Design of Abutment Bridge

8/3/2019 Design of Abutment Bridge

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Substructure Kondre (r

Dimension of slab and beams (brought from slab sheet) Check for correctness of data

Length of single span = 35.000 m

Carriageway width, Bc = 7.500 m

Overall width of Bridge, Bo = 10.700 m

No. of main girder = 5

Dimensions of the longitudinal girder is taken as follows:

Length (Lsi

) = 35.000 m

Breadth (B si ) = 0.550 m

Thickness(Tsi) = 1.680 m

Distance betn girders = 2.375 m

No. of cross beams = 15

For the design purposes the dimensions of the cross beam is taken as follows:

Length (L si ) = 2.375 m



Distance betn cross beams = 2.500 m

For the design purposes the dimensions of the internal slab is taken as follows:

Length (L si ) = 2.500 m Hence, effec.span in long.dir. = 2.360 m

Breadth (B si ) = 2.375 m Hence, effec.span in transv.dir. = 1.825 mThickness(Tsi) = 0.230 m

Avg. pavement cover over Slab(Tsp ) = 0.075 m

kerb/footpath

breadth = 0.225 m at bottom and 0.225 m at top

height = 1.000 m

For design purpose dimension of cantilever slab is taken as


Breadth (B si ) = 0.600 m from c/c of outer girder

Breadth (B si ) = 0.325 m from outer face of outer main girder

Breadth of carriageway = 1.000 m from c/c of outer girder

1.016 m from outer face of outer main girder

Thickness(Tsi) = 0.230 m at the endThickness(Tsi) = 0.230 m at outer face of main girder


2.2 DESIGN OF RIGHT ABUTMENT

The freebody diagram of the R.C.C Abutment is shown below:

i = 0 0.200

0.40 CG of Stem from inner vertical face:

A 1= 3.324

0.23 A2= 2.674

M.A.= 1.327

A1+A2 = 5.998

0.600 CG = 0.221

b = 90

Dimensions taken for the design of abutment:

Retaining wall Bottom level of base = 0.000

Length = 9.500 m Road level = 5.000 ######

Breadth = 0.600 m at bottom 0.23 m at top 1.25

Height = 14.454 m CG of retain. wall from point A = 0.221 m

Ballast wall

Length = 9.500 m

Breadth = 0.400 m

Breadth of ret. wall = 0.200 m

Height = 1.000 m height of superstr. plus bearing

Return wall without pav.cover

No. = 2

Breadth = 0.200 m Height of Return wall H1= 0.7

14.4

54

Pa

A

d


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Substructure Kondre (rig

Length = 6.000 m Height of Return wall H2= 0.4

Total height of abutment = 15.454 m

Approach slab

Length = 5.000 m 16.354

Breadth = 3.000 m -11.354

Thickness = 0.200 m

Foundation is open foundation

Length = 6.000 m

Width 2.750 m Heel 1.000 m 0.750Toe 1.000 m 0.750

Height of footing = 0.900 1.250

Volume = 14.850 m3 W = 356.40 KN

2.2.1 CALCULATION OF LOAD

STRESS DUE TO DEAD LOAD

S.N Elements Unit No. L(m) B(m) H(m) Volume Unit Wt Total e from

Dead Load of Superstructure

1 Hand rails 6.368 rm 4 35.60 142.40 0.06 8.54

2 Railing posts cum 18 0.15 0.15 1.00 0.41 24 9.72

3 Wearing course cum 1 35.60 7.50 0.08 20.03 22 440.55

4 Kerbs cum 2 35.60 0.23 1.00 16.02 24 384.48

5 Deck slab i) rectangle cum 1 35.60 2.93 0.23 23.95 24 574.80

ii)trapezoid cum 2 35.60 0.33 0.23 5.32 24 127.73

6 Main Girder i) fillets cum 8 35.60 0.30 0.15 6.41 24 153.79

ii) webs cum 5 35.60 0.60 1.45 154.86 24 ######

7 Internal Cross Girder i) fillets cum 0 1.83 0.15 0.15 0.00 24 0.00

ii) webs cum 0 1.83 0.14 0.07 0.00 24 0.00

8 End Cross Girder i) fillets cum 2 1.83 0.15 0.15 0.04 24 0.99

ii) webs cum 2 1.83 0.14 0.07 0.04 24 0.86

Total Dead load of the superstructure #####

Total DL coming to one abutment = 2709 KN

Dead Load of Abutment

retaining wall 9.500 0.415 14.454 56.98 24 1367.64 0.000 m

return walls 2 6.000 0.200 0.55 1.32 24 31.68 2.208 m

ballast wall 9.500 0.400 1.00 3.80 24 91.20 0.007 m

1490.52

STRESS DUE TO LIVE LOAD

Live load due to class A loading:

Maximum load on the given abutment is produced when the train of load is positioned as shown in the figure:

Axle Load,KN 27 27 114 114 68 68 68 68

x i , m 35.0 33.8 29.5 26.5 23.5 20.5

y i 1.0 1.0 0.8 0.8 0.7 0.6

load, KN 114.0 110.1 57.3 51.5 45.7 39.8

Maximum total live load = 418.4 KN

Load due to impact = 418.37714 x 0.110 = 45.919 KN

Live load including impact = 418.4 + 45.919 = 464.3 KN

Live load due to class AA wheeled loading:

Axle Load,KN 200 200

x i , m 19.0 17.8

y i 0.54 0.51

load, KN 108.6 101.71




Live load due to class AA tracked loading:


Load due to impact = 664 x 0.1 = 66.4 KN



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Out of above three loadings class AA tracked loading gives maximum load to abutment.

Maximum live load to abutment = 730.4 KN

CALCULATIONOF EARTH PRESSURE FOR RETAINING WALL

Horizontal earth pressure due to backfil

Corresponding to F = 30 of the backfill material

g = 18 KN/m3 g ' = 8 KN/m3

Ka = 0.2973 ( Coulomb's theory applicable )

Ep1 = Ko x g h2

/2 = 639.1 KN/m h = 15.454 m

Horizontal component of earth pressure = 639.1 cos 20 = 600.5 KN/m

which acts at 0.42 x h = 6.49 m from the top of footing base.

Vertical component of earth pressure = 639.1 x sin 20 = 218.570 KN/m

which acts at 0.221 m inward from the CG of wall.

Total Horizontal Thrust, E1 = Ep1 x L = 5704.9 KN

Total Vertical Thrust, E2= Ep2 x L = 2076.4 KN

WEIGHT OF BACKFILL BEHIND THE ABUTMENT

Wb = 9.1 x 1.000 x 15.5 x 18 = 2531.37 KN

STRESS DUE TO SEISMIC FORCES

Horizontal Forces due to seismicity are computed using the formula as shown in IS Code 1893 or IRC - 6

F eq = a x b x l x G

where a = 0.08 for the zone V of seismic intensityb = 1.2 for medium soil with isolated footingl = 1 importance factor

Thus seismic coefficient is taken as 0.096

a. Seismic force due to Superstructure = 0.096 x 2709.05 = 260.1 KN

height of action of the force = 14.45 + 0.025 + 1.68 / 2 =

= 15.319 m

b. Seismic force due to Abutment = 0.096 x 1491 = 143.1 KN

act at cg of abutment = 7.727 m

c. Seismic force due to Back fill = 0.096 x 2531.4 = 243.01 KN

acts at cg = 7.727 m

STRESS DUE TO WATER CURRENT

Horizontal Thrust due to water current is ignored, because it usually contributes to the stability

of the wall and the load case with seismic condition does not concide with HFL.

STRESS DUE TO LONGITUDINAL FORCES

Due to Tractive Effort or Braking Force

Braking Force due to Class AA tracked load = @20% of live load/2 = 1/2 to one abutment

= 55.4 KN

height of the line of action of the brake from the road level = 1.2 m

height of the bearing level from the top of base = 14.454 m

Due to Resistance in Bearing (Temperature)

The critical case will be when seismic force acts in the direction of traffic. But according to IRC 6-1966 (222.7) "The seismic force due to live loa

shall not be considered when acting in the direction of traffic,..".

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Severe moment is produced when the LL is within span. Assume frictional coeff. = 0.225 in free bearing

= 0.2 in other bearing

Frictional resistance at free bearing = 773.9 KN

Frictional resistance of fixed bearing = 687.9 KN

Unbalanced frictional force = 86.0 KN

Line of action of the force from the bottom of the abutment = 14.454 m

STRESS DUE TO WIND LOAD

Exposed height of structure = depth of the beams + thickness of slab +

+ height of kerb =

= 2.68 m

Exposed area contributing to wind pressure = 93.8 m

Avg. height of superstructure from the bed = 6.000 m

intensity of wind load = 0.73 KN/m

velocity of wind = 107 km/hr

The design is to be conducted for one span only loaded with class A train of vehicle

Length of Class A train = 20.4 m

a. Wind Force on Surface (SS) = 0.73 x 93.8 = 68.474 KN

b. Wind Force on Moving

vehicle Class A train = 20.4 x 3 = 61.2 KN

acts at 1.5 m above road level = -9.854 m

Maximum load is obtained in exposed area = 129.7 KN

For lateral direction

wind load = 129.7 KN

acts at = 15.454 m

SUMMARY OF LOADS FOR DESIGN OF ABUTMENTS

S.No. Load Vertical (KN) Horizontal (KN) Arm Moment (KN

Longit. dir Later. (m) Longit. Later.

1 Superstructure

1.1 Dead load of SS(D) 2709.1 0.48 1297.01.2 Live Load of SS (L) including impact 438.0 0.48 209.7

Substructure2 DL of abutment(E)

Retaining wall 1367.6 0.00

Return wall - Top section 31.7 -2.21 -69.9

Ballast wall 91.2 -0.01 -0.7

Backfill (BF) 2531 0.00 0.0

Earth pressure (EP) 35062.779

(EP1) 5704.9 6.49 ######

Due to overburden(EP2)

Frontal passive soil mass (EPp)

3 Buoyancy (B)

4 Longitudinal force4.1 Tractive force (T) 55.40 14.45 800.8

4.2 Due to temperature (F) 85.99 14.45 1242.8

4.3 Wind Load 129.67 15.45 2003.

4.4 Centrifugal force (CF)

4.5 Thrust due to water current (WC)

4.6 Seismic forces (FS)

on superstructure 130.03 260.07 260.07 15.32 3984.0 3983.

on abutment 71.54 143.09 143.09 7.73 1105.7 1105.

on live load 42.05 16.95 712.8

from backfill 121.51 243.01 243.011 7.727 1877.7 1877.

Load combination Group II (Normal+Temp. condition)

4638 5846 129.7 ##### 200

D + L + E + Ep1+ Ep2 + Ep3+ T + F + CF

Taking

Rankin's

earth

pressure

coefficient.


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Substructure Kondre (righ

Load combination Group IV (for Seismic dry condition)

D + L + E + Ep1+ Ep2 + T + F + CF + FS 7492 6492 688.2 ##### 7680

Note: Moment is taken at the critical section at the bottom of the ABUTMENT

2.2.2 STRUCTURAL DESIGN OF ABUTMENT RETAINING WALL

Design of the returning wall will be done based on normal loading condition, but checked on maximum loading

Design parameters

Vertical force = 4638 KNLong. Moment = 40508 KN-m

Substructures are designed using M20 concrete and Fe415 steel reinforcement

Parameters for RCC design

Conccrete M20 sck = 20 N/mm2

Steel sy = 415 N/mm2

(Under combination group IV the permissible increase of stress is

50%, hence Combination I governs the design longitudinal

moment)

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Substructure Kondre (right

Approximate Area of Steel reinforcement required is calculated considering singly reinforced beam theory

Ast = Mu/ (0.87xy x 0.7984 x d) = 234060 mm2

Provide 25 mm dia. bars Ast = 490.625 mm2

Required spacing = 0.9 mm c/c Provide at 100 mm c/c Decrease spacing !

Area of reinforcement provided, Ast = 45873.438 mm2

r = 0.8047971 % > 0.2 %

At the frontal face, the design moment is greatly reduced because of lack of earth pressure.

Therefore on this side provide only 50 % of the main reinforcement

Ast = 234060 x0.5 = 117030 mm2

Provide 20 mm dia. bars Ast = 314 mm2

Required spacing = 20 1.2 mm c/c Provide at 100 mm c/c Decrease spec

Area of reinforcement provided, Ast = 29359 mm2

Same spacing of reinforcement is provided on the side faces of the abutment.

Distribution re-bar

Providing 0.12% of cross sectional area = 0.0012 x 600 x 1000= 720 mm2


Required spacing = 314.0 mm c/c Provide at 250 mm c/c at both faces. ok

Check for stresses (For seismic condition) Parameters for RCC design


Normal Thrust , W = 7492018 N Steel sy = 415 N/mm2

Bending Moment = ######### N-mm

Ecentricity, e = 6336.850 mm > 128.125 mm

breadth, b = 9500 mm m = 0.0

Total depth , D = 600 mm

Effective depth, d = 512.5 mm

Compressive steel cover, dc = 75 mm Ac = 29359 mm2

Tensile steel cover, dt = 75 mm At = 45873.44 mm2

Ast = Mu/ (0.87xy x 0.7984 x d) 178050.265 mm2/m

x= 0.87*y*Ast 8928 mm 150 mm say

0.36*ck*b

Check for shear stress:

Critical secion for shear in case of abutment stem is considered at distance d = 512.5 mm from the junction

of base slab.

Horizontal force due to soil pressure = 5559.80 KN

Breaking force = 55.40 KN

Longitudinal force due to temperature effect = 85.99 KN

Shear force V = 5701.18 KN

Shear stress tc = V/(b d) = 0.744 N/mm2

Permissible shear stress without shear reinforcement tcmax =2.8 Mpa for M20 concrete

Where, As = Area of Steel provided in mm2

b = width of abutment vertical wall and d = thickness of abutment wall p= 4.05%

tc = basic values given for different p% of Steel provided = 0.28 Mpa for M20 concr < 0.744 N/mm2

From Table 19 of IS 456-200

The design of ballest wall and return wall will be as in left side abutment.

2.1.3 DESIGN OF BALLAST WALL

A

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RCC balast wall (as shown in the figure) monolithic to the abutment has been designed.

0.40 m

m

CALCULATION OF DEAD LOAD STRESSES

Dead load from the approach slab (triangular portion) = 36 KN

= 3.7894737 KN/m

Dead load of the Ballast wall = 91.2 KN

= 9.6 KN/m

Earth pressure from the back fill = ca x r x h2 /2 = 3.12 KN/mr = 21 KN/m3ca = 0.2973

LIVE LOAD STRESS CALCULATION

Class AA (tr) load component on wall , LL =

= 291.667 KN

= 30.702 KN/m

Add impact to live load @ , I = 0.1 = 3.070 KN/m

e = 0.15 m from axis of ballast wall

Horizontal Load due to braking of train of vehicle class A vehicle, T =(0.2 ) x(27+114*2) = 51.0 KN

(Front vehicle only considered) = 5.4 KN/m

Compaction load is maximum at the depth of 1.2m from the top.

P comp = 25.2 KN/m

Maximum shear force on section I-I

H = T or Pcopm+ Ep = 28.3 KN/mV = DL + LL + I = 47.2 KN/m

Moment given due to the forces on the critical section I - I at the base of the ballast wall

M = ( LL + I) x e + T x h + Ep x h /3 == 17.92 KN-m/m

Area of vertical steel reinforcement required

Ast = Mu/ (0.87xy x 0.7984 x d) 165.746387 mm2/m

Provide 12 mm dia bars @ 200 c/c for the continuation of bar from retaining wall

Area of steel provided = 565.2 mm2/m

Provide 12 mm dia bars @ 200 c/c on the opposite face

Area of steel provided = 565.2 mm2/mHorizontal links are provided of 12 mm dia. bars @ 200 c/c

Stresses at the section,

Shear stress,tmax = 0.083 N/mm2

Permissible shear stress without shear reinforcement tcmax = 2.8 N/mm2


p= 0.28% 0.28

tc = basic values given for different p% of Steel provided = 0.38 Mpa for M20 concr > 0.083 N/mm2

From Table 19 of IS 456-200 ok

2.1.4 DESIGN OF RETURN WALLS

1

.000


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These will be monolithic with the back wall. They are joined together through 300 mm x 300 mm fillets.

The loads acting on the wing wall (return wall) would be earth pressure and horizontal force to kerb as specified in the code.

STRESS DUE TO VERTICAL LOADS

All the stresses are calculated for the critical section I-I at the bottom face of the abutment

Dead load of the returnwall = 6.000 x 0.550 x 0.200 x 24 = 15.84 KN

= 2.64 KN/m

Vert. LL from Class A load = 168.72 KN 56.24 KN/m

Add Impact due to LL @ 0.25= 42.18 KN 14.06 KN/m

STRESSES DUE TO HORIZONTAL FORCES

Horizontal Thrust due to DL+LL+IL=q*L*ka = 24.070 KN/m

acts at h/2 0.28 m from the top of foundation

Moment due to hor. component of vertical loads = 433.2636 KN-m/r.m

Earth pressure from the backfill = ka x rs x h^2/2 = 0.94 KN/m

Moment due to earth pressure = 16.998 KN-m

Momt due to later. force = 7.5 KN/m for the curb as per code = 135.000 KN-m

Total horz. uniformly distributed load = 24.07 + 0.94 + 7.5 = 32.51 KN/m

Total Moment = 433.26 + 17.00 + 135.000 = 585.26

depth required = M /( 0.138ck b ) = 620.925 mmwhere b = ( 700 + 400 ) / 2 = 550 mm

Provide overall depth = 650 mm

Ast = Mu/ (0.87xy x 0.7984 x d) 3530.97 mm2/m

Tensile steel re-bar equired = 3530.97 mm2


Required spacing = 139.0 mm c/c Provide at 150 mm c/c Decrease specing !

2.2.3 DESIGN OF FOUNDATION

Foundation is designed as open foundation of the rectangular shape

Length = 6.00 m

Breadth = 2.75 m heel 1.000 m

toe 1.000 m

width of abutment 0.600 m

Height = 0.9 m

Concrete cover 75 mm

DEAD LOAD due to foundation = 356.4 KN

CALCULATIONOF EARTH PRESSURE FOR FOUNDATION

Horizontal earth pressure due to backfill

Ka = 0.2973 As calculated earlier for abutment stem. ( Coulomb's theory applicable )

Ep1 = Ka x g h2

/2 = 715.66 KN/m h = 16.354 m

Horizontal component of earth pressure = 715.7 x cos 20 = 672.499 KN/m

which acts at 0.42 x h = 6.87 m from the bottom of footing base.

Pressure due to surcharge: As approach slab has been provided no surcharge load is to be taken as per code.


Wb = 6.000 x 1.000 x 15.454 x 18.00 = 1669.0 KN

Wf = 6.000 x 1.000 x 0 x 18.00 = 0.0 KN

acts at 0.8 m from centre of abutment

Seismic force due to Foundation = 0.096 x 356.4 = 34.214 KN


Seismic force due to back fill = 0.096 x 1669.0 = 160.23 KN

act at cg of backfill = 8.627 m from the bottom of foundation base

SUMMARY OF LOADS FOR DESIGN OF FOUNDATION

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S.No. Load Vertical (KN) Horiz L. arm M[r] M(o)

(KN) (m) KN-m KN-m

1 Superstructure

1.1 Dead load of SS(D) 2709.05 0.8 2224.79

1.2 Live Load on SS including impact (L) 438.0 0.8 359.705

Substructure2 DL of abutment(E)

Retaining wall 1367.64 1.3 1777.93Return wall 31.68 3.5 111.118

Ballast wall 91.20 1.3 118.56

Backfill (BF) 1669.03 2.10 3504.97

Backfill (FR)

Earth pressure (EP)

Triangular dry(EP1) 6388.7 -6.9 -43882

Due to overburden(EP2) 0.00 0


DL of foundation 356.40 1.4 490.05

3 Buoyancy (B)

4 Longitudinal force

4.1 Breaking (breaking) force (T) 55.40 -15.4 -850.6

4.2 Due to temperature (F) 85.99 -15.4 -13204.3 Wind Load




on superstructure 130.03 260.1 -15.3 -1992

on abutment 71.54 143.1 -7.7 -552.8

on live load 0

on foundation 17.11 34.2 -0.5 -7.698

from backfill 80.11 160.2 -8.6 -691.1

Load combination Group I (N+T condition)

D + L + I + E + Ep1+ Ep2 +BF+ T + F + CF 6663.0 6530 8587.12 -46053


D + L + E + Ep1+ Ep2 +BF+ T + F + CF + FS 6961.8 7128 8587 -48446

Note: Moment is taken about the toe for check against overturning .

403.82

For Normal condition For Seismic condition

P = 6663 KN P = 6962 KN

Mnet = Mr-Mo= -37465.918 Mnet = Mr-Mo= -39859 KN-m

x = Mnet/P= -5.6229806 x = Mnet/P= -5.7254 m

e = b/2-x'= 6.9979806 > b/6, hence tesion e = b/2-x'= 7.10 > b/6, hence t

Pmax=P/A*(1+6*e/b) = 6569.444 KN/m2 Pmax=P/A*(1+6*e/b) = 6958.3 KN/m

Pmin=P/A*(1-6*e/b) = -5761.807 KN/m2 Pmin=P/A*(1-6*e/b) = -6114 KN/m

Where, A=b x d = 16.5 m2

CHECK AGAINST OVERTURNING

Case: Normal

Factor of Safety against overturning =Mres/Mov= 8587.1227 / 46053 = 0.2 > 2.0 Safe

Case: Seismic

Factor of Safety against overturning =Mres/Mov= 8587.1227 / 48446 = 0.2 >1.5 Safe

CHECK AGAINST SLIDING

Case : Normal

The minimum vertical force will be when there is no live load and the front backfill is washed out = ###### KN

The resisting force againest sliding = (vertical force) x (coefficient of friction b/w concrete and soil)

= 6225.00 x 0.45 = 2801.25 KN

The sliding force is the horizontal forces = 6474.72 KN

Hence, Factor of Safety against sliding =Pres/Psl= 2801.25 / 6474.7 = 0.433 >1.5 Hence Sa

( f = 0.45 soil/concrete)

Case: Seismic

The minimum vertical force will be when there is no live load and the front backfill is washed out = ######


= 6523.80 x 0.45 = 2935.71 KN


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Substructure Kondre (righ


Hence, Factor of Safety against sliding =Pres/Psl= 2935.71 / 7072.3 = 0.415 >1.25 Hence Sa

STRUCTURAL DESIGN OF FOUNDATION

Design of toe slab:

Wt of toe slab = 21.6 KN/m2 1412.739 -1277.72

1.000 m 0.600 m 1.000 m

2.75 m

Pressure distribution diagram

concrete grede : M 20

Steel grade : Fe 415

Moment about EF = 2414.47 KN-m 2414.47

d= M /( 0.138 564.0 mmAdopt D = 600 mm

d = 500 mm

Ast = Mu/ (0.87xy x 0.7 16752 mm2 Minimum reinforc. as per code = 1000 mm2

Using 16 mm dia bar, spacing = 12.0 mm

Adopt 16 mm dia bar, spacing = 150 mm c/c Area of re-bar provided = 1339.73

mm2

Check for shear stress:In toe critical section for shear is considered at a distance d = 500

from abutment stem.

Shear force at critical section V = 15776.00 KN b= 6.00 m 6548

3969

13

91

Shear stress = V/(b d) = 5.259 N/mm2


Where, k1 = 1.14 - 0.7 d >= 0.5 ( d in m ) 0.5 m


b = width of abutment vertical wall and d = thickness of abutment wall p= 0.037% 0.04

tc = basic values given for different p% of Steel provided = 0.280 N/mm2 < 5.259 N/mm2

From Table 19 of IS 456-200 Not ok. Revise section or provide shear re-bar

Design of heel slab:

Upward pressure on the heel slab varies from -5761.8073 Kn/m2 to -1277.72 Kn/m2

In addition to the upward pressure the heel slab is subjected to downward pressure due to the self weight of heel

slab and weight of earth fill on heel slab:

Down pressure due to self weight = 14.4 KN/m2

Down pressure due to backfill = 278.172 KN/m2

292.572 KN/m2 1570

6054

Net down pressure = 6054 KN/m2 max

1570 KN/m2 min 1.000 m

Moment about DG= 2279.841 KN-m

d= M /( 0.138ck b ) 908.862 mmAdopt D = 600 mm

d = 500 mm

Ast = Mu/ (0.87xy x 0.7984 x d) = 15818 mm2 Minimum reinforc. as per code = 1000 mm2

Using 16 mm dia bar, specing = 13 mm

Adopt 16 mm dia bar, specing = 150 mm c/c Area of re-bar provided = 1339.7 mm2

####

#

KN/m2

6547.8

4 ####

6569.4

4

A B

CDE

F G

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Distribution re-bar

0.12/100 x 900 = 1080 mm2/runnig meter

Using 12 mm dia bar, specing = 209.3 mm

Adopt 12 mm dia bar, specing = 200 mm c/c at top and bottom


Since the load on the heel is predominantly downward, it induces tensile reactio at its junction with stem.

Hence the critical section for shear for the heel slab is considered at the face of support.

Shear force V = 22874.002 KN b = 6.00 m


Permissible shear stress without shear reinforcement tcmax =2.8 N/mm2

Mpa for M20 concrete

Where, k1 = 1.14 - 0.7 d >= 0.5 ( d in m )


b = width of abutment vertical wall and d = thickness of abutment wall p= 0.037% 0.04

tc = basic values given for different p% of Steel provided =

From Table 19 of IS 456-200 0.28 N/mm2 < 7.6247 N/mm2 Not ok

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P 12


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Substructure Kondre (ri

i i

Ws

d

PaPa


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b b

Coulomb Rankin

i F b d

0 30 90 20 a

h = 30 Let us find out which theory is applicable:

a = 60 1.7321 m < 15.454 m hence Coulomb's theory applicable

Ka=

Coul. 0.75 0.9397 2.6845 0.2973

Rankin 1 0.75 0.5 1.5 0.3333

Hor pressure from Rankin's theory = 716.48 KN/m

Horizontal thrust = 6806.5 acts at 5.15 m from the top of foundation footing.

Weight of soil is also to

be considered in Rankin's

Theroy.

Weight of soil is not

to be considered in

Coulumb's Theroy.

90-b i

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P 15


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P 16


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Substructure Kondre (right)

g !

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P 18


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P 19


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n

P 20


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P 21


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vise section or provide shear re-bar

P 22


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Dimension of slab and beams (brought from slab sheet) Check for correctne

Length of single span = 8.000 m

Carriageway width, Bc = 3.000 m

Overall width of Bridge, Bo = 6.000 m

No. of main girder = 3

Dimensions of the longitudinal girder is taken as follows:




Distance betn girders = 2.400 m

No. of cross beams = 2

For the design purposes the dimensions of the cross beam is taken as follows:




Distance betn cross beams = 8.000 m

For the design purposes the dimensions of the internal slab is taken as follows:

Length (L si ) = 8.600 m Hence, effec.span in long.dir. =

Breadth (B si ) = 3.600 m Hence, effec.span in transv.dir. =



kerb/footpath

breadth = 1.000 m at bottom and 0.225 m at top

height = 0.500 m

For design purpose dimension of cantilever slab is taken as


Breadth (B si ) = 0.600 m from c/c of outer girder

Breadth (B si ) = 0.300 m from outer face of outer main girder

Breadth of carriageway = 2.400 m from c/c of outer girder

0.300 m from outer face of outer main girder

Thickness(Tsi) = 0.200 m at the endThickness(Tsi) = 0.200 m at outer face of main girder


2.2 DESIGN OF RIGHT ABUTMENT

The freebody diagram of the R.C.C Abutment is shown below:

0.20 i = 0 CG of Stem from inner vertical face:

A 1= 1.860

0.60 A2= 0.233

M.A.= 0.709d

Pa


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A1+A2 = 2.093

0.750 b = 90 CG = 0.339

Dimensions taken for the design of abutment:

Retaining wall Bottom level of base =

Length = 6.000 m Road level =Breadth = 0.750 m at bottom 0.60 m at top 1.25

Height = 3.100 m CG of retain. wall from point A = 0.339 m

Ballast wall

Length = 6.000 m

Breadth = 0.200 m

Height = 1.000 m height of superstr. plus bearing

Return wall without pav.cover

No. = 2

Breadth = 0.200 m Height of Return wall H1= 0.7

Length = 6.000 m Height of Return wall H2= 0.4

Total height of abutment = 4.100 m

Approach slab

Length = 5.000 m 5.000

Breadth = 3.000 m 0.000

Thickness = 0.200 m

Foundation is open foundation

Length = 6.000 m

Width 2.750 m Heel 1.000 m 0.750

Toe 1.000 m 0.750

Height of footing = 0.900 1.250

Volume = 14.850 m3 W = 356.40 KN

2.2.1 CALCULATION OF LOAD

STRESS DUE TO DEAD LOAD

S.N Elements Unit No. L(m) B(m) H(m) Volume

Dead Load of Superstructure

1 Hand rails 6.368 rm 4 8.60 34.40

2 Railing posts cum 5 0.15 0.15 1.00 0.11

3 Wearing course cum 1 8.60 3.00 0.07 1.81

4 Kerbs cum 2 8.60 0.61 0.50 5.27

5 Deck slab i) rectangle cum 1 8.60 3.00 0.20 5.16

ii)trapezoid cum 2 8.60 0.30 0.20 1.036 Main Girder i) fillets cum 4 8.60 0.30 0.15 0.77

ii) webs cum 3 8.60 0.60 0.55 8.51

7 Internal Cross Girder i) fillets cum 0 2.40 0.15 0.15 0.00

ii) webs cum 0 2.40 0.60 0.70 0.00

8 End Cross Girder i) fillets cum 2 2.40 0.15 0.15 0.05

ii) webs cum 2 2.40 0.60 0.70 2.02

Total Dead load of the superstructure

Total DL coming to one abutment =

3.1

00

A


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Dead Load of Abutment

retaining wall 6.000 0.675 3.100 12.56 24 301.32

return walls 2 6.000 0.200 0.55 1.32 24 31.68

ballast wall 6.000 0.200 1.00 1.20 24 28.80

361.80

STRESS DUE TO LIVE LOAD

Live load due to class A loading:

Maximum load on the given abutment is produced when the train of load is positioned as shown in the figure:

Axle Load,KN 27 27 114 114 68 68 68 68

x i , m 8.0 6.8 2.5 -0.5 -3.5 -6.5

y i 1.0 0.9 0.3 -0.1 -0.4 -0.8

load, KN 114.0 96.9 21.3 -4.3 -29.8 -55.3




Live load due to class AA wheeled loading:

Axle Load,KN 200 200

x i , m 19.0 17.8

y i 2.38 2.23

load, KN 475.0 445.00


Load due to impact = 920 x 0.18 = 165.6 KN


Live load due to class AA tracked loading:


Load due to impact = 542.5 x 0.1 = 54.25 KNLive load including impact = 542.5 + 54.25 = 596.8 KN

Out of above three loadings class AA tracked loading gives maximum load to abutment.

Maximum live load to abutment = 1085.6 KN

CALCULATIONOF EARTH PRESSURE FOR RETAINING WALL

Horizontal earth pressure due to backfil

Corresponding to F = 30 of the backfill material

g = 18 KN/m3 g ' = 8 KN/m3

Ka = 0.2973 ( Coulomb's theory applicable )Ep1 = Ko x g h

2/2 = 45.0 KN/m h = 4.100 m

Horizontal component of earth pressure = 45.0 cos 20 = 42.3

which acts at 0.42 x h = 1.72 m from the top of footing base.

Vertical component of earth pressure = 45.0 x sin 20 =

which acts at 0.339 m inward from the CG of wall.

Total Horizontal Thrust, E1 = Ep1 x L = 253.6 KN


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Total Vertical Thrust, E2= Ep2 x L = 92.3 KN


Wb = 5.6 x 1.000 x 4.1 x 18 = 413.28

STRESS DUE TO SEISMIC FORCES

Horizontal Forces due to seismicity are computed using the formula as shown in IS Code 1893 or IRC - 6

F eq = a x b x l x G

wher a = 0.08 for the zone V of seismic intensityb = 1.2 for medium soil with isolated footingl = 1 importance factor

Thus seismic coefficient is taken as 0.096

a. Seismic force due to Superstructure = 0.096 x 296.058

height of action of the force = 3.10 + 0.025 +

= 3.5 m

b. Seismic force due to Abutment = 0.096 x 362


c. Seismic force due to Back fill = 0.096 x 413.28 =

acts at cg = 2.05 m

STRESS DUE TO WATER CURRENT

Horizontal Thrust due to water current is ignored, because it usually contributes to the stability

of the wall and the load case with seismic condition does not concide with HFL.

STRESS DUE TO LONGITUDINAL FORCES

Due to Tractive Effort or Braking Force

Braking Force due to Class AA tracked load = @20% of live load/2 =

= 55.4 KN

height of the line of action of the brake from the road level = 1.2

height of the bearing level from the top of base = 3.100 m

Due to Resistance in Bearing (Temperature)

Severe moment is produced when the LL is within span. Assume frictional coeff. = 0.225

The critical case will be when seismic force acts in the direction of traffic. But according to IRC 6-1966 (222.7) "Theload shall not be considered when acting in the direction of traffic,..".


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= 0.2

Frictional resistance at free bearing = 310.9 KN

Frictional resistance of fixed bearing = 276.3 KN

Unbalanced frictional force = 34.5 KN

Line of action of the force from the bottom of the abutment = 3.100 m

STRESS DUE TO WIND LOAD

Exposed height of structure = depth of the beams + thickness of slab +

+ height of kerb =

= 1.25 m

Exposed area contributing to wind pressure = 10 m

Avg. height of superstructure from the bed = 6.000 m

intensity of wind load = 0.73 KN/m

velocity of wind = 107 km/hr

The design is to be conducted for one span only loaded with class A train of vehicle

Length of Class A train = 20.4 m

a. Wind Force on Surface (SS) = 0.73 x 10 =

b. Wind Force on Moving

vehicle Class A train = 20.4 x 3 =

acts at 1.5 m above road level = 1.500 m

Maximum load is obtained in exposed area = 68.5 KN

For lateral direction

wind load = 68.5 KN

acts at = 4.100 m

SUMMARY OF LOADS FOR DESIGN OF ABUTMENTS

S.N

o. Load Vertical (KN) Horizontal (KN)

Longit. dir Later.

1 Superstructure

1.1 Dead load of SS(D) 296.1

1.2 Live Load of SS (L) including impact 438.0

Substructure

2 DL of abutment(E)Retaining wall 301.3

Return wall - Top section 31.7

Ballast wall 28.8

Backfill (BF) 413

Earth pressure (EP) 413.526

(EP1) 253.6

Due to overburden(EP2)

Taking

Rankin's

earth

pressure


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3 Buoyancy (B)


4.1 Tractive force (T) 55.40

4.2 Due to temperature (F) 34.54

4.3 Wind Load 68.50

4.4 Centrifugal force (CF)4.5 Thrust due to water current (WC)


on superstructure 14.21 28.42 28.42

on abutment 17.37 34.73 34.73

on live load 42.05

from backfill 19.84 39.675 39.6749

Load combination Group II (Normal+Temp. condition)

1096 344 68.5


D + L + E + Ep1+ Ep2 + T + F + CF + FS 1561 446 144.9

Note: Moment is taken at the critical section at the bottom of the ABUTMENT

2.2.2 STRUCTURAL DESIGN OF ABUTMENT RETAINING WALL

Design of the returning wall will be done based on normal loading condition, but checked on maximum loading

Design parameters

Vertical force = 1096 KN

Long. Moment = 753 KN-m

Substructures are designed using M20 concrete and Fe415 steel reinforcement

Parameters for RCC design


Steel sy = 415 N/mm2Approximate Area of Steel reinforcement required is calculated considering singly reinforced beam theory

Ast = Mu/ (0.87xy x 0.7984 x d) = 3332 mm2


Required spacing = 88.3 mm c/c Provide at 100 mm c/c

Area of reinforcement provided, Ast = 28701.563 mm2

r = 0.6378125 % > 0.2 %

At the frontal face, the design moment is greatly reduced because of lack of earth pressure.

Therefore on this side provide only 50 % of the main reinforcement

Ast = 3332 x0.5 = 1666 mm2

coefficient.

D + L + E + Ep1+ Ep2 + Ep3+ T + F + CF

(Under combination group IV the permissible increase of

50%, hence Combination I governs the design longitudina

moment)


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Provide 20 mm dia. bars Ast = 314 mm2

Required spacing = 20 113.1 mm c/c Provide at 100 mm c/c

Area of reinforcement provided, Ast = 18369 mm2

Same spacing of reinforcement is provided on the side faces of the abutment.

Distribution re-bar

Providing 0.12% of cross sectional area = 0.0012 x 750 x 1000= 900



Check for stresses (For seismic condition) Parameter

Conccrete M20 sck =

Normal Thrust , W = 1560553 N Steel sy =

Bending Moment = ######### N-mm

Ecentricity, e = 643.948 mm > 165.625 mmbreadth, b = 6000 mm m = 0.0

Total depth , D = 750 mm

Effective depth, d = 662.5 mm

Compressive steel cover, dc = 75 mm Ac = 18369 mm2

Tensile steel cover, dt = 75 mm At = 28701.56 mm2

Ast = Mu/ (0.87xy x 0.7984 x 3768.77108 mm2/m

x= 0.87*y*Ast 189 mm 150 mm say

0.36*ck*b


Critical secion for shear in case of abutment stem is considered at distance d = 662.5 mm from t

of base slab.

Horizontal force due to soil pressure = 173.25 KN

Breaking force = 55.40 KN

Longitudinal force due to temperature effect = 34.54 KN

Shear force V = 263.19 KN

Shear stress tc = V/(b d) = 0.125 N/mm2




tc = basic values given for different p% of Steel provided = 0.28 Mpa for M20 concr


The design of ballest wall and return wall will be as in left side abutment.

A


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2.1.3 DESIGN OF BALLAST WALL

RCC balast wall (as shown in the figure) monolithic to the abutment has been designed.

0.20 m

m

CALCULATION OF DEAD LOAD STRESSES

Dead load from the approach slab (triangular portion) = 36 KN= 6 KN/m

Dead load of the Ballast wall = 28.8 KN

= 4.8 KN/m

Earth pressure from the back fill = ca x r x h2 /2 = 3.12 KN/mr = 21 KN/m3ca = 0.2973

LIVE LOAD STRESS CALCULATION

Class AA (tr) load component on wall , LL == 291.667 KN

= 48.611 KN/m

Add impact to live load @ , I 0.1 = 4.861 KN/m

e = 0.15 m from axis of ballast wall

Horizontal Load due to braking of train of vehicle class A vehicle, T =(0.2 ) x(27+114*2) =

(Front vehicle only considered) =

Compaction load is maximum at the depth of 1.2m from the top.

P comp = 25.2 KN/m

Maximum shear force on section I-IH = T or Pcopm+ Ep = 28.3 KN/mV = DL + LL + I = 64.3 KN/m

Moment given due to the forces on the critical section I - I at the base of the ballast wall

M = ( LL + I) x e + T x h + Ep x h /3 == 27.76 KN-m/m

Area of vertical steel reinforcement required

1.0

00


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Ast = Mu/ (0.87xy x 0.7984 x 256.816378 mm2/m

Provi 12 mm dia bars @ 200 c/c for the continuation of bar from retainin


Provi 12 mm dia bars @ 200 c/c on the opposite face


Horizontal links are provided of 12 mm dia. bars @ 200 c/c

Stresses at the section,

Shear stress,tmax = 0.202 N/mm2

Permissible shear stress without shear reinforcement tcmax = 2.8 N/mm2


p= 0.57% 0.57

tc = basic values given for different p% of Steel provided = 0.51 Mpa for M20 concr


2.1.4 DESIGN OF RETURN WALLS

These will be monolithic with the back wall. They are joined together through 300 mm x 300 mm fillets.

The loads acting on the wing wall (return wall) would be earth pressure and horizontal force to kerb as specified in th

STRESS DUE TO VERTICAL LOADS

All the stresses are calculated for the critical section I-I at the bottom face of the abutment

Dead load of the returnwall = 6.000 x 0.550 x 0.200

Vert. LL from Class A load = 168.72 KN 56.24 KN/m

Add Impact due to LL @ 0.25= 42.18 KN 14.06 KN/m

STRESSES DUE TO HORIZONTAL FORCES

Horizontal Thrust due to DL+LL+IL=q*L*ka = 24.070 KN/m

acts at h/2 0.28 m from the top of foundation

Moment due to hor. component of vertical loads = 433.2636 KN-m/r.m

Earth pressure from the backfill = ka x rs x h^2/2 = 0.94

Moment due to earth pressure = 16.998 KN-m

Momt due to later. force = 7.5 KN/m for the curb as per code = 135.000 KN-m

Total horz. uniformly distributed load = 24.07 + 0.94 + 7.5

Total Moment = 433.26 + 17.00 + 135.000 = 585.26

depth required = M /( 0.138ck b ) = 620.925 mmwhere b = ( 700 + 400 ) / 2 = 550 mm

Provide overall depth = 650 mm

Ast = Mu/ (0.87xy x 0.7984 x 3530.97 mm2/m


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Tensile steel re-bar equired = 3530.97 mm2



2.2.3 DESIGN OF FOUNDATION

Foundation is designed as open foundation of the rectangular shape

Length = 6.00 m

Breadth = 2.75 m heel 1.000 m

toe 1.000 m

width of abutment 0.750 m

Height = 0.9 m

Concrete cover 75 mm

DEAD LOAD due to foundation = 356.4 KN

CALCULATIONOF EARTH PRESSURE FOR FOUNDATION

Horizontal earth pressure due to backfill

Ka = 0.2973 As calculated earlier for abutment stem. ( Coulomb's theory applicable

Ep1 = Ka x g h2

/2 = 66.90 KN/m h = 5.000 m

Horizontal component of earth pressure = 66.9 x cos 20 =

which acts at 0.42 x h = 2.10 m from the bottom of footing base.

Pressure due to surcharge: As approach slab has been provided no surcharge load is to be taken as per code.


Wb = 6.000 x 1.000 x 4.100 x 18.00 = 442.8

Wf = 6.000 x 1.000 x 0 x 18.00 = 0.0

acts at 0.875 m from centre of abutment

Seismic force due to Foundation = 0.096 x 356.4


Seismic force due to back fill = 0.096 x 442.8

act at cg of backfill = 2.95 m from the b

SUMMARY OF LOADS FOR DESIGN OF FOUNDATION.

. Load Vertical (KN) Horiz L. arm

(KN) (m)

1 Superstructure

1.1 Dead load of SS(D) 296.06 1.2

1.2 Live Load on SS including impact (L) 438.0 1.2

Substructure

2 DL of abutment(E)

Retaining wall 301.32 1.4

Return wall 31.68 3.7

Ballast wall 28.80 1.4

Backfill (BF) 442.80 2.25

Backfill (FR)


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Earth pressure (EP)

Triangular dry(EP1) 377.2 -2.1

Due to overburden(EP2) 0.00


DL of foundation 356.40 1.4

3 Buoyancy (B)


4.1 Breaking (breaking) force (T) 55.40 -4.0

4.2 Due to temperature (F) 34.54 -4.0

4.3 Wind Load




on superstructure 14.21 28.4 -3.5

on abutment 17.37 34.7 -2.1

on live load

on foundation 17.11 34.2 -0.5

from backfill 21.25 42.5 -3.0

Load combination Group I (N+T condition)

D + L + I + E + Ep1+ Ep2 +BF+ T + F + CF 1895.1 467


D + L + E + Ep1+ Ep2 +BF+ T + F + CF + FS 1965.0 607

Note: Moment is taken about the toe for check against overturning .

For Normal condition For Seismic conditi

P = 1895 KN P = 1965

Mnet = Mr-Mo= 1797.12344 Mnet = Mr-Mo=

x = Mnet/P= 0.94832107 x = Mnet/P=e = b/2-x'= 0.42667893 ok e = b/2-x'=

Pmax=P/A*(1+6*e/b) = 221.772 KN/m2 Pmax=P/A*(1+6*e/

Pmin=P/A*(1-6*e/b) = 7.932 KN/m2 Pmin=P/A*(1-6*e/b

Where, A=b x d = 16.5 m2

CHECK AGAINST OVERTURNING

Case: Normal

Factor of Safety against overturning =Mres/Mov= 2948.9419 / 1151.8 =

Case: Seismic

Factor of Safety against overturning =Mres/Mov= 2948.9419 / 1086 =

CHECK AGAINST SLIDING

Case : NormalThe minimum vertical force will be when there is no live load and the front backfill is washed out =


= 1457.06 x 0.45 = 655.676


Hence, Factor of Safety against sliding =Pres/Psl= 655.68 / 411.7 =

( f = 0.45 soil/concr

Case: Seismic

The minimum vertical force will be when there is no live load and the front backfill is washed out =


34/51


= 1527.00 x 0.45 = 687.149


Hence, Factor of Safety against sliding =Pres/Psl= 687.15 / 551.6 =

STRUCTURAL DESIGN OF FOUNDATION

Design of toe slab:

Wt of toe slab = 21.6 KN/m2 144.012 85.692

1.000 m 0.750 m

2.75 m

Pressure distribution diagram

concrete grede :

Steel grade :

Moment about EF = 87.13 KN-m

d= M /( 0.138 107.1 mmAdopt D = 600 mm

d = 500 mm

Ast = Mu/ (0.87xy x 0.7 604 mm2 Minimum reinforc. as per cod

Using 16 mm dia bar, spacing = 201.1 mm

Adopt 16 mm dia bar, spacing = 150 mm c/c Area of r

Check for shear stress:In toe critical section for shear is considered at a distance d = 500

from abutment stem.

Shear force at critical section V = 542.20 KN b= 6.00 m #####

161.3



Where, k1 = 1.14 - 0.7 d >= 0.5 ( d in m ) 0.5 m



tc = basic values given for different p% of Steel provided = 0.280 N/mm2 >From Table 19 of IS 456-200 ok. Shear re-bar is not necessary.

200.1

7

KN/m2

221.7

7

####

A

DE

F G


35/51

Design of heel slab:

Upward pressure on the heel slab varies from 7.93215698 Kn/m2 to 85.69 Kn/m2

In addition to the upward pressure the heel slab is subjected to downward pressure due to the self weight of heel

slab and weight of earth fill on heel slab:

Down pressure due to self weight = 14.4 KN/m2

Down pressure due to backfill = 73.8 KN/m288.2 KN/m2 3

Net down pressure = 80 KN/m2 max

3 KN/m2 min 1.000 m

Moment about DG= 27.174 KN-m

d= M /( 0.138ck b ) 99.225 mmAdopt D = 600 mm

d = 500 mm

Ast = Mu/ (0.87xy x 0.7984 x d) = 189 mm2 Minimum reinforc. as per code =

Using 16 mm dia bar, specing = 201 mm

Adopt 16 mm dia bar, specing = 150 mm c/c Area of re-bar pro

Distribution re-bar0.12/100 x 900 = 1080 mm2/runnig meter

Using 12 mm dia bar, specing = 209.3 mm

Adopt 12 mm dia bar, specing = 200 mm c/c at top and bottom


Since the load on the heel is predominantly downward, it induces tensile reactio at its junction with stem.

Hence the critical section for shear for the heel slab is considered at the face of support.

Shear force V = 248.3274 KN b = 6.00 m


Permissible shear stress without shear reinforcement tcmax =2.8N/mm

2

Mpa for M20 concreteWhere, k1 = 1.14 - 0.7 d >= 0.5 ( d in m )



tc = basic values given for different p% of Steel provided =

From Table 19 of IS 456-200 0.28 N/mm2 >


36/51

ss of data

8.000 m

3.000 m


37/51

0.000

5.000 ######

Unit Wt Total e from c/c

0.06 2.06

24 2.70

22 39.73

24 126.42

24 123.84

24 24.7724 18.58

24 204.34

24 0.00

24 0.00

24 1.30

24 48.38

592.12

296.1 KN


38/51

0.000 m

2.338 m

0.238 m

i

Ws

b b

Coulomb

i F b d

0 30 90 20

h = 30 Let us find out which theory is appl

a = 60 1.7321 m < 4.100 m henceKa=

KN/m Coul. 0.75 0.9397 2.6845 0.2973

Rankin 1 0.75 0.5 1.5 0.3333

15.384 KN/m

Hor pressure from Rankin's theory = 50.43 KN/m

Horizontal thrust = 302.6 acts at 1.37 m from the top of foundation footin

Weight of soil is not

to be considered in

Coulumb's Theroy.

90-bd

Pa


39/51

KN

= 28.42 KN

0.75 / 2 =

= 34.73 KN

39.675 KN

1/2 to one abutment

m

in free bearing

seismic force due to live


40/51

in other bearing

7.3 KN

61.2 KN

Arm Moment (KN)

(m) Longit. Later.

0.16 47.7

0.16 70.6

0.00

-2.34 -74.1

-0.24 -6.8

0.00 0.0

1.72 436.7


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3.10 171.7

3.10 107.1

4.10 280.85

3.50 99.5 99.47549

2.05 71.2 71.20224

5.60 235.4688

2.05 81.3 81.3335

752.9 280.9

1004.9 487.5

Decrease spacing !

stress is

l


42/51

ok

mm2

at both faces. ok

for RCC design

20 N/mm2

415 N/mm2

he junction

> 0.125 N/mm2


43/51

51.0 KN

8.5 KN/m


44/51

wall

> 0.202 N/mm2

ok

code.

x 24 = 15.84 KN

= 2.64 KN/m

KN/m

= 32.51 KN/m


45/51

Decrease specing !

)

62.8613 KN/m

KN

KN

= 34.214 KN

= 42.509 KN

ottom of foundation base

M[r] M(o)

KN-m KN-m

359.382

531.683

414.315

117.612

39.6

996.3


46/51

-792.1

0

490.05

-221.6

-138.2

-49.74

-35.6

0

-7.698

-62.7

2948.94 -1152

2949 -1086

114.85

n

KN

1862.99 KN-m

0.94809 m0.43 ok

b) = 230.02 KN/m2

) = 8.1638 KN/m2

2.6 > 2.0 Safe

2.7 >1.5 Safe

###### KN

KN

1.593 >1.5 Hence Safe

te)

######


47/51

KN

1.246 >1.25 Hence Safe

1.000 m

M 20

Fe 415

87.1259

= 1000 mm2

-bar provided = 1339.733

mm2

122.4

0.04

0.181 N/mm2

7.9

B

C


48/51

80

1000 mm2

vided = 1339.7 mm2

0.04

0.0828 N/mm2

ok


49/51


50/51


51/51

i

Rankin

a

icable:

Coulomb's theory applicable

Weight of soil is also to

be considered in Rankin's

Theroy.

i

Pa

Documents

Design of Abutment Bridge