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

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.

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.

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.

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.

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.

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)

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.

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.

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.

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.

2/3 Fig 5.1 (a) Top storey displacement 5 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

(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.)

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.

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 :

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.

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.

Contd… 2/8 Fig 5.2

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

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

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)

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 :

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 1006.558 1006.658 0.221 CQC 0.9121 0.905 991.172 991.564 0.214 ABSSUM 0.9621 0.971 1134.546 1152.872 0.228 0.223 Time history 0.8921 0.893 980.098 983.332 0.197 0.198

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

3/2 Figure 5.4:

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 3.21. Use digitized values of the response spectrum of El centro earthquake ( Appendix 5A of the book). Results are obtained following the steps of section 5.3.4. Results are shown in Table 5.2. Solution :

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.

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.

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. 5.16 is established between and PSDF. Mean peak value of any response quantity r consists of two parts:

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:

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)

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:

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

3/11 Contd…

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.5.26 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.

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

3/14 Contd…

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

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.

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

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.

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 0.8635m. The time history analysis for the entire system (with C matrix for P-S system) is found as 0.9163m. Floor displacement response spectrum (Exmp. 5.6)

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.

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

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 .

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

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.

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

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

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

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

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 9@3m Fig5.8

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

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)

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

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

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)

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

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.

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

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.

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.

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

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

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.

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

6/7 Contd… Example 5.7: Seven storey frame shown in Fig 5.14 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.

A Seven storey-building frame for analysis 6/8 Contd… 5m 7@3m 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

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 4203 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

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

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