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)

Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI

WELCOME

to the

GROUP SABARIof

AEEs of 2008 BATCH

VIJAYAKUMAR SREEKANTAM.Tech;MHRM;

Master Trainer (GoI)FACULTY,WALAMTARI


OFF - TAKE SLUICE - IMPORTANCE

- DESIGN PRINCIPLES

by

VIJAYAKUMAR SREEKANTAM.Tech;MHRM;

Master Trainer (GoI)FACULTY,WALAMTARI


OFF –TAKE SLUICE- IRRIGATION SYSTEM

OT-L2

OT-L3

OT –R 1 OT-L1

HR

OT Channel

OT C

hannel

Left Main CanalRight Main Canal


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


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


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)


OFF- TAKE SLUICE – DESIGN DATA REQUIRED

-Hydraulic particulars of

Parent canal

Distributory

at the point of proposed OT location


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


OFF- TAKE SLUICE –

WORKED OUT EXAMPLE


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 +48.46 +48.46

HYDRAULIC PARTICULARS


OT DESIGN:


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 0.07m 0.07m The sill of the sluicesShould also be fixedSuch that Lower and lower as the location goes towards the end of the distributaries and minors.

10% to 15% 0.15m 0.07m 0.07m

5% to 10% 0.30m 0.15m 0.07m

2% to 5% 0.46m 0.30m 0.15m

1% to 2% 0.61m 0.46m 030m

0.5% to 1% 0.76m 0.61m 0.46m

Less than 0.5%

0.91m 0.76m 0.61m



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) 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


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


0.64 – 0.19Cot α = ------------------- = 0.99 0.455µ L t--- x ----- x cot α = --- x p1 2 2Where 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 mP = mean permissible stress at the crown of the arch section and is taken equal to

27.34 t/m2

µ 0.91 0.38--- x -------- x 0.99 = -------- x 27.341 2 2

0.38 x 27.34µ = -------------------- = 11.53 m

0.91 x 0.99Hence the floor of the barrel is safe against uplift head of 48.92 – 47.40 = 1.52 mc) Thickness of floor:Percentage of pressure at D/s head wall

(92-8)= 8 + ------- x 2.51 = 8 + 19.3 = 27.3%

10.97Considering buoyant weight of foundation concrete and 75% of theoretical head.


The thickness of floor required27.3 75 1

= 1.37 x ---------x --------- x ------- = 0.22 m00 100 1.25

as against 0.46m thick provided. Hence safeD) DESIGN OF SUB-STRUCTURE:

1. Design of upstream head wall:


S.No

Force

Particulars

Magni-tude

Lever arm (mrs)

Moment in t.m.

1. W1 0.52x415x1/100 = 0.2158 0.26 0.0561

2. W2 0.52x0.93x2243/1000 = 1.0847 0.26 0.2820

3. PV 0.0384[(1.54)2 – 0.61)2] x 2083/1000

=0.1600

------ -----

Total vertical load (V) 1.46054.

PH

0.134[(1.54)2 – (0.61)2] x 2083/1000

= 0.556 0.372 0.2072

Total Moments (M) 0.5453

Taking moments about point ‘A’W1 = 415 kg/m2 live load


M 0,5453L.A of the resultant load = --- = ------------- = 0.37 m V 1.4605

0.52Eccentricity = 0.37 - --------- = 0.11 m

2 1.4605 6x0.11Stresses = ------------ (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.


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

210.2 x 2403

Dead weight of slab = ------------------ 100

= 245 kg/m


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

32750Effective depth = ------------------ --= 6.319 cm

8.203 x 100


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)


(b) Check for Shear: 3410 x 0.98

Maximum shear at the support = ---------------------- = 1551 kg 2

1551Shear 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.

100Perimeters of bars = (----------- +1) π X 1/m width

2 x 15Bond 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


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


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 100B.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


0.79 x 100Spacing of 10 mm dia. Bars = ------------------------ = 19.91 cm

3.968Adopt a spacing of 15cm.

5.266Percentage steel = ---------- = 0.68

7.75(b)Check for shear:

Maximum shear force at support = 4203 x 0.98/2 = 2060 kg.

2060Actual shear stress = (-------------------)

7.75 x 100= 2.66 kg/cm2 < allowable shear stress as per tables = 3.26. Hence safe.


(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

060Bond stress = ------------------------------ = 22.32 kg/cm2

0.875 x 7.75 x 13.61

Provide 30cm of anchorage


4 Design of side walls for the barrel


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 0.497 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

(a) Stresses in masonry: Taking moments about point A.


W7 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)

6.060PH 0.134 (2.842 – 1.902)

2083/1000 1.244 0.372 0.46



2.563L.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 0.12Stresses = -----------(1±--------------------) = 9.934 (1±1.2) 0.61 0.61

Maximum stress (compressiove)= 9.934 x 2.2 = 21.86 t/m2

Minimum stress (tension) = 9.934 x 0.2 = 2.0 t/m2


Force Particulars Magnitude (t) Level arm (m)

Moment( tm)

W1 Same as force = 2.579 0.717 1.927

W2 “ = 0.159 0.717 0.114

W3 “ = 0.424 0.717 0.303

W4 “ = 1.009 0.493 0.497

W5 “ = 0.475 0.493 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


(b) Stress on soil:

Taking moments about point B.


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 0.053Stresses = -----------(1±--------------------) = 9.934 (1±1.2) 1.05 1.05

Maximum compressive stress = 8.039 x 1.3029 = 10.47 t/m2

Minimum compressive stress = 8.039 x 0.697 = 5.60 t/m2

&&&&&&&&&&&&&&


THANK YOU

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)

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