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

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

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

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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

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Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI 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

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Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI 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)

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

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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

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Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI OFF- TAKE SLUICE – WORKED OUT EXAMPLE

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Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI s.no ParticularsParental CanalOff-take 1 F.S. discharge 11.5 cumecs0.84 cumecs 2 Velocity 0.44 m / sec Section 10.5 mx 1.52 m2.44m x 0.68m 4 Surface fall 1/52801/ Banks L/R 3.66m/1.82m1.82m/1.82m 6 Half supply level Bed level F.S. Level T.B. Level Ground level HYDRAULIC PARTICULARS

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

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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 2.13 below 2.13 m to 1.22 m 1.22M 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%0.15m0.07m 5% to 10%0.30m0.15m0.07m 2% to 5%0.46m0.30m0.15m 1% to 2%0.61m0.46m030m 0.5% to 1%0.76m0.61m0.46m Less than 0.5% 0.91m0.76m0.61m

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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= 0.27 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 = √h h = Driving head = A = = m x √0.27

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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 m 2 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 q 2 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.

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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 /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 = Φ D ’ = 8% = 8/100 x 1.37 = m

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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. b)Uplift Pressure: Uplift head resisted by floor of the barrel: The thickness of floor under barrel = 0.38 m R 2 = (R-0.19) 2 = R 2 – 0.38 R = – 0.38 R R= 0.246/0.38 = 0.64

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Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI 0.64 – 0.19 Cot α = = 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/m 2 µ x x 0.99 = x x µ = = m 0.91 x 0.99 Hence the floor of the barrel is safe against uplift head of – = 1.52 m c)Thickness of floor: Percentage of pressure at D/s head wall (92-8) = x 2.51 = = 27.3% Considering buoyant weight of foundation concrete and 75% of theoretical head.

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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:

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Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI S. No Forc e Particulars Magni- tude Lever arm (mrs) Moment in t.m. 1.W1W1 0.52x415x1/100= W2W2 0.52x0.93x2243/1000= PVPV [(1.54) 2 – 0.61) 2 ] x 2083/1000 = Total vertical load (V) PHPH 0.134[(1.54) 2 – (0.61) 2 ] x 2083/1000 = Total Moments (M) Taking moments about point ‘ A ’ W 1 = 415 kg/m 2 live load

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Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI M 0,5453 L.A of the resultant load = --- = = 0.37 m V Eccentricity = = 0.11 m x0.11 Stresses = (1 ± ) = (1±0.66/0.52) 0.52 Stress = 6.33 t/m 2 (Compressive) & t/m 2 (tension) As there will be arch action due to abutting of side walls these stresses may be neglected.

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Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI 2.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/m Average loading = x 8000 = 3165 kg/m x 2403 Dead weight of slab = = 245 kg/m

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Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI Total uniformly distributed load = = 3410 kg x B.M. due to this U.D.L. = x = kg. cm. Adopt HYSD bars & M15 mix Effective depth = = cm x 100

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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 Area of steel required = = 3.22 cm x x 7.75 Area of 10mm. dia bar = 0.79 cm 2 Spacing of 10mm.dia bars 0.79 x 100 = = Adopt a spacing of 15cm centres (equal to the spacing of barrel slab reinforcement)

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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 Shear stress = = 2.00 kg/cm x 75 x 100 Percentage steel = allowable shear stress as per tables = 3.26 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 1551 = = kg/cm x 7.75 x π x 1 ( ) 2 x 15 Provide 50 Φ anchorage

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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

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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 Effective depth = √ = cm x 100 Adopt 7.75 cm. effective depth as assumed. Area of steel = ( ) = 3.96 cm x 7.75 x 0.875

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Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI 0.79 x 100 Spacing of 10 mm dia. Bars = = cm Adopt a spacing of 15cm Percentage steel = = (b)Check for shear: Maximum shear force at support = 4203 x 0.98/2 = 2060 kg Actual shear stress = ( ) 7.75 x 100 = 2.66 kg/cm 2 < allowable shear stress as per tables = Hence safe.

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Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI (c)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 Bond stress = = kg/cm x 7.75 x Provide 30cm of anchorage

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Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI 4 Design of side walls for the barrel

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Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI Force Particulars Magnitude (t) Level arm (m) Moment in t.m W1W X 1.90 X 2083/1000= W2W X 10.2/100 X 2403/1000 = W3W X ( ) X 2403/1000 = W4W X 1.90 X 2083/1000= W5W X 1.90 X 2243/1000= W6W X 2.02 X 2083/1000= (a) Stresses in masonry: Taking moments about point A.

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Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI W7W X 0.84/2 X 2083/1000 = W8W X 0.84/2 X 2243/1000 = PVPV ( – ) 2083/1000 = Total vertical load (V) PHPH ( – ) 2083/ Total Moments (M) 2.563

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Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI L.A. of the resultant = = 0.42 m Eccentricity = /2 = 0.12 m 1/6 th of base width = 0.61 / 6 = 0.10m x 0.12 Stresses = (1± ) = (1±1.2) Maximum stress (compressiove)= x 2.2 = t/m 2 Minimum stress (tension) = x 0.2 = 2.0 t/m 2

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Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI Force ParticularsMagnitude (t) Level arm (m) Moment ( tm) W1W1 Same as force= W2W2 “= W3W3 “= W4W4 “= W5W5 “= W6W6 “= W7W7 “= W8W8 “= W9W x 2.84 x2.83/1000= W x 0.38 x 2243/1000= PVPV ( – 1.902) 2083/1000= 0.541/ Total vertical load (V) PHPH ( – ) 2083/1000= Total Moments (M)4.881 (b)Stress on soil: Taking moments about point B.

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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/ 6 th of the base width = 1.05/6 = m x Stresses = (1± ) = (1±1.2) Maximum compressive stress = x = t/m 2 Minimum compressive stress = x = 5.60 t/m 2 &&&&&&&&&&&&&&

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Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI THANK YOU

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