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


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)


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


Types of tanks

Elevated tanks

Two parts:Container Staging (Supporting tower)


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


Use of tanks

Water distribution systems use ground supported and elevated tanks of RC & steel

Petrochemical industries use ground supported steel tanks


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



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



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


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


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



Hydrodynamic pressureHas curvilinear variation along wall heightIts direction is opposite to base motion

Base motion




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



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



Hydrostatic pressure Hydrodynamic pressure

Base motion

Net resultant force = zero Net resultant force ≠ zero

Rectangular tanks (Plan View)



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



Seismic design: Hydrodynamic pressure is considered

Net hydrodynamic force on the container is not zeroAffects design of container, staging and foundation



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


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)


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


Modeling of liquid

Impulsive Convective


Base motionBase motion


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


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


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


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


Modeling of liquid

D

L

Base motion

L

Base motion

ElevationPlan of Circular tank

Plan of Rectangular tank

h


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.


0

0.5

1

0 0.5 1 1.5 2h/D

mi/m

mc /m


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


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)


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


Modeling of liquid

mi/m

mc/m

Kch/mg

0

0.5

1

0 0.5 1 1.5 2h/D


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


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


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


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


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


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


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.


0

0.2

0.4

0.6

0.8

1

0 0.5 1 1.5 2h/D

hc/h

hi/h


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.


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


Modeling of liquid

Hydrostatic pressure on base

Base motion

Hydrodynamic pressure on base


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


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


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


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


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


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*


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


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


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;


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


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


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


Mechanical analogueor

spring mass model of tank


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


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



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


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


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


Modeling of liquid

Vi and Vc will cause Bending Moment (BM) in wall


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


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


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*


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


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


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


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

mailto:[email protected]

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


In this lecture

Seismic force evaluationProcedure in codesLimitations of IS 1893:1984


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



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.



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



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



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



Thus, base shear coefficient depends on:Severity of ground motionSoil conditionStructural characteristicsDesign philosophy



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


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


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


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


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


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


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)


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


IS 1893 (Part 1):2002

For 5% damping


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


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


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


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


IS 1893:1984


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


IS 1893:1984

Importance factor, IEnsures higher design seismic force for more important structures

IS 1893 (Part 1):2002, gives values only for buildings


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


IS 1893:1984


IS 1893:1984

Coefficient, CDepends on time period see next slide

Spectral acceleration, Sa/gDepends on time period and dampingSee next slide


IS 1893:1984

Graphs for C and Sa/g from IS 1893:1984

Natural Period (Sec) Natural Period (Sec)


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


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


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


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


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


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


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


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



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



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



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



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



Building with good ductility is chosenSay, buildings with SMRF

In IBC, for buildings with SMRF, R = 8.0Refer Table shown earlier



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



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



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



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



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



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



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



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



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



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



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



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



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



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



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



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



Impulsive and convective masses will have different time periods

Hence, will have different Sa/g valuesProcedure for finding time period in next lecture



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


Sesimic Design of Liquid Storage Tanks January 16 - February 6, 2006


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.


http://www.iitk.ac.in/nicee/IITK-GSDMA/EQ05.pdf

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


In this lecture

Modeling of tanksTime period of tanks


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


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


hc* = Location of convective mass


Mechanical analogueor

spring mass model of tank

Graphs and expression for these parameters are given in lecture 1.


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


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



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



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


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



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



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



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



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


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



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



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



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.



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.


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.


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


Modeling of elevated tanks

Elevated tank consists of container and staging

Elevated tank

Container

Staging

Wall

Roof slab

Floor slab



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



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



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


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.


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



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



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.



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



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



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



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.



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



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



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


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


Time period

Mathematical model of tank comprises of impulsive and convective componentsHence, time periods of impulsive and convective mode are to be obtained


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


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



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



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



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.



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



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



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



Such tanks are perhaps not used in India

Types of connections between tank wall and base slab



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



Impulsive mode time period of ground supported tanks likely to remain in this range

Sa/g



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



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.



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


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


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=



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



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



Time period of impulsive modes

sii K

mm2T += π

00020180135

2,

Ti

+π=

= 0.79 sec.


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



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



4 columns 6 columns 8 columns

9 columns 12 columns

Plan view of frame staging configurations



24 columns 52 columns



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



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.


Time period of convective mode

Convective mass is mc and stiffness is Kc

Time period of convective mode is:

c

cc K

mT π2=



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



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=



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



Fig. 5 For circular tanks

0

2

4

6

8

10

0 0.5 1 1.5 2

Ci

Cc

C

h/D



2

4

6

8

10

0 0.5 1 1.5 2h/L

Cc

Fig. 7 For rectangular tanks



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



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




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


In this lecture

Z, I, Sa/g and R values for tanks



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



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



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


(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


(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


(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


(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


(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


(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


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


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


Damping

IS 1893(Part 1), needs to have a re-look at the damping values

Accordingly, damping values for tanks can also be modified


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


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



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



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


Response reduction factor, R

R values for tanks are given in Table 2 of the Guideline

This is reproduced in next two slides



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



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.



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



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



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



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



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



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



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.



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


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



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


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



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



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)



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



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



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.



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 )



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)



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.


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


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


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



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


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


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


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


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



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



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



Base shear in convective mode

mc = mass of convective liquid

( ) gmAV cchc =



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



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.


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



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



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



For bending moment at the bottom of wall, effect of base pressure is not included

Hence, hi and hc are used



Bending moment at the bottom of wall

( ) ( )ghmhmhmAM ttwwiiihi ++=

( ) ghmAM ccchc )(=

Ground level

(Ah)imihi

(Ah)imw

hw

(Ah)imt

(Ah)cmchc

ht



Total bending moment at the bottom of wall

22ci MMM +=

SRSS rule used to combine impulsive and convective responses


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


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 )()( ** +=



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 +=


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


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 =+=+=


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 =+=+=


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 =+=+=


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


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 +=


( ) gmAV cchc =

Total base shear22

ci VVV +=


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



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



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



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



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



Total bending moment

22c

*i

** MMM +=

For shaft supported tanks, M* will be the design moment for shaft



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



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


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


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 .. +=


( ) gmAV cchc =

819x180x040 ..= kN6.70=

= 78.5 + 157 = 235.5 kN


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


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.


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.


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


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).


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


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


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



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



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.



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



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



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



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



Critical directions for 4 - column staging

Critical direction for shear force in brace

Critical direction for axial force in column

Bending Axis



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)




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




Critical direction for shear force and bending moment in braces

Critical direction for shear force, bending moment and axial force in columns

Bending Axis



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



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



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



Important to note that the earthquake directions are reversible

Hence, in 100%+30% rule, there are total eight load combinations



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



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



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


In this Lecture

Hydrodynamic pressure on wall and baseEffect of vertical ground accelerationSloshing wave heightAnchorage requirements for ground supported tanks



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



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



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


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



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



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



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’


x

Plan



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



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



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.



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


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



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


Walls perpendicular to direction of seismic force

LB



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



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



-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


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



(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



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



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



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



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



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.



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


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



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



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



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)


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



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



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



Equivalent linear impulsive pressure

Actual Impulsive pressure distribution

hi

qi

h ˜

Equivalent linearpressure distribution

hi

qi

ai

bi



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



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



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


gB

m)A(q cchc 2=



Linearised distribution can be treated as summation of two parts

Uniform pressureTriangular pressure

+=

bi ai - biai

bi



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.



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



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



( ) ( ) 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:



b) Convective mode


( ) ( ) 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



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


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



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



Simplified distribution in circumferential direction

φ

pmax

Actual pressure distribution Simplified pressure distribution

pmax


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



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



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



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



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


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


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


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 +++=


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


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


Example

Maximum pressure due to wall inertia, pww = (Ah)i t ρm g

= 0.23 x 0.2 x 25 = 1.15 kN/m2


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.


Sloshing wave height

Convective liquid undergoes sloshing motioni.e., liquid undergoes vertical motion

dmax



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



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



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


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



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



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



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


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


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


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


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


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


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.


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


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



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



Rigid link

Structural model of staging used for analysis

10.0



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



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



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 =


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 =×=∑



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



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



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)



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



⎥⎦

⎤⎢⎣

⎡+

=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



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



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



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.



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



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


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



Convective time period is 3.65 times impulsive time period

Hence, two DoF model can be treated as two uncoupled SDoF systems


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



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


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



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 +=



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



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



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



Location of forces as per Approach 2

Top of footing

h1 =19.46

280 kN



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



Tank empty conditionMoment = 211 x 19.18 = 4053 kNm

Top of footing

hcg =19.18

211 kN



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


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.



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=



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



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


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



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 ===



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


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

-----


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


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.


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



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



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



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



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



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



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



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


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


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


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 =



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 ×=×××

=



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



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



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



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 =



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



For hollow circular cross-sectionκ’ = 0.5

Hence, lateral stiffness of shaft staging, including shear deformations, will be

AG.L

EIL

Ks

503

13

+=



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



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



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



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



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



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



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 +=



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



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 +=



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


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


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 =

×==



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



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 =

×==



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


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

-----


Example 3: Comparison of results


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


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


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


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



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



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


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


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


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


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


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


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


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


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



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


In this Lecture

Following issues will be discussed Soil structure interactionBuried tanksP-Delta effectFlexibility of pipingBuckling of shellIS 11682:1985


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



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



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.



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



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



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



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 −=



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



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



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



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



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



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


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


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


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


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


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


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


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


Buried tanks

Dynamic soil pressure acts opposite to direction of seismic loading

Seismic force

Soil pressure


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.


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


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


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


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


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


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


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


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


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


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


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


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


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.


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


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


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 !!


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


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


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


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


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.


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


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


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


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


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


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.


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.


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.


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


© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 1 / slide 2


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.



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



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.


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.


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


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.=


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


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


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


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


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


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


© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Assignment 2/ slide 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



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.


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


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


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.


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


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


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


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





Read Clause 4.2.2, 4.2.3 and 4.3 of the Guidelines



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


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


© Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 3/ slide 2


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



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



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.


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


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,


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


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


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


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


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





Read Clause 4.4, 4.5 of the Guideline



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.


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.


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




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.



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.



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.


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


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)


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)


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


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


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)


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


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




Read Clause 4.6, 4.7 and 4.8 of the Guidelines



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.


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


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





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.



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.



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.



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 .


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


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


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


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


= 0.024 x 189 x 9.81

= 44.5 kN

Total base shear at bottom of wall is

( ) ( ) 398.3kN44.5395.8VVV 222c

2i =+=+=


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


( ) ( ) m-868.5kN110.4861.5MMM 222c

2i =+=+=


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.


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

* =+=+=


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


= 0.037 x 112 x 9.81 = 40.7 kN

Total base shear at bottom of wall is

( ) ( ) 532.8kN40.7531.2VVV 222c

2i =+=+=


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


( ) ( ) m-1097kN119.11090.3MMM 222c

2i =+=+=


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.


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

* =+=+=


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 =+=+=


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




Read Clause 4.9, 4.10, 4.11, 4.12 of the Guideline



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.


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



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.



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.



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


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


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


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 =×−×=−=


Linear impulsive pressure distribution

bi = 0.95 kN/m2

ai = 6.60 kN/m2


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 =×−×=−=


Linear convective pressure distribution ac = 0.57 kN/m2

bc = 1.08 kN/m2


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


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=


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 =×−×=−=


Linear impulsive pressure distribution

bi = 0.79 kN/m2

ai = 5.47 kN/m2


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 =×−×=−=


Linear convective pressure distribution ac = 0.035 kN/m2

bc = 0.72 kN/m2


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


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




Read Examples 2 and 3 of the Guideline



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.


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


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





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.


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

+=



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.


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


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.


Part III: Solutions

Rigid link

Brace

Radial beam

10 kN

Column


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


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 =


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


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


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


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


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


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


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


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


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.


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−=−×+−×=+=


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


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.


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+

=


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)


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

==


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.

is 1893 - tanks & water retaining

Documents