Design of a Cross Drainage Structure for D 20 Distributary in A Tadipudi Lift Irrigation
Scheme
Ch. Apparao,Rishu kumar,N.Sambasivarao
College of Agricultural Engineering, Bapatla
Acharya N.G. Ranga Agricultural University
ABSTRACT
The present study a methodology was developed for the Design of Well Syphon, a cross drainage
structure for D 20 Distributary in A Tadipudi Lift Irrigation Scheme(TLIS) on the river Godavari
under Jalayagnam project. The objectives of this study was to i) to develop methodology for the
design of cross drainage work for canal to cross two drains and a highway ii) to design well siphon
system as an appropriate cross drainage structure
The maximum flood discharge for the drains was estimated by using Ryve’s formula. The maximum
scour depth was estimated to be 2.22 m. The structural stability of well was estimated by considering
two conditions (i) when the well was full, and (ii) when the well is empty (worst condition) with the
help of Lame’s analogy. The maximum compressive stress developed in concrete structure was
15.526 t/m2 which was less than the maximum compressive strength of concrete. The maximum
compressive stress developed in the foundation of well was 7.85 t/m2 which is less than the bearing
capacity of soil. The maximum compressive stress developed in profile wall was estimated as 18.913
t/m2 which is less than the compressive stress of concrete.
Key words: Well Syphon, lift irrigation, cross drainage structure
1.INTRODUCTION
Investment in irrigation sector has led not only to a substantial increase in agricultural growth, income
and development, but also to an increase in the gross National product. About 40% of the Andhra
Pradesh state’s gross cropped area is irrigated, and irrigation contribution to State agricultural
production is about 60%. The Government has, therefore, given highest priority for the completion of
on-going irrigation projects. For this purpose, in the year 1996 Government has evolved a threefold
strategy namely; (i) to achieve maximum irrigation by completing on-going irrigation projects; (ii) to
rehabilitate and modernize existing irrigation schemes to bridge the gap ayacut in tail-end area; and
(iii) to hand over the management and maintenance of all irrigation schemes in the State to farmers’
organizations to ensure reliable and timely supply of water. Twenty-six major and medium irrigation
projects costing Rs. 46,000 crores (Revised to Rs. 67, 823 crores) are taken up for execution.
Tadipudi Lift Irrigation Scheme (TLIS) on the river Godavari is an on-going project under
Jalayagnam (Fig. 1) that has been planned to utilize 14.47 TMC s of Godavari river water to irrigate
2,06,572 acres.
80.45 KM
TADIPUDI LIFT IRRIGATION SCHEME
L1
KM 5
GODAVARI
RIVER
L2
KM 15.90
L3
KM 33.50
L4
KM 80
0.00 KM
Lift Height
Pumps
each
Discharge
Agreement Date
Present Stage
Length of the canal
Ayacut
Power required
Tender Amount
1st Pump House, Pressure Mains, Cistern : Progress
2nd Pump House, Pressure Mains
: Progress
Main Canal
: Progress
2nd Lift Pumps, Pump Houses, Pres. Mains : Designs
&Drawings
Under
Approval
Utilisation : 14.40 TMC
Ayacut
: 2,06,600 Acres
PUMP
HOUSE
DELIVERY
CISTERN
MAIN
: 27.00 Mts
: 6 Nos. of 4,000 HP
:
:
:
:
:
:
:
700 Cusecs.
23-1-2004
work in progress
80.45 Km.
2,06,572 Acres
28 MW
Rs. 302.64 Cr.
CANAL
PSC PRESSURE
MAINS (2 x 6
Nos.)
L1
KM 5
L2
KM 15.90
L3
KM 33.50
L4
KM 80
PROGRESS
4 Nos. of 3,000 HP each
For each pump house (2 Pump houses)
80.45 KM
TADIPUDI (V)
(WEST GODAVARI (DT))
0.00 KM
GODAVARI
RIVER
Fig. 1: Tadapudi Lift Irrigation Scheme
2. MATERIALS AND METHODS
The study area (Fig.2) was located near national highway joining Tadepalligudem and
Nallajerla . it was about 6 km from Tadepalligudem.
Fig. 2 : Location of Cross drainage structure on D20 distributary
2.1 Collection of Hydraulic Data
The following hydraulic data was collected for the study
a. Canal
1. Full supply discharge
2. Bed width
3. Full supply depth
4. Bed level
5. Bed slope
6. Full supply level
7. Top of bank level
8. Cross section of canal
9. Nature of bed material and value of Manning’s coefficient (n)
10. Velocity in canal
11. Scour depth
b. Drainage channel
1. Extent and nature of drainage area ( catchment area )
2. Maximum flood level
3. Silt factor
4. Allowable afflux in view of water spread, upstream of the proposed work
5. Reported or observed scour depth for any nearby structure on the same drainage channel
6. Bearing capacity of the foundation strata
c. Structural Data
1. Empty self-weight of the structure
2. Surcharge loads, if any
3. Full internal water pressure
4. Soil reaction and up-lift pressure
5. Earth loads on sides
2.2 Criteria for Selection of Particular Cross Drainage Work
Table 1: ..................................................
Particulars
Bed level, m
Water level, m
Discharge, cumecs
Drain
19.290
20.290 (HFL)
5.73
Canal
19.665
20.045 (FSL)
0.113
Since, HFL was at higher level than FSL as well as the discharge from canal was very less compare to
drain. Therefore, the canal bed was depressed and syphoned. Since, canal was passing two drains and
a national highway, therefore, well syphon was proposed.
2.3 Estimation of maximum flood discharge and High Flood Level(HFL)
Maximum flood discharge (MFD) was estimated by using Rvye’s formula.
𝑄 = 𝐶₁𝐴₁2/3
...(1)
Where,
Q = Maximum flood discharge in cusecs and A₁ is in square miles.
𝐶₁= 750 (Adopted from I & CAD Dept., Govt. of AP)
2.4 Design of drainage channel
Drainage channel was designed based on Kennedy theory with Manning’s equation. The estimation of
various parameters like drain area, hydraulic radius and wetted perimeter were given below.
i.
Calculation of area of drain
Area, A= (B+ZD) D,
...(2)
Where,
A= Area (m2), B= Base width (m), D= Drain depth( m), Z= Side Slope
ii.
Calculation of Perimeter:
Perimeter, P= B+ 2D√ (1+Z²)
...(3)
Where,
P= Perimeter (m), B= Base width (m), D= Drain depth (m), Z= Side Slope
iii.
Calculation of Hydraulic Radius:
Hydraulic Radius, R=A/P
...(4)
Where,
R= Hydraulic radius (m), A= Area (m2), P= Perimeter in meter
iv.
Calculation of Velocity:
Velocity, v= Q/A
...(5)
Where,
v= Velocity in m/s, Q= Discharge in cumec, A= Area in square meter
v.
Calculation of Bed Slope:
Bed Slope, S= (nv/R⅔)²
...(6)
Where,
n= Manning’s constant, v= Velocity in m/s, R= Hydraulic radius in meter
2.5 Determination of scour depth
Scour depth was estimated by using Lacey’s formula .
𝑄2
1⁄
3
𝑑 = 1.34 ( 𝑓 )
...(7)
Where,
d= normal depth of scour in meters below the flood level corresponding to the value of Q adopted
Q= the discharge adopted for design, in cubic meter per second per meter width
f= the silt factor for a representative sample of the bed material;
=1.76√m where m is the mean diameter in millimetre = 1.0 for silty soils
To calculate maximum scour depth, normal scour depth was calculated by using Lacey’s formula and
was multiplied by a factor of safety 1.5.
2.6 Head loss through syphon barrel
Head loss through barrel was calculated by using Unwin’s formula.
𝐿 𝑉2
𝑅 2𝑔
ℎ = (1 + 𝑓1 + 𝑓2 )
−
𝑉𝑎2
2𝑔
...(8)
Where,
h= head loss through the syphon, in m
L= length of barrel, in m
R= hydraulic mean radius of the barrel, in m
V= velocity of flow through the barrel, in m/s
𝑉𝑎 = velocity of approach, and is generally neglected
𝑓1= coefficient of head loss at entry = 0.505 for unshaped mouth= 0.08 for bell mouth
0.3𝑏
𝑓2 = 𝑎 (1 +
) = coefficient for the loss of head by the friction through the syphon barrel,
𝑅
a and b are constants. The constants 𝑎= 0.00316 and b = 0.1 were assumed.
2.7 Intensity of active earth pressure
Intensity of active earth pressure was estimated by using Rankine’s formula.
𝑊𝐻
𝑃𝑎 =
2
1−sin ∅
𝑥 1+sin ∅
where
...(9)
Where,
𝑃𝑎 = Earth Pressure, t/m2, 𝑊= Unit Weight of Saturated Soil, t/m3,𝐻= Height of Backfill, m
∅= Angle of repose of the soil= 29˚ (Anonymous, 1998)
2.8 Design of inlet and outlet well:
i.
Thickness of well:
The thickness of well was estimated by using anology of Lame’s theorem of thick cylinder.
𝑏
𝑃𝑥 =
𝑥2
−𝑎
...(10)
Where,
𝑃𝑥 = Intensity of Pressure at a distance x. a and b are arbitrary constants.
𝑏
𝜎𝑥 =
𝑥2
+𝑎
...(11)
Where,
𝜎𝑥 = Intensity of Hoop Stress induced
ii.
Stress developed in well
The stress developed in well was computed by estimating loads acting of the structure, lever arm,
eccentricity and maximum and minimum stress developed in concrete and foundation soil of well.
a. Lever arm
Lever arm was estimated by using the following formula.
𝑍=
∑𝑀
...(12)
∑𝑉
Where,
∑𝑀= Total moment, ∑𝑉= Total load
b. Eccentricity
To calculate eccentricity the following formula is used.
𝑏
𝑒 =2−𝑍
...(13)
Where,
b= Base width, Z= Lever arm
c. Stress developed at base point
To calculate maximum stress developed at base point the following formula is used.
𝜎=
∑V
𝑏
Where,
(1 ±
6𝑒
𝑏
)
...(14)
𝜎= Stress at base point, ∑𝑉= Total load
b= Base width, 𝑒= Eccentricity
2.9 Design of head wall
To design the head wall, we have to calculate stress developed in head wall. To calculate stress
developed in head wall, we need to calculate lever arm, eccentricity and maximum and minimum
stress developed in concrete and foundation soil of head wall by using above formulae.
3. RESULT AND DISCUSSIONS
The hydraulic particulars of the canal are given in Table.2
Table 2. Hydraulic Particulars of Canal
S. NO.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
Description
Ayacut
Discharge
Bed Width
FSD
Free Board
Side Slope
Bed Fall
Value of n
Velocity
Top Width of Bank
CBL
FSL
TBL
GL
Unit
Ac
Cumec
m
m
m
m/s
m
m
m
m
m
Particulars @ km 6.204
629.17
0.113
0.6
0.38
0.3
1.5:1
1:2300
0.025
0.309
1.5
19.665
20.045
20.345
19.290
3.1 Hydraulic Particulars of Drain
3.1.1 Drain @ 6.204 km
Bed level of the drain= 19.290 m
Maximum flood depth= 1.00 m
Maximum flood level of the drain= 20.290
Drain bed width= 5.00 m
Estimated catchment area of the drain= 0.363 km² = 0.1399 miles²
a. Estimation of maximum flood discharge
C= 750
M= 0.1399 miles²
From Ryves’ formula
Q= 202.1187 cusecs = 5.73 cumec
b. Estimation of scour depth
q= 5.73/5= 1.146 cumecs/m
f= Silt factor = 1.00
Therefore,
1.1462
𝑅 = 1.34( 1 )1/3= 1.467m
c. Maximum Scour Depth, Rmax= 1.5R= 1.5 x 1.467= 2.21m
3.1.2 Drain @ 6.256 km
Bed level of the drain= 19.345 m
Maximum flood depth= 0.700 m
Maximum flood level of the drain= 20.045 m
Drain bed width= 0.900 m
Estimated catchment area of the drain= 0.026 km² = 0.0100 miles²
a.Estimation Maximum flood discharge: 0.988 cumecs
Normal Scour depth (𝑅) : 1.426m
Maximum Scour Depth (Rmax): 2.139m
Particulars@ km 6.256
629.17
0.113
0.6
0.38
0.3
1.5:1
1:2300
0.025
0.309
1.5
19.642
20.022
20.322
19.345
b.Decision on foundation depth
Scour depth for canal: 0.43
Maximum Scour Depth (Rmax):0.659 m
Estimated scour depths for drain @ 6.204 km= 2.21 m
Estimated scour depths for drain @ 6.204 km= 2.139 m
The maximum of above three calculated scour depth is considered as the depth of foundation. Hence,
the top level of the foundation should be at least 2.21m below the ground level.
3.2 Design of Syphon Barrel
Discharge in Canal, Q₁ = 0.113 m³/s
Velocity allowed in the Barrel, v₁= 0.6 m/s (assume, double the velocity in the channel)
Cross-sectional area of Barrel, A₁= Q₁/v₁= 0.113/0.6=0.1883 m/s
If d is the diameter of barrel, then
(∏/4)d² = 0.1883
Or, d= 0.4898m
Considering a diameter of pipe about 0.450m or 450mm.
Therefore, Actual cross-sectional area of barrel, A₁´ = (∏/4) d²= (∏/4) x (0.450)²= 0.1589m²
Actual velocity through barrel, v₁´= Q₁/A₁´= 0.113/0.1589= 0.711m/s
Since, the velocity in barrel is almost twice the velocity in the canal. Hence, its OK.
Hence, NP3 class hume pipes 450 mm ø conforming to BIS 458/2003, was selected as barrel.
3.3 Details of NP3 class hume pipe
Diameter= 450 mm, Thickness= 75 mm, Length= 2.5 m, Proposed length of the barrel= 25m
Hence, 10 numbers of NP3 hume pipes should be provided.
3.4 Collar dimensions
Caulking space= 19mm, Minimum thickness= 35 mm, Minimum length= 200 mm
3.5 Details of the craddle bedding
As per IS: 783-1985, minimum thickness of the bedding is 1/4th of pipe internal diameter.
Thickness of the bedding= 0.3m, Minimum earth cushion over pipes= 0.5m
Offset on either sides of pipe= 0.15m, Spacing between pipes= 0.3m
3.6 Profile wall
Provide profile wall of wide= 300 mm on U/S and D/S.
3.7 Head loss through syphon
L= Length of barrel= 25m
V= Velocity in barrel= 0.711m/s
Vₐ= Velocity of approach (velocity of approach was considered as zero)
R= Hydraulic mean radius= d/4= 0.450/4= 0.1125m
𝑓1= 0.505 for unshaped mouth
0.3𝑏
𝑓2 = 𝑎 (1 + 𝑅 )= 0.00403 (where 𝑎=0.00316, 𝑏=0.1)
From UNWIN’s formula:
h= 0.062m≈0.1m
Therefore, the water level in the U/S well would head=20.022+ 0.1= 20.122m
3.8 Hydraulics of Inlet and Outlet Well
a. Check for thickness of the well for safety against hoop tension
W= 2.10T/m³, H= 3.05m, ∅= 29˚
From Rankine’s active earth pressure,
𝑃𝑎 = (2.10x3.05/2) x ((1-sin29˚)/ (1+sin29˚)) = 1.111 T/m²
External radial pressure on the outer periphery of the well wall can be estimated by Lame’s theory of
thick cylinder. For that we need to calculate a and b which are constant and can be calculated by
considering two different situations.
i.
External earth pressure due to soil
ii.
Internal pressure due to water
Let, r₁= internal radius of well= 900mm= 0.900m
Fig.3.1. Radial Earth Pressure working on wall periphery
r₂= outer radius of well= (1800+2x177)/2 mm= 1077mm = 1.077m
Case1: External earth pressure due to soil:
𝜎𝑥 = 1.111 T/m², x = r₂= 1.077m
Then by Lame’s theory
1.111= (b/ (1.077)²) + a………………….. (i)
Case 2: Internal pressure due to water:
The worst condition will be when there is no water in the well. Hence,
𝜎𝑥 =0 and x= r₁= 0.900m
0= (b/ (0.900)²) + a……………………… (ii)
Solving equations (i) and (ii), we will get:
b= -2.98, a= 3.679
Therefore,
Hoop compression at the outer periphery:
𝑏
𝑃𝑥 = 𝑎 − 𝑥 2 = 3.679 – (- 2.98/1.077²)= 6.248 T/m²
Hoop compression at the inner periphery:
𝑏
𝑃𝑥 = 𝑎 − 𝑥 2 = 3.679 – (-2.98/0.90²)= 7.358 T/m²
This showed that no tension was developing in the well wall and the maximum compressive stress
was within the permissible limits of masonry. Hence section assumed was quite safe.
3.9 Stability calculation of well
Assumptions:
Unit weight of soil= 2.10 T/m³
Unit weight of concrete= 2.40 T/m³
Unit weight of water= 1.00 T/m³
Unit weight of saturated soil= 2.20 T/m³
Horizontal coefficient of soil= 0.134
Vertical coefficient of soil= 0.0384
Height of surcharge on top of the abutment= 0m
Equivalent height of surcharge= 0/2.40 =0m
Total height of soil on foundation= 3.750m
Stability of well checked for two conditions
i.
When well is full
ii.
When well is empty
Well is made of RCC hume pipe of diameter 1800mm and thickness of 177mm.
a. Stress developed in concrete (taking moment about A):
Case I: When well is full
Fig 3.2. Forces acting on the wall
Table: Stress developed in concrete of Well (Taking moment about A)
Lever arm= (∑M/∑V) = (26.390/20.126)= 1.311m
Eccentricity= │ (d/2 – LA) │= │ (2.15/2) – 1.311│= 0.236m
Allowable limit= 0.358
∑V
6𝑒
Maximum stress developed, 𝜎𝑚𝑎𝑥 = 𝑏 (1 + 𝑏 ) = 15.526 T/m²
∑V
6𝑒
Minimum stress developed, 𝜎𝑚𝑖𝑛 = 𝑏 (1 − 𝑏 ) = 3.19 T/m²
Case I: When well is empty:
Lever arm= (∑M/∑V) = (19.121/13.376)= 1.43 m
Eccentricity= │ (d/2 – LA) │= │ (2.15/2) – 1.43│= 0.355m
Allowable limit= 0.358
∑V
6𝑒
Maximum stress developed, 𝜎𝑚𝑎𝑥 = (1 + ) = 12.356/m²
𝑏
𝑏
∑V
6𝑒
(1 − )
𝑏
𝑏
Minimum stress developed, 𝜎𝑚𝑖𝑛 =
= 0.057 T/m²
b. Stress developed in soil (taking moment about B):
Case I: When well is full:
Lever arm= (∑M/∑V) = (65.819/26.926)= 2.444 m
Eccentricity = │ (d/2 – LA) │= │ (4.45/2) – 2.444) │= 0.219 m
Allowable limit = 0.742
∑V
6𝑒
Maximum stress developed, 𝜎𝑚𝑎𝑥 = 𝑏 (1 + 𝑏 ) = 7.85 T/m²
S.
No.
Force
Description
Minimum stress developed, 𝜎𝑚𝑖𝑛 =
Density
∑V
6𝑒
(1 − 𝑏 )
𝑏
When well is full
Weight
L.A.
T
M
= 4.26 T/m²
Table3................................................................................
Moment
T.m
When Well is empty
Weight
L.A.
Moment
T
m
T.m
1
W1
2
W2
3
W3
4
W4
½
5
W5
½
6
Pv
0.0384
Total load, ∑V=
7
Ph
0.134
total moment, ∑M=
x
x
x(
0.18
1.8
0.18
1.15
1.15
3.75
x
x
x
x
x
²) -(
3.75
3.75
3.75
3.75
3.75
0
x(
3.75
²) -(
0
X
X
X
X
X
²)x
2.4
1
2.4
2.1
2.1
2.1
²)x
2.1
1.593
6.750
1.593
4.528
4.528
1.134
20.126
3.957
0.089
1.077
2.066
2.537
-0.383
0.141
7.270
3.290
11.489
-1.736
0
1.5
5.936
26.390
1.593
0.000
1.593
4.528
4.528
1.134
13.376
3.957
0.089
0.000
2.066
2.537
-0.383
0.141
0.000
3.290
11.489
-1.736
0.000
1.500
5.936
19.121
Table 4. Stress developed in soil (taking moment about B)
Force Description
S.
No.
0.18
1
W1
1.8
2
W2
0.18
3
W3
1/2
X
1.15
4
W4
1/2
X
1.15
5
W5
0.6
6
W6
0.0384 x ( 4.35
7
Pv
Total load, ∑V=
8
Ph
0.134
x ( 4.35
total moment, ∑M=
Density
x
x
x
x
x
x
²) -(
3.75
3.75
3.75
3.75
3.75
4.45
0
x
x
x
x
x
x
²)x
2.4
1
2.4
2.1
2.1
2.4
2.1
²) -(
0
²)x
2.1
When well is full
Weight L.A.
Moment
T
m
T.m
1.593
1.239 1.973
6.750
2.227 15.032
1.593
3.216 5.122
4.528
3.687 16.697
4.528
0.767 3.472
6.408
2.225 14.258
1.526
0.000
26.926
5.325
1.740 9.265
65.819
When Well is empty
Weight L.A.
Moment
T
m
T.m
1.593
1.239 1.973
0.000
0.000 0.000
1.593
3.216 5.122
4.528
3.687 16.697
4.528
0.767 3.472
6.408
2.225 14.258
1.526
0.000
20.176
5.325
1.740 9.265
50.786
Case II: When well is empty:
Lever arm= (∑M/∑V) = (50.786/20.176) = 2.52 m
Eccentricity = │ (d/2 – LA) │= │ (4.45/2) – 2.52 │= 0.295m
Allowable limit = 0.742
∑V
6𝑒
Maximum stress developed, 𝜎𝑚𝑎𝑥 = 𝑏 (1 + 𝑏 ) = 6.33 T/m²
Minimum stress developed, 𝜎𝑚𝑖𝑛 =
∑V
6𝑒
(1 − 𝑏 )
𝑏
= 2.73 T/m²
Table5. Stress table
Concrete
Soil
When well is full
Maximum Stress, T/m²
Minimum Stress, T/m²
When well is empty
Maximum Stress, T/m²
Minimum Stress, T/m²
15.526
7.85
3.19
4.26
12.356
6.33
0.057
2.73
3.10 Structural Design of U/S & D/S Head Wall:
Assumptions:
Unit weight of soil= 2.10 T/m³
Unit weight of concrete= 2.40 T/m³
Unit weight of water= 1.00 T/m³
Unit weight of saturated soil= 2.20 T/m³
Horizontal coefficient of soil= 0.158
Vertical coefficient of soil= 0.0397
Height of surcharge on top of the abutment= 0m
Equivalent height of surcharge= 0/2.40 =0m
Total height of soil on foundation= 3.750m
a. Stress developed in concrete (taking moments about A)
Lever arm= (∑M/∑V) = (33.317/24.628)= 1.353M
Eccentricity = │ (d/2 – LA) │= │ (2.20/2) – 1.353) │= 0.253m
Allowable limit = 0.367
∑V
6𝑒
Maximum stress developed, 𝜎𝑚𝑎𝑥 = (1 + ) = 18.91 T/m²
Minimum stress developed, 𝜎𝑚𝑖𝑛 =
S.
No
1
2
3
4
5
6
7
Force
Description
W1
W2
1/2
W3
1/2
W4
W5
Pv
0.0397
Total load, ∑V=
Ph
0.1584
total moment, ∑M=
𝑏
𝑏
∑V
6𝑒
(1 − 𝑏 )
𝑏
= 3.47 T/m²
x(
0.3
1.9
1.9
0.3
0.3
3.75
x
x
x
x
x
²) - (
3.75
3.75
3.75
3.75
3.75
0
x
x
x
x
x
²) x
Density
2.4
2.4
2.1
2.1
2.1
2.1
x(
3.75
²) -(
0
²) x
2.1
X
X
Weight
2.700
8.550
7.481
2.363
2.363
1.172
24.629
4.678
L.A.
2.050
1.267
0.633
-0.150
2.350
Moment
5.535
10.830
4.738
-0.354
5.552
0.000
1.500
7.017
33.317
Fig 3.3. Forces acting on the Head wall
Table: Stress developed in concrete of Head wall (Taking moment about A)
b. Table 6.Stress developed in soil (taking moments about B):
Table7. Stress developed in soil (Taking moment about B)
S. No
1
2
3
4
5
6
7
Force
W1
W2
W3
W4
W5
W6
Pv
Description
1/2
1/2
x
x
0.0397
x(
0.3
1.9
1.9
0.3
0.3
0.6
4.35
X
X
X
X
X
X
²) - (
3.75
3.75
3.75
3.75
3.75
2.8
0
x
x
x
x
x
x
²) x
Density
2.4
2.4
2.1
2.1
2.1
2.4
2.1
Weight, T
2.700
8.550
7.481
2.363
2.363
4.032
1.578
L.A., m
2.350
1.567
0.933
0.150
2.650
1.400
Moment,
T.m
6.345
13.395
6.983
0.354
6.261
5.645
0.000
8
Total load, ∑V=
Ph
0.158
total moment, ∑M=
x(
4.35
²) -(
0
²) x
29.066
6.278
2.1
1.740
Lever arm= (∑M/∑V) = (49.907/29.068)= 1.717 M
Eccentricity = │ (d/2 – LA) │= │ (2.80/2) – 1.717) │= 0.317 m
Allowable limit =0.467
∑V
6𝑒
Maximum stress developed, 𝜎𝑚𝑎𝑥 = 𝑏 (1 + 𝑏 ) = 17.433T/m²
Minimum stress developed, 𝜎𝑚𝑖𝑛 =
Table8. Stress table:
Concrete
Soil
∑V
6𝑒
(1 − )
𝑏
𝑏
Maximum Stress, T/m²
18.913
17.433
= 3.329 T/m²
Minimum Stress, T/m²
3.47
3.329
Fig. 3.4 Design of well- Siphin(Side view)
10.925
49.907
`
Fig 3.5. Plan of Well- siphon (Top View)
4CONCLUSION
1. The maximum pressure exerted on the foundation soil within the bearing capacity of the soil
2. No tension developed anywhere in the structure
3. The structure was safe against sliding and overturning.
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