1. WELCOME to the GROUP SABARI of AEEs of 2008 BATCH
VIJAYAKUMAR SREEKANTA
M.Tech;MHRM;
Master Trainer (GoI)
FACULTY,WALAMTARI
Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

2. OFF - TAKE SLUICE - IMPORTANCE - DESIGN PRINCIPLES by
VIJAYAKUMAR SREEKANTA
M.Tech;MHRM;
Master Trainer (GoI)
FACULTY,WALAMTARI
Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

3. OFF –TAKE SLUICE- IRRIGATION SYSTEMHR
Right Main Canal
OT –R 1
OT-L1
Left Main Canal
OT-L2
OT-L3
OT Channel
OT Channel
Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

4. Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI
OFF TAKE SLUICE
It is the main structure in an irrigation system
Draws a specified amount of water from parent canal to the distributory
It is at the head of a distributory
It passes the required designed discharge
It is to organize water delivery in a planned way in an irrigation system
It can be of barrel or Pipe
Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

5. OFF- TAKE SLUICE COMPONENTS
Vent way
Barrel , Pipe
Head walls / Wings & Returns
Up stream & Down Stream
Hoist
Shutters & Hoist Equipment
Upstream & Down Stream Bed Levels
Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

6. OFF- TAKE SLUICE - DESIGN FEATURES
Flow condition for which the OT vent way is to be designed (Full supply / Half supply)
Fixation of sill of Off Take sluice in reference to parent canal Bed level (atio of ‘q / Q’)
Using appropriate formula for Vent way design (Barrel / Pipe) from DRIVING HEAD point of view
Provision of control arrangements; Hoist etc
Floor thickness ( Uplift conditions)
Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

7. OFF- TAKE SLUICE – DESIGN DATA REQUIRED
-Hydraulic particulars of
Parent canal
Distributory
at the point of proposed OT location
Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

8. Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI
So to ensure delivery of required quantity of water in the irrigation channel ….
we need to
Design an Off take sluice at the head of every distributory / channel
Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

9. OFF- TAKE SLUICE – WORKED OUT EXAMPLE
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10. Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI
HYDRAULIC PARTICULARS
s.no
Particulars
Parental Canal
Off-take 1
F.S. discharge
11.5 cumecs
0.84 cumecs 2
Velocity
0.44 m / sec 3
Section
10.5 mx 1.52 m
2.44m x 0.68m 4
Surface fall
1/5280
1/3000 5
Banks L/R
3.66m/1.82m
1.82m/1.82m 6
Half supply level
+48.46
----- 7
Bed level
+47.40
+47.55 8
F.S. Level
+48.92
+48.23 9
T.B. Level
+49.83
+48.84 10
Ground level
Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

11. Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI
OT DESIGN:
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12. Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI
Perecentage of off take discharge to parent canal discharge
Height of sill above the bed of parent canal when the F.S.D. in the parent canal is
Above below
2.13 m to 1.22 m M
Remarks
15% and above
0.07m
The sill of the sluices
Should also be fixed
Such that Lower and lower as the location goes towards the end of the distributaries and minors.
10% to 15%
0.15m
5% to 10%
0.30m
2% to 5%
0.46m
1% to 2%
0.61m
030m
0.5% to 1%
0.76m
Less than 0.5%
0.91m
Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

13. Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI
A) Vent Way Calculations: ( Design of Off-take is being done for a range of
Full supply - Half supply in the Major )
Half supply level in major =
Water surface level at D =
Driving head available = m
Discharge to pass through vent way
Q = 2.86 A√h
Where Q = Discharge through vent = 0.84 cumecs. Q
A = Area of vent way required =
2.86√h
h = Driving head = 0.27
0.84
A = = m2
2.86 x √0.27
Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

14. Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI
Hence a vent way of 0.91 m x 0.65 m is provided giving an area of
0.59 m2 The dimensions of the shutter may be 1.06m x0.71 m
B ) Scour depth Calculations:
a)Scour depth at the entrance
q = discharge per meter width = 11.5/10.81 = cumecs
(Average width= /2 = m)
f = silt factor , equal to 1 q2
R = depth of scour below water surface = (-----)1/3 f
As this is only a normal reach without any obstruction, no factor of safety is
Considered and R = x /3 = m. below F.S.L.
(against 1.52 m FSD)
However 0.46m. deep cut off is provided.
Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

15. Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI
b) Scour depth at the end of Downstream wings:
q = 0.84/2.44 = or cumecs
f = silt factor equal to 1
R (with a factor of safety of 1.5) = x 1.5 x 0.342/3= 0.99 m
Depth below B.L. = 0.99 – 0.68 = 0.31 m
Floor thickness itself is 0.46 m
No cut off is therefore provided.
C) Exit gradient (GE), Uplift pressures and Thickness of floor Calculations:
a) Exit Gradient:
The total effective horizontal length of floor b = 10.97m.
d = depth of downstream cut off = 0.46 m
Head acting H = – = 1.37 m
1/α = D/B = 0.46/10.97 = 0.417
Φ D’ = 8%
= 8/100 x 1.37 = m
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16. Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI
GE = 0.84 x /0.46 = 0.20 < 0.3
Which is less than 0.3, hence safe.
Uplift Pressure:
Uplift head resisted by floor of the barrel:
The thickness of floor under barrel = 0.38 m
R2 = (R-0.19)2 = R2 – 0.38 R
= – 0.38 R
R = 0.246/0.38 = 0.64
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17. Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI
0.64 – 0.19
Cot α = = 0.99
0.455
µ L t
--- x x cot α = --- x p
Where µ = the maximum safe uplift pressure head taken by arch action.
L= span of arch = 0.91 m (width of barrel)
T = thickness of floor = 0.38 m
P = mean permissible stress at the crown of the arch section and is taken equal to t/m2 µ
--- x x 0.99 = x 27.34
0.38 x 27.34
µ = = m
0.91 x 0.99
Hence the floor of the barrel is safe against uplift head of
48.92 – = 1.52 m
c) Thickness of floor:
Percentage of pressure at D/s head wall
(92-8)
= x = = 27.3%
10.97
Considering buoyant weight of foundation concrete and 75% of theoretical head.
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18. Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI
The thickness of floor required
= 1.37 x x x = 0.22 m
as against 0.46m thick provided.
Hence safe
D) Design of sub-structure:
1. Design of upstream head wall:
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19. Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI
Taking moments about point ‘A’
W1 = 415 kg/m2 live load S. No
Force
Particulars
Magni-tude
Lever arm (mrs)
Moment in t.m. 1. W1
0.52x415x1/100 =
0.26
0.0561 2. W2
0.52x0.93x2243/1000 =
0.2820 3. PV
0.0384[(1.54)2 – 0.61)2]
x 2083/1000
=0.1600
------
-----
Total vertical load (V)
1.4605 4. PH
0.134[(1.54)2 – (0.61)2] x 2083/1000
= 0.556
0.372
0.2072
Total Moments (M)
0.5453
Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

20. Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI
L.A of the resultant load = = = m V
0.52
Eccentricity = = m 2
x0.11
Stresses = (1 ± )
1.4605
= (1±0.66/0.52)
Stress = 6.33 t/m2 (Compressive) & t/m2 (tension)
As there will be arch action due to abutting of side walls these stresses may
be neglected.
Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

21. Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI
Design of Lintel under upstream head wall:
(a) Main reinforcement : Slab in proximity to earth or moisture
The clear Span = 0.91
Thickness of slab assumed = 10.2 cm (overall)
Effective depth = 7.75 cm (assumed)
Effective span = 0.98 m.
Maximum compressive stress = 6.33 t/m2
6.33
Average loading = x 8000 = 3165 kg/m2 2
10.2 x 2403
Dead weight of slab =
100
= kg/m
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22. Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI
Total uniformly distributed load = = 3410 kg.
3410 x 0.982
B.M. due to this U.D.L. = x 100 8
= kg. cm.
Adopt HYSD bars & M15 mix
32750
Effective depth = = cm
8.203 x 100
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23. Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI
However adopt 7.75 cm as assumed
Using 10 mm.dia.bars
Total depth = = or 10.2 cm
32750
Area of steel required = = 3.22 cm2
500 x x 7.75
Area of 10mm. dia bar = 0.79 cm2
Spacing of 10mm.dia bars
0.79 x 100
= = 24.54
3.22
Adopt a spacing of 15cm centres (equal to the spacing of barrel slab reinforcement)
Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

24. Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI
(b) Check for Shear:
3410 x 0.98
Maximum shear at the support = = 1551 kg 2
1551
Shear stress = = 2.00 kg/cm2
1.0 x 75 x 100
Percentage steel = 0.68.
allowable shear stress as per tables = kg /cm2
Check for Bond:
Maximum shear force at the support = 1551 kg.
100
Perimeters of bars = ( ) π X 1/m width
2 x 15
Bond stress, for M 15 alternate bars cranked
= = kg/cm2
0.875 x 7.75 x π x 1 ( )
Provide 50 Φ anchorage
Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

25. Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI
3. Design of slab over barrel:
(a) Main reinforcement:
Clear span = 0.91 m
The thickness of slab = 10.2 cm
Using 10 mm.dia. bars and a clear cover of 1.92 cm
The effective depth = 10.2 – 0.50 – 1.92 = 7.78 cm or 7.75 cm.
Effective span = = 0.98 m
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26. Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI
Dead weight of slab/metre width = 10.2 x 2403/100 = 245 kg.
Weight of earth (including live load) = 1.90 x 2.83 x 1.0 = 3958 kg .
Total U.D.L = 4203 kg.
Assuming partial fixity
4203 x x 100
B.M = = kg.cm 10
40366
Effective depth = √ = cm
8.203 x 100
Adopt 7.75 cm. effective depth as assumed.
Area of steel
= ( ) = cm 2
1500 x 7.75 x 0.875
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27. Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI
0.79 x 100
Spacing of 10 mm dia. Bars = = cm
3.968
Adopt a spacing of 15cm.
5.266
Percentage steel = = 0.68
7.75
Check for shear:
Maximum shear force at support = 4203 x 0.98/2 = 2060 kg.
2060
Actual shear stress = ( )
7.75 x 100
= 2.66 kg/cm2 < allowable shear stress as per tables = 3.26.
Hence safe.
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28. Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI
Check for Bond:
Maximum shear force = 2069 kg.
Perimeter of 50% bars per metre width, alternate bar cracked.
100
= ( ) π X 1 = cm
2 x 15
060
Bond stress = = kg/cm2
0.875 x 7.75 x 13.61
Provide 30cm of anchorage
Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

29. Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI
Design of side walls for the barrel
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30. Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI
(a) Stresses in masonry:
Taking moments about point A.
Force
Particulars
Magnitude (t)
Level arm (m)
Moment in t.m W1
0.68 X 1.90 X 2083/1000
= 2.679
0.497
1.331 W2
0.68 X 10.2/100 X 2403/1000
= 0.159
0.079 W3
0.225 X ( ) X 2403/1000
= 0.424
0.211 W4
0.225 X 1.90 X 2083/1000
= 1.009
0.273
0.275 W5
0.225 X 1.90 X 2243/1000
= 0.475
0.129 W6
0.16 X 2.02 X 2083/1000
= 0.667
0.08
0.053
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31. Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI
0.16 X 0.84/2 X 2083/1000
= 0.140
0.053
0.008 W8
0.16 X 0.84/2 X 2243/1000
= 0.151
0.11
0.017 PV
0.384 (2.542 – 1.902) 2083/1000
= 0356
Total vertical load (V)
6.060 PH
0.134 (2.842 – 1.902) 2083/1000
1.244
0.372
0.46
Total Moments (M)
2.563
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32. Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI
2.563
L.A. of the resultant = = 0.42 m
6.060
Eccentricity = /2 = 0.12 m
1/6th of base width = 0.61 / 6 = 0.10m
x 0.12
Stresses = (1± ) = (1±1.2)
Maximum stress (compressiove)= x 2.2 = t/m2
Minimum stress (tension) = x 0.2 = t/m2
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33. Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI
(b) Stress on soil:
Taking moments about point B.
Force
Particulars
Magnitude (t)
Level arm (m)
Moment
( tm) W1
Same as force
= 2.579
0.717
1.927 W2 “
= 0.159
0.114 W3
= 0.424
0.303 W4
= 1.009
0.493
0.497 W5
= 0.475
0.234 W6
= 0.667
0.30
0.200 W7
= 0.140
0.273
0.038 W8
= 0.151
0.33
0.050 W9
0.22 x 2.84 x2.83/1000
= 1.301
0.11
0.143
W10
1.05 x 0.38 x 2243/1000
= 0.895
0.525
0.470 PV
0.384 (3.222 – 1.902) 2083/1000
= 0.541/8.441
Total vertical load (V)
8.441 PH
0.134 (2.842 – 1.902) 2083/1000
= 1.886
0.48
0.905
Total Moments (M)
4.881
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34. Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI
L.A of the resultant = 4.881/8.441 = m
Eccentricity = – 0525 = m
1/ 6th of the base width = 1.05/6 = m
x 0.053
Stresses = (1± ) = (1±1.2)
Maximum compressive stress = x = t/m2
Minimum compressive stress = x = 5.60 t/m2
&&&&&&&&&&&&&&
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35. Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI
THANK YOU
Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI