Check Dam Prog

8/8/2019 Check Dam Prog

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1.0 Name of Village Surauli Bujurg

2.0 Name of Block Sumerpur

3.0 Name of District Hamirpur

4.0 Name of river /nalla Baredi Nala

5.0 Catchment Area 12 Sq.Km.

6.0 Nature of Catchment Average

7.0 Average annual rainfall 1000 mm

8.0 Bed level of river 98.4 M

9.0 Height of crest 2.15 M

10.0 Ground Level of left Bank 102.93 M

11.0 Ground Level of right Bank 100.95 M

12.0 Bed width of river/nalla 5.70 M

13.0 Top width of river/nalla 12.00 M14.0 Bed slope of river/nalla 0.66 M/Km.

15.0 H.F.L. of river/nalla 101.9 M

16.0 Discharge factor 14

17.0 Manning's rugosity coefficient 0.03

18.0 Silt factor 1

19.0 Exit gradient 0.167

20.0 Specific gravity of concrete 2.24

21.0 X-Sectional area at H.F.L. 32.74 M

22.0 Wetted Perimeter at H.F.L. 16.80 M

6.31


2/46


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WIDTH OF CREST 1.250 FALSE M

PROVIDE WIDTH OF CREST 1.250 M

Total waterway 14.000 M

14.000 MCheck Dam

ENTER TRIAL VALUE OF D2 - 2.88O.K.

VALUE OF D1 CORRESPONDING D2 0.80

PROVIDE LE


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1.00


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Design of Check Dam

Data :

Name of Village Surauli Bujurg

Name of Block Sumerpur

Name of District Hamirpur

Name of river /nalla Baredi Nala

Catchment Area 12.00 Sq.K

Nature of Catchment Average

Average annual rainfall 1000 MM

Bed level of river 98.395 M

Height of crest 2.15 M

Ground Level of left Bank 102.930 M

Ground Level of right Bank 100.950 M

Bed width of river/nalla 5.70 M

Top width of river/nalla 12.00 M

Bed slope of river/nalla 0.66 M/KmH.F.L. of river/nalla 101.895 M

Discharge factor 14

Manning's rugosity coefficient 0.03

Silt factor 1

Exit gradient 0.17

Specific gravity of concrete 2.24

X-Sectional area at H.F.L. 32.740

Wetted Perimeter at H.F.L. 16.800

A Yield from Catchment

Design rainfall at 60 percent dependability

= 1000 60 100 600 mm

Catchment Type - 2 Average

From Strange's Table,

For Design rain fall = 600 mm,

Rainfal Yield = 14.69 %

Yield/Sq.Km.= 0.6 14.69 100 = 0.09 MCM

Yield From The Catchment = 12.00 0.09 = 1.06 MCM

B. Design Flood Discharge

As per Dicken's formula

= 14 12.00 = 90.26 cumec

As per Manning's Equation

1/N = 1 0.03 = 40

A = 32.740

P = 16.800

R = A/P = 32.740 16.800 = 1.95

1.95 = 1.560.00066 = 0.03

H Q 40 32 740 1 56 0 03 53 12

Q = C A 3/4

Q = (1/N) A (R)2/3 S1/2

R2/3 =S1/2 =

x / =

x /x

x ( )3/4

/

/

2/3( )1/2( )

x


6/46

Design of Check Dam

Hence, adopt maximum of two i.e. 90.26 Cumec

C. Water Way

Hence, L = 4.83 90.26 = 45.89 m

As the top width of river at crest level = 12.00 m

Adopt length of crest = top width of river at crest + 1 m on each bank

= 12.00 2.00 14.00 m

Hence , Length of crest = 14.00 m

D. Scour Depth

Where,

q = Discharge per unit length = Q/L

90.26 14.00 = 6.45 cumec

f = silt factor = 1

Thus, R = 1.35 6.45 1.00 4.68 m

Design Scour Depth = 1.5 x R = 7.020 m

H.F.L. before construction of check dam - 101.895 mPermissible afflux = 1.00 m

Thus, u.s. H.F.L. = H.F.L. + afflux =

101.895 1.00 102.895 m

Velocity of approach = q / R = 6.45 4.68 1.38 m/s

= 1.38 1.38 9.81 0.10m

u/s T.E.L. = u/s H.F.L + velocity head = 102.895 0.1 102.995

d/s T.E.L. = d/s H.F.L. + velocity head= 101.895 0.1 101.995

Crest Level of Check dam = 100.545 m

Head over crest = 102.995 100.545 = 2.450 m

Head Loss (HL) = u/s T.E.L - d/s T.E.L. = 102.995 101.995 1

For broad crested weir, discharge passing over crest

Q =Cd x L x H 3/2

1.71 14.00 2.450 91.54 cumec > 90

Hence the assumed waterway and the crest level is in order.

Assuming free board = 0.50 m

Top of Bund Level (TBL) = 102.995 0.50 = 103.

E. Shape of crest

Adopt , Top width ; 1.25m

u/s slope ; 1 : 8

Waterway as per Lacey's, L = 4.83 x Q 1/2

) 1/2

Scour Depth, R = 1.35 x ( q2 / f )1/3

Velocity head = V2 / 2g

3/2

x ( 2 / ) 1/3 =

/

x (

+

+ =

+ =

/ =

/ ( 2 xx ) =

+ =

+

- =

-

=x

x=


7/46

Design of Check Dam

F. Depth of Cistern

Depth of cistern = dc/3

Where, dc = 6.45 9.81 1Hence, Depth of cistern = 1.62 3.0 0

But limiting depth of cistern = 0.50 m

Hence, Provide depth of cistern = 0.50 m

G. Level and Length of Downstream floor

u.s. Bed level = 98.395m

d.s. Bed level = 98.395m

Here for ,

q = 6.45 cumec/m

and 1.00 mFrom Blench curve, Ef2 = 3.14

R.L. of d/s floor = d/s H.F.L. -Ef2

101.895 3.14 98.755 m

Adopt cistern level = 98.395m

3.14 1.00 4.14 m

Values of D1 & D2 corresponding to Ef1 & Ef2,

D1 = 0.80 m

D2 = 2.88 m

Cistern length required = 5 (D2 - D1)

2.88 0.80 10.40 m

= 3.8 x 1.619 0.42 1.00 = 7.57 m

Hence, provide length of cistern = 10.40m

Say - 10.40 m

H. Vertical Cut-offs

normal scour depth, R = 4

Below Bed Level = 4.68 102.895 98.395 0.

(I) Depth of u/s cut-off:as per scour depth consideration = 1.1 R 5.

Below Bed Level = 5.148 102.895 98.395 0.

as per water depth consideration =

Depth of u/s cutoff, d1 = d/3 + 0.60

4.50 2

Provide, 0.5 m wide and 2.10 m deep cutoff.

R.L. of bottom of cut-off = 98.395 2.10 96.

(II) Depth of d/s cut-off:

as per scour depth consideration = 1.25R 5.85m

Below Bed Level = 5.850 101.895 98.395 2.

as per water depth consideration =Depth of d/s cutoff, d2 = d/2 + 0.60

3 50 2 35 m

(q2

/g)1/3

HL =

Again Ef1 = Ef2 + HL

Lp= 3.8 x dc + 0.415 + H

L

2

/ )1/3

== (/ =

- ==

= +

= 5 x( - ) =

= / 3 + 0.60

/ 2 + 0 60

- ( - ) =

- ( - ) =

- ( - ) =

- =

+ +


8/46

100.545 98.395 2.150 m

Design of Check Dam

Hence, provide d/s cutoff of 0.5 m wide & 2.35 m dee

R.L. of bottom of cut-off = 98.395 2.35 96.

I. Total Floor Length and Exit Gradient

u.s. F.S.L. = 100.545 m

d.s. Floor level = 98.395 m

Max. static head (Hs) = 100.545 98.395 2.15 m

Depth of d/s cut-off, d2 = 2.35 m

Now, GE = (1 / ).Hs / d2

(1/) = GE.d2/Hs

Here,

GE = 0.167

(1/) = 0.167 2.35 2.15 0.18

From Khosla,s curve, = 5.02 m

say - 5.00 m

Total floor length required = d2

5.00 2.35 11.75 m

Provide Total floor length, b = 11.80m

Width of crest = 1.25 m

slope of u/s face 1 :

slope of d/s face 1 :

The length will be provided as under :

Length below the toe of weir = 10.40 m

Length of d/s weir =

0.33 101.895 98.395 1.170 m

Width of crest = 1.25 m

Length of u/s weir =

0.13 101.895 98.395 0.440 m

u/s floor balance = -1.020m

Provide u/s floor length = 1.800 m

Hence, Total floor length provided = 15.060 m

Assume u/s floor thickness near cut-off = 0.60 m

Assume d/s floor thickness near cut-off = 0.60 m

L. Pressure calculation

(i) Upstream cutoff

d1 = 2.10 m

b = 15.060m

= b/d1 7.17

4.12

Y = (-1)/ 0.76

12.90

87.10%

18.79

81.21%5.89%

C ti i 1 f d th 1 68%

=(1+1+2)/2

D = 1/cos-1(Y)

D1 =

100 -D

E = 1/cos-1((-2)/)

C1 =100 -ED1

-c1

=

= - =

- =

- =

x( ) / =

= x =

= x ( - ) =

= x ( - ) =


9/46

Design of Check Dam

(ii) Downstream cutoff

d = 2.35 m

b = 15.060 m

= b/d1 6.41

3.74

Y = (-1)/ 0.73

13.72

19.82

D2 - E2 = -6.10 % (-)ve

Correction in E2 for depth = -1.557% (-)ve

Corrected E2 =E2 + correction 18.26%

(iii)Toe of glacis

pressure = E2+((c1-E2)xLp/b) 75.22 %

M. Impervious floor thickness

u/s floor:

Floor thickness in u/s 0.60 m

(ii) Next point of unbalanced head from the downward end =

2.60

% pressure = 32.50 %

Thickness required = 0.56 m

Provide thickness 0.60 m

(iii)Next point of unbalanced head from the downward end =

5.20 m




(iv) Next point of unbalanced head from the downward end =

7.80 m



Provide thickness 1.10 m (v) At the toe of weir:

Next point of unbalanced head from the downward end =

10.40 m




Design of Check Dam

N. U/S protection

U/s scour depth, R 4.68 m

Anticipated scour = 1.5 R = 1.5 4.68 7.02 m

U/s scour level = 102.895 -7.020 = 95.875 m

=(1+1+2)/2

D2 = 1/cos-1(Y)

E2 = 1/cos-1((-2)/)

x =


10/46

98.395 -95.875 = 2.520 m

(i) Block protection

volume = d1 2.52 cum./

Thickness of protection = 1.20 m

Length of protection = 2.52 1.20 2.10 m

Length of block = 0.80 mNos. of row = 2.10 0.80 2.63

say- 3.00

Length to be provided = 3.00 0.80 2.40 m

Hence Provide 1.20 m thick and 2.40 m long

block protection.

(ii) Lanching apron.

volume = 2.25xd1

2.25 2.52 5.67 cum./


Length of protection = 4.73 m

Hence Provide 1.20 m thick and 5.00 m long

launching apron.

O. d/S protection

D/s scour depth, R 4.68 m

Anticipated scour = 2 R = 2.0 4.68 9.36 m

d/s scour level = 101.895 -9.360 = 92.535 m

Scour depth 'D' below d/s floor =

98.395 -92.535 = 5.860 m

(i) (a) Inverted filter

Volume = d2 5.86 cum./


Design Length = 5.86 1.2 5.00 m

Minimum Length required = 5.90m

Length of block with 100 mm thick jhirri filled with bajri = 0.90

Nos. of row = 5.90 / 0.90 = 6.56

Say- 7

Hence provide length of filter = 6.20m

(i) (b) Toe wall of concrete at the end of filter:width = 0.40 m

depth = 1.2 + 0.30 = 1.50 m

(ii) Luanching apron

Volume = 2.25 d2

2.25 5.86 13.19 cum./

Design of Check Dam

Minimum Length = 1.5D

1.50 5.86 8.79 m

Design Thickness = 13.19 8.79 1.50 mProvide- 1.50 m

Length of launching apron =

Provide 0.8 m x 0.8 m x 0.6 m C.C. block over 600 mm thick invertedfilter.

Provide 0.8 m x 0.8 m x 0.6 m C.C. block over 600 mm thickinverted filter.

/ =

/ =

x

= x =

x =

/ =

= x =

= x =

/ =


11/46

say- 8.80 m


12/46


13/46

m

m


14/46

m

m

m

m

m

m

m

m


15/46

m

8

3


16/46

###


17/46


18/46


19/46

Design of Check Dam

STABILITY CHECK FOR BODY WALL

Case 1. Empty Condition

No.

Particulars

Force L.A.Toe MomentL D Density Constant

W1 1.25 2.15 1.90 5.110 1.8 9.172W2 0.440 2.15 1.90 0.50 0.900 2.567 2.310

W3 1.170 2.15 1.90 0.50 2.390 0.780 1.864

PV = 8.400

H = 0 M = 13.35

Factor of safety against rupture from tension

Excess Moment, Me = M1 -M2 = 13.35 0 = 13.35 T-M

V = Total Vertical Force = 8.400 T

X = Me / V = 13.35 8.400 = 1.59 m

e = Eccentricity = b/2 - X = 2.860 1.59 = -0.16 mb/ 6 = 2.860 6 = 0.48 m

e < b/6, Hence safe.

Factor of safety against Compression or crushing

Pmax = [V / b] [1+6e/b]

Here, [1+6e/b] = 1 -0.16 2.86 0.66 T/m2

V / b = 8.400 2.86 = 2.94

Pmax = 2.94 0.66 = 1.94 T/m2 < 300, Hence safe.

Pmin = [V / b] [1-6e/b]

Here, [1-6e/b] = 1 -0.16 2.86 1.34 T/m2Pmin = 2.94 1.34 = 3.940 T/m2 < 300, Hence safe.

b

W3

(b-a-c)ac

W1

W2

H

FSL

1:3

1:8

-

/

/ 2 -/

[+ 6 x / ] =

x

[ - 6 x ] =/

/

x


20/46

Design of Check Dam

Case 2. FSL Condition

No.

Particulars


W1 1.250 2.15 1.90 5.110 1.795 9.172

W2 0.440 2.15 1.90 0.50 0.900 2.567 2.310

W3 1.170 2.15 1.90 0.50 2.390 0.780 1.864

P 2.150 2.15 1.00 0.50 2.310 0.720 1.663

U 2.860 2.15 1.00 0.50 3.070 1.910 5.864V = 5.330 M1 = 13.346H = 2.310 M2 = 7.527

Factor of safety against overturning

Restoring Moment, M1 = 13.35 T-m

Overturnig moment, M2 = 7.53 T-m

FOS = M1 / M2 = 13.35 7.53 = 1.77

> 1, Hence safe.


Excess Moment, Me = M1 -M2 = 13.35 7.53 = 5.82 T-M


X = Me / V = 5.82 5.33 = 1.09 m

e = Eccentricity = b/2 - X = 2.86 1.09 = 0.34 m

b/ 6 2 86 6 0 48

HbU

P

b

(b-a-c)ac

W1

W2

H

FSL

1:31:8

-

// 2 -/

/


21/46

Design of Check Dam


Pmax = [V / b] [1+6e/b]

Here, [1+6e/b] = 1 0.34 2.86 1.71V / b = 5.330 2.86 = 1.86


Pmin = [V / b] [1-6e/b]

Here, [1-6e/b] = 1 0.34 2.86 0.29

Pmin = 1.86 0.29 = 0.54 T/m2 < 300, Hence safe.

Factor of safety against Sliding

= Coefficient of sliding = 0.67

FOS sliding = V/H = 0.67 5.33 2.31 1.55> 1, Hence safe.

Case 2. Afflux HFL Condition

[+ 6 x / ] =

x

[- 6 x ] =/

/

x

/x =

b

(b-a-c)ac

W1

W2

H b

HFL

1:31:8

(Hb + H +h) U 1U 2

(Hb + H )

W3

W4

W5

P1

P2

P3

P4

H(H+h) w7

w6w8

Afflux HFL


22/46

Design of Check Dam

No.

Particulars


W1 1.250 2.150 1.90 5.110 1.795 9.172

W2 0.440 2.150 1.90 0.50 0.900 2.567 2.310

W3 1.170 2.150 1.90 0.50 2.390 0.780 1.864

W4 1.170 2.150 1.00 0.50 1.260 0.390 0.491

W5 0.440 2.150 1.00 0.50 0.470 2.713 1.275

W6 1.170 1.350 1.00 1.580 0.585 0.924

W7 1.250 2.350 1.00 2.940 1.795 5.277

W8 0.440 2.350 1.00 1.030 2.640 2.719

P1 2.350 2.150 1.00 5.050 1.075 5.429

P2 2.150 2.150 1.00 0.50 2.310 0.717 1.656P3 1.350 2.150 1.00 2.900 1.075 3.118

P4 2.150 2.150 1.00 0.50 2.310 0.720 1.663

U1 2.860 4.500 1.00 0.50 6.440 1.910 12.300

U2 2.860 3.500 1.00 0.50 5.010 0.950 4.760V = 4.230 M1 = 6.972

H = 2.150 M2 = 2.304




FOS = M1 / M2 = 6.97 2.3 = 3.030

> 1, Hence safe.




X = Me / V = 4.67 4.23 = 1.10 m


b/ 6 = 2.86 6 = 0.48 m



Pmax = [V / b] [1+6e/b]

Here, [1+6e/b] = 1 0.33 2.86 1.69

V / b = 4.23 2.86 = 1.48


Pmin = [V / b] [1-6e/b]

Here, [1-6e/b] = 1 0.33 2.86 0.31

Pmin = 1.48 0.31 = 0.46 T/m2



FOS sliding = V/H = 0.67 4.23 2.15 1.32

-

// 2 -/

/

[+ 6 x / ] =

x

[- 6 x ] =/

/

x

/x =


23/46

Design of Check DamDESIGN AND STABILITY CHECK FOR ABUTMENT

Case 1 : Simply check dam

(A) Height of abutment ; H = 103.5 98.245 5.25 m

Base width of abutment, b = 0.75 H = 0.70 5.25 3.680 m

Top width of abutment, a = 0.69 m

C = 0.44 m

Z = 2.300 m

Y = 2.950 m

(b-a) = 2.990 m

c = 0.44 m

No.

Particulars


W1 0.438 5.250 1.90 0.50 2.180 0.292 0.637

W2 0.690 5.250 1.90 6.880 0.783 5.387

W3 2.990 5.250 1.90 0.50 14.910 2.125 31.684

W4 1.680 2.950 1.75 0.50 4.340 2.248 9.756

W5 1.310 2.950 1.75 6.760 3.463 23.410

W6 1.310 2.300 2.00 0.50 3.010 2.553 7.685

U 3.680 2.300 1.00 0.50 -4.230 2.453 -10.376

P4 2.950 2.950 1.75 0.165 2.510 3.283 8.240

=x =

H

b

W2

W3

(b-a)a

W1

FSL

P4

TBL

-

W4

cA

U Z

P1

P2 P3

W5

W6

Z

Y


24/46

P2 2.300 2.300 1.00 0.165 0.870 0.767 0.667

P3 2.300 2.300 1.00 0.50 2.650 0.767 2.032V = 33.850 MR = 68.183

H = 9.950 Mo = 15.447

Design of Check Dam




FOS = M1 / M2 = 68.18 15.45 = 4.41

> 1, Hence safe.




X = Me / V = 52.74 33.85 = 1.56 m


b/ 6 = 3.68 6 = 0.61 m



Pmax = [V / b] [1+6e/b]

Here, [1+6e/b] = 1 0.28 3.68 1.46

V / b = 33.85 3.68 = 9.2


Pmin = [V / b] [1-6e/b]

Here, [1-6e/b] = 1 0.28 3.68 0.54

Pmin = 9.2 0.54 = 4.97 T/m2



FOS sliding = V/H = 0.67 33.85 9.95 2.28

> 1, Hence safe.

-

// 2 -/

/

[+ 6 x / ] =

x

[- 6 x ] =/

/

x

/x =


25/46

Design of Check DamDesign of Downstream wing wall

1.0 Data

Top of wing wall = 102.495 m

Junction level of the wingwall with concrete = 98.245 m

Height of wing wall = 4.250 m

Pond level = 100.545 m

Foundation level = 97.095 m

2.0 Condition of testing

The wing wall setion shall be tested at the junction level and the foundation level for the following conditions.

3.0 Tentative section of wing wall

Front face batter = 1 : 12

Top width of wing wall = 0.50 m

Base width of junction with concrete = 0.6 H

= 2.55 m

Base with of foundation level = 2.550 1.200 3.75 m

a = 0.50 m

Here, b = 2.55 m

d = 0.35 m

c = 1.700 mH = 4.250 m

Z = 2.300 m

" No water on the front of face and earth fill saturated upto the pond level and wet between pond level and topwing wall."

+

=

U

b

c

W2

W5

P1

P2P3

P4

W3W1

W4

a

H

Y

Z

W6

X

W7

A


26/46

Density of masonary 1.90 T/m3

Density of wet earth = 1.75 T/m3

Density of saturated earth = 2.00 T/m3

Cp = 0.33

Density of water = 1.00 T/m3

Density of concrete = 2.40 T/m3

Design of Check Dam

No.

Particulars Force

L.A.Toe MomentL D Density Constant V H

W1 0.350 4.250 1.900 0.500 1.410 - 0.117 0.

W2 0.500 4.250 1.900 4.040 - 0.600 2.4

W3 1.700 4.250 1.900 0.500 6.860 - 1.417 9.

W4 1.700 2.300 2.000 0.500 3.910 - 1.983 7.

W5 0.558 1.125 1.750 0.500 0.550 - 1.222 0.

W6 1.142 1.125 1.750 2.250 - 1.979 4.4

W7 1.700 0.825 1.750 2.450 - 1.700 4.

U 2.550 2.300 1.000 0.500 -2.930 - 1.700 -4.

P4 0.825 0.825 1.750 0.165 - 0.200 2.950 0.

P1 0.825 2.300 1.750 0.330 - 1.100 1.150 1.2

P2 2.300 2.300 1.000 0.165 - 0.870 0.767 0.

P3 2.300 2.300 1.000 0.500 - 2.650 0.767 2.V = 18.540 MR = 24.

H = 4.820 Mo = 4.


Restoring Moment, M1 = 24.373 T-mOverturnig moment, M2 = 4.555 T-m

FOS = M1 / M2 = 24.37 4.56 = 5.35

> 1.5, Hence safe.




X = Me / V = 19.82 18.54 = 1.07 m


b/ 6 = 2.55 6 = 0.43 m



Pmax = [V / b] [1+6e/b]

Here, [1+6e/b] = 1 0.21 2.55 1.48

V / b = 18.54 2.55 = 7.27


Pmin = [V / b] [1-6e/b]

Here, [1-6e/b] = 1 0.21 2.55 0.52

Pmin = 7.27 0.52 = 3.78 T/m2 < 10, Hence safe.



FOS sliding = V/H = 0.65 18.54 4.82 2.5

-

// 2 -/

/

[ + 6 x / ] =

x

[ - 6 x ] =/

/

x

/x =

---

/


27/46

Design of Check Dam

(ii) Checking Stability of wing wall section at foundation level.

Parameters -

a = 0.50 m Z' = 0.450 m

b = 3.75 m Z= 2.300 m

c = 2.05 m Y = 1.125 m

d = 0.60 m X = 0.825 m

e = 0.60 m Z+Z' = 2.750 m

Density of masonary 1.90 T/m3

Density of wet earth = 1.75 T/m3

Density of saturated earth = 2.00 T/m3

Cp = 0.33Density of water = 1.00 T/m3

Density of concrete = 2.40 T/m3

P1

P2P3

P4

W3W1

W4

aa

Z

Y

W6

X

W7

A

a c

b

e

d

Z'

H

W2

W5

UZ+Z'

W8

W9

W10


28/46

Design of Check Dam

No.

Particulars Force

L.A.Toe MomentL D Density Constant V H

W1 0.350 4.250 1.900 0.50 1.410 - 0.717 1.

W2 0.500 4.250 1.900 0.00 4.040 - 1.200 4.

W3 1.700 4.250 1.900 0.50 6.860 - 2.017 13.

W4 1.700 2.300 2.000 0.50 3.910 - 2.583 10.

W5 0.558 1.125 1.750 0.50 0.550 - 1.822 1.W6 1.142 1.125 1.750 - 2.250 - 2.579 5.

W7 1.700 0.825 1.750 - 2.450 - 2.300 5.

W8 0.600 1.950 1.750 - 2.050 - 3.800 7.

W9 0.600 2.300 1.750 - 2.420 - 3.800 9.

W10 3.750 0.450 2.400 - 4.050 - 1.875 7.

U 3.750 2.750 1.000 0.50 -5.160 - 2.500 -12.

P4 0.825 0.825 1.750 0.17 - 0.200 3.400 0.

P1 0.825 2.300 1.750 0.33 - 1.100 1.600 1.

P2 2.300 2.300 1.000 0.17 - 0.870 1.217 1.

P3 2.300 2.300 1.000 0.50 - 2.650 1.217 3.2V = 24.830 MR = 53.

H = 4.820 Mo = 6.




FOS = M1 / M2 = 53.916 / 6.724 = 8.018

> 1.5, Hence safe.


Excess Moment, Me = M1 -M2 = 53.916 6.724 = 47.192 T-m


X = Me / V = 47.192 / 24.830 = 1.9 m

e = Eccentricity = b/2 - X = 3.75 1.9 = -0.03 m

b/ 6 = 3.75 / 6 = 0.63 m

e < b/6, Hence no tension will develop at the base.


Pmax = [V / b] [1+6e/b]

Here, [1+6e/b] = -0.03 3.75 0.96

V / b = 24.83 / 3.75 = 6.62

Pmax = 6.62 x 0.96 6.34 T/m2 < 30, Hence safe.

Pmin = [V / b] [1-6e/b]Here, [1-6e/b] = -0.03 3.75 1.04

Pmin = 6 62 x 1 04 6 9 T/m2 < 10 Hence safe

-

/ 2 -

[1+ 6 x / ] =

=

[1+ 6 x

[ 1 - 6 x / ] =


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Designed and developed by Er. Rajeev Kumar, Executive Engineer, Irrigation Department of U.P.

1 of 3

Design of Check Dam

Data :

Name of Village : Surauli BujurgName of Block : Sumerpur

Name of District : Hamirpur

Name of river /nalla : Baredi Nala

Catchment Area : 12.00 Sq.Km.

Nature of Catchment : Average

Average annual rainfall : 1000.00 MM

Bed level of river : 98.40 M

Height of crest : 2.15 M

Ground Level of left Bank : 102.93 M

Ground Level of right Bank : 100.95 MBed width of river/nalla : 5.70 M

Top width of river/nalla : 12.00 M

Bed slope of river/nalla : 0.66 M/Km.

H.F.L. of river/nalla : 101.90 M

Discharge factor : 14.00

Manning's rugosity coefficient : 0.03

Silt factor : 1.00

Exit gradient : 0.17

Specific gravity of concrete : 2.24

X-Sectional area at H.F.L. : 32.74

Wetted Perimeter at H.F.L. : 16.80

Design Report :1.0 Design Flood Discharge : 90.26 cumec

2.0 Water Way : 45.89 m

3.0 Length of crest adopted : 14.00m

4.0 Normal Scour Depth : 4.68 m

5.0 u/s T.E.L. : 103 m

6.0 d/s T.E.L. : 102 m

7.0 Crest Level of Check dam = : 100.55 m

8.0 Head over crest = : 2.45 m

9.0 Discharge passing over crest : 91.54 cumec

10.0 Top of Bund Level (TBL) = : 103.5m

11.0 Shape of crest

Top width ; 1.25 m

u/s slope ; 1 : 8

d/s slope ; 1 : 3


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2 of 3

Design of Check Dam

11.0 Cistern Level : 98.4m12.0 Length of cistern : 10.40m

13.0 R.L. of bottom of u/s cut-off (0.50 m) : 96.3m

14.0 R.L. of bottom of d/s cut-off (0.50 m) : 96.05m

15.0 d.s. Floor level : 98.4m

16.0 u.s. floor level : 98.4m

17.0 Total Floor Length - : 15.060 m

Length below the toe of weir : 10.40m

Length of d/s weir : 1.170m

Width of crest : 1.250m

Length of u/s weir : 0.440m

u/s floor length : 1.800m18.0 Floor thickness in u/s : 0.60m

19.0 Toe of crest : #REF!m

20.0 Toe of glacis : 1.30m

21.0 2.60 m from the toe of crest : 0.60m



24.0 U/S protection

(i) Block protection

1.20 m thick and 2.40 m long

(ii) Lanching apron.


25.0 d/S protection

(i) Inverted filter

Provide 0.8 m x 0.8 m x 0.6 m C.C. block over 600 mm thick inverted filter.

6.20 m long.

(ii) Toe wall of concrete at the end of filter:

0.40 m wide and 1.50 m depth.

(iii) Luanching apron


26.0 Abutment

(i) Top Width of abutment ; : 0.69m

(ii) : 3.680m

Base width of Wing Wall at the foundation: 4.680

(iii)Height of abutment : 5.250m

(iv)Length of abutment : 2.860m

Provide 0.8 m x 0.8 m x 0.6 m C.C. block over 600 mm thickinverted filter.

Base width of abutment at the junction ofconcrete


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27.0 Wing Wall

(A)D/S Wing Wall

(i) Top Width of Wing Wall ; 0.50m

3 of 3

Design of Check Dam

(ii)2.55m

(iii)Base width of Wing Wall at the foundation

3.15m

(iv) Length of wing wall 10.40 m

(B)U/S Wing Wall

(i) Top Width of Wing Wall ; 0.50m

(ii)3.15m

(iii)Base width of Wing Wall at the foundation

3.75m(iv) Length of wing wall 1.800 m

Base width of Wing Wall at the junctionof concrete

Base width of Wing Wall at the junctionof concrete


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1.43

0.63

1.17

1.8

0.44

1.07

2.86


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Documents

Check Dam Prog