Sunday, 17 February 2013
Design For Water Tank
193
Water Tanks
WATER TANKS
Structure
10.1
Introduction
Objectives
10.2
Design and Detailing Specifications of Water Tanks
10.2.1
Specifications
10.2.2
Design Procedure for Overground Cylindrical Tanks with Flexible Base10.2.3
Design Procedure for Underground Cylindrical Tanks with Flexible Base
10.2.4
Design Procedure for Overhead Intze Type Tanks
10.3
Design Specifications for Circular Shaft Type Staging
10.4
Design of Annular Footing
10.5
Summary
10.6
Answers to SAQs
10.1 INTRODUCTION
A water tank is used for storage and distribution of water over itscommand area
(Figures 10.1 to 10.3). Tanks may be of rectangular, cylindrical, polygonal or of
any other shape depending upon their capacities, economy and architectural
requirements. Tanks may be :
(a)
Underground water tanks,
(b)
Surface tanks resting on ground, and
(c)
Overhead water tanks.
Figure 10.1 : Overground Tank
Figure 10.2 : Underground Tank
The supportfor an overhead tank may be a group of columns with intermediate
bracings. The bracings are provided for resisting horizontal forces as well as for
reducing effective lengthof columns. For larger capacity tanks, staging may be in
the form of single shaft circular or polygonal in plan and may be tapering. The
foundation may be isolated footing, or annular footing with or without beam, orraft (Figure 10.3). For simplicity ofanalysis and design, only those types of tanks
and staging have been described which may be analysed by membrane theory.
194
Theory of StructuresII
(a)
(b)
Figure10.3 : Overhead Tanks with different Types of Staging
Objectives
After studying this unit, you should be able to
?
describe the water tanks and its components,
?
classify the water tanks, and
?
explain the design steps of the cylindricaltanks resting on ground
with flexible base, underground cylindrical tanks with flexible base,and overhead intze type tank with circular shaft staging.
10.2 DESIGN AND DETAILING SPECIFICATIONS
OF WATER TANKS
10.2.1 Specifications
Water tanks must be leak proof. The designed sections of its components must be
crack resistant, and even if the cracks are formed they must be very fine and
evenly distributed, so that the full section may be effective to prevent leak and to
protect reinforcement from corroding.
To meet above requirements IS : 3370 (Part I to IV), Code of Practice for
Concrete Structures for Storage of Liquids, prescribes the following:
(a)
The elements shall be designed by Working Stress Method assuming
full cross section (including cover) uncracked and effective by
allowing limited tensile strength ofconcrete
?
ct
(Table 10.1).
195
Water Tanks
Table 10.1 : Permissible Concrete Stress in Calculation Relating to
Resistance to Cracking
Grade ofConcrete
(N/mm )
2
Permissible Stresses
Direct Tension
M 20
1.2
M25
1.3
M 30
1.5
M 35
1.6
M 40
1.7
(b)
Reduced tensile strength ofreinforcement (Table 10.2) are taken to
indirectly offset the effect of corrosion.
Table 10.2 : Permissible Stresses in Steel Reinforcement
for Strength Calculation
Permissible Stresses in N/mm
2
Sl.
No.
Type of Stress in Steel
Reinforcement
Plain Round Mild Steel
Bars Conforming to
Grade I of IS : 432
(Part I ) � 1966
High Yield Strength
Deformed Bars
Conforming to
IS : 17861966
IS : 11391966
(1) (2)
(3)
(4)
1.Tensile stress in members
under direct tension
115 150
(c)
Increased nominal covers than those as per IS : 456 � 2000 are
provided so that the surface cracks may not reachup to reinforcement.
For waterfaces of parts of members either in contact with the water
or enclosing the space above the water (such as inner faces of roof
slab), the minimum cover to all reinforcement should be 25 mm or the
diameter of the main bar, whichever is greater.
In the presence of sea water, soils and water of corrosive character the
cover should be increased by 12 mm but this additional cover shall
not be taken into account for design calculations.
For faces away from the water and for parts of the structure neither in
contact with the water on any face nor enclosing the space above the
water, the cover shall conform to the requirementsof IS : 4562000
(Refer Unit 1).In case of mild steel the minimum reinforcement in walls, floors androofs in each of two directions atright angles shall have an area of0.3 percent of the concrete section in that direction for sections upto
100 mm thick. Forsections of thickness greater than 100 mm and less
than 450mm the minimum reinforcement in each of the two
directions shallbe linearly reduced from 0.3 percent for 100 mm thick
section to 0.2 percent for 450 mmthick section. For sections of
thickness greater than 450 mm, minimum reinforcement in each of the
two directions shall be keptat 0.2 percent. In concrete section of
thickness 225 mm or greater, two layers of reinforcingsteel shall be
placed one near each face of the section to make up the minimum
reinforcement specified in this clause.
196 In case of high yield strength deformed bars the minimum
reinforcement specified above may be decreased by 20 percent.Theory of StructuresII
In no case the percentage of reinforcement in any member shallbe
less than that specified in IS: 4562000.
Other design and detailing requirements such as compressive stresses,
bar size, development length etc. shall comply with the provisions of
IS: 456  2000 (Refer Unit 1).
10.2.2 Design Procedure for Overground Cylindrical Tanks withFlexible Base
Size of Tank
Volume, V =
4
?
D H
2
where
D = Inside diameter of tank, and
H = Total depth of tank.
Since D and H are both unknowns, therefore, either of the two is assumed to
get the other. Here H = Depth of water + Free board.
Design of Wall
Hoop Reinforcement (A )
s
Maximum Hoop Tension, T = w H
2
D
where w = Density of water (kN/m ), and
3
T = Hoop tension/m height.
EquatingApplied Force, T = Resisting force (tensile steel A only
s
taken)T =
?
st
A
s
or, Hoop Reinforcement/m, A =
s
st
T
?
Thickness of Wall (t)
Equating Applied Force, T = Resisting force (whole cross section including
A is taken).
s
or,
T =
ct
?
{b t + (m � 1) A }
s
where
= Permissible direct tensile stress inconcrete (Table 10.1),
ct
?
b = breadth of section in vertical direction (usually taken 1 m),
t= thickness of section, and
m = modular ratio.
Thickness of wall is usually not less than 150 mm.
Minimum Reinforcement (A
s,min)
The hoop reinforcement as well as temperature and shrinkage reinforcement
in the vertical direction shall not be less than A
st,min
(10.2.1).
197
Water Tanks
Curtailment of Hoop Reinforcement
Theoretically hoop reinforcement requirement gradually reduces and
approaches to zero at the top. Hence curtailment becomes necessary
specially where the tank is deep. This is explained through Example 10.1.
Design ofBase Slab
The pressure on the ground due to water, wt. of wall and self weight of base
slab is almost uniformly distributed over the whole base area and it is equal
and opposite to the uniformly distributed reaction. Hence, a minimum
thickness of slab with minimum reinforcement is provided for base slab. To
make the joint between wall and base slab flexible, such that the wall may
deform at the junctionwithout any restraint, the wall and the base are
separated from each other and the joint is filled with bitumen
(Figures 10.1and 10.2).
* In case of
aggressive soil or
injurious subsoil
water, M 15
concretewith
sulphate resisting
cement shall be
provided as thescreed.
A screed or concrete layer of 75 mm thick of concrete grade M 10* shall be
first laid on the ground and covered with a sliding layer of bitumen paper or
other suitable material to destroy the bond between the screed and base
slab.
Example 10.1
Design a circular water tank with dome as top cover resting over ground for
a capacity of 200,000 litres. Depthof the tank is to be 3.2 m including 0.2 m
free board. Use M 30 concrete and Fe 250 steel.
Solution
Size of Tank
Depth of water in the tank = 3.2 � 0.2 =3 m
3
10
3
m
6
10
1 litre = 1000 cc
If
D = inside diameter of tank, then
3
D
4
10
10
000
,200
2
6
3
?
?
?
or,
D = 9.213 m
Hence, provided D = 9.25 m.
Design of Wall
Hoop Reinforcement (A )
s
Maximum hoop tension, T = wH
2
D
or,
T = 10 � 3.2 �
2
25
.
9
= 148 kN
st
?
= 115 N/mm from Table 10.2
2
A required at the base level
s
A =
s
148
10
00
11
5
st
T
?
?
= 1287 mm /m
2
Hence provided
?
12 @ 85 mm c/c =1329 mm /m.
2
Thickness of Wall (t)
From equilibrium of forces,
T =
ct
?
{bt + (m� 1) A }
s
198
or,
148 � 10 =1.5 � {1000 � t + (9 �1) � 1329}
3
Theory of StructuresII
or,
t = 88.035
Provided thickness of wall, t = 150mm.
Temperature and Shrinkage reinforcement
p % = 0.3 �
t
)
100
150
(
)
100
450
(
)
2
.
0
3
.
0
(

?


= 0.286 %
?
A
s,min
=
100
286
.
0
� h = 429 mm
2
/m
Provided
?10 @ 180 mm c/c.
These vertical bars will also act as tie bars for main reinforcement.
Curtailment of Hoop Reinforcement
Half of the hoop reinforcement may be curtailed athalf the depth of
the tank, i.e. at 1.6 m depth as the variation ofA is proportional to H.
s
The curtailed bars, however will be extended by greater of
?
12and d
(Figure 10.4).
Design ofBase Slab
Provided total depth of slab = 150 mm and temperature and shrinkage
reinforcement same as that for wall of 150 mm thickness, i.e.
?10 @ 180 mm c/c.
The reinforcement detailing is shown in Figure 10.4.
Figure 10.4 : Reinforcement Detailing
10.2.3 Design Procedure for Underground Cylindrical Tanks
with Flexible Base
The cylindricalwall, in this case, will be subjected to
(a)
Water pressure only from inside when the tank is full and earth is not
filled from outside,
(b)
Earth pressure only from outside when the tank is empty, and
(c)
Water pressure from inside and earth pressure fromoutside
simultaneously in normal circumstances.
199
Water Tanks
The above two cases, (a) and (b) are generally critical and design of wall will be
guided by these two.
Other design and detailing specifications will remain the same discussed in
Section 10.2.2.
Example 10.2
Design the same water tank as in Example 10.1 but lying underground.
Angle of repose = 25 and soil density = 17 kN/m .
o
3
Solution
Size of Tank
Total height, H = 3.2 m, and inside diameter, D = 9.25 m.
Design of Wall
(a)
when water pressure is only considered, the design will be same
as done in Example 10.1.
(b)
when earth pressure only is acting.
Hoop Reinforcement
Maximum hoop compression = k ? H
a
2D
2
25
9
2
3
17
25
sin
1
25
sin
1
.
.
?
?
?
?

= 102.114 kN
Compressive Stress,
=
cc
??
t
?
?
1000
10
114
.
102
3
or,
t
?
?
1000
10
114
.
102
8
3
(for M30 concrete,
?
cc
= 8 N/mm )
2
or
t = 12.895 mm << 150 taken in Example 10.1. Hence, the design and detailing will remain same as for Example 10.1 (Figure 10.4). 10.2.4 Design Procedure for Overhead Intze Type Tanks Size of Tank Most economical and structurally efficient tank size is determined by the formula : V = 0.585 D o 3 where V = Total volume of the Intze tank (Figure 10.5), and D = Inside diameter of the tank. o All other dimensions are given in terms of Do As this type of tank isprovided only when the tank size is comparatively large. Flat slab for base as well as for top cover become costlier propositions with respect to those for domes including the cost for shuttering. 200 Theory of StructuresII Figure 10.5 : Fixing Size The shape of the bottom dome is conical so that the horizontal components of force at the junction at the base will be of opposite sign to that for the spherical dome base to make the base ring beam economical. Design ofTop Dome The section of dome is very thin in comparison to its diameter. From membrane analysis, the hoop as well as meridional stresses are compressive only up to 51.8 semicentral angle (Figure 10.6(a)). o Figure 10.6(a) : Semicentral Angle Therise of the dome is taken as 1/5 th to 1/8 th of the diameter of the tank. The thickness is taken between 60 mm and 150 mm depending upon the diameter of the tank. Once the diameter, rise and thickness of the dome arefixed, the other dimensional data such as semi central angle (?) andthe radius of the sphere (R) and the load data may be calculated. Figure 10.6(b) : Meridional Thrust and Hoop Force For Design (Figure 10.6(b)). Meridional thrust, = ? N ? ? cos 1 wR and Corresponding compressive stress = bt N? 201 Water Tanks where w = load per unit area on dome, and b = width of section (generally taken as 1 m). The meridional compressive stress is checked against permissiblecompressive stress. Generally permissible stress is greater thanthe actual one, hence only nominal reinforcement is provided along meridian. Next, Hoop thrust, = w R ? N ? ?? � � � ? ? ? ?  ? cos 1 1cos ? Hoop stress= bt N ? Generally, hoopstress is less than the permissible compressive stress and, therefore, only nominal hoop reinforcement is provided. Designof Top Ring Beam It is designed for the horizontal component of maridonial force at the bottom of the dome acting radially. The hoop force, T, developeddue to above radial force is given by the formula (Figure10.6(c)). T = (N cos ?) r ? 1 where r = Radius ofthe ring beam. Figure 10.6(c) : Components of Meridional Force The rest of the design procedure is the same as that fora tank wall. Design of Vertical Wall It is the same as that for vertical wall in Section 10.2.2. Design of Ring Beam at the Junction of Vertical and Conical Dome (Figure 10.7) It isdesigned for the horizontal component of the axial force in conical dome, P = W cot ? where W = Total Vertical load above the beam including its selfweight. The rest of the designprocedure is the same as that for wall of a cylindrical wall. Figure 10.7 : Force at the Junction of Vertical Wall and Conical Dome 202 Design of Conical Dome Theory of StructuresII An element of the conical dome will be designed for meridional force (i.e. axial force) as well as for lateral force perpendicular to the surface of thedome (Figure 10.8). Figure 10.8 : Forces for Design of ConicalDome Design of Bottom DomeIt is similar to top dome except that the load is due to weight of water above it including its self weight. Design of Bottom Ring Beam It is similar to the topring beam except that the designradial pressure is the difference of the horizontal component of forces for bottom spherical and conical domes. The whole design procedure has been explained through Example 10.3. SAQ 1 (a) Draw free hand sketches of elevated water tanks with (i) column type staging, and (ii) with circular shaft type staging. (b) Why water tanks are designed by working stress method?(c) Write short notes on specifications which are applicable to water tanks only. (d) Why the base of an underground or overground tank made flexible? How it is done? Explain with a sketch. (e) Why an Intze type tank becomes economical? Howeconomically efficient dimensions are fixed? (f) How a dome is designed? 203 Water Tanks 10.3 DESIGN SPECIFICATIONS FOR CIRCULAR SHAFT TYPE STAGING Thickness of Shaft (t) Thickness ofshaft shall not be less than 150 mm. If the internal diameter of the shaft is more than 6 m, t = 150 + 120 6000  D where D =Internal diameter of the shaft in mm. Reinforcement in Shaft Vertical Reinforcement The diameter of bars shall not be less than 10 mm. A minimum of 0.25% reinforcement in two layersshall be provided. The spacing of bars shall neither be more 2 t or 400 whichever is less. Circumferential Reinforcement A minimum of 0.2% reinforcement in two layers shall be provided. In no case the reinforcement shall be less than 400 mm /m height. The 2 spacing of bars shall neither be more than t nor more than 300, whichever is less. These reinforcement shall be nearer to the faces. Cover Inside nominal cover shall notbe less than 25, however that foroutside not less 40. Analysis of Shaft For analysis, the loads acting on the shaft are evaluatedas explained below : (a)Vertical loads include load of the supported tank, weight of water of its full capacity, self weight of shaft and imposed load on top dome of the tank. * Earthquake load has not been included here for simplicity. (b) Horizontal loadsmay be either due to wind or due to earthquake . * Wind pressure is evaluated as per IS : 8751987 Part III. The total horizontal load is equal to wind pressure multiplied by projected area of tank including staging.However, this load is modified by a factor called �Shape Factor� whose value depends upon the shape of the tank and its staging. For determining the critical stresses due to vertical and horizontal loads following combinations shall be considered : (a) All vertical (gravity) loads, (b) All vertical loads excluding imposed load with tank empty + wind load, and (b) All vertical loadsexcluding imposed load with tank full + wind load. Whole cross section of shaft shall be in compression if 204 Theory of StructuresII e 2 r = where e = The eccentricity of theload = W M = ionconsiderat under section above load vertical Total ionconsiderat under section atthe plane lin vertica Moment In such case, the maximum vertical compressive stress in concrete shall be given by r I M A W cv ? ? or, ? ? � � ? ? ? ? ? r e t r Wcv 2 1 2 where A = 2 p r t I = I = p r t x r 3 and = Maximum vertical stress in concrete at outside diameter of shaft shell in N/mm cv ? 2 . Permissible Stresses * Concrete The stress in concrete shall not exceed the following limits for various combination of loads : [ Note : If shell thickness is adequateto satisfy Cl. (e) of permissible stresses of concrete this requirement may be waived.] Combination of Loads Stress Limit (a) Dead load + wind load 0.38 cv ????? (b) Dead load + earthquake forces 0.40 cv ????? (c) Circumferential tensile stress in concrete due to wind induced ring moment 0.07 cv???????? Here = 28day ultimate cube strength of concretein N/mm cv ? 2 . Reinforcement The stresses in steel shall not exceed the following limits for various combination of loads : Combination of Loads Stress Limit (a) Dead load + Wind load 0.57sy ????? (b) Dead load+ Earthquake Loads 0.60 sy ?????????? *R efer IS : 116821985 : criteria for design of RCC staging for overhead tanks. 205 Water Tanks (c) Circumferential tensile steel due to wind induced ring moment 0.50 sy ?????????? Here = yield or proof stress of steel in N/mm sy ? 2 . The reinforcement detailing is shown in Figure 10.9. Figure 10.9 : ExplainingSpecifications for ReinforcementDetailing SAQ 2 Writeshort notes on specificationsfor the design of a circular shafttype staging. 10.4 DESIGN OF ANNULAR FOOTING Vertical or gravity load is the total load on footing including its self weight. Themoment due to horizontal load about ground level is alsoconsidered for the design. The maximum pressure (f BC ) on the soil is given by formula f BC = 2 . e D I M A W ? where D = External diameter ofannular footing for safety and stability, and e f BC (i.e. bearing capacity of soil). Bc ? =?? Example 10.3 An overhead Intze type water tank is to be designed for a capacity of 600 klfor a staging height above GL of 15 m. Design parameters are as follows : 206 Bearing capacity of soil = 200 kN/m ; f 2ck = 25 N/mm for tank portion and 2 20 N/mm for staging f = 250 N/mm for tank and f = 415 N/mm for 2 y 2 y 2 staging. Theory of StructuresII Solution Fixing Size Total capacity = 600kl 600 = 0.585 D 3 o or, D = 10.085 m o Provided D = 10 m oInside Surface Dimension of the Tank Refer Figures 10.5 and 10.10. Figure 10.10 : Dimension of the Tank Radius of cylindrical portion m 5 2 1 0 r D Rise of Top dome h = 2m 1 Rise of Bottom dome h = 1 m 2 Chord of Bottom dome= 0 8 5 D = 2 r 2 = 6 m Depth of cylindrical tank = 0 3 2 D = h = 7 m Height of conical dome = 016 3 D = h = 2 m 3 Capacity Calculation Total volume of water = Volume of cylindricalportion + Volume of conical portion � Volume of the bottom dome portion = V = Q + Q � Q 1 2 3 = p r 1 2 (h � 0.15) + p (r 1 2 + r r + r ) 1 2 2 2 6 ) 3 ( 3 2 22 2 2 3 h h r h ? ?  where free board = 0.15 m. 207 Water Tanks = p [(5 � (7 � 0.15) + (5 + 5 � 3 + 3 ) 2 2 2 32  (3 � 3 + 1 ) � 2 2 6 1 )] = 625.962 m > 600 m3
3
Hence, dimension chosen are OK.
Design of Top Dome
Let thickness of dome = 125 mm
Radius of dome = R =
t
2
2
)
2
5
(
2
2
2
1
2
1
2
1
?
?
?
h
h
r
= 7.25 m.
The surface area of the dome
S = 2 p R h = 2 p � 7.25 � 2 = 91.106 m
tt
1
2
sin ? =
25
.
7
5
= 0.6897
or
? = sin

1
0.6897 = 43.61
cos ? = cos 43.61 = 0.724
Load Calculation
Self wt., w = 0.125 � 1 � 1 � 25 = 3.125 kN/m
s
2
IL,
w = 0.75 kN/m
i
2
Total load on the dome/m
2
= 3.875 kN/m
2
Total IL = W = 2 p R h w = 2p � 7.25 � 2 � 0.75 = 68.34 kN
1
t
1
i
Total DL =W = 2p R h w = 2p � 7.25 � 2 �.125 = 284.7 kN
2
t
1
sTotal (IL + DL) on top dome
= 353.04 kN
Meridional Thrust
N =
?
?
?
cos
1
t
R
w
=
)
724
.
0
1(
25
.
7
875
.
3
??
= 16.296 kN/m
?
Compressive stress =
125
1010
296
.
16
1
3
3
?
?
?
?
t
N
= 0.13 N/mm << 6 N/mm
2
2
Hence, OK.
Hoop Thrust
N = w R
?
t
?
?
?
�
�
�
?
?
?
?

?
cos1
1
cos
= 3.875 � 7.25 �
?
?
?
��
�
?
?
?

)
724
.
0
1
(
1
724
.0
= 4.044 kN/m
Hoop stress =
125
10
10
044
.
4
3
3
?
?
= 0.03 N/mm<< 6 N/mm
2
2
Hence, OK.
Nominal Reinforcement
208 Theory of StructuresII
For 125thickness p% =
)
100
125
(
)
100
450
(
)
2
.
0
3
.
0
(
3
.
0

?



= 0.293%
125
1000
100
?
st
A
= 0.293
or,
A = 366.25 mm /m
st
2
Hence, provided
?
10 @ 210 mm c/c bothways.
Design of top ring beam (Figure 10.11).
Figure 10.11 : Component of Meridional Force at Base for Ring Beam Design
T = (N cos ?) r = 16.296 � 0.724 � 5 = 58.99 kN
?
1
A =
s
115
10
99
.
58
3
?
?
stT
= 513 mm
2
Provided 8
?
10 (A = 8 � 78 = 624 mm)
s
2
Assuming b = 250
for M 25, m =
cbc
?
3
280
=
11
98
.
10
5
.
8
3
280
�
?
T =
?
ct
(b D + (m � 1) A )
st
58.99 � 10 = 1.3 (250 �D + (11 � 1) � 624)
3
or,
D = 156.55 mm
Provided D = 250 mm and shear stirrups of
?
6
@ 300 mm c/c.
Wt. of ring beam = 2p
?
?
�
�
?
?
?
2
250
.
0
5
� 0.250 � 0.25 � 25
= 50.315 kN.
Design of Vertical Wall
Height of wall = 7 m
Hoop tension (maximum) = (wh) r = 10 �7 � 5 = 350 kN
1
Hoop Tension at 6 m = (wh) r = 10 � 6 � 5 =300 kN
1
?
Average force for 1 m depth, i.e. from 6.0 m to 7.0 m =
2
)
300
350
(
?
= 325 kN.
A =
s
115
10
325
3
?
= 2826 mm
2
Provided
?
16 @ 140 mm c/c on each face, i.e. A = 2871 mm
s
2
209
Water Tanks
T =
ct
?
(bD + (m � 1) A )
st
325 � 10 = 1.3 (1000 D + (11 � 1) � 2871)
3
or,
D = 221.29 mm
Provided
D = 225 mm
Nominal Reinforcement in Vertical Direction
For 225 thickness p% =
)
100
225
(
)
100
450
(
)
2
.
0
3
.
0(
3
.
0

?


= 0.264%
or,
225
1000100
?
st
A
= 0.264
or,A = 594 mm /m.
st
2
Hence,provided
?
10 @ 270 mm c/c on both faces.
Design of Middle Ring Beam
Taking cross sectional dimension = 500 � 500
wt.of top dome = 353.040 kN
wt. of top ring beam = 50.315 kN
wt. of cylindrical portion = 2p (5.113 � 0.225 � 7 � 25) = 1265 kNwt of ring beam at the junction of cylindrical and conical dome
= 2p � 5.25 � 0.5 � 0.5 �25 = 206.16 kN
?
Total load on top of conical dome, w = 1874.5 kN
?
Load/m run =
)
113
.
0
5
(
25
.
1874
?
?
= 58.34 kN/m.
T cos ? = P and T sin ? = Wor,
P = W cot ? =
1
34
.
58
?
= 58.34 kN (Taking ? =45�)
Hoop tension in ring beam= 58.34 � 5.113 = 298.29 kN
?A =
s
115
10
29
.
298
3
?
= 2593.82 mm
2
Hence, provided 16
?
16.
Hoop tension = 298.29 � 10 =
3ct
?
(bD + (m � 1) A )
s
or
D =
500
3216
)
1
11
(
3
.
1
10
29
.
298
3
?
?
?
?
?
?
?
?
?
?
?
?


?
= 394.58
Hence, Provided crosssectional area of Middle Ring Beam
= 500
?
500.
Design of Conical Dome
Taking thickness of conical dome = 250 mm
Depth at mid height = 7 + 1 = 8 m
Mean radius of ring at mid height = 5 � 1 = 4 m
Thickness of conical dome = 250 mm
Total vertical load upto the middle ring beam
= 1874.5 kN
210
weight of conical portion up to the ring of 1 m length
at mid heightof the conical dome
Theory of StructuresII
= 2p �
2
)
4
5
(
?
� 0.25 � 2
2
� 25 = 499.74 kN
Subtotal = 2374.24 kN
Weight of water of rectangular crosssection + that of
?
ar
crosssection
= 10 � p � 4.5 � 7 � 1 +
2
10
� p � 4.5 �1 � 1 = 1060.287
kN
Total = 34340.53 kN
Load/m =
5
.
4
53
.
3434
?
?
= 242.94 kN/m
?
T = 242.94 cosec ? = 242.94 �
2
2= 343.52 kN/m.
Horizontal component of the thrust
H = T cos? = 343.52 �
1
2
2
1
= 242.94 kN/m
?
Meridional stress =
250
1000
10
52
.
343
3
?
?
= 1.37 N/mm < 6 N/mm
2
2
Vertical pr. on the slab per unit horizontal area (m )
2
= weight of water + weight of slab component
= 10 � 8 + 0.25 � 1 �
2
� 25 = 88.83 kN/m
2
on horizontal area.
Normal pressure on unit length of the conical length
=
83
.
88
2
12
1
?
?
= 44.41 kN/m
2
Hoop Tension =
2
pD
= 44.41 � 4 = 177.66 kN
A =
s115
10
66
.
177
3
?= 1544.86 mm /m
2
Hence, Provided
?
12 @ 140 mm c/c (A = 1614 mm ) in two
s
2
layers.
Hoop Force = 177.66� 10 =
3
ct
?
{bD + (m � 1) A }
s
or,
D =
1000
1614
)
1
11
(
3
.
1
10
66
.
177
3
?
?
?
?
?
?
?
?
?
?
?
?


?
= 120.52 < 250.
Hence, provided thickness of conical dome slab = 250 mm.
Nominal Reinforcement in Axial direction
For 250 thicknessp% =
)
100
250
(
)
100
450
(
)
2
.
0
3
.
0
(
3
.
0

?


??
211
Water Tanks
= 0.26%
bd
A
s
100
= 0.26
or
A =
s
650
100
250
1000
26
.
0
?
?
Hence, provided
?
10 @ 240 mm c/o in each face.
Design of Bottom Dome
Half chord length, i.e.r = 3 m
2
Rise of dome h = 1 m
2
Thickness of dome = 150mm
Radius of dome R =
b
)1
3
(
1
2
1
)
(
2
1
2
2
2
2
2
2
2
?
?
?
h
rh
= 5 m
Self wt. = (2p R h )t
b
2
2
?
c
= 2p � 5 � 1 � 0.15 � 25 = 117.81 kN
Wt. of water on the dome =
?
?
?
?
?
?
?
'
6
)
3
(
2
2
2
2
2
2
2
h
h
r
h
r
p
?
=10 p
?
?
?
?
?
?
?
?

?
6
1
)
1
3
3
(
9
3
2
2
2
= 2398.08 kN
Total wt. on dome, w
= 25
nextprevious15.89 kN
Semicentral angle ? = sin

1
5
3
= 36.87
o
w =
2
2
kN/m
08
.
80
1
5
2
89
.
2515
2
area
Surface
?
?
?
?
h
R
W
W
b
Meridional thrust, N =
???2
kN/m
44
.
222
)
87
.
36
cos
1
(
508
.
80
)
cos
1
(
?
?
?
?
b
wR
Meridional stress =
150
1000
10
44
.
222
3
?
?
?
bt
N
= 1.48 N/m < 6 N/mm2
2
Hoop compression = wR
?
?
?
�
�
�
?
?
?
?

?
cos
1
1
cos
= 80.08 � 5 �
??
�
�
?
?
?

87
.
36
cos
1
1
87
.
36
cos
= 97.87 kN/m
2Hoop stress =
2
2
3
N/mm
6
N/mm
652
.
0
150
1000
10
87
.
97
?
?
?
Nominal Reinforcement
For 150 thickness p% =
%286
.
0
)
100
150
(
)
100
450
(
)
2
.0
3
.
0
(
3
.
0

?



286
.
0
100
bd
A
s
212
or,
A =
s
429
100150
1000
286
.
0
?
?
Theory of StructuresII
Hence, provided
?
10 @ 180 mmc/c both ways.
Reinforcement detailing for the tank has been shown in Figure 10.12.
Design of Circular Shaft Staging
Diameter ofshaft = 6 m
Height of shaft above GL = 15 m
Thickness of shaft wall above GL = 150 mm
Depth of shaft below GL = 2 m
Thickness of shaft below GL = 400 mmFigure 10.12 : Reinforcement Detailing of Tank
Load acting on Shaft at GL
Load of top dome including imposed load = 353.04 kNLoad of top ring beam
= 50.315 kN
Load of tank
= 1265 kN
Weight of middle ring beam
= 206.12 kN
Loadof conical dome
= 499.74 kN
Load of bottom spherical dome= 117.81 kN
Load of bottom Ring Beam
= 2p � 3 � 0.5 � 0.5 � 25
= 117.80 kN
213 Water Tanks
Self wt. of shaft from bottom ring beam
to GL = p � 6 � 15 � 0.15 � 25
= 1060.29 kN
Subtotal = 3670.11kN
Self wt. of shaft from GL to top of footing
p � 6 � 2 � 0.4 � 25
= 376.99 kN
Subtotal
= 4047.10 kN
wt. of water fromcapacity calculation
= 6259.62 kN
Total = 10306.72 kN
Wind Load
Shape Factor = 0.7
Sl.
No.
Segment Wind
Pr
kN/m
2
Area
m
2
Distance
of CG
from
GL(m)
Moment
About
GL (kN)
Distance of
CG
about
Foundation
Level
Moment
about
Foundation
Level
1. Top dome 1.0 13.7 24.834
238.16
26.834
257.34
2.
Cylindrica
l
Portion of
tank
1.0 70 20.5
1004.5
22.5
1102.5
3. Trapezium 1.0
16 16.083
180.13
18.083
202.53
4. Supporting
Cyl.shaft
upto GL
1.0 90 7.5
472.5
9.5
598.5
?
MGL
=1895.29 kN
?
MFDN
= 2160.87
Eccentricity, e, of the Load When Tank is Empty
e =
11
.
3670
29
.
1895
?
GL
GL
W
M
= 0.506 m < ?
?
�
�
?
?
m
5.
1
4
6
Hence there will be compressive stress on the whole cross
section in all cases as eccentricity (e) has been calculated for the
lightest load.Stress at GL When Tank is Emptycv
?
at GL when tank empty =
?
?
�
�
?
?
??
r
e
rt
W
2
1
2=
?
?
�
�
?
?
??
?
?
?
3
516
.
02
1
150
.
3
2
11
.
3670
6
3
10
10
?= 1.74 N/mm < (0.38
2
?
25 = 9.5 N/mm )
2
Eccentricitye of the Load When Tank is Full
214
Theory of StructuresII
e =
)
62
.
625911
.
3670
(
29
.
1895
?
CL
GL
W
M
= 0.19
cv
?
at when tank full =
6
3
10
10
3
19
.
0
2
1
15
.
0
3
2
)
62
.
6259
11
.
3670
(
?
?
?
�
�
?
?
?
?
?
?
?
?
= 3.96 < 6 N/mm
2
When Tank Emptyat Foundation Level
e =
?
?
�
�
?
?
?
m
5.
1
4
6
m
53
.
01
.
4047
87
.
2160
cv
?
at FDN Level when tank empty
=
6
3
10
10
3
53
.
0
2
1
4
.
0
3
2
40471
?
?
?�
�
?
?
?
?
?
??
= 0.726 < (0.38 � 25 = 9.5 N/mm )
2
When tank full at foundation Level
e =
209
.
0
72
.
10306
87
.
2160cv
?
at foundation level when tank is full
2
6
3
N/mm
6
56
.
1
10
10)
3
209
.
0
2
1
(
4
.
0
3
2
72
.
10306
?
?
?
?
?
?
The reinforcement detailing will be done as per Section 10.3.
Design of Foundation (Figure 10.13)Total load on footing = 10306.72kN
Self wt 10% = 1030.67 kN
Total wt = 11337.39 kN
f
Bc
=
5
)
1
10
(
64
87
.
2160
)
1
10
(
4
39
.
11337
2
.
4
4
2
2
?

?
?

?
?
DI
M
A
W
x
x
= 145.81 +22.013
= 167.82 kN/m <200 kN/m .
2
2
Design of Cantilever Slab
Taking 1m strip
Effective Span = 2 m
215
Water Tanks
Figure 10.13 : Plan of Foundation
Estimate of Total Depth
From Deflection Control
4
3
2
1
k
k
k
k
k
d
l
B
ef
=
where k = 7, k
B
B
1
= 1k for M 20 and Fe 415 = 1
2k = k = 1
3
4
d
1
11
1
7
10
2
3
43
2
1
?
?
?
?
?
=
k
k
k
k
k
l
Bef
= 285.714
Taking
D = 600 mm; d = 600 � 50 �
2
20= 540 mm
From Moment of Resistance Consideration
Loads
Loads from reaction = 167.82 � 0.6 � 1 � 1 � 25
= 152.82 kN/m
2
Design load = w = 1.5 � 152.82 = 229.23 kN/m
u
2
DesignMoment = M =
u
2
223
.
229
2
2
2
?ef
u
l
W
= 458.46 kN m/mM
u, lim
= 458.46 � 106
= 0.36 � 0.48 (1 � 0.42 � 0.48) 1000 � d � 20
2
d = 407.61 mm < 540 mm
Hence, OK.A
st
M = 0.87 f A d
u
y st
?
?
?
�
�
�?
?

ck
y
st
bdf
f
A
1
458.46 � 10 = 0.87 � 415 � A � 540
6
st
??
�
�
?
?
?
?
?

20
540
1000
415
1
st
A
458.46 � 10 = 194967 A � 7.492 A
6
st
st
2
or,
A
st
2
� 26023.35 A +61193272.8 = 0
st
216
Theory of StructuresII
or,
A =
st
2
8
.
61193724
35
.
26023
(
35
.
26023
2
?

?
= 2614.06 mm
2
Provided 20
?
@ 120 mm c/c
?
?
�
�?
?
?
2617
120
314
1000
Check for shear at d from face of support
V = w l =229.23 � (2 � 0.54)
u
n
ef= 334.67 kN
=
v
???
5401000
10
67
.
334
3
?
?
bd
V
u
= 0.62N/mm
2
540
1000
2617100
100
?
?
bd
A
st= 0.484%
Corresponding M 20 and
c
?
= 0.62
%1
100
bd
A
st
or,
A =
st
100
540
1000
?
= 5400 mm /m
2
Hence, provided?
20 @ 55 mm (Figure 10.14).Figure 10.14 : Reinforcement Detailing of Foundation
[ Note : Same design will be applicable for inside portion of footing as the
pressure will reduce substantially.]
10.5 SUMMARY
The design and detailing of simpletypes of water tanks � circular water tanks
with flexible base (both under and over ground) and Intze type overhead water
tank with cylindrical staging and annular foundation � have been described.
Only direct stresses (i.e. compression or tension) are developed in their
components due to applied forces and, hence, principles and specifications of
analysis and design are very simple. To make the tanks leak proof, corrosion
resistant and durable, working stress method of design with higher grades of
concrete, reduced permissible stresses in steel,larger amount of temperature and
shrinkage reinforcementand thicker cover to reinforcement have been prescribed
by theCodes.
217
Water Tanks
10.6 ANSWERS TO SAQs
SAQ 1
(a)
Refer Section 10.1.
(b)
Refer Section 10.2.
(c)
Refer Section 10.2.
(d)
Refer Section 10.2.
(e)
Refer Section 10.2.4.
(f)
Refer Section 10.2.4.
SAQ 2
Refer Section 10.3.
FURTHER READING
BIS : 4562000, Code of Practice for Plain and Reinforced Concrete, Bureau of
Indian Standards, Manak Bhawan, Bahadur Shah Zafar Marg, New Delhi.
Ashok K. Jain, Reinforced Concrete Limit StateDesign, New Chand and
Brothers, Roorkee.
S. K. Mallick and A. P. Gupta, Reinforced Concrete, Oxford and IBH Publishing
Company Private Limited.
S. N. Sinha, Reinforced Concrete Design, Tata McGrawHill Publishing
Company Limited, Asaf Ali Road, Delhi.
S. U. Pillai and D. Menon, Reinforced Concrete Design, Tata McGrawHill
Publishing Company Limited, West Petal Nagar, New Delhi.
218
Theory of StructuresII
219 Water Tanks
THEORY OFSTRUCTURESII
Construction has been an activity which has witnessed many civilizations.
Different construction materialsand techniques have been tried in the past. In
modern times, forconstruction of any type of structure generally the choice for
material is confined to either concrete or steel. Concrete, though strong in
compression, is extremely weak in tension.Steel, on the other hand, is verystrong in tension as well as in compression. Therefore, the combination of
concrete and steel has proved to be most suitable choice to withstand stresses.
The widespread use of reinforced concrete in a variety of structural members in
different type of structures has compelled a proper understanding of the design
and detailing procedures. All reinforced concrete structures need proper designing
taking into account tensile and compressive stresses, shears, creep and thermal
effect, etc. This course, entitled �Theory of StructuresII�, covers the key aspects
of design and detailing of different reinforced concrete structures.
This coursecomprises ten units.
In Unit 1, you will be introduced to the limit state method of design of reinforced
concrete structures or their elements and limit state of flexural collapse.
Unit 2 deals with the method of design ofbeams and slabs for shear and torsion.
It also discussesconcrete reinforcement and detailing.
In Unit 3, the principles of design and detailing have been applied for the design
and detailing of simply supported rectangular beam, cantilever beam and simply
supported flanged beams.
In Unit 4, the designand detailing of cantilever slab, oneway simply supported
slabs, twoway simply supported slabs and twoway restrained slabsare
described.
Unit 5 introduces you to the design and detailing of axially loaded rectangular
and circular columns.
220
In Unit 6, design and detailing of strip footings, isolated footings and combined
footingsare discussed.
Theory of StructuresII
Unit 7 deals with the planning as well as structured design and detailing of flights
with or without stringer beams of rectangular staircases.
Unit 8 discusses the design and detailingof reinforced concrete cantilever type of
retailing wall.
In Unit 9, working stress method of design for structures and theirelements is
explained.
Finally, the design and detailing of circular water tank with flexible base(both
under and overground) and Intze type overhead water tank with cylindrical
stagingand annular foundation are described in Unit 10.
A number of SelfAssessment Questions (SAQs) are given in each unit to help
you to self monitoryour own progress. You are advised to study the text
carefully. Try to solve the SAQs on your own and verify youranswers with those
given atthe end of each unit. This will definitely develop your confidence.At the end, we wish you all the best for your all educational endeavours.
previous
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