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Computation of Bed slope for U/s

U/s Bed level @ 1000 m = 165.900

Bed level at the proposed site = 164.200

Difference in elevation 165.9 - 164.2 = 1.700

Distance between the U/s and proposed site 1

= 1000

Bed fall = 1.7 / 1000 = 0.0017

1 in 588.2

say 1 in 600

Construction of Check dam

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m

m

Km

m


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Computation of Bed slope for D/s

Bed level at the proposed site = 164.200

D/s Bed level @ 1000 m = 162.200

Difference in elevation 164.2 - 162.2 = 2.000

Distance between the proposed site and D/s 1

= 1000

Bed fall = 2 / 1000 = 0.002

1 in 500.000

say 1 in 500



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m

m

m

Km

m


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CALCULATION OF REAR WATER LEVEL (RWL)

Base width of the canal b 20 m

Side slope of the canal 1 in 1.5 (z)

Bed level of the canal + 164.200 m

Bed slope of the canal 1 in 500

Rugosity coefficient n 0.025Rear Water Level 164.775 m

RWL = + 164.775

1 y = 0.572 m

1.5

+ 164.200

Area of flow A = (b+zy) y = ( 20 + 1.5 * 0.572 ) * 0.572

= 11.927 m2

Wetted Perimeter P = b+2y (1+z^2) 0.5 = 20 + 2 * 0.572 * ( 1 + 1.5 2 ) 0.5

= 22.062 m

Hydraulic Radius R = A / P = 11.927 / 22.062

= 0.541 m

Bed slope S = 1 / 500 = 0.0020

Velocity of flow V = 1 / n x R2/3

x S1/2

= 1/ 0.025 x 0.541 ^ (2/3) x 0.0020 ^ (1/2)

= 1.187 m /sec

Discharge Q = A * V = 11.927 x 1.187

= 14.158 cumecs OR

500 cusecs

Rear Water Level = Bed level + y = 164.2 + 0.572 = 164.775 m

20.00

Gwall/lnattar/rwl

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Fixing the FMFL (Using Drowned weir formula)

Design Discharge : 14.16 m3/sec 500 c/s

Crest Level : 165.100 m

Assume FMFL : 165.882 m 165.885 m

RWL (calculated from the regraded section of rive 164.625 mUpstream Bed Level : 164.200 m

Downstream Bed Level 164.200 m

Head over crest 165.882 - 165.1 : 0.78 m

Total width of river : 20 m

Length of check dam L 20 m

Width of check dam 0.45 m

1.5 times width of weir 0.675 m

Hence provide Narrow crested weir

Drowned weir formula

Q = L * h^0.5 ( 1.84* h + 3.54 *h1)

FMFL+165.885

h RWL+164.625

crest+165.1 h1

NC Weir bed level+164.2

Where L = effective length = 20 m

h = FMFL - RMFL = 165.882 - 164.625 = 1.257 m

h1 = depth of d/s water level above crest = 164.625 - 165.1 = -0.475 m

Q = L * h^0.5 ( 1.84* h + 3.54 *h1)

= 20 * 1.257 ^0.5 ( 1.84* 1.257 + 3.54 * -0.475 )

= 14.158 m3/sec

14.158 m3/sec

Required Discharge 14.158 m3/sec

Total discharge over checkdam

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Fixing the FMFL (Using Free flow formula)

Design Discharge : 14.16 m3/sec 500 c/s

Crest Level : 165.100 m

Assume FMFL : 165.332 m 165.335 mRWL (calculated from the regraded section of rive 164.775 m

Upstream Bed Level : 164.200 m

Downstream Bed Level 164.200 m

Head over crest 165.332 - 165.1 : 0.23 m

Length of check dam L 20 m

Width of check dam 0.45 m

1.5 times width of weir 0.675 m

Hence provide Broadcrested weir

Free flow formula

Q = 1.704 * L * h^1.5

FMFL+165.335

h

crest+165.1

RWL+164.775

BC Weir bed level+164.2

Where L = effective length 20 m

h = Head over crest 165.332 - 165.1 = 0.557 m

Q = 1.704 * L * h^1.5

= 1.704 * 20 * 0.557 ^1.5

= 14.158 m3/sec

14.158 m3/sec

Required Discharge 14.158 m3/sec

Total discharge over checkdam

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146912493.xlsx.ms_officeStability Analysis of bodywall (For Drowned weirs)

HYDRAULIC PARTICULARS

Maximum flood discharge 14.16 m3/sec 500 c/s

Top of crest + 165.100 m

FMFL + 165.885 m

RWL + 164.625 m

Upstream bed level + 164.200 m

Downstream Bed level + 164.200Down stream side slope 0.3 H to 1 V

Upstream side slope 0 H to 1 V

Top width 0.45 m

Unit weight of concrete 2.4 t/m

The stability of body wall of the anicut was checked for the following conditions

1 Reservoir empty

2 Reservoir at MWL,with tailwater

3 Reservoir at FRL, no tail water

165.885

0.785 164.625165.100 C

a

165.1 0.3

0 1

0.9 1

ii

O

0.9 0.785 0.000 0.45 0.270

0.4500.720

Stability analysis:

1.Reservoir empty

S.NO L.A

Co-eff Length Depth Area Unit wt. V H + + -

Weight of masonry

1 0.5 0.000 0.900 0.000 2.4 0 0.720 0.000

2 1 0.45 0.900 0.405 2.4 0.972 0.495 0.481

3 0.5 0.270 0.900 0.122 2.4 0.29 0.180 0.052

SV= 1.264 SM= 0.534

Base width b = 0.720 m 2/3 *b = 0.48 m 1/3 *b = 0.24 m

X= SM / SV = 0.534 / 1.264 = 0.422 mFalls within the middle third

Eccentricity e = b/2-X 0.72 /2 - 0.422 = -0.062 m

6e/b = 6 * -0.062 / 0.72 = 0.519 m

Maximum stress = SV / b * (1 + 6e/b)


DESCRIPTION FORCE MOMENT

164.2001

23

146912493.xlsx.ms_officeweir s tability (drown)

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146912493.xlsx.ms_office= 1.264 / 0.72 * ( 1 + 0.519 ) = 2.666 t/m2

Minimum stress = SV/ b * (1 - 6e/b)= 1.264 / 0.72 * ( 1 - 0.519 ) = 0.844 t/m

2

2.Reservoir at MWL, with tailwater

S.NO L.A

Co-eff Length Depth Area Unit wt. V H + -

Weight of masonry

1 0.5 0.000 0.900 0.000 1.4 0 0.720 0

2 1 0.45 0.900 0.405 1.4 0.567 0.495 0.281

3 0.5 0.270 0.900 0.122 1.4 0.17 0.180 0.031

Water Pressure

i 1 0.785 0.9 1 0.71 0.45 0.318

ii 0.5 0.9 0.9 1 0.41 0.3 0.122

SV= 0.737 SM= 0.311 0.439


X= SM / SV = ( 0.311 - 0.439 )/ 0.737 = -0.174 mNot safe

Eccentricity e = b/2-X 0.72 /2 - -0.174 = 0.534 m

6e/b = 6 * 0.534 / 0.72 = 4.449 m

Maximum stress = SV / b * (1 + 6e/b)= 1.264 / 0.72 * ( 1 + 4.449 ) = 9.563 t/m

2

Minimum stress = SV/ b * (1 - 6e/b)= 1.264 / 0.72 * ( 1 - 4.449 ) = -6.053 t/m

2

3.Reservoir at FRL, no tailwater0.720

0.900 Uplift

S.NO L.A


Weight of masonry

1 0.5 0.000 0.900 0.000 2.4 0 0.720 0.000

2 1 0.45 0.900 0.405 2.4 0.972 0.495 0.481

3 0.5 0.270 0.900 0.122 2.4 0.29 0.180 0.052weight of water

a 0.5 0.000 0.9 1 0.000 0.000 0.0000

Water Pressure

i 0.5 0.9 0.9 1 0.41 0.3 0.122

Uplift 0.5 0.720 0.900 0.324 1 -0.324 0.48 0.156

SV= 0.940 SM= 0.534 0.277

MOMENT

DESCRIPTION

FORCE

FORCE MOMENT

DESCRIPTION


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146912493.xlsx.ms_officeBase width b = 0.720 m 2/3 *b = 0.48 m 1/3 *b = 0.24 m

X= SM / SV = ( 0.534 - 0.277 ) / 0.94 = 0.273 mFalls within the middle third

Eccentricity e = b/2-X 0.72 /2 - 0.273 = 0.087 m

6e/b = 6 * 0.087 / 0.72 = 0.724 m



1 Empty condition

2 MWL condition

3 FRL condition

stress in t/m2

stress in t/m2

3.026

0.844

-6.053

0.484

9.563

2.666

Maximum MinimumConditionS.No


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146912493.xlsx.ms_office

max min

Cond. 1 empty 2.666 0.844

Cond. 2 MWL 9.563 -6.053

Cond. 3 FRL 3.026 0.484


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146912493.xlsx.ms_officeStability Analysis of bodywall ( for Free flow weirs)


Maximum flood discharge 14.16 m3/sec 500 c/s

Top of crest + 165.100 m

FMFL + 165.335 m

RWL + 164.775 m

Upstream bed level + 164.200 m

Downstream Bed level + 164.200Down stream side slope 0.4 H to 1 V

Upstream side slope 0 H to 1 V

Top width 0.45 m

Unit weight of concrete 2.4 t/m

The stability of body wall of the anicut was checked for the following conditions

1 Reservoir empty

2 Reservoir at MWL,with tailwater

3 Reservoir at FRL, no tail water

165.335

0.235165.100 C

x y 0.325

a z 164.775

165.1 0.4

0 1

0.9 1

ii

O

0.9 0.235 0.000 0.45 0.360

0.4500.810

Stability analysis:

1.Reservoir empty

S.NO L.A

Co-eff Length Depth Area Unit wt. V H + + -

Weight of masonry

1 0.5 0.000 0.900 0.000 2.4 0 0.810 0.000

2 1 0.45 0.900 0.405 2.4 0.972 0.585 0.569

3 0.5 0.360 0.900 0.162 2.4 0.39 0.240 0.093

SV= 1.361 SM= 0.662


X= SM / SV = 0.662 / 1.361 = 0.486 mFalls within the middle third


6e/b = 6 * -0.081 / 0.81 = 0.603 m



164.200


12

3

146912493.xlsx.ms_officeweir stability (Free)

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146912493.xlsx.ms_office= 1.361 / 0.81 * ( 1 + 0.603 ) = 2.693 t/m2


2

2.Reservoir at MWL, with tailwater

S.NO L.A


Weight of masonry

1 0.5 0.000 0.900 0.000 1.4 0 0.810 0

2 1 0.45 0.900 0.405 1.4 0.567 0.585 0.332

3 0.5 0.360 0.900 0.162 1.4 0.23 0.240 0.054

Add Weight of masonry

x 0.5 0 0.325 0.000 1 0 0.81 0.000

y 1 0.45 0.325 0.146 1 0.146 0.585 0.086

z 0.5 0.13 0.325 0.021 1 0.021 0.317 0.007

Water Pressure

i 1 0.235 0.9 1 0.21 0.45 0.095

ii 0.5 0.9 0.9 1 0.41 0.3 0.122

SV= 0.961 SM= 0.478 0.217


X= SM / SV = ( 0.478 - 0.217 )/ 0.961 = 0.272 mFalls within the middle third


6e/b = 6 * 0.133 / 0.81 = 0.983 m


= 1.361 / 0.81 * ( 1 + 0.983 ) = 3.332 t/m2


3.Reservoir at FRL, no tailwater

0.810

0.900 Uplift

S.NO L.A


Weight of masonry

1 0.5 0.000 0.900 0.000 2.4 0 0.810 0.000

2 1 0.45 0.900 0.405 2.4 0.972 0.585 0.569

3 0.5 0.360 0.900 0.162 2.4 0.39 0.240 0.093

weight of water

a 0.5 0.000 0.9 1 0.000 0.000 0.0000

Water Pressure

i 0.5 0.9 0.9 1 0.41 0.3 0.122




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146912493.xlsx.ms_officeUplift 0.5 0.810 0.900 0.3645 1 -0.3645 0.54 0.197

SV= 0.996 SM= 0.662 0.318


X= SM / SV = ( 0.662 - 0.318 )/ 0.996 = 0.345 mFalls within the middle third


6e/b = 6 * 0.06 / 0.81 = 0.445 m



1 Empty condition

2 MWL condition

3 FRL condition 2.428 0.932

2.693 0.667

3.332 0.028

S.No ConditionMaximum Minimum

stress in t/m2

stress in t/m2


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max min

Cond. 1 empty 2.693 0.667

Cond. 2 MWL 3.332 0.028

Cond. 3 FRL 2.428 0.932


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Construction Checkdam

Surface Flow Condition

Design Data:

Top of Check dam = 165.1 m

Rear Water Level = 164.775 mFMFL = 165.335 m

Maximum Flood Discharge = 500 c/s = 14.158 Cumecs

U/s bed level = 164.2 m

D/s bed level = 164.2 m

Total width of canal = 20 m

Length of Checkdam = 20 m

Concentration factor = 0 %

Exit gradient 1 in 6 = 0.17

Depth of U/s water (FMFL - U/s bed level) 1.135 m

Thickness of cut-off wall 0.45 m

Design of structure for surface flow considerations:

Discharge intensity/Unit discharge Total discharge / Length of check dam

q = 14.158 / 20 = 0.708 m3/s/m

Regime Width:

Regime width of the canal = 4.83 Q = 4.83 X sqrt ( 14.158 )

= 18.17 m say 19 m

Looseness factor = Existing Over all length of check dam/ Regime width

= 20 / 19 = 1.053 > 1

Hence, the scour depth may be calculated as follows

Scour depth:

Assume silt factor f = 1.25

Scour depth (R) = 0.475 (Q / f) ^(1/3)

= 0.475 X ( 14.158 / 1.25 )^(1/3)

1.06675 m say 1.067 m

Energy level:

Depth of water in front of weir 1.135 m

Velocity of approach (Va) = Q / A

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= 14.158 / ( 20 x 1.135 = 0.624 m/sec

Approach Velocity head (hva) = Va /2g 0.624 ^2 / (2 X 9.81)

= 0.019846 m say 0.020 m

U/s Energy Line(U/s E.L) = U/s F.S.L + hva

= 165.335 + 0.02 = 165.355 m

s ve oc y d = 1.187 m/sec (From RWL calculation)

D/s Velocity head (hvd) = Vd2/2g = 1.187 ^2 / ( 2 X 9.81)

= 0.07181 m say 0.072 m

D/s Energy Line (D/s E.L) =D/s W.L + h

vd

= 164.775 + 0.072 = 164.847 m

Fixation of Stilling basin level:

Hydraulic jump calculations:

Sl.No.

1 Discharge intensity (q) 0.708 m2/sec

2 D/s Water level in m 164.775 m

3 U/s Water level in m 165.335 m

4 D/s Total Energy Level ( D/s E.L) in m 164.847 m

5 U/s Total Energy Level ( U/s E.L) in m 165.355 m

6 Head Loss (HL) in m (U/s E.L - D/s E.L) 0.508 m

7 Postjump Depth D2 (Assume) 0.760 m 0.003

8 Velocity (V2) = q / D2 = 0.7079 / 0.76 0.931 m/sec

9 D/s specific energy Ef2 = D2 + V2 /2g

0.76 + 0.931 2 / (2 * 9.81) 0.804 m

10 Froude's Number F2 =V2 / sqrt(g D2)

= 0.931 / sqrt( 9.81 * 0.76 ) 0.341

11 Prejump Depth D1 = D2 / 2 ( -1+ sqrt(1+ 8 F2 ))

Item


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= 0.76 / 2 * ( -1 + sqrt (1+ 8 * 0.3411 2 ) ) 0.148 m

12 Velocity (V1 ) = q / D1 = 0.7079 / 0.148 4.786 m/sec

13 U/s specific energy Ef1 = D1 + V1 / 2g 1.315 m

14 Ef1 - Ef2 - HL = 0 0.003 ~ 0

15 Level at which jump would form ( D/s Total Energy Lvl - Ef2) 164.040 m

16 Length of concrete floor required beyond the jump 3.06 m

Floor length = 5 * (D2 - D1)

Existing Floor level = 164.2 m

The Stilling Basin level(Required) = 164.04 m

Depth of stilling basin = 0.16 m

Provide a depth of stilling basin = 0.2 m

The Stilling Basin level is = 164.000 m

Total horizontal floor length:

U/S floor level = 164.2 m

D/S floor level = 164.2 m

1.D/s floor length /Basin length+cutoff = 4.00 m

2. D/s glacis length with 0.4 : 1 slope 0.44

3. U/s glacis length with 0 : 1 slope 0.00

4.Width of the body wall = 0.45 m

5. Length of U/s floor ( Assumed ) + cutoff = 0.91 m

=

Total Floor length required = 5.80 m

However provide a length of 5.80 m


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Sub surface Flow Condition (Koshla's Theory)

Design Data:

Maximum Flood Discharge 14 Cumecs 500 Cusecs

Crest level 165.100 m

MWL 165.335 m

Rear Water Level 164.775 m

Exit gradient 1 in 6 0.17

U/s bed level 164.200 m

D/s bed level 164.200 m

Length of Check dam 20 m

Thickness of cut-off wall 0.45 m

Assume U/s floor thickness as 0.45 m

Assume D/s floor thickness as 0.6 m

Unit weight of the floor material 2.4 t/m3

Unit weight of water 1 t/mFor static condition the FTL has been taken as FMFL = 165.100 m

For dynamic condition the MWL has been taken as FMFL = 165.335 m

Normal Scour depth (R) = 1.067 m

U/s cut-off level :

U/S scour level = U/s Water level - 1.25 R

= 165.335 - 1.25 * 1.067 = 164 m

Depth of U/s cut off = U/s bed level - U/s scour level

= 164.2 - 164 = 0.2 m

Depth of U/s water Y u = 1.14 m

Minimum depth of U/s cutoff = Yu /3 + 0.6

= 1.135 / 3 + 0.6 = 0.978 m

Depth of U/s cut off to be provided= 1 m

Provide a depth of U/s cutoff as 1 m

Bottom of U/s cut off = U/s bed level - Depth of U/s cutoff

Bottom of U/s cut off = 164.2 - 1 = 163.200 m

D/s cut-off level :

D/s scour level = Rear Water level - 1.5 R

164.775 - 1.5 X 1.067 163.175 m

Depth of D/s cut off = D/s bed level - D/s scour level

164.2 - 163.175 = 1.025 m

Depth of D/s water Y d = 0.58 m


DESIGN OF CHECKDAM ON PERMABLE FOUNDATION

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Minimum depth of D/s cutoff = Yd /2 + 0.6

= 0.575 / 2 + 0.6 = 0.888 m

Depth of D/s cut off to be provided= 1.6 m

Provide a depth of D/s cutoff as 1.6 m

Bottom of D/s cut off = D/s bed level - Depth of D/s cutoff

Bottom of D/s cut off = 164.2 - 1.6 = 162.600 m

To draw Hydraulic Gradient Line

165.1

0 5.35

0.685 0.890 3.78

1 0.4

164.2 0.45 1 164.2

164

163.75 0.6

1 m 163.4

1.6 m

0.910 4.44 162.6

163.2 2

1

Proposed Floor

Floor Length Required From Exit Gradient Considerations

Exit gradient 0.17

Taking head upto MFL

Static head H = FRL - D/S Bed level = 165.1 - 164.2 = 0.900 m

d = Depth of d/scutoff 1.6 m

Ge = H 1

d

= 1.1541_____

a = (2 -1)2

-1 0.843474Length of floor required from exit gradie a d = 1.350 m

2 m

Length of floor required (Exit gradient criteria) 2 m

Length of floor provided (hydraulic jump criteria) 5.80 m

Hence provide a floor length of 5.80 m


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Uplift Pressures:

for Pile 1

Total length of floor (b) = 5.80 m

Depth of pile (d) = 1 m

a = b / da = 5.8 / 1a = 5.8

_____

=( 1 + 1+ a2 )/ 2 = ( 1 + sqrt(1 + 5.8 ^2) ) / 2 = 3.443

fc = 1 / p cos-

(( - 2)/)

fc = 36.235 %

fD = 1 / p cos- (( - 1)/)

fD = 24.890 %

fC1 = 100-fcfC1 = 63.765 %

fD1 = 100-fDfD1 = 75.110 %

fE1 = 100 %

fE1 = 100 %

fD1 = 75 %

fC1 = 74 % (add 10% cor correction)

for Pile 2

Total length of floor (b) = 5.80 m

Depth of pile (d) = 1.6 m

a = b / da = 5.8 / 1.6

a = 3.625 =( 1 + 1+ a )/ 2 = ( 1 + sqrt(1 + 3.625 ^2) ) / 2 = 2.3802

fE = 1 / p cos-

(( - 2)/)

fE = 44.894 %


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fD = 1 / p cos-

(( - 1)/)

fD = 30.310 %

fC2 = 0 %

fD2 = 30.310 %

fE2 = 44.894 %

fE2 = 45 %

fD2 = 30 %

fC2 = 0 %

Pile No 1 2

Upstream pile Downstream pile

Pressure fE1 = 100 fE2 = 45

at the pt fD1 = 75 fD2 = 30

fC1 = 74 fC2 = 0

Un- Elevation of HGL

U/S water D/S water balanced Pile1 Pile2

Level Level Head fE1 fD1 fC1 fE2 fD2 fC2

100 75 74 45 30 0

Static condition

165.100 164.200 0.900 0.9 0.68 0.66 0.40 0.27 0

165.1 164.88 164.86 164.60 164.47 164.2

Dymanic condition

165.335 164.775 0.560 0.56 0.42 0.41 0.25 0.17 0.00

165.34 165.20 165.19 165.03 164.94 164.78

165.19

164.86 Dynamic HGL

165.14

165.335 Static HGL

165.1 164.79 165.03

0.953 164.60


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0.587

164.2 0.45

0.148 164.2

164.04 164

0.6

1 m

0.02 1.6 m162.6

163.2 2

1

0.685 0.89 3.78

5.35

Proposed Floor

Uplift Pressures

74

100 45 0

64

45

2 30

75 1

FLOOR THICKNESS(Dynamic head)

At the point of formation of jump 3.80 m from the centre of d/s pile

Dynamic head unbalance 0.953 m

2/3 rd of head 0.635 m

Thickness required 0.454

0.454 m

Provide a depth of 0.500 m

(Static head)

the static head is calculated from d/s basin level.

Maximum ordinate for static condition is at toe of the anicut. 3.78 m

(a) Unbalanced head at 2 m from center of d/s pile

0.701

Thickness required 0.501 m


(d) Unbalanced head at 3.78 m from center of d/s pile

Thickness required 0.787

0.562 m


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Static dominates the flow

Provided thickness at toe 0.600 m


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Design of Flexible Protection for Check dam

Downstream Block Protection: ( As per clause 20.2. of IS 6966(part I) - 1989 )

Calculation for the length of D/s block protection required

D/s Bed level 164.200 m

U/s Bed level 164.200 m

Designed Front MFL 165.335 m

Designed Rear MFL 164.775 m

Scour depth ( R ) 1.067 m

Width of toe wall 0.450 m

The likely extent of scour for downstream flexible protection = 1.75 x R

Design Depth of scour below the floor level ( D ) = 1.292 m

Length of downstream block protectio 1.5 D = 1.938 < 2.1 m

Hence provide 1 rows of CC block of size 1.5 X 1.5 X 0.9 m

Downstream Launching Apron: ( As per clause 20.3 of IS 6966(part I) - 1989 )

Length of launching apron = 1.5 to 2.5 DWhere D = Design Depth of scour below the floor level

D = 1.292 m

Length of launching apron = 1.938 to 3.231

= 2 m say 2.00 m

Slope of the river = 1 in 500

River Slope in m/km 2.000 m

Restricted to 0.4

Thickness of pitching ( T 1000 mm (From Table 2 of IS 6966(part I) - 1989)

Thickness of pitching required for covering the launched slope= 1.25 T = 1.250 m

As per clause 20.3.3 of IS 6966, Provide a slope of

2.50 : 1

2

Length of launching = 2.154 m 1.10

Quantity of pitching required to be laid per m width in the launched

slope length o 2.154 m with thickness 1.25 m = 2.154

2.693 m3

2.5: 1

This should be laid in the length of 2 m

Quantity of pitching required for the thickness(T) 2.154 m3

Inner Thickness of loose apron = 1.077 m

Provide an inner thickness of 1.1 m

Balance quantity of pitching= 0.539 m3

This should be laid in the wedge portion

Thickness of wedge portion= 0.539 m 2

Outer thickness required= 1.639 m 1.10

Provide an outer thickness= 1.7 m

Quantity of pitching provided = 2.8 m3

< 2.693 m3

?

1.70

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Hence O.K

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DESIGN OF ABUTMENT


Top level of abutment = 166.085 m Free board = 0.75

Foundation level= 164.200 m

Height of abutment wall= 1.885 m

Top width= 0.450 mFront slope= 1 in 0

Front slope starting point= 166.085 m

Front batter 0 m

Rear slope starting point= 166.085 m

Rear slope= 1 in 0.6 7.72

Rear batter= 1.131 m 0.28

b3= Base width= 1.581 m

Height= 1.885 m

Unit weight of soil= 2 t/m3

Top of earthfill= 166.085 m

f= Angle of internal friction of soil 22 od= f/2= Angle of friction between the wall and earthfill 11 oa= Angle which earth face of wall makes with vertical 31 oi= Slope of earthfill 0Unit weight of concrete= 2.4 t/m

3

Unit weight of water= 1 t/m3

STRESS DUE TO EARTH PRESSURE (ACTIVE EARTH PRESSURE):

Cos^2(f-a)= 0.976Cos^2a= 0.735

Cos(d+a)= 0.744

Sin(f+d)= 0.545

Sin(f-i)= 0.375Cos(a-i)= 0.857

Cos(d+a)= 0.744

1+ Sin(f+d)*Sin(f-i) 0.5 = 1.566

Cos(a-i)*Cos(d+a)

1 2

1+ Sin(f+d)*Sin(f-i) 0.5 = 0.408Cos(a-i)*Cos(d+a)

Cos2( f-a) = 1.785Cos

2a * Cos(d+a)

Cos2( f-a) 1 2

Cos2a * Cos(d+a) 1+ Sin(f+d)*Sin(f-i) ^0.5

Cos(a-i)*Cos(d+a)

= 1.785 * 0.408 = 0.7280


Ca=

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Ca = 0.7280

Pa = Pa =1/2*w*h2*Ca = 0.5 * 2 * 1.885 2 * 0.728 = 2.587 t

Earth pressure:

Vertical component= Pa* Sin(a+d) = 2.587 * Sin ( 31 + 11 ) = 1.730 t

Horizontal componen Pa* Cos(a+d) 2.587 * Cos( 31 + 11 ) = 1.923 t

166.085

0.450 i0 a 0.6

1 d Pa 1

1.885

O

0 0.450 1.131

Forces(t) e Moments

V H (m) t-m

Weight of concrete

1 1 0.450 1.885 2.4 2.036 1.356 2.761

0.5 1 1.131 1.885 2.4 2.558 0.754 1.929

0.5 1 0 1.885 2.4 0.000 1.581 0.000

Static Earth pressure:

Vertical component= 1.730 0.377 0.652

Horizontal component= 1.923 0.628 1.209

6.324 1.923 6.550

SV= 6.324 SM= 6.550

Base width b = 1.581 m 2/3 *b = 1.054 m 1/3 *b = 0.527

X= SM / SV = 6.55 / 6.324 = 1.036 m

Falls within the middle third


6e/b = 6 * -0.245 / 1.581 = 0.931 m


2

Minimum stress = SV/ b * (1 - 6e/b)

= 6.324 / 1.581 * ( 1 - 0.931 ) = 0.276 t/m2

Maximum stress= 7.723 t/m2

Minimum stress= 0.276 t/m2

Unit.wt.Coeff. L B D

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DESIGN OF D/s WING WALL


Top level of D/s wing wall = 165.525 m Free board = 0.75

Foundation level= 164.200 m

Height of wall= 1.325 m

Top width= 0.450 mFront slope= 1 in 0

Front slope starting point= 165.525 m

Front batter 0 m

Rear slope starting point= 165.525 m

Rear slope= 1 in 0.5 5.41

Rear batter= 0.6625 m 0.34

b3= Base width= 1.113 m

Height= 1.325 m

Unit weight of soil= 2 t/m3

Top of earthfill= 165.525 m

f= Angle of internal friction of soil 22 od= f/2= Angle of friction between the wall and earthfill 11 oa= Angle which earth face of wall makes with vertical 27 oi= Slope of earthfill 0Unit weight of concrete= 2.4 t/m

3

Unit weight of water= 1 t/m3

STRESS DUE TO EARTH PRESSURE (ACTIVE EARTH PRESSURE):

Cos^2(f-a)= 0.994Cos^2a= 0.800

Cos(d+a)= 0.793

Sin(f+d)= 0.545

Sin(f-i)= 0.375Cos(a-i)= 0.894

Cos(d+a)= 0.793

1+ Sin(f+d)*Sin(f-i) 0.5 = 1.536

Cos(a-i)*Cos(d+a)

1 2

1+ Sin(f+d)*Sin(f-i) 0.5 = 0.424Cos(a-i)*Cos(d+a)

Cos2( f-a) = 1.567Cos

2a * Cos(d+a)

Cos2( f-a) 1 2

Cos2a * Cos(d+a) 1+ Sin(f+d)*Sin(f-i) ^0.5

Cos(a-i)*Cos(d+a)

= 1.567 * 0.424 = 0.6638


Ca=

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Ca = 0.6638

Pa = Pa =1/2*w*h2*Ca = 0.5 * 2 * 1.325 2 * 0.664 = 1.165 t

Earth pressure:

Vertical component= Pa* Sin(a+d) = 1.165 * Sin ( 27 + 11 ) = 0.710 t

Horizontal componen Pa* Cos(a+d) 1.165 * Cos( 27 + 11 ) = 0.924 t

165.525

0.450 i0 a 0.5

1 d Pa 1

1.325

O

check dam 0.9,500

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