HW14_Solution - Problem 13.7 System properties: m = 59—9-...

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Unformatted text preview: Problem 13.7 System properties: m = 59—9- : “399-” = 0.25881dpusec2/in. g 386.4 3 k 2 39.5.1. z W = 325,32 laps/in, I: (12 x 12) Vibration properties (from Probiem £0.11): '1 k 5 k w; = {2—fi)—; a): a 2 3—; mi = (2+J§)~—- I“ In ' m 1'} = 0.34185ec T2 = 0.12515ec T3 = 0.07165ec 05 —1 05 et a 0.866 63 = 0 a3 = -O.866 1 l 1 Medal properties (from Probiem 13.5): l"E = 1.244 I} — 0.3333 ll 13 = 0.0893 Part a The displacements DRU} and pseudo-acceleration Ana) of the three modal SDF systems (with 7:, given abcwe and {,1 =0.05} are calculated using the numerical procedure of Section 5.2 with At =0.0lsec. The results are shown in Figs. P13.7a—b. Part I) The modal static responses for the various response quantities are given in Table Pl3.‘7a (aiso see Problem 13.5}. Table P13,7a 0.132 x 10" 0.019x10" 12;; [m 2.3213 0.1667 0.01% viz/m 1.6993 -O.1667 00326 vlfi/m 0.6220 0.1667 0.0447 -0.1667 0.0242 Step 5c of Section 13.2.4 is implemented to determine the contribution of the nth mode to selected response quantities: M?) a 5,51 A30) where r;l and A” (I) are both known. These results for roof displacement u3 (t), base shear me, and base overturning moment Mb(r} are plotted in Figs. PiBJ’c-e where their peak values are noted. Part C The modai contributions to each response quantity are combined at each time instant to obtain Figs. P13.7c-e. Table Pi3.7b summarizesthe peak values of the total responses. Table P13.7b Overturning moment. 138.08 139.29 O. 2096 «1 Mode 3 0.0498 ‘1 W O 5 10 15 Time, sec Fig. P1372: 14 £13m in. Mode 3 H3, in. TEme, sec Fig. P13.7c V5.1, kips 11.92 200 Mode 3 V1,, kips Time, sec Fig. 19117:: 15 Mode 3 17.5 16 ?r0b1em 13.41 Initial calcularr'oru K Subatituting values of E = 29,000 1:51. I 2- 1400 in}. m = 100 kips/g and it = 12ft in the solution for Problem 13.11 gives: m1 = 10.57 I] = 0.595500 75 = 0.1825130 T3 = 0.106513: (0, = 34.56 0.73 “'1 From the design spectrum of Fig. 6.95. scaled to up = (1/3) g. the spectral ordinates are .01 m 3.13m. 03 = 0.292111. D3 = 0.03301. 5‘- 2 0.903 51 2 0-903 333— : 0.803 g s 0 Part 0: Floor displacements Substituting numerical values for 1“” and (0,, from Problem 13.11 and D“ values above in Eq. (13.8.10) gives the peak displacements due to each of the three modes" 03156 1335 11.: 1“, (11 DE =13513 0.7451 (3.13) = 3.152 in. 1 4230 «0.7409 0.110 01 = 1‘2 01 D; a “05083 «0.3572 (02.92) r 0.053 in. 1 "0.148 12546 0.017 u3 = 13¢; 03 = 0.1569 “1.2024 (0.033) = —0.017 in. t 0014 Combining modal displacements by the SRSS rule gives . ug : 110.335): + (0.110)2 + {0.017)2 = 1.34001. u; a 1{(3.152}2 + (0.053}2 + 0—0017)2 = 3.152111. 2 (4.230)2 «1— (m 0.148)2 + (0.014) = 4.233111. 03 = Observe that the first motle contributes essentially the entire floor displacements. Part 0: Element forces ’i‘o compute the element forces use the results from Problem 13.11 and replace Ana) by the Spectral values A”. The bending moments in a first-story column due to the first mode are M a] = 0.7526 mi: A1 0.7526 (100/ g) (12) 0.9121315,I a 81552 kip—rt M“ = 0.3014171111‘1i = = 326.60 kip-ft O.3014(100/g)(12} 0.903g These bending moments due so the second mode are Ma: :2 0.083811111111511 a 0.08381(100/g)(12)0.903g ¢ 90.82 100m Mb2 = 00682311112143 = 0.06823(100/g) (12)0.903g m 73.93 inp‘tt Due to the third mode the bending moments are MB3 2 0.020301th3 = 0.02030(100/s}(12)0.803g = 19.56 kip-1t Mb3 = 0023021102113 2 0.02302(100/g)(12)0.803g —. 22.18kip‘ft Combining modal responses by the SRSS rule gives Ma x (315.52)2 + {90.32)2 + (19.56}2 = 820.80 kip—ft Mb = 110,116.60)2 + (73.93)2 + (22.13)2 = 335.50 kip—ft The bending moments in the second—story beam due to each of me three modes are computed similarly to oetain ME] = Mbi 2 assess kip—ft Mal 3: M52 = kip‘ft M53 7': Mb; = Combining modal responses by the SRSS rule gives Ma = M 2 f- 586.66)1 + (61.44)2 -I- (2.68): b ":- 589.87 kip—ft Problem 13.42 Initial calculations Substituting values of e = 29.000 1:51. 1 = 1400 in“, m = 100 kips/g and I: = 12 ft in the solution for Problem 13.12 gives (01 = 8.67 red I sec 032 = 30.27 rad lsec 033 = 57.25 rad I see '1; r- 0725501: 13 3 0.207 sec '13 = 0.109 see From the design spectrum of Fig. 6.9.5, scaled to 1138 x (l I 3)g , the spectral ordinates are 295 ...
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This note was uploaded on 01/04/2010 for the course CE 225 taught by Professor Chopra during the Spring '09 term at University of California, Berkeley.

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HW14_Solution - Problem 13.7 System properties: m = 59—9-...

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