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

TNHW SAI Consulting Engineers Pvt. Ltd.,

Design if Deck Slab

Sailent fetures of super-structure:Effective span of the longitudinal girder(C/C of supports) = 18 mOverall depth of girder = 1.2 mWidth of web = 0.25 mThickness of slab = 0.2 m Spacing of longitudinal girders = 1.7 mNumber of longitudinal girders = 5 nos.Number of cross girders = 3 nos.Thickness of crosss girder = 0.3 mDepth of cross girder = 1 m (Incl. deck slab)

Materials adopted:Grade of Concrete =M 40

Grade of steel = 415

13.334

200n = 0.401j = 0.867

k = 2.318

Cross girders are provided at the supports and at the mid span

Distance between the C/L of the end and middle cross girders = 9 mDistance between the C/L of the end amd middle longitudinal girders = 1.7 mAspect ratio = (l/b) = 5.295Hence the deck slab shall be designed as one way slab

Load caculations:Load due to self weight of the deck slab = 0.2 x 25

= 5

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 = 75 mm (or) 0.075 m

Load due to wearing coat = 1.65

Live Load Calculations:Deck slab is designed for three lane traffic and the following loads were considered in the designClass -AA WheeledClass -AA Tracked Vehicle70R - Bogie loadingClass - A WheeledThe 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

scbc= N/mm2

sst = N/mm2

kN/m2

kN/m2

Fe

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

TNHW SAI Consulting Engineers Pvt. Ltd.,

Case-2 Vehicle is placed concentric with the Central longitudinal girderCase -3 Vehicle is placed concentric with the panel on hand rail side

1. Class - AA Wheeled vehicle:Case -1: With minimum eccentricity:For outer row tyresDistance between two tyres of the axle = 2.2 m (C/C of tyres)Spacing of Axle = 1.2 mLoad per axle = 75 kNLoad per tyre 37.5 kNWidth of tyre = 300 mmContact length of tyre = 150 mm

Aspct ratio = (b/lo) = 5.2952.6 (for continuous slab)

Effective width of the concentrated load =a = Distance of the center of gravity of the concentrated load from the nerear supportlo = Effective spanb1 = Breadth of the concentration area of load.(parallel to support)

For Tyres 1 & 3 (Fallign in cantilever portion)a = 0.6 mb1 = 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 + 200 ) = 850 mm

(or) 0.85 mAs per IRC:6, the impact factor of 25% shall be considered.

Intensity of loading = 37.5 x 1.25= 54.0661.02 x 0.85

For Tyres 2 & 4a = 0.1 mlo = 1.7 mb1 = 150 + (2 x 75 ) = 300 mm

2.6beff = 0.545 mThe effective width is overlapping, hence the beff = 0.545 m

Width of tyre along the span direction =300 + (2 x ( 75 + 200 ) = 850 mm

(or) 0.85 mAs per IRC:6, the impact factor of 25% shall be considered.

Intensity of loading = 37.5 x 1.25= 101.1880.545 x 0.85

For inner row tyresDistance between two tyres of the axle = 1 m (C/C of tyres)Spacing of Axle = 1.2 m

As the aspect ratio is grater than 2.00, a =

beff = a x a (1 - a/lo) + b1

kN/m2

a =

kN/m2

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

TNHW SAI Consulting Engineers Pvt. Ltd.,

Load per axle = 125 kNLoad per tyre 62.5 kNWidth of tyre = 300 mmContact length of tyre = 150 mm

Aspct ratio = (b/lo) = 5.2952.6 (for continuous slab)

Effective width of the concentrated load =a = Distance of the center of gravity of the concentrated load from the nerear supportlo = Effective spanb1 = Breadth of the concentration area of load.(parallel to support)

For Tyres 1 & 3a = 0 mlo = 1.7 mb1 = 150 + (2 x 75 ) = 300 mm

2.6beff = 0.3 mThe effective width is not overlapping , hence the beff = 0.3 m

Width of tyre along the span direction =300 + (2 x ( 75 + 200 ) = 850 mm

(or) 0.85 mAs per IRC:6, the impact factor of 25% shall be considered.

Intensity of loading = 37.5 x 1.25= 183.8240.3 x 0.85

For Tyres 2 & 4a = 0.7 mlo = 1.7 mb1 = 150 + (2 x 75 ) = 300 mm

2.6beff = 1.371 mThe effective width is overlapping, hence the beff = 2.571 m

Width of tyre along the span direction =300 + (2 x ( 75 + 200 ) = 850 mm

(or) 0.85 mAs per IRC:6, the impact factor of 25% shall be considered.

Intensity of loading = 75 x 1.25= 42.92.571 x 0.85

Case -2 :Vehicle is placed concentric with one of Central longitudinal girderFor outer row tyresDistance between two tyres of the axle = 2.2 m (C/C of tyres)Spacing of Axle = 1.2 mLoad per axle = 75 kNLoad per tyre 37.5 kNWidth of tyre = 300 mmContact length of tyre = 150 mm

As the aspect ratio is grater than 2.00, a =

beff = a x a (1 - a/lo) + b1

a =

kN/m2

a =

kN/m2

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

TNHW SAI Consulting Engineers Pvt. Ltd.,

Aspct ratio = (b/lo) = 5.2952.6 (for continuous slab)

Effective width of the concentrated load =a = Distance of the center of gravity of the concentrated load from the nerear supportlo = Effective spanb1 = Breadth of the concentration area of load.(parallel to support)

For Tyres 1, 2, 3 and 4a = 0.6 mlo = 1.7 mb1 = 150 + (2 x 75 ) = 300 mm

2.6beff = 1.31 mThe effective width is overlapping, hence the beff = 2.51 m

Width of tyre along the span direction =300 + (2 x ( 75 + 200 ) = 850 mm

(or) 0.85 mAs per IRC:6, the impact factor of 25% shall be considered.

Intensity of loading = 75 x 1.25= 43.9422.51 x 0.85

For inner row tyresDistance between two tyres of the axle = 1 m (C/C of tyres)Spacing of Axle = 1.2 mLoad per axle = 125 kNLoad per tyre 62.5 kNWidth of tyre = 300 mmContact length of tyre = 150 mm

Aspct ratio = (b/lo) = 5.2952.6 (for continuous slab)

Effective width of the concentrated load =a = Distance of the center of gravity of the concentrated load from the nerear support

lo = Effective spanb1 = Breadth of the concentration area of load.(parallel to support)

For Tyres 1, 2, 3 and 4a = 0.5 mlo = 1.7 mb1 = 150 + (2 x 75 ) = 300 mm

2.6beff = 1.218 mThe effective width is overlapping, hence the beff = 2.418 m

Width of tyre along the span direction =300 + (2 x ( 75 + 200 ) = 850 mm

(or) 0.85 m

As the aspect ratio is grater than 2.00, a =

beff = a x a (1 - a/lo) + b1

a =

kN/m2

As the aspect ratio is grater than 2.00, a =

beff = a x a (1 - a/lo) + b1

a =

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

TNHW SAI Consulting Engineers Pvt. Ltd.,

As per IRC:6, the impact factor of 25% shall be considered.

Intensity of loading = 75 x 1.25= 45.6142.418 x 0.85

Case -3 :Vehicle is placed concentric with the Slab pannelFor outer row tyresDistance between two tyres of the axle = 2.2 m (C/C of tyres)Spacing of Axle = 1.2 mLoad per axle = 75 kNLoad per tyre 37.5 kNWidth of tyre = 300 mmContact length of tyre = 150 mm

Aspct ratio = (b/lo) = 5.2952.6 (for continuous slab)

Effective width of the concentrated load =a = Distance of the center of gravity of the concentrated load from the nerear supportlo = Effective spanb1 = Breadth of the concentration area of load.(parallel to support)

For Tyres 1, 2, 3 and 4a = 0.25 mlo = 1.7 mb1 = 150 + (2 x 75 ) = 300 mm

2.6beff = 0.855 mThe effective width is not overlapping , hence the beff = 0.855 m

Width of tyre along the span direction =300 + (2 x ( 75 + 200 ) = 850 mm

(or) 0.85 mAs per IRC:6, the impact factor of 25% shall be considered.

Intensity of loading = 37.5 x 1.25= 64.50.855 x 0.85

For inner row tyresDistance between two tyres of the axle = 1 m (C/C of tyres)Spacing of Axle = 1.2 mLoad per axle = 125 kNLoad per tyre 62.5 kNWidth of tyre = 300 mmContact length of tyre = 150 mm

Aspct ratio = (b/lo) = 5.2952.6 (for continuous slab)

Effective width of the concentrated load =

kN/m2

As the aspect ratio is grater than 2.00, a =

beff = a x a (1 - a/lo) + b1

a =

kN/m2

As the aspect ratio is grater than 2.00, a =

beff = a x a (1 - a/lo) + b1

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

TNHW SAI Consulting Engineers Pvt. Ltd.,

a = Distance of the center of gravity of the concentrated load from the nerear supportlo = Effective spanb1 = Breadth of the concentration area of load.(parallel to support)

For Tyres 1, 2, 3 and 4a = 0.35 mlo = 1.7 mb1 = 150 + (2 x 75 ) = 300 mm

2.6beff = 1.023 mThe effective width is not overlapping , hence the beff = 1.023 m

Width of tyre along the span direction =300 + (2 x ( 75 + 200 ) = 850 mm

(or) 0.85 mAs per IRC:6, the impact factor of 25% shall be considered.

Intensity of loading = 37.5 x 1.25= 53.9081.023 x 0.85

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 kNLoad per tyre 350 kNWidth of tyre = 850 mmContact length of tyre = 3600 mm

Aspct ratio = (b/lo) = 5.2952.6 (for continuous slab)

Effective width of the concentrated load =a = Distance of the center of gravity of the concentrated load from the nerear supportlo = Effective spanb1 = 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 mlo = 1.7 mb1 = 3600 + (2 x 75 ) = 3750 mm

2.6beff = 4.266 mThe effective width is = 4.266 m

Width of tyre along the span direction =850 + (2 x ( 75 + 200 ) = 1400 mm

(or) 1.4 mAs per IRC:6, the impact factor of 25% shall be considered.

Intensity of loading = 350 x 1.25= 73.2544.266 x 1.4

For Track 2a = 0.07 mlo = 1.7 m

a =

kN/m2

As the aspect ratio is grater than 2.00, a =

beff = a x a (1 - a/lo) + b1

a =

kN/m2

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

TNHW SAI Consulting Engineers Pvt. Ltd.,

b1 = 3600 + (2 x 75 ) = 3750 mm2.6

beff = 3.925 mThe effective width is = 3.925 m

Width of tyre along the span direction =850 + (2 x ( 75 + 200 ) = 1400 mm

(or) 1.4 mAs per IRC:6, the impact factor of 25% shall be considered.

Intensity of loading = 350 x 1.25= 79.6183.925 x 1.4

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 kNLoad per tyre 350 kNWidth of tyre = 850 mmContact length of tyre = 3600 mm

Aspct ratio = (b/lo) = 5.2952.6 (for continuous slab)

Effective width of the concentrated load =a = Distance of the center of gravity of the concentrated load from the nerear supportlo = Effective spanb1 = Breadth of the concentration area of load.(parallel to support)

For Track 1 & 2a = 0.67 mlo = 1.7 mb1 = 3600 + (2 x 75 ) = 3750 mm

2.6beff = 4.806 mThe effective width is = 4.806 m

Width of tyre along the span direction =850 + (2 x ( 75 + 200 ) = 1400 mm

(or) 1.4 mAs per IRC:6, the impact factor of 25% shall be considered.

Intensity of loading = 350 x 1.25= 65.0234.806 x 1.4

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 kNLoad per tyre 350 kNWidth of tyre = 850 mmContact length of tyre = 3600 mm

a =

kN/m2

As the aspect ratio is grater than 2.00, a =

beff = a x a (1 - a/lo) + b1

a =

kN/m2

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

TNHW SAI Consulting Engineers Pvt. Ltd.,

Aspct ratio = (b/lo) = 5.2952.6 (for continuous slab)

Effective width of the concentrated load =a = Distance of the center of gravity of the concentrated load from the nerear supportlo = Effective spanb1 = Breadth of the concentration area of load.(parallel to support)

For Track 1 & 2a = 0.18 mlo = 1.7 mb1 = 3600 + (2 x 75 ) = 3750 mm

2.6beff = 4.169 mThe effective width is = 4.169 m

Width of tyre along the span direction =850 + (2 x ( 75 + 200 ) = 1400 mm

(or) 1.4 mAs per IRC:6, the impact factor of 25% shall be considered.

Intensity of loading = 350 x 1.25= 74.9594.169 x 1.4

3. 70R - Wheeled (Bogie) loading:Case -1: With minimum eccentricity:Distance between two tyres of the axle = 1.93 m (C/C of tyres)Spacing of Axle = 1.22 mLoad per axle = 200 kNLoad per tyre 100 kNWidth 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 (or) 0.5273Actual max. Load = 5 t (or) 50000 N

Contact width of tyre = 360 mmContact length of tyre = 264 mmContact length of tyre = 264 mm

Aspct ratio = (b/lo) = 5.2952.6 (for continuous slab)

Effective width of the concentrated load =a = Distance of the center of gravity of the concentrated load from the nerear supportlo = Effective spanb1 = Breadth of the concentration area of load.(parallel to support)

For Tyres 1, 4a = 0.42 mlo = 1.7 mb1 = 264 + (2 x 75 ) = 414 mm

2.6beff = 1.237 mThe effective width is overlapping, hence the beff = 2.457 m

As the aspect ratio is grater than 2.00, a =

beff = a x a (1 - a/lo) + b1

a =

kN/m2

kg/cm2 N/mm2

As the aspect ratio is grater than 2.00, a =

beff = a x a (1 - a/lo) + b1

a =

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

TNHW SAI Consulting Engineers Pvt. Ltd.,

Width of tyre along the span direction =860 + (2 x ( 75 + 200 ) = 1410 mm

(or) 1.41 mAs per IRC:6, the impact factor of 25% shall be considered.

Intensity of loading = 200 x 1.25= 72.1642.457 x 1.41

For Tyres 2, 3a = 0.19 mlo = 1.7 mb1 = 264 + (2 x 75 ) = 414 mm

2.6beff = 0.853 mThe effective width is overlapping, hence the beff = 0.853 m

Width of tyre along the span direction =860 + (2 x ( 75 + 200 ) = 1410 mm

(or) 1.41 mAs per IRC:6, the impact factor of 25% shall be considered.

Intensity of loading = 100 x 1.25= 103.9310.853 x 1.41

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 mLoad per axle = 200 kNLoad per tyre 100 kNWidth of tyre = 860 mmContact length of the tyre shall be arrived based on the max. tyre pressures specified in IRC:6-2000

Maximum tyre pressure = 5.273 (or) 0.5273Actual max. Load = 5 t (or) 50000 N

Contact width of tyre = 360 mmContact length of tyre = 264 mmContact length of tyre = 264 mm

Aspct ratio = (b/lo) = 5.2952.6 (for continuous slab)

Effective width of the concentrated load =a = Distance of the center of gravity of the concentrated load from the nerear supportlo = Effective spanb1 = Breadth of the concentration area of load.(parallel to support)

For Tyres 1, 2, 3, ,4a = 0.735 mlo = 1.7 mb1 = 264 + (2 x 75 ) = 414 mm

2.6beff = 1.499 mThe effective width is overlapping, hence the beff = 2.719 m

kN/m2

a =

kN/m2

kg/cm2 N/mm2

As the aspect ratio is grater than 2.00, a =

beff = a x a (1 - a/lo) + b1

a =

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

TNHW SAI Consulting Engineers Pvt. Ltd.,

Width of tyre along the span direction =860 + (2 x ( 75 + 200 ) = 1410 mm

(or) 1.41 mAs per IRC:6, the impact factor of 25% shall be considered.

Intensity of loading = 200 x 1.25= 65.212.719 x 1.41

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 mLoad per axle = 200 kNLoad per tyre 100 kNWidth of tyre = 860 mmContact length of the tyre shall be arrived based on the max. tyre pressures specified in IRC:6-2000

Maximum tyre pressure = 5.273 (or) 0.5273Actual max. Load = 5 t (or) 50000 N

Contact width of tyre = 360 mmContact length of tyre = 264 mmContact length of tyre = 264 mm

Aspct ratio = (b/lo) = 5.2952.6 (for continuous slab)

Effective width of the concentrated load =a = Distance of the center of gravity of the concentrated load from the nerear supportlo = Effective spanb1 = Breadth of the concentration area of load.(parallel to support)

For Tyres 1, 2, 3 & 4a = 0.115 mlo = 1.7 mb1 = 264 + (2 x 75 ) = 414 mm

2.6beff = 0.693 mThe effective width is not overlapping , hence the beff = 0.693 m

Width of tyre along the span direction =860 + (2 x ( 75 + 200 ) = 1410 mm

(or) 1.41 mAs per IRC:6, the impact factor of 25% shall be considered.

Intensity of loading = 100 x 1.25= 127.9260.693 x 1.41

4. Class - A (2 Nos.)Case -1: With minimum eccentricity:Distance between two tyres of the axle = 1.8 m (C/C of tyres)Spacing of Axle = 1.2 mLoad per axle = 114 kNLoad per tyre 57 kNWidth of tyre = 500 mmContact length of tyre = 250 mm

kN/m2

kg/cm2 N/mm2

As the aspect ratio is grater than 2.00, a =

beff = a x a (1 - a/lo) + b1

a =

kN/m2

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

TNHW SAI Consulting Engineers Pvt. Ltd.,

Aspct ratio = (b/lo) = 5.2952.6 (for continuous slab)

Effective width of the concentrated load =a = Distance of the center of gravity of the concentrated load from the nerear supportlo = Effective spanb1 = 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 ) = 400 mm

beff = 0.527 mThe effective width is overlapping, hence the beff = 0.527 m

Width of tyre along the span direction =500 + (2 x ( 75 + 200 ) = 1050 mm

(or) 1.05 mAs per IRC:6, the impact factor of 50% shall be considered.Intensity of loading = 57 x 1.5

= 154.5140.527 x 1.05

For Tyres 2 & 3 (First vehicle)a = 0.15 mlo = 1.7 mb1 = 250 + (2 x 75 ) = 400 mm

2.6beff = 0.756 mThe effective width is not overlapping , hence the beff = 0.756 m

Width of tyre along the span direction =500 + (2 x ( 75 + 200 ) = 1050 mm

(or) 1.05 mAs per IRC:6, the impact factor of 50% shall be considered.

Intensity of loading = 57 x 1.5= 107.710.756 x 1.05

For Tyres 1 & 4 (Second vehicle)a = 0.15 mlo = 1.7 mb1 = 250 + (2 x 75 ) = 400 mm

2.6beff = 0.756 mThe effective width is overlapping, hence the beff = 0.756 m

Width of tyre along the span direction =500 + (2 x ( 75 + 200 ) = 1050 mm

(or) 1.05 mAs per IRC:6, the impact factor of 50% shall be considered.

Intensity of loading = 57 x 1.5= 107.710.756 x 1.05

For Tyres 2 & 3 (Second vehicle)

As the aspect ratio is grater than 2.00, a =beff = a x a (1 - a/lo) + b1

kN/m2

a =

kN/m2

a =

kN/m2

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

TNHW SAI Consulting Engineers Pvt. Ltd.,

a = 0.25 mlo = 1.7 mb1 = 250 + (2 x 75 ) = 400 mm

2.6beff = 0.955 mThe effective width is not overlapping , hence the beff = 0.955 m

Width of tyre along the span direction =500 + (2 x ( 75 + 200 ) = 1050 mm

(or) 1.05 mAs per IRC:6, the impact factor of 50% shall be considered.

Intensity of loading = 57 x 1.5= 85.2660.955 x 1.05

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 mLoad per axle = 114 kNLoad per tyre 57 kNWidth of tyre = 500 mmContact length of tyre = 250 mm

Aspct ratio = (b/lo) = 5.2952.6 (for continuous slab)

Effective width of the concentrated load =a = Distance of the center of gravity of the concentrated load from the nerear supportlo = Effective spanb1 = Breadth of the concentration area of load.(parallel to support)

For Tyres 1 & 4 (First vehicle)a = 0.75 mlo = 1.7 mb1 = 250 + (2 x 75 ) = 400 mm

2.6beff = 1.49 mThe effective width is overlapping, hence the beff = 2.69 m

Width of tyre along the span direction =500 + (2 x ( 75 + 200 ) = 1050 mm

(or) 1.05 mAs per IRC:6, the impact factor of 50% shall be considered.

Intensity of loading = 114 x 1.5= 60.5422.69 x 1.05

For Tyres 2 & 3 (First vehicle)a = 0.85 mlo = 1.7 mb1 = 250 + (2 x 75 ) = 400 mm

2.6beff = 1.505 mThe effective width is overlapping, hence the beff = 2.705 m

a =

kN/m2

As the aspect ratio is grater than 2.00, a =beff = a x a (1 - a/lo) + b1

a =

kN/m2

a =

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

TNHW SAI Consulting Engineers Pvt. Ltd.,

Width of tyre along the span direction =500 + (2 x ( 75 + 200 ) = 1050 mm

(or) 1.05 mAs per IRC:6, the impact factor of 50% shall be considered.

Intensity of loading = 114 x 1.5= 60.2062.705 x 1.05

For Tyres 2 & 3 (Second vehicle)a = 0.85 mlo = 1.7 mb1 = 250 + (2 x 75 ) = 400 mm

2.6beff = 1.505 mThe effective width is overlapping, hence the beff = 2.705 m

Width of tyre along the span direction =500 + (2 x ( 75 + 200 ) = 1050 mm

(or) 1.05 mAs per IRC:6, the impact factor of 50% shall be considered.

Intensity of loading = 114 x 1.5= 60.2062.705 x 1.05

For Tyres 1 & 4 (Second vehicle)a = 0.75 mlo = 1.7 mb1 = 250 + (2 x 75 ) = 400 mm

2.6beff = 1.49 mThe effective width is overlapping, hence the beff = 2.69 m

Width of tyre along the span direction =500 + (2 x ( 75 + 200 ) = 1050 mm

(or) 1.05 mAs per IRC:6, the impact factor of 50% shall be considered.

Intensity of loading = 114 x 1.5= 60.5422.69 x 1.05

1 2 3 4 5 6 7

0.85 1.7 1.7 1.7 1.7 0.85

Descritisation Diagram

1 3 5 6 8 10 11 132 4 7 9 12 14

kN/m2

a =

kN/m2

a =

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.

1 432

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

TNHW SAI Consulting Engineers Pvt. Ltd.,

Ctitical sections for design

Summary of moments in various members

Load descriptionSection Numbers

1 2 3 4 5 6 7Self-weight 0.83 2.43 1.26 0.7 -0.38 0.18 0.6SDL 1.22 3.38 4.01 3.38 1.31 -0.21 -0.52Class-AA (case-1) 0 0 -0.47 -1.04 -2.37 10.96 20.45Class-AA (case-2) 0 0 0.39 0.85 2.64 4.42 4.89Class-AA (case-3) 0 0 -0.38 -0.83 -2.55 1.53 5.93Class-AA -T(case-1) 0 0 -2.37 -5.23 -13.26 -0.29 7.1Class-AA -T(case-2) 0 0 0.49 1.07 3.3 5.53 6.11Class-AA -T(case-3) 0 0 -1.29 -2.85 -8.76 -1.04 5.0770R-Bogie(case-1) 0 0 -2.1 -4.62 -11.54 2.13 9.6570R-Bogie(case-2) 0 0 0.49 1.09 3.35 5.62 6.2170R-Bogie(case-3) 0 0 -1.73 -3.82 -11.79 -0.79 9.05Class-A (case-1) 3.1 8.14 8.42 0.78 -2.08 2.53 6.43Class-A (case-2) 0 0 -2.39 -5.27 -10.8 2.97 9.58Max. Hogging BM 5.15 13.95 13.69 4.93 4.28 10.93 20.53Max. Sagging BM 0 0 0 -1.19 -12.33 -1.07 0

Load descriptionSection Numbers

8 9 10 11 12Self-weight 0.56 0.1 -0.64 0.26 0.77SDL -0.79 -0.8 -0.49 0.36 0.67Class-AA (case-1) 17.31 3.75 -29.15 -7.13 8.34Class-AA (case-2) 1.56 -2.56 -8.79 4.39 10.37Class-AA (case-3) 5.6 1.09 -5.09 0.87 5.32Class-AA -T(case-1) 10..66 7.55 -1.37 -0.56 3.93Class-AA -T(case-2) 1.27 -3.85 -9.82 5.57 11.91Class-AA -T(case-3) 7.06 3.35 0.71 2.55 6.0570R-Bogie(case-1) 11.95 7.22 -7.26 -2 5.0570R-Bogie(case-2) 1.32 -3.85 -10 5.39 12.3870R-Bogie(case-3) 11.54 4.7 -1.32 3.14 10.15Class-A (case-1) 4.53 -1.67 -4.53 1.74 6.07Class-A (case-2) 10.81 5.61 -5.97 2.19 6.49Max. Hogging BM 17.08 6.85 -0.42 6.01 13.35Max. Sagging BM 0 -4.55 -30.28 -6.51 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 mmOverall depth provided = 200 mmEffective depth available = 200 - 50 - 5

= 145 mm > 114.294 mmHence, the section adopted is ADEQUATE

1 4

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

TNHW SAI Consulting Engineers Pvt. Ltd.,

Design of reinforcement :For hogging moment:(at the face of the web)

Design hogging moment = 20.53 kNmOverall depth provided = 375 mmEffective depth availbale = 375 - 40 - 5

= 330 mm

= 358.779Hence, provide Y 10 bars @ 218 mm C/CMax. hogging moment at other locations = 10.93 kNm

= 409.309Hence, provide Y 10 bars @ 191 mm C/C

However provide @ 190 mm C/C

Area of steel provided = 413

For sagging moment:Design sagging moment = 30.28 kNm

E. Depth provided = 200 - 40 - 6= 154 mm C/C

= 1133.93Hence, provide Y 12 bars @ 99 mm C/CHence, provide Y 12 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 < 409.309Hence, OKAY

For distribution Reinforcement::Design moment = 9.084 kNm

(0.3 times the maximum BM considered)

= 361.294Hence, provide Y 10 bars @ 217 mm C/CHence, provide Y 8 bars @ 150 mm C/C

Area of steel required = BM/sst x j x d

mm2/m

Area of steel required = BM/sst x j x d

mm2/m

mm2/m

Area of steel required = BM/sst x j x d

mm2/m

mm2/m mm2/m

Area of steel required = BM/sst x j x d

mm2/m

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