115325841 design of syphon aqueduct

DESIGN OF TWO VENT BOX BARREL FOR SYPHON AQUEDUCT

Name of the work:-Elamanchili minor drain under Medpadu field channel @ 0/750 Km

Note on site conditions

at proposed site are as given below:-

Hydraulic Particulars Units Field channel

Chainage Km 0.500 ----

Discharge Cumecs 2.950 0.600

Bed level m 1.200 2.585

OFL m 2.210 2.910

MFL m 3.130 3.210

TBL m 4.130 3.510

RTL m 4.130 4.130

Bed width m 1.400 0.600

Top width m 7.260 0.600

Bed fall ---- 0.000250 0.000250

From the above hydraulic particulars,it can be observed that the MFL of Chilla minor drain is above

the bed level of field channel.It is proposed to construct Syphon Aqeduct with depressed floor

and 1V-RCC barrel of size 2.50mx1.50m for a length of 6m to accommodate the

existing village road after the field channel.

Further fluming of field channel is not proposed, hence existing width of the channel is not altered.

and raising of floor on down stream side in 1V to 5H to ensure clearing of sediments.

Wing walls are proposed both upstream&down stream sides of drain along with revetment for

slopes and bed protection.Cisterns are also proposed both U/S and D/S as per the requirement.Cut-off

walls are proposed both U/S and D/S sides of drain.

Similarly,wing walls along with bed and slope protections are proposed for field channel also

both U/S and D/S sides.

The hydraulic design of syphon aqueduct is carried out as per the guide lines stipulated in

IS 7784:Part1--1993.The structural design of the box barrel along with load combinations is carried out as

per the guide lines stipulated in IS 7784:Part2--5--1993.

In view of the above facts,the box barrel is designed as free flowing vented structure with all

As per the record the hydraulic particulars of Chilla minor drain&Field channel

Chilla Minor drain

The floor of the aqeduct is proposed to be lowered to a level of +0.485 with vertical drop on upstream side

possible combinations of loads.Further,the stability of the stucture is also ensured by providing

appropriate factors of safety against overturning and sliding.

As,it is not possible propose foundation of box culvert below Max.scour depth,protective

works are proposed for the bed of the minor drain as per the guidelines.

Note on site conditions

3.425

From the above hydraulic particulars,it can be observed that the MFL of Chilla minor drain is above -2.1

1.325

2.065

-1.325

Further fluming of field channel is not proposed, hence existing width of the channel is not altered.

Wing walls are proposed both upstream&down stream sides of drain along with revetment for

slopes and bed protection.Cisterns are also proposed both U/S and D/S as per the requirement.Cut-off

Similarly,wing walls along with bed and slope protections are proposed for field channel also

IS 7784:Part1--1993.The structural design of the box barrel along with load combinations is carried out as

In view of the above facts,the box barrel is designed as free flowing vented structure with all

with vertical drop on upstream side

Design of 1V Box Barrel

I)Design Parameters:-

= 6.00m

= 2.50m

= 1.50m

= 0.30m

= 0.30m

= 0.30m

= 3.10m

= 2.10m

Haunch width = 0.15Haunch depth = 0.15

= 0.075m

= 1.200m

= 25KN/cum

= 24KN/cum

= 18KN/Cum

= 10KN/Cum

= 30

= 90

= 0

= 15

= 1.20m

= 2.585m

= 1.200m

= 3.130m

Depressed floor level of barrel = 0.785m

= 0.485m

= 2.585m

= 3.510m

= 4.130m

Upstream side breast wall top level = 4.130m

Downstream side breast wall top level = 4.580m

Middle breast wall top level = 4.130m

Thickness of brest walls = 0.300m

Barrel length in Road portion = 4.500m

Barrel length in Channel portion = 0.600m

Over all width of the barrel (WL)Clear vent (b)Clear depth (d)Thickness of top slab (t1)Thickness of side wall (t2)Thickness of raft (t3)Outer span (B)Outer Depth (D)

Thickness of wearing coat (t4)

Height of railing (h2)

Unit weight of RCC (yrc)

Unit weight of PCC (ypc)

Density of back fill soil behind side walls of Box (y)

Unit weight of water (yw)

Angle of shearing resistance of back fill material(Q)

Angle of face of wall supporting earth with horizontal(In degrees)(in clock wise direction)(a)

Slope of back fill (b)

Angle of wall friction (q)

Height of surcharge considered (h3)

Barrel top level (BTL)

Drain bed level (DBL)

Drain High flood Level (HFL)

Barrel foundation level (BFL) Channel bed level(CBL)Channel top level(CTL)Road top level (RTL)

= 7.50t/sqm

= 20.00N/sqmm

= 415.00N/sqmm

Cover to reinforcement = 50.00mm

Details of the preliminary structure assumed is as given below :-

II)General loading pattern:-

As per IRC:6---2000,the following loadings are to be considered on the boxbarrel in road portion:-

1.Dead load2.Live load3.Impact load4.Wind load5.Water current6.Tractive,braking effort of vehicles&frictional resistance of bearings7.Buoyancy8.Earth pressure9.Seismic force10.Water pressure force11.Static water pressure due to water in canal

As per clause 202.3,the increase in permissible stresses is not permissible for theabove loading combination.

Further as per IS 7784-Part 1---1993,the structure should be designed for the following forces :-

12.Uplift pressure due to flowing water13.Uplift pressure due to subsurface flow

Safe Bearing Capacity of the soil (SBC)

Compressive strength of concrete for Box (fck)

Yield strength of steel (fy)

1545

SECTIONPLAN

6000

4500

300 300

300

300

ROAD PORTION

CA

NA

L P

OR

TIO

N

3100

300

600

1500

2500

2100

450

III)Loading on the box culvert :-

1.Dead Load:-

i)Self wieght of the top slab = 139.50KN

(3.1*6*0.3*25) =

ii)Self wieght of the bottom slab/Raft = 139.50KN

(3.1*6*0.3*25) =

iii)Self wieght of side walls = 135.00KN

(2*1.5*6*0.3*25) =

iv)Self weight of haunches = 6.75KN

(4*0.5*0.15*0.15*6*25) =

v)Self weight of wearing coat = 34.88KN

(3.1*6*0.075*25) =

vi)Self weight of U/S side brest wall = 35.92KN

(1*3.1*0.3*1.545*25) =

vii)Self weight of D/S side brest wall = 46.38KN

(1*3.1*0.3*1.995*25) =

viii)Self weight of middle brest wall = 35.92KN

(1*3.1*0.3*1.545*25) =

ix)Weight of earth on barrel in Road portion = 387.95KN

(18*3.1*4.5*1.545) =961.80KN

There is no need to consider snow load as per the climatic conditions

Taking moments of all loads about upstream end,we get

S.No Item Weight

1 Top slab 139.50KN 3.00 418.50

2 Bottom slab 139.50KN 3.00 418.50

3 Side walls 135.00KN 3.00 405.00

4 Haunches 6.75KN 3.00 20.25

5 Wearing coat 34.88KN 3.00 104.64

6 U/S brest wall 35.92KN 0.15 5.39

7 D/S brest wall 46.38KN 5.85 271.32

Distance of centroid from U/S end

Moment about upstream end in KN-m

8 Mid. brest wall 35.92KN 0.75 26.94

9 Earth fill 387.95KN 3.15 1222.04

961.80KN 2892.58

Distance of centroid of above dead load from U/S end = 3.007m

Eccentricity in x- direction = 0.007m

The position of resultant dead load is as shown below:-

2.Live Load:-

As per clause 201.1 of IRC:6--2000,the bridges and culverts of medium importance

GENERAL IRC Class-A loading Pattern

are to be designed for IRC Class A loading.

2.7t

2.7t

11.4

t

11.4

t

6.8t

6.8t

6.8t

6.8t

1.10

1.80

3.20 1.20 4.30 3.00 3.00 3.00

Y

X

7

3100

6000

considered as per clauses 207.1.3&207.4

The ground contact area of wheels for the above placement,each axle wise isgiven below:-

Axle load Ground Contact Area(Tonnes) B(mm) W(mm)

11.4 250 5006.8 200 3802.7 150 200

Assuming 0.3m allowance for guide posts and the clear distance of vehicle from

the edge of guide post being 0.15m as per clause 207.1,the value of 'f' shown in the figure will

be 0.45m

0.45m

1.75m

2.20m

The IRC Class A loading as per the drawing is severe for bearing stresses on soil and the same is to be

Hence,the width of area to be loaded with 5KN/m2 on left side is (f) =

Similarly,the area to be loaded on right side (k) =

2.7t

2.7t

11.4

t

11.4

t

6.8t

6.8t

6.8t

6.8t

1.10

1.80

3.20 1.20 4.30 3.00 3.00 3.00

1200

2750

b

u

Portion to be loaded withlive load of 5KN/sqm

Y

X

1800

0.45

Position of Live load forMax.Soil pressure

1525

3100

11

.4t

11

.4t

375

4500

1750

The total live load on the top slab composes the following components:-

1.Wheel loads----Point loads

(114+114)=

3.Live load in remaing portion(Left side)----UDL

(0.45*3.1*5)=

4.Live load in remaing portion(Right side)----UDL

(1.75*3.1*5)=

Resultant live load:-

Eccentricity of live load w.r.t y-direction(Along the direction of travel of vehicles)

Taking moments of all the forces w.r.t y-axis

S.No Distance from Y-axis

1 57 1.00m

2 57 1.00m

3 57 2.80m

4 57 2.80m

5 6.975 (0.3+0.45/2)= 0.525m

6 27.125 2.28m

262.100

Distance of centroid of forces from y-axis(498.57/262.1) =

= 1.902m

Eccentricity = 1.098m(1.902-6/2)=

Eccentricity of live load w.r.t x-direction(At right angle to the travel of vehicles)

Taking moments of all the forces w.r.t x-axis

Wheel Load/UDL in KN

(0.3+0.45+0.5/2)=

(0.3+0.45+0.5/2)

(0.3+0.45+0.5/2+1.8)

(0.3+0.45+0.5/2+1.8)

(0.3+0.45/2+1.75) =

S.No Load in KN Distance from X-axis

1 57 (3.1-0.38) 2.72m

2 57 (3.1-0.38) 2.72m

3 57 1.52m

4 57 1.52m

7 6.98KN (3.1/2)= 1.55m

8 27.13KN (3.1/2)= 1.55m

262.1

Distance of centroid of forces from x-axis(536.22/262.1) =

= 2.046m

Eccentricity = 0.496m(2.046-3.1/2)=

Location of the resultant of live load is as shown in the figure given below:-

The eccentricty of the line of action of live load wrt centroid in y-direction = 0.496m

The eccentricty of the line of action of live load wrt centroid in x-direction = 1.098m

3.Impact of vehicles:-

As per Clause 211 of IRC:6--2000,impact allowance shall be made by an increment

of live load by a factor 4.5/(6+L)

Hence,the factor is 0.495

[3.1-(0.38+1.2)]

=[3.1-

(0.38+1.2)] =

Y

X

3100

6000

1098496

[4.5/(6+3.1)]=

Further as per clause 211.7 of IRC:6--2000,the above impact factor shall be only

50% for calculation of pressure on piers and abutments just below the level of bed block.There

is no need to increase the live load below 3m depth.

As such,the impact allowance for the top 3m of box culvert will be (0.495/2)=

For the remaining portion,impact need not be considered.

4.Wind load:-

The deck system is located at height of (RTL-LBL) 3.38m[4.58-(1.20)]=

The Wind pressure acting on deck system located at that height is considered for design.

As per clause 212.3 and from Table .4 of IRC:6---2000,the wind pressure at that height is (52+1.38*(63-52)/2)=

59.59

Height of the deck system = 3.380

Breadth of the deck system =(3.1) = 3.10

The effective area exposed to wind force =HeightxBreadth = 10.478

Hence,the wind force acting at centroid of the deck system =(Taking 50% perforations) (0.5*59.59*10.478*10/1000)=

Further as per clause 212.4 of IRC:6---2000 ,300 Kg/m wind force is considered to be

acting at a hieght of 1.5m from road surface on live load vehicle.

Hence,the wind force acting at 1.5m above the road surface =(300*6.0*10/1000)=

The location of the wind force from the top of Raft slab of box culvert =(1.5+0.3+1.5+0.075) =

5.Water current force:-

Water pressure considered on square ended abutments as per clause 213.2 of IRC:6---2000 is

37.14

Kg/m2.

P = 52KV2 = Kg/m2.

(52*1.5*0.692)=

(where the value of 'K' is 1.5 for square endedside walls)

For the purpose of calculation of exposed area to water current force,only 1.0m

width of box is considered for full hieght upto HFL

Hence,the water current force = 0.51KN[37.14*1.0*(2.585-(1.20)*10/1000] =

Point of action of water current force from the top of RCC raft slab = [3.13-(1.20)]/3 =

6.Tractive,braking effort of vehicles&frictional resistance of bearings:-

The breaking effect of vehicles shall be 20% of live load acting in longitudinal

direction at 1.2m above road surface as per the clause 214.2 of IRC:6--2000.

As no bearings are assumed in the present case,50% of the above longitudinal

force can be assumed to be transmitted to the supports of simply supported spans resting on

stiff foundation with no bearings as per clause 214.5.1.3 of IRC:6---2000

Hence,the longitudinal force due to braking,tractive or frictional resistance of

bearings transferred to abutments is

(262.1*0.1)= 26.21KN

The location of the tractive force from the top of RCC raft slab =(1.2+0.075+0.3+1.50) =

7.Buoyancy :-

As per clause 216.4 of IRC:6---2000,for abutments or piers of shallow depth,the dead weight of the box culvert shall be reduced by wieght of equal volume of water upto HFL.

The above reduction in self wieght will be considered assuming that the back fill behind the box is scoured.

For the preliminary section assumed,the volume of box section is

i)Volume of Raft slab section = 5.580Cum

(186.0/25) =

ii)Volume of side walls upto MFL = 5.400Cum

(180/25) =

iii)Volume of haunches = 0.140Cum

(9/25) =

11.120Cum

Reduction in self wieght = 111.20KN(11.12*10)=

8.Earth pressure :-

As per clause 217.1 of IRC:6---2000,the abutments are to be designed for a

surcharge equivalent to a back fill of hieght 1.20m behind the abutment.

The coefficient of active earth pressure exerted by the cohesion less back fill on

the box as per the Coulomb's theory is given by

'2Ka = Sin(a+Q)

sina sin(a-q) sin(Q+q)sin(Q-b)

sin(a+b)

Sin(a+Q) = SIN[3.14*(90+30)/180] = 0.867Sin(a-q) = SIN[3.14*(90-15)/180] = 0.966Sina = SIN[3.14*(90)/180] = 1Sin(Q+q) = SIN[3.14*(30+15)/180] = 0.707Sin(Q-b) = SIN[3.14*(30-0)/180] = 0.5Sin(a+b) = SIN[3.14*(90+0)/180] = 1

From the above expression,

0.3

The hieght of box above GL,as per the preliminary section assumed =

Hence,maximum pressure at the base of the wall Pa =(0.3*1800*1.80)*10/1000 =

The pressure distribution along the height of the wall is as given below:-

Surcharge load = 6.48 KN/sqm(0.3*1800*1.2)*10/1000=

6.48

1.800

9.72 6.48

Ka =

Area of the rectangular portion = 6.48*1.80 = 11.66Area of the triangular portion = 0.5*9.72*1.8 = 8.75

20.41

Taking moments of the areas about the toe of the wall

S.No Description Area Lever arm Moment

1 Rectangular 11.66 0.9 10.4942 Triangular 8.75 0.6 5.25

20.41 15.744

Height from the bottom of the wall = 0.77m(15.744/20.41)

The active Earth pressure acts on the box culvert is as shown below:-

Inclination of earth pressure 15.00 Deg.force with horizontal

Total earth pressure acting on the wall P = 122.47KN(6.48*1.8+0.5*9.72*1.8)*8 =

Horizontal component of the earth pressure Ph =

(PCos150 )

Vertical component of the earth pressure Pv =

3100

150015°

770

P

Ph

Pv

Earth Pressure on Box culvert

Eccentricity of vertical component of earth pressure = (3.1/2-0.0) =

9.Siesmic force :-

As per clause 222.1 of IRC:6---2000,the bridges in siesmic zones I and II need not be

designed for siesmic forces.The location of the slab culvert is in Zone-I.Hence,there is no need to

design the bridge for siesmic forces.

10.Water pressure force:-

The water pressure distribution on the sidewall is as given below:-

HFL 3.130m

1.93

LBL 1.200m

19.30kn/sqm

Total horizontal water pressure force = 0.5*19.3*1.93*8 = 111.75KN

The above pressure acts at height of H/3 =1.93/3 = 0.64m

Static water pressure force (Vertical) = (1.93+0.16)*10 = 20.90kn/sqm

11.Static water pressure due to water in canal

Height of water in channel above bed level = 0.725m

The static water pressure on top slab of box barrel = 7.25kn/sqm

Total load due to static water pressure = 13.48KN

Eccentricity of the above load in x-direction = 2.40m

(PSin150 )

12.Uplift pressure due to flowing water:-

Level of bottom of the top slab of box barrel = +2.29m

MFL of the drain = +3.130m

Afflux = 0.160m

Datum head =Height of water above the bottom of top slab of box barrel = 1.005m

0.040m

Total head causing uplift on the top slab of box barrel = 1.045m

Hence uplift pressure on the top slab = 10.45kn/sqm

Total uplift force = 156.75KN

13.Uplift pressure due to subsurface flow

Afflux = 0.160m

The difference between subsoil hydraulic gradient line h' = 2.090and the bottom of floor just at entry sectionwhen the water is held up.

20.90kn/sqm

Uplift force due to sub surface water = 194.37KN

IV)Check for stability of Box Barrel:-

a)Load Envelope-I:-(The drain is dry,back fill intact with live load on span)

The following co-ordinates are assumed:-

a)x-Direction-----At right angle to the movement of vehicles

b)y-Direction-----In the direction of movement of vehicles

S.No Type of load

1 Self wieght of Box culvert 961.80KN 0.000

2 391.84KN -0.496

3 Vertical component of Active Earth pressure 31.68KN 1.550

1385.32KN

Kinetic head = v2/2g =

Hence,the theoritical max.uplift pressure = yw x h' =

Vertical load acting on the Box barrel (P) composes of the following components

Intensity in KN

Eccentricty about x-axis(m)

Live load with impact factor---(Wheel loads+UDL) [262.1*(1+0.495)] =

S.No Type of load Direction x or y

1 Wind load 18.00KN x-Direction

2 Tractive,Braking&Frictional resistance of bearings 26.21KN y-Direction

3 Horizontal Active Earth pressure force 118.30KN y-Direction

162.51KN

Check for stability against over turning(Assuming that the earth fill on toe side is scoured):-

Taking moments of all the overturning forces about toe of the box wrt x-axis,

Moment due to tractive,braking&frictional resistance of bearings =(26.21*3.38)=

Moment due to active earth pressure force =(118.3*0.77)=

Total overturning moment =

Taking moments of all the restoring forces about toe of the box wrt x-axis,,

Moment due to self weight of box =[961.8*(3.1/2+0.0)]=

Moment due to live load reaction on box =[391.84*(3.1/2-0.496)]=

Moment due to vertical component of active earth pressure =[31.68*(3.1/2+1.55)]=

Total Restoring moment =

Factor of safety = 11.6407156 > 2.0 Hence safe(2002/171.98)= (As per clause 706.3.4 of IRC:78-2000)

Check for stability against sliding:-

Coefficient of friction between concrete surfaces =

6.81957361 > 1.5 Hence safe

(0.8*1385.32/162.51)= (As per clause 706.3.4 of IRC:78-2000)

Horizontal load acting/transferred on the box (H) composes of the following components

Intensity in KN

Total vertical load acting on the base of the box culvert Vb =

Total sliding force,ie,horizontal load on the box Hb =

Factor of safety against sliding Fs =

b)Load Envelope-II:-(The Canal is running upto HFL with no live load on span)




S.No Type of load

1 Self wieght of box 961.80KN

-111.20KN

2 Net self wieght 850.60KN 0.000

3 Uplift force due to flowing water -156.75KN 0.000

4 Uplift force due to subsurface water -194.37KN 0.000

5 Static water force due to water in drain 388.74KN 0.000

6 Static water force due to water in channel 13.48KN 0.000

7 Vertical component of Active Earth pressure 31.68 1.550




3 Active Earth pressure force 118.30KN y-Direction

4 Force due to water pressure 111.75KN y-Direction

Check for stability against over turning:-

Taking moments of all the overturning forces about toe of the box wrt x-axis,

Moment due to tractive,braking&frictional resistance of bearings =

Moment due to active earth pressure force =(118.3*0.77)=

Total overturning moment =

Vertical load acting on the box (P) composes of the following components

Intensity in KN


Reduction in self weight due to buoyancy

Horizontal load acting/transferred on the box (H) composes of the following components

Intensity in KN

Taking moments of all the restoring forces about toe of the box wrt x-axis,

Moment due to self weight of box =[850.6*(3.1/2+0.0)]=

Force due to static water pressure need not be considered,as it acts on both side walls in opposite directions

Moment due to vertical component of active earth pressure =[31.68*(3.1/2+1.55)]=

Total Restoring moment =

Factor of safety = 15.5238491 > 2.0 Hence safe(As per clause 706.3.4 of IRC:78-2000)

Check for stability against sliding:-

(850.6+31.68) =

Coefficient of friction between concrete surfaces =

5.96634219 > 1.5 Hence safe

(0.8*882.28/118.3)= (As per clause 706.3.4 of IRC:78-2000)

V)Check for bearing pressure:-

a)Load Envelope-I:-(The Canal is dry, back fill intact with live load on span)

i)At the bottom of RCC raft slab




S.No Type of load

1 Self weight of box 961.80KN 0.000

Total vertical load acting on the base of the boxVb =

Total sliding force,ie,horizontal load on the box Hb =

Factor of safety against sliding Fs =

Vertical load acting on the Box (P) composes of the following components

Intensity in KN


2 159.84KN 0.00

3 391.84KN -0.496

4 Vertical component of earth pressure 31.68KN 1.550




3 Horizontal load due to earth pressure 118.30KN y-Direction

Safe bearing capacity SBC of the soil = 7.50t/sqm

Check for stresses:-

About x-axis:-

Breadth of footing b = 6.00m

Depth of footing d = 3.10m

Area of the footing = A = 18.6

Section modulus of bottom footing 9.61

about x-axis --Zx =

For RCC Strip footing permissible bearing pressure is 1.5xSBC = 113KN/sqm

No tension is allowed on soil as per clause 706.3.3.1 of IRC 78:2000

S.No Type of load

Vertical loads:-(Stress = P/A(1+6e/b)1 Self wieght of Box 961.80KN 0.002 159.84KN 0.003 391.84KN -0.496

4 Vertical component of Earth pressure 31.68KN 1.55Horizontal loads:- (Stress = M/Z)

1 Wind load 18.00KN 0.002 Tractive,Braking&Frictional resistance of bearings 26.21KN 3.083 Horizontal load due to earth pressure 118.30KN 1.07

Self weight of levelling concrete=(3.7*6*0.30*24)


Horizontal load acting/transferred on the abutment (H) composes of the following components

Intensity in KN

m2

(1/6)bd2 = m3

Intensity in KN (P)

Eccentricity/Lever arm

Self weight of levelling concreteLive load with impact factor---(Wheel loads+UDL) [262.1*(1+0.495)] =

S.No Type of load Eccentricity

Vertical loads:-(Stress = P/A(1+6e/b)1 Self wieght of Box 961.80KN 0.002 159.84KN 0.003 391.84KN 0.496

4 Vertical component of Earth pressure 31.68KN -1.55Horizontal loads:- (Stress = M/Z)


Stress at heel = P/A(1+6e/b)+M/Z = 53.67 KN/Sqm>0

Hence safe.

Stress at toe = P/A(1+6e/b)+M/Z = 112.47 KN/Sqm>113KN/sqm

Hence safe.

About y-axis:-

Breadth of footing b = 3.10mDepth of footing d = 6.00mArea of the footing = A = 18.6


about y-axis --Zy =



S.No Type of load

Vertical loads:-(Stress = P/A(1+6e/b)1 Self wieght of Box 961.80KN 0.0072 159.84KN 0.003 391.84KN -1.098



Intensity in KN (P)


m2

(1/6)bd2 = m3

Intensity in KN (P)




Vertical loads:-(Stress = P/A(1+6e/b)1 Self wieght of Box 961.80KN -0.0072 159.84KN 0.003 391.84KN 1.098



Stress at up stream side edge of abutment = P/A(1+6e/b)+M/Z = 35.73 KN/Sqm>0

Hence safe.Stress at down stream side edge of abutment = P/A(1+6e/b)+M/Z = 108.77 KN/Sqm<113KN/sqm

Hence safe.

Net bearing pressure for design = 52.87 KN/Sqm(72.97-(8+8.08))=

b)Load Envelope-III:-(The Canal is running upto HFL with live load on span)

i)At the bottom of RCC raft slab


a)x-Direction-----At right angle to the movement of vehiclesb)y-Direction-----In the direction of movement of vehicles

S.No Type of load

1 Self weight of box 850.60KN 0.000

2 159.84KN 0.00

Intensity in KN (P)


Vertical load acting on the Box (P) composes of the following components

Intensity in KN



3 391.84KN -0.496





8 Vertical component of earth pressure 31.68KN 1.550




3 Horizontal load due to earth pressure 118.30KN y-Direction

4 Force due to water pressure 111.75KN y-Direction

Safe bearing capacity SBC of the soil = 7.50t/sqm

Check for stresses:-

About x-axis:-



about x-axis --Zx =



S.No Type of load

Vertical loads:-(Stress = P/A(1+6e/b)1 Self wieght of Box 850.60KN 0.00

2 159.84KN 0.00

3 391.84KN -0.496


Horizontal load acting/transferred on the abutment (H) composes of the following components

Intensity in KN

m2

(1/6)bd2 = m3

Intensity in KN (P)









1 Wind load 18.00KN 0.00

2 Tractive,Braking&Frictional resistance of bearings 26.21KN 3.08

3 Horizontal load due to earth pressure 118.30KN 1.07

4 Force due to water pressure 111.75KN 0.00



2 159.84KN 0.00

3 391.84KN 0.496





8 Vertical component of Earth pressure 31.68KN -1.55

Horizontal loads:- (Stress = M/Z)1 Wind load 18.00KN 0.00




Stress at heel = P/A(1+6e/b)+M/Z = 50.44 KN/Sqm>0

Hence safe.

Stress at toe = P/A(1+6e/b)+M/Z = 109.24 KN/Sqm>113KN/sqm

Intensity in KN (P)



Hence safe.

About y-axis:-



about y-axis --Zy =



S.No Type of load


2 159.84KN 0.00

3 391.84KN -1.098





8 Vertical component of Earth pressure 31.68KN 0.00

Horizontal loads:- (Stress = M/Z)1 Wind load 18.00KN 3.38





Vertical loads:-(Stress = P/A(1+6e/b)

1 Self wieght of Box 850.60KN -0.01

2 159.84KN 0.00

m2

(1/6)bd2 = m3

Intensity in KN (P)




Intensity in KN (P)


3 391.84KN 1.098




7 Static water force due to water in channel 13.48KN -2.400

8 Vertical component of Earth pressure 31.68KN 0.00

Horizontal loads:- (Stress = M/Z)

1 Wind load 18.00KN 3.38



4 Force due to water pressure 111.75KN -0.64

Stress at up stream side edge of abutment = P/A(1+6e/b)+M/Z = 46.17 KN/Sqm>0

Hence safe.Stress at down stream side edge of abutment = P/A(1+6e/b)+M/Z = 98.4 KN/Sqm<113KN/sqm

Hence safe.

Net bearing pressure for design = 103.88 KN/Sqm(112.47-(8.59))=

Design Bearing Pressure = 103.88 KN/Sqm

VI)Analysis of Box culvert:-

a)Calculation of equivalent Uniformly Distributed Load(UDL) for IRC live load for Road portion:-

The position of IRC class A live load for maximum bending moment in top slab in road portion is as given below:-


1200

2750

b

u

Portion to be loadedwith live loadof 5KN/sqm

Y

X

1800

0.45

Position of Live load forMax.Bending moment ineach span

2800

11.4

t

11

.4t

4500

1750

500

800 800

Each axle load is 114 KN and hence wheel load = 57.00KN

Assuming 1:1 dispersion parallel to Y-axis,the effective width of concentrated load in that direction = [0.25+2(0.075+0.30)]=

There is no over lapping of areas of load dispersion in this direction.

As per the clause 24.3.2.1 of IS 456:2000,

L = Effective span x = Distance of centre of gravity of load from nearer support

k = A constant depending on the ratio (B/L),where B is the width of the slab(From Table 14 of IS 456:2000)

In the present case,

L = 2.80m x = 0.80m bw = 0.50m B/L = 1.89 k = 2.6

1.99m

The effective width be = kx[1-x/l]+bw

where be = The effective width of slab on which the load acts

bw = Breadth of concentration area of load

Hence be =

1200

2750

b

u


Y

X

1800

0.45


2800

11

.4t

11.4

t

4500

1750

500

800 800

500

500

Dispersion of Loads

500745

450

1300 500

745

From the above figure net effective width of dispersion of loads = 3.495m

For each axle of load,the area of dispersion = 3.495sqm(1.0*3.495)=

Total load of each axle with impact = 170.43KN[11.4(1+0.495)]

Average intensity of load = 48.76 KN/sqmAdd the UDL due to overlapping at certain area = 5.00 KN/sqm

53.76 KN/sqm

500

500

Dispersion of Loads

500745

450

1300 500

745

Design of 1V Box Barrel

1.545

0.45

Further as per IS 7784-Part 1---1993,the structure should be designed for the following forces :-

1545

SECTIONPLAN

6000

4500

300 300

300

300

ROAD PORTION

CA

NA

L P

OR

TIO

N

3100

300

600

1500

2500

2100

450

2.7t

2.7t

11.4

t

11.4

t

6.8t

6.8t

6.8t

6.8t

1.10

1.80

3.20 1.20 4.30 3.00 3.00 3.00

Y

X

7

3100

6000

4500-2750

-375-1200

5300-4250

is severe for bearing stresses on soil and the same is to be

2.7t

2.7t

11.4

t

11.4

t

6.8t

6.8t

6.8t

6.8t

1.10

1.80

3.20 1.20 4.30 3.00 3.00 3.00

1200

2750

b

u

Portion to be loaded withlive load of 5KN/sqm

Y

X

1800

0.45

Position of Live load forMax.Soil pressure

1525

3100

11

.4t

11

.4t

375

4500

1750

228.00KN

6.98KN

27.13KN

262.10KN

Moment

57.00KNm

57.00KNm

159.60KNm

159.60KNm

3.66KNm

61.71KNm

498.57KNm

Moment

155.04KNm

155.04KNm

86.64KNm

86.64KNm

10.81KNm

42.04KNm

536.22KN

Y

X

3100

6000

1098496

0.2475

3.12KN

18.00KN

3.38m

0.64m

3.08m

1.800m

9.72KN/sqm

118.30KN

31.68KN

3100

150015°

770

P

Ph

Pv

Earth Pressure on Box culvert

0.007

1.098

0.00

Eccentricty about y-axis(m)

3.38

3.08

0.77

Check for stability against over turning(Assuming that the earth fill on toe side is scoured):-

80.73Kn-m

91.26Kn-m

171.98Kn-m

1490.79Kn-m

413.00Kn-m

98.21Kn-m

2002.00Kn-m

(As per clause 706.3.4 of IRC:78-2000)

1385.32KN

162.51KN

0.80


Location(Ht.from the section considered).(m)

0.007

0.000

0.000

0.000

2.400

0.00

3.38

0.00

0.77

0.64

0.00Kn-m

91.26Kn-m

91.26Kn-m



1318.43Kn-m

Force due to static water pressure need not be considered,as it acts on both side walls in opposite directions

98.21Kn-m

1416.64Kn-m


882.28KN

118.30KN

0.80


0.007


0.00

1.098

0.000

3.38

3.08

1.07

51.718.59

10.62

4.34

0-8.40

-13.1953.67

composes of the following components


Stress at heelP/A(1+6e/b)

51.718.59

31.52

-0.94

08.4

13.19112.47

KN/Sqm>113KN/sqm

52.418.59-23.7

1.7

-3.270.00.0

35.73

Stress at toeP/A(1+6e/b)

Stress at U/S edgeP/A(1+6e/b)

51.018.5944.2

1.7

3.2700

108.77

KN/Sqm<113KN/sqm

0.007

0.00

Stress at D/S edgeP/A(1+6e/b)


1.098

0.000

0.000

0.000

2.400

0.000

3.38

3.08

1.07

0.64

45.73

8.59

10.62

composes of the following components


Stress at heelP/A(1+6e/b)

-8.43

-10.45

20.9

0.73

4.34

0

-8.40

-13.19

0

50.44

45.73

8.59

31.52

-8.43

-10.45

20.9

0.73

-0.94

0

8.4

13.19

0

109.24

KN/Sqm>113KN/sqm

Stress at toeP/A(1+6e/b)

46.35

8.59

-23.7

-8.43

-10.45

20.9

4.09

1.7

3.27

0.0

0.0

3.9

46.17

45.11

8.59

Stress at U/S edgeP/A(1+6e/b)

Stress at D/S edgeP/A(1+6e/b)

44.2

-8.43

-10.45

20.9

-2.64

1.7

3.27

0

0

-3.85

98.4

KN/Sqm<113KN/sqm

a)Calculation of equivalent Uniformly Distributed Load(UDL) for IRC live load for Road portion:-

The position of IRC class A live load for maximum bending moment in top slab in road portion is as given

1200

2750

b

u


Y

X

1800

0.45


2800

11.4

t

11

.4t

4500

1750

500

800 800

1.00m

450500

1200

2750

b

u


Y

X

1800

0.45


2800

11

.4t

11.4

t

4500

1750

500

800 800

500

500

Dispersion of Loads

500745

450

1300 500

745

1300500745

1990-5001490 745

500

500

Dispersion of Loads

500745

450

1300 500

745

b)Loading on the Box culvert:-

Effective span of the culvert = 2.80mEffective height of the culvert = 1.80m

Case-I---Drain is running full,with live load on span and approaches are intact

Load on Top slab :-Self weight of top slab = 7.50 KN/mSelf weight of wearing coat = 1.88 KN/mWeight of earth filling in Road portion = 27.81 KN/mAverage intensity of load due to IRC Class-A live load = 53.76 KN/m

90.95 KN/m

Load on bottom slab :-Upward bearing pressure on raft = 103.88 KN/mUpward bearing pressure due to weight of water = 19.30 KN/mDownward static pressure due static water pressure = -19.30 KN/mAdd uplift pressure due to water in Channel = 20.90

124.78 KN/m

Load on side walls from out side :-At top:-Earth pressure on side wall = 6.48 KN/mAt bottom:-Earth pressure on side wall = 16.20 KN/m

Load on side walls from in side :-At top:-Hydro static pressure from inside wall = 0.00 KN/mAt bottom:-Hydro static pressure from inside wall = 19.30 KN/m

The loading on slab culvert is shown below for the above case of loading

Loading on Box culvert

90.95 KN/sqm

144.08 KN/sqm19.30KN/sqm

6.48 KN/sqm6.48 KN/sqm


1800

2800

19.3 KN/sqm

A B

C D

c)Calculation of internal forces:-

i) Calculation of fixed end moments:-

Span AB:-

59.42 KN-m

-59.42 KN-m

Span BD:-

For Earth pressure from outside:-

For UDL part1.75 KN-m

-1.75 KN-m

For Triangular part1.57 KN-m

-1.05 KN-m

For Water pressure from inside:-

-3.13 KN-m

2.08 KN-m

Total :-

0.19 KN-m

-0.72 KN-m

Span CD:-

Net Upward pressure = 124.78 KN-m

-81.52 KN-m

81.52 KN-m

Span AC:-


For UDL part-1.75 KN-m

MFAB = wl2/12 =

MFBA = -wl2/12 =

MFBD = wl2/12 =

MFDB = -wl2/12 =

MFBD = wl2/20 =

MFDB = -wl2/30 =

MFBD = -wl2/20 =

MFDB = wl2/30 =

MFBD =

MFDB =

MFCD = -wl2/12 =

MFDC = wl2/12 =

MFAC = wl2/12 =


90.95 KN/sqm




1800

2800

19.3 KN/sqm

A B

C D

1.75 KN-m

For Triangular part-1.57 KN-m

1.05 KN-m


3.13 KN-m

-2.08 KN-m

Total :-

-0.19 KN-m

0.72 KN-m

ii) Moment distribution:-

A B C D

AC AB BA BD CA CD DC

0.391 0.609 0.609 0.391 0.609 0.391 0.391

-0.190 59.420 -59.420 0.190 0.720 -81.520 81.520

-23.159 -36.071 36.071 23.159 49.207 31.593 -31.593

24.604 18.036 -18.036 -24.604 -11.579 -15.796 15.796

-16.672 -25.967 25.967 16.672 16.672 10.704 -10.704

8.336 12.984 -12.984 -8.336 -8.336 -5.352 5.352

-8.336 -12.984 12.984 8.336 8.336 5.352 -5.352

4.168 6.492 -6.492 -4.168 -4.168 -2.676 2.676

-4.168 -6.492 6.492 4.168 4.168 2.676 -2.676

2.084 3.246 -3.246 -2.084 -2.084 -1.338 1.338

-2.084 -3.246 3.246 2.084 2.084 1.338 -1.338

1.042 1.623 -1.623 -1.042 -1.042 -0.669 0.669

-1.042 -1.623 1.623 1.042 1.042 0.669 -0.669

-15.417 15.417 -15.417 15.417 55.020 -55.020 55.020

MFCA = -wl2/12 =

MFAC = wl2/20 =

MFCA = -wl2/30 =

MFAC =

MFCA =

MFAC =

MFCA =

iii) Calculation of span moments for simply supported condition:-

Span AB:-

89.13 KN-m

Span BD:-

Due to earth pressure from out sideFor UDL:-

2.62 KN-m

For Triangular load:-

2.02 KN-m

Due to water pressure from in sideMax.moment for simply supported condition = -2.08 KN-m(Assuming that the max.moments occur at mid span(On safe side)

Max.moment at mid span 2.56 KN-m

Span CD:-

122.28 KN-m

Span AC:-


2.62 KN-m


2.02 KN-m

Due to water pressure from in sideMax.moment for simply supported condition = -2.08 KN-m(Assuming that the max.moments occur at mid span(On safe side)


iv) Calculation of Max.span moments for the box culvert:-

Span AB:-

The max.span moment = 89.13-(15.42+15.42)/2 = 73.71

Span BD:-

The max.span moment = 2.56-(15.42+55.02)/2 = -32.66

Max.moment for simply supported condition = wl2/8 =


Max.moment for simply supported condition = wl2/(9*30.5) =




There is no sagging moment

Span AC:-



Span CD:-


The bending moment diagram for the box culvert is as shown below:-

v) Calculation of shear force for the box culvert:-

Span AB:-

Shear force at A = (90.95*2.8/2)+((15.42-15.42)/2.8) =

Shear force at B = (90.95*2.8/2)+((15.42-15.42)/2.8) =

Span BD:-

A B

C D

15.42

15.42

15.4215.42

73.71

87.06

55.02

55.02

55.02

55.02

32.66 32.66

Shear force at B = (6.48*1.8/2)+(9.72*1.8)/6+(55.02-15.42)/1.8) =

Shear force at D =(6.48*1.8/2)+(9.72*1.8)/3+((55.02-15.42)/1.8) =

Span CD:-

Shear force at C = (124.78*2.8/2)+((55.02-55.02)/2.8) =

Shear force at D = (124.78*2.8/2)+((55.02-55.02)/2.8) =

Span AC:-

Shear force at A = (6.48*1.8/2)+(9.72*1.8)/6+(55.02-15.42)/1.8) =

Shear force at C =(6.48*1.8/2)+(9.72*1.8)/3+((55.02-15.42)/1.8) =

The Shear force diagram for the box culvert is as shown below:-

vi) Computation of axial forces for the box culvert:-

Top slab:-

127.33

127.33

30.7530.75

33.67 33.67

174.69

174.69

Axial force in top slab = Shear force at A or B for the spans BD&AC =

Bottom slab:-

Axial force in bottom slab = Shear force at C or D for the spans BD&AC =

Side walls:-

Axial force in side walls = Shear force at A or C for the spans AB&CD =

b)Loading on the Box culvert:-

Effective span of the culvert = 2.80mEffective height of the culvert = 1.80m

Case-I---Drain is running full,with live load on span and approaches are intact

Load on Top slab :-Self weight of top slab = 7.50 KN/mSelf weight of wearing coat = 1.88 KN/mWeight of earth filling in Road portion = 27.81 KN/m

Average intensity of load due to IRC Class-A live load = 53.76 KN/m90.95 KN/m

Load on bottom slab :-Upward bearing pressure on raft = 103.88 KN/mUpward bearing pressure due to weight of water = 0.00 KN/mDownward static pressure due static water pressure = 0.00 KN/mAdd uplift pressure due to water in Channel = 20.90

124.78 KN/m

Load on side walls from out side :-At top:-Earth pressure on side wall = 6.48 KN/mAt bottom:-Earth pressure on side wall = 16.20 KN/m

Load on side walls from in side :-At top:-Hydro static pressure from inside wall = 0.00 KN/mAt bottom:-Hydro static pressure from inside wall = 0.00 KN/m

The loading on slab culvert is shown below for the above case of loading

c)Calculation of internal forces:-

90.95 KN/sqm

124.78 KN/sqm



1800

2800

A B

C D

i) Calculation of fixed end moments:-

Span AB:-

59.42 KN-m

-59.42 KN-m

Span BD:-


For UDL part1.75 KN-m

-1.75 KN-m

For Triangular part1.57 KN-m

-1.05 KN-m


0.00 KN-m

0.00 KN-m

Total :-

3.32 KN-m

-2.80 KN-m

Span CD:-

Net Upward pressure = 124.78 KN-m

-81.52 KN-m

81.52 KN-m

Span AC:-


For UDL part-1.75 KN-m

1.75 KN-m

For Triangular part-1.57 KN-m

1.05 KN-m


0.00 KN-m

0.00 KN-m

MFAB = wl2/12 =

MFBA = -wl2/12 =

MFBD = wl2/12 =

MFDB = -wl2/12 =

MFBD = wl2/20 =

MFDB = -wl2/30 =

MFBD = -wl2/20 =

MFDB = wl2/30 =

MFBD =

MFDB =

MFCD = -wl2/12 =

MFDC = wl2/12 =

MFAC = wl2/12 =

MFCA = -wl2/12 =

MFAC = wl2/20 =

MFCA = -wl2/30 =

MFAC =

MFCA =

Total :-

-3.32 KN-m

2.80 KN-m

ii) Moment distribution:-

A B C D

AC AB BA BD CA CD DC

0.391 0.609 0.609 0.391 0.609 0.391 0.391

-3.320 59.420 -59.420 3.320 2.800 -81.520 81.520

-21.935 -34.165 34.165 21.935 47.940 30.780 -30.780

23.970 17.082 -17.082 -23.970 -10.968 -15.390 15.390

-16.052 -25.001 25.001 16.052 16.052 10.306 -10.306

8.026 12.501 -12.501 -8.026 -8.026 -5.153 5.153

-8.026 -12.501 12.501 8.026 8.026 5.153 -5.153

4.013 6.250 -6.250 -4.013 -4.013 -2.576 2.576

-4.013 -6.250 6.250 4.013 4.013 2.576 -2.576

2.006 3.125 -3.125 -2.006 -2.006 -1.288 1.288

-2.006 -3.125 3.125 2.006 2.006 1.288 -1.288

1.003 1.563 -1.563 -1.003 -1.003 -0.644 0.644

-1.003 -1.563 1.563 1.003 1.003 0.644 -0.644

-17.336 17.336 -17.336 17.336 55.825 -55.825 55.825

iii) Calculation of span moments for simply supported condition:-

Span AB:-

89.13 KN-m

Span BD:-


2.62 KN-m


2.02 KN-m

MFAC =

MFCA =




Due to water pressure from in sideMax.moment for simply supported condition = 0.00 KN-m(Assuming that the max.moments occur at mid span(On safe side)


Span CD:-

122.28 KN-m

Span AC:-


2.62 KN-m


2.02 KN-m

Due to water pressure from in sideMax.moment for simply supported condition = 0 KN-m(Assuming that the max.moments occur at mid span(On safe side)


iv) Calculation of Max.span moments for the box culvert:-

Span AB:-


Span BD:-



Span AC:-



Span CD:-


The bending moment diagram for the box culvert is as shown below:-




A B

C D

17.34

17.34

17.3417.34

71.79

85.70

55.83

55.83

55.02

55.83

31.94 31.94

v) Calculation of shear force for the box culvert:-

Span AB:-

Shear force at A = (90.95*2.8/2)+((17.34-17.34)/2.8) =

Shear force at B = (90.95*2.8/2)+((17.34-17.34)/2.8) =

Span BD:-

Shear force at B = (6.48*1.8/2)+(9.72*1.8)/6+(55.83-17.34)/1.8) =

Shear force at D =(6.48*1.8/2)+(9.72*1.8)/3+(55.83-17.34)/1.8) =

Span CD:-

Shear force at C = (124.78*2.8/2)+((55.83-55.83)/2.8) =

Shear force at D = (124.78*2.8/2)+((55.83-55.83)/2.8) =

Span AC:-

Shear force at A = (6.48*1.8/2)+(9.72*1.8)/6+(55.83-17.34)/1.8) =

A B

C D

17.34

17.34

17.3417.34

71.79

85.70

55.83

55.83

55.02

55.83

31.94 31.94

Shear force at C =(6.48*1.8/2)+(9.72*1.8)/3+(55.83-17.34)/1.8) =

The Shear force diagram for the box culvert is as shown below:-

vi) Computation of axial forces for the box culvert:-

Top slab:-

Axial force in top slab = Shear force at A or B for the spans BD&AC =

Bottom slab:-

Axial force in bottom slab = Shear force at C or D for the spans BD&AC =

Side walls:-

Axial force in side walls = Shear force at A or C for the spans AB&CD =

f)Design of Members:-

= =

Cover to reinforcement =

Compressive strength of concrete for Box (fck)

Yield strength of steel (fy)

127.33

127.33

30.1330.13

33.05 33.05

174.69

174.69

Limiting stress in steel for axial tension =

i) Design of top slab:-

Factored critical bending moment(Sagging) = 110.57KN-m(1.5*73.71)Factored critical bending moment(Hogging) 23.13KN-m(1.5*15.42)Factored axial force = 46.13KN(1.5*30.75)Factored shear force = 191.00KN(1.5*127.33)

Effective depth required d = 200.15mm

The over all depth required(Assuming 16mm dia bars) = 258.15mm

[200.15+50+8] =

However provide overall depth of = 300.00mm

Hence,effective depth = 242.00mm

[300-50-8]=

Bottom steel:-

1.888

From table 2 of SP 16,percentage of steel required = 0.641

Area of steel required for bending tension = 1551.2sqmm[0.641*242*1000/100]=

Area of steel required for axial force = 153.8sqmm

(46.13*1000/2*150) =

Total area of steel required = 1705.0sqmm

Hence provide 16mm dia bars at 110mm c/c at bottom,the area of reinforcement comes to

1826.9sqmm

Top steel:-

0.395

From table 2 of SP 16,percentage of steel required = 0.114(Minimum steel as per IS 456-2000 = 0.15


Mu/0.138fckb =

[110.57x106/(0.138*20*1000)]0.5

Mu/bd2 =

=(1000/110)*(3.14/4)*162 =

Mu/bd2 =


(46.13*1000/2*150) =


Hence provide 10mm dia bars at 125mm c/c at top,the area of reinforcement comes to

628.0sqmm

Distribution steel:-

Provide distribution reinforcement of 0.15% of cross sectional area of footing

Hence,the distribution reinforcement required = (0.15*300*1000/100) = 450.00sqmm

Adopting 10mm dia bars,the spacing required = 174.00mm

However provide 10mm dia bars at 150mm c/c spacing,as distribution reinforcement

Check for one way shear:-

The critical section for beam shear is at distance of 'd' from the face of the support

137.52KNat a distance D from the face of the support[191-(191/1400)*(150+242)] =

0.6N/sqmm <2.8 N/sqmm[137.52x1000/(1000*242)] = (As per Table 20 of 1S 456)

Hence,the section is safe from shear point of view

Assumed percentage area of the steel reinforcement = 0.75%[100*1826.9/(1000*242)] =

The design shear strength of concrete for the above steel percentage from Table 19 of IS 456 is

0.56 135.52KN

Considering the enhanced shear strength at supports as 1.5 times the shear strength,

0.84>0.6

Hence,no shear reinforcement is required

ii) Design of bottom slab:-

Factored critical bending moment(Sagging) = 130.59KN-m(1.5*87.06)

=(1000/125)*(3.14/4)*102 =

1000/[450/(3.14/4)*102]=

Hence,the factored design shear force VFd =

Nominal shear stress Tv =

Hence Vuc =

Factored critical bending moment(Hogging) 83.75KN-m(1.5*55.83)Factored axial force = 50.51KN(1.5*33.67)Factored shear force = 262.03KN(1.5*174.69)



[217.52+50+8] =



[300-50-8]=

Top steel:-

2.23




(50.51*1000/2*150) =


Hence provide 16mm dia bars at 100mm c/c at TOP,the area of reinforcement comes to

2009.6sqmm

Bottom steel:-

1.43




(50.51*1000/2*150) =


Hence provide 16mm dia bars at 150mm c/c at BOTTOM ,the area of reinforcement comes to

Mu/0.138fckb =

[130.59x106/(0.138*20*1000)]0.5

Mu/bd2 =

=(1000/100)*(3.14/4)*162 =

Mu/bd2 =

1339.7sqmm








188.67KNat a distance D from the face of the support262.04-(262.04/1400)*(150+242)] =





[0.56+{0.08*(0.62-0.56}/0.25] = 0.579 140.12KN

Considering the enhanced shear strength at supports as 1.5 times the shear strength,

0.87>0.80


iii) Design of side walls:-

Factored critical bending moment(Sagging) = 0.00KN-m

Factored critical bending moment(Hogging) 83.75KN-m(1.5*55.83)Factored axial force = 262.03KN(1.5*174.69)Factored shear force = 50.51KN(1.5*33.67)

=(1000/150)*(3.14/4)*162 =

1000/[450/(3.14/4)*102]=



Hence Vuc =



[174.19+50+8] =



[300-50-8]=

Steel on outer face :-

1.43




(262.04*1000/2*150) =


Hence provide 16mm dia bars at 100mm c/c on outer face,the area of reinforcement comes to

2009.6sqmm

Steel on inner face :-

Area of steel required for bending tension = 0.0sqmm


(262.04*1000/2*150) =


Hence provide 12mm dia bars at 125mm c/c on inner face,the area of reinforcement comes to

904.3sqmm





Mu/0.138fckb =

[83.75x106/(0.138*20*1000)]0.5

Mu/bd2 =

=(1000/100)*(3.14/4)*162 =

=(1000/125)*(3.14/4)*122 =

1000/[450/(3.14/4)*102]=




36.36KNat a distance 'd' from the face of the support50.51-(50.51/1400)*(150+242)] =





[0.56+{0.08*(0.62-0.56}/0.25] = 0.579 140.12KN

0.579>0.2




Hence Vuc =


90.95 KN/sqm




1800

2800

19.3 KN/sqm

A B

C D

0.3571 0.555556 0.912698 0.391304-1

D Joint

DB Member

0.609 Distribution Factor

-0.720 Fixed end moment

-49.207 Distribution

11.579 Carry over


8.336 Carry over

-8.336 Distribution

4.168 Carry over

-4.168 Distribution

2.084 Carry over

-2.084 Distribution

1.042 Carry over

-1.042 Distribution

-55.020 Final moments

KN-m

KN-m

KN-m

KN-m

127.33 KN

127.33 KN

A B

C D

15.42

15.42

15.4215.42

73.71

87.06

55.02

55.02

55.02

55.02

32.66 32.66

30.75 KN

33.67 KN

174.69 KN

174.69 KN

30.75 KN

33.67 KN

127.33

127.33

30.7530.75

33.67 33.67

174.69

174.69

30.75 KN

33.67 KN

174.69 KN

90.95 KN/sqm

124.78 KN/sqm



1800

2800

A B

C D

D Joint

DB Member

0.609 Distribution Factor

-2.800 Fixed end moment


10.968 Carry over


8.026 Carry over

-8.026 Distribution

4.013 Carry over

-4.013 Distribution

2.006 Carry over

-2.006 Distribution

1.003 Carry over

-1.003 Distribution

-55.825 Final moments

KN-m

KN-m

KN-m

KN-m

A B

C D

17.34

17.34

17.3417.34

71.79

85.70

55.83

55.83

55.02

55.83

31.94 31.94

127.33 KN

127.33 KN

30.13 KN

33.05 KN

174.69 KN

174.69 KN

30.13 KN

A B

C D

17.34

17.34

17.3417.34

71.79

85.70

55.83

55.83

55.02

55.83

31.94 31.94

33.05 KN

30.13 KN

33.05 KN

174.69 KN

20N/sqmm

415N/sqmm

50.00mm

127.33

127.33

30.1330.13

33.05 33.05

174.69

174.69

150N/sqmm

Hence provide 16mm dia bars at 110mm c/c at bottom,the area of reinforcement comes to

Hence provide 10mm dia bars at 125mm c/c at top,the area of reinforcement comes to



Hence provide 16mm dia bars at 100mm c/c at TOP,the area of reinforcement comes to

Hence provide 16mm dia bars at 150mm c/c at BOTTOM ,the area of reinforcement comes to

Hence provide 16mm dia bars at 100mm c/c on outer face,the area of reinforcement comes to

Hence provide 12mm dia bars at 125mm c/c on inner face,the area of reinforcement comes to

DESIGN OF CANTILEVER BREST WALL

Data:-

Height of wall above G.L=Height of wall below G.L=

Grade of concrete =Grade of steel =Ground water Table level =

(in clock wise direction)

(Assumed)Characteristic compressive strength =Tensile strength of steel =Unit weight of RCC =Unit weight of PCC =

Coefficient of active earth pressure by Coulomb's theory

Sin(a+Q)

sina sin(a-q) sin(Q+q)sin(Q-b)

sin(a+b)

Sin(a+Q) = SIN[3.14*(87.17+30)/180] = 0.89Sin(a-q) = SIN[3.14*(87.17-15)/180] = 0.952Sina = SIN[3.14*(87.17)/180] = 0.999Sin(Q+q) = SIN[3.14*(30+15)/180] = 0.707Sin(Q-b) = SIN[3.14*(30-0)/180] = 0.5Sin(a+b) = SIN[3.14*(87.17+0)/180] = 0.999

From the above expression,

0.32

Dimensions of the brest wall(Assumed for preliminary design):-

Thickness of stem at bottom =Thickness of stem at top =

Height of Retaining wall(h) =

Density of back fill soil&material in toe portion(y) =

Angle of shearing resistance of back fill material&material at toe portion(Q) = Angle of face of wall supporting earth with horizontal(a)(In degrees)

Slope of back fill(b) =Angle of wall friction (q) =Undrained Cohesion ( c) =Safe bearing capacity(SBC) =Surcharge over the back fill(s) =

Ka =

Ka =

1995

300

(0.32*1800*0.6) =

[0.32*1800*(1.995+0.6)] =

Pressure distribution is as shown below:-

345.6

1.995m

1149.1 345.6

Area of the rectangular portion = 345.6*1.995 = 689.47Area of the triangular portion = 0.5*1149.1*1.995= 1146.25

1835.72

Taking moments of the areas about the toe of the wall

S.No Description Area Lever arm Moment

1 Rectangular 689.47 0.9975 687.7462 Triangular 1146.25 0.665 762.256

Earth pressure at top including surcharge = Kays =

Earth pressure at bottom including surcharge = Kay(s+h) =

1995

300

1835.72 1450

Height from the bottom of the wall = 0.79m(1450/1835.72)

The active Earth pressure acts on the abutment as shown below:-

0.20

17.83

1.995m

0.79m

87.17

0.30m0.04

Total earth pressure acting on the wall per 1m length P = 2180.46Kg(345.6*1.995+0.5*1149.1*1.995)*1 =

Eccentricity of vertical component of earth pressure = (0.3/2-0.04) =

Total earth pressure =

It acts at a hieght offrom the base

Moment due to horizontal component of earth pressure =

Design of wall or stem:-

Factored bending moment Mu = 2459.51Kgm(1.5*1639.67) =


Over all depth provided = 300.00mm

Effective depth provided(Assuming 50mm cover) d = 242.00mm

0.42

Horizontal component of the earth pressure Ph =

(PCos20.150 )

Vertical component of the earth pressure Pv =

(PSin20.150 )

Mu/0.138fckb =

Mu/bd2 =

From table 2 of SP 16,percentage of steel required = 0.129(Minimum percentage of steel as per IS 456:2000) = 0.15

Area of steel required = 363.00sqmm

Hence provide 10mm dia HYSD bars@ 150mm c/c spacing

523.33sqmm

Check for shear:-


11.11KNat a distance 'd' from the face of the support





[0.28+0.07*(0.36-0.28)/0.10] = 0.336

0.336>0.02

Hence,no shear reinforcement is required.

Provide temperature re inforcement @ 0.15%

Area required = 375.00sqmm

Provide 1/3rd of above reinforcement on earthen side = 125.00sqmm

Provide 8mm dia @ 300mm c/c on earthen side

Provide 2/3rd of above reinforcement on other side = 250.00sqmm

Provide 8mm dia @ 200mm c/c on other side

Hence Ast provided =



DESIGN OF CANTILEVER BREST WALL

1.995m1.995m0.00m

1800Kg/CumM20

Fe415

30

87.17

015

0Kg/sqm7500Kg/sqm

0.60m

20N/sqmm415N/sqmm2500Kg/Cum2400Kg/Cum

2

0.30m0.20m

1995

300

345.6Kg/sqm

1494.7Kg/sqm

1995

300

2075.84Kg

667.32Kg

0.11m

2180.5Kg/m

0.79m

1639.67Kg/m

Hydraulic design

a)Hydraulic Particulars of drain:-

1.Maximum Flood Level 3.130

2.Ordinary Flood level 2.210

3.Lowest Bed level 1.200

4.Average bed slope 0.000250(Calculated from the profile of drain)

0.033(As per table 5 of IRC:SP 13)

6.Top of bund level 4.130

b)Hydraulic Particulars of field channel:-

7.Maximum Flood Level 3.210

8.Ordinary Flood level 2.910

9.Lowest Bed level 2.585

10.Average bed slope 0.000250(Calculated from the profile of drain)

0.033(As per table 5 of IRC:SP 13)

12.Top of bund level 3.510

c)Hydraulic Particulars of the proposed box barrel:-

13.Bed level 0.785

14.Bottom level of the top slab 2.285

15.Top level of the top slab 2.585 (Bed level of the field channel)

16.No.of vents provided 1.0

17.Breadth of the vent 2.50m

18.Depth of the vent 1.50m

19.Width of box barrel 6.000

Discharge Calculations:-

5.Rugosity Coefficient(n)

11.Rugosity Coefficient(n)

b

b+2h

Cross section of minor drain

h

LBL

MFL

1)Area Velocity method:-

Depth of flow w.r.t HFL = 1.930m(MFL-LBL)

1.40m

5.260m

6.43sqm1.93*(1.4+5.26)/2

6.86m

Hydraulic Radius R= Wetted area/Wetted perimeter = 0.94

0.46m/sec

Discharge Q = AXV = 2.96Cumecs

Discharge as per the record = 2.950Cumecs

Higher of the above two values is adopted for design of CD Work

Design Discharge = 2.960Cumecs

Design Velocity = 0.46m/sec

Ventway Calculations(H.F.L Condition):-

Assuming the minor drain to be truly alluvial,the regime width is equal to linear waterway required for the drain.

8.26m

But the actual top width is much less than the above regime width.Hence,the stream is no truly alluvial in nature.It is quasi-alluvial.As per IRC:SP--13,the ventway calculations for quasi-alluvial streams are as given below:-

Bed width(b) =

Assuming side slopes 1 :1 in clayey soils,top width at HFL (b+2h) =

Wetted Area(A) =

Wetted perimetre(P) =

1.4+2*[(1.4)2+(1.4)2]0.5

Velocity V = 1/nX(R2/3XS1/2) =

Hence,as per Lacey's silt theory,the regime width W = 4.8Q1/2 = 4.8*2.960.5 =

b

b+2h


h

LBL

MFL

The linear water way required = Actual width of the drain at HFL (W) = 5.260m

As per the clause 8.1.2.2 of IS 7784-Part1--1993,for the works with rigid floors,the waterway can be further flumed within the permissible limits of velocity for the particular type of floor.

The limiting velocity for the floor with concrete below M30 is = 4.00m/s

The area of cross section required = Q/A = 0.74sqm

Assuming 1V--box barrel, vent being 2.50mX1.50m,the area of cross-section provided = 3.75sqm

Hence,the flow velocity in the box barrel will be within the limits.

Calculation of Afflux at HFL condition:-

The area of un obstructed cross section A = 6.43sqm

The area of cross section provided a = 3.75sqm

0.650m

Linear water way of the Box barrel (L) = 2.500m

2.960Cumecs

For the ratio a/A = 0.58

0.87 h = afflux in m e = factor accounting for recovery of some velocity head as = 1

potential head v = average velocity in approach section

0.2233730292077

0.56

substituting h = (0.56/v)-0.65 and re-arranging the terms

solving for v,we get v = 0.69m/s

Hence,afflux h = 0.160m

Scour Depth Calculations:-

As per the clause 101.1.2 of IRC:5--1985,the design discharge should be increased by 30% to ensure adequate margin of safety for foundations and protection works

Depth of flow on down stream side Dd =

Adopting,orifice formula for discharge Q = C0 (2g)1/2 LDd[h+(1+e)v2/2g]1/2 =

where, C0 = Coefficient of discharge =

h+0.102v2 =

Further, as per discharge formula ,Q =W x (Dd+h) x v,hence v x (0.65+h) =

we get 0.56-0.87v+0.102v3 = 0

Hence,the discharge for design of foundations = (1.3*2.96)

Discharge per metre width of foundations = q =(3.94/2.5)

For quasi-alluvial streams,linear water way required is natural width of the channel at HFL . =

Linear water way provided =

Maximum scour depth Dm = 1.5XD =(For design of foundations,as per clause 110.1.4.2 of IRC:5--1985)

(1.5*2.85)

(4.28+1.2)

Bottom level of foundation = (3.13-5.48) =

Depth of foundation below low bed level = [1.2-(-2.35)] =

The Minimum Safe Bearing capacity of the soil is considered as 7.5 KN/m2

Box barrel of size 1v-2.00mx2.35m is proposed at a level of Hence,cut-off walls and aprons are required to protect the foundation&floor bed

Design of cut-off walls:-

a)From scour depth criteria:-

It is sufficient to take the D/S side cut-off upto 1.27 times the normal scour depth as per caluse 703.2.3.2 of IRC 78-1983 considering that the reach is straight.

1.27XNormal scour depth from HFL = 3.62m(1.27*2.85) =

Depth of bottom of Cut-off wall from HFL = 3.62m

Bottom level of cut-off wall = -0.49(3.13-3.62)=

Depth of bottom of D/S side Cut-off wall from LBL 1.69m[1.2-(-0.49)]=

1.30XDesign Discharge = Qdf =

Lacey's Silt factor for standard silt ' f ' = 1.76Xm1/2= = (Value taken from table 4 of SP13) =

Normal scour depth D = 1.34(q2/f)1/3 =

[1.34*(1.5762/1.0)0.33]

As per clause 12.1 of IRC SP:13,Increased scour depth due to contraction of linear water way D' = D(W/L)0.61 =

1.81(5.26/2.5)0.61 =

Depth of foundation = Dm + Max.of 1.2m or 1/3 Dm =

The value of B/d = 3.550

For the above value,1 = 0.07

The exit gradient for clayey soil = 0.25

Static head including afflux = 2.09

Hence,the depth of cut-off required = 0.59m

Design of LAUNCHING APRONS:-

Size of stones as per clause 5.3.7.2 of IRC 89-1985

0.020m

Say 0.23m

Weight of each stone considering the specific gravity of stone as 2.65

15.81Kg

Say 20Kg

Weight of each stone shall not be less than 20 Kg.

Dimensions of apron as per clause 5.3.5.2 of IRC 89-1985

0.095

Thickness of apron at inner edge(near raft) = 1.5 x T = 0.143m

Say 0.30m

Thickness of apron at outer edge = 2.25 x T = 0.214m

Say 0.60m

Hence provide a thickness of 0.30m at near end and 0.6m at the outer end

According to Khosla's theory for two dimensional flow,the exit gradient GE = (H/d) x 1

However, provide 2.40 depth cut-off wall

Diametre (d) = (Vmax/4.893)2 =

(0.92/4.893)2 =

Weight of stone = 4/3x x(d/2)3 x 2.65 x 1000 =

( T ) = 0.06 x (Qdf)1/3 =

0.06*(3.48)1/3 =

Hydraulic design

b

b+2h


h

LBL

MFL

But the actual top width is much less than the above regime width.Hence,the stream is no truly alluvial in nature.

b

b+2h


h

LBL

MFL

As per the clause 8.1.2.2 of IS 7784-Part1--1993,for the works with rigid floors,the waterway can be further flumed

0.003385

-0.006792

As per the clause 101.1.2 of IRC:5--1985,the design discharge should be increased by 30% to ensure adequate

3.94Cumecs

1.00

1.576

1.81m

5.26m

2.50m

2.85m

4.28m

5.48m

-2.35m

3.55m

0.485m

115325841 design of syphon aqueduct

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