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