1. RESPONSE SPECTRUM METHOD OF ANALYSIS
Chapters – 5 & 6
Chapter -5
RESPONSE SPECTRUM METHOD OF ANALYSIS

2. Introduction It is also useful for approximate evaluation
1/1
Response spectrum method is favoured by
earthquake engineering community because of:
It provides a technique for performing an
equivalent static lateral load analysis.
It allows a clear understanding of the
contributions of different modes of vibration.
It offers a simplified method for finding the
design forces for structural members for
earthquake.
It is also useful for approximate evaluation
of seismic reliability of structures.

3. Contd… The concept of equivalent lateral forces for earth-
1/2
Contd…
The concept of equivalent lateral forces for earth-
quake is a unique concept because it converts a
dynamic analysis partly to dynamic & partly to
static analysis for finding maximum stresses.
For seismic design, these maximum stresses are
of interest, not the time history of stress.
Equivalent lateral force for an earthquake is
defined as a set of lateral force which will
produce the same peak response as that
obtained by dynamic analysis of structures .
The equivalence is restricted to a single mode of
vibration.

4. Contd… developed using the following steps.
1/3
Contd…
The response spectrum method of analysis is
developed using the following steps.
A modal analysis of the structure is carried out
to obtain mode shapes, frequencies & modal
participation factors.
Using the acceleration response spectrum, an
equivalent static load is derived which will
provide the same maximum response as that
obtained in each mode of vibration.
Maximum modal responses are combined to
find total maximum response of the structure.

5. Contd… The first step is the dynamic analysis while , the
1/4
Contd…
The first step is the dynamic analysis while , the
second step is a static analysis.
The first two steps do not have approximations,
while the third step has some approximations.
As a result, response spectrum analysis is
called an approximate analysis; but applications
show that it provides mostly a good estimate of
peak responses.
Method is developed for single point, single
component excitation for classically damped
linear systems. However, with additional
approximations it has been extended for multi
point-multi component excitations & for non-
classically damped systems.

6. Development of the method
1/5
Development of the method
Equation of motion for MDOF system under
single point excitation
(5.1)
Using modal transformation, uncoupled sets of
equations take the form
is the mode shape; ωi is the natural frequency
is the more participation factor; is the
modal damping ratio.

7. Contd… mode Response of the system in the ith mode is (5.3)
1/6
Contd…
Response of the system in the ith mode is
(5.3)
Elastic force on the system in the ith mode
(5.4)
As the undamped mode shape satisfies
(5.5)
Eq 5.4 can be written as
(5.6)
The maximum elastic force developed in the ith
mode
(5.7)

8. Contd… Referring to the development of displacement response spectrum
1/7
Contd…
Referring to the development of displacement
response spectrum
(5.8)
Using , Eqn 5.7 may be written as
(5.9)
Eq 5.4 can be written as
(5.10)
is the equivalent static load for the ith mode
of vibration.
is the static load which produces structural
displacements same as the maximum modal
displacement.

9. Contd… Since both response spectrum & mode shape
1/8
Contd…
Since both response spectrum & mode shape
properties are required in obtaining , it is known
as modal response spectrum analysis.
It is evident from above that both the dynamic &
static analyses are involved in the method of
analysis as mentioned before.
As the contributions of responses from different
modes constitute the total response, the total
maximum response is obtained by combining modal
quantities.
This combination is done in an approximate manner
since actual dynamic analysis is now replaced by
partly dynamic & partly static analysis.

10. Contd… Modal combination rules
2/1
Contd…
Three different types of modal combination rules
are popular
ABSSUM
SRSS
CQC
Modal combination rules
ABSSUM stands for absolute sum of maximum
values of responses; If is the response quantity
of interest
(5.11)
is the absolute maximum value of response in the ith mode.

11. 2/2
Contd…
The combination rule gives an upper bound to the computed values of the total response for two
reasons:
It assumes that modal peak responses occur at
the same time.
It ignores the algebraic sign of the response.
Actual time history analysis shows modal peaks
occur at different times as shown in Fig. 5.1;further
time history of the displacement has peak value at
some other time.
Thus, the combination provides a conservative
estimate of response.

12. 2/3 Fig 5.1 (a) Top storey displacement5 10 15 20 25 30
-0.4
-0.2
0.2
0.4
Top floor displacement (m)
t=6.15
Time (sec)
First generalized displacement (m)
t=6.1
(a) Top storey displacement
(b) First generalized displacement
Fig 5.1

13. (c) Second generalized displacement
Contd…
2/3 5 10 15 20 25 30
-0.06
-0.04
-0.02
0.02
0.04
0.06
Time (sec)
Second generalized displacement (m)
t=2.5
(c) Second generalized displacement
Fig 5.1 (contd.)

14. Contd… SRSS combination rule denotes square root of sum
2/4
Contd…
SRSS combination rule denotes square root of sum
of squares of modal responses
For structures with well separated frequencies, it
provides a good estimate of total peak response.
When frequencies are not well separated, some
errors are introduced due to the degree of
correlation of modal responses which is ignored.
The CQC rule called complete quadratic
combination rule takes care of this correlation.

15. Contd… It is used for structures having closely spaced frequencies:
2/5
Contd…
It is used for structures having closely spaced
frequencies:
Second term is valid for & includes the effect
of degree of correlation.
Due to the second term, the peak response may be
estimated less than that of SRSS.
Various expressions for are available; here
only two are given :

16. Contd… (Rosenblueth & Elordy) (5.14) (Der Kiureghian) (5.15)
2/6
Contd…
(Rosenblueth & Elordy) (5.14)
(Der Kiureghian) (5.15)
Both SRSS & CQC rules for combining peak modal
responses are best derived by assuming
earthquake as a stochastic process.
If the ground motion is assumed as a stationary
random process, then generalized coordinate in
each mode is also a random process & there
should exist a cross correlation between
generalized coordinates.

17. Contd… Because of this, exists between two modal peak responses.
2/7
Contd…
Because of this, exists between two modal
peak responses.
Both CQC & SRSS rules provide good estimates of
peak response for wide band earthquakes with
duration much greater than the period of structure.
Because of the underlying principle of random
vibration in deriving the combination rules, the
peak response would be better termed as mean
peak response.
Fig 5.2 shows the variation of with frquency
ratio rapidly decreases as frequency ratio
increases.

18. Contd…
2/8
Fig 5.2

19. 2/9
Contd…
As both response spectrum & PSDF represent frequency contents of ground motion, a relationship
exists between the two.
This relationship is investigated for the smoothed
curves of the two.
Here a relationship proposed by Kiureghian is
presented

20. 2/10
Contd…
Example 5.1 : Compare between PSDFs obtained from the smoothed displacement RSP and FFT of Elcentro record. 10 20 30 40 50 60
0.01
0.02
0.03
0.04
0.05
Frequency (rad/sec)
PSDF of acceleration (m 2
sec - 3
/rad)
Unsmoothed PSDF from Eqn 5.16a
Raw PSDF from fourier spectrum 70 80 90
100
0.005
0.015
0.025
PSDFs of acceleration (m
Eqn.5.16a
Fourier spectrum of El Centro
Unsmoothed
5 Point smoothed
Fig5.3

21. Application to 2D frames
2/11
Degree of freedom is sway degree of freedom.
Sway d.o.f are obtained using condensation
procedure; during the process, desired response
quantities of interest are determined and stored in
an array R for unit force applied at each sway
d.o.f.
Frequencies & mode shapes are determined
using M matrix & condensed K matrix.
For each mode find (Eq. 5.2) & obtain Pei
(Eq. 5.9)

22. Contd… Solution : Obtain ; is the modal peak response vector.
2/12
Contd…
Obtain ; is the modal peak
response vector.
Use either CQC or SRSS rule to find mean peak
response.
Example 5.2 : Find mean peak values of top dis-placement, base shear and inter storey drift between 1st & 2nd floors.
Solution :

23. Base shear in terms of mass (m)
2/13
Contd…
Table 5.1
Approaches
Disp (m)
Base shear in terms of mass (m)
Drift (m)
2 modes
all modes
SRSS
0.9171
0.917
0.221
CQC
0.9121
0.905
0.214
ABSSUM
0.9621
0.971
0.228
0.223
Time history
0.8921
0.893
0.197
0.198

24. Application to 3D tall frames
3/1
Analysis is performed for ground motion applied to
each principal direction separately.
Following steps are adopted:
Assume the floors as rigid diaphragms & find
the centre of mass of each floor.
DYN d.o.f are 2 translations & a rotation; centers
of mass may not lie in one vertical (Fig 5.4).
Apply unit load to each dyn d.o.f. one at a
time & carry out static analysis to find
condensed K matrix & R matrix as for 2D frames.
Repeat the same steps as described for 2D
frame

25. 3/2
Figure 5.4:

26. 3/3
Contd…
Example 5.3 : Find mean peak values of top floor displacements , torque at the first floor & at the base of column A for exercise for problem Use digitized values of the response spectrum of El centro earthquake ( Appendix 5A of the book).
Results are obtained following the steps of
section
Results are shown in Table 5.2.
Solution :

27. Contd… TABLE 5.2 Results obtained by CQC are closer to those of
3/4
Contd…
TABLE 5.2
Approaches
displacement (m)
Torque (rad)
Vx(N)
Vy(N)
(1)
(2)
(3)
SRSS
0.1431
0.0034
0.0020
214547
44081
CQC
0.1325
0.0031
0.0019
207332
43376
Time history
0.1216
0.0023
0.0016
198977
41205
Results obtained by CQC are closer to those of
time history analysis.

28. RSA for multi support excitation
3/5
RSA for multi support excitation
Response spectrum method is strictly valid for
single point excitation.
For extending the method for multi support
excitation, some additional assumptions are
required.
Moreover, the extension requires a derivation
through random vibration analysis. Therefore, it is
not described here; but some features are given
below for understanding the extension of the
method to multi support excitation.
It is assumed that future earthquake is
represented by an averaged smooth response
spectrum & a PSDF obtained from an ensemble
of time histories.

29. Contd… Lack of correlation between ground motions at
3/6
Contd…
Lack of correlation between ground motions at
two points is represented by a coherence function.
Peak factors in each mode of vibration and the
peak factor for the total response are assumed to
be the same.
A relationship like Eqn is established
between and PSDF.
Mean peak value of any response quantity r consists of two parts:

30. Contd… Pseudo static response due to the displacements of the supports
3/7
Contd…
Pseudo static response due to the displacements of the supports
Dynamic response of the structure with respect to supports.
Using normal mode theory, uncoupled dynamic equation of motion is written as:

31. Contd… If the response of the SDOF oscillator to then
3/8
Contd…
If the response of the SDOF oscillator to
then
Total response is given by
are vectors of size m x s (for s=3 & m=2)

32. Contd… Assuming to be random processes, PSDF of is given by:
3/9
Contd…
Assuming to be random processes, PSDF of is given by:
Performing integration over the frequency range
of interest & considering mean peak as peak
factor multiplied by standard deviation,
expected peak response may be written as:

33. Contd… and are the correlation matrices whose elements are given by:
3/10
Contd…
and are the correlation matrices
whose elements are given by:

34. 3/11
Contd…

35. 3/12
Contd…
For a single train of seismic wave,
that is displacement response spectrum for a specified ξ ; correlation matrices can be obtained
if is additionally provided; can be determined from (Eqn 5.6).
If only relative peak displacement is required,third
term of Eqn is only retained.
Steps for developing the program in MATLAB is
given in the book.
Example 5.4 Example 3.8 is solved for El centro earthquake spectrum with time lag of 5s.

36. Contd… Solution :The quantities required for calculating the
3/13
Contd…
Solution :The quantities required for calculating the
expected value are given below:

37. 3/14
Contd…

38. Contd… Mean peak values determined are:
3/15
Contd…
Mean peak values determined are:
For perfectly correlated ground motion

39. Contd… Mean peak values of relative displacement
3/16
Contd…
Mean peak values of relative displacement
It is seen that’s the results of RHA & RSA match
well.
Another example (example 3.10) is solved for a time
lag of a 2.5 sec.
Solution is obtained in the same way and results
are given in the book. The calculation steps
are self evident.

40. Cascaded analysis Cascaded analysis is popular for seismic analysis
4/1
Cascaded analysis
Cascaded analysis is popular for seismic analysis
of secondary systems (Fig 5.5).
RSA cannot be directly used for the total system
because of degrees of freedom become
prohibitively large ; entire system becomes
nonclasically damped.
Secondary System xg .. k c m
xa = xf + xg
Secondary system mounted on a floor of a building frame
SDOF is to be analyzed for obtaining floor response spectrum xf
Fig 5.5

41. Contd… In the cascaded analysis two systems- primary
4/2
Contd…
In the cascaded analysis two systems- primary
and secondary are analyzed separately; output of
the primary becomes the input for the secondary.
In this context, floor response spectrum of the
primary system is a popular concept for
cascaded analysis.
The absolute acceleration of the floor in the figure is
Pseudo acceleration spectrum of an SDOF is
obtained for ; this spectrum is used for RSA of
secondary systems mounted on the floor.

42. Floor displacement response spectrum (Exmp. 5.6)
4/3
Contd…
Example 5.6 For example 3.18, find the mean peak displacement of the oscillator for El Centro earthquake.
for secondary system = 0.02 ; for the main
system = 0.05 ;floor displacement spectrum shown in the Fig5.6 is used
Solution 5 10 15 20 25 30 35 40
0.5 1
1.5
Frequency (rad/sec)
Displacement (m)
Using this spectrum, peak displacement of the secondary system with T=0.811s is m.
The time history analysis for the entire system (with C matrix for P-S system) is found as m.
Floor displacement response spectrum (Exmp. 5.6)

43. Approximate modal RSA For nonclassically damped system, RSA cannot
4/4
For nonclassically damped system, RSA cannot
be directly used.
However, an approximate RSA can be performed.
C matrix for the entire system can be obtained
(using Rayleigh damping for individual systems
& then combining them without coupling terms)
matrix is obtained considering all d.o.f. &
becomes non diagonal.
Ignoring off diagonal terms, an approximate
modal damping is derived & is used for RSA.

44. Seismic coefficient method
4/5
Seismic coefficient method uses also a set of
equivalent lateral loads for seismic analysis of
structures & is recommended in all seismic codes
along with RSA & RHA.
For obtaining the equivalent lateral loads, it uses
some empirical formulae. The method consists of
the following steps:
Using total weight of the structure, base
shear is obtained by
is a period dependent seismic coefficient

45. Contd… Base shear is distributed as a set of lateral
4/6
Contd…
Base shear is distributed as a set of lateral
forces along the height as
bears a resemblance with that for the
fundamental mode.
Static analysis of the structure is carried out
with the force
Different codes provide different recommendations
for the values /expressions for

46. Distribution of lateral forces can be written as
4/7
Distribution of lateral forces can be written as

47. Computation of base shear is based on first mode.
4/8
Computation of base shear is based on first mode.
Following basis for the formula can be put forward.

48. Seismic code provisions
5/1
Seismic code provisions
All countries have their own seismic codes.
For seismic analysis, codes prescribe all three
methods i.e. RSA ,RHA & seismic coefficient
method.
Codes specify the following important factors for
seismic analysis:
Approximate calculation of time period for
seismic coefficient method.
plot.
Effect of soil condition on

49. Contd… Reduction factor for obtaining design forces
5/2
Contd…
Seismicity of the region by specifying PGA.
Reduction factor for obtaining design forces
to include ductility in the design.
Importance factor for structure.
Provisions of a few codes regarding the first three
are given here for comparison. The codes include:
IBC – 2000
NBCC – 1995
EURO CODE – 1995
NZS 4203 – 1992
IS 1893 – 2002

50. 5/3
Contd…
IBC – 2000
for class B site,
for the same site, is given by

51. Contd… T may be computed by can have any reasonable distribution.
5/4
Contd…
T may be computed by
can have any reasonable distribution.
Distribution of lateral forces over the height
is given by

52. Contd… Distribution of lateral force for nine story frame is
5/5
Contd…
Distribution of lateral force for nine story frame is
shown in Fig5.8 by seismic coefficient method . 2 4 1 3 5 6 7 8 9
Storey force
Storey
T=2sec
T=1sec
T=0.4sec W
Fig5.8

53. Seismic response factor S
5/6
Contd…
NBCC – 1995
is given by
For U=0.4 ; I=F=1, variations of with T
are given in Fig 5.9.
0.5 1
1.5 2
2.5 3
3.5 4
4.5
Time period (sec)
Seismic response factor S
Fig5.9

54. Contd… For PGV = 0.4ms-1 , is given by T may be obtained by
5/7
Contd…
For PGV = 0.4ms-1 , is given by
T may be obtained by
S and Vs T are compared in Fig 5.10 for
v = 0.4ms-1 , I = F = 1; (acceleration and
velocity related zone)

55. 5/8
Contd…
0.5 1
1.5 2
2.5 3
3.5 4
0.2
0.4
0.6
0.8
1.2
1.4
Time period (sec)
A/g S
S or A/g
Fig5.10

56. 5/9
Contd…
Distribution of lateral forces is given by

57. Contd… EURO CODE 8 – 1995 Base shear coefficient is given by
5/10
Contd…
EURO CODE 8 – 1995
Base shear coefficient is given by
is given by
Pseudo acceleration in normalized form is given
by Eqn 5.58 in which values of Tb,Tc,Td
are
(A is multiplied by 0.9)

58. 5/11
Contd…
Pseudo acceleration in normalized form ,
is given by

59. Rayleigh's method may be used for calculating T.
5/12
Rayleigh's method may be used for calculating T.
Distribution of lateral force is
Variation of are shown in
Fig 5.11.

60. Contd… 5/13 .. Fig 5.11 Ce /ug0 or A/ug0 Time period (sec) 0.5 1 1.5 2
0.5 1
1.5 2
2.5 3
3.5 4
Time period (sec)
A/ug0
Ce/ug0
Ce /ug0 or A/ug0 ..
Fig 5.11

61. Contd… NEW ZEALAND CODE ( NZ 4203: 1992)
6/1
Contd…
NEW ZEALAND CODE ( NZ 4203: 1992)
Seismic coefficient & design response curves
are the same.
For serviceability limit,
is a limit factor.
For acceleration spectrum, is replaced by T.

62. Contd… Lateral load is multiplied by 0.92. Fig5.12 shows the plot of
6/2
Contd…
Lateral load is multiplied by 0.92.
Fig5.12 shows the plot of
Distribution of forces is the same as Eq.5.60
Time period may be calculated by using
Rayleigh’s method.
Categories 1,2,3 denote soft, medium and hard.
R in Eq 5.61 is risk factor; Z is the zone factor;
is the limit state factor.

63. Contd… 6/3 Fig5.12 Cb Time period (sec) 0.5 1 1.5 2 2.5 3 3.5 4 0.2
0.5 1
1.5 2
2.5 3
3.5 4
0.2
0.4
0.6
0.8
1.2
Time period (sec)
Category 1
Category 2
Category 3 Cb
Fig5.12

64. Contd… IS CODE (1893-2002) Time period is calculated by empirical
6/4
IS CODE ( )
Time period is calculated by empirical
formula and distribution of force is given by:
are the same; they are given by:

65. Contd… For the three types of soil Sa/g are shown in Fig 5.13
6/5
Contd…
For the three types of soil Sa/g are shown in Fig
5.13
Sesmic zone coefficients decide about the PGA
values.

66. Variations of (Sa/g) with time period T
6/6
Contd…
0.5 1
1.5 2
2.5 3
3.5 4
Time period (sec)
Hard Soil
Medium Soil
Soft Soil
Spectral acceleration coefficient (Sa/g)
Variations of (Sa/g) with time period T
Fig 5.13

67. 6/7
Contd…
Example 5.7: Seven storey frame shown in Fig is analyzed with
For mass: 25% for the top three & rest 50% of live load are considered.
Solution:
First period of the structure falls in the falling
region of the response spectrum curve.
In this region, spectral ordinates are different
for different codes.

68. A Seven storey-building frame for analysis
6/8
Contd… 5m
All beams:-23cm  50cm
Columns(1,2,3):-55cm  55cm
Columns(4-7):-:-45cm  45cm
A Seven storey-building frame for analysis
Fig 5.14

69. 1st Storey Displacement (mm) Top Storey Displacement (mm)
6/9
Contd…
Table 5.3: Comparison of results obtained by different codes
Codes
Base shear (KN)
1st Storey Displacement (mm)
Top Storey Displacement (mm)
SRSS
CQC 3
all
IBC
33.51
33.66
33.52
33.68
0.74
10.64
NBCC
35.46
35.66
35.68
0.78
11.35 NZ
37.18
37.26
37.2
37.29
0.83
12.00
Euro 8
48.34
48.41
48.35
48.42
1.09
15.94
Indian
44.19
44.28
44.21
44.29
0.99
14.45

70. Comparison of displacements obtained by different codes
6/10
Contd… 2 4 6 8 10 12 14 16 1 3 5 7
Displacement (mm)
Number of storey
IBC
NBCC
NZ 4203
Euro 8
Indian
Comparison of displacements obtained by different codes
Fig 5.15

71. Lec-1/74