Lecture 5
January 31, 2006
In this Lecture


Impulsive and convective base shear
Critical direction of seismic loading
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 2
Base shear

Previous lectures have covered

Procedure to find impulsive and convective
liquid masses


Procedure to obtain base shear coefficients in
impulsive and convective modes


This was done through a mechanical analog model
This requires time period, damping, zone factor, importance
factor and response reduction factor
Now, we proceed with seismic force or base
shear calculations
 Sudhir K. Jain, 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
Base shear

Now, we evaluate impulsive and convective
weights



Or, impulsive and convective masses
Earlier we have obtained impulsive and
convective liquid mass
Now, we consider structural mass also
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 5
Base shear : Ground supported tanks

Impulsive liquid mass is rigidly attached to
container wall


Hence, wall, roof and impulsive liquid vibrate
together
In ground supported tanks, total impulsive mass
comprises of



Mass of impulsive liquid
Mass of wall
Mass of roof
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 6
Base shear : Ground supported tanks

Hence, base shear in impulsive mode
Vi   A h i mi  mw  mt  g
mi = mass of impulsive liquid
 mw = mass of container wall
 mt = mass of container roof
 g = acceleration due to gravity

 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 7
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
Base shear : Ground supported tanks

Base shear in convective mode
Vc   A h

c mc g
mc = mass of convective liquid
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 9
Base shear : Ground supported tanks

Total base shear, V is obtained as:
V 


Vi 2  Vc2
Impulsive and convective base shear are
combined using Square Root of Sum of Square
(SRSS) rule
Except Eurocode 8, all international codes use
SRSS rule

Eurocode 8 uses absolute summation rule

i.e, V = Vi + Vc
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 10
Base shear : Ground supported tanks

In the latest NEHRP recommendations (FEMA
450), SRSS rule is suggested

Earlier version of NEHRP recommendations (FEMA
368) was using absolute summation rule


FEMA 450, 2003, “NEHRP recommended provisions for
seismic regulations for new buildings and other structures”,
Building Seismic Safety Council, National Institute of Building
Sciences,, USA.
FEMA 368, 2000, “NEHRP recommended provisions for
seismic regulations for new buildings and other structures”,
Building Seismic Safety Council, National Institute of Building
Sciences,, USA.
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 11
Bending moment:Ground supported tanks


Next, we evaluate bending or overturning
effects due to base shear
Impulsive base shear comprises of three parts



(Ah)i x mig
(Ah)i x mwg
(Ah)i x mtg
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 12
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
mi acts at hi*


If base pressure effect is included
Recall hi and hi* from Lecture 1
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 13
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. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 14
Bending moment:Ground supported tanks

For bending moment at the bottom of wall,
effect of base pressure is not included

Hence, hi and hc are used
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 15
Bending moment:Ground supported tanks

Bending moment at the bottom of wall
(Ah)imt
(Ah)cmc
hc
hi
(Ah)imi
(Ah)imw
ht
hw
Ground level
M i  A h i mi hi  mw hw  mt ht  g
M c   A h c (mc hc ) g
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 16
Bending moment:Ground supported tanks

Total bending moment at the bottom of wall
M  M M
2
i

2
c
SRSS rule used to combine impulsive and
convective responses
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 17
Overturning moment:Ground supported tanks

Overturning moment


This is at the bottom of base slab
Hence, must include effect of base pressure

hi* and hc* will be used
Ground level
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 18
Overturning moment:Ground supported tanks

Overturning moment in impulsive mode
*

mi ( hi  tb )  mw hw  tb   
*
M i   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, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 19
Bending moment:Ground supported tanks

Overturning moment is at the bottom of base
slab



Hence, lever arm is from bottom of base slab
Hence, base slab thickness, tb is added to
heights measured from top of the base slab
Total overturning moment
M  M
*
 Sudhir K. Jain, IIT Kanpur
*2
i
M
*2
c
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 20
Example

Example: A ground-supported circular tank is shown
below along with some relevant data. Find base shear
and bending moment at the bottom of wall. Also find
base shear and overturning moment at the bottom of
base slab.
Roof slab 150 mm thick
mw = 65.3 t, mt =33.1 t,
10 m
4m
Wall 200
mm thick
Base slab 250 mm thick
 Sudhir K. Jain, IIT Kanpur
mi = 141.4 t; mc = 163.4 t
mb = 55.2 t,
hi =1.5 m, hi* = 3.95 m,
hc = 2.3 m, hc* = 3.63 m
(Ah)i = 0.225, (Ah)c = 0.08
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 21
Example
Solution:
Impulsive base shear at the bottom of wall is
Vi = (Ah)i (mi + mw + mt) g
= 0.225 x (141.4 + 65.3 + 33.1) x 9.81
= 529.3 kN
Convective base shear at the bottom of wall is
Vc = (Ah)c mc g
= 0.08 x 163.4 x 9.81
= 128.2 kN
Total base shear at the bottom of wall is
V  Vi2  Vc2 
 Sudhir K. Jain, IIT Kanpur
529.3   128.2 
2
2
 544.6kN
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 22
Example
For obtaining bending moment, we need height of CG of roof slab
from bottom of wall, ht.
ht = 4.0 + 0.075 = 4.075 m
Impulsive bending moment at the bottom of wall is
Mi = (Ah)i (mihi + mwhw + mtht) g
= 0.225 x (141.4 x 1.5 + 65.3 x 2.0 + 33.1 x 4.075) x 9.81
= 1054 kN-m
Convective bending moment at the bottom of wall is
Mc = (Ah)c mc hc g
= 0.08 x 163.4 x 2.3 x 9.81
= 295 kN-m
Total bending moment at bottom of wall is
M  Mi2  M2c 
 Sudhir K. Jain, IIT Kanpur
10542  2952
 1095kN - m
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 23
Example
Now, we obtain base shear at the bottom of base slab
Impulsive base shear at the bottom of base slab is
Vi = (Ah)i (mi + mw + mt + mb) g
= 0.225 x (141.4 + 65.3 + 33.1 + 55.2) x 9.81
= 651.1 kN
Convective base shear at the bottom of base slab is
Vc = (Ah)c mc g
= 0.08 x 163.4 x 9.81
= 128.2 kN
Total base shear at the bottom of base slab is
V  Vi2  Vc2 
 Sudhir K. Jain, IIT Kanpur
651.1   128.2 
2
2
 663.6kN
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 24
Example
Impulsive overturning moment at the bottom of base slab
Mi* = (Ah)i [mi (hi* + tb) + mw(hw + tb) + mt(ht +tb) + mb tb/2]g
= 0.225 x [141.4(3.95 + 0.25) + 65.3(2.0 + 0.25) +
33.1(4.075 + 0.25) + 55.2 x 0.25/2] x 9.81
= 1966 kN-m
Convective overturning moment at the bottom of base slab
Mc* = (Ah)c mc (hc* + tb) g
= 0.08 x 163.4 x (3.63 + 0.25) x 9.81
= 498 kN-m
Total overturning moment at bottom of base slab
*
M 
M   M 
* 2
i
* 2
c

1966
2
 498  2028 kN - m
2
Notice that this value is substantially larger that the value at the
bottom of wall (85%)
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 25
Base shear : Elevated tanks


In elevated tanks, base shear at the bottom of
staging is of interest
Ms is structural mass

Base shear in impulsive mode
Vi   A h i mi  ms  g

Base shear in convective mode
Vc   A h

c mc g
Total base shear
V 
 Sudhir K. Jain, IIT Kanpur
Vi 2  Vc2
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 26
Bending moment:Elevated tanks

Bending moment at the bottom of staging


Impulsive base shear comprises of two parts



Bottom of staging refers to footing top
(Ah)i x mig
Ah)i x msg
Convective base shear has only one part

(Ah)c x mcg
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 27
Bending moment:Elevated tanks


mi acts at hi*
mc acts at hc*


Bending moment at bottom of staging is being
obtained
Hence, effect of base pressure included and hi*
and hc* are used
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 28
Bending moment:Elevated tanks

Structural mass, ms comprises of mass of empty
container and 1/3rd mass of staging


ms is assumed to act at CG of empty container
CG of empty container shall be obtained by
considering roof, wall, floor slab and floor beams
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 29
Bending moment:Elevated tanks

Bending moment at the bottom of staging
 

M i   Ah i mi hi  hs  ms hcg
*
M



g
  Ah c mc h  hs g
*
c
hs = staging height


*
c
*
Measured from top of footing to bottom of wall
hcg = distance of CG of empty container from
bottom of staging
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 30
Bending moment:Elevated tanks

Bending moment at the bottom of staging
(Ah)i mig
(Ah)c mcg
(Ah)i msg
hi*
hc *
hs
hs
hcg
Top of footing
 


M i   Ah i mi hi  hs  ms hcg g
*
*
 Sudhir K. Jain, IIT Kanpur
M
*
c
*



 Ah c mc hc  hs  g
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 31
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. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 32
Bending moment:Elevated tanks



For analysis of frame staging, two approaches
are possible
Approach 1: Perform analysis in two steps
Step 1:



Step 2:



Analyze frame for (Ah)imig + (Ah)imsg
Obtain forces in columns and braces
Analyze the frame for (Ah)cmcg
Obtain forces in columns and braces
Use SRSS rule to combine the member forces
obtained in Step 1 and Step 2
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 33
Bending moment:Elevated tanks


Approach 2:
Apply horizontal force V at height h1 such that



V x h1 = M*
V and M* are obtained using SRSS rule as
described in slide nos. 26 and 32
In this approach, analysis is done in single step

Simpler and faster than Approach 1
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 34
Example

Example: An elevated tank on frame staging is shown
below along with some relevant data. Find base shear
and bending moment at the bottom of staging.
A is CG of empty container
A
2.8 m
mi = 100t; mc = 180 t
Mass of container = 160 t
Mass of staging = 120 t
hs = 15 m
hi* = 3 m, hc* = 4.2 m
(Ah)i = 0.08, (Ah)c = 0.04
GL
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 35
Example
Structural mass, ms = mass of container + 1/3rd mass of staging
= 160 + 1/3 x 120 = 200 t
Base shear in impulsive mode
Vi   A h i mi  ms  g
 0.08x100  200 x9.81
= 78.5 + 157 = 235.5 kN
Base shear in convective mode
Vc   A h  c mc g
 0.04 x180 x 9.81  70.6kN
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 36
Example
Total base shear
V 

Vi 2  Vc2
235.52  70.6 2  245.8kN
Now, we proceed to obtain bending moment at the bottom staging
Distance of CG of empty container from bottom of staging,
hcg = 2.8 + 15 = 17.8 m
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 37
Example
Base moment in impulsive mode
Mi
*
 

  Ah i mi hi  hs  ms hcg
*

g
 0.08100x3.0  15  200x 17.8x9.81
= 78.5 x 18 + 157 x 17.8
= 4207 kNm
Note: 78.5 kN of force will act at 18.0m and 157 kN of force will act at
17.8 m from top of footing.
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 38
Example
Base moment in convective mode


M c   Ah c mc hc  hs g
*
*
 0.04x180x4.2  15x9.81
= 70.6 x 19.2
= 1356 kNm
Note: 70.6 kN of force will act at 19.2 m from top of footing.
Total base moment
M* 
2
2
M *i  M *c
 42072  13562
 4420 kNm
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 39
Example

Now, for staging analysis, seismic forces are to be
applied at suitable heights

There are two approaches


Refer slide no 33
Approach 1:



Step 1: Apply force of 78.5 kN at 18 m and 157 kN at 17.8
m from top of footing and analyze the frame
Step 2: Apply 70.6 kN at 19.2 m from top of footing and
analyze the frame
Member forces (i.e., BM, SF etc. in columns and braces) of
Steps 1 and 2 shall be combined using SRSS
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 40
Example

Approach 2:

Total base shear, V = 245.8 kN will be applied at height h1,
such that
V x h1 = M*
245.8 x h1 = 4420
 h1 = 17.98 m

Thus, apply force of 245.8 kN at 17.98 m from top of footing
and get member forces (i.e., BM, SF in columns and
braces).
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 41
Elevated tanks:Empty condition

Elevated tanks shall be analysed for tank full as
well as tank empty conditions


Design shall be done for the critical condition
In empty condition, no convective liquid mass


Hence, tank will be modeled using single
degree of freedom system
Mass of empty container and 1/3rd staging mass
shall be considered
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 42
Elevated tanks:Empty condition


Lateral stiffness of staging, Ks will remain same
in full and empty conditions
In full condition, mass is more


Hence, time period of empty tank will be less



In empty condition mass is less
Recall, T = 2Π M
K
Hence, Sa/g will be more
Usually, tank full condition is critical

However, for tanks of low capacity, empty
condition may become critical
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 43
Direction of seismic force


Let us a consider a vertical cantilever with
rectangular cross section
Horizontal load P is applied




First in X-Direction
Then in Y-direction (see Figure below)
More deflection, when force in Y-direction
Hence, direction of lateral loading is important !!
P
Y
X
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
P
Lecture 5/ Slide 44
Direction of seismic force

On the other hand, if cantilever is circular



Direction is not of concern
Same deflection for any direction of loading
Hence, it is important to ascertain the most critical
direction of lateral seismic force

Direction of force, which will produce maximum
response is the most critical direction

In the rectangular cantilever problem, Y-direction is the most
critical direction for deflection
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 45
Direction of seismic force

For frame stagings consisting of columns and braces, IS
11682:1985 suggests that horizontal seismic loads shall be
applied in the critical direction

IS 11682:1985, “Criteria for Design of RCC Staging for Overhead
Water Tanks”, Bureau of Indian Standards, New Delhi
Clause 7.1.1.2 Horizontal forces – Actual forces and moments resulting
from horizontal forces may be calculated for critical direction and used
in the design of the structures. Analysis may be done by any of the
accepted methods including considering as space frame.
Clause 7.2.2 Bending moments in horizontal braces due to horizontal
loads shall be calculated when horizontal forces act in a critical
direction. The moments in braces shall be the sum of moments in the
upper and lower columns at the joint resolved in the direction of
horizontal braces.
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 46
Direction of seismic force


Section 4.8 of IITK-GSDMA Guidelines contains
provisions on critical direction of seismic force for
tanks
Ground-supported circular tanks need to be
analyzed for only one direction of seismic loads


These are axisymmetric
Hence, analysis in any one direction is sufficient
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 47
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, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 48
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, analysis in two directions will be necessary
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 49
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, in a 4 - column staging



Seismic loading along the length of the brace is
critical for braces
Seismic loading in diagonal direction gives
maximum axial force in columns
See next slide
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 50
Direction of seismic force

Critical directions for 4 - column staging
Bending Axis
Critical direction for
shear force in brace
 Sudhir K. Jain, IIT Kanpur
Critical direction for
axial force in column
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 51
Direction of seismic force

For 6 – column and 8 – column staging, critical
directions are given in Figure C-6 of the
Guideline


See next two slides
More information available in Sameer and Jain
(1994)

Sameer, S. U., and Jain, S. K., 1994, “Lateral load analysis
of frame staging for elevated water tanks”, Journal of
Structural Engineering, ASCE, Vol.120, No.5, 1375-1393.
(http://www.nicee.org/ecourse/Tank_ASCE.pdf)
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 52
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 and
bending moment in braces and axial
force in columns
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 53
Direction of seismic force

Critical directions for 8 - column staging
Bending Axis
Critical direction for
shear force and bending
moment in braces
 Sudhir K. Jain, IIT Kanpur
Critical direction for shear force,
bending moment and axial force
in columns
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 54
Direction of seismic force

As an alternative to analysis in the critical
directions, following two load combinations can
be used

100 % + 30% rule


Also used in IS 1893(Part 1) for buildings
SRSS rule
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 55
Direction of seismic force

100%+30% rule implies following combinations




ELx + 0.3 ELY
ELY + 0.3 ELx
ELx is response quantity when seismic loads are
applied in X-direction
ELY is response quantity when seismic loads are
applied in Y-direction
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 56
Direction of seismic force

100%+30% rule requires

Analyze tank with seismic force in X-direction;
obtain response quantity, ELX




Response quantity means BM in column, SF in brace, etc.
Analyze tank with seismic force in Y-direction;
obtain response quantity, ELY
Combine response quantity as per 100%+30%
rule
Combination is on response quantity and not
on seismic loads
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 57
Direction of seismic force

Important to note that the earthquake
directions are reversible

Hence, in 100%+30% rule, there are total eight
load combinations
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 58
Direction of seismic force
SRSS rule implies following combination

EL2x  EL2y

Note:



ELx is response quantity when seismic loads are
applied in X-direction
ELY is response quantity when seismic loads are
applied in Y-direction
Hence, analyze tank in two directions and use
SRSS combination of response quantity
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 59
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, 100%+30% or SRSS rule be used
 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks/ January 2006
Lecture 5/ Slide 60