Deck

Design of 20m span superstructure
(Design of Deck slab)
Design if Deck Slab
Sailent fetures of super-structure:
Effective span of the longitudinal girder(C/C of supports)
Overall depth of girder
Width of web
Thickness of slab
Spacing of longitudinal girders
Number of longitudinal girders
Number of cross girders
Thickness of crosss girder
Depth of cross girder
Materials adopted:
Grade of Concrete =M 40
Grade of steel = Fe
cbc=
13.334 N/mm2
st =
200 N/mm2
n=
0.401
j=
0.867
k=
2.318

=
=
=
=
=
=
=
=
=

18
1.2
0.25
0.2
1.7
5
3
0.3
1

m
m
m
m
m
nos.
nos.
m
m (Incl. deck slab)

415

Cross girders are provided at the supports and at the mid span
Distance between the C/L of the end and middle cross girders =
Distance between the C/L of the end amd middle longitudinal girders =
Aspect ratio =
(l/b) =
5.295
Hence the deck slab shall be designed as one way slab
Load caculations:
Load due to self weight of the deck slab =
0.2
=
5 kN/m2

x

9 m
1.7 m

25

Load due toCrashbarrier:
0.31 x
24 =
7.44 kN/m
And the same is assumed to be acting as UDL over a width of 500mm
I.e
14.88 kN/m (over a width of 500mm)
For calculation purpose load due to crash barrier is applied on median side also.

Load due to wearing coat:
Thickness of wearing coat =
Load due to wearing coat =

75 mm (or)
1.65 kN/m2

0.075 m

Live Load Calculations:
Deck slab is designed for three lane traffic and the following loads were considered in the design
Class -AA Wheeled
Class -AA Tracked Vehicle
70R - Bogie loading
Class - A Wheeled
The following live load positions were studied for maximum effects of the live load on the deck slab
Case-1
When vehicle is placed with minimum eccentricity condition

TNHW

SAI Consulting Engineers Pvt. Ltd.,

Design of 20m span superstructure
(Design of Deck slab)
Case-2
Case -3

Vehicle is placed concentric with the Central longitudinal girder
Vehicle is placed concentric with the panel on hand rail side

1. Class - AA Wheeled vehicle:
Case -1: With minimum eccentricity:
For outer row tyres
Distance between two tyres of the axle =
Spacing of Axle =
1.2 m
Load per axle =
75 kN
Load per tyre
37.5 kN
Width of tyre =
300 mm
Contact length of tyre =
150 mm

2.2 m (C/C of tyres)

Aspct ratio = (b/lo) =
5.295
As the aspect ratio is grater than 2.00,  =

2.6 (for continuous slab)

beff =  x a (1 - a/lo) + b1
Effective width of the concentrated load =
a = Distance of the center of gravity of the concentrated load from the nerear support
lo = Effective span
b1 = Breadth of the concentration area of load.(parallel to support)
For Tyres 1 & 3 (Fallign in cantilever portion)
a=
0.6 m
b1 =
150 + (2 x
75
)=
300 mm
beff =
1.02 m (as the load is faling in the cantleverportion beff. = 1.2 a +b1)
The effective width is not overlapping , hence the beff =
1.02 m
Width of tyre along the span direction =
300 + (2 x (
75
+
(or)
0.85
m
As per IRC:6, the impact factor of 25% shall be considered.
Intensity of loading =

37.5 x 1.25
1.02 x 0.85

=

For Tyres 2 & 4
a=
0.1 m
lo =
1.7 m
b1 =
150 + (2 x
75
=
2.6
beff =
0.545 m
The effective width is overlapping, hence the beff =

54.066

37.5 x 1.25
0.545 x 0.85

For inner row tyres
Distance between two tyres of the axle =
Spacing of Axle =
1.2 m

TNHW

=

850 mm

kN/m2

)=

300 mm

0.545 m

Width of tyre along the span direction =
300 + (2 x (
75
+
(or)
0.85
m
As per IRC:6, the impact factor of 25% shall be considered.
Intensity of loading =

200 ) =

200 ) =

101.188

850 mm

kN/m2

1 m (C/C of tyres)

SAI Consulting Engineers Pvt. Ltd.,

Design of 20m span superstructure
(Design of Deck slab)
Load per axle =
Load per tyre
Width of tyre =
Contact length of tyre =

125 kN
62.5 kN
300 mm
150 mm

Aspct ratio = (b/lo) =
5.295
As the aspect ratio is grater than 2.00,  =

2.6 (for continuous slab)

beff =  x a (1 - a/lo) + b1
Effective width of the concentrated load =
a = Distance of the center of gravity of the concentrated load from the nerear support
lo = Effective span
b1 = Breadth of the concentration area of load.(parallel to support)
For Tyres 1 & 3
a=
0 m
lo =
1.7 m
b1 =
150 + (2 x
75
)=
=
2.6
beff =
0.3 m
The effective width is not overlapping , hence the beff =
Width of tyre along the span direction =
300 + (2 x (
75
+
(or)
0.85
m
As per IRC:6, the impact factor of 25% shall be considered.
Intensity of loading =

37.5 x 1.25
0.3 x 0.85

=

For Tyres 2 & 4
a=
0.7 m
lo =
1.7 m
b1 =
150 + (2 x
75
=
2.6
beff =
1.371 m
The effective width is overlapping, hence the beff =

75 x 1.25
2.571 x 0.85

=

0.3 m

200 ) =

183.824

850 mm

kN/m2

)=

Width of tyre along the span direction =
300 + (2 x (
75
+
(or)
0.85
m
As per IRC:6, the impact factor of 25% shall be considered.
Intensity of loading =

300 mm

300 mm

2.571 m

200 ) =

42.9

850 mm

kN/m2

Case -2 :Vehicle is placed concentric with one of Central longitudinal girder
For outer row tyres
Distance between two tyres of the axle =
2.2 m (C/C of tyres)
Spacing of Axle =
1.2 m
Load per axle =
75 kN
Load per tyre
37.5 kN
Width of tyre =
300 mm
Contact length of tyre =
150 mm

TNHW

SAI Consulting Engineers Pvt. Ltd.,

Design of 20m span superstructure
(Design of Deck slab)

Aspct ratio = (b/lo) =
5.295
As the aspect ratio is grater than 2.00,  =

2.6 (for continuous slab)

beff =  x a (1 - a/lo) + b1
Effective width of the concentrated load =
a = Distance of the center of gravity of the concentrated load from the nerear support
lo = Effective span
b1 = Breadth of the concentration area of load.(parallel to support)
For Tyres 1, 2, 3 and 4
a=
0.6 m
lo =
1.7 m
b1 =
150 + (2 x
75
=
2.6
beff =
1.31 m
The effective width is overlapping, hence the beff =

)=

2.51 m

Width of tyre along the span direction =
300 + (2 x (
75
+
(or)
0.85
m
As per IRC:6, the impact factor of 25% shall be considered.
Intensity of loading =

75 x 1.25
2.51 x 0.85

=

200 ) =

43.942

For inner row tyres
Distance between two tyres of the axle =
Spacing of Axle =
1.2 m
Load per axle =
125 kN
Load per tyre
62.5 kN
Width of tyre =
300 mm
Contact length of tyre =
150 mm
Aspct ratio = (b/lo) =
5.295
As the aspect ratio is grater than 2.00,  =

300 mm

850 mm

kN/m2

1 m (C/C of tyres)

2.6 (for continuous slab)

beff =  x a (1 - a/lo) + b1
Effective width of the concentrated load =
a = Distance of the center of gravity of the concentrated load from the nerear support
lo = Effective span
b1 = Breadth of the concentration area of load.(parallel to support)
For Tyres 1, 2, 3 and 4
a=
0.5 m
lo =
1.7 m
b1 =
150 + (2 x
75
=
2.6
beff =
1.218 m
The effective width is overlapping, hence the beff =
Width of tyre along the span direction =
300 + (2 x (
75
(or)

TNHW

+
0.85

)=

300 mm

2.418 m

200 ) =

850 mm

m

SAI Consulting Engineers Pvt. Ltd.,

Design of 20m span superstructure
(Design of Deck slab)
As per IRC:6, the impact factor of 25% shall be considered.
Intensity of loading =

75 x 1.25
2.418 x 0.85

=

45.614

kN/m2

Case -3 :Vehicle is placed concentric with the Slab pannel
For outer row tyres
Distance between two tyres of the axle =
2.2 m (C/C of tyres)
Spacing of Axle =
1.2 m
Load per axle =
75 kN
Load per tyre
37.5 kN
Width of tyre =
300 mm
Contact length of tyre =
150 mm
Aspct ratio = (b/lo) =
5.295
As the aspect ratio is grater than 2.00,  =

2.6 (for continuous slab)

beff =  x a (1 - a/lo) + b1
Effective width of the concentrated load =
a = Distance of the center of gravity of the concentrated load from the nerear support
lo = Effective span
b1 = Breadth of the concentration area of load.(parallel to support)
For Tyres 1, 2, 3 and 4
a=
0.25 m
lo =
1.7 m
b1 =
150 + (2 x
75
)=
=
2.6
beff =
0.855 m
The effective width is not overlapping , hence the beff =
0.855 m
Width of tyre along the span direction =
300 + (2 x (
75
+
(or)
0.85
m
As per IRC:6, the impact factor of 25% shall be considered.
Intensity of loading =

37.5 x 1.25
0.855 x 0.85

=

For inner row tyres
Distance between two tyres of the axle =
Spacing of Axle =
1.2 m
Load per axle =
125 kN
Load per tyre
62.5 kN
Width of tyre =
300 mm
Contact length of tyre =
150 mm
Aspct ratio = (b/lo) =
5.295
As the aspect ratio is grater than 2.00,  =
Effective width of the concentrated load =

TNHW

300 mm

200 ) =

64.5

850 mm

kN/m2

1 m (C/C of tyres)

2.6 (for continuous slab)
beff =  x a (1 - a/lo) + b1

SAI Consulting Engineers Pvt. Ltd.,

Design of 20m span superstructure
(Design of Deck slab)
a = Distance of the center of gravity of the concentrated load from the nerear support
lo = Effective span
b1 = Breadth of the concentration area of load.(parallel to support)
For Tyres 1, 2, 3 and 4
a=
0.35 m
lo =
1.7 m
b1 =
150 + (2 x
75
)=
=
2.6
beff =
1.023 m
The effective width is not overlapping , hence the beff =
1.023 m
Width of tyre along the span direction =
300 + (2 x (
75
+
(or)
0.85
m
As per IRC:6, the impact factor of 25% shall be considered.
Intensity of loading =

37.5 x 1.25
1.023 x 0.85

=

300 mm

200 ) =

53.908

850 mm

kN/m2

2. Class - AA Track vehicle:
Case -1: With minimum eccentricity:
Distance between two tyres of the axle =
2.05 m (C/C of tyres)
Load per axle =
700 kN
Load per tyre
350 kN
Width of tyre =
850 mm
Contact length of tyre =
3600 mm
Aspct ratio = (b/lo) =
5.295
As the aspect ratio is grater than 2.00,  =

2.6 (for continuous slab)

beff =  x a (1 - a/lo) + b1
Effective width of the concentrated load =
a = Distance of the center of gravity of the concentrated load from the nerear support
lo = Effective span
b1 = Breadth of the concentration area of load.(parallel to support)
beff for cantilever portion = 1.2 x a + b1
For Track 1
a=
0.43 m
lo =
1.7 m
b1 =
3600 + (2 x
75
)=
3750 mm
=
2.6
beff =
4.266 m
The effective width is =
4.266 m
Width of tyre along the span direction =
850 + (2 x (
75
+
(or)
1.4
m
As per IRC:6, the impact factor of 25% shall be considered.
Intensity of loading =
For Track 2
a=
lo =

TNHW

350 x 1.25
4.266 x 1.4

=

200 ) =

73.254

1400 mm

kN/m2

0.07 m
1.7 m

SAI Consulting Engineers Pvt. Ltd.,

Design of 20m span superstructure
(Design of Deck slab)
b1 =
=
beff =
3.925 m
The effective width is =

3600
2.6

+ (2 x

75

)=

3.925 m

Width of tyre along the span direction =
850 + (2 x (
75
+
(or)
1.4
m
As per IRC:6, the impact factor of 25% shall be considered.
Intensity of loading =

3750 mm

350 x 1.25
3.925 x 1.4

=

200 ) =

79.618

1400 mm

kN/m2

Case -2 :Vehicle is placed concentric with the Central longitudinal girder
Distance between two tyres of the axle =
2.06 m (C/C of tyres)
Load per axle =
700 kN
Load per tyre
350 kN
Width of tyre =
850 mm
Contact length of tyre =
3600 mm
Aspct ratio = (b/lo) =
5.295
As the aspect ratio is grater than 2.00,  =

2.6 (for continuous slab)

beff =  x a (1 - a/lo) + b1
Effective width of the concentrated load =
a = Distance of the center of gravity of the concentrated load from the nerear support
lo = Effective span
b1 = Breadth of the concentration area of load.(parallel to support)
For Track 1 & 2
a=
lo =
b1 =
=
beff =
4.806 m
The effective width is =

0.67 m
1.7 m
3600 + (2 x
2.6

75

)=

4.806 m

Width of tyre along the span direction =
850 + (2 x (
75
+
(or)
1.4
m
As per IRC:6, the impact factor of 25% shall be considered.
Intensity of loading =

3750 mm

350 x 1.25
4.806 x 1.4

=

200 ) =

65.023

1400 mm

kN/m2

Case -3 :Vehicle is placed concentric with slab pannel
Distance between two tyres of the axle =
2.05 m (C/C of tyres)
Load per axle =
700 kN
Load per tyre
350 kN
Width of tyre =
850 mm
Contact length of tyre =
3600 mm

TNHW

SAI Consulting Engineers Pvt. Ltd.,

Design of 20m span superstructure
(Design of Deck slab)
Aspct ratio = (b/lo) =
5.295
As the aspect ratio is grater than 2.00,  =

2.6 (for continuous slab)

beff =  x a (1 - a/lo) + b1
Effective width of the concentrated load =
a = Distance of the center of gravity of the concentrated load from the nerear support
lo = Effective span
b1 = Breadth of the concentration area of load.(parallel to support)
For Track 1 & 2
a=
lo =
b1 =
=
beff =
4.169 m
The effective width is =

0.18 m
1.7 m
3600 + (2 x
2.6

75

)=

4.169 m

Width of tyre along the span direction =
850 + (2 x (
75
+
(or)
1.4
m
As per IRC:6, the impact factor of 25% shall be considered.
Intensity of loading =

3750 mm

350 x 1.25
4.169 x 1.4

3. 70R - Wheeled (Bogie) loading:
Case -1: With minimum eccentricity:
Distance between two tyres of the axle =
Spacing of Axle =
1.22 m
Load per axle =
200 kN
Load per tyre
100 kN
Width of tyre =
860 mm

=

200 ) =

74.959

1400 mm

kN/m2

1.93 m (C/C of tyres)

Contact length of the tyre shall be arrived based on the max. tyre pressures specified in IRC:6-2000
Maximum tyre pressure =
5.273 kg/cm2
(or)
0.5273 N/mm2
Actual max. Load =
5 t (or)
50000 N
Contact width of tyre =
360 mm
Contact length of tyre =
264 mm
Contact length of tyre =
264 mm
Aspct ratio = (b/lo) =
5.295
As the aspect ratio is grater than 2.00,  =

2.6 (for continuous slab)

beff =  x a (1 - a/lo) + b1
Effective width of the concentrated load =
a = Distance of the center of gravity of the concentrated load from the nerear support
lo = Effective span
b1 = Breadth of the concentration area of load.(parallel to support)
For Tyres 1, 4
a=
0.42 m
lo =
1.7 m
b1 =
264 + (2 x
75
=
2.6
beff =
1.237 m
The effective width is overlapping, hence the beff =

TNHW

)=

414 mm

2.457 m

SAI Consulting Engineers Pvt. Ltd.,

Design of 20m span superstructure
(Design of Deck slab)

Width of tyre along the span direction =
860 + (2 x (
75
+
(or)
1.41
m
As per IRC:6, the impact factor of 25% shall be considered.
Intensity of loading =

200 x 1.25
2.457 x 1.41

=

For Tyres 2, 3
a=
0.19 m
lo =
1.7 m
b1 =
264 + (2 x
75
=
2.6
beff =
0.853 m
The effective width is overlapping, hence the beff =

72.164

100 x 1.25
0.853 x 1.41

=

1410 mm

kN/m2

)=

414 mm

0.853 m

Width of tyre along the span direction =
860 + (2 x (
75
+
(or)
1.41
m
As per IRC:6, the impact factor of 25% shall be considered.
Intensity of loading =

200 ) =

200 ) =

103.931

1410 mm

kN/m2

Case -2: With vehicle is concentric with the central girder:
Distance between two tyres of the axle =
1.93 m (C/C of tyres)
Spacing of Axle =
1.22 m
Load per axle =
200 kN
Load per tyre
100 kN
Width of tyre =
860 mm
Contact length of the tyre shall be arrived based on the max. tyre pressures specified in IRC:6-2000
Maximum tyre pressure =
5.273 kg/cm2
(or)
0.5273 N/mm2
Actual max. Load =
5 t (or)
50000 N
Contact width of tyre =
360 mm
Contact length of tyre =
264 mm
Contact length of tyre =
264 mm
Aspct ratio = (b/lo) =
5.295
As the aspect ratio is grater than 2.00,  =

2.6 (for continuous slab)

beff =  x a (1 - a/lo) + b1
Effective width of the concentrated load =
a = Distance of the center of gravity of the concentrated load from the nerear support
lo = Effective span
b1 = Breadth of the concentration area of load.(parallel to support)
For Tyres 1, 2, 3, ,4
a=
0.735 m
lo =
1.7 m
b1 =
264 + (2 x
75
=
2.6
beff =
1.499 m
The effective width is overlapping, hence the beff =

TNHW

)=

414 mm

2.719 m

SAI Consulting Engineers Pvt. Ltd.,

Design of 20m span superstructure
(Design of Deck slab)

Width of tyre along the span direction =
860 + (2 x (
75
+
(or)
1.41
m
As per IRC:6, the impact factor of 25% shall be considered.
Intensity of loading =

200 x 1.25
2.719 x 1.41

=

200 ) =

65.21

1410 mm

kN/m2

Case -3: With vehicle is concentric with slab pannel :
Distance between two tyres of the axle =
1.93 m (C/C of tyres)
Spacing of Axle =
1.22 m
Load per axle =
200 kN
Load per tyre
100 kN
Width of tyre =
860 mm
Contact length of the tyre shall be arrived based on the max. tyre pressures specified in IRC:6-2000
Maximum tyre pressure =
5.273 kg/cm2
(or)
0.5273 N/mm2
Actual max. Load =
5 t (or)
50000 N
Contact width of tyre =
360 mm
Contact length of tyre =
264 mm
Contact length of tyre =
264 mm
Aspct ratio = (b/lo) =
5.295
As the aspect ratio is grater than 2.00,  =

2.6 (for continuous slab)

beff =  x a (1 - a/lo) + b1
Effective width of the concentrated load =
a = Distance of the center of gravity of the concentrated load from the nerear support
lo = Effective span
b1 = Breadth of the concentration area of load.(parallel to support)
For Tyres 1, 2, 3 & 4
a=
0.115 m
lo =
1.7 m
b1 =
264 + (2 x
75
)=
=
2.6
beff =
0.693 m
The effective width is not overlapping , hence the beff =
0.693 m
Width of tyre along the span direction =
860 + (2 x (
75
+
(or)
1.41
m
As per IRC:6, the impact factor of 25% shall be considered.
Intensity of loading =

100 x 1.25
0.693 x 1.41

4. Class - A (2 Nos.)
Case -1: With minimum eccentricity:
Distance between two tyres of the axle =
Spacing of Axle =
1.2 m
Load per axle =
114 kN
Load per tyre
57 kN
Width of tyre =
500 mm
Contact length of tyre =
250 mm

TNHW

=

414 mm

200 ) =

127.926

1410 mm

kN/m2

1.8 m (C/C of tyres)

SAI Consulting Engineers Pvt. Ltd.,

Design of 20m span superstructure
(Design of Deck slab)

Aspct ratio = (b/lo) =
5.295
As the aspect ratio is grater than 2.00,  =
2.6 (for continuous slab)
beff
=

x a (1 - a/lo) + b1
Effective width of the concentrated load =
a = Distance of the center of gravity of the concentrated load from the nerear support
lo = Effective span
b1 = Breadth of the concentration area of load.(parallel to support)
For Tyres 1 & 4 (First vehicle)
a=
0.05 m (From the face of support)
b1 =
250 + (2 x
75
)=
beff =
0.527 m
The effective width is overlapping, hence the beff =
0.527 m

400 mm

Width of tyre along the span direction =
500 + (2 x (
75
+
200 ) =
(or)
1.05
m
As per IRC:6, the impact factor of 50% shall be considered.
Intensity of loading =
57 x 1.5
=
154.514
kN/m2
0.527 x 1.05
For Tyres 2 & 3 (First vehicle)
a=
0.15 m
lo =
1.7 m
b1 =
250 + (2 x
75
)=
=
2.6
beff =
0.756 m
The effective width is not overlapping , hence the beff =
0.756 m
Width of tyre along the span direction =
500 + (2 x (
75
+
(or)
1.05
m
As per IRC:6, the impact factor of 50% shall be considered.
Intensity of loading =

57 x 1.5
0.756 x 1.05

=

For Tyres 1 & 4 (Second vehicle)
a=
0.15 m
lo =
1.7 m
b1 =
250 + (2 x
75
=
2.6
beff =
0.756 m
The effective width is overlapping, hence the beff =

57 x 1.5
0.756 x 1.05

=

1050 mm

kN/m2

)=

400 mm

0.756 m

Width of tyre along the span direction =
500 + (2 x (
75
+
(or)
1.05
m
As per IRC:6, the impact factor of 50% shall be considered.
Intensity of loading =

400 mm

200 ) =

107.71

1050 mm

200 ) =

107.71

1050 mm

kN/m2

For Tyres 2 & 3 (Second vehicle)

TNHW

SAI Consulting Engineers Pvt. Ltd.,

Design of 20m span superstructure
(Design of Deck slab)
a=
lo =
b1 =
=

0.25 m
1.7 m
250 + (2 x
2.6

75

)=

beff =
0.955 m
The effective width is not overlapping , hence the beff =

0.955 m

Width of tyre along the span direction =
500 + (2 x (
75
+
(or)
1.05
m
As per IRC:6, the impact factor of 50% shall be considered.
Intensity of loading =

57 x 1.5
0.955 x 1.05

=

400 mm

200 ) =

85.266

1050 mm

kN/m2

Case -2: With vehicle is concentric with the central girder:
Distance between two tyres of the axle =
1.8 m (C/C of tyres)
Spacing of Axle =
1.2 m
Load per axle =
114 kN
Load per tyre
57 kN
Width of tyre =
500 mm
Contact length of tyre =
250 mm
Aspct ratio = (b/lo) =
5.295
As the aspect ratio is grater than 2.00,  =
2.6 (for continuous slab)
beff
=

x a (1 - a/lo) + b1
Effective width of the concentrated load =
a = Distance of the center of gravity of the concentrated load from the nerear support
lo = Effective span
b1 = Breadth of the concentration area of load.(parallel to support)
For Tyres 1 & 4 (First vehicle)
a=
0.75 m
lo =
1.7 m
b1 =
250 + (2 x
75
=
2.6
beff =
1.49 m
The effective width is overlapping, hence the beff =

)=

2.69 m

Width of tyre along the span direction =
500 + (2 x (
75
+
(or)
1.05
m
As per IRC:6, the impact factor of 50% shall be considered.
Intensity of loading =

114 x 1.5
2.69 x 1.05

=

For Tyres 2 & 3 (First vehicle)
a=
0.85 m
lo =
1.7 m
b1 =
250 + (2 x
75
=
2.6
beff =
1.505 m
The effective width is overlapping, hence the beff =

TNHW

400 mm

200 ) =

60.542

)=

1050 mm

kN/m2

400 mm

2.705 m

SAI Consulting Engineers Pvt. Ltd.,

Design of 20m span superstructure
(Design of Deck slab)
Width of tyre along the span direction =
500 + (2 x (
75
+
(or)
1.05
m
As per IRC:6, the impact factor of 50% shall be considered.
Intensity of loading =

114 x 1.5
2.705 x 1.05

=

60.206

For Tyres 2 & 3 (Second vehicle)
a=
0.85 m
lo =
1.7 m
b1 =
250 + (2 x
75
=
2.6
beff =
1.505 m
The effective width is overlapping, hence the beff =

114 x 1.5
2.705 x 1.05

=

400 mm

200 ) =

60.206

1050 mm

kN/m2

)=

400 mm

2.69 m

Width of tyre along the span direction =
500 + (2 x (
75
+
(or)
1.05
m
As per IRC:6, the impact factor of 50% shall be considered.
114 x 1.5
2.69 x 1.05

kN/m2

2.705 m

For Tyres 1 & 4 (Second vehicle)
a=
0.75 m
lo =
1.7 m
b1 =
250 + (2 x
75
=
2.6
beff =
1.49 m
The effective width is overlapping, hence the beff =

Intensity of loading =

1050 mm

)=

Width of tyre along the span direction =
500 + (2 x (
75
+
(or)
1.05
m
As per IRC:6, the impact factor of 50% shall be considered.
Intensity of loading =

200 ) =

=

200 ) =

60.542

1050 mm

kN/m2

Continuous beam analysis has been carried-out by using the structural analysis programe (STAAD)
by inputting the various loads calculated above on the beam members. The analysis results were
appended for ready reference.
1

2

3

0.85

1.7

4

5

1.7

1.7

6

7

1.7

0.85

Descritisation Diagram
1

3
2

1

TNHW

5

6

4

8
7

2

10
9

11

13
12

3

14
4

SAI Consulting Engineers Pvt. Ltd.,

Design of 20m span superstructure
(Design of Deck slab)
4

1
Ctitical sections for design

Summary of moments in various members
Load description
Self-weight
SDL
Class-AA (case-1)
Class-AA (case-2)
Class-AA (case-3)
Class-AA -T(case-1)
Class-AA -T(case-2)
Class-AA -T(case-3)
70R-Bogie(case-1)
70R-Bogie(case-2)
70R-Bogie(case-3)
Class-A (case-1)
Class-A (case-2)
Max. Hogging BM
Max. Sagging BM

1
0.83
1.22
0
0
0
0
0
0
0
0
0
3.1
0
5.15
0

2
2.43
3.38
0
0
0
0
0
0
0
0
0
8.14
0
13.95
0

Load description
Self-weight
SDL
Class-AA (case-1)
Class-AA (case-2)
Class-AA (case-3)
Class-AA -T(case-1)
Class-AA -T(case-2)
Class-AA -T(case-3)
70R-Bogie(case-1)
70R-Bogie(case-2)
70R-Bogie(case-3)
Class-A (case-1)
Class-A (case-2)
Max. Hogging BM
Max. Sagging BM

8
0.56
-0.79
17.31
1.56
5.6
10..66
1.27
7.06
11.95
1.32
11.54
4.53
10.81
17.08
0

9
0.1
-0.8
3.75
-2.56
1.09
7.55
-3.85
3.35
7.22
-3.85
4.7
-1.67
5.61
6.85
-4.55

Section Numbers
3
4
5
1.26
0.7
-0.38
4.01
3.38
1.31
-0.47
-1.04
-2.37
0.39
0.85
2.64
-0.38
-0.83
-2.55
-2.37
-5.23
-13.26
0.49
1.07
3.3
-1.29
-2.85
-8.76
-2.1
-4.62
-11.54
0.49
1.09
3.35
-1.73
-3.82
-11.79
8.42
0.78
-2.08
-2.39
-5.27
-10.8
13.69
4.93
4.28
0
-1.19
-12.33

Section Numbers
10
11
-0.64
0.26
-0.49
0.36
-29.15
-7.13
-8.79
4.39
-5.09
0.87
-1.37
-0.56
-9.82
5.57
0.71
2.55
-7.26
-2
-10
5.39
-1.32
3.14
-4.53
1.74
-5.97
2.19
-0.42
6.01
-30.28
-6.51

6
0.18
-0.21
10.96
4.42
1.53
-0.29
5.53
-1.04
2.13
5.62
-0.79
2.53
2.97
10.93
-1.07

7
0.6
-0.52
20.45
4.89
5.93
7.1
6.11
5.07
9.65
6.21
9.05
6.43
9.58
20.53
0

12
0.77
0.67
8.34
10.37
5.32
3.93
11.91
6.05
5.05
12.38
10.15
6.07
6.49
13.35
0

Note: Values with negative sign are hogging moments
Check for depth requirement:
Maximum BM =
30.28 kNm
(at the face of web)
Efective depth required = sqrt(BM/Q x b) =
114.294 mm
Overall depth provided =
200 mm
Effective depth available =
200
50
=
145 mm
>
Hence, the section adopted is
ADEQUATE

TNHW

5
114.294 mm

SAI Consulting Engineers Pvt. Ltd.,

Design of 20m span superstructure
(Design of Deck slab)
Design of reinforcement :
For hogging moment:(at the face of the web)
Design hogging moment =
20.53 kNm
Overall depth provided =
375 mm
Effective depth availbale =
375 40 =
330 mm
Area of steel required = BM/st x j x d
=
358.779 mm2/m
Hence, provide Y
10
bars @
218 mm C/C
Max. hogging moment at other locations =
10.93 kNm
Area of steel required = BM/st x j x d
Hence, provide

Y

=
10

However provide @
Area of steel provided =

409.309 mm2/m
bars @
191 mm C/C
190 mm C/C
413 mm2/m

For sagging moment:
Design sagging moment =
E. Depth provided
=
200
=
Area of steel required = BM/st x j x d
Hence, provide
Hence, provide

Y
Y

=
12
12

5

30.28 kNm
40
154 mm C/C

-

6

1133.93 mm2/m
bars @
99 mm C/C
bars @
95 mm C/C

Check for minimum Reinforcement:
As per IRC:21-2000, minimum reinforcemrnt in slabs shall be
0.12 %
Area of steel =
240 mm2/m
<
409.309 mm2/m
Hence,
OKAY
For distribution Reinforcement::
Design moment =
9.084 kNm
(0.3 times the maximum BM considered)
Area of steel required = BM/st x j x d
=
361.294 mm2/m
Hence, provide Y
10
bars @
217 mm C/C
Hence, provide Y
8
bars @
150 mm C/C
Note: As per the IRC:21-2000, Shear need not be checked if the live load dispersions are taken
as per the dispersion width formula

TNHW

SAI Consulting Engineers Pvt. Ltd.,