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PROJECT: DOCUMENT NO. DATE
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INDEX
PG.NO
1.0 INTRODUCTION 1
2.0 DESIGN OF OHT 2
3.0 ANNEXURE -1-CALCULATION OF WIND PRESUURE 32
4.0 ANNEXURE -2-CALCULATION OF DESIGN CRACK WIDTH 33
5.0 ANNEXURE -3-CALCULATION OF SCOUR DEPTH 34
Arugula Rajaram Guthpa Lift Irrigation Project, Nizamabad
Dist., A.P
Design of OST for Surge protection-Stage 1
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1
1.0 INTRODUCTION:
1.1 General
Individual Surge tank of diameter - 10m is proposed for each pipe line as per surge analysis report.
Tank detils are proposed as per the details given in Table - 1 of surge analysis report.
This document covers the Design of over head water tank for Surge protection of stage 1 -
Pipe line.The design of OHT includes the design of Cover Slab , tank walls , Base slab ,
Ring beam , Columns, Tie beams and Foundation.
Calculation of Wind Pressure,Crack Width Check and scour depth are given in Annexure I,II and III.
1.2 Material Specifications
a) Grade of Concrete M25
b) Grade of steel - High yield deformed bars with yield stress of 415 N/mm2
1.3 References
a) IS:456- 2000 - Plain & reinforced concrete - Code of Practice (Fourth revision)
b) SP: 16;
c) IS:3370 - 1965 - Code of practice for concrete structures for the Storage of Liquids
Part I - General Requirements
Part II - Reinforced Concrete Structures
Part IV - Design Tables
e) IS:1893-1984 - Criteria for Earthquake resistant Design of Structures (Fourth revision)
f) IS:1893-2002 - Criteria for Earthquake resistant Design of Structures (Part 1) (Fifth revision)
g) Reinforced Concrete Sructures (Vol 2 ) by B.C.Punmia
Argula Rajaram Guthpa Lift Irrigation Project, Nizamabad
Dist., A.P
Design of OST for Surge protection-Stage 1
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ANNEXURE - III
5.0 CALCULATION OF SCOUR DEPTH
Scour Depth calculations:
Discharge = 1600000 Cusecs( SRSP Dam discharge )
= 45307.8099 Cumecs
M.W.L at Pump house Location = 337.00 M
Scour Depth, R = 0.473 x ( Q / f ) 1/3
Taking f = 4.75
\Scour Depth, R = 0.473 x ( 45307.81 / 4.75 ) 1/3
= 10.031 m
Scour Depth, R proposed = 1.27 R
= 1.27 x 10.031
= 12.740 m
say 12.750 m
Scour Level = 337 - 12.75
= 324.25 M
Hence the foundation Level is proposed at = + 324.25 m
( Since Morrum is available at
anticipated scour level. )
Arugula Rajaram Guthpa Lift Irrigation Project, Nizamabad
Dist., A.P
Design of OST for Surge protection-Stage 1
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1.0 DESIGN OF ONE WAY SURGE TANK
1.1 Basic Data
Design capacity of tank V =10.0 m
Free board h' = 0.6 m
Storage depth of water tank H = 4.0 m
Proposed Diameter of water tank D = 10.0 m
Ground level =
Height of staging = 20.0 m
Bottom of water tank =
Full tank level = GL
Clear Height of tank walls H = 4.60 m
Maximum depth of water = 4.00 m 4
Grade of Concrete = M25
Grade of steel = Fe415
Characteristic strength of concrete Fck= 25.0
Characteristic strength of steel Fy= 415
Unit weight of concrete Wc= 25
Unit weight of water w = 10
1.2 Assumptions
Thickness of Cover Slab = 0.20 m
Overall thickness of wall = 0.275 m
Thickness of Base Slab = 0.375 m
Dimensions of ring beam = 0.50 m x 0.70 m
Mean diameter of ring beam = 8.0 m
No of Columns = 8
Dimensions of Column below base Slab = 0.50 m x 0.50 m
Dimensions of Braces = 0.40 m x 0.50 m
Argula Rajaram Guthpa Lift Irrigation Project, Nizamabad Dist., A.P
Design of OST for Surge protection-Stage 1
EL. 352.3 m
315000 Lit
EL. 328.3 m
EL. 348.30 m
20.0
N/mm2
kN/m3
N/mm2
kN/m3
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Argula Rajaram Guthpa Lift Irrigation Project, Nizamabad Dist., A.P
Design of OST for Surge protection-Stage 1
Let the approximate length of brace be (pd/n) = 3.14 m
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Argula Rajaram Guthpa Lift Irrigation Project, Nizamabad Dist., A.P
Design of OST for Surge protection-Stage 1
Load due to floof finish = 0.5Sq. m
1.3 Calculation of Seismic Forces
1.3.1 Calculation of Seismic Weight
Case 1 : When Tank is Empty
IS1893:1984-
Cl5.2.4Seismic Weight of Structure = Wt of Empty tank+1/3 of wt of Staging
Weight of Cover Slab : pi( )/4 * 10.55 ^ 2 * 0.2 * 25 = 437 kN
Weight of Tank Walls : p( )/4*(10.55 ^ 2 - 10 ^ 2) * 4.6 * = 1020 kN
Weight of Base Slab : p( )/4 * 10.55 ^ 2 * 0.375 * 25 = 819 kN
Weight of Finishes : p( )/4 * 10.55 ^ 2 * 0.5 * 2 = 87 kN
Weight of Ring Beam : 220 kN
Total Weight of Tank Portion = 2583 kN
Weight of Columns : 25 * 8 * 0.5 * 0.5 * 20 = 1000 kN
Weight of Braces : (25 * 0.4 * 0.5 * 3.14 * 8 ) * 4 = 503 kN
Total Weight of Staging : 1000 + 503 = 1503 kN
Total Seismic Weight W: 2583 + 1/3 * 1503 = 3084 kN
Case 2 : When Tank is Full
Weight of Water : p( ) / 4 * 10 * 10 * 4 * 10 = 3140 kN
Total Seismic Weight W 3084 + 3140 = 6224 kN
1.3.2 Calculation of Seismic Co-efficient (ah)
Seismic Co-efficient (ah) = b I F
oS
a/g
IS1893:1984-
Tab3For raft Foundations b = 1.00
IS1893:1984-
Tab4Importance Factor I = 1.50
IS1893:2002-
Tab2Zone Factor Fo for Zone III = 0.16
Time Period T = 2pSqrt(D/g)
Percentage of Damping = 5.00
To find total Stiffness of the entire Columns
p( )/4*(8.5^ 2 - 7.5 ^ 2 ) * 0.7 * 25 =
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Argula Rajaram Guthpa Lift Irrigation Project, Nizamabad Dist., A.P
Design of OST for Surge protection-Stage 1
I1 = (0.5x0.5^3 / 12) x 2 = 0.010 m4
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Argula Rajaram Guthpa Lift Irrigation Project, Nizamabad Dist., A.P
Design of OST for Surge protection-Stage 1
I2 = [(0.5x0.5^3 / 12)+( 0.5x0.5) x(8/4)^2]4 = 4.005
I3 = [(0.5x0.5^3 / 12)+( 0.5x0.5) x(8/2)^2]2 = 8.005
Total Moment of Inertia of the Section I = 0.01+4.005+8.005
= 12.021
Distance of C.G of tank from Ground Level = 20 + 4.6 / 2 + 0.7 + 0.375
= 23.4 m
Stiffness of the Section = K = 12EI/L3
= (12 * 5000 * SQRT( 25 ) * 12.02 * 10 ^ 12 / ( 23.375 * 1000 ) ^ 3 )
= 282358.6
When tank is Empty
Total Deflection D = W / K = 3084.37 x 10^3 / 282358.57 = = 10.92
Time Period ( T ) = 2pSQRT(D/g) = (2*3.141592*SQRT((10.92*10 -3/9.81)))
= 0.21
IS1893:1984-
FIG 2Average Acceleration Co-eeficient Sa/g = 0.20
ah = 1 x 1.5 x 0.16 x 0.2 = 0.048
Lateral Force ah * W = 0.048 x 3084.37 = 148 kN
When tank is Full
Total Deflection D = W / K = 6224.37 x 10^3 / 282358.57 = = 22.04
Time Period ( T ) = 2pSQRT(D/g) = (2*3.141592*SQRT((22.04*10 -3/9.81)))
= 0.30
IS1893:1984-
FIG 2Average Acceleration Co-efficient Sa/g = 0.20
ah = 1 x 1.5 x 0.16 x 0.2 = 0.048
Lateral Force ah * W = 0.048 x 6224.37 = 299 kN
Hence Critical Condition will be when the tank is full and the above corresponding lateral force
is applied at the C.G of tank and analysed
m4
m4
m4
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Argula Rajaram Guthpa Lift Irrigation Project, Nizamabad Dist., A.P
Design of OST for Surge protection-Stage 1
1.3.3 Calculation of Hydrodynamic Pressure
When a tank Containing fluid vibrates the fluid exerts impulsive and Convective pressure on the tank
As per Clause 5.2.7.3 of IS1893-1984
Pressure on the wall of Circular Container Pw
Pw = ahwh*sqrt(3)Cosf'[(y/h)-0.5(y/h)2]Tanh*Sqrt(3)(R/h)
Substituting the values we get
y 0.2 H 0.4 H 0.6 H 0.8 H 1.0 H
Pw(KN/m2) 0.657 1.168 1.533 1.753 1.826
Pressure at the bottom of the tank Pb = ahwh*Sqrt(3/2)[(SinhSqrt(3)(x/h)) /CoshSqrt(3)(R/h)]
y 0 0.2 0.4 0.6 0.8 1.0
Pb(KN/m2) 0 0.219 0.470 0.788 1.219 1.826
Maximum Pressure on the wall = 1.826
Maximum Pressure at the base = 1.826
1.4 Critical Load Condition
Case-1: Water is full inside & no wind pressure on outer side of the tank
Water pressure on wall(Static Pressure) pw= 4.6 * 10 = 46.0
Maximum Hydro dynamic Pressure on the wall pw' = 1.83
Maximum water pressure on wall = 46 + 1.83 = 47.83
1.4.1 Determination of Hoop tension and Bending Moment
Maximum water pressure on wall = 47.83
kN/m2
kN/m2
Reservoir Bed47.83
kN/m2
EL. 352.3 m
EL. 348.30 m
kN/m2
kN/m2
kN/m2
R R R RRR
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Argula Rajaram Guthpa Lift Irrigation Project, Nizamabad Dist., A.P
Design of OST for Surge protection-Stage 1
TYPICAL CROSS SECTION OF TANK
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Argula Rajaram Guthpa Lift Irrigation Project, Nizamabad Dist., A.P
Design of OST for Surge protection-Stage 1
Considered that the wall panel is fixed at bottom & sides and hinged at top.
5.00 m 5.00 m
From IS:3370 Moment and shear coefficients
Thickness of wall (tentatively) 3H + 5 cm = 3 * 4.6 + 5
= 18.8 cm
Assume thickness of wall = 0.275 m
H2/Dt = 4.6^2 / 10x0.275 = 7.695
IS3370:1967-
Part4The ring tension at any height T = coefficient x w x H R
IS3370:1967-
Part4The Bending moment at any height M = coefficient x w x H
3
R = 5.00 m
H = 4.60 m
w = 10.0
w' = 1.826 / 4.6
= 0.40
Free
FixedX
Y
RR
H =
4.60 m
kN/m3
kN/m3
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Argula Rajaram Guthpa Lift Irrigation Project, Nizamabad Dist., A.P
Design of OST for Surge protection-Stage 1
As per Table 9 and 10 of IS:3370
Coefficient Tension (KN) CoefficientMoment
(KN.m)
0.0 H 0 0.00190182 0.45 0.00000 0.000
0.1 H 0.46 0.10629091 25.42 0.00002 0.015
0.2 H 0.92 0.22044364 52.71 0.00013 0.132
0.3 H 1.38 0.33637455 80.44 0.00029 0.295
0.4 H 1.84 0.44269455 105.86 0.00097 0.980
0.5 H 2.3 0.52941818 126.60 0.00184 1.866
0.6 H 2.76 0.56568364 135.27 0.00307 3.112
0.7 H 3.22 0.51732364 123.71 0.00400 4.046
0.8 H 3.68 0.36878182 88.19 0.00290 2.935
0.9 H 4.14 0.14504364 34.68 -0.00249 -2.520
1.0 H 4.6 0 0.00 -0.01523 -15.409
Maximum Tension = ######
Max. Bending moment =
1.5 Design of side wall for Hoop tension
1.5.1 Check for Thickness of wall
As per IS:337 (Designed as Uncracked Section)
Concrete(N/Sq.mm)
m K = 0.38
1.8 52 J = 0.87
Q = 0.30
150
Depth x below
Water surface
Bending momentHoop tension
15.41 KN.m
Allowable bending stresses Design factors
Steel(N/Sq.mm)
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Argula Rajaram Guthpa Lift Irrigation Project, Nizamabad Dist., A.P
Design of OST for Surge protection-Stage 1
For members with thickness greater than 225mm
Allowable tensile stress on face away for liquid = 190
Thickness of wall required SQRT(6M / b scbt) = SQRT(15.41x10^6 x 6 / (1000 x 1.8 ))
=
Nominal cover to reinforcement = 25 mm (on water face side)
Diameter of reinforcement bar = 16 mm
Total thickness of wall required = 226.64+25+ 16 / 2
=
Provide overall thickness of wall = 275 mm
Effective thickness of wall provided =
1.5.2 Design of Hoop reinforcement
Maximum tesnion T = 135.27 KN
Area of steel required near water face = 135.27x 10^3 / 150 = 902
Area of steel reqd away from water face =135.27x 10^3 / 190 = 712
Minimum percentage = 0.20 % (on both faces)
Minimum area of reinforcement = 0.2 / ( 2 * 100 ) * 1000 * 275 )
= 275 (on each face)
Required area of steel near water face = 902
Required area of steel away from water face = 712
Required near water face
Provide near water face
Required away from water face
Provide away from water face
1.5.3 Design of Vertical Reinforcement
Maximum Moment M =
Area of steel required = 15.41x10^6 / (150x0.87x242) = 487
Minimum percentage = 0.20 % (on both faces)
275 mm c/c as
15.41 KN.m
16 mm dia bars at 200 mm c/c as
16 mm dia bars at 282 mm c/c as
16 mm dia bars at
226.6 mm
259.6 mm
242 mm
16 mm dia bars at 223 mm c/c as
mm2
mm2
mm2
mm2
N/mm2
mm2
mm2
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Design of OST for Surge protection-Stage 1
Minimum area of reinforcement = 275 (on each face)mm2
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Argula Rajaram Guthpa Lift Irrigation Project, Nizamabad Dist., A.P
Design of OST for Surge protection-Stage 1
Required area of reinforcement = 487
Required
Provide Vertical reinforcement
1.5.4 Provision of Haunches
It is compulsory to provide 150mm x 150mm haunches at the junction of the wall and the base.
A haunch reinforcement of 8mm dia bars at 200mm c/c may be provided.
1.6 Design of Base slab
Assume thickness of base slab T =
Self weight of base slab = 375x 10^ -3 x25 = 9.38
Assuming load due to finishes = 0.5
Water pressure 4.6 * 10 = 46.0
Hydro dynamic pressure at the base slab = 1.83
Total UDL on base slab = 9.375+0.5+46+1.83 P = 57.7
Weight of walls = Pi /4 (10.55^2 - 10^2 ) x4.6 x 25 1021 KN
Diameter of cover slab = 10 + 2 x 275x10^ -3 =
Live load on cover slab = 1.5
Assuming load due to finishes = 0.5
Assume thickness of cover slab =
Self weight of cover slab = 200x 10^ -3 x25 = 5.00
Total UDL on cover slab = 1.5+0.5+5 = 7.00
Total load on walls from cover slab = Pi /4 x 10.55^2 x7 = 612 KN
Total load on base slab from walls = 1021 +612 = 1633 KN
Diameter of supporting circular ring beam = 8.0 m
10 mm dia bars at 150 mm c/c as
375 mm
10.550 m
200 mm
10 mm dia bars at 161 mm c/c as
mm2
KN/m2
KN/m2
KN/m2
KN/m2
KN/m2
KN/m2
KN/m2
KN/m2
KN/m2
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Design of OST for Surge protection-Stage 1
4 4
a = 5.138 a = 5.138
Total Load on base slab W = 1633+ ((PI()/4) * ( 10 + 275 * 10 ^ -3 ) ^ 2 * 57.7 )
= 6417 KN
Radial deflection of the wall due to hoop tension is "0" for wall is monolithick with base slab
Hence the clockwise slope y'w= 1/E { (prD2/4TH) (1-a H) + 0.5 M/aI }
Where y'w= 1/E { k1 + k2 M } Eq. 1.0
k1 = (prD2/4TH) (1-aH)
k1 = ((47.83*10^2)/(4*0.275*4.6))*(1-1.12*4.6) = -3934.6
k2 = 0.5/aI
k2 = 0.5 / ( 1.122 * 0.002 ) = 257.05
pr= (w+w' ) * H
= 47.83
D =
Thickness of wall T = 0.275 m
H = 4.6 m
Moment of inertia I = 1.T3/12
I = (1 * 0.275 3 ) / 12 = 0.0017
a4= T/ I D
2
a4= 0.275 / ( 0.0017 * 10 2 ) = 1.5868
a2= 1.2597
a= 1.122
Clockwise moment per unit circumferential
length of wall at its bottom edge. = M
10.0 m
57.7 KN/m2
1633 KN1633 KN
KN/m2
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Design of OST for Surge protection-Stage 1
Clockwise slope of slab at its edges
ys= 1/E{(1.5Pa3/T
3)-(0.95493*W/T
3)((a
2-b
2)/a)-(12aM/T
3)}
Where ys= 1/E { c1 - c2 - c3 M } Eq. 2.0
c1 = (1.5Pa3/T
3)
c1 = (1.5 * 57.7 * 5.1375 ^ 3 ) / ( 0.375 ^ 3 ) =
c2 = (0.95493*W/T3)((a
2-b
2)/a)
c2 = ((0.954929658*6417.23)/(0.052734375))*(((5.1375^2)-( =
c3 = 12a/T3
c3 = (12 * 5.1375 ) / 0.053 = 1169.1
a = 5.1375 m
b = 4.000 m
Thickness of base slab T = 0.375 m
T3= 0.0527
W = 6417 KN
P = 57.70
By equating equation 1 & 2
1/E { k1 + k2 M } = 1/E { c1 - c2 - c3 M }
k1 + k2 M = c1 - c2 - c3 M
M = (c1-c2-k1) / (c3+k2)
Bending moment = (222552.74-235099.91--3934.57)/(1169.07+257.05)
=
A. Determination of Radial and circumferential BM
BM due to UDL p = 57.70
a = 5.1375 m
(Mr)c=
(Mr)e=
r - Varies from 0 to a (Mr) =
0.00 kN.m
(3/16) *p (a2-r
2)
-6.04 kN.m
222552.7
235099.9
(3/16) *p a2
KN/m2
KN/m2
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Design of OST for Surge protection-Stage 1
(Mq)c= (3/16) *p a2
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Design of OST for Surge protection-Stage 1
(Mq)e=
(Mq) =
The values of Mr& Mqat various locations are tabulated below: (KN.m)
r (m) Mr=(3/16) *p (a2-r
2)
0.0 285.6 285.6
2.0 242.3 271.1
4.00 112.5 227.9
5.1375 0.0 190.4
B. Determination of moment due to upward load W
Total upward load W =
For r < b (Mr) = (Mr)b= (Mq) = (Mq)0=
(Mr) = (- W/8p) *{2ln (a/b) + 1 - (b/a)2}
For r > b (Mr) = (- W/8p) *{2ln (a/r) - (b/a)2+ (b/r)
2}
(Mq) = (- W/8p) *{2ln (a/r) - (b/r)2+ 2 - (b/a)
2}
The values of Mr& Mqat various locations are tabulated below: (KN.m)
r (m) Mr Mq
0.0 -228.4 -228.4
2.0 -228.4 -228.4
4.00 -228.4 -228.4
5.138 0.0 -201.1
C. Bending moment due to M
Hogging BM at the ends there will be constant BM of
(Mr) =
(2/16) *p a2
Mq= (1/16) *p (3a2-r
2)
(1/16) *p (3a2-r
2)
6417 kN
-6.04 kN.m
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Design of OST for Surge protection-Stage 1
(Mq) = -6.04 kN.m
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Design of OST for Surge protection-Stage 1
D. Net Bending moments
The net moments will be algebraic sum of the three, and are tabulated below:
r (m) Mr Mq
0.0 51.16 51.16
2.0 7.88 36.73
4.00 -121.95 -6.54
5.138 -6.04 -16.77
E. Determination of Shear force
Maximum shear force occurs at the outer edge of the ring beam and its magnitude is
F = (pb/2) - (W/2pb)
p = 57.70
W =
b = 4.00 m
F = (57.7x4 / 2) - (6417.23 /2x3.14x4 ) =
1.6.1 Check for Base slab thickness
Slab is to be designed for maximum BM M = 121.950 (Hogging)
Designed as limited Crack Width Section
Concrete(N/Sq.mm)
m K = 0.38
8.5 11 J = 0.87
Q = 1.40
Thickness of Slab required = SQRT(121.95 x 10^6 / (1000 x 1.4 )
=
Nominal cover to reinforcement = 25 mm (on water face side)
Diameter of reinforcement bar = 20 mm
Total thickness of slab required = 295.14+25+20 / 2
=
295.1 mm
6417 kN
-139.9 kN
Allowable bending stresses Design factors
Steel(N/Sq.mm)
150
330.1 mm
KN/m2
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Design of OST for Surge protection-Stage 1
Provide overall thickness of slab = 375 mm
Effective thickness of slab provided = 340.0 mm
Area of Radial reinforcement = 121.95x10^6 / (150x0.87x340) = 2742
Minimum percentage = 0.06 % (on each face)
Minimum area of reinforcement = 225
Required area of reinforcement = 2742
Hence provide 25 mm dia bars at 175 mm c/c as radial reinforcement at top
Radial moment is zero at r = SQRT(((0.1875*57.7*5.14^2)+-228.36--6.04)/(0.1875*57.7))
= 2.42 m
The radial moment is found to be zero at a radiu 'r' given by
r Mr1 Mr2 Mr3 SMri
2.42 222.3 -228.4 -6.04 -12.1
Therefor, the hogging BM eixists from r = a to r = 2.42 m
with its maximum value at r
Hence the above radial bars may be completely curtailed at r = 2.42 m
providing hooks there.
Radial & Circumferential Reinforcement at centre of the base slab (at bottom)
At the centre of the slab radial moment is positive and the circumferential moment is also positive.
M =
Effective depth of slab =
Area of steel for Radial reinforcement Ast= 1000000 * 51.16 / ( 315 * 0.87 * 150 )
= 1242
Minimum percentage = 0.06 % (on each face)
Minimum area of reinforcement = 0.06 / 100 * 375 * 1000
= 225
Required area of reinforcement = 1242
51.2 kN.m
= b i.e 4.00 m
315.0 mm
mm2
mm2
mm2
mm2
mm2
mm2
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Design of OST for Surge protection-Stage 1
Hence provide Circumferential and radials16 mm dia bars at 150 mm c/c as
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Design of OST for Surge protection-Stage 1
Circumferential reinforcement at outer edge of the slab (at top)
M =
Effective depth of slab = 340 - 16 =
Area of steel for Circumferential reinforcement Ast= 1000000 * 16.77 / ( 324 * 0.87 * 150 )
= 396
Minimum percentage = 0.06 % (on each face)
= 0.06 / 100 * 375 * 1000
Minimum area of reinforcement = 225
Required area of reinforcement = 396
Required Circumferential
Hence Provide 10 mm dia bars at 175 mm c/c as circumferential rft at top
1.7 Design of Circular beam
Vertical load on beam W =
Mean radius of curved beam R = 4.00 m
UDL on circular beam = 6417.23 / (2x3.14x4) =
Assume size of beam as D = 0.7 m
B = 0.5 m
Self weight of beam = 0.7x0.5x25 Wsb =
Total load on beam = 255.33 + 8.75 ( w )=
Let us support the beam on 8 equally spaced columns
324.0 mm
10 mm dia bars at
6417 kN
198 mm c/c as
16.77 kN.m
255.3 kN/m
8.8 kN/m
264.08 kN/m
mm2
mm2
mm2
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Design of OST for Surge protection-Stage 1
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Design of OST for Surge protection-Stage 1
From Table 1.1 (B.C. Punmia)
For 8 nos of supports
2q C1 C2 C3 fm
45 0.066 0.03 0.005 9.5
w R22q = 264.08x4^2 x (3.14/180) x45 =
Maximum -ve B.M at Support Mo = C1w R22q
Mo = 0.066 * 3319 =
Maximum +ve BM at mid span M1 = C2w R22q
M1 = 0.03 * 3319 =
Maximum torsional moment Mtm = C3w R
22q
Mtm = 0.005 * 3319 =
Maximum shear force at support Fo = w R q
Fo = 264.08 * 4 * ( 45 / 2 ) * 3.141592 / 180 =
Shear force at any point is given by F= w R(q- f)
At f= fm F= w R(q- fm)
264.08x4x((45x3.14 / (2x180)) - (9.5x3.14/180)) = F =
Bending moment at the point of maximum torsional moment
=
As per IS:456-2000
Concrete(N/Sq.mm)
m K = 0.38
8.5 11 J = 0.87
Q = 1.42
1.7.1 Check for Depth of beam
Maximum moment M =
Width of beam B =
Required effective depth of beam d = SQRT(219.03x10^6 / 1.42x500)
0.0 kN.m
Design factors
Steel(N/Sq.mm)
150
500 mm
Allowable bending stresses
239.7 kN
414.8 kN
219.03 kN.m
100 kN.m
3319 kN.m
219 kN.m
16.6 kN.m
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Design of OST for Surge protection-Stage 1
=
Nominal cover to reinforcement = 25 mm
Diameter of reinforcement bar = 25 mm
Total depth of beam required =
Provide overall depth of beam = 700 mm
Effective depth of beam provided = 700 - 25 - 25 / 2
=
1.7.2 Main and Longitudinal reinforcement
T = Mt
m = 16.6 kN.m
M = 0.0 kN.m
Met = MT+ M
MT = T ((1+D/B)/1.7)
16.6[(1+(700/500)/1.7)] = 23.4 kN.m
Met = 23.4 kN.m
Area of steel for Radial reinforcement Ast1 = 1000000 * 23.43 / ( 150 * 0.87 * 662.5 )
= 270
Minimum percentage = 0.20 % (on each face)
Minimum area of reinforcement = 0.2 / 100 * 700 * 500
= 700
Required area of reinforcement = 700
Required 2 Nos. as Torsion reinforcement
Met2
= MT- M
= 23.4 kN.m
Hence provide 2 Nos. as at top and bottom as torsional rft
Maximum hogging BM at support Mo =
Mto = 0.0 kN.m
Area of steel for Radial reinforcement Ast1 = 219.03 * 10^6 / ( 150 * 0.87 * 662.5 ))
662.5 mm
25 mm dia bars at
219.03 kN.m
25 mm dia bars at
555.4 mm
592.9 mm
mm2
mm2
mm2
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Design of OST for Surge protection-Stage 1
= 2528
Minimum percentage = 0.20 % (on each face)
Minimum area of reinforcement = 0.2 / 100 * 700 * 500
= 700
Required area of reinforcement = 2528
Required 6 Nos. as Main longitudinal reinforcement
Provide 6 Nos. as longitudinal reinforcement at top
Maximum Sagging BM at mid sapn M1 =
Mt
c = 0.0 kN.m
Area of steel for Radial reinforcement Ast1 = 1000000 * 99.56 / ( 150 * 0.87 * 662.5)
= 1149
Minimum percentage = 0.20 % (on each face)
= 0.2 / 100 * 700 * 500
Minimum area of reinforcement = 700
Required area of reinforcement = 1149
Required 6 Nos. as Main longitudinal reinforcement
Hence provide 6 Nos. as longitudinal reinforcement at bottom
1.7.3 Transverse Reinforcement
At point of maximum torsional moment F = 239.67 kN
Ve= F+1.6T/B
Ve = 239.67+(1.6x16.59/500x10^ -3)= 292.77 kN
Shear stress tve
= Ve/ B d
tve= 292.77x10^6 / 500x662.5 = 0.884
IS 456:2000
Table 24< 1.900 Hence OK
Percentage of steel provided at bottom 6 Nos. of 16 mm dia bars
100 Ast/Bd = 100 * 6 * 0.78539 * 16 ^ 2 / ( 500 * 662.5 ) = 0.364 %
IS 456:2000
Table 23Allowable shear stress from IS:456-2000 tc= 0.267
Hence Shear Reinforcement Required
25 mm dia bars at
99.6 kN.m
25 mm dia bars at
16 mm dia bars at
16 mm dia bars at
mm2
mm2
mm2
mm2
mm2
mm2
N/mm2
N/mm2
N/mm2
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Design of OST for Surge protection-Stage 1
Asv / Sv = = T/(b1d1ssv) + V/(2.5 d1ssv)
b1 = 500 - 2 * 25 - 25 =
d1 = 700 - 2 * 25 - 25 =
T = = 16.6 kN.m
V = = 239.7 kN
Asv / Sv = 16.59*10^6/(425*625*150))+((239.67*1000)/(2.5*625*150))
= 1.439
Minimum Asv / Sv > b. (tc- tv) /ssv = 500 * ( 0.884 - 0.267 ) / 150
= 2.058
Use stirrups of 12 mm dia 2 Legged Asv = 226.2
Spacing of vertical stirrups (Sv) = 226.19 / 2.06 =
Hence Provide stirrups of 12 mm dia 2 Legged 100 mm c/c
Shear reinforcement at the point of maximum shear
Fo = 414.8 kN
Shear stress tve= Fo/ B d
tve= 414.82 * 1000 / ( 662.5 * 500 ) = 1.252
IS 456:2000
Table 24< 1.900 Hence OK
Percentage of steel provided at support 2 Nos. of 25 mm dia bars
100 Ast/Bd = 100*2*0.78539*25^2/(500*662.5) = 0.296 %
IS 456:2000
Table 23Allowable shear stress from IS:456-2000 tc= 0.245
Hence Shear Reinforcement Required
Balance shear force ( Vs ) = (1.252 - 0.245)500x662.5/1000 = 334 kN
Use stirrups of 16 mm dia 2 Legged Asv = 402.1
Spacing of vertical stirrups (Sv) =402x150x662.5 / (334x10^3 ) =
Hence Provide stirrups of 16 mm dia 2 Legged 100 mm c/c
Shear reinforcement at the Mid span
Minimum Asv /b Sv > 0.4 / 0.87 fy = 0.001
120 mm
425.0 mm
625.0 mm
110 mm
mm2
N/mm2
N/mm2
N/mm2
mm2
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Argula Rajaram Guthpa Lift Irrigation Project, Nizamabad Dist., A.P
Design of OST for Surge protection-Stage 1
Use stirrups of 8 mm dia 2 Legged Asv = 100.5
Spacing of vertical stirrups Sv =
Maximum spacing 0.75 x d =
Hence Provide stirrups of 8 mm dia 2 Legged 175 mm c/c
1.7.4 Side face reinforcement
For beam depth is more thatn 750mm, provide reinforcement at 0.1% as side face reinfrocemetn.
Ast = 0.1 * 500 * 700 / 100 = 350
Required reinforcement on each face
1.8 Design of Columns
Tank is supported on 8 columns symmetrically placed on a circle of 8m mean dia. Height of
staging above ground level is 20m. Divided the total height in to 4 panels, Ht. of which is 5.0m .
Let the column be connected to the raft foundation by means of a ring beam, the top of which is
provided 1m below the ground level. So that the actual height of bottom panel is 5m or vertical
loads on column.
Load calculation
UDL on the beam =
Total load on the beam = 264.08x3.14x8 = 6637 KN
Load on each column = 6637.14 / 8 = 829.64 kN
Assume the column size as 500 mm x 500 mm
Self weight of column per meter height = 500x500x10^-6x25 = 6.25 kN
Assume the size of brace as 400 mm x 500 mm
Length of each brace [(R.Sin 2p/n)/Cos p/n] = 4*(SIN((2*180/8)*3.141592/180)))/(COS((180/8)*3.141592/180))
= 3.06 m
Clear distance of brace = 3.06 - (500x10^-3) = 2.56 m
Self weight of brace = 400x500x10^-6x2.56x25 = 12.81 kN
Height of Staging = 20 m
No of levels of braces above ground level = 3
16 mm dia bars
497 mm
181 mm
2 Nos. as side face
264.1 kN/m
mm2
mm2
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Design of OST for Surge protection-Stage 1
Total load on the column = 829.64+ 6.25x20+ 12.81x3 = 993.07 kN
1.8.1 Analysis of Wind Load
Wind pressure = 1.76
Coefficient for Buildings Circular in Plan = 0.7
Outer diameter of tank = 10.55 m
Outer diameter of ring beam = 8+ 500x10^-3 = 8.50 m
Overall depth of ring beam = 700 mm
Wind load on wall and bottom of ring beam
(10.55x4.6+10.55x375x10^-3+8.5x700x10 -3)x0.7x1.76 =71.94 KN
C.G of wind load above bottom of ring beam = 4.6/2+700*10^-3+375*10^-3
= 3.375 m
WL on each 5 m Panel
( 8x5x500x10^-3 +3.06x500x10^-3) x1.76x0.7 = 29.86 kN
WL on top 5 m Panel
( 8x5x500x10^-3 )x1.76x0.7x0.5 = 12.31 KN
71.94 KN
12.31 KN3.375 m
5 m O4
29.86 kN
5 m O3
29.86 kN
5 m O2
29.86 kN
KN/m2
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Argula Rajaram Guthpa Lift Irrigation Project, Nizamabad Dist., A.P
Design of OST for Surge protection-Stage 1
5 m O1
29.86 kNGL
4.00
Qw Mw
At level O1 173.8 kN 2389.1 kN.m
At level O2 144.0 kN 1594.6 kN.m 71.94*(3.375+5+5+5/2)+12.31*(5+5+5/2)+29.86*(5+5/2)+29.86*(5/2)
At level O3 114.1 kN 949.4 kN.m 71.94*(3.375+5+5/2)+12.31*(5+(5/2))+29.86*5/2
At level O4 84.3 KN 453.5 kN.m 71.94*(3.375+5/2)+12.31*5/2
At O1
Axial thrust Vmax = 0.05 Mw 0.05*2389. =
Shear force Smax = 0.25 Qw 0.25*173.8 = 43.46 KN
Bending Moment Mmax = Smax*h/2 43.46*5/2 =
1.8.2 Analysis of Seismic Load
Critical Condition is when the tank is full
Lateral Force under critical Condition = 298.77 KN
Distance of C.G of tank above ground level = 23.38 m
Moment Ms = 298.77*23.38 =
Axial thrust Vsmax = 0.05 Ms = 0.05 x 6983.74 = 349 kN
Shear force Ssmax = 0.25 Qs = 0.25 x 298.77 = 74.69 KN
Bending Moment Msmax = Ssmax*h/2 = 74.69x5/2 =
Seismic force is governing
Hence the Column is to be designed for loads and moments due to seismic force
1.8.3 Main reinforcement
6983.74 kN.m
71.94*(3.375+5+5+5+5/2))+12.31*(5+5+5+5/2)+29.86*(5+5+5/2)+2
9.86*(5+5/2)+29.86*(5/2)
119.45 kN
108.64 kN.m
186.73 kN.m
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Design of OST for Surge protection-Stage 1
Adopt a Load factor of 1.5
Assuming M25 Concrete and Fe 415 Steel
Fck = 25 fy = 415
Ultimate moment Mu = 1.5x186.73 = 280 kN.m
Ultimate load Pu = 1.5x993.07 = 1489.6 KN
Pu / FckbD = 1489.6x10^3 / (25x500x500) = 0.24
Mu / FckbD2= 280.1x10^6 / (25x500x500^2) = 0.09
Assume Clear cover = 40 mm
Diameter of bars = 20 mm
d' / D = (40+20/2)/500
= 0.100
SP 16 Chart
44p / Fck = 0.040
Percentage of Steel required = 0.04x25 = 1.000
Area of steel required = 1x500x500/100 = 2500
Minimum percentage of Steel = 0.80 %
Minimum Area of steel = 0.8*500*500/100 = 2000
Provided Area of Steel = 2500
No of 20 mm dia bars = 7.958
Provide 8 Nos. of 20 mm dia bars as main reinforcement
Percentage of Steel provided ppro= 1.00
Transverseve reinforcement
Diameter of ties = 5 But not < 5mm
Provide 8 mm Diameter ties
Pitch
Least lateral dimension = 500 mm
16 times dia of main bar = 320 mm
48 times of ties = 384 mm
Hence provide 8 mm dia ties at 250 mm c/c as transverse reinforcement
mm2
mm2
mm2
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Design of OST for Surge protection-Stage 1
1.9 Design of Braces
1.9.1 Main Reinforcement
Moment M1 = (Qw1*h1+Qw2*h2)*Cos2q* Sin(q+p/n)
n.Sin(2p/n)
Qw1 =
Qw2 =
h1 = 5.0 m
h2 = 5 m
n = 8
q = 22.5
M1 =
= 169.53 kN.m
Twisting moment T = 0.05*M1
= 0.05 * 169.53
=
Met = MT+ M
MT = T ((1+D/B)/1.7)
Depth of beam D = 500 mm
Width of beam B = 400 mm
MT = 8.48*((1+(500/400))/1.7)
=
Met
= 11.22+169.53
=
Use diameter of the main reinforcement = 20 mm
Nominal cover to the reinforcement d' = 25 mm
Effective depth of beam d = 465 mm
Mu /B d2= 1.5*180.75*10^6/(400*465*465) =
SP 16:Table
3For d'/D = (25+20/2)/465 0.08 pt = 1.05
((173.82*5)+(143.97*5))*COS(22.5*3.141592/180)^2*SIN((22.5+180/8)*3.14
1592/180)/(8* SIN((2*180/8)*3.141592/180))
11.22 kN.m
3.13
8.48 kN.m
180.75 kN.m
173.82 kN.m
143.97 kN.m
N/mm2
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Design of OST for Surge protection-Stage 1
Assuming minimum % of compression steel pc = 0.2
Area of Tension steel Ast =
Area of compression steel Asc =
Minimum percentage of steel p = 0.20
Minimum area of steel Asmin =
Hence provide 4 Nos. at bottom
Hence provide 4 Nos. at top
1.9.2 Shear Reinforcement
Max.Shear force V = (Qw1x h1+Qw2h2) 2Cos2p/n x Sin (2p/n)
L n Sin (2p/n)
Vmax =
= 110.75 KN
Veq = V + 1.6 T/B
T = 8.48 kN.m
Veq = 110.75+(1.6*8.48/(400*0.001)) = 144.66 KN
Shear stress developed t=Vu/Bd
t= 1.5*144.66*1000/(400*465) = 1.17
From table 61 of SP 16 for pt = 1.05
tc = 0.650
Balance shear force = (1.17-0.65)*400*465/1000 Vus = 96.1 KN
Vus/d = 96.09/46.5 =
Hence provide 2 legged stirrups of 10mm dia at 250mm c/c.
Also provide 150x150 haunches at junction of braces with columns and reinforced with 10mm dia
tor steel bars.
1.10 Design of Foundation
Adopt a Load factor of 1.5
((173.82*5)+(143.97*5))*2*(COS((180/8)*3.141592/180)^2)*(SIN(((2*180/8)*3.141592/18
0)))/(3.06*8*SIN(((2*180/8)*3.141592/180)))
25 mm dia bars
12 mm dia bars
2.07 kN/cm
1961
372
400
mm2
mm2
mm2
N/mm2
N/mm2
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Design of OST for Surge protection-Stage 1
Assuming M25 Concrete and Fe 415 Steel
Fck = 25 fy = 415
Total vertical load from filled tank and column = 993.07x8 = 7944.5 KN
Weight of water: PI()/4 * 10 2 * 4 * 10 = 3141.6 KN
Vertical load of empty tank and columns = 7944.5 - 3141.6 = 4802.9 KN
Maximum shear force due to Wind = 119.45x8 = 955.62 KN
Percentage of shear force w.r.t self wt. =(955.6 / 4802.9)x100 = 19.90 < 33.33 %
Assume self weight of foundation = 0.1 x W
0.1x7944.5 = 794.5 KN
Total load on subsoil 7944.5+794.5 =
Safe bearing pressure at 4.0m depth below EGL = 250
Area of foundation required 8739 / 250 = 34.96
Centre diameter of the foundation = 8.0 m
Width of foundation required 34.96 / (3.14x8) = 1.39 m
Provide foundation width(B) as = 2.0 m
Hence adopt raft slab having inner dia (D i) = 8 - 2 = 6.00 m
Outer diameter of the foundation slab (Do) = 8 + 2 = 10.00 m
1.10.1 Design of circular girder of raft slab
Load / m run of girder (w) = 8738.97 / (3.14 x8 ) = 347.71
Maximum - ve moment at support = C1w R22q
0.066*347.71*4^2*45*3.141592/180 =
Maximum + ve moment at mid span =C2w R
2
2q
0.03*347.71*4^2*45*3.141592/180 =
Maximum torsional moment at 9.5ofrom either support
= C3w R22q
0.005*347.71*4^2*45*3.141592/180 =
Shear force at support section = w R q
347.71*4*22.5*3.141592/180 = 546.19 KN
21.85 kN.m
131.08 kN.m
288.39 kN.m
8738.97 KN
KN/m2
m2
KN
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Design of OST for Surge protection-Stage 1
Shear force at maximum torsion, V
347.71*4*((22.5*3.141592/180)-(9.5*3.141592/180)) = 315.57 KN
Assume width of the section b = 500 mm
Mu = 0.1386 Fck B d2
Effective depth of beam required, d = Mu / sqrt ( 0.1386 Fck B )
SQRT((1.5*288.39*10^6)/(0.1386*20*500)) = 559 mm
Provide overall depth of beam D = 750 mm
Nominal cover to the reinforcement = 50 mm
Dia of the reinforcement bar = 16 mm
Effective depth of beam provided = 750 - 50 - 8
= 692 mm
Reinforcement at Support
Mu = 1.5 * 288.39 =
Mu /B d2= 432.58 * 10 6 / ( 500 * 692 2 ) = 1.81
From table 3 of SP 16 the required percentage of rinforcement
pt = 0.55
Area of steel required 0.55/100*(500*692) Ast = 1907
Hence Provide 25 mm dia bars of 4 Nos. at supports
Percentage of steel provided pprov= 0.57 %
Shear stress t=Vu/Bd
t= 1.5*546.19*1000/(500*692) = 2.37
From table 61 of SP 16 for pt = 0.55
tc = 0.51
Balance shear force Vus = ( 2.37 - 0.51)*(500*692)/1000
= 642.64 KN
Vus/d = 642.64 / 69.2 =
From table 62, Provide 4L - 12dia strps at 150mm c/c.
Mu = 1.5 * 131.08 = 196.63 kN.m
432.58 kN.m
9.3 kN/cm
KN/m2
mm2
N/mm2
N/mm2
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Design of OST for Surge protection-Stage 1
Mu /B d2= (196.63 * 10 6 )/(500 * 692 2) = 0.821
From table 3 of SP 16 the required percentage of rinforcement
pt = 0.237
Area of steel required 0.237/100*(500*692) Ast = 821
Hence Provide 16 mm dia bars of 5 Nos. at bottom
Percentage of steel provided pprov= 0.29 %
Torsional Reinforcement
The section is subjected to maximum Torsional moment and shear should be designed for the
following forces.
D = 750 mm T =
B = 500 mm V = 315.57 KN
d = 692 mm M=
Mt = 21.85* ( 1 + ( 750 / 500)) / 1.7 =
Me = 32.13 + 0 =
Ve = 315.57 + 1.6 * ( 21.85 / 0.5 ) = 385.49 KN
Mu = 1.5 * 32.13 =
Mu /B d2= 48.19 * 10 6 / ( 500 * 692 2 ) = 0.201
From table 3 of SP 16 the required percentage of rinforcement
pt = 0.084
Area of steel required 0.084/100*(500*692) Ast = 291
Hence Provide 12 mm dia bars of 3 Nos. as torsional reinforcement
Percentage of steel provided pprov= 0.10 %
Shear stress t =Vu/Bd
t = 1.5 * 385.49 * 1000 / ( 500 * 692 ) = 1.67
From table 61 of SP 16 for pt = 0.10
tc = 0.30
Balance shear force Vus = ( 1.67 - 0.3)*(500*692)/1000
48.19 kN.m
21.85 kN.m
32.13 kN.m
0.0 kN.m
32.13 kN.m
KN/m2
mm2
KN/m2
mm2
N/mm2
N/mm2
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PROJECT: DOCUMENT NO. DATE
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TITLE: DESIGNED CHECKED PAGE
Argula Rajaram Guthpa Lift Irrigation Project, Nizamabad Dist., A.P
Design of OST for Surge protection-Stage 1
= 474.43 KN
Vus/d = 474.43 / 69.2 = 6.9 kN/cm
From table 62, Provide 4L - 10dia strps at 125mm c/c.
1.10.2 Design of Raft slab
Maximum projection of raft slab from face of column = (B-b)/2
(2 - 500x10^-3) / 2 = 0.750 m
Soil pressure = W/[(p/4)*(Do2-Di
2)]
8739 / Pi/4 (10^2 - 6^2) = 173.86
Maximum bending moment = 173.86x0.75^2 / 2 =
Mu = 0.1386 Fck B d2
Effective depth of beam required, d = Mu / sqrt ( 0.1386 Fck B )
SQRT((1.5*48.9*10^6)/(0.1386*25* 1000)) = 145 mm
Nominal cover to the reinforcement = 50 mm
Dia of the reinforcement bar = 16 mm
Provide overall depth of slab D = 450 mm
Effective depth of beam provided = 450 - 50 - 8
= 392 mm
Reinforcement
Mu = 1.5 * 48.9 =
Mu /B d2= 73.35 * 10 6 /(1000* 392 2 ) = 0.477
From table 3 of SP 16 the required percentage of rinforcement
pt = 0.135
Area of steel required Ast = 530
Hence required 12 mm dia bars at 200 mm c/c
Hence Provide 12 mm dia bars at 200 mm mm c/c as main reinforcement
Distribution steel pt = 0.12
Area of steel required Ast = 470
73.35 kN.m
48.9 kN.m
KN/m
2
N/mm2
mm2
mm2
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PROJECT: DOCUMENT NO. DATE
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TITLE: DESIGNED CHECKED PAGE
Argula Rajaram Guthpa Lift Irrigation Project, Nizamabad Dist., A.P
Design of OST for Surge protection-Stage 1
Hence required 10 mm dia bars at 167 mm c/c
Provide 10 mm dia bars at 150 mm c/c at top and bottom
1.11 Design of Cover Slab
Assuming M25 Concrete and Fe 415 Steel
Fck = 25 fy = 415
Assume LL on the Slab = 1.500
Assume thickness of Slab = 200 mm
Self weight of Slab = 200 * 10^-3 * 25 = 5.00
Load due to finishes = 0.50
Total Load on Cover Slab= 1.5 + 5 + 0.5 = 7.00
Diameter of Cover slab = 10.55 m
Check for Effective Depth
Radial moment at centre = 3/16*w*r2
3 / 16 * 7 * 5.275 2 ) =
Effective depth d = SQRT(36.52 * 10 6 / (1000*1.42))
=
Nominal cover to the reinforcement = 25 mm
Dia of the reinforcement bar = 16 mm
Overall Depth required 160.37 + 25 + 8 = 193.4 mm
Provide overall depth of slab D = 200 mm
Effective depth of slab provided= 200 - 25 - 8 = 167 mm
Calculation of Radial reinforcement
Radial moment at centre =
Area of Steel required = 36.52*10^6/(150*0.87*167) = 1672
Minimum percentage of Steel = 0.12 %
36.52 kN.m
36.52 kN.m
160.37 mm
KN/m2
KN/m2
KN/m2
KN/m2
mm2
KN/m2
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PROJECT: DOCUMENT NO. DATE
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TITLE: DESIGNED CHECKED PAGE
Argula Rajaram Guthpa Lift Irrigation Project, Nizamabad Dist., A.P
Design of OST for Surge protection-Stage 1
Minimum Area of Steel= 0.12 * 167 * 1000/100 = 200
Provided Area of Steel = 1672
Spacing of 20 mm dia bars = 188 mm
Provide 20 mm dia bars at 175 mm c/c as radial reinforcement
Calculation of Circumferential reinforcement
Circumferential moment at Support = 2/16*w*r2
2/16 * 7 * 5.275 2 =
Area of Steel required 24.35*10^6/(150*0.87*167) = 1115
Spacing of 16 mm dia bars = 180 mm
Hence Provide 16 mm dia bars at 175 mm c/c at Support
Circumferential moment at Centre = 3/16*w*r2
= 36.5 kN.m
Area of Steel required = 1672
Spacing of 20 mm dia bars = 188 mm
Hence Provide 20 mm dia bars at 175 mm c/c at centre
24.35 kN.m
mm2
mm2
mm2
mm2
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PROJECT: DOCUMENT NO. DATE
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TITLE: DESIGNED CHECKED PAGE
Argula Rajaram Guthpa Lift Irrigation Project, Nizamabad Dist., A.P
Design of OST for Surge protection-Stage 1
SUMMARY
Component
Cover Slab Thickness 200 mm 20 mm dia as radial rft
20 mm dia as Circumfl at centre
16 mm dia as Circfl at support
Tank walls Thickness 275 mm 16 mm dia
16 mm dia away from water fac
10 mm dia as vertical rft
Base Slab Thickness 375 mm 25 mm dia as radial at top
16 mm dia as radial at bottom
16 mm dia as Circumfl at bottom
10 mm dia as circumflat top
Ring beam Width 500 mm 25 mm dia 2 Nos. as torsional rft at top and bottom
Depth 700 mm 25 mm dia 6 Nos. as main reinforcement at top
16 mm dia 6 Nos. as main reinforcement at bottom
8 mm dia 2 Legged as shear rft
16 mm dia 2 Nos. as
Columns Width 500 mm 20 mm dia 8 Nos. as main reinforcement
Depth 500 mm 8 mm dia as transverse rft
Braces Width 400 mm 25 mm dia 4 Nos. as main reinforcement (top)
Depth 500 mm 12 mm dia 4 Nos. as
8 mm dia 2 Legged as shear rft
Footing
Raft Beam Width 500 mm 25 mm dia 4 Nos. as main reinforcement at top
Depth 750 mm 16 mm dia 5 Nos. as main reinforcement at tottom.
12 mm dia 3 Nos. as torsional rft at top and bottom
12 mm dia 4legged at 150mm c/c as shear rft
Raft Slab Depth 450 mm 12 mm dia as main Rft
at 175 mm c/c
Dimensions Reinforcement
at 175 mm c/c
at 175 mm c/c
at 175 mm c/c
at 200 mm c/cas hoop rft near
water face
at 250 mm c/c
main reinforcement (bottom)
at 250 mm c/c
at 200 mm c/c
at 275 mm c/c
at 150 mm c/c
as side face reinforceemnt
at 175 mm c/c
at 150 mm c/c
at 150 mm c/c
at 175 mm c/c
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PROJECT: DOCUMENT NO. DATE
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TITLE: DESIGNED CHECKED PAGE
Argula Rajaram Guthpa Lift Irrigation Project, Nizamabad Dist., A.P
Design of OST for Surge protection-Stage 1
10 mm distribution bars150 mm c/c as
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32
ANNEXURE - I
3.0 CALCULATION OF WIND PRESSURE
Design Wind Speed (Vz) = VbK1K2K3
App-A of IS875
Part3Basic Wind Speed Vbfor Nizamabad = 44
IS875 Part3-
Table1Risk Co-efficient (K1) = 1.07
(Class:All ganeral Buildings with Design life 50yrs)
IS875 Part3-
Table 2Terrain, Height & Structure factor (K2) = 1.15
(Height=10m and Terrain Category 2 with Class A)IS875 Part3-
Cl:5.3.3Topography factor (K3) = 1.00
Design Wind Speed (Vz) = 54.142
IS875 Part3-
Cl:5.4Design Wind Pressure (pz) = 0.6Vz
2
= 1.76
Arugula Rajaram Guthpa Lift Irrigation Project,
Nizamabad Dist., A.P
Design of OST for Surge protection-Stage1
m/s
m/s
KN/m2
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ANNEXURE - II
4.0 CALCULATION OF CRACK WIDTH FOR BASE SLAB
As per Annexe F of IS 456 2000
Design Surface Crack Width(Wcr)
Wcr = (3acrem)/ 1+ [2(acr- Cmin) / h - x]
em = e1-[b(h-x)(a-x) / 3Es As(d-x)]
e1 = eh = (h-x / d-x)es
es = a M (d-x) / Es Ic
a = Es / Ec
Calculation of Neutral Axis Depth (x)
xu/d = 0.87fyAst / 0.36fckbd
Ast = pi()/4*25^ 2 * 1000 / 175 = 2803.57
xu = 0.87*415 * 2803.57/ ( 0.36 * 25 * 1000 )
= 112.5 mm
Check for Crack Width
a = 200*1000/(5000*SQRT(25)) = 8
es = 8*(121.95*10^6*((340-112.5))*12)/(200*10^3*1000*340^3) = 0.000339
e1 = 0.000339*(375-112.5)/(340-112.5) = 0.000391
em == 0.000211
acr = SQRT((25 2 + (175 / 2 ) 2 )) = 91.00
Wcr = 3*91.001*0.00021/(1+((2*(91-25))/(375-112.47))) 0.038 m
Arugula Rajaram Guthpa Lift Irrigation Project, Nizamabad
Dist., A.P
Design of OST for Surge protection-Stage 1
0.000391 - ((1000 * (375-112.5) ^ 2/
(3*200*10^3 * 2803.57 * (340-112.5))