This preview shows pages 1–5. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 pseudoacceleration
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’ce where their
peak values are noted. Part C The modai contributions to each response quantity are
combined at each time instant to obtain Figs. P13.7ce.
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 0903 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 ﬁrststory 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 kipft 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 kip1t 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 ...
View
Full
Document
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.
 Spring '09
 chopra

Click to edit the document details