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WELCOME to the GROUP SABARI of AEEs of 2008 BATCH. VIJAYAKUMAR SREEKANTA M.Tech;MHRM; Master Trainer (GoI) FACULTY,WALAMTARI. OFF - TAKE SLUICE - IMPORTANCE - DESIGN PRINCIPLES by. VIJAYAKUMAR SREEKANTA M.Tech;MHRM;

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Welcome to the group sabari of aees of 2008 batch

WELCOMEto the GROUP SABARIofAEEs 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

OFF - TAKE SLUICE - IMPORTANCE - DESIGN PRINCIPLESby

VIJAYAKUMAR SREEKANTA

M.Tech;MHRM;

Master Trainer (GoI)

FACULTY,WALAMTARI

Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI


Off take sluice irrigation system
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


Off take sluice
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
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
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
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
So to ensure delivery of required quantity of water in the irrigation channel ….

we need to

Design an Off take sluice at the headof every distributory / channel

Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI


Off take sluice worked out example
OFF- TAKE SLUICE – irrigation channel ….WORKED OUT EXAMPLE

Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI


HYDRAULIC PARTICULARS irrigation channel ….

Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI


OT DESIGN irrigation channel ….:

Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI



  • A) Vent Way Calculations irrigation channel ….:( 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)

However 0.46m. deep cut off is provided.

Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI


  • b) Scour depth at the end of Downstream wings area of :

  • 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 area of = 0.84 x 0.1096/0.46 = 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 = 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 area of – 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/m2

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 area of

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 area of ‘A’

W1 = 415 kg/m2 live load

Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI


M 0,5453

L.A of the resultant load = --- = ------------- = 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


  • Design of Lintel under upstream head wall: 0,5453

  • (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. 0,5453

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 0,5453

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) 0,5453Check 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. 0,5453Design 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. 0,5453

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

  • 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


  • Check for Bond: 100

  • 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 100Design of side walls for the barrel

Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI


(a) 100Stresses in masonry:

Taking moments about point A.

Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI



2.563 100

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±--------------------) = 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

Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI


(b) Stress on soil: 100

Taking moments about point B.

Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI


L.A of the resultant = 4.881/8.441 = 0.578 m 100

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±--------------------) = 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

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

Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI


Thank you
THANK YOU 100

Presentation of S.VIJAYA KUMAR,Faculty WALAMTARI


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