Transcript Tanks_lecture_3 - Civil Technocrats
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
K c /2 Rigid m c m i K c /2 h i (h i * ) h c (h c * )
m i = Impulsive liquid mass m c = Convective liquid mass K c = Convective spring stiffness h i = Location of impulsive mass (without considering overturnig caused by base pressure) h c = Location of convective mass (without considering overturning caused by base pressure) Mechanical analogue or spring mass model of tank h i * = Location of impulsive mass (including base pressure effect on overturning) h c * = 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 m i equal to total liquid mass, m and m c This difference may be about 2 to 3 % may not be Difference arises due to approximations in the derivation of these expressions More about it, later If this difference is of concern, then First, obtain m c from the graph or expression Obtain m i = m – m c 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 m hydrodynamic forces.
3 . Diameter of container at top level of liquid is 16 m. Find dimensions of equivalent circular container for computation of Equivalent circular container will have diameter of 16 m and volume of 1000 m 3.
Height of liquid, h can be obtained as : /4 x 16 2 x h = 1000 h = 1000 x 4/( x 16 2 ) = 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 m i , m c obtained using h/D = 0.31
etc.) are to be 16 m 16 m 4.97 m Equivalent circular container Intze container volume = 1000 m 3 Sudhir K. Jain, IIT Kanpur 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 4 m 12 m Elevation Hollow shaft of 2 m diameter Plan Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 12
Effect of obstructions inside tank Solution: Net volume of container = /4x(12 2 –2 2 )x4 = 439.8 m 3 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 /4xD 2 x4 = 439.8 m 3 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 m i , m c 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) 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.
Effect of wall flexibility on impulsive pressure depends on 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 t w h/D = 0.5
/ D = 0.0005
t w / D = 0.0005
t w is wall thickness
z h
Rigid tank Impulsive pressure on wall From Veletsos (1984) Sudhir K. Jain, IIT Kanpur 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
m i and m c 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 m i and m c 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: Obtain various parameters of mechanical model These include, m i , m c , K c , h i , h c , h i * Step 2: and h c * Calculate mass of tank wall (m w ), mass of roof (m t ) and mass of base slab (m b )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
Sudhir K. Jain, IIT Kanpur
Elevated tank
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, m i and m c All other parameters such as h i , h c , etc, shall be obtained as described earlier Lateral stiffness, K s , 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
h i K c /2
m c
m i K c /2 h c m c K c m i + m s h s K s
Spring mass model Two degree of freedom system OR Two mass idealization of elevated tanks Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 23
Modeling of elevated tanks m s 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 (T 1 and T 2 ) are well separated.
T 1 generally may exceed 2.5 times T 2.
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 m i +m s Second representing m c and K s and K c Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 26
Modeling of elevated tanks
m c K c m i + m s m i + m s K s K s m c K c
Two degree of freedom system Sudhir K. Jain, IIT Kanpur Two uncoupled single degree of freedom systems when T 1 ≥ 2.5 T 2 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 m i , m c , K c , h i , h c , h i * and h c * Other tank shapes and obstructions inside the container shall be handled as described earlier Step 2: 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 i + m s m +m s m = Total liquid mass K s m c K c
As per the Guideline
K s
As per IS 1893:1984 Sudhir K. Jain, IIT Kanpur 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 D 2 x h = /4 x 11.3
2 x 3 = 300.9 m 3 . 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: m i /m = 0.31, m c /m = 0.65 and K c h/mg = 0.47
m i m c K c = 0.31 x m = 0.31 x 300.9 = 93.3 t = 0.65 x m = 0.65 x 300.9 = 195.6 t = 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, m s = mass of container +1/3 rd mass of staging = 180 +1/3 x 100 = 213.3 t Lateral stiffness of staging, K s = 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 i + m s m + m s K s m c K c K s
m i K s = 93.3 t, m s = 213.3 t, m c = 20,000 kN/m, K c = 195.6 t, = 462.5 kN/m Model of tank as per the Guideline m = 300.9 t, m s K s = 213.3 t, = 20,000 kN/m Model of tank as per IS 1893:1984 Sudhir K. Jain, IIT Kanpur 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 (T i ) 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
T i for ground-supported circular tanks Ground supported circular tanks Time period of impulsive mode, T i is given by:
T i
C i h t/D ρ E C i
h/D
0.46
1
0.3h/D
0.067(h/D) 2
= 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
T i for ground-supported circular tanks C i can also be obtained from Figure 5 of the Guidelines 10 8 6 4 2 C i C c 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
T i 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 1/3 rd 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
T i 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
T i 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 m i and m c 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
T i 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
T i 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
T i 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
T i 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 T i will usually be less than 0.4 seconds In this short period range, spectral acceleration, S a /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
T i for ground-supported circular tanks 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
T i 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/m 3 .
For steel, Young’s modulus, E = 2x10 8 kN/m 2 .
h/D = 25/15 = 1.67. From Figure 5, C i = 5.3
Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 49
T i for ground-supported circular tanks Time period of impulsive mode,
T i
C i h t/D ρ E T i
5.3
25 0.015/15
= 0.30 sec
1.0
2x10 8
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
T i 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
T i for ground-supported
rectangular
tanks
T i 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 This will not be repeated here Time period is likely to be very low and S a /g will remain constant 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
T
i
for Elevated tanks
For elevated tanks, flexibility of staging is important Time period of impulsive mode, T i is given by:
T i
2
m i
m s K s
OR
T
2
g
m i m s = Impulsive mass of liquid = Mass of container and one-third mass of staging K s = Lateral stiffness of staging = Horizontal deflection of center of gravity of tank when a horizontal force equal to (m i center of gravity of tank + m s )g is applied at the Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 53
T i 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
T i 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, m i /m = 0.54; m i = 0.54 x 250 = 135 t Structural mass of tank, m s = mass of container + 1/3 rd 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
T i for Elevated tanks Time period of impulsive mode
T i
2
m i
m s K s
T i 2 135 180 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, K
s Lateral stiffness of staging, K s is force required to be applied at CG of tank to cause a corresponding unit horizontal deflection
CG P
K s = P/
Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 57
Lateral stiffness of staging, K s 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, K s Plan view of frame staging configurations 4 columns 6 columns 8 columns 9 columns Sudhir K. Jain, IIT Kanpur 12 columns E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 59
Lateral stiffness of staging, K s 24 columns 52 columns Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 60
Lateral stiffness of staging, K s 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, K s 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 K s = 3EI/L 3 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 m c and stiffness is K c Time period of convective mode is:
T c
2
m c K c
Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 63
Time period of convective mode m c and K c 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 m c and K c are substituted in the formula for T c Then one gets, For circular tanks:
T c
C c D/g C c
2 3 .
68
tanh
( 3 .
68
h
/
D
) For rectangular tanks:
T c
C c L/g C c
2 3 .
16
tanh
( 3 .
16 (
h
/
L
)) Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 65
Time period of convective mode Graphs for obtaining C c and 7 of the Guidelines are given in Figures 5 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 i 4 C c 2 0 0 0.5
h/D
1 Fig. 5 For circular tanks 1.5
Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 2 Lecture 3 / Slide 67
Time period of convective mode 10 8 6 4 2 0 0.5
h/L
1 Fig. 7 For rectangular tanks 1.5
Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 2 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, C c = 3.6
Time period of convective mode,
T
c T c
C c D/g
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