Lecture 3
January 23, 2006
In this lecture


Modeling of tanks
Time period of tanks
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 2
Modeling of tanks

As seen in Lecture 1 liquid may be replaced by
impulsive and convective mass for calculation of
hydrodynamic forces

See next slide for a quick review
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 3
Modeling of tanks
mi = Impulsive liquid mass
Kc/2
mc
mc = Convective liquid mass
Kc/2
Kc = Convective spring stiffness
Rigid m
i
hi
(hi*)
hc
(hc*)
Mechanical analogue
or
spring mass model of tank
hi = Location of impulsive mass
(without considering overturnig
caused by base pressure)
hc = Location of convective mass
(without considering overturning
caused by base pressure)
hi* = Location of impulsive mass
(including base pressure effect on
overturning)
hc* = Location of convective mass
(including base pressure effect on
overturning)
Graphs and expression for these parameters are given in lecture 1.
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 4
Approximation in modeling

Sometimes, summation of mi and mc may not be
equal to total liquid mass, m


This difference may be about 2 to 3 %
Difference arises due to approximations in the
derivation of these expressions


More about it, later
If this difference is of concern, then


First, obtain mc from the graph or expression
Obtain mi = m – mc
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 5
Tanks of other shapes

For tank shapes such as Intze, funnel, etc. :

Consider equivalent circular tank of same
volume, with diameter equal to diameter at the
top level of liquid
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 6
Tanks of other shapes
Example:
An Intze container has volume of 1000 m3. Diameter of
container at top level of liquid is 16 m. Find dimensions of
equivalent circular container for computation of
hydrodynamic forces.
Equivalent circular container will have diameter of 16 m
and volume of 1000 m3. Height of liquid, h can be obtained
as :
/4 x 162 x h = 1000
 h = 1000 x 4/( x 162) = 4.97 m
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 7
Tanks of other shapes
Thus, for equivalent circular container,
h/D = 4.97/16 = 0.31
All the parameters (such as mi, mc etc.) are to be
obtained using h/D = 0.31
16 m
16 m
Intze container
volume = 1000 m3
 Sudhir K. Jain, IIT Kanpur
4.97 m
Equivalent circular container
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 8
Effect of obstructions inside tank

Container may have structural elements inside



For example: central shaft, columns supporting
the roof slab, and baffle walls
These elements cause obstruction to lateral
motion of liquid
This will affect impulsive and convective masses
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 9
Effect of obstructions inside tank

Effect of these obstructions on impulsive and
convective mass is not well studied


A good research topic !
It is clear that these elements will reduce
convective (or sloshing) mass

More liquid will act as impulsive mass
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 10
Effect of obstructions inside tank

In the absence of detailed analysis, following
approximation may be adopted:


Consider a circular or a rectangular container of
same height and without any internal elements
Equate the volume of this container to net
volume of original container


This will give diameter or lateral dimensions of container
Use this container to obtain h/D or h/L
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 11
Effect of obstructions inside tank
Example: A circular cylindrical container has internal
diameter of 12 m and liquid height of 4 m. At the center
of the tank there is a circular shaft of outer diameter of
2 m. Find the dimensions of equivalent circular
cylindrical tank.
12 m
4m
12 m
Elevation
 Sudhir K. Jain, IIT Kanpur
Hollow shaft of
2 m diameter
Plan
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 12
Effect of obstructions inside tank
Solution:
Net volume of container = /4x(122 –22)x4 = 439.8 m3
Equivalent cylinder will have liquid height of 4 m and its
volume has to be 439.8 m3.
Let D be the diameter of equivalent circular cylinder, then
/4xD2x4 = 439.8 m3
 D = 11.83 m
Thus, for equivalent circular tank, h = 4 m, D = 11.83m
and h/D = 4/11.83 = 0.34.
This h/D shall be used to find parameters of mechanical
model of tank
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 13
Effect of wall flexibility

Parameters mi, mc etc. are obtained assuming
tank wall to be rigid

An assumption in the original work of Housner
(1963a)



Housner, G. W., 1963a, “Dynamic analysis of fluids in
containers subjected to acceleration”, Nuclear Reactors and
Earthquakes, Report No. TID 7024, U. S. Atomic Energy
Commission, Washington D.C.
RC tank walls are quite rigid
Steel tank walls may be flexible

Particularly, in case of tall steel tanks
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 14
Effect of wall flexibility

Wall flexibility affects impulsive pressure distribution


It does not substantially affect convective pressure
distribution
Refer Veletsos, Haroun and Housner (1984)



Effect of wall flexibility on impulsive pressure depends
on



Veletsos, A. S., 1984, “Seismic response and design of liquid
storage tanks”, Guidelines for the seismic design of oil and gas
pipeline systems, Technical Council on Lifeline Earthquake
1Engineering, ASCE, N.Y., 255-370, 443-461.
Haroun, M. A. and Housner, G. W., 1984, “Seismic design of liquid
storage tanks”, Journal of Technical Councils of ASCE, Vol. 107,
TC1, 191-207.
Aspect ratio of tank
Ratio of wall thickness to diameter
See next slide
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 15
Effect of wall flexibility

Effect of wall flexibility on impulsive pressure distribution
h/D = 0.5
tw / D = 0.0005
tw / D = 0.0005
tw is wall thickness
z
h
Rigid
tank
Impulsive pressure on wall
 Sudhir K. Jain, IIT Kanpur
From Veletsos (1984)
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 16
Effect of wall flexibility



If wall flexibility is included, then mechanical
model of tank becomes more complicated
Moreover, its inclusion does not change seismic
forces appreciably
Thus, mechanical model based on rigid wall
assumption is considered adequate for design.
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 17
Effect of wall flexibility

All international codes use rigid wall model for
RC as well as steel tanks

Only exception is NZSEE recommendation
(Priestley et al., 1986)


Priestley, M J N, et al., 1986, “Seismic design of storage
tanks”, Recommendations of a study group of the New
Zealand National Society for Earthquake Engineering.
American Petroleum Institute (API) standards,
which are exclusively for steel tanks, also use
mechanical model based on rigid wall
assumption

API 650, 1998, “Welded storage tanks for oil storage”,
American Petroleum Institute, Washington D. C., USA.
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 18
Effect of higher modes

mi and mc described in Lecture 1, correspond
to first impulsive and convective modes




For most tanks ( 0.15 < h/D < 1.5) the first
impulsive and convective modes together
account for 85 to 98% of total liquid mass
Hence, higher modes are not included
This is also one of the reasons for summation of
mi and mc being not equal to total liquid mass
For more information refer Veletsos (1984) and
Malhotra (2000)

Malhotra, P. K., Wenk, T. and Wieland, M., 2000, “Simple
procedure for seismic analysis of liquid-storage tanks”,
Structural Engineering International, 197-201.
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 19
Modeling of ground supported tanks

Step 1:



Step 2:


Obtain various parameters of mechanical model
These include, mi, mc, Kc, hi, hc, hi* and hc*
Calculate mass of tank wall (mw), mass of roof
(mt) and mass of base slab (mb)of container
This completes modeling of ground supported
tanks
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 20
Modeling of elevated tanks

Elevated tank consists of container and staging
Roof slab
Wall
Container
Floor slab
Staging
Elevated tank
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 21
Modeling of elevated tanks

Liquid is replaced by impulsive and convective
masses, mi and mc


All other parameters such as hi, hc, etc, shall be
obtained as described earlier
Lateral stiffness, Ks, of staging must be
considered


This makes it a two-degree-of-freedom model
Also called two mass idealization
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 22
Modeling of elevated tanks
mc
Kc/2
hi
Kc/2
mc
Kc
hc
mi
mi + m s
hs
Spring mass model
 Sudhir K. Jain, IIT Kanpur
Ks
Two degree of freedom system
OR
Two mass idealization of elevated tanks
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 23
Modeling of elevated tanks

ms is structural mass, which comprises of :



Mass of container, and
One-third mass of staging
Mass of container includes



Mass of roof slab
Mass of wall
Mass of floor slab and beams
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 24
Two Degree of Freedom System

2-DoF system requires solution of a 2 × 2 eigen
value problem to obtain




Two natural time periods
Corresponding mode shapes
See any standard text book on structural
dynamics on how to solve 2-DoF system
For most elevated tanks, the two natural time
periods (T1 and T2) are well separated.

T1 generally may exceed 2.5 times T2.
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 25
Two Degree of Freedom System

Hence the 2-DoF system can be treated as two
uncoupled single degree of freedom systems


One representing mi +ms and Ks
Second representing mc and Kc
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 26
Modeling of elevated tanks
mc
Kc
mi + m s
mi + m s
Ks
mc
Kc
Ks
Two degree of freedom system
Two uncoupled
single degree of freedom systems
when T1 ≥ 2.5 T2
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 27
Modeling of elevated tanks


Priestley et al. (1986) suggested that this
approximation is reasonable if ratio of two time
periods exceeds 2.5
Important to note that this approximation is
done only for the purpose of calculating time
periods


This significantly simplifies time period calculation
Otherwise, one can obtain time periods of 2-DoF
system as per procedure of structural dynamics.
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 28
Modeling of elevated tanks


Steps in modeling of elevated tanks
Step 1:


Obtain parameters of mechanical analogue
These include mi, mc, Kc, hi, hc, hi* and hc*


Step 2:


Other tank shapes and obstructions inside the container shall
be handled as described earlier
Calculate mass of container and mass of staging
Step 3:

Obtain stiffness of staging
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 29
Modeling of elevated tanks

Recall, in IS 1893:1984, convective mass is not
considered



It assumes entire liquid will act as impulsive mass
Hence, elevated tank is modeled as single
degree of freedom ( SDoF) system
As against this, now, elevated tank is modeled
as 2-DoF system

This 2-DoF system can be treated as two
uncoupled SDoF systems
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 30
Modeling of elevated tanks

Models of elevated tanks
m +ms
mi + m s
Ks
mc
m = Total liquid mass
Ks
Kc
As per the Guideline
 Sudhir K. Jain, IIT Kanpur
As per IS 1893:1984
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 31
Modeling of elevated tanks
Example: An elevated tank with circular cylindrical
container has internal diameter of 11.3 m and water
height is 3 m. Container mass is 180 t and staging mass
is 100 t. Lateral stiffness of staging is 20,000 kN/m.
Model the tank using the Guideline and IS 1893:1984
Solution:
Internal diameter, D = 11.3 m, Water height, h = 3 m.
Container is circular cylinder,
 Volume of water = /4 x D2 x h
=  /4 x 11.32 x 3 = 300.9 m3.
 mass of water, m = 300.9 t.
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 32
Modeling of elevated tanks
h/D = 3/11.3 = 0.265
From Figure 2 of the Guideline, for h/D = 0.265:
mi/m = 0.31, mc/m = 0.65 and Kch/mg = 0.47
mi = 0.31 x m = 0.31 x 300.9 = 93.3 t
mc = 0.65 x m = 0.65 x 300.9 = 195.6 t
Kc = 0.47 x mg/h
= 0.47 x 300.9 x 9.81/3 = 462.5 kN/m
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 33
Modeling of elevated tanks
Mass of container = 180 t
Mass of staging = 100 t
Structural mass of tank, ms
= mass of container +1/3rd mass of staging
= 180 +1/3 x 100
= 213.3 t
Lateral stiffness of staging, Ks = 20,000 kN/m
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 34
Modeling of elevated tanks
m + ms
mi + m s
Ks
mc
Ks
Kc
mi = 93.3 t, ms = 213.3 t, mc = 195.6 t,
Ks = 20,000 kN/m, Kc = 462.5 kN/m
Model of tank as per the Guideline
 Sudhir K. Jain, IIT Kanpur
m = 300.9 t, ms = 213.3 t,
Ks = 20,000 kN/m
Model of tank as per IS 1893:1984
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 35
Time period


What is time period ?
For a single degree of freedom system, time
period (T ) is given by
T  2




M
K
M is mass and K is stiffness
T is in seconds
M should be in kg; K should be in Newton per
meter (N/m)
Else, M can be in Tonnes and K in kN/m
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 36
Time period


Mathematical model of tank comprises of
impulsive and convective components
Hence, time periods of impulsive and convective
mode are to be obtained
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 37
Time period of impulsive mode

Procedure to obtain time period of impulsive
mode (Ti) will be described for following three
cases:



Ground supported circular tanks
Ground supported rectangular tanks
Elevated tanks
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 38
Ti for ground-supported circular tanks


Ground supported circular tanks
Time period of impulsive mode, Ti is given by:
Ti  Ci
h ρ
t/D E

1
Ci  
2
 h/D 0.46  0.3h/D 0.067( h/D)







 = Mass density of liquid
E = Young’s modulus of tank material
t = Wall thickness
h = Height of liquid
D = Diameter of tank
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 39
Ti for ground-supported circular tanks

Ci can also be obtained from Figure 5 of the
Guidelines
10
8
C
6
C
i
4
C
c
2
0
0
 Sudhir K. Jain, IIT Kanpur
0.5
h/D
1
1.5
E-Course on Seismic Design of Tanks/ January 2006
2
Lecture 3 / Slide 40
Ti for ground-supported circular tanks

This formula is taken from Eurocode 8


Eurocode 8, 1998, “Design provisions for earthquake
resistance of structures, Part 1- General rules and Part 4 –
Silos, tanks and pipelines”, European Committee for
Standardization, Brussels.
If wall thickness varies with height, then
thickness at 1/3rd height from bottom shall be
used

Some steel tanks may have step variation of wall
thickness with height
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 41
Ti for ground-supported circular tanks


This formula is derived based on assumption
that wall mass is quite small compared to liquid
mass
More information on time period of circular
tanks may be seen in Veletsos (1984) and
Nachtigall et al. (2003)

Nachtigall, I., Gebbeken, N. and Urrutia-Galicia, J. L., 2003, “On the
analysis of vertical circular cylindrical tanks under earthquake
excitation at its base”, Engineering Structures, Vol. 25, 201-213.
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 42
Ti for ground-supported circular tanks

It is important to note that wall flexibility is
considered in this formula


For tanks with rigid wall, time period will be zero
This should not be confused with rigid wall
assumption in the derivation of mi and mc


Wall flexibility is neglected only in the evaluation
of impulsive and convective masses
However, wall flexibility is included while
calculating time period
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 43
Ti for ground-supported circular tanks

This formula is applicable to tanks with fixed
base condition


i.e., tank wall is rigidly connected or fixed to the
base slab
In some circular tanks, wall and base have
flexible connections
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 44
Ti for ground-supported circular tanks

Ground supported tanks with flexible base are
described in ACI 350.3 and AWWA D-110




ACI 350.3, 2001, “Seismic design of liquid containing
concrete structures”, American Concrete Institute,
Farmington Hill, MI, USA.
AWWA D-110, 1995, “Wire- and strand-wound circular,
prestressed concrete water tanks”, American Water Works
Association, Colorado, USA.
In these tanks, there is a flexible pad between
wall and base
Refer Figure 6 of the Guideline
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 45
Ti for ground-supported circular tanks
Types of connections between tank wall and base slab

Such tanks are perhaps not used in India
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 46
Ti for ground-supported circular tanks

Impulsive mode time period of ground
supported tanks with fixed base is generally
very low




These tanks are quite rigid
Ti will usually be less than 0.4 seconds
In this short period range, spectral acceleration,
Sa/g has constant value
See next slide
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 47
Ti for ground-supported circular tanks
Sa/g
Impulsive mode time period of ground supported
tanks likely to remain in this range
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 48
Ti for ground-supported circular tanks
Example: A ground supported steel tank has water
height, h = 25 m, internal diameter, D = 15 m and
wall thickness, t=15 mm. Find time period of
impulsive mode.
Solution: h = 25 m, D = 15 m, t = 15 mm.
For water, mass density,  = 1 t/m3.
For steel, Young’s modulus, E = 2x108 kN/m2.
h/D = 25/15 = 1.67. From Figure 5, Ci = 5.3
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 49
Ti for ground-supported circular tanks
Time period of impulsive mode, Ti  Ci
Ti  5.3
h ρ
t/D E
25 1.0
0.015/15 2x108
= 0.30 sec
 Important to note that, even for such a slender tank of
steel, time period is low.
 For RC tanks and other short tanks, time period will be
further less.
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 50
Ti for ground-supported circular tanks


In view of this, no point in putting too much
emphasis on evaluation of impulsive mode time
period for ground supported tanks
Recognizing this point, API standards have
suggested a constant value of spectral
acceleration for ground supported circular steel
tanks

Thus, users of API standards need not find
impulsive time period of ground supported tanks
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 51
Ti for ground-supported rectangular tanks


Ti for ground-supported rectangular tanks
Procedure to find time period of impulsive mode
is described in Clause no. 4.3.1.2 of the
Guidelines


Time period is likely to be very low and Sa/g will
remain constant


This will not be repeated here
As described earlier
Hence, not much emphasis on time period
evaluation
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 52
Ti for Elevated tanks


For elevated tanks, flexibility of staging is important
Time period of impulsive mode, Ti is given by:
mi  ms
Ti  2
Ks
OR
T  2

g
mi = Impulsive mass of liquid
ms = Mass of container and one-third mass of staging
Ks = Lateral stiffness of staging
= Horizontal deflection of center of gravity of tank when a
horizontal force equal to (mi + ms)g is applied at the
center of gravity of tank
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 53
Ti for Elevated tanks

These two formulae are one and the same


Expressed in terms of different quantities
Center of gravity of tank refers to combined
mass center of empty container plus impulsive
mass of liquid
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 54
Ti for Elevated tanks
Example: An elevated tank stores 250 t of water. Ratio of
water height to internal diameter of container is 0.5.
Container mass is 150 t and staging mass is 90 t.
Lateral stiffness of staging is 20,000 kN/m. Find time
period of impulsive mode
Solution: h/D = 0.5, Hence from Figure 2a of the
Guideline, mi/m = 0.54;
 mi = 0.54 x 250 = 135 t
Structural mass of tank, ms
= mass of container + 1/3rd mass of staging
= 150 + 90/3 = 180 t
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 55
Ti for Elevated tanks
Time period of impulsive mode Ti  2
Ti  2
mi  ms
Ks
135  180
20,000
= 0.79 sec.
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 56
Lateral stiffness of staging, Ks

Lateral stiffness of staging, Ks is force required
to be applied at CG of tank to cause a
corresponding unit horizontal deflection
CG
 Sudhir K. Jain, IIT Kanpur
P

Ks = P/ 
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 57
Lateral stiffness of staging, Ks

For frame type staging, lateral stiffness shall be
obtained by suitably modeling columns and
braces

More information can be seen in Sameer and
Jain (1992, 1994)



Sameer, S. U., and Jain, S. K., 1992, “Approximate methods
for determination of time period of water tank staging”,
The Indian Concrete Journal, Vol. 66, No. 12, 691-698.
Sameer, S. U., and Jain, S. K., 1994, “Lateral load analysis
of frame staging for elevated water tanks”, Journal of
Structural Engineering, ASCE, Vol.120, No.5, 1375-1393.
Some commonly used frame type staging
configurations are shown in next slide
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 58
Lateral stiffness of staging, Ks
Plan view of frame staging configurations
4 columns
6 columns
9 columns
 Sudhir K. Jain, IIT Kanpur
8 columns
12 columns
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 59
Lateral stiffness of staging, Ks
24 columns
 Sudhir K. Jain, IIT Kanpur
52 columns
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 60
Lateral stiffness of staging, Ks

Explanatory handbook, SP:22 has considered
braces as rigid beams






SP:22 – 1982, Explanatory Handbook on Codes for
Earthquake Engineering, Bureau of Indian Standards, New
Delhi
This is unrealistic modeling
Leads to lower time period
Hence, higher base shear coefficient
This is another limitation of IS 1893:1984
Using a standard structural analysis software,
staging can be modeled and analyzed to
estimate lateral stiffness
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 61
Lateral stiffness of staging, Ks


Shaft type staging can be treated as a vertical
cantilever fixed at base and free at top
If flexural behavior is dominant, then



Its stiffness will be Ks = 3EI/L3
This will be a good approximation if height to
diameter ratio is greater than two
Otherwise, shear deformations of shaft would
affect the stiffness and should be included.
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 62
Time period of convective mode


Convective mass is mc and stiffness is Kc
Time period of convective mode is:
m
T  2
K
c
c
c
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 63
Time period of convective mode

mc and Kc for circular and rectangular tanks can
be obtained from graphs or expressions


These are described in Lecture 1
Refer Figures 2 and 3 of the Guidelines
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 64
Time period of convective mode


For further simplification, expressions for mc
and Kc are substituted in the formula for Tc
Then one gets,
For circular tanks:
Tc  Cc D/g
Cc 
2
3.68 tanh(3.68h / D)
For rectangular tanks:
Tc  Cc L/g
 Sudhir K. Jain, IIT Kanpur
Cc 
2
3.16 tanh(3.16(h / L))
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 65
Time period of convective mode

Graphs for obtaining Cc are given in Figures 5
and 7 of the Guidelines




These are reproduced in next two slides
Convective mass and stiffness are not affected
by flexibility of base or staging
Hence, convective time period expressions are
common for ground supported as well as
elevated tanks
Convective mode time periods are usually very
large

Their values can be as high as 10 seconds
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 66
Time period of convective mode
10
8
6
C
C
i
4
C
c
2
0
0
0.5
h/D
1
1.5
2
Fig. 5 For circular tanks
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 67
Time period of convective mode
10
8
Cc
D
6
4
2
0
0.5
h/L
1
1.5
2
Fig. 7 For rectangular tanks
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 68
Time period of convective mode

Example: For a circular tank of internal diameter, 12 m
and liquid height of 4 m. Calculate time period of
convective mode.
Solution: h = 4 m, D = 12 m,
 h/D = 4/12 = 0.33
From Figure 5 of the Guidelines, Cc = 3.6
Time period of convective mode,
Tc  Cc D/g
Tc  3.6 12/9.81
= 3.98 sec
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 69
At the end of Lecture 3



Based on mechanical models, time period for
impulsive and convective modes can be
obtained for ground supported and elevated
tanks
For ground supported tanks, impulsive mode
time period is likely to be very less
Convective mode time period can be very large
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 3 / Slide 70