rcc oht alternate

8/13/2019 Rcc Oht Alternate

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PROJECT: DOCUMENT NO. DATE

30.8.06TITLE : DESIGNED CHECKED PAGE

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



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TITLE : DESIGNED CHECKED PAGE

34

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. )


Dist., A.P



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TITLE: DESIGNED CHECKED PAGE

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


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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Let the approximate length of brace be (pd/n) = 3.14 m


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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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I1 = (0.5x0.5^3 / 12) x 2 = 0.010 m4


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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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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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TYPICAL CROSS SECTION OF TANK


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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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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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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

226.6 mm

259.6 mm

242 mm


mm2

mm2

mm2

mm2

N/mm2

mm2

mm2


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Minimum area of reinforcement = 275 (on each face)mm2


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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


375 mm

10.550 m

200 mm


mm2

KN/m2

KN/m2

KN/m2

KN/m2

KN/m2

KN/m2

KN/m2

KN/m2

KN/m2


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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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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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(Mq)c= (3/16) *p a2


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(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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(Mq) = -6.04 kN.m


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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)


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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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


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 area of reinforcement = 0.06 / 100 * 375 * 1000

= 225


51.2 kN.m

= b i.e 4.00 m

315.0 mm

mm2

mm2

mm2

mm2

mm2

mm2


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Hence provide Circumferential and radials16 mm dia bars at 150 mm c/c as


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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


= 0.06 / 100 * 375 * 1000



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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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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=

Nominal cover to reinforcement = 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



= 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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= 2528



= 700


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


= 0.2 / 100 * 700 * 500



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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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


Spacing of vertical stirrups (Sv) =402x150x662.5 / (334x10^3 ) =


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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Spacing of vertical stirrups Sv =

Maximum spacing 0.75 x d =


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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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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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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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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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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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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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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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


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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Mu /B d2= (196.63 * 10 6 )/(500 * 692 2) = 0.821


pt = 0.237


Hence Provide 16 mm dia bars of 5 Nos. at bottom


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


pt = 0.084


Hence Provide 12 mm dia bars of 3 Nos. as torsional reinforcement


Shear stress t =Vu/Bd

t = 1.5 * 385.49 * 1000 / ( 500 * 692 ) = 1.67


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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= 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



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


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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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


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))

=



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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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


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


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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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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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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TITLE : DESIGNED CHECKED PAGE

33

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


Dist., A.P


0.000391 - ((1000 * (375-112.5) ^ 2/

(3*200*10^3 * 2803.57 * (340-112.5))

rcc oht alternate

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