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**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

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**OFF - TAKE SLUICE - IMPORTANCE - DESIGN PRINCIPLES by**

VIJAYAKUMAR SREEKANTA M.Tech;MHRM; Master Trainer (GoI) FACULTY,WALAMTARI Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

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**OFF –TAKE SLUICE- IRRIGATION SYSTEM**

HR 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

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**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

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**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

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**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

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**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

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**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

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**OFF- TAKE SLUICE – WORKED OUT EXAMPLE**

Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

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**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

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**Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI**

OT DESIGN: Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

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**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

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**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

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**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

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**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 Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

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**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 Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

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**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. Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

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**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: Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

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**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

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**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

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**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 Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

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**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 Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

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**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

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**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

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**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 Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

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**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 Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

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**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. Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

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**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

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**Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI**

Design of side walls for the barrel Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

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**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 Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

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**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 Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

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**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 Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

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**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 Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

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**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 &&&&&&&&&&&&&& Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

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**Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI**

THANK YOU Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

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