Aqueduct Main Project Modify

8/14/2019 Aqueduct Main Project Modify

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Design of Syphon Aqueduct

Sri Sunflower College of Engineering & technology 1

AQUEDUCT

Chapter 1: Introduction 3-10

Chapter 2: Hydraulic Particulars 11

Chapter 3: Selection of type of aqueduct 12

Chapter 4: Design of canal trough 13

Chapter 5: Design of drainage water way 14-15

Chapter 6: Check for loss of head in the canal due to 16-19

Fluming of canal water way through the

Trough

Chapter 7: Fixing of M.F.L of drainage 20-22

Chapter 8: Design of side walls of canal trough 23-26

Chapter 9: Design of bottom slab of canal trough 27-30

Chapter 10: Design of tail channel 31-32

Chapter 11: Design of canal transitions 33-36

Chapter 12: Design of abutments 37-40

Chapter 13: Design of piers 41-44

Chapter 14: Design of wing walls 45-48

Chapter 15: Design of return walls 49-50

Chapter 16: Design of canal aprons 51-52


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Chapter 17: Checking the depth of foundation of 53

Drainage returns by scour depth

Chapter 18: Design of inspection track 54

Chapter 19: Design of pier cap 55-58

Chapter 20: Design of pile foundation 59

Chapter 21: Design of foundation of abutment and piers 60-61

Chapter 22: Design of anchorage arrangements 62

Chapter 23: Drawings 63


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Chapter 1

Introduction

Cross Drainage Works

Definition:

A cross drainage work is a structure carrying the discharge from a natural stream across a canalintercepting the stream.

Canal comes across obstructions like rivers, natural drains and other canals.

The various types of structures that are built to carry the canal water across the above mentioned

obstructions or vice versa are called cross drainage works.

It is generally a very costly item and should be avoided by

Diverting one stream into another.

Changing the alignment of the canal so that it crosses below the junction of two streams.

Types of cross drainage works

Depending upon levels and discharge, it may be of the following types:

(a) Cross drainage works carrying canal across the drainage:

the structures that fall under this type are:

1. An Aqueduct

2. Siphon Aqueduct


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Aqueduct:

When the HFL of the drain is sufficiently below the bottom of the canal such that the drainagewater flows freely under gravity, the structures known as Aqueduct. An aqueduct is a water

supply or navigable channel constructed to convey water. In modern engineering, the term is

used for any system of pipes, ditches, canals, tunnels, and other structures used for this purpose.In a more restricted use, aqueduct (occasionally water bridge) applies to any bridge or viaduct

that transports water instead of a path, road or railway across a gap. Large navigable aqueducts

are used as transport links for boats or ships. Aqueducts must span a crossing at the same level asthe watercourses on each end. The word is derived from the Latinaqua ("water") and ducere ("to

lead").

In this, canal water is carried across the drainage in a trough supported on piers.

Bridge carrying water

Provided when sufficient level difference is available between the canal and natural andcanal bed is sufficiently higher than HFL.


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Classification of aqueduct and siphon aqueduct:

Depending upon the nature of the sides of the aqueduct or siphon aqueduct it

may be classified under three headings:

Type I:

Sides of the aqueduct in earthen banks with complete earthen slopes. The

length of culvert should be sufficient to accommodate both, water section of

canal, as well as earthen banks of canal with aqueduct slope.

Sides of the aqueduct in earthen banks, with other slopes supported by

masonry wall. In this case, canal continues in its earthen section over the

drainage but the outer slopes of the canal banks are replaced by retaining wall,

reducing the length of drainage culvert.

Type II:

Sides of the aqueduct made of concrete or masonry. Its earthen section of the

canal is discontinued and canal water is carried in masonry or concrete trough,

canal is generally flumed in this section.

Siphon Aqueduct:

In case of the siphon Aqueduct, the HFL of the drain is much higher above the canal bed, andwater runs under siphonic action through the Aqueduct barrels.

The drain bed is generally depressed and provided with pucci floors, on the upstream side, the

drainage bed may be joined to the pucca floor either by a vertical drop or by glacis of 3:1. The

downstrean rising slope should not be steeper than 5:1. When the canal is passed over thedrain, the canal remains open for inspection throughout and the damage caused by flood is rare.

However during heavy floods, the foundations are succeptible to scour or the waterway of drain

may get choked due to debris, tress etc.


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The structures that fall under this type are:

Super passage

Canal siphon or called syphon only.

Super passage:

The hydraulic structure in which the drainage is passing over the irrigation canal is

known as super passage. This structure is suitable when the bed level of drainage is

above the flood surface level of the canal. The water of the canal passes clearly below thedrainage

A super passage is similar to an aqueduct, except in this case the drain is over the canal.

The FSL of the canal is lower than the underside of the trough carrying drainage water.Thus, the canal water runs under the gravity.

Reverse of an aqueduct


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Canal Syphon:

If two canals cross each other and one of the canals is siphoned under the other, then the

hydraulic structure at crossing is called canal siphon. For example, lower Jhelum canalis siphoned under the Rasul-Qadirabad (Punjab, Pakistan) link canal and the

crossing structure is called L.J.C siphon In case of siphon the FSL of the canal is much above the bed level of the drainage trough,

so that the canal runs under the siphonic action.

The canal bed is lowered and a ramp is provided at the exit so that the trouble of silting is

minimized.

Reverse of an aqueduct siphon

In the above two types, the inspection road cannot be provided along the canal and a

separate bridge is required for roadway. For economy, the canal may be flumed but the

drainage trough is never flumed.

Selection of suitable site for cross drainage works:

The factors which affect the selection of suitable type of cross drainage works are:

Relative bed levels and water levels of canal and drainage

Size of the canal and drainage.

The following considerations are important


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When the bed level of the canal is much above the HFL of the drainage, an

aqueduct is the obvious choice.

When the bed level of the drain is well above FSL of canal, super passage is

provided.

The necessary headway between the canal bed level and the drainage HFL can

be increased by shifting the crossing to the downstream of drainage. If,

however, it is not possible to change the canal alignment, a siphon aqueduct

may be provided.

When canal bed level is much lower, but the FSL of canal is higher than the

bed level of drainage, a canal siphon is preferred.

When the drainage and canal cross each other practically at same level, a level

crossing may be preferred. This type of work is avoided as far as possible.

Factors which influence the choice / Selection of Cross Drainage Works

1. The considerations which govern the choice between aqueduct and siphon aqueduct are:

2. Suitable canal alignment

3. Suitable soil available for bank connections4. Nature of available foundations

5. Permissible head loss in canal

6. Availibility of funds

Compared to an aqueduct a super passage is inferior and should be avoided whenever possible.

Siphon aqueduct is preferred over siphon unless large drop in drainage bed is required.

Uses:

Historically, agricultural societies have constructed aqueducts to irrigate crops. Archimedes

invented the water screw to raise water for use in irrigation of croplands.

Another use for aqueducts is to supply large cities with drinking water. Some of the Roman

aqueducts still supply water to Rome today. In California, United States, three large aqueducts


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supply water over hundreds of miles to the Los Angeles area. Two are from the Owens River

area and a third is from the Colorado River.

In more recent times, aqueducts were used for transportation purposes to allow canalbarges to

cross ravines or valleys. During the Industrial Revolution of the 18th century, aqueducts were

constructed as part of the boom in canal-building.

In modern civil engineering projects, detailed study and analysis of open channel flow is

commonly required to support flood control, irrigation systems, and large water supply systems

when an aqueduct rather than a pipeline is the preferred solution.

In the past, aqueducts often had channels made of earth or other porous materials but significant

amounts of water are lost through such unlined aqueducts. As water gets increasingly scarce,

these canals are being lined with concrete, polymers or impermeable soil. In some cases, a new

aqueduct is built alongside the old one because it cannot be shut down during construction.


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DESIGN STEPS:

1) Hydraulic particulars of canal & Drainage

2) Selection of type of aqueduct

3) Design of canal trough

4) Design of drainage water way

5) Check for loss of head in the canal due to fluming of canal water way

Through the trough

6) Fixing the M.F.L of the drainage

7) Design of side walls of canal trough

8) Design of bottom slab of canal trough

9) Design of tail channel

10) Design of canal transitions

11) Design of abutments

12) Design of piers

13) Design of wing walls

14) Design of return walls

15) Design of aprons

16) Checking the depth of foundations of drainage returns by scour depth

17) Design of inspection track

18) Design of pier cap

19) Reinforcement in pier

20) Pier foundation

21)Design of pier foundation

22)Design of foundations of abutments and piers


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Chapter -2

Hydraulic particulars

Canal:

Discharge = 35 m3/s

Bed width = 20 m

Bed level = +40.00

F.S.L = 42.00

Ultimate bed level = 39.75

Ultimate full supply level = 42.50

Top width of left bank = 5m

Top width of right bank = 2m

T.B.L = +43.50

Slope of canal banks = 2:1

Drain:

Catchment area = 8 km2

Maximum discharge = 60 m3/s

Bed level = 38.00

Average ground level = +38.00

Maximum flood level of drain at the site of crossing = +39.75

Hard soil for foundation is available at = +37.00


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Chapter-3

Selection of type of aqueduct

The above data of hydraulic particulars, a type-3 aqueduct is designed. It is only purely from an

economical aspect that we go in for the type-3 aqueduct. For major drains it will beuneconomical to go in for a type-3 aqueduct as the extra cost of barrel with a large number of

vents will be more than the cost of other works necessary for a type-3 aqueduct.

So, whenever an aqueduct are to be actually constructed, comparative costs are to be

worked out for a type-2 and a type-3, and whichever is economical is to be chosen and adopted.

In the case of a type-3 aqueduct, the canal will be flumed and taken through a masonry or

reinforced concrete trough supported on piers and abutments. The maximum velocity through the

trough is generally taken as twice the normal velocity or 1.5m/s whichever is less.


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Chapter - 4

Design of canal trough

Discharge, Q = 35 m3/s

Average velocity = 0.83 m3/s

Design velocity = 2% average velocity = 2*0.83=1.66 m/s

But the maximum design velocity = 1.5 m/s

Adopt design velocity = 1.5 m/s

Q= A x v

35= Ax1.5

A=23.3 m2

Depth of flow, y= F.S.D= F.S.L B.L= 42.00-40.00=2m

Bottom level = Ultimate bed level of canal = 39.75

Top level = Ultimate F.S.L + 0.5 = 42.50 + 0.50 = 43.00


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Chapter - 5

Design of drainage water way

Let us provide 3 vents of 2.5 m wide length of water way, l=3X2.50=7.5 m

Sill level of canal trough = 39.75 m (given)

Thickness of bottom slab = 250 mm (assume)

Thickness of wearing coat = 80 mm (assume)

Bottom level of the canal trough = sill level- thickness of wearing coat-

Thickness of bottom slab

= 39.75-0.08-0.025=39.64 m

Bottom level of slab = 39.75-0.25 = 39.50

Average bed level of the drain = 38.00

Since the M.F.L of the drain = ultimate bed level of the canal

Let us adopt depressed bed level of the drain = 37.00

Depth of water = y1 = Bottom level of the canal trough slab- depressed B.L

Of drain=39.45-37=2.45 m

Let the design velocity in the drain, V1= 3.25 m/s (assume)

Discharge in the drain, Q1 = 60 m3/s

Q1=A1 X V1,

60= A1 X 3.25

A1=18.46 m2

A1=L X y

1

18.46 = L X 2.45

L = 7.54 m

Length of barrel = B+ 2X (thickness of side wall)

= 12 + 2X (0.3) =12.60 m


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Sri Sunflower College of Engineeri

Check for loss of head

In a type-1 or type-2 aqueduct,

it is. Hence, there is no loss of h

However, in case of a

resulting in an increase in veloci

before entry and after exit, attain

In aqueduct of short leng

the loss of head may be very s

assumption is that the upstream

extent required to drive the flow

However, in large and longer a

trough, we may be forced to inc

the trough is an additional fact

head, which will have to be p

Structures constructed ignoring t

In the present case, to illustrate t

Consider section A-A

Canal bed level =

Full supply level =

Average velocity, V1 =

Velocity head =

Design of

g & technology

Chapter - 6

in the canal due to fluming of canal

through the trough

the canal water way is not reduced and is taken

ad in the canal.

type-3 aqueduct, the canal water-way is flu

y through the trough. Unless there is a differen

ment of increased velocity in the trough is not p

hs, by limiting the velocity to twice the norm

all or almost negligible and hence it is gener

ater surface will in course of time assume a fl

through the trough with that bit of extra velocit

ueducts it is not so. In order to economies in

rease the velocity through the trough. In additi

r. These two factors combine to indicate a si

ovided for, while formulating the canal hydr

his aspect will not function properly.

is aspect, the loss of head in the canal is comp

40.00

42.00

.83 m/s

12/2g = 0.83

2/2x9.81 = 0.037 m

Syphon Aqueduct

16

ater way

over the drain as

ed or reduced,

e in water levels

ossible.

l canal velocity,

lly ignored. The

atter slope to the

.

cost of the canal

on, the length of

gnificant loss of

aulic particulars.

ted.


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Total energy line at A-A = F.S.L + (V12/2g) = 42.00+0.037 = 42.037

Consider section B-B

Canal width = 20 m

Depth of water, y =F.S.L- Bed level = 42.00 40.00 = 2.00

At B-B the canal is rectangle in cross section, area of flow a= 20x2 = 40 m2

Discharge, Q = 35 m3/s

Velocity of flow, V2 = Q/a = 35/40 = 0.87 m/s

Head loss from A-A to B-B due to change in the velocity V1 to V2

Head loss = (V22-V1

2)/2g

= (0.8752-0.832)/2x9.81

= 0.004

On the U/S end the transition is abrupt and not smooth. So the entire eddy

Loss is taken in to consideration.

The T.E.L at B-B = T.E.L at A-A eddy loss = 42.035 0.004 = 42.031

Section at the entrance of trough C-C

Width of the canal = 12 m (assume)

Depth of canal, y = 2 m

A3 = 12x2 = 24 m2

Velocity, V3 =

3 =3524 = 1.46 m/s

There is a gradual change in c/s from B-B to C-C. There is a loss of head from

B-B to C-C due to change in velocity.

Head loss = 0.25 x ( V32- V2

2)/2g = 0.25 x (1.46

2-0.875

2)/2 x 9.81

= 0.018

T.E.L at C-C with reference to section B-B = 42.035 0.018 = 42.017


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Consider section D-D

From C-C to D-D, there is a uniform velocity. The loss of head in the trough is only

friction loss which manifests itself as surface fall to sustain the velocity.

This loss of head is calculated using mannings formula, using the value of 0.014 for n

Length of R.C trough = (3x 2.5) + (2x1) + (2x0.5) = 10.5 m

Sectional area in the trough A4 = 12x2 = 24 m2

Velocity developing =3524 = 1.46 m/s

Wetted perimeter = p = (2x2) + 12 = 16 m

Hydraulic mean depth = R =4 =

2416 = 1.5 m

Mannings formula, V =1 (R)

2/3 (S)

1/2

1.46 =1

0.014 (1.5)2/3

(S)1/2

S =1

4110

Head loss = S x length =1

4110 x 10.5 = 0.003 m

T.E.L at D-D = T.E.L at C-C Head loss = 42.017 0.003 = 42.014 m

Consider section E-E

Neglecting the frictional loss in the exit transition the eddy loss in the transition is

calculated as follows

Q = 35 m3/s

A5 = (B +n y) y

Here B = 20 m, y = 2 m, n =12

A5 = (20 +12 2) 2 = 42 m

2

V5 =

5 =3542 = 0.83 m/s

There is a gradual change in the section from D-D to E-E. There is a change in the

velocity.


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Head loss = 0.25 x (V42- V5

2)/2g = 0.25 x (1.46

2- 0.83

2)/2 x 9.81= 0.018 m

T.E.L at E-E = T.E.L at D-D Head loss = 42.014 0.018 = 41.99 m

Velocity head = V52/2g = 0.83

2/ 2x 9.81 = 0.0351 m/s

T.E.L at E-E = H.F.L at TEE + Velocity head

41.996 = HFL at TEE + 0.0351

HFL at TEE = 41.961 m

F.S.L at A-A = 42.00 m

Total loss of head = loss of head from A-A to E-E

= TEL at A-A TEL at E-E

= 42.037 41.961 = 0.039 m

Total head loss from A to E = 0.004 + 0.0175 + 0.003 + 0.018 = 0.0425

The total head loss is very small. Hence it can be neglected.


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Chapter - 7

Fixing the M.F.L of the drainage

The MFL of the drain in rear of the siphon barrel is 39.75. So, the barrel flows full under

maximum flow conditions. The necessary afflux required to push through 60 m3/s with a velocityof 3.25 m/s is calculated by unwins inverted syphon formula,

Afflux, d = (1+f1 +f2 )

22

Here v= velocity in the drain = 3.25 m/s

g= 9.81 m/s2

L = length of drain = 12.60 m

C/S area = 3 x 2.5 x 2.45 = 18.375 m2

Wetted perimeter = p = 3 x (2 x (2.5+2.45)) = 29.7 m

R= hydraulic radius = =

18.37529.7 = 0.62 m

f1= 0.0505

f2 = a (1+ 0.3) = 0.003 (1+ 0.3

0.10.62 ) = 0.003145

d = (1+0.505+ 0.00315

12.600.62 ) (

3.25 3.252 9.81 ) = 0.85 m

M.F.L on U/S = D/S M.F.L + afflux = 39.75 + 0.85 = 40.60 m


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Afflux at the drop of bed

Bed level of drainage = 38.00

At crossing with canal, bed level of drainage = 37.00

Drop in the bed level = 38 37 = 1

M.F.L of the drain = 39.75 m

Bottom level of the canal trough = 39.45 m

Afflux required = 39.75 39.45 = 0.3 m

Here drop is 1 m then it is treating it as a drowned weir (submerged weir)

For submerged weir, q = 3.54 y2 d10.5

+ 1.77 d13/2

y2 = depth of water = 2.60 m

q =

2 =6011 = 5.45 m

2/s

B2 = width of drainage b/w wing walls

5.45 = 3.45 (2.60) d10.5

+ 1.77 d13/2

5.45 = 8.97 d10.5

+ 1.77 d1 d10.5

5.45 = d10.5

(8.97+1.77 d1)

5.452

= d1 (8.97+1.77 d1)2

29.70 = d1 (80.46+3.14 d12+31.75 d1)

3.14 d13

+ 31.75 d12

+ 80.46 d1 29.70 = 0

d1= 0.326

M.F.L over the drop = U/S M.F.L + afflux = 40.60 + 0.32 = 40.92 m


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Chapter - 8

Design of side walls of canal trough

Bottom level of the side wall = bottom level of the canal trough

Top level of the side wall = U/S F.S.L + 0.5 = 42.50 + 0.5 = 43.00

Depth of water h = U/S F.S.L Sill level = 42.50 39.75 = 2.75 m

Let top thickness = 2 m

It is designed as a cantilever wall

of water = 10 KN/m3

Water pressure = p = x h = 10 x 2.75 = 27.5 KN/m2

Let us consider 1 m length of side wall

Total pressure, p= Area of pressure diagram x length of wall

=12 x 27.5 x 2.75 x 1 = 37.81 KN

Centre of pressure,y = 3 = 2.753 = 0.917 m

Bending moment, m = p x y = 37.81 x 0.917 = 35KN-m

Mu = 1.5 x 35 = 52.17 KN-m

Adopting M-20 grade concrete & Fe-415 grade steel


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= 0.48

D = 300 mm, b = 1000 mm, d1

= 40 mm, d = D - d1

d = 300 40 = 260 mm

Mu, limit = 0.36

bd2

fck(1 0.42

)

= 0.36 x 0.46 x 1000 x 2602

(1 0.42 x 0.48)x20

= 186.6 KN-m

Mu< Mu, limit (O.K)

Design of steel reinforcement

(a) Main steel :

1) Minimum area of steel,

Ast, min = 0.12% of gross area

=0.12100 x 1000 x 300

= 360 mm2

2) Maximum area of steel ,

Ast, max = 4% of gross area

=4

100 x 1000 x 300

= 12,000 mm2

3) Mu1 = 0.87 fyAst d (1

)

52.17 x 106

= 0.87x415x260x Ast1x

(1-- 415

1000 260 20)

555.75 = Ast1- Ast12

(7.98 x 10-6

)

Ast1 = 582.86 mm2

, 11947.25 mm2

Ast1< Ast1, max (O.K)

Ast1 > Ast1, min (O.K)

For Ast1 = 11947.25 mm2

=

0.87 0.36 =

0.87 415 11947.250.36 20 1000 260 = 2.304

= 0.48

>

(Not O.K)


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For Ast1 = 583 mm2

=

0.87 415 5830.36 20 1000 260 = 0.112

<

(O.K)

Ast1 = 583 mm2

Let us adopt 12 mm dia bars, 1 = 12 mm

Area of one bar, A =3.14

4 x 122

= 113 mm2

Number of bars, n1 =Ast1A =

583113 = 5.16

Spacing, S1 =1000

n1 = 193.8 mm

Adopt 12 mm @ 180 mm C/C

Check for shear design

Nominal shear force, v =

=37.81 1031000 260 = 0.145 N/mm

2

Percentage of steel provided =100

=100 583

1000 260 = 0.224%

As per IS 456 2000

For, M-20, 100 = 0.224% then c = 0.81 N/mm2

v


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2001000 =

0.40.87 415

Sv = 180 mm

2) 0.75 X d = 0.75 x 260 = 195 mm

3) 300 mm

Sv = 180 mmAdopt 4 legd vertical stirrups @ 8 mm dia @ 150 mm C/C

Distribution steel

Providing steel on both faces

1) Area of steel , Ast2 =

2 =360

2 = 180 mm2

2) Let us adopt dia of bar, 2 = 8 mm

Area of one bar, A 2 =3.14

4 x 82

= 50 mm2

Number of bars, n2 =Ast2A =18050 = 3.6

Spacing, s2 =1000

n2 =1000

3.6 = 277.7 mm

Adopt 8mm dia @ 250 mm C/C


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Chapter - 9

Design of bottom slab of canal trough

It is designed as a continuous slab.

Let us consider 1m wide slab

Thickness of wearing coat = 8 cm

Weight of wearing coat =8

100 x 24 = 1.92 KN/m2

Thickness of slab = 25 cm

Weight of slab =25

100 x 25 = 6.25 KN/m2

Depth of water = 2.75 m

Weight of water = 2.75 x 10 = 27.5 KN/m2

Total load on the slab = 1.92 + 6.25 + 27.5

= 35 KN/m2

Effective span (l) = clear water way + effective thickness of slab

= 2.5 + 0.26 = 2.76 m

Maximum B.M = M3 = 210 =35 2.76 2.76

10 = 26.67 KN-m

Mu3 = 1.5 x M1 = 40 KN-m

Mu, limit = 0.36

bd2

fck(1 0.42

)

= 0.36x0.48x (1 0.42 x 0.48) x1000x260 x 20

= 186.6 KN-m

Mu3


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Design reinforcement

a) Main steel

1) Minimum area of steel , Ast, min = 0.12% of gross area

=0.12

100x 1000 x 300

= 360 mm2

2) Maximum area of steel ,Ast, max = 4% of gross area

=4

100 x 1000 x 300

= 12,000 mm2

3) Mu3 = 0.87 fyAst d (1

)

40 x 106

= 0.87 x 415 x Ast3 x 260 x (13 415

1000 260 20)

426.10 = Ast3- Ast32

(7.98 x 10-5

)

Ast3 = 441.66 mm2

= 12089 mm2

Ast3 = 442 mm2

Ast3Ast,min (O.K)

Let us adopt 12mm dia bars, 3 = 12 mm

Area of one bar, A 3 =3.14

4 x 122

= 113 mm2

Spacing, s3 =1000

n3 =10003.91 = 255.6 mm

Adopt 12 mm dia bars @ 250 mm c/c

b) Distribution steelsProviding steel on both faces

Area of steel, Ast4 =

2 =360

2 = 180 mm2

Adopt 8mm dia @ 250 mm c/c.

Check for shear design

Shear force, V =2 =

35 2.762 = 48.3 KN

Nominal shear force, v =

=48.3 10 10 10

10 1010260

= 0.185 N/mm2

As per IS-456-2000, for M-20 grade , c, max =2.8 N/mm2

v


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As per IS-456-2000, for M-20 grade,100

= 0.17% then

c = 0.3 N/mm2

v


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Chapter -10

Design of tail channel

Tail channel will be always straight, its length will be 50 to 60m on either side of roads

Top level = M.F.L of drain

= 39.75m

Bottom level = bed level of the drain at crossing = 37.00m

Depth of flow, Y3 = 39.75-37.00 = 2.75m

Let us assume velocity of flow, V = 1.5m/sec

Q = AV

60 =1.5A

A = 40m2

Slope =12H: 1V (assume)

A = (B+ny) y

40 = (B+0.52.75)2.75

B = 13.17m

Adopt B = 14m

Slope of channel,

Mannings formula, V =1 (R)

2/3(S)

1/2

Here, V = 1.5m/sec, N = 0.015 (assume)

A = (B+ny)y

= (14+0.52.75)

A = 42.3m2

Wetted perimeter, P = B+2(2 + ) .y

= 14+2((.)2 + )2.75


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= 20.15m

R = = = 2.1m

V = (R)2/3

(S)1/2

1.5 = (2.1)2/3

(S)1/2

S =

Design of

g & technology

Syphon Aqueduct

32


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Chapter - 11

Design of canal transitions

Transitions are wing walls

Upstream side transitions for canal:

The bottom of foundation is same as that of the abutments. Assuming 60cm depth of concrete

foundation. The top of foundation is kept at 37.00m

Top level = ultimate full supply level + 0.5m

= 42.50 + 0.50 = 43.00m

Bottom level = hard soil level = 37.00m

Height of the wall = 43.00 37.00

= 6.00m

Thickness of foundation = 0.6m

Top thickness = 0.5m

From top level to the ultimate bed level of the canal the front face is vertical

Height of wing wall from ultimate bed level=43.00-39.75=3.25m.

Width of wing wall @ ultimate bed level=0.4Xheight

=0.4X3.25

=1.3m

Let us provide width ofwing wall @ ultimate bed level=1.75m


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From ultimate bed level to t

Height=top level-bottom lev

=43.00-37.00

=6.00m

Bottom width=0.4Xheight

=0.4X6

=2.4m

Let us provide widt

Design of

g & technology

p of foundation is vertical on earth phase

el

of wing wall=2.5m

Syphon Aqueduct

34


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Downstream side transition for canal:

In order to minimize the eddy losses, a suitable exit transition is necessary after the R.C trough

Top level = ultimate F.S.L + 0.5m

= 42.50m + 0.5m

= 43.00m

Bottom level = 37.00m


Bottom level of foundation = 37.00 0.6

= 36.40m

Height = top level bottom level

= 43 37

= 6m

Top width = 0.50m

The downstream side wing wall is sloping at12 : 1 up to the bed level, then it is vertical

Height of wing wall from ultimate bed level=43.00-39.75=3.25m.

Width of wing wall @ ultimate bed level=0.4Xheight

=0.4X3.25

=1.3m

Let us provide width ofwing wall @ ultimate bed level=2.125m

From ultimate bed level to top of foundation is vertical on earth phase

Height=top level-bottom level

=43.00-37.00

=6.00m

Bottom width=0.4Xheight

=0.4X6


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=2.4m

Let us provide width ofwing wal

Design of

g & technology

l=2.125m

Syphon Aqueduct

36


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Top level = bottom level of cent

Bottom level = hard soil level =

Bearing of canal tough = 0.5m

Top width = 2 length of bearin

= 2 0.5 = 1m

Height of wall = top level bott

= 39.42 37.00

= 2.42m

Bottom width = 0.4 height

= 0.4 2.42

= 0.968m

Adopt bottom width = 2m

Front batter = 1 in 8

Design of

g & technology

Chapter -12

Design of abutments

al tough = 39.42m

7.00m

m level

Syphon Aqueduct

37


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Check for stability of abut

Let us consider 1m length of wal

Unit weight of RCC = 24 KN/m

Weigh of triangular portion 1, V

X1 = ( + 0.5) = 0.2m from point

M1 = V1X1 = 8.71 0.2 = 1.74

Weigh of triangular portion 2, V

Design of

g & technology

ment:

l

1 = unit weight volume

= 24 (0.5bhl)

= 24 (0.50.32.421)

= 8.71KN

B

KN-m

= unit weight volume

= 24 (bhl)

= 24 (12.421)

= 58.08 KN

Syphon Aqueduct

38


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X2 = (0.3 + 0.5) = 0.8m from point B

M2 = V2X2 = 58.08 0.8 = 46.46 KN-m

Weigh of triangular portion 3, V3 = unit weight volume

= 24 (0.5bhl)

= 24 (0.50.72.421)

= 20.32 KN

X3 = (0.3+1+13 .) = 1.53m from point B

M3 = V3X3 = 20.32 1.53 = 31.15 KN-m

Reaction from canal tough, V4 =

2=

35.62.5

2 = 44.58 KN

X4 = (0.3+12 .) = 0.55m from point B

M4 = V4X4 = 44.58 0.55 = 24.51 KN-m

Total vertical force, V = V1+V2+V3+V4

= 8.71+58.08+20.32+44.58

= 131.69KN

Earth pressure, p = KaH

Ka =13, of earth = 20 KN/m

3

p =13 202.42 = 16.13 KN/m

2

Total pressure P = area of pressure diagram length

= (12 16.13 2.42 1)

= 19.51KN

Y =13 2.42 = 0.81m

M5 = P Y


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= 19.51 0.81

= 15.8 KN-m

Overturning moment, Mo = 15.8 KN-m

Net moment, M = MR MO = 103.86 15.8 = 88.06 KN-m

Lever arm, a =MV =

88.06131.69 = 0.66m

Eccentricity, e = (b2 - a ) = (

22 0.66 ) = 0.34m

Max stress, max =VB ( 1 +

6eB )

=131.69

2 ( 1 +60.34

2 )

= 133 KN/m2

Min stress, min =VB ( 1 +

6eB )

=131.69

2 ( 1 -60.34

2 )

= -1.31N/mm2

(tension occurs)

Factor of safety against overturning =MrMo =

103.8615.8

= 6.57 > 1.5 ( ok )

Factor of safety against sliding =VP =

0.6131.6919.51 = 4.04 > 1 ( ok )

Allowable compressive stress in foundation concrete = 4 N/mm2

(for M-15 grade)

= 4000 KN/m2

maxis less than 4000 KN/m2

( ok )


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Top level = bottom level of cent

Bottom level = hard soil level =

Bearing of canal tough = 0.5m

Top width = 2 length of bearin

= 2 0.5 = 1m


Height of wall = top level bott

= 39.42 37.00

= 2.42m

Bottom width = 0.4 height

= 0.4 2.42

= 0.968m

Adopt bottom width = 2m

Front batter = 1 in 8

Design of

g & technology

Chapter -13

Design of piers

al tough = 39.42m

7.00m

m level

Syphon Aqueduct

41


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Check for stability of piers:

Width of pier = 1m

Length of pier = width of canal tough = 12m

Clear spacing between piers = 2.5m

Length of canal tough on one pier = 2.5 + 1 = 3.5m

On pier there is water pressure from 3.5m wide.

Weight of pier, V1 = unit weight volume

= (24122.421)

= 696.96 KN

It is acting at a distance, X1 = 0.5 12 = 6m from D

M1=V1X1 = 696.96 6 = 4181.76 KN-m

Weight of canal tough, V2 = Wwidth of canal toughlength of canal trough in consideration

= 35.67123.5

= 1498.14 KN

It is acting at a distance, X2 = 0.5 12 = 6m from D


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M1=V2X2 = 1498.14 6 = 8988.84 KN-m

Total vertical force, V = V1 + V2

= 696.96 + 1498.14

= 2195.11 KN

Resisting moment, MR= M1 + M2 = 4181.76 + 8988.84

= 13170.6 KN-m

Weight of water =w = 10KN/m3

Water pressure, p =H =102.42 = 24.2 KN/m2

Total pressure, P = area of pressure diagram width of water way

= (12 24.22.42) (3.5)

= 102.5 KN

It is acting at a distance = 13 .

= .

M3 = = . . =

Overturning moment, mo= m3= Net moment , m = ( mr mo) = 13170.6 83

= 13087.6 KN-M

Lever arm, a = =

13087.62195.1 = 5.9 m

Eccentricity, e = ( 2 ) = 122 . = .

max = +

6 =

2195.1121 +

60.112

= . / 2

min = 6 =

2195.112

60.112 = . /

2

Factor of safety against over turning = =13170.6

83


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= . ()

Factor of safety against sliding = =0.62195.1

102.5

=.()

Allowable compressive stress in foundation concrete = / 2

= / 2

max < 4000 KN/m2,

min >0 (ok)


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DESIG

U/S Wing wall :

The wing wall as two portions

1) Slopping wing wall

2) Level wing wall

The slopping wing wall is sloppi

Let us assume M.F.L of drain =

1) Slopping wing wall :

a) Wing wall at junctioTop of wing wall (or)

Top of foundation (o

Height = top level

= 43 - 37 = 6

Thickness of foundati

Bottom level of foun

Top width = 0.5 m

Bottom thickness

Front batten

Design of

g & technology

Chapter 14

OF DRAINAGE WING WALL

ng from top of wing wall for canal to the level

1.50 m

of trough : top level = 43.00

) bottom level = 37.00

ottom level

on = 0.6 m

ation = 37.00 - 0.6

= 36.40 m

Syphon Aqueduct

45

f M.F.L.


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b) Wing wall at + 41.50

Top level = 41.50

Bottom level = 37.00

Thickness of foundatiBottom level of foun

Height = top level

= 41.50 37.

= 4.50 m

Top width = 0.5 m

Bottom width

Front batten

2) Level wing wall :

Top level =


Thickness of foundatiBottom level of foun

Height = top level

= 41.50 37.

= 4.50 m

Top width = 0.5 m

Design of

g & technology

level :

on = 0.6 ation = 37 0.6

= 36.40 m

ottom level

00

41.50

on = 0.6 ation = 37 0.6

= 36.40 m

ottom level

00

Syphon Aqueduct

46


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Bottom width

Front batten

D/S Wing wall :

The sloping wing wall is slop

1) Slopping wing wall :

a) Wing wall at junction

Top of wing wall

Top of foundation (o

Height = top level

= 43 - 37 = 6



Top width = 0.5 m

Bottom thickness

Front batten

Design of

g & technology

ping from top of wing wall for canal to level of

f trough

(or) top level = 43.00

) bottom level = 37.00

ottom level

on = 0.6 m

ation = 37.00 - 0.6

= 36.40 m

Syphon Aqueduct

47

40.50 m


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2) Level wing wall :

Top wing wall = 40.50 m

Bottom level = 37.00 m

Height of wall = 3.50 m

Thickness of foundation =

Top width = 0.5 m

Bottom width

Front batten = 1 in 8

Design of

g & technology

0.60 m

Syphon Aqueduct

48


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U/S Return wall :

Top level = 41.50




Height = top level

= 41.50 37.

= 4.50 m

Top width = 0.5 m

Bottom width

Front batten

Design of

g & technology

Chapter -15

Design of return wall

on = 0.6

ation = 37 0.6

= 36.40 m

ottom level

00

Syphon Aqueduct

49


50/64


D/S return wall :

Top wing wall =

Bottom level =

Height of wall =

Thickness of foTop width = 0.5

Bottom width

Front batten = 1

Design of

g & technology

40.50 m

7.00 m

3.50 m

ndation = 0.60 m m

in 8

Syphon Aqueduct

50


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Chapter -16

Design of canal apron

In between the canal wings, concrete is laid as a apron to prevent water percolating by the side

of the abutment exert pressure under the drainage apron. The uplift is maximum when the canalis full and drainage is empty.

In this case, the gross head causing uplift at F.S.L

Before the water creeps to the bottom of the drainage aprons, some head is lost .assuming that

the canal apron is laid sloping from the top of trough slab +39.75 to a level +38.00 (ground level

) for a length of L meters, as shown in fig the vertical creep is neglected.

The canal wing upstream and downstream will be splayed such that by the time the end of apron

is reached, the distance between their faces is equal to the theoretical bed width of

canal,i.e.,20.00 meters. This decides the splay of canal wings.

The apron is to be laid from canal sill level to natural ground level

Gross uplift head = canal F.S.L-foundation level

= 42.00-37.00

= 5m

Let us provide the canal apron sloping from top of the trough slab +39.75 to a level (ground

level) = 38.00


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Let us provide 60cm thick mass concrete for drainage floor

Thickness = residual head/(P-1)

0.6 = residual head/(2.40-1)

Residual head =0.84m

Net uplift head = gross uplift head -residual head

= 5.0-0.84

= 4.16

Let us assume exit gradient

GE= 1 in 4 = =

14

4.16 =

14

Therefore L = 17m

Length of apron(L)=17m.


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Chapter -17

Checking the depth of foundation of drainage returns by scour depth

Upstream side

Maximum flood discharge = 60 m3/sec

Maximum flood level = 40.94m

Length of apron retaining wall = 14m (assume)

Discharge per meter length of apron retaining wall

q = 60/14 = 4.3m2/sec

Depth of scour = 1.374 (q2/f)1/3

= 1.374 ((4.3)2/1)

1/3

= 3.562m

The foundations of returns and apron retaining wall are to be taken down to 40.36-3.56

= 37.38

However the foundations have been taken down to +36.40 and are quite safe.

On D/s side also, the foundations are taken down to +36.40 and as the distance between the

returns is also more or less same . The foundations adopted are safe.


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Chapter -18

INSPECTION TRACK

The inspection track on the left embankment of canal also has to be taken across the drain by

means of a bridge just by the side of the canal trough.

The width of roadway between kerbs may be kept large enough to meet the demands of traffic

proposed on the canal banks. In this case, it is kept as 3.65 meters (12 feet wide). Depending on

the traffic, the road bridge may be designed. In the present case, the roadway is carried over plain

concrete arches since there is enough headroom above water level. The springings of arches are

kept above water level in the drain.

Semi-circular arches or reinforced concrete deck slabs can be adopted.

In this case the thickness of 50 cms adopted for the arches is quite enough and the detailed

design of arch is not attempted.

Keep the springing level of arches a little above the rear M.F.L., i.e., +40.00.

Bottom level of arch (intrados) is 40.00 +.

=..

Thickness of arch = 0.50 m.

Therefore, top of arch (extrados) = 41.75.

This top level of road surface may be kept at +43.00. The space in between the road level and

top of arch is covered with earth to act as cushion over the arches.

The road width between kerbs is kept at 3.65 meters with suitable parapets. The inspection

track over the arches is suitably connected to the canal banks by canal transitions.


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Chapter -19

Design of pier-cap

i. Thickness of cap = 600mm (assume)

ii. Projection on each side = 600mm (assume)

Reinforcement detailing

1. Minimumarea of steel, ast1min =0.12% gross area

= (0.12/100)*600*1000

= 720mm2

2. Maximum area of steel, ast1max = 4% of gross area

= (4/100)*600*1000

= 24,000mm2

Load on the pier V =2195.11 KN

Number of pier = 2

Load on each pile = 2195/2

V= 1097.5 KN


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Shear force v = 2 1097.5 =

1.82

W = 1220 KN/m

Here the beam is considerd as a continuous beam

Max bending moment =212

=1220 1.8 1.8

12

= 329.25 KN-m

Factored moment = Mu = 1.5 X M = 1.5 X 329.25 = 493.875 KN-m

Check for thickness

Mu, limit = 0.36

bd2

fck(1 0.42

)

for Fe-415 grade steel

= 0.48

Here b= 1000 mm

D= 600 mm

d1= 50 mm

d = D - d1

= 600 50 = 550 mm

fck= 20 N/mm2

Mu, limit = 0.36 X 0.48 X 1000 X 5502 X 20 (1 0.42 X 0.48)


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= 834 KN-m

Mu < Mu, limit (O.K)

Thickness is satisfied.

(a)Main steel

Mu= 438 KNm

Mu=0.87fyastd (1- (Ast fy/ b d f c k ) )

= .

= . ((()/()))

= 2(. )

Ast = 6396 mm2

Ast = 20109 mm2

Adopt Ast = 6396 mm2

Xu/d = 0.87fyAst/0.36fckbd = 0.42 ;

(xumax/d) > (xu/d) (ok)

Ast = 20109 mm2

Xu/d =2.96 (not ok)

Ast = 6396 mm2

Let us adopt =16 mm

A =

4 * 162

= 201mm2

= AstA==2240201 =

= = .

Adopt 16mm @75mm c/c on both sides

(b)Distribution steel:-

Ast= Ast1in2 =760

2 = 380mm2


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Adopt = 10mm

A =

4 * 102

= 78.53mm2

= Ast

A

== 380

78.53

= .8

s = 206mm

Adopt 10mm @ 200mm c/c on both sides


59/64



Chapter -20

Design of pile foundation

Pier cap size 2.4 wide *4.2m length

Adopt 6 numbers 450mm piles

Pier cap thickness 1.2m

20 mm dia @ 150mm c/c in top and bottom


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Chapter -21

Design of foundation of abutment and piers

Foundation of piers and abutments are to be taken down to hard ground. As per data, hard

ground is available below +37.00.

So the foundation of piers level of foundation is +37.00.

As seen in the plan the drainage sill is kept at +37.00 and with a depth of drainage floor as 60

cms, the bottom level of the drainage apron also is at +36.40.

So, the drainage apron, foundation of piers and abutments will all be laid as one block of

concrete of 60 cms. thick (for the drainage barrel portion) as shown in the plan.

This distributes the load of the structure evenly on the soil below and drainage apron will also be

capable of acting as an inverted arch to take care of the extra uplift pressure.

Where the soil of enough bearing capacity is met with at a deeper level, the foundation will be

taken deeper and the drainage apron will be at higher levels. in such cases, the actual pressure

under the foundation of abutments will have to be checked and verified so that they do not

exceed the safe bearing capacity.

The drainage apron in such cases, not being monolithic with the foundation of the abutments and

piers will not be able to take care of the any uplift pressure by arch action. The uplift pressure

that can be resisted is only due to the weight of the concrete apron.

In the present case, for safety the thickness of the pier adopted is 1.00 meter. Abutment under the

road arches has a bottom width of 1.75 meters and top width of 1.00 meter .the abutment under

the road arches has a bottom width of 2.25 meters and a top width of 1.25 meters. The abutment

has a uniform face better 1 in 8.

These abutments, in actual construction, will have to be carefully checked for the stability taking

into account the earth pressures, surcharge, etc. acting on them. The maximum pressures on the

soil have to be checked so that they do not exceed the safe bearing pressure on the foundation

soil.


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Arc lengths to fix the lengths of drainage wings

Downstream side of drain:

Canal F.S.L +42.00

Drainage bed +37.00

Difference +5.00 m

Arc length =5 meters.

Upstream of drain:

Canal F.S.L +42.00

Drainage bed +38.00

Difference +4.00 m

Arc length along which creep may occur =4X4=16 meters.

Since the downstream transitions of canal will be lined with masonry even beyond the canal

wings, the creep length is fixed from the end of canal aprons upstream and the arch length along

which creep occurs is shown in the plan.

In the drawing, keeping the distance between the returns as 14 meters both upstream anddownstream and keeping the lengths till the hydraulic gradient cuts the proposed drain bed level,

the actual arc lengths along which the creep occurs are more than the required. Hence the

proposed splays as in the drawing are adopted.

Solid apron in the drainage bed will be provided up to the end of the end of the drainage wings.

The length of the drainage wings is limited by a hydraulic gradient as shown in dotted line in the

drawing.


62/64



Chapter -22

Design of anchorage arrangements

Since the barrel is flowing full and the drain M.L.F. on both sides is above the bottom level of

the roof slab there will be an upward thrust acting on the roof slab there will be an upward thrustacting on the roof slab.so the roof slab have to be well anchored to the piers and abutments to

prevent the upward movement of the R.C. slab.

The uplift is maximum when the barrel is full and canal empty. The worst condition is at

upstream end of roof slab. M.F.L. just upstream of R.C. trough

Bottom level of the trough

Difference =1.14

Thickness of roof slab

Roof slab will counteract

meters of uplift head

Net uplift head

Say 0.50 meters.

So, necessary anchoring arrangements are provided as holding-down bolts fixed on the piers

through R.C slab.

Design of anchor bolts:

Clear span of the slab =2.50m

Upward thrust acting on one span along the entire width of slab

= 2.5 X 12.60 X 0.5 X 1000 = 15750 kg.

Assuming 20 mm dia.rod, the thrust that can be registered by one holding-down bolt

= 1260 X 3.14 = 3956 kg.

No bolts required =157503956 = 4 Nos (approx.)

So provide 4 holding-down bolts on each pier.it is enough to provide half the number on the

abutment, but in this case provide 3 bolts, one at each end of trough and one in the middle.


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DRAWINGS


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Documents

Aqueduct Main Project Modify