IS 1893 - TANKS & WATER RETAINING

Lecture 1January 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) © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 3 Types of tanks Ground supported tanks Shape: Circular or Rectangular Material : RC, Prestressed Concrete, Steel These are ground supported vertical tanks Horizontal tanks are not considered in this course © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 4 Types of tanks Elevated tanks Two parts: Container Staging (Supporting tower) © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 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 © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 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 © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 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 © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 8 Indian Codes on Tanks 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 © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 9 Indian Codes on Tanks 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 © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 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 © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 11 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 E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 12 Hydrodynamic pressure Hydrodynamic pressure Has curvilinear variation along wall height Its direction is opposite to base motion Hydrodynamic pressure Base motion © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 13 IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 14 .Hydrodynamic pressure Summation of pressure along entire wall surface gives total force caused by liquid pressure Net hydrostatic force on container wall is zero Net hydrodynamic force is not zero © Sudhir K. Jain. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 15 . hydrodynamic is asymmetric © Sudhir K.Hydrostatic pressure is axisymmetric.Hydrodynamic pressure Circular tanks (Plan View) Hydrostatic pressure Hydrodynamic pressure Base motion Net resultant force = zero Net resultant force ≠ zero Note:. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 16 . Jain.Hydrodynamic pressure Rectangular tanks (Plan View) Hydrostatic pressure Hydrodynamic pressure Base motion Net resultant force = zero Net resultant force ≠ zero © Sudhir K. causes no overturning moment on foundation or staging Thus. hydrostatic pressure affects container design only and not the staging or the foundation © Sudhir K. Jain.Hydrodynamic pressure 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. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 17 . staging and foundation © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 18 . Jain.Hydrodynamic pressure Seismic design: Hydrodynamic pressure is considered Net hydrodynamic force on the container is not zero Affects design of container. TID 7024. Nuclear Reactors and Earthquakes. Atomic Energy Commission.Hydrodynamic pressure Procedure for hydrodynamic pressure & force: Very simple and elegant Based on classical work of Housner (1963a) Housner. U. We need not go in all the details Only basics and procedural aspects are explained in next few slides © Sudhir K. 1963a. Washington D. W. Report No. “Dynamic analysis of fluids in containers subjected to acceleration”. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 19 .. G.C. S. Jain. Jain.Modeling of liquid Liquid in bottom portion of the container moves with wall 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) This is called impulsive liquid Impulsive liquid (moves with tank wall) © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 20 . rigidly attached Has same acceleration as wall Convective liquid 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 © Sudhir K.Modeling of liquid Impulsive liquid Moves with wall. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 21 . Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 22 .Modeling of liquid Impulsive Base motion Convective Base motion Hydrodynamic pressure © Sudhir K. Modeling of liquid At this point. we will not go into details of hydrodynamic pressure distribution Rather. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 23 . we will first find hydrodynamic forces Impulsive force is summation of impulsive pressure on entire wall surface Similarly. Jain. convective force is summation of convective pressure on entire wall surface © Sudhir K. 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 E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 24 © Sudhir K. mi Convective liquid mass. IIT Kanpur . gets divided into two parts: Impulsive liquid mass.Modeling of liquid Total liquid mass. m. Jain. Jain.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. mi as fraction of m is more For short tanks. mc as fraction of m is more © Sudhir K. 2a and 3a of Guidelines Also see next slide For taller tanks (h/D or h/L higher). IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 25 . 5 mc /m 0.5 mc /m 0 0 0. Jain.5 h/D 1 1. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 26 .5 2 0 0 0.5 h/L 1 1. and L © Sudhir K. D.5 2 For circular tanks For rectangular tanks See next slide for definition of h.Modeling of liquid 1 mi/m 1 mi/m 0. Jain.Modeling of liquid h D Elevation Base motion Plan of Circular tank L L Base motion Plan of Rectangular tank © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 27 . mass density of water is 1000 kg/m3. D = 8 m.0 = 150.Modeling of liquid Example 1: A circular tank with internal diameter of 8 m. h = 3 m ∴ h/D = 3/8 = 0. Solution: Total volume of liquid = π/4 x 82 x 3 = 150. Jain. © Sudhir K. m = 150.81 x 1000 = 9810 N/m3.375. weight density of water is 9. stores 3 m height of water.8 x 1. Find impulsive and convective water mass.8 t Note:. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 28 .8 m3 ∴ Total liquid mass. 5 mc /m 0 0 0. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 29 .5 2 © Sudhir K. Jain.5 h/D 1 1.1 mi/m 0. 3 t and mc = 0.8 = 63.Modeling of liquid From graph. for h/D = 0.42 x 150.42 and mc/m = 0.56 x 150. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 30 .8 = 84.375 mi/m = 0.56 mi = 0.5 t © Sudhir K. Jain.Modeling of liquid Impulsive liquid is rigidly attached to wall Convective liquid moves relative to wall As if. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 31 . attached to wall with springs Kc/2 Kc/2 mc Convective liquid (moves relative to wall) Rigid m i Impulsive liquid (moves with wall) © Sudhir K. Modeling of liquid Stiffness associated with convective mass. Kc Kc depends on aspect ratio of tank Can be obtained from graph Refer Fig. 2a. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 32 . Jain. 3a of guidelines See next slide © Sudhir K. 5 1 1.Modeling of liquid 1 Kch/mg mi/m 0.5 2 h/D © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 33 . Jain.5 mc/m 0 0 0. for h/D = 0. m = 150.Modeling of liquid Example 2: A circular tank with internal diameter of 8 m.8 t (from Example 1) = 150. Jain.375. Solution: Total liquid mass. h = 3m ∴ h/D = 3/8 = 0. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 34 . From graph.65 © Sudhir K.8 x 1000 = 150800 kg g = acceleration due to gravity = 9.375. Kc h/mg = 0. Find Kc. stores 3 m height of water.81 m/sec2 D = 8 m. hence unit of Kc is N/m.65 x150800 x 9. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 35 .Modeling of liquid 1 Kch/mg mi/m 0.5 mc/m 0 0 0.65 mg/h ∴ Kc = 0. then unit of Kc will be kN/m.4 N/m Note: . If we take m in ton.0 = 320. Jain.5 1 1.81/3.525. © Sudhir K.5 2 h/D Kc = 0.Unit of m is kg. 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.Modeling of liquid Now. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 36 . convective mass © Sudhir K. we know liquid masses mi and mc Next. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 37 .Modeling of liquid Impulsive mass acts at centroid of impulsive pressure diagram Location of centroid: Obtained by dividing the moment due to pressure distribution by the magnitude of impulsive force Similarly. Jain. location of convective mass is obtained See next slide © Sudhir K. 2b. Jain. 3b of guidelines See next slide E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 38 © Sudhir K. IIT Kanpur . hc can be obtained from graphs They also depend on aspect ratio.Modeling of liquid Resultant of impulsive pressure on wall Resultant of convective pressure on wall hc hi hi. h/D or h/L Refer Fig. 4 0.8 hc/h 0.6 1.8 2 0 0 0.4 0.2 0 0 0.5 1 h/L 1.2 0.6 0. Jain.4 1.8 1 h/D 1.8 0.Modeling of liquid 1 1 0.6 hc/h hi/h hi/h 0.4 0.6 0.2 1.2 0. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 39 .5 2 For circular tanks For rectangular tanks © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 40 . Find hi and hc. h = 3m ∴ h/D = 3/8 = 0. stores 3 m height of water. Solution: D = 8 m.375.Modeling of liquid Example 3: A circular tank with internal diameter of 8 m. Jain. © Sudhir K. Jain.6 0.4 hi/h 0.8 hc/h 0.5 1 h/D 1.1 0.2 0 0 0.5 2 © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 41 . hi/h = 0. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 42 .375 hi = 0.55 hc = 0. Jain. hc > hi.Modeling of liquid From graph. © Sudhir K.375.65 m Note :.Since convective pressure is more in top portion.375 x 3 = 1.125 m and hc/h = 0. for h/D = 0.55 x 3 = 1. Jain.Modeling of liquid Hydrodynamic pressure also acts on base Under static condition. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 43 . liquid exerts nonuniform pressure on base This is in addition to the hydrostatic pressure on the base See next slide © Sudhir K. base is subjected to uniformly distributed pressure Due to base motion. Jain.Modeling of liquid Base motion Hydrostatic pressure on base Hydrodynamic pressure on base © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 44 . 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 © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 45 . Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 46 . Jain.Both the overturning effects are in the same direction © Sudhir K.Modeling of liquid hi Overturning effect due to wall pressure Overturning effect due to base pressure Note:. hi* and hc*.Modeling of liquid Total overturning effect of wall and base pressure is obtained by applying resultant of wall pressure at height. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 47 . Jain. • In place of hi and hc discussed earlier For overturning effect due to wall pressure alone. see next slide © Sudhir K. resultant was applied at hi For hi and hi*. 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 © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 48 . Jain. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 49 . 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 © Sudhir K.Modeling of liquid Similarly. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 50 .Modeling of liquid hi and hi* are such that Moment due to impulsive pressure on walls only = Impulsive force x hi Moment due to impulsive pressure on walls and base = Impulsive force x hi* hc and hc* are such that Moment due to convective pressure on walls only = Convective force x hc Moment due to convective pressure on walls and base = Convective force x hc* © Sudhir K. 2b & 3b of guidelines Also see next slide Please note. Jain. hc. C-1 of the Guidelines hi* & hc* depend on aspect ratio Graphs to obtain hi. hi* and hc* can be greater than h © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 51 . hc* are provided Refer Fig. hi*.Modeling of liquid hi* is greater than hi hc* is greater than hc Refer Fig. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 52 .5 1 1.5 1 0.5 2 hc/h hi/h hi*/h h/D © Sudhir K.Modeling of liquid 2.5 0 0 0. Jain.5 hc*/h 2 1. 375.Modeling of liquid Example 4: A circular tank with internal diameter of 8 m. Solution: D = 8 m. From graph. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 53 . © Sudhir K.375. Jain. h = 3m ∴ h/D = 3/8 = 0. stores 3 m height of water. for h/D = 0. Find hi* and hc*. 1 x 3 = 3. Jain.0 x 3 = 3.5 2 1. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 54 .1 Hence hi* = 1.5 1 hc/h 0.5 hi /h 0 0 0. hc*/h = 1.3 m Similarly.5 2 hi */h hc*/h h/D hi*/h = 1.5 1 1.Modeling of liquid 2. hc* = 1.0 m © Sudhir K.0 Hence. mi & mc This is called mechanical analogue or spring mass model for tank See next slide © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 55 . Jain.Modeling of liquid This completes modeling of liquid Liquid is replaced by two masses. Jain.Modeling of liquid mi = Impulsive liquid mass Kc/2 mc Kc/2 mc = Convective liquid mass Kc = Convective spring stiffness hi = Location of impulsive mass (without considering overturning caused by base pressure) hc = Location of convective mass (without considering overturning caused by base pressure) hi* = Location of impulsive mass (including base pressure effect on overturning) hc* = Location of convective mass (including base pressure effect on overturning) © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 56 Rigid m i hi (hi*) hc (hc*) Mechanical analogue or spring mass model of tank . IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 57 . Jain. hi. 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 © Sudhir K. Kc. hc. mc.Modeling of liquid mi. 75 hi * = h 0. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 mg h⎞ ⎛ tanh 2 ⎜ 3. 09375 h / D for h / D > 0 .23 h m D h⎞ ⎛ cosh ⎜ 3.68 ⎟ D D⎠ ⎝ = 0.68 sinh⎜ 3.68 sinh ⎜ 3.68 ⎟ − 1. 375 h = 0 . 75 0 .836 © Sudhir K.0 hc D⎠ ⎝ = 1− h h⎞ h ⎛ 3.5 − h⎞ ⎛ tanh ⎜ 3.125 for h / D ≤ 1.33 h⎞ ⎛ cosh⎜ 3.45 for h / D > 1.68 ⎟ mc D⎠ ⎝ = 0.0.01 hc * D⎠ ⎝ =1− h h h⎞ ⎛ 3. Jain.866 ⎟ h⎠ ⎝ D h .68 ⎟ − 2.68 ⎟ D D⎠ ⎝ for h / D ≤ 0 .Modeling of liquid For circular tanks D⎞ ⎛ tanh ⎜ 0 .866 D⎞ ⎛ 2 tanh⎜ 0.33 K c = 0. 866 ⎟ mi h ⎠ ⎝ = D m 0 .68 ⎟ h D⎠ ⎝ Lecture 1 / Slide 58 . 866 h hi = 0 . 33 for h / L > 1.16 ⎟ − 2. 75 hi * = h 0.264 h m L h⎞ ⎛ cosh ⎜ 3.16 sinh⎜ 3.16 ⎟ L L⎠ ⎝ = 0 .866 L h L⎞ ⎛ 2 tanh⎜ 0.33 h⎞ ⎛ cosh⎜ 3.Modeling of liquid For rectangular tanks L⎞ ⎛ tanh⎜ 0.16 ⎟ − 1. 375 h = 0 .833 mg h⎞ ⎛ tanh 2 ⎜ 3.5 − 0 .16 ⎟ h L⎠ ⎝ Lecture 1 / Slide 59 © Sudhir K.0 hc L⎠ ⎝ = 1− h h⎞ h ⎛ 3. Jain.16 ⎟ mc L⎠ ⎝ = 0.16 sinh ⎜ 3.866 ⎟ h⎠ ⎝ − 0.866 h h⎞ ⎛ tanh⎜ 3. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 .01 hc * L⎠ ⎝ = 1− h h h⎞ ⎛ 3.125 for h / L ≤ 1.866 ⎟ mi h⎠ ⎝ = L m 0. 75 for h / L > 0 . 09375 h/L for h / L ≤ 0 .45 K c = 0.16 ⎟ L L⎠ ⎝ hi = 0 . IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 60 .Modeling of liquid Note. in Table C-1 of the Guideline. there shall be minus sign before 0.75” For circular tank. in the expression for hi*/h. Jain. there are two typographical errors in these expressions For circular tank. first expression for hi/h shall have limit as “for h/D ≤ 0.125 These two errors have been corrected in the expressions given in previous two slides © Sudhir K. Vi = mi x acceleration Convective force.Modeling of liquid mi and mc are needed to find impulsive and convective forces Impulsive force. Jain. Vc = mc x acceleration Kc/2 Kc/2 mc Vc Vi Rigid m i © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 61 . Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 62 .Modeling of liquid Vi and Vc will cause Bending Moment (BM) in wall © Sudhir K. 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. Jain. later Kc/2 mc Kc/2 Vc hc hi © Sudhir K. IIT Kanpur Rigid mi Vi E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 63 . Jain.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 © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 64 . Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 65 .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 mc Kc/2 Vc hc* hi* Rigid mi Vi © Sudhir K. hc.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. hi. later Use of mi. mc. hi* and hc* in next example Acceleration values are assumed © Sudhir K. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 66 . D = 8 m.3 t mc = 84.5 t hi = 1. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 67 .1g = 0. Assuming impulsive mass acceleration of 0.98 m/sec2 © Sudhir K. ai = 0.65 m hi* = 3. Jain. ac = 0. find seismic forces on tank.81 = 0.3 x 9.125 m hc = 1.1 x 9.3 m hc* = 3. h = 3m From previous examples: mi = 63.Modeling of liquid Example 5: A circular tank with internal diameter of 8 m.3g and convective mass acceleration of 0.81 = 2.0 m Impulsive acceleration.3g = 0.94 m/sec2 Convective acceleration. stores 3 m height of water.1g. Solution: Geometry of tank is same as in previous examples. 3 = 614.4 kN-m © Sudhir K. Jain.0 = 248.94 = 186.4 kN-m Bending moment at bottom of wall due to Vc = Vc x hc = 82. Vi = mi x ai = 63.65 = 136.8 kN Bending moment at bottom of wall due to Vi = Vi x hi = 186..8 x 1.1 x 3.3 x 2.125 = 209.1 kN Convective force.6 kN-m Overturning moment at bottom of base due to Vi = Vi x hi* = 186.1 kN-m Overturning moment at bottom of base due to Vc = Vc x hc* = 82.98 = 82.1 x 1.5 x 0.Modeling of liquid Example 5 (Contd. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 68 . Vc = mc x ac = 84.) Impulsive force.8 x 3. Jain.At the end of Lecture 1 In seismic design. mechanical analogue of tanks are used. wherein. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 1 / Slide 69 . 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) © Sudhir K. Depending on convenience. table C-1 of guidelines as pointed in lecture1/slide57 3. mi/m. show to calculate hydrodynamic pr. Expressions for hi* and hc* for circular as well as rectangular tanks are also given in Table C-1. K. Umesh H. Expressions for these quantities are given in Table C-1 of the Guideline. fig 2a.parameters for tapering sections of wall/container . This is known as hydrostatic pressure. liquids apply a pressure on the bodies in contact. Hydrostatic pressure induces hoop forces and bending of the wall. Patil ([email protected] (through Distance Learning Mode) on Seismic Design of Liquid Storage Tanks January 16 . Direction of this pressure is always normal (perpendicular) to the surface of the body.something about hydrostatic design Under static conditions. Kukreja (guduja@rediffmail. IS 3370:1967 gives these forces for circular and straight walls. which is Figure 2b of the Guideline. one can find these quantities either from graphs or from expressions. mc/m and Kch/mg are plotted as a function of h/D.3a of guidelines as pointed in lecture1/slide25 2. In the Lecture we have given graphs of hi* and hc* only for circular tanks (Slide 54). Housner graphs to find h* are not given for rectangular tanks In Figure 2a of the Guideline. 2.in) 1. pl. 2006 Response to Questions and Comments on Lecture 1 Question from Mr. Refer standard textbooks for more information.com) 1. graphs plotted in these figures are based on these expressions only. and it varies linearly with height. Question from Mr. clarify following 1. This is for circular tanks and corresponding figure for rectangular tanks is given in Figure 3a. This pressure is a product of unit weight of liquid and depth of the liquid at the point under consideration.co. Basically. which has not been reproduced in the Lecture.February 6. Corresponding figure for rectangular tank is Figure 3b of the Guideline. S. expressions are also given. i. Another good example is heating greasers in bathrooms. Railway tankers carrying petroleum products are also horizontal tanks. some are of horizontal type and some are of vertical type !!. its thickness varies along the height. however. longitudinal axis of tank is horizontal is termed as horizontal tank. 4. conical. hydrodynamic pressure is obtained considering wall as rigid. However. then there will not be any change in hydrodynamic pressure. etc. H.. 1) What exactly does Horizontal Tank and Vertical Tank mean A circular tank kept horizontally on ground. 3.. If cross-section of wall is tapered. Abdul Gani (gani@igcar. 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. effect of top covered slab on hydrodynmic pressure Top slab will impart more stiffness to wall and make it more rigid. Classification as horizontal and vertical tank is more appropriate for circular tanks only.ernet. then hydrodynamic pressure distribution will be affected. This is not going to affect hydrodynamic pressure distribution.Lecture 3 has covered information on tanks of other shapes such as Intze. in which longitudinal axis of tank is vertical.in) The following are my questions on Lecture 1.e. Question from Mr.e. In petrol pumps. The difference. may not be significant from engineering view point. why the sum of impulsive and convective masses (147=2E8kg) is less than the total mass of water (150.11 of the Guideline. Recall. there is difference in answers from graph & equations Reading the value from graph is likely to induce some approximation and hence. steel tanks used for storing petrol/diesel are usually horizontal tanks. A vertical tank is the one. i. I. if sloshing liquid touches roof slab. 2) It was shown that the hydrodynamic pressure is curvilinear in shape. Please refer Lecture 6 and Commentary to Clause 4.8 kg)? This also has been addressed in Lecture 3 . This increase or decrease in pressure is the hydrodynamic effect. which in turn will change hydrostatic pressure on wall. liquid is divided into two parts. then. 9) Why effect of base pressure is considered along with both convective mass and impulsive mass instead of convective mass only? In our modeling. The decrease in the pressure is like an upward pressure. how to find impulsive and convective mass? Information on effect of vertical excitation is covered in Lecture 6. Also read Clause 4. Vertical excitation will cause change in weight density of liquid. liquid exerts uniform pressure on base. 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. both impulsive and convective liquids exert non-uniform pressure on base. masses shall be lumped at h. how to find impulsive and convective masses? Any simplified methods available Please refer Lecture 3 8) During vertical acceleration. one that vibrates along with wall (impulsive mass) and other which moves relative to wall (convective mass). During lateral excitation. it is not clear how pressure can develop in the upward direction? Can you please explain Under static condition.5) In the hydrodynamic pressure profile in the 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. If overturning moment at the bottom of base slab is to . h or h*? If bending moment at the bottom of wall is to be obtained. During lateral excitation. the liquid increases vertical pressure on some portion of slab and reduces pressure on other portion.10 of the Guideline. in) 1. This can also be observed in the pressure distributions shown in Lecture 6.078 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. How convective pressure is distributed and upto what depth ? We need to recognize that we are dividing liquid in two parts.078/2. 2. we have not included this in the course contents. and convective and impulsive pressures are present over the entire height of liquid.com) I didnt understood the following pl explain mi/m=3D tanh(. Question from (cshekhar66@rediffmail. R. 12) For load application in a mathematical model. 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. only C.078/2. Murugan (rmurugan@igcar. The derivation of this formula is given in Housner’s paper (1963a) and you may go through the same. then masses shall be lumped at h*. mass must be lumped at h*. It is an idealization. For design of staging and foundation also.866*12/5)/.ernet. which is not present in the expressions given in Lectures and in Table C-1 of the Guideline. Since it is a bit mathematical. It comes from potential equation of liquid or Laplace equation.obtained. (h) of the impulsive/ convective force is known. Question from Mr. 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 .G.078 mi/m=3Dtan 5*2.866*12/5 now in this expression mi/m=3Dtanh 2.39 and then dividing it by 2.078 is it finding tan 10. 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. You are also interested in knowing how this formula is arrived at. due to presence of roof slab. Generally. wherein. and we have sent addendum to it. The formulas and graph derived to find hi. Please refer Lecture 3 and commentary of Clause 4. Effect of flexibility of wall on pressure distribution has been studied by Veletsos (1984). tank with cover slab or without cover slab) Actually these are derived for open tanks. mi × g) . . this height can become larger than h. In this context. for design purpose we use formulae derived for rigid walls. hc. effect of moment due to base pressure is included. one can say that a tank with h/D less than 0. For example. Since. we can see that for h/D = 0. This was error in our solution. in short tank mi is less only See Example-1 in Lecture -1. provided sloshing liquid does not touch roof slab. Thus. is always less than h. However. 3.e.2 is squat tank and a tank with h/D greater than 1..0 is slender tank.2 is false in Assignment i. 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. On the other hand hi. impulsive mass is less.2 5.e.2. 4. 6. pressure distribution on wall is not going to change. slender tank? Slender tank is one in which h/D or h/L is quite high.11 of the Guideline. we say that in slender tanks. about 80% mass contributes to convective mode and for h/D = 1. from Figure 2a of the Guideline. How Question 1. There is no fixed limit on these values at which this demarcation begins.0 about 80% mass contributes to impulsive mode. these terminologies are used for impulsive and convective masses. which includes effect of impulsive pressure only on the walls. mi.moment due to mi (actually. However. What is mean by squat tank.1.2. The formulae derived for the above tanks are applicable for open tank or closed tank ? (i. These issues have been addressed in Lecture 6 and in Clause 4.. In squat tank h/D or h/L is less. question of liquid touching roof slab does not arises. If it touches roof slab then some additional issues will come up. convective mass is less and in squat tanks. 75 is to be given. Abdul Gani (gani@igcar. Nearly 3% of the liquid mass is unaccounted. 60). while the actual one furnished reads.. 8. far we learnt about impulsive mass. This also has been pointed out in Lecture 1 (Slide no. Why is it so? Is it not correct to include the missing mass in impulsive liquid component? This has been discussed in Lecture 3.5*x* h. 9.969.ernet. Question from Mr. impulsive pressure distribution as below: Impulsive Let us assume mi/m =x: If the pressure distribution is as shown in the above figure. So. hi should come less than or equal to 0. which we have pointed out in the Lecture 1. convective mass and it is centeroid. Table below compares the hi expected versus actual computed .503. i. Table C1-Expression for parameters of spring mass model.7. When we do the modeling in computer. when both are added the value is 0. How impulsive mass is distributed i.5x. hi/h = 0. hi/h should always be lower than 0. There is another typographical error in the expression for hi* for circular tanks. mc/m = 0. h/d greater than 0.e.375 for h/D less than or equal to 0.466.in) It is explained in lecture 1. Why mi + mc is not coming to m ? In solution 1.75 This is a typographical error in Table C-1.e how applied and upto what depth? Distribution of impulsive and convective pressure along wall height is covered in Lecture 6. mi/m = 0. 27 Less than or equal to 0. Similarly.72/2 0.375 0. This question arose because of the feeling that physically liquid got divided into two parts.Problem no. hi/h is always less than 0.5. but more liquid from upper portion participates in convective mode. Impulsive liquid is present throughout the height.54/2 0. Similarly. however. Please refer Figure 2b and 3b of the Guideline.4 From above computations.36 hi/h computed 0.466 0. hc/h is always greater than 0.54 hi/h expected Less than or equal to 0. there is no physical line.5h. convective liquid is also distributed along the entire height. Since more liquid from bottom portion is participating in impulsive mode.2 Rectangular tank in shorter direction mi/m 0.375 0. . more liquid from bottom portion participates in impulsive mode. 1. As has been explained in the answer to earlier question. which demarcates impulsive and convective portions of liquid.2 Rectangular tank in longer direction 1. You are thinking that liquid below ( x ) × (h ) height is impulsive and liquid above this height is convective.72 0.233 Less than or equal to 0.1 Circular tank 1. it appears that the pressure distribution for impulsive loading explained in lecture 1 appears to be incorrect. Kindly explain. 2006 .Lecture 2 January 19. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 2 .In this lecture Seismic force evaluation Procedure in codes Limitations of IS 1893:1984 © Sudhir K. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 3 .Seismic force evaluation During base excitation Structure is subjected to acceleration From Newton’s second law Force = mass x acceleration Hence. Jain. seismic force acting on structure = Mass x acceleration © Sudhir K. 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. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 4 . we need maximum seismic force Hence. © Sudhir K. PGA Maximum acceleration of rigid structure is same as PGA.Seismic force evaluation For design. Jain. Jain. IIT Kanpur Lecture 2 / Slide 5 .Seismic force evaluation Seismic force = mass x maximum acceleration Can be written as: Force = (maximum acceleration/g) x (mass x g) = (maximum acceleration/g) x W 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 E-Course on Seismic Design of Tanks/ January 2006 © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 6 .Seismic force evaluation Maximum acceleration of structure depends on Severity of ground motion Soil conditions Structural characteristics These include time period and damping More about time period. base shear coefficient. will also depend on these parameters © Sudhir K. later Obviously. Ah. Jain. Seismic force evaluation Seismic design philosophy is such that. base shear coefficient would also have a parameter associated with design philosophy © Sudhir K. Jain. 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. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 7 . Jain. base shear coefficient depends on: Severity of ground motion Soil condition Structural characteristics Design philosophy © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 8 .Seismic force evaluation Thus. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 9 .Seismic force evaluation Let us examine how following codes have included these parameters in base shear coefficient IS 1893 (Part 1): 2002 IS 1893:1984 International Building code (IBC) 2003 from USA Study of IBC provisions will help us understand the present international practice © Sudhir K. Jain. IS 1893 (Part 1):2002 Ah = (Z/2). (Sa/g) Z is zone factor I is importance factor R is response reduction factor Sa/g is spectral acceleration coefficient © Sudhir K. (I/R). IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 10 . Jain. 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. Jain.1 for zone II and Z = 0.IS 1893 (Part 1):2002 Zone factor.36 for zone V © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 11 . 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.IS 1893 (Part 1):2002 Importance factor. Jain. values will be given in Part 2 © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 12 . IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 13 .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. Jain. This depends on Ductility Redundancy Overstrength See next slide © Sudhir K. IS 1893 (Part 1):2002 Total Horizontal Load Linear Elastic Response Due to Ductility Non linear Response Due to Redundancy Due to Overstrength Δ Maximum force if structure remains elastic Fel Total Horizontal Load Maximum Load Capacity Fy Load at First Yield Fs First Significant Yield Design force Fdes Figure: Courtesy Dr. Jain. C V R Murty 0 Δw Δy Δmax Roof Displacement (Δ) Response Reduction Factor = © Sudhir K. IIT Kanpur Maximum Elastic Force (Fel) Design Force (Fdes) Lecture 2 / Slide 14 E-Course on Seismic Design of Tanks/ January 2006 . . R values will be given in IS:1893 (Part 2) © Sudhir K. building with SMRF has good ductility and has R = 5.) A structure with good ductility.5 for unreinforced masonry building which does not have good ductility Table 7 gives R values for buildings For tanks. redundancy and overstrength is designed for smaller seismic force and has higher value of R For example. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 15 .0 as against R = 1.IS 1893 (Part 1):2002 Response reduction factor (contd. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 16 .IS 1893 (Part 1):2002 Spectral acceleration coefficient. Sa/g Depends on structural characteristics and soil condition Structural characteristics include time period and damping Refer Fig. Jain. 2 and Table 3 of IS:1893 See next slide © Sudhir K. IS 1893 (Part 1):2002 For 5% damping © Sudhir K. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 17 . for higher damping.90 For higher damping.40 5 7 10 0.20 2 1.IS 1893 (Part 1):2002 For other damping. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 18 . Jain.60 25 0. Sa/g values are to be multiplied by a factor given in Table 3 of IS:1893 Table 3 is reproduced below % 0 damping Factor 3.70 20 0. multiplying factor is less Hence.50 1.00 0.55 30 0.80 15 0. Sa/g is less © Sudhir K. 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 © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 19 . Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 20 .IS 1893:1984 Ah= KCβIαo Seismic Coefficient Method (SCM) = KβIFoSa/g Response Spectrum Method (RSM) K is performance 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 © Sudhir K. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 21 . Jain.IS 1893:1984 Seismic coefficient. Fo Depends on severity of ground motion Used in response spectrum method © Sudhir K. αo Depends on severity of ground motion Used in seismic coefficient method Zone factor. IS 1893:1984 © Sudhir K. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 22 . type of foundation does not have any influence on base shear coefficient © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 23 .IS 1893:1984 β is soil foundation coefficient Depends on type of soil and foundation In IS 1893:2002. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 24 . I Ensures higher design seismic force for more important structures IS 1893 (Part 1):2002. Jain.IS 1893:1984 Importance factor. gives values only for buildings © Sudhir K. Jain. K Depends on ductility of structure Similar to response reduction factor of IS1893(Part 1):2002 K is in numerator whereas.IS 1893:1984 Performance factor.6 Thus.0 For ordinary buildings. R is in denominator For buildings with good ductility. a building with good ductility will have lower value of base shear coefficient than ordinary building © Sudhir K. K = 1. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 25 . K = 1. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 26 .IS 1893:1984 © Sudhir K. 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 © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 27 IS 1893:1984 Graphs for C and Sa/g from IS 1893:1984 Natural Period (Sec) Natural Period (Sec) © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 28 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 © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 29 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 © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 30 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 © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 31 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 © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 32 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 © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 33 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 © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 34 IBC 2003 Type of structure Building with special reinforced concrete moment resisting concrete frames Building with intermediate reinforced concrete moment resisting concrete frames Building with ordinary reinforced concrete moment resisting concrete frames Building with special steel concentrically braced frames Elevated tanks supported on braced/unbraced legs Elevated tanks supported on single pedestal Tanks supported on structural towers similar to buildings Flat bottom ground supported anchored steel tanks Flat bottom ground supported unanchored steel tanks Ground supported reinforced or prestressed concrete tanks with reinforced nonsliding base © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 R 8.0 5.0 3.0 8.0 3.0 2.0 3.0 3.0 2.5 2.0 Lecture 2 / Slide 35 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 © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 36 Base shear coefficient Important to note that: IS codes specify base shear coefficient at working stress level For limit state design, these are to be multiplied by load factors to get factored loads IBC specifies base shear coefficient at ultimate load level For working stress design, seismic forces are divided by a factor of 1.4 © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 37 Base shear coefficient 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 © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 38 Base shear coefficient 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 © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 39 4. I = 1.0. Jain.0.36. Zone V β = 1.0.0 Soil type D.6 I = 1. R = 8. Zone V I = 1. SD1 = 0. R = 5. Fo = 0.0 K = 1. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 40 . equivalent to soft soil of IS codes 5% damping They represent similar seismic hazard level © Sudhir K.0.Base shear coefficient Comparison for buildings Following parameters are chosen IS 1893 (Part 1): 2002 Z = 0. raft foundation 5% damping IBC2003 SDs = 1.08.0 Soft soil 5% damping IS 1893: 1984 αo = 0.0 Soft soil. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 41 . for buildings with SMRF.0 Refer Table shown earlier © Sudhir K. R = 8. buildings with SMRF In IBC.Base shear coefficient Building with good ductility is chosen Say. 08 IBC 2003 Ah = SDSI/(1.09 IS 1893:1984 C = 1.08 = 0.08 RSM: Ah = KβIFoSa/g = 1.3 sec IS 1893(Part 1):2002 Sa/g =2.0/5.0x 0.089 © Sudhir K.Sa/g = (0.5 Ah = Z/2. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 42 .0 x1.0) x 2.0/(1.Base shear coefficient For building with T = 0.0 x1.2 SCM: Ah = KCβIαo = 1.0 x 1. Jain.4 x 8.5 = 0.0 x1.0) = 0.I/R.0 x1.0x0.0 and Sa/g = 0.0 x 1.2 = 0.36/2 )x (1.4x0.4xR) = 1. 06 C = 0.11 = 0.0x1.6x1.0x1.36/2 )x(1.67 Ah = Z/2.0x1.67 = 0.0x0.4xRxT) = 0. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 43 . Jain.53x1.08 = 0.4 x 8.054 IBC 2003 © Sudhir K.Sa/g = (0.0 x0.044 Ah = SD1I/(1.11 SCM: Ah = KCβIαo = 1.0/5.0)x1.I/R.0x0.4x0.0 x1.042 RSM: Ah = KβIFoSa/g = 1.0/(1.53 and Sa/g = 0.Base shear coefficient For building with T = 1 sec IS 1893(Part 1):2002 IS 1893:1984 Sa/g = 1.0) = 0. 4 and Sa/g = 0.4 x 8. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 44 .Sa/g = (0.11 = 0. Jain.Base shear coefficient For building with T = 1.36/2 )x(1.0x1.0/5.0x1.032 RSM: Ah = KβIFoSa/g =1.0x1.5 sec IS 1893(Part 1):2002 IS 1893:1984 Sa/g = 1.4RT) = 0.078 SCM: Ah = KCβIαo = 1.0x0.11 Ah = Z/2.078 = 0.4x1.0 x0.0/(1.6x1.0)x1.036 IBC 2003 © Sudhir K.4x0.08 = 0.040 C = 0.0x0.I/R.5) = 0.031 Ah = SD1I/(1.0 x1. this value is higher Graphical comparison on next slide © Sudhir K.032 0.06 0.08 0.08 0.089 0.054 0.024 RSM 0.5 2.040 0. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 45 .0 1.0314* IBC2003 * Due to lower bound.024 0.031 0.Base shear coefficient Base shear coefficients for four time periods T (Sec) 0.042 0.3 1.044 0. Jain.036 0.03 IS 1893: 1984 SCM 0.09 0.0 IS 1893 (Part 1): 2002 0. 5 Time Period (S) 2 2.Base shear coefficient Comparison of base shear coefficient (Buildings) 0. SCM Base shear coefficient 0.1 IS 1893(Part 1):2002 IBC 2003 IS 1893:1984.5 3 © Sudhir K.075 Note the lower bound of IBC 0.05 0. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 46 .025 0 0 0. Jain.5 1 1. RSM IS 1893:1984. Base shear coefficient We have seen that: Codes follow similar strategy to obtain design base shear coefficient In similar seismic zones. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 47 . base shear coefficient for buildings matches reasonably well from these three codes © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 48 . let us compare design base shear coefficients for tanks From IS1893:1984 and IBC 2003 IS 1893(Part 1):2002 is only for buildings Hence.Base shear coefficient Similarly. can’t be used for tanks Only elevated tanks will be considered 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 © Sudhir K. Jain. base shear coefficient will be same for all types of elevated tanks © Sudhir K. Hence.0).Base shear coefficient In IBC I = 1.0 for tanks on shaft or pedestal In 1893:1984 I = 1. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 49 .5 for tanks K is not present in the expression for base shear coefficient (implies k=1.0 for tanks on frame staging (braced legs) R = 2. Jain.25 for tanks R = 3. 2 = 0.25.0 Ah = SDSI/(1.0 x1.4 x 2.298 For shaft staging.2 Ah = βIFoSa/g = 1.5.0 x1.25/(1. R = 2.Base shear coefficient For tank with T = 0. I = 1.4xR) = 1.4x0. Sa/g = 0.0) = 0.4xR) = 1.4 x 3.25/(1.0 x 1. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 50 . R = 3. Jain. I = 1.0 Ah = SDSI/(1.0) = 0.25.12 This value is common for frame and shaft staging IBC 2003 For frame staging.3 sec IS 1893:1984 I = 1.446 © Sudhir K.5 x 0. 0x1.4 x 3.5x0.Base shear coefficient For tank with T = 1 sec IS 1893:1984 IBC 2003 I = 1.4 x 2.4xRxT) = 0. Sa/g = 0. R = 2. R = 3.6x1. I = 1.6x1.4xRxT) = 0.268 © Sudhir K. I = 1.11 Ah = βIFoSa/g =1.11 = 0.0x1.25/(1.0) = 0.5.4x0. Jain.25/(1.25.25.066 For frame staging. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 51 .0) = 0.0x1.0 Ah = SD1I/(1.178 For shaft staging.0 Ah = SD1I/(1. 066 0.0 IS 1893:1984* IBC 2003 Frame staging 0.268 * Base shear coefficient values are common for frame and shaft staging Graphical comparison on next slide © Sudhir K.Base shear coefficient Base shear coefficients for tanks T (Sec) 0.178 Shaft staging 0.446 0. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 52 .3 1. Jain.298 0.12 0. All types of elevated tanks 2 2.4 IBC 2003.2 0.5 IBC 2003.5 Time period (S) IS 1893:1984.5 3 Lecture 2 / Slide 53 © Sudhir K. Jain.1 0 0 0. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 .3 0. Tanks on shaft staging Base shear coefficient 0.5 1 1. Tanks on frame staging 0.Base shear coefficient Comparison of base shear coefficient (Tanks) 0. IS 1893:1984 and IBC have much better comparison © Sudhir K.5 times for shaft staging than that from IS1893:1984 Recall.Base shear coefficient 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. Jain. for buildings. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 54 . 0 for tanks as against R = 8. Clearly . Jain. Same as for buildings with good ductility.0 for tanks. This is a major limitation of IS 1893:1984 © Sudhir K.elevated tanks do not have same ductility. redundancy and overstrength as buildings.Base shear coefficient Reason for lower values in IS 1893:1984 IBC uses R = 2. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 55 .0 for buildings with good ductility IS 1893:1984 uses K = 1.0 and R = 3. Base shear coefficient Another limitation of IS 1893:1984 In Lecture 1. these limitations have been removed © Sudhir K. rigidly attached to wall In IITK-GSDMA guidelines. Jain. we have seen. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 56 . does not consider convective mass It assumes entire liquid mass will act as impulsive mass. liquid mass gets divided into impulsive and convective masses IS 1893:1984. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 57 . 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).Base shear coefficient Let us now. redundancy and overstrength Sa/g will depend on time period © Sudhir K. 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. (I/R). Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 58 . Jain. will have different Sa/g values Procedure for finding time period in next lecture © Sudhir K.Base shear coefficient Impulsive and convective masses will have different time periods Hence. Ah. Redundancy and overstrength IS 1893:1984 has some serious limitations in design seismic force for tanks © Sudhir K. depends on Seismic Zone Soil type Structural characteristics Ductility.At the end of Lecture 2 Seismic force = (Ah) X (W) Base shear coefficient. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 2 / Slide 59 . A very stiff or rigid structure will have very low time period. Partial safety factors on materials also provide overstrength. materials may have strength greater than their characteristic strength.1 sec. there is a rising portion from T = 0.E-Course (through Distance Learning Mode) on Sesimic Design of Liquid Storage Tanks January 16 . For such rigid structures. strictly speaking.February 6. live loads and earthquake loads. In such rigid structures. 1.iitk.in) The following are my questions on Lecture 2. spectrum is flat (or horizontal) from T = 0. use of R for these structures is not appropriate. in design we do not include strength provided by these sources. actually it is present in the structure. Please note. whereas. IS 1893(Part 1) should in fact remove this rising portion.in/nicee/IITK-GSDMA/EQ05. codes make spectrum flat in short period range and allow the use of R values. for tanks. More discussion on this aspect is available in the following document: Proposed draft provisions and commentary on IS 1893(Part 1):2002 (www. Abdul Gani ([email protected] and greater than 1. . In order to address this issue. For limit state design of RC structures. we use partial safety factor of 1. As per fig C-4a of the Guidelines. 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 2000.5 on gravity loads. In IBC 2000. This additional strength comes from factors on gravity loads. Sa/g and base shear coefficient will be less.0 sec itself. Moreover.0 up to T = 0. this rising portion has been removed in the IITKGSDMA Guidelines.ernet. Nonstructural elements and special ductile detailing also constitute another source of overstrength. 2006 Response to Questions and Comments on Lecture 2 Question from Mr. 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. in IS 1893.pdf) Please note. which may fall in this range.ac. Thus. reduction in seismic forces on account of ductility is not allowed.7s. 2. What exactly is meant by ‘Overstrength’? Overstrength means additional strength which structure possess over and above the design strength. however. However. For example.7 sec also there is difference between IBC and IS 1893(Part 1). response reduction factors are different for impulsive and convective modes. then also this lower limit acts as a safeguard. design seismic force will remain same. in IITK-GSDMA Guideline. i. At present. in some international codes (like. This makes calculations simple and does not cause significant error.e.3 and AWWA D-115 use R = 1. 3. as of now there is no such lower limit. which means to say that. IS 1893 spectrum has 1/T variation for all time periods. For convective mode. On the other hand. At present. Pearson Education. That means. first Indian reprint. Please note. This is due to the fact that IBC has given a lower limit on its base shear coefficient. R is kept same for impulsive and convective modes and same spectrum is used for impulsive and convective modes. convective forces are like static forces and hence.3 and AWWA D-115).0 for convective mode. Why response reduction factor is different for convective and impulsive masses? In IITK-GSDMA Guideline. as per IS 1893:2002. convective forces would vary slowly with time. 2003). These codes argue that convective mode is low frequency mode (large time period) and hence. response spectrum has 1/Tc2 variation in long period range. How to find base shear coefficient based on IS 1893:2003 for two identical buildings designed for 50 and 100 years respectively. ACI 350. In probabilistic analysis. beyond certain time period. IBC specifies maximum considered earthquake with 2% probability of being exceeded in 50 years (2500 year return period). Moreover. it is important to note that.For time period greater than 1. 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. PGA is specified with certain probability of exceedance in given number of years. ACI 350. base shear coefficient will be same for both the buildings. reduction due to ductility shall not be applicable to them. IS 1893 does not have such lower limit and for tanks also in the IITK-GSDMA Guideline. response reduction factor is same for impulsive and convective modes. these codes use different response spectrum for convective and impulsive modes. However. if time period is incorrectly estimated on higher side. 4. This is done to ensure certain minimum strength against lateral loads. they do not allow any reduction in convective seismic forces. Indian . if design forces have to depend on expected life of building. which is usually having very large time period. ridges valleys etc. With the availability of more reliable data. it does not really matter. One may assume that this period will be less than approximately 0. working stress method is still quite common. TecDoc 1347) provision is made to account for local site effects (Site amplification etc). The spectral amplification factors are obtained from recorded data of ground motion in various soil conditions. Unlike this. Hence. the spectral amplifications are same for both rock and soil sites. In some of the international codes (eg. SD1 is specified at 1. As of now.0 sec time period. are currently not included in the design force calculations. while IBC specifies at Ultimate stress level? In India. Thus. Why is it so? But in IS 1893:2003. low frequency (long time period) waves get amplified. the spectral amplification factor for hard rock is higher followed by medium and soft soils. We use load factors to arrive at loads to be used for limit state design. such as hill slopes. 5. In IS 1893:2002 why no such provision is made? Site effects in IS 1893 are included by specifying different design spectra for rock. In IBC. medium and soft soils. in long time period range.codes have not yet adopted such an approach for quantifying seismic hazards. In some of the standards (eg. However. 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. 6. New Zealand and French codes). spectral amplification factor will be higher for hard soil. Why IS code specifies base shear coefficients at working stress level. spectra in short period range may be modified for different soil conditions. Other topographic effects. 7. IS 1893 (Part 1) has kept same spectral amplification in short period range for all types of soils. . 8.1 sec. IS 1893 has given design spectrum at working stress level. spectral amplification for soft soil will be higher. Similarly. as available information is not sufficient for reliable estimation of their effects.7 to arrive at forces for working stress design. in low time period range. Can please explain why? In hard soil. high frequency (low time period) waves get amplified and in soft soil. especially accounting for the higher ground velocities observed in soft soils. Users of IBC use a factor of 0. 2006 .Lecture 3 January 23. In this lecture Modeling of tanks Time period of tanks © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 2 . Jain. Modeling of tanks As seen in Lecture 1 liquid may be replaced by impulsive and convective mass for calculation of hydrodynamic forces See next slide for a quick review © Sudhir K. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 3 . Modeling of tanks mi = Impulsive liquid mass Kc/2 Kc/2 mc = Convective liquid mass Kc = Convective spring stiffness hi = Location of impulsive mass (without considering overturnig caused by base pressure) hc = Location of convective mass (without considering overturning caused by base pressure) hi* = Location of impulsive mass (including base pressure effect on overturning) hc* = Location of convective mass (including base pressure effect on overturning) Lecture 3 / Slide 4 mc Rigid m i hi (hi*) hc (hc*) Mechanical analogue or spring mass model of tank Graphs and expression for these parameters are given in lecture 1. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 . © Sudhir K. Jain. summation of mi and mc may not be equal to total liquid mass. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 5 . 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. obtain mc from the graph or expression Obtain mi = m – mc © Sudhir K. then First.Approximation in modeling Sometimes. Tanks of other shapes For tank shapes such as Intze. with diameter equal to diameter at the top level of liquid © Sudhir K. Jain. etc. : Consider equivalent circular tank of same volume. funnel. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 6 . Equivalent circular container will have diameter of 16 m and volume of 1000 m3.Tanks of other shapes Example: An Intze container has volume of 1000 m3.97 m © Sudhir K. Diameter of container at top level of liquid is 16 m. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 7 . h can be obtained as : π/4 x 162 x h = 1000 ∴ h = 1000 x 4/(π x 162) = 4. Height of liquid. Jain. Find dimensions of equivalent circular container for computation of hydrodynamic forces. h/D = 4. for equivalent circular container.31 16 m 16 m 4. Jain.Tanks of other shapes Thus. IIT Kanpur Equivalent circular container E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 8 .97/16 = 0.31 All the parameters (such as mi. mc etc.97 m Intze container volume = 1000 m3 © Sudhir K.) are to be obtained using h/D = 0. columns supporting the roof slab. and baffle walls These elements cause obstruction to lateral motion of liquid This will affect impulsive and convective masses © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 9 . Jain.Effect of obstructions inside tank Container may have structural elements inside For example: central shaft. Jain.Effect of obstructions inside tank Effect of these obstructions on impulsive and convective mass is not well studied A good research topic ! It is clear that these elements will reduce convective (or sloshing) mass More liquid will act as impulsive mass © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 10 . Effect of obstructions inside tank In the absence of detailed analysis. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 11 . Jain. following approximation may be adopted: Consider a circular or a rectangular container of same height and without any internal elements Equate the volume of this container to net volume of original container This will give diameter or lateral dimensions of container Use this container to obtain h/D or h/L © Sudhir K. 4m 12 m 12 m Elevation Hollow shaft of 2 m diameter Plan © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 12 . Find the dimensions of equivalent circular cylindrical tank. Jain. At the center of the tank there is a circular shaft of outer diameter of 2 m.Effect of obstructions inside tank Example: A circular cylindrical container has internal diameter of 12 m and liquid height of 4 m. Jain.8 m3.34.8 m3 ∴ D = 11. Let D be the diameter of equivalent circular cylinder.83 m Thus.Effect of obstructions inside tank Solution: Net volume of container = π/4x(122 –22)x4 = 439. then π/4xD2x4 = 439. for equivalent circular tank. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 13 . h = 4 m.8 m3 Equivalent cylinder will have liquid height of 4 m and its volume has to be 439.83m and h/D = 4/11. This h/D shall be used to find parameters of mechanical model of tank © Sudhir K.83 = 0. D = 11. U. “Dynamic analysis of fluids in containers subjected to acceleration”.C.Effect of wall flexibility Parameters mi. S. Nuclear Reactors and Earthquakes. Washington D. mc etc. RC tank walls are quite rigid Steel tank walls may be flexible Particularly. Atomic Energy Commission. G. Report No. W.. TID 7024. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 14 . in case of tall steel tanks © Sudhir K. 1963a. are obtained assuming tank wall to be rigid An assumption in the original work of Housner (1963a) Housner. Haroun. A. IIT Kanpur It does not substantially affect convective pressure distribution Refer Veletsos. N. A. “Seismic design of liquid storage tanks”. G.. 191-207. Vol. Journal of Technical Councils of ASCE. ASCE. W. TC1. Guidelines for the seismic design of oil and gas pipeline systems. Technical Council on Lifeline Earthquake 1Engineering. “Seismic response and design of liquid storage tanks”. 443-461. S.Effect of wall flexibility Wall flexibility affects impulsive pressure distribution Effect of wall flexibility on impulsive pressure depends on See next slide © Sudhir K.. Aspect ratio of tank Ratio of wall thickness to diameter E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 15 . and Housner. 107. Haroun and Housner (1984) Veletsos. Jain. 255-370.Y. 1984. M.. 1984. Jain.Effect of wall flexibility Effect of wall flexibility on impulsive pressure distribution h/D = 0. IIT Kanpur From Veletsos (1984) Lecture 3 / Slide 16 E-Course on Seismic Design of Tanks/ January 2006 .0005 tw is wall thickness z h Rigid tank Impulsive pressure on wall © Sudhir K.5 tw / D = 0.0005 tw / D = 0. then mechanical model of tank becomes more complicated Moreover.Effect of wall flexibility If wall flexibility is included. mechanical model based on rigid wall assumption is considered adequate for design. © Sudhir K. its inclusion does not change seismic forces appreciably Thus. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 17 . Jain. “Seismic design of storage tanks”. USA. C.. also use mechanical model based on rigid wall assumption API 650.Effect of wall flexibility All international codes use rigid wall model for RC as well as steel tanks Only exception is NZSEE recommendation (Priestley et al. 1986. © Sudhir K... “Welded storage tanks for oil storage”. Recommendations of a study group of the New Zealand National Society for Earthquake Engineering. et al. American Petroleum Institute (API) standards. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 18 . Jain. Washington D. which are exclusively for steel tanks. American Petroleum Institute. 1998. M J N. 1986) Priestley. Jain. T. Structural Engineering International. P. 2000. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 19 .. correspond to first impulsive and convective modes For most tanks ( 0. Wenk. 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. K. © Sudhir K. M. “Simple procedure for seismic analysis of liquid-storage tanks”..5) the first impulsive and convective modes together account for 85 to 98% of total liquid mass Hence.15 < h/D < 1. 197-201.Effect of higher modes mi and mc described in Lecture 1. and Wieland. mc.Modeling of ground supported tanks Step 1: Obtain various parameters of mechanical model These include. hi* and hc* Step 2: Calculate mass of tank wall (mw). mi. hi. mass of roof (mt) and mass of base slab (mb)of container This completes modeling of ground supported tanks © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 20 . hc. Kc. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 21 . Jain.Modeling of elevated tanks Elevated tank consists of container and staging Roof slab Wall Container Floor slab Staging Elevated tank © Sudhir K. Jain. etc. Ks. of staging must be considered This makes it a two-degree-of-freedom model Also called two mass idealization All other parameters such as hi. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 22 . shall be obtained as described earlier © Sudhir K.Modeling of elevated tanks Liquid is replaced by impulsive and convective masses. mi and mc Lateral stiffness. hc. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 . Jain.Modeling of elevated tanks mc Kc/2 hi Kc/2 mc mi hc Kc mi + ms hs Ks Spring mass model Two degree of freedom system OR Two mass idealization of elevated tanks Lecture 3 / Slide 23 © Sudhir K. which comprises of : Mass of container. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 24 .Modeling of elevated tanks ms is structural mass. Jain. and One-third mass of staging Mass of container includes Mass of roof slab Mass of wall Mass of floor slab and beams © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 25 .5 times T2. Jain. the two natural time periods (T1 and T2) are well separated.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. T1 generally may exceed 2. © Sudhir K. Two Degree of Freedom System Hence the 2-DoF system can be treated as two uncoupled single degree of freedom systems One representing mi +ms and Ks Second representing mc and Kc © Sudhir K. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 26 . Jain.5 T2 © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 27 .Modeling of elevated tanks mc Kc mi + ms Ks Ks mc Kc mi + ms Two degree of freedom system Two uncoupled single degree of freedom systems when T1 ≥ 2. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 28 .Modeling of elevated tanks Priestley et al. (1986) suggested that this approximation is reasonable if ratio of two time periods exceeds 2. one can obtain time periods of 2-DoF system as per procedure of structural dynamics. Jain. © Sudhir K.5 Important to note that this approximation is done only for the purpose of calculating time periods This significantly simplifies time period calculation Otherwise. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 29 .Modeling of elevated tanks Steps in modeling of elevated tanks Step 1: Obtain parameters of mechanical analogue These include mi. mc. hi. hi* and hc* Other tank shapes and obstructions inside the container shall be handled as described earlier Step 2: Calculate mass of container and mass of staging Step 3: Obtain stiffness of staging © Sudhir K. Kc. hc. elevated tank is modeled as 2-DoF system This 2-DoF system can be treated as two uncoupled SDoF systems © Sudhir K. convective mass is not considered It assumes entire liquid will act as impulsive mass Hence.Modeling of elevated tanks Recall. elevated tank is modeled as single degree of freedom ( SDoF) system As against this. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 30 . now. in IS 1893:1984. Jain. Jain. IIT Kanpur As per IS 1893:1984 Lecture 3 / Slide 31 E-Course on Seismic Design of Tanks/ January 2006 .Modeling of elevated tanks Models of elevated tanks mi + ms m +ms m = Total liquid mass Ks mc Kc Ks As per the Guideline © Sudhir K. h = 3 m.Modeling of elevated tanks Example: An elevated tank with circular cylindrical container has internal diameter of 11. © Sudhir K.000 kN/m. ∴ Volume of water = π/4 x D2 x h = π /4 x 11. Container is circular cylinder. Lateral stiffness of staging is 20. Jain. Container mass is 180 t and staging mass is 100 t. Model the tank using the Guideline and IS 1893:1984 Solution: Internal diameter.9 m3.3 m and water height is 3 m.3 m. Water height.9 t. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 32 .32 x 3 = 300. ∴ mass of water. D = 11. m = 300. 47 x 300.6 t Kc = 0.265 From Figure 2 of the Guideline. for h/D = 0.9 = 93.31 x 300.Modeling of elevated tanks h/D = 3/11.65 and Kch/mg = 0. mc/m = 0. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 33 .265: mi/m = 0.47 mi = 0.3 t mc = 0.3 = 0.9 = 195.31.65 x m = 0.47 x mg/h = 0.31 x m = 0.9 x 9.5 kN/m © Sudhir K.81/3 = 462.65 x 300. 000 kN/m © Sudhir K. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 34 . Ks = 20.3 t Lateral stiffness of staging.Modeling of elevated tanks Mass of container = 180 t Mass of staging = 100 t Structural mass of tank. ms = mass of container +1/3rd mass of staging = 180 +1/3 x 100 = 213. ms = 213. ms = 213. Ks = 20.6 t.000 kN/m.5 kN/m Model of tank as per the Guideline © Sudhir K. IIT Kanpur m = 300.3 t. Ks = 20. mc = 195.3 t.9 t.000 kN/m Model of tank as per IS 1893:1984 Lecture 3 / Slide 35 E-Course on Seismic Design of Tanks/ January 2006 . Kc = 462.Modeling of elevated tanks mi + ms m + ms Ks mc Kc Ks mi = 93. Jain.3 t. K should be in Newton per meter (N/m) Else. Jain. 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. M can be in Tonnes and K in kN/m © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 36 .Time period What is time period ? For a single degree of freedom system. time periods of impulsive and convective mode are to be obtained © Sudhir K. Jain.Time period Mathematical model of tank comprises of impulsive and convective components Hence. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 37 . Jain.Time period of impulsive mode Procedure to obtain time period of impulsive mode (Ti) will be described for following three cases: Ground supported circular tanks Ground supported rectangular tanks Elevated tanks © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 38 . 3h/D + 0. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 39 .46 − 0. Jain. Ti is given by: Ti = Ci h ρ t/D E ⎛ 1 ⎜ Ci = ⎜ h/D 0.Ti for ground-supported circular tanks Ground supported circular tanks Time period of impulsive mode.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. 5 2 © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 40 . Jain.5 h/D 1 1.Ti for ground-supported circular tanks Ci can also be obtained from Figure 5 of the Guidelines 10 8 6 C C i 4 C c 2 0 0 0. “Design provisions for earthquake resistance of structures. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 41 .Ti for ground-supported circular tanks This formula is taken from Eurocode 8 Eurocode 8. Part 1. If wall thickness varies with height. 1998. tanks and pipelines”. European Committee for Standardization. Brussels. Jain.General rules and Part 4 – Silos. then thickness at 1/3rd height from bottom shall be used Some steel tanks may have step variation of wall thickness with height © Sudhir K. Gebbeken. “On the analysis of vertical circular cylindrical tanks under earthquake excitation at its base”. 2003.. (2003) Nachtigall. N. J. Jain. © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 42 . I. 25. L.Ti for ground-supported circular tanks This formula is derived based on assumption that wall mass is quite small compared to liquid mass More information on time period of circular tanks may be seen in Veletsos (1984) and Nachtigall et al. Engineering Structures. 201-213. Vol.. and Urrutia-Galicia. wall flexibility is included while calculating time period © Sudhir K. 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. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 43 . Jain.Ti for ground-supported circular tanks It is important to note that wall flexibility is considered in this formula For tanks with rigid wall. wall and base have flexible connections © Sudhir K.Ti for ground-supported circular tanks This formula is applicable to tanks with fixed base condition i. tank wall is rigidly connected or fixed to the base slab In some circular tanks.. Jain.e. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 44 . “Seismic design of liquid containing concrete structures”.3. MI.Ti for ground-supported circular tanks Ground supported tanks with flexible base are described in ACI 350. American Concrete Institute.and strand-wound circular. Farmington Hill. Jain. USA. 1995. “Wire. American Water Works Association.3 and AWWA D-110 ACI 350. there is a flexible pad between wall and base Refer Figure 6 of the Guideline © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 45 . In these tanks. USA. AWWA D-110. prestressed concrete water tanks”. 2001. Colorado. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 46 .Ti for ground-supported circular tanks Types of connections between tank wall and base slab Such tanks are perhaps not used in India © Sudhir K. Sa/g has constant value See next slide © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 47 . Jain.4 seconds In this short period range.Ti for ground-supported circular tanks Impulsive mode time period of ground supported tanks with fixed base is generally very low These tanks are quite rigid Ti will usually be less than 0. spectral acceleration. IIT Kanpur Sa/g E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 48 . Jain.Ti for ground-supported circular tanks Impulsive mode time period of ground supported tanks likely to remain in this range © Sudhir K. Young’s modulus.67. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 49 . internal diameter. For water. t=15 mm. h = 25 m. For steel. Ci = 5.Ti for ground-supported circular tanks Example: A ground supported steel tank has water height. From Figure 5. Jain. ρ = 1 t/m3. h/D = 25/15 = 1. mass density. Find time period of impulsive mode. E = 2x108 kN/m2. D = 15 m.3 © Sudhir K. Solution: h = 25 m. t = 15 mm. D = 15 m and wall thickness. Ti for ground-supported circular tanks Time period of impulsive mode. time period is low. time period will be further less. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 50 .0 0.30 sec Important to note that. Ti = Ci h ρ t/D E Ti = 5. © Sudhir K. For RC tanks and other short tanks. even for such a slender tank of steel.015/15 2x10 8 = 0. Jain.3 25 1. Jain. API standards have suggested a constant value of spectral acceleration for ground supported circular steel tanks Thus.Ti for ground-supported circular tanks In view of this. no point in putting too much emphasis on evaluation of impulsive mode time period for ground supported tanks Recognizing this point. users of API standards need not find impulsive time period of ground supported tanks © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 51 . 2 of the Guidelines This will not be repeated here Time period is likely to be very low and Sa/g will remain constant As described earlier Hence.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.1. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 52 .3. not much emphasis on time period evaluation © Sudhir K. Jain. 4. flexibility of staging is important Time period of impulsive mode. Jain. Ti is given by: Ti = 2π mi + ms 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. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 53 .Ti for Elevated tanks For elevated tanks. Ti for Elevated tanks These two formulae are one and the same Expressed in terms of different quantities Center of gravity of tank refers to combined mass center of empty container plus impulsive mass of liquid © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 54 . Jain. 5. Hence from Figure 2a of the Guideline. Container mass is 150 t and staging mass is 90 t. ms = mass of container + 1/3rd mass of staging = 150 + 90/3 = 180 t © Sudhir K.000 kN/m.54. mi/m = 0. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 55 .54 x 250 = 135 t Structural mass of tank.5. Ratio of water height to internal diameter of container is 0. ∴ mi = 0. Find time period of impulsive mode Solution: h/D = 0.Ti for Elevated tanks Example: An elevated tank stores 250 t of water. Jain. Lateral stiffness of staging is 20. © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 56 .Ti for Elevated tanks Time period of impulsive mode Ti = 2π Ti = 2π mi + ms Ks 135 + 180 20.000 = 0. Jain.79 sec. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 57 . Jain. 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 © Sudhir K.Lateral stiffness of staging. 12.5. 1375-1393.. K.Lateral stiffness of staging. “Lateral load analysis of frame staging for elevated water tanks”. “Approximate methods for determination of time period of water tank staging”. Sameer. Vol. Some commonly used frame type staging configurations are shown in next slide © Sudhir K. and Jain. Journal of Structural Engineering. 1994) Sameer. ASCE. lateral stiffness shall be obtained by suitably modeling columns and braces More information can be seen in Sameer and Jain (1992. The Indian Concrete Journal. S. 691-698. No.. S.120. S. and Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 58 . 1992.. U. No. Vol. S. 66. Jain. K. Ks For frame type staging.. U. 1994. IIT Kanpur 12 columns Lecture 3 / Slide 59 E-Course on Seismic Design of Tanks/ January 2006 .Lateral stiffness of staging. Jain. Ks Plan view of frame staging configurations 4 columns 6 columns 8 columns 9 columns © Sudhir K. Lateral stiffness of staging. Jain. Ks 24 columns 52 columns © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 60 . Explanatory Handbook on Codes for Earthquake Engineering. SP:22 has considered braces as rigid beams SP:22 – 1982. Ks Explanatory handbook. staging can be modeled and analyzed to estimate lateral stiffness © Sudhir K. higher base shear coefficient This is another limitation of IS 1893:1984 Using a standard structural analysis software.Lateral stiffness of staging. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 61 . Bureau of Indian Standards. New Delhi This is unrealistic modeling Leads to lower time period Hence. Jain. Jain. shear deformations of shaft would affect the stiffness and should be included. then Its stiffness will be Ks = 3EI/L3 This will be a good approximation if height to diameter ratio is greater than two Otherwise.Lateral stiffness of staging. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 62 . © Sudhir K. Ks Shaft type staging can be treated as a vertical cantilever fixed at base and free at top If flexural behavior is dominant. 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 © Sudhir K. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 63 . IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 64 .Time period of convective mode mc and Kc for circular and rectangular tanks can be obtained from graphs or expressions These are described in Lecture 1 Refer Figures 2 and 3 of the Guidelines © Sudhir K. Jain. expressions for mc and Kc are substituted in the formula for Tc Then one gets. Jain.16(h / L)) Lecture 3 / Slide 65 E-Course on Seismic Design of Tanks/ January 2006 .68 tanh (3.16 tanh (3. IIT Kanpur Cc = 2π 3. For circular tanks: Cc = 2π 3.Time period of convective mode For further simplification.68h / D) Tc = Cc D/g For rectangular tanks: Tc = Cc L/g © Sudhir K. Jain.Time period of convective mode Graphs for obtaining Cc are given in Figures 5 and 7 of the Guidelines These are reproduced in next two slides Convective mass and stiffness are not affected by flexibility of base or staging Hence. convective time period expressions are common for ground supported as well as elevated tanks Convective mode time periods are usually very large Their values can be as high as 10 seconds © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 66 . 5 h/D 1 1. Jain.5 2 Fig. 5 For circular tanks © Sudhir K.Time period of convective mode 10 8 6 C C i 4 C c 2 0 0 0. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 67 . 5 2 Fig. 7 For rectangular tanks © Sudhir K.Time period of convective mode 10 8 6 Cc 4 2 0 0. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 68 .5 h/L 1 1. Jain. Calculate time period of convective mode.Time period of convective mode Example: For a circular tank of internal diameter.81 = 3.6 12/9. Solution: h = 4 m.98 sec © Sudhir K. 12 m and liquid height of 4 m. D = 12 m. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 69 .33 From Figure 5 of the Guidelines. Jain.6 Time period of convective mode. ∴ h/D = 4/12 = 0. Tc = Cc D/g Tc = 3. Cc = 3. impulsive mode time period is likely to be very less Convective mode time period can be very large © Sudhir K. Jain.At the end of Lecture 3 Based on mechanical models. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 70 . time period for impulsive and convective modes can be obtained for ground supported and elevated tanks For ground supported tanks. R (rmurugan@igcar.. This has been answered in a subsequent question. Graph with solid line is for tw/D = 0.in ) • In Slide-16: Please correct the tw/D ratio. How to combine them? Response from these two uncoupled SDOF systems is combined using SRSS rule.ernet. • In Slide-62: Kindly provide the formula for stiffness of staging based on shear deformation. second SDOF). This has been discussed in Lecture 7 .e. 2006 Response to Questions and Comments on Lecture 3 Question from Mr. first SDOF) and convective mode (i. Both dotted line and thick line are showing the same ratio of 0. Murugan. Please note..0005.February 6. You are right.E-Course (through Distance Learning Mode) on Sesimic Design of Liquid Storage Tanks January 16 . • 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. total base shear is SRSS combination of base shear in impulsive mode (i.e.005 • In Slide-26: We are treating 2-DOF system as two uncoupled SDOF system. between wall and base there is a flexible pad. This has been pointed out in the Commentary to clause 4. H. I.2. How to find the damping if impulsive and convective time periods are not well separated? If two time periods are not well separated. 1. ACI 350. Wr. this question will not arise. Time period. 3.5. Sa/g graph has been made horizontal in this low period range. Ww. Please refer Clause 4. roof and impulsive liquid respectively. 2.3 and are reproduced in Figure 6 of IITK-GSDMA Guideline. Even though the time period of impulsive mode of ground supported tanks with flexible base is low. then we have to solve 2-DoF system. wherein.ernet. Flexible base does not affect convective mode time.4.2. hence. How the flexibility of pad affects the impulsive mode time period? This question is answered in your first question . 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.2 of the Guideline. D is diameter of tank and Ka is stiffness of base pad in the units of force/m2. How to account this reduced sa/g values in design? For tanks. How to find the time period of ground supported tanks with flexible connection? Tanks with flexible connections are those wherein. the sa/g values as per IS 1893 is low in this region.3 suggests following formula for impulsive mode time period of circular tanks on flexible base Ti = 8π (Ww + Wr + Wi ) gDK a Where. Abdul Gani (gani@igcar. 4. damping is different for both the DoF. These types of tanks are described in ACI 350.Question from Mr. and Wi are weights of wall.in ) The following are my doubts regarding lecture 3. This will be nonclassically damped system. one finds that in the time period only one-third mass of spring contributes. Priestley’s criterion may not work well. In slide 16. there may be some extreme cases wherein. 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. Some discussion on this topic is included in Lecture 7. a) On the graph. In the case of ground supported rectangular tanks impulsive and convective time periods will be different in both directions. which is for secondary systems like equipments.kc (apart from Priestly's method). Ti and Tc). there is a range of various parameters (mi. You need to be cautious while using decoupling criterion of ASCE4-98. 9. 6. what does 'z' represent? b) The curves are presented for h/D = 0. Assuming that spring deflection varies linearly along its length. For most of the tanks. You consider a SDoF system and include mass of the spring also. Which value is to be adopted for the design? In rectangular tanks. You will notice that time period of 2-DoF systems depends on ratio of two masses and ratio of two uncoupled time periods.3.2. However. Staging acts like a lateral spring. What is the criteria for decoupling mi+ms. For other values are curves available? . total lateral force and bending moment acting on that wall is required. forces arising due to seismic load in the perpendicular direction shall be considered.ks and mc.5. 8. for the design of a particular wall. for wall design. How the remaining mass is accounted? This has been explained in the commentary to Clause 4. Interestingly.2. one can derive this criterion. 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. Can ASCE4-98 guideline for decoupling equipment masses be used here? For elevated tanks used in practice.5. only 1/3rd of the mass of staging is considered. mc. 7. the criterion suggested by Preistley is OK. While decoupling. Hence. Slide 62. If the tank container is believed to be more stiff (or rigid). 107..com) The formula specified by the code to consider lateral stiffness of shaft type staging considers Ks = 3*EI /L^3. M. the symbol "L" being referred to in the above formula should be identified as distance of CG of container from top of Footing. Page 72)). More studies on effect of wall flexibility are available in the literature. It is to be recognized that we need to model the stiffness of staging properly. 1984. it will be grossly on conservative side. Journal of Technical Councils of ASCE. Note h is total depth of liquid and z/h varies from 0 to 1. Please correct.You can also try your hands on literature on pressure vessels. W. Some discussion on this issue is given in Lecture 7. 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. 191-207. G. In frame staging base beams. In the simple formula (k = 3 E I /Lcg^3) for lateral stiffness of staging. Refer paper Haroun. as the case may be).Course on Tanks) As in the case of Tanks supported on framed stagings. TC1. 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. 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). Of course. “Seismic design of liquid storage tanks”. one should consider Lcg being the height of cg of container from the base (with or without water. They deal with flexible walls. Vol. and Housner. Question from Mr. Rushikesh Trivedi (rushikeshtrivedi@yahoo. what will be the value of tw/D? z represents the height at which pressure is to be obtained. (Lecture 3.0005 is represented by both bold and dotted curves. Hence. which are comparatively more rigid. This figure is taken from Veletsos (1984). E. A.0. Kindly opine on the same.c) tw/D = 0. d) For rigid tank. here also we should consider stiffness through a force applied at the CG of container. are also modeled and diaphragm effect is included to account for in-plane . This rigid link does not undergo much higher deflection than staging top.rigidity of slab. Lcg L1 W θ ∆ θ T = 2π Δ g . The force at CG is applied by putting a rigid link from staging top to CG. 2006 .Lecture 4 January 30. Jain. Sa/g and R values for tanks © Sudhir K. I. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 2 .In this lecture Z. Base shear coefficient Seismic force V = (Ah) x (W) Ah is base shear coefficient Ah = ⎛Z⎞ ⎛ I ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ 2 ⎠ ⎝ R⎠ Sa g Structural characteristics Depends on time period and damping Zone Depends on severity of ground motion Design philosophy © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 3 . Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 4 . respectively © Sudhir K. Jain.Base shear coefficient Tanks have two modes Impulsive Convective Seismic force In impulsive mode. Vc = (Ah)c x convective weight (Ah)i and (Ah)c are base shear coefficient in impulsive and convective modes. Vi = (Ah)i x impulsive weight In convective mode. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 5 . Jain. (Sa/g)i and (Sa/g)c © Sudhir K. R has been used in (Ah)i as well as (Ah)c Zone factor. Z As per Table 2 of IS 1893(Part1):2002 I. (Sa/g)i and (Sa/g)c will be discussed here First.Base shear coefficient Impulsive base shear coefficient (Ah)i = (Z/2) x (I/R) x (Sa/g)i Convective base shear coefficient (Ah)c = (Z/2) x (I/R) x (Sa/g)c Note. R. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 6 .(Sa/g)i and (Sa/g)c (Sa/g)i is average response acceleration for impulsive mode Depends on time period and damping of impulsive mode (Sa/g)c is average response acceleration for convective mode Depends on time period and damping of convective mode © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 7 . Jain.(Sa/g)i and (Sa/g)c Sa/g is obtained from design spectra Figure 2 of IS 1893(Part 1):2002 See next slide These spectra are slightly modified for tanks © Sudhir K. 1 sec is usually disallowed by the codes.(Sa/g)i and (Sa/g)c Modifications are: The rising portion in short period range from (0 to 0. 1/T variation is retained © Sudhir K. Spectra is extended beyond 4 sec Since convective time period may be greater than 4 sec.1 sec) has been made constant Very stiff structures have time period less than 0. actual time period may be slightly higher As the structure gets slightly damaged.1 sec There may be modeling errors. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 8 . Jain. its natural period elongates Ductility does not help in reducing response of very stiff structures Hence. Beyond 4 sec. rising portion in the range 0 to 0. (Sa/g)i and (Sa/g)c Sa/g Sa/g Sa/g Spectra of IS 1893 (Part 1):2002 Modified spectra For 5% damping © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 9 . Jain. 10 = 2.40 ≤ T ≤ 4.00 ≤ T < 0.0 © Sudhir K.0 For medium soil sites Sa/g = 1 + 15 T 0.40 = 1.55 = 1.10 ≤ T < 0.67 = 1.67 = 1.00 / T 0.67 / T 0.50 0.36 / T 0.(Sa/g)i and (Sa/g)c Expressions for design spectra at 5% damping Expressions for spectra of IS 1893(Part 1):2003 For hard soil sites 0.67 / T T ≥ 0.50 0.50 T < 0. IIT Kanpur Expressions for spectra for tanks For hard soil sites Sa/g = 2.67 ≤ T ≤ 4. Jain.40 For medium soil sites T < 0.10 ≤ T < 0.55 Sa/g = 2.10 = 2.40 = 1.5 T< 0.36 / T T ≥ 0.10 ≤ T < 0.0 / T T ≥ 0.00 ≤ T < 0.00 ≤ T < 0.50 0.55 ≤ T ≤ 4.10 Sa/g = 1 + 15 T = 2.67 E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 10 .50 = 1.0 For soft soil sites Sa/g = 1 + 15 T 0.55 For soft soil sites Sa/g = 2. Jain.4. higher damping reduces base shear coefficient or design seismic forces Multiplying factor =1. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 11 .0 for 5% damping Multiplying factor = 0.8 for 10% damping This multiplier is not used for PGA © Sudhir K.(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. for 2% damping Multiplying factor = 1. 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 7.2. IBC 2003 suggests 5% damping for all tanks It suggests 5% damping for all types of buildings also © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 12 .8.1 of IS 1893(Part 1) suggests 5% damping for RC and 2% damping for steel buildings However. Jain. one fixed value of damping is used in our analysis © Sudhir K. Jain. while the real behavior is non-linear Hence.Damping Damping depends on material and level of vibration Higher damping for stronger shaking Means that during the same earthquake. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 13 . damping will increase as the level of shaking increases We are performing a simple linear 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 © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 14 Damping Damping for convective mode 0.5% of critical for all types of tanks Convective mode damping does not depend on material of tank or type of liquid stored In Table 3 of IS 1893(Part 1):2002 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 © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 15 Importance factor, I Importance factor, I for tanks is given in Table 1 of the Guideline This Table is reproduced here Type of liquid storage tank 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. All other tanks with no risk to life and with negligible consequences to environment, society and economy. NOTE: Values of importance factor, I given in IS 1893 (Part 4) may be used where appropriate © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 16 I 1.5 1.0 Importance factor, I 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 © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 17 Importance factor, I Footnote below this Table is given to avoid conflict with I values of IS1893(Part 4) IS 1893(Part 4) will deal with industrial structures Not yet published Some industries assign very high importance factor to tanks storing hazardous materials Depending on their own requirements For such tanks, Importance factor (I) will be as per part 4 © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 18 Response reduction factor, R R values for tanks are given in Table 2 of the Guideline This is reproduced in next two slides © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 19 Response reduction factor, R Elevated tank 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 Tank supported on RC shaft RC shaft with two curtains of reinforcement, each having horizontal and vertical reinforcement 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# # R 1.3 1.5 1.8 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 E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 20 © Sudhir K. Jain, IIT Kanpur Response reduction factor, R Ground supported tank 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 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) Steel tank a) Unanchored base b) Anchored base Underground RC and steel tank+ + R 1.3 1.5 2.0 2.5 1.5 2.0 2.5 4.0 For partially buried tanks, values of R can be interpolated between ground supported and underground tanks based on depth of embedment. © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 21 Response reduction factor, R R values for tanks are smaller than buildings This is in line with other international codes As discussed earlier, R depends on Ductility Redundancy Overstrength Tanks possess low ductility, redundancy and overstrength © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 22 Response reduction factor, R 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 © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 23 Response reduction factor, R 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 © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 24 Response reduction factor, R Moreover in buildings, non-structural elements, such as infill walls, contribute significantly to overstrength Staging are bare frames In view of this, for staging with SMRF, R = 2.5 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 © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 25 Response reduction factor, R Comparison for frame staging Zone and soil parameters are same used in Lecture 2 0.5 0.4 0.3 0.2 0.1 0 0 0.5 1 1.5 Time period (sec) © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 26 IBC 2003; Frame staging, R = 3.0 Guideline; Frame staging, R = 2.5 IS 1893:1984; All types of staging, K = 1.0 Base shear coefficient 2 2.5 3 IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 27 . Jain. 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 © Sudhir K.Response reduction factor. R Let us now consider. 1990. Earthquake Spectra. 2002. USA. Boston. M J N. © Sudhir K. “Inelastic seismic demand on circular shaft type staging for elevated tanks”. Vol. 18 No. Rao M L N. R Studies on ductility of shaft Zanh F A. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 28 . 2002. “Flexural strength and ductility of circular hollow reinforced concrete columns without reinforcement on inside face”. of Roorkee. 2000. “Effect of confinement on ductility of RC hollow circular columns”.Response reduction factor. 7th National Conf. a Master’s thesis submitted to Dept. Paper No. Univ. Park R. ACI Journal 87 (2). 91. EERI. Jain. 156-166. “Retrofitting of shaft type staging for elevated tanks”. of Earthquake Engineering. and Priestley. India. on Earthquake Engrg. 745760. 4. Rai D C and Yennamsetti S. Rai D C. R 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 © Sudhir K. Jain.Response reduction factor. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 29 . Jain. P = axial load on shaft fc’ = characteristic strength of concrete Ag = gross area of concrete Hollow circular section © Sudhir K.Effect of Axial Load on Ductility Figure from Rai (2002) Ast/Ag = ratio longitudinal reinforcement to concrete area. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 30 . 25 Now. Jain.94. If ratio of axial load (P) to ultimate load (fck.02 This value reduces to 3 for P/ (f’c.Response reduction factor.Ag) is 0. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 31 . curvature ductility is about 9 for Ast/Ag = 0. R In this figure.1 then. let us see some results on effect of shaft thickness © Sudhir K.Ag) of 0. curvature ductility is plotted as a function of longitudinal reinforcement These results are for inner (Di) to outer (Do) diameter ratio of 0. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 32 .05 Very low axial force ratio © Sudhir K.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. 8 Thus. curvature ductility is 12 For longitudinal steel ratio Ast/Ag = 0. thickness has significant effect on ductility A thick shaft has reasonably good ductility © Sudhir K. Jain. R For thin shaft with Di/Do = 0. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 33 .02 This value increases to about 25 for thick shaft with Di/Do = 0.Response reduction factor.95. Jain. Roorkee..Response reduction factor. Proc. 2002. R These analytical studies clearly indicate that thin RC hollow sections possess very low ductility Issues connected with poor ductility of shaft. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 34 . “Review of code design forces for shaft supported elevated water tanks”. Ed.pdf) © Sudhir K. pp 1407 -1418. inadequate provisions of IS 1893:1984. D K Paul et al.org/ecourse/12_symp_tanks. (http://www.nicee. and their correlation to behavior during recent earthquakes is discussed in following paper: Rai D C.of 13th Symposium on Earthquake Engineering . base shear coefficient for shaft supported tanks matches well with international codes Comparison with IBC 2003 on next slide © Sudhir K.Response reduction factor. R = 1. Jain.8 for shaft supported tanks With this value of R. R Based on all these considerations. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 35 . 3 0.8 IS 1893:1984.5 Time period (Sec) 2 2. All types of staging.5 Base shear coefficient IBC 2003.0 0. Shaft staging.2 0.0 Guideline. Shaft staging. R Comparison for shaft staging Zone and soil parameters are same as used in Lecture 2 0.5 3 © Sudhir K.Response reduction factor. K = 1. Jain.5 1 1. R = 1.4 0. R = 2. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 36 .1 0 0 0. Response reduction factor. Jain. and construction of concrete-pedestal water Towers”. © Sudhir K. Farmington Hill. USA. design . American Concrete Institute. 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. “ Guide for the analysis. R Some useful information on RC shaft is given in ACI 371-98 ACI 371-98 . IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 37 . MI. 1998. C. C. (www. O. “Codal provisions on design seismic forces for liquid storage tanks: a review”. “Codal provisions on seismic analysis of liquid storage tanks: a review” Report No.. S. D. Kanpur. Indian Institute of Technology Kanpur.in/nicee/IITK-GSDMA/EQ04.K. Rai.iitk. S. R. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 38 .ac. 2004a. IITK-GSDMA-EQ-01-V1. D. Jain. Rai. and Jain.. O.iitk. IITK-GSDMA-EQ-04-V1.Response reduction factor.. (www.in/nicee/IITK-GSDMA/EQ01.0.0.K. 2004b.pdf ) © Sudhir K.ac. Report No. R We have seen comparison with IBC 2003 Comparison with other international codes is available in following documents: Jaiswal. and Jain.pdf ) Jaiswal. Indian Institute of Technology Kanpur. R. Kanpur. Response reduction factor. following international codes are reviewed and compared: IBC 2000 (now. (1986) © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 39 . 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. R In the above two documents. Jain. Response reduction factor. R Now we know Z. © Sudhir K. I. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 40 . R and Sa/g for tanks One can now obtain base shear coefficient for impulsive and convective modes An example follows. respectively.Example Example: An elevated water tank has RC frame staging detailed for ductility as per IS: 13920 and is located in seismic zone IV.2 sec and 4.24 From Table 2 of IS 1893 (PART I):2002. Solution: Zone: IV ∴ Z = 0. Jain. Impulsive and convective time periods are 1.0 sec. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 41 . Site of the tank has soft soil. Obtain base shear coefficient for impulsive and convective mode.5 From Table 1 of the Guideline R = 2.5 for RC frame with good ductility (SMRF) From Table 2 of the Guideline © Sudhir K. I = 1. 731 © Sudhir K.Example on (Ah)i and (Ah)c Impulsive time period. Tc = 4.75 = 0.3 of the Guideline) Convective mode time period.67/Ti = 1.0 x 1.67/1.2 = 1.5% damping (Clause 4.67/Tc) x 1.2 sec. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 42 .75 is to be used for scaling up (Sa/g) for 0. and soil is soft.5.67/4.0 sec and soil is soft Damping = 0.5.4 of the Guideline) Factor 1.75 = 1. Damping = 5% (RC Frame) ∴ (Sa/g)i = 1.4 of the Guideline) ∴ (Sa/g)c = (1.5% (Clause 4. Jain.392 (Clause 4. Ti = 1. 24/2 x 1.5 x 1.5/2. Jain.24/2 x 1.5 x 0.10 Base shear coefficient for convective mode (Ah)c = (Z/2) x (I/R) x (Sa/g)c = 0.392 = 0. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 43 .053 © Sudhir K.5/2.731 = 0.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. slight modifications are recommended for design spectrum of IS 1893(Part1) Damping for convective mode may be taken as 0. and Observed behavior of tanks For tanks.The basis for this is Analytical studies Provisions of international codes.5% for all types of tanks © Sudhir K.At the end of Lecture 4 R values for tanks are less than those for buildings. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 4 / Slide 44 . Jain. Lecture 5 January 31. 2006 . IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 2 .In this Lecture Impulsive and convective base shear Critical direction of seismic loading © Sudhir K. Jain. importance factor and response reduction factor Now. we proceed with seismic force or base shear calculations © Sudhir K. Jain. damping.Base shear Previous lectures have covered Procedure to find impulsive and convective liquid masses This was done through a mechanical analog model Procedure to obtain base shear coefficients in impulsive and convective modes This requires time period. zone factor. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 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 © Sudhir K. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 4 . we evaluate impulsive and convective weights Or. we consider structural mass also © Sudhir K. impulsive and convective masses Earlier we have obtained impulsive and convective liquid mass Now. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 5 .Base shear Now. Jain. Jain. total impulsive mass comprises of Mass of impulsive liquid Mass of wall Mass of roof © Sudhir K. wall.Base shear : Ground supported tanks Impulsive liquid mass is rigidly attached to container wall Hence. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 6 . roof and impulsive liquid vibrate together In ground supported tanks. 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 © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 7 .Base shear : Ground supported tanks Hence. Jain. Base shear : Ground supported tanks 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 © Sudhir K. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 8 . Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 9 .Base shear : Ground supported tanks Base shear in convective mode Vc = ( A h )c m c g mc = mass of convective liquid © Sudhir K. e. all international codes use SRSS rule Eurocode 8 uses absolute summation rule i. V = Vi + Vc © Sudhir K. Jain.Base shear : Ground supported tanks 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. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 10 . © Sudhir K. USA.Base shear : Ground supported tanks In the latest NEHRP recommendations (FEMA 450). National Institute of Building Sciences. National Institute of Building Sciences. USA. SRSS rule is suggested Earlier version of NEHRP recommendations (FEMA 368) was using absolute summation rule FEMA 450. “NEHRP recommended provisions for seismic regulations for new buildings and other structures”. Building Seismic Safety Council.. 2003. Building Seismic Safety Council. “NEHRP recommended provisions for seismic regulations for new buildings and other structures”. FEMA 368. Jain.. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 11 . 2000. Bending moment:Ground supported tanks Next. Jain. 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 © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 12 . Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 13 .Bending moment:Ground supported tanks mw acts at CG of wall mt acts at CG of roof mi acts at height hi from bottom of wall If base pressure effect is not included If base pressure effect is included Recall hi and hi* from Lecture 1 mi acts at hi* © Sudhir K. Jain.Bending moment:Ground supported tanks 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 © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 14 . effect of base pressure is not included Hence. hi and hc are used © Sudhir K.Bending moment:Ground supported tanks For bending moment at the bottom of wall. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 15 . Bending moment:Ground supported tanks Bending moment at the bottom of wall (Ah)imt (Ah)cmc hc hi (Ah)imw ht hw Ground level (Ah)imi Mi = ( Ah )i (mi hi + mw hw + mt ht ) g M c = ( A h )c (mc hc ) g © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 16 . Jain. Bending moment:Ground supported tanks Total bending moment at the bottom of wall 2 M = M i2 + M c SRSS rule used to combine impulsive and convective responses © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 17 . Jain. must include effect of base pressure hi* and hc* will be used Ground level © Sudhir K. Jain.Overturning moment:Ground supported tanks Overturning moment This is at the bottom of base slab Hence. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 18 . IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 19 .Overturning moment:Ground supported tanks Overturning moment in impulsive mode ⎡mi ( hi* + tb ) + mw (hw + tb ) +⎤ * Mi = ( Ah )i ⎢ ⎥g ⎢mt (ht + tb ) + mb tb / 2 ⎥ ⎣ ⎦ Overturning moment in convective mode * M c = ( Ah ) c mc ( hc + tb ) g * tb = thickness of base slab © Sudhir K. Jain. base slab thickness. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 20 .Bending moment:Ground supported tanks Overturning moment is at the bottom of base slab Hence. lever arm is from bottom of base slab Hence. tb is added to heights measured from top of the base slab Total overturning moment M = M * *2 i +M *2 c © Sudhir K. Jain. 95 m. hc = 2. Find base shear and bending moment at the bottom of wall.225.3 t. mt =33.4 t mw = 65.08 Lecture 5/ Slide 21 Base slab 250 mm thick © Sudhir K.3 m. (Ah)c = 0.1 t. mc = 163. hc* = 3. Jain. hi =1. Roof slab 150 mm thick 10 m 4m Wall 200 mm thick mi = 141.5 m.63 m (Ah)i = 0. Also find base shear and overturning moment at the bottom of base slab.Example Example: A ground-supported circular tank is shown below along with some relevant data. hi* = 3.4 t. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 . mb = 55.2 t. Example Solution: Impulsive base shear at the bottom of wall is Vi = (Ah)i (mi + mw + mt) g = 0.81 = 529.2 kN Total base shear at the bottom of wall is V = Vi2 + Vc2 = © Sudhir K.225 x (141.3 ) + (128.3 kN Convective base shear at the bottom of wall is Vc = (Ah)c mc g = 0.4 + 65.81 = 128. Jain.4 x 9.3 + 33. IIT Kanpur (529.2 ) 2 2 = 544.6kN Lecture 5/ Slide 22 E-Course on Seismic Design of Tanks/ January 2006 .08 x 163.1) x 9. 5 + 65.4 x 2.81 = 1054 kN-m Convective bending moment at the bottom of wall is Mc = (Ah)c mc hc g = 0.075 = 4.075 m Impulsive bending moment at the bottom of wall is Mi = (Ah)i (mihi + mwhw + mtht) g = 0.0 + 0.3 x 9.225 x (141.m Lecture 5/ Slide 23 E-Course on Seismic Design of Tanks/ January 2006 .0 + 33.81 = 295 kN-m Total bending moment at bottom of wall is M = Mi2 + M2 = c © Sudhir K.Example For obtaining bending moment. IIT Kanpur (1054 )2 + (295 )2 = 1095kN .3 x 2. we need height of CG of roof slab from bottom of wall.075) x 9.1 x 4.4 x 1.08 x 163. ht. ht = 4. Jain. Example Now. IIT Kanpur (651.81 = 128.1 kN Convective base shear at the bottom of base slab is Vc = (Ah)c mc g = 0.4 x 9. Jain.225 x (141.2 kN Total base shear at the bottom of base slab is V = Vi2 + Vc2 = © Sudhir K.6kN Lecture 5/ Slide 24 E-Course on Seismic Design of Tanks/ January 2006 .3 + 33. 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 = 0.2 ) 2 2 = 663.1 ) + (128.4 + 65.1 + 55.08 x 163.81 = 651.2) x 9. Jain.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 [141.25/2] x 9.0 + 0. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 25 .2 x 0.63 + 0.95 + 0.4(3.81 = 1966 kN-m Convective overturning moment at the bottom of base slab Mc* = (Ah)c mc (hc* + tb) g = 0.m Notice that this value is substantially larger that the value at the bottom of wall (85%) © Sudhir K.81 = 498 kN-m Total overturning moment at bottom of base slab * M = (M ) + (M ) * 2 i * 2 c = (1966 )2 + (498 )2 = 2028 kN .1(4.25) x 9.4 x (3.08 x 163.3(2.075 + 0.25) + 33.25) + 55.25) + 65. 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 = © Sudhir K. IIT Kanpur V i 2 + V c2 Lecture 5/ Slide 26 E-Course on Seismic Design of Tanks/ January 2006 . Jain.Base shear : Elevated tanks In elevated tanks. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 27 . Jain.Bending moment:Elevated tanks Bending moment at the bottom of staging Bottom of staging refers to footing top Impulsive base shear comprises of two parts (Ah)i x mig Ah)i x msg Convective base shear has only one part (Ah)c x mcg © Sudhir K. effect of base pressure included and hi* and hc* are used © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 28 .Bending moment:Elevated tanks mi acts at hi* mc acts at hc* Bending moment at bottom of staging is being obtained Hence. Jain. Jain. floor slab and floor beams © Sudhir K. wall. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 29 . 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.Bending moment:Elevated tanks Structural mass. Bending moment:Elevated tanks Bending moment at the bottom of staging M i = ( Ah )i mi hi + hs + ms hcg * * [ ( M c = ( Ah )c m c hc + hs g * * ( ) ) ]g hs = staging height Measured from top of footing to bottom of wall hcg = distance of CG of empty container from bottom of staging © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 30 . Jain. IIT Kanpur [ ( ) ] M c = ( Ah )c mc hc + hs g * * Lecture 5/ Slide 31 ( ) E-Course on Seismic Design of Tanks/ January 2006 . Jain.Bending moment:Elevated tanks Bending moment at the bottom of staging (Ah)i mig (Ah)i msg hi* (Ah)c mcg hc* hcg hs Top of footing hs M = ( Ah )i mi h + hs + ms hcg g * i * i © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 32 . Jain.Bending moment:Elevated tanks Total bending moment M = * M *2 i +M *2 c For shaft supported tanks. M* will be the design moment for shaft © Sudhir K. Bending moment:Elevated tanks For analysis of frame staging. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 33 . Jain. two approaches are possible Approach 1: Perform analysis in two steps Step 1: Analyze frame for (Ah)imig + (Ah)imsg Obtain forces in columns and braces Step 2: Analyze the frame for (Ah)cmcg Obtain forces in columns and braces Use SRSS rule to combine the member forces obtained in Step 1 and Step 2 © Sudhir K. Bending moment:Elevated tanks Approach 2: Apply horizontal force V at height h1 such that V x h1 = M* V and M* are obtained using SRSS rule as described in slide nos. analysis is done in single step Simpler and faster than Approach 1 © Sudhir K. Jain. 26 and 32 In this approach. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 34 . Example Example: An elevated tank on frame staging is shown below along with some relevant data. hc* = 4. A is CG of empty container A 2.2 m (Ah)i = 0. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 35 . Find base shear and bending moment at the bottom of staging. (Ah)c = 0. Jain.08.8 m mi = 100t.04 GL © Sudhir K. mc = 180 t Mass of container = 160 t Mass of staging = 120 t hs = 15 m hi* = 3 m. 81 = 70 . Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 36 . 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 .Example Structural mass.5 + 157 = 235.5 kN Base shear in convective mode Vc = (A h ) c m c g = 0 . 6 kN © Sudhir K. 04 x 180 x 9 . 81 = 78. we proceed to obtain bending moment at the bottom staging Distance of CG of empty container from bottom of staging.6 2 = 245.8 m © Sudhir K.Example Total base shear V = V i 2 + V c2 = 235 . hcg = 2.8kN Now.8 + 15 = 17. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 37 . Jain.5 2 + 70. 0 + 15 = 78.0m and 157 kN of force will act at 17. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 38 .5 x 18 + 157 x 17.8 = 4207 kNm )+ 200 x 17 .5 kN of force will act at 18. 8 ]x 9 .Example Base moment in impulsive mode M * i = ( A h )i [m i (h i * + h s + m s h cg ) ] g = 0 . 81 Note: 78.8 m from top of footing. 08 [ 100 x (3 . © Sudhir K. Jain. Jain. 81 = 70. 04 x180 x (4 . 2 + 15 )x 9 .Example Base moment in convective mode M c = ( Ah )c mc hc + hs g * * ( ) = 0 .2 = 1356 kNm Note: 70.2 m from top of footing.6 x 19. Total base moment M * = M * 2 i + M *2 c 2 = 4207 2 + 1356 = 4420 kNm © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 39 .6 kN of force will act at 19. 5 kN at 18 m and 157 kN at 17.Example Now.2 m from top of footing and analyze the frame Member forces (i. SF etc. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 40 .8 m from top of footing and analyze the frame Step 2: Apply 70. BM.6 kN at 19. 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. Jain. for staging analysis.e.. in columns and braces) of Steps 1 and 2 shall be combined using SRSS © Sudhir K. V = 245.8 x h1 = 4420 ∴ h1 = 17. BM. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 41 .98 m from top of footing and get member forces (i.Example Approach 2: Total base shear. apply force of 245.e. SF in columns and braces). such that V x h1 = M* 245. © Sudhir K. Jain.98 m Thus.8 kN will be applied at height h1..8 kN at 17. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 42 .Elevated tanks:Empty condition Elevated tanks shall be analysed for tank full as well as tank empty conditions Design shall be done for the critical condition In empty condition. tank will be modeled using single degree of freedom system Mass of empty container and 1/3rd staging mass shall be considered © Sudhir K. Jain. no convective liquid mass Hence. Ks will remain same in full and empty conditions In full condition. empty condition may become critical © Sudhir K.Elevated tanks:Empty condition Lateral stiffness of staging. time period of empty tank will be less Recall. T = 2 Π M K Hence. Sa/g will be more Usually. tank full condition is critical However. mass is more In empty condition mass is less Hence. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 43 . Jain. for tanks of low capacity. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 P Lecture 5/ Slide 44 . when force in Y-direction Hence.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. direction of lateral loading is important !! P Y X © Sudhir K. Jain.Direction of seismic force On the other hand. it is important to ascertain the most critical direction of lateral seismic force Direction of force. Y-direction is the most critical direction for deflection © Sudhir K. which will produce maximum response is the most critical direction In the rectangular cantilever problem. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 45 . if cantilever is circular Direction is not of concern Same deflection for any direction of loading Hence. Analysis may be done by any of the accepted methods including considering as space frame. IS 11682:1985 suggests that horizontal seismic loads shall be applied in the critical direction IS 11682:1985.2. “Criteria for Design of RCC Staging for Overhead Water Tanks”.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. © Sudhir K.1.Direction of seismic force For frame stagings consisting of columns and braces.1. Bureau of Indian Standards.2 Bending moments in horizontal braces due to horizontal loads shall be calculated when horizontal forces act in a critical direction. New Delhi Clause 7. 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. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 46 . Clause 7. Jain. 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.Direction of seismic force Section 4. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 47 . Jain. analysis in any one direction is sufficient © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 48 .Direction of seismic force Ground-supported rectangular tanks shall be analyzed for two directions Parallel to length of the tank Parallel to width of the tank Stresses in a particular wall shall be obtained for seismic loads perpendicular to that wall © Sudhir K. Jain. Direction of seismic force RC circular shafts of elevated tanks are also axisymmetric Hence. analysis in one direction is sufficient If circular shaft supports rectangular container Then. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 49 . Jain. analysis in two directions will be necessary © Sudhir K. Direction of seismic force 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. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 50 . Jain.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 © Sudhir K. in a 4 . Jain. IIT Kanpur Critical direction for axial force in column Lecture 5/ Slide 51 E-Course on Seismic Design of Tanks/ January 2006 .column staging Bending Axis Critical direction for shear force in brace © Sudhir K.Direction of seismic force Critical directions for 4 . U.Direction of seismic force For 6 – column and 8 – column staging.pdf) © Sudhir K.org/ecourse/Tank_ASCE. and Jain.120.5.. critical directions are given in Figure C-6 of the Guideline See next two slides More information available in Sameer and Jain (1994) Sameer. 1375-1393. K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 52 . “Lateral load analysis of frame staging for elevated water tanks”.. 1994. Journal of Structural Engineering.nicee. (http://www. S. S. Jain. ASCE. No. Vol. IIT Kanpur Critical direction for shear force and bending moment in braces and axial force in columns Lecture 5/ Slide 53 E-Course on Seismic Design of Tanks/ January 2006 .Direction of seismic force Critical directions for 6 .column staging Bending Axis Critical direction for shear force and bending moment in columns © Sudhir K. Jain. IIT Kanpur Critical direction for shear force. bending moment and axial force in columns Lecture 5/ Slide 54 E-Course on Seismic Design of Tanks/ January 2006 .column staging Bending Axis Critical direction for shear force and bending moment in braces © Sudhir K. Jain.Direction of seismic force Critical directions for 8 . following two load combinations can be used 100 % + 30% rule Also used in IS 1893(Part 1) for buildings SRSS rule © Sudhir K. Jain.Direction of seismic force As an alternative to analysis in the critical directions. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 55 . 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 © Sudhir K. Jain.3 ELY ELY + 0. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 56 .Direction of seismic force 100%+30% rule implies following combinations ELx + 0. ELX Analyze tank with seismic force in Y-direction. ELY Response quantity means BM in column. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 57 . Combine response quantity as per 100%+30% rule Combination is on response quantity and not on seismic loads © Sudhir K.Direction of seismic force 100%+30% rule requires Analyze tank with seismic force in X-direction. Jain. etc. SF in brace. obtain response quantity. obtain response quantity. there are total eight load combinations © Sudhir K.Direction of seismic force Important to note that the earthquake directions are reversible Hence. Jain. in 100%+30% rule. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 58 . Direction of seismic force SRSS rule implies following combination EL2 + EL2 x y 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. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 59 . analyze tank in two directions and use SRSS combination of response quantity © Sudhir K. At the end of Lecture 5 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. Jain. 100%+30% or SRSS rule be used © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 60 . this issue is discussed. if liquid is highly viscous. Why damping in convective mode is independant of type of liquid stored? For highly viscous liquids. that in highly viscous liquid. we have referred Newmark and Hall (1982).75 was arrived at for 0. I have been attending the ecourse on the design of liquid storage structures. I. The formulation used for obtaining hydrodynamic pressure. 2006 Questions and Comments on Lectures Question from Mr. 3. 2. In the commentary to Clause 4. thereby implying that more liquid will participate in impulsive mode.in ) Lecture 4 1. considers liquid to be non-viscous. Since we deal mostly with petroleum tank storages I am more interested in learning about the same. The potential equation is written with these assumptions on liquid properties.10 of IS 1893(Part 1):2002 Question from Mr. N.5% damping? Was logarithmic interpolation used? Yes. One can guess. but the convective mass will also change.5. homogeneous and incompressible. not only the damping. How the multiplicative factor of 1. does not viscosity of the liquid affect damping? This is an interesting question. then it would be an irregular building. less mass will undergo sloshing motion. wherein. Abdul Gani ([email protected] (through Distance Learning Mode) on Sesimic Design of Liquid Storage Tanks January 16 . logarithmic interpolation is used.February 6.There were a few points I would like . Please refer Clause 7. If a building has soft storey.ernet. Importance factor is not related to soft storey. Devanarayanan ([email protected]. Now. For multistoreyed buildings with "soft storey" what is the importance factor to be adopted? This is a question on buildings. this needs to be substantiated with in-depth study. H.co.in) Dear Sir. However. there will be a RC ring beam (or circular pile cap). first find the impulsive time period using formula given in Lecture 3 (also refer Clause 4.1.1 of the Guideline). Lecture 5 1.clarification on. the latest NEHRP recommendation of 2003 (FEMA 450) has used SRSS rule. Then. Tank wall and some portion of base plate rest on this ring beam. which is supported on compacted stone columns. For circular steel tanks. Hence. This tank will be considered as ground-supported tank and not as elevated tank. NEHRP recommendation of 2000 (FEMA 368) had suggested absolute summation. In the first case. For this case also. which supports rest of the base plate. 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. . However. find the convective mode time period and proceed with the evaluation of base shear coefficient. Till sometime back. 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. and will the piles/stone columns act as a staging thus making the whole structure behave like an elevated tank. In the second case. which is supported on piles. the underlying starta below the sand pad foundations are strngthened with stone columns. Inside the ring beam there will be sand filling. In such cases do we have to take the effect of seismic forces and design the tanks using the two degree of freedom.3. some recent numerical simulation studies have indicated that absolute summation rule overestimates the response. these modal responses may not reach their maximum values simultaneously. Hence. The two degree of freedom model of elevated tank is not applicable to these tanks. the entire base plate rests on sand fill. absolute summation is not used. tank will be treated as ground supported tank. 2006 .Lecture 6 February 7. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 2 .In this Lecture Hydrodynamic pressure on wall and base Effect of vertical ground acceleration Sloshing wave height Anchorage requirements for ground supported tanks © Sudhir K. Jain. emphasis was on lateral forces on tanks These lateral forces comprised of Impulsive component Convective component Impulsive component has two parts One due to impulsive liquid mass Second due to structural mass of tank Convective component has one part Due to convective liquid mass © Sudhir K.Hydrodynamic pressure So far. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 3 . Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 4 . we will see procedure to obtain Impulsive and convective pressure distribution on wall and base © Sudhir K.Hydrodynamic pressure Recall from Lecture 1 Impulsive force is summation of impulsive hydrodynamic pressure on wall Convective force is summation convective hydrodynamic pressure on wall Now. New Delhi © Sudhir K. Jain. Bureau of Indian Standards. “Code of Practice for Concrete Structures for the storage of Liquids”. 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.Hydrodynamic pressure Pressure distribution on wall is needed to obtain Hoop forces. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 5 . 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.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 . IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 6 . Jain.866 ⎢1 − ⎜ ⎟ ⎥ tanh ⎜ 0 . Jain. pressure will be maximum At φ = 900.Impulsive pressure:Circular tanks Note that impulsive pressure varies with circumferential angle. φ At φ = 00. 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 © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 7 . 0 0.4 1.5 0.8 Y/h 0.6 0.Impulsive pressure:Circular tanks Qiw(y) can also be read from Figure 9(a) of the Guideline This Figure is reproduced here 1 0.25 0 0 0.2 h/D =2 or h/L 1.4 Q iw 0.5 0. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 8 .6 0.8 1 © Sudhir K.2 0. Jain. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 9 .Impulsive pressure:Circular tanks Impulsive pressure on base of circular tanks changes in radial as well as circumferential direction A strip of length l’ is considered Refer plan of circular tank shown in Figure 8(a) of the Guideline l’ x D/2 φ O Direction of seismic force Plan © Sudhir K. 866 ⎝ = 0 . from the reference axis at the center of tank © Sudhir K. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 10 . 866 ( Ah )i ρ g h ⎛ cosh ⎜ 0 . 866 ⎜ ⎝ x⎞ ⎟ h⎠ l' ⎞ ⎟ h ⎟ ⎠ p ib x = horizontal distance in the direction of seismic force.Impulsive pressure:Circular tanks Impulsive pressure on this strip is given by ⎛ sinh ⎜ 0 . of a point on base slab. and water height of 5 m.866 [ 1. Impulsive base shear coefficient is 0. (Ah)i = 0. Jain. piw = Qiw (y) (Ah)i ρ g h cos φ Where.Impulsive pressure:Circular tanks Example: A circular tank has internal diameter of 12 m.(y /h)2] tanh(0.23 Impulsive pressure on wall is given by. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 11 .23. Find the impulsive pressure distribution on wall Solution: D = 12 m.866 D/h) Maximum pressure will occur at φ = 0 © Sudhir K. h = 5 m. Qiw (y)= 0. © Sudhir K.866 tanh(0. at φ = 0. ρ = 1000 kg/m3 ∴ piw (y =0) = 0.84 x 0. Some values are tabulated. piw = Qiw (y) (Ah)i ρ g h Mass density of water. piw is similarly calculated.5 N/m2 = 9.84 Now.866 tanh(0.866 x 12/5) = 0. Jain.81 x 5 = 9476. and a distribution plotted on the next slide. Qiw = 0. y = 0.866 D/h) = 0.Impulsive pressure:Circular tanks At base of wall.23 x 1000 x 9.48 kN/m2 For different values of y. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 12 . 84 9.25 0. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 13 .17 5 0 0 Qiw (y) piw (y) (kN/m2) h = 5m At φ = o 9.91 3.79 8.48 kN/m2 Impulsive pressure distribution on wall © Sudhir K.37 4. Jain.75 0.48 1.Impulsive pressure:Circular tanks y (m) 0 0. 866 ⎟ ⎢ h ⎠ ⎥ h ⎠ ⎝ ⎝ ⎢ ⎣ ⎦ Except following. 866 2 ⎡ L ⎞ ⎛ y ⎞ ⎤ ⎛ 1 − ⎜ ⎟ ⎥ tanh ⎜ 0 . Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 14 .Impulsive pressure:Rectangular tanks Impulsive pressure on walls of rectangular tanks is given by p iw = Q iw ( y ) ( A h ) i ρ g h Q iw ( y ) = 0 . 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 © Sudhir K. Impulsive pressure:Rectangular tanks This pressure is on walls perpendicular to direction of seismic force Pressure remains same along the length of these perpendicular walls However. Jain. it changes with wall height Walls perpendicular to direction of seismic force B L Direction of seismic force © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 15 . Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 16 .Impulsive pressure:Rectangular tanks Impulsive pressure on base of rectangular 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. 866 ⎝ ( x ) = ⎛ cosh ⎜ 0 . IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 17 . 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 © Sudhir K.Impulsive pressure:Rectangular tanks Impulsive pressure is given by pib = Qib ( x ) ( Ah )i ρ g h Q ib ⎛ sinh ⎜ 0 . Jain. Impulsive pressure:Rectangular tanks Coefficient of impulsive pressure on base Qib(x) 1.5 1.2 2.2 Figure 9(b) of the Guideline Lecture 6/ Slide 18 © Sudhir K.0 -0.8 0.4 -0 .4 -0. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 .25 = h/L -1.0 1. Jain.0 0 0 x/L 0.0 1.2 0.5 0.5 0.5 0.4 2.4 Qib -0 .2 0.8 Qib(x) 1.25 0. Jain.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 ⎝ y⎞ ⎟ D⎠ h⎞ ⎟ D⎠ © Sudhir K.5625 ⎛ cosh ⎜ 3.674 ⎝ Qcw ( y ) = 0. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 19 . Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 20 .Convective pressure:Circular tanks (Ah)c = Convective base shear coefficient Qcw = Coefficient of convective pressure on wall ρ = Mass density of liquid φ = Circumferential angle y = Vertical distance of a point on wall from the bottom of wall Qcw can be obtained from Figure 10(a) of the Guideline See next slide © Sudhir K. 0 1.4 h/D=0. Jain.Convective pressure:Circular tanks Coefficient of convective pressure on wall. Qcw 1 0.3 0.4 0.0 0.2 Q cw 0.25 0.5 Y/h 0.8 2.5 1. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 21 .1 0.6 Figure 10(a) of the Guideline © Sudhir K.6 0.2 0 0 0.5 0. Jain.Convective pressure:Circular tanks 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. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 22 .674 ⎟ D⎠ ⎝ ⎢D 3 ⎝ D ⎠ ⎥ ⎦ ⎣ Qcb can also be obtained from Figure 10(b) of the Guideline See next slide © Sudhir K.125⎢ − ⎜ ⎟ ⎥ sech⎜ 3. 2 -0.2 h/D=0.5 0 x/D 0.3 h/D=0.Convective pressure:Circular tanks Coefficient of convective pressure on base. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 23 .0 0.2 0.1 0.1 Q b c 0.0 0 1. Jain.5 0.4 -0.25 0.2 Qcb 0.25 -0. Qcb 0.75 1.4 -0.3 Figure 10(b) of the Guideline © Sudhir K.75 -0. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 24 . Convective base shear coefficient is 0. h = 5 m. and water height of 5 m.1/3 cos 2φ] cos φ Where. pcw = Qcw (y) (Ah)c ρ g D [1.5625 cosh(3. (Ah)c = 0.Convective pressure:Circular tanks Example: A circular tank has internal diameter of 12 m. Jain.674 h/D) Maximum pressure will occur at φ = 0 © Sudhir K. Find the convective pressure distribution on wall Solution: D = 12 m.07. Qcw (y)= 0.674 y/D)/cosh(3.07 Convective pressure on wall is given by. 5625/cosh(3. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 25 . Jain. y = 0. ρ = 1000 kg/m3 ∴ pcw (y =0) = 0.Convective pressure:Circular tanks At base of wall.674 x 5/12) = 0.674 h/D) = 0.26 kN/m2 For different values of y. © Sudhir K.1/3] = 1263. pcw is similarly calculated.07 x 1000 x 9. pcw = Qcw (y) (Ah)c ρ g D [1.23 Now. Qcw (y =0) = 0. at φ = 0.1/3 ] Mass density of water.23 x 0. Pressure distribution with wall height is shown in the next slide.81 x 12 x [1.5 N/m2 = 1.5625/cosh(3. 40 2.75 0.20 5 0.25 1.09 Qcw (y) pcw (y) (kN/m2) 3.25 0.26 1. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 26 .23 1.Convective pressure:Circular tanks y (m) 0 0.37 3.09 kN/m2 h = 5m At φ = o 1.5625 3.26 kN/m2 Convective pressure distribution on wall © Sudhir K. Jain. 162 ⎝ y⎞ ⎟ L⎠ h⎞ ⎟ L⎠ Qcw can also be obtained from Figure 11(a) of the Guideline See next slide © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 27 .4165 ⎛ cosh ⎜ 3 .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 . Jain.162 ⎝ Qcw ( y ) = 0 . 4 h/L=0. Qcw 1 0.5 0.1 0.0 0.Convective pressure:Rectangular tanks Coefficient of convective pressure on wall.6 2. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 28 .2 Q cw 0.8 Y/h 0.3 0. Jain.2 0 0 0.0 1.2 0.5 1.4 0.5 Figure 11(a) of the Guideline © Sudhir K. 162 ⎟ L⎠ ⎝ ⎢L 3 ⎝ L ⎠ ⎥ ⎣ ⎦ Qcb can be obtained from Figure 11(b) of the Guideline However.Convective pressure:Rectangular tanks 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 . IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 29 . Figure 11 (b) of the Guideline is incorrect Correct Figure is shown in next slide © Sudhir K.25 ⎢ − ⎜ ⎟ ⎥ sec h ⎜ 3 . Jain. 4 -0.3 0.2 -0.2 h/L = 0.1 0.25 Figure 11(b) of the Guideline (corrected) © Sudhir K.2 -0.5 Qcb 0.4 0.50 x/L h/L = 0. Jain.4 0.3 -0.Convective pressure:Rectangular tanks Coefficient of convective pressures on base.5 1. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 30 .0 0 0.2 0.50 0.4 0. Qcb 0.3 0.1 -0.1 0 -0.3 -0.0 -0.1 -0.25 0.75 1.75 0. 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. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 31 . refer IS 3370 (Part IV):1967 © Sudhir K. Jain. equivalent linearised pressure distribution is suggested For impulsive as well as convective pressure This linearisation is such that. base shear and bending moment at the bottom of wall remains same © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 32 . Jain.Linearised Pressure Distribution Hence. for convective pressure © Sudhir K.Linearised Pressure Distribution Equivalent linear distribution for impulsive pressure will be such that Total base shear and bending moment due to linear distribution will be same as that due to curvilinear distribution Similarly. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 33 . Jain. Linearised Pressure Distribution Equivalent linear impulsive pressure bi qi h hi Actual Impulsive pressure distribution qi hi ai Equivalent linear pressure distribution ˜ © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 34 . Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 . Jain.Linearised Pressure Distribution The ordinates of linear distribution are ai = qi (4 h − 6 hi ) 2 h bi = qi (6 h i − 2 h 2 h ) Both for circular and rectangular tanks Both for qi is different for circular and rectangular tanks ( A h )i m i qi = g π D / 2 ( A h )i m i qi = g 2B Lecture 6/ Slide 35 For circular tanks For rectangular tanks © Sudhir K. Linearised Pressure Distribution 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 © Sudhir K. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 36 . IIT Kanpur ( A h )c m c = g 2B Lecture 6/ Slide 37 E-Course on Seismic Design of Tanks/ January 2006 . equivalent linear convective pressure bc qc h qc hc ˜ hc qc b c = 2 (6 h c − 2 h h qc a c = 2 (4 h − 6 h c ) h For circular tanks ) Actual convective pressure distribution ac Equivalent linear pressure distribution qc = ( Ah )c mc g π D/2 For rectangular tanks qc © Sudhir K. Jain.Linearised Pressure Distribution Similarly. IIT Kanpur + bi ai . Jain.Linearised Pressure Distribution Linearised distribution can be treated as summation of two parts Uniform pressure Triangular pressure bi = ai © Sudhir K.bi Lecture 6/ Slide 38 E-Course on Seismic Design of Tanks/ January 2006 . Linearised Pressure Distribution Hoop forces and BM in wall due to uniform and triangular pressure can be obtained from IS 3370(Part IV):1967 Linearised pressure distributions discussed here are taken from ACI 350. MI. Farmington Hill.3 ACI 350.3. © Sudhir K. 2001. “Seismic design of liquid containing concrete structures”. American Concrete Institute. Jain. USA. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 39 . 07 Capacity of tank = π x 122 /4 x 5 = 565. Find linearised impulsive and convective pressure distribution on wall.47.65.500 kg For h/D = 5/12 = 0. h = 5 m. respectively. (Ah)c = 0. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 40 .5 x 1000 = 5.07.23 and 0.23.5 m3 ∴ Mass of water = 565. Solution: D = 12 m. Jain. Impulsive and convective base shear coefficients are 0.Linearised Pressure Distribution Example: A circular tank has internal diameter of 12 m.47 x 565500 = 2. and water height of 5 m.800 kg © Sudhir K.42. ∴ mi = 0.65. from Figure 2a of the Guideline: mi/m = 0. (Ah)i = 0. 9 m a) Impulsive mode ( Ah ) i mi 0 .81kN/m π D /2 π × 12 / 2 Pressure at bottom & top is given by qi 31 . hc = 0.375 x 5 = 1.82.13 kN/m a i = 2 (4 h − 6 hi ) = 2 h 5 © Sudhir K.58.5. hi = 0.50 x 565500 = 2.875 ) = 11 .58 x 5 = 2.Linearised Pressure Distribution mc/m = 0.800 kg hi /h = 0. mc = 0.875 m hc /h = 0. 81 (4 × 5 − 6 × 1 .800 × 9 . Jain.375.81 qi = g= = 31 .23 × 265 . IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 2 Lecture 6/ Slide 41 . IIT Kanpur 9. Jain.59 kN/m2 h = 5m ai = 11.281 (6 × 1. Linear and actual pressure distributions are shown below: bi = 1.For the same tank.875 − 2 × 5) = 1.48 kN/m2 Actual pressure distribution ( Obtained in earlier example) Lecture 6/ Slide 42 E-Course on Seismic Design of Tanks/ January 2006 . in earlier example.13 kN/m2 Linear pressure distribution © Sudhir K.Linearised Pressure Distribution bi = qi (6hi − 2h ) = 31. we had obtained actual distribution of impulsive pressure.59kN / m 2 5 h2 Note:. 3 (6 × 2 . 07 kN/m h 5 qc 10 . 9 ) = 1 . 3 a c = 2 (4 h − 6 h c ) = 2 (4 × 5 − 6 × 2 . Jain. 9 − 2 × 5 ) = 3 .800 × 9 .81 qc = g= = 10 . IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 43 .Linearised Pressure Distribution b) Convective mode ( Ah ) c m c 0 .3kN/m π D/2 π × 12 / 2 Pressure at bottom & top is given by qc 10 .07 × 282 . 05 kN/m b c = 2 (6 h c − 2 h ) = 2 h 5 2 © Sudhir K. Linear and actual pressure distributions are shown below: bi = 3. in earlier example we had obtained actual distribution of convective pressure.26 kN/m2 Actual convective pressure distribution (Obtained in earlier example) © Sudhir K.For the same tank.Linearised Pressure Distribution Note:.07 kN/m2 Linear pressure distribution 1. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 44 .05 kN/m2 3. Jain.09 kN/m2 h = 5m ai = 1. Circumferential distribution In circular tanks. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 45 . pressure varies along the circumference of the wall Recall there is cosφ term in the expression for pressure on wall φ © Sudhir K. Jain. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 46 . 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 © Sudhir K.Circumferential distribution Hoop forces and BM in wall for such varying pressure are also not readily available Hence. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 47 .Circumferential distribution Simplified distribution in circumferential direction pmax pmax φ Actual pressure distribution Simplified pressure distribution © Sudhir K. Jain. weight of liquid and tank will increase or decrease depending on direction of acceleration If acceleration is downward. weight will increase Increase in weight = vertical acceleration x mass © Sudhir K. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 48 . weight will decrease If acceleration is upward.Effect of vertical acceleration Due to vertical acceleration. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 49 . 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 © Sudhir K.Effect of vertical acceleration Increase in liquid weight would lead to increase in hydrostatic pressure exerted by liquid on wall Decrease in weight will reduce hydrostatic pressure. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 pv y Lecture 6/ Slide 50 . Jain. pv due to vertical acceleration pv = ( Av ) ρ g h (1 − y h) Av = 2 3 ⎛Z S ⎞ I ⎜ × × a⎟ ⎜ 2 R g ⎟ ⎠ ⎝ ρ = Mass density of liquid h = Total liquid height g = Acceleration due to gravity Z = Zone factor I = Importance factor R = Response reduction factor © Sudhir K.Effect of vertical acceleration Pressure. Effect of vertical acceleration Design acceleration spectrum in vertical direction is taken as 2/3rd of design acceleration spectrum in horizontal direction This is consistent with IS 1893 (Part 1):2002 For finding the value of Sa/g One needs to know time period of tank in vertical mode © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 51 . Jain. 3 sec for all tanks This implies. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 52 . value of Sa/g will be 2. Jain.Effect of vertical acceleration Usually tanks are quite rigid in vertical direction Hence. time period in vertical mode may be taken as as 0.5 Vertical mode of vibration for circular tanks is called breathing mode More information on time period of breathing mode is available in Eurocode 8 and ACI 350.3 © Sudhir K. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 53 . 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 g = thickness of tank wall = acceleration due to gravity pww (Ah)i = impulsive base shear coefficient © Sudhir K. during lateral excitation.Pressure due to wall inertia Mass of wall is distributed along its length and height Hence. 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. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 54 . pressure due to wall inertia may be significant © Sudhir K. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 55 . pww Total pressure on wall is obtained as p = ( piw + pww ) + p + pv 2 2 cw 2 © Sudhir K. pv Pressure due to wall inertia. Jain. pcw Pressure due to vertical acceleration. piw Convective hydrodynamic pressure.Total pressure on wall Total pressure on wall comprises of Impulsive hydrodynamic pressure. R = 2. Find maximum pressure due to vertical excitation and wall inertia. I = 1. It is ground supported RC tank with fixed base. Z= 0.Example A ground supported circular RC water tank has internal diameter of 12 m and liquid height is 5 m. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 56 . Wall thickness = 200 mm Zone IV. Wall thickness is 200 mm. D = 12 m. Solution: h = 5 m.23. It is located in zone IV and has fixed base.0 © Sudhir K. Jain. (Ah)i = 0.24.5. 0 x 2.0/5) = 7.81 x 5 x (1.3 sec for all tanks ∴ Sa/g = 2. I/R . Sa/g = 2/3 x 0. Jain.y/h) = 0.e. i.24/2 x 1. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 57 .5/2.5 = 0.15 Maximum pressure due to vertical acceleration will occur at y = 0. at the base of wall.36 kN/m2 © Sudhir K. T = 0.15 x 1000 x 9.Example For vertical mode.5 Av = 2/3 . time period is taken as. Z/2. ∴ pv = (Av) ρ g h (1. Jain.2 x 25 = 1.15 kN/m2 © Sudhir K. pww = (Ah)i t ρm g = 0.23 x 0.Example Maximum pressure due to wall inertia. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 58 . Hence. Jain.48 + 1. from the previous examples. Pv = 7. Pcw = 1. total pressure p = ( piw + pww) + p + pv 2 2 cw 2 = ( 9. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 59 . © Sudhir K.26 kN/m2.36 2 = 13 kN/m2 Total hydrostatic pressure at base = ρ g h = 49 kN/m2. Thus.262 + 7.48 kN/m2.Example For this tank. Pww = 1.36 kN/m2.15 kN/m2.15 )2 + 1. at the bottom of wall (y = 0) we have Piw = 9. total hydrodynamic pressure of 13 kN/m2 is 26% of hydrostatic pressure. Sloshing wave height Convective liquid undergoes sloshing motion i. Jain. liquid undergoes vertical motion dmax © Sudhir K.. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 60 .e. Sloshing wave height Maximum sloshing wave height is given by d max D = ( A h )c R 2 L 2 For circular tanks For rectangular tanks d max = ( A h )c R (Ah)c = convective base shear coefficient R D = response reduction factor = diameter of circular tank L = Length of rectangular tank in the direction of seismic force © Sudhir K. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 61 . Sloshing wave height Multiplication with R is due to the fact that design forces are reduced by factor R Recall. Jain. (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 © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 62 . 1185-1192 © Sudhir K.Sloshing wave height 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. 21. Vol. 4. sloshing liquid mass will also change Refer Malhotra (2005) for more information Malhotra P K (2005). Earthquake Spectra. No. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 63 . Jain. liquid will exert pressure on roof In the process. “Sloshing loads in liquid-storage tanks with insufficient free board”. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 64 . Jain. if overturning moment due to lateral force is more than stabilizing moment © Sudhir K.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. one can say. diameter D is to be replaced by L © Sudhir K. Jain. overturning will occur if h > D h = height of liquid D = diameter of tank 1 (A h )i (Ah)i= impulsive base shear coefficient For rectangular tanks.Anchorage requirement As a quick estimate. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 65 . 12 of the Guideline © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 66 . Jain.Anchorage requirement 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. At the end of Lecture 6 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 © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 6/ Slide 67 . Jain. 2006 .Lecture 7 February 16. we will discuss only important points These tanks will also be analysed using IS 1893:1984 Comparison of forces obtained from the Guideline and IS 1893:1984 will be presented © Sudhir K. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 2 .In this Lecture Examples 2 & 3 of IITK-GSDMA Guideline These examples have been sent separately Here. base shear and base moment at the base of staging are obtained © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 3 . Jain.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. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 4 . Jain. all the parameters of spring-mass model are obtained Step 4: Lateral stiffness of staging Using standard structural analysis software.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. stiffness is obtained More about this later © Sudhir K. I. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 5 . and damping values Step 7: Base shear and base moment Step 8: Analysis of empty tank © Sudhir K. Jain.Example 2 Step 5: Impulsive convective mode time period Step 6: Design horizontal seismic coefficient (Ah)i and (Ah)c are obtained using Z. R. 6 t © Sudhir K.6 t Structural mass. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 6 . ms = 160.6 + 196 = 451. Jain.7 t Mass of staging = 105.7 + 105.Example 2 Mass of water = 255.6/3 = 196 t ms = empty container + 1/3rd mass of staging Total mass = 255.6 t Mass of empty container = 160. mi = 140.4 m. h/D = 0. about 55% of total water mass mc =110 t. This was discussed in Lecture 3. about 43% of total water mass mi + mc is about 2% less than total water mass. hi. 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. mc. Jain. hence.6 t. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 7 .51 This h/D is used to find mi. hc etc. © Sudhir K.Example 2 Parameters of spring-mass model Here. Example 2 hi = 1. hi* = 3. Hence.65 m. hi* and hc* are same in this case.43 m Incidentally.43 m hc = 2.3 times hc Indicates that impulsive pressure at the base is more significant than the convective pressure at the base. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 8 . hc* = 3.68 m. hi* is significantly larger than hi hc* is not much larger than hc © Sudhir K. hc* is only 1. Jain. hi* is twice that of hi However. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 9 . In this example. mi. Jain.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. CG of empty container is treated as CG of tank A reasonable approximation for design purpose © Sudhir K. 88 + 0. force is applied at 2.Example 2: Lateral stiffness CG of empty container is at a distance of 2. a rigid link of length 3.18 m from center-line of circular ring beam 0.3 = 3.3 m is half-depth of ring beam Since only staging is modeled.88 m from top of circular ring beam In the model.18 m is provided to apply force at that height Model is shown in next slide © Sudhir K. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 10 . center-line dimensions are used Hence. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 11 .0 Rigid link Structural model of staging used for analysis © Sudhir K.Example 2: Lateral stiffness 10. Jain. lateral Stiffness of Staging Ks = 10/(5.Example 2: Lateral stiffness A force of 10 kN is applied at CG Deflection of CG = 5.616x10-4) = 17.800 kN/m Here.616 x10-4 m Hence. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 12 . computer software is used for analysis Traditionally. designers use manual methods Present frame is a polygon frame © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 13 . braces are considered as rigid beams © Sudhir K. general practice in India has been to use method suggested by SP 22 SP:22 – 1982. Bureau of Indian Standards.Example 2: Lateral stiffness For finding stiffness of such frames. Explanatory Handbook on Codes for Earthquake Engineering. New Delhi In this method. Jain. Example 2: Lateral stiffness If braces are considered as rigid beams. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 14 . 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 © Sudhir K. Jain. Jain. IIT Kanpur C = 6 × 36730 = 220. Kc K C =12EI =12 ×22360 ×10 33×8.400 kN/m E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 15 .730 kN/m L3 4.76 ×10 9 =36. there are 6 columns Hence.0 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ In each panel. stiffness of each panel ∑K © Sudhir K.Example 2: Analysis as per IS 1893:1984 For the frame in the present example Stiffness of each column in a panel. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 16 .090 kN/m 4 © Sudhir K. Jain.Example 2: Analysis as per IS 1893:1984 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. 800 kN/m from computer analysis Ks = 55. stiffness of frame Ks = 17. Jain.090 kN/m from method of SP22 Thus. braces should not be treated as rigid beams Their flexibility should be included in the analysis © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 17 .Example 2: Analysis as per IS 1893:1984 Thus. SP 22 grossly overestimates the stiffness Clearly. 1992.. S. K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 18 . and Jain. 691-698.. “Approximate methods for determination of time period of water tank staging”.nicee. Jain. Vol. 66.pdf) © Sudhir K. U. 12.Example 2: Analysis as per IS 1893:1984 Jain and Sameer (1992) have developed a simple method for obtaining stiffness of polygon frames This method considers flexibility of braces Sameer. The Indian Concrete Journal. No. (http://www. S.org/ecourse/Tank_ICJ. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 .Example 2: Lateral stiffness As per Jain and Sameer (1992) 1 1 1 = + K s K flexure K axial 1 K flexure = ∑K i =1 Np 1 panel Top most panel Kpanel = stiffness of each panel Np = number of panels Bottom most panel Lecture 7/ Slide 19 © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 20 . Jain.Example 2: Lateral stiffness K panel ⎤ Ib L 12 EI c N c ⎡ = ⎢ ⎥ h3 Ib L + 2 Ic h ⎦ ⎣ = For intermediate panels K panel ⎤ For top most panel 12 EI c N c ⎡ I cbr L ⎢ ⎥ h3 I cbr L + I c h ⎦ ⎣ For bottom most panel K panel ⎤ 12 EI c N c ⎡ Ib L = ⎥ ⎢ h3 Ib L + Ic h ⎦ ⎣ 1 K axial 2 = N c Ac E R 2 ∑ i =1 Np Hi h 2 © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 21 .Example 2: Lateral stiffness E = Modulus of elasticity Ic = Moment of inertia of column Nc = Number of columns h = Height of each panel L = Length of each brace Ib = Moment of inertia of brace beam Icbr = Moment of inertia of circular ring beam Hi = Distance of lateral load from inflection point of ith panel © Sudhir K. Jain. Jain. 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 © Sudhir K. point of inflection is at mid height of panel For the top most panel.Example 2: Lateral stiffness Hi = Distance of lateral load from inflection point of ith panel For intermediate panels. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 22 . 4 x 109 mm4 Icbr = 500 x (600)3 / 12 = 9 x 109 mm4 h = 4. Nc = 6.180 mm. Jain.330 mm Note:.141 mm Ic = π x (650)4/ 64 = 8. L = 3. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 23 .180 mm.800 mm2. and distances Hi are from the load.Load is applied at CG of container.31.141 mm R = 3. Ac = π x (650)2/ 4 = 3. H2 = 9. H3 = 13. © Sudhir K.76 x 109 mm4 Ib = 300 x (600)3 / 12 = 5.000 mm H1 = 6030 mm.Example 2: Lateral stiffness For 6–column staging of this Example E = 22.360 N/mm2. H4 = 16. we get Stiffness of staging. we have Ks = 17. Which is very reasonable Recall.Example 2: Lateral stiffness Using this method.090 kN/m Which is on much higher side © Sudhir K.440 kN/m From computer analysis. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 24 . Ks = 16. Ks = 55.800 kN/m Difference is about 7%. Jain. as per SP 22. usually in practice. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 25 .800 kN/m is used © Sudhir K. For further calculations here. method of SP 22 is used Also this method has been used in a number of text books in India.Example 2: Lateral stiffness Note. Ks = 17. Jain. 14 sec © Sudhir K.51. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 26 . Tc Tc = Cc D/g For h/D = 0. Ti Ti = 2π mi + ms Ks = 0. Tc = 3.Example 2: Time period Tank full condition Impulsive mode time period. Jain. from Figure 5 of the Guideline. Cc = 3.35 Hence.86 Sec Convective time period. Example 2: Time period Convective time period is 3.65 times impulsive time period Hence, two DoF model can be treated as two uncoupled SDoF systems © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 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 © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 28 Example 2: Base shear Design Horizontal Coefficient (Ah)i = Z/2 . I/R . (Sa/g)i = 0.084 (Ah)c = Z/2 . I/R . (Sa/g)c = 0.040 (Ah)c is less than (Ah)i Convective mass is subjected to much smaller acceleration © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 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 © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 30 Example 2:Base shear V = Vi 2 +Vc2 Total Base shear, = 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 © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 31 Example 2:Base shear 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 © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 32 Example 2:Base shear Location of forces as per Approach 1 116 kN 161 kN hi*= 3.43 m 43 kN hc*=3.43 m hcg = 19.18 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 E-Course on Seismic Design of Tanks/ January 2006 Example 2:Base shear Approach 2: Apply base shear, V at height h1 such that V x h1 = M* V = 280 kN and M* = 5448 kN-m Hence, h1 = 5448/280 = 19.46 m Thus, a force of 280 kN to be applied at 19.46 m © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 34 Example 2:Base shear Location of forces as per Approach 2 280 kN h1 =19.46 Top of footing © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 35 Example 2:Base shear 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 © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 36 Example 2:Base shear Tank empty condition Moment = 211 x 19.18 = 4053 kNm 211 kN hcg =19.18 Top of footing © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 37 Example 2:Base shear Base shear and moment are more in tank full condition Hence, design will be governed by tank full condition These forces are to be applied in suitable directions to get critical design forces in columns and braces Recall, the issue of critical direction for frame staging © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 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. © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 39 5 Fo= 0.Example 2: Analysis as per IS 1893:1984 Zone and soil parameters from IS 1893:1984 β = 1. T M T = 2π K M is total Mass This is sum of structural mass and total liquid mass K is Stiffness of Staging © Sudhir K.0 I = 1.25 (Zone IV) Time Period. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 40 . Ks = 55.Example 2: Analysis as per IS 1893:1984 Total Mass. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 41 .090 kN/m This we have seen earlier © Sudhir K.6 t Lateral stiffness Ks Braces are treated as as rigid beams as per prevailing practice Hence. Jain. M = Structural mass + liquid mass = 196 +255.6 = 451. Example 2: Analysis as per IS 1893:1984 Time period T = 2π M 451.86 sec calculated earlier Because.155 From Figure 2 of IS 1893 : 1984 © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 42 . Jain.6 = 2π = 0. stiffness of braces is taken infinite For 5% damping & T = 0.57 Sec Sa/g = 0.57 Sec K 55090 This time period is less than the impulsive time period of 0. 929 kNm © Sudhir K.81) = 257 kN Base Shear. M = 257 x 19.058 x (451.25 x 0.18 = 4. Jain.8 % of seismic weight of 4429 kN This will act at the CG of container CG of container is at 19. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 43 .0 x 1.18 m from footing top Hence base moment.5 x 0. V = αh x W Base shear is 5.155 = 0.6 x 9. αh αh = βIFo(Sa/g) = 1.058 = 0.Example 2: Analysis as per 1893:1984 Design horizontal seismic coefficient. 2 As per Figure 2 of IS 1893:1984 Design horizontal seismic coefficient. Jain.Example 2: Analysis as per 1893:1984 Analysis for empty condition Structural mass.2 = 0.0 x 1. M =196 t T = 2π M 196 = 2π = 0.075 © Sudhir K.25 x 0. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 44 .5 x 0.375 Sec Sa/g = 0.375 Sec K 55090 For 5% damping & T = 0. αh αh = βIFo(Sa/g) = 1. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 45 . M = 144 x 19. we compare seismic forces obtained from the Guideline and IS 1893:1984 © Sudhir K. V = αh x W = 0. Jain.18 = 2762 kNm Next.075 x (196 x 9.81) = 144 kN Moment at the bottom of staging.Example 2: Analysis as per 1893:1984 Base Shear. IIT Kanpur Convective mode.089 kN/m Impulsive mode. Tank empty ( Ti ) Tank full ( Ti) 0.66 sec 0.14 sec 0.11 0.075 0. 0. Jain.86 sec 3. Tank full ( Tc) Design seismic horizontal coefficient Impulsive mode Tank empty ( Ah)i Tank full ( Ah)i Tank full ( Ah)c © Sudhir K.058 ----Lecture 7/ Slide 46 E-Course on Seismic Design of Tanks/ January 2006 .Example 2:Comparison of forces Comparison of results from the Guideline and IS 1893:1984 Lateral stiffness of staging Time period Guideline 17.57 sec ----- Convective mode.375 sec 0.040 0.084 0.800 kN/m IS 1893:1984 55. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 47 .Example 2:Comparison of forces Comparison of results from the Guideline and IS 1893:1984 Guideline Total base shear (V) Tank empty Tank full Bending Moment Tank empty Tank full 211 kN 280 kN 4053 kN 5448 kN IS 1893:1984 144 kN 257 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 © Sudhir K. 0. These two effects compensate each other Hence.Example 2:Comparison of forces Two points may be noted IS 1893:1984 implied K = 1. Jain. there is not much difference in design forces from the Guideline and IS 1893:1984 for elevated tanks on frame type staging.which increases seismic coefficient. the practice was to overestimate stiffness. which leads to smaller seismic forces However. © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 48 . Alternatively. it is a space frame © Sudhir K. moment distribution method and Kani’s method In this example. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 49 .Example 2: Member forces To obtain design forces in columns and braces Staging frame is to be analysed for lateral seismic force One can use standard structural analysis programs. staging has polygon frame It is not a plane frame Strictly speaking. one may use approximate analysis methods. Plane frame analysis methods are well known For example. Jain. Eighth Edition.. 1980. Nem Chand Brothers. New Delhi These methods make simplifying assumptions Accuracy of these methods depends on validity of these assumptions © Sudhir K. Jain. 2. 1986. Roorkee Dayaratnam P. Oxford & IBH Publishing Co. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 50 . Vol. “Design of Reinforced Concrete Structures”. “Plain and reinforced Concrete”.Example 2: Member forces Some text books describe methods for analysis of such polygon frames Jain and Jai Krishna (1980) and Dayaratnam (1986) Jain O P and Krishna J. OR Jain and Jai Krishna distribute lateral shear similar to that in a cross-section of a cantilever column © Sudhir K.Example 2: Member forces Major assumptions in these methods are: 1) Axial force in column due to lateral force is proportional to its distance from bending 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. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 51 . Example 2: Member forces Sameer and Jain (1994) have critically examined these assumptions Sameer. and Jain. “Lateral load analysis of frame staging for elevated water tanks”. Vol. K. Journal of Structural Engineering. assumption 3) in the previous slide. 1994. ASCE. No.. Jain. (http://www.120. U. Column shear is obtained by moment equilibrium in the plane of bending © Sudhir K. 1375-1393. S. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 52 .pdf) They pointed out that assumption regarding shear distribution is not necessary That is.. S.5.org/ecourse/Tank_ASCE.nicee. Jain and Sameer (1994) have developed a new approach to obtain design forces in columns and braces © Sudhir K. Based on these observations. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 53 .Example 2: Member forces Point of inflection for the top and bottom panels is not at mid-span Sameer and Jain obtained point of inflection for these panels more accurately. . design forces for columns and braces are obtained Total lateral force = 280 kN acting at 19.Example 2: Member forces For the present example.46 m from footing top (i. Jain. 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 © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 54 .e. Jain.Example 2: Member forces 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) Braces Bending moment (kNm) Shear force (kN) 492 150 66 175 112 494 160 70 187 119 519 187 93 280 178 519 94 47 187 119 © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 55 . Example 2: Member forces 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 © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 56 . Jain. 28 m Height of shaft above GL = 15 m Hard soil.15 m Centre line diameter of shaft = 6. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 57 .Example 3:Elevated Tank on Shaft Example 3 of the Guideline Elevated tank on RC shaft is considered Container is same as in Example 2 Thickness of shaft = 0. Zone IV © Sudhir K. Jain. 7/3 = 201.. shaft = π x 6. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 58 . structural mass ms ms = 160.28 x 0.Example 3:Elevated tank on Shaft Weight of staging.15 x 16.e.9 t mi = 140. i.4 x 25 = 1.213 kN Mass of staging = 123.7 + 123.7 t Hence. hence.6 t mc =110 t Container is same as in Example 2. © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 59 .Example 3:Lateral stiffness Lateral stiffness of staging Shaft is considered as a cantilever of length 16. E = Modulus of elasticity = 22. Jain.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.36 x 106 kN/m2 © Sudhir K. 36 ×10 6 ×14.Example 3:Lateral stiffness I = Moment of inertia of shaft cross section = π x (6.22 ×10 5 kN/m 16.59 KS = = 2. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 60 .43 © Sudhir K.134)/64 = 14.4 m Thus.434 – 6. Jain. 3 × 22.59 m4 L = Height of shaft = 16. in frame staging. stiffness was obtained by applying force at the CG of empty container In frame staging. there is not much difference in stiffness Rigid link portion does not have much higher deflection than deflection at staging top © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 61 . Jain.Example 3:Lateral stiffness Height of shaft is taken up to ring beam only and not up to CG of empty container Recall. 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. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 62 . then. if we wish to consider height up to CG. Jain. 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.Example 3:Lateral stiffness In shaft. It will undergo only rigid body rotation Deflection at CG will not be much higher than that at staging top © Sudhir K. Jain.Example 3:Lateral stiffness Hence. staging height taken up to ring beam is adequate for design This will be on the conservative side Also recall time period varies as square root of stiffness A 15% error in stiffness will give about 7% error in time period There is another approximation in stiffness of shaft Shear deformations © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 63 . IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 64 . P Tip deflection due to shear deformation PL δs = ' κ AG G = Shear modulus A = Cross-sectional area κ’ = Shape factor Depends on shape of cross-section © Sudhir K.Example 3:Lateral stiffness Effect of shear deformation Consider a cantilever subjected to tip load . Jain. Example 3:Lateral stiffness Thus. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 65 . 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 © Sudhir K. Jain. Example 3:Lateral stiffness For hollow circular cross-section Hence. Jain.5 © Sudhir K.5 AG κ’ = 0. lateral stiffness of shaft staging. including shear deformations. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 66 . will be Ks = 1 L3 L + 3EI 0. Jain.22 x 105 kN/m is used This value is also used in the Guideline However. it should have been 1. if shear deformations are included.22 x 105 kN/m By including shear deformations. stiffness has reduced by 20%.Example 3:Lateral stiffness In the present example.77 x 105 kN/m Recall. For further calculations. IIT Kanpur Lecture 7/ Slide 67 . Ks=2. Ks = 2.77 x 105 kN/m E-Course on Seismic Design of Tanks/ January 2006 © Sudhir K. then Ks = 1. without shear deformation. 25 Sec Shafts are quite rigid. low time period Recall. Ti Ti = 2π mi + ms Ks = 0.Example 3: Time period Analysis for tank full condition Impulsive mode time period. Ti = 0. for frame staging. hence. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 68 . Jain.86 sec © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 69 . the assumption of two uncoupled SDoF can be used © Sudhir K.Example 3: Time period Convective time period. Jain. Tc = 3.0 Impulsive and convective time periods are quite well separated Hence.14 sec Same as in Example 2 Ratio of Tc and Ti is about 12. 5. Jain.Example 3: Base shear Design horizontal seismic coefficient Zone IV.5 From Clause 4. then time period Ti would have been 0. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 70 . Z = 0.3 of the Guideline Note: If we include the effect of shear deformation.28 sec And. (Sa/g)i = 2.5 R = 1. hence.5 © Sudhir K.8 For Ti = 0.24 I = 1. (Sa/g)i will remain 2.25 sec and hard soil. I/R .75 is for 0. I/R .318 x 1. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 71 .5% damping Design Horizontal Coefficient.25 (Ah)c = Z/2 .14. (Sa/g)i = 0. Ah (Ah)i = Z/2 .75 = 0.56 Multiplication by 1.Example 3: Base shear For Tc = 3. Jain. (Sa/g)c = 0.06 (Ah)c is much less than (Ah)i Convective mass is subjected to much smaller acceleration © Sudhir K. (Sa/g)c = 0. 6 + 201.06 x (110) x 9.8) x 9.25 x (140.Example 3:Base shear Base Shear. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 72 . V Impulsive mode Vi = (Ah)i (mi + ms)g Vi = 0.81 = 345 + 495 = 840 kN Convective mode Vc = (Ah)c mc g = 0. Jain.81 = 65 kN Convective forces are much smaller than impulsive forces © Sudhir K. = 843 kN Base shear is about 19 % of total seismic weight of 4.Example 3:Base shear V = Vi 2 +Vc2 Total Base shear. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 73 .488 kN (tank is located in seismic zone IV) © Sudhir K. Example 3:Base shear 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 © Sudhir K. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 74 . 43 + 17) = 1322 kNm Total base moment M * = M * 2 i +M * 2 c = 16. IIT Kanpur Lecture 7/ Slide 75 E-Course on Seismic Design of Tanks/ January 2006 .00 kNm Base moment in convective mode Mc* = 65 x (3. Jain.Example 3:Base shear Base moment in impulsive mode Mi* = 345 x (3.43 + 17) + 495 x 19.940 kNm © Sudhir K.88 = 169. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 76 .19 Sec Base Shear. Jain. V = 495 kN Base Moment.Example 3:Base shear Tank Empty Condition Time Period Ti = 0.842 kN-m Seismic base shear and moment are higher in tank full condition Hence. M* = 9. tank full condition will govern the design © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 77 . M = Mass of container + Mass of water + 1/3rd Mass of staging = 160.7 = 457.6 + 1/3 x 123.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. Jain.7 + 255.5 t © Sudhir K. 22 x 105 kNm Time period T = 2π M 457. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 78 .075 © Sudhir K. Jain.2 = 0.25 x 0.0 x 1.5 = 2π = 0.22 ×10 Base shear coefficient αh = βIFo(Sa/g) = 1.285 Sec 5 K 2. Ks = 2.Example 3: Analysis using IS 1893:1984 Stiffness of staging.5 x 0. 075 x 4488 = 336.88 = 6.692 kN-m © Sudhir K. V = αh x W = 0.6 x 19. Jain.Example 3: Analysis using IS 1893:1984 Base Shear. M = V x hcg = 336.6 kN Base Moment. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 79 . 9 = 2π = 0.7 + 1/3 x 123.2 = 0.22 ×10 Design horizontal seismic coefficient.25 x 0. αh αh = βIFo(Sa/g) = 1. Jain.5 x 0.075 © Sudhir K.9 T T = 2π M 201. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 80 .7 = 201.Example 3: Analysis using IS 1893:1984 Analysis for Tank Empty Structural mass.0 x 1. M M = Mass of container + 1/3rd Mass of staging = 160.19 Sec 5 K 2. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 81 .5 kN Base Moment. V = αh x W = 0. Jain.075 x 1980 = 148.5 x 19.Example 3: Analysis using IS 1893:1984 Base Shear.88 = 2952 kN-m Let us compare results from the Guideline and IS 1893:1984 See next slide © Sudhir K. M = V x hcg = 148. Tank empty ( Ti ) Tank full ( Ti) 0.19 sec 0.Example 3: Comparison of results Comparison of results from the Guideline and IS 1893:1984 Idealization of Tank Lateral stiffness of staging Time period Guideline IS 1893:1984 2.075 Convective mode.19 sec 0.060 E-Course on Seismic Design of Tanks/ January 2006 ----Lecture 7/ Slide 82 . Jain.075 0.22 x 105 kN/m Impulsive mode. Tank full ( Ah)c © Sudhir K. IIT Kanpur 0.14 sec 0.285 sec ----- Convective mode.25 0.22 x 105 kN/m 2.25 sec 3.25 0. Tank full ( Tc) Design seismic horizontal coefficient Impulsive mode Tank empty ( Ah)i Tank full ( Ah)i 0. 940 kN-m IS 1893:1984 148.952 kN-m 6. Jain.5 kN 336. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 83 . seismic forces from the Guideline are on much higher side Base moment is 150% higher in full condition © Sudhir K.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 495 kN 843 kN 9.842 kN-m 16.6 kN 2.692 kN-m For shaft supported tanks. 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. Jain. increase in base moment is of particular interest © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 84 . Analysis using IS 1893:1984 Main reason for this increase is: IS 1893:1984 has K = 1. redundancy and overstrength © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 85 . Jain.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. 0 Hence. 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.Analysis using IS 1893:1984 In frame staging. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 86 . in frame staging net increase in forces is not very significant © Sudhir K. Jain. cost of shaft and foundation is obtained for one tank See next example © Sudhir K. Jain.Analysis using IS 1893:1984 For shaft staging. increase in design seismic forces will not necessarily cause proportionate increase in cost of the tank Since lateral forces are higher. one will have to suitably choose the dimensions of shaft A wider shaft will help in achieving more economical design To explain this point. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 87 . Jain.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. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 88 . SBC = 25 t/m2 Solid raft foundation is used © Sudhir K. shaft of 7 m diameter is considered Then. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 89 . 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 © Sudhir K.Issue of cost In the first instance. shaft IS 1893:1984 150 m3 7t 135 m3 9t Guideline 150 m3 9t 340 m3 12 t * Design of 7 m shaft is not possible for forces as per the Guideline © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 90 . Shaft IS 1893:1984 Guideline Shaft Concrete Steel Foundation Concrete Steel 90 m3 6t 200 m3 8t * * 540 m3 21 t 12 m dia. Jain.Issue of cost Material quantities required 7 m dia. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 91 .Issue of cost Thus. by increasing shaft diameter from 7 m to 12 m About 200 m3 of concrete and 9 t of steel in foundation could be saved This example is shown to emphasize the importance of proportioning of dimensions in achieving economy in design © Sudhir K. Jain. 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 © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 92 . Jain. design of container does not get much affected. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 93 .3% This would usually accommodate the hydrodynamic pressure © Sudhir K. since hydrodynamic pressure is about 30 to 40% of hydrostatic pressure In working stress design.Issue of cost Note. permissible stresses are increased by 33. Jain. wind forces used to govern the design © Sudhir K. in many elevated tanks. so far.Issue of cost It is to be recognized that. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 94 . Jain. simplistic design approaches prevailed And above discussed options were not needed and were not attempted Moreover. seismic design forces for tanks were rather low Hence. in India. and Good seismic performances © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 95 .Issue of cost In view of new design forces There is a need to try some of these options To achieve better economy. Jain. 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. © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 7/ Slide 96 . will discuss some miscellaneous issues.At the end of Lecture 7 In Part B of the Guideline. Jain. 2006 .Lecture 8 February 21. Jain.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 © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 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. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 3 . medium and soft soil Spectrum for soft soil is higher than hard soil Except in short period range © Sudhir K. Jain. Jain.Soil Structure Interaction Effect of soil on structure’s response is included in soil structure interaction Soil-structure interaction is a complex problem Soil properties are frequency dependent Based on research. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 4 . simplified methods have been proposed © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 5 .. 255-370. A.. 1986) Veletsos. 443-461.Soil Structure Interaction We will briefly discuss method suggested by Veletsos (1984) This has been used in Eurocode 8 and NZSEE Recommendations (Priestley. N. Technical Council on Lifeline Earthquake Engineering.Y. “Seismic response and design of liquid storage tanks”. © Sudhir K. ASCE. 1984. S. Jain. Guidelines for the seismic design of oil and gas pipeline systems. Jain. it imparts flexibility and increases time period Soil flexibility affects impulsive mode time period It does not affect convective mode time period © Sudhir K.Soil Structure Interaction 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. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 6 . it is treated as fixed at base If soil is soft. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 7 .Soil Structure Interaction 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 © Sudhir K. Jain. mi Ti = Impulsive mode time period of tank with fixed base © Sudhir K. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 8 .Soil Structure Interaction Veletsos (1984) suggested following expression for impulsive time period of tank on soft soil Ti = Ti 1 + − K f Kx ⎡ K xh2 f 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. Soil Structure Interaction 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 © Sudhir K. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 9 . time period of impulsive mode.3.1.Soil Structure Interaction For elevated tanks Stiffness of impulsive mode.1 and 4.2 of the Guideline © Sudhir K.1.3. Kf is lateral stiffness of staging. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 10 . Jain. Ti is known Recall expression for Ti for circular tanks from Lecture 3 Also refer Clause 4. Ks For ground supported tanks We have considered circular and rectangular tanks For both types of tanks. Kf for ground supported tanks can be obtained from mi Ti = 2π Kf mi being impulsive mass is also known © Sudhir K. stiffness of impulsive mode.Soil Structure Interaction Hence. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 11 . Soil Structure Interaction tank In ground supported tanks hf = hi + tb hf is distance of impulsive mass from base of In elevated tanks Total impulsive mass = mi + ms Hence. hf = hcg Recall. we assume that entire mass is lumped at CG of empty container © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 12 . Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 13 . Jain.Soil Structure Interaction Veletsos (1984) included effect of soil damping as follows: Effective damping ratio of structure is ξ = ξS + − ξ ⎛ − ⎞ ⎜ Ti / T i ⎟ ⎝ ⎠ 3 ξs ξ Ti − ξ − = 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. Hence.Soil Structure Interaction It is clear that soft soil would have lower values of lateral stiffness Kx and rocking stiffness. soil structure interaction would be important for structures on soft soil © Sudhir K. Kθ. damping of structure will be more Thus. damping in soft soil is more compared to hard soil Hence. more effect on impulsive time period Similarly. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 14 . Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 15 .Soil Structure Interaction Increase in time period and damping would reduce seismic forces This is also called beneficial effects of soil Before using these beneficial effects of soil. one needs to properly ascertain the soil properties Note. soil properties are frequency dependent © Sudhir K. a detailed soil-structure interaction study is needed to find earthquake induced soil pressure Specialised literature shall be referred to find dynamic earth pressure and its distribution © Sudhir K.Buried tanks In buried tanks. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 16 . Jain. effect of soil pressure on wall shall be included During lateral excitation. soil induces dynamic pressure on wall Strictly speaking. IIT Kanpur σx = 0. He Idealised pressure distribution Tank base © Sudhir K.5 (Ah)i γs He Lecture 8/ Slide 17 E-Course on Seismic Design of Tanks/ January 2006 .5 (Ah)i γs He Ground surface PE Embedment depth.Buried tanks NZSEE has suggested following simplified distribution for earthquake induced soil pressure Tank wall σx = 1. Jain. Jain.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 © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 18 . 5 Buried tanks are quite rigid Convective time period will be same as that for ground supported tanks © Sudhir K. remains same Time period of impulsive mode will be very less. (Sa/g)i = 2. hence. mc etc. procedure to find mi. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 19 ..e. Jain.Buried tanks Seismic analysis procedure for buried tanks is same as that for ground supported tanks i. Buried tanks For buried tanks. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 20 . R value shall be linearly interpolated Depending on depth of embedment © Sudhir K. R = 4. then.0 This is for fully buried tank Refer Table 2 of the Guideline If tank is partially buried. response reduction factor. Jain. then.33 For a steel tank with anchored base R = 2.5 = 3.5 + (4. R = 2.5 for ground supported tank If tank is half buried.0+(4.0 for fully buried tank If tank is half buried.Buried tanks For a RC tank with fixed base R = 2.67 = 3.0 – 2.0 – 2.0 + (4.0 for ground supported tank R = 4. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 21 .0 – 2.5) x 0.5 = 3. then.0) x 0.0)x 0.25 © Sudhir K. R = 2. Jain.0 For 2/3rd buried tank. R = 2. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 22 . Jain. soil pressure will act over buried portion of shaft © Sudhir K. shaft diameter could be more than 10 m Foundation for shaft is usually 2 to 3 m below ground Hence.Buried tanks Dynamic soil pressure would also act on shaft staging of elevated tanks For large tanks. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 23 . Jain.Buried tanks Dynamic soil pressure acts opposite to direction of seismic loading Seismic force Soil pressure © Sudhir K. © Sudhir K. net moment on foundation would reduce Codes do not allow this reduction for design purpose Due to uncertainties in compactness of soil etc.Buried tanks Soil pressure would induce moment on the foundation This moment would be opposite to moment due to seismic forces Thus. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 24 . column sizes are quite small Compared to shaft diameter Members with such small width may not be able to mobilize dynamic soil pressure Hence.Buried tanks In frame staging. no dynamic soil pressure on embedded length of columns © Sudhir K. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 25 . M = Q x L P Q L © Sudhir K.P-Delta effect Consider a vertical cantilever subjected to lateral force Q and vertical force P Bending moment at the base. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 26 . Jain. total moment at the base = QxL+PxΔ © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 27 .P-Delta effect Due to force Q. let deflection be Δ Point of application of P shifts And. additional moment p x Δ will act on the cantilever P Q Δ L Now. lateral deflection would further increase Which. in turn would cause more moment © Sudhir K. vertical loads interact with lateral displacements And. Jain. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 28 .P-Delta effect Thus. induce additional forces on structure This is termed as P-Delta effect Due to additional moment of P x Δ. Jain.e.P-Delta effect P-Delta effect can be minimised by limiting the lateral deflection For buildings.4% i. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 29 . IS 1893(Part 1) requires that storey drift shall be less than 0. total lateral deflection will be always less than h/250. h is total height of the building © Sudhir K.. P-Delta effect will be more severe in elevated tanks In buildings.P-Delta effect Compared to buildings. Jain. almost entire vertical load is at the top © Sudhir K. vertical loads are distributed over the entire height In elevated tanks. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 30 . IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 31 .. in building limit is h/250 i.e. total lateral deflection shall be less than 0. Jain.P-Delta effect Thus.2% of height © Sudhir K. 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. Jain. 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 © Sudhir K.Flexibility of piping In past earthquakes. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 32 . IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 33 .Flexibility of piping All the international codes emphasize on flexible piping 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 © Sudhir K. Jain. AWWA D-100 are exclusively for ground supported steel tanks These codes give provisions for safety against buckling © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 34 . Jain. many steel tanks have suffered buckling failure during earthquakes API 650.Buckling of shell Ground supported circular steel tanks are sensitive to buckling These are thin shell structures In the past. 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 © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 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 © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 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. © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 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 © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 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 © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 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 !! © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 40 Buckling of shell ACI 371 specifies limits on construction tolerances; Some of these are Variation in wall thickness: -3% and + 5% of wall thickness Variation in plumb of shaft: 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 © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 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 © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 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 © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 43 IS 11682:1985 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 Increase in permissible stresses is allowed for columns and not for braces i.e., braces will be stronger than columns Whereas, in seismic design we need strong column-weak beam combination © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 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. © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 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 © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 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 © Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 47 limit state design is compulsory for members subjected to combined direct load and flexural Clause B-4.3 of IS 456:2000 Hence. provisions on limit state design of shaft are needed in IS 11682 © Sudhir K.IS 11682:1985 Clause 8. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 48 .2 suggests formulae for stresses in shaft These are based on working stress design method As per IS 456:2000. Jain.5. Jain. This would help designers in analysis and design of staging © Sudhir K.IS 11682:1985 There are many more such modifications needed in IS 11682 An urgent revision of this code is needed. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 49 . 2 and 3 have already been sent to you Please send your remaining questions Thank You……… © Sudhir K.And finally…. 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. IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 8/ Slide 50 .. With this Lecture. Jain. Assignment 1 Date of assignment: 18 January 2006 E-course on Seismic Design of Tanks 1 . IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Assignment 1 / slide 2 .What is expected? Part I: Reading assignments are to be completed as soon as possible Best if you finish the reading assignments on the same day It will help you appreciate the next lecture © Sudhir K. Jain. Feel free to go back and review the lecture while answering the questions. © Sudhir K. Please do not send the solutions to us. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Assignment 1 / slide 3 . Some questions may not have very specific answers This is in line with our engineering profession wherein certain questions have no specific answers.What is expected? Parts II and III To be completed by 19 January. Jain. Compare your answers with the solutions we will send. before we send out the solutions of these to you. 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. © Sudhir K. 4...2. 4. 1.Part I: Reading Assignment 1 Read carefully Preface and Clauses 0.2. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Assignment 1 / slide 4 . & 4. 2. Jain..1. 2)Impulsive mass is less than convective mass for short tanks.1)Hydrodynamic pressure varies linearly with liquid height.7)Impulsive and convective masses depend only on height of liquid. 1. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Assignment 1 / slide 5 . Jain.3)Net hydrodynamic force on the container wall is zero.4)Hydrodynamic pressure on base causes overturning moment on tank.Part II: True / False Identify the following statements as True or False 1. 1. 1. © Sudhir K.6)With the inclusion of base pressure effect. 1. 1.5)Convective mass acts at lower height than impulsive mass. 1. overturning moment on tank reduces. hi* and hc* in both the directions.Part III: Questions 1. 1. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Assignment 1 / slide 6 . hc.2) A rectangular tank has internal plan dimensions of 10 m x 16 m. Kc. Obtain mi. © Sudhir K.1)A circular tank has internal diameter of 12 m and water height of 5 m. Obtain mi. mc. hi. Water height is 8 m. mc. hi* and hc*. hc. Jain. hi. Solution 1 Date of assignment: 18 January 2006 Date of solution: 20 January 2006 E-course on Seismic Design of Tanks 1 . False Hydrodynamic pressure has curvilinear variation along the height. Jain. Hydrostatic pressure varies linearly with height. impulsive liquid mass is more.2)Impulsive mass is less than convective mass for short tanks. 1. In tall tanks.Part II: True / False Identify the following statements as True or False 1. In fact.1)Hydrodynamic pressure varies linearly with liquid height. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 1 / slide 2 . © Sudhir K. convective liquid or liquid undergoing sloshing motion is more. False In short tanks. Note the graphs for hi and hc in figure 2b and 3b of guidelines © Sudhir K.Part II: True / False 1. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 1 / slide 3 .4)Hydrodynamic pressure on base causes overturning moment on tank. 46 of Lecture 1 1. Jain. hence. convective mass is always located at higher height than impulsive mass. False Rather. undergoes convective or sloshing motion. net hydrostatic force on wall is zero 1. False Liquid in upper portion. True Refer slide no.5)Convective mass acts at lower height than impulsive mass.3)Net hydrodynamic force on the container wall is zero. overturning moment increases. © Sudhir K. 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. with the inclusion of base pressure effect. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 1 / slide 4 .6)With the inclusion of base pressure effect. h. Jain. Hence. False Impulsive and convective masses depend on aspect ratio (h/D or h/L) rather than only on height.Part II: True / False 1.7)Impulsive and convective masses depend only on height of liquid. 1. This can be seen from the fact that hi* is always greater than hi and hc* is always greater than hc. m = 565. these formulae will be used to obtain various parameters. from Figures 2a and 2b of the Guideline. mc. Kc. © Sudhir K. h = 5 m ∴ h/D = 5/12 = 0. values of various parameters can be read. hc.1)A circular tank has internal diameter of 12 m and water height of 5 m. Obtain mi. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 1 / slide 5 . Here.5 t D = 12 m. hi* and hc*. Jain.5 x 1. hi.Part III: Solutions 1. One can also use formulae given in Table C-1 of the Guideline. Solution: Total volume of water = π/4 x 122 x 5 = 565.5 m3 ∴ Total water mass.0 = 565.417 For this value of h/D. 866 ⎟ mi h⎠ ⎝ = D m 0 .68 ⎟ 12 12 ⎠ ⎝ = 0.68 ⎟ mc D⎠ ⎝ = 0.68 sinh⎜ 3.68 sinh⎜ 3.466 5⎞ ⎛ tanh⎜ 3.503 Since. 866 ⎟ 5 ⎠ ⎝ = 12 0 . h/D is less than 0.23 5 12 = 0.68 ⎟ − 1.0 12 ⎠ ⎝ = 1− 5 5⎞ ⎛ 3.375 h⎞ ⎛ cosh⎜ 3. hi/h = 0.68 ⎟ 12 ⎠ ⎝ = 0.Part III: Solutions D⎞ ⎛ tanh ⎜ 0 .579 © Sudhir K. 866 5 = 0.75.0 hc D⎠ ⎝ = 1− h h⎞ h ⎛ 3. Jain. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 1 / slide 6 .23 h m D 12 ⎞ ⎛ tanh ⎜ 0 .68 ⎟ − 1.68 ⎟ D D⎠ ⎝ 5⎞ ⎛ cosh⎜ 3. 866 h h⎞ ⎛ tanh⎜ 3. 68 ⎟ − 2. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 1 / slide 7 .68 ⎟ D D⎠ ⎝ 5⎞ ⎛ cosh⎜ 3.866 D⎞ ⎛ 2 tanh⎜ 0.01 12 ⎠ ⎝ = 1− 5 5⎞ ⎛ 3.878 K c = 0. Jain.125 = 0.Part III: Solutions D h hi * = h 0.68 ⎟ h D⎠ ⎝ ∴ 5⎞ ⎛ K c h / mg = 0.68 sinh⎜ 3.125 = 0.694 © Sudhir K.68 ⎟ 12 12 ⎠ ⎝ = 0.01 hc * D⎠ ⎝ = 1− h⎞ h h ⎛ 3.947 h⎞ ⎛ cosh⎜ 3.68 sinh⎜ 3.836 mg h⎞ ⎛ tanh 2 ⎜ 3.836 tanh 2 ⎜ 3.68 ⎟ 12 ⎠ ⎝ = 0.68 ⎟ − 2.866 ⎟ 5⎠ ⎝ 12 5 .866 ⎟ h⎠ ⎝ - 0.866 12 ⎞ ⎛ 2 tanh⎜ 0.0. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 1 / slide 8 .947 hc*/h = 0.375 x 5 ∴hc = 0.878 Kch/mg = 0.5 t ∴mc = 0.5 x 9.579 hi*/h = 0.375 hc/h = 0.579 x 5 ∴hi* = 0.466 x 565.88 m = 2.Part III: Solutions Thus.694mg/h = 0.39 m © Sudhir K.5 = 263.735 m = 4.947 x 5 ∴hc* = 0. Jain.90 m = 4.878 x 5 = 1.81/5.503 x 565.694 Kc = 0.503 hi/h = 0. mi/m = 0. we get.694 x 565.5 = 284.5 t ∴hi = 0.0 = 770 kN/m ∴mi = 0.466 mc/m = 0. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 1 / slide 9 X . hc.2) A rectangular tank has internal plan dimension of 10 m x 16 m. Solution: Y 16 m 10 m Plan view Total volume of water = 10 x16 x 8 = 1280 m3 Total mass of water. Jain. Water height is 8 m. hi* and hc* in both the directions. mc. m = 1280 x 1.0 = 1280 t © Sudhir K.Part III: Solutions 1. hi. Obtain mi. 87 x 8 = 3. consider base motion in X-direction: L = 16 m.4 m = 7.375 hc/h = 0.87 ∴mi = 0.8 x 8 ∴hc* = 0. 3b of guidelines. Jain.48 x 1280 = 614.6 m = 6.57 x 8 ∴hi* = 0. for h/L = 0.0 m © Sudhir K.54 mc/m = 0.54 x 1280 = 691.57 hi*/h = 0.4 t ∴hi = 0. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 1 / slide 10 .375 x 8 ∴hc = 0.0 m = 4.Part III: Solutions First.48 hi/h = 0.80 hc*/h = 0. h = 8 m ∴ h/L = 8/16 = 0.2 t ∴mi = 0.5 From Figure 3a.5: mi/m = 0. 73 ∴mi = 0. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 1 / slide 11 .72 x 1280 = 921. Jain. 3b of guidelines.58 x 8 ∴hc* = 0.72 mc/m = 0.73 x 8 = 3.58 hc*/h = 0.40 hc/h = 0.Part III: Solutions Now.2 m = 4.2 m = 5.8 From Figure 3a.8: mi/m = 0. consider base motion in Y-direction: L = 10 m.32 x 1280 = 409.32 hi/h = 0.6 t ∴hi = 0. h = 8 m ∴ h/L = 8/10 = 0.40 x 8 ∴hc = 0.64 m = 5.65 x 8 ∴hi* = 0.6 t ∴mi = 0. for h/L = 0.65 hi*/h = 0.84 m © Sudhir K. h/L = 0. in Y-direction. For shallow tanks.8. wherein. © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 1 / slide 12 . for which h/L = 0. in x-direction. Jain. Whereas.5. impulsive and convective masses are almost same.Part III: Solutions Thus. inclusion of base pressure significantly changes the height . Also note the difference in hi and hi* values ( or hc and hc*) for two cases. more liquid contributes to impulsive mass. Additional Slide for Solution 1 Insert after slide 1 of Solution 1 1 . 2 1.2)Impulsive mass is less than convective mass for short tanks. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 1 / slide 2 .Correction in Part II True/False . It is printed as False It should be printed as TRUE © Sudhir K. Jain.2 Notice that there is a correction in Part II True/False – 1.1. Assignment 2 Date of assignment: 20 January E-course on Seismic Design of Tanks 1 . Jain. 4.4.2.1.Part I: Reading Assignment 2 Read the following: Clause 4.1.4 of IS 1893(Part 1):2002 Clause 3. and 5.5 of the Guidelines Clause 6.6 of IS 1893:1984 © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Assignment 2/ slide 2 .2. base shear coefficient depends on type of foundation. 2. 2. 2. 2.4) In IS 1893:1984.Part II: True / False Identify the following statements as True or False 2. there are five seismic zones.2) In IS 1893(Part 1):2002.3) In IBC 2003. 2.5) Importance factor for tanks is higher than for buildings.6)Damping of structure does not affect base shear coefficient. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Assignment 2/ slide 3 . © Sudhir K. performance factor. K for buildings with good ductility is more than that for elevated tanks. 2.7) In IS 1893(Part 1):2002. response modification factor for buildings with good ductility is lower than that for elevated tanks on frame type staging.1) Seismic force depends on mass of the structure. Jain. 4.1 sec. consider 5% damping) 1) A building with good ductility and time period of 1. 2.2) For following elevated tanks.1 sec. 2% and 5% © Sudhir K.1)As per IS 1893:1984. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Assignment 2/ slide 4 .1 sec. damping of 0%. damping of 0%.4 sec. obtain base shear coefficient as per IS 1893:1984 (Fo = 0. β = 1.Part III: Questions 2.0) 1) Time period = 0.4. Jain. 2) A building with low ductility and time period of 1. β = 1. obtain base shear coefficient for following structures using response spectrum method (Fo = 0.5 sec. 3) An elevated tank with time period of 1.0. 2% and 5% 2) Time period = 1. Solution 2 Date of assignment: 20 January Date of solution: 23 January E-Course on seismic design of tanks 1 . 3) In IBC 2003.0 for elevated tanks on frame staging © Sudhir K.Part II: True/ False Identify the following statements as True or False 2. response modification factor.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. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 2 / Slide 2 . there are five seismic zones. 2. True 2.0 as against.1) Seismic force depends on mass of structure. Jain. numbered from II to V.2) In IS 1893(Part 1):2002. R = 3. R = 8. False There are four zones. 5) Importance factor for tanks is higher than for buildings. True 2. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 2 / Slide 3 . for a building with good ductility and elevated tanks.4) In IS 1893:1984. False See Table 3 of IS 1893(Part 1):2002 and Fig.performance factor. K for buildings with Good ductility is more than that for elevated tanks False Performance factor. K is same (K = 1. 2. Jain.Part II: True/ False (contd…) 2.0).6) Damping of structure does not affect base shear coefficient. 2 of IS 1893:1984 © Sudhir K. in IS 1893:1984.Part II: True/ False (contd…) 2. © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 2 / Slide 4 . Jain. base shear coefficient depends on type of foundation. False In IS 1893(Part 1):2002. Rather.7) In IS 1893(Part 1):2002. base shear coefficient does not depend on type of foundation. it use to depend on type of foundation. 1 sec.4. β = 1. Jain.1 sec.Part III: Solutions 2. Fo = 0. 3) An elevated tank with time period of 1. 2) A building with low ductility and time period of 1. consider 5% damping) 1) A building with good ductility and time period of 1.0 © Sudhir K.4. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 2 / Slide 5 .1 sec.0.1)As per IS 1893:1984. obtain base shear coefficient for following structures using response spectrum method (Fo = 0. base shear coefficient is expressed as: Ah = KβIFoSa/g for buildings Ah = βIFoSa/g for elevated tanks We have. Solution: In response spectrum method. β = 1. from Figure 2 of IS 1893:1984.5 Ah = βIFoSa/g = 1.1 sec and damping of 5%.0 Ah = KβIFoSa/g = 1.0 x 1.1 = 0.6.0 x 1.0 Ah = KβIFoSa/g = 1.4 x 0.06 © Sudhir K.6 x 1.1 = 0.5 x 0.0.0 x 0.1 For building with good ductility: K = 1.064 For elevated tank: I = 1.04 For building with low ductility: K = 1.1 = 0.0 x 1.4 x 0.0 x 0. Jain. and hence.Part III: Solutions All three structures have time period of 1. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 2 / Slide 6 .4 x 0. Sa/g = 0. I = 1. I = 1.0 x 1. 0) 1) Time period = 0. β = 1. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 2 / Slide 7 . obtain base shear coefficient as per IS 1893:1984 (Fo = 0.Part III: Solutions 2. 2% and 5% 2) Time period = 1. damping of 0%.4.0 and Importance factor.4 sec. I = 1. 2% and 5% Solution: Base shear coefficient for tank is: Ah = βIFoSa/g We have. β = 1. Jain. damping of 0%.2) For following elevated tanks.4.5 sec. Fo = 0. 2 of IS 1893:1984 © Sudhir K.5 Values of Sa/g are to be obtained from Fig. 08 Ah = βIFoSa/g 1.15 0.4 x 0.0 x 1. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 2 / Slide 8 .096 1.5 x 0.048 0.300 1.5 x 0.15 = 0.25 0.0 x 1.5 x 0.50 0.4 2 5 0.5 x 0.5 x 0.150 1.5 x 0.4 x 0.105 = 0.0 x 1.4 x 0.0 x 1.5 2 5 0 1.4 x 0.0 x 1.Part III: Solutions Time period Damping Sa/g (Sec) (%) 0 0.090 1.0 x 1. Jain.105 1.5 = 0.4 x 0.16 0.063 Note the effect of damping on base shear coefficient © Sudhir K.4 x 0.25 = 0.16 = 0.08 = 0. Assignment 3 Date of assignment: 24 January E-course on Seismic Design of Tanks 1 . 2. Jain.3 of the Guidelines © Sudhir K.3 and 4.2. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Assignment 3/ slide 2 . 4.Part I: Reading Assignment 3 Read Clause 4.2. 4 seconds. convective mode time period depends on lateral stiffness of staging. 3. braces shall be treated as rigid beams © Sudhir K.3)Impulsive mode time period of ground supported tanks is usually less than 0.Part II: True / False Identify the following statements as True or False 3. Jain. 3. 3. 3.1)Time period does not depend on mass of the structure.2)Time period increases with stiffness. 3.4)Convective mode time period depends only on aspect ratio of the container.6)While finding lateral stiffness of frame staging. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Assignment 3/ slide 3 .5)In elevated tanks. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Assignment 3/ slide 4 . © Sudhir K.000 kN and weight of staging is 1. Weight of container is 3. Liquid height is 4m. Jain. Lateral stiffness of staging is 50.700 kN.Part III: Questions 3. 3.2) An elevated tank with circular cylindrical container has internal diameter of 12 m and water height is 4. Obtain two mass model and find time period of impulsive and convective modes.000 kN/m.1) A ground-supported rectangular tank has length of 12 m and width of 6 m.5 m. Find time period of convective mode in the two horizontal directions. Solution 3 Date of assignment: 24 January Date of solution: 27 January E-course on Seismic Design of Tanks 1 . T = 2π M/K © Sudhir K. Jain. False Time period decreases with stiffness and is inversely proportional to square root of stiffness.1)Time period does not depend on mass of the structure.2)Time period increases with stiffness. False 3. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 3/ slide 2 .Part II: True / False 3. Part II: True / False 3.5)In elevated tanks.4)Convective mode time period depends only on aspect ratio of the container. See formula for Tc in slide no. 3. 65. True 3.3)Impulsive mode time period of ground supported tanks is usually less than 0. convective mode time period depends on lateral stiffness of staging.4 seconds. Jain. False It depends not only on aspect ratio but also on diameter or lateral dimension of container. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 3/ slide 3 . False © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 3/ slide 4 . braces shall be treated as rigid beams. False For proper evaluation of lateral stiffness. © Sudhir K. Jain.6)While finding lateral stiffness of frame staging.Part II: True / False 3. it is necessary that flexibility of brace is included in the model. Jain.1) A ground-supported rectangular tank has length of 12 m and width of 6 m. Liquid height is 4m. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 3/ slide 5 . Solution: Y 12 m 6m Plan view X © Sudhir K.Part III: Solutions 3. Find time period of convective mode in the two horizontal directions. ∴ h/L = 4/6 = 0.33.0 Time period of convective mode. TC = CC Lg TC = 4. for h/L = 0.65 Time period of convective mode. h = 4.67 From Figure 7 of guideline. L = 6. T C = C C Lg Tc = 3.33 From Figure 7 of guideline. Jain.65 (6.0 m. Cc = 3. h = 4. for h/L = 0. one gets Cc = 4.81) = 2. © Sudhir K.0 m.0 m ∴ h/L = 4/12 = 0.Part III: Solutions 1) Analysis in X-direction: Length. L = 12.0 9.81) = 4.85 Sec.42 Sec.0 (12. 2) Analysis in Y-direction: Length. Height of liquid. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 3/ slide 6 . one gets.0 m.67.0 9. 5/12 = 0.5 m.Part III: Solutions 3. Container is circular cylinder. Weight of container is 3. h/D = 4.000 kN/m. mc/m = 0. Jain.2) An elevated tank with circular cylindrical container has internal diameter of 12 m and water height is 4.545 and Kch/mg = 0. mass of water.700 kN.665 © Sudhir K.5 m. Lateral stiffness of staging is 50. h = 4.375 From Figure 2 of guideline. Obtain two mass model and find time period of impulsive and convective modes.43.375: mi/m = 0. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 3/ slide 7 . Water height.000 kN and weight of staging is 1. Solution: Internal diameter. for h/D = 0.5 = 509 m3. m = 509 t. ∴ Volume of water = π/4 x D2 x h = π /4 x 122 x 4. D = 12 m. 8 + 1/3 x 173.9 kN/m Weight of container = 3.43 x 509 = 218.Part III: Solutions mi = 0.665 x 509 x 9.5 = 737. Ks = 50.8 t Weight of staging = 1.4 t Kc = 0.43 x m = 0.665 x m g/h = 0.3 t Structural mass of tank. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 3/ slide 8 . ms = mass of container +1/3rd mass of staging = 305.6 t Lateral stiffness of staging.9 t mc = 0.81 = 305.545 x 509 = 277.545 x m = 0. Jain.700 kN ∴ Mass of staging = 1700/9.000 kN/m © Sudhir K.3 = 363.81 = 173.000 kN ∴ Mass of container = 3000/9.81/4. 9 kN/m and Ks = 50000 kN/m Two mass model: mc Kc mi + ms Ks 277. Kc = 737. Jain. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 3/ slide 9 .6t 50000 kN/m © Sudhir K.9 t. we have: mi =218.Part III: Solutions Thus.9t + 363.6 t. ms = 363.4 t. mc = 277.4 t 737.9 kN/m 218. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 .5 t 218. this two mass model will be treated as two uncoupled single degree of freedom system: 277. Jain.9 kN/m 582.4 t 737.9t + 363.Part III: Solutions For evaluating time period.6t 50000 kN/m 277.9 kN/m (a) Two mass idealization 50000 kN/m (b) Uncoupled system Solution 3/ slide 10 © Sudhir K.4 t 737. Ti = 2π mi + ms Ks 218.000 = 0. Jain.6 Ti = 2π 50.Part III: Solutions Now.678 Sec.4 737.85 Sec.9 + 363. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 3/ slide 11 . Impulsive Time Period.9 = 3. mc Convective Time Period. © Sudhir K. Tc = 2π Kc Tc = 2π 277. Assignment 4 Date of assignment: 30 January E-course on Seismic Design of Tanks 1 . 5 of the Guideline © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Assignment 4/ slide 2 .4.Part I: Reading Assignment 4 Read Clause 4. Jain. 4. 6)Response reduction factor for tanks depends on importance of tank. 4.Part II: True / False Identify the following statements as True or False 4. Jain. 4.2)Frame staging has less redundancy than shaft staging. © Sudhir K. 4.5)Thin shafts have more ductility than thick shafts. 4. damping in impulsive mode depends on material of container. 4.7)In elevated tanks.1) R values for elevated tanks with SMRF staging are lower than for buildings with SMRF. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Assignment 4/ slide 3 .3)Damping in convective mode depends on material of tank. 4.4)Elevated tanks are inverted pendulum type structures. Ti = 0. Find impulsive and convective base shear coefficients in both the directions. Jain.1) A ground-supported rectangular RC water tank with fixed base is located in zone III on medium soil.0 sec. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Assignment 4/ slide 4 .4 sec and Tc = 3. Along the length direction. © Sudhir K.Part III: Questions 4.2 sec and Tc = 5. Ti = 0.0 sec. Along the width direction. Find the base shear coefficient in impulsive and convective mode for following soil conditions. Jain.5 sec and Tc = 4. i) Hard ii) Medium iii) Soft.2) An elevated water tank on ductile RC frame staging is located in zone V. Ti = 1.0 sec. © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Assignment 4/ slide 5 .Part III: Questions 4. Solution 4 Date of assignment: 30 January 2006 Date of solution: 01 February 2006 E-course on Seismic Design of Tanks 1 . Jain. R = 5. © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 4/ slide 2 .5% damping for all types of tank materials.0 4.5.2)Frame staging has less redundancy than shaft staging False 4. whereas for buildings with SMRF.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.Part II: True / False 4.3)Damping in convective mode depends on material of tank False Convective mode has 0. I. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 4/ slide 3 . Jain. False Response reduction factor depends on ductility.4)Elevated tanks are inverted pendulum type structures True 4. © Sudhir K. Importance of tank decides the value of importance factor.5)Thin shafts have more ductility than thick shafts False 4.6)Response reduction factor for tanks depends on importance of tank. redundancy and overstrength.Part II: True / False 4. 7)In elevated tanks. damping in impulsive mode depends on material of staging and not of container. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 4/ slide 4 .Part II: True / False 4. Jain. © Sudhir K. damping in impulsive mode depends on material of container False In elevated tanks. Along the length direction.Part III: Solutions 4. Ti = 0. Along the width direction. Find impulsive and convective base shear coefficients in both the directions. Jain.1) A ground supported rectangular RC water tank with fixed base is located in zone III on medium soil. Ti = 0.0 sec.4 sec and Tc = 3. Solution: Width Direction Length Direction Plan of tank © Sudhir K.2 sec and Tc = 5. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 4/ slide 5 .0 sec. R = 2. hence I = 1. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 4/ slide 6 . Z = 0.16 (As per Table 2 of IS 1893 (Part I): 2002) Tank is being used for water storage.Part III: Solutions Zone III. Jain.5 (As per Table 1 of the Guideline) For RC ground supported tank with fixed base.0 (As per Table 2 of the Guideline) © Sudhir K. Tc = 5.15 Convective mode time period.75 is for 0.5.5 (Clause 4.75 x 1.5/2.0 (Clause 4.0 x 2.Part III: Solutions 1) Analysis in Length Direction: Impulsive mode time period.0 sec Damping in convective mode = 0.2 sec.5.4 of the Guideline) ∴ (Sa/g)c = 1.36/5.5 = 0. Damping in impulsive mode = 5% (RC tank) ∴ (Sa/g)i = 2.5% (Clause 4.4 of the Guideline) © Sudhir K. factor 1. Ti = 0. (Ah)i = Z/2 x I/R x (Sa/g)i = 0.3 of the Guideline) Impulsive base shear coefficient.476 Note.16/2 x 1. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 4/ slide 7 .3 of the Guideline) = 0. Jain.5.5% damping (Clause 4. Damping in impulsive mode = 5% (RC tank) ∴ (Sa/g)i = 2.16/2 x 1.5/2.0 x 0.5.3 of the Guideline) Impulsive base shear coefficient.4 Sec.5/2.15 and (Ah)c = 0.029 2) Analysis in Width Direction: Impulsive mode time period. Ti = 0.5 (Clause 4.Part III: Solutions Convective base shear coefficient (Ah)c= Z/2 x I/R x (Sa/g)c = 0.0 x 2. Jain.5 = 0.16/2 x 1. (Ah)i = Z/2 x I/R x (Sa/g)i = 0.029 Hence. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 4/ slide 8 . (Ah)i = 0.15 © Sudhir K.476 = 0. in length direction. 048 © Sudhir K.793 Convective base shear coefficient (Ah)c= Z/2 x I/R x (Sa/g)c = 0.5.16/2 x 1.4 of the Guideline) ∴ (Sa/g)c = 1.0 x 0. Tc = 3. (Ah)i = 0. Jain.Part III: Solutions Convective mode time period. in width direction.3 of the Guideline) = 0.5% (Clause 4.75 x 1.0 sec Damping in convective mode = 0. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 4/ slide 9 .048 Hence.793 = 0.5/2.36/3.15 and (Ah)c = 0.0 (Clause 4. Ti = 1.5 (As per Table 1 of the Guideline) Staging is ductile RC frame. i) Hard ii) Medium iii) Soft. hence. Z = 0.5 sec and Tc = 4. hence.Part III: Solutions 4.36 (As per Table 2 of IS 1893 (Part I): 2002) Tank stores water. Jain. I = 1.0 sec. Find the base shear coefficient in impulsive and convective mode for following soil conditions. Solution: Zone V.5 (As per Table 2 of the Guideline) © Sudhir K.2) An elevated water tank on ductile RC frame staging is located in zone V. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 4/ slide 10 . R = 2. 5 sec Damping is 5% (RC staging) Values of (Sa/g)i and impulsive base shear coefficient . Jain.5 x 1.67 1.91 1.098 = 0.67/Ti = 1.36/2 x 1.5/2.5 x 0. (Ah)i for different types of soil are given below: Soil type (Sa/g)i Hard Medium Soft 1/Ti = 0.36/Ti = 0.11 = 0. Ti = 1.91 = 0.072 = 0.5/2.5/2.Part III: Solutions Time period of impulsive mode.67 = 0.36/2 x 1. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 4/ slide 11 .11 (Ah)i = Z/2 x I/R x (Sa/g)i = 0.120 © Sudhir K.5 x 0.36/2 x 1. 67/Tc x 1. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 4/ slide 12 .5% Values of (Sa/g)c and convective base shear coefficient .438 = 0.595 = 0. (Ah)c for different types of soil are given below: Soil type (Sa/g)c Hard Medium Soft 1/Tc x1.0 sec Damping is 0.36/2 x 1.047 = 0.438 1.5/2.064 = 0.36/2 x 1. Tc = 4.75 = 0.5 x 0.36/Tc x 1.731 (Ah)c = Z/2 x I/R x (Sa/g)c = 0.36/2 x 1.75 = 0.595 1.5/2. Jain.5 x 0.079 © Sudhir K.75 = 0.5 x 0.5/2.731 = 0.Part III: Solutions Time period of convective mode. Assignment 5 Date of assignment: 3 February Date of solution: 5 February E-course on Seismic Design of Tanks 1 . 7 and 4.Part I: Reading Assignment 5 Read Clause 4. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Assignment 5/ slide 2 .8 of the Guidelines © Sudhir K.6. Jain. 4. 5)In IITK-GSDMA Guideline.7)In elevated tanks. acts at CG of empty container. 5. 5.4)In elevated tanks. impulsive mass of liquid.1)In ground supported tanks. © Sudhir K. base shear at the bottom of container wall. Jain. base moment at the top of footing includes effect of base pressure.3)For calculating bending moment at the bottom of wall. will not depend on mass of staging. mi. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Assignment 5/ slide 3 .Part II: True / False 5. base shear at the bottom of wall is same as base shear at the bottom of base slab. 5. 5. impulsive and convective base shear are combined using absolute summation rule. effect of base pressure is included. 5.2)In elevated tanks.6)Critical direction of seismic loading is always same for braces and columns of frame staging. 5. 024 0.15 (Ah)c 0. Jain.69 1.1 m projection along the periphery 12 m Y X 6m 12 m 4.69 hi * (m) 4.Part III: 5. bending moment at the bottom of wall and overturning moment in X and Y directions.5 m Plan mi (t) 138 230 hi (m) 1.07 3.15 0. Thickness of wall is 250 mm. roof slab is 120 mm thick and base slab is 250 mm thick. Obtain base shear.29 (Ah)i 0.1) Plan and elevation of a rectangular RC tank are shown below along with some relevant data.61 mc (t) 189 112 Elevation hc (m) 2. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Assignment 5/ slide 4 .73 2.93 hc* (m) 5. 0.48 2.037 Earthquake Direction X Y © Sudhir K. hi* = 3.0 m from top of floor slab.15 and of convective mode is 0. Impulsive and convective mode parameters are: mi = 140 t.43. mc = 110 t. Jain. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Assignment 5/ slide 5 .2) An elevated water tank has CG of empty container at 3. Height of staging from footing top to top of floor slab is 12 m. ms = 225 t. © Sudhir K.Part III: 5. Find base shear and base moment at the bottom of staging. hc* = 3.06.43 m. Base shear coefficient of impulsive mode is 0. Solution 5 Date of assignment: 3 February Date of solution: 6 February E-course on Seismic Design of Tanks 1 . 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.1)In ground supported tanks. base shear at the bottom of wall is same as base shear at the bottom of base slab.Part II: True / False 5. True Since staging is below the base slab (or floor slab). Jain. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 5/ slide 2 .2)In elevated tanks. base moment at the top of the footing includes effect of hydrodynamic pressure at the base. base moment in staging includes the bending effect caused by hydrodynamic pressure on base slab. 5. © Sudhir K. base pressure affects overturning moment. mi. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 5/ slide 3 . acts at CG of empty container. Rather. mi. is considered to act at hi*. ms.4)In elevated tanks.3)For calculating bending moment at the bottom of wall. impulsive mass of liquid. Structural mass of tank. which is obtained at the bottom of base slab. acts at CG of empty container. False Base pressure does not affect bending moment in wall. Jain. impulsive mass of liquid. © Sudhir K. effect of base pressure is included. 5. False In elevated tanks.Part II: True / False 5. Part II: True / False 5. False They are combined using Square Root of Sum of Squares (SRSS) rule.5)In IITK-GSDMA Guideline. False Columns and braces of frame staging can have different critical directions.6)Critical direction of seismic loading is always same for braces and columns of frame staging. Jain. 5. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 5/ slide 4 . impulsive and convective base shear are combined using absolute summation rule. © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 5/ slide 5 .Part II: True / False 5. Since staging is below the container wall. False: Mass of staging affects the natural period of the tank which decides the seismic coefficient (Ah)i . it is not going to affect base shear at the bottom of container. base shear at the bottom of container wall. True and False True: Base shear at any section depends on all the horizontal forces acting above that section.7)In elevated tanks. will not depend on mass of staging. Base shear at the bottom of staging depends on mass of staging. © Sudhir K. Jain. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 5/ slide 6 .07 3.69 1.15 0.024 0.Part III: Solutions 5.5 m 12 m 6 m periphery 12 m Y X Plan mi (t) 138 230 hi (m) 1.73 2.1 m projection along the 4.29 (Ah)i 0. Thickness of wall is 250 mm.93 hc* (m) 5. bending moment at the bottom of wall and overturning moment in X and Y directions.1) Plan and elevation of a rectangular RC tank are shown below along with some relevant data.69 hi * (m) 4. Obtain base shear.037 Earthquake Direction X Y © Sudhir K.61 mc (t) 189 112 Elevation hc (m) 2. 0. Jain.48 2. roof slab is 120 mm thick and base slab is 250 mm thick.15 (Ah)c 0. Jain.1 t © Sudhir K.8 kN and mass of roof slab.81 = 106. mw = 2 x (12.12 x 25 = 243.Part III: Solutions Solution: First we calculate mass of roof slab.5 x 25/9.25 x 4.5 m.81 = 24.25) x 0.5 x 0.25 + 6.5m ∴ Weight of roof slab = 12. mt = 243. wall. Weight density of concrete = 25 kN/m3 Thickness of roof slab= 120 mm Plan dimensions of roof slab = 12.9 t Thickness of wall = 250 mm Internal dimensions are 12m x 6m and height is 4. ∴ Mass of wall.5m x 6. and base slab. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 5/ slide 7 .8/9.5 x 6. 48 m.15.69 m. Analysis in X.direction In X-direction. hc = 2.7 m x 6.024 © Sudhir K.Part III: Solutions Thickness of base slab. (Ah)i = 0. mi = 138 t.7 m ∴ Mass of base slab = 12. it will have different seismic forces in X. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 5/ slide 8 . hi*= 4.81 = 54. tb = 250 mm Plan dimensions of base slab are 12.7 x 0.and Y-direction. Jain.2 t Since tank is rectangular in plan.73 m. we have.07 m. mc = 189 t. hc*= 5. hi = 1.7 x 6. (Ah)c = 0.25 x 25/9. Part III: Solutions Impulsive base shear at the bottom of wall is Vi = (Ah)i (mi + mw + mt) g = 0.81 = 395.024 x 189 x 9.9) x 9.5 ) 2 2 = 398. Jain.15 x (138 + 106. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 5/ slide 9 .81 = 44.1 + 24.3kN © Sudhir K.5 kN Total base shear at bottom of wall is V = Vi2 + Vc2 = (395.8 ) + (44.8 kN Convective base shear at the bottom of wall is Vc = (Ah)c mc g = 0. 81 = 110.9 x 4.5 kN-m Convective bending moment at the bottom of wall is Mc = (Ah)c mc hc g = 0.4 ) = 868.81 = 861.48 x 9.25 + 24.56) x 9. Jain.4 kN-m Total bending moment at bottom of wall is M = Mi2 + M2 = c (861.15 x (138 x 1.Part III: Solutions Impulsive bending moment at the bottom of wall is Mi = (Ah)i (mihi + mwhw + mtht) g = 0.024 x 189 x 2.69 + 106.5 ) 2 + (110.5kN .1 x 2.m 2 © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 5/ slide 10 . 5/2 = 434. © Sudhir K. Jain. Hence. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 5/ slide 11 .Part III: Solutions This is total bending moment and will be shared by two walls which are perpendicular to the direction of seismic force.25 kNm. bending moment at the bottom of one wall = 868. 81 = 1588 kN-m Convective overturning moment at the bottom of base slab is Mc* = (Ah)c mc (hc* + tb) g = 0.07+ 0.15 x [138(4.25) + 24.m © Sudhir K.7 kN-m Total overturning moment at bottom of base slab is M* = (M ) + (M ) * 2 i * 2 c = (1588 )2 + (236.9(4. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 5/ slide 12 .7 )2 = 1605kN .73 + 0.56 + 0. Jain.25) x 9.024 x 189 x (5.25) + 106.25/2] x 9.1(2.2 x 0.25) + 54.81 = 236.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 = 0.25 + 0. 93 m.61 m.7 kN Total base shear at bottom of wall is V = Vi2 + Vc2 = (531. mi = 230 t. (Ah)i = 0. hi*= 2.037 x 112 x 9. Jain. hc = 2.9) x 9.direction In Y – direction we have.8kN Solution 5/ slide 13 © Sudhir K.69 m.15. hi = 1.7 ) 2 2 = 532. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 . (Ah)c = 0.1 + 24.29 m. mc = 112 t.2 ) + (40.15 x (230 + 106.81 = 40.037 Impulsive base shear at the bottom of wall is Vi = (Ah)i (mi + mw + mt) g = 0. hc*= 3.81 = 531.Part III: Solutions Analysis in Y.2 kN Convective base shear at the bottom of wall is Vc = (Ah)c mc g = 0. 25 + 24.Part III: Solutions Impulsive bending moment at the bottom of wall is Mi = (Ah)i (mihi + mwhw + mtht) g = 0.81 = 119.69 + 106.3 )2 + (119. Jain.m © Sudhir K.1 x 2.15 x (230 x 1.037 x 112 x 2.81 = 1090 kN-m Convective bending moment at the bottom of wall is Mc = (Ah)c mc hc g = 0.9 x 4. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 5/ slide 14 .1 )2 = 1097kN .93 x 9.56) x 9.1 kN-m Total bending moment at bottom of wall is M = Mi2 + M2 = c (1090. 5 kN-m. Hence.Part III: Solutions This is total bending moment and will be shared by two walls which are perpendicular to the direction of seismic force. bending moment at the bottom of one wall = 1097/2 = 548. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 5/ slide 15 . Jain. © Sudhir K. 25) + 24.25/2] x 9.81 = 143.9(4.25) + 54.29+ 0.25) + 106.9 )2 = 1551kN .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 = 0.037 x 112 x (3.61 + 0.56 + 0.25 + 0. Jain.15 x [230(2.m © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 5/ slide 16 .1(2.2 x 0.25) x 9.81 = 1544 kN-m Convective overturning moment at the bottom of base slab is Mc* = (Ah)c mc (hc* + tb) g = 0.9 kN-m Total overturning moment at bottom of base slab is M = * (M ) + (M ) * 2 i * 2 c = (1544 )2 + (143. 0 m from top of floor slab.15 x (140 + 225) x 9.1) + (64. Solution: Impulsive base shear at the bottom of staging is Vi = (Ah)i (mi + ms) g = 0.06. Impulsive and convective mode parameters are: mi = 140 t.43.06 x 110 x 9. hc* = 3.0kN Solution 5/ slide 17 © Sudhir K.7 ) 2 2 = 541. mc = 110 t. Jain.2) An elevated water tank has CG of empty container at 3. hi* = 3.43 m.81 = 64. Height of staging from footing top to top of floor slab is 12 m.81 = 537.Part III: Solutions 5.1 kN Convective base shear at the bottom of staging is Vc = (Ah)c mc g = 0. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 . ms = 225 t. Base shear coefficient of impulsive mode is 0.15 and of convective mode is 0.7 kN Total base shear at bottom of staging is V = Vi2 + Vc2 = (537. Find base shear and base moment at the bottom of staging. 0 m Impulsive base moment at the bottom of staging is Mi* = (Ah)i [mi (hi* + hs) + ms hcg] g = 0.43 + 12) x 9.0 kN & Base moment = 8206 kN-m © Sudhir K.15 x [140 (3. base shear = 541.43 + 12) + 225 x 15. Jain.0] x 9.Part III: Solutions Distance of CG of container from bottom of staging.0 = 15.0 + 12.m Thus.06 x 110 x (3. hcg = 3.81 = 8145 kN-m Convective base moment at the bottom of staging is Mc* = (Ah)c mc (hc* + hs) g = 0. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 5/ slide 18 .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 . Assignment 6 Date of assignment: 8 February 2006 E-course on Seismic Design of Tanks 1 . 12 of the Guideline © Sudhir K. 4. 4.Part I: Reading Assignment 6 Read Clause 4.11. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Assignment 6/ slide 2 .9. 4. Jain.10. 6. impulsive pressure on wall varies in circumferential direction. 6.5)In rectangular tanks. convective pressure on a particular wall varies along the length of that wall. 6.1)Convective pressure varies linearly along wall height.2)In circular tanks. 6. © Sudhir K.4)Due to vertical ground acceleration.Part II: True / False 6. lateral pressure exerted by the liquid on the wall changes. Jain.6)Impulsive pressure is maximum at the bottom of wall. 6. hydrodynamic pressure acts on walls parallel to the direction of seismic force.7)Convective hydrodynamic force is more significant for tall tanks. 6.3)In rectangular tanks. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Assignment 6/ slide 3 . Tank is fixed at base.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 12 m Y X 6m Plan © Sudhir K. (Ah)c =0.15. Tank is located on hard soil in seismic zone III. Find in X.037.024. In Y – direction: (Ah)i = 0.and Y. In X – direction: (Ah)i = 0. Wall has uniform thickness of 200 mm.Part III: Questions 6. (Ah)c = 0.1) A ground supported rectangular RC water tank has plan dimensions of 12 x 6 m and water height of 4. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Assignment 6/ slide 4 .15.5 m. Jain. Solution 6 Date of assignment: 8 February 2006 Date of solution: 10 February 2006 E-course on Seismic Design of Tanks 1 . convective pressure on a particular wall varies along the length of that wall. 6. convective pressure on a wall remains constant along the length of that wall. © Sudhir K. This assumes that the direction of shaking is perpendicular to the wall. impulsive pressure on wall varies in circumferential direction. Jain. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 6/ slide 2 .3)In rectangular tanks.2)In circular tanks.Part II: True / False 6. True Impulsive pressure has cosφ variation in circumferential direction. False Convective pressure has curvilinear distribution along wall height. False In rectangular tanks. 6.1)Convective pressure varies linearly along wall height. 6. weight density of liquid increases or decreases depending on direction of vertical acceleration. hydrostatic pressure which depends on weight density of liquid.5)In rectangular tanks. © Sudhir K. False In rectangular tanks. Hence. This additional pressure is called pressure due to vertical excitation.4)Due to vertical ground acceleration. hydrodynamic pressure acts on walls perpendicular to the direction seismic force. lateral pressure exerted by the liquid on the wall changes. hydrodynamic pressure acts on walls parallel to the direction of seismic force. Jain. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 6/ slide 3 . True when subjected to vertical excitation. will also increase or decrease.Part II: True / False 6. Jain. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 6/ slide 4 .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).6)Impulsive pressure is maximum at the bottom of wall. True 6.Part II: True / False 6. Refer Figures 10(a) and 11(a) of Guidelines © Sudhir K. In X – direction: (Ah)i = 0. (Ah)c = 0. (Ah)c =0. Tank is fixed at base.15.5 m.1) A ground supported rectangular RC water tank has plan dimensions of 12 x 6 m and water height of 4. Tank is located on hard soil in zone III. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 6/ slide 5 . Find in X.15. Jain.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 12 m Y X 6m Plan © Sudhir K.Part III: Solution 6.and Y. Wall has uniform thickness of 200 mm.037.024. In Y – direction: (Ah)i = 0. hi = 0. mc = 0.5 = 2.375. tw = 200 mm Weight density of concrete = 25 kN/m3 Analysis in X.425.200 kg hi/h = 0.551.5/12 = 0. 584.375.000 = 137.Part III: Solution Solution: Capacity of tank = 12 x 6 x 4. L = 12 m.375 x 4.69 m hc/h =0. from Figure 3 of the Guideline.5 x 1000 = 324. B = 6 m.551 x 4. Jain. 584 x 324.5 m.024 h = 4.700 kg mc/m = 0. h/L = 4.direction: (Ah)i = 0. mi = 0. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 6/ slide 6 . (Ah)c = 0.425 x 324.5 = 1. For h/L = 0.15.000 = 189.375.5 = 324 m3 ∴ Mass of water = 12 x 6 x 4. hc = 0.000 kg Thickness of wall.48 m © Sudhir K. we have: mi/m = 0. 15 × 137 .60 kN/m2 Solution 6/ slide 7 © Sudhir K.5) = 0.60 kN/m a i = 2 (4 h − 6 hi ) = 2 4 . 90 qi (4 × 4 .95kN/m 2 bi = 2 (6hi − 2h ) = h 4.69 ) = 6 .5 − 6 × 1 .69 − 2 × 4. 81 g = = 16 .5 h qi 16. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 .95 kN/m2 i 2 Linear impulsive pressure distribution ai = 6.52 b = 0. 90 kN/m 2B 2×6 Pressure at bottom & top is given by qi = 16 .700 × 9 . Jain.Part III: Solution a) Impulsive mode ( A h )i m i 0 .90 (6 × 1. 57 kN/m 2 4 . IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 .52 h2 qc 3.52 2 bc = 1.48 ) = 0.71 (6 × 2. 81 = 3 .71 (4 × 4.48 − 2 × 4.08kN/m 2 bc = 2 (6hc − 2h ) = h 4.5 − 6 × 2. 024 × 189 .Part III: Solution b) Convective mode qc = ac = ( Ah ) c m c 0 . 71 kN/m g = 2B 2×6 Pressure at bottom & top is given by qc (4 h − 6 hc ) = 3.57 kN/m2 Solution 6/ slide 8 © Sudhir K.08 kN/m Linear convective pressure distribution ac = 0. Jain.5) = 1. 200 × 9 . 42 kN/m2 d) Pressure due to wall inertia.. pww = (Ah)i t ρm g = 0.16/2 x 1.15 x 0.e. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 6/ slide 9 . i.y/h) = 0.5 x (1. R = 2.5) = 4. Z/2.81 x 4. at the base of wall. Z = 0.10 Maximum pressure due to vertical acceleration occurs at y=0.0/4.5 Av = 2/3 .10 x 1000 x 9.Part III: Solution c) Pressure due to vertical acceleration Zone III. hence.5/2. I/R .0 For vertical mode. Sa/g = 2/3 x 0. ∴ Sa/g = 2.5 = 0. Jain.16 I = 1. ∴ pv = (Av) ρ g h (1. T = 0.3 Sec.75 kN/m2 © Sudhir K.5 It is ground supported RC tank with fixed base.0 x 2.2 x 25 = 0. 100 Kg hi/h = 0.59 kN/m 2 2 (6 .375.5/6 = 0. 60 Analysis in Y.15.5 = 1. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 6/ slide 10 .65 x 4.71 x 324. 57 2 + 4 . For h/L = 0.93 m © Sudhir K. 75. 75 ) + 0 . mc = 0. hi = 0. 346 x 324. 42 2 = 8.69 m hc/h =0. 346.037 h = 4. L = 6 m.5 m. B = 12 m.000 = 230.Part III: Solution Total pressure.000 = 112. (Ah)c = 0.direction: (Ah)i = 0. Jain.551. we have: mi/m = 0.375.000 Kg mc/m = 0. hc = 0. h/L = 4.5 = 2. mi = 0.375 x 4. p= (p iw + p ww ) 2 + p cw + p v 2 2 = + 0 . from Figure 3 of the Guideline.71. Jain.5) = 0.Part III: Solution a) Impulsive mode ( A h )i m i 0 . 81 g = = 14 .040 × 9 .52 b = 0.47 kN/m2 Solution 6/ slide 11 © Sudhir K.69 ) = 5 . 10 kN/m 2B 2 × 12 Pressure at bottom & top is given by qi = ai = qi (4 h − 6 hi ) = 14 .5 2 h2 qi 14.5 − 6 × 1 .47 kN/m 4 .79kN/m 2 bi = 2 (6hi − 2h ) = h 4. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 .10 (6 × 1. 15 × 230 .69 − 2 × 4.10 (4 × 4 .79 kN/m2 i 2 Linear impulsive pressure distribution ai = 5. 035 kN/m2 Solution 6/ slide 12 © Sudhir K.5) = 0.93 − 2 × 4. 70 kN/m 2B 2 × 12 Pressure at bottom & top is given by qc = ac = qc (4 h − 6 hc ) = 1 .72kN/m 2 bc = 2 (6hc − 2h ) = h 4.5 − 6 × 2 .035 kN/m 4 .93 ) = 0 .70 (6 × 2. 037 × 112 .70 (4 × 4 .52 2 bc = 0.Part III: Solution b) Convective mode ( A h )c m c 0 . IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 .72 kN/m 2 Linear convective pressure distribution ac = 0.104 × 9 . 81 g = = 1 . Jain.5 2 h2 qc 1. 42 kN/m2 d) Pressure due to wall inertia. ∴ Sa/g = 2.5/2. at the base of wall. Z = 0.5 Av = 2/3 .75 kN/m2 © Sudhir K.81 x 4.15 x 0.0 For vertical mode.16/2 x 1.. T = 0. pww = (Ah)i t ρm g = 0.0 x 2.Part III: Solution c) Pressure due to vertical acceleration Zone III. I/R .5) = 4.16 As it stores water. ∴ pv = (Av) ρ g h (1.5 It is ground supported RC tank with fixed base.10 x 1000 x 9. i.5 = 0. Sa/g = 2/3 x 0. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 6/ slide 13 . Z/2. R = 2.2 x 25 = 0. I = 1.3 Sec.0/4.y/h) = 0.5 x (1.10 Maximum pressure due to vertical acceleration is occur at y=0. Jain.e. 035 © Sudhir K.42 pww 1.42 4. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 6/ slide 14 .15 Total pressure 8. Jain. 75 ) 2 + 0 .15 1.59 7.Part III: Solution Total pressure p= (p iw + p ww ) 2 + p cw + p v 2 2 = (5 .6 pcw 0. 42 2 = 7.62 kN/m 2 At the base of wall pressure in kN/m2 are: piw X-direction Y-direction 6.57 pv 4.62 5. 035 2 + 4 .47 2 + 0 .47 0. Assignment 7 Date of assignment: 17 February E-course on Seismic Design of Tanks 1 . IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Assignment 7/ slide 2 .Part I: Reading Assignment 7 Read Examples 2 and 3 of the Guideline © Sudhir K. Jain. Part II: True / False 7.1) If braces are treated as rigid beams. then seismic forces are overestimated .2) Inclusion of shear deformation. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Assignment 7/ slide 3 .5) Empty tank does not have convective mode of vibration.2 times. 7. 7. 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.4) If stiffness decreases by 1. Jain. time period will increase by 1. © Sudhir K.44 times then. 7. and (c) approximate analysis using the approach of Sameer and Jain (1992). (b) approximate analysis considering braces as rigid beams as per SP:22 approach. Find the lateral stiffness by (a) computer analysis of a 3-dimensional model. Compare and discuss your results. All columns have 400 mm diameter.1) A eight column RC frame staging has four brace levels. Top ring beam is 400 mm wide and 750 mm deep.Part III: Questions 7. Grade of concrete is M 25. braces are 300 mm wide and 400 mm deep. CG of tank is at a distance of 3m from top ring beam (Refer Figures below). 3m 3m 4m 4m 4m 4m 4m Plan Elevation Assignment 7/ slide 4 © Sudhir K. Jain. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 . 20 m and 25 m.2. Jain. © Sudhir K. Find the lateral stiffness of shaft with and without considering shear deformation for following three staging heights: L = 15 m. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Assignment 7/ slide 5 .Part III: Questions 7. Take Poisson’s ratio as 0. Grade of concrete is M 25.2) A RC shaft has inner diameter of 6m and wall thickness of 200 mm. Solution 7 Date of assignment: 17 February Date of solution: 20 February E-course on Seismic Design of Tanks 1 . they do not allow any rotation at joints of columns and braces. higher seismic forces. increases. then seismic forces are overestimated . IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 7/ slide 2 . 7. False With the inclusion of shear deformation. This leads to significant overestimation of stiffness. Jain.1) If braces are treated as rigid beams. © Sudhir K.2) Inclusion of shear deformation reduces time period of shaft staging.Part II: True / False 7. True If braces are treated as rigid. lateral stiffness decreases and time period which is inversely proportional to square root of stiffness. This reduces time period and hence. . contribution of flexural deformation increase as L3 whereas contribution of shear deformation increase linearly. L. Jain. False The lateral stiffness is given by Ks = 1 L L + 3 EI 0 . i.2 of part III of this assignment. 7. in shafts with more height. Flexural contribution varies as L3. 5 AG 3 Here.Part II: True / False 7. L3/3EI is flexural contribution and L/0. Hence. This is also explained in the solution of problem no. © Sudhir K. flexural deformation governs the stiffness and effect of shear deformation is low.3) Effect of shear deformation on lateral stiffness is more important in shafts with large height-to-diameter ratio.5AG is shear contribution. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 7/ slide 3 . L is more. whereas. shear contribution varies linearly with length. Assuming same cross section. if height.e. True In empty tank.5 = 1. © Sudhir K.5. Note: (1.44. no water.2 times.4) If stiffness decreases by 1.2 times. then T will increase by 1. hence no convective mode.2 7. If K decreases to K/1. time period will increase by 1.44)0. True Time period T = 2 π (M/K)0.44 times then.Part II: True / False 7. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 7/ slide 4 . Jain.5) Empty tank does not have convective mode of vibration. 3m 3m 4m 4m 4m 4m 4m Plan Elevation Solution 7/ slide 5 © Sudhir K. All columns have 400 mm diameter. (b) approximate analysis considering braces as rigid beams as per SP:22 approach. CG of tank is at a distance of 3m from top ring beam (Refer Figures below).Part III: Solutions 7. Top ring beam is 400 mm wide and 750 mm deep. Grade of concrete is M 25. Find the lateral stiffness by (a) computer analysis of a 3-dimensional model.1) A eight column RC frame staging has four brace levels. and (c) approximate analysis using the approach of Sameer and Jain (1992). IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 . Compare and discuss your results. Jain. braces are 300 mm wide and 400 mm deep. a vertical rigid link of 3.0 m from circular ring beam. Computer model is shown on next slide. Columns are considered to be fixed at the base. a rigid link of 3. Jain. eight radial beams of same size as circular beams are also modeled. In order to have a node at the center. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 7/ slide 6 . On this rigid link. columns and circular ring beams.0 m length is required at the center of staging. a lateral force of 10 kN is applied. From the central node. Lateral load is to be applied at a height of 3. © Sudhir K.Part III: Solutions Solution: a) Stiffness from computer analysis Staging is modeled using SAP 2000.0 m length is put. Frame elements are used to model braces. For this purpose. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 7/ slide 7 .Part III: Solutions 10 kN Rigid link Column Brace Radial beam © Sudhir K. Jain. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 7/ slide 8 .00267 = 3.Part III: Solutions For lateral force of 10 kN.745 kN/m Deflection of staging © Sudhir K.00267 m ∴ Lateral stiffness of staging. Ks = 10/0. Jain. deflection = 0. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 7/ slide 9 . I = Moment of inertia and L = Length of panel. Jain.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. © Sudhir K. 4 m ∴ Moment of inertia.2 6 × 10 = 5 906 .26 x 10-3 m4 Hence.4)4/64 = 1. Hence.3 = 47250 kN/m Stiffness of one panel = 47. there are 8 columns. stiffness of each panel Kp = ∑ K C = 8 × 5906.5 = 25000 N/mm2 = 25 x 106 kN/m2 Column diameter = 400 mm = 0.Part III: Solutions In the present example. E = 5000 (25)0.3 kN/m 3 L3 ⎛ 4. 6 -3 K C = 12EI = 12 × 25 × 10 × 1 . Jain.250 kN/m © Sudhir K. Grade of concrete is M 25 Hence.0 ⎞ ⎜ ⎟ ⎝ ⎠ In each panel. I = π (0. These eight columns act like eight springs in parallel. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 7/ slide 10 . Hence.e.450 kN/m © Sudhir K. All panels have same stiffness and they act like five springs in series. Jain.. Ks = Kp/5 = 47.250/5 = 9.Part III: Solutions Staging comprises of five such panels. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 7/ slide 11 . Stiffness of staging. Stiffness of staging Ks is given by 1 1 1 1 1 1 = + + + + Ks Kp Kp Kp Kp Kp i. 1 K flexure = ∑K i =1 Np 1 panel . IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 7/ slide 12 .0 for end panels and α = 2. K panel ⎤ ⎡ 12EI c N c ⎢ Ib L ⎥ = 3 − ⎥ ⎢ h ⎣Ib L + α Ic h ⎦ α = 1.0 for intermediate panels 1 K axial 2 = N c A cE R 2 ∑ Hi h 2 i =1 Np © Sudhir K. Jain.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. 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 © Sudhir K. Jain. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 7/ slide 13 . For end panels.Part III: Solutions For intermediate panels. © Sudhir K. hence. For the bottom most panel. y from the restrained end. point of inflection is at a distance. point of inflection is at mid-height of panel. y is measured from bottom end. 3I y = b + Ic L h h 6I b L Top most panel Panel 1 Panel 2 Intermediate panels Panel 3 Panel 4 Bottom most panel Panel 5 For the top most panel. y is measured from top end.restrained end is at top ring beam. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 7/ slide 14 . Jain. Part III: Solutions For this Example. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 7/ slide 15 .50) x 2 = 2. we have: E = Modulus of elasticity = 5000 (25)0.000 mm Ic = Moment of inertia of column = π x (400)4/ 64 = 1. Jain.26 x 109 mm4 © Sudhir K.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 (22.296 mm R = Radius of staging system = 3. 699 mm © Sudhir K.41 x 1010 mm4 h = Height of each panel = 4.Part III: Solutions Ib = Moment of inertia of brace beam = 300 x(400)3/12 = 1.000 mm H4 = 3000 + 12000 + 2000 = 17.000 mm H5 = 3000 + 20000 – 2301 = 20.6x109 mm4 Icbr = Moment of inertia of circular ring beam = 400 x (750)3 / 12 = 1. Jain.6 × 10 9 1.000 mm H3 = 3000+ 8000 + 2000 = 13.000 mm 3I 3 × 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 = 9.26 × 10 9 b + Ic + 2296 4000 h h = L y = × 4000 = 2301 mm 9 6I 6 × 1. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 7/ slide 16 . 98 x1012 = 3.52 x10 − 5 mm / N © Sudhir K. Jain.84 x10 − 18 8 x125664 x 25000 x (3000) 2 2 ∑ Hi h = ((5301) 2 + (9000) 2 + (13000) 2 + (17000) 2 + (20699) 2 ) x 4000 = 3 . 98 x 10 1 12 ∴ K axial = 8.84 x10 −18 x 3. Kaxial is given by 1 K axial 2 Nc A c E R 2 Np i =1 N 2 2 = ∑H h Nc AcE R 2 i=1 i p = 2 = 8.Part III: Solutions Now. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 7/ slide 17 . 0 ) K panel = ⎤ 12 EI c N c ⎡ Ib L ⎢ ⎥ = 3.0 ) K panel = 12 EI c N c h3 ⎡ ⎤ Ib L ⎢ ⎥ = 2. for bottom most panel.5 x 104 N/mm. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 7/ slide 18 . for intermediate 3 panels. ( α = 1.48 x 104 N/mm.25 x 104 N/mm.0 ) K panel ⎤ 12 EI c N c ⎡ I cbr L = ⎥ = 4. Jain. ⎢ Ib L + α Ic h ⎥ ⎣ ⎦ © Sudhir K.Part III: Solutions Now. ( α = 2. ( α = 1. let us find Kpanel For the upper most panel. 3 h ⎢ Ib L + α Ic h ⎥ ⎣ ⎦ Similarly. ⎢ 3 h ⎢ I cbr L + α I c h ⎥ ⎦ ⎣ Similarly. Part III: Solutions 1 K flexure = ∑K i =1 Np 1 panel 1 K flexure = 1 1 2 2 2 + + + + 4.302 x10 − 4 mm / N ∴ Ks = 3. Jain.029 kN/m.48 ×10 4 2.48 ×10 4 = 2.25 ×10 4 2.95 x 10-4 mm/N Now. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 7/ slide 19 . © Sudhir K.48 ×10 4 2.95 × 10 − 4 + 3. Ks is obtained as 1 = Ks K 1 flexure + 1 K axial = 2.52 × 10 −5 = 3.029 N/mm = 3.5 ×10 4 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 © Sudhir K.Part III: Solutions Thus. Jain. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 7/ slide 20 . stiffness of staging using these three approaches is a) From computer analysis of 3-D model Ks = 3. © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 7/ slide 21 . stiffness is almost 2. if braces are treated as rigid beams. we have put radial beams. some of the difference in Sameer and Jain method is due to model error of computer Analysis.Part III: Solutions Thus. which impart rotational rigidity to columns at top level. The stiffness obtained by approach suggested by Sameer and Jain is about 19% lower than that from computer analysis. Jain. In computer analysis. as suggested by SP 22.5 times higher than that obtained computer analysis of 3-D model of staging. Hence. Find the lateral stiffness of shaft with and without considering shear deformation for following three staging heights: L = 15 m. Take Poisson’s ratio as 0. then Ks = 1 L3 L + 3EI κ' AG Solution 7/ slide 22 © Sudhir K. Grade of concrete is M 25. 20 m and 25 m. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 .2) A RC shaft has inner diameter of 6m and wall thickness of 200 mm. Solution: Lateral stiffness of RC shaft without considering shear deformation Ks = 3EI L3 If shear deformations are considered. Jain.2.Part III: Solutions 7. 02)/4 = 3.Part III: Solutions Grade of concrete is M 25. Jain.5 = 25. Hence.2 ∴ G = E/(2 x (1 + ν)) = 25000 x 103 /(2 x (1 + 0.000 N/mm2. = 25000 x 103 kN/m2 G = Shear modulus = E/(2 x (1 + ν)) where. ν = Poisson’s ratio = 0. E = 5000 x (25)0.04)/64 = 18.417 x103 kN/m2 I = Moment of inertia of shaft cross section = π x (6.5 (for hollow circular cross section) © Sudhir K.44 – 6.90 m2 κ’ = 0.2)) = 10.74 m4 A = Area of cross section of shaft = π x (6. Modulus of elasticity.42 – 6. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 7/ slide 23 . 74 0. Jain.9x10 417x10 3 3 = 318 .Part III: Solutions Now.4 x10 3 kN / m 153 Ks with considering shear deformation Ks = 1 L3 L + 3EI κ' AG = 1 15 15 + 3x25000x10 3 x18 .74 = = 416. for shaft with L = 15 m: Ks without considering shear deformation Ks = 3EI L3 3x25000x10 3 x18.5 x10 3 kN / m © Sudhir K. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 7/ slide 24 .5x3. Part III: Solutions Thus. Similarly.5%.8 % 10 % Note: Thus. for shafts with large height-to-diameter ratio. © Sudhir K. Ks (kN/m) of shaft diameter % Reduction without shear with shear ratio deformation deformation 15 m 20 m 25 m 2. Jain.5 3.7x103 81x103 23. IIT Kanpur E-Course on Seismic Design of Tanks / January 2006 Solution 7/ slide 25 . lateral stiffness is obtained for L = 20 m and 25 m and results are given in Table below Height Height-toLateral stiffness.5x103 149.5 % 14. 7x103 90x103 318.3 4.2 416. Ks reduces by 23.4x103 175. shear deformation has less effect on the lateral stiffness. with the inclusion of shear deformation.