sarada well
TRANSCRIPT
-
7/29/2019 Sarada Well
1/46
DISCHARGE CALCULATIONS.
Discharge by Catchment area method:
a) By Ryve's formula = 860 cumecs
b) By Dicken's formula = 1980 cumecs
c) By Rainfall Method = 1160 cumecs
Discharge by A-V Method:
a) = 1140 cumecsb) = 1140 cumecs
c) = 1100 cumecs
d) = 820 cumecs
e) = 1140 cumecs
The Discharge by A.V method and rainfall method are coinciding.
Hence discharge of 1140 cumecs is taken as design discharge.
At 100m D/S
At 20m U/S
At 100m U/S
At Site of crossing.
CONSTRUCTION OF HLB ACROSS RIVER VEGAVATHI NEAR VANTHARAM
@ KM 11/8-10 ON PANUKUVALASA - BALIJIPETA ROAD IN
VIJAYANAGARAM DISTRICT.
At 200m D/S
1
-
7/29/2019 Sarada Well
2/46
CONSTRUCTION OF HLB ACROSS RIVER VEGAVATHI NEAR VANTHARAM
@ KM 11/8-10 ON PANUKUVALASA - BALIJIPETA ROAD IN
M.F.L = +13.06
L.B.L = +7.58
V = 2.66 m/sec
Design discharge Q = 3064.2 Cumecs
D = 13.06 - 7.58 = 5.48 m
d = 0.15 m
ha = = 0.34167 m
= 0.49167 m
Vv = = 2.80 m/sec
Linear Water way required = Q/D*Vv = 200.04 m
14 vents of 16.5 m = 231 m
However provide 14 vents of 16.5m effective with footpaths from bank to bank.
Total Head
Velocity through Vents
Velocity of flow at M.F.L
Depth of flow at M.F.L
Assuming afflux
LINEAR WATERWAY CALCULATIONS.
Head due to velocity at approach V2.D
2/ 2g ( d+D)
2
Cd2gh
2
-
7/29/2019 Sarada Well
3/46
3
-
7/29/2019 Sarada Well
4/46
4
-
7/29/2019 Sarada Well
5/46
Design Discharge 3983.46 cumecs
M.F.L +13.06
L.B.L +7.58
Sill Level +7.58
Vertical Clearance 1.65 m
Thickness of Deckslab D 1.675 m
Wearing coat thickness 0.1 m
R.C.L = MFL + AFFLUX +VERT. CLEAR +D ( DECK) +WEAR.COAT = +16.64
Linear water way required = 200.04 m
Span arrangement provided 14 vents of 16.50 m eff. With footpaths.
Mean Scour Depth = 9.476 m
F.L for Pier as per Scour depth = -+12.146
F.L for Abutment as per Scour depth = -+2.946
Based on the soil report the following F.L' s are proposed with about 1.50m embedment into S.D.R
A1 , A2 -+2.95
P2,P5,P9,P11,P12 -+12.146P3,P4,P6,P7 +88.000
P1,P8,P10,P13 +89.350
Length of flywing = RCL-SILL LEVEL - BASE WIDTH
= 7.000 M
PROVIDING 350MM THICK FLYWING
3.5 m
0.45 m
Weight of Flywing : 3.5 m
(3.5x.45+0.5x3.52)0.35x2.40 = 6.468 t
C.G. of flywing:A X AX
1 1.575 1.75 2.756
2 6.125 1.167 7.146
Total 7.7 9.902
C.G = 1.286 m
CONSTRUCTION OF HLB ACROSS RIVER VEGAVATHI NEAR VANTHARAM @ KM
ON PANUKUVALASA - BALIJIPETA ROAD IN VIJAYANAGARAM DISTRICT.
PRELIMINARY DATA
-
7/29/2019 Sarada Well
6/46
6
11.00
Thickness of Wall steining h = Kd = 0.657
Provide 600mm thick for Pier also.
75mm 700mmCURB HEIGHT
300mm
30
100mm
Angle = 30
Height = 0.775 + 0.3 = 1.643
Tan300
Provide 1.7 m curb height.
L
-
7/29/2019 Sarada Well
7/46
1/8-10
-
7/29/2019 Sarada Well
8/46
DESIGN OF BACKING WALL & BED BLOCK:
0.30
CONCRETE MIX = M20 0.10
RCL 14.282
R.C.L = 14.282
= 35 deg.1.7
Deck thickness = 1.7 m
Wearing coat thk. = 0.10 m
Ka = 0.246
0.211
= 1.80 t/cum
0.30
Width of backing wall = 0.30 m
Live load surcharge = 1.20
Thickness of Bed = 0.30 m
block
Width of Bed block = 1.05 m
(h) Height of Backing wall h = 0.10 + 1.7 0.211 = 2.01 m
a) Moment due to Earth Pressure:
Pa = 1/2 Ka h2
= 0.895 t/sqm
Lever arm = 0.42 X 2.01 = 0.845 m
Moment = 0.895 X 0.845 = 0.76 t-m
3
B.M due to fluid pressure of 480 kg/sqm = 480 2.011 = 0.651 t-m
1000 6
< 0.756 t-m
b) Moment due to Live load surcharge:
Intensity of Live load surcharge at RCL = 0.246 1.80 1.20 = 0.531 t/sqm
2Moment = 0.531 X 2.01 2 = 1.074 t-m
c) Moment due to Breaking force:
Max.axle load is 20.00 tonne for class 70r wheeled load.
Breaking force is 20% of axle load I.e = 20.00 X 20 = 4.000 t
100
BACKING WALL:
-
7/29/2019 Sarada Well
9/46
Breaking force under each wheel = 2.00 t
a) Dispersion width at top of bed block = 0.86 2 2.01 = 4.88 m
(Tyre width = 0.86 m) (For second wheel)
b) Dispersion width at top of bed block
(For first wheel)= 0.475 + 1.2 + 0.860 + 2.01 = 4.55 m
Intensity of second Wheel load / RM = 2.00 4.88 = 0.410 t/m
Intensity of first Wheel load for overlapping portion = 2.00 4.55 = 0.44 t/m
Total Wheel Intensity = 0.41 + 0.44 = 0.85 t/m
Lever arm = 2.01 + 1.20 = 3.21 m
Moment due to Breaking force = 0.850 X 3.211 = 2.728 t-m
Total Moment = 0.76 + 1.074 + 2.728 = 4.56 t-m/m
For V.R.C.C M20 mix: Q = 7.702 j = 0.916
d = 4.56 * 105
= 24.33 cm
100 * 7.70
d available = 30.00 5 0.6 = 24.40 cm
> 24.33 cm
A st = 4.56 * 105
= 10.20 Cm2
2000 0.916 24.40
Spacing of 12 mm dia tor bars = 1.131 X 1000 = 111 Cm c/c
10.20
Spacing of 12 mm dia tor bars @ 110 mm c/c vertically in the form of Stirrups.
Distribution steel:
A st = 0.12 X 100 X 30.00 = 3.6 Cm2 /m
100
Spacing of 10 mm dia bars = 3.142 1.0 X 100 = 21.819 cm
4 X 3.6
Provide 10 mm dia tor bars @ 200 mm c/c horizontally on both faces.
Width of bed block = 105 cm
Steel to be provided = 105 X 22.5 100 = 23.625 Cm2
BED BLOCK DESIGN:
-
7/29/2019 Sarada Well
10/46
Longitudinal bars at top & bottom = 23.6 4 = 5.906 Cm2
Provide 6 nos of 12 mm dia tor bars @ top & bottom equally spaced in a
width of 1.05 m (Steel provided = 6 X 1.131 = 6.785 Cm2)
Steel in Transverse direction = 23.6 2 = 11.813 Cm2
Volume of steel per 1m length = 11.81 X 100 = 1181 Cm3
Length of each Stirrup = 100 + 25 2 = 250 cm
Using 10 mm dia bars volume of each stirrup = 250.0 0.786 = 196.500 Cm3
No of stirrups = 1181 196.5 = 6.011 no
Spacing of Stirrups = 100 6.011 = 16.63 cm
Provide 10 mm dia stirrups @ 150 mm c/c
12 mm dia tor bars @
110 mmc/c
10 mm dia tor bars @
2.01 m 200 mm c/c
6 nos of
12 mm dia each at top &bottom
0.30 m 10 mm dia stirrups @
150 mm c/c
1.05 m
-
7/29/2019 Sarada Well
11/46
Width of bed block 90 cm
Steel to be provided 90 X 22.5 100 20.250 Cm2
Longitudinal bars at top & bottom 20.3 4 5.063 Cm2
Provide 6 nos of 12 mm dia tor bars @ top & bottom equally spaced in a
width of 0.75 m (Steel provided 6 X 1.131 6.785 Cm2)
Steel in Transverse direction 20.3 2 10.125 Cm2
Volume of steel per 1m length 10.13 X 100 1013 Cm3
Length of each Stirrup 85 + 25 2 220 cm
Using 10 mm dia bars volume of each stirrup = 220.0 0.786 172.920 Cm3
No of stirrups 1013 172.9 5.855 no
Spacing of Stirrups 100 5.855 17.08 cm
Provide 10 mm dia stirrups @ 160 mm c/c
BED BLOCK DESIGN:
-
7/29/2019 Sarada Well
12/46
-
7/29/2019 Sarada Well
13/46
DESIGN OF R.C.C CIRCULAR PIER.
a) Checking of Circular Pier design using P- body curves:
P = 455.50 t CONCRETE MIX = M 20
ML = 105.19 t-mMT = 125.88 t-m
MR = ML2 + MT2 164.04 t-m say 164.04 t-m
D = 2.00
Checking of Pier section based on actual area required & min. % of steel
for direct load & moment.
With 0.8 % steel .
Asc = 0.008 Ag m = 10
Ac = 3.142* D*D 0.008 = 0.84 D2
4 4
Dm = 200.00 2 5 2.25 = 185.5 cmD = 200.00
Dm = 185.5 D = 0.9275 D
200
Ic = 3.142 x D4 +(m-1)* 0.008 * 3.142* D2*Dm2
64 4 8
= 0.0491 9.0 0.008 3.142 0.860 D4
4 8
= 0.0552 D4
c = P + M* y
Ac Ic
For M 20 c = 66.67 Kg/Cm2 M = Mr 164.04
66.67 = 455.5 1000 + 164.04 10^5 X D
0.84 D2
0.0552 D4
2
= 540938 + 1.49E+08
D2
D3
L.H.S = 66.666667
(by trail&error)
R.H.S = 540937.89 + 1.49E+08 D = 171.0 cm
D2
D3
Required
= 48.229596 200.00 cmAvailable D = 2.00 m
Hence Safe
Asc in Column
Pt = 0.8 %
2
Ast = 0.8 3.142 X 171.0 = 183.75 Cm2
100 4
AST REQD OF ALL CASES(A,B,C) = 183.75 Cm2
Provide 32 nos of 28 mm dia tor bars 197.06624 Cm2
Spacing of 28 mm dia steel
= 19.63 cm C/C Hence O.K
+(m-1) * 3.142 * D2
-
7/29/2019 Sarada Well
14/46
Transverse reinforcement:
Provide 10 mm dia tor stirrups @ 250 mm c/c
B) Checking of Pier section on actual area required and min. % of steel for direct load.
With 0.8% steel
Asc = 0.008Ag m = 10
Ac = 3.142* D2 = 0.84 D2
4 4
c = P
Ac
For M 20 c = 66.67 Kg/Cm2
P = 455.5 t
66.6667 = 455.5 1000 D = 90.08 cm
0.842 D2
Actual Area required for direct load = 0.84 D2
= 6832.95 Cm2
Ast min = 0.8 X 6832.95 = 54.66 Cm2
100 < 0.3% of steel
C) Asc in Column
Pt = 0.3 % OF GROSS AREA
2
Ast = 0.3 3.142 X 200.0 = 94.26100 4 Cm2
% STEEL PROVIDED = (183.750444*100*4)/(3.142*200*200) = 0.5848
d) Check by Rational formulae:
Equivalent Concrete Area = 0.83 X 4 = 3.307 m2
Direct Stress = P/A = 455500 / 33073.75 = 13.772 Kg/Cm2
Bending Stress = M x y = ####### X 100.00 = 19.150 Kg/Cm2
Ie 0.0535 X #########
Direct stress + Bending sress < 1Perm.comp.str Per.Bend.Strs
13.772 + 19.15 = 0.563 < 1
50.00 66.67
Hence o.k
+(m-1)* 3.142 *D2*0.008
-
7/29/2019 Sarada Well
15/46
a) Checking of Circular Pier design using P- body curves:
P = 455.5 t CONCRETE MIX = M 20
ML = 105.19 t-mMT = 125.88 t-m
MR = ML2 + MT2 164.04 t-m
R = 1.00
D = 2.00
Cover = 5.00 cm
m = 10.00
Eccentricity e = MR = 164.04 = 0.3601 m
P 455.5
e = 36.01 / 100 = 0.3601
R
r = 100 5 1 1.4 = 92.60 cm
r = 0.93
R
From Pea - body curves
mp = 10 0.58482 = 0.058
100
For e/R = 0.360 , r = 0.926
R
r/R e/R mp c k
For 0.950 0.360 0.075 1.880 0.420
0.900 0.360 0.075 1.900 0.450
0.926 0.360 0.075 1.890 0.434
For 0.950 0.360 0.050 1.980 0.480
0.900 0.360 0.050 2.020 0.480
0.926 0.360 0.050 1.999 0.480
C = 1.999 k = 0.480
fc = CMr = 1.999 164.04 10^5 = 32.80 Kg/Cm2
R3
1003
< 66.66667 Kg/Cm2
Fs = m k fc = 10.00 0.480 32.80 = 157.42 Kg/Cm2
< 2000 Kg/Cm2
-
7/29/2019 Sarada Well
16/46
Name of work;- Construction of ROB at KOLAKALURU IN GUNTUR Dist (pier)
P = 427 t
ML = 138 t-m
MT = 174 t-m
MR = ML2
+ MT2
= 222.08 t-m say 223.00 t-m
External dia. of well = 6.00 m
Internal dia of well = 4.20 m
Pier top width = 2.00 m
Pier bottom width = 2.00 m
Verticle loads = 427.00 t
Concrete mix =M 20
C = 66.60
n1 = 10.00 = 0.250
10.00 + 2000
66.60j = 1.00 - 0.250 = 0.917
3
Q = 0.50 x 0.917 x 0.250 66.6 = 7.63
Providing 1.25 m thick Well cap
Intensity of slf weight of well cap = t*2.40 = 3.000 t/sqm
The wellcap is most critical against shear which governs he design.
Hence the cap is designed for shear and checked for bending.
*Dia of wellcap is reduced for design purpose for a thickness of 'ts' 175 mm
considering false steining.
Ex.dia = 6.00 m
45 Degrees dispersion of load from pier is taken upto C.G. of
steel I.e upto a depth of 1.50-0.075-0.016-0.016/2 'd' = 1.151 m (effective depth)
Dispersion length l(ds) = Pier bot.width + 2*disp.width = 4.302 m < 6.00 m
2
Area under UDL = 4.302 = 14.54 sqm
4
UdL. for shear = vrt.load/area +slf wt = 32.38 t/sqm
( 427.00 + 3.000 )
14.54Shear will be calculated at a distance of 'a' i.e 'd'
from inner face of steining
Where a = Int.dia of well/2 -d = 0.949 M
1.151 0.949
V = q*a/2 = 15.363 t/m
DESIGN OF PIER WELL CAP
-
7/29/2019 Sarada Well
17/46
( 32.38 X 0.949 /2)
4.20 m
(Int.dia)
Shear due to overturning moment
Max. SF per 'm' run = po d/4
po = pd/r
d = Dia.of pier r =Ext dia of well/2
Where p = 4*MR R3
= 10.47 t/m
= 4 x 223.00 3
3.00
po = pd/r 10.47 2.00 3 = 6.98 t/m
Shear force = po d/4 = 3.49 t
Total Shear force = v + po d/4 = 18.853 t
Shear stress = 18.853 1.151 16.38 t/sqm < 22.91225 t/sqm
(Permissible shear stress)
or 1.64 Kg/Cm2
< 2.29 Kg/Cm2
Hence shear reinforcement not reqd.)
% of steel M20 M25For 0.150 1.8 Kg/Cm2 1.9 Kg/Cm2
0.250 2.2 Kg/Cm2 2.3 Kg/Cm2
0.500 3 Kg/Cm2 3.1 Kg/Cm2
0.750 3.5 Kg/Cm2 3.6 Kg/Cm2
For M20 Grade concrete for As/bd= 0.273
Permissible Shear stress 2.29 Kg/Cm2
Check for flexure:
For U.D.L Sagging Moment at centre = w*D2/30
Hogging moment at edge = w*D2/30
Where 'D' is c/c of steining D = 6.00 + 4.20 = 5.10 m
2
or Int.dia+Eff.depth = 4.20 + 1.151 = 5.35 m
Effective Span = 5.10 m
For over turning Moments Mu = BM/6a
-
7/29/2019 Sarada Well
18/46
o = (dia. Of pier + well eff.depth)/2 = 1.576
a = (Ex.dia )/2 = 3
o a = 1.576 = 0.5252
3
o a0.3 2.947
0.4 2.062
0.5 1.489
0.6 1.067
0.7 0.731
For simply supported case p 368 case 20
= 1.489 - 1.489 - 1.067 * 0.5252 - 0.50
0.6 - 0.5
= 1.390
For fixed case
For fixed case
o a
0.3
0.4 1.73
0.5 1.146
0.6 0.749
0.7 0.467
= 1.146 - 1.146 - 0.749 * 0.525 - 0.50
0.6 - 0.5
= 1.050
Considering 50% partial fixidity final moments have been considered as one of both the cases.
Case A :
w = 25.62 t/m
D = 6 m
Sagging moment = w*D2/30 = 38.85 t-m
Hogging moment = w*D2/30 = 38.85 t-m
Case B :Due to overturning moments
MR = 223.00 t-m
Moment due to shift of well cap top
= 427 X 0.15 = 64.05 t-m
`
`
-
7/29/2019 Sarada Well
19/46
Total moment = MR + 64.05 = 287.05 t-m
Max. radial moment = S.S + FIX X 287.05 = 19.46 t-m
2 6 X a
Sagging Moment = Sag.Mom + Max. Rad.Mom. = 58.31 t-m
Hogging Moment = Hog.Mom + Max. Rad.Mom. = 58.31 t-m
d eff required = Sag.Mom = 87.44 Cm < 1.151 m
7.63 100
Bottom Reinforcement = Sag.Moment * 1/100 = 27.63 Cm2/m
Ast provided = 30 Cm2/m width
Using 20 10.47 Cm
Provide 20 mm dia. Tor bars both ways both at top & bottom @ . 100 mm c/c
% of Steel = 100 X 3.14 X 100 = 0.273
10 100 1.151 X 100
Steel = 100 + 100 * 2 * 1.578 = 6.312 Kg
100 100
Per Cum = 38.25 Kg Say 40 Kg
1450 Kg per Pier cap.
t * j * d
mm dia. Tor bars , Spacing =
-
7/29/2019 Sarada Well
20/46
-
7/29/2019 Sarada Well
21/46
-
7/29/2019 Sarada Well
22/46
`
-
7/29/2019 Sarada Well
23/46
Name of work;- Construction of ROB at TADEPALLIGUDEM
in West Godavari District.
P = 475.5 t
ML = 155.19 t-m
MT = 255.88 t-m
MR = ML + MT
= 299.26 t-m say 300.00 t-m
External dia. of well = 6.00 m
Internal dia of well = 4.20 m
Pier top width = 2.00 m
Pier bottom width = 2.00 m
Verticle loads = 475.50 t
Concrete mix =M 20
C = 66.60
n1 = 10.00 = 0.250
10.00 + 2000
66.60
j = 1.00 - 0.250 = 0.9173
Q = 0.50 x 0.917 x 0.250 66.6 = 7.63
Providing 1.25 m thick Well cap
Intensity of slf weight of well cap = t*2.40 = 3.000 t/sqm
The wellcap is most critical against shear which governs he design.
Hence the cap is designed for shear and checked for bending.
*Dia of wellcap is reduced for design purpose for a thickness of 'ts' 175 mm
considering false steining.
Ex.dia = 6.00 m
45 Degrees dispersion of load from pier is taken upto C.G. of
steel I.e upto a depth of 1.50-0.075-0.016-0.016/2 ' d' = 1.151 m (effective depth)
Dispersion length l(ds)= Pier bot.width + 2*disp.width = 4.302 m < 6.00 m
2
Area under UDL = 4.302 = 14.54 sqm
4
UdL. for shear = vrt.load/area +slf wt = 35.71 t/sqm
( 475.50 + 3.000 )
14.54
Shear will be calculated at a distance of 'a' i.e 'd'
from inner face of steining
Where a = Int.dia of well/2 -d = 0.949 M
1.151 0.949
V = q*a/2 = 16.946 t/m
( 35.71 X 0.949 /2)
4.20 m
(Int.dia)
DESIGN OF PIER WELL CAP
-
7/29/2019 Sarada Well
24/46
Shear due to overturning moment
Max. SF per 'm' run = po d/4
po = pd/r
d = Dia.of pier r =Ext dia of well/2
Where p = 4*MR R3
= 14.11 t/m
= 4 x 300.00 3
3.00
po = pd/r 14.11 2.00 3 = 9.41 t/m
Shear force = po d/4 = 4.70 t
Total Shear force = v + po d/4 = 21.650 t
Shear stress = 21.650 1.151 18.81 t/sqm < 24.12472 t/sqm(Permissible shear stress)
or 1.88 Kg/Cm2
< 2.41 Kg/Cm2
Hence shear reinforcement not reqd.)
% of steel M20 M25
For 0.150 1.8 Kg/Cm2 1.9 Kg/Cm2
0.250 2.2 Kg/Cm2 2.3 Kg/Cm2
0.500 3 Kg/Cm2 3.1 Kg/Cm2
0.750 3.5 Kg/Cm2 3.6 Kg/Cm2
For M20 Grade concrete for As/bd= 0.303
Permissible Shear stress 2.41 Kg/Cm2
Check for flexure:
For U.D.L Sagging Moment at centre = w*D2/30
Hogging moment at edge = w*D2/30
Where 'D' is c/c of steining D = 6.00 + 4.20 = 5.10 m
2
or Int.dia+Eff.depth = 4.20 + 1.151 = 5.35 m
Effective Span = 5.10 m
For over turning Moments Mu = BM/6a
o = (dia. Of pier + well eff.depth)/2 = 1.576
a = (Ex.dia )/2 = 3
o a = 1.576 = 0.5252
3
o a
0.3 2.947
0.4 2.062
0.5 1.489
0.6 1.067
0.7 0.731
For simply supported case p 368 case 20
-
7/29/2019 Sarada Well
25/46
= 1.489 - 1.489 - 1.067 * 0.5252 - 0.50
0.6 - 0.5
= 1.390
For fixed caseFor fixed case
o a
0.3
0.4 1.73
0.5 1.146
0.6 0.749
0.7 0.467
= 1.146 - 1.146 - 0.749 * 0.525 - 0.50
0.6 - 0.5
= 1.050
Considering 50% partial fixidity final moments have been considered as one of both the cases.
Case A :
w = 25.62 t/m
D = 6 m
Sagging moment = w*D2/30 = 42.86 t-m
Hogging moment = w*D2/30 = 42.86 t-m
Case B :
Due to overturning moments
MR = 300.00 t-m
Moment due to shift of well cap top
= 475.5 X 0.15 = 71.325 t-m
Total moment = MR + 71.325 = 371.33 t-m
Max. radial moment = S.S + FIX X 371.33 = 25.17 t-m
2 6 X a
Sagging Moment = Sag.Mom + Max. Rad.Mom. = 68.02 t-m
Hogging Moment = Hog.Mom + Max. Rad.Mom. = 68.02 t-m
d eff required = Sag.Mom = 94.44 Cm < 1.151 m
7.63 100
Bottom Reinforcement = Sag.Moment * 1/100 = 32.23 Cm2/m
Ast provided = 33 Cm2/m width
Using 20 9.52 Cm
Provide 20 mm dia. Tor bars both ways both at top & bottom @ . 90 mm c/c
t * j * d
mm dia. Tor bars , Spacing =
`
`
-
7/29/2019 Sarada Well
26/46
% of Steel = 100 X 3.14 X 100 = 0.303
9 100 1.151 X 100
Steel = 100 + 100 * 2 * 1.578 = 7.013 Kg
90 90
Per Cum = 38.25 Kg Say 40 Kg
1450 Kg per Pier cap.
-
7/29/2019 Sarada Well
27/46
Vertical Load P = 352.44 t
Long.Moment ML = 180.27 t
Trans.Moment MT = 107.38 t-m
Resultant Moment MR = 209.828 t-m
Thickness of well cap = 1.50 m
External dia. Of well = 8.00 m
Internal dia of well = 6.50
Steining Thickness = 0.75 m
Steining height = 4.00 m
Curb height = 1.50 m
Straight length of Abutment 11.05 m
Assuming Reinforcement dia. as 2 cm
Cover = 7.5 cm
B(C/C ABUT TO EXTREEM END C/C STEINING) = 6.4875 M
Base width of Abutment at Sill level = 1.9 m
R.C.L = 100.470
SILL = 95.000
effective depth d = 139.5 Cm
(For Long.direction)
dp, in perpendicular dimension 141.5 Cm
(For Tran.direction)
Assume 1:1 dispersion for vertical load through well cap
Width of dispersion (2.50+1.395) = 3.295 m
Length of dispersion L = 11.05 m
Intensity of load load = 9.68 t/sqm
unit area
The wellcap is assumed as partially fixed alround its circumstance.
i) Effective span S.S case = 7.895 m
(Int.dia + w.cap thik -cover -2-2/2)c/c length = Int.dia + st.thk) = 7.25 m
Eff.span = 7.25
Effective span fixed case = 6.50 m
Effective span for partially fixed case D 6.88 m
For simplicity of analysis the intensity of direct load coming
on wellcap is assumed to act over entire area of wellcap .
From Reynold's handbook
Ave. B.M at centre of partially fixed circular slab = wD2/32
Average B.M around edge = -wD2/32
Self weight of wellcap = 3.60 t/sqm
Load intensity = 9.68 t/sqm
Total w = 13.28 t/sqm
Moment in longitudinal direction /m = 15.39 t-m/m
ml ML/disp.length = MLXB/D*L
Moment in transverse direction /m = 32.59 t-m/m
MT/disp.width
B.M due to direct load wD2/32 = 19.615 t-m/m
MOVEMENT IN LONGITUDINAL DIRECTION + = 15.39 t-m/m
= (MLXB)/(DXL)
Moment at support IN L.direct = = 3.85 t-m/m
DESIGN OF ABUTMANT WELL CAP
(ML*B)/(4*D*L)
-
7/29/2019 Sarada Well
28/46
Assuming 50% for externally applied moments IN TRANSVERSE DIRECTION
Moment at centre = M/2
Moment at support = M/8
Final B.M at centre of wellcap
I) AT CENTRE OF WELL CAP:
Longitudinal B.M +ve = wD2/32 +(MLXB)/(D*L) = 35.01 t-m/m (Bottom)
Longitudinal B.M -ve = 0 +(ML*B)/(D*L) = 15.39 t-m/m (Top)
Transverse B.M +ve = wD /32 +MT/2 = 35.91 t-m/m (Bottom)
Transverse B.M -ve = 0 +MT/2 = 16.29 t-m/m (Top)
II) AT SUPPORT:
Longitudinal B.M -ve = w X D2 + (MLXB) = 23.46 t-m/m (Bottom)
32 (4*D*L)
Transverse B.M -ve = w X D2 + MT = 23.69 t-m/m
32 8.00
Max. B.M = 35.91 t-m/m
CHECK FOR DEPTH: CONCRETE MIX = M 20
STEEL Fe 415
C = 66.66 kg/sq.Cm Q = 7.630
t = 2000 kg/sq.Cm j = 0.917
eff.depth required = 35.91 * 105
= 68.60 Cm
100 * 7.63
< 139.5 Cm
Keeping Transverse Reinforcement nearer to Surface.
Hence O.KReinforcement:
1 Longitudinal B.M +ve = 35.01 t-m/m
Ast = Longitudinal B.M +ve = 13.68 Cm2
2000 X 0.917 X 139.50
>
Ast minimum
Ast minimum =0.12*b*D/100 = 18.00 Cm2
Using 20 mm dia.bars
Spacing required = 4.91*100/18 = 17.45 Cm
However provide 20 mm dia.bars at 150 mm c/c at bottom in
Longitudinal direction.
2 Longitudinal B.M -ve at Support (top) = 23.46 t-m/m
Ast = Longitudinal B.M -ve = 9.17 Cm2
2000 X 0.917 X 139.50