - 7.41
8.325 4.58 2.64 0.18 - - - 7.40

9.25 4.63 2.48 0.08 - - - 7.18
Coeff. ( Loading 2 ) sum Coef. Maximum MLL+IM
A B C/4 D E F/4 A+B+C/4+D+E+F/4 (kNm/m)
- - - - - - - -
- 0.88 0.166 - - - 1.045 348.28
- 1.67 0.309 - - - 1.974 652.76
- 2.36 0.428 - - - 2.787 913.43
- 2.96 0.525 - - - 3.485 1130.29
0.24 3.47 0.598 - - - 4.311 1303.34
0.88 3.89 0.649 - - - 5.409 1432.59
1.41 4.21 0.676 - - - 6.298 1518.04
1.86 4.44 0.680 - - - 6.980 1559.67
2.21 4.58 0.661 - - - 7.453 1568.81
2.48 4.63 0.619 - - - 7.719 1624.66

When the live load is Design Tandem
Design Tandem Load
x Coefficients sum Coef. MLL+IM
(m) A B C D ( A + B + C ) (kNm/m)
0.000 - - - - - -
0.925 0.88 0.82 0.07 0.01 1.775 283.42
1.850 1.67 1.55 0.04 - 3.255 519.75
2.775 2.36 2.18 - - 4.538 724.53
3.700 2.96 2.72 - - 5.680 906.96
4.625 3.47 3.17 - - 6.638 1059.85
5.550 3.89 3.53 - - 7.410 1183.20
6.475 4.21 3.79 - - 7.998 1277.01
7.400 4.44 3.96 - - 8.400 1341.28
8.325 4.58 4.04 - - 8.618 1376.01
9.250 4.63 4.03 - - 8.650 1381.20
c-2) Interior Girder
Design Truck: VLL + IM = (1.00+IM )*(Dist. Factor)*(Ptruck)* Max(Sum Coeff's) IM = 33%

Design Tandem: VLL + IM = (1.00+IM )*(Dist. Factor)*(Ptandem)*(A+B) L = 18.50
Dist. Factor =1.024
P truck=72.50
P tandem=55.00
When the live load is Design Truck
x Coef. ( Loading 1 ) sum Coef.
(m) A B C/4 D E F/4 A+B+C/4+D+E+F/4
0 - - - - - - -
0.925 0.88 0.66 0.11 - - - 1.65
1.85 1.67 1.24 0.20 - - - 3.10
2.775 2.36 1.71 0.27 - - - 4.34
3.7 2.96 2.10 0.31 - - - 5.37
4.625 3.47 2.39 0.33 - - - 6.19
5.55 3.89 2.60 0.33 - - - 6.81
6.475 4.21 2.70 0.30 - - - 7.21
7.4 4.44 2.72 0.25 - - - 7.41
8.325 4.58 2.64 0.18 - - - 7.40
9.25 4.63 2.48 0.08 - - - 7.18

Coef. ( Loading 2 ) Sum Coef. Maximum,MLL+IM
A B C/4 D E F/4 A+B+C/4+D+E+F/4 (kNm/m)
- - - - - - - -
- 0.88 0.166 - - - 1.045 163.364
- 1.67 0.309 - - - 1.974 306.180
- 2.36 0.428 - - - 2.787 428.448
- 2.96 0.525 - - - 3.485 530.168
0.24 3.47 0.598 - - - 4.311 611.341
0.88 3.89 0.649 - - - 5.409 671.966
1.41 4.21 0.676 - - - 6.298 712.043
1.86 4.44 0.680 - - - 6.980 731.573
2.21 4.58 0.661 - - - 7.453 735.862
2.48 4.63 0.619 - - - 7.719 762.055

When the live load is Design Tandem
Design Tandem Load
Maximum, MLL+IM
x Coefficients Sum Coef.
(m) A B C D ( A + B + C+D )
0 - - - - - -
0.925 0.88 0.82 0.07 0.01 1.78 132.94
1.85 1.67 1.55 0.04 - 3.26 243.79
2.775 2.36 2.18 - - 4.54 339.84
3.7 2.96 2.72 - - 5.68 425.41
4.625 3.47 3.17 - - 6.64 497.13
5.55 3.89 3.53 - - 7.41 554.99
6.475 4.21 3.79 - - 8.00 598.99
7.4 4.44 3.96 - - 8.40 629.13
8.325 4.58 4.04 - - 8.62 645.42
9.25 4.63 4.03 - - 8.65 647.86
6.1.3 Seismic Force Effects

Earthquake zones: EBCS Zone -4
Site Coefficient: Type I = 1.5
Acceleration coefficient(A): =
The horizontal seismic force is the product of the site coefficient, the acceleration coefficient and the permanent load
Computation of permanent loads:
Railing and posts = 131.81 kN
Sidewalks = 869.50 kN
Top Slab = 638.25 kN
Girders= 777.00 kN
Diaphrams= 135.00 kN
Asphalt = 303.86 kN
Sum Wp = 2855.43 kN
Horizontal earhquake force FH = site coeff.*A* Wp = 428.31
This force is transferred to the substructure through the bearings located at ends of the bridge
Earthquake force transferred to one support = 1/2*FH = 214.16
This force is applied horizontally to the bearings
Total Dead Load reaction at one support = 1427.71
Assume coef. of friction between bearing and concrete seat at abut, µ= 0.50
The frictional resistance force developed between the interface = µ*Dead load Rxn
= 713.86
So Earthquake effects are negligible!!

6.1.4 FACTORED LOADS
Load Factors and Load Combinations
The load factors and load combinations are according to ERA's Bridge Design Manual 2002, section 3.3
The load combination to be used for design is Strenght - I Limit state, Table 3-1.
Factored Load = gb *DL + 1.75* ( LL + IM ) = 1.25*DL+1.75*(LL+IM) gb =1.25
Factored Shear Forces

x
Exterior Girder Interior Girder
Design
S.F.
VDL Vlane VLL+I
1.25VDL+
1.75VLL+IM+
1.75Vlane VDL Vlane VLL+I
1.25VDL+
1.75VLL+IM+
1.75Vlane Vmax
(m) (kN) (kN) (kN) (kN) (kN) (kN) (kN) (kN) (kN)
0.00 330.07 41.58 253.72 929.36 330.07 63.09 281.9 1016.33 1016.33
0.925 297.62 37.42 238.71 855.26 297.62 56.78 265.2 935.55 935.55
1.85 265.18 33.26 223.70 781.16 265.18 50.47 248.6 854.76 854.76
2.775 232.74 29.11 208.69 707.05 232.74 44.16 231.9 773.98 773.98
3.7 200.29 24.95 193.67 632.95 200.29 37.85 215.2 693.19 693.19
4.63 167.85 20.79 178.66 558.85 167.85 31.54 198.5 612.40 612.40
5.55 135.40 16.63 163.65 484.74 135.40 25.23 181.8 531.62 531.62
6.475 102.96 12.47 148.64 410.64 102.96 18.93 165.2 450.83 450.83
7.4 70.51 8.32 133.62 336.54 70.51 12.62 148.5 370.05 370.05
8.325 38.07 4.16 118.61 262.43 38.07 6.31 131.8 289.26 289.26
9.25 5.63 0.00 103.60 188.33 5.63 0.00 115.1 208.47 208.47
Factored Bending Moments

Exterior Girder Interior Girder
Design
B.M.
x MDL Mlane MLL+IM
1.25MDL+
1.75MLL+IM+
1.75Mlane MDL Mlane MLL+IM
1.25MDL+
1.75MLL+IM+
1.75Mlane Mmax
(m) (kNm) (kNm) (kNm) (kNm) (kNm) (kNm) (kNm) (kNm) (kNm)
0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.0 0.00 0.00
0.925 295.51 36.54 348.28 1042.82 295.51 55.44 163.4 752.29 1042.82
1.85 561.01 69.23 652.76 1964.74 561.01 105.04 306.2 1420.89 1964.74
2.775 796.50 98.07 913.43 2765.75 796.50 148.80 428.4 2005.81 2765.75
3.7 1001.98 123.07 1130.29 3445.86 1001.98 186.73 530.2 2507.05 3445.86
4.625 1177.45 144.23 1303.34 4005.05 1177.45 218.83 611.3 2924.60 4005.05
5.55 1322.90 161.53 1432.59 4443.35 1322.90 245.09 672.0 3258.47 4443.35
6.475 1438.35 174.99 1518.04 4760.74 1438.35 265.51 712.0 3508.65 4760.74
7.4 1523.78 184.61 1559.67 4957.22 1523.78 280.10 731.6 3675.15 4957.22
8.325 1579.20 190.38 1568.81 5052.59 1579.20 288.85 735.9 3767.25 5052.59
9.25 1604.61 192.30 1381.20 4759.39 1604.61 291.77 762.1 3849.96 4759.39
6.2. DESIGN FOR FLEXURE
a) Design Loads
The factored loads in the above tables are the design loads and the structure shall be
designed to carry the expected design loads as calculated above

b) Effective Compression Flange Width, beff, ( AASHTO Art. 4.6.2.6.1)
Exterior Girder
For exterior beams, the effective flange width may be taken as one-half the effective width of the
adjacent interior beam, plus the least of:
a. 1/8 effective span length = 2.31m L = 18.5 m
b.
6 times the average thickness of the slab, plus the
greater of half the web thickness or one-quarter of
the width of the top flange of the basic girder = 1.75m hf=200 mm
c. The width of the overhang = 1.00m Bw=400 mm
beff < 1/2*Effective width of interior girder + minimum value of
= 2.10m
Overhand =1.0 m
Clear girder spac. =
1.8 m

Interior Girder
The effective compression flange width, beff, is the minimum of the following:
a. 1/4 span = 4.63 m
b.
12 times the average thickness of the slab, plus the greater of web thickness or one-half
of the width of the top flange of the girder = 2.80
c. The average spacing of the adjacent beam = 2.20
Effective compression flange width, beff = 2.20
Use compression flange width, beff = 2.10
c) Spacing Limits for Reinforcements ( AASHTO 2004 - Art. 5.10.3.1.1)
Horizontal
In cast-in-place concrete, the clear distance between parallel bars in a layer shall not be less than:
1.5 times bar diameter = 48 mm Assume d=32
1.5 times the
max. size of
coarse
aggregate
(i.e.25.4mm)
= = 38 mm
38mm
= 38 mm
Hence the minimum clear distance between parallel bars in
a layer with no lapping is = 48 mm =

Assuming vertical lapping of bars, one over the other, The minimum clear distance between parallel bars in a layer
is = 48 mm
Therefore, the minimum center to center spacing of bars in a layer is = 80 mm
Use center to center spacing of bars in a layer = 80 mm
Therefore, the minimum center to center spacing of bars in a layer with overlap is = 112 mm
Use center to center spacing of bars in a layer = 120 mm
Vertical
Minimum clear distance between two layers of bars is 1 inch or dia of bar = 32 mm
Thus, minimum center to center spacing of bars when there is lap = 96 mm
Use vertical center to center spacing of bars b/n layers = 100 mm
d) Section Capacity and Reinforcement
The maximum design moment is:
Mmax = 5,052.59 kNm Cover(mm)= 40
Try No of bars ф of bars ai di(m) ai*di
G0 1st row = 4 32 3217 0.072 232
G1 2nd row = 4 32 3217 0.080 257
G2 3rd row = 4 32 3217 0.088 283
G3 4th row = 4 32 3217 0.096 309
16 12868 1081
D = 1.400 m

d' = sum ai*di = 0.084 m
sum ai
d = D - d' = 1.316 m
b = 2.100 m
Assume rectangular beam analysis
As= 12,868 mm2
b = 2.100 m
d = 1.316 m
fy = 400.0 MPa
fc' = 24.00 Mpa
bw = 0.400 m
hf = 0.20 m
Ø = 0.90 m
r= As/bd = 0.00466
a = As*fy/(0.85*fc'*b) 120.15 mm
It is rectangular beam as assumed
ØMn =Ø*As*fy*(d-a/2) 5,818.03 kNm
The following computations for T-beam analysis do not apply
Asf = 0.85*fc'*(b-bw)*hf/fy = - mm2
M1=Asf*fy*(d-hf/2) = - kNm

a = (As - Asf)*fy/(0.85*fc'*bw) = - mm
M2 =(As - Asf)*fy*(d-a/2) = - kNm
Mu = M1 + M2 = - kNm
ØMu = Ø ( M1 + M2 ) = - kNm
Therefore, the
section capacity is
ØMn = 5818.03 kN-m >> Mmax = 5,05,59 kN - m
e) Checking maximum steel area
rf = Asf/b'wd = 0.00211 b' bot-flange reinf.dist- width = 2.10 m
rmax = 0.75*rb = 0.75*0.85*b1*(fc'/fy)*(600/(600+fy)+rf) b1=0.85
= 0.0195 fc' = 24 MPa
rprovided = (As-Asf)/bwd = 0.0047 << rmax fy =400 MPa
OK !!
f) Check for minimum reinforcement (ERA BDM Section 9.4.5)
i. The minimum reinforcement criteria: - AASHTO 2004, Art. 5.7.3.3.2
ØMn > = minimum of 1.2Mcr. or 1.33 Mu

ØMn > = 1.2Mcr.
btop= 2.10 m
hf= 0.20 m
bw= 0.40 m
D= 1.40 m
hbot= 0.15 m
bbot= 2.10 m
length accommodating the reinforcement is
taken
Mcr = fr * Ig / yt
fr = 0.63 * SQRT( fc' ) = 3.086 N/mm2
Centroid of cross section
y1=Sum(Ai*yi)/(sumAi) =
Sum Ai*yi = 0.853 m3
Sum Ai = 1.155 m2
Yt=Sum(Ai*yi)/(sumAi) = 0.739 m
Gross moment of inertia, Ig, about centroid
I1=1/12*btop*hf
3
+ A1*d1
2
= 0.1338 m4
I2=1/12*bw*(D - hf -hbot)3
+ A2*d2
2
= 0.0403 m4
I3=1/12*bbot*(hbot )3
+ A3*d3
2
= 0.1393 m4
Sum Ig = 0.3134 m4

The crackimg moment, Mcr,
Mcr = fr * Ig / yt = 1,309.37 KNm
1.2Mcr = 1,571.25 KNm
Mmax = 5,052.59 KNm >>
OK!
ii. The minimum reinforcement criteria: - AASHTO 2004, Art. 5.7.3.3.2
rmin = 0.03*fc'/fy = 0.0018
r = As/bd = 0.0244 b= bw = 0.40
OK!
g) Serviceability Requirements
Fatigue stress limits ( ERA BDM Section 9.6.2 ) or (AASHTO LRFD Art. 5.5.3.2)
Fatigue stress limits will be checked for the service load conditions. The permissible stress range is given by Eq. 9.19.
ff = 145 - 0.33 fmin + 55 (r/h)
k = [rn+0.5(ts/d)2
] / (rn+(ts/d) = 0.258
k*d = 0.339 m
T-beam
If it is rectangular beam

j = 1 - k/3 = -
If it is T-beam
j = 6 - 6(hf/d)+2(hf/d)2
- (hf/d)3
/(2pn) = 0.935
6 - 3 (hf/d)
The minimum stress, fmin , is : (the minimum stress is created when there is no vehicle moving on the brid
fmin = MDL/(As*j*d) = 101.39 N/mm2
The maximim stress, fmax, is caused by the total load ( MDL+LL+IM ) (the maximum stress is created when the
vehicle is moving on the bridge, i.e., due to the self weight + moving load)
fmax = (MDL+LL+IM) /(As*j*d) = 200.51 N/mm2
The actual stress range, delta ff, is:
Delta ff = fmax - fmin = 99.12 N/mm2
The fatigue stress limit, ff, is
ff = 145 - 0.33* fmin + 55 (r/h) = 128.04 N/mm2 >>
OK !!
Control of Cracking by Distribution of Reinforcement (Sec. 9.4.5)
To control flexural cracking of the concrete, tension reinforcement shall be well distributed within maximum flexural zones.
Components shall be so proportioned that the tensile stress in the steel reinforcement at service limit state, fs, does not exceed
fsa = Z / (dc*A)1/3
< = 0.6*fy

A = Effective tension area/ No. of bars=bw*(2y')/N= 0.00420, y’ = d’ = 0.084 m
dc = distance measured from extreme tension fiber to
center of the closest bar = 0.072; bw = 0.40
Z = crack width parameter, assumed = 30.00 ; No of bars = 16
Therefore, for crack control the maximum allowable stress is
fsa = Z / (dc*A)1/3 <=0.6fy = 240.00 fsa = 240 MPa
The maximim stress, fmax, at service load is
fmax = (MDL+MLL+IM) /(As*j*d) = 200.51
OK
h) Bar Cutting
Development of Reinforcement (ERA BDM Sec. 9.4.3)or AASHTO, Art. 5.11.1.2.2
Positive moment reinforcement: Atleast one-third of the positive moment reinforcement in simple span
members shall extend along the same face of the member beyond the centerline of the support.
The basic development length, ldb, in mm is
For bars diam. 35 and smaller, ldb = 0.02Ab*fy/sqrt(fc') >=0.06db*fy db=
ldb = 0.02Ab*fy/sqrt(fc') = 1313 mm fy= 400
fc'= 24.0
The tension development length,ld, is 0.06db*fy= 768.0
ld = ldb * modification factor = 1313 mm mod. Factor = 1.00

Modification factor 2 taken to account for clear spacing between bars
Lap splices of Reinforcement in Tension
The length of lap for tension lap splices shall not be less than either 300mm or the 1.3 times the development
length
Lap splices for diam. 32 bar = 1,707 mm
Flexural Reinforcement Extension Length (Sec. 9.4.5)
Except at supports of simple spans and at the free ends of cantilevers, reinforcement shall be extended beyond the point at
which it is no longer required to resist flexure for a distance not less than:
a) the effective depth of the member = 1.32 m
b) 15 times the nominal diameter of bar = 0.48 m
c) 1/20 of the clear span = 0.90 m
Therefore, extension length is max. of the above
values and development length ld = 1.32 m
x
(m)
Mx
(KNm) 4Æ32 8Æ32 12Æ32 16Æ32 0Æ32
0.00 0.000 1520.59 2997.12 4415.70 5780.97 0.00
0.93 1042.82 1520.59 2997.12 4415.70 5780.97 0.00

1.85 1964.74 1520.59 2997.12 4415.70 5780.97 0.00
2.78 2765.75 1520.59 2997.12 4415.70 5780.97 0.00
3.70 3445.86 1520.59 2997.12 4415.70 5780.97 0.00
4.63 4005.05 1520.59 2997.12 4415.70 5780.97 0.00
5.55 4443.35 1520.59 2997.12 4415.70 5780.97 0.00
6.48 4760.74 1520.59 2997.12 4415.70 5780.97 0.00
7.40 4957.22 1520.59 2997.12 4415.70 5780.97 0.00
8.33 5052.59 1520.59 2997.12 4415.70 5780.97 0.00
9.25 4759.39 1520.59 2997.12 4415.70 5780.97 0.00

D = 1.40 m
beff= 2.10 m
bw= 0.40 m
bw'= 2.10 m
hf= 0.20 m
f= 0.90
fy= 400 Mpa
f'c= 24 Mpa
n(no.of bottom bar) 4

dc= 72 mm
Rectangular Analysis T Beam Analysis
fM= fAs*fy(d-a/2) Asf=.85*f'c*(beff-bw)*hf/fy
a=(As-Asf)fy/(.85*f'c*bw)
M1=Asf*fy(d-hf/2)
M2=(As-Asf)fy*(d-a/2)
fM= f(M1+M2)
No. of bars
Total No.
of bars Diam.(mm) As(mm2)
y
bar(mm)
d=D-
ybar(mm) r = As/bd
a=As*fy/
(.85*f'cbef
f) R or T fMr
4 4 32 3217 72 1328.00 0.001 30.037 R 1520.59
4 8 32 6434 80 1324.00 0.002 60.075 R 2997.12
4 12 32 9651 88 1316.00 0.003 90.112 R 4415.70
4 16 32 12868 96 1308.00 0.005 120.149 R 5780.97
x
(m)
Mx
(KNm)

0.00 0.000
0.925 1042.823
1.850 1964.740
2.775 2765.751
3.700 3445.856
4.625 4005.055
5.550 4443.348
6.475 4760.736
7.400 4957.218
8.325 5052.592
9.25 4759.392
covered dis
Type No. of bars R or T Resisting Moment
x for
moment From to length of bar
G0 4 R 1,520.59 1.502 0 1.502 18.9
G1 4 R 2,997.12 3.28 1.502 3.28 15.50
G2 4 R 4,415.70 5.754 3.28 5.754 11.94
G3 4 R 5,780.97 10.25 3.28 10.25 11.94

a = 1/2*circumf.+4*db = 630mm
e = 4*db = 4*32mm = 128mm
d
a
d = 10*db =320mm
Standard Hook Detail
e
Standard hook length = 0.63 M
4db = 128 mm
5db = 160 mm

length=
Length of Flexural Reinforcement bars
resistance length total length
The bar
start at
x= d
Bar- G0 = 14* Diam. 32
bars 1520.59 9.24 19.74 m 0.050 1.328
Bar- G1 =4*diam.32 bars 2997.12 7.75 18.13 m 0.136 1.324
Bar- G2 =4*diam.32 bars 4415.70 5.97 14.57 m 1.914 1.324
Bar- G3 = 4*diam.32 bars 5780.97 5.97 14.57 m 1.914 1.324
Bar- G4 =4*diam.32 bars 0.00 0.00 2.63 m 7.884 1.324
i ) Skin Reinforcement
If the effective depth, de, of reinforced concrete member exceeds 900mm, longitudinal skin reinforcement shall be uniformly distributed
along both side faces of the member for a distance d/2 nearest the flexural tension reinforcement.
The area of skin reinforcement, Ask, in mm2/m of height on each side face shall not be less than
Ask > = 0.001*(de - 760) <= As/1200
Effective depth de = D - d' = 1050 mm > 900mm
Therefore, Skin Reinf. shall be provided!
Ask > = 0.001*(de - 760) = 290 mm2/m d = 1.322
Spacing = as/As * 1000 = 690 mm diam. = 16
Maximum spacing, Smax, lesser of d/6 or 300mm = 220 mm
Use diam. 16 mm bars c/c 200 mm G4

6.3 DESIGN FOR SHEAR
Shear strength
Design of cross sections subject to shear shall be based on
Vu < = ØVn = Ø (Vc + Vs ) Ø = 0.900
where Vu = factored shear forces at the section
Vn = the nominal shear strength, determined as the lesser of Vn = Vc + Vs or Vn=0.25*fc'*bv*dv
Vc = the nominal shear strength provided by the concrete, determined by:
Vc = 0.083*b*SQRT(fc')*bv*dv, where b= 2.00
Vs = the nominal shear strength provided by the shear reinforcement
Vs = Av*fv*d / s
Shear strength provided by concrete
Shear stress provided by concrete
vc = 0.083*b*sqrt(fc') = 0.813 (MPa )
The shear strength carried by concrete,Vc=vc*bw*d= 0.813 *bw*d (N)
Shear strength provided by shear reinforcement
Where the factored shear force, Vu, exceeds shear strength ØVc, shear reinforcement shall be provided to satisfy
the equation, Vu < = ØVn = Ø ( Vc + Vs )
When shear reinforcement perpendicular to the axis of the member is used
Vs = Vu/Ø - Vc = Av*fy*d / s
There fore, spacing of shear reinforcement, s, is :
S = Av*fy*d / Vs = Av*fy*d/(Vu/Ø-Vc)=Av*fy*d/(Vu/Ø - 0.813*bw*d )
Minimum shear reinforcement ( Eq.12.34 )
Av = 0.083*sqrt(fc')*bw*S / fy fc' = 24MPa

Smax = Av*fy / ( 0.083*sqrt(fc')*bw ) fy = 300MPa
Maximum spacing of shear reinforcement Av =diam.12= 226mm2
If Vu < 0.10*fc' * bw * d, S <= 0.8d <= 600mm 0.10*fc' * bw= 960N
If Vu >= 0.10*fc' * bw * d, S <= 0.4d <= 300mm
Sections located less than a distance 'd' from support may be designed for the same shear as that computed at a distance d
x
(m)
Vx
(KN)
0.00 1016.33
0.93 935.55
1.85 854.76
2.78 773.98
3.70 693.19
4.63 612.40
5.55 531.62
6.48 450.83
7.40 370.05
8.33 289.26
9.25 208.47

Shear
reinforcement
x(m) Vu(KN) dv(m) Vc(KN) Vs(KN)
0.1*f'c*bw*
dv (KN)
Smax1
(mm)
Smax2
(mm) Scal.(mm)
Sprov.
(mm)
Actual/Recomm
ended Spacing
(mm)
0.00 1016.33 1.328 431.988 697.27 1274.88 600 417 129 129 80
0.93 935.55 2.394 778.750 260.75 2298.24 600 417 623 417 300
1.85 854.76 2.394 778.750 170.99 2298.24 600 417 950 417 300
2.78 773.98 2.394 778.750 81.22 2298.24 600 417 2000 417 300
3.70 693.19 2.369 770.617 0.00 2274.24 600 417 0 417 300
4.63 612.40 2.369 770.617 0.00 2274.24 600 417 0 417 300
5.55 531.62 2.369 770.617 0.00 2274.24 600 417 0 417 300
6.48 450.83 2.369 770.617 0.00 2274.24 600 417 0 417 300
7.40 370.05 2.344 762.485 0.00 2250.24 600 417 0 417 300
8.33 289.26 2.344 762.485 0.00 2250.2 600 417 0 417 300
9.25 208.47 2.344 762.485 0.00 2250.2 600 417 0 417 300
G5
6.4 DEFLECTION (Serviceabilty Requirement (AASHTO-Art. 5.7.3.6.2))

6.4.1 Computation of Gross Moment of Inertia
A) Centroid of cross section
For simplicity of calculations, the slab surface is assumed level, i.e., without crossfall.
The center of gravity is calculated from bottom of girder
Part
Area, Ai
(m2)
Centroid, yi
(m)
Ai*yi
(m3) Centroid of area
Top Slab 1.80 1.30 2.340 y1=sum Ai*yi/(sum Ai)= 1.00
Bottom slab
Girder 1.68 0.68 1.134
Sum 3.48 3.47
B) Gross moment of
Inertia
Part
Area, Ai
(m2)
yi
(m)
ycg = yi-y1
(m)
Ig
(m4)
Ai*(ycg)2
(m3)
Ig+Ai*(ycg)2
(m4)
Top Slab 1.80 1.30 0.30 0.006000 0.164 0.1699
Bottom slab
Girder 1.68 0.68 0.32 0.154350 0.176 0.3299
Sum Ig = 0.4998 m4
C) Computation of Effective Moment of Inertia

Ie = [ Mcr/Ma ]3
*Ig + [ 1 - (Mcr/Ma)3
] Icr < = Ig Mcr = cracking moment (N-mm)
fr = 0.63*sqrt(fc') = 3.09 N/mm2 fr = modulus of rupture of concrete as specified in
Mcr = fr*Ig / yt = 1545.19 kNm Article 5.4.2.6 (MPa)
yt = distance from the neutral axis to the extreme
D) Weight of superstructure tension fiber (mm)
Part
W
(KN)
W
(KN/m) Ma = maximum moment in a component at the stage
Top Slab 855.00 45.00 for which deformation is computed (N-mm)
Bottom Slab
Girder 798.00 42.00
Cross Girder 19.13 1.01
Sum W (KN/m) 88.01
Ma(Total Slab) = W*L2
/8 = 3,971
Ma = W*L2
/8
=
MDl ext.
=
W*L2
/8
= 6418.459
Ma = 6418.459 KNM
E) Transformed section

fc
T
jd
c
C
fs
kd
d
nAs
kd
d
beff
bw
k*d = 0.339 m (previous calculation)
beff= 8.40 m
bw= 1.60 m
n= 8
As= 51,472 mm2
d= 1.316 m
d'= 0.084 m
D= 1.400 m
hf= 0.200 m

j= 0.935 m
Centroid of cross section ( y1 measured from bottom)
Part
Area, Ai
(m2)
yi
(m)
Ai*yi
(m3)
Top Slab 1.80 1.30 2.34 y1=sum Ai*yi/(sum Ai)= 0.77
0.0000
Girder 6.72 0.68 4.536
Transformed steel,
nAs 0.41 0.08 0.035
Sum 8.93 6.91
Computation of Moment of Inertia of Cracked section, Icr
Part
Area, Ai
(m2)
ycg = yi-y1
(m)
Icg
(m4)
Ai*(ycg)2
(m3)
Ig+Ai*(ycg)2
(m4)
Slab 1.80 0.5263 0.004867 0.4986 0.5034
Girder 6.72 0.0987 0.154350 0.0655 0.2198
Transformed
steel, nAs 0.41 0.6897 - 0.1959 0.1959
Sum = 0.9191 m4
Effective moment of Inertia, Ie

Ie = [ Mcr / Ma]3
* Ig + [ 1 - (Mcr/Ma)3
] Icr = 0.9133 m4 <= Ig = 0.50
Ie= 0.500
Computation of Dead Load Deflection using effective moment of Inertia, Ie.
i. Instantaneous deflection
The maximum deflection is calculated by using the formula;
Deflection, max. = 5*W*L4
/(384*E*Ie) = 18.5mm E = Ec=24,768 MPa
ii. Long term deflection W =150.03 kN/m
Long term deflection = Instantaneous deflection * factor L=18.5 m
The factor is :
If the instantaneous deflection is based on Ig : = 3
If the instantaneous deflection is based on Ie :3.00-1.2(As'/As)>=1.6 2
Therefore, the Long term deflection = 55 mm
Provide camber 60 mm at mid span of the bridge
Calculation of live load deflection using the effective moment of inertia
The live load deflection at mid -span for different loading conditions that is analogous to the moving
load pattern is given as follows:

P
P P P/4
- When investigating the maximum absolute deflection, all design lanes should be loaded; and all
supporting components should be assumed to deflect equally.
- The live load portion of load combination Service I of Table 3-3( i.e, load factors for both live
and dead load equal to 1.0) with dynamic load allowance factor should be applied.
P with IM = 385.7 kN
**two
lane
loads
take
n
w lane load = 18.6 N/mm
def. Lane =5wL4
/(384EIe) = 0.57 mm
According to Chapter 3,
- The live load deflection should be taken as the larger of :

* That resulting from the design truck alone, or
* That resulting from 25 percent of the design truck together with the design lane
load.
a x x1 x2 x3
delta 1
(mm)
delta 2
(mm)
delta 3
(mm)
total truck
load
deflection
Maximum
deflection (max
(truck,0.25*truc
k+lane)
0 0.000 9.900 14.20 18.50 0.00 0.00 0.00 0.00 0.57
0.05 0.925 8.975 13.28 17.58 0.61 0.57 0.15 1.33 1.33
0.1 1.850 8.050 12.35 16.65 1.15 1.19 0.53 2.87 2.87
0.15 2.775 7.125 11.43 15.73 1.58 1.82 1.07 4.46 4.46
0.2 3.700 6.200 10.50 14.80 1.87 2.38 1.68 5.93 5.93
0.25 4.625 5.275 9.58 13.88 2.01 2.85 2.31 7.17 7.17
0.3 5.550 4.350 8.65 12.95 1.98 3.19 2.90 8.07 8.07
0.35 6.475 3.425 7.73 12.03 1.80 3.38 3.40 8.58 8.58
0.4 7.400 2.500 6.80 11.10 1.46 3.41 3.79 8.66 8.66
0.45 8.325 1.575 5.88 10.18 1.00 3.27 4.03 8.30 8.30
0.5 9.250 0.650 4.95 9.25 0.43 2.98 4.11 7.53 7.53
Therefore, the maximum live load deflection is the
maximum of the tabulated values above.
Def. Live load Max = 8.66 mm
Allowable Live Load Deflection - AASHTO Art. 2.5.2.6.2
Allowable deflection = Span /800 = 23.13 mm
OK
!

Chap-5-T-Girder Example-1.pdf

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