Design Of Syphon Aqueduct 3zw4t

DESIGN OF TWO VENT BOX BARREL FOR SYPHON AQUEDUCT Name of the work:-Elamanchili minor drain under Medpadu field channel @ 0/750 Km

F TWO VENT BOX BARREL FOR SYPHON AQUEDUCT

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

Note on site conditions As per the record the hydraulic particulars of Chilla minor drain&Field channel at proposed site are as given below:-

Hydraulic Particulars Units

Chilla Minor drain

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

0.000250

0.000250

Bed fall

----

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

3.425

of Chilla minor drain is above

of the channel is not altered.

ertical drop on upstream side

along with revetment for

er the requirement.Cut-off

d for field channel also

ombinations is carried out as

ented structure with all

-2.1 1.325 2.065 -1.325

Design of 1V Box Barrel I)Design Parameters: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) Haunch width Haunch depth Thickness of wearing coat (t4)

= = = = = = = = = = =

6.00m 2.50m 1.50m 0.30m 0.30m 0.30m 3.10m 2.10m 0.15 0.15 0.075m

Height of railing (h2)

=

1.200m

Unit weight of RCC (yrc)

=

25KN/cum

Unit weight of PCC (ypc)

=

24KN/cum

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

=

18KN/Cum

Unit weight of water (yw)

=

10KN/Cum

Angle of shearing resistance of back fill material(Q)

=

30

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

=

90

Slope of back fill (b)

=

0

Angle of wall friction (q)

=

15

Height of surcharge considered (h3)

=

1.20m

Barrel top level (BTL)

=

2.585m

Drain bed level (DBL)

=

1.200m

Drain High flood Level (HFL) Depressed floor level of barrel

= =

3.130m 0.785m

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

= = = =

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

Safe Bearing Capacity of the soil (SBC)

=

7.50t/sqm

Compressive strength of concrete for Box (fck)

=

20.00N/sqmm

Yield strength of steel (fy) Cover to reinforcement

=

415.00N/sqmm

=

50.00mm

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

300

450

PLAN

1545

300

ROAD PORTION

2500 2100

1500

4500

300

CANAL PORTION

SECTION

300 6000 600

II)General loading pattern:-

As per IRC:6---2000,the following loadings are to be considered on the box barrel in road portion:1.Dead load 2.Live load 3.Impact load 4.Wind load 5.Water current 6.Tractive,braking effort of vehicles&frictional resistance of bearings 7.Buoyancy 8.Earth pressure 9.Seismic force 10.Water pressure force 11.Static water pressure due to water in canal As per clause 202.3,the increase in permissible stresses is not permissible for the above 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 water 13.Uplift pressure due to subsurface flow

III)Loading on the box culvert :1.Dead Load:-

300

3100

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

Distance of centroid from U/S end

Moment about upstream end in KN-m

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

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:-

6000

X

7 3100

Y 2.Live Load:As per clause 201.1 of IRC:6--2000,the bridges and culverts of medium importance are to be designed for IRC Class A loading. GENERAL IRC Class-A loading Pattern

Position of Live load for Max.Soil pressure 3100

Portion to be loaded with

6.8t

X

The IRC Class A loading as per the drawing is severe for bearing stresses on soil and the same is to be considered as per clauses 207.1.3&207.4

6.8t

3.00

3.00

6.8t

11.4t

3.00

6.8t

4.30

1.20

11.4t

3.20

2.7t

1.80

2.7t

1.10

X

Position of Live load for Max.Soil pressure 3100

Portion to be loaded with live load of 5KN/sqm

1750

4500 1525

11.4t

11.4t

1200 1800

2750

Y

u 0.45

b 375

The ground area of wheels for the above placement,each axle wise is given below:Axle load (Tonnes) 11.4 6.8 2.7

Ground Area B(mm)

250 200 150

W(mm)

500 380 200

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

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

0.45m

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

1.75m 2.20m

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

Wheel Load/UDL in KN

1

57

2

57

3

Distance from Y-axis

(0.3+0.45 +0.5/2)= (0.3+0.45 +0.5/2)

1.00m

57

(0.3+0.45 +0.5/2+1.8)

2.80m

4

57

2.80m

5

6.975

(0.3+0.45 +0.5/2+1.8) (0.3+0.45/2)=

6

27.125

1.00m

(0.3+0.45/2 +1.75) =

0.525m 2.28m

262.100 Distance of centroid of forces from y-axis (498.57/262.1) = = 1.902m Eccentricity = (1.902-6/2)=

1.098m

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 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

[3.1(0.38+1.2)] =

1.52m

4

57

[3.1(0.38+1.2)] =

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 = (2.046-3.1/2)=

0.496m

Location of the resultant of live load is as shown in the figure given below:6000

X

3100 496 1098

Y 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 [4.5/(6+3.1)]=

0.495

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) [4.58-(1.20)]=

3.38m

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 Kg/m2. Height of the deck system =

3.380

Breadth of the deck system =(3.1) =

3.10

The effective area exposed to wind force =HeightxBreadth = 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 2

P = 52KV = 2 (52*1.5*0.69 )=

37.14 Kg/m2.

(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 = [37.14*1.0*(2.585-(1.20)*10/1000] =

0.51KN

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

10.478

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 s of simply ed 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 = (11.12*10)=

111.20KN

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

'2 Ka = sina

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

sin(a-q)

sin(a+b)

Sin(a+Q) = Sin(a-q) = Sina = Sin(Q+q) = Sin(Q-b) = Sin(a+b) =

SIN[3.14*(90+30)/180] = SIN[3.14*(90-15)/180] = SIN[3.14*(90)/180] = SIN[3.14*(30+15)/180] = SIN[3.14*(30-0)/180] = SIN[3.14*(90+0)/180] =

0.867 0.966 1 0.707 0.5 1

From the above expression, Ka =

0.3

The hieght of box above GL,as per the preliminary section assumed = Hence,maximum pressure at the base of the wall (0.3*1800*1.80)*10/1000 =

Pa =

The pressure distribution along the height of the wall is as given below:Surcharge load = (0.3*1800*1.2)*10/1000=

6.48 KN/sqm

6.48

1.800

9.72

6.48

Area of the rectangular portion = Area of the triangular portion =

6.48*1.80 = 0.5*9.72*1.8 =

11.66 8.75 20.41

Taking moments of the areas about the toe of the wall S.No 1 2

Description

Area

Rectangular Triangular

11.66 8.75

Lever arm Moment 0.9 0.6

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

10.494 5.25 15.744

0.77m

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

Earth Pressure on Box culvert 3100

Pv

P 15°

1500

Ph 770

Inclination of earth pressure force with horizontal

15.00 Deg.

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

122.47KN

Horizontal component of the earth pressure Ph = (PCos150 ) Vertical component of the earth pressure Pv = (PSin150 ) 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 = The above pressure acts at height of H/3 =1.93/3 = Static water pressure force (Vertical) = (1.93+0.16)*10 =

111.75KN 0.64m 20.90kn/sqm

11.Static water pressure due to water in canal Height of water in channel above bed level = The static water pressure on top slab of box barrel = Total load due to static water pressure = Eccentricity of the above load in x-direction =

0.725m 7.25kn/sqm 13.49KN 2.40m

12.Uplift pressure due to flowing water:Level of bottom of the top slab of box barrel = MFL of the drain =

+2.29m +3.130m

Afflux =

0.160m

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

1.005m

Kinetic head = v2/2g =

0.040m

Total head caulift on the top slab of box barrel =

1.045m

Hence uplift pressure on the top slab = Total uplift force =

10.45kn/sqm 156.75KN

13.Uplift pressure due to subsurface flow Afflux =

0.160m

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

2.090

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

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 Vertical load acting on the Box barrel (P) composes of the following components S.No

Type of load

1

Self wieght of Box culvert

2

Live load with impact factor---(Wheel loads+UDL) [262.1*(1+0.495)] = Vertical component of Active Earth pressure

3

Intensity in KN

Eccentricty about xaxis(m)

961.80KN

0.000

391.84KN

-0.496

31.68KN

1.550

1385.32KN

Horizontal load acting/transferred on the box (H) composes of the following components S.No

Type of load

Intensity in KN

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 = (2002/171.98)=

11.6407156

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

Check for stability against sliding:Total vertical load acting on the base of the box culvert Vb = Total sliding force,ie,horizontal load on the box Hb = Coefficient of friction between concrete surfaces = Factor of safety against sliding Fs = (0.8*1385.32/162.51)=

6.81957361 > 1.5 Hence safe (As per clause 706.3.4 of IRC:78-2000)

b)Load Envelope-II:-(The Canal is running upto HFL with no 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 Vertical load acting on the box (P) composes of the following components S.No

1

Type of load

Intensity in KN


Self wieght of box

961.80KN

Reduction in self weight due to buoyancy

-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

13.49KN

0.000

7

Vertical component of Active Earth pressure

31.68

1.550

Horizontal load acting/transferred on the box (H) composes of the following components S.No

Type of load

Intensity in KN

Direction x or y

1

Wind load

18.00KN

x-Direction

2


0.00KN

y-Direction

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 = 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

Check for stability against sliding:Total vertical load acting on the base of the boxVb = (850.6+31.68) =

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

Total sliding force,ie,horizontal load on the box Hb = Coefficient of friction between concrete surfaces = Factor of safety against sliding Fs = (0.8*882.28/118.3)=

5.96634219 > 1.5 Hence safe (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

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 Vertical load acting on the Box (P) composes of the following components S.No

Type of load

Intensity in KN


1

Self weight of box

961.80KN

0.000

2

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

159.84KN

0.00

3

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

4

Vertical component of earth pressure

391.84KN 31.68KN

-0.496 1.550

Horizontal load acting/transferred on the abutment (H) composes of the following components S.No

Type of load

Intensity in KN

Direction x or y

1

Wind load

18.00KN

x-Direction

2


26.21KN

y-Direction

3

Horizontal load due to earth pressure Safe bearing capacity SBC of the soil =

Check for stresses:About x-axis:-

118.30KN 7.50t/sqm

y-Direction

Breadth of footing b =

6.00m

Depth of footing d =

3.10m 18.6 m2

Area of the footing = A = Section modulus of bottom footing about x-axis --Zx =

(1/6)bd2 =

9.61 m3

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

1 2 3 4 1 2 3

Type of load

Vertical loads:-(Stress = P/A(1+6e/b) Self wieght of Box Self weight of levelling concrete Live load with impact factor---(Wheel loads+UDL) [262.1*(1+0.495)] = Vertical component of Earth pressure Horizontal loads:- (Stress = M/Z) Wind load Tractive,Braking&Frictional resistance of bearings Horizontal load due to earth pressure

S.No

1 2 3 4 1 2 3

Intensity in KN (P)

Type of load

Stress at heel =

961.80KN 159.84KN 391.84KN

0.00 0.00 -0.496

31.68KN

1.55

18.00KN 26.21KN 118.30KN

0.00 3.08 1.07

Intensity in KN (P)


P/A(1+6e/b)+M/Z =

Eccentricity/Lever arm

Eccentricity

961.80KN 159.84KN 391.84KN

0.00 0.00 0.496

31.68KN

-1.55

18.00KN 26.21KN 118.30KN

0.00 3.08 1.07

53.67 KN/Sqm>0

Hence safe. Stress at toe =

P/A(1+6e/b)+M/Z =

Hence safe.

About y-axis:-

112.47 KN/Sqm>113KN/sqm

Breadth of footing b = Depth of footing d = Area of the footing = A =

3.10m 6.00m 18.6 m2

Section modulus of bottom footing about y-axis --Zy =

(1/6)bd2 =

18.60 m3


113KN/sqm


1 2 3 4 1 2 3

S.No

1 2 3 4 1 2 3

Type of load

Intensity in KN (P)


Type of load

Stress at up stream side edge of abutment =

961.80KN 159.84KN 391.84KN

0.007 0.00 -1.098

31.68KN

0.00

18.00KN 26.21KN 118.30KN

3.38 0.00 0.00

Intensity in KN (P)


P/A(1+6e/b)+M/Z =


Eccentricity

961.80KN 159.84KN 391.84KN

-0.007 0.00 1.098

31.68KN

0.00

18.00KN 26.21KN 118.30KN

3.38 0.00 0.00

35.73 KN/Sqm>0

Hence safe. Stress at down stream side edge of abutment =

P/A(1+6e/b)+M/Z =

Hence safe.

108.77 KN/Sqm<113KN/sqm

Net bearing pressure for design = (72.97-(8+8.08))=

52.87 KN/Sqm

b)Load Envelope-III:-(The Canal is running upto HFL with live load on span) i)At the bottom of RCC raft slab

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 Vertical load acting on the Box (P) composes of the following components S.No

Type of load

Intensity in KN


1

Self weight of box

850.60KN

0.000

2


159.84KN

0.00

3


4

391.84KN

-0.496


-156.75KN

0.000

5


-194.37KN

0.000

6


388.74KN

0.000

7


13.49KN

0.000

8

Vertical component of earth pressure

31.68KN

1.550

Horizontal load acting/transferred on the abutment (H) composes of the following components S.No

Type of load

Intensity in KN

Direction x or y

1

Wind load

18.00KN

x-Direction

2


26.21KN

y-Direction

3

Horizontal load due to earth pressure

118.30KN

y-Direction

4


111.75KN

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 = Area of the footing = A = Section modulus of bottom footing about x-axis --Zx =

3.10m 18.6 m2 2

9.61 m3

(1/6)bd =


113KN/sqm


1

Type of load

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

Intensity in KN (P)


850.60KN

0.00

2


159.84KN

0.00

3

391.84KN

-0.496

4

Live load with impact factor---(Wheel loads+UDL) [301.5*(1+0.413)] = Uplift force due to flowing water

-156.75KN

0.000

5


-194.37KN

0.000

6


388.74KN

0.000

7


13.49KN

0.000

8

31.68KN

1.55

1

Vertical component of Earth pressure Horizontal loads:- (Stress = M/Z) Wind load

18.00KN

0.00

2


26.21KN

3.08

3


118.30KN

1.07

4


111.75KN

0.00

S.No

1

Type of load

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

Intensity in KN (P)

Eccentricity

850.60KN

0.00

2


159.84KN

0.00

3


391.84KN

0.496

4


-156.75KN

0.000

5


-194.37KN

0.000

6


388.74KN

0.000

7


13.49KN

0.000

8

Vertical component of Earth pressure

31.68KN

-1.55

1

Horizontal loads:- (Stress = M/Z) Wind load

18.00KN

0.00

2


26.21KN

3.08

3


118.30KN

1.07

4


111.75KN

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

Hence safe.

About y-axis:Breadth of footing b = Depth of footing d = Area of the footing = A = Section modulus of bottom footing about y-axis --Zy =

3.10m 6.00m 18.6 m2 (1/6)bd2 =

18.60 m3


113KN/sqm


1 2 3

Type of load

Vertical loads:-(Stress = P/A(1+6e/b) Self wieght of Box Self weight of levelling concrete =(5.5*12*0.30*24) Live load with impact factor---(Wheel loads+UDL) [301.5*(1+0.413)] =

Intensity in KN (P)


850.60KN

0.01

159.84KN

0.00

391.84KN

-1.098

4


-156.75KN

0.00

5


-194.37KN

0.00

6


388.74KN

0.00

7


13.49KN

2.40

8


31.68KN

0.00

1

Horizontal loads:- (Stress = M/Z) Wind load

18.00KN

3.38

2


26.21KN

0.00

3


118.30KN

0.00

4


111.75KN

0.64

S.No

Type of load

Intensity in KN (P)

Eccentricity

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

Self wieght of Box

850.60KN

-0.01

2


159.84KN

0.00

3


391.84KN

1.098

4


-156.75KN

0.000

5


-194.37KN

0.000

6


388.74KN

0.000

7


13.49KN

-2.400

8


31.68KN

0.00

Horizontal loads:- (Stress = M/Z) 1

Wind load

18.00KN

3.38

2


26.21KN

0.00

3


118.30KN

0.00

4


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 = (112.47-(8.59))=

103.88 KN/Sqm

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:-

X

Position of Live load for Max.Bending moment in each span 2800

Portion to be loaded with live load of 5KN/sqm

1750

4500

11.4t

800

11.4t

1200

800

2750

u

500

Y

1800

0.45

b

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, The effective width be = kx[1-x/l]+bw where be = The effective width of slab on which the load acts L = Effective span x = Distance of centre of gravity of load from nearer bw = Breadth of concentration area of load 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 = x = bw = B/L = k =

2.80m 0.80m 0.50m 1.89 2.6

Hence be =

1.99m

Dispersion of Loads 500

500

1300

745

500

500

745

450 From the above figure net effective width of dispersion of loads =

3.495m

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

3.495sqm

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

170.43KN

Average intensity of load = Add the UDL due to overlapping at certain area =

48.76 KN/sqm 5.00 KN/sqm 53.76 KN/sqm

1.545 0.45

CANAL PORTION

N

300

0

or the following forces :-

600

300

3100

3100

Y

6.8t

6.8t

X

ses on soil and the same is to be

6.8t

3.00

3.00

X 1750

2750

4500 -2750 -375 -1200 5300 -4250

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

3100

Y

0.2475

3.12KN

18.00KN

3.38m

0.64m

3.08m

1.800m 9.72KN/sqm

ox culvert

1500

770

118.30KN

31.68KN

1.55m

Eccentricty about yaxis(m) 0.007 1.098 0.00

Location(Ht.from the section considered).(m) 3.38 3.08 0.77

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

e 706.3.4 of IRC:78-2000)

1385.32KN 162.51KN 0.80

e 706.3.4 of IRC:78-2000)

Eccentricty about yaxis(m)

0.007 0.000

0.000 0.000 2.400 0.00

Location(Ht.from the section considered).(m) 3.38 0.00 0.77 0.64

0.00Kn-m 91.26Kn-m

91.26Kn-m

1318.43Kn-m

both side walls in opposite directions

98.21Kn-m

1416.64Kn-m

e 706.3.4 of IRC:78-2000)

882.28KN

118.30KN 0.80

e 706.3.4 of IRC:78-2000)

Eccentricty about yaxis(m) 0.007 0.00 1.098 0.000

e following components Location(Ht.from the section considered).(m) 3.38 3.08 1.07

Stress at heel P/A(1+6e/b)

51.71 8.59 10.62 4.34 0 -8.40 -13.19 53.67

Stress at toe P/A(1+6e/b)

51.71 8.59 31.52 -0.94 0 8.4 13.19 112.47

KN/Sqm>113KN/sqm

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

52.41 8.59 -23.7 1.7 -3.27 0.0 0.0 35.73

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

51.01 8.59 44.2 1.7 3.27 0 0 108.77

KN/Sqm<113KN/sqm

Eccentricty about yaxis(m) 0.007 0.00 1.098 0.000 0.000 0.000 2.400 0.000

e following components Location(Ht.from the section considered).(m) 3.38 3.08 1.07 0.64

Stress at heel P/A(1+6e/b)

45.73 8.59 10.62 -8.43 -10.45 20.9 0.73 4.34 0 -8.40 -13.19 0 50.44

Stress at toe P/A(1+6e/b)

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 U/S edge P/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

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

45.11 8.59 44.2 -8.43 -10.45 20.9 -2.64 1.7

3.27 0 0 -3.85 98.4

KN/Sqm<113KN/sqm

live load for Road portion:-

X

op slab in road portion is as given

1750

2750

1.00m

450 500 1300 500 745

5

1990 -500 1490

745

b)Loading on the Box culvert:Effective span of the culvert = Effective height of the culvert =

2.80m 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 = Self weight of wearing coat = Weight of earth filling in Road portion = Average intensity of load due to IRC Class-A live load =

Load on bottom slab :Upward bearing pressure on raft = Upward bearing pressure due to weight of water = Downward static pressure due static water pressure = Add uplift pressure due to water in Channel =

Load on side walls from out side :At top:Earth pressure on side wall = At bottom:Earth pressure on side wall =

7.50 1.88 27.81 53.76 90.95

KN/m KN/m KN/m KN/m KN/m

103.88 19.30 -19.30 20.90 124.78

KN/m KN/m KN/m KN/m

6.48 KN/m 16.20 KN/m

Load on side walls from in side :At top:Hydro static pressure from inside wall = At bottom:Hydro static pressure from inside wall =

0.00 KN/m 19.30 KN/m

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

Loading on Box culvert 2800 90.95 KN/sqm 6.48 KN/sqm

6.48 KN/sqm

B

A

1800 19.3 KN/sqm

1800 19.3 KN/sqm

D

C 16.20 KN/sqm

16.20 KN/sqm

19.30KN/sqm

144.08 KN/sqm

c)Calculation of internal forces:i) Calculation of fixed end moments:Span AB:MFAB = wl2/12 = 2

MFBA = -wl /12 =

59.42 KN-m -59.42 KN-m

Span BD:For Earth pressure from outside:For UDL part MFBD = wl2/12 = 2

MFDB = -wl /12 = For Triangular part MFBD = wl2/20 = 2

MFDB = -wl /30 =

1.75 KN-m -1.75 KN-m

1.57 KN-m -1.05 KN-m

For Water pressure from inside:MFBD = -wl2/20 =

-3.13 KN-m

MFDB = wl2/30 =

2.08 KN-m

Total :MFBD =

0.19 KN-m

MFDB =

-0.72 KN-m

Span CD:Net Upward pressure =

Span AC:-

124.78 KN-m

MFCD = -wl2/12 =

-81.52 KN-m

MFDC = wl2/12 =

81.52 KN-m

For Earth pressure from outside:For UDL part MFAC = wl2/12 = MFCA = -wl2/12 = For Triangular part MFAC = wl2/20 = 2

MFCA = -wl /30 =

-1.75 KN-m 1.75 KN-m

-1.57 KN-m 1.05 KN-m

For Water pressure from inside:MFAC =

3.13 KN-m

MFCA =

-2.08 KN-m

Total :MFAC =

-0.19 KN-m

MFCA =

0.72 KN-m

ii) Moment distribution:-

A

B

C

D

AC

AB

BA

BD

CA

CD

DC

DB

0.391

0.609

0.609

0.391

0.609

0.391

0.391

0.609

-0.190

59.420

-59.420

0.190

0.720

-81.520

81.520

-0.720

-23.159

-36.071

36.071

23.159

49.207

31.593

-31.593

-49.207

24.604

18.036

-18.036

-24.604

-11.579

-15.796

15.796

11.579

-16.672

-25.967

25.967

16.672

16.672

10.704

-10.704

-16.672

8.336

12.984

-12.984

-8.336

-8.336

-5.352

5.352

8.336

-8.336

-12.984

12.984

8.336

8.336

5.352

-5.352

-8.336

4.168

6.492

-6.492

-4.168

-4.168

-2.676

2.676

4.168

-4.168

-6.492

6.492

4.168

4.168

2.676

-2.676

-4.168

2.084

3.246

-3.246

-2.084

-2.084

-1.338

1.338

2.084

-2.084

-3.246

3.246

2.084

2.084

1.338

-1.338

-2.084

1.042

1.623

-1.623

-1.042

-1.042

-0.669

0.669

1.042

-1.042

-1.623

1.623

1.042

1.042

0.669

-0.669

-1.042

-15.417

15.417

-15.417

15.417

55.020

-55.020

55.020

-55.020

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

Span AB:Max.moment for simply ed condition = wl2/8 =

89.13 KN-m

Span BD:Due to earth pressure from out side For UDL:Max.moment for simply ed condition = wl2/8 =

2.62 KN-m

For Triangular load:Max.moment for simply ed condition = wl2/(9*30.5) =

2.02 KN-m

Due to water pressure from in side Max.moment for simply ed condition = (Assuming that the max.moments occur at mid span(On safe side) Max.moment at mid span

-2.08 KN-m

2.56 KN-m

Span CD:Max.moment for simply ed condition = wl2/8 =

122.28 KN-m

Span AC:Due to earth pressure from out side For UDL:Max.moment for simply ed condition = wl2/8 =

2.62 KN-m


2.02 KN-m


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

-2.08 KN-m

2.56 KN-m

Span AB:The max.span moment = 89.13-(15.42+15.42)/2 =

73.71 KN-m

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

-32.66 KN-m

Span BD:-

There is no sagging moment

Span AC:The max.span moment = 2.56-(15.42+55.02)/2 =

-32.66 KN-m

There is no sagging moment Span CD:The max.span moment = 122.28-(55.02+55.02)/2 =

87.06 KN-m

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

15.42 15.42

15.42

A

15.42

B 73.71

32.66

32.66 87.06

C 55.02 55.02

D 55.02 55.02

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) =

127.33

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

127.33

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

30.75

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

33.67

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

174.69

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

174.69

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

30.75

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

33.67

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

127.33 30.75

30.75 127.33

174.69 33.67

33.67 174.69

174.69 33.67

33.67 174.69

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 =

30.75

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

33.67

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

174.69

b)Loading on the Box culvert:Effective span of the culvert = Effective height of the culvert =

2.80m 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 = Self weight of wearing coat = Weight of earth filling in Road portion = Average intensity of load due to IRC Class-A live load =

Load on bottom slab :Upward bearing pressure on raft = Upward bearing pressure due to weight of water = Downward static pressure due static water pressure = Add uplift pressure due to water in Channel =

Load on side walls from out side :At top:Earth pressure on side wall = At bottom:Earth pressure on side wall =

7.50 1.88 27.81 53.76 90.95

KN/m KN/m KN/m KN/m KN/m

103.88 0.00 0.00 20.90 124.78

KN/m KN/m KN/m KN/m

6.48 KN/m 16.20 KN/m

Load on side walls from in side :At top:Hydro static pressure from inside wall = At bottom:Hydro static pressure from inside wall =

0.00 KN/m 0.00 KN/m

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

2800 90.95 KN/sqm 6.48 KN/sqm

6.48 KN/sqm

A

B

1800

90.95 KN/sqm 6.48 KN/sqm

6.48 KN/sqm

A

B

1800

D

C 16.20 KN/sqm

16.20 KN/sqm

124.78 KN/sqm

c)Calculation of internal forces:i) Calculation of fixed end moments:Span AB:MFAB = wl2/12 =

59.42 KN-m

MFBA = -wl2/12 =

-59.42 KN-m

Span BD:For Earth pressure from outside:For UDL part MFBD = wl2/12 = MFDB = -wl2/12 = For Triangular part MFBD = wl2/20 = 2

MFDB = -wl /30 =

1.75 KN-m -1.75 KN-m

1.57 KN-m -1.05 KN-m

For Water pressure from inside:MFBD = -wl2/20 = 2

Span CD:-

0.00 KN-m

MFDB = wl /30 =

0.00 KN-m

Total :MFBD =

3.32 KN-m

MFDB =

-2.80 KN-m

Net Upward pressure =

124.78 KN-m

MFCD = -wl2/12 =

-81.52 KN-m

MFDC = wl2/12 =

81.52 KN-m

Span AC:For Earth pressure from outside:For UDL part MFAC = wl2/12 = MFCA = -wl2/12 = For Triangular part MFAC = wl2/20 = 2

MFCA = -wl /30 =

-1.75 KN-m 1.75 KN-m

-1.57 KN-m 1.05 KN-m

For Water pressure from inside:MFAC =

0.00 KN-m

MFCA =

0.00 KN-m

Total :MFAC =

-3.32 KN-m

MFCA =

2.80 KN-m

ii) Moment distribution:-

A

B

C

D

AC

AB

BA

BD

CA

CD

DC

DB

0.391

0.609

0.609

0.391

0.609

0.391

0.391

0.609

-3.320

59.420

-59.420

3.320

2.800

-81.520

81.520

-2.800

-21.935

-34.165

34.165

21.935

47.940

30.780

-30.780

-47.940

23.970

17.082

-17.082

-23.970

-10.968

-15.390

15.390

10.968

-16.052

-25.001

25.001

16.052

16.052

10.306

-10.306

-16.052

8.026

12.501

-12.501

-8.026

-8.026

-5.153

5.153

8.026

-8.026

-12.501

12.501

8.026

8.026

5.153

-5.153

-8.026

4.013

6.250

-6.250

-4.013

-4.013

-2.576

2.576

4.013

-4.013

-6.250

6.250

4.013

4.013

2.576

-2.576

-4.013

2.006

3.125

-3.125

-2.006

-2.006

-1.288

1.288

2.006

-2.006

-3.125

3.125

2.006

2.006

1.288

-1.288

-2.006

1.003

1.563

-1.563

-1.003

-1.003

-0.644

0.644

1.003

-1.003

-1.563

1.563

1.003

1.003

0.644

-0.644

-1.003

-17.336

17.336

-17.336

17.336

55.825

-55.825

55.825

-55.825

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

Span AB:Max.moment for simply ed condition = wl2/8 =

89.13 KN-m

Span BD:Due to earth pressure from out side For UDL:Max.moment for simply ed condition = wl2/8 =

2.62 KN-m


2.02 KN-m


0.00 KN-m

4.64 KN-m

Span CD:Max.moment for simply ed condition = wl2/8 =

122.28 KN-m

Span AC:Due to earth pressure from out side For UDL:Max.moment for simply ed condition = wl2/8 =

2.62 KN-m


2.02 KN-m


0 KN-m

4.64 KN-m

iv) Calculation of Max.span moments for the box culvert:Span AB:The max.span moment = 89.13-(17.34+17.34)/2 =

71.79 KN-m

The max.span moment = 4.64-(17.34+55.83)/2 =

-31.94 KN-m

Span BD:-

There is no sagging moment

Span AC:The max.span moment = 4.64-(17.34+55.83)/2 =

-31.94 KN-m

There is no sagging moment Span CD:The max.span moment = 122.28-(55.83+55.83)/2 =

85.7 KN-m

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

17.34 17.34

17.34

A

17.34

B 71.79

31.94

31.94 85.70

C 55.02 55.83

D 55.83 55.83

55.02

55.83

55.83

55.83

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) =

127.33

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

127.33

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

30.13

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

33.05

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

174.69

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

174.69

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

30.13

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

33.05

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

127.33 30.13

30.13 127.33

174.69 33.05

33.05

174.69 33.05

33.05 174.69

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 =

30.13

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

33.05

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

174.69

f)Design of :Compressive strength of concrete for Box (fck)

= =

Yield strength of steel (fy) Cover to reinforcement Limiting stress in steel for axial tension

= =

i) Design of top slab:Factored critical bending moment(Sagging) = (1.5*73.71) Factored critical bending moment(Hogging) (1.5*15.42) Factored axial force = (1.5*30.75) Factored shear force = (1.5*127.33) Effective depth required d =

Mu/0.138fckb =

6

[110.57x10 /(0.138*20*1000)]

110.57KN-m 23.13KN-m 46.13KN 191.00KN

200.15mm

0.5

The over all depth required(Assuming 16mm dia bars) = [200.15+50+8] =

258.15mm

However provide overall depth of =

300.00mm

242.00mm

Hence,effective depth = [300-50-8]= Bottom steel:Mu/bd2 =

1.888

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

Area of steel required for bending tension = [0.641*242*1000/100]= Area of steel required for axial force = (46.13*1000/2*150) = Total area of steel required =

0.641

1551.2sqmm

153.8sqmm

1705.0sqmm

Hence provide 16mm dia bars at 110mm c/c at bottom,the area of reinforcement comes to =(1000/110)*(3.14/4)*162 =

1826.9sqmm

Top steel:Mu/bd2 =

0.395

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

0.114 0.15

Area of steel required for bending tension = [0.15*242*1000/100]=

363.0sqmm

Area of steel required for axial force = (46.13*1000/2*150) =

153.8sqmm

Total area of steel required =

516.8sqmm

Hence provide 10mm dia bars at 125mm c/c at top,the area of reinforcement comes to =(1000/125)*(3.14/4)*102 =

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) = Adopting 10mm dia bars,the spacing required = 1000/[450/(3.14/4)*102]=

450.00sqmm 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 Hence,the factored design shear force VFd = at a distance D from the face of the [191-(191/1400)*(150+242)] = Nominal shear stress Tv = [137.52x1000/(1000*242)] =

137.52KN

0.6N/sqmm <2.8 N/sqmm (As per Table 20 of 1S 456)

Hence,the section is safe from shear point of view

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

0.75%

The design shear strength of concrete for the above steel percentage from Table 19 of IS 456 is 0.56 Hence Vuc = 135.52KN Considering the enhanced shear strength at s 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) = (1.5*87.06) Factored critical bending moment(Hogging) (1.5*55.83) Factored axial force = (1.5*33.67) Factored shear force = (1.5*174.69) Effective depth required d =

Mu/0.138fckb =

6

[130.59x10 /(0.138*20*1000)]

130.59KN-m 83.75KN-m 50.51KN 262.04KN

217.52mm

0.5


275.52mm


300.00mm

Hence,effective depth =

242.00mm

[300-50-8]= Top steel:Mu/bd2 =

2.23


Area of steel required for bending tension = [0.707*242*1000/100]= Area of steel required for axial force = (50.51*1000/2*150) = Total area of steel required =

0.707

1710.9sqmm

168.4sqmm

1879.3sqmm

Hence provide 16mm dia bars at 100mm c/c at TOP,the area of reinforcement comes to =(1000/100)*(3.14/4)*162 =

2009.6sqmm

Bottom steel:Mu/bd2 =

1.43

From table 2 of SP 16,percentage of steel required = Area of steel required for bending tension = [0.457*242*1000/100]= Area of steel required for axial force = (50.51*1000/2*150) = Total area of steel required =

0.457 1105.9sqmm

168.4sqmm

1274.3sqmm

Hence provide 16mm dia bars at 150mm c/c at BOTTOM ,the area of reinforcement comes to =(1000/150)*(3.14/4)*162 =

1339.7sqmm

Distribution steel:Provide distribution reinforcement of 0.15% of cross sectional area of footing Hence,the distribution reinforcement required = (0.15*300*1000/100) = Adopting 10mm dia bars,the spacing required = 1000/[450/(3.14/4)*102]=

450.00sqmm 174.00mm


Check for one way shear:The critical section for beam shear is at distance of 'd' from the face of the Hence,the factored design shear force VFd = at a distance D from the face of the 262.04-(262.04/1400)*(150+242)] = Nominal shear stress Tv = [188.67x1000/(1000*242)] =

188.67KN




0.83%

The design shear strength of concrete for the above steel percentage from Table 19 of IS 456 is [0.56+{0.08*(0.62-0.56}/0.25] = 0.579 Hence Vuc = 140.12KN Considering the enhanced shear strength at s as 1.5 times the shear strength, 0.87>0.80 Hence,no shear reinforcement is required

iii) Design of side walls:Factored critical bending moment(Sagging) = Factored critical bending moment(Hogging) (1.5*55.83) Factored axial force = (1.5*174.69) Factored shear force = (1.5*33.67) Effective depth required d =

Mu/0.138fckb =

6

[83.75x10 /(0.138*20*1000)]

0.00KN-m 83.75KN-m 262.04KN 50.51KN

174.19mm

0.5


230.19mm


300.00mm

Hence,effective depth = [300-50-8]=

242.00mm

Steel on outer face :-

Mu/bd2 =

1.43


0.457

Area of steel required for bending tension = [0.457*242*1000/100]=

1105.9sqmm

873.5sqmm


1979.4sqmm


Hence provide 16mm dia bars at 100mm c/c on outer face,the area of reinforcement comes to =(1000/100)*(3.14/4)*162 =

2009.6sqmm

Steel on inner face :Area of steel required for bending tension =

0.0sqmm


873.5sqmm


873.5sqmm

Hence provide 12mm dia bars at 125mm c/c on inner face,the area of reinforcement comes to =(1000/125)*(3.14/4)*122 =

904.3sqmm

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 174.00mm

Adopting 10mm dia bars,the spacing required = 1000/[450/(3.14/4)*102]=


Check for one way shear:The critical section for beam shear is at distance of 'd' from the face of the Hence,the factored design shear force VFd = at a distance 'd' from the face of the 50.51-(50.51/1400)*(150+242)] =

36.36KN

Nominal shear stress Tv = [36.36x1000/(1000*242)] =




0.83%

The design shear strength of concrete for the above steel percentage from Table 19 of IS 456 is [0.56+{0.08*(0.62-0.56}/0.25] = 0.579 Hence Vuc = 140.12KN 0.579>0.2 Hence,no shear reinforcement is required

6.48 KN/sqm

1800

1800

16.20 KN/sqm

0.3571 0.555556

t Member Distribution Factor Fixed end moment Distribution Carry over Distribution Carry over Distribution Carry over Distribution Carry over Distribution Carry over

0.912698 0.391304 -1

Distribution Final moments

15.42

6

55.02

2

KN KN

KN KN

KN KN

KN KN

KN

KN

KN

6.48 KN/sqm

1800

6.48 KN/sqm

1800

16.20 KN/sqm

t Member Distribution Factor Fixed end moment Distribution Carry over Distribution Carry over Distribution Carry over Distribution Carry over

Distribution Carry over Distribution Final moments

7.34

4

5.83

5.83

KN KN

KN KN

KN KN

KN KN

KN

KN

KN

20N/sqmm 415N/sqmm 50.00mm 150N/sqmm

nt comes to

nt comes to

DESIGN OF CANTILEVER BREST WALL Data:Height of Retaining wall(h) = Height of wall above G.L= Height of wall below G.L= Density of back fill soil&material in toe portion(y) = Grade of concrete = Grade of steel = Ground water Table level = Angle of shearing resistance of back fill material&material at toe portion(Q) = Angle of face of wall ing earth with horizontal(a)(In degrees) (in clock wise direction) Slope of back fill(b) = Angle of wall friction (q) = Undrained Cohesion ( c) = Safe bearing capacity(SBC) = Surcharge over the back fill(s) = (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 Ka =

Sin(a+Q) sina

sin(a-q)

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

Sin(a+Q) = Sin(a-q) = Sina = Sin(Q+q) = Sin(Q-b) = Sin(a+b) =

SIN[3.14*(87.17+30)/180] = SIN[3.14*(87.17-15)/180] = SIN[3.14*(87.17)/180] = SIN[3.14*(30+15)/180] = SIN[3.14*(30-0)/180] = SIN[3.14*(87.17+0)/180] =

From the above expression, Ka =

0.32

Dimensions of the brest wall(Assumed for preliminary design):Thickness of stem at bottom = Thickness of stem at top =

0.89 0.952 0.999 0.707 0.5 0.999

1995

300

Earth pressure at top including surcharge = Kays = (0.32*1800*0.6) = Earth pressure at bottom including surcharge = Kay(s+h) = [0.32*1800*(1.995+0.6)] = Pressure distribution is as shown below:345.6

1.995m

1149.1 Area of the rectangular portion = Area of the triangular portion =

345.6 345.6*1.995 = 0.5*1149.1*1.995=

689.47 1146.25 1835.72

Taking moments of the areas about the toe of the wall S.No 1 2

Description

Area

Lever arm

Moment

Rectangular Triangular

689.47 1146.25

0.9975 0.665

687.746 762.256

1835.72

1450

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

0.79m

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

0.20

17.83

1.995m 0.79m 87.17

0.30m 0.04 Total earth pressure acting on the wall per 1m length P = (345.6*1.995+0.5*1149.1*1.995)*1 =

2180.46Kg

Horizontal component of the earth pressure Ph = (PCos20.150 ) Vertical component of the earth pressure Pv = (PSin20.150 ) Eccentricity of vertical component of earth pressure = (0.3/2-0.04) =

Total earth pressure = It acts at a hieght of from the base Moment due to horizontal component of earth pressure =

Design of wall or stem:Factored bending moment Mu = (1.5*1639.67) = Effective depth required d = Over all depth provided =

2459.51Kgm

Mu/0.138fckb = 300.00mm

Effective depth provided(Assuming 50mm cover) d = 2

Mu/bd =

94.40mm

242.00mm

0.42

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

0.129 0.15

Area of steel required =

363.00sqmm

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

523.33sqmm

Check for shear:The critical section for beam shear is at distance of 'd' from the face of the Hence,the factored design shear force VFd = at a distance 'd' from the face of the

Nominal shear stress Tv = [11.11x1000/(1000*242)] =

11.11KN




0.22%

The design shear strength of concrete for the above steel percentage from Table 19 of IS 456 is [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 = Provide 8mm dia @ 200mm c/c on other side

250.00sqmm

BREST WALL

1.995m 1.995m 0.00m 1800Kg/Cum M20 Fe415 30 87.17 0 15 0Kg/sqm 7500Kg/sqm 0.60m 20N/sqmm 415N/sqmm 2500Kg/Cum 2400Kg/Cum

2

0.30m 0.20m

345.6Kg/sqm

1494.7Kg/sqm

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 (Calculated from the profile of drain)

0.000250

5.Rugosity Coefficient(n) (As per table 5 of IRC:SP 13)

0.033

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 (Calculated from the profile of drain)

0.000250

11.Rugosity Coefficient(n) (As per table 5 of IRC:SP 13)

0.033

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 (Bed level of the field channel)

2.585

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:-

Cross section of minor drain b+2h

MFL

1:1

e

p slo

slo pe

1:1 h

Cross section of minor drain b+2h

MFL

1:1 pe

1:1

slo p

e

slo h

LBL

b

1)Area Velocity method:-

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

1.930m

Bed width(b) =

1.40m

Assuming side slopes 1 :1 in clayey soils,top width at HFL (b+2h) = Wetted Area(A) = 1.93*(1.4+5.26)/2

5.260m

6.43sqm

Wetted perimetre(P) = 1.4+2*[(1.4)2+(1.4)2]0.5

6.86m

Hydraulic Radius

R=

Wetted area/Wetted perimeter =

Velocity V =

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

Discharge Q =

AXV =

Discharge as per the record =

0.94 0.46m/sec 2.96Cumecs

2.950Cumecs

Higher of the above two values is adopted for design of CD Work Design Discharge = Design Velocity =

2.960Cumecs 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.

Hence,as per Lacey's silt theory,the regime width W = 4.8Q 1/2 = 4.8*2.960.5 =

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: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 = The area of cross section required =

4.00m/s

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

Depth of flow on down stream side Dd =

0.650m

Linear water way of the Box barrel (L) =

2.500m

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

2.960Cumecs

0.58

where, C0 = Coefficient of discharge = h = afflux in m e = factor ing for recovery of some velocity head as = potential head v = average velocity in approach section h+0.102v2 =

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

0.87 1

0.223373029

0.56

substituting h = (0.56/v)-0.65 and re-arranging the we get 0.56-0.87v+0.102v3 = 0 solving for v,we get v = Hence,afflux h =

0.69m/s 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 Hence,the discharge for design of foundations = 1.30XDesign Discharge = Qdf = (1.3*2.96) Lacey's Silt factor for standard silt ' f ' = 1.76Xm1/2= = (Value taken from table 4 of SP13) =

Discharge per metre width of foundations = q = (3.94/2.5) Normal scour depth D = 1.34(q2/f)1/3 = [1.34*(1.5762/1.0)0.33] For quasi-alluvial streams,linear water way required is natural width of the channel at HFL . = Linear water way provided = 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 =

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) Depth of foundation = Dm + Max.of 1.2m or 1/3 Dm = (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 = (1.27*2.85) =

3.62m

Depth of bottom of Cut-off wall from HFL =

3.62m

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

-0.49

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

1.69m

According to Khosla's theory for two dimensional flow,the exit gradient G E = (H/d) x 1 The value of B/d = For the above value,1

3.550 =

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

However, provide 2.40 depth cut-off wall

Design of LAUNCHING APRONS:Size of stones as per clause 5.3.7.2 of IRC 89-1985 Diametre (d) = (Vmax/4.893)2 =

0.020m

(0.92/4.893)2 = Say

0.23m

Weight of each stone considering the specific gravity of stone as 2.65 Weight of stone = 4/3x x(d/2)3 x 2.65 x 1000 = Say

15.81Kg 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 ( T ) = 0.06 x (Qdf)1/3 = 0.06*(3.48)

1/3

0.095

=

Thickness of apron at inner edge(near raft) = 1.5 x T = Say Thickness of apron at outer edge = 2.25 x T = Say

0.143m 0.30m 0.214m 0.60m

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

MFL

MFL

LBL

m is no truly alluvial in nature.

an be further flumed

0.003385

-0.00679

d 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

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