Canal Syphon Programme

Office of Director DamI.D.R., Jaipur

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Program by- Sunil K Goyal Hydraulic 1 Checked by - P.P.Pareek

DESIGN OF A CANAL SYPHON

NAME OF PROJECT : Case study, Design example 9.4/p394

DESIGN INPUT DATA :

(a) Canal

(i) Full supply discharge of canal 40.00 Cumec

(ii) Bed width of canal 18.00 m

(iii) Full supply depth of canal 2.10 m

(iv) Bed level of canal (C.B.L.) at D/S 250.00 m

(v) Side slope of canal (s) 1.50 :1

(vi) Free board of canal 0.75 m

(b) Drain

(i) Max. observed flood discharge 100.00 cumec

(ii) Bank level 254.00 m

(iii) Bed level 251.80 m

(iv) Highest Flood Level (H.F.L.) 253.25 m

(v) slope 1/600

HYDRAULIC DESIGN :

(1) Section of the drainage channel

According to Lacey's formula

P = 4.83 X= 4.83 X 10= 48.3

Provide bed width of the drain at the crossing = 44.50 m.

(2) Canal waterway

Bed width of canal = 18.00 m.

Normal X-area of the channel = BD +(A) = 44.42 Sq.m.

Velocity in the normal section = Q/A= 0.90 m/sec

Width = 3.00 m , Wall thickness = 0.30 m

Q 1/2

sD 2

Adopt size of the barrel as


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Height = 2.50 m , No. of barrels = 2 No.


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Reduce the canal waterway from 18.00 m to 6.30 m

Velocity through the barrels =40

= 2.67 m/sec15 < 6.00 m/sec

The velocity is within the recommended rangeThe size of the barrel is, therefore okayIt should be checked that the flow is subcritical in the barrel, .i.e., Froude number (F) shouldbe less than unity

Now F =Where,

V = 2.67 m/secg = 9.81 m/secd = 2.50 m

Therefore,F = 0.538

Since the value of F is less than 1 the flow will be subcritical in the barrel

(3) Head loss and bed levels at different sections :

Width of canal in the flumed portion = 6.3 mProvide 2 in 1splay in contraction and 3 in 1 splay in expansion transition

11.70Length of contraction transition = -------- X 2 = 11.7 m

211.70

Length of expansion transition = -------- X 3 = 17.55 m 2

Assume,Length of the barrels in the flumed portion= 70.0 m > 68.50 massumed length is O.K.

In the transitions, the side slopes of the section shall be warped from 1.50 : 1 to vertical.

1 2 3 4

0.3 m Thick wall

Canal

18.0 6.3 18.0

R.C.C. barrels

11.70 70.00 17.55

1 2 3 4

Canal waterway (All dim in meter)

V/(gd)1/2


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At section 4-4

Area of section = 44.42 Sq.m.(Normal channel section)Velocity = Q/A = 0.901 m/sec

Velocity head = 0.0414 mR.L. of bed = 250.00 m (given)R.L. of water surface = 250.00 + 2.1

= 252.10 mR.L. of T.E.L. = 252.10 + 0.0414

= 252.141 m

At section 3-3

Water depth at the entry and exit of the barrel should be kept slightly higher than the depth of the barrels so as to keep the ends of the barrel submerged for proper syphoningProvide water depth equal to 3.00 mArea of section = 3.00 X 6.3

= 18.9 Sq.m.Velocity = Q/A = 2.116 m/sec

0.228 mLoss of head in expansion from section 3-3 to section 4-4

== 0.056 m

Hence elevation of T.E.L. at section 3-3= 252.141 + 0.056= 252.197 m

R.L. of water surface = 252.197 - 0.228= 251.969 m

R.L. of bed = 251.969 - 3.00= 248.969 m

From section 3 - 3 to section 2 - 2, area and velocity are constant.

Head loss through barrels

Head loss through barrels is given by

=

= 0.080 ,for bell mouthed syphon

= a(1+b/R)

Where, a = 0.00316 , b = 0.10000

L = 70.0 mR = A/P

= 0.682 mHence,

= 0.00362Therefore loss of head in barrels = 0.526 m

V2/2g =

Velocity head = V2/2g =

0.3 (V22 - V1

2) /2g

( 1 + f1 + f2 xL/R )V 2/2g

Where, f1

f2

f2


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At section 2-2

R.L. of T.E.L. = T.E.L. at section 3-3 + head loss through barrels= 252.197 + 0.526= 252.724 m

R.L.of water surface = R.L. of T.E.L. - Velocity head at section 3-3= 252.724 - 0.228= 252.495 m

R.L. of bed = 252.495 - 3.00= 249.495 m

At section 1-1

Loss of head in contraction transition from section 1-1 to section 2-2

== 0.037 m

R.L. of T.E.L. = T.E.L. at section 2-2 + head loss in contraction transition= 252.724 + 0.037= 252.76 m

R.L. of water surface = 252.76 - 0.041 = 252.72 m

R.L. of bed = 252.72 - 2.10= 250.62 m

(4) Transitions

The general method of Hinds shall be applied for designing the transitions, as the water depths inthe transitions vary from 2.10 m to 3.00 m

(a) Contraction transitionw.s.profile

1

R.L. 252.72 2

0.112

5.85 R.L. 252.50

1

11.70 2

(All dim in m)

Contraction transition

=Water level at section 1-1 - Water level at section 2-2

2

=252.72 - 252.50

2= 0.1122

= Length of contraction transition2

=11.70

2= 5.85 m

0.2 (V22 - V1

2) /2g

y1 =

x1=

y1

x1


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C = =0.112

= 0.0032834.223

y = 0.00328 (equation of water surface profile in contraction transition)

The details of the contraction transition have been worked out in Table 1

(b) Expansion transitionw.s.profile 4

3 R.L. 252.10

0.065

R.L. 251.97

8.775

17.55 4

3

Expansion transition

=Water level at section 4-4 - Water level at section 3-3

2

=252.10 - 251.97

2= 0.0654 m

=Length of expansion transition

2

=17.55

2= 8.775 m

Hence,

C = =0.06541

= 0.0008577.0006

y = 0.00085 (equation of water surface profile in expansion transition )

The details of expansion transition have been worked out in Table 1

y1

x12

x2

y1 =

x1=

y1

x1

y1

x12

x2


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Table 1

water Elevation Velocity Velocity side Area A= Bed Depth(D) Bed

Distance surface of slope 's' Q/V Level Col.(3) width

elevation T.E.L. (4) - (3) -Col(9) B = A/D

m m m m -sD

1 2 3 4 5 6 7 8 9 10 11

From section 1-1 to 2-2 CONTRACTION TRANSITION

0 0.0000 252.720 252.761 0.041 0.901 1.500 :1 44.392 250.62 2.10 18.00

3.0 0.0295 252.690 252.752 0.062 1.099 1.115 :1 36.406 250.34 2.35 12.86

5.85 0.1122 252.608 252.742 0.135 1.627 0.750 :1 24.593 250.06 2.55 7.73

8.7 0.0295 252.525 252.733 0.208 2.021 0.385 :1 19.794 249.78 2.75 6.30

11.7 0.0000 252.495 252.724 0.228 2.116 0.00 :1 18.900 249.50 3.00 6.30

From section 3-3 to 4-4 EXPANSION TRANSITION

0 0.0000 251.969 252.197 0.228 2.116 0 :1 18.900 248.969 3.00 6.30

3 0.0076 251.977 252.188 0.211 2.035 0.256 :1 19.656 249.145 2.83 6.30

6 0.0306 252.000 252.178 0.179 1.872 0.513 :1 21.372 249.322 2.68 6.61

8.775 0.0654 252.035 252.169 0.135 1.627 0.750 :1 24.593 249.485 2.55 7.73

11.55 0.0306 252.069 252.161 0.091 1.337 0.987 :1 29.913 249.648 2.42 9.96

14.55 0.0076 252.092 252.151 0.059 1.072 1.244 :1 37.300 249.824 2.27 13.62

17.55 0.0000 252.100 252.141 0.041 0.901 1.500 :1 44.392 250.00 2.10 18.00

(5) Invert level

Bed level of drain = 251.80 m

Provide 0.30 m thick concrete slab and 0.60 m thick earth fill over the slab

Invert level of the concrete = 251.80 - ( 0.6 + 0.3 + 2.5 )= 248.40 m

Invert level at the entrance and exit of the barrel shall be the same as the bed levels alreadyworked out at sections 2-2 and 3-3 respectively.

Thus the invert level at the entry = 249.495 mThe invert level at the exit = 248.969 m

The invert level of the barrel would be kept at 248.40 m in a length of 44.5 m(under base of drain) after which it would meet the respective bed levels at the entrance andexit, so as to obtain a slope of about 1in 15 in the barrel at either side.Thus, Length of barrel upstream = 16.00 m

Length of barrel downstream = 8.00 m

The length of the pucca floor on either end should be adequate to provide safe hydraulic gradientand its thickness sufficient to counterbalance the total uplift pressure by gravity. The barrel shallbe made of reinforced concrete box construction and its structural design is given subsequently.

y=cx2 head (hV) V= Ö2ghV


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(6) Pucca Floor

Provide pucca floor in half the transition length in the upstream and 3/4 th the length of expansiontransition in the downstream.Length of pucca floor upstream = 1/2 X 11.7

= 5.85 msay 6.00 m

Length of pucca floor downstream = 3/4 X 17.55= 13.16 m= say 13.00 m

The floor shall be subjected to static uplift and seepage head; it is maximum when high flood ispassing through the drain and there is no flow in the barrel. The seepage head would becalculated by Bligh's theory.

(7) Uplift pressures on the barrel floor and pucca floor

(a) Static pressure

At bottom of barrel floor

Deepest invert level of the barrel = 248.40 mThe thickness of the barrel is = 0.30 mThe bottom level of the barrel floor = 248.10 mAssuming the sub-soil water level upto the bed level i.e.at R.L. 250.00 m, the maximumstatic head = 250.00 - 248.10

= 1.90 m

At the downstream end of barrel

Floor level at d/s end of barrel = 248.969 mAssuming floor thickness at this point 2.00 m,The bottom level of pucca floor = 248.969 - 2.00

= 246.969 m Hence,

Static head = 250.00 - 246.969= 3.031 m

At the upstream end of barrel

Floor level at u/s end of barrel = 249.495 mAssuming floor thickness at this point 1.50 m,The bottom level of pucca floor = 249.495 - 1.50

= 247.995 m Hence,

Static head = 250.62 - 247.995= 2.624 m

(b) Seepage head

The seepage head will be maximum when the drain is running full and there isno flow in the canal. Thus total seepage head =

= H.F.L. in the drain - Bed level of canal


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= 253.25 - 250.00= 3.25 m


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At bottom of barrel floor

The residual seepage head at point 'a' in the centre of the first barrel has been calculated byBligh's theory. The seepage line would follow the path indicated by the line x a b y. Its total length(neglecting floor depression or thickness) is the sum of the following

(i) Half the barrel span = 1.5 m(ii) Length of barrel in indicated portion = 8.00 m(iii) Length of pucca floor = 13.00 m

Thus total creep length = 22.50 m and creep length upto point 'a' i.e. centre of firstbarreel= 1.5 m

8 13.00

1.5

Barrel Plan

Residual seepage head point 'a' = 3.033 m

Thus total uplift in the barrel = 1.90 + 3.033= 4.93 m

say 4.93

At the downstream end of barrel floor

total creep length upto the end of barrel floor i.e. at point b= 1.5 + 8.0= 9.5 m

Hence , Residual seepage head at this point = 1.88 m

Thus total uplift = Static uplift + residual seepage head= 3.031 + 1.88= 4.909 m

4.909The floor thickness required (sp.gr.=2.22) = ----------- = 2.211 m

2.22Say 2.20 m

Provide 2.20 m thick c.c.floor d/s and reduce it to thickness 0.90 m at the end of floor

t/m2

c.c.cutoff

c.c.floor

R.C.C.Barrel

x

a by


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At the upstream end of barrel floor

total creep length upto the end of barrel floor = 1.5 + 16.00= 17.5 m

Hence , Residual seepage head at this point = 0.83 m

Thus total uplift = Static uplift + residual seepage head= 2.624 + 0.83= 3.454 m

3.454The floor thickness required (sp.gr.=2.22) = ----------- = 1.556 m

2.22Say 1.60 m

Provide 1.60 m thick c.c.floor u/s and reduce it to thickness 0.70 m at the end of floor

**********************

Office of Director DamI.D.R.,Jaipur

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Program by - Sunil K Goyal Drawing 12 Checked by- P P Pareek

DR

AIN

4.28TRANSITION WING TRANSITION WING

4.28B

R.C.C. BARRELS 0.30 M THICK

A A

18.0

0 3.00

18.0

0

CANAL 3.00

4.28 16.00 44.50B

8.00 4.28

CONTRACTION TRANSITION EXPANSION TRANSITION

11.7 68.50 17.55

PLAN

0.60 TH. EARTH FILL

TOP OF WING WALL TOP OF WING WALL

R.L. 253.47R.L. 254.00 H.F.L. 253.25

254.00 R.L. 252.85

U/S F.S.L. 252.72 D/S F.S.L. 252.10

CANAL

U/S BED R.L. 250.62249.50 2.50 248.97

D/S R.L. 250.00

248.40

U/S TOE WALL U/S CUT OFF C.C. BLOCK 0.30 TH. R.C.C. BARREL LEAN CONCRETE C.C. BLOCK D/S CUT OFF D/S TOE WALL

DRY BRICK PITCHING 5.70 6.00 16.00 44.50 8.00 13.00 4.55 D.B. PITCHING

11.7 68.50 17.55

TOP OF BANK 254.00 SECTION AT A-AH.F.L. OF DRAIN 253.25 DETAILS OF PROTECTION WORKS

BED LEVEL OF DRAIN 251.80 TOP OF BARREL ROOF 1 U/S TOE WALL 0.40 X 0.80

R.L. 251.20 2 U/S CUT OFF 0.50 X 1.00

R.L. 248.403.10 3 C.C. BLOCK 0.50 X 0.50

4 D/S CUT OFF 0.50 X 1.50

6.90 5 D/S TOE WALL 0.40 X 1.00


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SECTION AT B-B 6 LEAN CONCRETE 0.15 M THICK

7 D.B.PITCHING 0.40 M THICK

DETAILS OF TRANSITION WINGS

4.275

TRANSITION WING TRANSITION WING

4.275

18.0

0

18.0

0

###

7.73

6.30

6.30

6.30

6.61

7.73

9.96

13.6

2

6.00

3.00 8.77

4.275

5.85 11.55

4.2758.70 14.55

11.70 17.55

CONTRACTION TRANSITION EXPANSION TRANSITION

DETAILS OF PUCCA FLOOR

0.70

1.15

1.60

2.20 1.

70 1.30

0.90

2.00

4.00 3.25 3.25 3.25 3.25

6.00

13.00

U/S PUCCA FLOOR D/S PUCCA FLOOR


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Program by- Sunil K Goyal Structural 15 Checked by - P P Pareek

STRUCURAL DESIGN OF A CANAL SYPHON

NAME OF PROJECT : Case study, Design example 9.4/p394

DESIGN DATA :

1 UNIT WEIGHT OF DRY EARTH 1.60

2 UNIT WEIGHT OF SATURATED EARTH 2.00

3 UNIT WEIGHT OF SUBMERGED EARTH 1.00

4 UNIT WEIGHT OF CONCRETE 2.40

5 30 Degree

6 GRADE OF STEEL Fe 415

7 GRADE OF CONCRETE M 20

6 DIAMETER OF REINF. BARS Main 16 FDist 12 F

7 NUMBER OF BARRELS 2 Nos.

8 WIDTH OF EACH BARREL 3.00 m

9 HEIGHT OF EACH BARREL 2.50 m

10 THICKNESS OF BARREL 0.30 m

11 BANK LEVEL 254.00 m

11 DRAIN H.F.L. 253.25 m

12 LEVEL AT TOP OF BARREL 251.20 M

13 UPLIFT AT BASE OF BARREL 4.93

t/m3

t/m3

t/m3

t/m3

ANGLE OF INTERNAL FRICTION ( f )

t/m2


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(1) Critical section of the barrel

The barrel shall be subjected to maximum loading under the bank at the lowest possiblelevel of the barrel as indicated below in the following section:-

Bank level 254.00 m

Saturation line 253.25 m

R.L. 251.20 m

2.50

R.L. 248.40 m

R.L. 248.10 m

6.90

(2) Design features

Following cosiderations have been made in the design of barrels :(i) Bottom slab : This design is for uplift pressure and reaction from soil resulting from

the loading on the foundations. Theoretically, the soil reaction is not uniform but for simplification it is assumed uniform.

(ii) side walls : Side walls would be tested in the following two critical conditions(a) High flood in the drain while barrels are empty(b) No water in the drain while barrels are full

(iii) Partition walls : The partition walls are subjected to equal pressures on either side, andtherefore,no reinforcement is required. Nominal reinforcement is, however,provided to take care of contingency arising due to unequal pressuresresulting from chocking up of any of the barrels.

(iv) Top slab : The loads considered for design of top slab are :(a) Earth load,(b) Weight of water below saturation line

As there is no roadway along the drain, no live load due to traffic shall be considered.

(3) Design

As the barrels are rigidly joined, they should be designed as a continuous structure. Hardy Crossmethod of moment distribution shall be used for design.The effective length of horizontal member = 3.30 mThe effective length of vertical member = 2.80 m

Distribution factors

At joint A

For member AB =2.8

= 0.462.8 + 3.3

For member AD =3.3

= 0.542.8 + 3.3


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At joint D

For member DA =3.3

= 0.542.8 + 3.3

For member DC =2.8

= 0.462.8 + 3.3

(i) Dead loads

Consider one meter length of barrelDepth of dry earth over barrel = 254.00 - 253.25 = 0.75 mDepth of saturated earth = 253.25 - 251.20 = 2.05 m

Weight of dry and saturard earth = 0.75 X 1.60 + 2.05 X 2.00= 5.3 t/m2

Weight of top slab = 0.30 X 2.40 = 0.72

Weight on the top slab including its own weight = 5.3 + 0.72

= 6.02

Weight of the barrels per metre of length=( 4 X 3.3 + 3 X 2.8 ) X 0.30 X 2.40

= 15.55 t

Total dead load/m length of barrels = 15.55 + 5.3 X 6.90= 52.12 t

Uplift/m length = 4.93 X 6.90 = 34.04 t

Net vertical load acting on foundation = 52.122 - 34.04= 18.08 t

Pressure on foundation soil =18.082

= 2.626.90

Pressure acting on the base slab = Soil reaction + uplift= 2.62 + 4.93

= 7.55

Net upward pressure on the base slab = 7.55 - 0.72

= 6.83

(ii) Earth pressure

The earth pressure shall comprise of the following :(a) dry earth pressure above saturation line from R.L. 253.25 to 254.00(b) saturated earth pressure from R.L. 253.25 to 248.25

t/m2

t/m2

t/m2

t/m2

t/m2


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= 30 degree , = 1/3The pressure at A= Surcharge due to dry earth + Surcharge due to saturated earth

= 0.75 + X X 2.20+ w X 2.20

= 1/3 X 1.60 X 0.75 + 1/3 X 1.0 X 2.20+ 1 X 2.20

= 3.33

Pressure at D = 3.33 + X X 2.8 + w X 2.8= 3.33 + 1/3 X 1.00 X 2.8 + 1 X 2.8

= 7.07

Loads acting on different members are shown below :

6.02

3.33 3.33

3.30 3.30

2.80

7.07 7.07

6.83

(iii) Fixing moments

6.02 X 3.3(i) On span AB = 12 = 12

= 5.46 t-m

6.83 X 3.3(ii) On span CD = 12 = 12

= 6.20 t-m

(iii) On span AD fixed end moments in the wall at each end due to rectangular portion

3.33 X 2.8= 12 = 12

= 2.178 t-m

Fixed end moments due to triangular portion

3.73 X 2.8

= 30= 0.9756 t-m

= 3.73 X 2.820

Corresponding to f Cp

Cp X wd X Cp ws

t/m2

Cp ws

t/m2

t/m2

t/m2 t/m2

t/m2 t/m2

t/m2

wl2 2

wl2 2

wl2 2

2

MAD

MDA 2

B

D

A

CD

E

F


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= 1.4635 t-m


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Total fixed end moments at A = 2.178 + 0.976= 3.15 t-m

Total fixed end moments at D = 2.178 + 1.463= 3.64 t-m

(iv) Distribution of moments

Joints C D A BDistribution 0.46 0.54 0.54 0.46factorsFixed end -6.20 6.20 -3.64 3.15 -5.46 5.46momentsBalance -1.18 -1.39 1.25 1.06Carry over -0.59 0.62 -0.69 0.53Balance -0.29 -0.34 0.37 0.32Carry over -0.14 0.19 -0.17 0.16Balance -0.09 -0.10 0.09 0.08Carry over -0.04 0.05 -0.05 0.04Balance -0.02 -0.02 0.03 0.02

Total -6.98 4.63 -4.63 3.98 -3.98 6.19

(v) Net moments at centre and face

Span AB

6.02 X 3.3 6.02 X 0.15Sagging moments at face = ------------------- X 0.15 - -----------------------------

2 2

= 1.4222 t-m

Fixing moments at face= 3.98 +3.15

( 6.19 - 3.98 )3.3

= 6.0907 t-m

Net fixing moments at face = 6.0907 - 1.422= 4.6685 t-m

Sagging moments at centre = 6.02 X 3.38

= 8.1947 t-m

Fixing moments at centre = 3.98 + 6.192

= 5.0876 t-m

Net sagging moments at centre = 8.195 - 5.088= 3.107 t-m

2

2


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Span CD

6.83 X 3.3 6.83 X 0.15Sagging moments at face = -------------------- X 0.15 - ----------------------------

2 2

= 1.6145 t-m

Fixing moments at face=4.63 +

3.15( 6.98 - 4.63 )3.3

= 6.8693 t-m


Sagging moments at centre = 6.83 X 3.38

= 9.3027 t-m

Fixing moments at centre = 6.98 + 4.632

= 5.8043 t-m


Span AD

(a) Due to rectangular portion

=3.33 X 2.8

X 0.15 -3.33 X 0.15

2 2

= 0.6625 t-m

(a) Due to triangular portion

=3.73 X 2.8

X1

X 0.15 -0.2 X 0.15

X 0.052 3 2

= 0.2606 t-mTotal sagging moments at face = 0.662 + 0.261

= 0.92 t-m

Fixing moments at face = 3.98 + 2.65 ( 4.63 - 3.98 )2.8

= 4.598 t-m

2

2

2


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Sagging moments at centre

(a) Due to rectangular portion = 3.33 X 2.808

= 3.27 t-m

(b) Due to triangular portion = 3.73 X 2.80X 2.8 X

12 7.81

= 1.87 t-m

Total sagging moments at centre = 3.27 + 1.87= 5.14 t-m

Fixing moments at centre =4.63 + 3.98

2

= 4.31 t-m


The net moments at face and centre of the different spans are tabulated below :

SpanMOMENTS

At face At centreAB 4.67 t-m (hogging) 3.11 t-m (sagging)CD 5.25 t-m (hogging) 3.50 t-m (sagging)AD 3.67 t-m (hogging) 0.83 t-m (sagging)

(vi) Thickness of members

The maximum moment in the barrels is 5.25 t-m

For M 20 grade concrete and steel of Fe 415

70 1500.00m = 13k = 0.378j = 0.874

Q = 11.552The minimum effective thickness (d) required for bending moment

d = ÖM= Ö 5.25 X

bQ 100 X 11.55

= 21.33 cmAdopt overall thickness of all the members = 30.0 cmThen effective thickness = 25.20 cm

2

scbc = Kg/cm2 sst = Kg/cm2

105


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(clear cover = 4.0 cm )


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The reinforcement required at various points is given below :

(vii) Reinforcement

Span AB & BEMoments in t-m Reinforcement required

At face 4.67 t-m (hogging) = 14.13Provide 16 mm bars @ 14.0 cm c/c

At centre 3.11 t-m (sagging) = 9.40Provide 16 mm bars @ 21.0 cm c/c

Span CD & CFMoments in t-m Reinforcement required



Span AD & EFMoments in t-m Reinforcement required



Nominal steel shall also be provided on the outer face of the walls for the condition when there isno water in the drain and barrels are running full. The details of reinforcement are shown below

16 14.0 cm c/c

16 14.0 cm c/c

12 25.0 cm c/c

16 21.0 cm c/c

C.C. M - 20 16 25.0 cm c/c 12 25.0 cm c/c

16 18.0 cm c/c

16 12.0 cm c/c clear cover = 4.00 cm

16 18.0 cm c/c

0.30 3.00 m 0.30 3.00 0.30

DETAILS OF REINFORCEMENT IN SYPHON BARRELS

A t Cm2

A t Cm2

A t Cm2

A t Cm2

A t Cm2

A t Cm2

mm f @

mm f @

mm f @

mm f @

mm f @ mm f @

mm f @

mm f @

mm f @


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*************

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