Report

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 OFF - TAKE SLUICE - IMPORTANCE - DESIGN PRINCIPLES by VIJAYAKUMAR SREEKANTA M.Tech;MHRM; Master Trainer (GoI) FACULTY,WALAMTARI Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI OFF –TAKE SLUICEIRRIGATION SYSTEM HR OT –R 1 OT-L1 OT-L2 OT-L3 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 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 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 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 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 OFF- TAKE SLUICE – WORKED OUT EXAMPLE Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI HYDRAULIC PARTICULARS s.no Particulars 1 F.S. discharge 2 Velocity Parental Canal Off-take 11.5 cumecs 0.84 cumecs 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 +48.46 +48.46 Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI OT DESIGN: Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI Perecentage of off take discharge to parent canal discharge 15% and above 10% to 15% 5% to 10% 2% to 5% 1% to 2% 0.5% to 1% Less than 0.5% Height of sill above the bed of parent canal when the F.S.D. in the parent Remarks canal is Above 2.13 below 2.13 m to 1.22 m 1.22M 0.07m 0.07m 0.07m The sill of the sluices Should also be 0.15m 0.07m 0.07m fixed 0.30m 0.15m 0.07m Such that Lower 0.46m 0.30m 0.15m and lower as the 0.61m 0.46m 030m location goes 0.76m 0.61m 0.46m towards the end of the 0.91m 0.76m 0.61m distributaries and minors. 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 = +48.46 Water surface level at D = +48.19 ------------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 = ----------2.86√h h = Driving head = 0.27 0.84 A = ------------------- = 0.5652 m2 2.86 x √0.27 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 = 1.063 cumecs (Average width= 10.05 + 1.52/2 = 10.81 m) f = silt factor , equal to 1 q2 R = depth of scour below water surface = 1.346 (-----)1/3 f As this is only a normal reach without any obstruction, no factor of safety is Considered and R = 1.346 x 1.0632/3 = 1.403 m. below F.S.L. (against 1.52 m FSD) Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI However 0.46m. deep cut off is provided. b) Scour depth at the end of Downstream wings: q = 0.84/2.44 = 0.3481 or 0.348 cumecs f = silt factor equal to 1 R (with a factor of safety of 1.5) = 1.343 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 = 48.92 – 47.55 = 1.37 m 1/α = D/B = 0.46/10.97 = 0.417 Φ D’ = 8% = 8/100 x 1.37 = 0.1096 m Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI GE = 0.84 x 0.1096/0.46 = 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 R2 = 0.4552 + (R-0.19)2 = 0.21 + R2 – 0.38 R + 0.036 = 0.246 – 0.38 R R = 0.246/0.38 = 0.64 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 1 2 2 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 27.34 t/m 2 µ 0.91 0.38 --- x -------x 0.99 = -------- x 27.34 1 2 2 0.38 x 27.34 µ = -------------------- = 11.53 m 0.91 x 0.99 Hence the floor of the barrel is safe against uplift head of 48.92 – 47.40 = 1.52 m c) Thickness of floor: Percentage of pressure at D/s head wall (92-8) = 8 + ------- x 2.51 = 8 + 19.3 = 27.3% 10.97 Considering buoyant weight of foundation concrete and 75% of theoretical head. Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI The thickness of floor required 27.3 75 1 = 1.37 x ---------x --------- x ------- = 0.22 m 00 100 1.25 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 Taking moments about point ‘A’ W1 = 415 kg/m2 live load S. Forc No e Particulars Magnitude 1. W1 0.52x415x1/100 2. 3. W2 PV 0.52x0.93x2243/1000 = 1.0847 0.0384[(1.54)2 – 0.61)2] x 2083/1000 =0.1600 Total vertical load (V) 4. PH = 0.2158 Lever arm (mrs) 0.26 Moment in t.m. 0.26 0.2820 ------ ----- 0.372 0.2072 0.0561 1.4605 0.134[(1.54)2 – (0.61)2] = 0.556 x 2083/1000 Total Moments (M) Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI 0.5453 L.A of the resultant load = M 0,5453 --- = ------------- = 0.37 m V 1.4605 0.52 Eccentricity = 0.37 - --------- = 0.11 m 2 1.4605 6x0.11 Stresses = ------------ (1 ± -------- ) 0.52 0.52 1.4605 = --------------- (1±0.66/0.52) 0.52 Stress = 6.33 t/m2 (Compressive) & 0.725 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 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/m2 6.33 Average loading = --------- x 8000 = 3165 kg/m2 2 10.2 x 2403 Dead weight of slab = -----------------100 = 245 kg/m Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI Total uniformly distributed load = 3165 + 245 = 3410 kg. 3410 x 0.982 B.M. due to this U.D.L. = ---------------------- x 100 8 = 32750 kg. cm. Adopt HYSD bars & M15 mix 32750 Effective depth = ------------------ --= 6.319 cm 8.203 x 100 Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI However adopt 7.75 cm as assumed Using 10 mm.dia.bars Total depth = 7.75 + 0.50 + 1.92 = 10.17 or 10.2 cm 32750 Area of steel required = -------------------- = 3.22 cm2 500 x 0.875 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 (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 = 3.26 kg /cm2 Check for Bond: Maximum shear force at the support = 1551 kg. 100 Perimeters of bars = (----------- +1) π X 1/m width 2 x 15 Bond stress, for M 15 alternate bars cranked 1551 = ---------------------------------------------- = 16.82 kg/cm2 100 0.875 x 7.75 x π x 1 (----------- +1) 2 x 15 Provide 50 Φ anchorage 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 = 0.91 + 0.0775 = 0.98 m 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 0.982 x 100 B.M = ------------------------- = 40366 kg.cm 10 40366 Effective depth = √-------------------- = 7.015 cm 8.203 x 100 Adopt 7.75 cm. effective depth as assumed. Area of steel 40366 = ( --------------------------- ) = 3.96 cm 2 1500 x 7.75 x 0.875 Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI 0.79 x 100 Spacing of 10 mm dia. Bars = ------------------------ = 19.91 cm 3.968 Adopt a spacing of 15cm. 5.266 Percentage steel = ---------- = 0.68 7.75 (b)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 (c) Check for Bond: Maximum shear force = 2069 kg. Perimeter of 50% bars per metre width, alternate bar cracked. 100 = (----------- +1) π X 1 = 13.61 cm 2 x 15 060 Bond stress = ------------------------------ = 22.32 kg/cm2 0.875 x 7.75 x 13.61 Provide 30cm of anchorage Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI 4 Design of side walls for the barrel Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI (a) Stresses in masonry: Taking moments about point A. Force Particulars Magnitude Level arm (t) (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 0.497 0.079 W3 0.225 X (0.76 + 0.08) X 2403/1000 = 0.424 0.497 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 0.273 0.129 W6 0.16 X 2.02 X 2083/1000 = 0.667 0.08 0.053 Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI W7 W8 PV 0.16 X 0.84/2 X 2083/1000 0.16 X 0.84/2 X 2243/1000 0.384 (2.542 – 1.902) 2083/1000 Total vertical load (V) PH 0.134 (2.842 – 1.902) 2083/1000 = 0.140 0.053 0.008 = 0.151 0.11 0.017 = 0356 ----------- ----------- 0.372 0.46 6.060 1.244 Total Moments (M) Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI 2.563 2.563 L.A. of the resultant = ---------------- = 0.42 m 6.060 Eccentricity = 0.42 - 0.61/2 = 0.12 m 1/6th of base width = 0.61 / 6 = 0.10m 6.060 6 x 0.12 Stresses = -----------(1±--------------------) = 0.61 0.61 9.934 (1±1.2) Maximum stress (compressiove)= 9.934 x 2.2 = 21.86 t/m2 Minimum stress (tension) = 9.934 x 0.2 = Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI 2.0 t/m2 (b) Stress on soil: Taking moments about point B. Force W1 W2 W3 W4 W5 W6 W7 W8 W9 W10 PV Particulars Same as force “ “ “ “ “ “ “ 0.22 x 2.84 x2.83/1000 1.05 x 0.38 x 2243/1000 0.384 (3.222 – 1.902) 2083/1000 Total vertical load (V) PH 0.134 (2.842 – 1.902) 2083/1000 Magnitude (t) = 2.579 = 0.159 = 0.424 = 1.009 = 0.475 = 0.667 = 0.140 = 0.151 = 1.301 = 0.895 = 0.541/8.441 Level arm (m) 0.717 0.717 0.717 0.493 0.493 0.30 0.273 0.33 0.11 0.525 ----------- Moment ( tm) 1.927 0.114 0.303 0.497 0.234 0.200 0.038 0.050 0.143 0.470 ----------- 8.441 = 1.886 0.48 0.905 Total Moments (M) Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI 4.881 L.A of the resultant = 4.881/8.441 = 0.578 m Eccentricity = 0.578 – 0525 = 0.053 m 1/ 6th of the base width = 1.05/6 = 0.175 m 8.441 6 x 0.053 Stresses = -----------(1±--------------------) 1.05 1.05 = 9.934 (1±1.2) Maximum compressive stress = 8.039 x 1.3029 = 10.47 t/m2 Minimum compressive stress = 8.039 x 0.697 &&&&&&&&&&&&&& Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI = 5.60 t/m2 THANK YOU Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI