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COMPARA COMPARATIVE STUD S TUDY Y OF DESIGN OF WATER WATER TANKS TANKS WITH WI TH REFERENCE REF ERENCE TO IS : 3370 Thesis · November 2011
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2 authors: Bharat Bhushan Jindal
Dhirendra Singhal
Maharishi Markandeshwar University
Deenbandhu Chhotu Chhotu Ram Universi ty of Sc ienc e and Technology, …
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COMPARATIVE STUDY OF DESIGN OF WATER TANKS WITH REFERENCE TO IS : 3370 THESIS SUBMITTED IN PARTIAL FULFILLMENT OF THE REQUIREMENT FOR THE AWARD OF THE DEGREE OF
MASTER OF TECHNOLOGY
(Structural Engineering) Engineering) SUBMITTED BY
BHARAT BHUSHAN JINDAL November2011
PUNJAB TECHNICAL UNIVERSITY
JALANDHAR, INDIA
COMPARATIVE STUDY OF DESIGN OF WATER TANKS WITH REFERENCE TO IS : 3370 THESIS
SUBMITTED IN PARTIAL FULFILLMENT OF THE REQUIREMENT FOR THE AWARD OF THE DEGREE OF
MASTER OF TECHNOLOGY
(Structural Engineering )
SUBMITTED BY BHARAT BHUSHAN JINDAL Registration no. 71601003
G.Z.S. COLLEGE OF ENGINEERING & TECHNOLOGY BATHINDA – 151001 NOVEMBER 2011
PUNJAB TECHNICAL UNIVERSITY
JALANDHAR, INDIA
CANDIDATE’S DECLARATION
I hereby certify that the work being presented in the thesis entitled “COMPARATIVE STUDY OF DESIGN OF WATER TANKS WITH REFERENCE TO IS : 3370 ” in partial fulfilment of the requirement for the award of degree of M.Tech. (Structural Engineering) submitted in department of Civil Engineering at G.Z.S. College of Engineering & Technology, Bathinda under PUNJABTECHNICAL UNIVERSITY, JALANDHAR is an authentic record of my own work carried out during a period from January 2011 to November 2011 under the supervision of Dr. Dhirendra Singhal. The matter presented in this thesis has not been submitted by me in any other University / Institute for the award of M. Tech. Degree.
(BHARAT BHUSHAN JINDAL) This is to certify that the above statement made by the candidate is correct to the best of my knowledge.
Dhirendra Singhal Associate Professor & Head of Department Civil Engg. Department GZSCET, Bathinda
The M.Tech Viva-Voce Examination of BHARAT BHUSHAN JINDAL has been held on ____________ and accepted. Signature of Supervisor ________________________________ Signature of Head of Department _________________________ Signature of External Examiner __________________________
i
ABSTRACT
Limit state method is widely used at present in comparison to working stress method with the following advantages: i)
Materials are treated according to their properties.
ii)
Loads are treated according to their nature.
iii)
Structures generally fail when they reach their limit state, not their elastic state.
However, when structures reach to their limit state, the cracking width in the structure may be significantly higher comparative to a structure designed by working stress method at the same stage. IS: 3370 i.e. the Indian Standard specifications for construction of liquid retaining structures did not adopt limit state design method for long. However, IS:3370 has adopted the limit state design method after considering checks over the cracking width. It has been recently adopted in the new version of IS 3370-2009 concrete structures for storage of liquids – code of practice, while going through IS 3370 – 2009 it can be found that three methods of design are available. i)
Working Stress Method.
ii)
Limit State Design Method – Crack width check.
iii)
Limit State Design Method – Deemed to satisfy.
Objectives of the study was to : 1.
To compare the design of water retaining structures done by WSM & LSM in reference to
2.
IS 3370 – 1965 and IS 3370 – 2009 ( new version ).
To analyze which method is more economical and efficient.
ii
Three types of water tanks were chosen for a comparative study of provisions of IS:3370. In case of OHSR, Intz types of water tank was chosen since it is widely used for large capacity and the comparison of provisions can thus be better highlighted through Intz Type of water tank. In other case, a square and a rectangular tank were taken as the forces and moments totally depend on L/B ratio. These tanks generally used for moderate capacity were supported on ground. A thorough study through both the versions of IS:3370 reveals the following four methods of designs: 1)
Working stress method in accordance IS 3370 (1965).
2)
Working stress method in accordance IS 3370 (2009).
3)
Designing by Ultimate Limit State and then checking cracking width by limit state of serviceability IS 3370 (2009).
4)
Limit state design method by limiting steel stresses in accordance IS 3370 ( 2009 ) and checking cracking width under serviceability not required (Deemed to be satisfied).
All the water tanks as discussed above were designed by the above four methods. M30 grade of concrete as required now by IS:3370 for reinforced concrete has been considered in all the above problems and methods. Also, the steel used is deformed steel which can be considered here as TOR or TMT. It has been observed that a)
The size of members remained same for working stress method by IS:3370(1965) and IS:3370 (2009). However, the requirement of area of steel increased in IS:3370 (2009)
iii
for Intz type and rectangular water tanks as the allowable stresses in steel were lower. The steel required in square tank was approximately same in both the cases. However, the change in the clause of requirement of minimum steel decreased the steel required in bottom spherical dome in Intz type of tank. b)
The size of members remained same for limit state design methods by IS:3370(2009) in limit state of collapse as well in deemed to satisfy criteria for all the three tank designs. However, the requirement of area of steel increased in IS:3370 (2009) ) in serviceability design method as well in deemed to satisfy criteria for all the three tank designs as the allowable stresses in steel were lower.
c)
The size of members as well as the requirement of steel decreased for limit state design method by IS:3370 (2009) in comparison to working stress design method of both IS : 3370 (1965 ) and IS : 3370 (2009) provisions for all three type of tanks taken in study.
It was found that the provisions of reinforcement through the surface zones in IS:3370(2009) provides economical and more effective reinforcement. However, it was also felt that IS:3370 (2009) should have provided direct tensile stress and compressive stress under bending and limit state.
iv
ACKNOWLEDGEMENT
I express my profound sense of gratitude to my guide Dr. Dhirender Singhal, Associate Professor & head of department of Civil Engineering, G.Z.S.
College of Engineering&
Technology, Bathinda for providing me prudent guidance and encouragement at every stage of this work without which it could not have been possible for me to complete it.
I thank the central library of Maharishi Markandeshwar University, Ambala for providing me the literature required for this thesis work.
I thank my family for the support and inspiration, they provided to me for the completion of this work.
Finally I want to express my sincere appreciation and gratitude to all whosoever has contributed in this thesis work at GZSCET, BATHINDA.
BHARAT BHUSHAN JINDAL
v
CONTENTS
Candidate’s declaration
i
Abstract
ii
Acknowledgement
v
Chapter 1: Introduction
1
1.1
Design Methods
2
1.1.1
Limit State Design Method
2
1.1.2
Working Stress Method
3
1.2
Objective of Study
4
1.3
Problem Formulation
4
1.4
Design Methods Used
6
1.5
Organisation of Thesis
6
Chapter 2: Literature Review
7
2.1
Tanks Resting on Ground
7
2.1.1
Circular Tank with Flexible Joint at the Base
7
2.1.1.1 Wall Design
7
2.1.1.2 Base Slab
8
2.1.2
Circular Tanks with Restraint at Base and Roof
8
2.2
Underground Tanks
10
2.3
OHSR Tanks
10
2.3.1
Intz Type Water Tank
11
2.4
Design of Water Retaining Structures
12
2.4.1
Basis of Design
13
2.4.2
Design Methods
14
2.4.3
Design Requirements of Concrete
15
2.5
General Design Requirement as per IS : 3370-2009
17
vi
2.5.1
Basis of Design
17
2.5.2
Permissible Stresses in Concrete
17
2.5.3
Permissible Stresses in Steel
18
2.4.3.1 For resistance to cracking
18
2.5.3.2 For strength calculations
18
2.5.4. Stresses due to drying Shrinkage or Temperature Change
19
2.5.5. Floors
19
2.5.5.1 Provision of movement joints
19
2.5.5.2 Floors of tanks resting on ground
19
2.5.5.3 Floor of tanks resting on supports
20
2.5.6
20
Walls
2.5.6.1 Provision of joints
20
2.5.6.2 Pressure on walls
20
2.5.6.3 Wall or tanks rectangular or polygon in plan
21
2.5.6.4 Walls of cylindrical tanks
22
2.5.7
22
Roofs
2.5.7.1 Provision of movement joints
22
2.5.7.2 Water Tightness
23
2.5.8
23
Reinforcement detailing
2.5.8.1 Minimum Reinforcement
23
2.5.9
24
Crack width due to temperature and moisture
2.5.9.1 Minimum Reinforcement
24
2.5.9.2 Crack Spacing
25
2.6
Comparative Study of IS 3370-1965 & IS 3370 –2009
26
2.6.1
Permissible Stresses in Steel Reinforcement for Strength Calculations
26
2.6.2
Reinforcement Detailing
27 vii
Chapter 3 : Identification of Problem
3.1
Identification of Problem
28
3.2
Water tank design problems
28
3.3
Design of INTZ Tank
29-44
3.3.1
Comparative result of INTZ Type Water Tank
45
3.4
Rectangular Water Tank Design
46-76
3.4.1
Comparative result of Rectangular Water Tank situated on ground
76
3.5
Square Water Tank Design
77-87
3.5.1. Comparative result of Square Water Tank situated on ground
87
Chapter 4 : Results and Discussion
4.1
Comparative result of INTZ Type Water Tank
88
4.2
Comparative result of Rectangular Water Tank situated on ground
89
4.3
Comparative result of Square Water Tank situated on ground
89
Chapter 5 : Conclusions and Future Scope of the Study
5.1
Conclusions and Future Scope of the Study
Reference
90
91-92
viii
CHAPTER 1 Introduction
Storage reservoirs and overhead tanks are used to store water, liquid petroleum, petroleum products and similar liquids. In general there are three kinds of water tanks: i) ii) iii)
Tanks resting on ground. Underground tanks. Elevated tanks.
The tanks resting on ground like clear water reservoirs, settling tanks, aeration tanks etc. are supported on the ground directly. The walls of these tanks are subjected to pressure and the base is subjected to weight of liquid and upward soil pressure. The tanks may be covered on top.
From design point of view, the tanks may be classified as per their shape as foll owing: i) ii)
Rectangular tanks Circular tanks
iii)
Over Head Service Reservoir (OHSR)
iv)
Intz Tank i.e. OHSR for large capacity.
Rectangular tanks are provided for smaller to moderate capacity. For small capacities, circular tanks prove uneconomical as the formwork for circular tanks is very costly. The rectangular tanks should be preferably square in plan from point of view of economy. It is desirable that longer side should not be greater than twice the smaller side.
1
1.1
Design Methods:
Three methods of design has been adopted so far i.e. 1. Working Stress Method 2. Ultimate Load Method 3. Limit State Method However, Ultimate Load Method has become obsolete these days. Other tw o methods are generally used and have been discussed here. 1.1.1
Limit State Design Method
Limit state design method, though semi-empirical approach, has been found to be the best for the design of reinforced concrete structures over the elastic theory of design where the level of stresses in concrete and steel are limited so that stress-deformations are taken to be linear. There are two limit states: i) ii)
Limit state of collapse. Limit state of serviceability serviceability which includes includes a) Deflection b) Cracking
The structure is first designed under Limit State of Collapse and then checked under serviceability. Because of its superiority to other two methods , IS 456:2000 has been thoroughly updated in its fourth revision in 2000 taking into consideration the rapid development in the field of concrete technology and incorporating important aspects like durability etc. This standard has put greater emphasis to limit state method of design by presenting it in a full section (section 5), while the working stress method has been given in Annex B of the same standard. Accordingly, structures or structural elements shall normally be designed by limit state method.
2
It is important to point out here that a structure designed through limit state method when fails, the failure will be in plastic stage and not in elastic stage. Therefore, the cracking and cracking width can be significant at the failure stage. 1.1.2
Working Stress Method
This method of design, considered as the method of earlier times, has several limitations. However, in situations where limit state method cannot be conveniently applied, working stress method can be employed as an alternative. It is expected that in the near future the working stress method will be completely replaced by the limit state method. Presently, this method is put in Annex B of IS 456:2000. Though the choice of the method of design is still left to the designer as per cl. 18.2 of IS 456:2000, the superiority of the limit state method is evident from the emphasis given to this method by presenting it in a full section (Section 5), while accommodating the working stress method in Annex B of IS 456:2000, from its earlier place of section 6 in IS 456:1978. It is expected that a gradual change over to the limit state method of design will take place in the near future after overcoming the inconveniences of adopting this method in some situations. In this method of design, the structure is not checked for serviceability perhaps with the fact that elastic limit is not crossed. Therefore, the structure designed by working stress method will not have significant cracking. Therefore, liquid retaining structures, are still preferred to be designed by working stress method as seepage of liquid is the also the main criteria. Limited cracking in the structure designed by working stress method was the main reason why the Indian Standard IS: 3370 (1965) did not adopt the limit state design method even after adoption by IS; 456 – 1978.
3
However, with the following advantages of Limit State Design method, IS:3370 adopted Limit State Design Method in 2009. i)
Limit State Design Method considers the materials according to their properties
ii)
Limit State Design Method considers the load according to their nature
iii)
The structures also fails mostly under limit state and not in elastic state
iv)
Limit State Method also checks for serviceability
IS:3370-2009 adopts Limit State Design Method with precautions. It adopts the criteria for limiting crack width when the structures are designed by considering ultimate limit state and restricts the stresses to 130MPa in steel so that cracking width is not exceeded. These precaution ensures cracking width to be less than 0.2 mm i.e. fit for liquid storage. This also specifies clearly how a liquid storage structure differs with other structures. 1.2
Objective Of Study 1. To compare the design of RCC water retaining structures done by WSM & LSM in
reference to IS 3370 – 1965 and IS 3370 – 2009 ( new version ). 2. To analyze which method is more economical and efficient.
1.3
Problem Formulation
To do the comparative study of provisions in IS 3370 (1965) and IS 3370 (2009), three type of problems of water tank were taken. One OHSR water tank problem and other ground level water tank of rectangular and square section were taken in this study. In case of OHSR, Intz types of water tank was chosen since it is widely used for large capacity and the comparison of provisions can thus be better highlighted through Intz type of water tank. In other case, a square and a rectangular tank were taken as the forces and moments totally depend on L/B ratio. These tanks generally used for moderate capacity were supported on 4
ground. A thorough study through both the versions of IS:3370 reveals the following four methods of designs: 1.
Working stress method in accordance IS 3370 (1965).
2.
Working stress method in accordance IS 3370 (2009).
3.
Designing by Ultimate Limit State and then checking cracking width by limit state of serviceability IS 3370 (2009).
4.
Limit state design method by limiting steel stresses in accordance IS 3370 (2009) and checking cracking width under serviceability not required (Deemed to b e satisfied).
Water tank design problems were as follows:
1)
Design an Intz type water tank of 1 million litres capacity, supported on an elevated lower comprising of 8 columns. The base of the tank is 16 m above ground level. Depth of foundation 1 m below ground level. Adopt M30 grade of concrete and Fe-415 grade of tor steel. The design of the tank should conform to the stresses specified in IS 3370 and IS 456.
2)
1
A rectangular water tank is to be designed to store 2500 kl water. The tank is to be made 2
just above the ground level and the safe bearing capacity of the soil is 75 kN/m . Adopt M30 grade of concrete and Fe-415 grade of tor steel. The design of the tank should conform to the stresses specified in IS 3370 and IS 456. 3)
2
Design a square water tank having inner dimensions of 7.5 × 7.5 × 2.65 m high with walls fixed at the bottom and free at the top. The tank is directly supported on the earth. The floor slab is monolithic with the walls. The free board is 15 cm. Use M30 grade of concrete and Fe 415 grade H.S.D. bars.
3
5
1.4 Design Methods used
As per discussion above, the three water tank design problems are designed by the following four design methods. 1. Working stress method in accordance IS 3370 (1965). 2. Working stress method in accordance IS 3370 (2009). 3. Limit state design method with crack width calculations and check in accordance IS 3370 (2009). 4. Limit state design method deemed to satisfy (limiting steel stresses) in accordance IS 3370 (2009).
1.5
Organization of Thesis
The thesis has been organized in the following Chapters:Chapter: 1
Deals with the introduction of the title and identifies the objectives.
Chapter: 2
Scans through the literature available in brief.
Chapter: 3
Identification of Problem
Chapter: 4
Results and discussions.
Chapter: 5
Conclusions and future scope of the study.
6
CHAPTER 2 Literature Review
Liquid storage tanks are commonly used in industries for storing chemicals, petroleum products, etc. and for storing water in public water distribution s ystems. A reinforced concrete tank is a very useful structure for the storage of water, sewage sedimentation and for other similar purposes. The usual types of water tanks are the following: (i)
Tanks situated on the ground.
(ii)
Tanks situated underground.
(iii)
Tanks situated above the ground level.
The tanks may be either open or roofed over and they may be circular or rectangular in plan. 2.1
Tanks Resting On Ground
2.1.1
Circular Tank with Flexible Joint at the Base
2.1.1.1 Wall Design
The walls of circular tank having flexible joint at the base are designed as vertical cylinders subjected to hydrostatic pressure. The intensity of water pressure at any depth ‘h’is equal to ‘w h’ units per unit area. The corresponding tension/unit height at this level is given by
, where D = internal diameter of the tank
The area of steel per unit height required to resist the above tension is calculated as
,
This hoop reinforcement may be provided preferably near both faces with a minimum clear cover of 25 mm. However, the minimum amount of steel provided should not be less than as specified in 2.5.8.
7
Thickness of wall should be computed to resist the hoop tension in the wall such that Tensile stress in wall concrete =
. , should be less than permissible direct
tensile stress in concrete as per table 1 . Where
t
= thickness of wall
m = modular ratio 2.1.1.2
Base Slab
The base slab shall be designed for the difference of pressure due to weight of water acting downward and the upward pressure due to soil. This difference is generally small and nominal thickness (minimum 100 mm) with minimum steel is provided. 2.1.2
Circular Tank with Restraint at Base and Roof
Due to restraint at base and roof, the walls are subjected to horizontal shear, hoop tension and bending moment. In this case the wall will resist the water pressure partly by hoop action and partly by cantilever action. Elastic behaviour of tank walls show that for a certain height from base there will be predominant cantilever action and at higher levels, there will be predominant hoop action. a)
Dr Reissner’s Method Dr Reissner developed a method to compute restraint moment M, and ring tension T and its position for rectangular and triangular wall sections, depending upon the value of a parameter.
. Where
H = height of water D = internal diameter of the tank t = thickness of wall’
The values of Mr and T are given in table 4 and table 5 respectively.
8
Table 4 : Values of Restraint Moment Mr . ( p = w H ) K
Rectangular wall section
Triangular wall section
0
0.167 p H
2
0.167 p H
2
10
0.110 p H
2
0.140 p H
2
100
0.0582 p H
1000
0.024 p H
10000
0.0085 p H
2
0.0707 p H
2 2
0.026 p H
2
0.009 p H
2
0
∞
2
0
Table 5 : Ring Tension in Tank Wall ( p = w H ) Rectangular wall section
K
b)
Max. Tension
Rectangular wall section
Height from base
Max. Tension
Height from base
0
0
---
0
----
10
0.18 p D /2
0.1 H
0.09 p D /2
0.65 H
100
0.27 p D /2
1.0 H
0.31 p D /2
0.58 H
1000
0.47 p D /2
0.47 H
0.52 p D /2
0.44 H
10000
0.67 p D /2
0.31 H
0.70 p D /2
0.30 H
∞
1.0 p D /2
0
1.0 p D /2
0
Carpenter’s Method Carpenter simplified Dr Reissner’s method. The maximum cantilever moment and the position and magnitude of maximum ring tension are given by the following relations. i)
Position of maximum ring tension = K . H from base Where K is a coefficient depending upon H / D and H / t
ii) Maximum hoop tension at K . H. above base = w ( H – K H ) . D/2
= w H D ( 1 – K ) /2
iii) Maximum cantilever bending moment
= F w H
3
Where F is a coefficient depending upon values of H / D and H / t The coefficient K and F can be obtained from table 6 9
Table 6 : Coefficients F and K Factors
F
H/t
Values of H/D
2.2
K
10
20
30
40
10
20
30
40
0.2
0.046
0.028
0.022
0.015
---
0.50
0.45
0.40
0.3
0.032
0.019
0.014
0.010
0.55
0.43
0.38
0.33
0.4
0.024
0.014
0.010
0.007
0.50
0.39
0.35
0.30
0.5
0.020
0.012
0.009
0.006
0.45
0.37
0.32
0.27
1.0
0.012
0.006
0.005
0.003
0.37
0.28
0.24
0.21
2.0
0.006
0.003
0.002
0.002
0.30
0.22
0.19
0.16
4.0
0.004
0.002
0.002
0.001
0.27
0.20
0.17
0.14
Underground Tanks
The walls of underground water tank should be investigated for both internal water pressure and external earth pressure. The external pressure may be due to dry earth or due to a combination of earth and ground water i.e. saturated soil. The design principles for such tanks are same as that for tanks resting on ground. In case of water tank built below ground with earth covering the roof slab, there will be a trapezoidal lateral earth pressure on the wall. Whenever there is a possibility of ground water table to rise above the base slab, not only walls are to be designed for saturated soil up to the extent of water above the base slab, but also the base slab is to be designed for the net uplift pressure of water ( less weight of slab for tank empty case ). In addition, check has to be applied for stability of the tank as a whole against uplift.
2.3
OHSR Tanks :
The elevated tanks may be rectangular or circular in shape. These tanks are elevated so that once they are filled, the water flows under gravity to the area it is to serve. The tanks are supported on staging which may be simple slab or a spherical dome with or without opening in 10
the centre, latter being mostly used in practice. Where the size of the tank is large, a network of beams is provided over columns to support the base slab. If columns are provided in staging, these are braced by ring beams not only at top and bottom but also at a number of places along the height. In such a case, the effective length of columns taken as the distance between centres of adjacent bracings. The staging and the tank is subjected to wind pressure and seismic forces depending upon the location of the tank. 2.3.1
Intz Type Water Tank
In case of large diameter elevated circular tanks, thicker floor slabs are required which will make the water tank design uneconomical. In such cases, Intz type tank with conical and bottom spherical domes provide economical solution. The properties of the conical and the spherical bottom domes are selected so that the outward thrust from the bottom dome balances the inward thrust due to the conical domes part of the tank floor. Structural elements of Intz tank are:
1. Top spherical dome 2. Top ring beam 3. Circular side vertical walls 4. Bottom ring beam 5. Conical dome 6. Bottom spherical dome 7. Bottom circular girder 8. Tower with columns and braces 9. Foundations Top spherical dome is designed for maximum meridional thrust and circumferential stress due
to water pressure and subsequently reinforcement is designed. 11
Top ring beam is designed for hoop tension which depends upon meridional thrust and
dimensions of tank. The cross sectional area of ring beam is determined by limiting the tensile stress in ring beam as per IS: 456-2000 specifications depending upon the grade of concrete used. Side Walls of tanks are designed for hoop tension developed due to the water pressure in the
tank. A minimum thickness of 150 to 200 mm is provided at the top of the tank and the thickness at the base of the vertical wall is designed by limiting the tensile stress. The spacing of the hoop reinforcement is gradually increased towards the top of the tank. Distribution and temperature reinforcement of 0.3% of the gross section is provided in the vertical direction. Bottom ring beam reinforcement is designed to resist the hoop tension and the section is
designed by limiting the tensile stresses in concrete. Conical dome reinforcement is designed for hoop tension and Meridional thrust. Bottom spherical dome: The floor of the dome is designed similar to the top dome. The design
load for the dome included the self-weight of the dome and the weight of water column above the dome. The dome reinforcement is designed for Meridional thrust and circumferential forces. The detailed analysis and design is done in design of Intz water tank chapter. 2.4
Design of Water Retaining Structures
A structure that is designed to retain liquids must fulfil the requirements for normal structures in having adequate strength, durability, and freedom from excessive cracking or deflection. In addition, it must be designed so that the liquid is not allowed to leak or percolate through the concrete structure. In the design of normal building structures, the most critical aspect of the 12
design is to ensure that the structure retains its stability under the imposed loads. In the design of structures to retain liquids, it is usual to find that, if the structure has been proportioned and reinforced so that the liquid is retained without leakage. Further, the strength should be retained with time in presence of the liquid stored. The requirements for ensuring a reasonable service life for the structure without undue maintenance are more onerous for liquid retaining structures than for normal structures, and adequate concrete cover to the reinforcement is essential to protect it from corrosion. Equally concrete itself must be of good quality, and be properly compacted, good workmanship during construction is critical. Further, the stresses in steel should limited to that extent only so that concrete is not over strained . Potable water from moorland areas may contain free carbon dioxide or dissolved salts from the gathering grounds which attack normal concrete. Similar difficulties may occur with tanks which are used to store sewage or industrial liquids. If investigation reveals aggressive exposure, it may be necessary to increase the cement content of the concrete mix or use special (4,5,6)
cements or use a special lining to the concrete tank
Concrete to be used in water tank has to be watertight. This can be achieved by proper mixing and placing and curing of concrete. Well graded cement in richer proportion shall be used. 2.4.1
Basis of Design:
There are two important factors to be noted in the design of an R.C.C. tank to have necessary strength as well as imperviousness. It is necessary to prevent shrinkage cracks in the tank walls which will be possible by adopting a distribution reinforcement amounting to at least 0.3% of the gross area of the wall. Further, slabs should be concreted in lengths not greater than 7.5 m. It is also necessary to provide contraction joints between adjacent slab units. In order to accommodate any additional thermal
13
displacement, expansion joints shall also be provided at intervals of 30 m. For floors it is necessary that the ground should be first covered with a 80 mm thick layer of plain concrete. 2.4.2
Design Methods
Historically, the design of structural concrete has been based on elastic theory, with specified maximum design stresses in the materials at working loads. More recently, limit state philosophy has been introduced, providing a more logical basis for determining factors of safety. In ultimate design, the working or characteristic loads are enhanced by being multiplied by a partial safety factor. The enhanced or ultimate loads are then used with the failure strengths of the materials to design the structure. Limit state design methods are now widely used throughout the world for normal structural design. Formerly, the design of liquid-retaining structures was designed on the based on the use of elastic design, with material stresses so low that no flexural tensile cracks developed. This led to the use of thick concrete sections with copious quantities of mild steel reinforcement. The probability of shrinkage and thermal cracking was not dealt with on a satisfactory basis, and nominal quantities of reinforcement were specified in most codes of practice. More recently, analytical procedures have been developed to enable flexure crack width to be estimated and compared with specified maxima. A method of calculating the effects of thermal 6
and shrinkage strains has also been published . These two developments enable limit state methods to be extended to the design of liquid-retaining structures so that water tank can be designed under limit state of collapse and then checked for serviceability. Limit state design methods enable the possible modes of failure of a structure to be identified and investigated so that a particular premature form of failure may be prevented. Limit state may be ultimate’ or ‘serviceability’. 14
In UK, limit state design has been used successfully for over 20 years for the design of liquid retaining structures. The former BS 5337 allowed a designer to choose between elastic design and limit state design. However, as nearly all designers decided to use limit state design, BS 8007 solely recommended limit state design. In Australia and the USA, design methods based on elastic theory are specified in the national codes. Elastic design is a simpler process, but with the widespread use of computer facilities, there is no difficulty in preparing limit state 4
designs . Structural design is often governed by a code of practice appropriate to the location of the structure. Whilst the basic design objectives are similar in each code, the specified stresses and factors of safety may vary. It is important to consider the climatic conditions at the proposed site, and not to use a code of practice written for temperature zones in parts of world with more extreme weather conditions. 2.4.3
Design Requirements of Concrete
In water retaining structure a dense impermeable concrete is required therefore, proportion of fine and course aggregates to cement should be such as to give high quality concrete. Minimum grade of concrete is M 20 for plain cement concrete, M 30 for reinforced concrete . For small capacity tanks up to 50 m
3
at locations where there is difficulty in providing M 30
grade of concrete, the minimum grade of concrete may be taken as M 25. However this 8
exception shall not apply in coastal areas . The minimum quantity of cement in the concrete 3
mix shall be not less than 250 kg/m for plain concrete and 320 kg/m
3
for reinforced concrete.
Maximum free water cement ratio is 0.50 for plain concrete and 0.45 for reinforced concrete. The design of the concrete mix shall be such that the resultant concrete is sufficiently impervious. Efficient compaction preferably by vibration is essential. The permeability of the thoroughly compacted concrete is dependent on water cement ratio. Increase in water cement ratio increases permeability, while concrete with low water cement ratio is difficult t o 15
compact. Other causes of leakage in concrete are defects such as segregation and honey combing. All joints should be made water-tight as these are potential sources of leakage. Design of liquid retaining structure is different from ordinary R.C.C, structures as it requires that concrete should not crack and hence tensile stresses in concrete should be within permissible limits. A reinforced concrete member of liquid retaining structure is designed on the usual principles ignoring tensile resistance of concrete in bending. Additionally it should be ensured that tensile stress on the liquid retaining face of the equivalent concrete section does not exceed the permissible tensile strength of concrete as given in table 1. Table 1 Permissible Concrete Stresses in Calculations Relating to Resistance to Cracking
Sl. No.
Grade of Concrete
i) ii) iii) iv) v) vi)
M 25 M 30 M 35 M 40 M 45 M 50
Permissible Concrete Stresses, 2 N/mm Tension Due Direct Tension to Bending 1.3 1.8 1.5 2.0 1.6 2.2 1.8 2.4 2.0 2.6 2.1 2.8
For calculation purposes the cover is also taken into concrete area. Cracking may be caused due to restraint to shrinkage, expansion and contraction of concrete due to temperature or shrinkage and swelling due to moisture effects. Such restraint may be caused by. (i) The interaction between reinforcement and concrete during shri nkage due to drying. (ii) The boundary conditions. (iii) The differential conditions prevailing through the large thickness of massive concrete. Use of small size bars placed properly, leads to closer cracks but of smaller width. The risk of cracking due to temperature and shrinkage effects may be minimi zed by limiting the changes in moisture content and temperature to which the structure as a whole is subjected. The risk of cracking can also be minimized by reducing the restraint on the free expansion of the structure with long walls or slab founded at or below ground level, restraint can be minimized by the 16
provision of a sliding layer. This can be provided by founding the structure on a flat layer of concrete with interposition of some material to break the bond and facilitate movement. In case length of structure is large it should be subdivided into suitable lengths separated by movement joints, especially where sections are changed the movement joints should be provided. Where structures have to store hot liquids, stresses caused by difference in temperature between inside and outside of the reservoir should be taken into account. -6
The coefficient of expansion due to temperature change is taken as 10 x 10 /° C
2.5
General Design Requirements As Per IS : 3370 – 2009
2.5.1
Basis of Design :
12
.
The structural members shall be designed based on adequate resistance to cracking and adequate strength. In calculation of stresses, for both flexural and direct tension or combination of both relating to resistance to cracking, the whole section of concrete including the cover together with the reinforcement can be taken into account provided the tensile stress in concrete is limited to Table 1.
2.5.2. Permissible Stresses in Concrete. (a) For resistance to cracking. For calculations relating to the resistance of members to
cracking, the permissible concrete stresses shall confirm to the values specified in Table 1. Although cracks may develop in practice, compliance with the assumption stated above ensures that these cracks are not excessive. (b) For strength calculations. In strength calculations the permissible concrete stresses shall
be in accordance with Table 2.
17
Table 2 Permissible Stresses in Concrete for Strength Calculations
Sl. No.
Grade of Concrete
i)
M 25
ii)
Permissible Stresses in Compression 2 N/mm Bending Direct
Permissible Stress in Bond ( Avg. ) for Plain Bars in Tension
6.0
0.9
M 30
10.0
8.0
1.0
iii)
M 35
11.5
9.0
1.1
iv)
M 40
13.0
10.0
1.2
v)
M 45
14.5
11.0
1.3
vi)
M 50
16.0
12.0
1.4
2.5.3
8.5
Permissible Stresses in Steel
2.5.3.1 For resistance to cracking.
The tensile stress in steel will necessarily be limited by the requirement that the permissible tensile stress in concrete is not exceeded; so the tensile stress in steel shall be equal to the product of the modular ratio of steel and concrete, and the corresponding permissible tensile stress in concrete. 2.5.3.2 For strength calculations.
In strength calculations the permissible stress in steel shall confirm to the values specified in shall confirm to Table 3. Table 3 Permissible Stresses in Steel Reinforcement for Strength Calculations
Sl. No.
i)
ii)
Type of Stress in Steel Reinforcement
Tensile stress in members under direct tension, bending and shear Compressive stress in columns subjected to direct load
Permissible Stresses, 2 N/mm Plain Round High Strength Mild Steel Bars Deformed Bars
115
130
125
140
18
2.5.4. Stresses due to drying Shrinkage or Temperature Change.
No separate calculation is required for stresses due to moisture or temperature change in the concrete provided that : a)
The reinforcement provided is not less than that specified in 2.5.8.1 .
b)
The recommendations of the standard with regard to the provision of movement of joints and for a suitable sliding layer beneath the tank given in IS 3370 ( Part 1 ) are complied with,
c)
The tank is to be used only for the storage of water or aqueous liquids at or near ambient temperature and the concrete never dries out, and
d)
Adequate precautions are taken to avoid cracking of the concrete during the construction period and until the tank is put into use.
e)
-6
For shrinkage stresses, shrinkage coefficient 300×10 may be assumed.
2.5.5. Floors 2.5.5.1 Provision of movement joints.
Movement joints should be provided in accordance with IS 3370 ( Part 1 ) 2.5.5.2 Floors of tanks resting on ground.
Where walls or floors are rested on ground, a layer of lean concrete not less than 75 mm thick shall be placed over the ground. In normal circumstances this flat layer of concrete may be weaker than that used in other parts of the structure, nut not weaker than M 15 as specified in IS 456, where injurious soils or aggressive water are expected, the concrete layer shall be of grade not weaker than M 20 and if necessary a sulphate resisting or other special cement should be used. Following a layer of lean concrete, the floor shall be cast in a single layer. A separating layer of 2
polyethylene sheet of mass 1 kg/m should be provided in between the floor slab and the layer of lean concrete. 19
2.5.5.3
Floor of tanks resting on supports
(a) If the tank is supported on walls or other similar supports the floor slab shall be designed as floor in buildings for bending moments due to water load and self-weight. The worst case of loading should not exceed the provisions of IS 456 -2000 as follows : (i) Design dead load on all spans with full design imposed load on two adjacent spans; and (ii) Design dead load on all spans with full design imposed load on alternate spans. (iii)When design imposed load does not exceed three-fourths of the design deal load, the load arrangement may be design dead load and design imposed load on all the spans. (b) When the floor is rigidly connected to the walls (as is generally the case) the bending moments at the junction between the walls and floors shall be taken into account in the design of floor together with any direct forces transferred to the floor from the walls or from the floor to the wall due to suspension of the floor from the wall. 9
(c) Sexena et al. (1987) presented the minimum cost design of Intz reinforced concrete water tanks based on the Indian and ACI ("building" 1969) codes using the heuristic flexible tolerance method (Himmelblau 1972). The cost function included the material costs of concrete, steel, and the formwork. They concluded that a large percentage in cost savings can be achieved for water tanks with large capacities . 2.5.6.
Walls
2.5.6.1 Provision of joints
Where it is desired to allow the walls to expand or contract separately from the floor, or to prevent moments at the base of the wall owing to fixity to the floor, sliding joints may be employed. 2.5.6.2 Pressure on Walls.
(a) In liquid retaining structures with fixed or floating covers the gas pressure developed above liquid surface shall be added to the liquid pressure.
20
(b) When the wall of liquid retaining structure is built in ground, or has earth embanked against it, the effect of earth pressure shall be taken into account as per IS 3370 (Part1)
2.5.6.3 Walls or Tanks Rectangular or Polygonal in Plan.
While designing the walls of rectangular or polygonal concrete tanks, the following points should be taken care of : (a) In plane walls, the liquid pressure is resisted by both vertical and horizontal bending moments. An estimate should be made of the proportion of the pressure resisted by bending moments in the vertical and horizontal planes. The direct horizontal tension caused by the direct pull due to water pressure on the end walls, should be added to that resulting from horizontal bending moments. (b) On liquid retaining faces, the tensile stresses due to the combination of direct horizontal tension and bending action shall satisfy the following condition:
1 where
= calculated direct tensile stress in concrete
= permissible direct tensile stress in concrete (Table 1) = calculated tensile stress due to bending in concrete.
= permissible tensile stress due to bending in concrete.
(c) At the vertical edges where the walls of a reservoir are ri gidly joined, horizontal reinforcement and haunch bars should be provided to resist the horizontal bending moments even if the walls are designed to with stand the whole load as vertical beams or cantilever without lateral supports.
21
(d) In the case of rectangular or polygonal tanks, the side walls act as two way slabs, whereby the wall is continued or restrained in the horizontal direction, fixed or hinged at the bottom and hinged or free at the top. The walls thus act as thin plate subjected triangular loading and with boundary conditions varying between full restraint and free edge. The analysis of moment and forces may be made on the basis of any recognized method. However , moment coefficients, for boundary conditions of wall panel for some common cases are given in IS 3370 (Part 4 ) for general guidance.
2.5.6.4 Walls of Cylindrical Tanks.
While designing walls of cylindrical tanks the following points should be borne in mind: (a) Walls of cylindrical tanks are either cast monolithically with the base or are set in grooves and key ways (movement joints). In either case deformation of wall under influence of liquid pressure is restricted at and above the base. (b) Unless the extent of fixity at the base is established by analysis with due consideration to the dimensions of the base slab, the type of joint between the wall and slab and the type of soil supporting the base slab, it is advisable to assume wall to be fully fixed at the base. Coefficient of ring tension and vertical moments for different conditions of the walls for some common cases are given in IS 3370 (Part 4) for general guidance.
2.5.7. Roofs 2.5.7.1 Provision of Movement joints.
To avoid the possibility of sympathetic cracking it is important to ensure that movement joints in the roof correspond with those in the walls, if roof and walls are monolithic. It, however, provision is made by means of a sliding joint for movement between the roof and the wall, Correspondence of joints is not so important.
22
2.5.7.2 Water tightness.
In case of tanks intended for the storage of water for domestic purpose, the roof must be made water-tight. This may be achieved by limiting the stresses as for the rest of the tank, or by the use of the covering of the waterproof membrane or by providing slopes to ensure adequate drainage. 2.5.8
Reinforcement Detailing
2.5.8.1
Minimum Reinforcement
(a) The minimum reinforcement in walls, floors and roofs in each of two directions at right angles shall have an area of 0.35 per cent of the surface zone, cross section as shown in fi g. 1 and fig. 2 for high strength deformed bars and not less than 0.64 percent for mild steel reinforcement bars. The minimum reinforcement can be further reduced to 0.24 percent for deformed bars and 0.40 percent for plain round bars for tanks having any dimension not more than 15 m. In wall slabs less than 200 mm in thickness, the calculated amount of reinforcement may all be placed in one face as near as possible to the upper surface consistent with the nominal cover. Bar spacing should generally not exceed 300 mm or the thickness of the section, whichever is less. �
NOTE - For D < 500 mm, assume each reinforcement face controls D/2 depth of concrete. For D > 500 mm assume each reinforcement face controls 250 mm depth of concrete, ignoring any central core beyond this surface depth. Fig. 1 Surface Zones : Walls and Suspended Slabs
23
2.5.9
Crack Width due to Temperature and Moisture
Calculation of minimum reinforcement, crack spacing and crack width in relation to temperature and moisture effects in thin section as per procedure given in 2.5.9.1 & 2.5.9.2.
2.5.9.1 Minimum Reinforcement
To be effective in distributing cracking, the amount of reinforcement provided needs to be at least as great as that given by formula :
/ =
where,
….. (1)
= critical steel ratio , that is , the minimum ratio, of steel area to the gross area of the
whole concrete section, required to distribute the cracking.
direct tensile strength of the immature concrete which is as given below :
24
Grade of concrete
2
, N/mm
M25
M30
M35
M40 M45 M50
1.15
1.3
1.45
1.6
1.7
1.8
characteristic strength of the reinforcement.
For ground slabs less than 200 mm thick the minimum reinforcement may be assessed on the basis of thickness of 100 mm and placed wholly in the top surface with cover not exceeding 50 mm. The top surface zone for ground slab from 200 to 500 mm thick may be assesses on half the thickness of the slab. For ground slabs over 500 mm thick, consider them as ‘thick’ sections with the bottom surface zone only 100 mm thick.
2.5.9.2 Crack Spacing
When sufficient reinforcement is provided to distribute cracking the likely maximum spacing of crack Smax shall be given by the formula:
=
where
…… (2)
ratio of the tensile strength of the concrete (
to the average bond strength
between concrete and steel.
= size of each reinforcing bar , and = steel ratio based on the gross concrete section.
For immature concrete, the value of
may be taken as unity for plain round bars and 2/3 for
deformed bars. The value of W max ( estimated maximum crack width ) for cooling to ambient from the peak hydration temperature may be assumed to be :
Wmax = where
α
= Coefficient of thermal expansion of mature concrete )
T1 = fall in temperature between the hydration peak and ambient. 25
The value of T1 depends on the temperature of concreting, cement content, thickness of the 0
member and material for shutters. As per IS 3370-2009 , it is recommended to use T 1 = 30 C 0
for concreting in summer and 20 C during winter, when steel shutters are used. 0
We have taken T 1 = 40 C in designs taking into consideration the Indian ambient conditions. 2.6
Comparative Study of IS 3370-1965 & IS 3370-2009
2.6.1
Permissible Stresses in Steel Reinforcement for Strength Calculations
Sl. No.
Type of Stress in Steel Reinforcement
i)
Tensile stress in members under direct tension
Permissible Stresses, as per IS 3370 -1965 2 N/mm Plain Round High Strength Mild Steel Deformed Bars Bars
150
150
Tensile stress in members under direct bending : a) On retaining face ii)
liquid
b) On face away from liquid for members less than 225 mm c) On face away from liquid for members more than 225 mm
iii)
Compressive stress in columns subjected to direct load
150
150
150
150
125
190
125
175
26
Permissible Stresses, as per IS 3370 -2009 2 N/mm Plain Round High Strength Mild Steel Deformed Bars Bars
115
130
115
130
125
140
2.6.2
Reinforcement Detailing
:
IS : 3370 – 1965
IS : 3370 -2009
Minimum Reinforcement Minimum Reinforcement A ) The minimum reinforcement in walls, A ) The minimum reinforcement in walls, floors and roofs in each of two directions at floors and roofs in each of two directions at right angles shall have an area of right angles, within each surface zone shall not be less than
a) 0.3 % of cross sectional area of sections thickness < 100 mm
a) 0.35 % of surface zone as shown in fig. 1 and fig. 2 for HYSD bars.
b) Linearly varying from 0.3 % to 0.2% for thickness 100 mm to 450 mm.
b) 0.64 % of surface zone as shown in fig. 1 and fig. 2 for mild steel bars
c) mm
0.2 % for section of thickness > 450
d) In concrete sections of thickness > = 225 mm ,two layers of reinforcement be placed one near each face . B ) The minimum reinforcement specified above may be decreased by 20 % in case of HYSD bars.
B ) The minimum reinforcement can be further reduced to a) 0.24 % for HYSD bars. c) 0.40 % for mild steel bars. for tanks having any dimension not more than 15 m. C ) In wall slabs less than 200 mm in thickness, the reinforcement may be placed in one face.
27
CHAPTER 3 3.1
Identification of Problem
To do the comparative study of provisions in IS 3370 (1965) and IS 3370 (2009), three type of problems of water tank were taken. One OHSR water tank problem and other ground level water tank of rectangular and square section were taken in this study. The following four methods of designs: 1.
Working stress method in accordance IS 3370 (1965).
2.
Working stress method in accordance IS 3370 (2009).
3.
Designing by Ultimate Limit State and then checking cracking width by limit state of serviceability IS 3370 (2009).
4.
Limit state design method by limiting steel stresses in accordance IS 3370 (2009) and checking cracking width under serviceability not required (Deemed to b e satisfied).
3.2
Water tank design problems were as follows:
1.)
Design an Intz type water tank of 1 million litres capacity, supported on an elevated lower comprising of 8 columns. The base of the tank i s 16 m above ground level. Depth of foundation 1 m below ground level. Adopt M30 grade of concrete and Fe-415 grade of tor steel. The design of the tank should conform to the stresses specified in Is 3370 and IS 456.
2)
A rectangular water tank is to be designed to store 2500 kl water. The tank is to be 2
made just above the ground level and the safe bearing capacity of the soil is 75 kN/m . Adopt M30 grade of concrete and Fe-415 grade of tor steel. The design of the tank should conform to the stresses specified in IS 3370 and IS 456. 3)
Design a square water tank having inner dimensions of 7.5 × 7.5 × 2.65 m high with walls fixed at the bottom and free at the top. The tank is directly supported on the earth. The floor slab is monolithic with the walls. The free board is 15 cm. Use M30 grade of concrete and Fe 415 grade H.S.D. bars.
28
3.3
DESIGN OF INTZ TANK
Design an Intz-type water tank of capacity 1 million litres, supported on an elevated tower comprising 8 columns. The base of the tank is 16 m above ground level and the depth of foundations is 1 m below ground level. Adopt M30 grade concrete and Fe 415 grade HYSD bars. The design of tank should confirm to the stresses specified in IS : 456-2000. 1.
Data
Capacity of tank = 1 million litres = 1000 m
3
Height of supporting tower = 16 m Number of columns = 8 Depth of foundations = 1 m below ground level. 2.
Permissible Stresses
M – 30 grade concrete and Fe-415 grade HYSD steel, for calculations relating to resistance to cracking as per (IS: 3370-1965)
= 1.5 N/mm
2
= 2.0 N/mm
2
= 2.0 N/mm
2
= 150 N/mm
2
= 130 N/mm
2
as per (IS: 3370-2009)
= 1.5 N/mm
2
For strength calculations the stresses in concrete and steel are same as that recommended in IS: 456-2000
= 8.0 N/mm
3.
2
= 10.0 N/mm
m = 9.34 2
Q = 1.3
j = 0.9
Dimensions of Tank
Let D = Inside diameter of the tank. Assuming the average depth = 0.75 D, we have,
0.75 1000
3
m
Height of cylindrical portion of tank = 8 m 29
i.e. D = 12 m
Depth of conical dome = 2 m Diameter of the supporting tower = 8 m Spacing of bracings = 4 m 4.
Design of Top Dome
Thickness of dome slab = t
= 100 mm
Live load on dome
= 1.5 kN/m
Self-weight of dome = (0.1 × 24)
= 2.4 kN/m
Finishes
= 0.10 kN/m
Total Load w
= 4.0 kN/m
if R
2
2
2
= Radius of the dome
D
= Diameter at base = 12 m
r
= central rise = (1/6 × 12) = 2 m
6 2 /2 2 2 2 10 cos
Semi central angle Therefore,
2
= (8/10) = 0.8
0
= 36 – 50’
22.22 / . 410 . . .
Meridional Thrust =
Circumferential Force = Meridional stress = Hoop stress =
=
2
2
= 0.22 N/mm <
= 0.10 N/mm
The stresses are within safe limits.
30
= 10 kN/m
8 N/mm
2
< 8 N/mm
2
a)
Working Stress Method Reinforcement : As per IS 3370-1965 provisions
Providing nominal reinforcement of 0.30% Ast =
. 300
Provide 8 mm φ at 160 mm centers both circumferentially and meridionally. As per IS 3370-2009 provisions
Providing nominal reinforcements of 0.35% of surface zone
. 175 A = st
Provide 8 mm φ at 200 mm centers both circumferentially and meridionally. b)
Limit State Design Method
Meridional Thrust = T1 = 22.22 kN/m Thickness of the dome required
Using , Factored Thrust =
.. 0.36 . . 0.1728 . .
; x = 0.48 d for Fe 415
=
3
i.e. 22.22 × 1.5 × 10 = 0.1728 × 30 × 1000 ×d Provide
d = 100
mm
Area of Steel Required :
... . Astm= 120 mm 5.
. . .
2
Design of Top Ring Beam
Assume 260 mm by 260 mm size top ring beam. a) Working Stress Method
. . . . 106.6
Hoop Tension = F t =
31
2
= 100 mm / m
As per IS 3370-1965 provisions
Area of steel required =
. .
= 711 mm
Min. steel required ( 0.3 %) =
2
= 780 mm
2
As per IS 3370-2009 provisions
. Area of steel required = = 820 mm . Min. steel required (0.24 %) = = 312 mm 2
2
2
Provide 8 bars of 12 mm φ ( Ast = 904 mm ) Cross – sectional area of Ring Beam ( Ac)
1.50 . .
; Ac=
62623.0 mm
2
Provide 260 mm by 260 mm size top ring beam, with 8 bars of 12 mm φ as main reinforcement and 6 mm φ stirrups at 200 mm centres. b)
Limit State Design Method : Method 1 : Calculation of Crack Width
Hoop Tension
=
Ft
= 106.60 kN
Area of Cross Section of Ring Beam :
Let
b = 150 mm & d = 230 mm Ac = 34,500 mm
2
Inclusion of direct tensile stress under the limit state is required at this stage of IS 3370-2009. It has been taken as 1.55
4
Area of Steel Required :
... .
. . .
= 445 mm
2
Provide 4 bars of 12 mm φ as main reinforcement and 6 mm φ stirrups at 200 mm 2
centres. ( 453 mm )
32
Check for crack width :
Minimum Reinforcement :
= critical steel ratio , that is , the minimum ratio, of steel area to the
gross area of the whole concrete section, required to distribute the cracking.
/
direct tensile strength of the immature concrete.
characteristic strength of the reinforcement.
/
= 453/34500 = 0.013 ,
= 1.3/415 = 0.003
>
Check for max. crack width: Maximum spacing of crack (
,
where = ratio of the tensile strength of the concrete to the average bond strength between concrete and steel which can be taken as 2/3 in this case.
.
= size of each reinforcing bar , and = steel ratio based on the gross concrete section = 307
Width of fully developed crack, Wmax Wmax =
where
α
= 1
-50
-1
10 C
( Coefficient of thermal expansion of
0
concrete ) , T = 40 C -5
Wmax = 307 × 10 × 40/2 = 0.0614 mm c)
< 0.2 mm ( permissible limit )
Method 2 : Limit State ( deemed to satisfy ) condition
Area of steel required
. =
= 820 mm
2
Provide 8 bars of 12 mm φ as main reinforcement and 6 mm φ stirrups at 200 mm 2
centres. ( 904 mm )
33
6.
Design Of Cylindrical Tank Wall
Maximum hoop tension at base of wall Where, w = density of water
and
= Ft =
h = depth of water
480 / 360 / 240 / 120 /
a) Hoop Tension at base of wall, Ft =
b) Hoop Tension at a height of 2 m from base, F t =
c) Hoop Tension at a height of 4 m from base, F t =
d) Hoop Tension at a height of 6 m from base , F t = a)
Working Stress Method As per IS 3370-1965 provisions Area of Steel Required
At base of wall
Ast =
At a height of 2m from base of wall Ast = At a height of 4m from base of wall Ast = At a height of 6m from base of wall Ast =
2
= 3200mm /m 2
= 2400 mm /m 2
= 1600 mm /m 2
= 800 mm /m
Min. steel required as per IS 3370-1965 ( 0.3 %)
. 1050 mm . 600 mm
At base of wall Astm = At top of wall Astm =
As per IS 3370-2009 provisions Area of steel required
At base of wall
Ast =
At a height of 2 m from base of wall Ast =
34
2
= 3700 mm /m 2
= 2770 mm /m
At a height of 4 m from base of wall Ast =
2
= 1850 mm /m
At a height of 6 m from base of wall Ast =
2
= 925 mm /m
Min. steel required as per IS 3370-2009 ( 0.35 %)
420 mm . / 240 mm . /
Min. steel required at base of wall = Astm = Min. steel required at top of wall =
Astm =
2
Provide 20 mm φ at 170 mm centres on each face (A st = 3750 mm ) steel area required 2
at 2 m below top is A st = (2/8 x 3750) = 937 mm provide 10 mm φ at 180 mm centres on each face
If t = thickness of side wall at bottom
. 1.5
,
t = 285 mm
Adopt 350 mm thick walls at bottom gradually reducing to 200 mm at top. Distribution Steel
At bottom,
Ast =
. 700 mm
Provide 10 mm φ at 100 mm centres on both faces. At top
Ast =
. 600 mm
Provide 10 mm φ at 250 mm centres on both faces. The details of reinforcements provided in the cylindrical tank walls at different heights are as follows: Distance From top (m)
Main hoop steel each face ( mm c/c )
Vertical distribution Steel, each face(mm c/c )
0-2
10 – 180
10 - 250
2–4
16 – 200
10 – 250
4–8
20 – 180
10 – 180 35
b)
Limit State Design Method :
Method 1 : Calculation of Crack Width
Max Hoop Tension at the base of wall = F t = 480 kN/m Thickness of the wall required at base of wall :
Using , Factored hoop tension =
.. 0.1728 . . 3
i.e. 480 × 1.5 × 10 = 0.1728 × 30 × 1000 ×d d = 140
mm
at top of wall :
Using , Factored hoop tension =
.. 0.1728 . . 3
i.e. 120 × 1.5 × 10 = 0.1728 × 30 × 1000 ×d d = 35
mm
Provide Minimum Thickness d = 100 mm
Area of Steel Required : at base of wall
.. . .
. .
= 1995 mm
2
. .
= 1496 mm
2
at 2 m from base of wall
.. . . at 4m from base of wall
... .
. .
= 998 mm
2
... .
. .
= 500 mm
2
at 6m from base of wall
Adopt 350 mm thick walls at bottom gradually reducing to 200 mm at top. 36
Provide 20 mm dia HYSD bars at 150 mm c/c. The details of reinforcements provided in the cylindrical tank walls at different heights are as follows:
Distance From top (m)
Main hoop steel each face ( mm c/c )
Vertical distribution Steel, each face ( mm c/c )
2
12 – 100
10 - 250
4
18 – 150
10 – 250
8
22 – 120
10 – 180
Check for Crack Width at each section Distance From top (m)
Crack Width
0-2
0.16
2–4
0.19
4–8
0.16
c) Method 2 : Limit State ( deemed to satisfy ) condition
Area of steel required
7.
=
= 3700 mm
2
Design Of Bottom Ring Beam
Loads on ring beam
1.
Load due to top dome
=
13.30 kN/m
2.
Due to top ring beam 0.260 × 0.260 × 24
=
1.6224 kN/m
3.
Load due to wall=
4.
Self weight 1.2 × 0.60 × 24
. . 8 24 52.8
kN/m
Total Vertical Load = V1 37
=
17.28 kN/m
=
85.00 kN/m
Horizontal Force = H =
cot
= 85 ×
cot45 =
Hoop tension due to vertical load is given by = Hoop tension due to water pressure = Total hoop tension
=
85 kN / m = 510.00 kN
. = 288.00 kN
510.00 + 288.00
=
798.00 KN
a) Working Stress Method Area of Steel Required As per IS 3370-1965 provisions
Area of steel required =
. 2160 mm =
5320 mm
2
Min. steel required ( 0.3 %) =
As per IS 3370-2009 provisions
= 6140 mm Area of steel required = . 2520 mm Min. steel required ( 0.35 %) = 2
2
Provide 18 bars of 22 mm diameter ( A st = 6842 mm ) . Maximum tensile stress =
.
2
1.01 N/mm < 1.5 N/mm
2
Provide a ring beam of 1200 mm wide and 600 mm deep with 18 bars of 22 mm diameter and 10 mm diameter stirrups at 180 mm c/c. b)
Limit State Design Method :
Method 1 : Calculation of Crack Width
Hoop Tension = F t = 798.00 KN Area of Cross Section : Assume 900 × 600 beam then Max Tensile Stress = 1.332 < 1.50 safe Ac = 540000 Area of Steel Required :
A .F. T. . 38
.. .
= 3315 mm
2
Since this is a case of deep beam as per clause 32.5 ( P-62 ) IS 456-2000 Additional side face reinforcement = Ast1 / Ac = 0.0025 Ast1 = 0.0025 × 540000 = 1350 m Total steel required = 3315 + 1350 = 4665 mm
2
2
2
Provide 16 bars of 20 mm φ as main reinforcement ( 5026 mm ) and 6 mm φ stirrups at 200 mm centres. Check for Crack Width : Minimum Reinforcement :
ρ
= 5026/540000 = 0.0093 ,
f /f
= 1.3/415 = 0.003
Check for max. crack width:
Maximum spacing of crack (
SM SM ,
SM .
= 717
Width of fully developed crack, Wmax
SM �
Wmax =
where
α
= 1
10
-50
-1
C
( Coefficient of thermal expansion of
concrete ) 0
T = 40 C -5
Wmax = 717 × 10 × 40/2 = 0.1434 mm
< 0.2 mm ( permissible )
Provide a ring beam of 900 mm wide by 600 mm deep with 16 bars of 20 mm dia. as main 2
reinforcement ( 5026 mm ) and 6 mm φ stirrups at 200 mm centres Method 2 : Deemed to Satisfy Case :
Area of steel required =
= 6140 mm
39
2
8.0
Design Of Conical Dome
Average diameter of conical dome
=
10.00 m
Average depth of water
=
9.0 m
Weight of water above conical dome =
109210
= 5655 kN
Assuming 600 mm thick slab.
Self weight of slab
=
π
× 10 × 2.83 × 0.60 × 24
= 1280.0 kN
Load from top dome, top ring beam, cylindrical wall & bottom ring beam =
π
12
Total load at base of conical slab =
Meridional Thrust
=
Meridional Stress
=
=
=
Load/ unit length
85
=
3205.0 kN
5655 + 1280 + 3205 =
10140 kN
404.00 kN/m 0
404.00 × cosec 45
= 0.954 N/mm
= 2
572.00 kN/m 2
< 8 N/mm
Hoop tension in the conical dome will remain maximum at the top of the conical dome slab since diameter D is maximum at this section.
cot /2
Hoop tension =
H =
Water Pressure =
P
Hoop tension =H=
108 450.6024cot4512/2
a)
= 10 × 8 = 80 kN/m
2
Working Stress Method
Area of Steel Required As per IS 3370-1965 provisions
= 5100 mm . = 1800 mm 2
Area of steel required = Ast minimum =
2
As per IS 3370 -2009 provisions
Area of steel required
=
=
5890 mm
2
40
= 765.0 kN
Minimum Steel Required =Astm=
. = 875 mm
2
2
Provide 25 mm at 190 mm c/c ( Ast = 5034.0 mm ) on both faces of slab. Distribution steel =
.
= 1200 mm
2
Provide 10 mm diameter at 130 mm centres on both faces along the meridians. Max. Tensile stress b)
= .
1.182 < 1.5 N/mm
2
Limit State Design Method :
Method 1 : Calculation of Crack Width Hoop Tension = Ft = 765.00 kN Area of Cross Section of Slab :
Using , Factored hoop tension =
�.�. 0.1728 . . 3
i.e. 765 × 1.5 × 10 = 0.1728 × 30 × 1000 × d d = 260 mm Provide d = 500 mm, D = 540 mm Area of Steel Required :
. . . . . . . = 3180 mm . = 875 mm Minimum Steel Required =A =
2
2
stm
2
Provide 12 bars of 20 mm φ as main reinforcement ( 3770 mm ) and 6 mm φ 200 mm centres.
Check for Crack Width :
/ .
= 3770/(540×1000) = 0.00698 ,
Maximum spacing of crack (
= 1.3/415 = 0.003
,
= 955
Width of fully developed crack, W max
41
stirrups at
W max =
where
α
= 1
-50
-1
10 C ( Coefficient of thermal expansion of
concrete ) 0
T = 40 C -5
W max = 955 × 10 × 40/2 = 0.191 mm
< 0.2 mm ( permissible )
Method 2 : Limit State ( deemed to satisfy ) condition
Area of steel required =
. = 5885 mm
2
2
Provide 12 bars of 20 mm φ as main reinforcement ( 3770 mm ) and 6 mm φ stirrups at 200 mm centres.
Max. Tensile stress 9.
=1.33 < 1.5 N/mm .
2
Design Of Bottom Spherical Dome
Thickness of dome slab assumed as 300 mm Diameter at base
=
Central rise
r
=
D = (1/5 × 8)
=
8m
=
1.6 m
2
R = 5.8 m
If R = radius of the dome 2
(2R – r) r = ( D/2 )
( 2R – 1.6 ) × 1.6 = 4 Self weight of dome slab =
2 5.8 1.6 0.3 24 420
Volume of water above the dome is =
4 8 2 . .
. .
∴
Weight of water = (460 × 10 ) = 4600 kN
∴
Total load on dome = (420+4600) = 5020 kN
Load/unit area = w = Meridional thrust = =
=
100 kN / m
2
42
460
3
m
4.2 = 0.724 5.8
cos θ =
θ = 44.5
o
1 cos 1100 0.75.248 337 / 1.123 cos 2
Meridional stress =
N/mm
Circumferential ( hoop ) force =
= 100 x 5.8
Circumferential ( hoop ) force =
= 100 × 5.8 × ( 0.724 - 1/(1+0.724) ) = 83.5 kN/m
83.5 ×103 2 Hoop stress = = 0.278 N/mm (safe) 300 ×1000 a)
Working Stress Method
Provide nominal reinforcement of 0.3% as per IS 3370-1965 provisions
Ast =
. 900
Provide nominal reinforcement of 0.35 % as per IS 3370-2009 provisions
. A = 525 st
2
Provide 12 mm diabars at 120 mm centres ( 942 mm ) circumferentially and along the meridians. b)
Limit State Design Method :
Method 1 : Calculation of Crack Width
Hoop Tension = F t = 83.50 KN/m Area of Cross Section of Slab :
Using , Factored hoop tension =
�.�. 0.1728 . .
3
i.e. 83.5 × 1.5 × 10 = 0.1728 × 30 ×1000 ×d , d = 25 mm Provide
D = 200 mm 43
Area of Steel Required :
.. .
.. . .
= 347 mm
2
2
Provide 12 mm diabars at 150 mm centres ( 753×2 = 1506 mm ) circumferentially and meridionally. . Check for Crack Width :
/ 2
= 1506/(200×1000) = 0.007530,
= 1.3/415 = 0.003
Maximum spacing of crack (
.
= 885
Width of fully developed crack, W max W max =
where
α
= 1
-50
-1
10 C ( Coeffiecient of thermal expansion of
concrete ) 0
T = 40 C -5
W max = 885× 10 × 40/2 = 0.177 mm
< 0.2 mm ( not permissible )
Method 2 : Limit State ( deemed to satisfy ) condition
. Area of steel required = = 642 mm
2
2
Provide 12 mm diameter bars at 150 mm centres ( 753×2 = 1506 mm ) on both faces of the slab.
44
3.3.1 Comparative Result of INTZ Type Water Tank Working Stress Method Structural Element
IS 3370 – 1965
Limit State Design Method
IS 3370 - 2009
Deemed to
Crack Theory
Satisfy
TOP DOME
Area of Steel Required
300 mm
2
175 mm
2
2
120 mm
%age Change
---
Thickness Required
100 mm
100 mm
100 mm
100 mm
%age Change
--
Nil
Nil
Nil
-
41.66 %
-----
60%
-
TOP RING BEAM
Area of Cross Section
62623 mm
%age Change
--
Area of Steel Reqd.
780 mm
%age Change
-----
2
62623 mm
2
34500 mm
Nil 2
820 mm
2
2
34500 mm
-45% 2
445 mm
+ 5.12 %
-45% 2
820 mm
43%
-
2
+ 5.12 %
CYLINDRICAL TANK WALL
Base Level Thickness
350 mm
350 mm
%age Change
-----
0%
Steel at base
3200 mm
%age Change
-----
+15.6 %
Top Level Thickness
200 mm
200 mm
%age Change
-----
+0 %
Steel at top
800 mm
%age Change
-----
2
925 mm
2
140 mm
60%
-
3700 mm
2
140 mm 1995 mm
- 60% 2
3700 mm
37.65%
-
+15.6 %
100 mm
2
500 mm
+15.62 %
100 mm
50%
-
-50%
2
923 mm
37.5%
-
2
2
+15.37%
BOTTOM RING BEAM
Area of C/S
720000 mm
%age Change
-----
Steel
5320 mm
%age Change
-----
2
720000 mm
2
+0 % 2
540000 mm 25%
-
6140 mm
2
+15.41 %
540000 mm -25%
3315 mm
2
37.68%
-
2
6140 mm
+15.41 %
CONICAL DOME
Thickness
600 mm
%age Change
-----
600 mm
500 mm
+0 %
Steel
5100 mm
%age Change
-----
2
5890 mm
+15.5 %
500 mm
20%
2
3180 mm
- 20% 2
37.64%
-
5885 mm
2
+15.40 %
BOTTOM SPHERICAL DOME
Thickness
300 mm
300 mm
%age Change
-----
+0 %
Steel
900 mm
%age Change
-----
2
525 mm
2
-41.66 %
45
200 mm
200 mm
33.33%
1506 mm +67.33%
2
2
-33.33%
642 mm
2
-28.66%
2
3.4
RECTANGULAR TANK
A rectangular water tank is to be designed to store 2500 kl water. The tank is to be made just 2
above the ground level and the safe bearing capacity of the soil is 75 kN/m . Design data & main dimensions of the tank : 3
2
Capacity of tank = Q = 2500 m , Safe bearing capacity of soil = p a = 75 kN / m , Free Board = 0.15 m. The allowable stresses are :
M – 30 grade concrete and Fe-415 grade tor steel, for calculations relating to resistance to cracking (IS: 3370)
σ ct = 1.5 N/mm 2
σ cb = 2.0 N/mm2
σ st = 150 N/mm2
For strength calculations the stresses in concrete and steel are same as that recommended in IS: 456.
σ cc = 8.0 N/mm 2
m = 9.34
σ cbc = 10.0 N/mm 2
Q = 1.3
J = 0.9
Assuming a clear height of the tank as 5.25 m, Clear base
=
..
490
2
m
Assuming the clear dimensions of the tank as followings:
Height = 5.4 m , length = 24.6 m , width = 19.6 m
Let the roof slab is supported by columns spaced 5 m apart in both directions. Let the center to center distance between the walls and the head of water be : Lx = 24.6+0.4 = 25 m
, L y == 19.6+0.4 = 20.0 m , H = 5.4-0.15 = 5.25 m
Assuming the thickness of the wall at the base as 400 mm. 46
Column
Let the size of the column be assumed as 440 mm square. The capacity of the tank is ( gross volume ) = 24.6 × 19.6 ×5.25 = 2531.3 m
2
Net Volume = Gross volume – volume of 12 columns = 2531.3 - ( 12×5.25×0.44×0.44 ) 1.
= 2518.3 m
3
Design of Roof Slab :
The roof slab is designed as a flat slabwith columns spaced at 5 m apart. Let the thickness of the slab be 240 mm for self-weight purpose and the loads on the slab are : Live load
=
1.5 kN/m
2
Surface finish load
=
2.0 kN/m
2
Self weight = 0.24 ×25
=
6.0 kN/m
2
Total load = w
=
9.5 kN/m
2
Centre to centre of panel L = 5 m Clear panel L0 = 5 - .44
= 4.56 m
Total design load on the panel is = W = w L L0 = 9.5 × 5 × 4.56
= 216.6 kN
a) Working Stress Method
The sum of the magnitudes of the positive and negative bending moments in a panel is M0 = 1/8 W . L0
= 216.6 × 4.56 / 8 = 123.5kNm
The effective height of the wall be Lw=5.4 m
The relative stiffness of the wall is , Kw =
= . = 8 × 10 .
-4
Where the average thickness of the wall is taken as 220 mm. The relative stiffness of the panel slab by assuming the thickness 220 mm is
47
. = 8.9 × 10
Ks = =
-4
The ratio of the relative stiffness of the wall and the roof slab panel is ;
αc =
K = = 0.9 K .
the ratio of live load to the dead load is ,
=
. 0.1875 .
The designed bending moments in the flat slab are computed using the moment coefficients 3
given in table 4.6 and 4.7 .
1
= 2.111
3
Using table 4.7 , c1 = 0.308 ,c2 = 0.702 ,c3 = 0.497 & cl = 0.65 The magnitude of the bending moments (in kNm) in a panel are:
3
= c1 M0= 0.308 × 123.5 × 10 = 38 ,
= c2 M0= 87 ,
= c3 M0 = 62 ,
= ci M0 = 81
Where 1 : the inside face of the outer wall 2 : the face at the inner column of the outer panel 3 : the middle point of the outer panel i : the interior panel The bending moments on the panel are distributed between the column and the middle strips as per table 4.6. The bending moments in the column strip which are indicated by a subscript are : Mnc1 = Mn1 = 38 kNm Mpc3 = 0.6 Mp3 = 37.2 kNm 48
Mnc2 = 0.75 Mn2 = 65.25 kNm Mnci= 0.75Mni = 60.75 kNm (100 % of the negative bending moment in the end support, 75% of the negative BM in the interior supports and 60% of the negative BM are assigned to the column strip ). In case of an end wall support 75% can be assigned to the column strip sectional. The effective depth of the section is calculated from the maximum negative bending moment, and is ; Using M = Q . b . d
2
. .. 224 provide d = 250 mm ( Roof slab has been designed as a cracked section as it is not in contact with the water ) Check for Shear Stress :
The allowable shear stress based on diagonal tension i s
0.16
= 0.16
√ 3 0
0.876
The critical shear plane is the peripheral plane which is at a distance 0.5 d from the face of the column. The length of the plane is b0 = 4 ( a+d ) = 4 ( 0.44+0.25) = 2.76 m 2
2
2
2
the shear force on the plane is , V = w ( L – (a+d) ) = 9.5 ( 5 – 0.69 ) = 233 kN
The nominal shear stress on the plane is,
=
. 0.337 ( 0.876 ) .. 49
Thus shear stress is within the allowable limits, therefore the depth is adequate against shear stress. The overall depth of the slab can be ;
t = d + 0.03 = 0.28 m
Design of reinforcement in the column strip
The area of reinforcement, Ast =
. .
The required areas of the reinforcement at different sections are ;
. .
734 mm .
. .
. 1260 mm .
. .
. 719 mm .
. .
. 1175 mm .
2
2
2
2
The minimum area of steel required is 0.20% , Astm= 0.12 b t / 100 = 0.12 ×2500×300/100 = 900 mm
2
There is direct tension in the slab as it supports the vertical wall.
The tension force is =
.= 34.5 kN/m
Direct tension per 2.5 m panel is, T = 34.5×2.5 = 86.25 kN The area of tensile reinforcement for direct tension is,
. =
375 mm
2
50
At any given section the total area of the reinforcement is equal to the sum of the area needed for bending and direct tension. The direct tensile stress caused in the concrete is,
0.115
Which is very small.
Design of reinforcement in the middle strip
The positive bending moment on the middle strip in the exterior panel is Mmp3= 0.35 × 0.25 × M 0 = 0.35×0.25×123.50 = 10.80 kNm The area of the reinforcement is ,
Ast =
. .
=
. .
= 208 mm
2
b) Limit State Design Method
Total factored design load on the panel is = 1.5 × 216.6 kN = 325 kN The sum of the magnitudes of the positive and negative bending moments in a panel is M0 = 1/8 W . L0
= 325 × 4.56 / 8 = 185.25 kNm
2
Using table 4.7 , c1 = 0.308 ,c2 = 0.702 ,c3 = 0.497 & cl = 0.65 The magnitude of the bending moments (in kNm) in a panel are :
3
= c1 M0= 0.308 × 18525 × 10 = 57 ,
= c2 M0= 130 ,
= c3 M0 = 92.02 ,
= ci M0 = 120.4
Where 1 : the inside face of the outer wall 2 : the face at the inner column of the outer panel 3 : the middle point of the outer panel i : the interior panel 51
The bending moments on the panel are distributed between the column and the middle strips as per table 4.6. The bending moments in the column strip which are indicated by a subscript c are: Mnc1 = Mn1 = 57 kNm Mpc3 = 0.6 Mp3 = 55.2 kNm Mnc2 = 0.75 Mn2 = 97.50 kNm Mnci= 0.75Mni = 90.3 kNm ( 100 % of the negative bending moment in the end support, 75% of the negative BM in the interior supports and 60% of the negative BM are assigned to the column strip ). In case of an end wall support 75% can be assigned to the column strip sectional. The effective depth of the section is calculated from the maximum negative bending moment, and is ; M = 0.138 f ck . b .d
2
9
0.0975 ×10 = 0.138 ×30 ×1000 × d
2
d = 154 mm provide 200 mm ( Roof slab has been designed as a cracked section as it is not in contact with the water ) Check for Shear Stress :
The allowable shear stress based on diagonal tension i s
0.25
= 0.25
√ 3 0
1.3693
The critical shear plane is the peripheral plane which is at a distance 0.5 d from the face of the column. The length of the plane is
52
b0 = 4 ( a+d ) = 4 ( 0.44+0.20) = 2.56 m 2
2
2
2
the shear force on the plane is , V = w ( L – (a+d) ) = 9.5 ( 5 – 0.64 ) = 233.6 kN
The nominal shear stress on the plane is,
=
. 0.456 ( 1.3693 ) ..
Thus shear stress is within the allowable limits; therefore the depth is adequate against shear stress. The overall depth of the slab can be ;
t = d + 0.03 = 0.23 m
Design of reinforcement in the column strip
0.87 0.420.48
using
= 0.7
The area of reinforcement, Ast =
.
The required areas of the reinforcement at different sections are ;
. 981 mm .
.
. 1678 mm .
.
. 950 mm .
.
. 1554 mm .
.
2
2
2
2
The minimum area of steel required is 0.12% , Astm = 0.12 b t / 100 = 0.12 ×2500×200/100 = 600 mm
53
2
There is direct tension in the slab as it supports the vertical wall.
The tension force is =
.= 34.5 kN/m
Direct tension per 2.5 m panel is, T = 34.5×2.5 = 86.25 kN Factored Tension = F.T. = 1.5 ×86.25 = 130 kN The area of tensile reinforcement for direct tension is,
... .
. = .
540 mm
2
At any given section the total area of the reinforcement is equal to the sum of the area needed for bending and direct tension. The direct tensile stress caused in the concrete is,
0.15
Which is very small.
Design of reinforcement in the middle strip
The positive bending moment on the middle strip in the exterior panel is Mmp3= 0.35 × 0.25 × M 0 = 0.35×0.25×185.25 = 16.2 kNm The area of the reinforcement is ,
Ast
2.
= .
. = .
= 280 mm
2
Design Of Column( M30 Grade concrete )
The columns are spaced at 5 m and are subjected to axial force only. The load from the roof slab on each column is, 2
Ps= w L = 9.5 × 25 =
237.50 kN
54
Let self wt.
=
Total load
P
=
32.5 kN 270 kN
a) Working Stress Method : As per IS 3370 - 1965
Assume only 0.8% of reinforcement in the column, the capacity of the column is then given by:
P 10.008 0.008 0.99280.008175 9.336 =
=
≥
≥
3
P ( = 270×10 )
2
28920 mm
Size of column required is = a =
√ 28920
= 170 mm
Provide 350 mm square column. Area Provided = 122500 mm
2
Slenderness ratio = g = L e / a = 3.9 / .35 = 11.14 Area of Reinforcement= 0.8 % × 122500 = 980 mm
2
Provide 350 mm size of column with 6 bars of 16 mm dia in each column. Also provide 6 mm ties at 200 mm spacings. A column head of 440 mm square is to be provided so as to match with the assumptions made in the design of the roof slab.
55
As per IS 3370 - 2009
Assume only 0.8% of reinforcement in the column, the capacity of the column is then given by
P 10.008 0.008 0.99280.008140 9.056 =
=
≥
≥
3
P ( = 270×10 )
2
28920 mm
Size of column required is = a =
√ 28920
= 170 mm
Provide 350 mm square column. Area Provided = 122500 mm
2
Slenderness ratio = g = L e / a = 3.9 / .35 = 11.14 Area of Reinforcement= 0.8 % × 122500 = 980 mm
2
Provide 350 mm size of column with 6 bars of 16 mm dia in each column. Also provide 6 mm ties at 200 mm spacings. A column head of 440 mm square is to be provided so as to match with the assumptions made in the design of the roof slab. b) Limit State Design Method
Factored Load = 1.5 × 270 = 405 kN Assume only 0.8% of reinforcement in the column, the capacity of the column is then given by:
P 0.4 0.67 3
2
2
405 ×10 = 0.4 × 30 × (1-.008) a + 0.67 × 415 × 0.008 a 2
a = 28665 ; square column of 200 mm size. 56
Area of steel provided = 1206 mm
2
Provide 350 mm size of column with 6 bars of 16 mm dia in each column. Also provide 6 mm ties at 200 mm spacings.
The slenderness factor for the column is , i)
g = L e / a = 3.9 / 0.35 = 11.14
Check for crack width :
Minimum Reinforcement :
=
Critical
steel
ratio,
that
is,
the
minimum
ratio,
of
steel
area
to
the
gross area of the whole concrete section, required to distribute the cracking.
/
direct tensile strength of the immature concrete.
characteristic strength of the reinforcement.
= 1206/(350×350) = 0.00984 ,
/
= 1.3/415 = 0.003
>
Maximum spacing of crack (
, where
=
ratio of the tensile strength of the concrete to the average
bond strength between concrete and steel which can be taken as 2/3 in this case.
.
= size of each reinforcing bar , and = steel ratio based on the gross concrete section
= 542
57
Width of fully developed crack, Wmax
Wmax =
where
α
= 1
10
-50
-1
C
( Coefficient of thermal expansion of
concrete ) 0
T = 40 C -5
Wmax = 542 × 10 × 40/2 = 0.1084 mm ii)
< 0.2 mm ( permissible
Method 2 : Limit State ( deemed to satisfy ) condition
Column has not been designed under this state as the stresses in concrete under limit state at steel stress level 140 MPa is not given. 3.
Design Of Vertical Wall :
a) Working Stress Method
The lengths of the walls (20 m and 25 m) are larger when compared with the height. The walls will act as cantilevered from the base with some end constraints. The bending moment coefficients are taken from table 3 of IS 3370 (Part IV)-1965 R-1999 cxa = -.02 , cyb = -0.063, cyc = -0.03 The corresponding bending moments are :
. . 0.02105.25 28.94 / . . 0.063105.25 91.16 / . . 0.03105.25 43.41 / The axial load coming on the wall from the roof slab for the 2.5 m width is Ps = w L/2 = 9.5 × 2.5/2 58
= 11.875 kN/m
Let the self weight of the wall for 300 mm average thickness ( assumed ) be Pg = 5.4 (1)(0.3)(25) = Total axial force Use
40.5 kN/m
P = Ps + Pg = 52.375 kN/m
P = 55 kN/m
At the bottom of the wall the bending causes tension in the concrete on the water face, so the design of the section is controlled by
Or
The bending moment and the axial force at the base are M = ( -Myb ) = 91.16 kNm/m P = 55 kN/m Therefore, substituting the above quantities in the governing equation, we get
. or
0.515
2.0
, Provide 520 mm thickness
The effective thickness of the wall is also obtained for comparison by the cracked theory. J = 0.90
, K = 0.29
Neglecting the effect of the axial force, we have
264 mm d QM. .. =
, Provide 520 mm thickness
59
As per IS 3370-1965
Then the area of the tension reinforcement is
Ast =
.. ..
Minimum Steel Required = 0.20% =
1300
2
mm /m
. = 1100 mm /m 2
As per IS 3370-2009
Ast =
.. ..
1500
2
mm /m
Minimum Steel Required = 0.35 % of surface zone =
. = 1925 mm /m 2
Provide 18 mm bars at 120 mm spacing on the inner face at the bottom of the wall. As the bending moment at the mid height of the tank is positive, all the bars can be curtailed or cranked at mid height . b) Limit State Design Method : 2
6
M = 0.138 f ck . b .d , 91.16 ×10 = 0.138 ×30 ×1000 × d
2
d = 148 mm provide 200 mm Providing a clear cover of about 30 mm, the overall thickness of the wall needed is about 230 mm . However, the smaller value is selected. Let
t = 230 mm and d= 200 mm
Area of Steel Required :
Ast =
.
=
. 1570 . 2
provide 16 mm dia bars @ 125 mm c/c . (1610 mm )
60
i)
check for crack width :
Minimum Reinforcement :
= Critical steel ratio , that is , the minimum ratio, of steel area to the
gross area of the whole concrete section, required to distribute the cracking.
/
direct tensile strength of the immature concrete.
characteristic strength of the reinforcement.
/
= 1610/200×1000 = 0.00805 ,
= 1.3/415 = 0.003
>
Maximum spacing of crack (
, where
=
ratio of the tensile strength of the concrete to the average
bond strength between concrete and steel which can be taken as 2/3 in this case.
.
= size of each reinforcing bar , and = steel ratio based on the gross concrete section
= 662
Width of fully developed crack, Wmax
Wmax =
where α = 1
-50
-1
10 C (Coefficient of thermal expansion of concrete)
0
T = 40 C -5
Wmax = 662 × 10 × 40/2 = 0.1324 mm
< 0.2 mm ( permissible
61
ii)
Method 2 : Limit State ( deemed to satisfy ) condition
Area of steel required =
. = 701 mm
2
Reinforcement At Mid-Height : a) Working Stress Method
The bending moment on the vertical fibres at mid-height of the tank is Myc = 43410 Nm/m The thickness of the wall at the top is assumed to be 150 mm and it is increased to 230 mm at the base. So the thickness at the mid height is 0.5 (150+230) = 190 mm, the corresponding effective depth to the reinforcement is 190-35 = 155 mm, therefore tc = 190 mm , d = 155mm As per IS 3370-1965
Then the area of the tension reinforcement is
Ast =
.. ..
Minimum Steel Required = 0.27 % =
2062
2
mm /m
. = 4185 mm /m 2
As per IS 3370-2009
Ast =
.. ..
4830
2
mm /m
Minimum Steel Required = 0.35 % of surface zone =
. / = 333 mm /m 2
b) Limit State Design Method 2
M = 0.138 f ck . b .d ,
6
43.140 ×10 = 0.138 ×30 ×1000 × d d = 102 mm provide 150 mm 62
2
Providing a clear cover of about 30 mm, the overall thickness of the wall needed is about 180 mm . However, the smaller value is selected. Let
t = 180 mm and d = 150 mm
Area of Steel Required :
Ast = .
. 825 mm = . 2
Provide 12 mm dia bars @ 125 mm c/c .( 904 mm ) i)
Check for crack width :
Minimum Reinforcement :
=
Critical
steel
ratio,
that
is,
the
minimum
ratio,
of
steel
area
to
the
gross area of the whole concrete section, required to distribute the cracking.
direct tensile strength of the immature concrete. characteristic strength of the reinforcement.
= 904/180×1000 = 0.0050 ,
>
/
= 1.3/415 = 0.003
/
Maximum spacing of crack (
, where
= ratio of the tensile strength of the concrete to the average
bond strength between concrete and steel which can be taken as 2/3 in this case.
= size of each reinforcing bar, and = steel ratio based on the gross concrete section
63
.
= 800
Width of fully developed crack, Wmax
Wmax =
where
α
= 1
10
-50
-1
C
( Coefficient of thermal expansion of
concrete ) 0
T = 40 C -5
Wmax = 800 × 10 × 40/2 = 0.16 mm ii)
< 0.2 mm ( permissible
Method 2 : Limit State ( deemed to satisfy ) condition
Area of steel required =
. = 331 mm
2
Design of the horizontal reinforcement :
The positive bending moment on the horizontal fibres of the wall is zero; therefore, only nominal reinforcement be provided. However, there is negative bending moment at the corners of the wall. Its value at about the mid height of the wall is Mxa = -28940 Nm/m There is an axial tension coming from the hydrostatic force acting on the walls normal to this wall. This force at mid height is of H/2 width of the wall. Therefore, the tension force can be taken approximately as
. 13.8kN/m
The tensile stress due to the combined action on the vertical plane of the wall is
The thickness of the wall at mid height is 190 mm so 64
4.809 0.061 4.87
> 2 MPa
The axial tension contribution is negligible but the tension caused by bending moment exceeds the allowable value of 2.0 MPa. Therefore, a fillet be provided at the corners so as to reduce the tension. Provide 150 mm thick haunch at the corners of the wall. The overall thickness at the joint including fillet is 190+150 = 340 mm and the
Bending stress is =
1.5 2.0
The effective depth of the section at this point can be taken as 340-100=240 mm. a) Working Stress Method As per IS 3370-1965
The area of the tension reinforcement is
Ast =
.. .
900
2
mm /m
The minimum reinforcement for 190 mm thickness is about
Astm =
. 532 mm /m 2
2
Provide 12 mm bars at 125 mm spacing on the inner face of the wall. (Astp = 905 mm ) As per IS 3370-2009
The area of the tension reinforcement is
Ast =
.. .
1040
2
mm /m
The minimum reinforcement for 190 mm thickness is about 65
Astm =
./ 333 mm /m 2
2
Provide 16 mm bars at 150 mm spacing on the inner face of the wall. (Astp = 1340 mm ) b) Limit State Design Method
Mxa = -28940 Nm/m Factored Moment = 1.5 ×28940 = 43410 Nm/m Thickness Required : M = 0.138 f ck ck . b .d
2
3
43410 ×10 = 0.138 ×30 ×1000 × d
2
d = 102 mm provide 150 mm Area of Steel Required :
Ast =
.
=
997 . 997 2
Provide 16 mm dia bars @ 150 mm c/c .( 1340 mm ) i)
Check For Crack Crack Width :
Minimum Reinforcement Reinforcement :
=
Critical
steel
ratio,
that
is,
the
minimum
ratio,
of
steel
gross area of the whole concrete section, required to distribute the cracking.
direct tensile strength of the t he immature concrete. characteristic strength of the reinforcement. reinforcement.
= 1340/150×1000 1340/150×1000 = 0.00893 ,
/
= 1.3/415 = 0.003,
66
/ >
area
to
the
Check for max. crack width: Maximum spacing spacing of crack crack (
, where
= ratio of the tensile strength of the concrete concrete to the average average
bond strength between concrete and steel which can be taken as 2/3 in this case.
.
= size of each reinforcing bar, and = steel ratio based on the gross concrete section
= 597
Width of fully developed crack, crack, Wmax
Wmax =
where
α
= 1
10
-50
-1
C
( Coefficient Coefficient of thermal expansion expansion of of
concrete ) 0
T = 40 C -5
Wmax = 597 × 10 × 40/2 = 0.1194 mm ii)
Method 2 : Limit State ( deemed to satisfy ) condition
. = 335 mm
Area of steel steel required = 4.
< 0.2 mm ( permissible
2
Design Of Base Slab
The bottom slab is resisting on soil and it supports the water and the columns. The weight of the water is directly transferred to the soil. Therefore, the bearing capacity has to be checked. The load from the column is transferred through the bottom slab. The slab need to be designed for tank empty condition as a two way slab subjected to net pressure from the soil. First the bearing capacity of the soil is computed.
67
Load from the roof
=
9.5 kN/m
2
Load from the columns
=
1.5 kN/m
2
Load from the water
=
52.5 kN/m
Load from the bottom slab
=
6.0 kN/m
Total load
=
69.5 kN/m
2
2
2
2
2
The bearing pressure is 69.5 kN/m as against the safe bearing capacity of 75 kN/m . Safe Structural design of bottom slab :
The net bearing pressure on the soil in the tank empty condition is that due to the roof and the column load and it is w = 9.5 + 1.5 = 11 kN/m
2
The bottom slab is designed as a flat slab subjected to the soil pressure and supported by the columns spaced at 5 m apart. Assume a widening of the column as shown in fig. The widened size of the column base be same as the column head and it is 440 mm square. Efective span of the slab L = 5 m Clear Span = L 0 = L – a = 4.56 m Total design load on the panel is = W = w L L0 = 11 × 5 × 4.56
= 250.8 kN
The sum of the magnitudes of the positive and negative bending moments in a panel is M0 = 1/8 1/8 W . L0
= 250.8 × 4.56 / 8 = 143 kNm
The effective height of the wall be Lw=5.4 m
The relative stiffness of the wall is , Kw =
= . = 8 × 10 . 68
-4
Where the average thickness of the wall is taken as 220 mm. The relative stiffness of the panel slab by assuming the thickness 220 mm is
. = 8.9 × 10
Ks = =
-4
The ratio of the relative stiffness of the wall and the roof slab panel is ;
αc =
K = = 0.9 K .
the ratio of live load to the dead load is ,
=
. 0.1875 .
The designed bending moments in the flat slab are computed using the moment coefficients given in table 4.6 and 4.7.
1
= 2.111
2
Using table 4.7 , c1 = 0.308 ,c2 = 0.702 ,c3 = 0.497 & cl = 0.65 The magnitude of the bending moments ( in kNm ) in a panel are :
= c1 M0= 0.308 × 143 = 45 ,
= c2 M0= 100.4 ,
= c3 M0 = 72 ,
= ci M0 = 93
Where 1 : the inside face of the outer wall 2 : the face at the inner column of the outer panel 3 : the middle point of the outer panel i : the interior panel
69
The bending moments on the panel are distributed between the column and the middle strips as per table 4.6. The bending moments in the column strip which are indicated by a subscript c are: Mnc1 = Mn1 = 45 kNm Mpc3 = 0.6 Mp3 = 43.2 kNm Mnc2 = 0.75 Mn2 = 75.3 kNm Mnci= 0.75Mni = 69.75 kNm ( 100 % of the negative bending moment in the end support, 75% of the negative BM in the interior supports and 60% of the negative BM are assigned to the column strip ). In case of an end wall support 75% can be assigned to the column strip sectional. The effective depth of the section is calculated from the maximum negative bending moment, and is ; a) Working Stress Method:
. .. =
= 152 mm Provide 200 mm thickness with overall depth = D = 230 mm Check for Shear Stress :
The allowable shear stress based on diagonal tension i s
0.16
= 0.16
√ 3 0
0.876
The critical shear plane is the peripheral plane which is at a distance 0.5 d from the face of the column. The length of the plane is b0 = 4 ( a+d ) = 4 ( 0.44+0.2) = 2.56 m
70
2
2
2
2
the shear force on the plane is , V = w ( L – (a+d) ) = 11 ( 5 – 0.64 ) = 270.50 kN
The nominal shear stress on the plane is,
=
. 0.528 ( 0.876) ..
Thus shear stress is within the allowable limits, therefore the depth is adequate against shear stress. The overall depth of the slab can be ;
t = d + 0.03 = 0.23 m
Design of Reinforcement in The Column Strip
The area of reinforcement ,Ast =
. .
The required areas of the reinforcement at different sections are ; As per IS 3370-1965
1667 mm . . .
. 2790 mm . . .
. 1600 mm . . .
. 2583 mm . . .
2
2
2
2
The minimum area of steel required is 0.27% , Astm= 0.27 b t / 100 = 0.27 ×2500×230/100 = 1552 mm As per IS 3370-2009
1923 mm . . .
2
71
2
. 3220 mm . . .
. 1850 mm . . .
. 2980 mm . . .
2
2
2
The minimum area of steel required is 0.35% ,
Astm =
.
= 1007 mm
2
There is direct tension in the slab as it supports the vertical wall.
The tension force is =
.= 34.5 kN/m
Direct tension per 2.5 m panel is, T = 34.5×2.5 = 86.25 kN The area of tensile reinforcement for direct tension is, AS per IS 3370-1965
. = 575 mm
2
AS per IS 3370-2009
. = 664 mm
2
At any given section the total area of the reinforcement is equal to the sum of the area needed for bending and direct tension. The direct tensile stress caused in the concrete is,
86250 0.15 2500230 Which is very small. 72
Design of reinforcement in the middle strip
The positive bending moment on the middle strip in the exterior panel is Mmp3= 0.35 × 0.25 × M 0 = 0.35×0.25×215 = 0.0188 MNm The area of the reinforcement is, AS per IS 3370-1965
Ast
= . .
. = .
2
= 697 mm
AS per IS 3370-2009
. .
= 805 mm
2
t = 0.200 + cover to the centre of the steel = 0.200 + 0.065 = 0.265 m The reinforcement of the bottom slab is also proportioned with respect to an allowable stress of 150 MPa and the depth. The bearing pressure on the soil with 280 mm thickness of the bottom slab is within the limits. The wall is resting on the slab with a cantilever of 500 mm; therefore the bending moment from the wall is to be distributed between the cantilever and inside of the slab. For all practical purpose, the slab can be treated as fixed because of the load of the water on it with equal soil reaction. Therefore, additional reinforcement must be provided at the edges to resist the bending moment and the tension from the hydrostatic force.
105.25 207
kN/m
Area of the tension steel in the bottom slab is 73
1380
2
mm / m
b) Limit State Design of Base Slab
M = 0.138 f ck . b .d
2
6
113.25 ×10 = 0.138 ×30 ×1000 × d
2
d = 165 mm provide 200 mm Check for Shear Stress :
The allowable shear stress based on diagonal tension i s
0.25
= 0.25
√ 3 0
1.3693
The critical shear plane is the peripheral plane which is at a distance 0.5 d from the face of the column. The length of the plane is b0 = 4 ( a+d ) = 4 ( 0.44+0.20) = 2.56 m 2
2
2
2
the shear force on the plane is , V = w ( L – (a+d) ) = 9.5 ( 5 – 0.64 ) = 233.6 kN
The nominal shear stress on the plane is,
=
. 0.456 (1.3693) ..
Thus shear stress is within the allowable limits, therefore the depth is adequate against shear stress. The overall depth of the slab can be ;
t = d + 0.03 = 0.23 m
Design Of Reinforcement In The Column Strip
0.87 0.420.48
using
The area of reinforcement ,Ast =
= 0.7
. 74
The required areas of the reinforcement at different sections are ;
.
. 1140 mm .
.
. 1950 mm .
.
. 1104 mm .
.
. 1805 mm .
2
2
2
2
The minimum area of steel required is 0.20% , Astm= 0.20 b t / 100 = 0.27 ×2500×230/100 = 1552 mm
2
Deemed to satisfy condition
.
. 4137mm .
2
There is direct tension in the slab as it supports the vertical wall.
The tension force is =
.= 34.5 kN/m
Direct tension per 2.5 m panel is, T = 34.5×2.5 = 86.25 kN Factored Tension = F.T. = 1.5 ×86.25 = 130 kN The area of tensile reinforcement for direct tension is,
... .
. = .
540 mm
2
At any given section the total area of the reinforcement is equal to the sum of the area needed for bending and direct tension. The direct tensile stress caused in the concrete is, 75
0.226 �Pa σ �T
Which is very small.
Design of Reinforcement In The Middle Strip
The positive bending moment on the middle strip in the exterior panel is Mmp3= 0.35 × 0.25 × M 0 = 0.35×0.25×215 = 0.01881 MNm The area of the reinforcement is ,
Ast =
.
=
. .
= 323 mm
2
3.4.1 Comparative result of Rectangular Water Tank situated on ground Working Stress Method Structural Element
IS 3370 – 1965
IS 3370 - 2009
ROOF SLAB 250 mm -0 %
Limit State Design Method Deemed to Crack Theory Satisfy
Thickness %age Change
250 mm -----
Steel
1260 mm
Area of Cross Section %age Change Area of Steel Reqd. %age Change
COLUMNS 122500 mm 122500 mm -----0 % 980 mm 980 mm -----0 % VERTICAL WALL
40000 mm - 67.34% 1206 mm + 23 %
40000 mm - 67.34% 2387 mm + 143 %
Wall Thickness bottom
520 mm
230 mm
230 mm
at
2
%age Change ----Wall Thickness at mid 190 mm height %age Change ----Steel at Base 1300 mm %age Change ----Steel at Mid Height 4185 mm %age Change -----
1260 mm
2
520 mm -0 %
190 mm -0 % 1925 mm +48 % 4830 mm +15.4 % BASE SLAB 230 mm
154 mm - 38.4% Not Applicable
- 55.76 %
154 mm - 38.4% Not Applicable
- 55.76 %
180 mm
180 mm
-6% 1570 mm +21 % 904 mm - 78 %
-6% 3900 mm 200 % 4830 mm +15.4 %
230 mm
230 mm
Thickness
230 mm
%age Change
-----
+0 %
�� �
�� �
Steel %age Change
2790 mm -----
3220 mm +15.4 %
1950 mm - 30.1 %
4137 mm +48.2 %
76
3.5
SQUARE WATER TANK DESIGN
Design a square water tank having inner dimensions of 7.5 × 7.5 × 2.65 m high with walls fixed at the bottom and free at the top. The tank is directly supported on the earth. The floor slab is monolithic with the walls. The free board is 15 cm. Use M30 grade of concrete and Fe 415 grade H.S.D. bars. Solution : Capacity of water tank
=
3
7.5 × 7.5 ×2.5 = 140 m = 140 kilo litre
Let the thickness of the vertical wall be 230 mm at the top and increases to 280 mm at the bottom of the tank . Let the centre to centre dimensions of the tank be H = 2.65 - 0.15 = 2.50 m L = 7.5 m + 0.23 m = 7.73 m , Therefore, L / H = 7.73/2.5 = 3.10 The bending moment can be obtained from Table 3 of IS : 3370 – Part 4 for L/H = 3.0. A negative bending moment means tension on the water face Table 3 of IS : 3370 – Part 4 for L/H = 3.0 1.
TANK WALL DESIGN
a)
Working Stress Method
Vertical Direction : ( Bending Moment )
Maximum Bending Moment in vertical direction at the bottom of wall =
where
= density of water = 10 kN/m
3
3
B.M. at the bottom of wall = -0.126 × 10 × 2.5 = -19.70 kNm/m
77
0126
3
B.M. at the mid of the wall = = + 0.005 × 10 × 2.5 = 0.78125 kNm/m Horizontal Direction : ( Bending Moment )
Maximum Bending Moment in horizontal direction at the ends of wall = -0.082 × 10 × 2.5
3
= -12.81 kNm/m 3
B.M. at the mid height of the wall = = -0.055 × 10 × 2.5 = -8.60 kNm/m For no cracking in the wall, near bottom
6 Or
Thickness of the wall required, t =
t
≥
. .
= 244 mm
For no cracking in the wall near top,
Thickness of wall required, t =
. .
= 196 mm
Let us adopt a wall thickness of 280 mm at the base and taper it to 230 mm at the top in a height of 2.65 m. Shear Force
IS : 3370 code does not give shear force coefficient for a wall fixed at the base and free at the top. Let us adopt the same coefficients as given for a wall hinged at the base and free at the top being conservative. 78
Maximum S.F. at mid point of bottom of the vertical edge of wall =
0.45
2
= 0.45 × 10 × 2.5 = 28.12 kN/m Maximum S.F. at mid point and top of the vertical edge of wall =
0.406
2
= 0.406 × 10 × 2.5 = 25.37 kN/m The S.F. in the wall on the vertical edge will cause tension in the adjacent wall. While the S.F. at bottom edge will cause tension in the base slab. Check the section:
Let us check the section of the wall on the vertical edge near top , t = 500 mm
. . 0.2291.5 .
.
Provide d = 500 mm and overall depth D = 530 mm at bottom of wall and taper to 300 mm at top. Vertical Reinforcement in wall , i)
At base of wall
B.M. at the bottom causes tension on the water face = 19.70 kNm/m
σ cb =10, σ st =150 M Pa , m = 9.33,σ bt
=
2.0, j = 0.90 , D = 530 mm , d = 500 mm
As per IS 3370-1965
Ast =
.. .
Minimum steel required =
292 . 1000500 1000 / 2
mm /m
79
As per IS 3370-2009
Ast =
.. .
Minimum steel required =
ii)
336
2
mm /m
. 1000500/2 875 /
At mid height of wall from top
As per IS 3370-1965
Ast =
. .. .
Minimum steel required =
15
2
mm /m
. 1000400 1000 /
As per IS 3370-2009
Ast =
. .. .
Minimum steel required =
20
2
mm /m
. 1000 700 / 2
Provide 12 mm dia bars @ 125 mm c/c (Ast = 904 mm ) on the water face at bottom of wall 2
Provide 10 mm dia bars @ 125 mm c/c ( Ast = 628 mm ) on the outer face of wall.
Horizontal Reinforcement in wall i) Near top of wall at corners :
M = 12.81 kNm/m
, T = 25.37 kN/m
80
As per IS 3370-1965
. 275 mm /m . . 1000200 600 / Minimum steel required = 2
Ast required =
As per IS 3370-2009
. 320 mm /m . . 1000400/2 700 / Minimum steel required = 2
Ast required = =
ii) Near top of wall in the middle :
M = 7.5 kNm/m
Provide minimum reinforcement 2
Provide 10 mm dia bars @ 125 mm c/c ( Ast = 628 mm / m ) . �� Limit State Design Method
Thickness of wall required: i) At bottom of wall : 2
using FM = 0.138 f ck b d
0.0155 √ 0.0155 1.519.7010 84.25 ii) At top of wall :
0.0155 1.512.8110 68 Provide wall thickness of 130 mm .
81
Reinforcement required
D = 160 mm, d = 130 mm
a) Vertical reinforcement in wall : �����
0.87 0.420.48
at base of wall :
Ast = .
. . 783 = .
. Minimum Steel Required = 156 At mid of wall
B.M. =
0.78125 /
, D = 100 mm & d = 70 mm
Area of steel required, Ast =
.
=
. . 10 2/ .
. 700 2/
Minimum Steel Required =
2
Provide 12 mm dia bars @ 125 mm c/c (Ast = 904 mm ) . CHECK FOR CRACK WIDTH :
Minimum Reinforcement :
= Critical steel ratio , that is , the minimum ratio, of steel area to the
gross area of the whole concrete section, required to distribute the cracking.
/
direct tensile strength of the immature concrete.
characteristic strength of the reinforcement.
= 904/(160×1000) = 0.00565 ,
/
= 1.3/415 = 0.003
>
82
Check for max. crack width: Maximum spacing of crack (
= ratio of the tensile strength of the concrete to the average
, where
bond strength between concrete and steel which can be taken as 2/3 in this case.
.
= size of each reinforcing bar, and = steel ratio based on the gross concrete section
= 707
Width of fully developed crack, Wmax
Wmax =
where
α
= 1
-50
-1
10 C
(Coefficient of thermal expansion of
concrete ) 0
T = 40 C -5
Wmax = 707 × 10 × 40/2 = 0.1414 mm
< 0.2 mm ( permissible
b) Method 2 : Limit State ( deemed to satisfy ) condition At bottom of wall
Area of steel required = A st =
a)
.
=
.
1665
2
mm /m
Horizontal reinforcement in wall :
Near top of wall at corners
.. = 105 mm / m . . 192 Minimum Steel Required =
... .
using ,
83
2
2
Provide 12 mm dia bars @ 150 mm c/c (Ast = 753 mm ) . CHECK FOR CRACK WIDTH :
Minimum Reinforcement :
= Critical steel ratio , that is , the minimum ratio, of steel area to the gross area of the
whole concrete section, required to distribute the cracking.
/
direct tensile strength of the immature concrete.
characteristic strength of the reinforcement.
= 753/(160×1000) = 0.0047 ,
/
= 1.3/415 = 0.003
>
Check for max. crack width: Maximum spacing of crack (
, where
= ratio of the tensile strength of the concrete to the average
bond strength between concrete and steel which can be taken as 2/3 in this case.
.
= size of each reinforcing bar , and = steel ratio based on the gross concrete section
= 851
84
Width of fully developed crack, Wmax
Wmax =
where
α =
1
-50
-1
10 C (Coefficient of thermal expansion of concrete)
0
T = 40 C -5
Wmax = 851 × 10 × 40/2 = 0.1702 mm
< 0.2 mm ( permissible)
b) Method 2 : Limit State ( deemed to satisfy ) condition At bottom of wall
Area of steel required = A st =
. 1082 mm /m . . . . 2
2. BASE SLAB DESIGN : a) Working Stress Method
M = 19.70 kNm/m, T = 28.12 kN/m , D = 280 mm , d = 250 mm , e = M/T = (19.70 ×1000)/28.12 = 700 mm As per IS 3370-1965
Taking moment about the c.g. of compression zone,
28.1210 7001402500.87250 1500.87250 Or
. . A = . t
2
At = 700 mm /m
Minimum steel required =
. 1000250 875 /
85
As per IS 3370-2009
.. = 810 mm /m . . 1000250/2 300 / Minimum steel required = 2
Ast required = At =
Provide 10 mm bars @ 100 mm c/c at top face of the slab near all edges. 50% bars may be curtailed at 1000 mm and the remaining bars at 2000 mm from the edges. Reduce the thickness of the base slab to 100 mm at 1000 mm from the ends. Provide 8 mm bars @ 200 mm c/c in each direction near the bottom of the slab. b) Limit state design method :
Using
Ast =
0.87 0.420.48 ... 468 . . =
Minimum steel required =
. 1000280/2490 / 2
Provide 12 mm diameter bars at 100 mm c/c ( Ast = 1130 mm )as per IS : 3370-2009 guidelines. Check for crack width :
Minimum Reinforcement :
= critical steel ratio , that is , the minimum ratio, of steel area to the
gross area of the whole concrete section, required to distribute the cracking.
/
direct tensile strength of the immature concrete.
characteristic strength of the reinforcement.
= 1130/(1000 × 280 ) = 0.00403,
/
= 1.3/415 = 0.003
>
86
Check for max. crack width: Maximum spacing of crack (
where
=
2
ratio of the tensile strength of the concrete to the average bond strength between
concrete and steel which can be taken as 2/3 in this case.
.
= size of each reinforcing bar, and
= steel ratio based on the gross concrete section = 992
Width of fully developed crack, Wmax Wmax =
where
α
= 1
-50
-1
10 C
( Coefficient of thermal expansion of
0
concrete ) , T = 40 C -5
Wmax = 992 × 10 × 40/2 = 0.1983 mm
< 0.2 mm ( permissible limit )
c) Method 2 : Limit State ( deemed to satisfy ) condition
Ast =
.
=
. 1236 .
3.5.1. Comparative result of Square Water Tank situated on ground Structural Element
Thickness
Working Stress Method IS 3370 – IS 3370 - 2009 1965 TANK WALL 530 mm 530 mm
%age Change
-----
Nil
Steel
1000 mm
%age Change
-----
2
875 mm
2
- 12.5 %
Limit State Design Method Deemed to Crack Theory Satisfy
160 mm - 69.8%
160 mm
783 mm
1082 mm
- 21.7%
+ 8.2 %
- 69.8% 2
BASE SLAB
Thickness
280 mm
280 mm
150 mm
150 mm
%age Change
-----
Nil
- 46.42%
- 46.42%
Steel
875 mm
810 mm
1130 mm
1236 mm
%age Change
-----
- 7.42 %
+ 29.15 %
+41.25 %
2
87
2
2
CHAPTER 4 Results and Discussions
4.1 Comparative Result of INTZ Type Water Tank Working Stress Method Structural Element
IS 3370 – 1965
IS 3370 - 2009
Limit State Design Method Deemed to Crack Theory Satisfy
TOP DOME
Area of Steel Required %age Change Thickness Required %age Change Area of Cross Section %age Change Area of Steel Reqd. %age Change Base Level Thickness %age Change Steel at base %age Change Top Level Thickness %age Change Steel at top %age Change Area of C/S %age Change Steel %age Change Thickness %age Change Steel %age Change Thickness %age Change Steel %age Change
300 mm
2
--100 mm --
175 mm
2
2
120 mm
41.66 % 60% 100 mm 100 mm Nil Nil TOP RING BEAM 62623 mm 62623 mm 34500 mm -Nil -45% 780 mm 820 mm 445 mm ----+ 5.12 % 43% CYLINDRICAL TANK WALL 350 mm 350 mm 140 mm ----0% 60% 3200 mm 3700 mm 1995 mm ----+15.6 % 37.65% 200 mm 200 mm 100 mm ----+0 % 50% 800 mm 925 mm 500 mm ----+15.62 % 37.5% BOTTOM RING BEAM 720000 mm 720000 mm 540000 mm ----+0 % 25% 5320 mm 6140 mm 3315 mm ----+15.41 % 37.68% CONICAL DOME 600 mm 600 mm 500 mm ----+0 % 20% 5100 mm 5890 mm 3180 mm ----+15.5 % 37.64% BOTTOM SPHERICAL DOME 300 mm 300 mm 200 mm ----+0 % 33.33% 900 mm 525 mm 1506 mm -----41.66 % +67.33%
-----
-
88
100 mm Nil 34500 mm -45% 820 mm + 5.12 % 140 mm - 60% 3700 mm +15.6 % 100 mm -50% 923 mm +15.37% 540000 mm -25% 6140 mm +15.41 % 500 mm - 20% 5885 mm +15.40 % 200 mm -33.33% 642 mm -28.66%
4.2 Comparative Result of Rectangular Water Tank Situated on ground Working Stress Method Structural Element
IS 3370 – 1965
IS 3370 - 2009
ROOF SLAB 250 mm -0 %
Limit State Design Method Deemed to Crack Theory Satisfy
Thickness %age Change
250 mm -----
Steel
1260 mm
Area of Cross Section %age Change Area of Steel Reqd. %age Change
COLUMNS 122500 mm 122500 mm -----0 % 980 mm 980 mm -----0 % VERTICAL WALL
40000 mm - 67.34% 1206 mm + 23 %
40000 mm - 67.34% 2387 mm + 143 %
Wall Thickness bottom
520 mm
230 mm
230 mm
at
2
%age Change ----Wall Thickness at mid 190 mm height %age Change ----Steel at Base 1300 mm %age Change ----Steel at Mid Height 4185 mm %age Change -----
1260 mm
2
520 mm -0 %
154 mm - 38.4% Not Applicable
- 55.76 %
190 mm -0 % 1925 mm +48 % 4830 mm +15.4 % BASE SLAB 230 mm
154 mm - 38.4% Not Applicable
- 55.76 %
180 mm
180 mm
-6% 1570 mm +21 % 904 mm - 78 %
-6% 3900 mm 200 % 4830 mm +15.4 %
230 mm
230 mm
Thickness
230 mm
%age Change
-----
+0 %
�� �
�� �
Steel %age Change
2790 mm -----
3220 mm +15.4 %
1950 mm - 30.1 %
4137 mm +48.2 %
4.3 Comparative Result of Square Water Tank Situated on ground Structural Element
Thickness % age change Steel %age Change Thickness
Working Stress Method Limit State Design Method IS 3370 – 1965 IS 3370 - 2009 Crack Theory Deemed to Satisfy TANK WALL 530 mm 530 mm 160 mm 160 mm ----Nil -69.8% -69.8% 1000 mm 875 mm 783 mm 1082 mm ----- 12.5 % - 21.7% + 8.2 % BASE SLAB 150 mm 280 mm 280 mm 150 mm
%age Change
-----
Nil
Steel
875 mm
810 mm
%age Change
-----
- 7.42 %
2
89
2
- 46.42%
- 46.42%
1130 mm
1236 mm
+ 29.15 %
+41.25 %
2
CHAPTER 5 Conclusions And Future Scope Of Study
Based on the results and discussions following conclusions are arrived at : 1. The size of members remained same for working stress method by IS:3370 (1965) and IS:3370 (2009). However, the requirement of area of steel increased in IS:3370 (2009) for Intz type and rectangular water tanks as the allowable stresses in steel were lower. The steel required in square tank was approximately same in both the cases. However, the change in the clause of requirement of minimum steel decreased the steel required in bottom spherical dome in intz type of tank. 2. The size of members remained same for limit state design methods by IS:3370 (2009) in limit state of collapse as well in deemed to satisfy criteria for all the three tank designs. However, the requirement of area of steel increased in IS:3370 (2009) in serviceability design method as well in deemed to satisfy criteria for all the three tank designs as the allowable stresses in steel were lower. 3. The size of members as well as the requirement of steel decreased for limit state design method by IS: 3370 (2009) in comparison to working stress design methods of both IS : 3370 (1965 ) and IS : 3370 (2009) provisions for all the three type of tanks taken in study.
It was found that the provisions of reinforcement through the surface zones in IS : 3370 (2009) provides economical and more effective reinforcement. However, it was also felt that IS:3370(2009) should have provided direct tensile stress and compressive stress under bending and limit state.
90
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1.
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rd
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3.
th
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4.
nd
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5.
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6.
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7.
th
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8.
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Saxena, K.C. and Adeli, H., 1987. " Cost Optimization of Intz Tanks on Shafts Using Nonlinear Programming ", Eng. Optimization, Vol.10, No.4.
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