IS 1893 - TANKS & WATER RETAINING

Lecture 1

January 17, 2006

In this lecture

Types of tanks IS codes on tanks Modeling of liquid

© Sudhir K. Jain, IIT Kanpur

E-Course on Seismic Design of Tanks/ January 2006

Lecture 1 / Slide 2

Types of tanks

Two categories

Ground supported tanks

Also called at-grade tanks; Ground Service Reservoirs (GSR)

Elevated tanks

Also called overhead tanks; Elevated Service Reservoirs (ESR)



Lecture 1 / Slide 3

Types of tanks


Shape: Circular or Rectangular

Material : RC, Prestressed Concrete, Steel

These are ground supported vertical tanks

Horizontal tanks are not considered in this course



Lecture 1 / Slide 4

Types of tanks

Elevated tanks Two parts:

Container Staging (Supporting tower)



Lecture 1 / Slide 5

Types of tanks

Elevated tanks Container:

Material: RC, Steel, Polymer Shape : Circular, Rectangular, Intze, Funnel, etc.

Staging:

RC or Steel frame RC shaft Brick 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



Lecture 1 / Slide 6

Use of tanks

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



Lecture 1 / Slide 7

Indian Codes on Tanks

IS 3370:1965/1967 (Parts I to IV)

For concrete (reinforced and prestressed) tanks Gives design forces for container due to hydrostatic loads Based on working stress design BIS is considering its revision



Lecture 1 / Slide 8


IS 11682:1985

For RC staging of overhead tanks Gives guidelines for layout & analysis of staging More about this code later

IS 803:1976

For circular steel oil storage tanks



Lecture 1 / Slide 9


IS 1893:1984

Gives seismic design provisions Covers elevated tanks only Is under revision More about other limitations, later

IS 1893 (Part 1):2002 is for buildings only

Can not be used for tanks



Lecture 1 / Slide 10

Hydrodynamic Pressure

Under static condition, liquid applies pressure on container.

This is hydrostatic pressure

During base excitation, liquid exerts additional pressure on wall and base.

This is hydrodynamic pressure This is in additional to the hydrostatic pressure




Hydrodynamic pressure

Hydrostatic pressure

Varies linearly with depth of liquid Acts normal to the surface of the container At depth h from liquid top, hydrostatic pressure = γh h γh

Hydrostatic pressure © Sudhir K. Jain, IIT Kanpur





Has curvilinear variation along wall height Its direction is opposite to base motion

Hydrodynamic pressure Base motion © Sudhir K. Jain, IIT Kanpur




Summation of pressure along entire wall surface gives total force caused by liquid pressure

Net hydrostatic force on container wall is zero Net hydrodynamic force is not zero





Circular tanks (Plan View) Hydrostatic pressure


Base motion

Net resultant force = zero

Net resultant force

≠ zero

Note:- Hydrostatic pressure is axisymmetric; hydrodynamic is asymmetric © Sudhir K. Jain, IIT Kanpur




Rectangular tanks (Plan View) Hydrostatic pressure


Base motion

Net resultant force = zero


Net resultant force


≠ zero



Static design: Hydrostatic pressure is considered

Hydrostatic pressure induces hoop forces and bending moments in wall IS 3370 gives design forces for circular and rectangular tanks Net hydrostatic force is zero on container wall Hence, causes no overturning moment on foundation or staging Thus, 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 zero Affects design of container, staging and foundation





Procedure for hydrodynamic pressure & force:

Very simple and elegant Based 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 details Only 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 liquid

Liquid 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 liquid

Convective liquid

Moves with wall; rigidly attached Has same acceleration as wall Also called sloshing liquid Moves relative to wall Has different acceleration than wall

Impulsive & convective liquid exert pressure on wall

Nature of pressure is different See next slide




Modeling of liquid

Impulsive

Convective Base motion

Base motion

Hydrodynamic pressure © Sudhir K. Jain, IIT Kanpur



Modeling of liquid

At this point, we will not go into details of hydrodynamic pressure distribution

Rather, we will first find hydrodynamic forces Impulsive force is summation of impulsive pressure on entire wall surface Similarly, 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 acceleration Convective force = mc x acceleration

mi & mc experience different accelerations Value of accelerations will be discussed later First we will find mi and mc




Modeling of liquid

Housner suggested graphs for mi and mc

mi and mc depend on aspect ratio of tanks Such graphs are available for circular & rectangular tanks

See Fig. 2a and 3a of Guidelines Also see next slide

For taller tanks (h/D or h/L higher), mi as fraction of m is more For short tanks, mc as fraction of m is more




Modeling of liquid

1

1

mi/m

mi/m

0.5

0.5

mc /m

mc /m 0

0 0

0.5

h/D

1

1.5

For circular tanks

2

0

0.5

h/L

1

1.5

2

For rectangular tanks

See next slide for definition of h, D, and L




Modeling of liquid

D

h

Plan of Circular tank

Elevation Base motion

L

L

Plan of Rectangular tank © Sudhir K. Jain, IIT Kanpur

Base motion



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.




1 mi/m

0.5 mc /m

0 0


0.5

h/D

1

1.5

2



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 wall Convective liquid moves relative to wall

As if, attached to wall with springs

Kc/2

Kc/2 mc

Rigid m i


Convective liquid (moves relative to wall)

Impulsive liquid (moves with wall)



Modeling of liquid

Stiffness associated with convective mass, Kc

Kc depends on aspect ratio of tank Can be obtained from graph

Refer Fig. 2a, 3a of guidelines See next slide




Modeling of liquid

1 Kch/mg mi/m

0.5 mc/m

0 0

0.5

1

1.5

2

h/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 1 Kch/mg mi/m

0.5 mc/m

0 0

0.5

1

1.5

2

h/D 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. © Sudhir K. Jain, IIT Kanpur



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 floor Location 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 diagram Location of centroid:

Obtained by dividing the moment due to pressure distribution by the magnitude of impulsive force

Similarly, location of convective mass is obtained

See next slide




Modeling of liquid Resultant of impulsive pressure on wall

Resultant of convective pressure on wall

hc hi

hi, hc can be obtained from graphs

They also depend on aspect ratio, h/D or h/L Refer Fig. 2b, 3b of guidelines See next slide




Modeling of liquid

1

1

0.8

0.8

hc/h

hc/h 0.6

0.6

hi/h

hi/h

0.4

0.4

0.2

0.2

0

0

0

0.2

0.4

0.6

0.8

1 h/D 1.2

1.4

For circular tanks


1.6

1.8

2

0

0.5

1

h/L

1.5

2

For rectangular tanks



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.




1

0.8

hc/h

0.6

hi/h

0.4

0.2

0 0


0.5

1 h/D

1.5

2



Modeling of liquid

From graph, for h/D = 0.375; hi/h = 0.375 hi = 0.375 x 3 = 1.125 m and 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 base

Under static condition, base is subjected to uniformly distributed pressure Due to base motion, liquid exerts nonuniform pressure on base

This is in addition to the hydrostatic pressure on the base

See next slide




Modeling of liquid

Base motion Hydrostatic pressure on base


Hydrodynamic pressure on base



Modeling of liquid

Impulsive as well as convective liquid cause nonuniform pressure on base

Nonuniform pressure on base causes overturning effect This will be in addition to overturning effect of hydrodynamic pressure on wall See next slide




Modeling of liquid

hi

Overturning effect due to wall pressure

Overturning effect due to base pressure

Note:- Both the overturning effects are in the same direction © Sudhir K. Jain, IIT Kanpur



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 earlier For overturning effect due to wall pressure alone, resultant was applied at hi * For hi and hi , see next slide




Modeling of liquid

h*i hi

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

h*c hc

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 hc

Refer Fig. C-1 of the Guidelines

hi* & hc* depend on aspect ratio Graphs to obtain hi, hc, hi*, hc* are provided

Refer Fig. 2b & 3b of guidelines Also see next slide Please note, hi* and hc* can be greater than h




Modeling of liquid 2.5 hc*/h 2 1.5 1 hc/h

0.5

hi*/h hi/h

0 0

0.5

1

1.5

2

h/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 2.5 hc*/h

2 1.5 1 hc/h

hi */h

0.5 hi /h 0 0

0.5

1

1.5

2

h/D

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 © Sudhir K. Jain, IIT Kanpur



Modeling of liquid

This completes modeling of liquid

Liquid is replaced by two masses, mi & mc This is called mechanical analogue or spring mass model for tank See next slide




Modeling of liquid mi = Impulsive liquid mass Kc/2

mc

Kc/2

Rigid m i

mc = Convective liquid mass Kc = Convective spring stiffness

hi (hi*)

hc (hc*)

Mechanical analogue or spring mass model of tank

hi = Location of impulsive mass (without considering overturning caused by base pressure) hc = Location of convective mass (without considering overturning caused by base pressure) hi* = Location of impulsive mass (including base pressure effect on overturning) hc* = Location of convective mass (including base pressure effect on overturning)




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 Guidelines These are reproduced in next two slides




Modeling of liquid For circular tanks h⎞ ⎛ tanh ⎜ 3.68 ⎟ mc D⎠ ⎝ = 0.23 h m D

D⎞ ⎛ tanh ⎜ 0 . 866 ⎟ mi h ⎠ ⎝ = D m 0 . 866 h hi = 0 . 375 h = 0 .5 −

hi * = h

for h / D ≤ 0 . 75 0 . 09375 h / D

0.866

D h

D⎞ ⎛ 2 tanh⎜ 0.866 ⎟ h⎠ ⎝

for h / D > 0 . 75

- 0.125

= 0.45

for h / D ≤ 1.33

h⎞ ⎛ cosh ⎜ 3.68 ⎟ − 1.0 hc D⎠ ⎝ = 1− h h⎞ h ⎛ 3.68 sinh ⎜ 3.68 ⎟ D D⎠ ⎝

h⎞ ⎛ cosh⎜ 3.68 ⎟ − 2.01 hc * D⎠ ⎝ =1− h h h⎞ ⎛ 3.68 sinh⎜ 3.68 ⎟ D D⎠ ⎝

for h / D > 1.33

K c = 0.836 © Sudhir K. Jain, IIT Kanpur


mg h⎞ ⎛ tanh 2 ⎜ 3.68 ⎟ h D⎠ ⎝ Lecture 1 / Slide 58

Modeling of liquid For rectangular tanks h⎞ ⎛ tanh⎜ 3.16 ⎟ mc L⎠ ⎝ = 0.264 h m L

L⎞ ⎛ tanh⎜ 0.866 ⎟ mi h⎠ ⎝ = L m 0.866 h

hi = 0 . 375 h = 0 .5 −

hi * = h

for h / L ≤ 0 . 75 0 . 09375 h/L

0.866

L h

L⎞ ⎛ 2 tanh⎜ 0.866 ⎟ h⎠ ⎝

for h / L > 0 . 75

− 0.125

= 0 .45


for h / L ≤ 1.33 for

h / L > 1.33

h⎞ ⎛ cosh ⎜ 3.16 ⎟ − 1.0 hc L⎠ ⎝ = 1− h h⎞ h ⎛ 3.16 sinh ⎜ 3.16 ⎟ L L⎠ ⎝

h⎞ ⎛ cosh⎜ 3.16 ⎟ − 2.01 hc * L⎠ ⎝ = 1− h h h⎞ ⎛ 3.16 sinh⎜ 3.16 ⎟ L L⎠ ⎝

K c = 0.833

mg h⎞ ⎛ tanh 2 ⎜ 3.16 ⎟ h L⎠ ⎝



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 0.125 These 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 acceleration Convective force, Vc = mc x acceleration Kc/2

mc

Rigid m i


Kc/2

Vc Vi



Modeling of liquid

Vi and Vc will cause

Bending Moment (BM) in wall




Modeling of liquid

BM at bottom of wall

BM due to Vi = Vi x hi BM due to Vc = Vc x hc

Total BM is not necessarily Vi X hi+ Vc X hc

More about this, later

Kc/2

Kc/2

hc

Rigid

mc

Vc

mi

Vi

hi © Sudhir K. Jain, IIT Kanpur



Modeling of liquid

Overturning of the container is due to pressure on wall and base

Pressure on base does not cause BM in wall

Overturning Moment (OM) at tank bottom

OM is at bottom of base slab Hence, includes effect of pressure on base Note the difference between bottom of wall and bottom of base slab




Modeling of liquid

OM at bottom of base slab

OM due to Vi = Vi x hi* BM due to Vc = Vc x hc*

Kc/2

Kc/2

Rigid

hc*

mc

Vc

mi

Vi

hi*




Modeling of liquid

mi and mc will have different accelerations We 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 example

Acceleration 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 = 3m From previous examples: mi = 63.3 t mc = 84.5 t hi = 1.125 m hc = 1.65 m hi* = 3.3 m hc* = 3.0 m Impulsive acceleration, ai = 0.3g = 0.3 x 9.81 = 2.94 m/sec2 Convective acceleration, ac = 0.1g = 0.1 x 9.81 = 0.98 m/sec2 © Sudhir K. Jain, IIT Kanpur



Modeling of liquid Example 5 (Contd..)

Impulsive force, Vi = mi x ai = 63.3 x 2.94 = 186.1 kN Convective 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-m Bending moment at bottom of wall due to Vc = Vc x hc = 82.8 x 1.65 = 136.6 kN-m Overturning moment at bottom of base due to Vi = Vi x hi* = 186.1 x 3.3 = 614.1 kN-m Overturning moment at bottom of base due to Vc = Vc x hc* = 82.8 x 3.0 = 248.4 kN-m © Sudhir K. Jain, IIT Kanpur



At the end of Lecture 1

In seismic design, mechanical analogue of tanks are used, wherein, liquid is replaced by impulsive & convective masses These masses and their points of application depend on aspect ratio Graphs and expressions are available to find all these quantities

These are based on work of Housner (1963a)




E-Course (through Distance Learning Mode) on

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

Response to Questions and Comments on Lecture 1 Question from Mr. S. K. Kukreja ([email protected]) 1. clarify following 1. fig 2a,3a of guidelines as pointed in lecture1/slide25 2. table C-1 of guidelines as pointed in lecture1/slide57 3. Housner graphs to find h* are not given for rectangular tanks In Figure 2a of the Guideline, mi/m, mc/m and Kch/mg are plotted as a function of h/D. This is for circular tanks and corresponding figure for rectangular tanks is given in Figure 3a. Expressions for these quantities are given in Table C-1 of the Guideline. Basically, graphs plotted in these figures are based on these expressions only. Depending on convenience, one can find these quantities either from graphs or from expressions. In the Lecture we have given graphs of hi* and hc* only for circular tanks (Slide 54), which is Figure 2b of the Guideline. Corresponding figure for rectangular tank is Figure 3b of the Guideline, which has not been reproduced in the Lecture. Expressions for hi* and hc* for circular as well as rectangular tanks are also given in Table C-1. Question from Mr. Umesh H. Patil ([email protected]) 1. pl.something about hydrostatic design Under static conditions, liquids apply a pressure on the bodies in contact. This pressure is a product of unit weight of liquid and depth of the liquid at the point under consideration. Direction of this pressure is always normal (perpendicular) to the surface of the body. This is known as hydrostatic pressure, and it varies linearly with height. Hydrostatic pressure induces hoop forces and bending of the wall. IS 3370:1967 gives these forces for circular and straight walls. Refer standard textbooks for more information. 2. show to calculate hydrodynamic pr.parameters for tapering sections of wall/container

Lecture 3 has covered information on tanks of other shapes such as Intze, conical, etc. If cross-section of wall is tapered, i.e., its thickness varies along the height, then there will not be any change in hydrodynamic pressure. Recall, hydrodynamic pressure is obtained considering wall as rigid. 3. effect of top covered slab on hydrodynmic pressure Top slab will impart more stiffness to wall and make it more rigid. This is not going to affect hydrodynamic pressure distribution. However, if sloshing liquid touches roof slab, then hydrodynamic pressure distribution will be affected. Please refer Lecture 6 and Commentary to Clause 4.11 of the Guideline. 4. there is difference in answers from graph & equations Reading the value from graph is likely to induce some approximation and hence, expressions are also given. The difference, however, may not be significant from engineering view point. Question from Mr. H. I. Abdul Gani ([email protected]) The following are my questions on Lecture 1. 1) What exactly does Horizontal Tank and Vertical Tank mean A circular tank kept horizontally on ground, i.e., longitudinal axis of tank is horizontal is termed as horizontal tank. In petrol pumps, steel tanks used for storing petrol/diesel are usually horizontal tanks. Railway tankers carrying petroleum products are also horizontal tanks. Another good example is heating greasers in bathrooms, some are of horizontal type and some are of vertical type !!. A vertical tank is the one, in which longitudinal axis of tank is vertical. Classification as horizontal and vertical tank is more appropriate for circular tanks only. 2) It was shown that the hydrodynamic pressure is curvilinear in shape. Does the pressure profile depend on the stiffness of the tank walls? 3) How the stiffness of the wall affects i) Convective mass and ii) impulsive mass These have been addressed in Lecture 3 4) In example 1, why the sum of impulsive and convective masses (147=2E8kg) is less than the total mass of water (150.8 kg)? This also has been addressed in Lecture 3

5) In the hydrodynamic pressure profile in the base, it is not clear how pressure can develop in the upward direction? Can you please explain Under static condition, liquid exerts uniform pressure on base. During lateral excitation, the liquid increases vertical pressure on some portion of slab and reduces pressure on other portion. This increase or decrease in pressure is the hydrodynamic effect. The decrease in the pressure is like an upward pressure. 6) How to find height of sloshing liquid? Does it depend on acceleration This is addressed in Lecture 6 7) For tanks with arbitrary shapes like triangular, hexagonal shapes etc, how to find impulsive and convective masses? Any simplified methods available Please refer Lecture 3 8) During vertical acceleration, how to find impulsive and convective mass? Information on effect of vertical excitation is covered in Lecture 6. Also read Clause 4.10 of the Guideline. Vertical excitation will cause change in weight density of liquid, which in turn will change hydrostatic pressure on wall. 9) Why effect of base pressure is considered along with both convective mass and impulsive mass instead of convective mass only? In our modeling, liquid is divided into two parts, one that vibrates along with wall (impulsive mass) and other which moves relative to wall (convective mass). During lateral excitation, both impulsive and convective liquids exert non-uniform pressure on base. 10) How to check adequacy of freeboard in a water tank during dynamic excitation? This issue is addressed in Lecture 6 11) For lumping mass in mathematical model for impulsive and convective mass which height is to be used, h or h*? If bending moment at the bottom of wall is to be obtained, then, masses shall be lumped at h. If overturning moment at the bottom of base slab is to

obtained, then masses shall be lumped at h*. For design of staging and foundation also, mass must be lumped at h*. 12) For load application in a mathematical model, only C.G. (h) of the impulsive/ convective force is known. How to find the area and distribution of load (or mass) with respect to height of the tank? Details about distribution of impulsive and convective pressure along wall are covered in Lecture 6. Question from ([email protected]) I didnt understood the following pl explain mi/m=3D tanh(.866*12/5)/.866*12/5 now in this expression mi/m=3Dtanh 2.078/2.078 mi/m=3Dtan 5*2.078/2.078 is it finding tan 10.39 and then dividing it by 2.078 all the other thing i understood and also since i have referred charts from the guidlines i also got the answers correct Abut only i didnt understood the formulae part how the value arrived thanking you chandrshekhar The expression which you have given above contains 3D term, which is not present in the expressions given in Lectures and in Table C-1 of the Guideline. You are also interested in knowing how this formula is arrived at. The derivation of this formula is given in Housner’s paper (1963a) and you may go through the same. It comes from potential equation of liquid or Laplace equation. Since it is a bit mathematical, we have not included this in the course contents. Question from Mr. R. Murugan ([email protected]) 1. How convective pressure is distributed and upto what depth ? We need to recognize that we are dividing liquid in two parts. That does not mean that physically liquid gets divided in two parts and liquid only up to certain depth participates in convective mode and liquid below it participates in impulsive mode. It is an idealization, and convective and impulsive pressures are present over the entire height of liquid. This can also be observed in the pressure distributions shown in Lecture 6. 2. In slide 54 How hi* is higher than the tank height? hi* is that height where mi should act so that moment due to impulsive pressure on wall and moment due to impulsive pressure on base is equal to

moment due to mi (actually, mi × g) . Since, effect of moment due to base pressure is included, this height can become larger than h. On the other hand hi, which includes effect of impulsive pressure only on the walls, is always less than h.

3. The formulae derived for the above tanks are applicable for open tank or closed tank ? (i.e. tank with cover slab or without cover slab) Actually these are derived for open tanks, wherein, question of liquid touching roof slab does not arises. However, due to presence of roof slab, pressure distribution on wall is not going to change, provided sloshing liquid does not touch roof slab. If it touches roof slab then some additional issues will come up. These issues have been addressed in Lecture 6 and in Clause 4.11 of the Guideline. 4. The formulas and graph derived to find hi, hc, mi, mc etc is applicable for RC tank or steel tank? The formulae are irrespective of materials or the materials proportion will change the co-efficient? This issue has been addressed in Lecture 3. Effect of flexibility of wall on pressure distribution has been studied by Veletsos (1984). However, for design purpose we use formulae derived for rigid walls. Please refer Lecture 3 and commentary of Clause 4.2.1.2 5. What is mean by squat tank, slender tank? Slender tank is one in which h/D or h/L is quite high. In squat tank h/D or h/L is less. There is no fixed limit on these values at which this demarcation begins. Generally, these terminologies are used for impulsive and convective masses. For example, we say that in slender tanks, convective mass is less and in squat tanks, impulsive mass is less. In this context, from Figure 2a of the Guideline, we can see that for h/D = 0.2, about 80% mass contributes to convective mode and for h/D = 1.0 about 80% mass contributes to impulsive mode. Thus, one can say that a tank with h/D less than 0.2 is squat tank and a tank with h/D greater than 1.0 is slender tank. 6. How Question 1.2 is false in Assignment i.e.., in short tank mi is less only See Example-1 in Lecture -1. This was error in our solution, and we have sent addendum to it.

7. So, far we learnt about impulsive mass, convective mass and it is centeroid. When we do the modeling in computer. How impulsive mass is distributed i.e how applied and upto what depth? Distribution of impulsive and convective pressure along wall height is covered in Lecture 6. 8. Why mi + mc is not coming to m ? In solution 1, mi/m = 0.466; mc/m = 0.503; when both are added the value is 0.969; Nearly 3% of the liquid mass is unaccounted. Why is it so? Is it not correct to include the missing mass in impulsive liquid component? This has been discussed in Lecture 3. 9. Table C1-Expression for parameters of spring mass model, hi/h = 0.375 for h/D less than or equal to 0.75 is to be given, while the actual one furnished reads, h/d greater than 0.75 This is a typographical error in Table C-1, which we have pointed out in the Lecture 1. There is another typographical error in the expression for hi* for circular tanks. This also has been pointed out in Lecture 1 (Slide no. 60). Question from Mr. Abdul Gani ([email protected]) It is explained in lecture 1, impulsive pressure distribution as below:

Impulsive Let us assume mi/m =x: If the pressure distribution is as shown in the above figure, hi should come less than or equal to 0.5*x* h, i.e., hi/h should always be lower than 0.5x. Table below compares the hi expected versus actual computed

Problem no.

mi/m

1.1 Circular tank

0.466

1.2 Rectangular tank in longer direction 1.2 Rectangular tank in shorter direction

0.54

0.72

hi/h expected Less than or equal to 0.233 Less than or equal to 0.54/2 0.27 Less than or equal to 0.72/2 0.36

hi/h computed 0.375 0.375

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 ( x ) × (h ) 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, hi/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 evaluation Procedure in codes Limitations of IS 1893:1984



Lecture 2 / Slide 2

Seismic force evaluation

During base excitation

From Newton’s second law

Structure is subjected to acceleration Force = mass x acceleration

Hence, seismic force acting on structure = Mass x acceleration



Lecture 2 / Slide 3


For design, we need maximum seismic force Hence, maximum acceleration is required

This refers to maximum acceleration of structure This is different from maximum acceleration of ground Maximum ground acceleration is termed as peak ground acceleration, PGA Maximum acceleration of rigid structure is same as PGA.



Lecture 2 / Slide 4


Seismic force = mass x maximum acceleration

Force = (maximum acceleration/g) x (mass x g) = (maximum acceleration/g) x W

Can be written as:

W is weight of the structure g 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 shear Ah is base shear coefficient



Lecture 2 / Slide 5


Maximum acceleration of structure depends on

Severity of ground motion Soil conditions Structural characteristics

These include time period and damping More about time period, later

Obviously, base shear coefficient, Ah, will also depend on these parameters



Lecture 2 / Slide 6


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



Lecture 2 / Slide 7


Thus, base shear coefficient depends on:

Severity of ground motion Soil condition Structural characteristics Design philosophy



Lecture 2 / Slide 8


Let us examine how following codes have included these parameters in base shear coefficient

IS 1893 (Part 1): 2002 IS 1893:1984 International Building code (IBC) 2003 from USA

Study of IBC provisions will help us understand the present international practice



Lecture 2 / Slide 9

IS 1893 (Part 1):2002

Ah = (Z/2). (I/R). (Sa/g)

Z is zone factor I is importance factor R is response reduction factor Sa/g is spectral acceleration coefficient




IS 1893 (Part 1):2002

Zone factor, Z

Depends on severity of ground motion India is divided into four seismic zones (II to V) Refer Table 2 of IS 1893(part1):2002 Z = 0.1 for zone II and Z = 0.36 for zone V




IS 1893 (Part 1):2002

Importance factor, I

Ensures higher design seismic force for more important structures Values for buildings are given in Table 6 of IS :1893

Values for other structures will be given in respective parts For tanks, values will be given in Part 2




IS 1893 (Part 1):2002

Response reduction factor, R

Earthquake resistant structures are designed for much smaller seismic forces than actual seismic forces that may act on them. This depends on

Ductility Redundancy Overstrength

See next slide




IS 1893 (Part 1):2002

Δ Total Horizontal Load

Total Horizontal Load

Maximum force if structure remains elastic Fel Linear Elastic Response

Maximum Load Capacity Fy Load at First Yield

Due to Ductility Non linear Response Due to Redundancy

First Significant Yield

Fs

Due to Overstrength

Design force Fdes

0

Δw

Δy

Figure: Courtesy Dr. C V R Murty

Δmax Roof Displacement (Δ)

Response Reduction Factor = © Sudhir K. Jain, IIT Kanpur

Maximum Elastic Force (Fel) Design Force (Fdes)



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 buildings

For 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:1893 See 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

% 0 damping

2

5

Factor

1.40

1.00 0.90

3.20

7

10

15

20

25

30

0.80

0.70

0.60

0.55

0.50

For higher damping, multiplying factor is less Hence, for higher damping, Sa/g is less




IS 1893:1984

Let us now look at the provision of IS 1893:1984 IS 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 factor C is a coefficient which depends on time period β is soil-foundation system coefficient I is importance factor αo is seismic coefficient Fo is zone factor Sa/g is average acceleration coefficient




IS 1893:1984

Seismic coefficient, αo

Depends on severity of ground motion Used in seismic coefficient method

Zone factor, Fo

Depends on severity of ground motion Used in response spectrum method




IS 1893:1984




IS 1893:1984

β is soil foundation coefficient

Depends 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


Ensures higher design seismic force for more important structures

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




IS 1893:1984

Performance factor, K

Depends on ductility of structure

Similar to response reduction factor of IS1893(Part 1):2002 K is in numerator whereas, R is in denominator

For buildings with good ductility, K = 1.0 For ordinary buildings, K = 1.6 Thus, a building with good ductility will have lower value of base shear coefficient than ordinary building




IS 1893:1984




IS 1893:1984

Coefficient, C

Depends on time period see next slide

Spectral acceleration, Sa/g

Depends on time period and damping See 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 considered

For elevated tanks, it suggests Ah = βIFoSa/g

Performance factor, K is not present Implies, 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 IS1893:1984

More about it, later




IBC 2003

International Building Code (IBC) 2003

In 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 coefficient Ah = SD1 I/(R T) ≤ SDS I/R Ah 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 Ah It ensures minimum design seismic force This 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 seconds SDS and SD1 are design spectral accelerations in short period and at 1 sec. respectively

SDS and SD1 depend on seismic zone and soil

I is importance factor and R is response modification factor

IBC suggests I = 1.0, 1.25 and 1.5 for different types of structures Values of R will be discussed later




IBC 2003

More about SDS and SD1

SDS = 2/3 Fa SS and SD1 = 2/3 Fv S1 SS is mapped spectral acceleration for short period S1 is mapped spectral acceleration for 1-second period SS 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 coefficients

Their values for are given for different soil types




IBC 2003

Response modification factor, R

IS 1893(Part 1):2002 calls it response reduction factor Values of R for some selected structures are given in next slide




IBC 2003 R

Type of structure Building with special reinforced concrete moment resisting concrete frames

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 Ah = (Z/2).(I/R).(Sa/g)

IS 1893: 1984 SCM: Ah = KCβIαo RSM: Ah = KβIFoSa/g For tanks: Ah = βIFoSa/g


IBC2003 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. W This is same as force = mass x acceleration





Let us compare base shear coefficient values from these codes

Comparison will be done at working stress level

IBC 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 code This comparison is first done for buildings





Comparison for buildings

Following parameters are chosen

IS 1893 (Part 1): 2002 Z = 0.36; Zone V I = 1.0; R = 5.0 Soft soil 5% damping

IS 1893: 1984 αo = 0.08; Fo = 0.4; Zone V β = 1.0 K = 1.0; I = 1.0 Soft soil, raft foundation 5% damping

IBC2003 SDs = 1.0; SD1 = 0.6 I = 1.0; R = 8.0 Soil type D, equivalent to soft soil of IS codes 5% damping

They represent similar seismic hazard level





Building with good ductility is chosen

Say, buildings with SMRF

In IBC, for buildings with SMRF, R = 8.0

Refer Table shown earlier





For building with T = 0.3 sec IS 1893(Part 1):2002

IS 1893:1984

Sa/g =2.5 Ah = Z/2.I/R.Sa/g = (0.36/2 )x (1.0/5.0) x 2.5 = 0.09 C = 1.0 and Sa/g = 0.2 SCM: Ah = KCβIαo = 1.0 x 1.0 x1.0 x1.0x0.08 = 0.08 RSM: Ah = KβIFoSa/g = 1.0 x 1.0 x1.0x 0.4x0.2 = 0.08

IBC 2003

Ah = SDSI/(1.4xR) = 1.0 x1.0/(1.4 x 8.0) = 0.089





For building with T = 1 sec IS 1893(Part 1):2002

IS 1893:1984

Sa/g = 1.67 Ah = Z/2.I/R.Sa/g = (0.36/2 )x(1.0/5.0)x1.67 = 0.06 C = 0.53 and Sa/g = 0.11 SCM: Ah = KCβIαo = 1.0 x0.53x1.0 x1.0x0.08 = 0.042 RSM: Ah = KβIFoSa/g = 1.0x1.0x1.0x0.4x0.11 = 0.044

IBC 2003

Ah = SD1I/(1.4xRxT) = 0.6x1.0/(1.4 x 8.0x1.0) = 0.054





For building with T = 1.5 sec IS 1893(Part 1):2002

IS 1893:1984

Sa/g = 1.11 Ah = Z/2.I/R.Sa/g = (0.36/2 )x(1.0/5.0)x1.11 = 0.040 C = 0.4 and Sa/g = 0.078 SCM: Ah = KCβIαo = 1.0 x0.4x1.0 x1.0x0.08 = 0.032 RSM: Ah = KβIFoSa/g =1.0x1.0x1.0x0.4x0.078 = 0.031

IBC 2003

Ah = SD1I/(1.4RT) = 0.6x1.0/(1.4 x 8.0x1.5) = 0.036





Base shear coefficients for four time periods T (Sec)

IS 1893 (Part 1): 2002

IS 1893: 1984 SCM

IBC2003

RSM

0.3

0.09

0.08

0.08

0.089

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*

* Due to lower bound, this value is higher

Graphical comparison on next slide





Comparison of base shear coefficient (Buildings)

0.1

IS 1893(Part 1):2002


IBC 2003 0.075

IS 1893:1984; RSM

Note the lower bound of IBC

IS 1893:1984; SCM 0.05

0.025

0 0

0.5

1

1.5

2

2.5

3

Time Period (S)





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 2003 IS 1893(Part 1):2002 is only for buildings

Only elevated tanks will be considered

Hence, can’t be used for tanks IS 1893:1984 has provisions for elevated tanks only

Zone and soil parameters will remain same as those considered for buildings Importance factor for tanks are different than those for buildings





In IBC

I = 1.25 for tanks R = 3.0 for tanks on frame staging (braced legs) R = 2.0 for tanks on shaft or pedestal

In 1893:1984

I = 1.5 for tanks K 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 sec IS 1893:1984

I = 1.5, Sa/g = 0.2 Ah = βIFoSa/g = 1.0 x 1.5 x 0.4x0.2 = 0.12 This value is common for frame and shaft staging

IBC 2003

For frame staging, I = 1.25, R = 3.0 Ah = SDSI/(1.4xR) = 1.0 x1.25/(1.4 x 3.0) = 0.298 For shaft staging, I = 1.25, R = 2.0 Ah = SDSI/(1.4xR) = 1.0 x1.25/(1.4 x 2.0) = 0.446





For tank with T = 1 sec IS 1893:1984

I = 1.5, Sa/g = 0.11 Ah = βIFoSa/g =1.0x1.5x0.4x0.11 = 0.066

IBC 2003

For frame staging, I = 1.25, R = 3.0 Ah = SD1I/(1.4xRxT) = 0.6x1.25/(1.4 x 3.0x1.0) = 0.178 For shaft staging, I = 1.25, R = 2.0 Ah = SD1I/(1.4xRxT) = 0.6x1.25/(1.4 x 2.0x1.0) = 0.268





Base shear coefficients for tanks T (Sec)

IS 1893:1984*

IBC 2003 Frame staging

Shaft staging

0.3

0.12

0.298

0.446

1.0

0.066

0.178

0.268

* Base shear coefficient values are common for frame and shaft staging

Graphical comparison on next slide





Comparison of base shear coefficient (Tanks) 0.5


IBC 2003; Tanks on shaft staging 0.4

IBC 2003; Tanks on frame staging 0.3

IS 1893:1984; All types of elevated tanks

0.2 0.1 0 0

0.5

1

1.5

2

2.5

3

Time period (S) © Sudhir K. Jain, IIT Kanpur




Base shear coefficient for elevated tanks from IS1893:1984 is on much lower side than IBC 2003 IBC 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 1893:1984

IBC uses R = 2.0 and R = 3.0 for tanks as against R = 8.0 for buildings with good ductility IS 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 1893:1984

In Lecture 1, we have seen, liquid mass gets divided into impulsive and convective masses IS 1893:1984, does not consider convective mass It 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 tanks Design base shear coefficient is to be expressed in terms of parameters of IS 1893(Part 1):2002

Ah = (Z/2). (I/R). Sa/g Z will be governed by seismic zone map of Part 1 I 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 values Procedure for finding time period in next lecture





Seismic force = (Ah) X (W) Base shear coefficient, Ah, depends on

Seismic Zone Soil type Structural characteristics Ductility, 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

Response to Questions and Comments on Lecture 2 Question from Mr. Abdul Gani ([email protected]) The following are my questions on Lecture 2. 1. What exactly is meant by ‘Overstrength’? Overstrength means additional strength which structure possess over and above the design strength. This additional strength comes from factors on gravity loads, live loads and earthquake loads. For limit state design of RC structures, we use partial safety factor of 1.5 on gravity loads. Partial safety factors on materials also provide overstrength. Moreover, materials may have strength greater than their characteristic strength, Nonstructural 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 IITKGSDMA Guidelines.

For time period greater than 1.7 sec also there is difference between IBC and IS 1893(Part 1). This is due to the fact that IBC has given a lower limit on its base shear coefficient, which means to say that, beyond certain time period, design seismic force will remain same. This is done to ensure certain minimum strength against lateral loads. Moreover, if time period is incorrectly estimated on higher side, then also this lower limit acts as a safeguard. IS 1893 does not have such lower limit and for tanks also in the IITK-GSDMA Guideline, as of now there is no such lower limit. 3. Why response reduction factor is different for convective and impulsive masses? In IITK-GSDMA Guideline, response reduction factor is same for impulsive and convective modes. However, in some international codes (like, ACI 350.3 and AWWA D-115), response reduction factors are different for impulsive and convective modes. These codes argue that convective mode is low frequency mode (large time period) and hence, convective forces would vary slowly with time. That means, convective forces are like static forces and hence, reduction due to ductility shall not be applicable to them. ACI 350.3 and AWWA D-115 use R = 1.0 for convective mode, i.e. they do not allow any reduction in convective seismic forces. However, it is important to note that, these codes use different response spectrum for convective and impulsive modes. For convective mode, which is usually having very large time period, response spectrum has 1/Tc2 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 tanks Time period of tanks



Lecture 3 / Slide 2

Modeling of tanks

As seen in Lecture 1 liquid may be replaced by impulsive and convective mass for calculation of hydrodynamic forces

See next slide for a quick review



Lecture 3 / Slide 3

Modeling of tanks mi = Impulsive liquid mass Kc/2

mc

mc = Convective liquid mass

Kc/2

Rigid m i

Kc = Convective spring stiffness

hi (hi*)

hc (hc*)

Mechanical analogue or spring mass model of tank

hi = Location of impulsive mass (without considering overturnig caused by base pressure) hc = Location of convective mass (without considering overturning caused by base pressure) hi* = Location of impulsive mass (including base pressure effect on overturning) hc* = Location of convective mass (including base pressure effect on overturning)

Graphs and expression for these parameters are given in lecture 1. © Sudhir K. Jain, IIT Kanpur


Lecture 3 / Slide 4

Approximation in modeling

Sometimes, summation of mi and mc may not be equal to total liquid mass, m

This difference may be about 2 to 3 % Difference arises due to approximations in the derivation of these expressions

More about it, later

If this difference is of concern, then

First, obtain mc from the graph or expression Obtain mi = m – mc



Lecture 3 / Slide 5

Tanks of other shapes

For tank shapes such as Intze, funnel, etc. :

Consider equivalent circular tank of same volume, with diameter equal to diameter at the top level of liquid



Lecture 3 / Slide 6


Example:

An Intze container has volume of 1000 m3. Diameter of container at top level of liquid is 16 m. Find dimensions of equivalent circular container for computation of hydrodynamic forces. Equivalent circular container will have diameter of 16 m and volume of 1000 m3. Height of liquid, h can be obtained as : π/4 x 162 x h = 1000 ∴ h = 1000 x 4/(π x 162) = 4.97 m © Sudhir K. Jain, IIT Kanpur


Lecture 3 / Slide 7


Thus, for equivalent circular container, h/D = 4.97/16 = 0.31 All the parameters (such as mi, mc etc.) are to be obtained using h/D = 0.31

16 m

16 m

Intze container volume = 1000 m3 © Sudhir K. Jain, IIT Kanpur

4.97 m

Equivalent circular container


Lecture 3 / Slide 8

Effect of obstructions inside tank

Container may have structural elements inside

For example: central shaft, columns supporting the roof slab, and baffle walls

These elements cause obstruction to lateral motion of liquid This will affect impulsive and convective masses



Lecture 3 / Slide 9


Effect of these obstructions on impulsive and convective mass is not well studied

A good research topic !

It is clear that these elements will reduce convective (or sloshing) mass

More liquid will act as impulsive mass





In the absence of detailed analysis, following approximation may be adopted:

Consider a circular or a rectangular container of same height and without any internal elements Equate the volume of this container to net volume of original container

This will give diameter or lateral dimensions of container

Use this container to obtain h/D or h/L





Example: A circular cylindrical container has internal diameter of 12 m and liquid height of 4 m. At the center of the tank there is a circular shaft of outer diameter of 2 m. Find the dimensions of equivalent circular cylindrical tank.

12 m

4m

12 m Elevation


Hollow shaft of 2 m diameter

Plan




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, then π/4xD2x4 = 439.8 m3

∴ D = 11.83 m Thus, for equivalent circular tank, h = 4 m, D = 11.83m and h/D = 4/11.83 = 0.34. This h/D shall be used to find parameters of mechanical model of tank © Sudhir K. Jain, IIT Kanpur



Effect of wall flexibility

Parameters mi, mc etc. are obtained assuming tank wall to be rigid

An assumption in the original work of Housner (1963a)

Housner, G. W., 1963a, “Dynamic analysis of fluids in containers subjected to acceleration”, Nuclear Reactors and Earthquakes, Report No. TID 7024, U. S. Atomic Energy Commission, Washington D.C.

RC tank walls are quite rigid Steel tank walls may be flexible

Particularly, in case of tall steel tanks





Wall flexibility affects impulsive pressure distribution

Effect of wall flexibility on impulsive pressure depends on

It does not substantially affect convective pressure distribution Refer Veletsos, Haroun and Housner (1984) Veletsos, A. S., 1984, “Seismic response and design of liquid storage tanks”, Guidelines for the seismic design of oil and gas pipeline systems, Technical Council on Lifeline Earthquake 1Engineering, ASCE, N.Y., 255-370, 443-461. Haroun, M. A. and Housner, G. W., 1984, “Seismic design of liquid storage tanks”, Journal of Technical Councils of ASCE, Vol. 107, TC1, 191-207.

Aspect ratio of tank Ratio of wall thickness to diameter

See next slide





Effect of wall flexibility on impulsive pressure distribution h/D = 0.5 tw / D = 0.0005 tw / D = 0.0005 tw is wall thickness

z h

Rigid tank

Impulsive pressure on wall © Sudhir K. Jain, IIT Kanpur

From Veletsos (1984)




If wall flexibility is included, then mechanical model of tank becomes more complicated Moreover, its inclusion does not change seismic forces appreciably Thus, mechanical model based on rigid wall assumption is considered adequate for design.





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 mass Hence, higher modes are not included This is also one of the reasons for summation of mi and mc being not equal to total liquid mass

For more information refer Veletsos (1984) and Malhotra (2000)

Malhotra, P. K., Wenk, T. and Wieland, M., 2000, “Simple procedure for seismic analysis of liquid-storage tanks”, Structural Engineering International, 197-201.




Modeling of ground supported tanks

Step 1:

Step 2:

Obtain various parameters of mechanical model These include, mi, mc, Kc, hi, hc, hi* and hc* Calculate mass of tank wall (mw), mass of roof (mt) and mass of base slab (mb)of container

This completes modeling of ground supported tanks




Modeling of elevated tanks

Elevated tank consists of container and staging Roof slab Wall

Container

Floor slab Staging

Elevated tank © Sudhir K. Jain, IIT Kanpur




Liquid is replaced by impulsive and convective masses, mi and mc

All other parameters such as hi, hc, etc, shall be obtained as described earlier

Lateral stiffness, Ks, of staging must be considered

This makes it a two-degree-of-freedom model Also called two mass idealization




Modeling of elevated tanks mc Kc/2 hi

Kc/2 mc

Kc hc

mi

mi + ms

hs

Spring mass model


Ks

Two degree of freedom system OR Two mass idealization of elevated tanks




ms is structural mass, which comprises of :

Mass of container, and One-third mass of staging

Mass of container includes

Mass of roof slab Mass of wall Mass of floor slab and beams




Two Degree of Freedom System

2-DoF system requires solution of a 2 × 2 eigen value problem to obtain

Two natural time periods Corresponding mode shapes

See any standard text book on structural dynamics on how to solve 2-DoF system For most elevated tanks, the two natural time periods (T1 and T2) are well separated.

T1 generally may exceed 2.5 times T2.




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





mc Kc

mi + ms

mi + ms Ks

mc Kc

Ks

Two degree of freedom system

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 2.5 Important to note that this approximation is done only for the purpose of calculating time periods

This significantly simplifies time period calculation Otherwise, one can obtain time periods of 2-DoF system as per procedure of structural dynamics.





Steps in modeling of elevated tanks Step 1:

Obtain parameters of mechanical analogue These include mi, mc, Kc, hi, hc, hi* and hc*

Step 2:

Other tank shapes and obstructions inside the container shall be handled as described earlier

Calculate mass of container and mass of staging

Step 3:

Obtain stiffness of staging





Recall, in IS 1893:1984, convective mass is not considered

It assumes entire liquid will act as impulsive mass Hence, elevated tank is modeled as single degree of freedom ( SDoF) system

As against this, now, elevated tank is modeled as 2-DoF system

This 2-DoF system can be treated as two uncoupled SDoF systems





Models of elevated tanks m +ms

mi + ms

Ks

mc

m = Total liquid mass

Ks

Kc

As per the Guideline © Sudhir K. Jain, IIT Kanpur

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. © Sudhir K. Jain, IIT Kanpur




h/D = 3/11.3 = 0.265 From Figure 2 of the Guideline, for h/D = 0.265: mi/m = 0.31, mc/m = 0.65 and Kch/mg = 0.47 mi = 0.31 x m = 0.31 x 300.9 = 93.3 t mc = 0.65 x m = 0.65 x 300.9 = 195.6 t Kc = 0.47 x mg/h = 0.47 x 300.9 x 9.81/3 = 462.5 kN/m





Mass of container = 180 t Mass of staging = 100 t Structural mass of tank, ms = mass of container +1/3rd mass of staging = 180 +1/3 x 100 = 213.3 t Lateral stiffness of staging, Ks = 20,000 kN/m





m + ms

mi + ms

Ks

mc

Ks

Kc

mi = 93.3 t, ms = 213.3 t, mc = 195.6 t, Ks = 20,000 kN/m, Kc = 462.5 kN/m Model of tank as per the Guideline © Sudhir K. Jain, IIT Kanpur

m = 300.9 t, ms = 213.3 t, Ks = 20,000 kN/m Model of tank as per IS 1893:1984



Time period

What is time period ? For a single degree of freedom system, time period (T ) is given by T = 2π

M K

M is mass and K is stiffness T is in seconds M should be in kg; K should be in Newton per meter (N/m) Else, M can be in Tonnes and K in kN/m




Time period

Mathematical model of tank comprises of impulsive and convective components Hence, 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 tanks Ground supported rectangular tanks Elevated tanks




Ti for ground-supported circular tanks

Ground supported circular tanks Time period of impulsive mode, Ti is given by: Ti = Ci

h ρ t/D E

⎛ 1 Ci = ⎜ ⎜ h/D 0.46 − 0.3h/D + 0.067(h/D)2 ⎝

(

)

⎞ ⎟ ⎟ ⎠

ρ = Mass density of liquid E = Young’s modulus of tank material t = Wall thickness h = Height of liquid D = Diameter of tank © Sudhir K. Jain, IIT Kanpur




Ci can also be obtained from Figure 5 of the Guidelines 10

8

C

6 C

i

4

C

c

2

0 0


0.5

h/D

1

1.5


2



This formula is taken from Eurocode 8

Eurocode 8, 1998, “Design provisions for earthquake resistance of structures, Part 1- General rules and Part 4 – Silos, tanks and pipelines”, European Committee for Standardization, Brussels.

If wall thickness varies with height, then thickness at 1/3rd height from bottom shall be used

Some steel tanks may have step variation of wall thickness with height





This formula is derived based on assumption that wall mass is quite small compared to liquid mass More information on time period of circular tanks may be seen in Veletsos (1984) and Nachtigall et al. (2003)

Nachtigall, I., Gebbeken, N. and Urrutia-Galicia, J. L., 2003, “On the analysis of vertical circular cylindrical tanks under earthquake excitation at its base”, Engineering Structures, Vol. 25, 201-213.





It is important to note that wall flexibility is considered in this formula

For tanks with rigid wall, time period will be zero

This should not be confused with rigid wall assumption in the derivation of mi and mc

Wall flexibility is neglected only in the evaluation of impulsive and convective masses However, wall flexibility is included while calculating time period





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 base Refer Figure 6 of the Guideline





Types of connections between tank wall and base slab

Such tanks are perhaps not used in India





Impulsive mode time period of ground supported tanks with fixed base is generally very low

These tanks are quite rigid Ti will usually be less than 0.4 seconds

In this short period range, spectral acceleration, Sa/g has constant value See next slide





Sa/g

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





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





Time period of impulsive mode, Ti = Ci

Ti = 5.3

h ρ t/D E

25 1.0 0.015/15 2x10 8

= 0.30 sec Important to note that, even for such a slender tank of steel, time period is low. For RC tanks and other short tanks, time period will be further less. © Sudhir K. Jain, IIT Kanpur




In view of this, no point in putting too much emphasis on evaluation of impulsive mode time period for ground supported tanks Recognizing this point, API standards have suggested a constant value of spectral acceleration for ground supported circular steel tanks

Thus, users of API standards need not find impulsive time period of ground supported tanks




Ti for ground-supported rectangular tanks

Ti for ground-supported rectangular tanks Procedure to find time period of impulsive mode is described in Clause no. 4.3.1.2 of the Guidelines

Time period is likely to be very low and Sa/g will remain constant

This will not be repeated here

As described earlier

Hence, not much emphasis on time period evaluation




Ti for Elevated tanks

For elevated tanks, flexibility of staging is important Time period of impulsive mode, Ti is given by: mi + ms Ti = 2π Ks

OR

T = 2π

Δ g

mi = Impulsive mass of liquid ms = Mass of container and one-third mass of staging Ks = Lateral stiffness of staging Δ= Horizontal deflection of center of gravity of tank when a horizontal force equal to (mi + ms)g is applied at the center of gravity of tank © Sudhir K. Jain, IIT Kanpur




These two formulae are one and the same

Expressed in terms of different quantities

Center of gravity of tank refers to combined mass center of empty container plus impulsive mass of liquid





Example: An elevated tank stores 250 t of water. Ratio of water height to internal diameter of container is 0.5. Container mass is 150 t and staging mass is 90 t. Lateral stiffness of staging is 20,000 kN/m. Find time period of impulsive mode Solution: h/D = 0.5, Hence from Figure 2a of the Guideline, mi/m = 0.54; ∴ mi = 0.54 x 250 = 135 t Structural mass of tank, ms = mass of container + 1/3rd mass of staging = 150 + 90/3 = 180 t © Sudhir K. Jain, IIT Kanpur




Time period of impulsive mode Ti = 2π Ti = 2π

mi + ms Ks

135 + 180 20,000

= 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


P

δ

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





Plan view of frame staging configurations

4 columns

6 columns

9 columns © Sudhir K. Jain, IIT Kanpur

8 columns

12 columns




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 modeling Leads to lower time period Hence, higher base shear coefficient This is another limitation of IS 1893:1984

Using a standard structural analysis software, staging can be modeled and analyzed to estimate lateral stiffness





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:

m T = 2π K

c

c


c




mc and Kc for circular and rectangular tanks can be obtained from graphs or expressions

These are described in Lecture 1 Refer Figures 2 and 3 of the Guidelines





For further simplification, expressions for mc and Kc are substituted in the formula for Tc Then one gets, For circular tanks:

Tc = Cc D/g

Cc =

2π 3.68 tanh (3.68h / D)

For rectangular tanks:

Tc = Cc L/g © Sudhir K. Jain, IIT Kanpur

Cc =

2π 3.16 tanh (3.16(h / L))




Graphs for obtaining Cc are given in Figures 5 and 7 of the Guidelines

These are reproduced in next two slides

Convective mass and stiffness are not affected by flexibility of base or staging Hence, convective time period expressions are common for ground supported as well as elevated tanks Convective mode time periods are usually very large

Their values can be as high as 10 seconds





10

8

6

C

C

i

4

C

c

2

0 0

0.5

h/D

1

1.5

2

Fig. 5 For circular tanks © Sudhir K. Jain, IIT Kanpur




10

8

Cc

6

4

2

0

0.5

h/L

1

1.5

2

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 Time period of convective mode, Tc

= Cc D/g

Tc = 3.6 12/9.81 = 3.98 sec © Sudhir K. Jain, IIT Kanpur




Based on mechanical models, time period for impulsive and convective modes can be obtained for ground supported and elevated tanks For ground supported tanks, impulsive mode time period is likely to be very less Convective mode time period can be very large






Response to Questions and Comments on Lecture 3 Question from Mr. Murugan.R ([email protected] ) •

In Slide-16: Please correct the tw/D ratio. Both dotted line and thick line are showing the same ratio of 0.0005. You are right. Graph with solid line is for tw/D = 0.005

•

In Slide-26: We are treating 2-DOF system as two uncoupled SDOF system. How to combine them? Response from these two uncoupled SDOF systems is combined using SRSS rule. Please note, total base shear is SRSS combination of base shear in impulsive mode (i.e., first SDOF) and convective mode (i.e., second SDOF).

•

Kindly furnish the formula for time period of a ground supported tanks with finned base? Finned base? You are perhaps asking formula for tanks with flexible base. This has been answered in a subsequent question.

•

In Slide-62: Kindly provide the formula for stiffness of staging based on shear deformation. This has been discussed in Lecture 7

Question from Mr. H. I. Abdul Gani ([email protected] ) The following are my doubts regarding lecture 3. 1. How to find the time period of ground supported tanks with flexible connection? 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

Ti =

8π (Ww + Wr + Wi ) gDKa

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 nonclassically 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 W

L1

∆

θ

θ T = 2π

Δ g

Lecture 4

January 30, 2006

In this lecture

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



Lecture 4 / Slide 2


Seismic force V = (Ah) x (W) Ah is base shear coefficient

Ah =

⎛Z⎞ ⎛ I ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ 2 ⎠ ⎝ R⎠

Zone Depends on severity of ground motion


Sa g

Design philosophy

Structural characteristics Depends on time period and damping


Lecture 4 / Slide 3


Tanks have two modes

Seismic force

Impulsive Convective 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



Lecture 4 / Slide 4


Impulsive base shear coefficient

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

(Ah)i = (Z/2) x (I/R) x (Sa/g)i

As per Table 2 of IS 1893(Part1):2002

I, R, (Sa/g)i and (Sa/g)c will be discussed here

First, (Sa/g)i and (Sa/g)c



Lecture 4 / Slide 5

(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



Lecture 4 / Slide 6

(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 tanks

See next slide



Lecture 4 / Slide 7

(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 sec There may be modeling errors; actual time period may be slightly higher As the structure gets slightly damaged, its natural period elongates Ductility does not help in reducing response of very stiff structures Hence, rising portion in the range 0 to 0.1 sec is usually disallowed by the codes.

Spectra is extended beyond 4 sec

Since convective time period may be greater than 4 sec. Beyond 4 sec, 1/T variation is retained



Lecture 4 / Slide 8

Sa/g

Sa/g Sa/g

(Sa/g)i and (Sa/g)c

Spectra of IS 1893 (Part 1):2002

Modified spectra

For 5% damping © Sudhir K. Jain, IIT Kanpur


Lecture 4 / Slide 9

(Sa/g)i and (Sa/g)c Expressions for design spectra at 5% damping Expressions for spectra of IS 1893(Part 1):2003

Expressions for spectra for tanks

For hard soil sites 0.00 ≤ T < 0.10 Sa/g = 1 + 15 T = 2.50 0.10 ≤ T < 0.40 = 1.00 / T 0.40 ≤ T ≤ 4.0

For hard soil sites Sa/g = 2.50 T < 0.40 = 1.0 / T T ≥ 0.40

For medium soil sites Sa/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 medium soil sites T < 0.55 Sa/g = 2.50 = 1.36 / T T ≥ 0.55

For soft soil sites Sa/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 soft soil sites Sa/g = 2.5 T< 0.67 = 1.67 / T T ≥ 0.67




(Sa/g)i and (Sa/g)c

Sa/g values also depend on damping

Multiplying 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% damping Multiplying factor = 1.0 for 5% damping Multiplying factor = 0.8 for 10% damping This multiplier is not used for PGA




Damping

Damping for impulsive mode

5% of critical for RC tanks 2% of critical for steel tanks These 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 shaking Means that during the same earthquake, damping will increase as the level of shaking increases We are performing a simple linear analysis, while the real behavior is non-linear Hence, 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 mode

In Table 3 of IS 1893(Part 1):2002

0.5% of critical for all types of tanks Convective mode damping does not depend on material of tank or type of liquid stored Multiplying factor for 0.5% damping is not given Values are given for 0% and 2% damping Linear interpolation shall not be done

Multiplying factor = 1.75, for 0.5% damping

In Eurocode 8 this multiplying factor is 1.673 In ACI 350.3, this factor is 1.5





Importance factor, I for tanks is given in Table 1 of the Guideline

This Table is reproduced here Type of liquid storage tank

I

Tanks used for storing drinking water, non-volatile material, low inflammable petrochemicals etc. and intended for emergency services such as fire fighting services. Tanks of post earthquake importance.

1.5

All other tanks with no risk to life and with negligible consequences to environment, society and economy.

1.0

NOTE: Values of importance factor, I given in IS 1893 (Part 4) may be used where appropriate © Sudhir K. Jain, IIT Kanpur




I = 1.5, is consistent with IS 1893(Part 1)

IS 1893(Part 1):2002 suggests, I = 1.5 for

Hospital buildings Schools Fire 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 structures

Some industries assign very high importance factor to tanks storing hazardous materials

Not yet published

Depending on their own requirements

For such tanks, Importance factor (I) will be as per part 4





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

This is reproduced in next two slides




Response reduction factor, R Elevated tank

R

Tank supported on masonry shafts a) Masonry shaft reinforced with horizontal bands * b) Masonry shaft reinforced with horizontal bands and vertical bars at corners and jambs of openings

1.3 1.5

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) Tank supported on steel frame#

1.8 2.5 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

#




Response reduction factor, R Ground supported tank

R

Masonry tank a) Masonry wall reinforced with horizontal bands* b) Masonry wall reinforced with horizontal bands and vertical bars at corners and jambs of openings

1.3 1.5

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)

2.0 2.5 1.5

Steel tank a) Unanchored base b) Anchored base

2.0 2.5

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 buildings

As discussed earlier, R depends on

This is in line with other international codes Ductility Redundancy Overstrength

Tanks possess low ductility, redundancy and overstrength





First let us consider, elevated tanks on frame type staging Staging frames are different than building frames

Hence, following footnote to Table 2

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.

Staging frames are non-building frames and are different than building frames





There are critical differences between building frames and non-building frames International 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 top These 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 = 2.5 as against R = 5.0 for buildings with SMRF With R = 2.5, base shear coefficient for elevated tanks on frame staging matches well with other international codes

See next slide





Comparison for frame staging

Zone and soil parameters are same used in Lecture 2 IBC 2003; Frame staging, R = 3.0


0.5 0.4

Guideline; Frame staging, R = 2.5

0.3

IS 1893:1984; All types of staging, K = 1.0

0.2 0.1 0 0

0.5

1

1.5

2

2.5

3

Time period (sec) © Sudhir K. Jain, IIT Kanpur




Let us now consider, elevated tanks on RC shaft They possess less redundancy and have single load path RC shafts are usually thin shell and possess low ductility There 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 shaft

Zanh 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, 745760. 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 shaft Longitudinal 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 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

Hollow circular section © Sudhir K. Jain, IIT Kanpur




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 = 0.02 This 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) = 0.05 Very 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 = 0.8 Thus, 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 ductility Issues 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





Comparison for shaft staging Zone and soil parameters are same as used in Lecture 2

IBC 2003; Shaft staging, R = 2.0


0.5 0.4

Guideline; Shaft staging, R = 1.8

0.3

IS 1893:1984; All types of staging, K = 1.0

0.2 0.1 0 0

0.5

1

1.5

2

2.5

3

Time period (Sec) © Sudhir K. Jain, IIT Kanpur




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 2003 It gives information on:

minimum steel construction tolerances safety against buckling shear design etc.





We have seen comparison with IBC 2003 Comparison 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 350.3 ACI 371 AWWA D-110 and AWWA D-115 AWWA D-100 and AWWA D-103 API 650 and API 620 Eurocode 8 NZSEE recommendations (From New Zealand)

Priestley et al. (1986)





Now we know Z, I, R and Sa/g for tanks One 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 Guideline R = 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 soft Damping = 0.5% (Clause 4.4 of the Guideline) Factor 1.75 is to be used for scaling up (Sa/g) for 0.5% damping (Clause 4.5.4 of the Guideline) ∴ (Sa/g)c = (1.67/Tc) x 1.75 = 1.67/4.0 x 1.75 = 0.731 © Sudhir K. Jain, IIT Kanpur



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

In this Lecture

Impulsive and convective base shear Critical direction of seismic loading



Lecture 5/ Slide 2

Base shear

Previous lectures have covered

Procedure to find impulsive and convective liquid masses

Procedure to obtain base shear coefficients in impulsive and convective modes

This was done through a mechanical analog model

This requires time period, damping, zone factor, importance factor and response reduction factor

Now, we proceed with seismic force or base shear calculations



Lecture 5/ Slide 3

Base shear

Seismic force in impulsive mode (impulsive base shear) Vi = (Ah)i x impulsive weight Seismic 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



Lecture 5/ Slide 4

Base shear

Now, we evaluate impulsive and convective weights

Or, impulsive and convective masses Earlier we have obtained impulsive and convective liquid mass Now, we consider structural mass also



Lecture 5/ Slide 5

Base shear :

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 liquid Mass of wall Mass of roof



Lecture 5/ Slide 6

Base shear : Ground supported tanks

Hence, base shear in impulsive mode

Vi = ( A h )i (mi + mw + mt ) g mi = mass of impulsive liquid mw = mass of container wall mt = mass of container roof g = acceleration due to gravity



Lecture 5/ Slide 7


This is base shear at the bottom of wall Base shear at the bottom of base slab is : Vi’ = Vi + (Ah)i x mb

mb is mass of base slab

Base shear at the bottom of base slab may be required to check safety against sliding



Lecture 5/ Slide 8


Base shear in convective mode Vc =

( A h )c m c

g

mc = mass of convective liquid



Lecture 5/ Slide 9


Total base shear, V is obtained as: V =

V i 2 + V c2

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 rule

i.e, V = Vi + Vc



Lecture 5/ Slide 10


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.



Lecture 5/ Slide 11

Bending moment:Ground supported tanks

Next, we evaluate bending or overturning effects due to base shear Impulsive base shear comprises of three parts

(Ah)i x mig (Ah)i x mwg (Ah)i x mtg



Lecture 5/ Slide 12


mw acts at CG of wall mt acts at CG of roof mi acts at height hi from bottom of wall

If base pressure effect is not included

mi acts at hi*

If base pressure effect is included Recall hi and hi* from Lecture 1



Lecture 5/ Slide 13


Bending moment at the bottom of wall

Due to impulsive base shear M i = ( A h )i (mi hi + mw hw + mt ht ) g

Due to convective base shear

M c = ( A h )c (mc hc ) g

hi = location of mi from bottom of wall hc = location of mc from bottom of wall hw = height of CG of wall ht = height of CG of roof



Lecture 5/ Slide 14


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

Hence, hi and hc are used



Lecture 5/ Slide 15


Bending moment at the bottom of wall (Ah)imt (Ah)cmc hc hi

(Ah)imi

(Ah)imw ht hw Ground level

Mi = ( Ah )i (mi hi + mw hw + mt ht ) g

M c = ( A h )c (mc hc ) g



Lecture 5/ Slide 16


Total bending moment at the bottom of wall

M = M i2 + M c2

SRSS rule used to combine impulsive and convective responses



Lecture 5/ Slide 17

Overturning moment:Ground supported tanks

Overturning moment

This is at the bottom of base slab Hence, must include effect of base pressure

hi* and hc* will be used

Ground level © Sudhir K. Jain, IIT Kanpur


Lecture 5/ Slide 18

Overturning moment:Ground supported tanks

Overturning moment in impulsive mode * ⎡ mi ( hi + tb ) + mw (hw + tb ) +⎤ * Mi = ( Ah )i ⎢ ⎥g ⎢⎣mt (ht + tb ) + mb tb / 2 ⎥⎦

Overturning moment in convective mode

M

*

= ( Ah ) c mc ( hc + tb ) g *

c

tb = thickness of base slab



Lecture 5/ Slide 19


Overturning moment is at the bottom of base slab

Hence, lever arm is from bottom of base slab Hence, base slab thickness, tb is added to heights measured from top of the base slab

Total overturning moment

M = M *


*2 i

+M

*2 c


Lecture 5/ Slide 20

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. Roof slab 150 mm thick

mw = 65.3 t, mt =33.1 t,

10 m 4m

Wall 200 mm thick

Base slab 250 mm thick © Sudhir K. Jain, IIT Kanpur

mi = 141.4 t; mc = 163.4 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


Lecture 5/ Slide 21

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 is Vc = (Ah)c mc g = 0.08 x 163.4 x 9.81 = 128.2 kN Total base shear at the bottom of wall is

V = Vi2 + Vc2 = © Sudhir K. Jain, IIT Kanpur

(529.3 ) + (128.2 ) 2

2

= 544.6kN


Lecture 5/ Slide 22

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 is Mi = (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 is Mc = (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 = Mi2 + M2c = © Sudhir K. Jain, IIT Kanpur

(1054 )2 + (295 )2

= 1095kN - m


Lecture 5/ Slide 23

Example Now, we obtain base shear at the bottom of base slab Impulsive base shear at the bottom of base slab is Vi = (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 is Vc = (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

V = Vi2 + Vc2 = © Sudhir K. Jain, IIT Kanpur

(651.1 ) + (128.2 ) 2

2

= 663.6kN


Lecture 5/ Slide 24

Example Impulsive overturning moment at the bottom of base slab Mi* = (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-m Convective 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 *

M =

(M ) + (M ) * 2 i

* 2 c

=

(1966 )

2

+ (498 ) = 2028 kN - m 2

Notice that this value is substantially larger that the value at the bottom of wall (85%)



Lecture 5/ Slide 25

Base shear :

Elevated tanks

In elevated tanks, base shear at the bottom of staging is of interest Ms is structural mass

Base shear in impulsive mode Vi = ( A h )i (mi + ms ) g

Base shear in convective mode V c = (A h

)c m c g

Total base shear V =


V i 2 + V c2


Lecture 5/ Slide 26

Bending moment:Elevated tanks

Bending moment at the bottom of staging

Impulsive base shear comprises of two parts

Bottom of staging refers to footing top (Ah)i x mig Ah)i x msg

Convective base shear has only one part

(Ah)c x mcg



Lecture 5/ Slide 27


mi acts at hi* mc acts at hc*

Bending moment at bottom of staging is being obtained Hence, effect of base pressure included and hi* and hc* are used



Lecture 5/ Slide 28


Structural mass, ms comprises of mass of empty container and 1/3rd mass of staging

ms is assumed to act at CG of empty container CG of empty container shall be obtained by considering roof, wall, floor slab and floor beams



Lecture 5/ Slide 29



[ (

)

M i = ( Ah )i mi hi + hs + ms hcg *

*

(

)

]g

M c = ( Ah )c m c hc + hs g *

hs = staging height

*

Measured from top of footing to bottom of wall

hcg = distance of CG of empty container from bottom of staging



Lecture 5/ Slide 30



(Ah)i mig

(Ah)c mcg

(Ah)i msg

hi*

hc*

hs

hs

hcg

Top of footing

[ (

)

]

M = ( Ah )i mi h + hs + ms hcg g * i

* i


(

)

M c = ( Ah )c mc hc + hs g *


*

Lecture 5/ Slide 31


Total bending moment

M = *

M

*2 i

+M

*2 c

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



Lecture 5/ Slide 32


For analysis of frame staging, two approaches are possible Approach 1: Perform analysis in two steps Step 1:

Step 2:

Analyze frame for (Ah)imig + (Ah)imsg Obtain forces in columns and braces 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



Lecture 5/ Slide 33


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 32 In this approach, analysis is done in single step

Simpler and faster than Approach 1



Lecture 5/ Slide 34

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. A is CG of empty container A 2.8 m

mi = 100t; mc = 180 t Mass of container = 160 t Mass of staging = 120 t hs = 15 m

hi* = 3 m, hc* = 4.2 m (Ah)i = 0.08, (Ah)c = 0.04

GL



Lecture 5/ Slide 35

Example Structural mass, ms = mass of container + 1/3rd mass of staging = 160 + 1/3 x 120 = 200 t Base shear in impulsive mode

Vi = ( A h )i (mi + ms ) g

= 0 .08 x (100 + 200 ) x 9 . 81

= 78.5 + 157 = 235.5 kN Base shear in convective mode

Vc =

(A h ) c m c g

= 0 . 04 x 180 x 9 . 81 = 70 . 6 kN © Sudhir K. Jain, IIT Kanpur


Lecture 5/ Slide 36

Example

Total base shear

V =

=

V i 2 + V c2

235 .5 2 + 70.6 2 = 245.8kN

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



Lecture 5/ Slide 37

Example Base moment in impulsive mode

M

* i

=

( A h )i [m i (h i

*

)

+ h s + m s h cg

= 0 . 08 [100 x (3 . 0 + 15

)+

]

g

200 x 17 . 8 ]x 9 . 81

= 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.



Lecture 5/ Slide 38

Example Base moment in convective mode

(

)

M c = ( Ah )c mc hc + hs g *

*

= 0 . 04 x180 x (4 . 2 + 15 )x 9 . 81 = 70.6 x 19.2 = 1356 kNm Note: 70.6 kN of force will act at 19.2 m from top of footing. Total base moment

M

*

=

M

* 2 i

+ M

*2 c

= 4207 2 + 1356 = 4420 kNm © Sudhir K. Jain, IIT Kanpur

2


Lecture 5/ Slide 39

Example

Now, for staging analysis, seismic forces are to be applied at suitable heights

There are two approaches

Refer 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 frame Step 2: Apply 70.6 kN at 19.2 m from top of footing and analyze the frame Member forces (i.e., BM, SF etc. in columns and braces) of Steps 1 and 2 shall be combined using SRSS



Lecture 5/ Slide 40

Example

Approach 2:

Total base shear, V = 245.8 kN will be applied at height h1, such that V x h1 = M* 245.8 x h1 = 4420

∴ h1 = 17.98 m

Thus, 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).



Lecture 5/ Slide 41

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 condition

In empty condition, no convective liquid mass

Hence, tank will be modeled using single degree of freedom system Mass of empty container and 1/3rd staging mass shall be considered



Lecture 5/ Slide 42

Elevated tanks:Empty condition

Lateral stiffness of staging, Ks will remain same in full and empty conditions In full condition, mass is more

Hence, time period of empty tank will be less

In empty condition mass is less Recall, T = 2 Π MK Hence, Sa/g will be more

Usually, tank full condition is critical

However, for tanks of low capacity, empty condition may become critical



Lecture 5/ Slide 43

Direction of seismic force

Let us a consider a vertical cantilever with rectangular cross section Horizontal load P is applied

First in X-Direction Then in Y-direction (see Figure below) More deflection, when force in Y-direction Hence, direction of lateral loading is important !! P Y X



P Lecture 5/ Slide 44


On the other hand, if cantilever is circular

Direction is not of concern Same 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



Lecture 5/ Slide 45


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.



Lecture 5/ Slide 46


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 axisymmetric Hence, analysis in any one direction is sufficient



Lecture 5/ Slide 47


Ground-supported rectangular tanks shall be analyzed for two directions

Parallel to length of the tank Parallel to width of the tank Stresses in a particular wall shall be obtained for seismic loads perpendicular to that wall



Lecture 5/ Slide 48


RC circular shafts of elevated tanks are also axisymmetric

Hence, analysis in one direction is sufficient

If circular shaft supports rectangular container

Then, analysis in two directions will be necessary



Lecture 5/ Slide 49


For elevated tanks on frame staging

Critical direction of seismic loading for columns and braces shall be properly ascertained

Braces and columns may have different critical directions of loading For example, in a 4 - column staging

Seismic loading along the length of the brace is critical for braces Seismic loading in diagonal direction gives maximum axial force in columns See next slide



Lecture 5/ Slide 50


Critical directions for 4 - column staging

Bending Axis

Critical direction for shear force in brace © Sudhir K. Jain, IIT Kanpur

Critical direction for axial force in column


Lecture 5/ Slide 51


For 6 – column and 8 – column staging, critical directions are given in Figure C-6 of the Guideline

See next two slides

More 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)



Lecture 5/ Slide 52



Bending Axis

Critical direction for shear force and bending moment in columns © Sudhir K. Jain, IIT Kanpur

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


Lecture 5/ Slide 53



Bending Axis

Critical direction for shear force and bending moment in braces © Sudhir K. Jain, IIT Kanpur

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


Lecture 5/ Slide 54


As an alternative to analysis in the critical directions, following two load combinations can be used

100 % + 30% rule

Also used in IS 1893(Part 1) for buildings

SRSS rule



Lecture 5/ Slide 55


100%+30% rule implies following combinations

ELx + 0.3 ELY ELY + 0.3 ELx ELx is response quantity when seismic loads are applied in X-direction ELY is response quantity when seismic loads are applied in Y-direction



Lecture 5/ Slide 56


100%+30% rule requires

Analyze 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% rule Combination is on response quantity and not on seismic loads



Lecture 5/ Slide 57


Important to note that the earthquake directions are reversible

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



Lecture 5/ Slide 58


SRSS rule implies following combination

EL2x + EL2y

Note:

ELx is response quantity when seismic loads are applied in X-direction ELY is response quantity when seismic loads are applied in Y-direction

Hence, analyze tank in two directions and use SRSS combination of response quantity



Lecture 5/ Slide 59


This completes seismic force evaluation on tanks There are two main steps

Evaluation of impulsive and convective masses Evaluation of base shear coefficients for impulsive and convective modes

SRSS rule is used to combine impulsive and convective responses Critical direction of seismic loading shall be properly ascertained

Else, 100%+30% or SRSS rule be used



Lecture 5/ Slide 60



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 base Effect of vertical ground acceleration Sloshing wave height Anchorage requirements for ground supported tanks



Lecture 6/ Slide 2


So far, emphasis was on lateral forces on tanks These lateral forces comprised of

Impulsive component has two parts

Impulsive component Convective component One due to impulsive liquid mass Second due to structural mass of tank

Convective component has one part

Due to convective liquid mass



Lecture 6/ Slide 3


Recall from Lecture 1

Impulsive force is summation of impulsive hydrodynamic pressure on wall Convective force is summation convective hydrodynamic pressure on wall

Now, we will see procedure to obtain

Impulsive and convective pressure distribution on wall and base



Lecture 6/ Slide 4


Pressure distribution on wall is needed to obtain

Hoop 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



Lecture 6/ Slide 5

Impulsive pressure:Circular tanks

Impulsive pressure on wall of circular tanks is given by

p iw = Q iw ( y ) ( A h ) i ρ g h cos φ 2 ⎡ D⎞ ⎛ y⎞ ⎤ ⎛ Qiw ( y ) = 0 .866 ⎢1 − ⎜ ⎟ ⎥ tanh ⎜ 0 .866 ⎟ h⎠ ⎝ h ⎠ ⎥⎦ ⎝ ⎢⎣

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 © Sudhir K. Jain, IIT Kanpur


Lecture 6/ Slide 6


Note that impulsive pressure varies with circumferential angle, φ

At φ = 00, pressure will be maximum At φ = 900, pressure will be zero Recall following figure from Lecture 1 zero pressure at φ = 900

φ Direction of seismic force


maximum pressure at φ = 00


Lecture 6/ Slide 7


Qiw(y) can also be read from Figure 9(a) of the Guideline

This Figure is reproduced here

1

0.8

0.6

Y/h

0.4 1.5

0.5

1.0

0.25

h/D =2 or h/L

0.2

0 0


0.2

0.4

Q

iw

0.6

0.8


1

Lecture 6/ Slide 8


Impulsive pressure on base of circular tanks changes in radial as well as circumferential direction

A strip of length l’ is considered

Refer plan of circular tank shown in Figure 8(a) of the Guideline l’ x


D/2 φ O

Plan © Sudhir K. Jain, IIT Kanpur


Lecture 6/ Slide 9


Impulsive pressure on this strip is given by

p ib

⎛ sinh ⎜ 0 . 866 ⎝ = 0 . 866 ( Ah )i ρ g h ⎛ cosh ⎜⎜ 0 . 866 ⎝

x⎞ ⎟ h⎠ l' ⎞ ⎟⎟ h ⎠

x = horizontal distance in the direction of seismic force, of a point on base slab, from the reference axis at the center of tank



Lecture 6/ Slide 10


Example: A circular tank has internal diameter of 12 m, and water height of 5 m. Impulsive base shear coefficient is 0.23. Find the impulsive pressure distribution on wall Solution:

D = 12 m, h = 5 m, (Ah)i = 0.23 Impulsive pressure on wall is given by,

piw = Qiw (y) (Ah)i ρ g h cos φ Where, Qiw (y)= 0.866 [ 1- (y /h)2] tanh(0.866 D/h) Maximum pressure will occur at φ = 0 © Sudhir K. Jain, IIT Kanpur


Lecture 6/ Slide 11

Impulsive pressure:Circular tanks At base of wall, y = 0;

Qiw = 0.866 tanh(0.866 D/h) = 0.866 tanh(0.866 x 12/5) = 0.84 Now, at φ = 0,

piw = Qiw (y) (Ah)i ρ g h Mass 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.



Lecture 6/ Slide 12


y (m)

0

1.25

3.75

5

Qiw (y)

0.84

0.79

0.37

0

piw (y) (kN/m2)

9.48

8.91

4.17

0

h = 5m

At φ = o 9.48 kN/m2 Impulsive pressure distribution on wall © Sudhir K. Jain, IIT Kanpur


Lecture 6/ Slide 13

Impulsive pressure:Rectangular tanks

Impulsive pressure on walls of rectangular tanks is given by p iw = Q iw ( y ) ( A h ) i ρ g h Q iw ( y ) = 0 . 866

2 ⎡ y L ⎞ ⎛ ⎞ ⎤ ⎛ ⎟ ⎢1 − ⎜ ⎟ ⎥ tanh ⎜ 0 . 866 h h ⎝ ⎠ ⎦⎥ ⎝ ⎠ ⎣⎢

Except following, this expression is same as that for circular tanks

D/h is replaced by L/h

Angle φ is not present

Qiw(y) can be read from Figure 9(a) of the Guideline



Lecture 6/ Slide 14


This pressure is on walls perpendicular to direction of seismic force Pressure remains same along the length of these perpendicular walls

However, it changes with wall height Walls perpendicular to direction of seismic force B

L

Direction of seismic force © Sudhir K. Jain, IIT Kanpur


Lecture 6/ Slide 15


Impulsive pressure on base of rectangular tanks Impulsive pressure will be defined for a point at a distance x from central axis

As shown in Figure below

Refer plan of rectangular tank in Figure 8 (b) of the Guideline L

Direction of Seismic Force x

Plan © Sudhir K. Jain, IIT Kanpur


Lecture 6/ Slide 16


Impulsive pressure is given by

pib = Qib ( x ) ( Ah )i ρ g h Q

ib

⎛ sinh ⎜ 0 . 866 ⎝ ( x ) = ⎛ cosh ⎜ 0 . 866 ⎝

x h L h

⎞ ⎟ ⎠ ⎞ ⎟ ⎠

Qib(x) is coefficient of impulsive pressure on base It can be read from Figure 9(b) of the Guideline See next slide



Lecture 6/ Slide 17


Coefficient of impulsive pressure on base Qib(x) 1.2 0.25 0.5

Qib(x)

0.8

1.0 1.5

0.4

2.0

Qib -0 .4

2.0

-0 .2

0

0

x/L

0.2

0.4

-0.4

1.5 1.0

-0.8 0.5 0.25 = h/L

-1.2


Figure 9(b) of the Guideline


Lecture 6/ Slide 18

Convective pressure:Circular tanks

Convective pressure on wall of circular tanks is given by

⎡ 1 2 ⎤ pcw = Qcw( y )( Ah ) c ρ g D⎢1 - cos φ ⎥ cosφ ⎣ 3 ⎦ ⎛ cosh ⎜ 3.674 ⎝ Qcw ( y ) = 0.5625 ⎛ cosh ⎜ 3.674 ⎝


y⎞ ⎟ D⎠ h⎞ ⎟ D⎠


Lecture 6/ Slide 19

Convective pressure:Circular tanks (Ah)c = Convective base shear coefficient Qcw = Coefficient of convective pressure on wall

ρ = Mass density of liquid φ = Circumferential angle y = Vertical distance of a point on wall from the bottom of wall

Qcw can be obtained from Figure 10(a) of the Guideline

See next slide



Lecture 6/ Slide 20


Coefficient of convective pressure on wall, Qcw 1

0.8

2.0

Y/h

1.5 1.0

0.6

0.5 0.4 h/D=0.25

0.2

0 0

0.1

0.2

Q

cw

0.3

0.4

0.5

0.6

Figure 10(a) of the Guideline © Sudhir K. Jain, IIT Kanpur


Lecture 6/ Slide 21


Convective pressure on base of circular tanks is given by

p cb = Q cb ( x ) ( Ah

)c ρ

g D

⎡ x 4 ⎛ x ⎞3 ⎤ h⎞ ⎛ Qcb ( x ) = 1.125⎢ − ⎜ ⎟ ⎥ sech⎜ 3.674 ⎟ D⎠ ⎝ ⎢⎣ D 3 ⎝ D ⎠ ⎥⎦

Qcb can also be obtained from Figure 10(b) of the Guideline

See next slide



Lecture 6/ Slide 22


Coefficient of convective pressure on base, Qcb 0.3 h/D=0.25

Qcb

0.2

0.5 0.1 0.75

b c

Q

1.0

0 1.0 0.75

-0.4

-0.2

0

x/D

0.2

0.4

-0.1 0.5

-0.2 h/D=0.25 -0.3

Figure 10(b) of the Guideline © Sudhir K. Jain, IIT Kanpur


Lecture 6/ Slide 23


Example: A circular tank has internal diameter of 12 m, and water height of 5 m. Convective base shear coefficient is 0.07. Find the convective pressure distribution on wall Solution:

D = 12 m, h = 5 m, (Ah)c = 0.07 Convective pressure on wall is given by,

pcw = Qcw (y) (Ah)c ρ g D [1- 1/3 cos 2φ] cos φ Where,

Qcw (y)= 0.5625 cosh(3.674 y/D)/cosh(3.674 h/D) Maximum pressure will occur at φ = 0 © Sudhir K. Jain, IIT Kanpur


Lecture 6/ Slide 24

Convective pressure:Circular tanks At base of wall, y = 0; Qcw (y =0) = 0.5625/cosh(3.674 h/D) = 0.5625/cosh(3.674 x 5/12) = 0.23 Now, at φ = 0, pcw = Qcw (y) (Ah)c ρ g D [1- 1/3 ] Mass density of water, ρ = 1000 kg/m3 ∴ pcw (y =0) = 0.23 x 0.07 x 1000 x 9.81 x 12 x [1- 1/3] = 1263.5 N/m2 = 1.26 kN/m2 For different values of y, pcw is similarly calculated. Pressure distribution with wall height is shown in the next slide.



Lecture 6/ Slide 25


y (m)

0

1.25

3.75

5

Qcw (y)

0.23

0.25

0.40

0.5625

pcw (y) (kN/m2)

1.26

1.37

2.20

3.09

3.09 kN/m2

h = 5m

At φ = o 1.26 kN/m2 Convective pressure distribution on wall © Sudhir K. Jain, IIT Kanpur


Lecture 6/ Slide 26

Convective pressure:Rectangular tanks

Convective pressure on wall of rectangular tank is given by

p cw = Q cw ( y )( Ah )c ρ g L ⎛ cosh ⎜ 3 .162 ⎝ Qcw ( y ) = 0 .4165 ⎛ cosh ⎜ 3 .162 ⎝

y⎞ ⎟ L⎠ h⎞ ⎟ L⎠

Qcw can also be obtained from Figure 11(a) of the Guideline

See next slide



Lecture 6/ Slide 27


Coefficient of convective pressure on wall, Qcw 1

0.8

Y/h

2.0

1.5

0.6

1.0 0.5

0.4 h/L=0.2 0.2

0 0

0.1

0.2

Q

cw

0.3

0.4

0.5

Figure 11(a) of the Guideline © Sudhir K. Jain, IIT Kanpur


Lecture 6/ Slide 28


Convective pressure on base of rectangular tanks is given by

pcb = Qcb( x )( Ah )c ρ g L ⎡ x 4 ⎛ x ⎞3 ⎤ h⎞ ⎛ Qcb ( x ) = 1 .25 ⎢ − ⎜ ⎟ ⎥ sec h ⎜ 3 .162 ⎟ L⎠ ⎝ ⎣⎢ L 3 ⎝ L ⎠ ⎦⎥

Qcb can be obtained from Figure 11(b) of the Guideline However, Figure 11 (b) of the Guideline is incorrect Correct Figure is shown in next slide



Lecture 6/ Slide 29


Coefficient of convective pressures on base, Qcb 0.4 0.3

h/L = 0.25

0.2

Qcb

1.0

-0.5

0.50 0.75 1.0

0.1 0

-0.4

-0.3

-0.2

-0.1

0

0.75

-0.1

0.50

-0.2

h/L = 0.25

0.1

0.2

0.3

0.4

0.5

x/L

-0.3 -0.4

Figure 11(b) of the Guideline (corrected) © Sudhir K. Jain, IIT Kanpur


Lecture 6/ Slide 30

Linearised Pressure Distribution

Impulsive and convective pressures on wall have curvilinear distribution Hoop 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



Lecture 6/ Slide 31


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



Lecture 6/ Slide 32


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



Lecture 6/ Slide 33


Equivalent linear impulsive pressure bi qi h hi

˜ hi

Actual Impulsive pressure distribution


qi

ai Equivalent linear pressure distribution


Lecture 6/ Slide 34


The ordinates of linear distribution are

qi a i = 2 (4 h − 6 hi ) h

qi b i = 2 (6 h i − 2 h h

)

Both for circular and rectangular tanks Both for qi is different for circular and rectangular tanks

For circular tanks

For rectangular tanks © Sudhir K. Jain, IIT Kanpur

qi =

( A h )i m i g π D / 2

qi =

( A h )i m i g 2B


Lecture 6/ Slide 35


Here

mi = impulsive liquid mass hi = height of impulsive mass h = Total liquid height (Ah)i = Impulsive base shear coefficient

All these quantities are known

Procedure to obtain these quantities is covered in previous lectures



Lecture 6/ Slide 36


Similarly, equivalent linear convective pressure bc

qc

hc

qc h

Actual convective pressure distribution

˜

hc

ac Equivalent linear pressure distribution

qc b c = 2 (6 h c − 2 h h qc a c = 2 (4 h − 6 h c ) h

)

For circular tanks

( Ah )c mc qc = g π D/2 For rectangular tanks

qc © Sudhir K. Jain, IIT Kanpur

( A h )c m c = g 2B


Lecture 6/ Slide 37


Linearised distribution can be treated as summation of two parts

Uniform pressure Triangular pressure bi

= ai © Sudhir K. Jain, IIT Kanpur

+ bi

ai - bi


Lecture 6/ Slide 38


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.



Lecture 6/ Slide 39


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 = 0.07 Capacity of tank = π x 122 /4 x 5 = 565.5 m3

∴ Mass of water = 565.5 x 1000 = 5,65,500 kg For h/D = 5/12 = 0.42, from Figure 2a of the Guideline: mi/m = 0.47, ∴ mi = 0.47 x 565500 = 2,65,800 kg © Sudhir K. Jain, IIT Kanpur


Lecture 6/ Slide 40

Linearised Pressure Distribution mc/m = 0.5, mc = 0.50 x 565500 = 2,82,800 kg hi /h = 0.375, hi = 0.375 x 5 = 1.875 m hc /h = 0.58, hc = 0.58 x 5 = 2.9 m a) Impulsive mode

( Ah ) i mi 0 .23 × 265 ,800 × 9 .81 qi = g= = 31 .81kN/m π D /2 π × 12 / 2 Pressure at bottom & top is given by

qi 31 . 81 (4 × 5 − 6 × 1 .875 ) = 11 .13 kN/m a i = 2 (4 h − 6 hi ) = 2 h 5 © Sudhir K. Jain, IIT Kanpur


2

Lecture 6/ Slide 41


qi 31.81 bi = 2 (6hi − 2h ) = 2 (6 × 1.875 − 2 × 5) = 1.59kN / m 2 5 h Note:- For the same tank, in earlier example, we had obtained actual distribution of impulsive pressure. Linear and actual pressure distributions are shown below: bi = 1.59 kN/m2

h = 5m

9.48 kN/m2

ai = 11.13 kN/m2 Linear pressure distribution © Sudhir K. Jain, IIT Kanpur

Actual pressure distribution ( Obtained in earlier example)


Lecture 6/ Slide 42

Linearised Pressure Distribution b) Convective mode

( Ah ) c m c 0 .07 × 282 ,800 × 9 .81 qc = g= = 10 .3kN/m π D/2 π × 12 / 2 Pressure at bottom & top is given by

ac =

qc 10 . 3 ( ) (4 × 5 − 6 × 2 . 9 ) = 1 . 07 kN/m 4 h − 6 h = c 2 2 h 5

qc 10 . 3 (6 × 2 . 9 − 2 × 5 ) = 3 . 05 kN/m b c = 2 (6 h c − 2 h ) = 2 h 5



2

Lecture 6/ Slide 43

Linearised Pressure Distribution Note:- For the same tank, in earlier example we had obtained actual distribution of convective pressure. Linear and actual pressure distributions are shown below: bi = 3.05

kN/m2

3.09 kN/m2

h = 5m

1.26 kN/m2

ai = 1.07 kN/m2 Linear pressure distribution


Actual convective pressure distribution (Obtained in earlier example)


Lecture 6/ Slide 44

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



Lecture 6/ Slide 45


Hoop forces and BM in wall for such varying pressure are also not readily available Hence, 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 pressure See next slide



Lecture 6/ Slide 46


Simplified distribution in circumferential direction pmax

pmax

φ

Actual pressure distribution


Simplified pressure distribution


Lecture 6/ Slide 47

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 decrease If acceleration is upward, weight will increase

Increase in weight = vertical acceleration x mass



Lecture 6/ Slide 48


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 height

Pressure due vertical acceleration would also vary linearly with height



Lecture 6/ Slide 49


Pressure, pv due to vertical acceleration

pv = ( Av ) ρ g h (1 − y h) 2 Av = 3

⎛Z Sa ⎞ I ⎟⎟ ⎜⎜ × × R g ⎠ ⎝ 2

ρ = Mass density of liquid

pv

h = Total liquid height g = Acceleration due to gravity y

Z = Zone factor I = Importance factor R = Response reduction factor © Sudhir K. Jain, IIT Kanpur


Lecture 6/ Slide 50


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):2002

For finding the value of Sa/g

One needs to know time period of tank in vertical mode



Lecture 6/ Slide 51


Usually tanks are quite rigid in vertical direction

Hence, time period in vertical mode may be taken as as 0.3 sec for all tanks This 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



Lecture 6/ Slide 52

Pressure due to wall inertia

Mass of wall is distributed along its length and height Hence, during lateral excitation, wall is subjected to lateral pressure Pressure due to wall inertia is given by

p ww = ( Ah )i t ρ m g

ρm = mass density of tank wall t

= thickness of tank wall

g

= acceleration due to gravity

pww

(Ah)i = impulsive base shear coefficient © Sudhir K. Jain, IIT Kanpur


Lecture 6/ Slide 53

Pressure due to wall inertia

Pressure due to wall inertia may not be significant in steel tanks

Steel tank walls have less mass

However for RC tanks, pressure due to wall inertia may be significant



Lecture 6/ Slide 54

Total pressure on wall

Total pressure on wall comprises of

Impulsive 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

p = ( piw + pww ) + p + pv 2


2 cw

2


Lecture 6/ Slide 55

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 mm Zone IV; Z= 0.24, I = 1.5, It is ground supported RC tank with fixed base, R = 2.0



Lecture 6/ Slide 56

Example For vertical mode, time period is taken as, T = 0.3 sec for all tanks

∴ Sa/g = 2.5 Av = 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 at y = 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



Lecture 6/ Slide 57

Example Maximum pressure due to wall inertia, pww = (Ah)i t ρm g = 0.23 x 0.2 x 25 = 1.15 kN/m2



Lecture 6/ Slide 58

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 2 p = ( piw + pww)2 + pcw + pv

2

= ( 9.48 + 1.15 ) + 1.26 + 7.36 2

2

2

= 13 kN/m2 Total hydrostatic pressure at base = ρ g h = 49 kN/m2. Thus, total hydrodynamic pressure of 13 kN/m2 is 26% of hydrostatic pressure. © Sudhir K. Jain, IIT Kanpur


Lecture 6/ Slide 59

Sloshing wave height

Convective liquid undergoes sloshing motion

i.e., liquid undergoes vertical motion dmax



Lecture 6/ Slide 60


Maximum sloshing wave height is given by d max

D = ( A h )c R 2

d max = ( A h )c R

L 2

For circular tanks For rectangular tanks

(Ah)c = convective base shear coefficient R

= response reduction factor

D

= diameter of circular tank

L = Length of rectangular tank in the direction of seismic force © Sudhir K. Jain, IIT Kanpur


Lecture 6/ Slide 61


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 forces Expression for sloshing wave height is taken from ACI 350.3



Lecture 6/ Slide 62


Sloshing wave height is useful in determining the free board to be provided

If loss of liquid needs to be prevented

If free board is less than sloshing wave height, liquid will exert pressure on roof

In the process, sloshing liquid mass will also change Refer 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



Lecture 6/ Slide 63

Anchorage requirement

Ground supported tanks have tendency to overturn during lateral base excitation

Particularly tall steel tanks

Anchorages are needed to ensure safety against overturning Tank will overturn, if overturning moment due to lateral force is more than stabilizing moment



Lecture 6/ Slide 64


As a quick estimate, one can say, overturning will occur if

h > D

1 (A h

)i

h = height of liquid D = diameter of tank (Ah)i= impulsive base shear coefficient

For rectangular tanks, diameter D is to be replaced by L



Lecture 6/ Slide 65


It is an approximate estimate for circular tanks

It assumes that entire tank mass and liquid mass is acting at the center of liquid height Read Clause 4.12 of the Guideline



Lecture 6/ Slide 66


Impulsive and convective pressure have curvilinear distribution along the wall height Equivalent linear distribution may be used

For obtaining hoop forces and BM in wall

Wall is also subjected to pressure due to vertical acceleration and wall inertia Sloshing wave height can be obtained using simple expression



Lecture 6/ Slide 67

Lecture 7

February 16, 2006

In this Lecture

Examples 2 & 3 of IITK-GSDMA Guideline

These examples have been sent separately Here, 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



Lecture 7/ Slide 2

Example 2

Elevated tank considered in Example 2 of the Guidelines has

RC Intze container of 250 m3 capacity RC Frame staging on six columns

Staging is assumed to have been designed for good ductility

Tank is located in seismic zone IV on hard soil Please refer the examples sent to you

In this example, base shear and base moment at the base of staging are obtained



Lecture 7/ Slide 3

Example 2

Analysis involves following calculations Step 1: Weight of various parts Step 2: Center of gravity of empty container Step 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 staging

Using standard structural analysis software, stiffness is obtained More about this later



Lecture 7/ Slide 4

Example 2

Step 5: Impulsive convective mode time period Step 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 moment Step 8: Analysis of empty tank



Lecture 7/ Slide 5

Example 2

Mass of water = 255.6 t Mass of empty container = 160.7 t Mass of staging = 105.6 t Structural mass, ms = 160.7 + 105.6/3 = 196 t

ms = empty container + 1/3rd mass of staging

Total mass = 255.6 + 196 = 451.6 t



Lecture 7/ Slide 6

Example 2

Parameters of spring-mass model Here, container is of Intze type Equivalent 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 = 0.51 This h/D is used to find mi, mc, hi, hc etc. mi = 140.6 t; about 55% of total water mass mc =110 t; about 43% of total water mass mi + mc is about 2% less than total water mass. This was discussed in Lecture 3.



Lecture 7/ Slide 7

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



Lecture 7/ Slide 8

Example 2: Lateral stiffness

Lateral stiffness of staging Staging is modeled using structural analysis software Force 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



Lecture 7/ Slide 9


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 used Hence, 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



Lecture 7/ Slide 10


10.0

Rigid link

Structural model of staging used for analysis © Sudhir K. Jain, IIT Kanpur


Lecture 7/ Slide 11


A force of 10 kN is applied at CG

Deflection of CG = 5.616 x10-4 m Hence, lateral Stiffness of Staging Ks = 10/(5.616x10-4) = 17,800 kN/m

Here, computer software is used for analysis Traditionally, designers use manual methods

Present frame is a polygon frame



Lecture 7/ Slide 12


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



Lecture 7/ Slide 13


If braces are considered as rigid beams, then

No rotation is allowed at joints of the columns Hence, stiffness of a column in a panel or between two brace levels is K C = 12EI L3

Where L is length of column in a panel



Lecture 7/ Slide 14

Example 2: Analysis as per IS 1893:1984

For the frame in the present example Stiffness of each column in a panel, Kc K C =12EI =12 ×22360 ×10 33×8.76 ×10 9 =36,730 kN/m L3 4.0 ⎛ ⎜ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎟ ⎠

In each panel, there are 6 columns Hence, stiffness of each panel

∑K © Sudhir K. Jain, IIT Kanpur

C

= 6 × 36730 = 220,400 kN/m


Lecture 7/ Slide 15


Stiffness of one panel = 220400 kN/m

There are four such panels Each having same stiffness They act as springs in series Hence, lateral stiffness of staging, Ks Ks

K ∑ =


4

C

=

220400 = 55,090 kN/m 4


Lecture 7/ Slide 16


Thus, stiffness of frame

Ks = 17,800 kN/m from computer analysis Ks = 55,090 kN/m from method of SP22

Thus, SP 22 grossly overestimates the stiffness

Clearly, braces should not be treated as rigid beams Their flexibility should be included in the analysis



Lecture 7/ Slide 17


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)



Lecture 7/ Slide 18


As per Jain and Sameer (1992) 1 1 1 = + K s K flexure K axial

1 K flexure

=

Np

∑K i =1

1 panel

Kpanel = stiffness of each panel Np = number of panels


Top most panel


Bottom most panel Lecture 7/ Slide 19


K

panel

K

⎤ Ib L 12 EI c N c ⎡ = ⎢ ⎥ h3 I L + I h 2 c ⎣ b ⎦

panel

K panel

1 K axial

For intermediate panels

⎤ For top most panel 12 EI c N c ⎡ I cbr L = ⎢ ⎥ h3 I L + I h c ⎣ cbr ⎦

⎤ 12 EI c N c ⎡ Ib L = ⎥ ⎢ h3 I L + I h c ⎦ ⎣ b

2 = N c Ac E R 2


For bottom most panel

Np

∑

2

Hi h

i =1


Lecture 7/ Slide 20


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



Lecture 7/ Slide 21


Hi = Distance of lateral load from inflection point of ith panel

For intermediate panels, point of inflection is at mid height of panel For the top most panel, point of inflection is at a distance y from circular ring beam y =

⎛ 3 I b + ⎜ L ⎝ ⎛ 6 I b ⎜ L ⎝

I c ⎞ ⎟ h ⎠ h ⎞ ⎟ ⎠

For the bottom most panel, distance y is measured from fixed end



Lecture 7/ Slide 22


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 mm R = 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 mm H1 = 6030 mm; H2 = 9,180 mm; H3 = 13,180 mm; H4 = 16,330 mm Note:- Load is applied at CG of container, and distances Hi are from the load. © Sudhir K. Jain, IIT Kanpur


Lecture 7/ Slide 23


Using this method, we get

From computer analysis, we have

Stiffness of staging, Ks = 16,440 kN/m Ks = 17,800 kN/m Difference is about 7%. Which is very reasonable

Recall, as per SP 22, Ks = 55,090 kN/m

Which is on much higher side



Lecture 7/ Slide 24


Note, usually in practice, method of SP 22 is used Also this method has been used in a number of text books in India. For further calculations here, Ks = 17,800 kN/m is used



Lecture 7/ Slide 25

Example 2: Time period

Tank full condition

Impulsive mode time period, Ti mi + ms Ti = 2π Ks

= 0.86 Sec

Convective time period, Tc

Tc = Cc D/g For h/D = 0.51, from Figure 5 of the Guideline, Cc = 3.35

Hence, Tc = 3.14 sec



Lecture 7/ Slide 26


Convective time period is 3.65 times impulsive time period

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



Lecture 7/ Slide 27

Example 2: Base shear

Design horizontal seismic coefficient

Zone 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 = 0.56

Multiplication by 1.75 is for 0.5% damping



Lecture 7/ Slide 28


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)i

Convective mass is subjected to much smaller acceleration



Lecture 7/ Slide 29

Example 2:Base shear

Base Shear, V

Impulsive mode Vi = (Ah)i (mi + ms)g Vi = 0.084 x (140.6 + 196) x 9.81 = 116 +161 = 277 kN Convective mode Vc = (Ah)c mc g = 0.04 x (110) x 9.81 = 43 kN

Convective forces are less than impulsive forces

Convective mass derives much smaller forces than impulsive mass



Lecture 7/ Slide 30


2 2 Total Base shear, V = Vi +Vc

= 280 kN

Base 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



Lecture 7/ Slide 31


Approach 1: Apply impulsive forces at their respective locations

Impulsive forces have two parts

Part1: 116 kN at hi* + hs

Part 2: 161 kN at hcg

Apply convective forces at its location

43 kN at hc* + hs



Lecture 7/ Slide 32


Location of forces as per Approach 1

116 kN 161 kN

hi*= 3.43 m

hcg = 19.18 m

43 kN

hc*=3.43 m

hs = 16.3 m

hs = 16.3 m

Top of footing Impulsive forces © Sudhir K. Jain, IIT Kanpur

Convective forces


Lecture 7/ Slide 33


Approach 2: Apply base shear, V at height h1 such that

V x h1 = M* V = 280 kN and M* = 5448 kN-m Hence, h1 = 5448/280 = 19.46 m Thus, a force of 280 kN to be applied at 19.46 m



Lecture 7/ Slide 34


Location of forces as per Approach 2

280 kN

h1 =19.46

Top of footing



Lecture 7/ Slide 35


Analysis for tank empty Condition

No convective mode Impulsive mass will be only structural mass, ms V = (Ah)i ms g = 211 kN

This will act at CG of empty container



Lecture 7/ Slide 36


Tank empty condition

Moment = 211 x 19.18 = 4053 kNm

211 kN

hcg =19.18

Top of footing © Sudhir K. Jain, IIT Kanpur


Lecture 7/ Slide 37


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



Lecture 7/ Slide 38

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 points

In IS 1893:1984 entire liquid mass is lumped as impulsive mass

Hence, no convective mode

While 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.



Lecture 7/ Slide 39


Zone and soil parameters from IS 1893:1984

β = 1.0 I = 1.5 Fo= 0.25 (Zone IV)

Time Period, T

M T = 2π K M is total Mass

This is sum of structural mass and total liquid mass

K is Stiffness of Staging



Lecture 7/ Slide 40


Total Mass, M = Structural mass + liquid mass = 196 +255.6 = 451.6 t Lateral stiffness Ks

Braces are treated as as rigid beams as per prevailing practice Hence, Ks = 55,090 kN/m

This we have seen earlier



Lecture 7/ Slide 41


Time period M 451.6 T = 2π = 2π = 0.57 Sec K 55090

This time period is less than the impulsive time period of 0.86 sec calculated earlier

Because, stiffness of braces is taken infinite

For 5% damping & T = 0.57 Sec

Sa/g = 0.155 From Figure 2 of IS 1893 : 1984



Lecture 7/ Slide 42

Example 2: Analysis as per 1893:1984

Design horizontal seismic coefficient, αh αh = βIFo(Sa/g) = 1.0 x 1.5 x 0.25 x 0.155 = 0.058

Base 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 kN This 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



Lecture 7/ Slide 43


Analysis for empty condition Structural mass, M =196 t M 196 T = 2π = 2π = 0.375 Sec K 55090

For 5% damping & T = 0.375 Sec

Sa/g = 0.2

As per Figure 2 of IS 1893:1984

Design horizontal seismic coefficient, αh

αh = βIFo(Sa/g) = 1.0 x 1.5 x 0.25 x 0.2 = 0.075



Lecture 7/ Slide 44


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



Lecture 7/ Slide 45

Example 2:Comparison of forces

Comparison of results from the Guideline and IS 1893:1984 Guideline 17,800 kN/m

IS 1893:1984 55,089 kN/m

Tank empty ( Ti ) Tank full ( Ti)

0.66 sec 0.86 sec

0.375 sec 0.57 sec

Tank full ( Tc) Design seismic horizontal coefficient

3.14 sec

-----

Tank empty ( Ah)i Tank full ( Ah)i

0.11 0.084

0.075 0.058

Tank full ( Ah)c

0.040

-----

Lateral stiffness of staging Time period

Impulsive mode,

Convective mode,

Impulsive mode

Convective mode,



Lecture 7/ Slide 46


Comparison of results from the Guideline and IS 1893:1984 Guideline

IS 1893:1984

Total base shear (V) Tank empty Tank full

211 kN 280 kN

144 kN 257 kN

Bending Moment Tank empty Tank full

4053 kN 5448 kN

2762 kNm 4929 kNm

In tank full condition, base shear and moment from the Guideline are about 10% more than those obtained using IS 1893:1984



Lecture 7/ Slide 47


Two points may be noted

IS 1893:1984 implied K = 1.0, which leads to smaller seismic forces However, the practice was to overestimate stiffness,which increases seismic coefficient.

These two effects compensate each other Hence, there is not much difference in design forces from the Guideline and IS 1893:1984 for elevated tanks on frame type staging.



Lecture 7/ Slide 48

Example 2: Member forces

To obtain design forces in columns and braces

Plane frame analysis methods are well known

Staging frame is to be analysed for lateral seismic force One can use standard structural analysis programs. Alternatively, one may use approximate analysis methods. For example, moment distribution method and Kani’s method

In this example, staging has polygon frame

It is not a plane frame Strictly speaking, it is a space frame



Lecture 7/ Slide 49


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, Roorkee Dayaratnam P, 1986, “Design of Reinforced Concrete Structures”, Oxford & IBH Publishing Co., New Delhi

These methods make simplifying assumptions

Accuracy of these methods depends on validity of these assumptions



Lecture 7/ Slide 50


Major assumptions in these methods are: 1) Axial force in column due to lateral force is proportional to its distance from bending axis 2)Point of inflection in columns and braces are at mid-span 3)Dayaratnam’s book assumes that lateral shear is equally distributed to all columns, OR Jain and Jai Krishna distribute lateral shear similar to that in a cross-section of a cantilever column



Lecture 7/ Slide 51


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



Lecture 7/ Slide 52


Point of inflection for the top and bottom panels is not at mid-span Sameer 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



Lecture 7/ Slide 53


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 Sameer Jain and Jai Krishna Dayaratnam



Lecture 7/ Slide 54


Comparison of design forces in columns and braces Computer Jain & Jain & Dayaratnam analysis Sameer JaiKrishna Columns Axial force (kN) Bending moment (kNm) Shear force (kN)

492 150 66

494 160 70

519 187 93

519 94 47

Braces Bending moment (kNm) Shear force (kN)

175 112

187 119

280 178

187 119



Lecture 7/ Slide 55


Jain & Jai Krishna’s approach overestimates bending moments and shear forces in columns and braces Dayaratnam’s approach underestimates bending moments and shear forces in columns The point here is that designer should properly assess the limitations of a particular method



Lecture 7/ Slide 56

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 m Centre line diameter of shaft = 6.28 m Height of shaft above GL = 15 m Hard soil; Zone IV



Lecture 7/ Slide 57

Example 3:Elevated tank on Shaft

Weight of staging, i.e., shaft

Hence, structural mass ms

= π x 6.28 x 0.15 x 16.4 x 25 = 1,213 kN Mass of staging = 123.7 t ms = 160.7 + 123.7/3 = 201.9 t

Container is same as in Example 2, hence,

mi = 140.6 t mc =110 t



Lecture 7/ Slide 58

Example 3:Lateral stiffness

Lateral stiffness of staging Shaft 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 beam Lateral stiffness is given by

3EI KS = 3 L

Where, E = Modulus of elasticity = 22.36 x 106 kN/m2



Lecture 7/ Slide 59


I = Moment of inertia of shaft cross section = π x (6.434 – 6.134)/64 = 14.59 m4 L = Height of shaft = 16.4 m Thus, 3 × 22.36 ×10 6 ×14.59 5 KS = = 2.22 × 10 kN/m 3 16.4



Lecture 7/ Slide 60


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



Lecture 7/ Slide 61


In shaft, if we wish to consider height up to CG, then, cantilever will have two different crosssections

One up to staging top Second 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 rotation Deflection at CG will not be much higher than that at staging top



Lecture 7/ Slide 62


Hence, staging height taken up to ring beam is adequate for design

Also recall time period varies as square root of stiffness

This will be on the conservative side

A 15% error in stiffness will give about 7% error in time period

There is another approximation in stiffness of shaft

Shear deformations



Lecture 7/ Slide 63


Effect of shear deformation

Consider a cantilever subjected to tip load , P Tip deflection due to shear deformation

δs =

PL κ ' AG

G = Shear modulus A = Cross-sectional area κ’ = Shape factor

Depends on shape of cross-section



Lecture 7/ Slide 64


Thus, total deflection of cantilever PL3 PL δ = + ' 3EI κ AG

First part is due to flexural deformation Second part is due to shear deformation

Hence, stiffness will be Ks =


1 L3 L + ' 3EI κ AG


Lecture 7/ Slide 65


For hollow circular cross-section

κ’ = 0.5

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


1 L3 L + 3EI 0.5 AG


Lecture 7/ Slide 66


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/m By 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 Guideline However, it should have been 1.77 x 105 kN/m



Lecture 7/ Slide 67


Analysis for tank full condition

Impulsive mode time period, Ti mi + ms Ti = 2π Ks

= 0.25 Sec

Shafts are quite rigid, hence, low time period

Recall, for frame staging, Ti = 0.86 sec



Lecture 7/ Slide 68


Convective time period, Tc = 3.14 sec

Same as in Example 2 Ratio of Tc and Ti is about 12.0

Impulsive and convective time periods are quite well separated

Hence, the assumption of two uncoupled SDoF can be used



Lecture 7/ Slide 69


Design horizontal seismic coefficient

Zone 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 sec And, (Sa/g)i will remain 2.5



Lecture 7/ Slide 70


For Tc = 3.14; (Sa/g)c = 0.318 x 1.75 = 0.56

Design Horizontal Coefficient, Ah

Multiplication by 1.75 is for 0.5% damping

(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)i

Convective mass is subjected to much smaller acceleration



Lecture 7/ Slide 71


Base Shear, V

Impulsive mode Vi = (Ah)i (mi + ms)g Vi = 0.25 x (140.6 + 201.8) x 9.81 = 345 + 495 = 840 kN Convective mode Vc = (Ah)c mc g = 0.06 x (110) x 9.81 = 65 kN

Convective forces are much smaller than impulsive forces



Lecture 7/ Slide 72


2 2 Total Base shear, V = Vi +Vc

= 843 kN

Base shear is about 19 % of total seismic weight of 4,488 kN (tank is located in seismic zone IV)



Lecture 7/ Slide 73


Impulsive forces have two parts

Part1: 345 kN at hi* + hs

Part 2: 495 kN at hcg

Apply convective forces at its location

65 kN at hc* + hs



Lecture 7/ Slide 74


Base moment in impulsive mode

Mi* = 345 x (3.43 + 17) + 495 x 19.88 = 169,00 kNm

Base moment in convective mode

Mc* = 65 x (3.43 + 17) = 1322 kNm

Total base moment M

*

=

M

* 2 i

+M

* 2 c

= 16,940 kNm © Sudhir K. Jain, IIT Kanpur


Lecture 7/ Slide 75


Tank Empty Condition

Time Period Ti = 0.19 Sec Base Shear, V = 495 kN Base 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



Lecture 7/ Slide 76

Example 3:Analysis using IS 1893:1984

This tank is also analysed using IS 1893:1984

Entire liquid mass is lumped as impulsive mass No convective mode

For tank full condition

Mass, M = Mass of container + Mass of water + 1/3rd Mass of staging = 160.7 + 255.6 + 1/3 x 123.7 = 457.5 t



Lecture 7/ Slide 77

Example 3: Analysis using IS 1893:1984

Stiffness of staging, Ks = 2.22 x 105 kNm Time period M 457.5 T = 2π = 2π = 0.285 Sec 5 K 2.22 ×10

Base shear coefficient αh = βIFo(Sa/g) = 1.0 x 1.5 x 0.25 x 0.2 = 0.075



Lecture 7/ Slide 78


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



Lecture 7/ Slide 79


Analysis for Tank Empty

Structural mass, M M = Mass of container + 1/3rd Mass of staging = 160.7 + 1/3 x 123.7 = 201.9 T T = 2π

M 201.9 = 2π = 0.19 Sec 5 K 2.22 ×10

Design horizontal seismic coefficient, αh αh = βIFo(Sa/g) = 1.0 x 1.5 x 0.25 x 0.2 = 0.075



Lecture 7/ Slide 80


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-m

Let us compare results from the Guideline and IS 1893:1984

See next slide



Lecture 7/ Slide 81

Example 3: Comparison of results

Comparison of results from the Guideline and IS 1893:1984

Idealization of Tank Lateral stiffness of staging

Guideline IS 1893:1984 2.22 x 105 kN/m 2.22 x 105 kN/m

Time period

Impulsive mode, Tank empty ( Ti ) Tank full ( Ti)

0.19 sec 0.25 sec

0.19 sec 0.285 sec

3.14 sec

-----

0.25 0.25

0.075 0.075

0.060

-----

Convective mode, Tank full ( Tc) Design seismic horizontal coefficient

Impulsive mode Tank empty ( Ah)i Tank full ( Ah)i

Convective mode, Tank full ( Ah)c © Sudhir K. Jain, IIT Kanpur


Lecture 7/ Slide 82

Example 3:Comparison of results

Comparison of results from the Guideline and IS 1893:1984 Idealization of Tank Total base shear (V) Tank empty Tank full Total base Moment (M) Tank empty Tank full

Guideline

IS 1893:1984

495 kN 843 kN

148.5 kN 336.6 kN

9,842 kN-m 16,940 kN-m

2,952 kN-m 6,692 kN-m

For shaft supported tanks, seismic forces from the Guideline are on much higher side

Base moment is 150% higher in full condition



Lecture 7/ Slide 83

Example 3:Comparison of results

Comparison of forces from the Guideline and IS 1893:1984 reveals: For frame staging

Design forces increase marginally However, this increase will vary for different tanks

For shaft staging

Design forces increase significantly In these tanks, increase in base moment is of particular interest



Lecture 7/ Slide 84

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



Lecture 7/ Slide 85


In frame staging, earlier practice was to overestimate lateral stiffness

Thereby increases design seismic forces To some extent, this compensates the lower values of design forces due to K = 1.0 Hence, in frame staging net increase in forces is not very significant



Lecture 7/ Slide 86


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 shaft A wider shaft will help in achieving more economical design To explain this point, cost of shaft and foundation is obtained for one tank See next example



Lecture 7/ Slide 87

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 mm Zone V, Medium soil, SBC = 25 t/m2 Solid raft foundation is used



Lecture 7/ Slide 88

Issue of cost

In the first instance, shaft of 7 m diameter is considered Then, shaft of 12 m diameter is considered Comparison of quantity of concrete and steel required in both the cases is given in next slide

Quantity for shaft and foundation are given



Lecture 7/ Slide 89

Issue of cost

Material quantities required 7 m dia. Shaft

12 m dia. shaft

IS 1893:1984 Guideline

IS 1893:1984

Guideline

Shaft Concrete Steel

90 m3 6t

* *

150 m3 7t

150 m3 9t

Foundation Concrete Steel

200 m3 8t

540 m3 21 t

135 m3 9t

340 m3 12 t

* Design of 7 m shaft is not possible for forces as per the Guideline © Sudhir K. Jain, IIT Kanpur


Lecture 7/ Slide 90

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



Lecture 7/ Slide 91

Issue of cost

Some more economy in foundation can perhaps be achieved by using different type of foundation system Some options are:

Solid mat foundation Annular mat foundation Strip foundation on piles Shell type foundation



Lecture 7/ Slide 92

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



Lecture 7/ Slide 93

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 prevailed And above discussed options were not needed and were not attempted Moreover, in many elevated tanks, wind forces used to govern the design



Lecture 7/ Slide 94

Issue of cost

In view of new design forces

There is a need to try some of these options To achieve better economy, and Good seismic performances



Lecture 7/ Slide 95


In Part B of the Guideline, six solved examples are given

These will be of help to the users of the Guideline

We are now almost at the end of this course

Next Lecture, which is the last one, will discuss some miscellaneous issues.



Lecture 7/ Slide 96

Lecture 8

February 21, 2006

In this Lecture

Following issues will be discussed

Soil structure interaction Buried tanks P-Delta effect Flexibility of piping Buckling of shell IS 11682:1985



Lecture 8/ Slide 2

Soil Structure Interaction

Soil condition is important in seismic analysis Type of soil affects the ground motion and structure’s response Effect on ground motion is reflected in design spectrum

Recall, there are different response spectra for hard, medium and soft soil Spectrum for soft soil is higher than hard soil

Except in short period range



Lecture 8/ Slide 3


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



Lecture 8/ Slide 4


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.



Lecture 8/ Slide 5


From Veletsos (1984): Soil structure interaction has two effects

It changes time period It changes total damping

Effect on time period

If structure is on hard soil, it is treated as fixed at base If 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



Lecture 8/ Slide 6


Effect on damping

Soil medium has higher damping Damping in soil comes from hysteric action and radial waves

Also called radial damping in soil

Damping of soil increases total damping of soilstructure system



Lecture 8/ Slide 7


Veletsos (1984) suggested following expression for impulsive time period of tank on soft soil −

Ti = Ti 1 +

K

f

Kx

⎡ K x h 2f ⎢1 + Kθ ⎢⎣

⎤ ⎥ ⎥⎦

Kx = Horizontal translational stiffness of soil Kθ = Rocking stiffness of soil Kf = Stiffness of impulsive mode hf = Distance from the tank base to the point of application of impulsive mass, mi Ti = Impulsive mode time period of tank with fixed base © Sudhir K. Jain, IIT Kanpur


Lecture 8/ Slide 8


KX

8 = Gs R0 2 −νs

8 3 Kθ = G s R0 3( 1 − ν s ) Gs = Shear modulus of elasticity of soil

νs = Poisson’s ratio of soil R0 = Radius of the foundation



Lecture 8/ Slide 9


For elevated tanks

Stiffness of impulsive mode, Kf is lateral stiffness of staging, Ks

For ground supported tanks

We have considered circular and rectangular tanks For both types of tanks, time period of impulsive mode, Ti is known

Recall expression for Ti for circular tanks from Lecture 3 Also refer Clause 4.3.1.1 and 4.3.1.2 of the Guideline



Lecture 8/ Slide 10


Hence, stiffness of impulsive mode, Kf for ground supported tanks can be obtained from mi Ti = 2π Kf

mi being impulsive mass is also known



Lecture 8/ Slide 11


hf is distance of impulsive mass from base of

tank In ground supported tanks

hf = hi + tb

In elevated tanks

Total impulsive mass = mi + ms Hence, hf = hcg Recall, we assume that entire mass is lumped at CG of empty container



Lecture 8/ Slide 12


Veletsos (1984) included effect of soil damping as follows:

Effective damping ratio of structure is −

ξ −

ξ

ξs ξ Ti

= ξS +

ξ ⎛ − ⎞ ⎜ Ti / T i ⎟ ⎝ ⎠

3

= Effective damping ratio of structure including soil effect = Damping ratio for soil = Damping ratio of structure = Impulsive mode time period with fixed base

−

Ti = Impulsive mode time period including soil effect © Sudhir K. Jain, IIT Kanpur


Lecture 8/ Slide 13


It is clear that soft soil would have lower values of lateral stiffness Kx and rocking stiffness, Kθ.

Similarly, damping in soft soil is more compared to hard soil

Hence, more effect on impulsive time period

Hence, damping of structure will be more

Thus, soil structure interaction would be important for structures on soft soil



Lecture 8/ Slide 14


Increase in time period and damping would reduce seismic forces

This is also called beneficial effects of soil

Before using these beneficial effects of soil, one needs to properly ascertain the soil properties

Note, soil properties are frequency dependent



Lecture 8/ Slide 15

Buried tanks

In buried tanks, effect of soil pressure on wall shall be included During lateral excitation, soil induces dynamic pressure on wall Strictly 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



Lecture 8/ Slide 16

Buried tanks

NZSEE has suggested following simplified distribution for earthquake induced soil pressure Tank wall σx = 1.5 (Ah)i γs He

Ground surface

Embedment depth, He

PE

Tank base © Sudhir K. Jain, IIT Kanpur

Idealised pressure distribution

σx = 0.5 (Ah)i γs He


Lecture 8/ Slide 17

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 condition

This earth pressure shall be used to check wall design This dynamic earth pressure shall not be relied upon to reduce dynamic effects due to liquid



Lecture 8/ Slide 18

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 same Time 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



Lecture 8/ Slide 19

Buried tanks

For buried tanks, response reduction factor, R = 4.0

This is for fully buried tank Refer Table 2 of the Guideline

If tank is partially buried, then, R value shall be linearly interpolated

Depending on depth of embedment



Lecture 8/ Slide 20

Buried tanks

For a RC tank with fixed base

R = 2.0 for ground supported tank R = 4.0 for fully buried tank If tank is half buried, then, R = 2.0 + (4.0 – 2.0) x 0.5 = 3.0 For 2/3rd buried tank, R = 2.0+(4.0 – 2.0)x 0.67 = 3.33

For a steel tank with anchored base

R = 2.5 for ground supported tank If tank is half buried, then, R = 2.5 + (4.0 – 2.5) x 0.5 = 3.25



Lecture 8/ Slide 21

Buried tanks

Dynamic soil pressure would also act on shaft staging of elevated tanks

For large tanks, shaft diameter could be more than 10 m Foundation for shaft is usually 2 to 3 m below ground Hence, soil pressure will act over buried portion of shaft



Lecture 8/ Slide 22

Buried tanks

Dynamic soil pressure acts opposite to direction of seismic loading Seismic force

Soil pressure



Lecture 8/ Slide 23

Buried tanks

Soil pressure would induce moment on the foundation

This moment would be opposite to moment due to seismic forces Thus, net moment on foundation would reduce

Codes do not allow this reduction for design purpose

Due to uncertainties in compactness of soil etc.



Lecture 8/ Slide 24

Buried tanks

In frame staging, column sizes are quite small

Compared 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



Lecture 8/ Slide 25

P-Delta effect

Consider a vertical cantilever subjected to lateral force Q and vertical force P Bending moment at the base, M = Q x L P Q L



Lecture 8/ Slide 26

P-Delta effect

Due to force Q, let deflection be Δ Point of application of P shifts

And, additional moment p x Δ will act on the cantilever P

Q

Δ

Now, total moment at the base = QxL+PxΔ

L



Lecture 8/ Slide 27

P-Delta effect

Thus, vertical loads interact with lateral displacements

And, induce additional forces on structure

This is termed as P-Delta effect Due to additional moment of P x Δ, lateral deflection would further increase

Which, in turn would cause more moment



Lecture 8/ Slide 28

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



Lecture 8/ Slide 29

P-Delta effect

Compared to buildings, P-Delta effect will be more severe in elevated tanks In buildings, vertical loads are distributed over the entire height

In elevated tanks, almost entire vertical load is at the top



Lecture 8/ Slide 30

P-Delta effect

Thus, to minimise P-Delta effect in elevated tanks

Permissible maximum lateral deflection shall be less than that for buildings h/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



Lecture 8/ Slide 31

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 junctions

Hence, piping shall have sufficient flexibility to accommodate seismic movement

Without causing damage to tank shell or base



Lecture 8/ Slide 32

Flexibility of piping

All the international codes emphasize on flexible piping system Flexibility in piping can be imparted by

Using flexible elbows Special coupling devices

These flexible devices need not be used at the junction of piping and tanks

They can be used at nearby joints also



Lecture 8/ Slide 33

Buckling of shell

Ground supported circular steel tanks are sensitive to buckling

These are thin shell structures In 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



Lecture 8/ Slide 34

Buckling of shell

During lateral base excitation, tank wall is subjected to excessive compressive force

This may trigger buckling Buckling 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



Lecture 8/ Slide 35

Buckling of shell

For safety against buckling

Codes restrict maximum allowable axial stresses in steel tanks

RC shaft staging may also be thin shell structure

They are not as thin as steel tanks Nevertheless, safety against buckling should be ensured



Lecture 8/ Slide 36

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 shaft This 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.



Lecture 8/ Slide 37

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 shaft Cw = Wall strength coefficient = 0.55 fck = Specified compressive strength of concrete Ac = Gross area of concrete βw = Wall slenderness coefficient



Lecture 8/ Slide 38

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



Lecture 8/ Slide 39

Buckling of shell

Geometrical imperfections are invariably present in shaft staging

Stringent construction tolerances should be followed to minimize these imperfections Recall, buckling load is sensitive to imperfections

Effect of these imperfections on static response, dynamic response and buckling should be studied

A good research topic !!



Lecture 8/ Slide 40

Buckling of shell

ACI 371 specifies limits on construction tolerances; Some of these are Variation in wall thickness:

Variation in plumb of shaft:

-3% and + 5% of wall thickness Less than 10 mm in any 1.6 m of height Less than 40 mm in any 16 m of height Maximum of 75 mm in total height

Variation in diameter of shaft:

0.4% of diameter and shall not exceed 75 mm



Lecture 8/ Slide 41

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 staging These should be included in IS 11682

In fact, there are many other limitations in IS 11682:1985 Some of these will be discussed briefly



Lecture 8/ Slide 42

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 load In seismic analysis of buildings, one applies reductions in live load. This confuses the design

Hence, this clause should be corrected



Lecture 8/ Slide 43

IS 11682:1985

Increase in permissible stresses is allowed for columns and not for braces

Clause 5.4: Increase in permissible stresses for column staging shall be as per IS 456-1978 Clause 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

i.e., braces will be stronger than columns

Whereas, in seismic design we need strong column-weak beam combination



Lecture 8/ Slide 44

IS 11682:1985

Clause 7.3.3. “Loss of contact in the soil under footing should not be allowed. In locations where the soil bearing capacity is high, loss of contact may be allowed provided it is safe against overturning and such other conditions that are to be fulfilled.”

This clause is for column footings It needs to be more specific regarding how much loss of contact can be allowed or how much tension in soil may be permitted Seismic loadings are momentary and like increase in SBC, some tension may also be permitted These should also be permitted for raft foundations.



Lecture 8/ Slide 45

IS 11682:1985

Clause 7.2.4. “For staging in seismic zones where seismic coefficient exceeds 0.05 twin diagonal vertical bracing of steel or RCC in addition to horizontal bracing may be provided….”

This clause is for frame staging Limit of 0.05 on seismic coefficient may be removed Suggestions on use of shear walls in addition to vertical diagonal bracing may be included



Lecture 8/ Slide 46

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 diameters Similarly, different types of foundation systems for shaft should be suggested

This has been done in ACI 371



Lecture 8/ Slide 47

IS 11682:1985

These are based on working stress design method As per IS 456:2000, limit state design is compulsory for members subjected to combined direct load and flexural

Clause 8.5.2 suggests formulae for stresses in shaft

Clause B-4.3 of IS 456:2000

Hence, provisions on limit state design of shaft are needed in IS 11682



Lecture 8/ Slide 48

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



Lecture 8/ Slide 49

And finally…..

With this Lecture, we conclude this E-course Please spare some time to go through the Lectures again We 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 questions Thank You………



Lecture 8/ Slide 50

Assignment 1 Date of assignment: 18 January 2006 E-course on Seismic Design of Tanks

1

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 day It will help you appreciate the next lecture

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

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 follow Also 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 False 1.1)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.


Solution 1 Date of assignment: 18 January 2006 Date of solution: 20 January 2006 E-course on Seismic Design of Tanks 1


Identify the following statements as True or False 1.1)Hydrodynamic pressure varies linearly with liquid height. False Hydrodynamic pressure has curvilinear variation along the height. In fact, Hydrostatic pressure varies linearly with height. 1.2)Impulsive mass is less than convective mass for short tanks. False In short tanks, convective liquid or liquid undergoing sloshing motion is more. In tall tanks, impulsive liquid mass is more.

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

Solution 1 / slide 2

Part II: True / False 1.3)Net hydrodynamic force on the container wall is zero. False Rather, net hydrostatic force on wall is zero 1.4)Hydrodynamic pressure on base causes overturning moment on tank. True Refer slide no. 46 of Lecture 1 1.5)Convective mass acts at lower height than impulsive mass. False Liquid in upper portion, undergoes convective or sloshing motion, hence, convective mass is always located at higher height than impulsive mass. Note the graphs for hi and hc in figure 2b and 3b of guidelines © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006


Part II: True / False 1.6)With the inclusion of base pressure effect, overturning moment on tank reduces. False Hydrodynamic 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. False Impulsive and convective masses depend on aspect ratio (h/D or h/L) rather than only on height, h. © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006


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 t D = 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 D⎞ ⎛ tanh ⎜ 0 . 866 ⎟ mi h⎠ ⎝ = D m 0 . 866 h h⎞ ⎛ tanh⎜ 3.68 ⎟ mc D⎠ ⎝ = 0.23 h m D

12 ⎞ ⎛ tanh ⎜ 0 . 866 ⎟ 5 ⎠ ⎝ = 12 0 . 866 5

= 0.466

5⎞ ⎛ tanh⎜ 3.68 ⎟ 12 ⎠ ⎝ = 0.23 5 12

= 0.503

Since, h/D is less than 0.75, hi/h = 0.375 h⎞ ⎛ cosh⎜ 3.68 ⎟ − 1.0 hc D⎠ ⎝ = 1− h h⎞ h ⎛ 3.68 sinh⎜ 3.68 ⎟ D D⎠ ⎝

5⎞ ⎛ cosh⎜ 3.68 ⎟ − 1.0 12 ⎠ ⎝ = 1− 5 5⎞ ⎛ 3.68 sinh⎜ 3.68 ⎟ 12 12 ⎠ ⎝

= 0.579



Part III: Solutions

hi * = h

0.866

D h

D⎞ ⎛ 2 tanh⎜ 0.866 ⎟ h⎠ ⎝

-

0.125

h⎞ ⎛ cosh⎜ 3.68 ⎟ − 2.01 hc * D⎠ ⎝ = 1− h⎞ h h ⎛ 3.68 sinh⎜ 3.68 ⎟ D D⎠ ⎝

K c = 0.836

∴

=

0.866

12 5

12 ⎞ ⎛ 2 tanh⎜ 0.866 ⎟ 5⎠ ⎝

= 0.947

- 0.125

5⎞ ⎛ cosh⎜ 3.68 ⎟ − 2.01 12 ⎠ ⎝ = 1− 5 5⎞ ⎛ 3.68 sinh⎜ 3.68 ⎟ 12 12 ⎠ ⎝

= 0.878

mg h⎞ ⎛ tanh 2 ⎜ 3.68 ⎟ h D⎠ ⎝

5⎞ ⎛ K c h / mg = 0.836 tanh 2 ⎜ 3.68 ⎟ 12 ⎠ ⎝

= 0.694



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* = 0.878 x 5

= 4.39 m

hc*/h = 0.878 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:

Y 16 m 10 m

X

Plan view Total volume of water = 10 x16 x 8 = 1280 m3 Total mass of water, m = 1280 x 1.0 = 1280 t © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006


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* = 0.87 x 8

= 7.0 m

hc*/h = 0.87



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* = 0.73 x 8

= 5.84 m

hc*/h = 0.73



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 .



Additional Slide for Solution 1 Insert after slide 1 of Solution 1

1

Correction in Part II True/False - 1.2

Notice that there is a correction in Part II True/False – 1.2 1.2)Impulsive mass is less than convective mass for short tanks.

It is printed as False It should be printed as TRUE



Assignment 2 Date of assignment: 20 January E-course on Seismic Design of Tanks

1


Read the following: Clause 4.5 of the Guidelines Clause 6.4 of IS 1893(Part 1):2002 Clause 3.4, 4.2.1.1, and 5.2.6 of IS 1893:1984


E-Course on Seismic Design of Tanks / January 2006 Assignment 2/ slide 2


Identify the following statements as True or False 2.1) 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%



Solution 2 Date of assignment: 20 January Date of solution: 23 January E-Course on seismic design of tanks 1

Part II: True/ False

Identify the following statements as True or False 2.1) Seismic force depends on mass of structure. True 2.2) In IS 1893(Part 1):2002, there are five seismic zones. False There 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 staging False For a building with good ductility, response modification factor, R = 8.0 as against, R = 3.0 for elevated tanks on frame staging


E-Course on Seismic Design of Tanks / January 2006

Solution 2 / Slide 2

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 False Performance factor, K is same (K = 1.0), for a building with good ductility and elevated tanks. 2.5) Importance factor for tanks is higher than for buildings. True 2.6) Damping of structure does not affect base shear coefficient. False See Table 3 of IS 1893(Part 1):2002 and Fig. 2 of IS 1893:1984 © Sudhir K. Jain, IIT Kanpur



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 buildings Ah = βIFoSa/g for elevated tanks We have, Fo = 0.4, β = 1.0 © Sudhir K. Jain, IIT Kanpur



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 = 1.0 Ah = KβIFoSa/g = 1.0 x 1.0 x 1.0 x 0.4 x 0.1 = 0.04 For building with low ductility: K = 1.6, I = 1.0 Ah = KβIFoSa/g = 1.6 x 1.0 x 1.0 x 0.4 x 0.1 = 0.064 For elevated tank: I = 1.5 Ah = 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 = 1.5 Values of Sa/g are to be obtained from Fig. 2 of IS 1893:1984




Part III: Solutions

Time period Damping Sa/g (Sec) (%) 0.5

1.4

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 = 0.063

5

0.08

1.0 x 1.5 x 0.4 x 0.08 = 0.048

Note the effect of damping on base shear coefficient © Sudhir K. Jain, IIT Kanpur




1


Read Clause 4.2.2, 4.2.3 and 4.3 of the Guidelines


Assignment 3/ slide 2


Identify the following statements as True or False 3.1)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.



Solution 3 Date of assignment: 24 January Date of solution: 27 January E-course on Seismic Design of Tanks 1

Part II: True / False 3.1)Time period does not depend on mass of the structure. False 3.2)Time period increases with stiffness. False Time period decreases with stiffness and is inversely proportional to square root of stiffness. T = 2π M/K


Solution 3/ slide 2

Part II: True / False 3.3)Impulsive mode time period of ground supported tanks is usually less than 0.4 seconds. True 3.4)Convective mode time period depends only on aspect ratio of the container. False It depends not only on aspect ratio but also on diameter or lateral dimension of container. See formula for Tc in slide no. 65. 3.5)In elevated tanks, convective mode time period depends on lateral stiffness of staging. False


Solution 3/ slide 3

Part II: True / False 3.6)While finding lateral stiffness of frame staging, braces shall be treated as rigid beams. False For proper evaluation of lateral stiffness, it is necessary that flexibility of brace is included in the model.


Solution 3/ slide 4

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:

Y 12 m 6m

X

Plan view


Solution 3/ slide 5

Part III: Solutions 1) Analysis in X-direction: Length, L = 12.0 m, Height of liquid, h = 4.0 m ∴ h/L = 4/12 = 0.33 From Figure 7 of guideline, for h/L = 0.33, one gets Cc = 4.0 Time period of convective mode, TC = CC

Lg

TC = 4.0 (12.0 9.81)

= 4.42 Sec. 2) Analysis in Y-direction: Length, L = 6.0 m, h = 4.0 m, ∴ h/L = 4/6 = 0.67 From Figure 7 of guideline, for h/L = 0.67, one gets, Cc = 3.65 Time period of convective mode, T C = C C

Lg

Tc = 3.65 (6.0 9.81) = 2.85 Sec. © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006

Solution 3/ slide 6

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 = 0.375 From Figure 2 of guideline, for h/D = 0.375: mi/m = 0.43, mc/m = 0.545 and Kch/mg = 0.665


Solution 3/ slide 7

Part III: Solutions mi = 0.43 x m = 0.43 x 509 = 218.9 t mc = 0.545 x m = 0.545 x 509 = 277.4 t Kc = 0.665 x m g/h = 0.665 x 509 x 9.81/4.5 = 737.9 kN/m Weight of container = 3,000 kN ∴ Mass of container = 3000/9.81 = 305.8 t Weight of staging = 1,700 kN ∴ Mass of staging = 1700/9.81 = 173.3 t Structural mass of tank, ms = mass of container +1/3rd mass of staging = 305.8 + 1/3 x 173.3 = 363.6 t Lateral stiffness of staging, Ks = 50,000 kN/m © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006

Solution 3/ slide 8

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


Solution 3/ slide 9

Part III: Solutions For evaluating time period, this two mass model will be treated as two uncoupled single degree of freedom system: 277.4 t 737.9 kN/m 582.5 t

50000 kN/m

218.9t + 363.6t 50000 kN/m

277.4 t 737.9 kN/m

(a) Two mass idealization

(b) Uncoupled system


Solution 3/ slide 10

Part III: Solutions

Now, Impulsive Time Period, Ti = 2π

Ti = 2π

mi + ms Ks 218.9 + 363.6 50,000

= 0.678 Sec.

mc Convective Time Period, Tc = 2π Kc 277.4 Tc = 2π 737.9 = 3.85 Sec.




1


Read Clause 4.4, 4.5 of the Guideline




Identify the following statements as True or False 4.1) 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: Questions 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.



Part III: Questions 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 4 Date of assignment: 30 January 2006 Date of solution: 01 February 2006 E-course on Seismic Design of Tanks 1

Part II: True / False 4.1) R values for elevated tanks with SMRF staging are lower than for buildings with SMRF True For elevated tanks with SMRF staging R = 2.5, whereas for buildings with SMRF, R = 5.0 4.2)Frame staging has less redundancy than shaft staging False 4.3)Damping in convective mode depends on material of tank False Convective mode has 0.5% damping for all types of tank materials.


Solution 4/ slide 2

Part II: True / False 4.4)Elevated tanks are inverted pendulum type structures True 4.5)Thin shafts have more ductility than thick shafts False 4.6)Response reduction factor for tanks depends on importance of tank. False Response reduction factor depends on ductility, redundancy and overstrength. Importance of tank decides the value of importance factor, I.


Solution 4/ slide 3

Part II: True / False 4.7)In elevated tanks, damping in impulsive mode depends on material of container False In elevated tanks, damping in impulsive mode depends on material of staging and not of container.


Solution 4/ slide 4

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: Width Direction

Length Direction Plan of tank © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006

Solution 4/ slide 5

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)


Solution 4/ slide 6

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 sec Damping 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)


Solution 4/ slide 7

Part III: Solutions

Convective base shear coefficient (Ah)c= Z/2 x I/R x (Sa/g)c = 0.16/2 x 1.5/2.0 x 0.476 = 0.029 Hence, in length direction, (Ah)i = 0.15 and (Ah)c = 0.029 2) Analysis in Width Direction: Impulsive mode time period, Ti = 0.4 Sec, Damping in impulsive mode = 5% (RC tank)

∴ (Sa/g)i = 2.5 (Clause 4.5.3 of the Guideline) Impulsive base shear coefficient, (Ah)i = Z/2 x I/R x (Sa/g)i = 0.16/2 x 1.5/2.0 x 2.5 = 0.15 © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006

Solution 4/ slide 8

Part III: Solutions Convective mode time period, Tc = 3.0 sec Damping 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


Solution 4/ slide 9

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 sec Damping 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 sec Damping 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



Assignment 5 Date of assignment: 3 February Date of solution: 5 February E-course on Seismic Design of Tanks 1


Read Clause 4.6, 4.7 and 4.8 of the Guidelines



Part II: True / False 5.1)In ground supported tanks, base shear at the bottom of wall is same as base shear at the bottom of base slab. 5.2)In elevated tanks, base moment at the top of footing includes effect of base pressure. 5.3)For calculating bending moment at the bottom of wall, effect of base pressure is included. 5.4)In elevated tanks, impulsive mass of liquid, mi, acts at CG of empty container. 5.5)In IITK-GSDMA Guideline, impulsive and convective base shear are combined using absolute summation rule. 5.6)Critical direction of seismic loading is always same for braces and columns of frame staging. 5.7)In elevated tanks, base shear at the bottom of container wall, will not depend on mass of staging.



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.

12 m

6m 12 m

Y X

Plan

4.5 m

0.1 m projection along the periphery

Elevation

Earthquake Direction

mi (t)

hi (m)

hi * (m)

mc (t)

hc (m)

hc* (m)

(Ah)i

(Ah)c

X

138

1.69

4.73

189

2.48

5.07

0.15

0.024

Y

230

1.69

2.61

112

2.93

3.29

0.15

0.037



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.



Solution 5 Date of assignment: 3 February Date of solution: 6 February E-course on Seismic Design of Tanks 1

Part II: True / False 5.1)In ground supported tanks, base shear at the bottom of wall is same as base shear at the bottom of base slab. False Base 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. True Since staging is below the base slab (or floor slab), base moment in staging includes the bending effect caused by hydrodynamic pressure on base slab.


Solution 5/ slide 2

Part II: True / False 5.3)For calculating bending moment at the bottom of wall, effect of base pressure is included. False Base 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. False In elevated tanks, impulsive mass of liquid, mi, is considered to act at hi*. Structural mass of tank, ms, acts at CG of empty container.


Solution 5/ slide 3


5.5)In IITK-GSDMA Guideline, impulsive and convective base shear are combined using absolute summation rule. False They 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. False Columns and braces of frame staging can have different critical directions.


Solution 5/ slide 4


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 .


Solution 5/ slide 5

Part III: Solutions 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. 0.1 m projection along the 4.5 m 12 m 6 m periphery 12 m Y X

Elevation

Plan

Earthquake Direction

mi (t)

hi (m)

hi * (m)

mc (t)

hc (m)

hc* (m)

(Ah)i

(Ah)c

X

138

1.69

4.73

189

2.48

5.07

0.15

0.024

Y

230

1.69

2.61

112

2.93

3.29

0.15

0.037


Solution 5/ slide 6

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 mm Plan 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 kN and mass of roof slab, mt = 243.8/9.81 = 24.9 t Thickness of wall = 250 mm Internal dimensions are 12m x 6m and height is 4.5 m.

∴ Mass of wall, mw = 2 x (12.25 + 6.25) x 0.25 x 4.5 x 25/9.81 = 106.1 t © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006

Solution 5/ slide 7

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


Solution 5/ slide 8

Part III: Solutions Impulsive base shear at the bottom of wall is Vi = (Ah)i (mi + mw + mt) g = 0.15 x (138 + 106.1 + 24.9) x 9.81 = 395.8 kN Convective base shear at the bottom of wall is Vc = (Ah)c mc g = 0.024 x 189 x 9.81 = 44.5 kN Total base shear at bottom of wall is V = Vi2 + Vc2 =

(395.8 ) + (44.5 ) 2

2

= 398.3kN


Solution 5/ slide 9

Part III: Solutions Impulsive bending moment at the bottom of wall is Mi = (Ah)i (mihi + mwhw + mtht) g = 0.15 x (138 x 1.69 + 106.1 x 2.25 + 24.9 x 4.56) x 9.81 = 861.5 kN-m Convective bending moment at the bottom of wall is Mc = (Ah)c mc hc g = 0.024 x 189 x 2.48 x 9.81 = 110.4 kN-m Total bending moment at bottom of wall is M = Mi2 + M2c =

(861.5 )

2

+ (110.4 ) = 868.5kN - m 2



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-m Convective overturning moment at the bottom of base slab is Mc* = (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 =

(M ) + (M ) * 2 i

* 2 c

=

(1588 )2 + (236.7 )2

= 1605kN - m



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 = 0.037 Impulsive base shear at the bottom of wall is Vi = (Ah)i (mi + mw + mt) g = 0.15 x (230 + 106.1 + 24.9) x 9.81 = 531.2 kN Convective base shear at the bottom of wall is Vc = (Ah)c mc g = 0.037 x 112 x 9.81 = 40.7 kN Total base shear at bottom of wall is

V = Vi2 + Vc2 =

(531.2 ) + (40.7 ) 2

2

= 532.8kN



Part III: Solutions Impulsive bending moment at the bottom of wall is Mi = (Ah)i (mihi + mwhw + mtht) g = 0.15 x (230 x 1.69 + 106.1 x 2.25 + 24.9 x 4.56) x 9.81 = 1090 kN-m Convective bending moment at the bottom of wall is Mc = (Ah)c mc hc g = 0.037 x 112 x 2.93 x 9.81 = 119.1 kN-m Total bending moment at bottom of wall is

M = Mi2 + M2c =

(1090.3 )2 + (119.1 )2

= 1097kN - m



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-m Convective 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 =

(M ) + (M ) * 2 i

* 2 c

=

(1544 )2 + (143.9 )2

= 1551kN - m



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 is Vc = (Ah)c mc g = 0.06 x 110 x 9.81 = 64.7 kN Total base shear at bottom of staging is V = Vi2 + Vc2 =

(537.1) + (64.7 ) 2

2

= 541.0kN



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 is Mi* = (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 is Mc* = (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

M* =

(M ) + (M ) * 2 i

* 2 c

=

(8145 )2 + (999.0 )2

= 8206kN - m

Thus, base shear = 541.0 kN & Base moment = 8206 kN-m © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006


Assignment 6 Date of assignment: 8 February 2006 E-course on Seismic Design of Tanks

1


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



Part II: True / False 6.1)Convective pressure varies linearly along wall height. 6.2)In circular tanks, impulsive pressure on wall varies in circumferential direction. 6.3)In rectangular tanks, convective pressure on a particular wall varies along the length of that wall. 6.4)Due to vertical ground acceleration, lateral pressure exerted by the liquid on the wall changes. 6.5)In rectangular tanks, hydrodynamic pressure acts on walls parallel to the direction of seismic force. 6.6)Impulsive pressure is maximum at the bottom of wall. 6.7)Convective hydrodynamic force is more significant for tall tanks.



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.

12 m Y X

Plan

6m

Find in X- and Y- directions: 1) Linearised impulsive and convective pressure distribution on the walls 2) Pressure due to vertical excitation 3) Pressure due to wall inertia 4) Total pressure at the bottom of wall



Solution 6 Date of assignment: 8 February 2006 Date of solution: 10 February 2006 E-course on Seismic Design of Tanks 1

Part II: True / False 6.1)Convective pressure varies linearly along wall height. False Convective pressure has curvilinear distribution along wall height. 6.2)In circular tanks, impulsive pressure on wall varies in circumferential direction. True Impulsive 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. False In 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.


Solution 6/ slide 2

Part II: True / False 6.4)Due to vertical ground acceleration, lateral pressure exerted by the liquid on the wall changes. True when 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. False In rectangular tanks, hydrodynamic pressure acts on walls perpendicular to the direction seismic force.


Solution 6/ slide 3

Part II: True / False 6.6)Impulsive pressure is maximum at the bottom of wall. True 6.7)Convective hydrodynamic force is more significant for tall tanks. False Convective 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


Solution 6/ slide 4

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.

12 m Y X

Plan

6m

Find in X- and Y- directions: 1) Linearised impulsive and convective pressure distribution on wall 2) Pressure due to vertical excitation 3) Pressure due to wall inertia 4) Total pressure at the bottom of wall


Solution 6/ slide 5

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 mm Weight density of concrete = 25 kN/m3

Analysis in X- direction: (Ah)i = 0.15, (Ah)c = 0.024 h = 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 kg mc/m = 0. 584, mc = 0. 584 x 324,000 = 189,200 kg hi/h = 0.375, hi = 0.375 x 4.5 = 1.69 m hc/h =0.551, hc = 0.551 x 4.5 = 2.48 m © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006

Solution 6/ slide 6

Part III: Solution

a) Impulsive mode ( A h )i m i 0 . 15 × 137 ,700 × 9 . 81 qi = g = = 16 . 90 kN/m 2B 2×6 Pressure at bottom & top is given by

16 . 90 qi ( ) (4 × 4 .5 − 6 × 1 .69 ) = 6 .60 kN/m 4 6 h − h = i 2 2 4 .5 h qi 16.90 2 ( ) bi = 2 (6hi − 2h ) = 6 × 1 . 69 − 2 × 4 . 5 = 0 . 95 kN/m h 4.52 b = 0.95 kN/m2

ai =

2

i

Linear impulsive pressure distribution

ai = 6.60 kN/m2


Solution 6/ slide 7

Part III: Solution

b) Convective mode ( Ah ) c m c 0 . 024 × 189 , 200 × 9 . 81 qc = = 3 . 71 kN/m g = 2B 2×6 Pressure at bottom & top is given by

3 .71 qc 2 ( ) a c = 2 (4 h − 6 hc ) = 4 4 . 5 6 2 . 48 0 . 57 × − × = kN/m 4 .52 h qc 3.71 2 ( ) bc = 2 (6hc − 2h ) = 6 × 2 . 48 − 2 × 4 . 5 = 1 . 08 kN/m h 4.52 2 bc = 1.08 kN/m

Linear convective pressure distribution

ac = 0.57 kN/m2


Solution 6/ slide 8

Part III: Solution c) Pressure due to vertical acceleration Zone III, Z = 0.16 I = 1.5 It is ground supported RC tank with fixed base, hence, R = 2.0 For vertical mode, T = 0.3 Sec,

∴ Sa/g = 2.5

Av = 2/3 . Z/2. I/R . Sa/g = 2/3 x 0.16/2 x 1.5/2.0 x 2.5 = 0.10 Maximum pressure due to vertical acceleration occurs at y=0, i.e., at the base of wall, ∴ pv = (Av) ρ g h (1- y/h) = 0.10 x 1000 x 9.81 x 4.5 x (1- 0/4.5) = 4.42 kN/m2 d) Pressure due to wall inertia, pww = (Ah)i t ρm g = 0.15 x 0.2 x 25 = 0.75 kN/m2 © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006

Solution 6/ slide 9

Part III: Solution Total pressure,

p=

=

(p

iw

+ p ww

(6 . 60

)

2

+ p cw + p v 2

2

+ 0 . 75 ) + 0 . 57 2 + 4 . 42 2 = 8.59 kN/m 2 2

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


Part III: Solution

a) Impulsive mode ( A h )i m i 0 . 15 × 230 ,040 × 9 . 81 qi = g = = 14 . 10 kN/m 2B 2 × 12 Pressure at bottom & top is given by

14 . 10 qi (4 × 4 .5 − 6 × 1 .69 ) = 5 .47 kN/m a i = 2 (4 h − 6 hi ) = 2 4 .5 h qi 14.10 2 ( ) bi = 2 (6hi − 2h ) = 6 × 1 . 69 − 2 × 4 . 5 = 0 . 79 kN/m h 4.52 b = 0.79 kN/m2

2

i

Linear impulsive pressure distribution

ai = 5.47 kN/m2



Part III: Solution

b) Convective mode ( A h )c m c 0 . 037 × 112 ,104 × 9 . 81 qc = g = = 1 . 70 kN/m 2B 2 × 12 Pressure at bottom & top is given by

1 . 70 qc (4 × 4 .5 − 6 × 2 .93 ) = 0 .035 kN/m a c = 2 (4 h − 6 h c ) = 2 4 .5 h qc 1.70 2 ( ) bc = 2 (6hc − 2h ) = 6 × 2 . 93 − 2 × 4 . 5 = 0 . 72 kN/m h 4.52 2

2

bc = 0.72 kN/m

Linear convective pressure distribution

ac = 0.035 kN/m2



Part III: Solution c) Pressure due to vertical acceleration Zone III, Z = 0.16 As it stores water, I = 1.5 It is ground supported RC tank with fixed base, R = 2.0 For 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 is occur at y=0, i.e., at the base of wall, ∴ pv = (Av) ρ g h (1- y/h) = 0.10 x 1000 x 9.81 x 4.5 x (1- 0/4.5) = 4.42 kN/m2 d) Pressure due to wall inertia, pww = (Ah)i t ρm g = 0.15 x 0.2 x 25 = 0.75 kN/m2 © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006


Part III: Solution Total pressure

p=

=

(p

+ p ww

iw

(5 .47

2

)

+ p cw + p v

2

2

+ 0 . 75

)

2

+ 0 . 035

2

2

+ 4 . 42 2

= 7.62 kN/m 2 At the base of wall pressure in kN/m2 are: 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



Assignment 7 Date of assignment: 17 February E-course on Seismic Design of Tanks

1


Read Examples 2 and 3 of the Guideline



Part II: True / False 7.1) If braces are treated as rigid beams, then seismic forces are overestimated . 7.2) Inclusion of shear deformation, reduces time period of shaft staging. 7.3) Effect of shear deformation on lateral stiffness is more important in shafts with large height-to-diameter ratio. 7.4) If stiffness decreases by 1.44 times then, time period will increase by 1.2 times. 7.5) Empty tank does not have convective mode of vibration.



Part III: Questions 7.1) 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. 3m 4m

3m

4m 4m 4m

Plan

4m

Elevation



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.



Solution 7 Date of assignment: 17 February Date of solution: 20 February E-course on Seismic Design of Tanks 1

Part II: True / False 7.1) If braces are treated as rigid beams, then seismic forces are overestimated . True If 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. False With the inclusion of shear deformation, lateral stiffness decreases and time period which is inversely proportional to square root of stiffness, increases.


Solution 7/ slide 2

Part II: True / False 7.3) Effect of shear deformation on lateral stiffness is more important in shafts with large height-to-diameter ratio. False The lateral stiffness is given by Ks =

1 L3 L + 3 EI 0 . 5 AG

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 and effect of shear deformation is low. This is also explained in the solution of problem no. 7.2 of part III of this assignment.


Solution 7/ slide 3

Part II: True / False 7.4) If stiffness decreases by 1.44 times then, time period will increase by 1.2 times. True Time 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. True In empty tank, no water, hence no convective mode.


Solution 7/ slide 4

Part III: Solutions 7.1) 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. 3m 4m

3m

4m 4m 4m 4m

Plan

Elevation


Solution 7/ slide 5

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 circular ring 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.


Solution 7/ slide 6

Part III: Solutions 10 kN Rigid link

Radial beam

Column Brace


Solution 7/ slide 7

Part III: Solutions

For lateral force of 10 kN, deflection = 0.00267 m ∴ Lateral stiffness of staging, Ks = 10/0.00267 = 3,745 kN/m

Deflection of staging © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006

Solution 7/ slide 8

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

KC = 12EI L3 E = Modulus of elasticity, I = Moment of inertia and L = Length of panel.


Solution 7/ slide 9

Part III: Solutions In the present example, Grade of concrete is M 25 Hence, 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,

12 × 25 × 10 6 × 1 .2 6 × 10 - 3 = 5 906 .3 kN/m K C = 12EI = 3 L3 ⎛ 4.0 ⎞ ⎜ ⎟ ⎝

⎠

In each panel, there are 8 columns. These eight columns act like eight springs in parallel. Hence, stiffness of each panel

Kp = ∑ K C = 8 × 5906.3 = 47250 kN/m Stiffness of one panel = 47,250 kN/m © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006


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 1 1 1 1 1 1 = + + + + Ks Kp Kp Kp Kp Kp

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

1 1 1 = + K s K flexure K axial Where, 1 K flexure

=

Np

∑K i =1

1

,

K panel

panel

⎤ ⎡ 12EI c N c ⎢ Ib L ⎥ = 3 − ⎥ ⎢ h I L I h + α c ⎦ ⎣ b

α = 1.0 for end panels and α = 2.0 for intermediate panels 1 K axial

2 = N c A cE R 2

Np

∑H i =1

2 i

h



Part III: Solutions

E = Modulus of elasticity Ac = Area of one column Nc = Number of columns Np = Number of panels L = Length of each brace beam R = Radius of staging system Ic = Moment of inertia of column Ib = Moment of inertia of brace beam Icbr = Moment of inertia of circular ring beam h = Height of each panel Hi = 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 the restrained end. 3I y =

b + Ic L h h 6I b L

For 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.

Top most panel

Panel 1 Panel 2

Intermediate panels

Panel 3 Panel 4

Bottom most panel


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 = 5 L = Length of each brace beam = 3000 x sin (22.50) x 2 = 2,296 mm R = 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 3I 3 × 1.6 × 10 9 1.26 × 10 9 b + Ic + 2296 4000 h L h = y = × 4000 = 2301 mm 9 6I 6 × 1.6 × 10 b L 2296

Distance of point of inflection of each panel from the load H1 = 3000 + 2301 = 5,301 mm H2 = 3000 + 4000 + 2000 = 9,000 mm H3 = 3000+ 8000 + 2000 = 13,000 mm H4 = 3000 + 12000 + 2000 = 17,000 mm H5 = 3000 + 20000 – 2301 = 20,699 mm © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006


Part III: Solutions Now, Kaxial is given by

1 K axial 2 Nc A c E R 2

N 2 2 = H ∑ 2 i =1 i h Nc AcE R p

=

2 − 18 = 8 . 84 x 10 8 x125664 x 25000 x (3000) 2

Np

2 ∑ Hi h = ((5301) 2 + (9000) 2 + (13000) 2 + (17000) 2 + (20699) 2 ) x 4000 i =1

∴

= 3 . 98 x 10 1

K axial

12

= 8.84 x10 −18 x 3.98 x1012 = 3.52 x10 − 5 mm / N



Part III: Solutions Now, let us find Kpanel For the upper most panel, ( α = 1.0 ) K panel

⎤ 12 EI c N c ⎡ I cbr L = ⎥ = 4.5 x 104 N/mm. ⎢ 3 h ⎢⎣ I cbr L + α I c h ⎥⎦

Similarly, for bottom most panel, ( α = 1.0 ) K panel

⎤ 12 EI c N c ⎡ Ib L = ⎢ ⎥ = 3.25 x 104 N/mm. 3 h ⎢⎣ I b L + α I c h ⎥⎦

Similarly, for intermediate 3 panels, ( α = 2.0 ) K panel =

12 EI c N c h3

⎡ ⎤ Ib L ⎢ ⎥ = 2.48 x 104 N/mm. ⎢⎣ I b L + α I c h ⎥⎦



Part III: Solutions 1

=

K flexure

1 K flexure

=

Np

1

i =1

panel

∑K

1 1 2 2 2 + + + + 4.5 ×10 4 3.25 ×10 4 2.48 ×10 4 2.48 ×10 4 2.48 ×10 4

= 2.95 x 10-4 mm/N Now, Ks is obtained as 1 = Ks K

1 flexure

+

1 K

= 2.95 × 10 − 4 + 3.52 × 10 −5 = 3.302 x10 − 4 mm / N

axial

∴ Ks = 3,029 N/mm = 3,029 kN/m. © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006


Part III: Solutions

Thus, stiffness of staging using these three approaches is a) From computer analysis of 3-D model

Ks = 3,745 kN/m b) From approximate analysis considering braces as rigid beams

Ks = 9,450 kN/m c) 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 computer analysis 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 Ks =

3EI L3

If shear deformations are considered, then Ks =

1 L3 L + 3EI κ' AG



Part III: Solutions Grade of concrete is M 25. Hence, Modulus of elasticity, E = 5000 x (25)0.5 = 25,000 N/mm2. = 25000 x 103 kN/m2 G = Shear modulus = E/(2 x (1 + ν)) where, ν = Poisson’s ratio = 0.2 ∴ G = E/(2 x (1 + ν)) = 25000 x 103 /(2 x (1 + 0.2)) = 10,417 x103 kN/m2 I = Moment of inertia of shaft cross section = π x (6.44 – 6.04)/64 = 18.74 m4 A = Area of cross section of shaft = π x (6.42 – 6.02)/4 = 3.90 m2 κ’ = 0.5 (for hollow circular cross section) © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006


Part III: Solutions Now, for shaft with L = 15 m: Ks without considering shear deformation Ks =

3EI L3

3x25000x10 3 x18.74 3 = = 416 . 4 x 10 kN / m 3 15

Ks with considering shear deformation Ks =

=

1 L3 L + 3EI κ' AG

1 3

15 15 + 3x25000x10 3 x18 .74 0.5x3.9x10 417x10 3

= 318 .5 x10 3 kN / m



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 Height Height-toLateral stiffness, Ks (kN/m) of shaft diameter % Reduction without shear with shear ratio deformation deformation 15 m

2.5

416.4x103

318.5x103

23.5 %

20 m

3.3

175. 7x103

149.7x103

14.8 %

25 m

4.2

90x103

81x103

10 %

Note: Thus, for shafts with large height-to-diameter ratio, shear deformation has less effect on the lateral stiffness. © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006


IS 1893 - TANKS & WATER RETAINING

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