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Intze tank design

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intze tank design

Intze tank design

  1. 1. Intze tank MAIN PROJECT REPORT ON DESIGN AND ESTIMATION OF INTZE TANK Submitted in partial fulfillment of the Requirements for the award of the degree of Bachelor of Technology in Civil Engineering By M.LOKESH 09241A0175 K.NAGA RAJU 09241A0178 R.RAJASHEKAR 09241A0188 J.RAJEEV 09241A0190 Under the esteemed guidance of G.V.V SATYA NARAYANA (Associate professor of Civil Engineering Department) DEPARTMENT OF CIVIL ENGINEERING GOKARAJU RANGARAJU INSTITUTE OF ENGINEERING AND TECHNOLOGY (Affiliated to JNTU)
  2. 2. Intze tank ABSTRACT Due to enormous need by the public, water has to be stored and supplied according to their needs. Water demand is not constant throughout the day. It fluctuates hour to hour. In order to supply constant amount of water, we need to store water. So to meet the public water demand, water tank need to be constructed. Storage reservoirs and overhead tanks are used to store water, liquid petroleum, petroleum products and similar liquids. The force analysis of the reservoirs or tanks is about the same irrespective of the chemical nature of the product. All tanks are designed as crack free structures to eliminate any leakage. This project gives in brief, the theory behind the design of liquid retaining structure (Elevated circular water tank with domed roof and conical base) using working stress method. Elements are design in working stress method.
  3. 3. Intze tank ACKNOWLEDGEMENT We would like to express our gratitude to all the people behind the screen who helped us to transform an idea into a real application. We would like to express our heart-felt gratitude to our parents without whom we would not have been privileged to achieve and fulfill our dreams. We are grateful to our principal Dr.JandyalaN.Murthi who most ably run the institution and has had the major hand in enabling us to do our project. We profoundly thank Dr. G.Venkataramana, Head of the Department of CIVIL ENGINEERING who has been an excellent guide and also a great source of inspiration to our work. We would like to thank our internal guide Sri. G.V.V.Satyanarayana Associate Professor for his technical guidance, constant encouragement and support in carrying out our project at college. The satisfaction and euphoria that accompany the successful completion of the task would be great but incomplete without the mention of the people who made it possible with their constant guidance and encouragement crowns all the efforts with success. In this context, We would like thank all the other staff members, both teaching and non-teaching, who have extended their timely help and eased our task. M.LOKESH 09241A0175 K.NAGA RAJU 09241A0178 R.RAJASHEKAR 09241A0188 J.RAJEEV 09241A0190
  6. 6. Intze tank 1. SYMBOLS A = Total area of section Ab = Equivalent area of helical reinforcement. Ac = Equivalent area of section Ah = Area of concrete core. Am = Area of steel or iron core. Asc = Area of longitudinal reinforcement (comp.) Ast = Area of steel (tensile.) Al = Area of longitudinal torsional reinforcement. Asv= Total cross-sectional are of stirrup legs or bent up bars within distance Sv Aw =Area of web reinforcement. AФ= Area of cross –section of one bars. a = lever arm. ac = Area of concrete. B =flange width of T-beam. b = width. br =width of rib. C =compressive force. c = compressive stress in concrete. c’= stress in concrete surrounding compressive steel. D = depth d = effective depth dc = cover to compressive steel ds= depth of slab dt= cover to tensile steel
  7. 7. Intze tank e = eccentricity. dc/d = compressive steel depth factor F =shear characteristic force. Fd= design load Fr = radial shear force. f= stress (in general) fck = characteristic compressive stress of concrete. Fy = characteristic tensile strength of steel. H = height. I = moment of inertia. Ie=equivalent moment of intertia. j= lever arm factor. Ka=coefficient of active earth pressure. Kp =coefficient of passive earth pressure. k = neutral axis depth factor (n/d). L=length. Ld =devolopment length. l = effective length of column or length or bond length. M = bending moment or moment. Mr=moment of resistance or radial bending moment. Mt=torsional moment. Mu=ultimate bending moment Mθ=circumferential bending moment m = modular ratio. n = depth of neutral axis.
  8. 8. Intze tank nc=depth of critical neutral axis. Pa=active earth pressure. Pp= passive earth pressure. Pu= ultimate axial load on the member(limit state design). P = percentage steel. P’= reinforcement ratio. Pa=active earth pressure indencity. Pe=net upward soil pressure. Q= shear resistance. ߬ = shear stress. q’=shear stress due to torsion R= radius. s= spacing of bars. sa= average bond stress. sb= local bond stress. T=tensile force. Tu=ultimate torsional moment. ߪst or t= tensile stress in steel. tc= compressive stress in compressive steel. Vu=ultimate shear force due or design load. Vus=shear carried by shear reinforcement. W= point load. X= coordinate. xu= depth of neutral axis. Z= distance.
  9. 9. Intze tank α = inclination. β = surcharge angle. γ = unit weight of soil γf = partial safety factor appropriate to the loading. γm = partial safety factor appropriate to the material. σcc = permissible stress in concrete. σcbc = permissible compressive stress in concrete due to bending. σsc = permissible compressive stress in bars. σst = permissible stress in steel in tension. σst = permissible tensile strss in shear reinforcement. σsy = yield point compressive stress in steel. µ = co efficient of friction.
  10. 10. Intze tank 2. INTRODUCTION A water tank is used to store water to tide over the daily requirement. In the construction of concrete structure for the storage of water and other liquids the imperviousness of concrete is most essential .The permeability of any uniform and thoroughly compacted concrete of given mix proportions is mainly dependent on water cement ratio .The increase in water cement ratio results in increase in the permeability .The decrease in water cement ratio will therefore be desirable to decrease the permeability, but very much reduced water cement ratio may cause compaction difficulties and prove to be harmful also. Design of liquid retaining structure has to be based on the avoidance of cracking in the concrete having regard to its tensile strength.Cracks can be prevented by avoiding the use of thick timber shuttering which prevent the easy escape of heat of hydration from the concrete mass the risk of cracking can also be minimized by reducing the restraints onfree expansion or contraction of the structure.
  11. 11. Intze tank 1. Objective: 1. To make a study about the analysis and design of water tanks. 2. To make a study about the guidelines for the design of liquid retaining Structure according to is code. 3. To know about the design philosophy for the safe and economical design of water tank. 4. To develop programs for the design of water tank of flexible base and rigid base and the under ground tank to avoid the tedious calculations. 5. In the end, the programs are validated with the results of manual calculation given in concrete Structure. 2.1 Sources of water supply: The various sources of water can be classified into two categories: Surface sources, such as 1. Ponds and lakes, 2. Streams and rivers, 3. Storage reservoirs, and 4. Oceans, generally not used for water supplies, at present. Sub-surface sources or underground sources, such as 1. Springs, 2. Infiltration wells, and 3. Wells and Tube-wells.
  12. 12. Intze tank 3. WATER DEMAND 3.1 Water Quantity Estimation: The quantity of water required for municipal uses for which the water supply scheme hasto be designed requires following data:Water consumption rate (Per Capita Demand in litres per day per head)Population to be served. Quantity= Per demand x Population 3.2 Water Consumption Rate: It is very difficult to precisely assess the quantity of water demanded by the public, sincethere are many variable factors affecting water consumption. The various types of waterdemands, which a city may have, may be broken into following class Water Consumption for Various Purposes: Types of Consumption Normal Range (lit/capita/day) Average % 1 Domestic Consumption 65-300 160 35 2 Industrial and Commercial Demand 45-450 135 30 3 Public including Fire Demand Uses 20-90 45 10 4 Losses and Waste 45-150 62 25 3.3 Fire Fighting Demand:The per capita fire demand is very less on an average basis but the rate at which the wateris required is very large. The rate of fire demand is sometimes treated as a function ofpopulation and is worked out from following empirical formulae: Authority Formula (P in thousand) Q for 1 lakh Population) 1 American InsuranceAssociation Q(L/min)=4637P(1-0.01 ÖP) 41760 2 Kuchling'sFormula Q(L/min)=3182 P 31800 3 Freeman'sFormula Q(L/min)=1136.5(P/5+10) 35050 4 Ministry ofUrbanDevelopmentManual Formula Q(kilo liters/d)=100P for P>50000 31623
  13. 13. Intze tank 3.4 Factors affecting per capita demand: • Size of the city: Per capita demand for big cities is generally large as compared tothat for smaller towns as big cities have sewered houses. • Presence of industries. • Climatic conditions. • Habits of economic status. • Quality of water: If water is aesthetically $ people and their . Medically safe, the consumption will increase as people will not resort to privatewells, etc. • Pressure in the distribution system. • Efficiency of water works administration: Leaks in water mains and services;and un authorised use of water can be kept to a minimum by surveys. • Cost of water. • Policy of metering and charging method: Water tax is charged in two different ways on the basis of meter reading and on the basis of certain fixed monthly rate. 3.5 Fluctuations in Rate of Demand: Average Daily Per Capita Demand = Quantity Required in 12 Months/ (365 x Population) If this average demand is supplied at all the times, it will not be sufficient to meet thefluctuations. •Seasonal variation:The demand peaks during summer.Firebreak outs are generally more in summer, increasing demand. So,there is seasonal variation .• Daily variation depends on the activity. People draw out more water on Sundaysand Festival days, thus increasing demand on these days. • Hourly variations are very important as they have a wide range. During activehousehold working hours i.e. from six to ten in the morning and four to eight inthe evening, the bulk of the daily requirement is taken. During other hours therequirement is negligible. Moreover, if a fire breaks out, a huge quantity of
  14. 14. Intze tank wateris required to be supplied during short duration, necessitating the need for amaximum rate of hourly supply.So, an adequate quantity of water must be available to meet the peak demand. To meet allthe fluctuations, the supply pipes, service reservoirs and distribution pipes must beproperly proportioned. The water is supplied by pumping directly and the pumps anddistribution system must be designed to meet the peak demand. The effect of monthlyvariation influences the design of storage reservoirs and the hourly variations influencesthe design of pumps and service reservoirs. As the population decreases, the fluctuationrate increases. Maximum daily demand = 1.8 x average daily demand Maximum hourly demand of maximum day i.e. Peak demand = 1.5 x average hourly demand = 1.5 x Maximum daily demand/24 = 1.5 x (1.8 x average daily demand)/24 = 2.7 x average daily demand/24 = 2.7 x annual average hourly demand
  15. 15. Intze tank 4. POPULATION FORECAST 4.1 Design Periods & Population Forecast: This quantity should be worked out with due provision for the estimated requirements ofthe future. The future period for which a provision is made in the water supply scheme isknown as the design period. Design period is estimated based on the following: • Useful life of the component , considering obsolescence, wear, tear, etc. • Expandability aspect. • Anticipated rate of growth of population, including industrial, commercial developments& migration-immigration. • Available resources. • Performance of the system during initial period. 4.2 Population Forecasting Methods: The various methods adopted for estimating future populations are given below. Theparticular method to be adopted for a particular case or for a particular city dependslargely on the factors discussed in the methods, and the selection is left to the discrectionand intelligence of the designer. 1. Incremental Increase Method 2. Decreasing Rate of Growth Method 3. Simple Graphical Method 4. Comparative Graphical Method 5. Ratio Method 6. Logistic Curve Method 7. Arithmetic Increase Method 8. Geometric Increase Method
  16. 16. Intze tank 5. WATER TANKS 5.1 CLASSIFICATIONS: Classification based on under three heads: 1. Tanks resting on ground 2. Elevated tanks supported on stagging 3. Underground tanks. Classification based on shapes 1. Circular tanks 2. Rectangular tanks 3. Spherical tanks 4. Intze tanks 5. Circular tanks with conical bottom
  17. 17. Intze tank 6. DESIGN REQUIREMENT OF CONCRETE (I. S. I) In water retaining structure a dense impermeable concrete is requiredtherefore,proportion of fine and course aggregates to cement should besuch as to give highqualityconcrete.Concrete mix lesser than M20 is not used. The minimum quantity ofcement in the concrete mix shall be not less than 30 kN/m3 .The design of the concretemix shall be such that the resultant concrete issu efficiently impervious. Efficientcompaction preferably by vibration is essential. The permeability of the thoroughlycompacted concrete is dependent on water cement ratio. Increase in water cement ratioincreases permeability, while concrete with low water cement ratio is difficult to 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 ofliquid retaining structure is different from ordinary R.C.C. structures as it requires thatconcrete should not crack and hence tensile stresses in concrete should be withinpermissible limits. A reinforced concrete member of liquid retaining structure is designedon the usual principles ignoring tensile resistance of concrete in bending. Additionally itshould be ensured that tensile stress on the liquid retaining ace of the equivalent concretesection does not exceed the permissible tensile strength of concrete as given in table 1. 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 shrinkage 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 minimized 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 provision of a sliding layer. This can be provided by founding the structure on a flat layer ofconcrete with interposition of some material to break the bond and facilitate movement.Incaselength of structure is large it should be subdivided into suitable lengths separated by movement joints, especially where sections are changed the
  18. 18. Intze tank movement joints should be provided.Where structures have to store hot liquids, stresses caused by difference in temperature between insideand outside of the reservoir should be taken into account.The coefficient of expansion due to temperature change is taken as 11 x 10-6 /° C and coefficient of shrinkage may be taken as 450 x 10-6 for initial shrinkage and 200 x 10-6 for drying shrinkage. 6.1 JOINTS IN LIQUID RETAINING STRUCTURES: 6.1.1 MOVEMENT JOINTS. There are three types of movement joints. (i)Contraction Joint. It is a movement joint with deliberate discontinuity without initial gap between the concrete on either side of the joint. The purpose of this joint is to accommodate contraction of the concrete. The joint is shown in Fig. (a) Fig (a) A contraction joint may be either complete contraction joint or partial contraction joint. A complete contraction joint is one in which both steel and concrete areinterrupted and a partial contraction joint is one in which only the concrete is interrupted, the reinforcing steel running through as shown in Fig.(b)
  19. 19. Intze tank Fig (b) (ii)Expansion Joint. It is a joint with complete discontinuity in both reinforcing steel and concrete and it is to accommodate either expansion or contraction of the structure. A typical expansion joint is shown in Fig.(c) Fig(c) This type of joint is provided between wall and floor in some cylindrical tank designs. 6.1.2 CONTRACTION JOINTS: This type of joint is provided for convenience in construction. This type of joint requires the provision of an initial gap between thead joining parts of a structure which by closing or opening accommodates the expansion or contraction of the structure.
  20. 20. Intze tank Fig (d) (iii) Sliding Joint. It is a joint with complete discontinuity in both reinforcement and concrete and with special provision to facilitate movement in plane of the joint. A typical joint is shown in Fig. This type of joint is provided between wall and floor in some cylindrical tank designs. Fig (e) 6.1.3 TEMPORARY JOINTS: A gap is sometimes left temporarily between the concrete of adjoining parts of a structurewhich after a suitable interval and before the structure is put to use, is filled with mortaror concrete completely with suitable jointing materials. In the first case width of the gap should be sufficient to allow the sidesto be prepared before filling.Figure (g) Fig (g)
  21. 21. Intze tank 7. GENERAL DESIGN REQUIREMENTS (I.S.I) 7.1 Plain Concrete Structures: Plain concrete member of reinforced concrete liquid retaining structure may be designed against structural failure by allowing tension in plain concrete as per the permissible limits for tension in bending. This will automatically take care of failure due to cracking. However, nominal reinforcement shall be provided, for plain concrete structural members. 7.2. Permissible Stresses in Concrete: (a) For resistance to cracking: For calculations relating to the resistance of members to cracking, the permissible stresses in tension (direct and due to bending) and shear shall confirm to the values specified in Table 1.The permissible tensile stresses due to bending apply to the face of the member in contact with the liquid. In members less than 225mm ∅ thick and in contact with liquid on one side these permissible stresses in bending apply also to the face remote from the liquid. (b) For strength calculations: In strength calculations the permissible concrete stresses shall be in accordance with Table 1. Where the calculated shear stress in concrete alone exceeds the permissible value, reinforcement acting in conjunction with diagonal compression in the concrete shall be provided to take the whole of the shear. 7.3 Permissible Stresses in Steel: (a) For resistance to cracking. When steel and concrete are assumed to act together for checking the tensile stress in concrete for avoidance of crack, the tensile stress in steel will be limited by the requirement that the permissible tensile stress in the concrete is not exceeded so the tensile stress in steel shall be equal to the product of modular ratio of steel and concrete, and the corresponding allowable tensile stress in concrete. (b) For strength calculations: In strength calculations the permissible stress shall be as follows: a) Tensile stress in member in direct tension 1000 kg/cm2 .
  22. 22. Intze tank b) Tensile stress in member in bending on liquid retaining face of members or face away from liquid for members less than 225mm thick 1000 kg/cm2 . c) On face away from liquid for members 225mm or more in thickness 1250 kg/cm2 . d) Tensile stress in shear reinforcement For members less than 225mm thickness 1000 kg/cm2 for members 225mm or more in thickness 1250 kg/cm2 . (v)Compressive stress in columns subjected to direct load 1250 kg/cm2 . 7.4 Stresses due to drying Shrinkage or Temperature Change: (i)Stresses due to drying shrinkage or temperature change may be ignored provided that . (a) The permissible stresses specified above in (ii) and (iii) are not otherwise exceeded. (b) Adequate precautions are taken to avoid cracking of concrete during the construction period and until the reservoir is put into use. (c) Recommendation regarding joints given in article 8.3 and for suitable sliding layer beneath the reservoir are complied with, or the reservoir is to be used only for the storageof water or aqueous liquids at or near ambient temperature and the circumstances aresuch that the concrete will never dry out. (ii)Shrinkage stresses may however be required to be calculated in special cases, when ashrinkage co-efficient of 300× 10ି଺ may be assumed. (iii) When the shrinkage stresses are allowed, the permissible stresses,tensile stresses to concrete (direct and bending) as given in Table 1 may be increased by 33.33 per cent. 7.5 Floors: (i) Provision of movement joints. Movement joints should be provided as discussed in article 3. (ii) Floors of tanks resting on ground. If the tank is resting directly over ground, floor may be constructed of concrete with nominal percentage of reinforcement provided that it is certain that the ground will carry the load without appreciable subsidence in any part and that the concrete floor is cast inpanels with sides not more than 4.5m.with contraction or expansion joints between. Insuch cases a screed or concrete layer less than 75mm thick shall
  23. 23. Intze tank first be placed on theground and covered with a sliding layer of bitumen paper or other suitable material todestroy the bond between the screed and floor concrete. In normal circumstances the screed layer shall be of grade not weaker than M10,where injurious soils or aggressivewater are expected, the screed layer shall be of grade not weaker than M15 and if necessary a sulphate resisting or other special cement should be used. (iii) Floor of tanks resting on supports (a) If the tank is supported on walls or other similar supports the floor slab shall bedesigned as floor in buildings for bending moments due to water load and self weight. (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 to gether with any direct forces transferred to the floor from the walls orfrom the floor to the wall due to suspension of the floor from the wall.If the walls are non-monolithic with the floor slab, such as in cases, where movement joints have been provided between the floor slabs and walls, the floor shall be designed only for the vertical loads on the floor. (c) In continuous T-beams and L-beams with ribs on the side remote from the liquid, the tension in concrete on the liquid side at the face of the supports shall not exceed the permissible stresses for controlling cracks in concrete. The width of the slab shall be determined in usual manner for calculation of the resistance to cracking of T-beam, L beam sections at supports. (d)The floor slab may be suitably tied to the walls by rods properly embedded in both the slab and the walls. In such cases no separate beam (curved or straight) is necessary under the wall, provided the wall of the tank itself is designed to act as a beam over the supports under it. (e)Sometimes it may be economical to provide the floors of circular tanks,in the shape of dome. In such cases the dome shall be designed for the vertical loads ofthe liquid over it and the ratio of its rise to its diameter shall be so adjusted that the stresses in the dome are, as far as possible, wholly compressive. The dome shall be supported at its bottom on the ring beam which shall be designed for resultan tcircumferential tension in addition to vertical loads.
  24. 24. Intze tank 7.6 Walls: (i)Provision of joints (a)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. (b)The spacing of vertical movement joints should be as discussed in article 3.3 while the majority of these joints may be of the partial or complete contraction type, sufficient joints of the expansion type should be provided to satisfy the requirements given in article (ii) 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. (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. (iii) Walls or Tanks Rectangular or Polygonal in Plan. While designing the walls of rectangular or polygonal concrete tanks, the following points should be borne in mind. (a) In plane walls, the liquid pressure is resisted by both vertical and horizontal bendingmoments. 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. On liquid retaining faces, the tensile stressesdue to the combination of direct horizontal tension and bending action shall satisfy the following condition: (t./t )+ ( óc t . /óct ) ≤ 1 t. = calculated direct tensile stress in concrete t = permissible direct tensile stress in concrete (Table 1) óc t = calculated tensile stress due to bending in concrete. óc t = permissible tensile stress due to bending in concrete. (d)At the vertical edges where the walls of a reservoir are rigidly joined, horizontal reinforcement and haunch bars should be provided to resist the horizontal bending
  25. 25. Intze tank moments even if the walls are designed to withstand the whole load as vertical beams or cantilever without lateral supports. (c) In the case of rectangular or polygonal tanks, the side walls act as two way slabs,where by the wall is continued or restrained in the horizontal direction, fixed or hinged atthe bottom and hinged or free at the top. The walls thus act as thin plates subjected triangular loading and with boundary conditions varying between full restraint and freeedge. The analysis of moment and forces may be made on the basis of any recognizedmethod. (iv) 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. Consequently, only part ofthe triangular hydrostatic load will be carried by ring tension and part of the load at bottom will be supported by cantilever action. (b)It is difficult to restrict rotation or settlement of the base slab and it is advisable toprovide vertical reinforcement as if the walls were fully fixed at the base, in addition to the reinforcement required to resist horizontal ring tension for hinged at base, conditions of walls, unless the appropriate amount 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 , where applicable, the type of soil supporting the base slab. 7.7 Roofs; (i) 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. (ii) Loading: Field covers of liquid retaining structures should be designed for gravity loads, such asthe weight of roof slab, earth cover if any, live loads and mechanical equipment. They should also be designed for upward load if the liquid retaining structure is subjected to internal gas pressure. A superficial load sufficient to
  26. 26. Intze tank ensure safety with the unequalintensity of loading which occurs during the placing of the earth cover should be allowedfor in designing roo fs. The engineer should specify a loading under these temporaryconditions which should not be exceeded. In designing the roof, allowance should bemade for the temporary condition of some spans loaded and other spans unloaded, eventhough in the final state the load may be small and evenly distributed. (iii) 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 water proof membrane or by providing slopes to ensure adequate drainage. (iv) Protection against corrosion: Protection measure shall be provided to the underside of the roof to prevent it from corrosion due to condensation. 7.8 Minimum Reinforcement: (a)The minimum reinforcement in walls, floors and roofs in each of two directions atright angles shall have an area of 0.3 per cent of the concrete section in that direction for sections up to 100mm, thickness. For sections of thickness greater than 100mm, and lessthan 450mm the minimum reinforcement in each of the two directions shall be linearly reduced from 0.3 percent for 100mm thick section to 0.2 percent for 450mm, thicksections. For sections of thickness greater than 450mm, minimum reinforcement in eachof the two directions shall be kept at 0.2 per cent. In concrete sections of thickness225mm or greater, two layers of reinforcement steel shall be placed one near each faceof the section to make up the minimum reinforcement. (b)In special circumstances floor slabs may be constructed with percentage of reinforcement less than specified above. In no case the percentage of reinforcement inany member be less than 0.15% of gross sectional area of the member. 7.9 Minimum Cover to Reinforcement: (a)For liquid faces of parts of members either in contact with the liquid (such as innerfaces or roof slab) the minimum cover to all reinforcement should be 25mm or the diameter of the main bar whichever is grater. In the presence of the sea water and soil sand water of corrosive characters the cover should be increased by 12mm but thisadditional cover shall not be taken into account for design calculations. (b)For faces away from liquid and for parts of the structure neither in contact with theliquid on any face, nor enclosing the space above the liquid, the cover shall be as forordinary concrete member.
  27. 27. Intze tank 8. DOMES A dome may be defined as a thin shell generated by the revolution of a regular curve about one of its axes. The shape of the dome depends on the type of the curve and the direction of the axis of revolution. In spherical and conoidal domes, surface is described by revolving an arc of a circle. The centre of the circle may be on the axis of rotation (spherical dome) or outside the axis (conoidal dome). Both types may or may not have assymmetrical lantern opening through the top. The edge of the shell around its base isusually provided with edge member cast integrally with the shell. Domes are used in variety of structures, as in the roof of circular areas, in circular tanks, in hangers, exhibition halls, auditoriums, planetorium and bottom of tanks, bins andbunkers. Domes may be constructed of masonry, steel, timber and reinforced concrete.However, reinforced domes are more common nowadays since they can be constructed over large spans membrane theory for analysis of shells of revolution can be developed neglecting effectof bending moment, twisting moment and shear and assuming that the loads are carriedwholly by axial stresses. This however applies at points of shell which are removed somedistance away from the discontinuous edge. At the edges, the results thus obtained maybe indicated but are not accurate. The edge member and the adjacent hoop of the shells must have very nearly the same strain when they are cast integrally. The significance of this fact is usually ignored and the forces thus computed are, therefore, subject to certain modifications.Stresses in shells are usually kept fairly low, as effect of the edge disturbance, as mentiioned above is usually neglected. The shell must be thick enough to allow space and protection for two layers of reinforcement. From this point of view 80 mm is considered as the minimum thickness of shell.
  28. 28. Intze tank 9. MEMBRANE THEORY OF SHELLS OF REVOLUTION Fig shows a typical shell of revolution, on which equilibrium of an element, obtained by intersection of meridian and latitude, is indicated. Forces along the circumference are denotted by Nf and are called meridian stresses and forces at right angles to the meridian plane and along the latitude are horizontal and called the hoop stresses, denoted by N .Neglecting variations in the magnitudes of Nf and N , since they are very small.the state of stress in the element is shown in fig (b). Shell of Revolution.
  29. 29. Intze tank two forces N߮(rd ߠ) have the resultant N߮(rd ߠ)d߮ as shown in Fig.(c) and the resultant acts normal to the surface pointed towards the inner side. Forces Nߠ(r1d߮) again have horizontal resultant of magnitude N߮(r1 d߮) dߠ as shown in Fig (d). It has a component N߮(r1d߮)dߠsin߮ directed normally to the shell and pointing towards the inner side. These two forces and the external force normal to the surface and a magnitude Pr(rdߠ) must bein equilibrium. Thus,Nf (rd)df++N (r1df)dsinf+Pr(rd)(r1d )= 0 Combining and as r = r2 sinf from Fig. ((a) Nf /r1+N/r2 = -Pr = pressure normal to the surface In this equation pr is considered positive when acting towards the inner side and negative when acting towards the outerside of the shell.Value s and Nf and N will be positive when tensile andnegative compressive. The equation is valid not only for shells in thform of a surface of revolution, but may be apped to allshells, when the coordinate lines for ߮= constant and ߠ = constant, are the linesof curvature of the surface. Forces in shell Force Nf act tangentially to the surface aall around the circumference. Considering thequilibrium of a segment of shell cut along the parallel to latitude defined by the angle as shown in Fig 2prNf sin f + W= 0, Where W= total load in the vertical dirction on the surface of the shell above the cut. This gives, Nf = -W/2prsinf Eq. is readily solved for Nfand N may then be detrminedby Eq. This theory is applicable to a shell of any material as only the conditiions of equilibrium have been applied and no compatibility relationsships in terms of deformation have been introduced. It is, therefore, immaterial whetherr Hooke's law is applicable or not.
  30. 30. Intze tank 10. WATER TANK WITH SPHERICAL BOTTOM Referring to the tank in Fig.(a),supported along the circumference as shown,the magnitude of Na may be obtained from consideration of equilibrium. If it is required to obtain Na at section 1 - 1 from calculation of the total downward load, there are two possibilities. The downward load may be taken to be the weight of water and tank of the annular part i.e. W1 shown in Fig.(b) Fig (a) Fig (b) Fig. Water tank with spherical bottom. Alterrnatively, the downward load may be calculated from the weight of water and tank bottom of the part i.e W2 less upward reaction of the support as shown in Fig. For section which cuts the tank bottom inside the support, the reaction has to be considered with the weight of water and tank of the annular part. Simillar is the case with Intze reservoir as in Fig. (a), which combines a truncated dome with a spherical segment. Pattern of the two forces Nf1and Nf2 at point A are shown in Fig (b). To eliminate horizontal forces on the supporting ring girder, it is necessary that Nf1cos a1 = Nf2cos a2.
  31. 31. Intze tank 11. DESIGN OF REINFORCED CONCRETE DOMES The requirements of thickness of dome and reinforcement from the point of view of induced stresses are usually very small. However, a minimum of 80 mm is provided so as to accommodate two layers of steel with adequate cover. Similarly a minimu m of steel provided is 0.15% of the sectional area in each direction along the meridians as well as along the latitudes. This reinforcement will be in addition to the requirements for hoop tensile stresses. The reinforcement is provided in the middle of the thickness of the dome shell Near the edges usually some ring beam is provided for taking the horizontal component of the meridian stress. Some bending moment develops in the shell near the edges. As shown in Fig. it is normal to thicken the shell near the edges and provide increased curvature. Reinforcements near the top as well as near the bottom face of the shell are also provided. The size of the ring beam is obtained on basis of the hoop tension developed in the ring due to the horizontal component of the meridian stress. The concrete area is obtained so that the resulting tensile stress when concrete alone is considered does not exceed 1.1N/mm2 to 1.70 N/mm2 for direct tension and 1.5 N/mm2 to 2.40 N/mm2 for tension due to bending in liquid resisting structure depending on the grade of concrete. Reinforcement for the hoop stress is also provided with the allowable stress in steel as 115 N/mm2 (or 150N/mm2 ) in case of liquid retaining structures and 140 N/mm2 (or190 N/ mm2 ) in other cases. The ring should be provided so that the central line of the shell passes through the centroid of the ring beam. Reinforcement has to be provided in both the directions. If the reinforcement along the meridians is continued upto the crown, there will be congestion of steel there. Hence, from practical considerations, the reinforcement along the meridian is stopped below the crown and a separate mesh, as shown in Fig (a), is provided. Alternatively, the arrangement of the bars may be made as shown in plan in Fig (b) In case of domes with lantern opening with concentrated load acting there, ring beam has to be provided at the periphery of the opening. The edge beam there will, however, be subjected to hoop compression in place of hoop tension. Openings may be provided in the dome as required from other functional or architectural requirements. However, reinforcement has to be provided all around theopening as shown in Fig. (c). The meridian and hoop reinforcement reaching the opening should be well anchored to such reinforcement.
  32. 32. Intze tank The allowable stresss specified in IS 3370 for such tanks are as follows: Type of stresses: Permissible stress in N/mm2 High yield strength Plain bars confirming to deformed bars as per Grade-I of IS 432-1966. IS 1786-1966 or is 1139-1966. Tensile stress in members under no table of contents entries found direct load. Direct tensile stress in concrete a may be taken as 1.1 N/mm2 , 1.2. N/mm2 ,1.32 N/mm2 , 1.5 N/mm2 , 1.6N/mm2 and 1.7 N/mm2 for M15, M20, MM25, M30, M35and M40 respectively, the value in tension due to bending i.e.,being1.5N/mm2 ,1.7N/mm2 ,1.82N/mm2 ,2.0 N/mm2 ,2.2 N/mm2 and 2.4 N/mm2 . When steel and concrete are assumed to act together for checking thetensile stress in concrete for avoidance of cracks, the tensile streess in the steel will be limited by the requirements that the stress as mentioned above should not be exceeded. The
  33. 33. Intze tank tensill stress in steel will be modular ratio multiplied by the corresponding allowable tensile stress in concrete. Stresses due to shrinkage or temperature change may be ignored if the permissible stresses in concrete and steel are not exceeded and adequate precautions are taken to avoid cracking of concrete during construction period, until the reservoir is put into use and if it is assured that the concrete will never dry out. If it is required to calculate shrinkage stresses, a shrinkage strain of 300×10-6 may be assumed. When shrinkage stresses are considered, the permissible stresses may be increasedby 33 ଵ ଷ %. When shrinkage stresses are considered it is necessary to check the thickness for no crack. Minimum reinforcement of each of two directions at right angles shall have an areof 0.3% for 100 mm thick concrete to 0.2% for 450 mm thick concrete wall. In floor slabs, minimum reinforcement to be provided is 0.15%. The minimum reinforcement as specified above may be decreased by 20%), if high strength deformed bars are used. Minimum cover to reinforcement on the liquid face is 25 mm or diameter of the bar, whichever is larger and should be increased by 12 mm for tanks for sea water or liquid of corrosive character.
  34. 34. Intze tank 12. OVERHEAD WATER TANKS AND TOWERS Overhead water tanks of various shapes can be used as service reservoirs, as a balancing tank in water supply schemes and for replenishing the tanks for various purposes. Reinforced concrete water towers have distinct advantages as they are not affected by climatic changes, are leak proof, provide greater rigidity and are adoptable for all shapes. Components of a water tower consists of- (a) Tank portion with (1) Roof and roof beams (if any) (2) sidewalls (3) Floor or bottom slab (4) floor beams, including circular girder (b)Staging portion, consisting of (5) Columns (6) Bracings and (7)Foundations Types of water Tanks may be (a) Square open or with cover at top (b) Rectangular open or with cover at top (c) Circular open or with cover at which may be flat or domed. Among these the circular types are proposed for large capacities. Such circular tanks may have flat floors or domical floors and these are supported on circular girder. The most common type of circular tank is the one which is called an Intze Tank. In such cases, a domed cover is provided at top with a cylindrical and conical wall at bottom. A ring beam will be required to support the domed roof.A ring beam is also provided at the junction of the cylindrical and conical walls.The conical wall and the tank floor are supported on a ring girder which is supported on a number of columns. Usually a domed floor is shown in fig a result of which the ring girder supported on the columns will be relieved from the horizontal thrusts as the horizonal thrusts of the conical wall and the domed floor act in opposite direction.
  35. 35. Intze tank Sometimes, a vertical hollow shaft may be provided which may be supported on the domed floor. The design of the tank will involve the following. (1) The dome: at top usually 100 mm to 150 mm thick with reinforcement along themeridians and latitudes. The rise is usually l/5th of the span. (2) Ring beam supporting the dome: The ring beam is necessary to resist thehorizontal component of the thrust of the dome. The ring beam will bedesigned for the hoop tension induced. (3) Cylindrical walls: This has to be designed for hoop tension caused due tohorizontal water pressure. (4) Ring beam at the junction of the cylindrical walls and the conical wall:This ring beam is provided to resist the horizontal component of the reaction of the conical wall on the cylindrical wall.The ring beam will be designed for theinducedhoop tension. (5) Conical slab: This will be designed for hoop tension due to water pressure.The slab will also be designed as a slab spanning between the ring beam at top and the ring girder at bottom. (6)Floor of the tank.The floor may be circular or domed. This slab is supportedon the ring girder. (7) The ring girder: This will be designed to support the tank and its contents.Thegirder will be supported on columns and should be designed for resulting bending moment and Torsion. (8) Columns: These are to be designed for the total load transferred to them. The columns will bebraced at intervels and have to be designed for wind pressure or seismic loads whichever govern. (9)Foundations: A combined footing is usuals provided for all supporting columns. When this is done it is usual to make the foundation consisting of a ring girder and acircular slab. Suitable proportions for the Intze. for case(1) suggested by Reynolds. Total volume ~0.585D3 for case (2), the proportion was suggested by Grey and Total Volume is given by
  36. 36. Intze tank V1 = p݀ మ ర × ‫ܪ‬ = 0.39‫ܦ‬ଷ . for H = D/2. V2 = ௣.௛ ଵଶ ×(‫ܦ‬ଶ + ݀ଶ + ݀) = 0.102D3 . V3 = ௣௛భ ଺ (3‫ݎ‬ଶ + ℎଵ ଶ ) = 0.017D3 . With h1 = 3/25D and r = 0.0179D3 . Volume V = 0.4693D3 . With h1 = D/6 and r = 3/10D. Volume V = 0.493D3 .
  37. 37. Intze tank 13. DESIGN 13. DETAILS OF DESIGN: Design of tank: Design of an intze tank for a capacity of 300000 lts . Assuming height of tank floor above the ground level is 17.3m. Safe bearing capacity of soil 200kn/m2 Wind pressure as per IS875 1200N/m2 Assuming M20 concrete For which σcbe = 7N/mm2 , σcc = 5N/mm2 Direct tension σt = 5N/mm2 Tension in bending = 1.70 N/mm2 Modular ratio m = 13 For Steel stress, Tensile stress in direct tension =115 N/mm2 Tensile stress in bending on liquid face =115 N/mm2 for t < 225 mm and 125 N/mm2 for > 225 mm. Solution: Taking the volume as 0.585 D3 for proportion given in Fig. D = 9.0 m. The dimension of the Tank is shown in fig.
  38. 38. Intze tank Design of Roof Dome: Considering a rise of 1.80 m, radius of the roof dome is given from 1.80(2R-1.80) = (4.75)2 R = 6.525m. Sin φ = (4.5)/6.525= 0.7241 and φ= 43.36< 51.8° Hence no tension Assuming t = 100mm. Hoop stress @ level of springing: f = ௐோ ௧ [cos ߠ − ଵ ଵିୡ୭ୱ ఏ ] = ହଶହ଴×଺∙ହଶହ ଴∙ଵହ [0.72− ଵ ଵ.଻ଶ ] f =0.0298 N/mm2 Hoop stress @ Crown: ߠ=0°
  39. 39. Intze tank f = ସଽହ଴×଺.ହଶହ ଴.ଵହ [1− ଵ ଶ ] f =0.107 N/mm2 Meridional thrust @ level of sprining: T = ௐோ ଵାୡ୭ୱ ఏ = ସଽହ଴×଺.ହଶହ ଵା଴.଻ଶ =18778.34 N/m Compressive stress = ଵ଼଻଻଼.ଷସ ଵହ଴×ଵ଴଴଴ =0.125 N/mm2 provoide 8mm Ring beam @ top : Horizontal component of T= Tcos ߠ =13520.40 N/m Hoop stress in the ring beam =14339.82× ଽ ଶ =60841.82 Area of steel required = ଺ସହଶଽ.ଶ ଴.ଽଶଷ଴ =311.73 mm2 We have to provide 12 mm ⱷ,4 bars of 452.38 mm Size of the ring beam: Let the area of the ring beam section = A mm2 Equivalent concrete area = ‫ܣ‬௖+(m-1)‫ܣ‬௦௧ =‫ܣ‬௖+(13.33-1)452.38
  40. 40. Intze tank = ‫ܣ‬௖+5577.8454 Limiting tensile stress on the eqvivalent concrete area to 2 N/mm2 Cylindrical wall: Pressure intensity at the bottom of cylindrical wall = 4×9810 =39240 N/mm2 Consider bottom strip of the wall as 1 mm. Hoop tension = 39240× ଽ ଶ = 176580 N Ast= ଵ଻଺ହ଴ ଶଷ଴×଴.ଽ = 853.04 mm2 Provide 8 bars of 12 mm diametre of 142.85 mm distance. Thickness of the wall may be kept as 200 mm. Distribution steel = ଴∙ଶସ ଵ଴଴ [200×1000] = 480 mm2 Provide 8 mm diametre bars. = ସ଼଴ ଵ଺×ଷ.ଵଶ Provide 10 mm diametre bars of spacing 100 mm between them. Check for compressive stress at the bottom of the cylindrical wall. Vertical component of T1 = V1 = T1sin ߠ = 24917 × 0.68 = 17184.137 N/m. Weight of the wall = 0.2×4×25000 = 20000 N/m.
  41. 41. Intze tank Weight of ring beam = 0.2×0.2×25000 = 1000 N/m. Total load V2 =38184.137 N/m. Compressive stress = ଷ଼ଵ଼ସ.ଵଷ଻ ଶ଴଴×ଵ଴଴଴ = 0.19 N/mm2 Nominal vertical stress is equql to 0.24% of gross area. Vertical steel = ଴.ଶସ ଵ଴଴ × 200 × 100 = 480 mm2 Provide 10 bars of 8 mm diametre of spacing 100 mm. Ring beam at B : Let T2 be the thrust /m run exerted by the conical wall at the junction B. Resolving vertically at B T2sin ߙ= V2 tan ߙ= ଵ.ହ ଵ.ହ = 1 ߙ = 45°. T2 = ௏మ ୱ୧୬ఈ = ଷ଼ଵ଼ସ.ଵଷ଻ ୱ୧୬ସହ° = 54000.52 N/m. Resolving horizontally at B H2 =T2cos ߙ=54000.52× cos 45° = 38184.137 N/m
  42. 42. Intze tank This horizontal load H2 will produce a hoop tension in ring beam B Hoop tension due to H2 =H2× ௗ ଶ =38184.137× ଽ ଶ N =171828.6165N Let the rinmg beam be 500mm deep Water pressure on the ringh beam =9810× 4 × ହ଴଴ ଵ଴଴଴ =19620 N/m Hoop tension due to water = 19620× ଽ ଶ =88290 N Total hoop tension = 88290 +171828.61 = 260118.61 N Steel for hoop tension = ଶ଺଴ଵଵ଼.଺ଵ ଶଷ଴×଴.ଽ = 1256.611mm2 Provide 6 bars 18 mm ∅ Ast = 1526.81 mm2 . Let ‘A’ be the area of ring beam Equivalent concrete area = A+(m-1)Ast = A+(13.33-1)× 1526.81 = A+18825.61 Limiting the tensile stress on the equivalent concrete area to 2 N/mm2 ଶ଺଴ଵଵ଼.଺ଵ ஺ାଵହ଺଼଼.଴ଵସ = 2 Ac =11233.688 mm2
  43. 43. Intze tank Provide 250× 500 mm size Design of conical slab: Conical slab should be designed for a) Hoop tension b) Bending as it spans on a sloping slab from the ring beam @ B at the ring girder @ ‘c’ Design for hoop tension: ௪ೢା௪ೞ ଶగ + ௪ೢ ଶగ tan ߙ Where Ww= weight of water resting on the conical slab. Ws = weight of the conical slab. ߙ = inclination of the conical slab with the horizontal. Area of water section standing on the conical slab = ହ.ହାସ ଶ × 1.5 = 7.125 m2 . X = ଺ା[ మ.మఱ య ] ଻.ଵଶହ = 0.52 m. Weight of water resting on the conical slab Ww = 9810×7.125×2ߨ[3.52] = 1545882.24 N Length of conical slab = 2.121 m. Take thickness of the slab as 200 mm. Weight of the conical slab Ws = 0.2×2.121×25000×2ߨ[ ଻.ହ ଶ ] = 249874.42 N. Hoop tension = ଵହସହ଼଼ଶ.ଶସାଶସଽ଼଻ସ.ସଶାଵହସହ଼଼ଶ ଶగ =531838.349 N. Hoop steel on the entire section = ହଷଵ଼ଷ଼.ସଽ ଶଷ଴×଴.ଽ
  44. 44. Intze tank = 2569.267 mm2 . Provide 14 bars of 6 mm ∅ =14× ߨ ×64 = 2814.86 mm2 . Design for bending moment: Load per metre width of the conical slab = ௐೢାௐೞ ଶగ × ௠௘௔௡ ௥௔ௗ௜௨௦ = ଵହସହ଼଼ଶ.ଶସାଶସଽ଼଻ସ.ସଶ ଶగ ×ଷ.଻ହ = 76214.279 N. Maximum bending moment = ௐ௟ ଼ = ଻଺ଶଵସ.ଶ଻ଽ×ଵ.ହ ଼ = 14290.177 Nm. Axial compression V2 = T2sin ߙ = ଷ଼ଵ଼ସ.ଵଷ଻ ୱ୧୬ ସହ° = 54000.52 N. Providing 16 mm diametre bar at clear covers of spacing 25 mm. Effective depth = 200−25 − 8 = 167 mm. Distance between centre of section and centre of steel x = d− ௧ ଶ = 167−100 = 67 mm Resultant bending moment = M+T2.x =14290.177× 10ଷ + 54000 × 67 = 17908212.15 Nmm. Ast = ଵ଻ଽ଴଼ଶଵଶ.ଵହ ଵ଺଻×ଶଷ଴×଴.ଽ = 518.04 mm2 Spacing of 16 mm diameter bars = 333.33 mm and provide 3 bars.
  45. 45. Intze tank The bottom dome: Let R be the radius of the dome,then 32 = 1.2(2R−1.2) = 4.35 m. Let 2ߠ be the angle subtended by the dome. sin ߠ = ଷ ସ∙ଷହ = 43°36°° cos ߠ = 0.68 Thickness of dome = 200 mm. Loads: Dead load = 25000×0.2 = 5000 N/mm2 . Weight of water resting on the dome = ߛ௪[ߨ‫ݎ‬ଶ h− గ௛೎ ଷ (3R−ℎ௖)] =9810[155.508−17.869] = 1350234.872 Area of dome surface = 2ߨRh = 2ߨ × 4.315 × 1.2
  46. 46. Intze tank = 32.79 m2 . Load intensity due to weight of water = ଵଷହ଴ଶଷସ.଼଻ ଷଶ.଻ଽ = 41178.25 N/m2 . Total load intensity = 5000+41178.25 = 46178.25 N/m2 . Meridional thrust = ௐோ ଵାୡ୭ୱ ఏ = ସ଺ଵ଻଼.ଶହ×ସ.ଷହ ଵା଴.଻ଶ = 116788.016 N/m. Meridional compressive stress = ଵଵ଺଻଼଼.ଵ଺ ଶ଴଴×ଵ଴଴଴ = 0.583 N/mm2 . Hoop stress = ௐோ ௧ [cos ߠ − ଵ ଵାୡ୭ୱ ఏ ] = ସ଺ଵ଻଼.ଶହ×ସ.ଷହ ଴.ଶ଴଴ [0.72− ଵ ଵ.଻ଶ ] = 0.139 N/mm2 . Hoop stress at the crown ߠ = 0°. Maximum hoop stres = ௐோ ௧ [cos ߠ − ଵ ଵାୡ୭ୱ ఏ ] = 502188.46 = 0.502 N/mm2 . These stresses are low and hence provide nominal 0.3% steel. Provide 8 mm ∅ bars @100 mm spacing. Circular girder: The total load on the circular girder consists of the following; Total weight of water W1 = weight of water on conical slab + weight of water on dome. = 1545882.24+1350234.872 = 2896117.112. Weight of dome + cylindrical wall + ring beam at A W2 = 38184.137×2ߨ ×4.5 = 1079631.039 N. Weight of ring beam at B W3 = 0.25×0.5×25000× 2ߨ ×4.5 = 88357.29 N. Weight of conical wall W4 = 249874.42 N.
  47. 47. Intze tank Weight of lower dome W5 = 5000×32.79 = 163950 N. = 0.0075×4560396.668×3 = 102608.925 N. Torsion = 0.0015×W×r = 20521.785 N. (from table 2) Angular distance for maximum torsion = 12°44°°. Let us provide 8 coloumns. Bending momment at the support = 0.0083×W×r = 0.0083×4591027.197×3 = 114316.577 Nm. Bending moment at centre = 0.00416×W×r = 0.00416×4591027.197×3 =57296.01 Nm. Torsion = 0.0006×4591027.197×3 = 8263.84 Nm. Angular distance for maximum torsion = 9°33°°. Load at each support = ௐ ଼ = ସହଽସହ଺ଵ.ସସ଼଼ ଼ = 573878.39 N. Shear force at the support = ௐ ଶ , V = 286939.199. Design at support section: Equating moment of resistance to the bending moment at support 0.913bd2 = 114316.577×1000, 0.913×400×d2 = 114316.577×1000, Then d2 = 278458.26, d =560 mm. Let the clear cover be 40 mm. Over all depth of beam = 600 mm. Actual effective depth = 600 mm.
  48. 48. Intze tank Equivalent shear force = V+1.6 ் ௕ = 286939.199+1.6 ் ௕ . = 287160.093+( ଵ.଺×଼ଶ଻଴.ଶଵ×ଵ଴଴଴ ସ଴଴ ). Vc = 319994.559. Equivalent nominal shear stress ߬ve = ௏೐ ௕ௗ = ଷଵ଺ହ଺ହ.ଶ଼ ସ଴଴×ହ଺଴ = 1.42 N/mm2 . Maximum shear stress ߬max> ߬v. ߬max= 1.8 N/mm2 . ߬c< ߬v. Provide longitudinal and transverse reinforcement according to B-6.4 Longitudinal reinforcement: Me = M+Mt , Mt = ்(ଵା ವ ್ ) ଵ.଻ = ଼ଶ଺ଷ.଼ସ[ଵା లబబ రబబ ]×ଵ଴଴଴ ଵ.଻ = 12152705.88 Nmm. M = moment at crosssection. Mer = 1000×114316.577+12152705.88 = 126469282.9 Nmm. Ast= ெ೐ೝ ଶଷ଴×଴.ଽ×ହ଺଴ = ଵଶହଶଵହଷଵଶ ଶଷ଴×଴.ଽ×ହ଺଴ = 1080.187 mm2 . Transverse reinforcement: Asv= ்∙௦ೡ ௕భௗభఙೞೡ + ௏∙௦ೡ ଶ.ହௗభఙೞೡ , b1 = 400−80 = 320 mm , d1 = 600−80 = 520 mm. Asv = [ ଼ଶ଺ଷ.଼ସ×ଵ଴଴଴ ଷଶ଴×ହଶ଴×ଶଷ଴ + ଶ଼଺ଽ.ଵଽଽ ଶ.ହ×ହଶ଴×ଶଷ଴ ]Sv Providing 4 legged 10 mm stirrups. Asv= 315 mm2 , 315 = 1.175, Sv = 267.95 mm. Take Sv as 250 mm. [ ఛೡ೐ିఛ೎ ఙೞೡ ]b×Sv , ଵ.ସଶି଴.ଶ଼ ଶଷ଴ ×400×Sv = 315 , Sv= 158.88 mm.
  49. 49. Intze tank Provide 150 mm spacing. Steel for sagging moment = ହ଻ଶଽ଺.଴ଵ×ଵ଴଴଴ ଶଷ଴×଴.ଽ×ହ଺଴ = 494.27 mm2 . Provide 5 bars of 12 mm diameter. Ast = 565.48 mm2 . Hoop stress: Tc = thrust exerted by the conical slab on the girder. Tcsin ߙ ×2ߨr = Ww+Ws+weight of cylindrical wall and upper dome. Tcsin ߙ ×2ߨr = 154588.24+249874.42+1079631.039 Tcsin ߙ ×2ߨr = 2875387.699. Tc = ଶ଼଻ହଷ଼଻.଺ଽଽ ଶగ×ଷ×ୱ୧୬ ସହ° = 215729.87 N. Horizontal component of Tc = 215729.87× cos 45°, H1 = 152544.055 N. Horizontal component due to dome = 11678.016× cos 43°36′, H2 = 84574.59, H1−H2 = Net,Net = 67969.46 N. Hoop stress = 67969.46×3 = 203908.38 N. Hoop compressive stress = ଶ଴ଷଽ଴଼.ଷ଼ ସ଴଴×଺଴଴ = 0.849 N/mm2 . Coloumns: Coloumns should be designed for direct loads coming upon them and for the bending moments caused by wind load. Vertical load on one column at top = ସହଽଵ଴ଶ଻.ଵଽ଻ ଼ = 573878.399 N. Let ߙ be the inclination of the column with the vertical. tan ߙ= ଵ ଵ଴ , ߙ = 5°42′ , sin ߙ = 0.0995, cos ߙ = ଵ଴ √ଵ଴ଵ = 0.995. Actual length of column = √10ଶ + 1ଶ = 10.05 m. Providing 300 mm × 300 mm column.
  50. 50. Intze tank Wt. Of column =10×0.3×0.3×25000 = 22500 N Total vertical load = 573878.399+22500 N = 596378.399 N Corresponding axil load = ହଽ଺ଷ଻଼.ଷଽଽ ଴.ଽଽହ = 59375.2754 N When tank is full = 599375.2754 N Wt. Of water in tank = ଶ଼ଽ଺ଵଵ଻.ଵଵଶ ଼ =3620124.639 N on each column Vertical load on each column when tank is empty = 596378.399−362014.239 = 237361.036 N Corresponding axial load= ଶଶସ଻ଽ଼.ଶଶଶ ଴.ଽଽହ = 238553.805 N Ignoring wind load effect if the steel requirement is Asc Then cAc + tAsc =599375.275 N 5×Ac + 190 ×Asc =599375.275 5[400× 400 − ‫ܣ‬sc] +190×Asc =599375.275 Asc =807.433 mm2 . Min. Requirement of steel = 0.8% = ଴.଼ ଵ଴଴ [300×300] =720 mm2 Provide 6 bars of 20mm dia. =1884 mm2 More steel has been subjected since column is subjected to B.M caused by wind load.
  51. 51. Intze tank Analysis due to wind pressure: Wind pr. =1200 mm2 . Wind force on the top dome & cylindrical walls =(4+ ଵ.଼ ଶ )×9.4×1200 @Ht=13.95 =55272 N Wind force on the circular wall = ଽ.ସା଺.ସ ଶ ×1.5×01200 =14220 N Wind force on circular girder =0.6×6.4×1200 =4608 N Wind force on column & braces =5×0.3×10×1200+3× ଺ା଼ ଶ ×0.3×1200 =25560 N Total moment of wind pr. About the base =55272×13.95+14220×0.8+4608×10+25560×5 =10982500Nm. Vetrical load on any column due to wind load = ெ௫ ∑ ௫^ଶ ∑ ‫ݔ‬2 =2×42 +4( ସ √ଶ )2 =64m2 Max. Wind load force in the most leeward side &the most windward side. = ଵ଴ଽ଼ହ଴଴.ସ×ସ ଺ସ =68656.275 N Max. Wind force in columns marked 5 = ଵ଴ଽ଼ହ଴଴.ସ ଺ସ × ସ √ଶ =48547.317 N Consider the windword column 1 Vertical load due to dead +wind load =596378.399 +68656.275 N
  52. 52. Intze tank =665034.674 N. Corresponding axial load = ଺଺ହ଴ଷସ.଺଻ସ ଴.ଽଽହ଴ =668376.556 N Horizontal comoponent of the axial forces caused by wind action =2×68456.275×0.0995+4×48547.317× 0.0995 × ଵ √ଶ =27285.39 N. Aactualhorizontal force @ base = 55272+14220+4608+25560−27285.39 = 72374.61 Horizontal shear column = ଻ଶଷ଻ସ.଺ଵ ଼ = 9046.826 N. Maximum bending moment for the column = 9046.826× ଶ.ହ ଶ = 11308.532 N. Analysis of column section: Direct load = 668376.556 N. Bending moment = 11308.532 Nm. Provide 300×300 column. Provide 6 bars of 20 mm diameter at effective cover of 50 mm. Ast = 1884 mm2 , Equivalent concrete area = Ac+(m-1)Ast = (300×300)+(12.33×1884) = 113229.72 N Polar moment of inertia of the equivalent concrete section, = ௔ర ଵଶ +(mAst×effective depth fromcentre), = ଷ଴଴ర ଺ +1884×12.33[150-50]2 = 1.582×109 mm4 . Equivalent moment of inertia about full section = ଵ.ହ଼ଶ×ଵ଴వ ଶ = 791.14×106 mm4.
  53. 53. Intze tank Direct stress in concrete = ௗ௜௥௘௖௧ ௟௢௔ௗ ௘௤௨௜௩௔௟௘௡௧ ௖௢௡௖௥௘௧௘ ௔௥௘௔ = 5.9 mm2 . Bending stress in concrete = ଵହ଴×ଵଵଷ଴଼.ହଷଶ×ଵ଴଴଴ ଻ଽଵ.ଵସ×ଵ଴ల = 2.14 N/mm2 . Maximum stress = 5.9+2.14 = 8.04 N/mm2 . Design of braces: Moment in brace BC = 2×moment for the column× sec 45°, = 2×11308.532× √2 = 31985.358 Nm. Provide 300×300 mm bar section and a doubly reinforced beam with equal steel at top and bottom. Ast = Asc = ଷଵଽ଼ହ.ଷହ଼×ଵ଴଴଴ ଶଷ଴×ଶଶ଴×଴.ଽ = 702.357 mm2 . Provide 4 bars of 18 mmdiameter at top and equal amount at bottom. Shear force for brace = ௕௘௡ௗ௜௡௚ ௠௢௠௘௡௧ ௙௢௥ ௕௥௔௖௘ భ మ ௦௣௔௡ ௢௙ ௕௥௔௖௘ , Span of brace =2 × ଻ ଶ × sin 22°30′ = 2.678 m. Shesr force for brace = ଷଵଽ଼ହ.ଷହ଼ భ మ ×ଶ.଺଻଼ = 23887.49 N. Nominal shear stress ߬v = ௏ ௕ௗ = ଶଷ଼଼଻.ସଽ ଷ଴଴×ଶ଺଴ = 0.30 N/mm2 . Provide nomonal stirrups say 2 legged 10 mmdiameter stirrups at 200 mm clear cover.
  54. 54. Intze tank Design of foundation: Total load on the column = 599375.2754×8 = 4795002.203 N. Approximate weight of foundation is 10% of column loads. = 479500.22 N. Then total load is equal to 5274502.22 N. Safe bearing capacity of 200 KN/m2 , Area = ௟௢௔ௗ ௌ஻஼ = ହଶ଻ସହ଴ଶ.ସଶଷ ଶ଴଴×ଵ଴య = 26.37 m2 . Let us provide outer dia of 9.5 m and inner dia of 6.5 m. = గ ସ [9.52 −6.52 ] = 37.69 m2 . Net intensity = ହଶ଻ସହ଴ଶ.ସଶଷ ଷ଻.଺ଽ
  55. 55. Intze tank = 139.9 KN/m2 . 139.9 KN/m2 < 200 KN/m2 . Design of circular girder: Maximum bending moment occurs at support = 0.00416×W×r = 11508.005 Nm. Maximum bending moment occurs at support = 0.0083× 4795002.203×4 = 159194.073 Nm. Maximum torsion = 0.0006×W×r = 11508.005 Nm. Maximum shear force at support = ସ଻ଽହ଴଴ଶ.ଶ଴ଷ ଶ×଼ (from table 2) = 299687.63 N. Design at support section; Moment of resistence = maximum bending moment at support. 0.913bd2 = 159194.073×1000 , bd2 = 174363716.30 , d = 590 mm ,clear cover = 60 mm , D = 650 mm. Equivalent shear stress Vv = V+1.6 ் ௕ = 299687.63+1.6 ଵଵହ଴଼.଴଴ହ×ଵ଴଴଴ ହ଴଴ , = 336550.0176 N. Equivalent nominal shear ߬v = ௏ೡ ௕ௗ = 1.14 N/mm2 , but ߬c = 1.8N/mm2 , Hence ߬c< ߬v . Longitudinal reinforcement: Mel = M+Mt , Mt = ்(ଵା ವ ್ ) ଵ.଻ = ଵଵହ଴଼.଴଴ହ[ଵା లఱబ ఱబబ ]×ଵ଴଴଴ ଵ.଻ = 15569653.82 N , Mel = 1000[159194.073+15569.653] = 174763.726×1000 N. Ast = ଵ଻ସ଻଺ଷ.଻ଶ଺×ଵ଴଴଴ ଶଷ଴×଴.ଽ×ହଽ଴ = 1430.964 mm2 ,
  56. 56. Intze tank Provide 9 bars of 16 mm diameter bars. Hence area os steel required is Ast = 1809.55 mm2 . Transverse reinforcement: Asv= ்∙௦ೡ ௕భௗభఙೞೡ + ௏∙௦ೡ ଶ.ହௗభఙೞೡ , providing 4 legged 10 mm diamater of stirrups. Asv = 4ߨ ×52 = 314 mm2 , b1 = 500-80 = 420 mm , d1 = 650-120 = 530 mm, 314 = ଵଵହ଴଼.଴଴ହ×ଵ଴଴଴ ସଶ଴×ହଷ଴×ଶଷ଴ + ଶଽଽ଺଼଻.଼ଷ ଶ.ହ×ହଷ଴×ଶଷ଴ , 314 = Sr[0.224+0.983] , Sv = 260 mm. Let us provide 200 mm clear cover spacing. Steel for hogging mommentAst = ଻ଽ଻଼଼.଼ଷ଺×ଵ଴଴଴ ଶଷ଴×଴.ଽ×ହଽ଴ = 653.31 mm2 , Provide 4 bars of 16 mm diameter. Design of bottom slab: Provide a cantilever projection beyond the face of the beam = 0.6 m. Maximum bending moment for 1 m wide stirup = 139944.346× ଴.଺మ ଶ Nm , = 2518.98 Nm. Equating moment of resistence to bending moment , 0.913×bd2 = 25189.98×1000 , b = 1000 mm. Then d2 = 27590.339 , d = 166.1 mm. Let us provide 170 mm effective depth and 40 mm clear cover. D = 210 mm. Ast = ଶହଵ଼ଽ.ଽ଼×ଵ଴଴଴ ଶଷ଴×଴.ଽ×ଵ଻଴ = 715.82 mm2 . Provide 4 bars of 18 mm diameter. Ast = 1017.87 mm2 ,and spacing of the bars is 250 mm clear cover. Distribution steel:
  57. 57. Intze tank Provide 0.12 % steel and the steel required is = ଴.ଵଶ×ଶଵ଴×ଵ଴଴଴ ଵ଴଴ = 252 mm2 . Provide 6 bars of 8 mm diameter bars and spacing = ଵ଴଴଴ ଺ = 160 mm clear cover. Check for sliding: Total load on the foundation when tank is empty = 5274502.423-2896117.112 = 2378385.311 N Horizantal force on the base = 72374.61 N. Let coefficient of friction = 0.5 Fs= ଴.ହ×ଶଷ଻଼ଷ଼ହ.ଷଵଵ ଻ଶଷ଻ସ.଺ଵ = 16.43.
  58. 58. Intze tank 14. ESTIMATION 14.1 Detailed estimation: Detailed estimate is an accurate estimate and consists of working out the quantities of each item of works, and working the cost. The dimensions, length, breadth and height of each item are taken out correctly from drawing and quantities of each item are calculated, and abstracting and billing are done. The detailed estimate is prepared in two stages: Details of measurement and calculation of quantities. The details of measurements of each item of work are taken out correctly from plan and drawing and quantities under each item are calculated in a tabular form named as details of measurement form. Abstract of estimated cost: The cost of each item of work is calculated in a tabular form the quantities already computed and total cost is worked out in abstract estimate form. The rates of different items of work are taken as per schedule of rates or current workable rates for finished item of work. Detailed estimation: S. N o DECRIPTION OF WORK N OS L m B m A m2 Hor D (m) QTY m3 REMARKS 1 Earthwork in excavation 1 73.89 1 73.89 ‫ܣ‬ = ߨ݀ଶ /4 =ߨ × 9.72 /4 =73.89 Earthwork in filling 1 64.316 2 a)R.C.C work in foundation b)steel in foundation i )Longitudanal ii)Transverse 1 9 4 7.068 ߨ ×0.0082 0.0082 × ߨ 0.2 1.4136 0.045 0.02 = ߨ ×8
  59. 59. Intze tank 3 R.C.C in columns Steel in columns 8 8× 6 0.3 0.3 0.09 ߨ ×0.012 10.0 49 10.0 49 7.235 0.151 4 RCC in Bracings@2.5 Steel in Bracings @2.5m from G.L 8 8× 8 0.637 5 0.637 5 0.3 0.09 ߨ ×0.0092 0.3 =0.459 =0.01 A=ߨ ×0.0092 =0.000254 5 a) R.C.C in bracings @5m from G.L. b) Steel 8 8× 8 0.575 0.575 0.3 0.09 ߨ ×0.0092 0.3 0.414 0.0093 6 A=ߨ ×0.0092 =0.000254 6 a)RCC in bracings 7.5m b)steel 8 8× 8 0.45 0.45 ߨ ×0.0092 0.324 0.0072 A=ߨ ×0.0092 =0.000254 7 Top ring girder a)R.C.C b)steel longitudinal transeverse 1 5 125 ߨD 6ߨ 6ߨ 0.4 0.24 ߨ ×0.0092 0.6 4.52 0.02 0.066 8 Bottom dome a)RCC in dome b) steel 1 L=6.6 2 =22.619 0.067 0.2 4.523 0.443 A=2ߨrh =2ߨ ×3×1.2 =22.619 ‫ܮ‬ = ௫ ଷ଺଴° × 2ߨr =6.62 9 a)RCC conical slab b)steel steel for B.M. 1 14 3 23.56 23.56 23.56 0.2 ߨ ×0.0082 2.12 1 9.994 0.066 0.014 ߨ[‫1ܦ‬ + ‫]2ܦ‬ 2 = ߨ(9+6)/2 10 a)RCC ring beam @ B b)steel 1 6 28.27 0.25 ߨ ×0.0082 0.5 3.534 0.034 11 Cylindrical wall a)Main steel b)Distribution steel 1 20 4 4.32 28.27 0.2 ߨ ×0.0062 ߨ ×0.0042 4 22.619 0.098 0.0056 L=4+16d =4+16×0.012 =4.32 12 Ring beam @ A a)concrete 1 9 ߨ 0.2 0.04 0.2 1.13 L=ߨD =9 ߨ
  60. 60. Intze tank b)Steel 4 9 ߨ ߨ ×0.0062 0.012 13 Top dome R.C.C a) concrete b) Steel 1 100 9.93 2ߨrh =50.89 ߨ ×0.0042 0.15 0 7.63 0.05 A=2ߨrh = ‫ܮ‬ = ௫ ଷ଺଴° × 2ߨr 12 Total RCC work 63.795 6 13 Total steel 1.017 14 Plastering in CM (1:2) for Inner surface Of conical dome (12mm) 1 50.89 9.15 A=2ߨrh =50.89 15 Plastering in CM (1:6) for outer surface Of conical dome (12mm) 1 55.135 9.92 A=2ߨrh =55.135 16 Plastering in CM (1:2) for Inner surface Of cylindrical wall (12mm) 1 ߨ ×D 28.2 112.8 4 20.354 17 Plastering in CM (1:6) for outer surface Of cylindrical wall (12mm) 1 ߨ ×D 29.5 118.82 4 28.349 18 Plastering in CM (1:2) for Inner surface Of domed roof (12mm) 1 22.619 4.07 A=2ߨrh =2ߨ ×3×1.2 =22.619 19 Plastering in CM (1:6) for outer surface Of domed roof (12mm) 1 26.38 4.74 A=2ߨrh =2ߨ ×3×1.4 20 Plastering in CM (1:6) for columns (12mm) 8 0.3 0.3 0.09 17.28 21 Plastering in CM (1:2) for ring beam at top (12mm) 1 9 ߨ 0.2 5.65 1.01 22 Plastering in CM (1:2) for ring beam at bottom (12mm) 1 1.27
  61. 61. Intze tank 23 Plastering in CM (1:6) for bracings at 2.5m ht.(12mm) 1 0.27 24 Plastering in CM (1:6) for bracings at 5m ht.(12mm) 1 0.24 25 Plastering in CM (1:6) for bracings at 7.5m ht.(12mm) 1 0.19 26 Plastering in CM(1:2) for inner surface of conical slab(12mm) 1 4.239 ߨ[‫1ܦ‬ + ‫]2ܦ‬ 2 = ߨ(9+6)/2 27 Plastering in CM(1:6) for outer surface of conical slab(12mm) 1 4.46 28 Total plastering 105.53 3 29 Thick water proof cement painting for tank portion 85.278 30 white washing for columns 8 0.312 0.31 2 10.04 7.826 31 Total white washing 93.104
  62. 62. Intze tank ABSTRACT S.NO DESCRIPTION OF WORK QTY OR NOS RATE RS PS COST RS PS 1 Earth work in excavation 73.89 2 Beldars 13 250 3250 3 Mazdoors 11 250 2750 4 Total 6000 5 Earth work in Filling In foundation 64.316 6 Beldar 7 250 1750 7 Bhisthi 2 285 570 8 Mazdoors 5 250 1250 9 Total 3570 10 Total earth work in Filling 11 Disposal of surplus earth in a lead 30m 9.574 12 Mazdoor 4 250 1000 13 Total 1000 Total cost of earth work 10,570
  63. 63. Intze tank 14.2 DATA SHEET: RCC M- 20 Nominal mix (Cement:fine aggregate: coarse aggregate) corresponding to Table 9 of IS 456 using 20mm size graded machine crushed hard granite metal (coarse aggregate) from approved quarry including cost and conveyance of all materials likecement FOUNDATION A. MATERIALS UNIT QTY RATERS AMOUNT RS 20mm HBG graded metal Cum Cum 0.601 1076 646.676 Sand Cum 1.2 375 450 Cement Cum 0.4 1620 648 1st Class Mason Day 0.38 285 108.3 2nd Class Mason Day 1.03 285 293.55 Mazdoor (Both Men and Women) Day 2.33 250 582.5 Concrete Mixer 10/7 cf (0.2/0.8cum)capacity Hour 1 250 250 Cost of Diesel for Miller Liter 0.52 45 23.4 Cost of Petrol for Vibrator Liter 0.75 68 51 Water (including for curing) Ki 1.2 77.0 92.4 Add 20% in Labour (1st Floor) 629.16 Add MA 20% 629.16 Add TOT 4% 176.166 BASIC COST per 1 cum 4580.31
  64. 64. Intze tank COLUMNS A. MATERIALS UNIT QTY RATERS AMOUNT RS 20mm HBG graded metal Cum Cum 6.156 1076 6623.85 Sand Cum 3.078 375 1154.25 Cement Cum 2.052 1620 3324.24 1st Class Mason Day 1.99 285 567.15 2nd Class Mason Day 5.26 285 1499.1 Mazdoor (Both Men and Women) Day 11.96 250 2990 Concrete Mixer 10/7 cf (0.2/0.8cum) capacity Hour 1 250 250 Labour centering Cum 1 971 971 Material hire charges for centering Cum 1 89 89 Water (including for curing) Ki 1.2 77.0 92.4 Add 20% in Labour (1st Floor) 2912.198 Add MA 20% 2912.198 Add TOT 4% 582.43 BASIC COST per 1 cum 20967.816 RCC RING BEAM AT TOP A. MATERIALS UNIT QTY RATERS AMOUNT RS 20mm HBG graded metal Cum Cum 0.96 1076 1032.96 Sand Cum 0.48 375 180 Cement Cum 0.32 1620 518.4 1st Class Mason Day 0.31 285 88.35 2nd Class Mason Day 0.83 285 236.55 Mazdoor (Both Men and Women) Day 1.86 250 465 Concrete Mixer 10/7 cf (0.2/0.8cum) capacity Hour 0.26 250 65 Labour centering Cum 1 971 971 Material hire charges for centering Cum 1 89 89 Water (including for curing) Ki 1.2 77.0 92.4 Add 20% in Labour (1st Floor) 747.73 Add MA 20% 747.73 Add TOT 4% 149.54 BASIC COST per 1 cum 5383.66
  65. 65. Intze tank RCC DOMED ROOF 150mm THICK A. MATERIALS UNIT QTY RATERS AMOUNT RS 20mm HBG graded metal Cum Cum 6.48 1076 6972.48 Sand Cum 3.24 375 1215 Cement Cum 2.16 1620 3499.2 1st Class Mason Day 2.1 285 598.5 2nd Class Mason Day 5.6 285 1596 Mazdoor (Both Men and Women) Day 12.6 250 3150 Concrete Mixer 10/7 cf (0.2/0.8cum) capacity Hour 0.267 250 66.75 Labour centering Cum 10 971 9710 Material hire charges for centering Cum 10 89 890 Water (including for curing) Ki 1.2 77.0 92.4 Add 20% in Labour (1st Floor) 5558.33 Add MA 20% 5558.33 Add TOT 4% 1111.61 BASIC COST per 1 cum 40018.6 CONICAL SLAB 200mm THICK A. MATERIALS UNIT QTY RATERS AMOUNT RS 20mm HBG graded metal Cum Cum 8.49 1076 9135.24 Sand Cum 4.25 375 1593.75 Cement Cum 2.83 1620 4584.6 1st Class Mason Day 2.75 285 783.75 2nd Class Mason Day 7.34 285 2091.9 Mazdoor (Both Men and Women) Day 16.52 250 4130 Concrete Mixer 10/7 cf (0.2/0.8cum) capacity Hour 0.26 250 65 Labour centering Cum 5 971 4855 Material hire charges for centering Cum 5 89 445 Water (including for curing) Ki 1.2 77.0 92.4 Add 20% in Labour (1st Floor) 5555.328 Add MA 20% 5555.328 Add TOT 4% 1111.06 BASIC COST per 1 cum 39998.35
  66. 66. Intze tank RCC CYLINDRICAL WALL A. MATERIALS UNIT QTY RATERS AMOUNT RS 20mm HBG graded metal Cum Cum 19.23 1076 20691.48 Sand Cum 9.62 375 3607.5 Cement Cum 6.41 1620 10384.2 1st Class Mason Day 6.23 285 1775.55 2nd Class Mason Day 16.62 285 4736.7 Mazdoor (Both Men and Women) Day 37.39 250 9347.5 Concrete Mixer 10/7 cf (0.2/0.8cum) capacity Hour 0.26 250 65 Labour centering Cum 1 971 971 Material hire charges for centering Cum 1 89 89 Water (including for curing) Ki 1.2 77.0 92.4 Add 20% in Labour (1st Floor) 10352.066 Add MA 20% 10352.066 Add TOT 4% 2070.432 BASIC COST per 1 cum 74534.89 RCC RING BEAM AT BOTTOM OF CYLINDRICAL WALL A. MATERIALS UNIT QTY RATERS AMOUNT RS 20mm HBG graded metal Cum Cum 3 1076 3228 Sand Cum 1.5 375 562.5 Cement Cum 1 1620 1620 1st Class Mason Day 0.97 285 267.45 2nd Class Mason Day 2.59 285 738.15 Mazdoor (Both Men and Women) Day 5.84 250 1460 Concrete Mixer 10/7 cf (0.2/0.8cum) capacity Hour 0.26 250 65 Labour centering Cum 1 971 971 Material hire charges for centering Cum 1 89 89 Water (including for curing) Ki 1.2 77.0 92.4 Add 20% in Labour (1st Floor) 1818.7 Add MA 20% 1818.7 Add TOT 4% 363.74 BASIC COST per 1 cum 13094.64
  67. 67. Intze tank RCC CIRCULAR GIRDER A. MATERIALS UNIT QTY RATERS AMOUNT RS 20mm HBG graded metal Cum Cum 3.84 1076 4131.84 Sand Cum 1.92 375 720 Cement Cum 1.28 1620 2073.6 1st Class Mason Day 1.24 285 353.4 2nd Class Mason Day 3.32 285 946.2 Mazdoor (Both Men and Women) Day 7.47 250 1867.5 Concrete Mixer 10/7 cf (0.2/0.8cum) capacity Hour 0.26 250 65 Labour centering Cum 1 971 971 Material hire charges for centering Cum 1 89 89 Water (including for curing) Ki 1.2 77.0 92.4 Add 20% in Labour (1st Floor) 2262.588 Add MA 20% 2262.58 Add TOT 4% 452.517 BASIC COST per 1 cum 16290.61 RCC BRACING AT 2.5m HT. A. MATERIALS UNIT QTY RATERS AMOUNT RS 20mm HBG graded metal Cum Cum 0.39 1076 419.64 Sand Cum 0.19 375 71.25 Cement Cum 0.13 1620 210.6 1st Class Mason Day 0.125 285 35.625 2nd Class Mason Day 0.33 285 94.05 Mazdoor (Both Men and Women) Day 0.75 250 187.5 Concrete Mixer 10/7 cf (0.2/0.8cum) capacity Hour 0.26 250 65 Labour centering Cum 1 971 971 Material hire charges for centering Cum 1 89 89 Water (including for curing) Ki 1.2 77.0 92.4 Add 20% in Labour (1st Floor) 447.213 Add MA 20% 447.213 Add TOT 4% 89.44 BASIC COST per 1 cum 3219.93
  68. 68. Intze tank RCC BRACING AT 5m HT. A. MATERIALS UNIT QTY RATERS AMOUNT RS 20mm HBG graded metal Cum Cum 0.33 1076 355.08 Sand Cum 0.17 375 63.75 Cement Cum 0.11 1620 178.2 1st Class Mason Day 0.125 285 35.625 2nd Class Mason Day 0.33 285 94.05 Mazdoor (Both Men and Women) Day 0.75 250 187.5 Concrete Mixer 10/7 cf (0.2/0.8cum) capacity Hour 0.26 250 65 Labour centering Cum 1 971 971 Material hire charges for centering Cum 1 89 89 Water (including for curing) Ki 1.2 77.0 92.4 Add 20% in Labour (1st Floor) 426.32 Add MA 20% 426.32 Add TOT 4% 85.264 BASIC COST per 1 cum 3069.50 RCC BRACING 7.5m HT. A. MATERIALS UNIT QTY RATE RS AMOUNT RS 20mm HBG graded metal Cum Cum 0.27 1076 290.52 Sand Cum 0.13 375 48.75 Cement Cum 0.09 1620 145.8 1st Class Mason Day 0.08 285 22.8 2nd Class Mason Day 0.23 285 65.55 Mazdoor (Both Men and Women) Day 0.535 250 133.75 Concrete Mixer 10/7 cf (0.2/0.8cum) capacity Hour 0.26 250 65 Labour centering Cum 1 971 971 Material hire charges for centering Cum 1 89 89 Water (including for curing) Ki 1.2 77.0 92.4 Add 20% in Labour (1st Floor) 384.91 Add MA 20% 384.91 Add TOT 4% 76.98 BASIC COST per 1 cum 2771.37
  69. 69. Intze tank Plastering with CM(1:6)&(1:2),12 mm thick Cement Mortor 1:6 1:2 cum cum 105.533 65.44 40.09 552 780 36165 31673 Mason 1st class day 39 285 11115 Bhisthi day 14 285 3990 Mazdoor (unskilled) day 39 250 9750 Add MA 20% 18539 Add TOT 4% 3719 Grand Total 114951 Painting to new walls of tank portion with 2 coats of water proof cement paint of approved brand and shade over a base coat of approved cement primer grade I making making 3 coats in all to give an even shade after thourughly brushing the surface to remove all dirt and remains of loose powdered materials, including cost and conveyance of all materials to work site and all operational, incidental, labour charges etc. complete for finished item of work as per SS 912 for walls Epoxy primer for Hibond floor & protective coatings : Procoat SNP2 or Zoriprime EFC 2 Pack 26 548 14250 1st class painter Day 4 355 1420 Mazdoor Day 4 250 1000 cost of water proof cement paint Cum 50 35 1750 1st class painter Day 2 355 710 Mazdoor (unskilled) Day 2 250 500 Add MA 20% 3926 Add TOT 4% 786 Total cost 24342
  70. 70. Intze tank Painting to new columns of tank portion with 2 coats of water proof cement paint of approved brand and shade over a base coat of approved cement primer grade I making making 3 coats in all to give an even shade after thourughly brushing the surface to remove all dirt and remains of loose powdered materials, including cost and conveyance of all materials to work site and all operational, incidental, labour charges etc. complete for finished item of work as per SS 912 for walls Cost of cement primer Pack 18 100 1800 1st class painter Day 1 355 355 2nd class painter Day 1 250 250 cost of water proof cement paint Cum 6 35 210 1st class painter Day 1 355 355 Mazdoor (unskilled) Day 1 250 250 Add MA 20% 644 Add TOT 4% 129 Total cost 3993 Total cost of project: Total cost of R.C.C = 2,23,930 Total cost of steel = 5,18,924 Total cost of plastering = 1,14,951 Total cost of painting = 28,335 Total cost of earthwork = 10,570 8,96,710
  71. 71. Intze tank 15. CONCLUSION Storage of water in the form of tanks for drinking and washing purposes, swimming pools for exercise and enjoyment, and sewage sedimentation tanks are gaining increasing importance in the present day life. For small capacities we go for rectangular water tanks while for bigger capacities we provide circular water tanks. Design of water tank is a very tedious method. With out power also we can consume water by gravitational force. Intze tank is constructed to minimize the project cost why because lower dome in this construction resists the horizontal thrust.
  72. 72. Intze tank 16. REFERENCES Table 16.2. Coefficients for moment in cylindrical wall fixed at base (As Per IS3370) Moment = Coefficient (wH3 ) Nm/m H2 Co efficient at points DT 0.1 H0.2 H 0.3 H0.4 H 0.5 H 0.6 H 0.7 H 0.8H 0.4 + 0.0005 + 0.0014 + 0.0021 + 0.0007 - 0.0042 -0.0150 -0.0302-0.0529 0.8 + 0.0011 + 0.0037 + 0.0063 + 0.0080 + 0.0070 + 0.0023 + 0.0068 -0.0024 1.2 + 0.0012 + 0.0042 + 0.0077 + 0.0103 + 00112 + 0.0090 + 0.0022 -0.0108 1.6+ 0.0011 + 0.0041 + 0.0075 + 0.0107 + 0.0121 + 0.0111 + 0.0058 -0.0051 2.0+ 0.0010 + 0.0035 + 0.0068 + 0.0099 + 0.0120 + 0.0115 + 0.0075 -0.0021 3.0 + 0.0006 + 0.0024 + 0.0047 + 0.0071 + 0.0090 + 0.0097 + 0.0077 +0.0012 4.0 + 0.0003 + 0.0015 + 0.0028 + 0.0047 + 0.0066 + 0.0077 + 0.0069 +0.0023 5.0 + 0.0002 + 0.0008 + 0.0016 + 0.0029 + 0.0046 + 0.0059 + 0.0059 +0.0028 6.0 + 0.0001 + 0.0003 + 0.0008 + 0.0019 + 0.0032 + 0.0046 + 0.0051 +0.0029 8.0 0.0000 + 0.0001 + 0.0002 + 0.0008 + 0.0016 + 0.0028 + 0.0038 +0.0029 10.0 0.0000 + 0.0000 + 0 0001 + 0.0004 + 0.0007 + 0.0019 + 0.0029 +0.0028 12.0 0.0000 + 0.0000 + 0.0001 + 0.0002 + 0.0003 + 0.0013 + 0.0023 +0.0026 14.0 0.0000 0.0000 0.0000 0.0000 + 0.0001 + 0.0008 + 0.0019 +0.0023 16.0 0.0000 0.0000 -0.0001 - 0.0002 -0.0001 + 0.0004 + 0.0013 +0.0019 Table 1:
  73. 73. Intze tank Permissible stresses in concrete All values in N/mm2 Grade permissible stresses in compression permissible stress in bond Of concrete for plain bars in tension Bending Direct (average) ࣌cbc ࣌cc ࣎bd M 10 3.0 2.5 _ M 15 5.0 4.0 0.6 M 20 7.0 5.0 0.8 M 25 8.5 6.0 0.9 M 30 10.0 8.0 1.0 M 35 11.5 9.0 1.1 M 40 13.0 10.0 1.2 M 45 14.5 11.0 1.3 M 50 16.0 12.0 1.4 Table 1.1: Grade of M10 M15 M20 M25 M30 M35 M40 M45 M50 Concrete Tensile 1.2 2.0 2.8 3.2 3.6 4.0 4.4 4.8 5.2 Stress(N/mm2 ) Table 2: Moments for circular girders For 8 columns B.M@ B.M@ Torsion Support centre 0.0083Wr 0.00416Wr 0.0006Wr
  74. 74. Intze tank 17. REFERENCE BOOKS • I.S 456:2000 for RCC. • I.S 800:1984 for STEEL. • I.S 872 Part I and Part II. • I.S 3373 (Part IV-1967). • Reinforced concrete structures (M.Ramamrutham). • Element of environmental engineering (BIRIDI). . Estimating, costing and evaluation (B.N.Datta). . Standard schedule of rates (SSR)
  75. 75. Intze tank
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