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WELCOME to the GROUP SABARI of AEEs of 2008 BATCH

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Presentation on theme: "WELCOME to the GROUP SABARI of AEEs of 2008 BATCH"— Presentation transcript:

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 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

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
Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

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: Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

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 Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

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 Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

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. Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

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: Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

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 Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

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 Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

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 Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

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 Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

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. Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

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 Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

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 Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

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 Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

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 Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

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 Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

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

35 Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI
THANK YOU Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI


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