Abut Design

7/29/2019 Abut Design

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BASIC DATA1.5m

Skew Angle 0 degrees

Overall Span L 10.000 m

Overall Span in skew direction Lskew 10.000 m

Clear Carriageway Bcw 7.550 m

Width of Solid Slab Wsolid 10.50 m

Width of Solid Slab in skew direction Wsolid skew 10.500 m

Depth of Solid Slab Dslab 0.8 m

Unit weight of concrete wcon 2.4 t/m3

Weight of Crash barrier wrail 1.0 t/m

DEAD LOAD CALCULATION

Dead LoadTotal wei (10.5*10*0.8*2.4) = 201.6 t

S.I.D.L.

Span = 10.000 m Width of the Carraige-way= 7.55 m

Superimposed dead load from Crash-Barrier= (1.0*10) = 10.00 t

Superimposed dead load from Crash-Barrier= (1.0*10) = 10.00 t

Superimposed dead load from Railing = (0.5*10) = 5.00 t

Superimposed dead load from Wearing Coat = (0.2*7.55*10) = 15.10 t

Total SIDL = (10+10.00+5.00+15.10) = 40.10 t

Total dead load+sidl reaction = (201.6+40.1) = 241.70 T

For design of substructure increasing by 5% = 1.05*241.7 = 253.79 T

Total Dead Load + SIDL on Abutment due to Superstructure = (253.79/2) = 126.895 T

Say= 127 T

DEAD LOAD CALCULATION

7.550m

Dslab


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1) Total Length of the Bridge : 10.0 m, Exp jt to exp jt

2) Span Arrangement : 1*10 m

3) Overall Width of Superstructure : 10.5 m

4) Type of Superstructure : Solid Slab

5) Depth of Superstructure : 0.8 m

6) Type of Bearings : Tar paper

7) Type of Substructure

Abutment : cantilever type Abutment

8) Type of Foundations

Abutment : Open foundation

9) Design Data

* Formation Level : 105.791 m

* Safe Bearing Capacity

Abutment : 15 t/m2

10) Properties of earthfill behindabutments / returns

* Dry Density : 1.8 t/cu.m

* Submerged Density : 2.07 t/cu.m

* f : 31 degrees

* Cohesion 'C' : 0 t/sq.m

* m : 0.377

11) Founding Levels

*Abutments : 107.352 m

* return wall : 107.352 m

12) Return Walllength : 2.500 m

Abutment cap top Level : 104.991 m

SALIENT FEATURES


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Front Ground Level : 109.352 m

High Flood Level : 113.976 m


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Reaction has been calculated for the following cases

1. One lane of class 70-R(W)

2. One lane of class 70-R(T)

3. One lane of class - A

4. Two lane of class - A

(For Elastomeric Bearing) : 4

Horizontal force at each end including braking = Fh/2+Vr.Ltc

(As per Cl.214.5

Vr = Shear rating of the elastomeric bearings

Ltc = Movement of deck above beraing , other than that applied forces.

(As per IRC-83 Part-II)

Total Longitudinal Strain , d = 0.0005

Movement of the deck , Lrc = (0.5*0.0005*(9.6+0.2+0.2)) = 0.0025 m

GA (102.04*0.135)

h 0.048

G = 102.04 t/m2

A = 0.135 m2

[ Lo = 0.45 m Bo

h = nhi+2he = (3*0.012+2*0.006) = 0.04

hi = Thickness of individual internal layer of elastomer = 0.012

he = Thickness of top/bottom layer opf elastomer = 0.006

n = No. of internal elastomer layer = 3

= (286.99*0.0025)

= (no. of bearing*0

= (4*0.72)

= 2.87 t

Case 1 : One lane of class 70-R(W)

4.4 m Cg of 92t

0.2 m 9.60 m

Ra

Ra = 92*(9.6-4.4+0.2)/9.6 = 51.8

Rb = 92-51.8 = 40.3

Braking Force = 0.2*92/2+2.87 = 12.1

CL of 70-R CL of c/w

1.630

Transvers eccentricity = 3.120

Transvers moment = 3.12*51.8 = 161.5

CALCULATION OF LIVE LOAD REACTION FOR ABUTMENT

Horizontal force without any applied forces for one bearing, V.Lrc

Total horizontal force without any applied forces ,V.Lrc

28

No. of Bearing =

Vr = = =

9.5

3.120


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Case 2 : One lane of class-A

2.99 m Cg of 36.4 t braking Force

0.2 m 9.60 m

Ra

Ra = 36.4*(9.6-2.99+0.2)/9.6 = 25.

Rb = 36.4-25.8 = 10.

Braking Force = 0.2*36.4/2+2.87 = 6.5

CL class A(1L) CL of c/w

1.30

9.5 m


Transvers moment = 3.45*25.8 = 89.

Case 3 : Two lane of class-A

Ra = 2*25.8 = 51.

Rb = 2*10.6 = 21.

Braking Force(for 1 lane only) = 6.5

CL class A(2L) CL of c/w

3.05

9.5 m


Transvers moment = 1.7*51.6 = 87.

Case 4 : One lane of 70R Tracked

cg of 70t Load

2.285m 2.285m

0.200m 9.600m 0.200m

Ra Rb

Ra = 70.00*7.315/9.6 = 54.80t

Rb= 70-54.80 = 15.20

CL of carriageway

2.60 2.2

9.500m

transverse moment = 118.09t-m

braking force = 9.870t = 115

Summary of Loads

1L class 70 - R

From the above 70R tracked case has been considered in the design of Abutment

mMax. vertical reaction (t)

12.1

6.5

6.5

Load case

51.8

25.8

51.6

1L class - A

3.45

Braking force (t)

1.7

2L class - A

70R-Tracked 54.8 9.9


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DESIGN OF ABUTMENT STEM

Width of abutment 8.500 m

Thickness of abutment 1.000 m

Reinf. requred at base = 346.46 cm2

Reinf. Provided at base = 390.6 cm2

OK ( 42Nos 28 f ) @ 203.6585

= 0.46 % ( 42Nos 20 f ) @ 203.6585

Reinf. requred at the mid = 188.02 cm2

Reinf. Provided at the mid = 258.6 cm2

OK ( 42Nos 28 f ) @ 203.6585

= 0.30 % ( 0Nos 16 f ) @

Minimum Reinf. In Pier Column

Minimum Reinf. Reqd. ( 0.3% ) as per cl. 306.2.1 = 0.003*8.5*1* 10000

= 255.0 cm2

Transverse Reinforcement in Pier ColumnsAs per Cl. 306.3.2 of IRC:21

Min. Dia of trans. Reinf. = 1/4 * Largest bar dia. of main reinf.

= 1/4*28 = 7 mm

Bar Dia. Provided = 10 mm @ c/c 150 mm OK

As per Cl. 306.3.3 of IRC:21

Pitch = 12 * dia. of smallest bar of main reinf.

= 12*20

= 240 mm

The calculations for reinforcement requirement at various sections is presented here. A standard in house program has been

used for calculation of stresses in concrete and steel in a rectangular section subjected to combined axial and bending forces.

of IRC:21-2000


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The abutment cap has been designed as per Clause 716.2.1 , IRC :78- 2000 .

The abutment cap shall be reinforced with a total minimum of 1% steel distributed equally at both facesand in both directions assuming a cap thickness of 225 mm.

Length of abutment cap = 8.5 m

Width of abutment cap = 2.12 m

Depth of abutment cap assumed = 0.225 m

Reinforcement in the direction of length of abutment :-

Area of steel required (mm2) = 2385 mm

2

Providing steel by distributing equally at top & bottom = 16 nos. 12 f @ 131 mm

Area of steel provided (mm2) = 3619 mm

2 OK

Reinforcement in the direction of width of abutment :-

Area of steel required (mm2)/m = 1125 mm

2

Providing stirrups of 16 mm dia. 2 legged

Spacing if stirrups required = 357.4 mm

Therefore, providing 12 f , 2 legged stirrups @ 150 c/c .

Depth of abutment provided = 1000.00 mm

DESIGN OF ABUTMENT CAP


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0.92

Design values :

0.300 m g = 1.80 t/m22.160

1 2 ka = 0.279

1.085 t/m2 0.603 t/m2

1)Earth Pressure due to surcharge equivalent to 1.2m of earthfil l = ka *g*1.2

= 0.603 t/m2

2)Earth Pressure due to backfill of earth = k a *g*h

= 1.085 t/m2

Bending moment at the base of dirtwall due to earth pressure (1) = ka *g*1.2*h2/2

= 1.406 t-m/m

Bending moment at the base of dirtwall due to earth pressure (1) = ka *g*h3*0.42/2

= 1.063 t-m/m

Total bending moment at the base of dirtwall = 2.469 t-m/m

due to earth pressure

Calculation of force and moment due to the effect of braking :(cosidering 40t bogie loading)

2.245 10t 8.5 10t

2.79m

2.160

1.750 2.79 2.160

Effective width = 6.700 m

Braking force, 0.2*20 = 4 t

braking force including impact of 50%= 6 t

Braking force per metre width = 0.90 t

Bending moment at the base of dirtwall due to effect of braking = 1.93 t-m/m

Therefore total bending moment at the base of the dirtwall = 4.40 t-m/m

height of

dirtwall,h = 2.160 m

DESIGN OF DIRTWALL

Earth Pressure diagram

width of

dirtwall =0.300 m


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Basic Design Data:

Grade of conc. 35

Grade of steel 500

Dia of bar used 12Permissible stress in concrete 1190 t/m2

Permissible stress in steel 24000 t/m2m , Modulur ratio 10

K value for concrete 0.332

j value for concrete 0.889Q for concrete 175.55

Max. moment in dirtwall (t-m) 4.40

Effective depth required (mm) 158

Effective depth provided (mm) 244

Ast required (mm ) 845

Provide longitudinal reinforcement:

f16 @ 150 c/c

Ast provided (mm2) 1340 Thus OK

Therefore providing 12 f @ 250 c/c on the approach side and 10 f @ 200 c/c on the outer sidein the vertical direction . Also providing 10 f @ 150 c/c on both faces in the horizontal direction .


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DESIGN OF FOOTING SLAB

L

3 10.836t/m^2

4.075t/m^2 2

10.500m

T T

6.219t/m^2 1

4 13.784t/m^2

L

From the design of abutment we obtain the maximum and minimum base pressures

considering LWL case as the governing case

Axial load = 768.837t

Net longitudinal moment = 445.0900t-m

transverse moment at base = 161.5000t-m

longitudinal section modulus at base = 117.670m^3

transverse section modulus = 150.675m^3

area of base = 86.10m^2

base pressure at point 1 = 13.784t/m^2

base pressure at point 2 = 10.836t/m^2

base pressure at point 3 = 4.075t/m^2base pressure at point 4 = 6.219t/m^2

Design of toe slab

D

a

1.3 C

0.5 B

3.5 0.9 3.8

5.15t/m^2 a 12.310t/m^2

8.200m


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Average base pressure at point 1 = 12.310t/m^2

Average base pressure at point 4 = 5.147t/m^2

base pressure at junction-section a-a= 8.991t/m^2

Bending moment at face of stem

Depth of soil cover above footing = 2.000m

loadings Element Description Area factor Force (t) lever arm(m)

moment(t-

m)

B

rec portion of

conrete 1.000 4.56 1.9 8.664

C

triangular

porton of

concrete 0.500 3.648 1.27 4.63296

D

rectangular

portion of soil

cover 1.000 4.788 1.9 9.0972

E

triangular

porton of soilcover 0.500 2.736 2.53 6.9312

F

rectanguar

portion of

base

pressure -1.000 -46.78 1.9 -88.879413

G

triangular

portion of

base

pressure -0.500 6.31 1.27 7.98890989Total -24.74 -51.57

Bending moment at face stem = -51.57t-m

Grade of concrete = M30permissible stress in concrete = 1019 t/m^2

permissible stress in steel = 20400 t/m^2

m = 10

k = 0.333

j = 0.889

Q value for concrete = 150.963t/m^2

cover to substructure = 75.000mm

assuming a 16 dia bar,eff depth at junction= 1.217m

effective depth required at the junction = 0.584m

area of steel required = 23.36cm^2/m

minimum steel required 0.12% = 7.80cm^2/m

provide25 dia bar @190c/c = 25.8cm^2/m

safe

distribution steel for the tension face of toe slab,providing 0.25*(DL+LL) at the

junction = 5.84cm^2/m

minimum steel required = 7.80cm^2/m

provide16 dia bar @200c/c = 10.1cm^2/m

Downwardpressure

upwardpressure


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14/32

check for shear

deff section

1.22m

1.30m

0.50m

3.50m 0.90m 3.80m

effective depth distance from stem junction = 1.217m

depth at 'd eff' distance from stem junction = 1.044m

effective depth at 'd eff' distance = 0.961mtan b = 0.229

base pressure at effective depth distance from

junction = 10.1t/m^2

Depth of soil cover above footing = 2.000m

Loadings Element Description Area factor Force Lever arm Moment

A

rectangular

part of

concrete 1.00 3.10 1.29 4.00

B

trianglar part

of concrete 0.50 1.69 0.86 1.45

C

rectangular

part of soil

cover 1.00 4.45 1.29 5.74

D

trianglar part

of soil cover 0.50 0.70 1.72 1.21

E

rectangular

part of base

pressure -1.00 -31.80 1.29 -41.07

F

trianglar partof base

pressure -0.50 2.91 0.86 2.51Total = -18.95 -26.15

downwardpressure

upward

pressur


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shear force at critical section = 18.95t

bending moment at critical sction = 26.15t-m

net shear force at critical section = 12.73t

net shear stress at critical section = 13.25t/m^2

0.1325Mpa

percentage of reinforcement = 0.2689

permissible shear stress = 0.2424Mpa

safe in shear

summary of reinforcement in Toe slab

main steel 25 f @190c/c

dist. Steel 16 f @200c/c

main steel 16 f @200c/cdist. Steel 16 f @200c/c

Bottom

Top


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HEEL SLAB DESIGN FML= 9.965

BRL = 8

3.164 CTRL = 7.7

STRL = 6.7

D

4.9

C

1.8 B

1.8 0.000

2.05

2.85t/m^2 41t/m^2

3 a 2

M-T

8.5

M-L

4 1

a

From the design of abutment we obtain the maximum and minimum base pressures

5.100m

1.000

1.000

5.100m

A


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considering LWL case as the governing case

Axial Load = 961.00t

Net longitudinal mom = 712.00t-m

Transverse moment = 99.90t-m

Longitudinal section modulus at base = 36.8475 m^3Transverse section modulus at base = 61.4125 m^3

Area of the base = 43.35 m^2

summary of base pressures at points 1,2,3 & 4

Base pressure at point 1 = 43.12




Base pressure at junction a-a = 29.37 t/m^2

Summary of bending moment at the junction

due to soil and concrete

Element no Force Lever arm Moment

A 15.12 1.025 15.50

B 0.00 0.68 0.00

C 0.00 1.37 0.00

D 36.77 1.025 37.69

Total = 51.89 53.19

Summary of bending moment at the junction

due to base pressure

Element Force Lever arm moment

Description

rectangular portion of

concrete

triangular portion of

concretetriangular portion of

soilrectangular portion of

soil

Description


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1 -60.20 1.025 -61.71

2 27.18 1.37 37.2366

-33.02 -24.4734

Bending moment at face of stem = 28.72 t-m

Shear force at face of stem = 18.87 t

Grade of concrete = M35

Grade of steel = 500

m = 10

permissible stress in steel = 24000 t/m^2

permissible stress in concrete = 1189.26 t/m^2

k = 0.33

j = 0.89

Q = 175.26 t/m^2

Effective depth required at junction= 0.40 massuming dia of steel = 16 mm

cover to the substructure = 75 mm

Effective depth provided at junction= 1.717 m

safe

area of steel required = 33.23 cm^2

provide 32 f bar @120 = 67.02 cm^2

safe

Design for shear

Heel has been checked for shear at the junction of heel portion

Effective depth = 1.717

shear force at the junction = 33.02 t

shear stress at the junction = 19.23 t/m^2

= 0.19 Mpa

triangular portion of

base pressure

diagram

Total =

rectangular portion of

base pressure

diagram


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permissible shear stress

percentage of reinforcement = 0.39 %

permissible shear stress = 0.35 Mpa

safe


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ka = 0.279

g = 1.80

Live load Surcharge = 1.20 m

(SHt)

Q1=ka*g*SHt Q2=ka*g*b

The end return wall has been designed as a plate fixed on its two faces i.e. at the base and on one of its vertical sides.

( Refer Table 26 of Formula of Stress & Strain by Roark & Young ).

1 8.165 2.05 0.251 0.603 0.183 0.190 0.573 0.265

0.251 3.729 0.134 0.090 0.423 0.151

500 248 740 800 -6.77

bar diaspacing

reqd

spacing

providedbar dia

spacin

g reqd

spacing

provided

20 f 619 c/c 150 c/c 12 f @ 125 c/c 12 f 300 c/c 125 c/c @

Ast provided (vertic Ast provided (horizontal) =

CHECK FOR SHEAR

0.251 0.603 0.183 0.190 0.573 0.265

DESIGN OF RETURN WALL

12 f

2999.17 mm2 1809.56 mm2

Earth face

Outer face

376

(at x=0 &

z=0.4b) Ma2

(t-m/m)

(at x=a & z=0) Mb1(t-m/m)

thk. of

end

return

wall at

top

Design of end return wall

a = 2.050

b

=8.1

65

Q1

(t/m2)

b1

(at x=0 &

z=0.8b) Ma1

(t-m/m)

-1.27

b2

Outer face

Horizontal reinforcement

Uniform load due to live load surcharge over entire plate

Ast for vertical

reinforcement

(mm2)

507

Moment in

horizontal

direction(Mh)

-5.02

Ast for horizontal

reinforcement (mm2)

"a/b" (at x=a & z=0) Mb2(t-m/m)

-5.54

Varying load due to earth pressure over entire plate

-3.75

(at x=a & z=0) R1

(t-m/m2)

2.819

(at x=0 &

z=0.8b) R2

(t-m/m2)

1.305

Uniform load due to live load surcharge over entire plate

b1 g2

g1

For "a/b" Q1

(t/m2)

Earth face

Deff

provided

Deff

reqd.

g2

-1.23

b2

Moment in

vertical

direction(

Mv)

Vertical reinforcement

g2

S.NO.

Height

of

return

wall

(m) "b"

Length

of return

wall (m)

"a"

"a/b"

b2 g1

Q2

(t/m2)

b1

thk. of

end

return

wall at

bottom

g1


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0.251 3.729 0.134 0.090 0.423 0.151

At x = a & z = 0

mm

mm

V (t-m/m2)

tv N/mm2

r

As per IRC 21-2000(Cl 304.7.1.3.3),

Permissible Shear Stress (0.5% steel)

N/mm2

Increase in permissible stress (d)

N/mm2

tv tc Thus OkAt x = 0 & z = 0.4*b

mm

z=0.4b 0.4 mm

V (t-m/m2)

tv N/mm2

r

As per IRC 21-2000(Cl 304.7.1.3.3),Permissible Shear Stress (0.5% steel)

N/mm2

Incease in permissible stress (d)

N/mm2

tv tc Thus Ok

0.214

1.00

0.214

10.80 o.k.

Check For One way Shear :

According to Cl. 307.2.5.4 of IRC-21,2000. Pile Cap is not checked for shear if it is designed by truss analogy.

Pile Cap is checked for shear as per Cl. 307.2.5.4 of IRC-21,2000.

The pile cap is checked for one way shear. The critical section for shear is considerred as thesection located deff distance from face of abutment.

Distance of the critical section from the face of the abutment(at deff distance) = 1625.0 mm

Distance of the Maximum Reaction Force from the Face of the abutment = 1300 mm

Distance of the critical section from the Maximum Reaction Force = (1,300-1,625) mm

= -325 mm < 600mm

Width of Pile Cap = 8.700 mShear at deff distance from face of abutment cap = 154.7 tShear at deff distance from face of abutment cap/m = 17.78 t/m

Pile Cap is checked for shear as per Cl. 307.2.5.4 of IRC-21,2000.The pile cap is checked for one way shear. The critical section for shear is considerred as thesection located deff distance from face of abutment.

deff provided = 1.625 m

Shear Stress = 17.78/(1.625)= 10.94 t/m2

100*Ast/bd = 0.302

Permissible Shear Stress tc = 24.666 t/m2

Shear carried by concrete, Vc = 24.666*1.0*0.000= 40.1 t/m

Greater than Max. Design Shear,Thus no Shear Reinforcement is required,

(In One Layers at top in both directions)

one way shear check is required.


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But minimum shear reinforcement will be provided as per Cl.304.7.1.5 of IRC-21,2000.

Asw 0.4bs 0.87fy

Where, s = Spacing of the Bars = 1000 mm

b = Width of the abutmen = 8700 mm

fy = Yield Strrength of steel = 500 MPa

Shear Reinforcement, Asw = (0.4*1000*8700)/(0.87*500) mm^2

= 8000.00 mm2

= 80.000 cm2

Provide 10 dia. 18L Stirrups @125 Spacing

Ast provided = 113.10 cm2

>80.00cm^2 , Thus O.K.

For construction convenience, required stirrups may be provided in alternate la ers as follows

10 dia. 18L Stirrups @250 c/c

+10 dia. 18L Stirrups @250 c/c

Check For Punching Shear around pile :

750 mm

1413 mm

2825 mm

750 mm

Permissible Punching Shear Stress around the pile = 0.16sqrt(f ck)

= 0.16*sqrt(35)

= 0.947 Mpa

Maximum Reaction on one pile = 233.0 t

Perimeter of area on which punching shear acts (taken as the minimum of two possible cases) = 3.719 m

Distance of the critical section from the face of the abutment = 0.813 m

Punching Shear Sress around the pile = 233.03/(3.72*1.625)

= 38.56 t/m2

= 0.39 Mpa


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Check For Punching Shear around abutment column :

10125 mm

0 mm

2625 mm

Permissible Punching Shear Stress around the abutment = 0.16sqrt(f ck)

= 0.16*sqrt(35)

= 0.947 Mpa

Maximum Reaction on single abutment = 961 t

Self wt. of Pile Cap = 114.8 t

Total Load = 1076 t

Perimeter of area on which punching shear acts = 25.50 m

Distance of the critical section from the face of the abutme = 0.813 m

Punching Shear Sress around the abutment = 1,076/(25.50*1.625)

= 25.96 t/m2

= 0.260


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Calculation of Top R/F for Sagging Moment :

Width of pile cap = 5.1 m

Net bending Moment = 0.0 t-m

Moment per unit width = 0.0 t-m

CALCULATION OF DESIGN PARAMETERS


Grade of steel = Fe 500

Unit. Wt of Concrete= 2.4 t/m3

Permissible stresses:

sst =

sbc =

Basic Design Parameters:

r= sst/sbc = 20.17

m= 10

k = m/(m+r) = 0.331

j = 1-k/3 = 0.890

q = 0.5*scbc*k*j = 175t/m^2

deff reqd. = sqrt(0.01/(175.44*1.0))

= 0.008 m = 7.5 mm

deff provd. = {(2.25*1000)-150-16-16/2} = 2076.0 mm

Main Bar Dia. Provided = 16 f mm

Distribution Bar Dia. Provided = 16 f mm Clear Cover Provided (including p = 150 mm

Ast reqd = (0.0*1000)/(24,000.0*0.890*2,076.0) = 0.000000 m2

= 0 mm2

= 0.00 cm2/m

Provide 16 f @ 125 c/c0 f @ 0 c/c

Rein.provided = 16.1 cm2/m > 0.00 o.k.

Calculation of Bottom R/F for Support Moment :

Width of pile cap = 5.1 m

Net bending Moment = 0.0 t-m

Moment per unit width = 0.0 t-m

CALCULATION OF DESIGN PARAMETERS



Unit. Wt of Concrete= 2.4 t/m3

Permissible stresses:

sst =

sbc =

Basic Design Parameters:

r= sst/sbc = 20.57

m= 10

k = m/(m+r) = 0.327j = 1-k/3 = 0.891

q = 0.5*scbc*k*j = 170t/m^2

1167t/m^2

1190t/m^2

Thus O.K

(In One Layers at top)

24000t/m^2

DESIGN OF PILE CAP (TRANSVERSE DIRECTION)

24000t/m^2


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deff reqd. = sqrt(0.00/(170.01*1.0))

= 0.000 m = 0.0 mm

deff provd. = {(2.25*1000)-150-25-16/2} = 2067.0 mm

Main Bar Dia. Provided = 16 f mm

Distribution Bar Dia. Provided = 25 f mm Clear Cover Provi = 150 mm

Ast reqd = (0.0*1000)/(24,000.0*0.891*2,067.0) = 0.000000 m2

= 0 mm2

= 0.00 cm2/m

Provide 16 f @ 125 c/c In One La ers at bottom in transverse dir.0 f @ 125 c/c

Minimum Ast reqd. (as per IS :456) = 0.06% of Cross-sectional area.

= (0.0006*2.25*1.0*1000000)

= 1350 mm2

= 13.50 cm2

Rein.provided = 16.08 cm2/m > 0.00 o.k.

Thus O.K


32/32

So, we will follow here IS 2911 ,part 1/section 2

No. of Pile = 6

Pile diameter = 1.2 m

Pile length = 20.000 m

Normal Case :

Transverse Horizontal Force = 0 t

Longitudinal Horizontal Force = 284 t (from the output - 1 of ABUT2 fortran programme )

Resultant Horizontal Force = sqrt(0^2+284^2) = 284.00 t

Longitudinal Seismic Case :

Transverse Horizontal Force = 0 t

Longitudinal Horizontal Force = 358 t (from the output - 1-s of ABUT2 fortran programme )

(In house programme )

Resultant Horizontal Force = sqrt(0^2+358^2) = 358.00 t

Maximum Lateral load = 286.40 t

Q = Lateral Load in each Pile = (286.40/6 = 47.73 t

0.00

T = (E*I/K1)^(1/5) K 1 = 0.146 kg/cm3

(337,216.55*0.10*100000000/0.146)^0.2/100)

= 4.72 m

Q Q DRY Submerged

Loose sand 0.26 0.146

L1 L1 Medium sand 0.775 0.525

Dense sand 2.075 1.245Very loose sand under - 0.04

Le Lf repeated loading ornormally loading clays

L1 = 0.0

L1/T = 0.0

(From Fig 2 of IS:2911(Part 1, Sec-2)-AppendixB Lf/T = 2.2

Lf = (4.72*2.2) = 10.39 m

Grade of concrete = M35 Permissible Stress:

E = 57000*(fck)0.5

kg/cm2

Concrete scbc= 11.67 Mpa

= 57000*(35^0.5) 1167 t/sqm (In Normal Case)

= 337216.5 kg/cm2

= 3372165.5 t/m2

Steel sst= 240 Mpa

24000 t/s m (In Normal Case)


Dia. of pile = 1.20 m

I of pile = (3.14*1.2^4/64) = 0.1018m^4

For fixed headpile, the fixed end moment of the equivalent cantilever is computed using the following equation :Mf =Q*(L1+Lf)/2

= 47.73*(0.00+10.39)/2

= 248.01 t-m

0.0

(m) = 0.82 (From Fig 3, IS:2911(PartI/Sec2)

m*Mf = 203.37 t-m

DESIGN OF PILEBored Cast in-Situ Piles

Final check OK

VALUES OF CONSTANT K1 ( kg/cm* )

Reduction Factor (m) , for L1/t or L1/R =

The Actual Maximum Moment (M) =

Abut Design

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