1 1. SYMBOLS A = Total area of section A b = Equivalent area of helical reinforcement. A c = Equivalent area of section A h = Area of concrete core. A m = Area of steel or iron core. A sc = Area of longitudinal reinforcement (comp.) A st = Area of steel (tensile.) A l = Area of longitudinal torsional reinforcement. A sv = Total cross-sectional are of stirrup legs or bent up bars within distance Sv A w = Area of web reinforcement. A Ф = Area of cross section of one bars. a = lever arm. a c = Area of concrete. B = flange width
Transcript
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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
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dc = cover to compressive steel
ds = depth of slab
dt = cover to tensile steel
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 inertia.
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 = development length.
l = effective length of column or length or bond length.
M = bending moment or moment.
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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.
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 intensity.
Pe = net upward soil pressure.
Q = shear resistance.
τ = 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.
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Tu = ultimate torsional moment.
σst 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.
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 stress in shear reinforcement.
σsy = yield point compressive stress in steel.
µ = co efficient of friction.
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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 on free expansion or contraction of the structure.
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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 underground 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.
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3. WATER DEMAND
3.1 Water Quantity Estimation: The quantity of water required for municipal uses for which the water supply scheme has to 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, since there are many variable factors affecting water consumption. The various types of water demands, 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 water is required is very large. The rate of fire demand is sometimes treated as a function of population and is worked out from following empirical formulae:
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Authority Formula (P in thousand)
Q (for 1 lakh Population)
1
American Insurance Association
Q(L/min)=4637P(1-0.01 ÖP)
41760
2 Kuchling's Formula Q(L/min)=3182 P 31800 3 Freeman's Formula Q(L/min)=1136.5(P/5+10) 35050 4 Ministry
Of Urban Development Manual Formula
Q(kilo liters/d)=100P for P>50000
31623
3.4 Factors affecting per capita demand:
• Size of the city: Per capita demand for big cities is generally large as compared to that 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 private wells, etc.
• Pressure in the distribution system.
• Efficiency of water works administration: Leaks in water mains and services; and unauthorized 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.
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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 the fluctuations.
• 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 Sundays and Festival days, thus increasing demand on these days.
• Hourly variations are very important as they have a wide range. During active household working hours i.e. from six to ten in the morning and four to eight in the evening, the bulk of the daily requirement is taken. During other hours the requirement is negligible. Moreover, if a fire breaks out, a huge quantity of water is required to be supplied during short duration, necessitating the need for a maximum rate of hourly supply. So, an adequate quantity of water must be available to meet the peak demand. To meet all the fluctuations, the supply pipes, service reservoirs and distribution pipes must be properly proportioned. The water is supplied by pumping directly and the pumps and distribution system must be designed to meet the peak demand. The effect of monthly variation influences the design of storage reservoirs and the hourly variations influences the design of pumps and service reservoirs. As the population decreases, the fluctuation rate 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
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4. POPULATION FORECAST
4.1 Design Periods & Population Forecast:
This quantity should be worked out with due provision for the estimated requirements of the future. The future period for which a provision is made in the water supply scheme is known 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. The particular method to be adopted for a particular case or for a particular city depends largely on the factors discussed in the methods, and the selection is left to the discretion and intelligence of the designer.
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. Concrete mix lesser than M20 is not used. The minimum quantity of cement in the concrete mix shall be not less than 30 kN/m3.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 to compact. Other causes of leakage in concrete are defects such as segregation and honey combing. All joints should be made watertight 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 ace of the equivalent concrete section 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 of concrete with interposition of some material to break the bond and facilitate movement. Incase 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,
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stresses caused by difference in temperature between inside and 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 are interrupted and a partial contraction joint is one in which only the concrete is interrupted, the reinforcing steel running through as shown in Fig.(b)
Fig (b)
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(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 the adjoining parts of a structure which by closing or opening accommodates the expansion or contraction of the structure.
Fig (d)
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(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 structure which after a suitable interval and before the structure is put to use, is filled with mortar or concrete completely with suitable jointing materials. In the first case width of the gap should be sufficient to allow the sides to be prepared before filling. Figure (g)
Fig (g)
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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. 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.
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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.
e) 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 storage of water or aqueous liquids at or near ambient temperature and the circumstances are such that the concrete will never dry out.
(ii)Shrinkage stresses may however be required to be calculated in special cases, when a shrinkage co-efficient of 300× 10 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 in panels with sides not more than 4.5m.with contraction or expansion joints between. In such cases a screed or concrete layer less than 75mm thick shall first be placed on the ground and covered with a sliding layer of bitumen paper or other suitable material to destroy 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 aggressive
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water 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 be designed 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 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. 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 of the 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 resultant circumferential tension in addition to vertical loads.
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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 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. On liquid retaining faces, the tensile stresses due 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.
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(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 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 at the 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 free edge. The analysis of moment and forces may be made on the basis of any recognized method.
(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 of the 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 to provide 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 as the weight of roof slab, earth cover if any, live loads and mechanical equipment. They
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should also be designed for upward load if the liquid retaining structure is subjected to internal gas pressure. A superficial load sufficient to ensure safety with the unequal intensity of loading which occurs during the placing of the earth cover should be allowed for in designing roofs. The engineer should specify a loading under these temporary conditions which should not be exceeded. In designing the roof, allowance should be made for the temporary condition of some spans loaded and other spans unloaded, even though 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 at right 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 less than 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, thick sections. For sections of thickness greater than 450mm, minimum reinforcement in each of 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 face of 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 in any 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 inner faces or roof slab) the minimum cover to all reinforcement should be 25mm or the diameter of the main bar whichever is greater. In the presence of the sea water and soil sand water of corrosive characters the cover should be increased by 12mm but this additional cover shall not be taken into account for design calculations.
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(b) For faces away from liquid and for parts of the structure neither in contact with the liquid on any face, nor enclosing the space above the liquid, the cover shall be as for ordinary concrete member.
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 as symmetrical lantern opening through the top. The edge of the shell around its base is usually 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 and bunkers. 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 effect of bending moment, twisting moment and shear and assuming that the loads are carried wholly by axial stresses. This however applies at points of shell which are removed some distance 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 mentioned 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.
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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 denoted 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).
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Shell of Revolution
Two forces N ) have the resultant N )d as shown in Fig.(c) and the resultant acts normal to the surface pointed towards the inner side. Forces N (r1d ) again have horizontal resultant of magnitude N (r1 d ) d as shown in Fig (d). It has a component N (r1d )d sin 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(rd ) must be in 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 outer side of the shell. Value s and Nf and N will be positive when tensile and negative compressive.
The equation is valid not only for shells in the form of a surface of revolution, but may be applied to all shells, when the coordinate lines for = constant and = constant, are the lines of curvature of the surface.
Forces in shell Force Nf act tangentially to the surface all around the circumference. Considering the equilibrium 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 direction on the surface of the shell above the cut.
This gives,
Nf = -W/2Prsinf
Eq. is readily solved for Nf and N may then be determined by Eq. This theory is applicable to a shell of any material as only the conditions of equilibrium have been applied and no compatibility relationships in terms of deformation have been introduced. It is, therefore, immaterial whether Hooke's law is applicable or not.
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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.
Alternatively, 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. Similar 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
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Nf1cos a1 = Nf2cos a2.
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 minimum 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 up to 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.
27
Openings may be provided in the dome as required from other functional or architectural requirements. However, reinforcement has to be provided all around the opening as shown in Fig. (c). The meridian and hoop reinforcement reaching the opening should be well anchored to such reinforcement.
The allowable stresses 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, M25, M30, M35 and 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 the tensile stress in concrete for avoidance of cracks, the tensile stress in the steel will be limited by the requirements that the stress as mentioned above should not be exceeded. The tensile
28
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 increased by 33 %.
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 area of 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.
29
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
30
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 horizontal thrusts of the conical wall and the domed floor act in opposite direction. 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 the meridians and latitudes. The rise is usually l/5th of the span.
(2) Ring beam supporting the dome: The ring beam is necessary to resist the horizontal component of the thrust of the dome. The ring beam will be designed for the hoop tension induced.
(3) Cylindrical walls: This has to be designed for hoop tension caused due to horizontal 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 the induced hoop 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 supported on the ring girder.
(7) The ring girder: This will be designed to support the tank and its contents. The girder 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 be braced at intervals and have to be designed for wind pressure or seismic loads whichever govern.
31
(9) Foundations: A combined footing is usually provided for all supporting columns. When this is done it is usual to make the foundation consisting of a ring girder and a circular 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
V1 = pd24 × H =0.39 D3 . for H = D/2.
V2 = p . h12
× ( D2+d2+d )=¿ 0.102D3
V3 = p h1
6(3 r2+h1
2) = 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.
32
13. DESIGN
13. DETAILS OF DESIGN:
Design of tank:
Design of an Intze tank for a capacity of 900000 lts .
Assuming height of staging is 16 m.
Safe bearing capacity of soil 150 kN/m2 .
Assume Wind pressure as 1500 N/m2 .
Assuming M20 concrete
For which σcbc = 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.
33
Solution: 1. Dimensions of the tank :
Let the diameter of the cylindrical portion= D =14 m ; R=7 m.
Let the diameter of the ring beam B2 =D0 =10 m . Height h0 of conical dome = 2m.
Rise h1 = 1.8 m ; Rise h2 =1.6 m.
The radius R2 of the bottom dome is given by
1.6(2R2 - 1.6) = 52 or R2 =8.61 m.
sin φ2 =5
8.61 = 0.5807 ; φ2=35.50°
cos φ2 =0.8141 ; tan φ2 =0.7133 ; cot φ2 =1.4019.
Let ‘h’ be the height of cylindrical portion.
Capacity of tank is given by
V= π4 D2h + π
12h0(D2 + D02 + DD0) -
π3 h2
2 (3R2 - h2)
34
Required volume = 900,000 litres = 900 m3.
∴ 900 = π4 (14)2h + π × 2
12 (142 + 102 + 14×10) - π
3 (1.6)2 (3×8.61 – 1.6)
From which h = 4.78 m. Allowing for free board, keep h = 5 m.
For the top dome, the radius R1 is given by :
1.8(2R1 - 1.8) = 72 or R1 = 14.51 m.
sin φ1 =7
14.51 = 0.4824 ; φ1 =28.84° ; cos φ1 = 0.8760.
2. Design of top Dome :
Let thickness t1 = 100 m = 0.1 m.
Taking live load of 1500 N/m2, total p per sq. m. of dome
= 0.1 × 25000 + 1500 = 4000 N/m2.
Meridional thrust at edges =T1= p R1
1+cosφ 1
= 4000 ×14.511+0.8760 =30938 N/m.
∴ Meridional stress = 30938
100× 1000 =0.31 N/mm2 (safe)
Maximum hoop stress occurs at the centre and its magnitude is
However provide t = 300 mm at bottom and taper it to 200 mm at top.
Average t =300+2002 = 250 mm; % of distribution steel = 0.3[ 250−100
450−100 ]×0.1 =0.24
∴ Ash = 0.24 ×250 ×1000
100 = 650 mm2. Area of steel on each face = 325mm2
Spacing of 8 mm Ф bars = 1000× 50.3325 = 155 mm.
Hence provide 8 mm Ф bars @ 150 mm c/c on both faces.
Keep the clear cover of 25 mm. Extend the vertical bars of the outer face into the dome to take care of continuity effects.
To resist the hoop tension at 2 m below top, Ash = 25
× 2286 = 914.4 mm2
∴ Spacing of 12 mm Ф rings = 1000× 113914.4/2 = 247 mm. Hence provide the rings @ 240
mm c/c in the top 2m height.
At 3m below the top, Ash = 35
×2286 = 1372 mm2
37
Spacing of 12 mm Ф rings = 1000× 1131.372/2 = 164.7 mm.
Hence provide rings @ 160 mm c/c in the next 1m height. At 4m below the top,
Ash= 45
×2286 = 1829 mm2
Spacing of 12mm Ф rings = 1000× 1131829/2 = 123.6 mm
Hence provide rings @ 120 mm c/c for the next 1 m height. In the last 1m height (4m to 5m) provide rings 95mm c/c.
5. Design of ring beam B3 :
This ring beam connects the tank wall with the conical dome. The vertical load at the junction of the wall with conical dome is transferred to ring beam B3 by meridional thrust in the conical dome. The horizontal component of the thrust cause hoop tension at the junction. The ring beam is provided to take up this hoop tension. The load W transmitted through tank wall, at the top of conical dome consists of the following:
(i) Load of top dome = T1 sin φ1 =30938×0.4824 = 14924 N/m.(ii) Load due to the ring beam B1 =0.36×(0.4-0.2)×1×25000 =1800 N/m.
(iii) Load due to the tank wall = 5[0.2+0.32 ]×1×25000= 31250 N/m.
(iv) Self-load of the beam B3 (1m×0.6 m, say) = (1-0.3)×0.6×25000=10500 N/m.
∴ Total W =58474 N/m.Inclination of conical dome wall with vertical= φ0 = 45º
∴ sin φ0 = cos φ0 = 0.7071= 1
√ 2 ; tan φ0 =1 ;
∴Pw =W tan φ0 =58474×1 =58474 N/m
Pw = w h d3 =9800×5×0.6 =29400 N/m.
Hence hoop tension in the ring beam is given by
P3 = (PW + Pw) D2 = (58474+29400)14
2 =615118 N
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This to be resisted entirely by steel hoops, the area of which is
Ash = 615118/150 = 4100 mm2.
No. of 30 mm Ф bars = 4100/706.9 =5.8
Hence provide 6 rings of 30 mm Ф bars.
Actual Ash =4241.4
Stress in equivalent section = 615118
(1000× 600 )+12× 4241.4 = 0.95 N/mm2 < 1.2
Hence safe.
The 8 mm Ф distribution bars (vertical bars) provided in the wall @ 150 mm c/c should be taken round the above rings to act as stirrups.
6. Design of conical dome :
(a) Meridional thrust : The weight of water is given by
Area of minimum steel = 0.3- [ 250−100450−100 ]0.1 = 0.26 %
∴ As = 0.26 ×2500 = 650 mm2 in each direction.
Spacing of 10 mm Ф bars =100× 78.5650 =121 mm.
Hence provide 10 mm Ф bars @ 120 mm c/c in both the directions. Also, provide 16 mm Ф meridional bars @ 100 mm c/c near water face, for 1 m length, to take care of the continuity effect. The thickness of the dome may be increased from 250 mm to 280 mm gradually in 1 m length.
8. Design of bottom circular beam B2
Inward thrust from conical dome = T0 sinφ0 = 361437× 1√ 2 = 255613 N/m
Outward thrust from bottom dome = T2 cosφ0 = 290093×0.8141 = 236165 N/m
∴ Net inward thrust = 255613 – 236165 = 19448 N/m
Hoop compression in beam = 19448× 102 = 97240 N
Assuming the size of beam to be 600 mm×1200 mm
Hoop stress = 97240600× 1200 =0.135 N/mm2.This is extremely small.
Vertical load on beam, per metre run = T0 cosφ0 + T2 sinφ2
= 361437× 1√ 2 + 290093×0.5807 = 424070 N/m
Self weight = 0.6×1.20×1×25000=18000 N/m
∴ The load on beam = w= 424070 + 18000 = 442070 N/m.
Let us support the beam on 8 equally spaced columns at a mean diameter of 10 m.
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Mean radius of curved beam is R = 5m.
Table : Coefficient For B.M. And Twisting Moment In Circular Beams.
No. of Supports
2Ө C1 C2 C3 φ
4 90º 0.137 0.070 0.021 1910
4
5 72º 0.108 0.054 0.014 1510
4
6 60º 0.089 0.045 0.009 1230
4
7 5130
70.077 0.037 0.007 1030
4
8 45º 0.066 0.030 0.005 910
2
9 40º 0.060 0.027 0.004 810
2
10 36º 0.054 0.023 0.003 710
4
12 30º 0.045 0.017 0.002 610
4
For Table, 2Ө = 45º = π4 ; Ө = 22.5º =
π8 radians ;
C1 = 0.066 ; C2= 0.030 ; C3= 0.005 ; φm = 910
2
wR2(2 Ө) = 442070(5)2( π4 ) = 8680024 N-m
44
Maximum –ve B.M. at support = M0 = C1.wR2. 2Ө = 0.066×8680024 = 572882 N-m
Maximum +ve B.M. at support = Mc =C2.wR2. 2Ө =0.030×8680024 = 260401 N-m
Maximum torsional moment = Mmt = C3.wR2. 2Ө = 0.005×8680024 = 43400 N-m
For M-20 concrete (σcbc =7 N/mm2) and HYSD bars (σst =150 N/mm2) we have
Hence we have the following combination of B.M. and torsional moment:
(a) At the supports
M0 = 572882 N-m (hogging or negative) ;M0
t = zero.
(b) At mid-span :
Mc = 260401 N-m (sagging or positive) ;M0
t = zero.
(c) At the point of max. torsion ( φ = φm= 910
2 )
M φ = 1421 N-m (hogging or negative) ; Mm
t = 43400 N-m
Main and Longitudinal Reinforcement
(a)Section at point of maximum torsion
T = Mmaxt = 43400 N-m ; M φ = M =1421 ; Me1 = M+MT
Where MT = T[1+D /b1.7 ] = 43400[1+1.2/0.6
1.7 ] = 76588 N-m
∴ Me1 = 1421 + 76588 = 78009 N-m or,
46
Ast1 = M e 1
σst jd = 78009× 1000150× 0.874 ×1160 = 513 mm2
No. of 25 mm Ф bars = 513/491 = 1.05
Let us provide a minimum of 2 bars. Since Mt > M ,
Me2 =MT – M = 76588 – 1421 = 75167 N-m
Ast2 = 75167× 1000150× 0.874 ×1160 = 494.3 mm2
∴ No. of 25 mm Ф bars ≈ 1. However, provide a minimum of 2 bars. Thus, at the point of maximum torsion, provide 2-25 mm Ф bars each at the top and bottom.
(b)Section at max. hogging B.M. (support)
M0 = 572882 N-m = Mmax ; M0t = 0
Ast = 572882× 1000150× 0.874 ×1160 = 3767 mm2
∴ No. of 25 mm Ф bars = 3767/491 = 7.7 ≈ 8. Hence provide 6 Nos. of 25 mm Ф bars in one layer and 2 bars in the second layer. These will be provided at the top of the section, near supports.
(c)Section at max. Sagging B.M. (mid span)
Mc = 260401 N-m ; Mct = 0
∴ For positive B.M., steel will be to other face, where stress in steel (σst) can be taken as 190 N/mm2. The constants for M20 concrete having c =7 N/mm2 and m =13 will be
k = 0.324 ; j = 0.892 and R= 1.011 ; Ast = 260401 ×100190× 0.892× 1160 = 1325 mm2
47
No. of 25 mm Ф bars = 1325/490 = 2.7
At the supports, provide 6-25 mm Ф bars at the top layer and 2-25 mm Ф bars in the second layer. Continue these upto the section of maximum torsion (i.e. at φm = 9.5o =0.166 rad.) at a distance = Rφm = 5×0.166 = 0.83 m or equal to Ld = 52φ = 1300 mm from supports.
At this point, discontinue four bars while continue the remaining four bars. Similarly, provide 4 bars of 25 mm Ф at the bottom, throughout the length. These bars will take care of both the max. positive B.M. as well as maximum torsional moment.
Transverse Reinforcement
(a) At point of max. torsional moment :
At the point of max. torsion, V = 501512 N
Ve = V +1.6Tb where T = Mm
t = 43400 N-m ; b = 600 mm = 0.6 m
∴ Ve = 501512 + 1.6× 434000.6 = 617245 N
∴ τ ve=
V e
b d= 617245
600 × 1160 = 0.887 N/mm2
This is less than τ cmax = 1.8 N/mm2 for M20 concrete; Hence O.K.
100 A s
bd
= 100(4 ×491)600 ×1160 = 0.282
Hence from table ; τ c ≈ 0.23 N/mm2
Since τ cve> τc , shear reinforcement is necessary. The area of cross section Asv of the stirrups is given by
Asv = T . sv
b1 d1σsv + V . sv
2.5 d1σsv
48
Where b1 = 600 - (40×2) - 25 = 495 mm ;
d1 = 1200 - (40×2) - 25 = 1095 mm
∴ A sv
Sv = 43400 ×1000
495 ×1095 ×150 + 5015122.5× 1095× 150 = 1.755
Minimum transverse reinforcement is governed by A sv
Sv≥[ τve−τc
σ sv ]b∴
A sv
Sv = 0.887−0.23
150×600 = 2.628.
Hence depth A sv
Sv = 2.628
Using 12 mm Ф 4 lgd stirrups, Asv = 4×113 = 452 mm2
Or Sv =452/2.628 = 172 mm
However, the spacing should be not exceed the least of x1 , x1+x2
4 and 300 mm, where
x1 = short dimension of stirrup = 495 + 25+12 = 532 mm
y1 = long dimension of stirrup = 1095 + 25+ 12 = 1032 mm
x1+ y1
4=532+1032
4=391 mm
Hence provide 12 mm Ф 4 lgd stirrups @ 170 mm c/c.
(b) At the point of max. shear (supports).
At supports, Fo = 868002 N ; τ v=868002
600× 1160 = 1.25 N/mm2
49
At supports, 100 A s
b d = 100(8 × 491)600× 1160 = 0.564
τ c ≈ 0.31 N/mm2. Hence shear reinforcement is necessary.
Vc = 0.31 ×600×1160 = 215760 ∴ Vs =
Fo – Vc = 868002 – 215760 = 652242 N
The spacing of 10 mm Ф 4 lgd stirrups having Asv = 314 mm2 is given by
sv=σ sv . A sv . d
V s = 150× 314 ×1160
652242 = 83.8 mm
This is small. Hence use 12 mm Ф 4 lgd stirrups, having :
Asv = 4 × π4(12)2 = 452.39 mm2 at spacing sv = 150× 452.39 ×1160
652242 ≈ 120 mm
(c) At mid-span:
At the mid-span, S.F. is zero. Hence provide minimum/nominal shear reinforcement,
given by A sv
b . sv≥ 0.4
f y
Or A sv
sv=0.4 b
f y For HYSD bars, fy = 415 N/mm2
∴ A sv
sv=0.4 ×600
415 = 0.578
Choosing 10 mm Ф 4 lgd stirrups, Asv = 314 mm2.
Sv = 3140.578 = 543 mm
Max. permissible spacing = 0.75d = 0.75×1160 = 870 or 300 mm, whichever is less.
Hence provide 10 mm Ф 4 lgd stirrups @ 300 mm c/c.
50
Side Face Reinforcement: Since the depth is more than 450 mm, provide side face reinforcement @ 0.1 %.
∴ Al=0.1100
(600× 1200 ) = 720 mm2 .
Provide 3-16 mm Ф bars on each face, having total Al=6 ×201 = 1206 mm2.
9. Design of columns: The tank is supported on 8 columns, symmetrically placed on a circle of 10 m mean diameter. Height of staging above ground level is 16 m. Let us divide this height into four panels, each of 4 m height. Let the column be connected to raft foundation by means of a ring beam, the top of which is provided at 1 m below the ground level, so that the actual height of bottom panel is 5 m.
(a) Vertical load on columns
1. Weight of the water = Ww + Wo = 4392368+4751259 = 9143627 N 2. Weight of tank :
(i) Weight of top dome + cylindrical walls etc. = W = 58474× π ×14 = 2571821 N(ii) Weight of conical dome = Ws = 1066131 N(iii) Weight of bottom dome = 540982 N(iv) Weight of bottom ring beam = 18000× π ×10 = 565487 N
∴ Total weight of tank = 4744421 N Total superimposed load = 9143627 + 4744421 = 13888048 N
∴ Load per column = 13888048/8 ≈ 1736000 N.Let the column be of 700 mm diameter.Weight of column per metre height = π
4(0.7)2 ×1×25000 = 9620 N
Let the brace be of 300 mm × 600 mm size.
Length of each brace = L = R sin 2 π
n
cos πn
= 5× sin π
4
cos π8
= 3.83 m
51
Clear length of each brace = 3.83 – 0.7 = 3.13 m
∴ Weight of each brace = 0.3×0.6×3.13×25000 = 14085 N
Hence total weight of column just above each brace is tabulated below:Brace GH:W = 1736000+4×9620 = 1774480 NBrace EF:W = 1736000+8×9620+14085 = 1827045 NBrace CD:W = 1736000+12×9620+2×14085 = 1879610 NBottom of column: W = 1736000+17×9620+3×14085 = 1941795 N
(b) Wind loads: Intensity of wind pressure = 1500 N/m2. Let us take a shape factor of 0.7 for sections circular in plan.
52
Wind load on tank, domes and ring beam
= [ (5× 14.4 )+(14.2× 23
×1.9)+(2× 12.8 )+(10.6× 1.2)]×1500 × 0.7
= 134720 N
This may be assumed to act at about 5.7 m above the bottom of ring beam.
×23520=11760 N. The wind loads are shown in fig. The point of contra flexure O1 , O2 , O3 and O4 are assumed to be at the mid-height of each panel. The shear forces Qw and moment Mw due to wind at these planes are given below :
The axial thrust Vmax = 4 M w
n Do =
4 M w
8 ×10 = 0.05 Mw in the farthest leeward column, the shear
force Smax = 2Qw/n = 0.25Qw in the column on the bending axis at each of the above
levels and bending moment M = Smax × h2 in the columns are tabulated below:
Level Vmax Smax (N) M (N-m)
O4 53040 36620 73240
O3 85650 44890 89780
O2 124860 53150 106300
O1 170950 61420 153550
54
The farthest leeward column will be subjected to be superimposed axial load plus Vmax given above. The column on the bending axis, on other hand, will be subjected to superimposed axial load plus a bending moment M given above. These critical combinations for various panels of these columns are tabulated below:
Panel
Farthest Leeward column Column on Bending axis
Axial load (N) Vmax (N) Axial load (N) M (N-m)
O4 O4' 1774480 53040 1774480 73240
O3 O3' 1827045 85650 1827045 89780
O2 O2' 1879610 124860 1879610 106300
O1 O1' 1941795 170950 1941795 153550
According to I.S. Code, when effect of wind load is to be considered, the permissible stresses in the material may be increased by 331
3 %. For the farthest leeward column the axial thrust Vmax due to wind load is less than even 10% of the superimposed axial load. Hence the effect of wind is not critical for the farthest leeward column. However, column situated on the bending axis need be considered to see the effect of maximum B.M. of 153550 N-m due to wind, along with the superimposed axial load of 1941795 N at the lowest panel. Use M20 concrete, for which and σ cbc= 7 N/mm2 and σ cc= 5 N/mm2 For steel, σ st= 230 N/mm2. All of three can be increased by 331
3 % when taking into account wind action.
Diameter of column = 700 mm. Use 12 bars of 30 mm dia. at an effective cover of 40 mm.
A sc=π4(30)2 ×12 = 8482 mm2
Equivalent area of column = π4
(700)2+(13−1)8482 = 486629 mm2
Equivalent moment of inertia = π64
d2+ (n-1) A sc d ' 2
8, where
55
d = 700 mm ; d' = 700 - 2×40 = 620 mm
Ic = π64
(700)4+(13−1)8482 ×(629)2
8 = 1.66766×1010 mm4
∴ Direct stress in column = σ cc' = 1941795/486629 = 3.99 N/mm2
For the safety of the column, we have the condition,
σcc 'σcc
+σcbc 'σcbc
≥ 1 ∴ 3.991.33× 5
+ 3.221.33 ×7
<1
Or 0.60 + 0.35 < 1 or 0.95 < 1 Hence safe.
Use 10 mm Ф wire rings of 250 mm c/c to tie up the main reinforcement. Since the columns are of 700 mm diameter, increase the width of curved beam B2 from 600 mm to 700 mm.
10. Design of braces: The bending moment m1 in a brace is given by equation its maximum value being governed by:
tan(θ+ π
8 )=12
cot θ
Solving this graphically, we get Ө = 24.8º
(m1)max = Qw 1 . h1+Qw 2 . h2
n sin 2πn
cos2 θ sin(θ+ πn )
For the lowest junction C, h1 =5 m and h2 = 4 m
(m1)max = (245660× 5 )+(212600 × 4)
8 sin 2π8
cos2 24.8° sin (24.8°+ 180 °8 ) = 222540 N-m
56
The maximum shear force (Sb)max in a brace is given by equation, for Ө = π8 :
(Sb)max = (245660× 5 )+(212600 × 4)
3.93× 8 sin 2 π8
(2 cos¿¿2 π8 × sin 2π
8 )¿ = 112870 N
For Ө = π8 , the value of m1 is given by:
[ ( m1 ) ]θ =π
8 =
(245650× 5 )+(212600 × 4)
8 sin 2π8
cos2 π8 sin( π
8 +π8 ) = 221786 N-m
Twisting moment at Ө = π8 is M t = 0.05 m1 = 0.05×221786 = 11090 N-m
Thus the brace will be subjected to a critical combination of max. shear force (Sb)max and
a twisting moment (M t) when the wind blows parallel to it (i.e. when Ө = π8 ).
The brace is reinforced equally at top and bottom since the sign of moment (m1) will depend upon the direction of wind.
For M20 concrete, c = σ cbc= 7 N/mm2 , m=13, also σ st= t =230 N/mm2 ; k = 0.283 ; j = 0.906 and R= 0.897. Depth of N.A = 0.283d. Let Asc =Ast = p b d and dc = 0.1 d
Equating the moment of equivalent area about N.A.
12
b(0.283 d)2+(13−1 ) p bd (0.283 d−0.1 d )=13 p bd (d−0.283 d )
From which p = 0.0056
Since the brace is subjected to the both B.M. as well as twisting moment, we have
Me1 = M + MT where M = B . M = (m1)max = 222540 N-m
MT = T( 1+D /b1.7 ), where T = Mt = 11090 N-m
57
Let D = 700 mm. ∴ MT = 11090[1+700 /3001.7 ] = 21745 N-m
Me1 = 222540 + 21745 = 244285
In order to find the depth of the section, equate the moment of resistance of the section to the external moment.
b . n . c2 [d− n
3 ]+(mc−1 ) Asc .c ' (d−dc )=M e1
Here c = 1.33×7 = 9.31 N/mm2 ; mc = 1.5 m = 1.5×13 = 19.5 ;
c '=9.31(0.283−0.1)
0.283 = 6.02 N/mm2
Hence,
300 ×0.283 d × 9.312 [1−0.283
3 ]d+(19.5−1 ) (0.0056 × 300 d )6.02 (1−0.1 ) d
¿244285 ×103
Or 358 d2+168.4 d2=244285 ×103. From which d = 680 mm
Adopt D = 700 mm so that d = 700-25-10 = 665 mm
Asc = Ast = p bd = 0.0056×300×700 = 1176 mm2
No. of 20 mm Ф bars = 1176/314 = 3.74
Hence provide 4 Nos. of 20 mm Ф bars each at top and bottom.
100 A s
b d=100 × 4 × 491
300 ×700=0.94 %. Maximum shear = 112870 N
Ve = V + 1.6 T/b = 112870 + 1.6× 110900.3 = 172017 N
∴ τve=172017
300 ×700 =0.82 N/mm2
58
This is less than τ c. max but more than τ c= 0.37 N/mm2. Hence transverse reinforcement is necessary.
A sv=T . sv
b1d1 σsv+
V . sv
2.5 d1 σsv
Where b1 = 300 – (25×2)-20 = 230 mm ; d1 = 700-(25×2)-20 =630 mm.
Using 12 mm Ф 2 lgd stirrups, A sv=2 π4(12)2 = 226 mm2.
∴A sv
sv=[ 11090× 100
230 ×630 ×230+ 112870
2.5× 630× 230 ]=0.333+0.312=0.645
Minimum transverse reinforcement is given by
A sv
sv≥( τ ve−τc
σ sv)b ∴
A sv
sv=0.82−0.37
230×300=0.587
∴ sv=A sv
0.645= 226
0.645=350 mm
This spacing should not exceed least of x1 , x1+ y1
4 and 300 mm
Where x1 = short dimension of stirrup = 230+20+12=262 mm y1 =long dimension of stirrup = 630+20+12 = 662 mm.
∴
x1+ y1
4=262+662
4 =231 mm.
Hence provide 12 mm Ф 2 lgd stirrups at 230 mm c/c throughout. Since the depth of section exceeds 450 mm, provide the side face reinforcement @ 0.1 %.
Al=0.1100
(300 ×700) = 210 mm2 .
Provide 2-10 mm Ф bars at each face, giving total Al=4 × 78.5 = 314 mm2.
Provide 300 mm × 300 mm haunches at the junction of braces with columns and reinforce it with 10 mm Ф bars.
59
11. Design of raft foundation
Vertical load from filled tank and columns = 1941795×8 = 15534360 N
Weight of water = 9143627. Vertical load of empty tank and columns = 6390733 N
Vmax due of wind load = 170959×8 , which is less than 3313 % of the super imposed load.
Assume self weight etc. as 10% = 1553436 N
∴ Total load = 15534360 + 1553436 = 17087796 N
∴ Area of foundation required = 17087796/150000 = 113.9 m2
Circumference of column circle = π × 10 = 31.42 m ∴
Width of foundation = 113.931.32 = 3.64 m. Hence inner diameter = 10 – 3.64 = 6.36 m
Outer diameter = 10 + 3.64 = 13.64 m.
Area of annular raft =π4
(13.642−6.362) = 114.35 m2
60
Moment of inertia of slab about a diametrical axis = π64
[13.644−6.364 ]=1618.8 m4
Total load, tank empty = 6390733+1553436 = 7944169 N
∴ Stabilizing moment = 7944169× 13.642 = 54179233 N-m
Let the base of the raft be 2m below ground level.
∴ Mw at base = (134720×23.7) + (11760×18) + 33060(14+10+6) = 4396344 N-m
Hence the soil pressure at the edges along a diameter are
(a) tank full : 17087796114.35
± 43963441618.8
× 13.642 = 167956 N/m2 or 130912 N/m2
(b) tank empty :7944169114.35
± 43963441618.8
× 13.642 = 87994 N/m2 or 509551 N/m2
Under the wind load, the allowable bearing capacity is increased to 150×1.333 = 200 kN/m2, which is greater than the maximum soil pressure of 167.96 kN/m2. Hence the foundation raft will be designed only for super-imposed load.
The layout of the foundation is shown in fig. A ring beam of 700 mm width may be provided. The foundation will be designed for an average pressure p:
p = 15534360114.35
≈135849 N/m2
The overhang x of raft slab = 12 [ 1
2(13.64−6.36 )−0.7]= 1.47 m
B.M. = 135849(1.47)2
2=¿146778 N-m ; S.F. = 135849×1.47 = 199698 N
d =√ 146778 ×10001000 ×0.897
= 405 mm
Provide 450 mm thick slab with effective depth of 410 mm. Decrease the total depth of 250 mm at the edges.
A st=146778 ×1000
230× 0.906 × 410 = 1718 mm2
61
Spacing of 16 mm Ф bars = 1000× 2011718 = 117 mm. Hence provide 16 mm Ф radial bars
@ 110 mm c/c at the bottom slab.
Area of distribution steel = 0.15100
×1000 × 450 = 675 mm2
Spacing of 10 mm Ф bars = 1000× 78.5675 = 116 mm. Hence provide 10 mm Ф bars @ 110
mm c/c at the support. Increase this spacing to 200 mm at the edge.
Design of circular beam of raft: The design of circular beam of raft will be practically similar to the circular beam B2 provided at the top of the columns.
Design load = 15534360π ×10 = 494474 N/m
The circular beam B2 was designed for w = 442070 N/m. Hence the B.M. etc. will be increased in this ratio of 494474/442070 = 1.119
∴ Max. (-) B.M. at support = Mo = 572882×1.119 = 640790 N-m
Max. (+) B.M. at mid span = Mc =260401×1.119 = 291268 N-m
Max. torsional moment Mmt = 43400×1.119 = 48545 N-m
B.M. at the point of max. torsion = 1421×1.119 = 1589 N-m
At φ=φm=9 12
° , F = 501512×1.119 = 560960 N
Max. shear force at supports = 868002×1.119 = 970893 N
Use b = 700 mm = diameter of columns. Use M 20 concrete.
σ st = 230 N/mm2 . d = √ 640790 ×1000700 ×0.897
= 1010 mm
However, keep total depth of 1200 mm from shear point of view. Using an effective cover of 60 mm, d= 1140 mm
62
Main or longitudinal reinforcement:
(a) Section at point of maximum torsion
T = Mmaxt = 48545 N-m ; M φ= M = 1589 N-m ; Me1 = M + MT
Where MT = T[1+ D /b1.7 ] = 48545[1+ 1200/700
1.7 ] = 77509 N-m
∴ Me1 = 1589 + 77509 = 79098 N/m
A st=M e1
σ st jd= 79098 ×1000
230 ×0.906 ×1140 = 333 mm2 ; No. of 25 mm Ф bars = 333/491≈ 1
Since MT>M , Me2 = MT – M = 77509 -1589 = 75920
A st 2=75920 ×1000
230× 0.906 ×1140 = 319.6 mm2 ∴ No. of 25 mm Ф bars = 319.6/491≈ 1
However provide minimum of 2 bars each at top and bottom.
(b) Section at max. hogging B.M. (support)
Mo = 640760 N-m = Mmax ; Mot = 0
A st=640790× 1000
230× 0.906 ×1140 = 2697 mm2 ; No. of 25 mm Ф bars = 2697/491≈ 5.49
However, provide 6 bars of 25 mm Ф at the bottom of the section, near supports.
(c) Section at max. sagging B.M. (mid-span)
Mc = 291268 N-m ; Mct = 0
∴ A st=
291268× 1000230× 0.906 ×1140 = 1226 mm2
∴ No. of 25 mm Ф bars = 1226/491≈ 2.5
Hence the scheme of reinforcement along the span will be as follows :
63
At supports, provide 6-25 mm Ф bars at bottom of section. Continue these upto the section of maximum torsion (i.e. at φm= 9.5º = 0.116 rad.), at a distance = Rφm= 5×0.166
= 0.83 or equal to Ld =Ф . σst
4 τbd=Ф ×230
4 ×1.12=¿ 52 Ф = 52×25 = 1300 mm whichever is more.
Beyond this, discontinue 2 bars, while the remaining 4 bars may be continued throughout the length.
Similarly, provide 4-25 mm Ф bars at top, throughout the length. These bars will take care of both the maximum positive B.M. as well as maximum torsional moment.
Transverse reinforcement:
(a) At the point of minimum torsional moment
V = 560960 N-m
∴ Ve = V + 1.6Tb = 560960 + 1.6× 48545
0.7 = 672152 N
τ ve=672152
700× 1140 = 0.84 N/mm2
This is less than τ cmax = 1.8 N/mm2. Hence O.K.
100 A s
b d=
100 (3 ×491)700 ×1140
= 0.185
∴ τ c= 0.22 N/mm2. Hence shear reinforcement is necessary.
A sv=T . sv
b1d1 σ sv+
V . sv
2.5 d1 sv
where b1 = 700-(40×2)-25 = 595 mm
d1 = 1200-(40×2)-25 = 1095 mm
∴ A sv
sv=[ 48545 ×1000
595× 1095× 230+ 560960
2.5×1095 ×230 ]=1.215
64
Minimum transverse reinforcement is governed by
∴ A sv
sv≥( τ ve−τ c
σ sv)b ∴
A sv
sv=0.84−0.22
230×700 = 1.887
Hence adopt A sv
sv = 1.887. Using 12 mm Ф 4 lgd stirrups,
Asv = 4×113 = 452 mm2.
∴ sv = 452/1.887 ≈ 240 mm.
However, spacing should not exceed least of x1 , x1+ y1
4 and 300 mm, where
x1 = short dimension of stirrup = 595+25+12 = 632 mm
y1 = long dimension of stirrup = 1095+25+12 = 1132 mm
∴
x1+ y1
4=632+1132
4 = 441 mm.
Hence provide 12 mm Ф 4 lgd stirrups @ 240 mm c/c.
(b) At the point of max. shear (supports)
At supports, Fo = 970893 N
∴ τ v=970893
700× 1140 = 1.22 N/mm2. At supports,100 A s
b d=
100 (6× 491)700 × 1140
≈ 0.37 %
Hence τ c ≈0.26 N/mm2. Hence shear reinforcement is necessary.
Vc = 0.26×700×1140 = 207480 N
∴ Vs = Fo – Vc = 970893 – 207480 = 763413 N
The spacing of 12 mm Ф 4-lgd stirrups having A sv=π4
(12)2 = 452.4 mm2 is given by
65
sv=σ sv . A sv . d
V=230 × 452.4 ×1140
763413=¿155.3 mm
Hence provide 12 mm Ф 4-lgd stirrups @ 150 mm c/c.
(c) At mid-span: At the mid-span, S.F. is zero, Hence provide minimum/nominal shear reinforcement given by
A sv
b . sv≥ 0.4
f y or
A sv
sv=0.4 b
f y=0.4 ×700
415=0.675
Choosing 10 mm Ф 4-lgd stirrups, Asv =314 mm2.
∴ sv = 314/0.675 = 465 mm
Max. permissible spacing = 0.75d = 0.75×1140 = 855 or 300 mm, whichever is less.
Side Face Reinforcement: Since depth is more than 450 mm, provide side face reinforcement @ 0.1%
Al=0.1100
(700× 1200 )=¿840 mm2
Provide 3-16 mm Ф bars on each face, having total Al= 6×201=1206 mm2.
66
12. Details of reinforcement: Shown in figure.
67
68
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.
Beldars 26 250 6500 Mazdoors 20 250 5000 Total 115002 Earth work in
Filling In foundation
154.4
Beldars 15 250 3750 Bhisthi 5 285 1425 Mazdoors 10 250 2500 Total 76753 Disposal of
surplus earth in a lead 30m
228.7
Mazdoors 20 250 5000 Total 5000 Total cost of
earth work 24,175
14.2 DATA SHEET:
75
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 like cement
FOUNDATION
MATERIALS UNIT QTY RATERS
AMOUNT RS
20mm HBG graded metal Cum Cum 62.48 1405.04 87786.89Sand Cum 124.97 509.92 63724.70Cement Cum 41.58 1820 75675.601st Class Mason Day 10 285 28502nd Class Mason Day 20 285 5700Mazdoor (Both Men and Women) Day 25 250 6250Concrete Mixer 10/7 cft (0.2/0.8cum)capacity Hour 6.2 250 1550Cost of Diesel for Miller Liter 2.4 53 127.20Cost of Petrol for Vibrator Liter 3.7 67 247.90Water (including for curing) KL 1.5 77 115.50Add 20% in Labour (1st Floor) 48805.55Add MA 20% 48805.55Add TOT 4% 9761.11TOTAL COST 351400
COLUMNS MATERIALS UNIT QTY RATERS
AMOUNT RS
76
20mm HBG graded metal Cum Cum 28.61 1405.04 40198.19Sand Cum 14.30 509.92 7291.85Cement Cum 9.52 1820 17326.401st Class Mason Day 6 285 17102nd Class Mason Day 10 285 2850Mazdoor (Both Men and Women) Day 15 250 3750Concrete Mixer 10/7 cft (0.2/0.8cum) capacity Hour 2 250 500Labour centering Cum 2 971 1942Material hire charges for centering Cum 2 89 178Water (including for curing) KL 1.5 77 115.50Add 20% in Labour (1st Floor) 15172.38Add MA 20% 15172.38Add TOT 4% 3034.47TOTAL COST 109241.17
RING BEAM AT TOP MATERIALS UNIT QTY RATERS
AMOUNT RS
20mm HBG graded metal Cum Cum 3.45 1405.04 4847.38Sand Cum 1.72 509.92 877.06Cement Cum 1.15 1820 20931st Class Mason Day 1 285 2852nd Class Mason Day 2 285 570Mazdoor (Both Men and Women) Day 3 250 750Concrete Mixer 10/7 cft (0.2/0.8cum) capacity Hour 1 250 250Labour centering Cum 2 971 1942Material hire charges for centering Cum 2 89 178Water (including for curing) KL 1.5 77 115.5Add 20% in Labour (1st Floor) 2381.58Add MA 20% 2381.58
77
Add TOT 4% 476.31TOTAL COST 17147.41
DOMED ROOF MATERIALS UNIT QTY RATERS
AMOUNT RS
20mm HBG graded metal Cum Cum 8.96 1405.04 12589.15Sand Cum 4.48 509.92 2284.44Cement Cum 2.98 1820 5423.601st Class Mason Day 3 285 8552nd Class Mason Day 6 285 1710Mazdoor (Both Men and Women) Day 13 250 3250Concrete Mixer 10/7 cft (0.2/0.8cum) capacity Hour 1 250 250Labour centering Cum 15 971 14565Material hire charges for centering Cum 15 89 1335Water (including for curing) KL 1.5 77 115.5Add 20% in Labour (1st Floor) 8475.53Add MA 20% 8475.53Add TOT 4% 1695.10TOTAL COST 61023.85
CONICAL SLAB MATERIALS UNIT QTY RATERS
AMOUNT RS
78
20mm HBG graded metal Cum Cum 23.3 1405.04 32737.43Sand Cum 11.65 509.92 5940.56Cement Cum 7.75 1820 141051st Class Mason Day 4 285 11402nd Class Mason Day 10 285 2850Mazdoor (Both Men and Women) Day 19 250 4750Concrete Mixer 10/7 cft (0.2/0.8cum) capacity Hour 0.86 250 215Labour centering Cum 10 971 9710Material hire charges for centering Cum 10 89 890Water (including for curing) KL 1.5 77 115.5Add 20% in Labour (1st Floor) 14490.69Add MA 20% 14490.69Add TOT 4% 2898.13TOTAL COST 104333
CYLINDRICAL WALL MATERIALS UNI
T QTY RATER
S
AMOUNT RS
20mm HBG graded metal Cum Cum 30.04 1405.04 42207.40Sand Cum 15.02 509.92 7658.99Cement Cum 9.99 1820 18181.801st Class Mason Day 8 285 22802nd Class Mason Day 18 285 5130Mazdoor (Both Men and Women) Day 40 250 10000Concrete Mixer 10/7 cft (0.2/0.8cum) capacity
Hour
0.96
250 240
Labour centering Cum 2 971 1942Material hire charges for centering Cum 2 89 178Water (including for curing) KL 1.5 77 115.5Add 20% in Labour (1st Floor) 17586.73
79
Add MA 20% 17586.73Add TOT 4% 3517.34TOTAL COST 126624.49
RING BEAM AT BOTTOM MATERIALS UNIT QTY RATERS
AMOUNT RS
20mm HBG graded metal Cum Cum 10.31 1405.04 14485.96Sand Cum 5.15 509.92 2626.08Cement Cum 3.43 1820 6242.601st Class Mason Day 3 285 8552nd Class Mason Day 6 285 1710Mazdoor (Both Men and Women) Day 9 250 2250Concrete Mixer 10/7 cft (0.2/0.8cum) capacity Hour 0.96 250 240Labour centering Cum 2 971 1942Material hire charges for centering Cum 2 89 178Water (including for curing) KL 1.5 77 115.5Add 20% in Labour (1st Floor) 6129.02Add MA 20% 6129.02Add TOT 4% 1225.80TOTAL COST 44128.98
CIRCULAR GIRDER MATERIALS UNIT QTY RATERS
AMOUNT RS
20mm HBG graded metal Cum Cum 14.42 1405.04 20260.67Sand Cum 7.21 509.92 3676.52
80
Cement Cum 4.79 1820 8717.801st Class Mason Day 6 285 17102nd Class Mason Day 12 285 3420Mazdoor (Both Men and Women) Day 16 250 4000Concrete Mixer 10/7 cft (0.2/0.8cum) capacity
Hour
0.96
250 240
Labour centering Cum 2 971 1942Material hire charges for centering Cum 2 89 178Water (including for curing) KL 1.5 77 115.5Add 20% in Labour (1st Floor) 8852.09Add MA 20% 8852.09Add TOT 4% 1770.41TOTAL COST 63735.08
BRACING AT 4m from G.L. MATERIALS UNIT QTY RATERS
AMOUNT
RS 20mm HBG graded metal Cum Cum 0.36 1405.04 505.81Sand Cum 0.18 509.92 91.78Cement Cum 0.12 1820 218.401st Class Mason Day 0.18 285 51.302nd Class Mason Day 0.40 285 114Mazdoor (Both Men and Women) Day 0.98 250 245Concrete Mixer 10/7 cft (0.2/0.8cum) capacity Hour 0.38 250 95Labour centering Cum 2 971 1942Material hire charges for centering Cum 2 89 178Water (including for curing) KL 1.5 77 115.5Add 20% in Labour (1st Floor) 711.35Add MA 20% 711.35Add TOT 4% 142.27TOTAL COST 5121.76
81
BRACING AT 8m from G.L. MATERIALS UNIT QTY RATERS
AMOUNT RS
20mm HBG graded metal Cum Cum 0.36 1405.04 505.81Sand Cum 0.18 509.92 91.78Cement Cum 0.12 1820 218.401st Class Mason Day 0.18 285 51.302nd Class Mason Day 0.40 285 114Mazdoor (Both Men and Women) Day 0.98 250 245Concrete Mixer 10/7 cft (0.2/0.8cum) capacity
Hour
0.38
250 95
Labour centering Cum 2 971 1942Material hire charges for centering Cum 2 89 178Water (including for curing) KL 1.5 77 115.5Add 20% in Labour (1st Floor) 711.35Add MA 20% 711.35Add TOT 4% 142.27TOTAL COST 5121.76
BRACING AT 12m from G.L. MATERIALS UNIT QTY RATE
RS AMOUNT RS
20mm HBG graded metal Cum Cum 0.36 1405.04 505.81Sand Cum 0.18 509.92 91.78Cement Cum 0.12 1820 218.401st Class Mason Day 0.18 285 51.30
82
2nd Class Mason Day 0.40 285 114Mazdoor (Both Men and Women) Day 0.98 250 245Concrete Mixer 10/7 cft (0.2/0.8cum) capacity Hour 0.38 250 95Labour centering Cum 2 971 1942Material hire charges for centering Cum 2 89 178Water (including for curing) KL 1.5 77 115.5Add 20% in Labour (1st Floor) 711.35Add MA 20% 711.35Add TOT 4% 142.27TOTAL COST 5121.76
Plastering with CM(1:6) & (1:2) (12 mm thick) Material Unit Qty Rate Amount
RS
Cement Mortar 1:6 1:2
cum cum
22.29 13.71 8.58
1889 3200
25898.19 27456
Mason 1st class day 30 285 8550Bhisthi day 10 285 2850Mazdoor (unskilled) day 30 250 7500Add MA 20% 14450.83Add TOT 4% 2890.16Grand Total 89595.18
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 3 coats in all to give an even shade after thoroughly 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
83
charges etc. complete for finished item of work as per SS 912 for walls Material Unit Qty Rate Amount
RS
Epoxy primer for Hi bond floor & protective coatings : Procoat SNP2 or Zoriprime EFC 2
Pack 20 548 10960
1st class painter Day 4 355 1420Mazdoor Day 4 250 1000cost of water proof cement paint Cum 40 35 14001st class painter Day 2 355 710Mazdoor (unskilled) Day 2 250 500Add MA 20% 3198Add TOT 4% 639.60Total cost 19827.60
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 3 coats in all to give an even shade after thoroughly 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 Material Unit Qty Rate Amount
RSCost of cement primer Pack 15 100 15001st class painter Day 1 355 3552nd class painter Day 1 250 250cost of water proof cement paint Cum 6 35 2101st class painter Day 1 355 355Mazdoor (unskilled) Day 1 250 250Add MA 20% 584Add TOT 4% 116.8Total cost 3620.80
84
Total cost of project:
Total cost of R.C.C = 8,92,999
Total cost of steel = 20,53,898
Total cost of plastering = 89,595
Total cost of painting = 23,449
Total cost of earthwork = 24,175
Over all cost = 30,84,116
Add 15 % contractors profit = 4,62,617
Add 3 % contingency = 92,524
Total Cost 36,39,257
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. Without 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.
85
16. REFERENCES
Table 16.2. Coefficients for moment in cylindrical wall fixed at base(As per IS3370)
