This document provides a design example for a reinforced concrete T-beam bridge girder. It includes the design of the deck slab, longitudinal girders, and cross girders. The design uses Courbon's method to calculate live load bending moments and shear forces. Details are given for the design of an interior deck slab panel including reinforcement sizing. Design of the longitudinal girders includes calculating reaction factors and sizing reinforcement to resist bending moments and shear forces from dead and live loads.

1. Design of Deck Slab & Girders

2. Design Example:
Design a RCC T beam girder bridge to suit the following data
Clear width of roadway – 7.5 m
Span (c/c of bearings) – 16 m
Live load – IRC class AA tracked vehicle
Average thickness of wearing coat – 80 mm
Concrete mix – M25
Steel – Fe415 grade HYSD bars
Using Courbon’s method, compute the design moments and
shear forces and design the deck slab, main girder and cross
girder and sketch the typical details of reinforcements.
2

3. Design of Deck Slab
3

4. 1. Permissible stresses:
Permissible flexural compressive stress (σcb) = 8.3 Mpa (table 9 of IRC
21-2000)
Permissible stress for Fe415 (σst) = 200 Mpa (table 10 of IRC 21-2000)
The modular ratio can be adopted as m = 10
Lever arm factor (j) =
Moment factor (Q) = 0.5 σcb nj = 0.5x8.3x0.9x0.293 = 1.09
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3
n
1
9
.
0
3
293
.
0
1
j
293
.
0
3
.
8
x
10
200
1
1
m
1
1
n
cb
st
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5. 2. Cross section of the deck:
5
2.5 m
2.5 m
4.0 m 4.0 m 4.0 m 4.0 m
The depth of cross girder is taken as equal to the depth of main
girder to simplify the computations

6. 2. Cross section of the deck:
6
0.08 m thick wearing coat
0.2 m
0.6 m
1.85 m 2.5 m 2.5 m
7.5 m
0.3 m
1.0 m
1.4 m
0.3 m

7. 3. Design of interior slab panel:
Bending moments
Dead weight of slab = (1x1x0.2x24) = 4.8 kN/m2
Dead weight of wc = (1x1x0.08x22) = 1.76 kN/m2
Total dead load = 4.8+1.76 = 6.56 kN/m2
7
B=2.5
m
L=4.0 m
3.60 m
0.85
m
u=1.01
m
v=3.76 m

8. 3. Design of interior slab panel:
Bending moments (live load)
u/B = 1.01/2.5 = 0.404
v/L = 3.76/4.0 = 0.94
K=(B/L)=2.5/4.0=0.625
From Pigeaud’s curves m1=0.085 and m2=0.024
8
Pigeaud's curves are based on the elastic analysis of thin isotrpoic plates
with simple supports on all the four sides. However, a deck slab cannot be
idealised as simply supported panel. The process of reducing Pigeaud‘s
moment values by 20% and assuming equal and support moments is
highly conservative. The deck slab behaves as a fixed slab for the usual
beam sizes of a bridge.

9. 3. Design of interior slab panel:
Bending moments (live load)
Bending moment in the short span direction (MS)
MS=(m1+μm2)W=(0.085+0.15x0.024)350 = 31.01 kNm
Bending moment in the long span direction (ML)
ML=(m2+μm1)W=(0.024+0.15x0.085)350 = 12.845 kNm
The design bending moment over an intermediate support of a
continuous deck supported on bearings may be calculated by
equation.
M1 = (M – qa2/8) or 0.8M, whichever is greater,
where, M, = Design bending moment.
9

10. 3. Design of interior slab panel:
Bending moments (live load)
Design bending moment including impact in the short span
direction (MS) = 1.25x0.8x31.01 = 31.01 kNm
Design bending moment including impact in the longer span
direction (ML) = 1.25x0.8x12.845 = 12.845 kNm
Bending moments (dead loads)
Dead load = 6.56 kN/m2
Total load on panel = (4x2.5x6.56) = 65.6 kN
u/B = 1.0; v/L = 1.0 as the panel is loaded with uniformly
distributed load.
K=(B/L)=2.5/4.0=0.625 and 1/K = 1.6
From Pigeaud’s curves m1=0.049 and m2=0.015
Bending moment in the short span direction (MS)
MS=(m1+μm2)W=(0.049+0.15x0.015)65.6 = 3.36 kNm
Bending moment in the long span direction (ML)
ML=(m2+μm1)W=(0.015+0.15x0.049)65.6 = 1.468 kNm
10

11. 3. Design of interior slab panel:
Bending moments (dead load)
Design bending moment in the short span direction
(MS) = 0.8x3.36 = 2.688 kNm
Design bending moment in the longer span direction
(ML) = 0.8x1.468 = 1.174 kNm
Total design bending moments in the short span
(MS) = 31.01+2.688 = 33.698 kNm
Total design bending moments in the long span
(ML) = 12.845+1.174 = 14.019 kNm
11

12. 3. Design of interior slab panel:
Shear forces
Dispersion in the direction of span = 0.85+2(0.08+0.2) = 1.41 m
For max. shear, load is kept such that the whole dispersion is in
span. The load is kept at 1.41/2 = 0.705 m from the edge of
beam.
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13. 3. Design of interior slab panel:
Shear forces
Breadth of girder = 0.3 m
Clear length of panel (L’) = 4.0-2(0.15) = 3.70 m
Clear breadth of panel (B’) = 2.5-2(0.15) = 2.2 m
(B’/L’) = 2.2/3.7 = 0.6
‘K’ for continuous slab is 1.84 (IRC 21-2000)
Effective width of slab = 1.84x0.705(1-0.705/2.2)+3.76 = 4.64 m
Load per meter width = 350/4.64 = 75.4 kN
Shear force = 75.4(2.2-0.705)/2.2 = 51.23 kN
Shear force with impact = 1.25x51.23 = 64 kN
Dead load shear force = 6.56x2.2/2 = 7.21 kN
Total shear force = 64+7.21 = 71.21 kN
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14. 3. Design of interior slab panel:
Design of section
Effective depth
Adopt overall depth = 200 mm
Area of tension reinforcement is given by
Similarly calculate reinforcement in longer span direction
14
mm
175
1000
x
1
.
1
10
x
698
.
33
Qb
M
d
6

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
2
6
st
st mm
1170
175
x
9
.
0
x
200
10
x
698
.
33
jd
M
A 
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15. 3. Design of interior slab panel:
Check for shear stress
Nominal shear stress
15
2
mm
/
N
4
.
0
175
x
1000
1000
x
21
.
71
bd
V
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16. 3. Design of interior slab panel:
Check for shear stress
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Maximum permissible shear
stress
Permissible shear stress
Hence safe in permissible
limits
2
max mm
/
N
9
.
1
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2
c mm
/
N
36
.
0
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17. Design of Girders
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18. 4. Design of longitudinal girders:
Dead loads from slab for girder
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0.2 m
0.6 m
1.85 m 2.5 m 2.5 m
0.3 m
1.0 m
1.4 m
0.3 m
W1 W2
W
1.2+2(0.85)
=2.05m
Axis of
bridge
CG of
loads
e=1.1 m

19. 4. Design of longitudinal girders:
Dead loads from slab for girder
Weight of
1. Parapet = 2x0.7 = 1.4 kN/m
2. Wearing coat = 0.08x7.5x22 = 13.2 kN/m
3. Deck slab = 0.2x7.5x24 = 36 kN/m
4. Kerb = 2(0.5x0.6x24) = 14.4 kN/m
Total dead load = 1.4+13.2+36+14.4 = 65 kN/m
It is assumed that the dead load is equally shared by all girders.
Dead load per girder = 65/3 = 21.66 kN/m
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20. 4. Design of longitudinal girders:
Reaction factors
Using Courbon’s method, the reaction factors are calculated as
follows:
Where, Rx= Reaction factor for the girder under consideration
I = Moment of Inertia of each longitudinal girder
dx= distance of the girder under consideration from the central
axis of the bridge
W = Total concentrated live load
n = number of longitudinal girders
e = Eccentricity of live load with respect to the axis of the
bridge.
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d
I
d
I
1
n
W
R x
2
x
x

21. 4. Design of longitudinal girders:
Reaction factors
Reaction factor for outer girder
Reaction factor for inner girder
Since W1=0.5W; RA = 0.48W; RB = 0.33W;
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1
2
1
A W
96
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0
1
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1
x
5
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2
5
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2
Ix
3
I
3
1
3
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2
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B W
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W
2
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22. 4. Design of longitudinal girders:
Live load BMs
Bending moment = 4x700 = 2800 kN.m
Bending moment including impact factor and reaction factor =
Reaction factor =2800x1.25x0.48 = 1680 kNm (for outer girder)
Bending moment including impact factor and reaction factor =
Reaction factor =2800x1.25x0.33 = 1155 kNm (for inner girder)
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23. 4. Design of longitudinal girders:
Live load Shears
Reaction on girder B = (350x0.45)/2.5=63 kN
Reaction on girder A = (350x2.05)/2.5=287 kN
Maximum reaction on girder B = (63x14.2)/16 = 56 kN
Maximum reaction on girder A = (287x14.2)/16 = 255 kN
Considering impact factor = 56x1.25 = 70 kN (girder B)
Considering impact factor = 255x1.25 = 318.75 kN (girder A)
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24. 4. Design of longitudinal girders:
Dead load BM and SFs
The depth of the girder is assumed as 1600 mm.
Depth of rib = 1.4 m
Width of rib = 0.3 m
Weight of rib = 1x1.4x0.3x24 = 10.08 kN/m (per meter)
The cross section is assumed to have the same cross section
dimension of main girder.
Hence weight of cross girder = 10.08 kN/m (per meter)
Reaction on main girder = 10.08x2.5 = 25.2 kN
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10.08 kN/m

25. 25