Download - IS 1893 - TANKS & WATER RETAINING
Lecture 1
January 17, 2006
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 2
In this lecture
Types of tanksIS codes on tanksModeling of liquid
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 3
Types of tanks
Two categoriesGround supported tanks
Also called at-grade tanks; Ground Service Reservoirs (GSR)
Elevated tanks Also called overhead tanks; Elevated Service Reservoirs (ESR)
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 4
Types of tanks
Ground supported tanksShape: Circular or Rectangular Material : RC, Prestressed Concrete, Steel
These are ground supported vertical tanksHorizontal tanks are not considered in this course
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 5
Types of tanks
Elevated tanks
Two parts:Container Staging (Supporting tower)
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 6
Types of tanks
Elevated tanks
Container:Material: RC, Steel, PolymerShape : Circular, Rectangular, Intze, Funnel, etc.
Staging:RC or Steel frame RC shaftBrick or masonry shafts
Railways often use elevated tanks with steel frame staging Now-a-days, tanks on brick or stone masonry shafts are not constructed
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 7
Use of tanks
Water distribution systems use ground supported and elevated tanks of RC & steel
Petrochemical industries use ground supported steel tanks
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 8
Indian Codes on Tanks
IS 3370:1965/1967 (Parts I to IV)For concrete (reinforced and prestressed) tanksGives design forces for container due to hydrostatic loadsBased on working stress designBIS is considering its revision
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 9
Indian Codes on Tanks
IS 11682:1985For RC staging of overhead tanksGives guidelines for layout & analysis of stagingMore about this code later
IS 803:1976For circular steel oil storage tanks
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 10
Indian Codes on Tanks
IS 1893:1984Gives seismic design provisionsCovers elevated tanks onlyIs under revisionMore about other limitations, later
IS 1893 (Part 1):2002 is for buildings onlyCan not be used for tanks
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 11
Hydrodynamic Pressure
Under static condition, liquid applies pressure on container.
This is hydrostatic pressureDuring base excitation, liquid exerts additional pressure on wall and base.
This is hydrodynamic pressureThis is in additional to the hydrostatic pressure
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 12
Hydrodynamic pressure
Hydrostatic pressure Varies linearly with depth of liquidActs normal to the surface of the containerAt depth h from liquid top, hydrostatic pressure = γh
Hydrostatic pressure
h
γh
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 13
Hydrodynamic pressure
Hydrodynamic pressureHas curvilinear variation along wall heightIts direction is opposite to base motion
Base motion
Hydrodynamic pressure
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 14
Hydrodynamic pressure
Summation of pressure along entire wall surface gives total force caused by liquid pressure
Net hydrostatic force on container wall is zeroNet hydrodynamic force is not zero
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 15
Hydrodynamic pressure
Hydrostatic pressure Hydrodynamic pressure
Base motion
Circular tanks (Plan View)
Net resultant force = zero Net resultant force ≠ zero
Note:- Hydrostatic pressure is axisymmetric; hydrodynamic is asymmetric
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 16
Hydrodynamic pressure
Hydrostatic pressure Hydrodynamic pressure
Base motion
Net resultant force = zero Net resultant force ≠ zero
Rectangular tanks (Plan View)
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 17
Hydrodynamic pressure
Static design: Hydrostatic pressure is consideredHydrostatic pressure induces hoop forces and bending moments in wallIS 3370 gives design forces for circular and rectangular tanksNet hydrostatic force is zero on container wallHence, causes no overturning moment on foundation or stagingThus, hydrostatic pressure affects container design only and not the staging or the foundation
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 18
Hydrodynamic pressure
Seismic design: Hydrodynamic pressure is considered
Net hydrodynamic force on the container is not zeroAffects design of container, staging and foundation
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 19
Hydrodynamic pressure
Procedure for hydrodynamic pressure & force:Very simple and elegantBased on classical 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.
We need not go in all the detailsOnly basics and procedural aspects are explained in next few slides
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 20
Modeling of liquid
Liquid in bottom portion of the container moves with wall
This is called impulsive liquidLiquid in top portion undergoes sloshing and moves relative to wall
This is called convective liquid or sloshing liquid
Convective liquid(moves relative to tank wall)
Impulsive liquid(moves with tank wall)
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 21
Modeling of liquid
Impulsive liquidMoves with wall; rigidly attached Has same acceleration as wall
Convective liquidAlso called sloshing liquidMoves relative to wallHas different acceleration than wall
Impulsive & convective liquid exert pressure on wall
Nature of pressure is differentSee next slide
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 22
Modeling of liquid
Impulsive Convective
Hydrodynamic pressure
Base motionBase motion
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 23
Modeling of liquid
At this point, we will not go into details of hydrodynamic pressure distribution
Rather, we will first find hydrodynamic forcesImpulsive force is summation of impulsive pressure on entire wall surfaceSimilarly, convective force is summation of convective pressure on entire wall surface
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 24
Modeling of liquid
Total liquid mass, m, gets divided into two parts:
Impulsive liquid mass, mi
Convective liquid mass, mc
Impulsive force = mi x accelerationConvective force = mc x acceleration
mi & mc experience different accelerationsValue of accelerations will be discussed laterFirst we will find mi and mc
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 25
Modeling of liquid
Housner suggested graphs for mi and mcmi and mc depend on aspect ratio of tanks
Such graphs are available for circular & rectangular tanks
See Fig. 2a and 3a of GuidelinesAlso see next slide
For taller tanks (h/D or h/L higher), mi as fraction of m is moreFor short tanks, mc as fraction of m is more
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 26
Modeling of liquid
0
0.5
1
0 0.5 1 1.5 2h/D
0
0.5
1
0 0.5 1 1.5 2h/L
For circular tanks
mi/m
mc /m
mi/m
mc /m
For rectangular tanks
See next slide for definition of h, D, and L
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 27
Modeling of liquid
D
L
Base motion
L
Base motion
ElevationPlan of Circular tank
Plan of Rectangular tank
h
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 28
Modeling of liquid
Example 1:A circular tank with internal diameter of 8 m, stores 3 m height
of water. Find impulsive and convective water mass.Solution:
Total volume of liquid = π/4 x 82 x 3 = 150.8 m3
∴ Total liquid mass, m = 150.8 x 1.0 = 150.8 t
Note:- mass density of water is 1000 kg/m3; weight density of water is 9.81 x 1000 = 9810 N/m3.
D = 8 m, h = 3 m ∴ h/D = 3/8 = 0.375.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 29
0
0.5
1
0 0.5 1 1.5 2h/D
mi/m
mc /m
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 30
Modeling of liquid
From graph, for h/D = 0.375 mi/m = 0.42 and mc/m = 0.56
mi = 0.42 x 150.8 = 63.3 t and mc = 0.56 x 150.8 = 84.5 t
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 31
Modeling of liquid
Impulsive liquid is rigidly attached to wallConvective liquid moves relative to wall
As if, attached to wall with springs
Rigid
mc
Kc/2Kc/2
mi
Convective liquid(moves relative to wall)
Impulsive liquid(moves with wall)
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 32
Modeling of liquid
Stiffness associated with convective mass, KcKc depends on aspect ratio of tankCan be obtained from graph
Refer Fig. 2a, 3a of guidelinesSee next slide
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 33
Modeling of liquid
mi/m
mc/m
Kch/mg
0
0.5
1
0 0.5 1 1.5 2h/D
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 34
Modeling of liquid
Example 2:A circular tank with internal diameter of 8 m, stores 3 m height of water. Find Kc.Solution:
Total liquid mass, m = 150.8 t (from Example 1)= 150.8 x 1000 = 150800 kg
g = acceleration due to gravity = 9.81 m/sec2
D = 8 m, h = 3m ∴ h/D = 3/8 = 0.375. From graph, for h/D = 0.375;
Kc h/mg = 0.65
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 35
Modeling of liquid
Kc = 0.65 mg/h∴ Kc = 0.65 x150800 x 9.81/3.0 = 320,525.4 N/m
Note: - Unit of m is kg, hence unit of Kc is N/m. If we take m in ton, then unit of Kc will be kN/m.
mi/m
mc/m
Kch/mg
0
0.5
1
0 0.5 1 1.5 2h/D
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 36
Modeling of liquid
Now, we know liquid masses mi and mc
Next, we need to know where these are attached with the wall
Like floor mass in building acts at centre of gravity (or mass center) of floorLocation of mi and mc is needed to obtain overturning effects
Impulsive mass acts at centroid of impulsive pressure diagram
Similarly, convective mass
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 37
Modeling of liquid
Impulsive mass acts at centroid of impulsive pressure diagramLocation of centroid:
Obtained by dividing the moment due to pressure distribution by the magnitude of impulsive force
Similarly, location of convective mass is obtainedSee next slide
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 38
Modeling of liquid
hi
hi, hc can be obtained from graphsThey also depend on aspect ratio, h/D or h/LRefer Fig. 2b, 3b of guidelinesSee next slide
hc
Resultant of impulsive pressure on wall
Resultant of convective pressure on wall
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 39
Modeling of liquid
0
0.2
0.4
0.6
0.8
1
0 0.5 1 1.5 2h/L
hc/h
hi/h
For circular tanks For rectangular tanks
0
0.2
0.4
0.6
0.8
1
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2h/D
hc/h
hi/h
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 40
Modeling of liquid
Example 3:A circular tank with internal diameter of 8 m, stores 3 m
height of water. Find hi and hc.Solution:
D = 8 m, h = 3m ∴ h/D = 3/8 = 0.375.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 41
0
0.2
0.4
0.6
0.8
1
0 0.5 1 1.5 2h/D
hc/h
hi/h
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 42
Modeling of liquid
From graph, for h/D = 0.375; hi/h = 0.375 hi = 0.375 x 3 = 1.125 mand hc/h = 0.55 hc = 0.55 x 3 = 1.65 m
Note :- Since convective pressure is more in top portion, hc > hi.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 43
Modeling of liquid
Hydrodynamic pressure also acts on baseUnder static condition, base is subjected to uniformly distributed pressureDue to base motion, liquid exerts nonuniformpressure on base
This is in addition to the hydrostatic pressure on the base
See next slide
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 44
Modeling of liquid
Hydrostatic pressure on base
Base motion
Hydrodynamic pressure on base
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 45
Modeling of liquid
Impulsive as well as convective liquid cause nonuniform pressure on base
Nonuniform pressure on base causes overturning effectThis will be in addition to overturning effect of hydrodynamic pressure on wallSee next slide
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 46
Modeling of liquid
Overturning effect due to wall pressure
Overturning effect due to base pressure
hi
Note:- Both the overturning effects are in the same direction
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 47
Modeling of liquid
Total overturning effect of wall and base pressure is obtained by applying resultant of wall pressure at height, hi
* and hc*.
• In place of hi and hc discussed earlierFor overturning effect due to wall pressure alone, resultant was applied at hi
For hi and hi*, see next slide
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 48
Modeling of liquid
hi
h*i
Location of resultant of wall pressure when effect of base pressure is not included
Location of Resultant of wall pressure when effect of base pressure is also included
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 49
Modeling of liquid
Similarly, hc and hc* are defined
hc
h*c
Location of resultant of wall pressure when effect of base pressure is not included Location of Resultant of wall
pressure when effect of base pressure is also included
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 50
Modeling of liquid
hi and hi* are such that
Moment due to impulsive pressure on walls only = Impulsive force x hi
Moment due to impulsive pressure on walls and base = Impulsive force x hi*
hc and hc* are such that
Moment due to convective pressure on walls only = Convective force x hc
Moment due to convective pressure on walls and base = Convective force x hc*
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 51
Modeling of liquid
hi* is greater than hi
hc* is greater than hcRefer Fig. C-1 of the Guidelines
hi* & hc
* depend on aspect ratioGraphs to obtain hi, hc, hi
*, hc* are provided
Refer Fig. 2b & 3b of guidelinesAlso see next slidePlease note, hi
* and hc* can be greater than h
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 52
Modeling of liquid
hi/h
hi*/hhc/h
hc*/h
0
0.5
1
1.5
2
2.5
0 0.5 1 1.5 2h/D
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 53
Modeling of liquid
Example 4:A circular tank with internal diameter of 8 m, stores 3 m height of
water. Find hi* and hc
*.Solution:D = 8 m, h = 3m ∴ h/D = 3/8 = 0.375. From graph, for h/D = 0.375;
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 54
Modeling of liquid
hi*/h = 1.1 Hence hi* = 1.1 x 3 = 3.3 m Similarly, hc*/h = 1.0 Hence, hc* = 1.0 x 3 = 3.0 m
hi/h
hi*/hhc/h
hc*/h
0
0.5
1
1.5
2
2.5
0 0.5 1 1.5 2h/D
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 55
Modeling of liquid
This completes modeling of liquidLiquid is replaced by two masses, mi & mc
This is called mechanical analogue or spring mass model for tankSee next slide
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 56
Modeling of liquid
Rigid
mc
Kc/2Kc/2
mihi
(hi*)
hc(hc
*)
mi = Impulsive liquid mass
mc = Convective liquid mass
Kc = Convective spring stiffness
hi = Location of impulsive mass (without considering overturning 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)
Mechanical analogueor
spring mass model of tank
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 57
Modeling of liquid
mi, mc, Kc, hi, hc, hi* and hc
* can also be obtained from mathematical expressions:
These are given in Table C 1 of GuidelinesThese are reproduced in next two slides
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 58
hD
hD
mmi
866.0
866.0tanh ⎟⎠⎞
⎜⎝⎛
=
⎟⎠⎞
⎜⎝⎛=
Dh
hmgKc 68.3tanh836.0 2
Dh
Dh
mmc
⎟⎠⎞
⎜⎝⎛
=68.3tanh
23.0
375.0=hh i
Dh /09375.05.0 −=
for 75.0/ ≤Dh
75.0/ >Dhfor ⎟⎠⎞
⎜⎝⎛
−⎟⎠⎞
⎜⎝⎛
−=
Dh
Dh
Dh
hhc
68.3sinh68.3
0.168.3cosh1
For circular tanks
1250 -86602
8660.
hD.tanh
hD.
h*hi
⎟⎠⎞
⎜⎝⎛
= for 33.1/ ≤Dh
45.0= 33.1/ >Dhfor
⎟⎠⎞
⎜⎝⎛
−⎟⎠⎞
⎜⎝⎛
−=
Dh
Dh
Dh
hhc
68.3sinh68.3
01.268.3cosh1
*
Modeling of liquid
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 59
For rectangular tanks
125.0866.0tanh2
866.0*−
⎟⎠⎞
⎜⎝⎛
=
hL
hL
hhi
for 33.1/ ≤Lh
45.0= 33.1/ >Lhfor
⎟⎠⎞
⎜⎝⎛
−⎟⎠⎞
⎜⎝⎛
−=
Lh
Lh
Lh
hhc
16.3sinh16.3
01.216.3cosh1
*
375.0=h
h i for 75.0/ ≤Lh
Lh /09375.05.0 −= 75.0/ >Lhfor
⎟⎠⎞
⎜⎝⎛
−⎟⎠⎞
⎜⎝⎛
−=
Lh
Lh
Lh
hhc
16.3sinh16.3
0.116.3cosh1
⎟⎠⎞
⎜⎝⎛=
Lh
hmgKc 16.3tanh833.0 2
hL
hL
mmi
866.0
866.0tanh ⎟⎠⎞
⎜⎝⎛
=Lh
Lh
mmc
⎟⎠⎞
⎜⎝⎛
=16.3tanh
264.0
Modeling of liquid
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 60
Modeling of liquid
Note, in Table C-1 of the Guideline, there are two typographical errors in these expressions
For circular tank, first expression for hi/h shall have limit as “for h/D ≤ 0.75”For circular tank, in the expression for hi
*/h, there shall be minus sign before 0.125These two errors have been corrected in the expressions given in previous two slides
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 61
Modeling of liquid
mi and mc are needed to find impulsive and convective forces
Impulsive force, Vi = mi x accelerationConvective force, Vc = mc x acceleration
Rigid
mc
Kc/2Kc/2
mi Vi
Vc
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 62
Modeling of liquid
Vi and Vc will cause Bending Moment (BM) in wall
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 63
Modeling of liquid
BM at bottom of wallBM due to Vi = Vi x hi
BM due to Vc = Vc x hc
Total BM is not necessarily Vi X hi+ Vc X hcMore about this, later
Rigid
mc
Kc/2Kc/2
mi Vi
Vc
hi
hc
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 64
Modeling of liquid
Overturning of the container is due to pressure on wall and base
Pressure on base does not cause BM in wallOverturning Moment (OM) at tank bottom
OM is at bottom of base slabHence, includes effect of pressure on baseNote the difference between bottom of wall and bottom of base slab
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 65
Modeling of liquid
OM at bottom of base slabOM due to Vi = Vi x hi
*
BM due to Vc = Vc x hc*
Vi
Vc
Rigid
mc
Kc/2Kc/2
mi
hi*
hc*
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 66
Modeling of liquid
mi and mc will have different accelerationsWe yet do not know these accelerations
ai = acceleration of mi
ac = acceleration of mc
Procedure to find acceleration, later
Use of mi, mc, hi, hc, hi* and hc
* in next exampleAcceleration values are assumed
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 67
Modeling of liquid
Example 5:A circular tank with internal diameter of 8 m, stores 3 m height of
water. Assuming impulsive mass acceleration of 0.3g and convective mass acceleration of 0.1g, find seismic forces on tank. Solution:Geometry of tank is same as in previous examples.D = 8 m, h = 3mFrom previous examples:mi = 63.3 t mc = 84.5 thi = 1.125 m hc = 1.65 mhi
* = 3.3 m hc* = 3.0 m
Impulsive acceleration, ai = 0.3g = 0.3 x 9.81 = 2.94 m/sec2
Convective acceleration, ac = 0.1g = 0.1 x 9.81 = 0.98 m/sec2
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 68
Modeling of liquid
Example 5 (Contd..)
Impulsive force, Vi = mi x ai = 63.3 x 2.94 = 186.1 kNConvective force, Vc = mc x ac = 84.5 x 0.98 = 82.8 kN
Bending moment at bottom of wall due to Vi = Vi x hi
= 186.1 x 1.125 = 209.4 kN-mBending moment at bottom of wall due to Vc = Vc x hc
= 82.8 x 1.65 = 136.6 kN-mOverturning moment at bottom of base due to Vi = Vi x hi
*
= 186.1 x 3.3 = 614.1 kN-mOverturning moment at bottom of base due to Vc = Vc x hc
*
= 82.8 x 3.0 = 248.4 kN-m
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 69
At the end of Lecture 1
In seismic design, mechanical analogue of tanks are used, wherein, liquid is replaced by impulsive & convective massesThese masses and their points of application depend on aspect ratioGraphs and expressions are available to find all these quantities
These are based on work of Housner (1963a)
E-Course (through Distance Learning Mode) on
Seismic Design of Liquid Storage Tanks January 16 - February 6, 2006
Response to Questions and Comments on Lecture 1
Question from Mr. S. K. Kukreja ([email protected]) 1. clarify following
1. fig 2a,3a of guidelines as pointed in lecture1/slide25 2. table C-1 of guidelines as pointed in lecture1/slide57 3. Housner graphs to find h* are not given for rectangular tanks
In Figure 2a of the Guideline, mi/m, mc/m and Kch/mg are plotted as a function of h/D. This is for circular tanks and corresponding figure for rectangular tanks is given in Figure 3a. Expressions for these quantities are given in Table C-1 of the Guideline. Basically, graphs plotted in these figures are based on these expressions only. Depending on convenience, one can find these quantities either from graphs or from expressions. In the Lecture we have given graphs of hi* and hc* only for circular tanks (Slide 54), which is Figure 2b of the Guideline. Corresponding figure for rectangular tank is Figure 3b of the Guideline, which has not been reproduced in the Lecture. Expressions for hi* and hc* for circular as well as rectangular tanks are also given in Table C-1.
Question from Mr. Umesh H. Patil ([email protected])
1. pl.something about hydrostatic design
Under static conditions, liquids apply a pressure on the bodies in contact. This pressure is a product of unit weight of liquid and depth of the liquid at the point under consideration. Direction of this pressure is always normal (perpendicular) to the surface of the body. This is known as hydrostatic pressure, and it varies linearly with height.
Hydrostatic pressure induces hoop forces and bending of the wall. IS 3370:1967 gives these forces for circular and straight walls. Refer standard textbooks for more information.
2. show to calculate hydrodynamic pr.parameters for tapering sections of
wall/container
Lecture 3 has covered information on tanks of other shapes such as Intze, conical, etc. If cross-section of wall is tapered, i.e., its thickness varies along the height, then there will not be any change in hydrodynamic pressure. Recall, hydrodynamic pressure is obtained considering wall as rigid.
3. effect of top covered slab on hydrodynmic pressure
Top slab will impart more stiffness to wall and make it more rigid. This is not going to affect hydrodynamic pressure distribution. However, if sloshing liquid touches roof slab, then hydrodynamic pressure distribution will be affected. Please refer Lecture 6 and Commentary to Clause 4.11 of the Guideline.
4. there is difference in answers from graph & equations
Reading the value from graph is likely to induce some approximation and hence, expressions are also given. The difference, however, may not be significant from engineering view point.
Question from Mr. H. I. Abdul Gani ([email protected]) The following are my questions on Lecture 1. 1) What exactly does Horizontal Tank and Vertical Tank mean
A circular tank kept horizontally on ground, i.e., longitudinal axis of tank is horizontal is termed as horizontal tank. In petrol pumps, steel tanks used for storing petrol/diesel are usually horizontal tanks. Railway tankers carrying petroleum products are also horizontal tanks. Another good example is heating greasers in bathrooms, some are of horizontal type and some are of vertical type !!. A vertical tank is the one, in which longitudinal axis of tank is vertical. Classification as horizontal and vertical tank is more appropriate for circular tanks only.
2) It was shown that the hydrodynamic pressure is curvilinear in shape. Does the pressure profile depend on the stiffness of the tank walls?
3) How the stiffness of the wall affects i) Convective mass and ii) impulsive mass
These have been addressed in Lecture 3
4) In example 1, why the sum of impulsive and convective masses (147=2E8kg)
is less than the total mass of water (150.8 kg)?
This also has been addressed in Lecture 3
5) In the hydrodynamic pressure profile in the base, it is not clear how pressure
can develop in the upward direction? Can you please explain
Under static condition, liquid exerts uniform pressure on base. During lateral excitation, the liquid increases vertical pressure on some portion of slab and reduces pressure on other portion. This increase or decrease in pressure is the hydrodynamic effect. The decrease in the pressure is like an upward pressure.
6) How to find height of sloshing liquid? Does it depend on acceleration
This is addressed in Lecture 6
7) For tanks with arbitrary shapes like triangular, hexagonal shapes etc, how to find impulsive and convective masses? Any simplified methods available Please refer Lecture 3
8) During vertical acceleration, how to find impulsive and convective mass?
Information on effect of vertical excitation is covered in Lecture 6. Also read Clause 4.10 of the Guideline. Vertical excitation will cause change in weight density of liquid, which in turn will change hydrostatic pressure on wall.
9) Why effect of base pressure is considered along with both convective mass
and impulsive mass instead of convective mass only?
In our modeling, liquid is divided into two parts, one that vibrates along with wall (impulsive mass) and other which moves relative to wall (convective mass). During lateral excitation, both impulsive and convective liquids exert non-uniform pressure on base.
10) How to check adequacy of freeboard in a water tank during dynamic
excitation? This issue is addressed in Lecture 6
11) For lumping mass in mathematical model for impulsive and convective mass which height is to be used, h or h*? If bending moment at the bottom of wall is to be obtained, then, masses shall be lumped at h. If overturning moment at the bottom of base slab is to
obtained, then masses shall be lumped at h*. For design of staging and foundation also, mass must be lumped at h*.
12) For load application in a mathematical model, only C.G. (h) of the impulsive/
convective force is known. How to find the area and distribution of load (or mass) with respect to height of the tank?
Details about distribution of impulsive and convective pressure along wall are covered in Lecture 6.
Question from ([email protected]) I didnt understood the following pl explain mi/m=3D tanh(.866*12/5)/.866*12/5 now in this expression mi/m=3Dtanh 2.078/2.078 mi/m=3Dtan 5*2.078/2.078 is it finding tan 10.39 and then dividing it by 2.078 all the other thing i understood and also since i have referred charts from the guidlines i also got the answers correct Abut only i didnt understood the formulae part how the value arrived thanking you chandrshekhar
The expression which you have given above contains 3D term, which is not present in the expressions given in Lectures and in Table C-1 of the Guideline. You are also interested in knowing how this formula is arrived at. The derivation of this formula is given in Housner’s paper (1963a) and you may go through the same. It comes from potential equation of liquid or Laplace equation. Since it is a bit mathematical, we have not included this in the course contents.
Question from Mr. R. Murugan ([email protected])
1. How convective pressure is distributed and upto what depth ?
We need to recognize that we are dividing liquid in two parts. That does not mean that physically liquid gets divided in two parts and liquid only up to certain depth participates in convective mode and liquid below it participates in impulsive mode. It is an idealization, and convective and impulsive pressures are present over the entire height of liquid. This can also be observed in the pressure distributions shown in Lecture 6.
2. In slide 54 How hi* is higher than the tank height?
hi* is that height where mi should act so that moment due to impulsive pressure on wall and moment due to impulsive pressure on base is equal to
moment due to mi (actually, mi × g) . Since, effect of moment due to base pressure is included, this height can become larger than h. On the other hand hi, which includes effect of impulsive pressure only on the walls, is always less than h.
3. The formulae derived for the above tanks are applicable for open tank or closed tank ? (i.e. tank with cover slab or without cover slab)
Actually these are derived for open tanks, wherein, question of liquid touching roof slab does not arises. However, due to presence of roof slab, pressure distribution on wall is not going to change, provided sloshing liquid does not touch roof slab. If it touches roof slab then some additional issues will come up. These issues have been addressed in Lecture 6 and in Clause 4.11 of the Guideline.
4. The formulas and graph derived to find hi, hc, mi, mc etc is applicable for RC tank or steel tank? The formulae are irrespective of materials or the materials proportion will change the co-efficient?
This issue has been addressed in Lecture 3. Effect of flexibility of wall on pressure distribution has been studied by Veletsos (1984). However, for design purpose we use formulae derived for rigid walls. Please refer Lecture 3 and commentary of Clause 4.2.1.2
5. What is mean by squat tank, slender tank?
Slender tank is one in which h/D or h/L is quite high. In squat tank h/D or h/L is less. There is no fixed limit on these values at which this demarcation begins. Generally, these terminologies are used for impulsive and convective masses. For example, we say that in slender tanks, convective mass is less and in squat tanks, impulsive mass is less. In this context, from Figure 2a of the Guideline, we can see that for h/D = 0.2, about 80% mass contributes to convective mode and for h/D = 1.0 about 80% mass contributes to impulsive mode. Thus, one can say that a tank with h/D less than 0.2 is squat tank and a tank with h/D greater than 1.0 is slender tank.
6. How Question 1.2 is false in Assignment i.e.., in short tank mi is less only See Example-1 in Lecture -1.
This was error in our solution, and we have sent addendum to it.
7. So, far we learnt about impulsive mass, convective mass and it is centeroid. When we do the modeling in computer. How impulsive mass is distributed i.e how applied and upto what depth?
Distribution of impulsive and convective pressure along wall height is covered in Lecture 6.
8. Why mi + mc is not coming to m ? In solution 1, mi/m = 0.466; mc/m = 0.503; when both are added the value is 0.969; Nearly 3% of the liquid mass is unaccounted. Why is it so? Is it not correct to include the missing mass in impulsive liquid component?
This has been discussed in Lecture 3.
9. Table C1-Expression for parameters of spring mass model, hi/h = 0.375 for h/D less than or equal to 0.75 is to be given, while the actual one furnished reads, h/d greater than 0.75
This is a typographical error in Table C-1, which we have pointed out in the Lecture 1. There is another typographical error in the expression for hi* for circular tanks. This also has been pointed out in Lecture 1 (Slide no. 60).
Question from Mr. Abdul Gani ([email protected])
It is explained in lecture 1, impulsive pressure distribution as below: Impulsive Let us assume mi/m =x: If the pressure distribution is as shown in the above figure, hi should come less than or equal to 0.5*x* h, i.e., hi/h should always be lower than 0.5x. Table below compares the hi expected versus actual computed
Problem no. mi/m hi/h expected hi/h computed
1.1 Circular tank 0.466 Less than or equal to 0.233 0.375
1.2 Rectangular tank in longer direction
0.54 Less than or equal to 0.54/2 0.27
0.375
1.2 Rectangular tank in shorter direction
0.72 Less than or equal to 0.72/2 0.36
0.4
From above computations, it appears that the pressure distribution for impulsive loading explained in lecture 1 appears to be incorrect. Kindly explain.
This question arose because of the feeling that physically liquid got divided into two parts. You are thinking that liquid below height is impulsive and liquid above this height is convective. As has been explained in the answer to earlier question, there is no physical line, which demarcates impulsive and convective portions of liquid. Impulsive liquid is present throughout the height; however, more liquid from bottom portion participates in impulsive mode. Similarly, convective liquid is also distributed along the entire height, but more liquid from upper portion participates in convective mode. Since more liquid from bottom portion is participating in impulsive mode, h
( ) ( )hx ×
i/h is always less than 0.5h. Similarly, hc/h is always greater than 0.5. Please refer Figure 2b and 3b of the Guideline.
Lecture 2
January 19, 2006
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 2
In this lecture
Seismic force evaluationProcedure in codesLimitations of IS 1893:1984
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 3
Seismic force evaluation
During base excitationStructure is subjected to acceleration
From Newton’s second lawForce = mass x acceleration
Hence, seismic force acting on structure = Mass x acceleration
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 4
Seismic force evaluation
For design, we need maximum seismic force Hence, maximum acceleration is required
This refers to maximum acceleration of structureThis is different from maximum acceleration of groundMaximum ground acceleration is termed as peak ground acceleration, PGAMaximum acceleration of rigid structure is same as PGA.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 5
Seismic force evaluation
Seismic force = mass x maximum accelerationCan be written as:
Force = (maximum acceleration/g) x (mass x g)= (maximum acceleration/g) x W
W is weight of the structureg is acceleration due to gravity
Typically, codes express design seismic force as:V = (Ah) x (W)
V is design seismic force, also called design base shearAh is base shear coefficient
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 6
Seismic force evaluation
Maximum acceleration of structure depends on Severity of ground motionSoil conditionsStructural characteristics
These include time period and dampingMore about time period, later
Obviously, base shear coefficient, Ah, will also depend on these parameters
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 7
Seismic force evaluation
Seismic design philosophy is such that, design seismic forces are much lower than actual seismic forces acting on the structure during severe ground shaking
Base shear coefficient has to ensure this reduction in forces
Hence, base shear coefficient would also have a parameter associated with design philosophy
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 8
Seismic force evaluation
Thus, base shear coefficient depends on:Severity of ground motionSoil conditionStructural characteristicsDesign philosophy
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 9
Seismic force evaluation
Let us examine how following codes have included these parameters in base shear coefficient
IS 1893 (Part 1): 2002IS 1893:1984International Building code (IBC) 2003 from USA
Study of IBC provisions will help us understand the present international practice
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 10
IS 1893 (Part 1):2002
Ah = (Z/2). (I/R). (Sa/g)Z is zone factorI is importance factorR is response reduction factorSa/g is spectral acceleration coefficient
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 11
IS 1893 (Part 1):2002
Zone factor, ZDepends on severity of ground motionIndia is divided into four seismic zones (II to V)Refer Table 2 of IS 1893(part1):2002Z = 0.1 for zone II and Z = 0.36 for zone V
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 12
IS 1893 (Part 1):2002
Importance factor, IEnsures higher design seismic force for more important structuresValues for buildings are given in Table 6 of IS :1893
Values for other structures will be given in respective partsFor tanks, values will be given in Part 2
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 13
IS 1893 (Part 1):2002
Response reduction factor, REarthquake resistant structures are designed for much smaller seismic forces than actual seismic forces that may act on them. This depends on
DuctilityRedundancyOverstrength
See next slide
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 14
Design force
MaximumLoad Capacity
Tota
l Hor
izon
tal L
oad
Roof Displacement (Δ)
Non linear Response
First Significant
Yield
Linear Elastic Response
Δmax
Fy
Fs
Fdes
ΔyΔw
Fel
Load at First Yield
Due to Overstrength
Due to Redundancy
Due to Ductility
Maximum force if structure remains elastic
0
)(F Force Design)(F Force Elastic MaximumFactor Reduction Response
des
el=
Total Horizontal
Load
Δ
Figure: Courtesy Dr. C V R Murty
IS 1893 (Part 1):2002
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 15
IS 1893 (Part 1):2002
Response reduction factor (contd..)A structure with good ductility, redundancy and overstrength is designed for smaller seismic force and has higher value of R
For example, building with SMRF has good ductility and has R = 5.0 as against R = 1.5 for unreinforced masonry building which does not have good ductility
Table 7 gives R values for buildingsFor tanks, R values will be given in IS:1893 (Part 2)
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 16
IS 1893 (Part 1):2002
Spectral acceleration coefficient, Sa/g Depends on structural characteristics and soil condition
Structural characteristics include time period and damping
Refer Fig. 2 and Table 3 of IS:1893See next slide
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 17
IS 1893 (Part 1):2002
For 5% damping
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 18
IS 1893 (Part 1):2002
For other damping, Sa/g values are to be multiplied by a factor given in Table 3 of IS:1893
Table 3 is reproduced below
% damping
0 2 5 7 10 15 20 25 30
0.55 0.500.60Factor 3.20 1.40 1.00 0.90 0.80 0.70
For higher damping, multiplying factor is lessHence, for higher damping, Sa/g is less
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 19
IS 1893:1984
Let us now look at the provision of IS 1893:1984IS 1893:1984 suggests two methods for calculating seismic forces
Seismic coefficient method (SCM)Response spectrum method (RSM)
These have different expressions for base shear coefficient
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 20
IS 1893:1984
Ah= KCβIαo Seismic Coefficient Method (SCM)= KβIFoSa/g Response Spectrum Method (RSM)
K is performance factorC is a coefficient which depends on time periodβ is soil-foundation system coefficientI is importance factorαo is seismic coefficientFo is zone factorSa/g is average acceleration coefficient
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 21
IS 1893:1984
Seismic coefficient, αo Depends on severity of ground motionUsed in seismic coefficient method
Zone factor, FoDepends on severity of ground motionUsed in response spectrum method
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 22
IS 1893:1984
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 23
IS 1893:1984
β is soil foundation coefficientDepends on type of soil and foundation
In IS 1893:2002, type of foundation does not have any influence on base shear coefficient
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 24
IS 1893:1984
Importance factor, IEnsures higher design seismic force for more important structures
IS 1893 (Part 1):2002, gives values only for buildings
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 25
IS 1893:1984
Performance factor, KDepends on ductility of structure
Similar to response reduction factor of IS1893(Part 1):2002K is in numerator whereas, R is in denominator
For buildings with good ductility, K = 1.0For ordinary buildings, K = 1.6Thus, a building with good ductility will have lower value of base shear coefficient than ordinary building
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 26
IS 1893:1984
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 27
IS 1893:1984
Coefficient, CDepends on time period see next slide
Spectral acceleration, Sa/gDepends on time period and dampingSee next slide
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 28
IS 1893:1984
Graphs for C and Sa/g from IS 1893:1984
Natural Period (Sec) Natural Period (Sec)
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 29
IS 1893:1984
IS 1893:1984 has provisions for elevated tanks only
Ground supported tanks are not consideredFor elevated tanks, it suggests
Ah = βIFoSa/gPerformance factor, K is not presentImplies, K = 1.0 for all types of elevated tanks
Unlike buildings, different types of tanks do not have different values of K
This is one of the major limitation of IS1893:1984More about it, later
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 30
IBC 2003
International Building Code (IBC) 2003In IITK-GSDMA guidelines IBC 2000 is referred This is now upgraded to IBC 2003
In USA codes are regularly upgraded every three year
There is no change in the base shear coefficient expression from IBC 2000 to IBC 2003
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 31
IBC 2003
Base shear coefficientAh = SD1 I/(R T)
≤ SDS I/RAh shall not be less than 0.044 SDSI for buildings and not less than 0.14 SDSI for tanks
This is a lower limit on AhIt ensures minimum design seismic forceThis lower limit is higher for tanks than for buildings
Variation with time period is directly given in base shear coefficient
Hence, no need to have response spectrum separately
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 32
IBC 2003
T is time period in secondsSDS and SD1 are design spectral accelerations in short period and at 1 sec. respectively
SDS and SD1 depend on seismic zone and soilI is importance factor and R is response modification factor
IBC suggests I = 1.0, 1.25 and 1.5 for different types of structuresValues of R will be discussed later
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 33
IBC 2003
More about SDS and SD1SDS = 2/3 Fa SS and SD1 = 2/3 Fv S1
SS is mapped spectral acceleration for short periodS1 is mapped spectral acceleration for 1-second periodSS and S1 are obtained from seismic map
This is similar to zone map of our code It is given in contour form
Fa and Fv are site coefficientsTheir values for are given for different soil types
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 34
IBC 2003
Response modification factor, RIS 1893(Part 1):2002 calls it response reduction factorValues of R for some selected structures are given in next slide
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 35
IBC 2003
Type of structureBuilding with special reinforced concrete moment resisting concrete frames
R
8.0
Building with intermediate reinforced concrete moment resisting concrete frames 5.0
Building with ordinary reinforced concrete moment resisting concrete frames 3.0
Building with special steel concentrically braced frames 8.0
Elevated tanks supported on braced/unbraced legs 3.0
Elevated tanks supported on single pedestal 2.0
Tanks supported on structural towers similar to buildings 3.0
Flat bottom ground supported anchored steel tanks 3.0
Flat bottom ground supported unanchored steel tanks 2.5
Ground supported reinforced or prestressed concrete tanks with reinforced nonsliding base 2.0
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 36
Base shear coefficient
In summary,Base shear coefficient from these three codes are:
IS 1893 (Part 1): 2002 IS 1893: 1984 IBC2003
Ah = (Z/2).(I/R).(Sa/g) SCM: Ah = KCβIαo
RSM: Ah = KβIFoSa/g
For tanks:Ah = βIFoSa/g
Ah = SD1 I/(R T) ≤ SDS I/R
> 0.044 SDS I for buildings> 0.14 SDS I for tanks
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 37
Base shear coefficient
Important to note that:IS codes specify base shear coefficient at working stress level
For limit state design, these are to be multiplied by load factors to get factored loads
IBC specifies base shear coefficient at ultimate load level
For working stress design, seismic forces are divided by a factor of 1.4
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 38
Base shear coefficient
Once, base shear coefficient is known, seismic force on the structure can be obtained
Recall, seismic force, V = Ah. WThis is same as force = mass x acceleration
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 39
Base shear coefficient
Let us compare base shear coefficient values from these codes
Comparison will be done at working stress levelIBC values are divided by 1.4 to bring them to working stress level
This shall be done for similar seismic zone or seismic hazard level of each codeThis comparison is first done for buildings
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 40
Base shear coefficient
Comparison for buildingsFollowing parameters are chosen
IS 1893 (Part 1): 2002 IS 1893: 1984 IBC2003
Z = 0.36; Zone VI = 1.0; R = 5.0Soft soil5% damping
αo = 0.08; Fo = 0.4; Zone Vβ = 1.0K = 1.0; I = 1.0Soft soil, raft foundation5% damping
SDs = 1.0; SD1 = 0.6 I = 1.0; R = 8.0Soil type D, equivalent to soft soil of IS codes5% damping
They represent similar seismic hazard level
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 41
Base shear coefficient
Building with good ductility is chosenSay, buildings with SMRF
In IBC, for buildings with SMRF, R = 8.0Refer Table shown earlier
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 42
Base shear coefficient
For building with T = 0.3 secIS 1893(Part 1):2002
Sa/g =2.5Ah = Z/2.I/R.Sa/g = (0.36/2 )x (1.0/5.0) x 2.5 = 0.09
IS 1893:1984C = 1.0 and Sa/g = 0.2SCM: Ah = KCβIαo = 1.0 x 1.0 x1.0 x1.0x0.08 = 0.08RSM: Ah = KβIFoSa/g = 1.0 x 1.0 x1.0x 0.4x0.2 = 0.08
IBC 2003Ah = SDSI/(1.4xR) = 1.0 x1.0/(1.4 x 8.0) = 0.089
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 43
Base shear coefficient
For building with T = 1 secIS 1893(Part 1):2002
Sa/g = 1.67Ah = Z/2.I/R.Sa/g = (0.36/2 )x(1.0/5.0)x1.67 = 0.06
IS 1893:1984C = 0.53 and Sa/g = 0.11SCM: Ah = KCβIαo = 1.0 x0.53x1.0 x1.0x0.08 = 0.042RSM: Ah = KβIFoSa/g = 1.0x1.0x1.0x0.4x0.11 = 0.044
IBC 2003Ah = SD1I/(1.4xRxT) = 0.6x1.0/(1.4 x 8.0x1.0) = 0.054
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 44
Base shear coefficient
For building with T = 1.5 secIS 1893(Part 1):2002
Sa/g = 1.11Ah = Z/2.I/R.Sa/g = (0.36/2 )x(1.0/5.0)x1.11 = 0.040
IS 1893:1984C = 0.4 and Sa/g = 0.078SCM: Ah = KCβIαo = 1.0 x0.4x1.0 x1.0x0.08 = 0.032RSM: Ah = KβIFoSa/g =1.0x1.0x1.0x0.4x0.078 = 0.031
IBC 2003Ah = SD1I/(1.4RT) = 0.6x1.0/(1.4 x 8.0x1.5) = 0.036
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 45
Base shear coefficient
Base shear coefficients for four time periods
IS 1893: 1984T (Sec)
IS 1893 (Part 1):
2002 SCM RSM
1.0 0.06 0.042 0.044 0.054
1.5 0.040 0.032 0.031 0.036
2.0 0.03 0.024 0.024 0.0314*
0.09
IBC2003
0.3 0.08 0.08 0.089
* Due to lower bound, this value is higher
Graphical comparison on next slide
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 46
Base shear coefficient
Comparison of base shear coefficient (Buildings)
Note the lowerbound of IBC
0
0.025
0.05
0.075
0.1
0 0.5 1 1.5 2 2.5 3
Time Period (S)
Bas
e sh
ear c
oeffi
cien
t
IS 1893(Part 1):2002
IS 1893:1984; SCMIS 1893:1984; RSM
IBC 2003
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 47
Base shear coefficient
We have seen that:Codes follow similar strategy to obtain design base shear coefficient In similar seismic zones, base shear coefficient for buildings matches reasonably well from these three codes
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 48
Base shear coefficient
Similarly, let us compare design base shear coefficients for tanks
From IS1893:1984 and IBC 2003IS 1893(Part 1):2002 is only for buildings
Hence, can’t be used for tanks
Only elevated tanks will be consideredIS 1893:1984 has provisions for elevated tanks only
Zone and soil parameters will remain same as those considered for buildingsImportance factor for tanks are different than those for buildings
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 49
Base shear coefficient
In IBCI = 1.25 for tanksR = 3.0 for tanks on frame staging (braced legs)R = 2.0 for tanks on shaft or pedestal
In 1893:1984I = 1.5 for tanksK is not present in the expression for base shear coefficient (implies k=1.0). Hence, base shear coefficient will be same for all types of elevated tanks
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 50
Base shear coefficient
For tank with T = 0.3 secIS 1893:1984
I = 1.5, Sa/g = 0.2Ah = βIFoSa/g = 1.0 x 1.5 x 0.4x0.2 = 0.12This value is common for frame and shaft staging
IBC 2003For frame staging, I = 1.25, R = 3.0Ah = SDSI/(1.4xR) = 1.0 x1.25/(1.4 x 3.0) = 0.298For shaft staging, I = 1.25, R = 2.0Ah = SDSI/(1.4xR) = 1.0 x1.25/(1.4 x 2.0) = 0.446
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 51
Base shear coefficient
For tank with T = 1 secIS 1893:1984
I = 1.5, Sa/g = 0.11Ah = βIFoSa/g =1.0x1.5x0.4x0.11 = 0.066
IBC 2003For frame staging, I = 1.25, R = 3.0Ah = SD1I/(1.4xRxT) = 0.6x1.25/(1.4 x 3.0x1.0) = 0.178For shaft staging, I = 1.25, R = 2.0Ah = SD1I/(1.4xRxT) = 0.6x1.25/(1.4 x 2.0x1.0) = 0.268
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 52
Base shear coefficient
Base shear coefficients for tanks
IBC 2003T (Sec)
IS 1893:1984*
Frame staging Shaft staging
1.0 0.066 0.178 0.268
0.120.3 0.298 0.446
* Base shear coefficient values are common for frame and shaft staging
Graphical comparison on next slide
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 53
Base shear coefficient
0
0.1
0.2
0.3
0.4
0.5
0 0.5 1 1.5 2 2.5 3Time period (S)
Ba
se s
he
ar
coe
ffic
ien
t
Comparison of base shear coefficient (Tanks)
IS 1893:1984; All types of elevated tanks
IBC 2003; Tanks on shaft staging
IBC 2003; Tanks on frame staging
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 54
Base shear coefficient
Base shear coefficient for elevated tanks from IS1893:1984 is on much lower side than IBC 2003IBC value is about 2.5 times for frame staging and 3.5 times for shaft staging than that from IS1893:1984
Recall, for buildings, IS 1893:1984 and IBC have much better comparison
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 55
Base shear coefficient
Reason for lower values in IS 1893:1984IBC uses R = 2.0 and R = 3.0 for tanks as against R = 8.0 for buildings with good ductilityIS 1893:1984 uses K = 1.0 for tanks. Same as for buildings with good ductility.Clearly ,elevated tanks do not have same ductility, redundancy and overstrength as buildings.
This is a major limitation of IS 1893:1984
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 56
Base shear coefficient
Another limitation of IS 1893:1984In Lecture 1, we have seen, liquid mass gets divided into impulsive and convective massesIS 1893:1984, does not consider convective massIt assumes entire liquid mass will act as impulsive mass, rigidly attached to wall
In IITK-GSDMA guidelines, these limitations have been removed
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 57
Base shear coefficient
Let us now, get back to seismic force evaluation for tanksDesign base shear coefficient is to be expressed in terms of parameters of IS 1893(Part 1):2002
Ah = (Z/2). (I/R). Sa/gZ will be governed by seismic zone map of Part 1I and R for tanks will be different from those for buildings
R depends on ductility, redundancy and overstrength
Sa/g will depend on time period
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 58
Base shear coefficient
Impulsive and convective masses will have different time periods
Hence, will have different Sa/g valuesProcedure for finding time period in next lecture
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 59
At the end of Lecture 2
Seismic force = (Ah) X (W)Base shear coefficient, Ah, depends on
Seismic ZoneSoil typeStructural characteristicsDuctility, Redundancy and overstrength
IS 1893:1984 has some serious limitations in design seismic force for tanks
E-Course (through Distance Learning Mode) on
Sesimic Design of Liquid Storage Tanks January 16 - February 6, 2006
Response to Questions and Comments on Lecture 2
Question from Mr. Abdul Gani ([email protected]) The following are my questions on Lecture 2. 1. What exactly is meant by ‘Overstrength’?
Overstrength means additional strength which structure possess over and above the design strength. This additional strength comes from factors on gravity loads, live loads and earthquake loads. For limit state design of RC structures, we use partial safety factor of 1.5 on gravity loads. Partial safety factors on materials also provide overstrength. Moreover, materials may have strength greater than their characteristic strength, Non-structural elements and special ductile detailing also constitute another source of overstrength. Please note, in design we do not include strength provided by these sources, however, actually it is present in the structure.
2. As per fig C-4a of the Guidelines, the base shear coefficients from IS 1893 (Part 1):2003 is much less than that from IBC 2000 for structures with time period less than 0.1s and greater than 1.7s. How to account for this effect while designing a structure which falls in this time period zone?
Figure C-4a gives comparison of base shear coefficient for buildings from IS 1893(Part 1):2002 and IBC 2000. In IBC 2000, spectrum is flat (or horizontal) from T = 0.0 sec itself, whereas, in IS 1893, there is a rising portion from T = 0.0 up to T = 0.1 sec. A very stiff or rigid structure will have very low time period, which may fall in this range. For such rigid structures, Sa/g and base shear coefficient will be less. In such rigid structures, reduction in seismic forces on account of ductility is not allowed. Thus, strictly speaking, use of R for these structures is not appropriate. In order to address this issue, codes make spectrum flat in short period range and allow the use of R values. IS 1893(Part 1) should in fact remove this rising portion. More discussion on this aspect is available in the following document: Proposed draft provisions and commentary on IS 1893(Part 1):2002 (www.iitk.ac.in/nicee/IITK-GSDMA/EQ05.pdf)
Please note, for tanks, this rising portion has been removed in the IITK-GSDMA Guidelines.
For time period greater than 1.7 sec also there is difference between IBC and IS 1893(Part 1). This is due to the fact that IBC has given a lower limit on its base shear coefficient, which means to say that, beyond certain time period, design seismic force will remain same. This is done to ensure certain minimum strength against lateral loads. Moreover, if time period is incorrectly estimated on higher side, then also this lower limit acts as a safeguard. IS 1893 does not have such lower limit and for tanks also in the IITK-GSDMA Guideline, as of now there is no such lower limit.
3. Why response reduction factor is different for convective and impulsive masses?
In IITK-GSDMA Guideline, response reduction factor is same for impulsive and convective modes.
However, in some international codes (like, ACI 350.3 and AWWA D-115), response reduction factors are different for impulsive and convective modes. These codes argue that convective mode is low frequency mode (large time period) and hence, convective forces would vary slowly with time. That means, convective forces are like static forces and hence, reduction due to ductility shall not be applicable to them. ACI 350.3 and AWWA D-115 use R = 1.0 for convective mode, i.e. they do not allow any reduction in convective seismic forces. However, it is important to note that, these codes use different response spectrum for convective and impulsive modes. For convective mode, which is usually having very large time period, response spectrum has 1/Tc
2 variation in long period range. On the other hand, IS 1893 spectrum has 1/T variation for all time periods.
At present, in IITK-GSDMA Guideline, R is kept same for impulsive and convective modes and same spectrum is used for impulsive and convective modes. This makes calculations simple and does not cause significant error.
4. How to find base shear coefficient based on IS 1893:2003 for two identical
buildings designed for 50 and 100 years respectively.
At present, as per IS 1893:2002, base shear coefficient will be same for both the buildings. Please note, if design forces have to depend on expected life of building, then zone map or zone factor or PGA will have to be arrived at using probabilistic seismic hazard analysis (Refer “Geotechnical Earthquake Engineering” by Kramer S L; Pearson Education, first Indian reprint, 2003). In probabilistic analysis, PGA is specified with certain probability of exceedance in given number of years. For example, IBC specifies maximum considered earthquake with 2% probability of being exceeded in 50 years (2500 year return period). Indian
codes have not yet adopted such an approach for quantifying seismic hazards.
5. Why IS code specifies base shear coefficients at working stress level, while
IBC specifies at Ultimate stress level?
In India, working stress method is still quite common. Hence, IS 1893 has given design spectrum at working stress level. However, it does not really matter. We use load factors to arrive at loads to be used for limit state design. Users of IBC use a factor of 0.7 to arrive at forces for working stress design.
6. In some of the international codes (eg. New Zealand and French codes), the
spectral amplification factor for hard rock is higher followed by medium and soft soils. Why is it so? But in IS 1893:2003, the spectral amplifications are same for both rock and soil sites. Can please explain why?
In hard soil, high frequency (low time period) waves get amplified and in soft soil, low frequency (long time period) waves get amplified. Thus, in low time period range, spectral amplification factor will be higher for hard soil. Similarly, in long time period range, spectral amplification for soft soil will be higher. The spectral amplification factors are obtained from recorded data of ground motion in various soil conditions. As of now, IS 1893 (Part 1) has kept same spectral amplification in short period range for all types of soils. With the availability of more reliable data, spectra in short period range may be modified for different soil conditions.
7. In some of the standards (eg. TecDoc 1347) provision is made to account for
local site effects (Site amplification etc). In IS 1893:2002 why no such provision is made?
Site effects in IS 1893 are included by specifying different design spectra for rock, medium and soft soils, especially accounting for the higher ground velocities observed in soft soils. Other topographic effects, such as hill slopes, ridges valleys etc. are currently not included in the design force calculations, as available information is not sufficient for reliable estimation of their effects.
8. In IBC, what is the time period at which SDS is specified?
SDS is specified in short period range and there is no specific value of time period at which it is specified. One may assume that this period will be less than approximately 0.1 sec. Unlike this, SD1 is specified at 1.0 sec time period.
Lecture 3
January 23, 2006
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 2
In this lecture
Modeling of tanksTime period of tanks
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 3
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 4
Modeling of tanks
Rigid
mc
Kc/2Kc/2
mihi
(hi*)
hc(hc
*)
mi = Impulsive liquid mass
mc = Convective liquid mass
Kc = Convective spring stiffness
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)
Mechanical analogueor
spring mass model of tank
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 5
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, thenFirst, obtain mc from the graph or expressionObtain mi = m – mc
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 6
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 7
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 :
π/4 x 162 x h = 1000∴ h = 1000 x 4/(π x 162) = 4.97 m
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 8
Tanks of other shapes
Thus, for equivalent circular container, h/D = 4.97/16 = 0.31All the parameters (such as mi, mc etc.) are to be
obtained using h/D = 0.31
16 m
Intze containervolume = 1000 m3
16 m 4.97 m
Equivalent circular container
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 9
Effect of obstructions inside tank
Container may have structural elements insideFor example: central shaft, columns supporting the roof slab, and baffle walls
These elements cause obstruction to lateral motion of liquidThis will affect impulsive and convective masses
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 10
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) massMore liquid will act as impulsive mass
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 11
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 elementsEquate 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 12
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
Hollow shaft of 2 m diameter
12 m
ElevationPlan
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 13
Effect of obstructions inside tank
Solution: Net volume of container = π/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, thenπ/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 14
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 rigidSteel 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 15
Effect of wall flexibility
Wall flexibility affects impulsive pressure distributionIt does not substantially affect convective pressure distributionRefer 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 gaspipeline 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 16
Effect of wall flexibility
Effect of wall flexibility on impulsive pressure distribution
From Veletsos (1984)
hz Rigid
tank
tw / D = 0.0005
tw / D = 0.0005
Impulsive pressure on wall
tw is wall thickness
h/D = 0.5
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 17
Effect of wall flexibility
If wall flexibility is included, then mechanical model of tank becomes more complicatedMoreover, its inclusion does not change seismic forces appreciablyThus, 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 18
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 19
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 massHence, higher modes are not includedThis 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 20
Modeling of ground supported tanks
Step 1:Obtain various parameters of mechanical modelThese include, mi, mc, Kc, hi, hc, hi
* and hc*
Step 2: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 21
Modeling of elevated tanks
Elevated tank consists of container and staging
Elevated tank
Container
Staging
Wall
Roof slab
Floor slab
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 22
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 modelAlso called two mass idealization
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 23
Modeling of elevated tanks
hi
mc
mihc
hs
Spring mass model Two degree of freedom system OR
Two mass idealization of elevated tanks
Ks
mi + ms
mc
Kc
Kc/2 Kc/2
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 24
Modeling of elevated tanks
ms is structural mass, which comprises of :Mass of container, andOne-third mass of staging
Mass of container includesMass of roof slabMass of wallMass of floor slab and beams
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 25
Two Degree of Freedom System
2-DoF system requires solution of a 2 × 2 eigenvalue problem to obtain
Two natural time periodsCorresponding mode shapes
See any standard text book on structural dynamics on how to solve 2-DoF systemFor 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 26
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 27
Modeling of elevated tanks
Ks
mi + ms
mc
Kc
Two degree of freedom system
Ks
mi + ms
mc
Kc
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 28
Modeling of elevated tanks
Priestley et al. (1986) suggested that this approximation is reasonable if ratio of two time periods exceeds 2.5Important to note that this approximation is done only for the purpose of calculating time periods
This significantly simplifies time period calculationOtherwise, 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 29
Modeling of elevated tanks
Steps in modeling of elevated tanksStep 1:
Obtain parameters of mechanical analogueThese include mi, mc, Kc, hi, hc, hi
* and hc*
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 30
Modeling of elevated tanks
Recall, in IS 1893:1984, convective mass is not considered
It assumes entire liquid will act as impulsive massHence, 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 31
Modeling of elevated tanks
Models of elevated tanks
Ks
mi + ms
mc
Kc
Ks
m +ms m = Total liquid mass
As per the Guideline As per IS 1893:1984
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 32
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 = π/4 x D2 x h
= π /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 33
Modeling of elevated tanks
h/D = 3/11.3 = 0.265From 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 tmc = 0.65 x m = 0.65 x 300.9 = 195.6 tKc = 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 34
Modeling of elevated tanks
Mass of container = 180 tMass 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 35
Modeling of elevated tanks
Ks
mi + ms
mc
Kc
Ks
m + ms
Model of tank as per the Guideline Model of tank as per IS 1893:1984
mi = 93.3 t, ms = 213.3 t, mc = 195.6 t, Ks = 20,000 kN/m, Kc = 462.5 kN/m
m = 300.9 t, ms = 213.3 t, Ks = 20,000 kN/m
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 36
Time period
What is time period ?For a single degree of freedom system, time period (T ) is given by
KM
T π2=
M is mass and K is stiffnessT is in secondsM 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 37
Time period
Mathematical model of tank comprises of impulsive and convective componentsHence, 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 38
Time period of impulsive mode
Procedure to obtain time period of impulsive mode (Ti) will be described for following three cases:
Ground supported circular tanksGround supported rectangular tanksElevated tanks
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 39
Ti for ground-supported circular tanks
Ground supported circular tanksTime period of impulsive mode, Ti is given by:
Et/Dρh
CT ii = ( )⎟⎟⎠⎞
⎜⎜⎝
⎛
+−=
2i0.067(h/D)0.3h/D0.46h/D
1C
ρ = Mass density of liquidE = Young’s modulus of tank materialt = Wall thicknessh = Height of liquidD = Diameter of tank
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 40
Ti for ground-supported circular tanks
Ci can also be obtained from Figure 5 of the Guidelines
0
2
4
6
8
10
0 0.5 1 1.5 2
Ci
Cc
h/D
C
© 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 taken from Eurocode 8Eurocode 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 42
Ti for ground-supported circular tanks
This formula is derived based on assumption that wall mass is quite small compared to liquid massMore 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 earthquakeexcitation 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 43
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 zeroThis 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 massesHowever, 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 44
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 45
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 baseRefer Figure 6 of the Guideline
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 46
Ti for ground-supported circular tanks
Such tanks are perhaps not used in India
Types of connections between tank wall and base slab
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 47
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 rigidTi will usually be less than 0.4 seconds
In this short period range, spectral acceleration, Sa/g has constant valueSee next slide
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 48
Ti for ground-supported circular tanks
Impulsive mode time period of ground supported tanks likely to remain in this range
Sa/g
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 49
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, ρ = 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 50
Ti for ground-supported circular tanks
Et/Dρh
CT ii =Time period of impulsive mode,
8i2x100.015/15
1.0255.3T =
= 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 51
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 tanksRecognizing 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 52
Ti for ground-supported rectangular tanks
Ti for ground-supported rectangular tanksProcedure to find time period of impulsive mode is described in Clause no. 4.3.1.2 of the Guidelines
This will not be repeated hereTime period is likely to be very low and Sa/g will remain constant
As described earlierHence, not much emphasis on time period evaluation
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 53
Ti for Elevated tanks
For elevated tanks, flexibility of staging is importantTime period of impulsive mode, Ti is given by:
s
sii K
mm2T += π
mi = Impulsive mass of liquidms = Mass of container and one-third mass of stagingKs = Lateral stiffness of stagingΔ= 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
ORg
2T Δπ=
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 54
Ti for Elevated tanks
These two formulae are one and the sameExpressed 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 55
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 tStructural 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 56
Ti for Elevated tanks
Time period of impulsive modes
sii K
mm2T += π
00020180135
2,
Ti
+π=
= 0.79 sec.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 57
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 δP Ks = P/ δ
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 58
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 59
Lateral stiffness of staging, Ks
4 columns 6 columns 8 columns
9 columns 12 columns
Plan view of frame staging configurations
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 60
Lateral stiffness of staging, Ks
24 columns 52 columns
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 61
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 modelingLeads to lower time periodHence, higher base shear coefficientThis 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 62
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 63
Time period of convective mode
Convective mass is mc and stiffness is Kc
Time period of convective mode is:
c
cc K
mT π2=
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 64
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 1Refer Figures 2 and 3 of the Guidelines
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 65
Time period of convective mode
For further simplification, expressions for mcand Kc are substituted in the formula for Tc
Then one gets,
)/68.3(68.32
DhtanhCc
π=D/gCT cc =
L/gCT cc =
For circular tanks:
For rectangular tanks:
))/(16.3(16.32
LhtanhCc
π=
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 66
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 slidesConvective mass and stiffness are not affected by flexibility of base or stagingHence, convective time period expressions are common for ground supported as well as elevated tanksConvective 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 67
Time period of convective mode
Fig. 5 For circular tanks
0
2
4
6
8
10
0 0.5 1 1.5 2
Ci
Cc
C
h/D
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 68
Time period of convective mode
2
4
6
8
10
0 0.5 1 1.5 2h/L
Cc
Fig. 7 For rectangular tanks
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 69
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
D/gCT cc =Time period of convective mode,
12/9.81 3.6Tc == 3.98 sec
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 70
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 tanksFor ground supported tanks, impulsive mode time period is likely to be very lessConvective mode time period can be very large
E-Course (through Distance Learning Mode) on
Sesimic Design of Liquid Storage Tanks January 16 - February 6, 2006
Response to Questions and Comments on Lecture 3
Question from Mr. Murugan.R ([email protected] ) • In Slide-16: Please correct the tw/D ratio. Both dotted line and thick line are
showing the same ratio of 0.0005.
You are right. Graph with solid line is for tw/D = 0.005
• In Slide-26: We are treating 2-DOF system as two uncoupled SDOF system. How to combine them?
Response from these two uncoupled SDOF systems is combined using SRSS rule. Please note, total base shear is SRSS combination of base shear in impulsive mode (i.e., first SDOF) and convective mode (i.e., second SDOF).
• Kindly furnish the formula for time period of a ground supported tanks with
finned base?
Finned base? You are perhaps asking formula for tanks with flexible base. This has been answered in a subsequent question.
• In Slide-62: Kindly provide the formula for stiffness of staging based on shear
deformation.
This has been discussed in Lecture 7
Question from Mr. H. I. Abdul Gani ([email protected] ) The following are my doubts regarding lecture 3. 1. How to find the time period of ground supported tanks with flexible connection?
a
irwi gDK
WWWT )(8 ++=
π
Tanks with flexible connections are those wherein, between wall and base there is a flexible pad. These types of tanks are described in ACI 350.3 and are reproduced in Figure 6 of IITK-GSDMA Guideline. ACI 350.3 suggests following formula for impulsive mode time period of circular tanks on flexible base
Where, Ww, Wr, and Wi are weights of wall, roof and impulsive liquid respectively. D is diameter of tank and Ka is stiffness of base pad in the units of force/m2. Flexible base does not affect convective mode time.
2. Even though the time period of impulsive mode of ground supported tanks with flexible base is low, the sa/g values as per IS 1893 is low in this region. How to account this reduced sa/g values in design?
For tanks, Sa/g graph has been made horizontal in this low period range, hence, this question will not arise. Please refer Clause 4.5.2 of the Guideline.
3. How to find the damping if impulsive and convective time periods are not well separated?
If two time periods are not well separated, then we have to solve 2-DoF system, wherein, damping is different for both the DoF. This will be non-classically damped system. This has been pointed out in the Commentary to clause 4.2.2.4. Time period, mode shapes and modal superposition methods for non-classically damped systems have been developed and this information is available in some advanced textbooks on Structural Dynamics.
4. How the flexibility of pad affects the impulsive mode time period?
This question is answered in your first question
5. In the case of ground supported rectangular tanks impulsive and convective time periods will be different in both directions. Which value is to be adopted for the design?
In rectangular tanks, for wall design, total lateral force and bending moment acting on that wall is required. Hence, for the design of a particular wall, forces arising due to seismic load in the perpendicular direction shall be considered.
6. How we can incorporate the effect of soil flexibility to find the impulsive mode time period?
Usually codes suggest to refer specialized literature for soil structure interaction. Some discussion on this topic is included in Lecture 7.
7. What is the criteria for decoupling mi+ms,ks and mc,kc (apart from Priestly's method). Can ASCE4-98 guideline for decoupling equipment masses be used here?
For elevated tanks used in practice, there is a range of various parameters (mi, mc, Ti and Tc). For most of the tanks, the criterion suggested by Preistley is OK. However, there may be some extreme cases wherein, Priestley’s criterion may not work well. You need to be cautious while using decoupling criterion of ASCE4-98, which is for secondary systems like equipments.
Interestingly, one can derive this criterion. What you need to do is to solve 2-DoF system using standard modal analysis (undamped case) and find its time periods and compare them with time periods of decoupled systems. You will notice that time period of 2-DoF systems depends on ratio of two masses and ratio of two uncoupled time periods.
8. While decoupling, only 1/3rd of the mass of staging is considered. How the remaining mass is accounted?
This has been explained in the commentary to Clause 4.2.2.3. Staging acts like a lateral spring. You consider a SDoF system and include mass of the spring also. Assuming that spring deflection varies linearly along its length, one finds that in the time period only one-third mass of spring contributes.
9. In slide 16, a) On the graph, what does 'z' represent? b) The curves are presented for h/D = 0.5. For other values are curves available?
c) tw/D = 0.0005 is represented by both bold and dotted curves. Please correct. d) For rigid tank, what will be the value of tw/D?
z represents the height at which pressure is to be obtained. Note h is total depth of liquid and z/h varies from 0 to 1.0. This figure is taken from Veletsos (1984). More studies on effect of wall flexibility are available in the literature. Refer paper 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.You can also try your hands on literature on pressure vessels. They deal with flexible walls.
Question from Mr. Rushikesh Trivedi ([email protected]) The formula specified by the code to consider lateral stiffness of shaft type staging considers Ks = 3*EI /L^3. (Lecture 3, Slide 62, E- Course on Tanks) As in the case of Tanks supported on framed stagings, here also we should consider stiffness through a force applied at the CG of container. Hence, the symbol "L" being referred to in the above formula should be identified as distance of CG of container from top of Footing. If we consider "L" to be the distance from top of footing up to the bottom of container (as considered in Proposed Draft (Example 3, Page 72)), it will be grossly on conservative side. Kindly opine on the same.
Some discussion on this issue is given in Lecture 7. It is to be recognized that we need to model the stiffness of staging properly.
In the simple formula (k = 3 E I /Lcg^3) for lateral stiffness of staging, one should consider Lcg being the height of cg of container from the base (with or without water, as the case may be). Of course, this only considers flexural deformation of the staging and the approximation is generally acceptable if the relative stiffness of the tank container is within 50% of the relative stiffness of the tank staging.
If the tank container is believed to be more stiff (or rigid), then one should determine the lateral stiffness from the deflection of the entire structure acting as a cantilever beam of length Lcg subjected to a concentrated load at Lcg from the base for which the portion (say L1) above the tank staging is considered as rigid which undergoes only rigid body rotation (Refer Figure below).
In frame staging base beams, which are comparatively more rigid, are also modeled and diaphragm effect is included to account for in-plane
rigidity of slab. The force at CG is applied by putting a rigid link from staging top to CG. This rigid link does not undergo much higher deflection than staging top.
Lcg
θ
θ
L1
∆
W
gT Δ
= π2
Lecture 4
January 30, 2006
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 2
In this lecture
Z, I, Sa/g and R values for tanks
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 3
Base shear coefficient
Seismic force V = (Ah) x (W)Ah is base shear coefficient
Zone
Depends on severity of ground motion
Structural characteristics
Depends on time period and damping
⎟⎠
⎞⎜⎝
⎛2Z
⎟⎠
⎞⎜⎝
⎛RI
gSa=hA
Design philosophy
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 4
Base shear coefficient
Tanks have two modesImpulsiveConvective
Seismic force In impulsive mode, Vi = (Ah)i x impulsive weight
In convective mode, Vc = (Ah)c x convective weight
(Ah)i and (Ah)c are base shear coefficient in impulsive and convective modes, respectively
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 5
Base shear coefficient
Impulsive base shear coefficient(Ah)i = (Z/2) x (I/R) x (Sa/g)i
Convective base shear coefficient(Ah)c = (Z/2) x (I/R) x (Sa/g)c
Note, R has been used in (Ah)i as well as (Ah)c
Zone factor, Z As per Table 2 of IS 1893(Part1):2002
I, R, (Sa/g)i and (Sa/g)c will be discussed hereFirst, (Sa/g)i and (Sa/g)c
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 6
(Sa/g)i and (Sa/g)c
(Sa/g)i is average response acceleration for impulsive mode
Depends on time period and damping of impulsive mode
(Sa/g)c is average response acceleration for convective mode
Depends on time period and damping of convective mode
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 7
(Sa/g)i and (Sa/g)c
Sa/g is obtained from design spectra Figure 2 of IS 1893(Part 1):2002
These spectra are slightly modified for tanksSee next slide
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 8
(Sa/g)i and (Sa/g)c
Modifications are:The rising portion in short period range from (0 to 0.1 sec) has been made constant
Very stiff structures have time period less than 0.1 secThere may be modeling errors; actual time period may be slightly higherAs the structure gets slightly damaged, its natural period elongatesDuctility does not help in reducing response of very stiff structuresHence, rising portion in the range 0 to 0.1 sec is usually disallowed by the codes.
Spectra is extended beyond 4 secSince convective time period may be greater than 4 sec.Beyond 4 sec, 1/T variation is retained
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 9
(Sa/g)i and (Sa/g)c
Spectra of IS 1893 (Part 1):2002
Sa/g
Sa/g
Modified spectra
For 5% damping
Sa/g
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 10
(Sa/g)i and (Sa/g)c
Expressions for spectra of IS 1893(Part 1):2003
Expressions for spectra for tanks
For hard soil sitesSa/g = 1 + 15 T 0.00 ≤ T < 0.10
= 2.50 0.10 ≤ T < 0.40 = 1.00 / T 0.40 ≤ T ≤ 4.0
For medium soil sitesSa/g = 1 + 15 T 0.00 ≤ T < 0.10
= 2.50 0.10 ≤ T < 0.55 = 1.36 / T 0.55 ≤ T ≤ 4.0
For soft soil sitesSa/g = 1 + 15 T 0.00 ≤ T < 0.10
= 2.50 0.10 ≤ T < 0.67 = 1.67 / T 0.67 ≤ T ≤ 4.0
For hard soil sitesSa/g = 2.50 T < 0.40
= 1.0 / T T ≥ 0.40
For medium soil sitesSa/g = 2.50 T < 0.55
= 1.36 / T T ≥ 0.55
For soft soil sitesSa/g = 2.5 T< 0.67
= 1.67 / T T ≥ 0.67
Expressions for design spectra at 5% damping
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 11
(Sa/g)i and (Sa/g)c
Sa/g values also depend on dampingMultiplying factors for different damping are given in Table 3 of IS 1893(Part 1)
Recall from Lecture 2, higher damping reduces base shear coefficient or design seismic forces
Multiplying factor =1.4, for 2% dampingMultiplying factor = 1.0 for 5% dampingMultiplying factor = 0.8 for 10% dampingThis multiplier is not used for PGA
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 12
Damping
Damping for impulsive mode5% of critical for RC tanks2% of critical for steel tanksThese are kept in line with IS 1893(Part 1)
Clause 7.8.2.1 of IS 1893(Part 1) suggests 5% damping for RC and 2% damping for steel buildings
However, IBC 2003 suggests 5% damping for all tanks
It suggests 5% damping for all types of buildings also
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 13
Damping
Damping depends on material and level of vibration
Higher damping for stronger shakingMeans that during the same earthquake, damping will increase as the level of shaking increasesWe are performing a simple linear analysis, while the real behavior is non-linearHence, one fixed value of damping is used in our analysis
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 14
Damping
IS 1893(Part 1), needs to have a re-look at the damping values
Accordingly, damping values for tanks can also be modified
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 15
Damping
Damping for convective mode0.5% of critical for all types of tanksConvective mode damping does not depend on material of tank or type of liquid stored
In Table 3 of IS 1893(Part 1):2002Multiplying factor for 0.5% damping is not givenValues are given for 0% and 2% dampingLinear interpolation shall not be done
Multiplying factor = 1.75, for 0.5% dampingIn Eurocode 8 this multiplying factor is 1.673In ACI 350.3, this factor is 1.5
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 16
Importance factor, I
Importance factor, I for tanks is given in Table 1 of the Guideline
This Table is reproduced here
Type of liquid storage tank I
Tanks used for storing drinking water, non-volatile material, low inflammable petrochemicals etc. and intended for emergency services such as fire fighting services. Tanks of post earthquake importance.
1.5
All other tanks with no risk to life and with negligible consequences to environment, society and economy.
1.0
NOTE: Values of importance factor, I given in IS 1893 (Part 4) may be used where appropriate
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 17
Importance factor, I
I = 1.5, is consistent with IS 1893(Part 1)IS 1893(Part 1):2002 suggests, I = 1.5 for
Hospital buildingsSchoolsFire station buildings, etc.
Tanks are kept at same importance level
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 18
Importance factor, I
Footnote below this Table is given to avoid conflict with I values of IS1893(Part 4)
IS 1893(Part 4) will deal with industrial structuresNot yet published
Some industries assign very high importance factor to tanks storing hazardous materials
Depending on their own requirements
For such tanks, Importance factor (I) will be as per part 4
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 19
Response reduction factor, R
R values for tanks are given in Table 2 of the Guideline
This is reproduced in next two slides
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 20
Response reduction factor, R
Elevated tank R
Tank supported on masonry shaftsa) Masonry shaft reinforced with horizontal bands *b) Masonry shaft reinforced with horizontal bands and vertical bars at corners
and jambs of openings
1.31.5
Tank supported on RC shaft RC shaft with two curtains of reinforcement, each having horizontal and vertical reinforcement
1.8
Tank supported on RC frame#
a) Frame not conforming to ductile detailing, i.e., ordinary moment resisting frame (OMRF)
b) Frame conforming to ductile detailing, i.e., special moment resisting frame (SMRF)
1.8
2.5
Tank supported on steel frame# 2.5
# These R values are meant for liquid retaining tanks on frame type staging which are inverted pendulum type structures. These R values shall not be misunderstood for those given in other parts of IS 1893 for building
and industrial frames. * These tanks are not allowed in Zone IV and V
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 21
Response reduction factor, R
Ground supported tank R
Masonry tanka) Masonry wall reinforced with horizontal bands*
b) Masonry wall reinforced with horizontal bands and vertical bars at corners and jambs of openings
1.31.5
RC / prestressed tank a) Fixed or hinged/pinned base tank (Figures 6a, 6b, 6c)b) Anchored flexible base tank (Figure 6d)c) Unanchored contained or uncontained tank (Figures 6e, 6f)
2.02.51.5
Steel tanka) Unanchored base b) Anchored base
2.02.5
Underground RC and steel tank+ 4.0
+ For partially buried tanks, values of R can be interpolated between ground supported and underground tanks based on depth of embedment.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 22
Response reduction factor, R
R values for tanks are smaller than buildingsThis is in line with other international codes
As discussed earlier, R depends onDuctilityRedundancyOverstrength
Tanks possess low ductility, redundancy and overstrength
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 23
Response reduction factor, R
First let us consider, elevated tanks on frame type stagingStaging frames are different than building frames
Hence, following footnote to Table 2These R values are meant for liquid retaining tanks on frame type staging which are inverted pendulum type structures. These R values shall not be misunderstood for those given in other parts of IS 1893 for building and industrial frames.
Staging frames are non-building frames and are different than building frames
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 24
Response reduction factor, R
There are critical differences between building frames and non-building framesInternational codes clearly differentiate between these two types of frames
Building frames have rigid diaphragms at floor levels
Frames of staging do not have rigid diaphragms
In buildings, seismic weight is distributed along the height at each floor level
In elevated tanks, almost entire seismic weight is concentrated at the topThese are inverted pendulum type structures
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 25
Response reduction factor, R
Moreover in buildings, non-structural elements, such as infill walls, contribute significantly to overstrength
Staging are bare frames
In view of this, for staging with SMRF, R = 2.5as against R = 5.0 for buildings with SMRFWith R = 2.5, base shear coefficient for elevated tanks on frame staging matches well with other international codes
See next slide
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 26
Response reduction factor, R
Comparison for frame stagingZone and soil parameters are same used in Lecture 2
0
0.1
0.2
0.3
0.4
0.5
0 0.5 1 1.5 2 2.5 3
Time period (sec)
Bas
e sh
ear c
oeffi
cien
t
IBC 2003; Frame staging, R = 3.0
Guideline; Frame staging, R = 2.5
IS 1893:1984; All types of staging, K = 1.0
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 27
Response reduction factor, R
Let us now consider, elevated tanks on RC shaftThey possess less redundancy and have single load path RC shafts are usually thin shell and possess low ductilityThere are analytical and experimental studies on ductility of hollow circular sections used in RC shafts
Some references are given on next slide
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 28
Response reduction factor, R
Studies on ductility of shaftZanh F A, Park R, and Priestley, M J N, 1990, “Flexural strength and ductility of circular hollow reinforced concrete columns without reinforcement on inside face”, ACI Journal 87 (2), 156-166.Rai D C, 2002, “Retrofitting of shaft type staging for elevated tanks”, Earthquake Spectra, EERI, Vol. 18 No. 4, 745-760.Rai D C and Yennamsetti S, 2002, “Inelastic seismic demand on circular shaft type staging for elevated tanks”, 7th National Conf. on Earthquake Engrg, Boston, USA, Paper No. 91. Rao M L N, 2000, “Effect of confinement on ductility of RC hollow circular columns”, a Master’s thesis submitted to Dept. of Earthquake Engineering, Univ. of Roorkee, India.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 29
Response reduction factor, R
These studies have revealed that ductility of shaft depends on
Thickness of wall (ratio of outer to inner diameter)Axial force on shaftLongitudinal and transverse reinforcement
Some results from these studies on ductility of RC shafts are discussed in next few slides
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 30
Effect of Axial Load on Ductility
Hollow circular section
Figure from Rai (2002)
Ast/Ag = ratio longitudinal reinforcement to concrete area.
P = axial load on shaft
fc’ = characteristic strength of
concrete
Ag = gross area of concrete
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 31
Response reduction factor, R
In this figure, curvature ductility is plotted as a function of longitudinal reinforcement
These results are for inner (Di) to outer (Do) diameter ratio of 0.94.If ratio of axial load (P) to ultimate load (fck.Ag) is 0.1 then, curvature ductility is about 9 for Ast/Ag = 0.02This value reduces to 3 for P/ (f’c.Ag) of 0.25
Now, let us see some results on effect of shaft thickness
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 32
Effect of Shell Thickness on Ductility
Effect of ratio of inner to outer diameter (Di/Do) is shown This result corresponds to P/(f’c.Ag) = 0.05Very low axial force ratio
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 33
Response reduction factor, R
For thin shaft with Di/Do = 0.95, curvature ductility is 12
For longitudinal steel ratio Ast/Ag = 0.02 This value increases to about 25 for thick shaft with Di/Do = 0.8Thus, thickness has significant effect on ductility
A thick shaft has reasonably good ductility
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 34
Response reduction factor, R
These analytical studies clearly indicate that thin RC hollow sections possess very low ductilityIssues connected with poor ductility of shaft, inadequate provisions of IS 1893:1984, and their correlation to behavior during recent earthquakes is discussed in following paper:
Rai D C, 2002, “Review of code design forces for shaft supported elevated water tanks”, Proc.of 13th Symposium on Earthquake Engineering , Roorkee, Ed. D K Paul et al., pp 1407 -1418.
(http://www.nicee.org/ecourse/12_symp_tanks.pdf)
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 35
Response reduction factor, R
Based on all these considerations, R = 1.8 for shaft supported tanks With this value of R, base shear coefficient for shaft supported tanks matches well with international codes
Comparison with IBC 2003 on next slide
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 36
Response reduction factor, R
0
0.1
0.2
0.3
0.4
0.5
0 0.5 1 1.5 2 2.5 3
Time period (Sec)
Bas
e sh
ear
coef
ficie
ntComparison for shaft staging
Zone and soil parameters are same as used in Lecture 2
IBC 2003; Shaft staging, R = 2.0
Guideline; Shaft staging, R = 1.8
IS 1893:1984; All types of staging, K = 1.0
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 37
Response reduction factor, R
Some useful information on RC shaft is given in ACI 371-98
ACI 371-98 , 1998, “ Guide for the analysis, design , and construction of concrete-pedestal water Towers”, American Concrete Institute, Farmington Hill, MI, USA.
It exclusively deals with tanks on RC shaft It suggests same design forces as IBC 2003It gives information on:
minimum steelconstruction tolerancessafety against bucklingshear design etc.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 38
Response reduction factor, R
We have seen comparison with IBC 2003Comparison with other international codes is available in following documents:
Jaiswal, O. R. Rai, D. C. and Jain, S.K., 2004a, “Codal provisions on design seismic forces for liquid storage tanks: a review”, Report No. IITK-GSDMA-EQ-01-V1.0, Indian Institute of Technology Kanpur, Kanpur. (www.iitk.ac.in/nicee/IITK-GSDMA/EQ01.pdf )Jaiswal, O. R., Rai, D. C. and Jain, S.K., 2004b, “Codal provisions on seismic analysis of liquid storage tanks: a review” Report No. IITK-GSDMA-EQ-04-V1.0, Indian Institute of Technology Kanpur, Kanpur. (www.iitk.ac.in/nicee/IITK-GSDMA/EQ04.pdf )
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 39
Response reduction factor, R
In the above two documents, following international codes are reviewed and compared:
IBC 2000 (now, IBC 2003)ACI 350.3ACI 371AWWA D-110 and AWWA D-115AWWA D-100 and AWWA D-103API 650 and API 620Eurocode 8NZSEE recommendations (From New Zealand)
Priestley et al. (1986)
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 40
Response reduction factor, R
Now we know Z, I, R and Sa/g for tanksOne can now obtain base shear coefficient for impulsive and convective modes
An example follows.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 41
Example
Example: An elevated water tank has RC frame staging detailed for ductility as per IS: 13920 and is located in seismic zone IV. Site of the tank has soft soil. Impulsive and convective time periods are 1.2 sec and 4.0 sec, respectively. Obtain base shear coefficient for impulsive and convective mode. Solution:
Zone: IV∴ Z = 0.24 From Table 2 of IS 1893 (PART I):2002,
I = 1.5 From Table 1 of the GuidelineR = 2.5 for RC frame with good ductility (SMRF)
From Table 2 of the Guideline
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 42
Example on (Ah)i and (Ah)c
Impulsive time period, Ti = 1.2 sec, and soil is soft, Damping = 5% (RC Frame)∴ (Sa/g)i = 1.67/Ti = 1.67/1.2 = 1.392
(Clause 4.5.3 of the Guideline)
Convective mode time period, Tc = 4.0 sec and soil is softDamping = 0.5% (Clause 4.4 of the Guideline)
Factor 1.75 is to be used for scaling up (Sa/g) for 0.5% damping (Clause 4.5.4 of the Guideline)
∴ (Sa/g)c = (1.67/Tc) x 1.75 = 1.67/4.0 x 1.75 = 0.731
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 43
Example on (Ah)i and (Ah)c
Base shear coefficient for impulsive mode(Ah)i= (Z/2) x (I/R) x (Sa/g)i
= 0.24/2 x 1.5/2.5 x 1.392 = 0.10
Base shear coefficient for convective mode(Ah)c = (Z/2) x (I/R) x (Sa/g)c
= 0.24/2 x 1.5/2.5 x 0.731= 0.053
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 44
At the end of Lecture 4
R values for tanks are less than those for buildings.The basis for this is
Analytical studies Provisions of international codes, and Observed behavior of tanks
For tanks, slight modifications are recommended for design spectrum of IS 1893(Part1)Damping for convective mode may be taken as 0.5% for all types of tanks
Lecture 5
January 31, 2006
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 2
In this Lecture
Impulsive and convective base shearCritical direction of seismic loading
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 3
Base shear
Previous lectures have coveredProcedure to find impulsive and convective liquid masses
This was done through a mechanical analog model
Procedure to obtain base shear coefficients in impulsive and convective modes
This requires time period, damping, zone factor, importance factor and response reduction factor
Now, we proceed with seismic force or base shear calculations
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 4
Base shear
Seismic force in impulsive mode (impulsive base shear)
Vi = (Ah)i x impulsive weightSeismic force in convective mode (convective base shear)
Vc = (Ah)c x convective weight(Ah)i = impulsive base shear coefficient (Ah)c = convective base shear coefficient
These are described in earlier lectures
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 5
Base shear
Now, we evaluate impulsive and convective weights
Or, impulsive and convective massesEarlier we have obtained impulsive and convective liquid mass Now, we consider structural mass also
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 6
Base shear : Ground supported tanks
Impulsive liquid mass is rigidly attached to container wall
Hence, wall, roof and impulsive liquid vibrate together
In ground supported tanks, total impulsive mass comprises of
Mass of impulsive liquidMass of wallMass of roof
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 7
Base shear : Ground supported tanks
Hence, base shear in impulsive mode
( ) ( )gmmmAV twihi i++=
mi = mass of impulsive liquidmw = mass of container wallmt = mass of container roofg = acceleration due to gravity
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 8
Base shear : Ground supported tanks
This is base shear at the bottom of wallBase shear at the bottom of base slab is :
Vi’ = Vi + (Ah)i x mbmb is mass of base slab
Base shear at the bottom of base slab may be required to check safety against sliding
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 9
Base shear : Ground supported tanks
Base shear in convective mode
mc = mass of convective liquid
( ) gmAV cchc =
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 10
Base shear : Ground supported tanks
Total base shear, V is obtained as:
22ci VVV +=
Impulsive and convective base shear are combined using Square Root of Sum of Square (SRSS) rule
Except Eurocode 8, all international codes use SRSS rule
Eurocode 8 uses absolute summation rulei.e, V = Vi + Vc
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 11
Base shear : Ground supported tanks
In the latest NEHRP recommendations (FEMA 450), SRSS rule is suggested
Earlier version of NEHRP recommendations (FEMA 368) was using absolute summation rule
FEMA 450, 2003, “NEHRP recommended provisions for seismic regulations for new buildings and other structures”, Building Seismic Safety Council, National Institute of Building Sciences,, USA.FEMA 368, 2000, “NEHRP recommended provisions for seismic regulations for new buildings and other structures”, Building Seismic Safety Council, National Institute of Building Sciences,, USA.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 12
Bending moment:Ground supported tanks
Next, we evaluate bending or overturning effects due to base shearImpulsive base shear comprises of three parts
(Ah)i x mig(Ah)i x mwg(Ah)i x mtg
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 13
Bending moment:Ground supported tanks
mw acts at CG of wallmt acts at CG of roofmi acts at height hi from bottom of wall
If base pressure effect is not includedmi acts at hi
*
If base pressure effect is includedRecall hi and hi
* from Lecture 1
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 14
Bending moment:Ground supported tanks
Bending moment at the bottom of wallDue to impulsive base shear
( ) ( )ghmhmhmAM ttwwiiihi ++=
hi = location of mi from bottom of wallhc = location of mc from bottom of wallhw = height of CG of wallht = height of CG of roof
( ) ghmAM ccchc )(=
Due to convective base shear
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 15
Bending moment:Ground supported tanks
For bending moment at the bottom of wall, effect of base pressure is not included
Hence, hi and hc are used
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 16
Bending moment:Ground supported tanks
Bending moment at the bottom of wall
( ) ( )ghmhmhmAM ttwwiiihi ++=
( ) ghmAM ccchc )(=
Ground level
(Ah)imihi
(Ah)imw
hw
(Ah)imt
(Ah)cmchc
ht
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 17
Bending moment:Ground supported tanks
Total bending moment at the bottom of wall
22ci MMM +=
SRSS rule used to combine impulsive and convective responses
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 18
Overturning moment:Ground supported tanks
Overturning momentThis is at the bottom of base slabHence, must include effect of base pressure
hi* and hc
* will be used
Ground level
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 19
Overturning moment:Ground supported tanks
Overturning moment in impulsive mode
tb = thickness of base slab
( ) ( )( )
g/tmthm
thm)th(mAMbbbtt
bwwb*ii
ih*i
⎥⎥⎦
⎤
⎢⎢⎣
⎡
++++++
=2
Overturning moment in convective mode
gthmAM bccchc )()( ** +=
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 20
Bending moment:Ground supported tanks
Overturning moment is at the bottom of base slab
Hence, lever arm is from bottom of base slabHence, base slab thickness, tb is added to heights measured from top of the base slab
Total overturning moment
2*2**ci MMM +=
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 21
Example
Example: A ground-supported circular tank is shown below along with some relevant data. Find base shear and bending moment at the bottom of wall. Also find base shear and overturning moment at the bottom of base slab.
mi = 141.4 t; mc = 163.4 t
mw = 65.3 t, mt =33.1 t,
mb = 55.2 t,
hi =1.5 m, hi* = 3.95 m,
hc = 2.3 m, hc* = 3.63 m
(Ah)i = 0.225, (Ah)c = 0.08
Roof slab 150 mm thick
Base slab 250 mm thick
4 m
10 m
Wall 200 mm thick
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 22
Example
Solution:Impulsive base shear at the bottom of wall is
Vi = (Ah)i (mi + mw + mt) g = 0.225 x (141.4 + 65.3 + 33.1) x 9.81= 529.3 kN
Convective base shear at the bottom of wall isVc = (Ah)c mc g
= 0.08 x 163.4 x 9.81 = 128.2 kN
Total base shear at the bottom of wall is
( ) ( ) 544.6kN128.2529.3VVV 222c
2i =+=+=
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 23
Example
For obtaining bending moment, we need height of CG of roof slab from bottom of wall, ht.ht = 4.0 + 0.075 = 4.075 m
Impulsive bending moment at the bottom of wall isMi = (Ah)i (mihi + mwhw + mtht) g
= 0.225 x (141.4 x 1.5 + 65.3 x 2.0 + 33.1 x 4.075) x 9.81= 1054 kN-m
Convective bending moment at the bottom of wall isMc = (Ah)c mc hc g
= 0.08 x 163.4 x 2.3 x 9.81 = 295 kN-m
Total bending moment at bottom of wall is
( ) ( ) m-1095kN2951054MMM 222c
2i =+=+=
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 24
Example
Now, we obtain base shear at the bottom of base slab
Impulsive base shear at the bottom of base slab isVi = (Ah)i (mi + mw + mt + mb) g
= 0.225 x (141.4 + 65.3 + 33.1 + 55.2) x 9.81= 651.1 kN
Convective base shear at the bottom of base slab isVc = (Ah)c mc g
= 0.08 x 163.4 x 9.81 = 128.2 kN
Total base shear at the bottom of base slab is
( ) ( ) 663.6kN128.2651.1VVV 222c
2i =+=+=
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 25
Example
Impulsive overturning moment at the bottom of base slabMi
* = (Ah)i [mi (hi* + tb) + mw(hw + tb) + mt(ht +tb) + mb tb/2]g
= 0.225 x [141.4(3.95 + 0.25) + 65.3(2.0 + 0.25) + 33.1(4.075 + 0.25) + 55.2 x 0.25/2] x 9.81
= 1966 kN-mConvective overturning moment at the bottom of base slab
Mc* = (Ah)c mc (hc
* + tb) g = 0.08 x 163.4 x (3.63 + 0.25) x 9.81 = 498 kN-m
Total overturning moment at bottom of base slab
Notice that this value is substantially larger that the value at the bottom of wall (85%)
( ) ( ) ( ) ( ) m-kN 20284981966MMM 22*c
*i
* =+=+=22
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 26
Base shear : Elevated tanks
In elevated tanks, base shear at the bottom of staging is of interestMs is structural mass
Base shear in impulsive mode
( ) ( )gmmAV siihi +=
Base shear in convective mode
( ) gmAV cchc =
Total base shear22
ci VVV +=
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 27
Bending moment:Elevated tanks
Bending moment at the bottom of stagingBottom of staging refers to footing top
Impulsive base shear comprises of two parts(Ah)i x migAh)i x msg
Convective base shear has only one part(Ah)c x mcg
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 28
Bending moment:Elevated tanks
mi acts at hi*
mc acts at hc*
Bending moment at bottom of staging is being obtainedHence, effect of base pressure included and hi
*
and hc* are used
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 29
Bending moment:Elevated tanks
Structural mass, ms comprises of mass of empty container and 1/3rd mass of staging
ms is assumed to act at CG of empty containerCG of empty container shall be obtained by considering roof, wall, floor slab and floor beams
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 30
Bending moment:Elevated tanks
Bending moment at the bottom of staging
( ) ( )[ ] ghmhhmAM cgss*
iiih*
i ++=
( ) ( ) ghhmAM s*
ccch*
c +=
hs = staging heightMeasured from top of footing to bottom of wall
hcg = distance of CG of empty container from bottom of staging
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 31
Bending moment:Elevated tanks
Bending moment at the bottom of staging
Top of footing
(Ah)i msg
hs
hcg
hi*
hs
hc*
( ) ( ) ghhmAM s*
ccch*
c +=( ) ( )[ ] ghmhhmAM cgss*iiih
*i ++=
(Ah)i mig (Ah)c mcg
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 32
Bending moment:Elevated tanks
Total bending moment
22c
*i
** MMM +=
For shaft supported tanks, M* will be the design moment for shaft
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 33
Bending moment:Elevated tanks
For analysis of frame staging, two approaches are possibleApproach 1: Perform analysis in two stepsStep 1:
Analyze frame for (Ah)imig + (Ah)imsgObtain forces in columns and braces
Step 2:Analyze the frame for (Ah)cmcg Obtain forces in columns and braces
Use SRSS rule to combine the member forces obtained in Step 1 and Step 2
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 34
Bending moment:Elevated tanks
Approach 2:Apply horizontal force V at height h1 such that
V x h1 = M*
V and M* are obtained using SRSS rule as described in slide nos. 26 and 32In this approach, analysis is done in single step
Simpler and faster than Approach 1
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 35
Example
Example: An elevated tank on frame staging is shown below along with some relevant data. Find base shear and bending moment at the bottom of staging.
AA is CG of empty container
mi = 100t; mc = 180 t
Mass of container = 160 t
Mass of staging = 120 t
hi* = 3 m, hc
* = 4.2 m
(Ah)i = 0.08, (Ah)c = 0.04
GL
hs = 15 m
2.8 m
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 36
Example
Structural mass, ms = mass of container + 1/3rd mass of staging= 160 + 1/3 x 120 = 200 t
Base shear in impulsive mode
( ) ( )gmmAV sihi i+=
( ) 819x200100x080 .. +=
Base shear in convective mode
( ) gmAV cchc =
819x180x040 ..= kN6.70=
= 78.5 + 157 = 235.5 kN
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 37
Example
Total base shear 22ci VVV +=
22 6705235 .. += kN8.245=
Now, we proceed to obtain bending moment at the bottom staging
Distance of CG of empty container from bottom of staging, hcg = 2.8 + 15 = 17.8 m
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 38
Example
Base moment in impulsive mode
( ) ( )[ ] ghmhhmAM cgss*
iiih*
i ++=
( )[ ] 819x817x2001503x100080 .... ++=
= 78.5 x 18 + 157 x 17.8
= 4207 kNm
Note: 78.5 kN of force will act at 18.0m and 157 kN of force will act at 17.8 m from top of footing.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 39
Example
Base moment in convective mode
( ) ( ) ghhmAM s*
ccch*
c +=
( ) 819x1524x180x040 ... +=
Total base moment22c
*i
** MMM +=
22 13564207 +=kNm4420=
= 70.6 x 19.2= 1356 kNm
Note: 70.6 kN of force will act at 19.2 m from top of footing.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 40
Example
Now, for staging analysis, seismic forces are to be applied at suitable heights
There are two approachesRefer slide no 33
Approach 1:Step 1: Apply force of 78.5 kN at 18 m and 157 kN at 17.8 m from top of footing and analyze the frameStep 2: Apply 70.6 kN at 19.2 m from top of footing and analyze the frameMember forces (i.e., BM, SF etc. in columns and braces) of Steps 1 and 2 shall be combined using SRSS
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 41
Example
Approach 2:Total base shear, V = 245.8 kN will be applied at height h1, such thatV x h1 = M*245.8 x h1 = 4420∴ h1 = 17.98 mThus, apply force of 245.8 kN at 17.98 m from top of footing and get member forces (i.e., BM, SF in columns and braces).
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 42
Elevated tanks:Empty condition
Elevated tanks shall be analysed for tank full as well as tank empty conditions
Design shall be done for the critical conditionIn empty condition, no convective liquid mass
Hence, tank will be modeled using single degree of freedom systemMass of empty container and 1/3rd staging mass shall be considered
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 43
Elevated tanks:Empty condition
Lateral stiffness of staging, Ks will remain same in full and empty conditionsIn full condition, mass is more
In empty condition mass is lessHence, time period of empty tank will be less
Recall, T = Hence, Sa/g will be more
Usually, tank full condition is criticalHowever, for tanks of low capacity, empty condition may become critical
KM2 Π
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 44
Direction of seismic force
Let us a consider a vertical cantilever with rectangular cross sectionHorizontal load P is applied
First in X-DirectionThen in Y-direction (see Figure below)More deflection, when force in Y-directionHence, direction of lateral loading is important !!
P
P
X
Y
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 45
Direction of seismic force
On the other hand, if cantilever is circularDirection is not of concernSame deflection for any direction of loading
Hence, it is important to ascertain the most critical direction of lateral seismic force
Direction of force, which will produce maximum response is the most critical direction
In the rectangular cantilever problem, Y-direction is the most critical direction for deflection
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 46
Direction of seismic force
For frame stagings consisting of columns and braces, IS 11682:1985 suggests that horizontal seismic loads shall be applied in the critical direction
IS 11682:1985, “Criteria for Design of RCC Staging for Overhead Water Tanks”, Bureau of Indian Standards, New Delhi
Clause 7.1.1.2 Horizontal forces – Actual forces and moments resulting from horizontal forces may be calculated for critical direction and used in the design of the structures. Analysis may be done by any of the accepted methods including considering as space frame.
Clause 7.2.2 Bending moments in horizontal braces due to horizontal loads shall be calculated when horizontal forces act in a critical direction. The moments in braces shall be the sum of moments in the upper and lower columns at the joint resolved in the direction of horizontal braces.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 47
Direction of seismic force
Section 4.8 of IITK-GSDMA Guidelines contains provisions on critical direction of seismic force for tanks
Ground-supported circular tanks need to be analyzed for only one direction of seismic loads
These are axisymmetricHence, analysis in any one direction is sufficient
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 48
Direction of seismic force
Ground-supported rectangular tanks shall be analyzed for two directions
Parallel to length of the tankParallel to width of the tank
Stresses in a particular wall shall be obtained for seismic loads perpendicular to that wall
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 49
Direction of seismic force
RC circular shafts of elevated tanks are also axisymmetric
Hence, analysis in one direction is sufficient
If circular shaft supports rectangular containerThen, analysis in two directions will be necessary
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 50
Direction of seismic force
For elevated tanks on frame stagingCritical direction of seismic loading for columns and braces shall be properly ascertained
Braces and columns may have different critical directions of loadingFor example, in a 4 - column staging
Seismic loading along the length of the brace is critical for bracesSeismic loading in diagonal direction gives maximum axial force in columnsSee next slide
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 51
Direction of seismic force
Critical directions for 4 - column staging
Critical direction for shear force in brace
Critical direction for axial force in column
Bending Axis
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 52
Direction of seismic force
For 6 – column and 8 – column staging, critical directions are given in Figure C-6 of the Guideline
See next two slidesMore information available in Sameer and Jain (1994)
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. (http://www.nicee.org/ecourse/Tank_ASCE.pdf)
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 53
Direction of seismic force
Critical directions for 6 - column staging
Critical direction for shear force and bending moment in columns
Critical direction for shear force and bending moment in braces and axial force in columns
Bending Axis
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 54
Direction of seismic force
Critical directions for 8 - column staging
Critical direction for shear force and bending moment in braces
Critical direction for shear force, bending moment and axial force in columns
Bending Axis
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 55
Direction of seismic force
As an alternative to analysis in the critical directions, following two load combinations can be used
100 % + 30% ruleAlso used in IS 1893(Part 1) for buildings
SRSS rule
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 56
Direction of seismic force
100%+30% rule implies following combinationsELx + 0.3 ELY
ELY + 0.3 ELx
ELx is response quantity when seismic loads are applied in X-directionELY is response quantity when seismic loads are applied in Y-direction
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 57
Direction of seismic force
100%+30% rule requiresAnalyze tank with seismic force in X-direction; obtain response quantity, ELX
Response quantity means BM in column, SF in brace, etc.
Analyze tank with seismic force in Y-direction; obtain response quantity, ELY
Combine response quantity as per 100%+30% ruleCombination is on response quantity and not on seismic loads
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 58
Direction of seismic force
Important to note that the earthquake directions are reversible
Hence, in 100%+30% rule, there are total eight load combinations
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 59
Direction of seismic force
SRSS rule implies following combination
22yx ELEL +
Note:ELx is response quantity when seismic loads are applied in X-directionELY is response quantity when seismic loads are applied in Y-direction
Hence, analyze tank in two directions and use SRSS combination of response quantity
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 60
At the end of Lecture 5
This completes seismic force evaluation on tanksThere are two main steps
Evaluation of impulsive and convective massesEvaluation of base shear coefficients for impulsive and convective modes
SRSS rule is used to combine impulsive and convective responsesCritical direction of seismic loading shall be properly ascertained
Else, 100%+30% or SRSS rule be used
E-Course (through Distance Learning Mode) on
Sesimic Design of Liquid Storage Tanks January 16 - February 6, 2006
Questions and Comments on Lectures
Question from Mr. H. I. Abdul Gani ([email protected] ) Lecture 4 1. Why damping in convective mode is independant of type of liquid stored? For highly viscous liquids, does not viscosity of the liquid affect damping?
This is an interesting question. The formulation used for obtaining hydrodynamic pressure, considers liquid to be non-viscous, homogeneous and incompressible. The potential equation is written with these assumptions on liquid properties. Now, if liquid is highly viscous, not only the damping, but the convective mass will also change. One can guess, that in highly viscous liquid, less mass will undergo sloshing motion, thereby implying that more liquid will participate in impulsive mode. However, this needs to be substantiated with in-depth study.
2. How the multiplicative factor of 1.75 was arrived at for 0.5% damping? Was logarithmic interpolation used?
Yes, logarithmic interpolation is used. In the commentary to Clause 4.5.4, we have referred Newmark and Hall (1982), wherein, this issue is discussed.
3. For multistoreyed buildings with "soft storey" what is the importance factor to be adopted?
This is a question on buildings. Importance factor is not related to soft storey. If a building has soft storey, then it would be an irregular building. Please refer Clause 7.10 of IS 1893(Part 1):2002
Question from Mr. N. Devanarayanan ([email protected]) Dear Sir, I have been attending the ecourse on the design of liquid storage structures. Since we deal mostly with petroleum tank storages I am more interested in learning about the same.There were a few points I would like
clarification on. In certain areas due to the soil characteristics the storage tanks are some times rested on RCC pile foundations In certain cases the tanks are resten on sandpad foundations, the underlying starta below the sand pad foundations are strngthened with stone columns. In such cases do we have to take the effect of seismic forces and design the tanks using the two degree of freedom, and will the piles/stone columns act as a staging thus making the whole structure behave like an elevated tank.
In the first case, there will be a RC ring beam (or circular pile cap), which is supported on piles. Tank wall and some portion of base plate rest on this ring beam. Inside the ring beam there will be sand filling, which supports rest of the base plate. This tank will be considered as ground-supported tank and not as elevated tank. In the second case, the entire base plate rests on sand fill, which is supported on compacted stone columns. For this case also, tank will be treated as ground supported tank. The two degree of freedom model of elevated tank is not applicable to these tanks. For circular steel tanks, first find the impulsive time period using formula given in Lecture 3 (also refer Clause 4.3.1.1 of the Guideline). Then, find the convective mode time period and proceed with the evaluation of base shear coefficient.
Lecture 5 1. Why Vi and Vc (also Moi&Moc) are combined using SRSS rule? Wouldn’t absolute sum be more accurate as the effects of impulsive and convective mass are simultaneously present?
Though impulsive and convective masses (or modes) may be present simultaneously, these modal responses may not reach their maximum values simultaneously. Hence, absolute summation is not used. Till sometime back, NEHRP recommendation of 2000 (FEMA 368) had suggested absolute summation. However, some recent numerical simulation studies have indicated that absolute summation rule overestimates the response. Hence, the latest NEHRP recommendation of 2003 (FEMA 450) has used SRSS rule.
Lecture 6
February 7, 2006
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 2
In this Lecture
Hydrodynamic pressure on wall and baseEffect of vertical ground accelerationSloshing wave heightAnchorage requirements for ground supported tanks
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 3
Hydrodynamic pressure
So far, emphasis was on lateral forces on tanksThese lateral forces comprised of
Impulsive componentConvective component
Impulsive component has two partsOne due to impulsive liquid massSecond due to structural mass of tank
Convective component has one partDue to convective liquid mass
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 4
Hydrodynamic pressure
Recall from Lecture 1Impulsive force is summation of impulsive hydrodynamic pressure on wallConvective force is summation convective hydrodynamic pressure on wall
Now, we will see procedure to obtainImpulsive and convective pressure distribution on wall and base
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 5
Hydrodynamic pressure
Pressure distribution on wall is needed to obtainHoop forces, and Bending moment in wall
IS 3370 (Part IV):1967 gives hoop forces and bending moments due to hydrostatic pressure
Hydrostatic pressure has linear distribution along wall height
IS 3370(Part IV):1967, “Code of Practice for Concrete Structures for the storage of Liquids”, Bureau of Indian Standards, New Delhi
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 6
Impulsive pressure:Circular tanks
Impulsive pressure on wall of circular tanks is given by
⎟⎠⎞
⎜⎝⎛
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎠⎞
⎜⎝⎛−=
hD.tanh
hy.)y(Qiw 866018660
2
Qiw = Coefficient of impulsive pressure on wall
(Ah)i = Impulsive base shear coefficient
ρ = Mass density of liquid
φ = Circumferential angle (see next slide)
y = Vertical distance of a point on wall from the bottom of wall
( ) coshgA)y(Qp ihiwiw ρ= φ
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 7
Impulsive pressure:Circular tanks
Note that impulsive pressure varies with circumferential angle, φ
At φ = 00, pressure will be maximumAt φ = 900, pressure will be zeroRecall following figure from Lecture 1
φ
Direction of seismic forcemaximum pressure at φ = 00
zero pressure at φ = 900
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 8
Impulsive pressure:Circular tanks
Qiw(y) can also be read from Figure 9(a) of the Guideline
This Figure is reproduced here
0
0.2
0.4
0.6
0.8
1
0 0.2 0.4 0.6 0.8 1
1.51.0 0.5 0.25
h/D =2or h/L
Qiw
Y/h
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 9
Impulsive pressure:Circular tanks
Impulsive pressure on base of circular tanks changes in radial as well as circumferential direction
A strip of length l’ is consideredRefer plan of circular tank shown in Figure 8(a) of the Guideline
D/2
O
φ
l’
Direction of seismic force
x
Plan
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 10
Impulsive pressure:Circular tanks
Impulsive pressure on this strip is given by
( )⎟⎟⎠
⎞⎜⎜⎝
⎛
⎟⎠⎞
⎜⎝⎛
=
hlhx
hgAp ihib '
866.0cosh
866.0sinh866.0 ρ
x = horizontal distance in the direction of seismic force, of a point on base slab, from the reference axis at the center of tank
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 11
Impulsive pressure:Circular tanks
Example: A circular tank has internal diameter of 12 m, and water height of 5 m. Impulsive base shear coefficient is 0.23. Find the impulsive pressure distribution on wall
Solution: D = 12 m, h = 5 m, (Ah)i = 0.23
Impulsive pressure on wall is given by,piw = Qiw (y) (Ah)i ρ g h cos φ
Where,Qiw (y)= 0.866 [ 1- (y /h)2] tanh(0.866 D/h)
Maximum pressure will occur at φ = 0
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 12
Impulsive pressure:Circular tanks
At base of wall, y = 0;Qiw = 0.866 tanh(0.866 D/h)
= 0.866 tanh(0.866 x 12/5)= 0.84
Now, at φ = 0, piw = Qiw (y) (Ah)i ρ g hMass density of water, ρ = 1000 kg/m3
∴ piw (y =0) = 0.84 x 0.23 x 1000 x 9.81 x 5
= 9476.5 N/m2 = 9.48 kN/m2
For different values of y, piw is similarly calculated. Some values are tabulated, and a distribution plotted on the next slide.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 13
Impulsive pressure:Circular tanks
y (m) 0 1.25Qiw (y) 0.84 0.79 0.37 0
8.919.48
3.75 5
piw (y) (kN/m2) 4.17 0
9.48 kN/m2
Impulsive pressure distribution on wall
h = 5m
At φ = o
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 14
Impulsive pressure:Rectangular tanks
Impulsive pressure on walls of rectangular tanks is given by
( ) hgA)y(Qp ihiwiw ρ=
⎟⎠⎞
⎜⎝⎛
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎠⎞
⎜⎝⎛−=
hL.tanh
hy.)y(Q iw 866018660
2
Except following, this expression is same as that for circular tanks
D/h is replaced by L/hAngle φ is not present
Qiw(y) can be read from Figure 9(a) of the Guideline
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 15
Impulsive pressure:Rectangular tanks
This pressure is on walls perpendicular to direction of seismic forcePressure remains same along the length of these perpendicular walls
However, it changes with wall height
Direction of seismic force
Walls perpendicular to direction of seismic force
LB
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 16
Impulsive pressure:Rectangular tanks
Impulsive pressure on base of rectangular tanksImpulsive pressure will be defined for a point at a distance x from central axis
As shown in Figure belowRefer plan of rectangular tank in Figure 8 (b) of the Guideline
Plan
x
L
Direction ofSeismic Force
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 17
Impulsive pressure:Rectangular tanks
Impulsive pressure is given by
( ) hgA)x(Qp ihibib ρ=
⎟⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛
=
hL.cosh
hx.sinh
)x(Q ib
8660
8660
Qib(x) is coefficient of impulsive pressure on base It can be read from Figure 9(b) of the GuidelineSee next slide
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 18
Impulsive pressure:Rectangular tanks
-1.2
-0.8
-0.4
0
0.4
0.8
1.2
-0.4 -0.2 0 0.2 0.4x/L
Qib
2.0
1.5
1.0
0.50.25 = h/L
2.0
1.5
1.0
0.50.25
Qib(x)
Coefficient of impulsive pressure on base Qib(x)
Figure 9(b) of the Guideline
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 19
Convective pressure:Circular tanks
Convective pressure on wall of circular tanks is given by
( ) φφρ coscosDgA)y(Qp chcwcw ⎥⎦⎤
⎢⎣⎡= 2
31- 1
⎟⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛
=
Dh.cosh
Dy.cosh
.)y(Qcw
6743
674356250
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 20
Convective pressure:Circular tanks
(Ah)c = Convective base shear coefficient
Qcw = Coefficient of convective pressure on wall
ρ = Mass density of liquid
φ = Circumferential angle
y = Vertical distance of a point on wall from the bottom of wall
Qcw can be obtained from Figure 10(a) of the Guideline
See next slide
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 21
Convective pressure:Circular tanks
Coefficient of convective pressure on wall, Qcw
0
0.2
0.4
0.6
0.8
1
0 0.1 0.2 0.3 0.4 0.5 0.6
2.0
1.51.0
0.5
h/D=0.25
Qcw
Y/h
Figure 10(a) of the Guideline
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 22
Convective pressure:Circular tanks
Convective pressure on base of circular tanks is given by
( ) DgA)x(Qp chcbcb ρ=
⎟⎠⎞
⎜⎝⎛
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎠⎞
⎜⎝⎛−=
Dh.hsec
Dx
Dx.)x(Qcb 6743
341251
3
Qcb can also be obtained from Figure 10(b) of the Guideline
See next slide
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 23
Convective pressure:Circular tanks
Coefficient of convective pressure on base, Qcb
-0.3
-0.2
-0.1
0
0.1
0.2
0.3
-0.4 -0.2 0 0.2 0.4x/D
Qcb
1.00.75
0.5
h/D=0.25
h/D=0.25
0.5
0.751.0
Qcb
Figure 10(b) of the Guideline
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 24
Convective pressure:Circular tanks
Example: A circular tank has internal diameter of 12 m, and water height of 5 m. Convective base shear coefficient is 0.07. Find the convective pressure distribution on wall
Solution:D = 12 m, h = 5 m, (Ah)c = 0.07
Convective pressure on wall is given by,pcw = Qcw (y) (Ah)c ρ g D [1- 1/3 cos 2φ] cos φ
Where,Qcw (y)= 0.5625 cosh(3.674 y/D)/cosh(3.674 h/D)
Maximum pressure will occur at φ = 0
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 25
Convective pressure:Circular tanks
At base of wall, y = 0;Qcw (y =0) = 0.5625/cosh(3.674 h/D)
= 0.5625/cosh(3.674 x 5/12) = 0.23
Now, at φ = 0, pcw = Qcw (y) (Ah)c ρ g D [1- 1/3 ] Mass density of water, ρ = 1000 kg/m3
∴ pcw (y =0) = 0.23 x 0.07 x 1000 x 9.81 x 12 x [1- 1/3]= 1263.5 N/m2 = 1.26 kN/m2
For different values of y, pcw is similarly calculated. Pressure distribution with wall height is shown in the next slide.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 26
Convective pressure:Circular tanks
y (m) 0 1.25
Qcw (y) 0.23 0.25 0.40 0.5625
1.371.26
3.75 5
pcw (y) (kN/m2) 2.20 3.09
Convective pressure distribution on wall
1.26 kN/m2
h = 5m
3.09 kN/m2
At φ = o
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 27
Convective pressure:Rectangular tanks
Convective pressure on wall of rectangular tank is given by
Lg)A)(y(Qp chcwcw ρ=
⎟⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛
=
Lh.cosh
Ly.cosh
.)y(Qcw
1623
162341650
Qcw can also be obtained from Figure 11(a) of the Guideline
See next slide
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 28
Convective pressure:Rectangular tanks
Coefficient of convective pressure on wall, Qcw
0
0.2
0.4
0.6
0.8
1
0 0.1 0.2 0.3 0.4 0.5Qcw
2.0 1.5
1.0
0.5
h/L=0.2
Y/h
Figure 11(a) of the Guideline
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 29
Convective pressure:Rectangular tanks
Convective pressure on base of rectangular tanks is given by
Qcb can be obtained from Figure 11(b) of the GuidelineHowever, Figure 11 (b) of the Guideline is incorrectCorrect Figure is shown in next slide
Lg)A)(x(Qp chcbcb ρ=
⎟⎠⎞
⎜⎝⎛
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎠⎞
⎜⎝⎛−=
Lh.hsec
Lx
Lx.)x(Qcb 1623
34251
3
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 30
Convective pressure:Rectangular tanks
Coefficient of convective pressures on base, Qcb
-0.4
-0.3
-0.2
-0.1
0
0.1
0.2
0.3
0.4
-0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.5
x/L
Qcb
h/L = 0.25
0.50
0.751.0
h/L = 0.25
0.50
0.75
1.0
Figure 11(b) of the Guideline (corrected)
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 31
Linearised Pressure Distribution
Impulsive and convective pressures on wall have curvilinear distributionHoop forces and bending moment in wall due to such curvilinear pressure distributions are not readily available
Hoop forces and BM are available in codes only for linearly varying pressure
As mentioned earlier, refer IS 3370 (Part IV):1967
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 32
Linearised Pressure Distribution
Hence, equivalent linearised pressure distribution is suggested
For impulsive as well as convective pressure
This linearisation is such that, base shear and bending moment at the bottom of wall remains same
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 33
Linearised Pressure Distribution
Equivalent linear distribution for impulsive pressure will be such that
Total base shear and bending moment due to linear distribution will be same as that due to curvilinear distribution
Similarly, for convective pressure
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 34
Linearised Pressure Distribution
Equivalent linear impulsive pressure
Actual Impulsive pressure distribution
hi
qi
h ˜
Equivalent linearpressure distribution
hi
qi
ai
bi
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 35
Linearised Pressure Distribution
The ordinates of linear distribution are
( )ii
i hhhqa 642 −= ( )hh
hqb i
ii 262 −=
g/Dm)A(q iih
i 2π=For circular tanks
gB
m)A(q iihi 2=For rectangular tanks
Both for circular and rectangular tanks Both for qi is different for circular and rectangular tanks
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 36
Linearised Pressure Distribution
Heremi = impulsive liquid masshi = height of impulsive massh = Total liquid height
(Ah)i = Impulsive base shear coefficient
All these quantities are knownProcedure to obtain these quantities is covered in previous lectures
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 37
Linearised Pressure Distribution
Similarly, equivalent linear convective pressure
Actual convective pressure distribution Equivalent linear
pressure distribution
qc
hch ˜ hc
ac
qc
bc
g/Dm)A(q cch
c 2π=
( )cc
c hhhqa 642 −=
( )hhhqb c
cc 262 −=
For circular tanks
For rectangular tanks
gB
m)A(q cchc 2=
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 38
Linearised Pressure Distribution
Linearised distribution can be treated as summation of two parts
Uniform pressureTriangular pressure
+=
bi ai - biai
bi
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 39
Linearised Pressure Distribution
Hoop forces and BM in wall due to uniform and triangular pressure can be obtained from IS 3370(Part IV):1967
Linearised pressure distributions discussed here are taken from ACI 350.3
ACI 350.3, 2001, “Seismic design of liquid containing concrete structures”, American Concrete Institute, Farmington Hill, MI, USA.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 40
Linearised Pressure Distribution
Example: A circular tank has internal diameter of 12 m, and water height of 5 m. Impulsive and convective base shear coefficients are 0.23 and 0.07, respectively. Find linearised impulsive and convective pressure distribution on wall.
Solution:
D = 12 m, h = 5 m, (Ah)i = 0.23, (Ah)c = 0.07Capacity of tank = π x 122 /4 x 5 = 565.5 m3
∴ Mass of water = 565.5 x 1000 = 5,65,500 kg
For h/D = 5/12 = 0.42, from Figure 2a of the Guideline:mi/m = 0.47, ∴ mi = 0.47 x 565500 = 2,65,800 kg
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 41
Linearised Pressure Distribution
mc/m = 0.5, mc = 0.50 x 565500 = 2,82,800 kghi /h = 0.375, hi = 0.375 x 5 = 1.875 mhc /h = 0.58, hc = 0.58 x 5 = 2.9 m
a) Impulsive mode
Pressure at bottom & top is given by
( ) ( ) 2i
ii kN/mhh
hqa 13.11875.1654
581.3164 22 =×−×=−=
kN/mgD
mAq iihi 81.31
2/1281.9800,26523.0
2/)(
=×
××==
ππ
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 42
Linearised Pressure Distribution
( ) ( ) 222 5915287516
5813126 m/kN...hh
hqb i
ii =×−×=−=
bi = 1.59 kN/m2
ai = 11.13 kN/m2
Linear pressure distribution
9.48 kN/m2
Actual pressure distribution ( Obtained in earlier example)
h = 5m
Note:- For the same tank, in earlier example, we had obtained actual distribution of impulsive pressure. Linear and actual pressure distributions are shown below:
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 43
Linearised Pressure Distribution
b) Convective mode
Pressure at bottom & top is given by
( ) ( ) kN/mhhhqa c
cc 07.19.2654
53.1064 22 =×−×=−=
( ) ( ) 2c
cc kN/mhh
hqb 05.3529.26
53.1026 22 =×−×=−=
kN/mgD
mAq cchc 3.10
2/1281.9800,28207.0
2/)(
=×
××==
ππ
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 44
Linearised Pressure Distribution
Note:- For the same tank, in earlier example we had obtained actual distribution of convective pressure. Linear and actual pressure distributions are shown below:
bi = 3.05 kN/m2
ai = 1.07 kN/m2
Linear pressure distribution
1.26 kN/m2
Actual convective pressure distribution (Obtained in earlier example)
h = 5m
3.09 kN/m2
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 45
Circumferential distribution
In circular tanks, pressure varies along the circumference of the wall
Recall there is cosφ term in the expression for pressure on wall
φ
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 46
Circumferential distribution
Hoop forces and BM in wall for such varying pressure are also not readily availableHence, for convenience in stress analysis
Hydrodynamic pressure may be approximated by an outward axisymmetric pressure distribution Intensity of this axisymmetric distribution is equal to that of maximum pressureSee next slide
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 47
Circumferential distribution
Simplified distribution in circumferential direction
φ
pmax
Actual pressure distribution Simplified pressure distribution
pmax
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 48
Effect of vertical acceleration
Due to vertical acceleration, weight of liquid and tank will increase or decrease depending on direction of acceleration
If acceleration is downward, weight will decreaseIf acceleration is upward, weight will increase
Increase in weight = vertical acceleration x mass
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 49
Effect of vertical acceleration
Increase in liquid weight would lead to increase in hydrostatic pressure exerted by liquid on wall
Decrease in weight will reduce hydrostatic pressure, and this need not be considered for design purpose
Hydrostatic pressure varies linearly with heightPressure due vertical acceleration would also vary linearly with height
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 50
Effect of vertical acceleration
Pressure, pv due to vertical acceleration
( ) ( )hyhgAp vv −= 1ρ
y
pv
⎟⎟⎠
⎞⎜⎜⎝
⎛××=
gS
RIZA a
v 232
ρ = Mass density of liquidh = Total liquid heightg = Acceleration due to gravityZ = Zone factorI = Importance factorR = Response reduction factor
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 51
Effect of vertical acceleration
Design acceleration spectrum in vertical direction is taken as 2/3rd of design acceleration spectrum in horizontal direction
This is consistent with IS 1893 (Part 1):2002For finding the value of Sa/g
One needs to know time period of tank in vertical mode
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 52
Effect of vertical acceleration
Usually tanks are quite rigid in vertical directionHence, time period in vertical mode may be taken as as 0.3 sec for all tanksThis implies, value of Sa/g will be 2.5
Vertical mode of vibration for circular tanks is called breathing mode
More information on time period of breathing mode is available in Eurocode 8 and ACI 350.3
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 53
Pressure due to wall inertia
Mass of wall is distributed along its length and heightHence, during lateral excitation, wall is subjected to lateral pressurePressure due to wall inertia is given by
( ) gtAp mihww ρ=
ρm = mass density of tank wallt = thickness of tank wallg = acceleration due to gravity(Ah)i = impulsive base shear coefficient
pww
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 54
Pressure due to wall inertia
Pressure due to wall inertia may not be significant in steel tanks
Steel tank walls have less massHowever for RC tanks, pressure due to wall inertia may be significant
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 55
Total pressure on wall
Total pressure on wall comprises ofImpulsive hydrodynamic pressure, piw
Convective hydrodynamic pressure, pcw
Pressure due to vertical acceleration, pv
Pressure due to wall inertia, pww
Total pressure on wall is obtained as
222vcwwwiw pp)pp(p +++=
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 56
Example
A ground supported circular RC water tank has internal diameter of 12 m and liquid height is 5 m. It is located in zone IV and has fixed base. Wall thickness is 200 mm. Find maximum pressure due to vertical excitation and wall inertia.
Solution:
h = 5 m, D = 12 m, (Ah)i = 0.23, Wall thickness = 200 mmZone IV; Z= 0.24, I = 1.5, It is ground supported RC tank with fixed base, R = 2.0
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 57
Example
For vertical mode, time period is taken as, T = 0.3 sec for all tanks
∴ Sa/g = 2.5Av = 2/3 . Z/2. I/R . Sa/g
= 2/3 x 0.24/2 x 1.5/2.0 x 2.5= 0.15
Maximum pressure due to vertical acceleration will occur aty = 0, i.e. at the base of wall,
∴ pv = (Av) ρ g h (1- y/h) = 0.15 x 1000 x 9.81 x 5 x (1- 0/5)= 7.36 kN/m2
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 58
Example
Maximum pressure due to wall inertia, pww = (Ah)i t ρm g
= 0.23 x 0.2 x 25 = 1.15 kN/m2
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 59
Example
For this tank, from the previous examples, at the bottom of wall (y = 0) we have
Piw = 9.48 kN/m2; Pcw = 1.26 kN/m2;Pv = 7.36 kN/m2; Pww = 1.15 kN/m2;
Hence, total pressure
222vcwwwiw pp)pp(p +++= 222 367261151489 ..)..( +++=
= 13 kN/m2
Total hydrostatic pressure at base = ρ g h = 49 kN/m2.Thus, total hydrodynamic pressure of 13 kN/m2 is 26% of hydrostatic pressure.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 60
Sloshing wave height
Convective liquid undergoes sloshing motioni.e., liquid undergoes vertical motion
dmax
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 61
Sloshing wave height
Maximum sloshing wave height is given by
2DR)A(d chmax = For circular tanks
2LR)A(d chmax = For rectangular tanks
(Ah)c = convective base shear coefficientR = response reduction factorD = diameter of circular tankL = Length of rectangular tank in the direction of seismic force
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 62
Sloshing wave height
Multiplication with R is due to the fact that design forces are reduced by factor R
Recall, (Ah)c = (Z/2) (I/R) (Sa/g)c
Displacements will be R times higher than displacements due to design forcesExpression for sloshing wave height is taken from ACI 350.3
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 63
Sloshing wave height
Sloshing wave height is useful in determining the free board to be provided
If loss of liquid needs to be preventedIf free board is less than sloshing wave height, liquid will exert pressure on roof
In the process, sloshing liquid mass will also changeRefer Malhotra (2005) for more information
Malhotra P K (2005), “Sloshing loads in liquid-storage tanks with insufficient free board”, Earthquake Spectra, Vol. 21, No. 4, 1185-1192
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 64
Anchorage requirement
Ground supported tanks have tendency to overturn during lateral base excitation
Particularly tall steel tanksAnchorages are needed to ensure safety against overturningTank will overturn, if overturning moment due to lateral force is more than stabilizing moment
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 65
Anchorage requirement
As a quick estimate, one can say, overturning will occur if
( )ihADh 1
>
h = height of liquidD = diameter of tank
(Ah)i= impulsive base shear coefficient
For rectangular tanks, diameter D is to be replaced by L
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 66
Anchorage requirement
It is an approximate estimate for circular tanksIt assumes that entire tank mass and liquid mass is acting at the center of liquid heightRead Clause 4.12 of the Guideline
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 67
At the end of Lecture 6
Impulsive and convective pressure have curvilinear distribution along the wall heightEquivalent linear distribution may be used
For obtaining hoop forces and BM in wallWall is also subjected to pressure due to vertical acceleration and wall inertiaSloshing wave height can be obtained using simple expression
Lecture 7
February 16, 2006
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 2
In this Lecture
Examples 2 & 3 of IITK-GSDMA GuidelineThese examples have been sent separatelyHere, we will discuss only important points
These tanks will also be analysed using IS 1893:1984
Comparison of forces obtained from the Guideline and IS 1893:1984 will be presented
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 3
Example 2
Elevated tank considered in Example 2 of the Guidelines has
RC Intze container of 250 m3 capacityRC Frame staging on six columns
Staging is assumed to have been designed for good ductility
Tank is located in seismic zone IV on hard soilPlease refer the examples sent to you
In this example, base shear and base moment at the base of staging are obtained
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 4
Example 2
Analysis involves following calculationsStep 1: Weight of various partsStep 2: Center of gravity of empty containerStep 3: Parameters of spring-mass model
From h/D of equivalent circular container, all the parameters of spring-mass model are obtained
Step 4: Lateral stiffness of stagingUsing standard structural analysis software, stiffness is obtainedMore about this later
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 5
Example 2
Step 5: Impulsive convective mode time periodStep 6: Design horizontal seismic coefficient
(Ah)i and (Ah)c are obtained using Z, I, R, and damping values
Step 7: Base shear and base momentStep 8: Analysis of empty tank
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 6
Example 2
Mass of water = 255.6 tMass of empty container = 160.7 tMass of staging = 105.6 tStructural mass, ms = 160.7 + 105.6/3 = 196 t
ms = empty container + 1/3rd mass of stagingTotal mass = 255.6 + 196 = 451.6 t
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 7
Example 2
Parameters of spring-mass modelHere, container is of Intze typeEquivalent circular container of same volume and diameter is obtained
Equivalent circular container has D = 8.6 m and h = 4.4 m; hence, h/D = 0.51This h/D is used to find mi, mc, hi, hc etc. mi = 140.6 t; about 55% of total water massmc =110 t; about 43% of total water massmi + mc is about 2% less than total water mass. This was discussed in Lecture 3.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 8
Example 2
hi = 1.65 m, hi* = 3.43 m
hc = 2.68 m, hc* = 3.43 m
Incidentally, hi* and hc
* are same in this case.hi
* is twice that of hi
However, hc* is only 1.3 times hc
Indicates that impulsive pressure at the base is more significant than the convective pressure at the base. Hence,
hi* is significantly larger than hi
hc* is not much larger than hc
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 9
Example 2: Lateral stiffness
Lateral stiffness of stagingStaging is modeled using structural analysis softwareForce is to be applied at the CG of tank
CG of tank is combined CG of container and impulsive liquid mass, mi.In this example, CG of empty container is treated as CG of tank
A reasonable approximation for design purpose
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 10
Example 2: Lateral stiffness
CG of empty container is at a distance of 2.88 m from top of circular ring beam
In the model, center-line dimensions are usedHence, force is applied at 2.88 + 0.3 = 3.18 m from center-line of circular ring beam
0.3 m is half-depth of ring beam
Since only staging is modeled, a rigid link of length 3.18 m is provided to apply force at that height
Model is shown in next slide
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 11
Example 2: Lateral stiffness
Rigid link
Structural model of staging used for analysis
10.0
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 12
Example 2: Lateral stiffness
A force of 10 kN is applied at CGDeflection of CG = 5.616 x10-4 mHence, lateral Stiffness of Staging
Ks = 10/(5.616x10-4) = 17,800 kN/mHere, computer software is used for analysisTraditionally, designers use manual methods
Present frame is a polygon frame
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 13
Example 2: Lateral stiffness
For finding stiffness of such frames, general practice in India has been to use method suggested by SP 22
SP:22 – 1982, Explanatory Handbook on Codes for Earthquake Engineering, Bureau of Indian Standards, New Delhi
In this method, braces are considered as rigid beams
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 14
Example 2: Lateral stiffness
If braces are considered as rigid beams, then No rotation is allowed at joints of the columnsHence, stiffness of a column in a panel or between two brace levels is
Where L is length of column in a panel
3C L12EIK =
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 15
Example 2: Analysis as per IS 1893:1984
For the frame in the present exampleStiffness of each column in a panel, Kc
kN/m 36,7304.0
108.76102236012L
12EIK3
933C =××××==
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛
In each panel, there are 6 columnsHence, stiffness of each panel
kN/m 220,400367306K C =×=∑
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 16
Example 2: Analysis as per IS 1893:1984
Stiffness of one panel = 220400 kN/mThere are four such panelsEach having same stiffnessThey act as springs in series Hence, lateral stiffness of staging, Ks
kN/m 55,0904
2204004
KK
C=== ∑s
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 17
Example 2: Analysis as per IS 1893:1984
Thus, stiffness of frameKs = 17,800 kN/m from computer analysisKs = 55,090 kN/m from method of SP22
Thus, SP 22 grossly overestimates the stiffnessClearly, braces should not be treated as rigid beams Their flexibility should be included in the analysis
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 18
Example 2: Analysis as per IS 1893:1984
Jain and Sameer (1992) have developed a simple method for obtaining stiffness of polygon frames
This method considers flexibility of braces 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. (http://www.nicee.org/ecourse/Tank_ICJ.pdf)
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 19
Example 2: Lateral stiffness
As per Jain and Sameer (1992)
axialflexures KKK111
+=
∑=
=pN
i panelflexure KK 1
11
Kpanel = stiffness of each panelNp = number of panels
Top most panel
Bottom most panel
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 20
Example 2: Lateral stiffness
⎥⎦
⎤⎢⎣
⎡+
=hILI
LIh
NEIKcb
bccpanel 2
123 For intermediate panels
⎥⎦
⎤⎢⎣
⎡+
=hILI
LIh
NEIKccbr
cbrccpanel 3
12 For top most panel
⎥⎦
⎤⎢⎣
⎡+
=hILI
LIh
NEIKcb
bccpanel 3
12 For bottom most panel
221
REANK ccaxial= ∑
=
pN
ii hH
1
2
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 21
Example 2: Lateral stiffness
E = Modulus of elasticity
Ic = Moment of inertia of column
Nc = Number of columns
h = Height of each panel
L = Length of each brace
Ib = Moment of inertia of brace beam
Icbr = Moment of inertia of circular ring beam
Hi = Distance of lateral load from inflection point of ith panel
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 22
Example 2: Lateral stiffness
Hi = Distance of lateral load from inflection point of ith panel
For intermediate panels, point of inflection is at mid height of panelFor the top most panel, point of inflection is at a distance y from circular ring beam
For the bottom most panel, distance y is measured from fixed end
h
LI
hI
LI
yb
cb
⎟⎠⎞
⎜⎝⎛
⎟⎠
⎞⎜⎝
⎛ +=
6
3
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 23
Example 2: Lateral stiffness
For 6–column staging of this Example E = 22,360 N/mm2, Ac = π x (650)2/ 4 = 3,31,800 mm2, Nc = 6, L = 3,141 mmR = 3,141 mm Ic = π x (650)4/ 64 = 8.76 x 109 mm4
Ib = 300 x (600)3 / 12 = 5.4 x 109 mm4
Icbr = 500 x (600)3 / 12 = 9 x 109 mm4
h = 4,000 mmH1 = 6030 mm; H2 = 9,180 mm; H3 = 13,180 mm; H4 = 16,330 mm
Note:- Load is applied at CG of container, and distances Hi are from the load.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 24
Example 2: Lateral stiffness
Using this method, we getStiffness of staging, Ks = 16,440 kN/m
From computer analysis, we haveKs = 17,800 kN/m
Difference is about 7%. Which is very reasonableRecall, as per SP 22, Ks = 55,090 kN/m
Which is on much higher side
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 25
Example 2: Lateral stiffness
Note, usually in practice, method of SP 22 is usedAlso this method has been used in a number of text books in India.For further calculations here, Ks = 17,800 kN/m is used
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 26
Example 2: Time period
Tank full conditionImpulsive mode time period, Ti
s
sii K
mm2T += π
= 0.86 SecConvective time period, Tc
D/gCT cc =For h/D = 0.51, from Figure 5 of the Guideline,
Cc = 3.35
Hence, Tc = 3.14 sec
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 27
Example 2: Time period
Convective time period is 3.65 times impulsive time period
Hence, two DoF model can be treated as two uncoupled SDoF systems
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 28
Example 2: Base shear
Design horizontal seismic coefficientZone IV, hence, Z = 0.24 I = 1.5 Frame has ductile detailing, hence, R = 2.5 For Ti = 0.86 sec and hard soil; (Sa/g)i = 1.16
From Clause 4.5.3 of the Guideline
For Tc = 3.14; (Sa/g)c = 0.318 x 1.75 = 0.56Multiplication by 1.75 is for 0.5% damping
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 29
Example 2: Base shear
Design Horizontal Coefficient(Ah)i = Z/2 . I/R . (Sa/g)i = 0.084(Ah)c = Z/2 . I/R . (Sa/g)c = 0.040
(Ah)c is less than (Ah)iConvective mass is subjected to much smaller acceleration
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 30
Example 2:Base shear
Base Shear, VImpulsive mode
Vi = (Ah)i (mi + ms)gVi = 0.084 x (140.6 + 196) x 9.81
= 116 +161 = 277 kNConvective modeVc = (Ah)c mc g = 0.04 x (110) x 9.81 = 43 kN
Convective forces are less than impulsive forcesConvective mass derives much smaller forces than impulsive mass
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 31
Example 2:Base shear
Total Base shear, = 280 kNBase shear is about 6.3 % of total seismic weight of 4429 kN (tank is located in seismic zone IV)
Now, to obtain member forces, impulsive and convective base shear is to be applied at corresponding locations Recall from Lecture 5, there are two approaches to apply these forces
2c
2i VVV +=
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 32
Example 2:Base shear
Approach 1:
Apply impulsive forces at their respective locations
Impulsive forces have two partsPart1: 116 kN at hi
* + hs
Part 2: 161 kN at hcg
Apply convective forces at its location43 kN at hc
* + hs
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 33
Example 2:Base shear
Location of forces as per Approach 1
Top of footing
161 kN
hs = 16.3 mhcg = 19.18 m
hi*= 3.43 m
116 kN 43 kN
hs = 16.3 m
hc*=3.43 m
Impulsive forces Convective forces
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 34
Example 2:Base shear
Approach 2:Apply base shear, V at height h1 such that
V x h1 = M*
V = 280 kN and M* = 5448 kN-mHence, h1 = 5448/280 = 19.46 m
Thus, a force of 280 kN to be applied at 19.46 m
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 35
Example 2:Base shear
Location of forces as per Approach 2
Top of footing
h1 =19.46
280 kN
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 36
Example 2:Base shear
Analysis for tank empty ConditionNo convective modeImpulsive mass will be only structural mass, ms
V = (Ah)i ms g = 211 kNThis will act at CG of empty container
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 37
Example 2:Base shear
Tank empty conditionMoment = 211 x 19.18 = 4053 kNm
Top of footing
hcg =19.18
211 kN
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 38
Example 2:Base shear
Base shear and moment are more in tank full condition
Hence, design will be governed by tank full condition
These forces are to be applied in suitable directions to get critical design forces in columns and braces
Recall, the issue of critical direction for frame staging
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 39
Analysis as per IS 1893:1984
Now, this tank will be analysed as per IS 1893:1984
This will help us compare design forces as per the Guideline and IS 1893:1984
Let us recall two pointsIn IS 1893:1984 entire liquid mass is lumped as impulsive mass
Hence, no convective modeWhile finding the stiffness of staging, braces are treated as infinitely rigid beams
This was general practice as suggested by solved example in SP22.
Even though not suggested by IS:1893-1984.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 40
Example 2: Analysis as per IS 1893:1984
Zone and soil parameters from IS 1893:1984β = 1.0 I = 1.5Fo= 0.25 (Zone IV)
Time Period, T
M is total Mass This is sum of structural mass and total liquid mass
K is Stiffness of Staging
KMT π2=
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 41
Example 2: Analysis as per IS 1893:1984
Total Mass, M = Structural mass + liquid mass= 196 +255.6 = 451.6 t
Lateral stiffness KsBraces are treated as as rigid beams as per prevailing practiceHence, Ks = 55,090 kN/m
This we have seen earlier
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 42
Example 2: Analysis as per IS 1893:1984
Time period
Sec 0.5755090451.62π
KM2πT ===
This time period is less than the impulsive time period of 0.86 sec calculated earlier
Because, stiffness of braces is taken infiniteFor 5% damping & T = 0.57 Sec
Sa/g = 0.155From Figure 2 of IS 1893 : 1984
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 43
Example 2: Analysis as per 1893:1984
Design horizontal seismic coefficient, αhαh = βIFo(Sa/g) = 1.0 x 1.5 x 0.25 x 0.155
= 0.058Base Shear, V = αh x W
= 0.058 x (451.6 x 9.81)= 257 kN
Base shear is 5.8 % of seismic weight of 4429 kNThis will act at the CG of container
CG of container is at 19.18 m from footing top Hence base moment, M = 257 x 19.18 = 4,929 kNm
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 44
Example 2: Analysis as per 1893:1984
Analysis for empty conditionStructural mass, M =196 t
For 5% damping & T = 0.375 SecSa/g = 0.2
As per Figure 2 of IS 1893:1984
Design horizontal seismic coefficient, αhαh = βIFo(Sa/g) = 1.0 x 1.5 x 0.25 x 0.2 = 0.075
Sec 0.375550901962π
KM2πT ===
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 45
Example 2: Analysis as per 1893:1984
Base Shear, V = αh x W= 0.075 x (196 x 9.81) = 144 kN
Moment at the bottom of staging, M= 144 x 19.18 = 2762 kNm
Next, we compare seismic forces obtained from the Guideline and IS 1893:1984
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 46
Example 2:Comparison of forcesComparison of results from the Guideline and IS 1893:1984
Guideline IS 1893:1984Lateral stiffness of staging 17,800 kN/m 55,089 kN/m
Time periodImpulsive mode, Tank empty ( Ti )Tank full ( Ti)Convective mode,Tank full ( Tc)
0.66 sec0.86 sec
3.14 sec
0.375 sec0.57 sec
-----Design seismichorizontal coefficientImpulsive modeTank empty ( Ah)i
Tank full ( Ah)i
Convective mode,Tank full ( Ah)c
0.110.084
0.040
0.0750.058
-----
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 47
Example 2:Comparison of forces
Guideline IS 1893:1984
Total base shear (V)Tank empty Tank full
211 kN280 kN
144 kN257 kN
Bending MomentTank emptyTank full
4053 kN5448 kN
2762 kNm4929 kNm
Comparison of results from the Guideline and IS 1893:1984
In tank full condition, base shear and moment from the Guideline are about 10% more than those obtained using IS 1893:1984
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 48
Example 2:Comparison of forces
Two points may be notedIS 1893:1984 implied K = 1.0, which leads to smaller seismic forcesHowever, the practice was to overestimate stiffness,which increases seismic coefficient. These two effects compensate each otherHence, there is not much difference in design forces from the Guideline and IS 1893:1984 for elevated tanks on frame type staging.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 49
Example 2: Member forces
To obtain design forces in columns and bracesStaging frame is to be analysed for lateral seismic forceOne can use standard structural analysis programs.Alternatively, one may use approximate analysis methods.
Plane frame analysis methods are well knownFor example, moment distribution method andKani’s method
In this example, staging has polygon frameIt is not a plane frameStrictly speaking, it is a space frame
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 50
Example 2: Member forces
Some text books describe methods for analysis of such polygon frames
Jain and Jai Krishna (1980) and Dayaratnam(1986)
Jain O P and Krishna J, 1980, “Plain and reinforced Concrete”, Vol. 2, Eighth Edition, Nem Chand Brothers, RoorkeeDayaratnam P, 1986, “Design of Reinforced Concrete Structures”, Oxford & IBH Publishing Co., New Delhi
These methods make simplifying assumptionsAccuracy of these methods depends on validity of these assumptions
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 51
Example 2: Member forces
Major assumptions in these methods are:1) Axial force in column due to lateral force is
proportional to its distance from bending axis2)Point of inflection in columns and braces are at
mid-span3)Dayaratnam’s book assumes that lateral shear is
equally distributed to all columns, ORJain and Jai Krishna distribute lateral shear similar to that in a cross-section of a cantilever column
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 52
Example 2: Member forces
Sameer and Jain (1994) have critically examined these assumptions
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. (http://www.nicee.org/ecourse/Tank_ASCE.pdf)
They pointed out that assumption regarding shear distribution is not necessary
That is, assumption 3) in the previous slide.Column shear is obtained by moment equilibrium in the plane of bending
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 53
Example 2: Member forces
Point of inflection for the top and bottom panels is not at mid-spanSameer and Jain obtained point of inflection for these panels more accurately.Based on these observations, Jain and Sameer(1994) have developed a new approach to obtain design forces in columns and braces
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 54
Example 2: Member forces
For the present example, design forces for columns and braces are obtained
Total lateral force = 280 kN acting at 19.46 m from footing top (i.e., fixed end)
Design forces in columns and braces from following methods are compared in next slide
Computer analysis Jain and SameerJain and Jai Krishna Dayaratnam
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 55
Example 2: Member forces
Comparison of design forces in columns and braces
Computer analysis
Jain & Sameer
Jain & JaiKrishna
Dayaratnam
ColumnsAxial force (kN)Bending moment (kNm)Shear force (kN)
49215066
49416070
51918793
5199447
BracesBending moment (kNm)Shear force (kN)
175112
187119
280178
187119
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 56
Example 2: Member forces
Jain & Jai Krishna’s approach overestimates bending moments and shear forces in columns and bracesDayaratnam’s approach underestimates bending moments and shear forces in columnsThe point here is that designer should properly assess the limitations of a particular method
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 57
Example 3:Elevated Tank on Shaft
Example 3 of the Guideline Elevated tank on RC shaft is considered
Container is same as in Example 2 Thickness of shaft = 0.15 mCentre line diameter of shaft = 6.28 mHeight of shaft above GL = 15 mHard soil; Zone IV
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 58
Example 3:Elevated tank on Shaft
Weight of staging, i.e., shaft= π x 6.28 x 0.15 x 16.4 x 25 = 1,213 kNMass of staging = 123.7 t
Hence, structural mass msms = 160.7 + 123.7/3 = 201.9 t
Container is same as in Example 2, hence,mi = 140.6 tmc =110 t
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 59
Example 3:Lateral stiffness
Lateral stiffness of stagingShaft is considered as a cantilever of length 16.4 m
This is the height of shaft from top of footing upto bottom of circular ring beamLateral stiffness is given by
Where,E = Modulus of elasticity = 22.36 x 106 kN/m2
3S L3EIK =
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 60
Example 3:Lateral stiffness
I = Moment of inertia of shaft cross section= π x (6.434 – 6.134)/64 = 14.59 m4
L = Height of shaft = 16.4 mThus,
kN/m102.2216.4
14.591022.363K 53
6
S ×=×××
=
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 61
Example 3:Lateral stiffness
Height of shaft is taken up to ring beam only and not up to CG of empty container
Recall, in frame staging, stiffness was obtained by applying force at the CG of empty container
In frame staging, a rigid link was used from staging top to apply force at CG
The effect of larger rigidity of container portion is thus included
Even if load is applied at top of staging rather than CG, there is not much difference in stiffness
Rigid link portion does not have much higher deflection than deflection at staging top
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 62
Example 3:Lateral stiffness
In shaft, if we wish to consider height up to CG, then, cantilever will have two different cross-sections
One up to staging topSecond from staging top to CG
This will represent container portion up to CG
Due to larger dimensions in the second portion, It will undergo only rigid body rotationDeflection at CG will not be much higher than that at staging top
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 63
Example 3:Lateral stiffness
Hence, staging height taken up to ring beam is adequate for design
This will be on the conservative sideAlso recall time period varies as square root of stiffness
A 15% error in stiffness will give about 7% error in time period
There is another approximation in stiffness of shaft
Shear deformations
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 64
Example 3:Lateral stiffness
Effect of shear deformationConsider a cantilever subjected to tip load , PTip deflection due to shear deformation
G = Shear modulusA = Cross-sectional areaκ’ = Shape factor
Depends on shape of cross-section
AGPL's κ
δ =
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 65
Example 3:Lateral stiffness
Thus, total deflection of cantilever
AGL
EIL
K
'
s
κ+
=
3
13
AGPL
EIPL
'κδ +=
3
3
First part is due to flexural deformationSecond part is due to shear deformation
Hence, stiffness will be
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 66
Example 3:Lateral stiffness
For hollow circular cross-sectionκ’ = 0.5
Hence, lateral stiffness of shaft staging, including shear deformations, will be
AG.L
EIL
Ks
503
13
+=
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 67
Example 3:Lateral stiffness
In the present example, if shear deformations are included, then
Ks = 1.77 x 105 kN/m
Recall, without shear deformation, Ks = 2.22 x 105
kN/mBy including shear deformations, stiffness has reduced by 20%. For further calculations, Ks=2.22 x 105 kN/m is used
This value is also used in the GuidelineHowever, it should have been 1.77 x 105 kN/m
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 68
Example 3: Time period
Analysis for tank full conditionImpulsive mode time period, Ti
s
sii K
mm2T += π
= 0.25 Sec
Shafts are quite rigid, hence, low time periodRecall, for frame staging, Ti = 0.86 sec
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 69
Example 3: Time period
Convective time period, Tc = 3.14 secSame as in Example 2Ratio of Tc and Ti is about 12.0Impulsive and convective time periods are quite well separated
Hence, the assumption of two uncoupled SDoFcan be used
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 70
Example 3: Base shear
Design horizontal seismic coefficientZone IV, hence, Z = 0.24 I = 1.5 R = 1.8 For Ti = 0.25 sec and hard soil; (Sa/g)i = 2.5
From Clause 4.5.3 of the Guideline
Note:If we include the effect of shear deformation,
then time period Ti would have been 0.28 secAnd, (Sa/g)i will remain 2.5
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 71
Example 3: Base shear
For Tc = 3.14; (Sa/g)c = 0.318 x 1.75 = 0.56Multiplication by 1.75 is for 0.5% damping
Design Horizontal Coefficient, Ah(Ah)i = Z/2 . I/R . (Sa/g)i = 0.25(Ah)c = Z/2 . I/R . (Sa/g)c = 0.06
(Ah)c is much less than (Ah)iConvective mass is subjected to much smaller acceleration
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 72
Example 3:Base shear
Base Shear, VImpulsive mode
Vi = (Ah)i (mi + ms)gVi = 0.25 x (140.6 + 201.8) x 9.81
= 345 + 495 = 840 kNConvective modeVc = (Ah)c mc g = 0.06 x (110) x 9.81 = 65 kN
Convective forces are much smaller than impulsive forces
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 73
Example 3:Base shear
Total Base shear, = 843 kNBase shear is about 19 % of total seismic weight of 4,488 kN (tank is located in seismic zone IV)
2c
2i VVV +=
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 74
Example 3:Base shear
Impulsive forces have two partsPart1: 345 kN at hi
* + hs
Part 2: 495 kN at hcg
Apply convective forces at its location65 kN at hc
* + hs
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 75
Example 3:Base shear
Base moment in impulsive modeMi
* = 345 x (3.43 + 17) + 495 x 19.88= 169,00 kNm
Base moment in convective modeMc
* = 65 x (3.43 + 17) = 1322 kNm
Total base moment
= 16,940 kNm
2c
*2i
** MMM +=
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 76
Example 3:Base shear
Tank Empty ConditionTime Period Ti = 0.19 SecBase Shear, V = 495 kNBase Moment, M* = 9,842 kN-m
Seismic base shear and moment are higher in tank full condition
Hence, tank full condition will govern the design
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 77
Example 3:Analysis using IS 1893:1984
This tank is also analysed using IS 1893:1984Entire liquid mass is lumped as impulsive massNo convective mode
For tank full conditionMass, M = Mass of container + Mass of water
+ 1/3rd Mass of staging= 160.7 + 255.6 + 1/3 x 123.7 = 457.5 t
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 78
Example 3: Analysis using IS 1893:1984
Stiffness of staging, Ks = 2.22 x 105 kNmTime period
Base shear coefficient αh = βIFo(Sa/g) = 1.0 x 1.5 x 0.25 x 0.2
= 0.075
Sec0.285102.22
457.52πKM2πT 5 =
×==
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 79
Example 3: Analysis using IS 1893:1984
Base Shear, V = αh x W= 0.075 x 4488 = 336.6 kN
Base Moment, M = V x hcg= 336.6 x 19.88= 6,692 kN-m
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 80
Example 3: Analysis using IS 1893:1984
Analysis for Tank EmptyStructural mass, M M = Mass of container + 1/3rd Mass of staging
= 160.7 + 1/3 x 123.7 = 201.9 T
Design horizontal seismic coefficient, αh
αh = βIFo(Sa/g) = 1.0 x 1.5 x 0.25 x 0.2 = 0.075
Sec0.19102.22
201.92πKM2πT 5 =
×==
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 81
Example 3: Analysis using IS 1893:1984
Base Shear, V = αh x W= 0.075 x 1980 = 148.5 kN
Base Moment, M = V x hcg
= 148.5 x 19.88 = 2952 kN-mLet us compare results from the Guideline and IS 1893:1984
See next slide
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 82
Idealization of Tank Guideline IS 1893:1984Lateral stiffness of staging 2.22 x 105 kN/m 2.22 x 105 kN/mTime period
Impulsive mode, Tank empty ( Ti )Tank full ( Ti)Convective mode,Tank full ( Tc)
0.19 sec0.25 sec
3.14 sec
0.19 sec0.285 sec
-----Design seismic horizontal coefficient
Impulsive modeTank empty ( Ah)i
Tank full ( Ah)i
Convective mode,Tank full ( Ah)c
0.250.25
0.060
0.0750.075
-----
Comparison of results from the Guideline and IS 1893:1984
Example 3: Comparison of results
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 83
Idealization of Tank Guideline IS 1893:1984
Total base shear (V)Tank empty Tank full
495 kN843 kN
148.5 kN336.6 kN
Total base Moment (M)Tank empty Tank full
9,842 kN-m16,940 kN-m
2,952 kN-m6,692 kN-m
Comparison of results from the Guideline and IS 1893:1984
Example 3:Comparison of results
For shaft supported tanks, seismic forces from the Guideline are on much higher side
Base moment is 150% higher in full condition
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 84
Example 3:Comparison of results
Comparison of forces from the Guideline and IS 1893:1984 reveals:For frame staging
Design forces increase marginallyHowever, this increase will vary for different tanks
For shaft stagingDesign forces increase significantlyIn these tanks, increase in base moment is of particular interest
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 85
Analysis using IS 1893:1984
Main reason for this increase is:IS 1893:1984 has K = 1.0 for all types of tanks
Thereby implying that all elevated tanks have same ductility or energy absorbing capacity as that of a building with good ductility
IITK-GSDMA Guideline has specified different values of response reduction factors for different types of staging
Depending on their ductility, redundancy and overstrength
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 86
Analysis using IS 1893:1984
In frame staging, earlier practice was to overestimate lateral stiffness
Thereby increases design seismic forcesTo some extent, this compensates the lower values of design forces due to K = 1.0Hence, in frame staging net increase in forces is not very significant
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 87
Analysis using IS 1893:1984
For shaft staging, increase in design seismic forces will not necessarily cause proportionate increase in cost of the tank
Since lateral forces are higher, one will have to suitably choose the dimensions of shaftA wider shaft will help in achieving more economical designTo explain this point, cost of shaft and foundation is obtained for one tankSee next example
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 88
Issue of cost
Consider a 1000 m3 elevated tank on RC shaft with following data
Height of shaft = 20 m (from foundation top)Shaft thickness = 200 mmZone V, Medium soil, SBC = 25 t/m2
Solid raft foundation is used
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 89
Issue of cost
In the first instance, shaft of 7 m diameter is consideredThen, shaft of 12 m diameter is consideredComparison of quantity of concrete and steel required in both the cases is given in next slide
Quantity for shaft and foundation are given
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 90
Issue of cost
Material quantities required
7 m dia. Shaft 12 m dia. shaft
IS 1893:1984 Guideline IS 1893:1984 Guideline
ShaftConcrete Steel
90 m3
6 t**
150 m3
7 t150 m3
9 t
FoundationConcreteSteel
200 m3
8 t540 m3
21 t135 m3
9 t340 m3
12 t
* Design of 7 m shaft is not possible for forces as per the Guideline
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 91
Issue of cost
Thus, by increasing shaft diameter from 7 m to 12 m
About 200 m3 of concrete and 9 t of steel in foundation could be saved
This example is shown to emphasize the importance of proportioning of dimensions in achieving economy in design
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 92
Issue of cost
Some more economy in foundation can perhaps be achieved by using different type of foundation systemSome options are:
Solid mat foundationAnnular mat foundationStrip foundation on pilesShell type foundation
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 93
Issue of cost
Note, design of container does not get much affected, since hydrodynamic pressure is about 30 to 40% of hydrostatic pressure
In working stress design, permissible stresses are increased by 33.3%This would usually accommodate the hydrodynamic pressure
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 94
Issue of cost
It is to be recognized that, so far, in India, seismic design forces for tanks were rather low
Hence, simplistic design approaches prevailedAnd above discussed options were not needed and were not attemptedMoreover, in many elevated tanks, wind forces used to govern the design
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 95
Issue of cost
In view of new design forcesThere is a need to try some of these optionsTo achieve better economy, and Good seismic performances
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 96
At the end of Lecture 7
In Part B of the Guideline, six solved examples are given
These will be of help to the users of the GuidelineWe are now almost at the end of this course
Next Lecture, which is the last one, will discuss some miscellaneous issues.
Lecture 8
February 21, 2006
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 2
In this Lecture
Following issues will be discussed Soil structure interactionBuried tanksP-Delta effectFlexibility of pipingBuckling of shellIS 11682:1985
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 3
Soil Structure Interaction
Soil condition is important in seismic analysis Type of soil affects the ground motion and structure’s responseEffect on ground motion is reflected in design spectrum
Recall, there are different response spectra for hard, medium and soft soilSpectrum for soft soil is higher than hard soil
Except in short period range
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 4
Soil Structure Interaction
Effect of soil on structure’s response is included in soil structure interaction Soil-structure interaction is a complex problem
Soil properties are frequency dependent Based on research, simplified methods have been proposed
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 5
Soil Structure Interaction
We will briefly discuss method suggested by Veletsos (1984)
This has been used in Eurocode 8 and NZSEE Recommendations (Priestley, 1986)
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 Engineering, ASCE, N.Y., 255-370, 443-461.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 6
Soil Structure Interaction
From Veletsos (1984):Soil structure interaction has two effects
It changes time period It changes total damping
Effect on time periodIf structure is on hard soil, it is treated as fixed at baseIf soil is soft, it imparts flexibility and increases time period Soil flexibility affects impulsive mode time period
It does not affect convective mode time period
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 7
Soil Structure Interaction
Effect on dampingSoil medium has higher dampingDamping in soil comes from hysteric action and radial waves
Also called radial damping in soil
Damping of soil increases total damping of soil-structure system
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 8
Soil Structure Interaction
Veletsos (1984) suggested following expression for impulsive time period of tank on soft soil
⎥⎥⎦
⎤
⎢⎢⎣
⎡++=
−
θKhK
KK
TT fx
x
fii
2
11
Kx = Horizontal translational stiffness of soilKθ = Rocking stiffness of soilKf = Stiffness of impulsive modehf = Distance from the tank base to the point of application of
impulsive mass, mi
Ti = Impulsive mode time period of tank with fixed base
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 9
Soil Structure Interaction
Gs = Shear modulus of elasticity of soil
νs = Poisson’s ratio of soilR0 = Radius of the foundation
028 RGK s
sX ν−=
3013
8 RG)(
K ssν
θ −=
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 10
Soil Structure Interaction
For elevated tanksStiffness of impulsive mode, Kf is lateral stiffness of staging, Ks
For ground supported tanksWe have considered circular and rectangular tanksFor both types of tanks, time period of impulsive mode, Ti is known
Recall expression for Ti for circular tanks from Lecture 3Also refer Clause 4.3.1.1 and 4.3.1.2 of the Guideline
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 11
Soil Structure Interaction
Hence, stiffness of impulsive mode, Kf for ground supported tanks can be obtained from
f
ii K
mT π2=
mi being impulsive mass is also known
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 12
Soil Structure Interaction
hf is distance of impulsive mass from base of tankIn ground supported tanks
hf = hi + tb
In elevated tanksTotal impulsive mass = mi + ms
Hence, hf = hcg
Recall, we assume that entire mass is lumped at CG of empty container
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 13
Soil Structure Interaction
Veletsos (1984) included effect of soil damping as follows:
Effective damping ratio of structure is
3
⎟⎠
⎞⎜⎝
⎛+=
−
−
i
S
T/iT
ξξξ
ξs = Damping ratio for soil
ξ = Damping ratio of structureTi = Impulsive mode time period with fixed base
ξ−
= Effective damping ratio of structure including soil effect
−
iT = Impulsive mode time period including soil effect
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 14
Soil Structure Interaction
It is clear that soft soil would have lower values of lateral stiffness Kx and rocking stiffness, Kθ.
Hence, more effect on impulsive time periodSimilarly, damping in soft soil is more compared to hard soil
Hence, damping of structure will be moreThus, soil structure interaction would be important for structures on soft soil
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 15
Soil Structure Interaction
Increase in time period and damping would reduce seismic forces
This is also called beneficial effects of soilBefore using these beneficial effects of soil, one needs to properly ascertain the soil properties
Note, soil properties are frequency dependent
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 16
Buried tanks
In buried tanks, effect of soil pressure on wall shall be includedDuring lateral excitation, soil induces dynamic pressure on wallStrictly speaking, a detailed soil-structure interaction study is needed to find earthquake induced soil pressure
Specialised literature shall be referred to find dynamic earth pressure and its distribution
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 17
Buried tanks
NZSEE has suggested following simplified distribution for earthquake induced soil pressure
Em
bedm
ent
dept
h, H
e
Tank wall
σx = 1.5 (Ah)i γs He Ground surface
Idealised pressure distribution
Tank baseσx = 0.5 (Ah)i γs He
PE
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 18
Buried tanks
Total force per unit length due to this pressure is given by
PE = (Ah)i γsHe2
This is in addition to soil pressure in at-rest conditionThis earth pressure shall be used to check wall designThis dynamic earth pressure shall not be relied upon to reduce dynamic effects due to liquid
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 19
Buried tanks
Seismic analysis procedure for buried tanks is same as that for ground supported tanks
i.e., procedure to find mi, mc etc. remains sameTime period of impulsive mode will be very less, hence, (Sa/g)i = 2.5
Buried tanks are quite rigid
Convective time period will be same as that for ground supported tanks
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 20
Buried tanks
For buried tanks, response reduction factor, R = 4.0
This is for fully buried tankRefer Table 2 of the Guideline
If tank is partially buried, then, R value shall be linearly interpolated
Depending on depth of embedment
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 21
Buried tanks
For a RC tank with fixed baseR = 2.0 for ground supported tankR = 4.0 for fully buried tankIf tank is half buried, then, R = 2.0 + (4.0 – 2.0) x 0.5 = 3.0For 2/3rd buried tank, R = 2.0+(4.0 – 2.0)x 0.67 = 3.33
For a steel tank with anchored baseR = 2.5 for ground supported tankIf tank is half buried, then, R = 2.5 + (4.0 – 2.5) x 0.5 = 3.25
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 22
Buried tanks
Dynamic soil pressure would also act on shaft staging of elevated tanks
For large tanks, shaft diameter could be more than 10 mFoundation for shaft is usually 2 to 3 m below groundHence, soil pressure will act over buried portion of shaft
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 23
Buried tanks
Dynamic soil pressure acts opposite to direction of seismic loading
Seismic force
Soil pressure
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 24
Buried tanks
Soil pressure would induce moment on the foundation
This moment would be opposite to moment due to seismic forcesThus, net moment on foundation would reduce
Codes do not allow this reduction for design purpose
Due to uncertainties in compactness of soil etc.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 25
Buried tanks
In frame staging, column sizes are quite smallCompared to shaft diameter
Members with such small width may not be able to mobilize dynamic soil pressure
Hence, no dynamic soil pressure on embedded length of columns
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 26
P-Delta effect
Consider a vertical cantilever subjected to lateral force Q and vertical force PBending moment at the base, M = Q x L
Q
P
L
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 27
P-Delta effect
Due to force Q, let deflection be ΔPoint of application of P shifts
And, additional moment p x Δ will act on the cantilever
Q
P
Δ
L
Now, total moment at the base = Q x L + P x Δ
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 28
P-Delta effect
Thus, vertical loads interact with lateral displacements
And, induce additional forces on structureThis is termed as P-Delta effectDue to additional moment of P x Δ, lateral deflection would further increase
Which, in turn would cause more moment
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 29
P-Delta effect
P-Delta effect can be minimised by limiting the lateral deflection
For buildings, IS 1893(Part 1) requires that storey drift shall be less than 0.4%i.e., total lateral deflection will be always less than h/250.
h is total height of the building
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 30
P-Delta effect
Compared to buildings, P-Delta effect will be more severe in elevated tanksIn buildings, vertical loads are distributed over the entire height
In elevated tanks, almost entire vertical load is at the top
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 31
P-Delta effect
Thus, to minimise P-Delta effect in elevated tanksPermissible maximum lateral deflection shall be less than that for buildingsh/500 may be a reasonable limit
Recall, in building limit is h/250
i.e., total lateral deflection shall be less than 0.2% of height
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 32
Flexibility of piping
In past earthquakes, many tanks have suffered damages at the junctions of tank and piping These junctions are critical locations
Severe stress concentration occurs at junctionsHence, piping shall have sufficient flexibility to accommodate seismic movement
Without causing damage to tank shell or base
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 33
Flexibility of piping
All the international codes emphasize on flexible piping systemFlexibility in piping can be imparted by
Using flexible elbowsSpecial coupling devices
These flexible devices need not be used at the junction of piping and tanks
They can be used at nearby joints also
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 34
Buckling of shell
Ground supported circular steel tanks are sensitive to buckling
These are thin shell structuresIn the past, many steel tanks have suffered buckling failure during earthquakes
API 650, AWWA D-100 are exclusively for ground supported steel tanks
These codes give provisions for safety against buckling
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 35
Buckling of shell
During lateral base excitation, tank wall is subjected to excessive compressive force
This may trigger bucklingBuckling loads are very sensitive to initial imperfections in geometry
Particularly, for thin shells
NZSEE recommendations (Priestley et al., 1984) have also given information on buckling of steel tanks
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 36
Buckling of shell
For safety against bucklingCodes restrict maximum allowable axial stresses in steel tanks
RC shaft staging may also be thin shell structureThey are not as thin as steel tanksNevertheless, safety against buckling should be ensured
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 37
Buckling of shell
Not much significant information is available on buckling of RC shaft staging
ACI 371 (1998) has specified a limit on maximum axial force on shaftThis limit depends on slenderness coefficient or ratio of thickness to diameter
ACI 371 , (1998) “ Guide for the analysis, design , and construction of concrete-pedestal water Towers”, American Concrete Institute, Farmington Hill, MI, USA.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 38
Buckling of shell
As per ACI 371, in limit state design, for shafts under pure compression
P < 0.7 x Cw x fck x βwxAc
P = Axial load on shaftCw = Wall strength coefficient = 0.55fck = Specified compressive strength of concrete Ac = Gross area of concrete βw = Wall slenderness coefficient
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 39
Buckling of shell
Wall slenderness coefficient, βw is obtained from classical elastic buckling strength of cylinder and is given by
βw = 80 t/dw
where, t = Wall thickness and dw= Center-line diameter of shaft
βw shall be less than 1.0
For wider shafts with less thickness, βw will be less than 1.0
Which will reduce the permissible axial load
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 40
Buckling of shell
Geometrical imperfections are invariably present in shaft staging
Stringent construction tolerances should be followed to minimize these imperfectionsRecall, buckling load is sensitive to imperfections
Effect of these imperfections on static response, dynamic response and buckling should be studied
A good research topic !!
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 41
Buckling of shell
ACI 371 specifies limits on construction tolerances; Some of these areVariation in wall thickness:
-3% and + 5% of wall thicknessVariation in plumb of shaft:
Less than 10 mm in any 1.6 m of heightLess than 40 mm in any 16 m of heightMaximum of 75 mm in total height
Variation in diameter of shaft:0.4% of diameter and shall not exceed 75 mm
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 42
IS 11682:1985
In India, we have IS 11682:1985 exclusively for RC staging
However, it does not have any provisions on construction tolerances for shaft stagingThese should be included in IS 11682
In fact, there are many other limitations in IS 11682:1985Some of these will be discussed briefly
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 43
IS 11682:1985
Clause 3.2 of IS 11682: “…. Weight of water may be considered as live load for members directly containing the same….”
This implies that for container design, water can be treated as live loadIn seismic analysis of buildings, one applies reductions in live load.This confuses the design
Hence, this clause should be corrected
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 44
IS 11682:1985
Clause 5.4: Increase in permissible stresses for column staging shall be as per IS 456-1978Clause 5.4.1: “The increase in permissible stresses need not be allowed in the design of braces for forces as wind or earthquake, which are primary forces in them
Increase in permissible stresses is allowed for columns and not for braces
i.e., braces will be stronger than columns
Whereas, in seismic design we need strong column-weak beam combination
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 45
IS 11682:1985
Clause 7.3.3. “Loss of contact in the soil under footing shouldnot be allowed. In locations where the soil bearing capacity is high, loss of contact may be allowed provided it is safe againstoverturning and such other conditions that are to be fulfilled.”
This clause is for column footingsIt needs to be more specific regarding how much loss of contact can be allowed or how much tension in soil may be permittedSeismic loadings are momentary and like increase in SBC, some tension may also be permittedThese should also be permitted for raft foundations.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 46
IS 11682:1985
Clause 7.2.4. “For staging in seismic zones where seismic coefficient exceeds 0.05 twin diagonal vertical bracing of steelor RCC in addition to horizontal bracing may be provided….”
This clause is for frame stagingLimit of 0.05 on seismic coefficient may be removedSuggestions on use of shear walls in addition to vertical diagonal bracing may be included
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 47
IS 11682:1985
Clause 8.2 provides provisions on minimum thickness of RC shaft, reinforcement and detailing near openings.
These provisions should include suggestions on proportioning of shaft and container diametersSimilarly, different types of foundation systems for shaft should be suggested
This has been done in ACI 371
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 48
IS 11682:1985
Clause 8.5.2 suggests formulae for stresses in shaft
These are based on working stress design methodAs per IS 456:2000, limit state design is compulsory for members subjected to combined direct load and flexural
Clause B-4.3 of IS 456:2000
Hence, provisions on limit state design of shaft are needed in IS 11682
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 49
IS 11682:1985
There are many more such modifications needed in IS 11682
An urgent revision of this code is needed. This would help designers in analysis and design of staging
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 50
And finally…..
With this Lecture, we conclude this E-coursePlease spare some time to go through the Lectures againWe have received some very good questions
Discussion and answers to questions on Lectures 1, 2 and 3 have already been sent to you
Please send your remaining questionsThank You………
1
Assignment 1
Date of assignment: 18 January 2006
E-course on Seismic Design of Tanks
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Assignment 1 / slide 2
What is expected?
Part I: Reading assignments are to be completed as soon as possible
Best if you finish the reading assignments on the same dayIt will help you appreciate the next lecture
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Assignment 1 / slide 3
What is expected?
Parts II and III To be completed by 19 January; before we send out the solutions of these to you.Please do not send the solutions to us. Compare your answers with the solutions we will send.Feel free to go back and review the lecture while answering the questions.Some questions may not have very specific answers
This is in line with our engineering profession wherein certain questions have no specific answers.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Assignment 1 / slide 4
Part I: Reading Assignment 1
Read carefully Preface and Clauses 0., 1., 2., 4.1, 4.2, & 4.2.1 of the Guidelines
Mark whatever you find difficult to followAlso mark portions that you find interesting
It will help you review the materials later.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Assignment 1 / slide 5
Part II: True / False
Identify the following statements as True or False1.1)Hydrodynamic pressure varies linearly with liquid height.1.2)Impulsive mass is less than convective mass for short
tanks.1.3)Net hydrodynamic force on the container wall is zero.1.4)Hydrodynamic pressure on base causes overturning
moment on tank.1.5)Convective mass acts at lower height than impulsive
mass.1.6)With the inclusion of base pressure effect, overturning
moment on tank reduces.1.7)Impulsive and convective masses depend only on height
of liquid.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Assignment 1 / slide 6
Part III: Questions
1.1)A circular tank has internal diameter of 12 m and water height of 5 m. Obtain mi, mc, Kc, hi, hc, hi
* and hc*.
1.2) A rectangular tank has internal plan dimensions of 10 m x 16 m. Water height is 8 m. Obtain mi, mc, hi, hc, hi
* and hc* in both the
directions.
1
Solution 1
Date of assignment: 18 January 2006Date of solution: 20 January 2006
E-course on Seismic Design of Tanks
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 1 / slide 2
Part II: True / False
Identify the following statements as True or False1.1)Hydrodynamic pressure varies linearly with liquid height.
FalseHydrodynamic pressure has curvilinear variation along the height. In fact, Hydrostatic pressure varies linearly with height.
1.2)Impulsive mass is less than convective mass for short tanks.False
In short tanks, convective liquid or liquid undergoing sloshing motion is more. In tall tanks, impulsive liquid mass is more.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 1 / slide 3
Part II: True / False
1.3)Net hydrodynamic force on the container wall is zero.FalseRather, net hydrostatic force on wall is zero
1.4)Hydrodynamic pressure on base causes overturning moment on tank.TrueRefer slide no. 46 of Lecture 1
1.5)Convective mass acts at lower height than impulsive mass.
FalseLiquid in upper portion, undergoes convective or sloshing motion, hence, convective mass is always located athigher height than impulsive mass. Note the graphs for hiand hc in figure 2b and 3b of guidelines
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 1 / slide 4
Part II: True / False
1.6)With the inclusion of base pressure effect, overturning moment on tank reduces.FalseHydrodynamic pressure on the base produces overturning moment in the same direction as that due to hydrodynamic pressure on wall. Hence, with the inclusion of base pressure effect, overturning moment increases. This can be seen from the fact that hi
* is always greater than hi and hc
* is always greater than hc.
1.7)Impulsive and convective masses depend only on height of liquid.FalseImpulsive and convective masses depend on aspect ratio (h/D or h/L) rather than only on height, h.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 1 / slide 5
Part III: Solutions
1.1)A circular tank has internal diameter of 12 m and water height of 5 m. Obtain mi, mc, Kc, hi, hc, hi
* and hc*.
Solution:Total volume of water = π/4 x 122 x 5 = 565.5 m3
∴ Total water mass, m = 565.5 x 1.0 = 565.5 tD = 12 m, h = 5 m ∴ h/D = 5/12 = 0.417
For this value of h/D, from Figures 2a and 2b of the Guideline, values of various parameters can be read. One can also use formulae given in Table C-1 of the Guideline. Here, these formulae will be used to obtain various parameters.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 1 / slide 6
Part III: Solutions
hD.
hD.tanh
mm i
8660
8660 ⎟⎠⎞
⎜⎝⎛
=
5128660
5128660
.
.tanh ⎟⎠⎞
⎜⎝⎛
= = 0.466
Dh
Dh.tanh
.mmc
⎟⎠⎞
⎜⎝⎛
=683
230
125
125683
230⎟⎠⎞
⎜⎝⎛
=.tanh
. = 0.503
Since, h/D is less than 0.75, hi/h = 0.375
⎟⎠⎞
⎜⎝⎛
−⎟⎠⎞
⎜⎝⎛
−=
Dh.sinh
Dh.
.Dh.cosh
hhc
683683
016831
⎟⎠⎞
⎜⎝⎛
−⎟⎠⎞
⎜⎝⎛
−=
125683
125683
01125683
1.sinh.
..cosh= 0.579
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 1 / slide 7
Part III: Solutions
1250 - 86602
8660.
hD.tanh
hD.
h*hi
⎟⎠⎞
⎜⎝⎛
= 1250 -
51286602
5128660
..tanh
.
⎟⎠⎞
⎜⎝⎛
= = 0.947
⎟⎠⎞
⎜⎝⎛
−⎟⎠⎞
⎜⎝⎛
−=
Dh.sinh
Dh.
.Dh.cosh
h*hc
683683
0126831
⎟⎠⎞
⎜⎝⎛
−⎟⎠⎞
⎜⎝⎛
−=
125683
125683
012125683
1.sinh.
..cosh= 0.878
⎟⎠⎞
⎜⎝⎛=
Dh.tanh
hmg.Kc 6838360 2
⎟⎠⎞
⎜⎝⎛=
1256838360 2 .tanh.mg/hKc∴ 6940.=
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 1 / slide 8
Part III: Solutions
Thus, we get,
mi/m = 0.466 ∴mi = 0.466 x 565.5 = 263.5 t
mc/m = 0.503 ∴mc = 0.503 x 565.5 = 284.5 t
hi/h = 0.375 ∴hi = 0.375 x 5 = 1.88 m
hc/h = 0.579 ∴hc = 0.579 x 5 = 2.90 m
hi*/h = 0.947 ∴hi
* = 0.947 x 5 = 4.735 m
hc*/h = 0.878 ∴hc
* = 0.878 x 5 = 4.39 m
Kch/mg = 0.694
Kc = 0.694mg/h = 0.694 x 565.5 x 9.81/5.0 = 770 kN/m
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 1 / slide 9
Part III: Solutions
1.2) A rectangular tank has internal plan dimension of 10 m x 16 m. Water height is 8 m. Obtain mi, mc, hi, hc, hi
* and hc* in both the directions.
Solution:
X
Y
16 m
10 m
Total volume of water = 10 x16 x 8 = 1280 m3
Total mass of water, m = 1280 x 1.0 = 1280 t
Plan view
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 1 / slide 10
Part III: Solutions
First, consider base motion in X-direction:L = 16 m, h = 8 m∴ h/L = 8/16 = 0.5
From Figure 3a, 3b of guidelines, for h/L = 0.5:mi/m = 0.54 ∴mi = 0.54 x 1280 = 691.2 t
mc/m = 0.48 ∴mi = 0.48 x 1280 = 614.4 t
hi/h = 0.375 ∴hi = 0.375 x 8 = 3.0 m
hc/h = 0.57 ∴hc = 0.57 x 8 = 4.6 m
hi*/h = 0.80 ∴hi
* = 0.8 x 8 = 6.4 m
hc*/h = 0.87 ∴hc
* = 0.87 x 8 = 7.0 m
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 1 / slide 11
Part III: Solutions
Now, consider base motion in Y-direction:L = 10 m, h = 8 m∴ h/L = 8/10 = 0.8
From Figure 3a, 3b of guidelines, for h/L = 0.8:mi/m = 0.72 ∴mi = 0.72 x 1280 = 921.6 t
mc/m = 0.32 ∴mi = 0.32 x 1280 = 409.6 t
hi/h = 0.40 ∴hi = 0.40 x 8 = 3.2 m
hc/h = 0.65 ∴hc = 0.65 x 8 = 5.2 m
hi*/h = 0.58 ∴hi
* = 0.58 x 8 = 4.64 m
hc*/h = 0.73 ∴hc
* = 0.73 x 8 = 5.84 m
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 1 / slide 12
Part III: Solutions
Thus, in x-direction, wherein, h/L = 0.5, impulsive and convective masses are almost same. Whereas, in Y-direction, for which h/L = 0.8, more liquid contributes to impulsive mass.
Also note the difference in hi and hi* values ( or hc and hc
*) for two cases. For shallow tanks, inclusion of base pressure significantly changes the height .
1
Additional Slide for Solution 1
Insert after slide 1 of Solution 1
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 1 / slide 2
Notice that there is a correction in Part II True/False – 1.2
1.2)Impulsive mass is less than convective mass for short tanks.
It is printed asFalseIt should be printed asTRUE
Correction in Part II True/False - 1.2
1
Assignment 2
Date of assignment: 20 January
E-course on Seismic Design of Tanks
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Assignment 2/ slide 2
Part I: Reading Assignment 2
Read the following:Clause 4.5 of the GuidelinesClause 6.4 of IS 1893(Part 1):2002Clause 3.4, 4.2.1.1, and 5.2.6 of IS 1893:1984
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Assignment 2/ slide 3
Part II: True / False
Identify the following statements as True or False2.1) Seismic force depends on mass of the structure.2.2) In IS 1893(Part 1):2002, there are five seismic zones.2.3) In IBC 2003, response modification factor for buildings
with good ductility is lower than that for elevated tanks on frame type staging.
2.4) In IS 1893:1984, performance factor, K for buildings with good ductility is more than that for elevated tanks.
2.5) Importance factor for tanks is higher than for buildings.2.6)Damping of structure does not affect base shear
coefficient.2.7) In IS 1893(Part 1):2002, base shear coefficient depends
on type of foundation.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Assignment 2/ slide 4
Part III: Questions
2.1)As per IS 1893:1984, obtain base shear coefficient for following structures using response spectrum method (Fo = 0.4, β = 1.0, consider 5% damping)
1) A building with good ductility and time period of 1.1 sec.2) A building with low ductility and time period of 1.1 sec.3) An elevated tank with time period of 1.1 sec.
2.2) For following elevated tanks, obtain base shear coefficient as per IS 1893:1984 (Fo = 0.4, β = 1.0)1) Time period = 0.5 sec; damping of 0%, 2% and 5%2) Time period = 1.4 sec; damping of 0%, 2% and 5%
1
Solution 2
Date of assignment: 20 JanuaryDate of solution: 23 January
E-Course on seismic design of tanks
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 2 / Slide 2
Part II: True/ False
Identify the following statements as True or False2.1) Seismic force depends on mass of structure.
True2.2) In IS 1893(Part 1):2002, there are five seismic zones.
FalseThere are four zones, numbered from II to V.
2.3) In IBC 2003,response modification factor for buildings with good ductility is lower than that for elevated tanks on frame type stagingFalseFor a building with good ductility, response modification factor, R = 8.0 as against, R = 3.0 for elevated tanks on frame staging
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 2 / Slide 3
Part II: True/ False (contd…)
2.4) In IS 1893:1984,performance factor, K for buildings with Good ductility is more than that for elevated tanks
FalsePerformance factor, K is same (K = 1.0), for a building withgood ductility and elevated tanks.
2.5) Importance factor for tanks is higher than for buildings.True
2.6) Damping of structure does not affect base shear coefficient.False
See Table 3 of IS 1893(Part 1):2002 and Fig. 2 of IS 1893:1984
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 2 / Slide 4
Part II: True/ False (contd…)
2.7) In IS 1893(Part 1):2002, base shear coefficient depends on type of foundation.False
In IS 1893(Part 1):2002, base shear coefficient does not depend on type of foundation. Rather, in IS 1893:1984, it use to depend on type of foundation.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 2 / Slide 5
Part III: Solutions
2.1)As per IS 1893:1984, obtain base shear coefficient for following structures using response spectrum method (Fo = 0.4, β = 1.0; consider 5% damping)
1) A building with good ductility and time period of 1.1 sec.2) A building with low ductility and time period of 1.1 sec.3) An elevated tank with time period of 1.1 sec.
Solution: In response spectrum method, base shear coefficient is expressed as:Ah = KβIFoSa/g for buildingsAh = βIFoSa/g for elevated tanks
We have, Fo = 0.4, β = 1.0
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 2 / Slide 6
Part III: Solutions
All three structures have time period of 1.1 sec and damping of 5%, and hence, from Figure 2 of IS 1893:1984, Sa/g = 0.1
For building with good ductility:K = 1.0, I = 1.0Ah = KβIFoSa/g = 1.0 x 1.0 x 1.0 x 0.4 x 0.1 = 0.04For building with low ductility:K = 1.6, I = 1.0Ah = KβIFoSa/g = 1.6 x 1.0 x 1.0 x 0.4 x 0.1 = 0.064
For elevated tank:I = 1.5Ah = βIFoSa/g = 1.0 x 1.5 x 0.4 x 0.1 = 0.06
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 2 / Slide 7
Part III: Solutions
2.2) For following elevated tanks, obtain base shear coefficient as per IS 1893:1984 (Fo = 0.4, β = 1.0)
1) Time period = 0.5 sec; damping of 0%, 2% and 5%2) Time period = 1.4 sec; damping of 0%, 2% and 5%
Solution: Base shear coefficient for tank is:Ah = βIFoSa/g
We have, Fo = 0.4, β = 1.0 and Importance factor, I = 1.5Values of Sa/g are to be obtained from Fig. 2 of IS 1893:1984
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 2 / Slide 8
Part III: Solutions
Time period (Sec)
Damping (%)
Sa/g Ah = βIFoSa/g
0 0.50 1.0 x 1.5 x 0.4 x 0.5 = 0.300
2 0.25 1.0 x 1.5 x 0.4 x 0.25 = 0.150
5 0.16 1.0 x 1.5 x 0.4 x 0.16 = 0.096
0 0.15 1.0 x 1.5 x 0.4 x 0.15 = 0.090
2 0.105 1.0 x 1.5 x 0.4 x 0.105 = 0.0631.4
5 0.08 1.0 x 1.5 x 0.4 x 0.08 = 0.048
0.5
Note the effect of damping on base shear coefficient
1
Assignment 3
Date of assignment: 24 January
E-course on Seismic Design of Tanks
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Assignment 3/ slide 2
Part I: Reading Assignment 3
Read Clause 4.2.2, 4.2.3 and 4.3 of the Guidelines
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Assignment 3/ slide 3
Part II: True / False
Identify the following statements as True or False3.1)Time period does not depend on mass of the structure.3.2)Time period increases with stiffness.3.3)Impulsive mode time period of ground supported tanks is
usually less than 0.4 seconds. 3.4)Convective mode time period depends only on aspect
ratio of the container.3.5)In elevated tanks, convective mode time period
depends on lateral stiffness of staging.3.6)While finding lateral stiffness of frame staging, braces shall
be treated as rigid beams
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Assignment 3/ slide 4
Part III: Questions
3.1) A ground-supported rectangular tank has length of 12 m and width of 6 m. Liquid height is 4m. Find time period of convective mode in the two horizontal directions.
3.2) An elevated tank with circular cylindrical container has internal diameter of 12 m and water height is 4.5 m. Weight of container is 3,000 kN and weight of staging is 1,700 kN. Lateral stiffness of staging is 50,000 kN/m. Obtain two mass model and find time period of impulsive and convective modes.
1
Solution 3
Date of assignment: 24 JanuaryDate of solution: 27 January
E-course on Seismic Design of Tanks
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 3/ slide 2
Part II: True / False
3.1)Time period does not depend on mass of the structure.False
3.2)Time period increases with stiffness.False
Time period decreases with stiffness and is inversely proportional to square root of stiffness. T = 2π M/K
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 3/ slide 3
Part II: True / False
3.3)Impulsive mode time period of ground supported tanks is usually less than 0.4 seconds.True
3.4)Convective mode time period depends only on aspect ratio of the container.False
It depends not only on aspect ratio but also on diameter or lateral dimension of container. See formula for Tc in slide no. 65.
3.5)In elevated tanks, convective mode time period depends on lateral stiffness of staging.False
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 3/ slide 4
Part II: True / False
3.6)While finding lateral stiffness of frame staging, braces shall be treated as rigid beams.FalseFor proper evaluation of lateral stiffness, it is necessary thatflexibility of brace is included in the model.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 3/ slide 5
Part III: Solutions
3.1) A ground-supported rectangular tank has length of 12 m and width of 6 m. Liquid height is 4m. Find time period of convective mode in the two horizontal directions.
Solution:
X
Y
12 m6 m
Plan view
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 3/ slide 6
Part III: Solutions
1) Analysis in X-direction:Length, L = 12.0 m, Height of liquid, h = 4.0 m∴ h/L = 4/12 = 0.33From Figure 7 of guideline, for h/L = 0.33, one gets Cc = 4.0
gL CT CC =
9.81)(12.004TC .=
Time period of convective mode,
2) Analysis in Y-direction:Length, L = 6.0 m, h = 4.0 m, ∴ h/L = 4/6 = 0.67From Figure 7 of guideline, for h/L = 0.67, one gets, Cc = 3.65
9.81)(6.03.65Tc =
gL CT CC =
= 4.42 Sec.
= 2.85 Sec.
Time period of convective mode,
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 3/ slide 7
Part III: Solutions
3.2) An elevated tank with circular cylindrical container has internal diameter of 12 m and water height is 4.5 m. Weight of container is 3,000 kN and weight of staging is 1,700 kN. Lateral stiffness of staging is 50,000 kN/m. Obtain two mass model and find time period of impulsive and convective modes.
Solution:Internal diameter, D = 12 m, Water height, h = 4.5 m.Container is circular cylinder, ∴ Volume of water = π/4 x D2 x h = π /4 x 122 x 4.5 = 509 m3.
mass of water, m = 509 t.h/D = 4.5/12 = 0.375From Figure 2 of guideline, for h/D = 0.375:mi/m = 0.43, mc/m = 0.545 and Kch/mg = 0.665
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 3/ slide 8
Part III: Solutions
mi = 0.43 x m = 0.43 x 509 = 218.9 tmc = 0.545 x m = 0.545 x 509 = 277.4 tKc = 0.665 x m g/h = 0.665 x 509 x 9.81/4.5 = 737.9 kN/m
Weight of container = 3,000 kN∴ Mass of container = 3000/9.81 = 305.8 t
Weight of staging = 1,700 kN∴ Mass of staging = 1700/9.81 = 173.3 t
Structural mass of tank, ms = mass of container +1/3rd mass of staging= 305.8 + 1/3 x 173.3= 363.6 t
Lateral stiffness of staging, Ks = 50,000 kN/m
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 3/ slide 9
Part III: Solutions
Thus, we have:mi =218.9 t, ms = 363.6 t, mc = 277.4 t, Kc = 737.9 kN/m and Ks = 50000 kN/m Two mass model:
mc
Kc
mi + ms
Ks
277.4 t
737.9 kN/m
218.9t +363.6t
50000 kN/m
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 3/ slide 10
Part III: Solutions
277.4 t
737.9 kN/m
582.5 t
50000 kN/m 50000 kN/m277.4 t
737.9 kN/m
(a) Two mass idealization (b) Uncoupled system
For evaluating time period, this two mass model will be treated as two uncoupled single degree of freedom system:
218.9t +363.6t
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 3/ slide 11
Part III: Solutions
s
sii K
mmT += π2
c
cc K
mT π2=
Now, Impulsive Time Period,
000,506.3639.2182 +
= πiT
= 0.678 Sec.
Convective Time Period,
9.7374.2772π=cT
= 3.85 Sec.
1
Assignment 4
Date of assignment: 30 January
E-course on Seismic Design of Tanks
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Assignment 4/ slide 2
Part I: Reading Assignment 4
Read Clause 4.4, 4.5 of the Guideline
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Assignment 4/ slide 3
Part II: True / False
Identify the following statements as True or False4.1) R values for elevated tanks with SMRF staging are lower
than for buildings with SMRF.4.2)Frame staging has less redundancy than shaft staging.4.3)Damping in convective mode depends on material of
tank.4.4)Elevated tanks are inverted pendulum type structures.4.5)Thin shafts have more ductility than thick shafts.4.6)Response reduction factor for tanks depends on
importance of tank.4.7)In elevated tanks, damping in impulsive mode depends
on material of container.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Assignment 4/ slide 4
Part III: Questions4.1) A ground-supported rectangular RC water tank with
fixed base is located in zone III on medium soil. Along the length direction, Ti = 0.2 sec and Tc = 5.0 sec. Along the width direction, Ti = 0.4 sec and Tc = 3.0 sec. Find impulsive and convective base shear coefficients in both the directions.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Assignment 4/ slide 5
Part III: Questions4.2) An elevated water tank on ductile RC frame staging is
located in zone V. Ti = 1.5 sec and Tc = 4.0 sec. Find the base shear coefficient in impulsive and convective mode for following soil conditions.
i) Hard ii) Medium iii) Soft.
1
Solution 4
Date of assignment: 30 January 2006Date of solution: 01 February 2006
E-course on Seismic Design of Tanks
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 4/ slide 2
Part II: True / False
4.1) R values for elevated tanks with SMRF staging are lower than for buildings with SMRFTrueFor elevated tanks with SMRF staging R = 2.5, whereas forbuildings with SMRF, R = 5.0
4.2)Frame staging has less redundancy than shaft stagingFalse
4.3)Damping in convective mode depends on material of tankFalseConvective mode has 0.5% damping for all types of tank materials.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 4/ slide 3
Part II: True / False
4.4)Elevated tanks are inverted pendulum type structuresTrue
4.5)Thin shafts have more ductility than thick shaftsFalse
4.6)Response reduction factor for tanks depends on importance of tank.FalseResponse reduction factor depends on ductility, redundancy and overstrength. Importance of tank
decides the value of importance factor, I.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 4/ slide 4
Part II: True / False
4.7)In elevated tanks, damping in impulsive mode depends on material of containerFalse
In elevated tanks, damping in impulsive mode depends on material of staging and not of container.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 4/ slide 5
Part III: Solutions
4.1) A ground supported rectangular RC water tank with fixed base is located in zone III on medium soil. Along the length direction, Ti= 0.2 sec and Tc = 5.0 sec. Along the width direction, Ti = 0.4 sec and Tc = 3.0 sec. Find impulsive and convective base shear coefficients in both the directions.
Solution:
Length Direction
Plan of tank
Width Direction
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 4/ slide 6
Part III: Solutions
Zone III, Z = 0.16 (As per Table 2 of IS 1893 (Part I): 2002)Tank is being used for water storage, hence
I = 1.5 (As per Table 1 of the Guideline)For RC ground supported tank with fixed base,
R = 2.0 (As per Table 2 of the Guideline)
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 4/ slide 7
Part III: Solutions
1) Analysis in Length Direction:Impulsive mode time period, Ti = 0.2 sec,Damping in impulsive mode = 5% (RC tank)
∴ (Sa/g)i = 2.5 (Clause 4.5.3 of the Guideline)
Impulsive base shear coefficient, (Ah)i = Z/2 x I/R x (Sa/g)i
= 0.16/2 x 1.5/2.0 x 2.5 = 0.15
Convective mode time period, Tc = 5.0 secDamping in convective mode = 0.5% (Clause 4.4 of the Guideline)
∴ (Sa/g)c = 1.75 x 1.36/5.0 (Clause 4.5.3 of the Guideline)= 0.476
Note, factor 1.75 is for 0.5% damping (Clause 4.5.4 of the Guideline)
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 4/ slide 8
Part III: Solutions
Convective base shear coefficient (Ah)c= Z/2 x I/R x (Sa/g)c
= 0.16/2 x 1.5/2.0 x 0.476= 0.029
Hence, in length direction, (Ah)i = 0.15 and (Ah)c = 0.029
2) Analysis in Width Direction:Impulsive mode time period, Ti = 0.4 Sec,Damping in impulsive mode = 5% (RC tank)
∴ (Sa/g)i = 2.5 (Clause 4.5.3 of the Guideline)Impulsive base shear coefficient, (Ah)i = Z/2 x I/R x (Sa/g)i
= 0.16/2 x 1.5/2.0 x 2.5 = 0.15
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 4/ slide 9
Part III: Solutions
Convective mode time period, Tc = 3.0 secDamping in convective mode = 0.5% (Clause 4.4 of the Guideline)
∴ (Sa/g)c = 1.75 x 1.36/3.0 (Clause 4.5.3 of the Guideline)= 0.793
Convective base shear coefficient (Ah)c= Z/2 x I/R x (Sa/g)c
= 0.16/2 x 1.5/2.0 x 0.793= 0.048
Hence, in width direction, (Ah)i = 0.15 and (Ah)c = 0.048
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 4/ slide 10
Part III: Solutions
4.2) An elevated water tank on ductile RC frame staging is located in zone V. Ti = 1.5 sec and Tc = 4.0 sec. Find the base shear coefficient in impulsive and convective mode for following soil conditions. i) Hard ii) Medium iii) Soft.
Solution: Zone V; Z = 0.36 (As per Table 2 of IS 1893 (Part I): 2002)Tank stores water, hence, I = 1.5 (As per Table 1 of the Guideline)Staging is ductile RC frame, hence, R = 2.5
(As per Table 2 of the Guideline)
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 4/ slide 11
Part III: Solutions
Time period of impulsive mode, Ti = 1.5 secDamping is 5% (RC staging)Values of (Sa/g)i and impulsive base shear coefficient , (Ah)i for different types of soil are given below:
Soil type (Sa/g)i (Ah)i = Z/2 x I/R x (Sa/g)i
Hard 1/Ti = 0.67 = 0.36/2 x 1.5/2.5 x 0.67 = 0.072
Medium 1.36/Ti = 0.91 = 0.36/2 x 1.5/2.5 x 0.91 = 0.098
Soft 1.67/Ti = 1.11 = 0.36/2 x 1.5/2.5 x 1.11 = 0.120
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 4/ slide 12
Part III: Solutions
Time period of convective mode, Tc = 4.0 secDamping is 0.5% Values of (Sa/g)c and convective base shear coefficient , (Ah)c for different types of soil are given below:
Soil type (Sa/g)c (Ah)c = Z/2 x I/R x (Sa/g)c
Hard 1/Tc x1.75 = 0.438
= 0.36/2 x 1.5/2.5 x 0.438 = 0.047
Medium 1.36/Tc x 1.75 = 0.595
= 0.36/2 x 1.5/2.5 x 0.595 = 0.064
Soft 1.67/Tc x 1.75 = 0.731
= 0.36/2 x 1.5/2.5 x 0.731 = 0.079
1
Assignment 5
Date of assignment: 3 FebruaryDate of solution: 5 February
E-course on Seismic Design of Tanks
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Assignment 5/ slide 2
Part I: Reading Assignment 5
Read Clause 4.6, 4.7 and 4.8 of the Guidelines
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Assignment 5/ slide 3
Part II: True / False
5.1)In ground supported tanks, base shear at the bottom of wall is same as base shear at the bottom of base slab.
5.2)In elevated tanks, base moment at the top of footing includes effect of base pressure.
5.3)For calculating bending moment at the bottom of wall, effect of base pressure is included.
5.4)In elevated tanks, impulsive mass of liquid, mi, acts at CG of empty container.
5.5)In IITK-GSDMA Guideline, impulsive and convective base shear are combined using absolute summation rule.
5.6)Critical direction of seismic loading is always same for braces and columns of frame staging.
5.7)In elevated tanks, base shear at the bottom of container wall, will not depend on mass of staging.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Assignment 5/ slide 4
Part III:5.1) Plan and elevation of a rectangular RC tank are shown below along
with some relevant data. Thickness of wall is 250 mm, roof slab is 120 mm thick and base slab is 250 mm thick. Obtain base shear, bending moment at the bottom of wall and overturning moment in X and Y directions.
Plan
Earthquake Direction
mi
(t)hi
(m)hi
*
(m)mc
(t)hc
(m)hc
*
(m)(Ah)i
189 5.07
3.29112
0.15
0.15
(Ah)c
X 138 1.69 4.73 2.48 0.024
0.037Y 230 1.69 2.61 2.93
4.5 m12 m
Elevation
0.1 m projection along the periphery
6 m12 mY
X
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Assignment 5/ slide 5
Part III:5.2) An elevated water tank has CG of empty container
at 3.0 m from top of floor slab. Height of staging from footing top to top of floor slab is 12 m. Impulsive and convective mode parameters are: mi= 140 t, hi
* = 3.43, mc = 110 t, hc* = 3.43 m, ms =
225 t. Base shear coefficient of impulsive mode is 0.15 and of convective mode is 0.06. Find base shear and base moment at the bottom of staging.
1
Solution 5
Date of assignment: 3 FebruaryDate of solution: 6 February
E-course on Seismic Design of Tanks
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 5/ slide 2
Part II: True / False
5.1)In ground supported tanks, base shear at the bottom of wall is same as base shear at the bottom of base slab. FalseBase shear at the bottom of base slab is sum of base shear at the bottom of wall and shear due to mass of base slab.
5.2)In elevated tanks, base moment at the top of the footing includes effect of hydrodynamic pressure at the base.TrueSince staging is below the base slab (or floor slab), base moment in staging includes the bending effect caused by hydrodynamic pressure on base slab.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 5/ slide 3
Part II: True / False
5.3)For calculating bending moment at the bottom of wall, effect of base pressure is included.FalseBase pressure does not affect bending moment in wall. Rather, base pressure affects overturning moment, which is obtained at the bottom of base slab.
5.4)In elevated tanks, impulsive mass of liquid, mi, acts at CG of empty container.FalseIn elevated tanks, impulsive mass of liquid, mi, is considered to act at hi
*. Structural mass of tank, ms, acts at CG of empty container.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 5/ slide 4
Part II: True / False
5.5)In IITK-GSDMA Guideline, impulsive and convective base shear are combined using absolute summation rule.FalseThey are combined using Square Root of Sum of Squares (SRSS) rule.
5.6)Critical direction of seismic loading is always same for braces and columns of frame staging.FalseColumns and braces of frame staging can have different critical directions.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 5/ slide 5
Part II: True / False
5.7)In elevated tanks, base shear at the bottom of container wall, will not depend on mass of staging.True and False
True: Base shear at any section depends on all the horizontal forces acting above that section. Since staging is below the container wall, it is not going to affect base shear at the bottom of container. Base shear at the bottom of staging depends on mass of staging.
False: Mass of staging affects the natural period of the tank which decides the seismic coefficient (Ah)i .
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 5/ slide 6
Part III: Solutions
5.1) Plan and elevation of a rectangular RC tank are shown belowalong with some relevant data. Thickness of wall is 250 mm, roofslab is 120 mm thick and base slab is 250 mm thick. Obtain base shear, bending moment at the bottom of wall and overturning moment in X and Y directions.
Plan
Earthquake Direction
mi
(t)hi
(m)hi
*
(m)mc
(t)hc
(m)hc
*
(m)(Ah)i
189 5.07
3.29112
0.15
0.15
(Ah)c
X 138 1.69 4.73 2.48 0.024
0.037Y 230 1.69 2.61 2.93
4.5 m12 m
Elevation
0.1 m projection along the periphery 6 m12 m
Y
X
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 5/ slide 7
Part III: Solutions
Solution:First we calculate mass of roof slab, wall, and base slab.Weight density of concrete = 25 kN/m3
Thickness of roof slab= 120 mmPlan dimensions of roof slab = 12.5m x 6.5m
∴ Weight of roof slab = 12.5 x 6.5 x 0.12 x 25 = 243.8 kNand mass of roof slab, mt = 243.8/9.81 = 24.9 t
Thickness of wall = 250 mm Internal dimensions are 12m x 6m and height is 4.5 m.
∴ Mass of wall, mw = 2 x (12.25 + 6.25) x 0.25 x 4.5 x 25/9.81= 106.1 t
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 5/ slide 8
Part III: Solutions
Thickness of base slab, tb = 250 mm Plan dimensions of base slab are 12.7 m x 6.7 m∴ Mass of base slab = 12.7 x 6.7 x 0.25 x 25/9.81 = 54.2 t
Since tank is rectangular in plan, it will have different seismic forces in X- and Y-direction.
Analysis in X- direction
In X-direction, we have, mi = 138 t, mc = 189 t, hi = 1.69 m, hi
*= 4.73 m, hc = 2.48 m, hc
*= 5.07 m, (Ah)i = 0.15, (Ah)c = 0.024
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 5/ slide 9
Part III: Solutions
Impulsive base shear at the bottom of wall is Vi = (Ah)i (mi + mw + mt) g
= 0.15 x (138 + 106.1 + 24.9) x 9.81
= 395.8 kN
Convective base shear at the bottom of wall isVc = (Ah)c mc g
= 0.024 x 189 x 9.81
= 44.5 kN
Total base shear at bottom of wall is
( ) ( ) 398.3kN44.5395.8VVV 222c
2i =+=+=
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 5/ slide 10
Part III: Solutions
Impulsive bending moment at the bottom of wall is
Mi = (Ah)i (mihi + mwhw + mtht) g = 0.15 x (138 x 1.69 + 106.1 x 2.25 + 24.9 x 4.56) x 9.81
= 861.5 kN-m
Convective bending moment at the bottom of wall is
Mc = (Ah)c mc hc g = 0.024 x 189 x 2.48 x 9.81
= 110.4 kN-m
Total bending moment at bottom of wall is
( ) ( ) m-868.5kN110.4861.5MMM 222c
2i =+=+=
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 5/ slide 11
Part III: Solutions
This is total bending moment and will be shared by two walls which are perpendicular to the direction of seismic force. Hence, bending moment at the bottom of one wall = 868.5/2 = 434.25 kNm.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 5/ slide 12
Part III: Solutions
Impulsive overturning moment at the bottom of base slab is
Mi* = (Ah)i [mi (hi
* + tb) + mw(hw + tb) + mt(ht +tb) + mb tb/2]g
= 0.15 x [138(4.73 + 0.25) + 106.1(2.25 + 0.25) + 24.9(4.56 + 0.25) + 54.2 x 0.25/2] x 9.81
= 1588 kN-mConvective overturning moment at the bottom of base slab isMc
* = (Ah)c mc (hc* + tb) g
= 0.024 x 189 x (5.07+ 0.25) x 9.81
= 236.7 kN-m
Total overturning moment at bottom of base slab is
( ) ( ) ( ) ( ) m-1605kN236.71588MMM 222*c
2*i
* =+=+=
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 5/ slide 13
Part III: Solutions
Analysis in Y- direction In Y – direction we have, mi = 230 t, mc = 112 t, hi = 1.69 m, hi
*= 2.61 m,hc = 2.93 m, hc
*= 3.29 m, (Ah)i = 0.15, (Ah)c = 0.037Impulsive base shear at the bottom of wall is Vi = (Ah)i (mi + mw + mt) g
= 0.15 x (230 + 106.1 + 24.9) x 9.81 = 531.2 kN
Convective base shear at the bottom of wall isVc = (Ah)c mc g
= 0.037 x 112 x 9.81 = 40.7 kN
Total base shear at bottom of wall is
( ) ( ) 532.8kN40.7531.2VVV 222c
2i =+=+=
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 5/ slide 14
Part III: Solutions
Impulsive bending moment at the bottom of wall is
Mi = (Ah)i (mihi + mwhw + mtht) g = 0.15 x (230 x 1.69 + 106.1 x 2.25 + 24.9 x 4.56) x 9.81
= 1090 kN-m
Convective bending moment at the bottom of wall is
Mc = (Ah)c mc hc g = 0.037 x 112 x 2.93 x 9.81
= 119.1 kN-m
Total bending moment at bottom of wall is
( ) ( ) m-1097kN119.11090.3MMM 222c
2i =+=+=
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 5/ slide 15
Part III: Solutions
This is total bending moment and will be shared by two walls which are perpendicular to the direction of seismic force. Hence, bending moment at the bottom of one wall = 1097/2 = 548.5 kN-m.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 5/ slide 16
Part III: Solutions
Impulsive overturning moment at the bottom of base slab is
Mi* = (Ah)i [mi (hi
* + tb) + mw(hw + tb) + mt(ht +tb) + mb tb/2]g
= 0.15 x [230(2.61 + 0.25) + 106.1(2.25 + 0.25) + 24.9(4.56 + 0.25) + 54.2 x 0.25/2] x 9.81
= 1544 kN-mConvective overturning moment at the bottom of base slab is
Mc* = (Ah)c mc (hc
* + tb) g = 0.037 x 112 x (3.29+ 0.25) x 9.81
= 143.9 kN-m
Total overturning moment at bottom of base slab is
( ) ( ) ( ) ( ) m-1551kN143.91544MMM 222*c
2*i
* =+=+=
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 5/ slide 17
Part III: Solutions
5.2) An elevated water tank has CG of empty container at 3.0 m from top of floor slab. Height of staging from footing top to top of floor slab is 12 m. Impulsive and convective mode parameters are: mi = 140 t, hi
* = 3.43, mc = 110 t, hc* = 3.43
m, ms = 225 t. Base shear coefficient of impulsive mode is 0.15 and of convective mode is 0.06. Find base shear and base moment at the bottom of staging.
Solution: Impulsive base shear at the bottom of staging is
Vi = (Ah)i (mi + ms) g = 0.15 x (140 + 225) x 9.81 = 537.1 kN
Convective base shear at the bottom of staging isVc = (Ah)c mc g
= 0.06 x 110 x 9.81 = 64.7 kNTotal base shear at bottom of staging is
( ) ( ) 541.0kN64.7537.1VVV 222c
2i =+=+=
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 5/ slide 18
Part III: Solutions
Distance of CG of container from bottom of staging, hcg = 3.0 + 12.0 = 15.0 m
Impulsive base moment at the bottom of staging isMi
* = (Ah)i [mi (hi* + hs) + ms hcg] g
= 0.15 x [140 (3.43 + 12) + 225 x 15.0] x 9.81 = 8145 kN-m
Convective base moment at the bottom of staging isMc
* = (Ah)c mc (hc* + hs) g
= 0.06 x 110 x (3.43 + 12) x 9.81 = 999.0 kN-m
Total base moment at bottom of staging is
Thus, base shear = 541.0 kN & Base moment = 8206 kN-m
( ) ( ) ( ) ( ) m-8206kN999.08145MMM 222*c
2*i
* =+=+=
1
Assignment 6
Date of assignment: 8 February 2006
E-course on Seismic Design of Tanks
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Assignment 6/ slide 2
Part I: Reading Assignment 6
Read Clause 4.9, 4.10, 4.11, 4.12 of the Guideline
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Assignment 6/ slide 3
Part II: True / False
6.1)Convective pressure varies linearly along wall height.6.2)In circular tanks, impulsive pressure on wall varies in
circumferential direction.6.3)In rectangular tanks, convective pressure on a particular
wall varies along the length of that wall.6.4)Due to vertical ground acceleration, lateral pressure
exerted by the liquid on the wall changes.6.5)In rectangular tanks, hydrodynamic pressure acts on
walls parallel to the direction of seismic force.6.6)Impulsive pressure is maximum at the bottom of wall.6.7)Convective hydrodynamic force is more significant for
tall tanks.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Assignment 6/ slide 4
Part III: Questions
6.1) A ground supported rectangular RC water tank has plan dimensions of 12 x 6 m and water height of 4.5 m. Tank is fixed at base. Wall has uniform thickness of 200 mm. Tank is located on hard soil in seismic zone III. In X – direction: (Ah)i = 0.15, (Ah)c = 0.024. In Y – direction: (Ah)i = 0.15, (Ah)c =0.037.
Plan
6 m12 mY
X
Find in X- and Y- directions:1) Linearised impulsive and convective
pressure distribution on the walls2) Pressure due to vertical excitation3) Pressure due to wall inertia 4) Total pressure at the bottom of wall
1
Solution 6
Date of assignment: 8 February 2006Date of solution: 10 February 2006
E-course on Seismic Design of Tanks
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 6/ slide 2
Part II: True / False 6.1)Convective pressure varies linearly along wall height.
FalseConvective pressure has curvilinear distribution along wall height.
6.2)In circular tanks, impulsive pressure on wall varies in circumferential direction.TrueImpulsive pressure has cosφ variation in circumferential
direction.
6.3)In rectangular tanks, convective pressure on a particular wall varies along the length of that wall.FalseIn rectangular tanks, convective pressure on a wall remains constant along the length of that wall. This assumes that the direction of shaking is perpendicular to the wall.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 6/ slide 3
Part II: True / False
6.4)Due to vertical ground acceleration, lateral pressure exerted by the liquid on the wall changes.Truewhen subjected to vertical excitation, weight density of liquid increases or decreases depending on direction of vertical acceleration. Hence, hydrostatic pressure which depends on weight density of liquid, will also increase or decrease. This additional pressure is called pressure due to vertical excitation.
6.5)In rectangular tanks, hydrodynamic pressure acts on walls parallel to the direction of seismic force.FalseIn rectangular tanks, hydrodynamic pressure acts on walls perpendicular to the direction seismic force.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 6/ slide 4
Part II: True / False
6.6)Impulsive pressure is maximum at the bottom of wall.True
6.7)Convective hydrodynamic force is more significant for tall tanks.FalseConvective pressure in tall tanks (high value of h/D or h/L) is not as significant as for short tanks (low h/D or h/L ratio). Refer Figures 10(a) and 11(a) of Guidelines
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 6/ slide 5
Part III: Solution
6.1) A ground supported rectangular RC water tank has plan dimensions of 12 x 6 m and water height of 4.5 m. Tank is fixed at base. Wall has uniform thickness of 200 mm. Tank is located on hard soil in zone III. In X – direction: (Ah)i = 0.15, (Ah)c = 0.024, In Y – direction: (Ah)i = 0.15, (Ah)c =0.037.
Plan
6 m12 mY
X
Find in X- and Y- directions:1) Linearised impulsive and convective
pressure distribution on wall2) Pressure due to vertical excitation3) Pressure due to wall inertia 4) Total pressure at the bottom of wall
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 6/ slide 6
Part III: Solution
Solution:Capacity of tank = 12 x 6 x 4.5 = 324 m3
∴ Mass of water = 12 x 6 x 4.5 x 1000 = 324,000 kg
Thickness of wall, tw = 200 mmWeight density of concrete = 25 kN/m3
Analysis in X- direction:
(Ah)i = 0.15, (Ah)c = 0.024h = 4.5 m, L = 12 m, B = 6 m, h/L = 4.5/12 = 0.375.
For h/L = 0.375, from Figure 3 of the Guideline, we have: mi/m = 0.425, mi = 0.425 x 324,000 = 137,700 kgmc/m = 0. 584, mc = 0. 584 x 324,000 = 189,200 kghi/h = 0.375, hi = 0.375 x 4.5 = 1.69 mhc/h =0.551, hc = 0.551 x 4.5 = 2.48 m
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 6/ slide 7
Part III: Solution
a) Impulsive mode
kN/m..,.gB
m)A(q iihi 9016
62819700137150
2=
×××
==
( ) ( ) 2i
ii kN/mhh
hqa 60.669.165.44
5.490.1664 22 =×−×=−=
( ) ( ) 2i
ii kN/mhh
hqb 95.05.4269.16
5.490.1626 22 =×−×=−=
Pressure at bottom & top is given by
Linear impulsive pressure distribution
bi = 0.95 kN/m2
ai = 6.60 kN/m2
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 6/ slide 8
Part III: Solution
b) Convective mode
kN/mgB
mAq cchc 71.3
6281.9200,189024.0
2)(
=×
××==
( ) ( ) 2c
cc kN/mhh
hqa 57.048.265.44
5.471.364 22 =×−×=−=
( ) ( ) 2c
cc kN/mhh
hqb 08.15.4248.26
5.471.326 22 =×−×=−=
Pressure at bottom & top is given by
Linear convective pressure distribution ac = 0.57 kN/m2
bc = 1.08 kN/m2
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 6/ slide 9
Part III: Solution
c) Pressure due to vertical accelerationZone III, Z = 0.16I = 1.5It is ground supported RC tank with fixed base, hence, R = 2.0For vertical mode, T = 0.3 Sec, ∴ Sa/g = 2.5
Av = 2/3 . Z/2. I/R . Sa/g = 2/3 x 0.16/2 x 1.5/2.0 x 2.5 = 0.10
Maximum pressure due to vertical acceleration occurs at y=0, i.e., at the base of wall,
∴ pv = (Av) ρ g h (1- y/h) = 0.10 x 1000 x 9.81 x 4.5 x (1- 0/4.5)= 4.42 kN/m2
d) Pressure due to wall inertia, pww = (Ah)i t ρm g
= 0.15 x 0.2 x 25 = 0.75 kN/m2
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 6/ slide 10
Part III: Solution
Total pressure,
( ) 222vcwwwiw ppppp +++=
( ) 222 424570750606 .... +++=
Analysis in Y- direction:
(Ah)i = 0.15, (Ah)c = 0.037
h = 4.5 m, L = 6 m, B = 12 m, h/L = 4.5/6 = 0. 75. For h/L = 0.375, from Figure 3 of the Guideline, we have:
mi/m = 0.71, mi = 0.71 x 324,000 = 230,000 Kg
mc/m = 0. 346, mc = 0. 346 x 324,000 = 112,100 Kg
hi/h = 0.375, hi = 0.375 x 4.5 = 1.69 m
hc/h =0.551, hc = 0.65 x 4.5 = 2.93 m
2kN/m 95.8=
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 6/ slide 11
Part III: Solution
a) Impulsive mode
kN/m..,.gB
m)A(q iihi 1014
122819040230150
2=
×××
==
( ) ( ) 2i
ii kN/mhh
hqa 47.569.165.44
5.410.1464 22 =×−×=−=
( ) ( ) 2i
ii kN/mhh
hqb 79.05.4269.16
5.410.1426 22 =×−×=−=
Pressure at bottom & top is given by
Linear impulsive pressure distribution
bi = 0.79 kN/m2
ai = 5.47 kN/m2
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 6/ slide 12
Part III: Solution
b) Convective mode
kN/m..,.gB
m)A(q cchc 701
1228191041120370
2=
×××
==
( ) ( ) 2c
cc kN/mhh
hqa 035.093.265.44
5.470.164 22 =×−×=−=
( ) ( ) 2c
cc kN/mhh
hqb 72.05.4293.26
5.470.126 22 =×−×=−=
Pressure at bottom & top is given by
Linear convective pressure distribution ac = 0.035 kN/m2
bc = 0.72 kN/m2
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 6/ slide 13
Part III: Solution
c) Pressure due to vertical accelerationZone III, Z = 0.16As it stores water, I = 1.5It is ground supported RC tank with fixed base, R = 2.0For vertical mode, T = 0.3 Sec, ∴ Sa/g = 2.5Av = 2/3 . Z/2. I/R . Sa/g = 2/3 x 0.16/2 x 1.5/2.0 x 2.5 = 0.10
Maximum pressure due to vertical acceleration is occur at y=0, i.e., at the base of wall,
∴ pv = (Av) ρ g h (1- y/h) = 0.10 x 1000 x 9.81 x 4.5 x (1- 0/4.5)= 4.42 kN/m2
d) Pressure due to wall inertia, pww = (Ah)i t ρm g
= 0.15 x 0.2 x 25 = 0.75 kN/m2
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 6/ slide 14
Part III: Solution
Total pressure
( ) 222vcwwwiw ppppp +++=
( ) 2222 4240350750475 .... +++=2kN/m62.7=
piw pcw pv pww Total pressure
X-direction 6.6 0.57 4.42 1.15 8.59
Y-direction 5.47 0.035 4.42 1.15 7.62
At the base of wall pressure in kN/m2 are:
1
Assignment 7
Date of assignment: 17 February
E-course on Seismic Design of Tanks
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Assignment 7/ slide 2
Part I: Reading Assignment 7
Read Examples 2 and 3 of the Guideline
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Assignment 7/ slide 3
Part II: True / False
7.1) If braces are treated as rigid beams, then seismic forces are overestimated .
7.2) Inclusion of shear deformation, reduces time period of shaft staging.
7.3) Effect of shear deformation on lateral stiffness is more important in shafts with large height-to-diameter ratio.
7.4) If stiffness decreases by 1.44 times then, time period will increase by 1.2 times.
7.5) Empty tank does not have convective mode of vibration.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Assignment 7/ slide 4
Part III: Questions7.1) A eight column RC frame staging has four brace levels. All
columns have 400 mm diameter; braces are 300 mm wide and 400 mm deep. Grade of concrete is M 25. Top ring beam is 400 mm wide and 750 mm deep. CG of tank is at a distance of 3m from top ring beam (Refer Figures below). Find the lateral stiffness by (a) computer analysis of a 3-dimensional model, (b) approximate analysis considering braces as rigid beams as per SP:22 approach, and (c) approximate analysis using the approach of Sameer and Jain (1992). Compare and discuss your results.
3 m
Plan
4 m3 m
4 m
4 m
4 m
4 mElevation
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Assignment 7/ slide 5
Part III: Questions
7.2) A RC shaft has inner diameter of 6m and wall thickness of 200 mm. Grade of concrete is M 25. Find the lateral stiffness of shaft with and without considering shear deformation for following three staging heights: L = 15 m, 20 m and 25 m. Take Poisson’s ratio as 0.2.
1
Solution 7
Date of assignment: 17 FebruaryDate of solution: 20 February
E-course on Seismic Design of Tanks
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 7/ slide 2
Part II: True / False
7.1) If braces are treated as rigid beams, then seismic forces are overestimated .TrueIf braces are treated as rigid, they do not allow any rotation at joints of columns and braces. This leads to significant overestimation of stiffness. This reduces time period and hence, higher seismic forces.
7.2) Inclusion of shear deformation reduces time period of shaft staging.
FalseWith the inclusion of shear deformation, lateral stiffness decreases and time period which is inversely proportional to square root of stiffness, increases.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 7/ slide 3
Part II: True / False 7.3) Effect of shear deformation on lateral stiffness is more
important in shafts with large height-to-diameter ratio.FalseThe lateral stiffness is given by
Here, L3/3EI is flexural contribution and L/0.5AG is shear contribution.Flexural contribution varies as L3, whereas, shear contribution varies linearly with length, L. Assuming same cross section, if height, i.e., L is more, contribution of flexural deformation increase as L3 whereas contribution of shear deformation increase linearly. Hence, in shafts with more height, flexural deformation governs the stiffness andeffect of shear deformation is low. This is also explained in the solution of problem no. 7.2 of part III of this assignment.
AG.L
EIL
K s
503
13
+=
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 7/ slide 4
Part II: True / False
7.4) If stiffness decreases by 1.44 times then, time period will increase by 1.2 times.
TrueTime period T = 2 π (M/K)0.5. If K decreases to K/1.44, then T will increase by 1.2 times. Note: (1.44)0.5 = 1.2
7.5) Empty tank does not have convective mode of vibration.
TrueIn empty tank, no water, hence no convective mode.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 7/ slide 5
Part III: Solutions7.1) A eight column RC frame staging has four brace levels. All
columns have 400 mm diameter; braces are 300 mm wide and 400 mm deep. Grade of concrete is M 25. Top ring beam is 400 mm wide and 750 mm deep. CG of tank is at a distance of 3m from top ring beam (Refer Figures below). Find the lateral stiffness by (a) computer analysis of a 3-dimensional model, (b) approximate analysis considering braces as rigid beams as per SP:22 approach, and (c) approximate analysis using the approach of Sameer and Jain (1992). Compare and discuss your results.
3 m
Plan
4 m3 m
4 m
4 m
4 m
4 m
Elevation
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 7/ slide 6
Part III: Solutions
Solution:a) Stiffness from computer analysis
Staging is modeled using SAP 2000. Frame elements are used to model braces, columns and circular ring beams. Columns are considered to be fixed at the base. Lateral load is to be applied at a height of 3.0 m from circularring beam. For this purpose, a rigid link of 3.0 m length is required at the center of staging. In order to have a node at the center, eight radial beams of same size as circular beams are also modeled. From the central node, a vertical rigid link of 3.0 m length is put. On this rigid link, a lateral force of 10 kN is applied. Computer model is shown on next slide.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 7/ slide 7
Part III: Solutions
Rigid link
Brace
Radial beam
10 kN
Column
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 7/ slide 8
Part III: Solutions
For lateral force of 10 kN, deflection = 0.00267 m
∴ Lateral stiffness of staging, Ks = 10/0.00267= 3,745 kN/m
Deflection of staging
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 7/ slide 9
Part III: Solutions
b) Stiffness by considering braces as rigid beams
If braces are treated as rigid beams, then stiffness of column between two brace levels or stiffness of column in a panel is given by
E = Modulus of elasticity, I = Moment of inertia and L = Length of panel.
3C L12EIK =
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 7/ slide 10
Part III: Solutions
In the present example, Grade of concrete is M 25Hence, E = 5000 (25)0.5 = 25000 N/mm2 = 25 x 106 kN/m2
Column diameter = 400 mm = 0.4 m∴ Moment of inertia, I = π (0.4)4/64 = 1.26 x 10-3 m4
Hence,
kN/m 54.0
1061102512L
12EIK 33-6
3C 39062 .. =××××==⎟⎠⎞⎜
⎝⎛
In each panel, there are 8 columns. These eight columns act like eight springs in parallel. Hence, stiffness of each panel
kN/m 472505906.38K C =×== ∑Kp
Stiffness of one panel = 47,250 kN/m
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 7/ slide 11
Part III: Solutions
Staging comprises of five such panels. All panels have same stiffness and they act like five springs in series. Hence, Stiffness of staging Ks is given by
ppppps KKKKKK111111
++++=
i.e., Stiffness of staging, Ks = Kp/5 = 47,250/5 = 9,450 kN/m
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 7/ slide 12
Part III: Solutions
c) Stiffness as per approach of Sameer and Jain (1992)Stiffness of staging is given by
axialflexures K1
K1
K1
+=
Where,
∑=
=pN
1i panelflexure K1
K1
2ccaxial REAN2
K1
= ∑=
pN
1i
2i hH
⎥⎥
⎦
⎤
⎢⎢
⎣
⎡
α+=
−
hILI
LIh
N12EIK
cb
b3
ccpanel,
α = 1.0 for end panels and α = 2.0 for intermediate panels
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 7/ slide 13
Part III: Solutions
E = Modulus of elasticityAc = Area of one column Nc = Number of columns Np = Number of panelsL = Length of each brace beamR = Radius of staging systemIc = Moment of inertia of columnIb = Moment of inertia of brace beam Icbr = Moment of inertia of circular ring beamh = Height of each panelHi = Distance of point of inflection of ith panel from the load
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 7/ slide 14
Part III: Solutions
For intermediate panels, point of inflection is at mid-height of panel. For end panels, point of inflection is at a distance, y from therestrained end.
h
Lb6I
hcI
Lb3I
y+
=Top most panel
Bottom most panel
Intermediate panelsFor the top most panel,restrained end is at top ring beam, hence, y is measured from top end.
For the bottom most panel, y is measured from bottom end.
Panel 1
Panel 2
Panel 3
Panel 4
Panel 5
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 7/ slide 15
Part III: Solutions
For this Example, we have: E = Modulus of elasticity = 5000 (25)0.5 = 25,000 N/mm2
Ac = Area of one column = π x (400)2/ 4 = 125,664 mm2
Nc = Number of columns = 8 Np = Number of panels = 5L = Length of each brace beam
= 3000 x sin (22.50) x 2 = 2,296 mmR = Radius of staging system = 3,000 mm Ic = Moment of inertia of column
= π x (400)4/ 64 = 1.26 x 109 mm4
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 7/ slide 16
Part III: Solutions
Ib = Moment of inertia of brace beam = 300 x(400)3/12 = 1.6x109 mm4
Icbr = Moment of inertia of circular ring beam = 400 x (750)3 / 12 = 1.41 x 1010 mm4
h = Height of each panel = 4,000 mm
mm 23014000
2296
9101.664000
9101.262296
9101.63
h
Lb6I
hcI
Lb3I
y =×××
×+
××
=+
=
Distance of point of inflection of each panel from the loadH1 = 3000 + 2301 = 5,301 mmH2 = 3000 + 4000 + 2000 = 9,000 mmH3 = 3000+ 8000 + 2000 = 13,000 mmH4 = 3000 + 12000 + 2000 = 17,000 mmH5 = 3000 + 20000 – 2301 = 20,699 mm
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 7/ slide 17
Part III: Solutions
Now, Kaxial is given by
2ccaxial REAN2
K1
= ∑=
pN
1i
2
i hH
1822
cc
10x84.8)3000(x25000x125664x8
2REAN
2 −==
4000x))20699()17000()13000()9000()5301(( 22222 ++++=∑=
pN
1i
2i hH
1210x98.3=
N/mm10x52.310x98.3x10x84.8 51218 −− ==∴axialK1
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 7/ slide 18
Part III: Solutions
Now, let us find Kpanel
For the upper most panel, ( )01.=α
01.=α
⎥⎥⎦
⎤
⎢⎢⎣
⎡
α+=
hILILI
hNEI
Kccbr
cbrccpanel 3
12= 4.5 x 104 N/mm.
Similarly, for bottom most panel, ( )
⎥⎥⎦
⎤
⎢⎢⎣
⎡
α+=
hILILI
hNEI
Kcb
bccpanel 3
12= 3.25 x 104 N/mm.
0.2=αSimilarly, for intermediate 3 panels, ( )
⎥⎥⎦
⎤
⎢⎢⎣
⎡
α+=
hILILI
hNEIK
cb
bccpanel 3
12= 2.48 x 104 N/mm.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 7/ slide 19
Part III: Solutions
44444flexure
1022
1022
1022
103.251
1041
K1
×+
×+
×+
×+
×=
48.48.48.5.
Now, Ks is obtained as
∴ Ks = 3,029 N/mm = 3,029 kN/m.
= 2.95 x 10-4 mm/N
∑=
=pN
1i panelflexure K1
K1
N/mm10x302.351025.341095.2axialK1
flexureK
1
sK1 4−=−×+−×=+=
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 7/ slide 20
Part III: Solutions
Thus, stiffness of staging using these three approaches isa) From computer analysis of 3-D model
Ks = 3,745 kN/mb) From approximate analysis considering braces as rigid beams
Ks = 9,450 kN/mc) From approximate analysis of Sameer and Jain (1992)
Ks = 3,029 kN/m
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 7/ slide 21
Part III: Solutions
Thus, if braces are treated as rigid beams, as suggested by SP 22, stiffness is almost 2.5 times higher than that obtained computeranalysis of 3-D model of staging.
The stiffness obtained by approach suggested by Sameer and Jain is about 19% lower than that from computer analysis.
In computer analysis, we have put radial beams, which impart rotational rigidity to columns at top level, Hence, some of the difference in Sameer and Jain method is due to model error of computer Analysis.
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 7/ slide 22
Part III: Solutions
7.2) A RC shaft has inner diameter of 6m and wall thickness of 200 mm. Grade of concrete is M 25. Find the lateral stiffness of shaft with and without considering shear deformation for following three staging heights: L = 15 m, 20 m and 25 m. Take Poisson’s ratio as 0.2.
Solution: Lateral stiffness of RC shaft without considering shear deformation
If shear deformations are considered, then
3s L3EI
K =
AG'L
3EIL
1K
3s
κ+
=
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 7/ slide 23
Part III: Solutions
Grade of concrete is M 25. Hence, Modulus of elasticity, E = 5000 x (25)0.5 = 25,000 N/mm2.
= 25000 x 103 kN/m2
G = Shear modulus = E/(2 x (1 + ν)) where, ν = Poisson’s ratio = 0.2
∴ G = E/(2 x (1 + ν)) = 25000 x 103 /(2 x (1 + 0.2)) = 10,417 x103 kN/m2
I = Moment of inertia of shaft cross section = π x (6.44 – 6.04)/64 = 18.74 m4
A = Area of cross section of shaft = π x (6.42 – 6.02)/4 = 3.90 m2
κ’ = 0.5 (for hollow circular cross section)
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 7/ slide 24
Part III: Solutions
Now, for shaft with L = 15 m:
Ks without considering shear deformation
3s L3EI
K =
Ks with considering shear deformation
AG'L
3EIL
1K
3s
κ+
=
m/kN10x5.318
417x100.5x3.9x1015
74.18x3x25000x1015
1 3
33
3 =+
=
m/kN10x4.41615
x18.743x25000x10 33
3
==
© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 7/ slide 25
Part III: Solutions
Thus, with the inclusion of shear deformation, Ks reduces by 23.5%.
Similarly, lateral stiffness is obtained for L = 20 m and 25 m and results are given in Table below
Lateral stiffness, Ks (kN/m)Height of shaft
Height-to-diameter
ratio
15 m
20 m
25 m
2.5
3.3
4.2
without shear deformation
with shear deformation
416.4x103 318.5x103
149.7x103
81x103
175. 7x103
90x103
% Reduction
23.5 %
14.8 %
10 %
Note: Thus, for shafts with large height-to-diameter ratio, shear deformation has less effect on the lateral stiffness.