Seismic Analysis of Structures by TK Dutta, Civil Department, IIT Delhi, New Delhi.

Transcription

1 Seisic Analysis of Structures by K Dutta, Civil Departent, II Delhi, New Delhi. Module 5: Response Spectru Method of Analysis Exercise Probles : 5.8. or the stick odel of a building shear frae shown in igure.4, find the ean peak values of the base shear, base oent and top displaceent. ake noralized response spectru for pseudo rotational acceleration to be one fourth of that for translational one. 5% response spectru for El Centro earthquake is the basic input ground otion. ake response spectru for rotational coponent of ground otion as /6th of the translational ground otion or the shear frae, shown in igure.9, find the ean peak values of base shear, top displaceent, inter story drift between first and second floor, and colun oents at the base considering contributions fro odes only and all six odes. Use SRSS, ABSSUM and CQC rules of cobinations and copare the results. Use 5% response spectru of El Centro earthquake as basic input ground otion. Also, copare the results with those of the tie history analysis or the D frae, shown in igure.0, find the ean peak values of base shear, top floor displaceent and oent at the base of colun A. ake noralized design response spectru given in IBC (000) and assue that the angle of incidence of earthquake is 0 0 with the x-axis. Copare the results with those obtained for zero angle of incidence. Use CQC rule of cobination. 5.. or the siplified odel of a cable stayed bridge, shown in igure.8, obtain the ean peak values of the vertical displaceent of the centre of the deck, base oents of the piers and the axial forces in the central cables. Assue the tie lag between the supports as 5s, El Centro earthquake spectru ( ξ 5% ) as seisic input and correlation function given by Equation.94 to be valid. Copare the results with those of the tie history analysis. 5.. or the pipeline, shown in igure.5, find the ean peak values of displaceents of the supports for the El Centro earthquake response spectru. Assue tie lag between supports as 5s and use Equation.94 as correlation function. Copare the results with those of the tie history analysis.

2 5.. or the frae with a secondary syste, shown in igure.6, find the ean peak value of the displaceent of the secondary syste by cascaded analysis. ake El Centro earthquake response spectru as input excitation with a tie lag of 5s between two supports, and 5% and % dapings for priary and secondary systes, respectively. Copare the results with those of the tie history analysis or the sae secondary syste, as above, find the ean peak value of the displaceent of the secondary syste using an approxiate odal response spectru analysis. Use the El Centro earthquake response spectru. Copare the results with those of the tie history analysis or the frae, shown in igure.7, find the base shear, top storey displaceent and storey drift between the second and first floor by seisic coefficient ethod of analysis using the recoendations of IBC (000), NBCC (995), IS89 (00) and NZ 40 (99) and Euro 8. ake R, hard soil and PGA0.g or the frae, shown in igure.9, find the ean peak values of the base shear, top storey displaceent and oents at the botto of the second storey using the response spectru ethod of analysis. Copare the results between those obtained by using the above five codes. ake R, hard soil and PGA0.g. urther, copare the results with those obtained by the seisic coefficient ethod. ake the relevant figures fro the slides or fro the reference book

3 Module 5: Response Spectru Method of Analysis Exercise Solution : ERRAA OR HE EX BOOK pp 6, Second para, last line s, but ; n, but Equation 5.4a: β should be β pp 7, igure 5.7 is wrong plots of Equations 5.46 and he values of c h and A g should be obtained directly fro the given Equations (igure 5.7 should not be used) pp 8, In Equation 5.5b, replace U by v; U in Equation 5.5a is defined as a calibration factor. pp 9, In igure 5.9, top curve is for Za > Zv; iddle one is for Za Zv; last one is for Refer to the exercise proble 5.8 and igure.4. he ass, stiffness and daping atrices of the stick odel are (exercise proble) 0 M 0 [ ] 0.5, 5rads K k,, [ ] 0 rad s It is assued that translational and rotational ground otions are perfectly correlated and are acting together. he influence coefficient vector on the RHS of the equation of otion is 6 I Mode participation factors: λ.9 ; λ 0.; 0.4 g ; 0.76 g he lateral load vectors (using Equation 5.9) [ ] [ ] S a S a

4 SRSS rule is used to copute the ean peak values as: op disp. () 0.86 Base shear Base oent 5.67 Refer to the exercise proble 5.9 and igure.9. he ass, stiffness and daping atrices of the frae are (Exercise proble.0) M K k C he ode shapes and frequencies are Mode participation factor are: λ.5 λ λ 0..55rad s 4.485rad s 7.85rad s rad s 5.rad s 6.8rads λ λ λ S a 0.0 g ; S a 0.8 g ; S a 0.57 g ; S a g ; S a g ; S a g

5 he lateral load vectors are [ ] able 5.5: Peak responses in each ode of vibration Responses st Mode nd Mode rd Mode 4 th Mode 5 th Mode 6 th Mode Base shear (in ters of ) op displaceent () Inter storey drift () Colun base oent (in ters of ) able 5.6: Mean peak responses Mode results All ode results Absolute Response peak tie SRSS ABSSUM CQC SRSS ABSSUM CQC history Base shear (in ters of ) op displaceent () Drift () Colun base oent (in ters of ) Since the ode shapes are well separated, the results of SRSS and CQC are nearly the sae. urther, results by considering all odes copare well with the tie history analysis. Refer to the exercise proble 5.0 and igure.0. he stiffness and ass atrices of the frae are (exercise proble.):

6 K k M rad s, 6.58rads, 5 9.6rad s 5.4 rad s 4 8rad s rad s MI λi M i i i in which I [ c s o c s o] ; 0 c cos0 ; 0 s sin 0 λ.06, λ 0.69, λ 0.7 λ , λ , λ S a S a 0.77 g, S a 0.74 g, 0.6 g, S a g, g, g S a S a

7 he equivalent load vectors are ( ) P N ( ) P ( N ) [ ] P N ( ) P4 N ( ) P5 N ( ) P6 N aking the contributions of the four odes and using CQC rule: able5.7: Response for different angles of incident of earthquake 0 Responses θ 0 0 θ 0 V x 04.N 640N V y 875N 7588N ( ) δ x top ( y ) δ top θ top rad rad M xa N 90.75N M ya 954.8N 897.N he results of the analysis show that the direction of earthquake in asyetric building should be carefully considered. he directions coinciding with principal axes do not necessarily provide the worst effect. Refer to the exercise probles 5.,.9 and igure.8. he acceleration response spectru copatible PSD of the Elcentro ground otion is shown in igure 5.. his PSD and the correlation function given by Equation.9 are used to calculate the correlation atricesl uu, l uz and l zz. ollowing the steps -8 given in section 5.4., the quantities required for calculating the expected value of the responses are given below:

8 pier axial [ ] [ ] r rads ; 5.8rad s ; 5.45rad s ( vertical displaceent) [ ] a a a ( base oent at the left pier) 0 [ ] ( axial force in the left central cable) [ ] β [ verticaldisplaceent of thecentreof thedeck] [ ] β [ Base oent at the left pier] [ ] β [ axial force in the left central cable] [ ] β ( βki k 4; i arranged in atrix for) β D [ vertical displaceent of the centre of the deck] [ ]

9 β D [ Base oent at the left pier] [ ] β D [ axial force in the cable] [ ] 0.084g D D D D D.05rad s ( ) (.05) 4 0.g D D D D D 5.6rad s 0.0 ( ) ( 5.6) 4 0.g D D D D D 5.65rad s ( ) ( 5.65) 4 (, ) coh i j ρ ρ ρ ρ ρ ρ ; ρ ρ ρ ρ ρ ρ 5 ρ exp π 0 ρ exp π 5 ρ exp π u p is the sae for the all supports and is equal to. c, since the value of the peak displaceent of the El Centro record, for which the PSD is given in Appendix 4A, is known. herefore u pi is not required to be calculated. Otherwise, it is to be calculated following the step 6 given in section b ( for vertical displaceent) 0 [ ] b for base oent and axial force are siilarly obtained h i and PSD of the generalized displaceent Z ki or Z lj are obtained using the fundaental odal Equation 5.8 by reoving the suation sign. he area under the curve provides etc. Using Equations , the eleents of the correlation atrices uu, uz and zz are deterined. σ z ki he calculation of these atrices is left to the readers (see exaple proble 5.4 and 5.5)

10 inally, use of Equation 5.6 provides ean peak values of the response quantities of interest: Mean peak (total) vertical displaceent of the deck 0.0 Mean peak bending oent at the left tower base 59.7 Mean peak bending oent at the right tower base 59.5 Mean peak tension in the left cable Mean peak tension in the right cable Refer to the exercise probles 5., exaple proble.0 and igure.5. Like the previous exaple, the quantities required for the calculation of responses are: K C r M rad s ; 9.8rad s ; rad s a r β ( ki β, k ; i in atrix for) ( left support) [ ] β ( centralsupport) [ ] β

11 ( centralsupport) [ ] β D D ( left support) [ ] β ( ) D D D D ( ) D D D D ( ) D D D D ρ coh( i, j) ρ ρ ρ ρ ; ρ 5 exp π ; 0 ρ exp π he coputations of correlations atrices uu, uz and zz are left to the readers b in whichu u u u 4.c p p p p Mean peak value of the displaceents for the d.o.f are: 0.04 () 0.04 () () Refer to the exercise proble 5. and igure.6. he ass, stiffness and daping atrices of the frae without the secondary syste are: M K 4 k C is assued to be classically daped with ξ 5%. he ode shapes and frequencies are ; ; 6.5rads rad s α and β are calculated as α ; β

12 C αm + βk r [ ] or a tie lag of 5 sec between the two supports, the tie histories of excitations ( x and of duration 5s are constructed for the two supports as explained in Exaple proble.8. he tie history of absolute acceleration of the top floor of the frae is deterined by direct integration of the equation of otion. he tie history of the absolute acceleration is shown in igure 5.6. he pseudo acceleration response spectru for ξ % for the tie history of acceleration is obtained using the ethod described in Chapter. he acceleration spectru is shown in igure g x g ) Absolute Acceleration (s - ) ie (s) igure 5.6 ie history of absolute acceleration of top floor

13 0 5 Acceleration (s - ) ie Period (s) igure 5.7 op floor acceleration response spectru (ψ %) he tie period of the secondary syste is obtained as K s rad s or % ξ, spectral acceleration ( ) a π s 0.8s S for 0.8s is g he ean peak value of the displaceent of the secondary syste is therefore, x S a peak s ( 7.746) x (fro tie history analysis) peak s s Refer to the exercise proble 5.4. he stiffness and ass atrices for the three degree of freedo proble are: K k M 0.5

14 Undaped ode shapes and frequencies of the syste are ; [ ] ; [ ] ; 5.7rad s 9.rad s 7.87 rad s he daping atrix of the three degree of freedo syste is constructed as CP 0 C Cs in which C C P is the x daping atrix obtained in the previous proble and ξ he coupling ters of the daping atrix between the priary and s s s secondary systes are assued as zero (like probles of soil structure interaction discussed in Chapter 7). 0.8 C C After ignoring the off diagonal ters of the C atrix, approxiate odal daping for the priary - secondary syste is ξ ; ξ 0.08 ; ξ 0.05 Daping odifier (of.9) for Sa g value is used to find Sa g value for ξ 0.08 as given in IS Code 000. or ξ and ξ 0.05, the values for 5% daping are used. Assuing no tie lag, the response spectru analysis is perfored which results in the following quantities λ 0.6 ; λ 0.7 ; λ S a 0.5 g ; S a 0.59 g ; 0.77 g S a

15 ; ; Using SRSS rule of cobination, ean peak value of the top ass (secondary syste) 0.07 Refer to the exercise proble 5.5 and igure.7. In order to aintain unifority, all ultiplying factors are taken as unity. urther, the axiu value of the seisic co-efficient/spectral acceleration noralized with respect to g is taken as unity for different codes. his provides a unifor PGA 0f 0.4g for all codes except the Canadian code whose PGA turns out to be 0.65g. he frae under consideration has a tie period of.475s which corresponds to a flexible syste (like tall buildings). In this range of tie period, the seisic coefficient values differ significantly fro code to code. hus, this exercise proble illustrates the difference of base shear and hence, the lateral load coputed by different codes for flexible systes. he results are obtained for ediu soil (not for the hard soil, as specified in the proble), PGA 0.g and ξ 5% IBC (000) 0.6 8g 0.6 C h ; Vb 6.57 Assuing a story height of and using Equation 5.49 with k [ ] NBCC (995) with (U ) g Ch 0.0 ; Vb g Using Equations 5.54 and 5.55 ( ) [ ] Euro 8 (995)

16 Using Equations 5.56 and 5.57 (with q ) or igure Cs Ch ; Vb 6.05 Using Equation 5.60 [ ] New Zealand (NZ40:99) Using igure 5. (category ; μ ; z factor such that PGA 0.g) 0. Cb Ch 0. ; Vb Using Equation 5.60 and ultiplying by the factor 0.9 [ ] IS (89:00) Using Equation 5.6 (z factor is taken such that PGA 0.g) Ce Ch 0.; Vb 8.6 Using Equation 5.65, [ ] able 5.8: Values of response quantities of interest Codes Base Shear in ters of op story displaceent () Drift () IBC NBCC NZ Euro IS ro the table it is seen that there are soe variations of responses as deterined by different codes. IS code provides the axiu values of responses, while Euro code provides the

17 iniu values. he difference is priarily due to the difference in the values of tie periods. C h at higher Refer to the exercise proble 5.6 and igure.9. he results are obtained for ediu soil, PGA 0.g and ξ 5% he frequencies and ode shapes are taken fro Exercise proble.0 ( s).55rad s 7.8rads 4.5 ; ( s) 5.rad s 0.87 ; ( s) ; ( s) 4.48rad s rad s.4 ( s) 6.8rads ( s) able 5.9: Spectral accelerations obtained fro different codes Period IBC NBCC Euro NZ IS (4.sec) 0.048g 0.07g 0.05g 0.075g 0.066g (.4sec) 0.4g 0.09g 0.4g 0.75g 0.94g (0.87sec) 0.9g 0.5g 0.44g 0.6g 0.g 4 (0.66sec) 0.0g 0.98g 0.454g 0.5g 0.4g 5 (0.56sec) 0.57g 0.g 0.5g 0.5g 0.485g 6 (0.5sec) 0.9g 0.56g 0.5g 0.7g 0.5g able 5.0: Coparison of response quantities as obtained by different codes using different cobination rules Base shear in op displaceent Bending oent

18 IBC NBCC Euro NZ IS ters of () in ters of SRSS CQC ABSSUM SRSS CQC ABSSUM SRSS CQC ABSSUM SRSS CQC ABSSUM SRSS CQC ABSSUM able 5.: Contributions of different odes to responses for IS code. st nd rd 4th 5th 6th op displaceent () Base shear in ters of he sae proble is solved using the Seisic coefficient ethod. Assue the storey height as able 5.: Calculated force and base shears Variables IBC NBCC Euro NZ IS C h in ters of g Base shear in ters of

19 6 in ters of 5 in ters of 4 in ters of in ters of in ters of in ters of able 5.: Response quantities of interest Response IBC NBCC Euro NZ IS op displaceent () Bending oent in ters of It is seen fro the tables that there is again a significant variation in the responses deterined by different codes. NZ and IS codes provide coparable results (higher values), while NBCC code provides the iniu values. urther, the results show that the seisic coefficient ethod estiates higher values of the responses copared to the response spectru ethod.