sarada well

7/29/2019 Sarada Well

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DISCHARGE CALCULATIONS.

Discharge by Catchment area method:

a) By Ryve's formula = 860 cumecs

b) By Dicken's formula = 1980 cumecs

c) By Rainfall Method = 1160 cumecs

Discharge by A-V Method:

a) = 1140 cumecsb) = 1140 cumecs

c) = 1100 cumecs

d) = 820 cumecs

e) = 1140 cumecs

The Discharge by A.V method and rainfall method are coinciding.

Hence discharge of 1140 cumecs is taken as design discharge.

At 100m D/S

At 20m U/S

At 100m U/S

At Site of crossing.

CONSTRUCTION OF HLB ACROSS RIVER VEGAVATHI NEAR VANTHARAM

@ KM 11/8-10 ON PANUKUVALASA - BALIJIPETA ROAD IN

VIJAYANAGARAM DISTRICT.

At 200m D/S

1


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CONSTRUCTION OF HLB ACROSS RIVER VEGAVATHI NEAR VANTHARAM

@ KM 11/8-10 ON PANUKUVALASA - BALIJIPETA ROAD IN

M.F.L = +13.06

L.B.L = +7.58

V = 2.66 m/sec

Design discharge Q = 3064.2 Cumecs

D = 13.06 - 7.58 = 5.48 m

d = 0.15 m

ha = = 0.34167 m

= 0.49167 m

Vv = = 2.80 m/sec

Linear Water way required = Q/D*Vv = 200.04 m

14 vents of 16.5 m = 231 m

However provide 14 vents of 16.5m effective with footpaths from bank to bank.

Total Head

Velocity through Vents

Velocity of flow at M.F.L

Depth of flow at M.F.L

Assuming afflux

LINEAR WATERWAY CALCULATIONS.

Head due to velocity at approach V2.D

2/ 2g ( d+D)

2

Cd2gh

2


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3


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4


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Design Discharge 3983.46 cumecs

M.F.L +13.06

L.B.L +7.58

Sill Level +7.58

Vertical Clearance 1.65 m

Thickness of Deckslab D 1.675 m

Wearing coat thickness 0.1 m

R.C.L = MFL + AFFLUX +VERT. CLEAR +D ( DECK) +WEAR.COAT = +16.64

Linear water way required = 200.04 m

Span arrangement provided 14 vents of 16.50 m eff. With footpaths.

Mean Scour Depth = 9.476 m

F.L for Pier as per Scour depth = -+12.146

F.L for Abutment as per Scour depth = -+2.946

Based on the soil report the following F.L' s are proposed with about 1.50m embedment into S.D.R

A1 , A2 -+2.95

P2,P5,P9,P11,P12 -+12.146P3,P4,P6,P7 +88.000

P1,P8,P10,P13 +89.350

Length of flywing = RCL-SILL LEVEL - BASE WIDTH

= 7.000 M

PROVIDING 350MM THICK FLYWING

3.5 m

0.45 m

Weight of Flywing : 3.5 m

(3.5x.45+0.5x3.52)0.35x2.40 = 6.468 t

C.G. of flywing:A X AX

1 1.575 1.75 2.756

2 6.125 1.167 7.146

Total 7.7 9.902

C.G = 1.286 m

CONSTRUCTION OF HLB ACROSS RIVER VEGAVATHI NEAR VANTHARAM @ KM

ON PANUKUVALASA - BALIJIPETA ROAD IN VIJAYANAGARAM DISTRICT.

PRELIMINARY DATA


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6

11.00

Thickness of Wall steining h = Kd = 0.657

Provide 600mm thick for Pier also.

75mm 700mmCURB HEIGHT

300mm

30

100mm

Angle = 30

Height = 0.775 + 0.3 = 1.643

Tan300

Provide 1.7 m curb height.

L


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1/8-10


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DESIGN OF BACKING WALL & BED BLOCK:

0.30

CONCRETE MIX = M20 0.10

RCL 14.282

R.C.L = 14.282

= 35 deg.1.7

Deck thickness = 1.7 m

Wearing coat thk. = 0.10 m

Ka = 0.246

0.211

= 1.80 t/cum

0.30

Width of backing wall = 0.30 m

Live load surcharge = 1.20

Thickness of Bed = 0.30 m

block

Width of Bed block = 1.05 m

(h) Height of Backing wall h = 0.10 + 1.7 0.211 = 2.01 m

a) Moment due to Earth Pressure:

Pa = 1/2 Ka h2

= 0.895 t/sqm

Lever arm = 0.42 X 2.01 = 0.845 m

Moment = 0.895 X 0.845 = 0.76 t-m

3

B.M due to fluid pressure of 480 kg/sqm = 480 2.011 = 0.651 t-m

1000 6

< 0.756 t-m

b) Moment due to Live load surcharge:

Intensity of Live load surcharge at RCL = 0.246 1.80 1.20 = 0.531 t/sqm

2Moment = 0.531 X 2.01 2 = 1.074 t-m

c) Moment due to Breaking force:

Max.axle load is 20.00 tonne for class 70r wheeled load.

Breaking force is 20% of axle load I.e = 20.00 X 20 = 4.000 t

100

BACKING WALL:


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Breaking force under each wheel = 2.00 t

a) Dispersion width at top of bed block = 0.86 2 2.01 = 4.88 m

(Tyre width = 0.86 m) (For second wheel)

b) Dispersion width at top of bed block

(For first wheel)= 0.475 + 1.2 + 0.860 + 2.01 = 4.55 m

Intensity of second Wheel load / RM = 2.00 4.88 = 0.410 t/m

Intensity of first Wheel load for overlapping portion = 2.00 4.55 = 0.44 t/m

Total Wheel Intensity = 0.41 + 0.44 = 0.85 t/m

Lever arm = 2.01 + 1.20 = 3.21 m

Moment due to Breaking force = 0.850 X 3.211 = 2.728 t-m

Total Moment = 0.76 + 1.074 + 2.728 = 4.56 t-m/m

For V.R.C.C M20 mix: Q = 7.702 j = 0.916

d = 4.56 * 105

= 24.33 cm

100 * 7.70

d available = 30.00 5 0.6 = 24.40 cm

> 24.33 cm

A st = 4.56 * 105

= 10.20 Cm2

2000 0.916 24.40

Spacing of 12 mm dia tor bars = 1.131 X 1000 = 111 Cm c/c

10.20

Spacing of 12 mm dia tor bars @ 110 mm c/c vertically in the form of Stirrups.

Distribution steel:

A st = 0.12 X 100 X 30.00 = 3.6 Cm2 /m

100

Spacing of 10 mm dia bars = 3.142 1.0 X 100 = 21.819 cm

4 X 3.6

Provide 10 mm dia tor bars @ 200 mm c/c horizontally on both faces.

Width of bed block = 105 cm

Steel to be provided = 105 X 22.5 100 = 23.625 Cm2

BED BLOCK DESIGN:


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Longitudinal bars at top & bottom = 23.6 4 = 5.906 Cm2

Provide 6 nos of 12 mm dia tor bars @ top & bottom equally spaced in a

width of 1.05 m (Steel provided = 6 X 1.131 = 6.785 Cm2)

Steel in Transverse direction = 23.6 2 = 11.813 Cm2

Volume of steel per 1m length = 11.81 X 100 = 1181 Cm3

Length of each Stirrup = 100 + 25 2 = 250 cm

Using 10 mm dia bars volume of each stirrup = 250.0 0.786 = 196.500 Cm3

No of stirrups = 1181 196.5 = 6.011 no

Spacing of Stirrups = 100 6.011 = 16.63 cm

Provide 10 mm dia stirrups @ 150 mm c/c

12 mm dia tor bars @

110 mmc/c

10 mm dia tor bars @

2.01 m 200 mm c/c

6 nos of

12 mm dia each at top &bottom

0.30 m 10 mm dia stirrups @

150 mm c/c

1.05 m


11/46

Width of bed block 90 cm

Steel to be provided 90 X 22.5 100 20.250 Cm2

Longitudinal bars at top & bottom 20.3 4 5.063 Cm2

Provide 6 nos of 12 mm dia tor bars @ top & bottom equally spaced in a

width of 0.75 m (Steel provided 6 X 1.131 6.785 Cm2)

Steel in Transverse direction 20.3 2 10.125 Cm2

Volume of steel per 1m length 10.13 X 100 1013 Cm3

Length of each Stirrup 85 + 25 2 220 cm

Using 10 mm dia bars volume of each stirrup = 220.0 0.786 172.920 Cm3

No of stirrups 1013 172.9 5.855 no

Spacing of Stirrups 100 5.855 17.08 cm

Provide 10 mm dia stirrups @ 160 mm c/c

BED BLOCK DESIGN:


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13/46

DESIGN OF R.C.C CIRCULAR PIER.

a) Checking of Circular Pier design using P- body curves:

P = 455.50 t CONCRETE MIX = M 20

ML = 105.19 t-mMT = 125.88 t-m

MR = ML2 + MT2 164.04 t-m say 164.04 t-m

D = 2.00

Checking of Pier section based on actual area required & min. % of steel

for direct load & moment.

With 0.8 % steel .

Asc = 0.008 Ag m = 10

Ac = 3.142* D*D 0.008 = 0.84 D2

4 4

Dm = 200.00 2 5 2.25 = 185.5 cmD = 200.00

Dm = 185.5 D = 0.9275 D

200

Ic = 3.142 x D4 +(m-1)* 0.008 * 3.142* D2*Dm2

64 4 8

= 0.0491 9.0 0.008 3.142 0.860 D4

4 8

= 0.0552 D4

c = P + M* y

Ac Ic

For M 20 c = 66.67 Kg/Cm2 M = Mr 164.04

66.67 = 455.5 1000 + 164.04 10^5 X D

0.84 D2

0.0552 D4

2

= 540938 + 1.49E+08

D2

D3

L.H.S = 66.666667

(by trail&error)

R.H.S = 540937.89 + 1.49E+08 D = 171.0 cm

D2

D3

Required

= 48.229596 200.00 cmAvailable D = 2.00 m

Hence Safe

Asc in Column

Pt = 0.8 %

2

Ast = 0.8 3.142 X 171.0 = 183.75 Cm2

100 4

AST REQD OF ALL CASES(A,B,C) = 183.75 Cm2

Provide 32 nos of 28 mm dia tor bars 197.06624 Cm2

Spacing of 28 mm dia steel

= 19.63 cm C/C Hence O.K

+(m-1) * 3.142 * D2


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Transverse reinforcement:

Provide 10 mm dia tor stirrups @ 250 mm c/c

B) Checking of Pier section on actual area required and min. % of steel for direct load.

With 0.8% steel

Asc = 0.008Ag m = 10

Ac = 3.142* D2 = 0.84 D2

4 4

c = P

Ac

For M 20 c = 66.67 Kg/Cm2

P = 455.5 t

66.6667 = 455.5 1000 D = 90.08 cm

0.842 D2

Actual Area required for direct load = 0.84 D2

= 6832.95 Cm2

Ast min = 0.8 X 6832.95 = 54.66 Cm2

100 < 0.3% of steel

C) Asc in Column

Pt = 0.3 % OF GROSS AREA

2

Ast = 0.3 3.142 X 200.0 = 94.26100 4 Cm2

% STEEL PROVIDED = (183.750444*100*4)/(3.142*200*200) = 0.5848

d) Check by Rational formulae:

Equivalent Concrete Area = 0.83 X 4 = 3.307 m2

Direct Stress = P/A = 455500 / 33073.75 = 13.772 Kg/Cm2

Bending Stress = M x y = ####### X 100.00 = 19.150 Kg/Cm2

Ie 0.0535 X #########

Direct stress + Bending sress < 1Perm.comp.str Per.Bend.Strs

13.772 + 19.15 = 0.563 < 1

50.00 66.67

Hence o.k

+(m-1)* 3.142 *D2*0.008


15/46

a) Checking of Circular Pier design using P- body curves:

P = 455.5 t CONCRETE MIX = M 20

ML = 105.19 t-mMT = 125.88 t-m

MR = ML2 + MT2 164.04 t-m

R = 1.00

D = 2.00

Cover = 5.00 cm

m = 10.00

Eccentricity e = MR = 164.04 = 0.3601 m

P 455.5

e = 36.01 / 100 = 0.3601

R

r = 100 5 1 1.4 = 92.60 cm

r = 0.93

R

From Pea - body curves

mp = 10 0.58482 = 0.058

100

For e/R = 0.360 , r = 0.926

R

r/R e/R mp c k

For 0.950 0.360 0.075 1.880 0.420

0.900 0.360 0.075 1.900 0.450

0.926 0.360 0.075 1.890 0.434

For 0.950 0.360 0.050 1.980 0.480

0.900 0.360 0.050 2.020 0.480

0.926 0.360 0.050 1.999 0.480

C = 1.999 k = 0.480

fc = CMr = 1.999 164.04 10^5 = 32.80 Kg/Cm2

R3

1003

< 66.66667 Kg/Cm2

Fs = m k fc = 10.00 0.480 32.80 = 157.42 Kg/Cm2

< 2000 Kg/Cm2


16/46

Name of work;- Construction of ROB at KOLAKALURU IN GUNTUR Dist (pier)

P = 427 t

ML = 138 t-m

MT = 174 t-m

MR = ML2

+ MT2

= 222.08 t-m say 223.00 t-m

External dia. of well = 6.00 m

Internal dia of well = 4.20 m

Pier top width = 2.00 m

Pier bottom width = 2.00 m

Verticle loads = 427.00 t

Concrete mix =M 20

C = 66.60

n1 = 10.00 = 0.250

10.00 + 2000

66.60j = 1.00 - 0.250 = 0.917

3

Q = 0.50 x 0.917 x 0.250 66.6 = 7.63

Providing 1.25 m thick Well cap

Intensity of slf weight of well cap = t*2.40 = 3.000 t/sqm

The wellcap is most critical against shear which governs he design.

Hence the cap is designed for shear and checked for bending.

*Dia of wellcap is reduced for design purpose for a thickness of 'ts' 175 mm

considering false steining.

Ex.dia = 6.00 m

45 Degrees dispersion of load from pier is taken upto C.G. of

steel I.e upto a depth of 1.50-0.075-0.016-0.016/2 'd' = 1.151 m (effective depth)

Dispersion length l(ds) = Pier bot.width + 2*disp.width = 4.302 m < 6.00 m

2

Area under UDL = 4.302 = 14.54 sqm

4

UdL. for shear = vrt.load/area +slf wt = 32.38 t/sqm

( 427.00 + 3.000 )

14.54Shear will be calculated at a distance of 'a' i.e 'd'

from inner face of steining

Where a = Int.dia of well/2 -d = 0.949 M

1.151 0.949

V = q*a/2 = 15.363 t/m

DESIGN OF PIER WELL CAP


17/46

( 32.38 X 0.949 /2)

4.20 m

(Int.dia)

Shear due to overturning moment

Max. SF per 'm' run = po d/4

po = pd/r

d = Dia.of pier r =Ext dia of well/2

Where p = 4*MR R3

= 10.47 t/m

= 4 x 223.00 3

3.00

po = pd/r 10.47 2.00 3 = 6.98 t/m

Shear force = po d/4 = 3.49 t

Total Shear force = v + po d/4 = 18.853 t

Shear stress = 18.853 1.151 16.38 t/sqm < 22.91225 t/sqm

(Permissible shear stress)

or 1.64 Kg/Cm2

< 2.29 Kg/Cm2

Hence shear reinforcement not reqd.)

% of steel M20 M25For 0.150 1.8 Kg/Cm2 1.9 Kg/Cm2

0.250 2.2 Kg/Cm2 2.3 Kg/Cm2

0.500 3 Kg/Cm2 3.1 Kg/Cm2

0.750 3.5 Kg/Cm2 3.6 Kg/Cm2

For M20 Grade concrete for As/bd= 0.273

Permissible Shear stress 2.29 Kg/Cm2

Check for flexure:

For U.D.L Sagging Moment at centre = w*D2/30

Hogging moment at edge = w*D2/30

Where 'D' is c/c of steining D = 6.00 + 4.20 = 5.10 m

2

or Int.dia+Eff.depth = 4.20 + 1.151 = 5.35 m

Effective Span = 5.10 m

For over turning Moments Mu = BM/6a


18/46

o = (dia. Of pier + well eff.depth)/2 = 1.576

a = (Ex.dia )/2 = 3

o a = 1.576 = 0.5252

3

o a0.3 2.947

0.4 2.062

0.5 1.489

0.6 1.067

0.7 0.731

For simply supported case p 368 case 20

= 1.489 - 1.489 - 1.067 * 0.5252 - 0.50

0.6 - 0.5

= 1.390

For fixed case

For fixed case

o a

0.3

0.4 1.73

0.5 1.146

0.6 0.749

0.7 0.467

= 1.146 - 1.146 - 0.749 * 0.525 - 0.50

0.6 - 0.5

= 1.050

Considering 50% partial fixidity final moments have been considered as one of both the cases.

Case A :

w = 25.62 t/m

D = 6 m

Sagging moment = w*D2/30 = 38.85 t-m

Hogging moment = w*D2/30 = 38.85 t-m

Case B :Due to overturning moments

MR = 223.00 t-m

Moment due to shift of well cap top

= 427 X 0.15 = 64.05 t-m

`

`


19/46

Total moment = MR + 64.05 = 287.05 t-m

Max. radial moment = S.S + FIX X 287.05 = 19.46 t-m

2 6 X a

Sagging Moment = Sag.Mom + Max. Rad.Mom. = 58.31 t-m

Hogging Moment = Hog.Mom + Max. Rad.Mom. = 58.31 t-m

d eff required = Sag.Mom = 87.44 Cm < 1.151 m

7.63 100

Bottom Reinforcement = Sag.Moment * 1/100 = 27.63 Cm2/m

Ast provided = 30 Cm2/m width

Using 20 10.47 Cm

Provide 20 mm dia. Tor bars both ways both at top & bottom @ . 100 mm c/c

% of Steel = 100 X 3.14 X 100 = 0.273

10 100 1.151 X 100

Steel = 100 + 100 * 2 * 1.578 = 6.312 Kg

100 100

Per Cum = 38.25 Kg Say 40 Kg

1450 Kg per Pier cap.

t * j * d

mm dia. Tor bars , Spacing =


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22/46

`


23/46

Name of work;- Construction of ROB at TADEPALLIGUDEM

in West Godavari District.

P = 475.5 t

ML = 155.19 t-m

MT = 255.88 t-m

MR = ML + MT

= 299.26 t-m say 300.00 t-m

External dia. of well = 6.00 m

Internal dia of well = 4.20 m

Pier top width = 2.00 m

Pier bottom width = 2.00 m

Verticle loads = 475.50 t

Concrete mix =M 20

C = 66.60

n1 = 10.00 = 0.250

10.00 + 2000

66.60

j = 1.00 - 0.250 = 0.9173

Q = 0.50 x 0.917 x 0.250 66.6 = 7.63

Providing 1.25 m thick Well cap

Intensity of slf weight of well cap = t*2.40 = 3.000 t/sqm

The wellcap is most critical against shear which governs he design.

Hence the cap is designed for shear and checked for bending.

*Dia of wellcap is reduced for design purpose for a thickness of 'ts' 175 mm

considering false steining.

Ex.dia = 6.00 m

45 Degrees dispersion of load from pier is taken upto C.G. of

steel I.e upto a depth of 1.50-0.075-0.016-0.016/2 ' d' = 1.151 m (effective depth)

Dispersion length l(ds)= Pier bot.width + 2*disp.width = 4.302 m < 6.00 m

2

Area under UDL = 4.302 = 14.54 sqm

4

UdL. for shear = vrt.load/area +slf wt = 35.71 t/sqm

( 475.50 + 3.000 )

14.54

Shear will be calculated at a distance of 'a' i.e 'd'

from inner face of steining

Where a = Int.dia of well/2 -d = 0.949 M

1.151 0.949

V = q*a/2 = 16.946 t/m

( 35.71 X 0.949 /2)

4.20 m

(Int.dia)

DESIGN OF PIER WELL CAP


24/46

Shear due to overturning moment

Max. SF per 'm' run = po d/4

po = pd/r

d = Dia.of pier r =Ext dia of well/2

Where p = 4*MR R3

= 14.11 t/m

= 4 x 300.00 3

3.00

po = pd/r 14.11 2.00 3 = 9.41 t/m

Shear force = po d/4 = 4.70 t

Total Shear force = v + po d/4 = 21.650 t

Shear stress = 21.650 1.151 18.81 t/sqm < 24.12472 t/sqm(Permissible shear stress)

or 1.88 Kg/Cm2

< 2.41 Kg/Cm2

Hence shear reinforcement not reqd.)

% of steel M20 M25

For 0.150 1.8 Kg/Cm2 1.9 Kg/Cm2

0.250 2.2 Kg/Cm2 2.3 Kg/Cm2

0.500 3 Kg/Cm2 3.1 Kg/Cm2

0.750 3.5 Kg/Cm2 3.6 Kg/Cm2

For M20 Grade concrete for As/bd= 0.303

Permissible Shear stress 2.41 Kg/Cm2

Check for flexure:

For U.D.L Sagging Moment at centre = w*D2/30

Hogging moment at edge = w*D2/30

Where 'D' is c/c of steining D = 6.00 + 4.20 = 5.10 m

2

or Int.dia+Eff.depth = 4.20 + 1.151 = 5.35 m

Effective Span = 5.10 m

For over turning Moments Mu = BM/6a

o = (dia. Of pier + well eff.depth)/2 = 1.576

a = (Ex.dia )/2 = 3

o a = 1.576 = 0.5252

3

o a

0.3 2.947

0.4 2.062

0.5 1.489

0.6 1.067

0.7 0.731

For simply supported case p 368 case 20


25/46

= 1.489 - 1.489 - 1.067 * 0.5252 - 0.50

0.6 - 0.5

= 1.390

For fixed caseFor fixed case

o a

0.3

0.4 1.73

0.5 1.146

0.6 0.749

0.7 0.467

= 1.146 - 1.146 - 0.749 * 0.525 - 0.50

0.6 - 0.5

= 1.050

Considering 50% partial fixidity final moments have been considered as one of both the cases.

Case A :

w = 25.62 t/m

D = 6 m

Sagging moment = w*D2/30 = 42.86 t-m

Hogging moment = w*D2/30 = 42.86 t-m

Case B :

Due to overturning moments

MR = 300.00 t-m

Moment due to shift of well cap top

= 475.5 X 0.15 = 71.325 t-m

Total moment = MR + 71.325 = 371.33 t-m

Max. radial moment = S.S + FIX X 371.33 = 25.17 t-m

2 6 X a

Sagging Moment = Sag.Mom + Max. Rad.Mom. = 68.02 t-m

Hogging Moment = Hog.Mom + Max. Rad.Mom. = 68.02 t-m

d eff required = Sag.Mom = 94.44 Cm < 1.151 m

7.63 100

Bottom Reinforcement = Sag.Moment * 1/100 = 32.23 Cm2/m

Ast provided = 33 Cm2/m width

Using 20 9.52 Cm

Provide 20 mm dia. Tor bars both ways both at top & bottom @ . 90 mm c/c

t * j * d

mm dia. Tor bars , Spacing =

`

`


26/46

% of Steel = 100 X 3.14 X 100 = 0.303

9 100 1.151 X 100

Steel = 100 + 100 * 2 * 1.578 = 7.013 Kg

90 90

Per Cum = 38.25 Kg Say 40 Kg

1450 Kg per Pier cap.


27/46

Vertical Load P = 352.44 t

Long.Moment ML = 180.27 t

Trans.Moment MT = 107.38 t-m

Resultant Moment MR = 209.828 t-m

Thickness of well cap = 1.50 m

External dia. Of well = 8.00 m

Internal dia of well = 6.50

Steining Thickness = 0.75 m

Steining height = 4.00 m

Curb height = 1.50 m

Straight length of Abutment 11.05 m

Assuming Reinforcement dia. as 2 cm

Cover = 7.5 cm

B(C/C ABUT TO EXTREEM END C/C STEINING) = 6.4875 M

Base width of Abutment at Sill level = 1.9 m

R.C.L = 100.470

SILL = 95.000

effective depth d = 139.5 Cm

(For Long.direction)

dp, in perpendicular dimension 141.5 Cm

(For Tran.direction)

Assume 1:1 dispersion for vertical load through well cap

Width of dispersion (2.50+1.395) = 3.295 m

Length of dispersion L = 11.05 m

Intensity of load load = 9.68 t/sqm

unit area

The wellcap is assumed as partially fixed alround its circumstance.

i) Effective span S.S case = 7.895 m

(Int.dia + w.cap thik -cover -2-2/2)c/c length = Int.dia + st.thk) = 7.25 m

Eff.span = 7.25

Effective span fixed case = 6.50 m

Effective span for partially fixed case D 6.88 m

For simplicity of analysis the intensity of direct load coming

on wellcap is assumed to act over entire area of wellcap .

From Reynold's handbook

Ave. B.M at centre of partially fixed circular slab = wD2/32

Average B.M around edge = -wD2/32

Self weight of wellcap = 3.60 t/sqm

Load intensity = 9.68 t/sqm

Total w = 13.28 t/sqm

Moment in longitudinal direction /m = 15.39 t-m/m

ml ML/disp.length = MLXB/D*L

Moment in transverse direction /m = 32.59 t-m/m

MT/disp.width

B.M due to direct load wD2/32 = 19.615 t-m/m

MOVEMENT IN LONGITUDINAL DIRECTION + = 15.39 t-m/m

= (MLXB)/(DXL)

Moment at support IN L.direct = = 3.85 t-m/m

DESIGN OF ABUTMANT WELL CAP

(ML*B)/(4*D*L)


28/46

Assuming 50% for externally applied moments IN TRANSVERSE DIRECTION

Moment at centre = M/2

Moment at support = M/8

Final B.M at centre of wellcap

I) AT CENTRE OF WELL CAP:

Longitudinal B.M +ve = wD2/32 +(MLXB)/(D*L) = 35.01 t-m/m (Bottom)

Longitudinal B.M -ve = 0 +(ML*B)/(D*L) = 15.39 t-m/m (Top)

Transverse B.M +ve = wD /32 +MT/2 = 35.91 t-m/m (Bottom)

Transverse B.M -ve = 0 +MT/2 = 16.29 t-m/m (Top)

II) AT SUPPORT:

Longitudinal B.M -ve = w X D2 + (MLXB) = 23.46 t-m/m (Bottom)

32 (4*D*L)

Transverse B.M -ve = w X D2 + MT = 23.69 t-m/m

32 8.00

Max. B.M = 35.91 t-m/m

CHECK FOR DEPTH: CONCRETE MIX = M 20

STEEL Fe 415

C = 66.66 kg/sq.Cm Q = 7.630

t = 2000 kg/sq.Cm j = 0.917

eff.depth required = 35.91 * 105

= 68.60 Cm

100 * 7.63

< 139.5 Cm

Keeping Transverse Reinforcement nearer to Surface.

Hence O.KReinforcement:

1 Longitudinal B.M +ve = 35.01 t-m/m

Ast = Longitudinal B.M +ve = 13.68 Cm2

2000 X 0.917 X 139.50

>

Ast minimum

Ast minimum =0.12*b*D/100 = 18.00 Cm2

Using 20 mm dia.bars

Spacing required = 4.91*100/18 = 17.45 Cm

However provide 20 mm dia.bars at 150 mm c/c at bottom in

Longitudinal direction.

2 Longitudinal B.M -ve at Support (top) = 23.46 t-m/m

Ast = Longitudinal B.M -ve = 9.17 Cm2

2000 X 0.917 X 139.50

sarada well

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