DESIGN OF DOUBLE LANE ROAD BRIDGE ACROSS SURPLUS
of #REF!
I.a. HYDRAULIC PARTICULARS
1 Discharge ( R ) #REF! cumecs
2 Discharge( D ) #REF! cumecs
3 Normal Bed width 13.200 m
Extended bed width 13.200 m
4 FSD #REF! m
5 Free board #REF! m
6 Side slopes #REF! ###
7 Value of 'n' #REF!
8 Velocity #REF! m/sec
9 Bed fall #REF! #REF!
10 CBL + #REF! m
11 FSL + #REF! m
12 TBL + 242.740 m
13 Ground level / + #REF! m
Existing Road level
14 Proposed road level + #REF! m
15 skew angle #REF! Degrees
Loss of Head allowed 0.05 m
II.b ROAD PARTICULARS
3 Carriage way width propsed for the bridge = #REF! ( in normal direction)
II. VENT WAY :
Vertical clearance provided = #REF! m,
Width of canal at FSL + ### m i.e at level #REF! m
= 13.200 + #REF! = #REF! m
Provide #REF! vents of ### m clear span
Top width of pier = #REF! m
Width between the abutments at bed block level ( In normal direction)
= #REF! x #REF! + #REF! x #REF! = #REF! m
III HEIGHT OF PIER :
Bottom of bed block = #REF! - #REF! = #REF! m
Depth of top of foundation 1st footing below BL = #REF! m
Height of pier from bed block = #REF! - #REF!
= #REF! m
IV TOP WIDTH OF PIER AND ABUTMENT :
The top widths adopted for pier, abutment are as follows :
Pier = #REF! m
Abutment = #REF! m
V. SCOUR DEPTH :
Scour depth, R = 1.35 x( q^2 / f )^1/3
q = #REF! / 13.200 = #REF! cumecs/ m width
f = 2.00 ( For ALL SOILS )
R= 1.35 x( #REF! x #REF! / 2 )^1/3
= #REF! m
Max. scour depth = 2R = #REF! m
Max. scour level = FSL - #REF! m
= #REF! - #REF! = #REF! m
However the foundations of piers are taken to a level of + #REF! m
VI SUPER STRUCTURE
MOST Specifications are adopted for super structure and the thickness of slab with
varying thickness from #REF! to #REF! m is proposed in accordance with the
MOST Dwaing No. SD/111
VII SUB-STRUCTURE
The returns and abutments are proposed in V.C.C. M15 and piers in RCC M25 For computing earth pressure due
to earth fill, in case of abutments and returns, T.V.A. procedure is adopted, taking the angle of
internal friction as 32 degrees. While designing the pier breaking force, Live load eccentricity,
Fricitional resistance to bearings , water currents effect and wind force etc., are considered.
VII. FOUNDATIONS
The foundations for returns and abutments are proposed in VCC M15 grade and for pier RCC M25
SCOUR DEPTH CALCULATIONS
FOR CANAL:-
Discharge(in Cumecs), Q = 54.000 m3/sec
Bed Width(in metres), b = 15.000 m
Silt Factor, f = 4.750
F.S.L = 521.575
Scour Depth multiplication factor, x = 2.000
Discharge / metre width(in Cumecs/m), q = Q / b = 3.600 m3/sec/m
Normal scour depth R = 1.34(q 2 / f )1/3 = 1.872 m
Max scour depth = x. R = 3.745 m
Max scour level = F.S.L - scour depth = 517.830
DESIGN OF BED BLOCK OVER SLRB PIER
Vide IRC:78-1983 Cl.No.716.2.1
Reinforcement = 1% of Cross sectional Area
= 1% x30x100
30
= 30 cm 2
100
Steel at top & Bottom 15 cm 2
Ast. Required :
Assuming Dia of steel Rod = 16 mm
Area of steel Rod = 2.011 cm2
No.of 16mm bars required 7.457386
say 10 Nos.
Provide 16 mm ᴓ bars 5 Nos. at top & 5 Nos. at Bottom
20.11429 > 15
Assuming Dia of steel Rod = 12 mm
Area of steel Rod = 1.131 cm2
Steel required in the transverse direction = 15
Volume of steel/Rmt. = 1500 cm3
Lenth of each Stirrup = 2*(100-2*2.5+30-2*2.5)
Provide 12 mm 2*155+2*25 = 240 cm
Volume of each stirrup = 271.54286 cm3
No.of stirrups required = 5.5239899 Nos.
Spacing of stirrups = 181.02857 mm
say 180 mm
Provide 12mm ᴓ bars at 180 mm c/c
DESIGN OF DIRT WALL OVER DLRB ABUTMENT
Reinforcement = 1% of Cross sectional Area
= 1%x500x1000
1% 500 300
= 1500 mm2
On each side 750 mm2 50.00
30.000
Ast. Required :
Assuming Dia of steel Rod = 16 mm 100.00 c.m
Area of steel Rod = 201.143 mm2
No.of 12mm bars required 3.728693
say 6 Nos.
Provide 16 mm ᴓ bars 3 on each face
1206.857 > 750 mm2
Steel required in the transvers direction = 7.5 cm2
Provide 12mm ᴓ bars 2 legged stirrups
Volume of steel/Rmt. = 750 cm3 - per metre
Lenth of each Stirrup = 2*(30-2*2.5+50-2*2.5) = 140 cm
Provide 12 mm
Area of steel Rod = 1.131 cm2
Provide 12 mmᴓ stirrup = 158.4 cm3
No.of stirrups required = 4.7348485 say 7
Spacing of stirrups = 142.85714 mm
say 140 mm
Provide 12mm ᴓ bars at 140 mm c/c
DESIGN OF RETURN WALL
Width of w1 = 0.55 m 550 + #REF!
Width of w3 = 0.55 m
Width of w2 = 1.000 m 600
Width of w4 = 3.450 m
Width of w5 = 3.450 m W1 + #REF!
Width of W6 = 0.40 m
Thickness of foundation = 0.600 m
Found., top level = #REF!
Found., bottom level = #REF! W6
Height of Parapet wall = 0.60 m #REF! W5 W3
Density of RCC = 2.5 t/m3
Density of Concrete : 2.4 t/m3 W4 W2
Density of Soil = 2.1 t/m3 A
Live load = 1.2 m 400 3450 550 1000 400 #REF!
Surcharge 600 W7
Stress on Max Min B #REF!
concrete #REF! #REF! 5000
soil #REF! #REF! 5800
STRESSES ON CONCRETE : Taking moments about " A "
Load Description Load-t L.A -m Moment-tm
W1 0.600 x 0.550 x 2.400 0.792 3.725 2.950
W2 0.50 x 1.000 x #REF! x 2.400 #REF! 4.333 #REF!
W3 0.550 x #REF! x 2.400 #REF! 3.725 #REF!
W4 0.50 x 3.450 x #REF! x 2.400 #REF! 2.300 #REF!
W5 0.50 x 3.450 x #REF! x 2.100 #REF! 1.150 #REF!
Pv 0.0395 x ( #REF! ² - 1.200 ² ) x 2.100 #REF! 0.000 #REF!
Total Load #REF!
Ph 0.1580 x ( #REF! ² - 1.200 ² ) x 2.100 #REF! #REF! #REF!
Total Moment #REF!
Leverarm = #REF! / #REF! = #REF! m
Eccentricity = 2.500 - #REF! = #REF! m ###
MAX. STRESS = #REF! ( 1.000 + 6 x #REF! ) = #REF! t/m²
5.00 5.000
MIN STRESS = #REF! ( 1.000 - 6 x #REF! ) = #REF! t/m²
5.00 5.000
CALCULATION OF STRESSES ON SOIL
Taking moments about 'B'
Load Description Load-t L.A -m Moment-tm
W1 0.600 x 0.550 x 2.400 0.792 4.125 3.267
W2 0.5 x 1.000 x #REF! x 2.400 #REF! 4.733 #REF!
W3 0.550 x #REF! x 2.400 #REF! 4.125 #REF!
W4 0.5 x 3.450 x #REF! x 2.400 #REF! 2.700 #REF!
W5 0.5 x 3.450 x #REF! x 2.100 #REF! 1.550 #REF!
W6 0.400 x #REF! x 2.100 #REF! 0.200 #REF!
W7 5.800 x 0.600 x 2.400 8.352 2.900 24.221
Pv 0.0395 x ( #REF! ² - 1.200 ² ) x 2.100 #REF! 0.000 #REF!
Total Load #REF!
Ph 0.1580 x ( #REF! ² - 1.200 ² ) x 2.100 #REF! #REF! #REF!
Total Moment #REF!
Leverarm = #REF! / #REF! = #REF! m
Eccentricity = 2.900 - #REF! = #REF! m #REF!
MAX. STRESS = #REF! ( 1.000 + 6 x #REF! ) = #REF! t/m²
5.80 5.800
MIN STRESS = #REF! ( 1.000 - 6 x #REF! ) = #REF! t/m²
5.80 5.800
HNSS-PK-31-MC-DR-DLRB@KM-134.402 19
Design of Bed block
Vide IRC:78-1983 Cl.No.716.2.1, 100 cm
Reinforcement = 1 % of cross sectional Area
30cm
= 1% x 30.0 x 100.0
= 30.00 cm2
Say 30.0 cm2
Steel at top & Bottom = 30.0 = 15.0 cm2
i.e Long bars 2
.'. No. of Tor 16 mm bars req. = 15 = 7.460
2.011
Say 10 No.s
\ provide 16 mm f bars 5 No.s at top & 5 No.s at bottom
Steel in transverse direction = 15.0 cm2
Volume of steel / Rmt = 15 x 100 = 1500 cm3
Length of each Stirrup = 2 x 95 + 2 x 25 = 240 cm
Providing 12 mm dia bars
Volume of each stirrup = 240 x 1.131 = 271.4 cm3
No.of Stirrups = 1500 = 5.526 No.s
271.434
Spacing of Stirrups = 1000 = 180.96 mm
5.5 Say 180.00 mm c/c
\ provide 12 mm f bars at 180.00 mm c/c
DESIGN OF ABUTMENT
Live load surcharge = 1.2 m
VR
+ #REF! + #REF!
+ #REF! #REF!
+ #REF! W8 W6 #REF!
+ #REF!
Top width 1.20 m W1 + #REF! #REF!
Rear side batter 2.50 m
Front side batter 0.00 m W7
Offset on earth side 0.60 m
Offset on front side 0.60 m #REF!
Depth of foundation 0.60 W4 #REF!
#REF! m W5
Friction Coef. = 0.500
W9 W3 W2
0
600 2500 500 500 600
+ #REF! A
#REF!
+ #REF! W10 B
3500
4700
HYDRAULIC PARTICULARS OF Drain:
Max flood Discharge = 24.283 cu.m
Bed width = 15.200 m
Full Supply Depth = 2.230 m
Side slope = 1.5 : 1
Surface fall = 1 in 7500 Density of soil : 2.10
Coeff. Of rugosity = 0.025 Density of concrete : 2.40
Velocity = 0.666 m/sec Density of RCC : 2.50
Bed level = + 357.720 m Angle of Ǿ = 28
Full supply level = + 359.950 m
Top of bank level = + 360.700 m Hor.coefft of soil : 0.1584
Ground level = + 364.047 m Ver.coefft of soil : 0.0395
Proposed road level + 364.047 m
Skew angle = 23 deg.
Type of Bridge = 2 lane
1) REACTION ON ABUTMENT;
Dead loads : Total width of Roadway = 12.000 m
Skew factor = Sec ( 22.5 ) = = 1.082
Width of slab at bottom = 10.900 m
Top width of return wall = 0.500 m
Total Length of abutment = 13.633 m
Eeective Length of abutment = 13.633 m
Effective span 8.9 x 1.082 = 9.630 m
Dead Load on abutment as per M.O.T drawings = #REF! t
Load on abutment = #REF! = #REF! t
Load coming on to abutment per m length = #REF! / #REF! = #REF! t/ m
Live load : For class A train of vehicles
Maximum of Max. LL::
Live load maximum = 63.750 t ( From Live load calculations)
Impact factor = 4.5 = 4.5 = 0.288
6+L 6+ 9.630
As the impact factor reduces to zero at 3.00 m below bed block and 0.5 times the
factor at bottom of bed block, the above factor comes down to
= 0.288 x #REF! = #REF! on top of concrete
2 x 3.00
Live load including impact per m length = Double lane
= 63.750 x #REF! = #REF! t/m
12.00
Total load coming on abutment = #REF! + #REF! = #REF! t/m
For skew increase this load as follows ( as per page 106 of Jhonson Victor, Bridge design)
Skew angle in degrees load increase in percentage
0 - 20 0 - 50
20 - 50 50 - 90
for skew of 23 degrees percentage load increase = 53.33333 %
Load on abutment = #REF! x 1.53333 = #REF! t/m
2) Force due to braking effect ;
20 % of max train load on span
The Braking force for this loading is the maximum that is coming on the span
I.e., 92.00 t as shown in sketch in case- I while calculating Live load reactions
Braking force = ( 92 ) 20 / 100 = 18.400 t
Horizontal force normal to abutment = 18.4 x cos 22.5 = 16.998 t
acting at 1.20 m above road level I.e.,
Force per running m = 16.998 / 12.00 = 1.417 t/m
2) Force due to temparature stresses ;
No.of bearings on each side 3
Bearing size = 560 x 220 x 61 mm
Total force on abutment per bearing= 10.7148 t From Pier Design
Total force on abutment per metre run= 10.715 / 12.000 = 0.893 t/m
STRESS ON CONCRETE: ( Taking moments about 'A' )
S.NO Description Load LA Moment
in Tons in mts in T.m
1 VR : Vertical loads #REF! 0.250 #REF!
2 W1 : 0.500 x #REF! x 2.500 #REF! 0.250 #REF!
3 W2 : 0.500 x 0.000 x #REF! x 2.400 #REF! 0.000 #REF!
4 W3 : 0.500 x #REF! x 2.400 #REF! 0.250 #REF!
5 W4 : 0.500 x #REF! x 2.400 #REF! 0.750 #REF!
6 W5 : 0.500 x 2.500 x #REF! x 2.400 #REF! 1.833 #REF!
7 W6 : 2.500 x #REF! x 2.400 #REF! 2.250 #REF!
8 W7 : 0.500 x 2.500 x #REF! x 2.100 #REF! 2.667 #REF!
9 WL : 1.000 x 3.000 x 1.200 x 2.100 7.560 2.000 15.120
10 Pv : 0.0395 x ( #REF! 2
- 1.200 2
)x 2.100 #REF! 3.500 #REF!
V= #REF! #REF!
11 Ph : 0.1584 x ( #REF! 2
- 1.200 2
)x 2.100 #REF! #REF! (-) #REF!
12 Braking force 1.417 #REF! (-) #REF!
Frictional force 0.893 #REF! (-) #REF!
#REF! M= #REF!
Lever arm = M/V = #REF! / #REF! = #REF! m
Eccentricity = #REF! - 3.500 /2 = #REF! m
B/6 = 3.50 /6 = 0.583 ### #REF! m #REF!
Max. Stress = ( #REF! / 3.500 ) x { 1 + ( 6 x #REF! ) / 3.500 }
= #REF! T/ Sqm #REF!
Min. Stress = ( #REF! / 3.500 ) x { 1 - ( 6 x #REF! ) / 3.500 }
= #REF! T/ Sqm #REF!
STRESS ON SOIL: ( Taking moments about 'B' )
S.NO Description Load LA Moment
in Tons in mts in T.m
1 VR : Vertical loads #REF! 0.850 #REF!
2 W1 : As calculated above #REF! 0.850 #REF!
3 W2 : As calculated above #REF! 0.600 #REF!
4 W3 : As calculated above #REF! 0.850 #REF!
5 W4 : As calculated above #REF! 1.350 #REF!
6 W5 : As calculated above #REF! 2.433 #REF!
7 W6 : As calculated above #REF! 2.850 #REF!
8 W7 : As calculated above #REF! 3.267 #REF!
3 W8 : 1.000 x 0.600 x #REF! x 2.500 #REF! 4.400 #REF!
4 W9 : 1.000 x 0.600 x #REF! x 2.100 #REF! 4.400 #REF!
5 W10 : 1.000 x 4.700 x #REF! x 2.400 #REF! 2.350 #REF!
9 WL : 1.000 x 3.600 x 1.200 x 2.100 9.072 2.900 26.309
6 Pv : 0.0395 x ( #REF! 2
- 1.200 2
)x 2.100 #REF! 4.700 #REF!
V= #REF! #REF!
7 Ph : 0.1584 x ( #REF! 2
- 1.200 2
)x 2.100 #REF! #REF! (-) #REF!
8 Braking force 1.417 #REF! (-) #REF!
Frictional force 0.893 #REF! (-) #REF!
#REF! #REF!
M= #REF!
Lever arm = M/V = #REF! / #REF! = #REF! m
Eccentricity = #REF! - 4.700 /2 = #REF! m
B/6 = 4.700 / 6 = 0.783 ### #REF! m #REF!
Max. Stress = ( #REF! / 4.700 )x{1+(6x #REF! ) / 4.700 }
= #REF! T/ Sqm #REF!
Min. Stress = ( #REF! / 4.700 ) x { 1 - ( 6 x #REF! ) / 4.700 }
= #REF! T/ Sqm #REF!
OVER TURNING MOMENT : #REF! T-M
CHECK FOR OVER TURNING : #REF! = #REF! #REF! 2 (IRC:78-2000.cl.706.3.4)
#REF!
CHECK FOR SLIDING : 0.60 X #REF! = #REF! #REF! 1.5 (IRC:78-2000.cl.706.3.4)
#REF!
max min
on concrete #REF! #REF!
on soil #REF! #REF!
mm
mm
mm
mm
mm
t/m^3
t/m^3
t/m^3
º
