Design of Abutment Bridge

Substructure Kondre (right) Dimension of slab and beams (brought from slab sheet)

Check for correctness of data

Length of single span = Carriageway width, Bc = Overall width of Bridge, Bo =

35.000 m 7.500 m 10.700 m

No. of main girder = 5 Dimensions of the longitudinal girder is taken as follows: Length (L si ) = 35.000 m Breadth (B si )

=

0.550 m

Thickness(Tsi) Distance betn girders =

=

1.680 m 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 Breadth (B si ) = 0.140 m Thickness(Tsi) = 0.300 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. = Breadth (B si ) = 2.375 m Hence, effec.span in transv.dir. = Thickness(Tsi) = 0.230 m Avg. pavement cover over Slab(T sp ) = 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 Length (L si ) = 1.00 m 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 end Thickness(Tsi) = 0.230 m at outer face of main girder Avg. pavement cover over Slab(T sp ) = 0.08 m

2.2

2.360 m 1.825 m

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:

Pa

d

14.454

0.23

0.600

A 1=

3.324

A2=

2.674

M.A.= A1+A2 =

1.327 5.998

CG =

0.221

b = 90 A Dimensions taken for the design of abutment: Retaining wall Length = 9.500 Breadth = 0.600 Height = 14.454 Ballast wall Length = 9.500 Breadth = 0.400 Breadth of ret. wall = 0.200 Height = 1.000 Return wall No. = 2 Breadth = 0.200

m m at bottom

m m m m m

m

Bottom level of base = Road level = 0.23 m at top 1.25 CG of retain. wall from point A = 0.221 m

height of superstr. plus bearing without pav.cover Height of Return wall H1=

Page 1

0.7

0.000 5.000 ######

Substructure Kondre (right) Length = Total height of abutment = Approach slab Length = Breadth = Thickness = Foundation is open foundation Length = Width Height of footing = Volume = 2.2.1

6.000 m

Height of Return wall H2= 15.454 m

5.000 m 3.000 m 0.200 m

16.354 -11.354

6.000 m 2.750 m 0.900

0.4

Heel Toe

1.000 m 1.000 m

W =

356.40 KN

0.750 0.750

1.250

14.850 m3

CALCULATION OF LOAD

STRESS DUE TO DEAD LOAD S.N

Elements

Unit

Dead Load of Superstructure 1 Hand rails 6.368 2 Railing posts 3 Wearing course 4 Kerbs 5 Deck slab i) rectangle ii)trapezoid 6 Main Girder i) fillets ii) webs 7 Internal Cross Girder i) fillets ii) webs 8 End Cross Girder i) fillets ii) webs

Dead Load of Abutment retaining wall return walls ballast wall

2

rm cum cum cum cum cum cum cum cum cum cum cum

9.500 6.000 9.500

No.

L(m)

B(m)

H(m) Volume Unit Wt

4 35.60 18 0.15 0.15 1.00 1 35.60 7.50 0.08 2 35.60 0.23 1.00 1 35.60 2.93 0.23 2 35.60 0.33 0.23 8 35.60 0.30 0.15 5 35.60 0.60 1.45 0 1.83 0.15 0.15 0 1.83 0.14 0.07 2 1.83 0.15 0.15 2 1.83 0.14 0.07 Total Dead load of the superstructure Total DL coming to one abutment =

0.415 0.200 0.400

14.454 0.55 1.00

56.98 1.32 3.80

142.40 0.41 20.03 16.02 23.95 5.32 6.41 154.86 0.00 0.00 0.04 0.04

24 1367.64 24 31.68 24 91.20 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 xi , m 35.0 33.8 29.5 26.5 23.5 20.5 yi 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 xi , m yi load, KN Maximum total live load = Load due to impact = Live load including impact =

210.28571 210.3

Live load due to class AA tracked loading: Maximum total live load = Load due to impact = Live load including impact =

664 664.0

200 19.0 0.54 108.6

200 17.8 0.51 101.71 210.3 KN x 0.18 = + 37.851429 =

664.0 KN x +

Page 2

0.1 = 66.4 =

37.851 KN 248.1 KN

66.4 KN 730.4 KN

0.06 24 22 24 24 24 24 24 24 24 24 24

Total Wt e from c/c 8.54 9.72 440.55 384.48 574.80 127.73 153.79 ###### 0.00 0.00 0.99 0.86 ##### 2709 KN

0.000 m 2.208 m 0.007 m

Substructure Kondre (right) 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

Ka =

g'

(

0.2973

=

8

Coulomb's theory applicable

Ep1 = Ko x g h /2 = 639.1 KN/m h= 15.454 m Horizontal component of earth pressure = 639.1 cos 20 which acts at 0.42 x h = 6.49 m from the top of footing base. Vertical component of earth pressure = 639.1 x which acts at 0.221 m inward from the CG of wall.

KN/m3

)

2

Total Horizontal Thrust, E1 = Ep1 x L = Total Vertical Thrust, E2 = Ep2 x L =

= sin 20

600.5 KN/m =

218.570 KN/m

5704.9 KN 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 = where

a= b= l=

axbx l xG 0.08 for the zone V of seismic intensity 1.2 for medium soil with isolated footing 1 importance factor

Thus seismic coefficient is taken as

0.096

a.

Seismic force due to Superstructure height of action of the force

= = =

b.

Seismic force due to Abutment act at cg of abutment

= =

c.

Seismic force due to Back fill acts at cg

= =

0.096 14.45 + 15.319 m

x 2709.05 0.025 +

0.096 x 7.727 m 0.096 x 7.727 m

2531.4 =

= 260.1 KN 1.68 / 2 =

1491 =

143.1 KN

243.01 KN

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 load shall not be considered when acting in the direction of traffic,..". 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 = = height of the line of action of the brake from the road level = height of the bearing level from the top of base =

Due to Resistance in Bearing (Temperature)

Page 3

1/2 to one abutment 55.4 KN 1.2 m 14.454 m

Substructure Kondre (right) Severe moment is produced when the LL is within span. Assume frictional coeff. = = Frictional resistance at free bearing

=

773.9 KN

Frictional resistance of fixed bearing

=

687.9 KN

Unbalanced frictional force = Line of action of the force from the bottom of the abutment =

0.225 in free bearing 0.2 in other bearing

86.0 KN 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 m²

Avg. height of superstructure from the bed = intensity of wind load velocity of wind 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

b.

Wind Force on Moving vehicle Class A train acts at 1.5 m above road level

=

20.4 x =

Maximum load is obtained in exposed area = For lateral direction wind load acts at

6.000 m 0.73 KN/m² 107 km/hr

93.8 =

68.474 KN

3 = m

61.2 KN

-9.854 129.7 KN

= =

129.7 KN 15.454 m

SUMMARY OF LOADS FOR DESIGN OF ABUTMENTS S.No.

Load

Vertical (KN)

Horizontal (KN) Arm Later. (m)

Moment (KN) Longit. Later.

Longit. dir

1 1.1 1.2

Superstructure Dead load of SS(D) Live Load of SS (L) including impact Substructure DL of abutment(E) Retaining wall Return wall - Top section Ballast wall Backfill (BF) Earth pressure (EP) (EP1)

2

3 4 4.1 4.2 4.3 4.4 4.5 4.6

2709.1 438.0

0.48 1297.0 0.48 209.7

1367.6 31.7 91.2 2531 35062.779

Due to overburden(EP2) Frontal passive soil mass (EPp) Buoyancy (B) Longitudinal force Tractive force (T) Due to temperature (F) Wind Load Centrifugal force (CF) Thrust due to water current (WC) Seismic forces (FS) on superstructure on abutment on live load from backfill

Taking Rankin's earth pressure coefficient.

0.00 -2.21 -0.01 0.00 5704.9

6.49 ######

55.40 85.99

130.03 71.54 121.51

-69.9 -0.7 0.0

129.67

14.45 800.8 14.45 1242.8 15.45 2003.982

260.07 143.09 42.05 243.01 243.011

15.32 3984.0 3983.994 7.73 1105.7 1105.654 16.95 712.8818 7.727 1877.7 1877.746

260.07 143.09

Load combination Group II (Normal+Temp. condition) D + L + E + Ep1+ Ep2 + Ep3+ T + F + CF 4638

Page 4

5846

129.7

#####

2004.0

Substructure Kondre (right) Load combination Group IV (for Seismic dry condition) D + L + E + Ep1+ Ep2 + T + F + CF + FS

7492

6492

688.2

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 KN (Under combination group IV the permissible increase of stress is Long. Moment = 40508 KN-m 50%, hence Combination I governs the design longitudinal moment) Substructures are designed using M20 concrete and Fe415 steel reinforcement

Conccrete M20

Parameters for RCC design sck = 20 N/mm2

Steel

sy =

415 N/mm2

Page 5

#####

7680.3

Substructure Kondre (right) Approximate Area of Steel reinforcement required is calculated considering singly reinforced beam theory Ast = Mu/ (0.87xσy x 0.7984 x d) = Provide Required spacing =

234060 mm2

25 mm dia. bars 0.9 mm c/c

Ast = Provide at

Area of reinforcement provided, Ast =

490.625 mm2 100 mm c/c

45873.438 mm2 0.8047971 % >

r=

Decrease spacing !

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

=

Provide Required spacing =

117030 mm2

20 mm dia. bars 20 1.2 mm c/c

Ast = Provide at


314 mm2 100 mm c/c

Decrease specing !

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 = Provide Required spacing =

0.0012 x


Check for stresses

Ast = Provide at

600 x 1000=

113.04 mm2 250 mm c/c

(For seismic condition) 7492018 ######### 6336.850 9500 600 512.5 75 75

mm mm mm mm mm mm

at both faces.

ok

Parameters for RCC design Conccrete M20

Normal Thrust , W = Bending Moment = Ecentricity, e = breadth, b = Total depth , D = Effective depth, d = Compressive steel cover, dc = Tensile steel cover, dt =

720 mm2

N N-mm > m=

Ac = At =

Steel

sck = sy =

20 N/mm2 415 N/mm2

128.125 mm 0.0

29359 mm2 45873.44 mm2

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

x=

0.87*σy*Ast 0.36*σck*b

8928 mm

150 mm say

Check for shear stress: Critical secion for shear in case of abutment stem is considered at distance d = 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) =

A

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

512.5 mm from the junction

DESIGN OF BALLAST WALL

Page 6

p=

4.05%

0.28 Mpa for M20 concrete

<

2 0.744 N/mm

Substructure Kondre (right)

1.000 m

RCC balast wall (as shown in the figure) monolithic to the abutment has been designed. 0.40 m

CALCULATION OF DEAD LOAD STRESSES Dead load from the approach slab (triangular portion) = = Dead load of the Ballast wall

36 KN 3.7894737 KN/m

=

91.2 KN 9.6 KN/m

= Earth pressure from the back fill

= 21 KN/m3 0.2973

r= ca =

ca x r x h2 /2 =

3.12 KN/m

LIVE LOAD STRESS CALCULATION Class AA (tr) load component on wall , LL =

Add impact to live load @ , I =

291.667 30.702 3.070 0.15

= = = e=

0.1

KN KN/m KN/m 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)

=

51.0 KN 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 = V = DL + LL + I =

28.3 KN/m 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/ (0.87xσy x 0.7984 x d) =165.746387 mm2/m Provide

12 mm dia bars @ Area of steel provided = Provide 12 mm dia bars @ Area of steel provided = Horizontal links are provided of

200 c/c

for the continuation of bar from retaining wall 565.2 mm2/m 200 c/c on the opposite face 565.2 mm2/m 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 2

Where, As = Area of Steel provided in mm

2.1.4

p=

tc = basic values given for different p% of Steel provided = From Table 19 of IS 456-200 DESIGN OF RETURN WALLS

Page 7

N/mm2

0.28%

0.28 0.38 Mpa for M20 concrete

>

2 0.083 N/mm ok

Substructure Kondre (right) 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 Vert. LL from Class A load = Add Impact due to LL @ 0.25=

168.72 KN 42.18 KN

56.24 14.06

0.550

x

0.200

x 24 = =

15.84 KN 2.64 KN/m

KN/m KN/m

STRESSES DUE TO HORIZONTAL FORCES Horizontal Thrust due to DL+LL+IL=q*L*ka acts at h/2 0.28 m

= 24.070 KN/m from the top of foundation

Moment due to hor. component of vertical loads =

433.2636 KN-m/r.m ka x rs x h^2/2

Earth pressure from the backfill =

=

Moment due to earth pressure = Momt due to later. force =

16.998 KN-m 7.5 KN/m for the curb as per code = 135.000

Total horz. uniformly distributed load = Total Moment = 433.26 + depth required = where b = ( Provide overall depth =

17.00 + M /( 0.138σck b ) 700 + 650 mm

Ast = Mu/ (0.87xσy x 0.7984 x d) =

24.07

+ 135.000 =

= 400 ) / 2 =

0.94 KN/m

KN-m

0.94 +

7.5 =

32.51 KN/m

585.26 620.925 mm 550 mm

3530.97 mm2/m

Tensile steel re-bar equired = 3530.97 mm2 Provide 25 mm dia. bars Required spacing = 139.0 mm c/c

Ast = Provide at

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 toe width of abutment Height = 0.9 m Concrete cover 75 mm DEAD LOAD due to foundation =

490.87 mm2 150 mm c/c

Decrease specing !

1.000 m 1.000 m 0.600 m

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

=

)

672.499 KN/m

Pressure due to surcharge: As approach slab has been provided no surcharge load is to be taken as per code. WEIGHT OF BACKFILL BEHIND THE ABUTMENT Wb = Wf =

6.000 6.000 acts at

x x 0.8

1.000 1.000 m

x 15.454 x 0 from centre of abutment

Seismic force due to Foundation act at cg of abutment Seismic force due to back fill act at cg of backfill SUMMARY OF LOADS FOR DESIGN OF FOUNDATION

x x

= = = =

Page 8

18.00 = 18.00 =

0.096 x 0.45 m 0.096 x 8.627 m

1669.0 KN 0.0 KN

356.4 =

34.214 KN

1669.0 = 160.23 KN from the bottom of foundation base

Substructure Kondre (right) S.No.

Load

Vertical (KN)

Horiz (KN)

1 1.1 1.2

2

4.1 4.2 4.3

M(o) KN-m

Superstructure Dead load of SS(D) Live Load on SS including impact (L)

2709.05 438.0

0.8 2224.79 0.8 359.705

Substructure DL of abutment(E) Retaining wall Return wall Ballast wall

1367.64 31.68 91.20

1.3 1777.93 3.5 111.118 1.3 118.56

1669.03

2.10 3504.97

Backfill (BF) Backfill (FR) Earth pressure (EP) Triangular dry(EP1)

3 4

L. arm M[r] (m) KN-m

6388.7

Due to overburden(EP2) Frontal passive soil mass (EPp) DL of foundation Buoyancy (B) Longitudinal force Breaking (breaking) force (T) Due to temperature (F) Wind Load

-6.9

-43882

0.00 356.40

Centrifugal force (CF) Thrust due to water current (WC) Seismic forces (FS) on superstructure on abutment on live load on foundation from backfill Load combination Group I (N+T condition) D + L + I + E + Ep1+ Ep2 +BF+ T + F + CF Load combination Group IV (for Seismic dry condition) D + L + E + Ep1+ Ep2 +BF+ T + F + CF + FS

0 1.4

490.05

55.40 85.99

-15.4 -15.4

-850.6 -1320

130.03 71.54

260.1 143.1

-15.3 -7.7

17.11 80.11

34.2 160.2

-0.5 -8.6

-1992 -552.8 0 -7.698 -691.1

6663.0

6530

8587.12 -46053

6961.8

7128

8587 -48446

4.4 4.5 4.6

Note: Moment is taken

about the toe for check against overturning . 403.82

For Normal condition P= 6663 KN Mnet = Mr-Mo= -37465.918 x = Mnet/P= -5.6229806 e = b/2-x'= 6.9979806 > b/6, hence tesion Pmax=P/A*(1+6*e/b) = 6569.444 KN/m2 Pmin=P/A*(1-6*e/b) = -5761.807 KN/m2 Where, A=b x d = 16.5 m2 CHECK AGAINST OVERTURNING Case: Normal Factor of Safety against overturning =Mres/Mov= Case: Seismic Factor of Safety against overturning =Mres/Mov=

For Seismic condition P= 6962 KN Mnet = Mr-Mo= -39859 x = Mnet/P= -5.7254 e = b/2-x'= 7.10 Pmax=P/A*(1+6*e/b) = Pmin=P/A*(1-6*e/b) =

KN-m m > b/6, hence tesion 6958.3 KN/m2 -6114 KN/m2

8587.1227

/

46053 =

0.2

> 2.0 Safe

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 Safe (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 = ###### The resisting force againest sliding = (vertical force) x (coefficient of friction b/w concrete and soil) = 6523.80 x 0.45 = 2935.71 KN

Page 9

Substructure Kondre (right) The sliding force is the horizontal forces = Hence, Factor of Safety against sliding =Pres/Psl=

7072.32 2935.71

KN /

7072.3 =

0.415 >1.25 Hence Safe

STRUCTURAL DESIGN OF FOUNDATION

6569.44 KN/m2

Wt of toe slab =

21.6 KN/m2

D

F

G

1412.739

C B #####

A

Design of toe slab:

E

-1277.72

1.000 m

0.600 m 2.75 m

1.000 m

Pressure distribution diagram

####

6547.84 Moment about EF =

concrete grede : Steel grade :

2414.47 KN-m d= M /( 0.138σck b ) 564.0 mm Adopt D = 600 mm d= 500 mm Ast = Mu/ (0.87xσy x 0.7984 x d)16752 = mm2 Using 16 mm dia bar, spacing = Adopt 16 mm dia bar, spacing =

M 20 Fe 415

2414.47

Minimum reinforc. as per code = 1000 mm2 12.0 mm 150 mm c/c Area of re-bar provided = 1339.733 mm2

Check for shear stress:

Shear stress =

V/(b d) =

b=

6.00 m

1391

15776.00 KN

500 3969

Shear force at critical section V =

d =

6548

In toe critical section for shear is considered at a distance from abutment stem.

5.259 N/mm2

Permissible shear stress without shear reinforcement tcmax =2.8 Mpa for M20 concrete Where, k1 = 1.14 - 0.7 d >= 0.5 ( d in m ) Where, As = Area of Steel provided in mm2 b = width of abutment vertical wall and d = thickness of abutment wall p=

0.5 m

0.037% 0.04 2 0.280 N/mm2 < 5.259 N/mm Not ok. Revise section or provide shear re-bar

tc = basic values given for different p% of Steel provided = From Table 19 of IS 456-200

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 = Moment about DG= d= M /( 0.138σck b ) Adopt

6054 1570 2279.841 908.862

KN/m2 KN/m2 KN-m mm

max min

D= 600 mm d= 500 mm Ast = Mu/ (0.87xσy x 0.7984 x d) = 15818 mm2 Using 16 mm dia bar, specing = Adopt 16 mm dia bar, specing =

Page 10

1.000 m

Minimum reinforc. as per code = 1000 mm2 13 mm 150 mm c/c Area of re-bar provided = 1339.7 mm2

Substructure Kondre (right) Distribution re-bar 0.12/100 x

900 Using Adopt

=

1080 mm2/runnig meter 12 mm dia bar, specing = 209.3 mm 12 mm dia bar, specing = 200 mm c/c

at top and bottom

Check for shear stress: 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 Shear stress = V/(b d) = 7.625 N/mm2 Permissible shear stress without shear reinforcement tcmax =2.8 Where, k1 = 1.14 - 0.7 d >= 0.5 ( d in m )

N/mm2

Where, As = Area of Steel provided in mm2 b = width of abutment vertical wall and d = thickness of abutment wall tc = basic values given for different p% of Steel provided = From Table 19 of IS 456-200


p= 0.28 N/mm2

Page 11

0.037%

<

0.04 2 7.6247 N/mm

Not ok. Revise section or p


Page 12


i

i Pa

Weight of soil is not to be considered in Coulumb's Theroy.

d 90-b

Page 13

Ws

Pa i


d 90-b


i 0 h= a= Coul. Rankin

Pa i

b

b Coulomb

F 30

b 90

d 20

30

a Let us find out which theory is applicable:

60

1.7321 m <

0.75 0.9397 2.6845 1 0.75 0.5

Hor pressure from Rankin's theory = Horizontal thrust = 6806.5 acts at

Rankin Weight of soil is also to be considered in Rankin's Theroy.

1.5

15.454 m hence Coulomb's theory applicable Ka= 0.2973 0.3333

716.48 KN/m 5.15 m from the top of foundation footing.

Page 14


Page 15


Page 16


Decrease specing !

Page 17


Page 18


Page 19


> b/6, hence tesion

Page 20


Page 21


Not ok. Revise section or provide shear re-bar

Page 22

Dimension of slab and beams (brought from slab sheet) Length of single span = Carriageway width, Bc = Overall width of Bridge, Bo =

Check for correctness of data 8.000 m 3.000 m 6.000 m

No. of main girder = 3 Dimensions of the longitudinal girder is taken as follows: Length (L si ) = 8.000 m Breadth (B si ) Thickness(Tsi) Distance betn girders =

=

0.600 m

=

0.750 m 2.400 m

No. of cross beams = 2 For the design purposes the dimensions of the cross beam is taken as follows: Length (L si ) = 3.000 m Breadth (B si ) = 0.600 m Thickness(Tsi) = 0.900 m 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. = Thickness(Tsi) = 0.200 m Avg. pavement cover over Slab(T sp ) = 0.070 m 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 Length (L si ) = 1.00 m 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 end Thickness(Tsi) = 0.200 m at outer face of main girder Avg. pavement cover over Slab(T sp ) = 0.07 m

2.2

DESIGN OF RIGHT ABUTMENT

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

0.20

i=0 Pa

0.60 3.100

d

CG of Stem from inner vertical face: A 1=

1.860

A2=

0.233

M.A.=

0.709

3.100

A1+A2 =

0.750 A

b = 90

CG =

Dimensions taken for the design of abutment: Retaining wall Length = 6.000 Breadth = 0.750 Height = 3.100 Ballast wall Length = 6.000 Breadth = 0.200 Height = 1.000 Return wall No. = 2 Breadth = 0.200 Length = 6.000 Total height of abutment = Approach slab Length = Breadth = Thickness = Foundation is open foundation Length = Width Height of footing = Volume =

m m at bottom

m m m m

m m

Bottom level of base = Road level = 0.60 m at top 1.25 CG of retain. wall from point A = 0.339 m

height of superstr. plus bearing without pav.cover Height of Return wall H1= Height of Return wall H2= 4.100 m

5.000 m 3.000 m 0.200 m

0.7 0.4

5.000 0.000

6.000 m 2.750 m 0.900

2.093

0.339

Heel Toe

1.000 m 1.000 m

W =

356.40 KN

0.750 0.750

1.250

14.850 m3

2.2.1 CALCULATION OF LOAD STRESS DUE TO DEAD LOAD S.N Elements

Unit

Dead Load of Superstructure 1 Hand rails 6.368 2 Railing posts 3 Wearing course 4 Kerbs 5 Deck slab i) rectangle ii)trapezoid 6 Main Girder i) fillets ii) webs 7 Internal Cross Girder i) fillets ii) webs 8 End Cross Girder i) fillets ii) webs

rm cum cum cum cum cum cum cum cum cum cum cum

No.

L(m)

B(m)

H(m) Volume

4 8.60 5 0.15 0.15 1 8.60 3.00 2 8.60 0.61 1 8.60 3.00 2 8.60 0.30 4 8.60 0.30 3 8.60 0.60 0 2.40 0.15 0 2.40 0.60 2 2.40 0.15 2 2.40 0.60 Total Dead load of the superstructure Total DL coming to one abutment =

1.00 0.07 0.50 0.20 0.20 0.15 0.55 0.15 0.70 0.15 0.70

34.40 0.11 1.81 5.27 5.16 1.03 0.77 8.51 0.00 0.00 0.05 2.02

Dead Load of Abutment retaining wall return walls ballast wall

2

6.000 6.000 6.000

0.675 0.200 0.200

3.100 0.55 1.00

12.56 1.32 1.20

24 24 24

301.32 31.68 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 xi , m 8.0 6.8 2.5 -0.5 -3.5 -6.5 yi 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 Maximum total live load = 142.9 KN Load due to impact = 142.9 x 0.321 = 45.932 KN Live load including impact = 142.9 + 45.932 = 188.8 KN Live load due to class AA wheeled loading: Axle Load,KN xi , m yi load, KN Maximum total live load = Load due to impact = Live load including impact = Live load due to class AA tracked loading: Maximum total live load = Load due to impact = Live load including impact =

200 19.0 2.38 475.0 920 920.0

200 17.8 2.23 445.00 920.0 KN x +

0.18 = 165.6 =

165.6 KN 1085.6 KN

542.5 542.5

542.5 KN x +

0.1 = 54.25 =

54.25 KN 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

Ka =

0.2973

g'

(

=

8

Coulomb's theory applicable

Ep1 = Ko x g h /2 = 45.0 KN/m h= 4.100 m Horizontal component of earth pressure = 45.0 cos 20 = 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.

KN/m3

)

2

Total Horizontal Thrust, E1 = Ep1 x L =

253.6 KN

42.3 =

Total Vertical Thrust, E2 = Ep2 x L =

92.3 KN

WEIGHT OF BACKFILL BEHIND THE ABUTMENT 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 = wherea = b=

l=

axbx l xG 0.08 for the zone V of seismic intensity 1.2 for medium soil with isolated footing 1 importance factor

Thus seismic coefficient is taken as

0.096

a.

Seismic force due to Superstructure height of action of the force

= = =

b.

Seismic force due to Abutment act at cg of abutment

= =

c.

Seismic force due to Back fill acts at cg

= =

0.096 3.10 + 3.5 m

x 296.058 0.025 +

0.096 x 2.05 m 0.096 x 2.05 m

362

413.28 =

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 li load shall not be considered when acting in the direction of traffic,..". 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 = = height of the line of action of the brake from the road level = height of the bearing level from the top of base =

55.4 KN 1.2 3.100 m

Due to Resistance in Bearing (Temperature) Severe moment is produced when the LL is within span. Assume frictional coeff. =

0.225

= Frictional resistance at free bearing

=

310.9 KN

Frictional resistance of fixed bearing

=

276.3 KN

Unbalanced frictional force = Line of action of the force from the bottom of the abutment =

34.5 KN

0.2

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 = Avg. height of superstructure from the bed = intensity of wind load = velocity of wind = 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

b.

Wind Force on Moving vehicle Class A train acts at 1.5 m above road level

=

20.4 x =

Maximum load is obtained in exposed area = For lateral direction wind load = acts at =

10 m² 6.000 m 0.73 KN/m² 107 km/hr

10 =

1.500

3= m

68.5 KN 68.5 KN 4.100 m

SUMMARY OF LOADS FOR DESIGN OF ABUTMENTS S.N o.

Load

Vertical (KN)

Horizontal (KN) Longit. dir

1

2

Superstructure 1.1 Dead load of SS(D) 1.2 Live Load of SS (L) including impact Substructure DL of abutment(E) Retaining wall Return wall - Top section Ballast wall Backfill (BF) Earth pressure (EP) (EP1) Due to overburden(EP2)

296.1 438.0

301.3 31.7 28.8 413 413.526


253.6

Later.

3 4 4.1 4.2 4.3 4.4 4.5 4.6


Frontal passive soil mass (EPp) Buoyancy (B) Longitudinal force Tractive force (T) Due to temperature (F) Wind Load Centrifugal force (CF) Thrust due to water current (WC) Seismic forces (FS) on superstructure on abutment on live load from backfill

55.40 34.54 68.50

14.21 17.37

28.42 34.73

28.42 34.73 42.05 39.675 39.6749

19.84

Load combination Group II (Normal+Temp. condition) D + L + E + Ep1+ Ep2 + Ep3+ T + F + CF Load combination Group IV (for Seismic dry condition) D + L + E + Ep1+ Ep2 + T + F + CF + FS

1096

344

68.5

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 (Under combination group IV the permissible increase of stress is Long. Moment = 753 KN-m 50%, hence Combination I governs the design longitudinal moment) Substructures are designed using M20 concrete and Fe415 steel reinforcement

Conccrete M20

Parameters for RCC design sck = 20 N/mm2

sy = Steel 415 N/mm2 Approximate Area of Steel reinforcement required is calculated considering singly reinforced beam theory Ast = Mu/ (0.87xσy x 0.7984 x d) = Provide Required spacing =

3332 mm2


Area of reinforcement provided, Ast = r=

Ast = Provide at

490.625 mm2 100 mm c/c

28701.563 mm2 0.6378125 % >

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

0.2 %

Provide Required spacing =

20 mm dia. bars 20 113.1 mm c/c

Ast = Provide at


314 mm2 100 mm c/c

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 = Provide Required spacing =

0.0012 x


Check for stresses

Ast = Provide at

750 x 1000=

113.04 mm2 250 mm c/c

(For seismic condition)

Parameters for RCC design Conccrete M20

Normal Thrust , W = Bending Moment = Ecentricity, e = breadth, b = Total depth , D = Effective depth, d = Compressive steel cover, dc = Tensile steel cover, dt =

1560553 ######### 643.948 6000 750 662.5 75 75

mm mm mm mm mm mm

900

N N-mm > m=

Ac = At =

Steel

sck = sy =

165.625 mm 0.0

18369 mm2 28701.56 mm2

Ast = Mu/ (0.87xσy x 0.7984 x d)3768.77108 = mm2/m

x=

0.87*σy*Ast 0.36*σck*b

189 mm

150 mm say

Check for shear stress: Critical secion for shear in case of abutment stem is considered at distance d = 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) =

662.5 mm from the junction

A

0.125 N/mm2

Permissible shear stress without shear reinforcement tcmax =2.8


2

Where, As = Area of Steel provided in mm b = width of abutment vertical wall and d = thickness of abutment wall tc = basic values given for different p% of Steel provided = From Table 19 of IS 456-200 The design of ballest wall and return wall will be as in left side abutment.

p=

0.21%


2.1.3 DESIGN OF BALLAST WALL

1.000 m

RCC balast wall (as shown in the figure) monolithic to the abutment has been designed. 0.20 m

CALCULATION OF DEAD LOAD STRESSES Dead load from the approach slab (triangular portion) = = Dead load of the Ballast wall

36 KN 6 KN/m

=

28.8 KN 4.8 KN/m

= Earth pressure from the back fill

= 21 KN/m3 0.2973

r= ca =

ca x r x h2 /2 =

3.12 KN/m

LIVE LOAD STRESS CALCULATION Class AA (tr) load component on wall , LL =

Add impact to live load @ , I =

0.1

= = = e=

291.667 48.611 4.861 0.15

KN KN/m KN/m 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-I H = T or Pcopm+ Ep = V = DL + LL + I =

28.3 KN/m 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

=

Ast = Mu/ (0.87xσy x 0.7984 x d)256.816378 = mm2/m Provide 12 mm dia bars @ Area of steel provided = Provide 12 mm dia bars @ Area of steel provided = Horizontal links are provided of

200 c/c

for the continuation of bar from retaining wall 565.2 mm2/m 200 c/c on the opposite face 565.2 mm2/m 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 Where, As = Area of Steel provided in mm2

p=

N/mm2

0.57%

0.57

tc = basic values given for different p% of Steel provided = From Table 19 of IS 456-200 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 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 Vert. LL from Class A load = Add Impact due to LL @ 0.25=

168.72 KN 42.18 KN

56.24 14.06

x

0.200

KN/m KN/m

STRESSES DUE TO HORIZONTAL FORCES Horizontal Thrust due to DL+LL+IL=q*L*ka acts at h/2 0.28 m Moment due to hor. component of vertical loads =

433.2636 KN-m/r.m ka x rs x h^2/2

Earth pressure from the backfill = Moment due to earth pressure = Momt due to later. force =

= 24.070 KN/m from the top of foundation

16.998 KN-m 7.5 KN/m for the curb as per code = 135.000

=

KN-m

Total horz. uniformly distributed load = 24.07 + 0.94 + Total Moment = 433.26 + 17.00 + 135.000 = 585.26 depth required = M /( 0.138σck b ) = 620.925 mm where b = ( 700 + 400 ) / 2 = 550 mm Provide overall depth = 650 mm Ast = Mu/ (0.87xσy x 0.7984 x d) = 3530.97 mm2/m

0.94

7.5

Tensile steel re-bar equired = 3530.97 mm2 Provide 25 mm dia. bars Required spacing = 139.0 mm c/c

Ast = Provide at

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 toe width of abutment Height = 0.9 m Concrete cover 75 mm DEAD LOAD due to foundation =

490.87 mm2 150 mm c/c

1.000 m 1.000 m 0.750 m

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. WEIGHT OF BACKFILL BEHIND THE ABUTMENT Wb = Wf =

6.000 6.000 acts at

x x 0.875

1.000 1.000 m

x 4.100 x 0 from centre of abutment

Seismic force due to Foundation act at cg of abutment Seismic force due to back fill act at cg of backfill SUMMARY OF LOADS FOR DESIGN OF FOUNDATION S.No . Load

x x

= = = =

Vertical (KN)

18.00 = 18.00 =

442.8 0.0

0.096 x 0.45 m 0.096 x 2.95 m

356.4

Horiz (KN)

1

2

Superstructure 1.1 Dead load of SS(D) 1.2 Live Load on SS including impact (L) Substructure DL of abutment(E) Retaining wall Return wall Ballast wall Backfill Backfill

(BF) (FR)

442.8 from the bottom of foundation ba

L. arm (m)

296.06 438.0

1.2 1.2

301.32 31.68 28.80

1.4 3.7 1.4

442.80

2.25

Earth pressure (EP) Triangular dry(EP1)

377.2

Due to overburden(EP2) Frontal passive soil mass (EPp) DL of foundation 3 Buoyancy (B) 4 Longitudinal force 4.1 Breaking (breaking) force (T) 4.2 Due to temperature (F) 4.3 Wind Load

0.00 356.40

4.4 Centrifugal force (CF) 4.5 Thrust due to water current (WC) 4.6 Seismic forces (FS) on superstructure on abutment on live load on foundation from backfill Load combination Group I (N+T condition) D + L + I + E + Ep1+ Ep2 +BF+ T + F + CF Load combination Group IV (for Seismic dry condition) D + L + E + Ep1+ Ep2 +BF+ T + F + CF + FS Note: Moment is taken For Normal condition P= 1895 Mnet = Mr-Mo= x = Mnet/P= e = b/2-x'= Pmax=P/A*(1+6*e/b) = Pmin=P/A*(1-6*e/b) = Where, A=b x d =

-2.1

1.4

55.40 34.54

-4.0 -4.0

14.21 17.37

28.4 34.7

-3.5 -2.1

17.11 21.25

34.2 42.5

-0.5 -3.0

1895.1

467

1965.0

607

about the toe for check against overturning . For Seismic condition P= 1965 Mnet = Mr-Mo= x = Mnet/P= e = b/2-x'= Pmax=P/A*(1+6*e/b) = Pmin=P/A*(1-6*e/b) =

KN 1797.12344 0.94832107 0.42667893 ok 221.772 KN/m2 7.932 KN/m2 16.5 m2

CHECK AGAINST OVERTURNING Case: Normal Factor of Safety against overturning =Mres/Mov= Case: Seismic Factor of Safety against overturning =Mres/Mov=

2948.9419

/

1151.8 =

2948.9419

/

1086 =

CHECK AGAINST SLIDING Case : Normal The minimum vertical force will be when there is no live load and the front backfill is washed out = The resisting force againest sliding = (vertical force) x (coefficient of friction b/w concrete and soil) = 1457.06 x 0.45 = 655.676 The sliding force is the horizontal forces = 411.71 KN Hence, Factor of Safety against sliding =Pres/Psl= 655.68 / 411.7 = (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 =

The resisting force againest sliding = (vertical force) x (coefficient of friction b/w concrete and soil) = 1527.00 x 0.45 = 687.149 The sliding force is the horizontal forces = 551.59 KN Hence, Factor of Safety against sliding =Pres/Psl= 687.15 / 551.6 = STRUCTURAL DESIGN OF FOUNDATION

Wt of toe slab =

21.6 KN/m2

221.77 KN/m2

A

Design of toe slab:

E

D

F

G

144.012

85.692

1.000 m

0.750 m 2.75 m

Pressure distribution diagram

####

200.17 Moment about EF =

concrete grede : Steel grade :

87.13 KN-m d= M /( 0.138σck b ) 107.1 mm Adopt D = 600 mm d= 500 mm Ast = Mu/ (0.87xσy x 0.7984 x d) =604 mm2 Using 16 mm dia bar, spacing = Adopt 16 mm dia bar, spacing =

Minimum reinforc. as per code = 201.1 mm 150 mm c/c Area of re-bar provided =

Check for shear stress:

Shear stress =

V/(b d) =

542.20 KN

b=

500

6.00 m

161.3

Shear force at critical section V =

d =

#####

In toe critical section for shear is considered at a distance from abutment stem.

2

0.181 N/mm

Permissible shear stress without shear reinforcement tcmax =2.8 Mpa for M20 concrete Where, k1 = 1.14 - 0.7 d >= 0.5 ( d in m ) Where, As = Area of Steel provided in mm2 b = width of abutment vertical wall and d = thickness of abutment wall p= tc = basic values given for different p% of Steel provided = From Table 19 of IS 456-200

0.5 m

0.280 N/mm2 ok. Shear re-bar is not necessary.

0.037%

>

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/m2 88.2 KN/m2 3 Net down pressure = Moment about DG= d= M /( 0.138σck b ) Adopt

80 3 27.174 99.225

D= d= Ast = Mu/ (0.87xσy x 0.7984 x d) = Using 16 Adopt 16 Distribution re-bar 0.12/100 x 900 = Using 12 Adopt 12

KN/m2 KN/m2 KN-m mm

max min

600 mm 500 mm 189 mm2 mm dia bar, specing = mm dia bar, specing =

1.000 m

Minimum reinforc. as per code = 201 mm 150 mm c/c Area of re-bar provided =

1080 mm2/runnig meter mm dia bar, specing = 209.3 mm mm dia bar, specing = 200 mm c/c

at top and bottom

Check for shear stress: 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 Shear stress = V/(b d) = 0.083 N/mm2 Permissible shear stress without shear reinforcement tcmax =2.8 Where, k1 = 1.14 - 0.7 d >= 0.5 ( d in m )

N/mm2

Where, As = Area of Steel provided in mm2 b = width of abutment vertical wall and d = thickness of abutment wall tc = basic values given for different p% of Steel provided = From Table 19 of IS 456-200


p=

0.037%

0.28 N/mm2

>

k for correctness of data

8.000 m 3.000 m

0.000 5.000 ######

Unit Wt

Total Wt e from c/c

0.06 2.06 24 2.70 22 39.73 24 126.42 24 123.84 24 24.77 24 18.58 24 204.34 24 0.00 24 0.00 24 1.30 24 48.38 592.12 296.1 KN

0.000 m 2.338 m 0.238 m

i Pa

i 0 h= a= KN/m

Coul. Rankin

Ws

d 90-b


b

b Coulomb

F 30

b 90

d 20

30

Let us find out which theory is applicable:

60

1.7321 m <

0.75 0.9397 2.6845 1 0.75 0.5

1.5

4.100 m hence Ka= 0.2973 0.3333

15.384 KN/m Hor pressure from Rankin's theory = Horizontal thrust = 302.6 acts at

50.43 KN/m 1.37 m from the top of foundation footing.

KN

= 28.42 KN 0.75 / 2 =

=

34.73 KN

39.675 KN

6 (222.7) "The seismic force due to live

1/2 to one abutment m

in free bearing

in other bearing

7.3 KN

61.2 KN

Arm (m)

Moment (KN) Longit. Later. 0.16 0.16

47.7 70.6

0.00 -2.34 -0.24 0.00

-74.1 -6.8 0.0

1.72

436.7

3.10 3.10 4.10

3.50 2.05 5.60 2.05

171.7 107.1 280.85

99.5 99.47549 71.2 71.20224 235.4688 81.3 81.3335

752.9

280.9

1004.9

487.5

le increase of stress is gn longitudinal

Decrease spacing !

ok

mm2

at both faces.

ok

Parameters for RCC design 20 N/mm2 415 N/mm2

mm from the junction

>

2 0.125 N/mm

51.0 KN 8.5 KN/m

rom retaining wall

>

2 0.202 N/mm ok

specified in the code.

x 24 = =

15.84 KN 2.64 KN/m

KN/m

=

32.51 KN/m

Decrease specing !

)

62.8613 KN/m

KN KN

=

34.214 KN

= 42.509 KN from the bottom of foundation base

M[r]

M(o)

KN-m

KN-m

359.382 531.683

414.315 117.612 39.6 996.3

-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

eismic condition KN 1862.99 0.94809 0.43 =P/A*(1+6*e/b) = =P/A*(1-6*e/b) =

KN-m m ok 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 soil/concrete) ######

KN 1.246 >1.25 Hence Safe

C

7.9

B

1.000 m

M 20 Fe 415

87.1259

rc. as per code =

1000 mm2

Area of re-bar provided = 1339.733

122.4

mm2

0.04 2 0.181 N/mm

80

1000 mm2

of re-bar provided =

1339.7 mm2

0.04 2 0.0828 N/mm

ok

i Pa i Rankin Weight of soil is also to be considered in Rankin's Theroy. a theory is applicable: Coulomb's theory applicable

ndation footing.

Design of Abutment Bridge

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