25.875 OT

GUNDLAKAMMA RESERVOIR PROJECT 2. DESIGN OF OFF-TAKE CHANNEL Unlined canal

Offtake at Km

Discharge (required)

=

Full supply depth (d)

=

0.35

m

Bed width (b)

=

0.50

m

Side slopes

( Inner) ( Outer)

Bed fall Area

P

R

= = 1

in

=

0.500

+

=

0.359

0.50 +

=

1.76 m A

=

0.359 1.762

=

1 0.030

Discharge (provided) = V = Vo

2R 1.325

0.107 0.971

OK OK

Manning's rugosity coefficient (n) =

0.03

1.5 :1 2.0 :1

=

=

=

0.1020 cumecs

: :

1500

OR

0.35 x

0.00067

1.5

x

0.35

Sq.m

2 x

1.803 x

0.35

P

V

Q

=

0.204 m

X

=

0.298 m/sec

=

V

2/3

X

0.00067

0.5

x A

=

0.298

=

0.10 7 cumecs

Vo

=

0.530 x

Vo

= =

0.530 x

=

0.298

V Vo

0.204

x

0.35875

d 0.350

0.52

m/sec

0.52

0.307 m/sec

=

0.971

0.307

1 :

0.35 0.350 0 m 5 1 .

Bed width 0.50 m

3

245932249.xls.ms_office

DESIGN OF OFF-TAKE SLUICE FOR 12 L MAJOR @ Km 25.875 1) HYDRAULIC PARTICULARS

Sl

Description

Unit

Particulars of Main Canal

Ha

41100

Cumecs

19.383

O.T 12 L

1

Ayacut

400

2

Discharge (Required/Designed)

3

Bed width

m

10.10

0.70

4

Full Supply Depth

m

2.00

0.35

5

Free board

m

0.90

0.45

6

Side slopes (Inner) (Outer)

-

1.50 :1 2 :1

1.50 2

7

Bed fall

-

1 in

8

Velocity

m/s

9

Value of 'n'

/

19.540

0.1360

5000

1 in

0.746

0.3250

-

0.0250

0.0250

10 Top width of banks L/R

m

7.000

11 C.B.L.

m

+ 453.325

+ 453.925

12 F.S.L.

m

+ 455.325

+ 454.275

13 T.B.L.

m

+ 456.225

+ 454.725

14 G.L.

m

+ 456.065

+ 456.065

0.7 ( 0.7 +

x 0.350 1.500 0.42875 + 1.803 1.962 0.429 0.2185 0.3628 1/ 0.025 0.3245 0.4288 0.1391 HENCE O.K.

/

:1 :1

4.00

4.200

2000

/

2.00

Cd =

0.75

2) DESIGN OF O.T.CHANNEL Assume the section Area -

= Perimeter Hydraulic Radius (R) 2/3

R Velocity Discharge -

V0 = = V/V0 = =

0.7 = A/P = = V= = Q= = 0.53 0.30704

x 0.350

m x 0.35 )

x 0.350

2

m

x2 x m / m m x

0.3628 x

x >

0.3245 0.1360

0.350

1.962

0.022361

0.52

0.3245 / 0.3070 1.0569 (1.1 to 0.9)

3) DRIVING HEAD: FSL in the parent channel 3/4 of FSD 3/4 of FSL 3/4 FSL in the Offtake channel

= = = =

455.325 m 1.500 + 454.825 + 454.275 m

Driving head = So, Consider the driving head =

454.825 0.5500 m

454.275

4) AREA OF VENT WAY Q = Cd A Sqrt (2g h)

(for circular vent )

Q = Cd A Sqrt (2g h)

(for square or Rectangular type vent) Cd = 0.62

Where

3


Q h A Cd

= = = =

Driving head considered

OT Discharge

= 0.1360 = 0.5500

Area of vent way Coefficient of discharge

= 0.750

Cumecs m

In this case "Provide circular type vent" 0.136

Area of vent (A)

= = = =

A

=

The minimum dia . Of pipe

=

Dia meter of pipe required (d)

0.750 x 0.750 x 2.464 x 0.136 2.464 0.055 Sqm

A A A

x x

Sqrt (2 x 9.81 x Sqrt 10.791

0.5500 )

0.300 m

= Sqrt(4x0.055/pi) 0.265 m = = 265.20 mm 300 mm minimum with ferrule of 265 mm @ the first pipe joint.

Provide So, provide

1 vents of

Therefore the area of vent provided

=

0.300 m

0.071 Sqm

>

0.055 Sqm

HENCE OK

5 HOIST PLATFORM LEVEL (HPL) (A)

(B)

Hoist platform level

= Sill level + twice height of gate + 0.3 + 0.15

Height of gate Hoist plotform level

= =

Hoist plotform level

=

0.30 m 453.925 + + + 455.075 m

FSL in the parent channel

=

+ 455.325 m

Hoist plotform level

= = =

FSL + 455.325 + + 455.625 m

2 x 0.3 +

0.30 + 0.15

0.05

0.3 0.3

Therefore, Provide the hoist platform level (HPL) = + 455.63 m Provide Width of platform = 1.200 m 4 LENGTH OF BARREL Length of barrel Required = 0.60 + 1.5 (TBL/GL - HPL) + TW + 2/1.5 (TBL/GL - d/s Hwl) + 0.6/0.45 Assume Level of D/s head wall

=

Length of barrel

=

+ 455.7250 0.6

+

+ 2

x

= Length of barrel Provide Length of barrel provided

m 1.500 x

456.225 - 455.725 0.6 + + 10.0000 m

= 4 no.s of =

456.225 +

0.9 + 0.5

455.625

+ 7.000

0.5 7.000 +

1.000

2.500 m length pipes 10.000 m HENCE OK

5) PROTECTION WORK Provide 5 m length of Lining in CC M10 on D/S of of structure in the O.T Channel and the same for 5 m length on either side of U/S transition to the side of Parent channel

6) UPSTERAM TRANSITION

The level of U/S sloped wing

= = =

Platform level - (Platform width /side slope) + 455.625 1.200 / 1.500 + 454.825

4


The length of U/S transition

(or) The Horizontal length of U/S sloped wing = (level of U/S sloped wing - CBL of parent channel) * side slope = 1.500 x + 454.825 - 453.325 = 2.250 m

Providing length of transition 1 in

3 flare

Outer width of transition = barrel width +( 2 x horizontal length of U/s sloped wing / flare) = 0.300 + 2 x 2.250 3 = 1.800 m say 1.800 m 7) DOWNSTERAM TRANSITION Barrel / Pipes width Bed width of O.T channel

= =

Providing length of transition in 1 in The length of D/S transition

0.300 m 0.70 m

5 flare =

5 x

0.70

= 1.0 m Howev er provide , a cistern of 2.000 x 2.00 from which the offtake channel runs

5

2

0.3

m size


3.(ii). LINE DIAGRAM OF OFFTAKE SLUICE (CIRCULAR TYPE VENT) GL T BL of Parent canal Platform level FSL parent channel Sill of OT CBL of Parent channel GL

= = = = = = =

######## ######## ######## ######## ######## ######## ########

m m m m m m m

Dia of pipe Thickness of pipe Caulking space Collar thickness Collar length

length of barrel 0.900 1.200 0.45

10.00 m Bank Width 4.000 + 456.225 TBL

= = = = =

m

300 30.00 16.00 30.00 150.00

mm mm mm mm mm

= = = = =

0.3 0.030 0.016 0.030 0.150

4.000

m m m m m

0.5

0.6

+ 456.065 GL

0.15

2.000 1.5 : 1

2.0 : 1

Platform level #######

#######

2.27 FSL

####### Pial Wall #######

+ 454.725 T.B.L + 453.955

Parent channel

1

0.03

2

0.3

3

4 + 453.925 CBL

#######

CBL ######## 1.2

2.5

2.5

2.5

2.5

7


8. DESIGN OF HEAD WALLS, WINGS AND RETURNS

1. DESIGN OF HEAD WALL (U/S)

0.77

unit wt of concrete unit wt of earth

1.5:1 W6

+

= =

2.400 2.100

t t

455.625

W3 W4

W1 2.375 W2

0.30

1.150 0.600 0.30 1.750

+

0.50

W5 2.350

+ 452.750

Taking moments about A Load W1 W2 W3 W6

453.25

(Stresses in concrete)

Particulars 0.600 x 1.150 x 1.150 x 0 767 x

2.375 2.375 2.375 1 150

x x x x

2.400 2.400 x 2.100 x 2 100 x

0.500 0.500 0 500

Force V H 3.420 3.28 2.868 0 926

L.A.

Moment

1.450 0.767 0.383 0 383

4.959 2.513 1.1 0.355

8. DESIGN OF HEAD WALLS, WINGS AND RETURNS

1. DESIGN OF HEAD WALL (U/S)

0.77


1.5:1 W6

+

= =

2.400 2.100

t t

455.625

W3 W4

W1 2.375 W2

0.30

1.150 0.600 0.30 1.750

+

453.25 0.50

W5 2.350

+ 452.750

Taking moments about A Load W1 W2 W3 W6 Pv Ph


Particulars 0.600 1.150 1.150 0.767 0.1031 0.3596

x x x x x x

2.375 2.375 2.375 1.150 3.142 3.142

x x x x x x

2.400 2.400 2.100 2.100 3.142 3.142

x x x x x

0.500 0.500 0.500 2.100 2.100 S V

Lever arm

= M/V

=

Eccentricity

= 1.268

16.0 -

/

1.750

Permissible 'e' =

=

/2

=

1.75

Max. comp. Stress

Min. Stress

1.268

m

0.393

m

/6

= 0.292

S M

16.008

16.939

t/sq.m.

= 12.63 x ( 1.750

1 -

6

x 0.393 1.750

) =

-2.507

t/sq.m.

(Stresses on soil)

2.375 2.375 2.375 3.142 0.500 1.150 3.642 3.642

= M/V

=

Eccentricity

= 1.498

7.081

REVISE THE SECTIO

) =

Lever arm

x x x x x x x x

2.400 2.400 2.100 2.100 2.400 2.100 3.642 3.642

27.2 -

x x

0.50 0.50

x x x

0.50 2.100 2.100

/

18.16

2.350

Permissible 'e' =

Min. Stress

4.959 2.513 1.1 0.355

x 0.393 1.750

Sv

Max. comp. Stress

1.450 0.767 0.383 0.383 --0.950

6

Particulars x x x x x x x x

0.393

Moment

1 +

Load 0.600 1.150 1.150 0.300 2.350 0.767 0.1031 0.360

<

L.A.

= 12.63 x ( 1.750

Taking moments about B

W1 W2 W3 W4 W5 W6 Pv Ph

12.63

Force V H 3.420 3.28 2.868 0.926 2.137 7.453 12.63

=

/2

=

2.35

/6

Force V H 3.420 3.28 2.868 1.979 2.820 0.926 2.871 10.015 18.16

1.498

m

0.323

m

= 0.392

>

0.323

L.A.

Moment

1.750 1.067 0.683 0.15 1.175 0.683 --1.15

5.985 3.496 1.96 0.297 3.314 0.633 11.517

SM

27.202

HENCE OK

= 18.16 x ( 2.350

1 +

6

x 0.323 2.350

) =

14.102

t/sq.m.

= 18.16 x ( 2.350

1 -

6

x 0.323 2.350

) =

1.355

t/sq.m.

6


2. DESIGN OF WING WALLS (U/S) unit wt of concrete unit wt of earth

= =

2.400 t 2.100 t + 454.825

W3 W4

W1 1.575 W2

0.300

0.100 0.500 0.30 0.600

+ 453.25 0.50

W5 1.200

+ 452.75

Taking moments about A Load W1 W2 W3 Pv Ph

(Stresses in concrete) Force

Particulars 0.500 0.100 0.100 0.0384 0.134

x x x x x

1.575 1.575 1.575 1.575 1.575

x x x x x

2.400 2.400 x 2.100 x 1.57 x 1.57

V 1.9 0.2 0.2 0.2

0.500 0.500 2.100 2.100

0.7 SV

Lever arm

= M/V

= 1.121

Eccentricity

= 0.459

-

/

0.600

Permissible 'e' =

=

/2

=

0.60

Max. comp. Stress

Min. Stress

m

0.159

m

= 0.100

<

0.159

) =

10.552

t/sq.m.

=

2.44 x ( 0.600

1 -

6

x 0.159 0.600

) =

-2.404

t/sq.m.

1.575 1.575 1.575 1.575 0.500 2.075 2.075

= M/V

(Stresses on soil)

x x x x x x x

=

Eccentricity

= 0.672

2.400 2.400 x 2.100 x 2.100 2.400 2.075 x 2.075

3.4 -

0.44

REVISE THE SECTIO

x 0.159 0.600

Lever arm

/

0.50 0.50

2.100 2.100

5.02

1.200

Permissible 'e' =

Min. Stress

1.121

6

Sv

Max. comp. Stress

SM

1 +

Particulars x x x x x x x

0.662 0.013 0.006

2.44 x ( 0.600

Load 0.500 0.100 0.100 0.300 1.200 0.0384 0.134

0.350 0.067 0.033 --0.630

2.444

0.459

/6

Moment

=


W1 W2 W3 W4 W5 Pv Ph

2.44

L.A.

H

=

/2

=

1.20

/6

Force V H 1.890 0.19 0.165 0.992 1.440 0.347 1.212 5.024

0.672

m

0.072

m

= 0.200

L.A.

Moment

0.650 0.367 0.333 0.15 0.6 --0.83

1.229 0.07 0.056 0.149 0.864

SM

3.374

>

0.072

HENCE OK

=

5.02 x ( 1.200

1 +

6

x 0.072 1.200

) =

5.694

t/sq.m.

=

5.02 x ( 1.200

1 -

6

x 0.072 1.200

) =

2.679

t/sq.m.

7

1.006


3. DESIGN OF HEAD WALL (D/S)

0.40


2:1 W6

+

2.400 2.100

= =

t t

455.725

W3 W4

W1

1.980

W2 3.96 0.300

0.600 0.500 0.30 1.100

+ 453.745 0.50

W5 1.700

+ 453.245

Taking moments about A Load W1 W2 W3 W6 Pv Ph


Particulars 0.500 0.600 0.600 0.600 0.0623 0.2174

x x x x x x

1.980 1.980 1.980 0.400 2.380 2.380

x x x x x x

2.400 2.400 2.100 2.100 2.380 2.380

x x x x x

0.500 0.500 0.500 2.100 2.100 SV

Lever arm

= M/V

=

Eccentricity

= 0.818

4.9 -

/

1.100

Permissible 'e' =

=

/2

=

1.10

Max. comp. Stress

Min. Stress

0.818

m

0.268

m

/6

= 0.183

SM

4.941

REVISE THE SECTIO

) =

13.522

t/sq.m.

=

6.04 x ( 1.100

1 -

6

x 0.268 1.100

) =

-2.537

t/sq.m.

1.980 1.980 1.980 2.380 0.500 0.400 2.480 2.480

= M/V

(Stresses on soil)

x x x x x x x x

=

Eccentricity

= 0.957

2.400 2.400 2.100 2.100 2.400 2.100 2.480 2.480

9.2 -

2.049

0.268

x 0.268 1.100

Lever arm

x x

0.50 0.50

x x x

0.50 2.100 2.100

/

9.65

1.700

Permissible 'e' =

Min. Stress

2.02 0.571 0.25 0.051

6

Sv

Max. comp. Stress

0.850 0.400 0.200 0.200 --0.792

1 +

Particulars x x x x x x x x

Moment

6.04 x ( 1.100

Load 0.500 0.600 0.600 0.300 1.700 0.600 0.0623 0.217

<

L.A.

=


W1 W2 W3 W4 W5 W6 Pv Ph

6.04

Force V H 2.376 1.43 1.247 0.252 0.741 2.586 6.042

=

/2

=

1.70

/6

Force V H 2.376 1.43 1.247 1.499 2.040 0.252 0.805 2.808 9.645

0.957

m

0.107

m

= 0.283

L.A.

Moment

1.150 0.700 0.500 0.15 0.85 0.500 --0.992

2.733 0.998 0.624 0.225 1.734 0.126

SM

9.226

>

0.107

HENCE OK

=

9.65 x ( 1.700

1 +

6

x 0.107 1.700

) =

7.816

t/sq.m.

=

9.65 x ( 1.700

1 -

6

x 0.107 1.700

) =

3.531

t/sq.m.

8

2.786


4. DESIGN OF WING & RETURN WALLS (D/S) unit wt of concrete unit wt of earth

= =

2.400 t 2.100 t + 455.725

W3 W4

W1 1.980 W2

0.300

0.300 0.500 0.30 0.800

+ 453.745 0.50

W5 1.400

+ 453.245

Taking moments about A Load W1 W2 W3 Pv Ph


Particulars 0.500 0.300 0.300 0.0384 0.134

x x x x x

1.980 1.980 1.980 1.980 1.980

x x x x x

2.400 2.400 x 2.100 x 1.980 x 1.980

0.500 0.500 2.100 2.100 SV

Lever arm

= M/V

=

Eccentricity

= 0.593

2.4 -

/

0.800

Permissible 'e' =

=

/2

=

0.80

Max. comp. Stress

Min. Stress

0.593

m

0.193

m

/6

= 0.133 <

Eccentricity

=

0.193

t/sq.m.

=

4.03 x ( 0.800

1 -

6

x 0.193 0.800

) =

-2.254

t/sq.m.

(Stresses on soil)

x x x x x x x

=

2.400 2.400 x 2.100 x 2.100 2.400 2.480 x 2.480

5.7 -

0.874

REVISE THE SECTIO

12.325

/

0.50 0.50

2.100 2.100

7.14

1.400

Permissible 'e' =

Min. Stress

2.387

) =

0.8

Max. comp. Stress

SM

x 0.193 0.800

Sv

= M/V

1.307 0.143 0.063

6

1.980 1.980 1.980 1.980 0.500 2.480 2.480

Lever arm

0.550 0.200 0.100 --0.792

1 +

Particulars x x x x x x x

Moment

4.03 x ( 0.800

Load 0.500 0.300 0.300 0.300 1.400 0.0384 0.134

L.A.

=


W1 W2 W3 W4 W5 Pv Ph

4.03

Force V H 2.376 0.71 0.624 0.316 1.103 4.029

=

/2

=

1.40

/6

Force V H 2.376 0.71 0.624 1.247 1.680 0.496 1.731 7.136

0.8

m

0.100

m

= 0.233

L.A.

Moment

0.850 0.500 0.400 0.15 0.7 --0.992

2.02 0.357 0.25 0.188 1.176

SM

5.708

>

0.100

HENCE OK

=

7.14 x ( 1.400

1 +

6

x 0.100 1.400

) =

7.281

t/sq.m.

=

7.14 x ( 1.400

1 -

6

x 0.100 1.400

) =

2.913

t/sq.m.

9

1.717


Sl no

STRESS TABLE

Concrete t / Sqm

Soil t / Sqm

1 HEAD WALL U/S

Max. comp. Stress Min. Stress

16.939 -2.507

14.102 1.355

2 WING WALL U/S


10.552 -2.404

5.694 2.679

3 HEAD WALLS D/S


13.522 -2.537

7.816 3.531

4 WING & RETURN


12.325 -2.254

7.281 2.913

WALLS

10


DESIGN OF OFF-TAKE (12L) @ km 17.350

S.NO I II

III

DESCRIPTION NOTE ON DESIGN FEATURES

PAGE NO 1

DESIGN CALCULATIONS 1. HYDRAULIC PARTICULARS OF CANAL/OT

3

2. VENT WAY

3

3. HOIST PLATFORM LEVEL

4

4. LENGTH BARREL

4

5. PROTECTION WORK

4

6. UPSTREAM TRANSITION

4

7. DOWNSTREAM TRANSITION

5

8. DESIGN HEADWALLS, WINGS & RETURNS

6

DRAWINGS GENERAL PLAN AND SECTION

25.875 OT

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