Class13 - é‘iéWMwQ—fi afggmmaé ', {VIEW}...

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Unformatted text preview: é‘iéWMwQ—fi afggmmaé ', {VIEW} vgvm‘fm‘fi , Hm”... . -‘_. ,a {69%ng , x Q f ' ngmm m, WfQKL/W‘DEFWWQQ fing {2;} F“? ’ ” wig.) "’ aw“ m c . - ,, m _ c: ,g [afimwfl ’19; IE?— Z‘AEW“? 1K x MW: e; ’ A a H wig?” 3%?fi?§m s 2&1 ‘3 {Ag “’ a? é” gig? $313533»; t3; 4% Cwb g 9? z‘ WWW Mamas (:Wéfimm firmw‘) Asawmm : Awwwmfiw Aggwgs 2% w:ng m @nga? mama Ligggg $ng 53' " ‘ A'ifi W A/ v11 WWW V mm, ® u I t f» g ~ .. w g Aaemg wt): *3 W @945 ‘ikfié fiwfl é> 3-; v1 2 _ \ VVVVVVVVV i VVVVVVVVVVVVVVVVVVVVVV d u y W__fl;,,_._._@@g$mf 3 jg?" tg<£4£fi£ 3“ “E” g ‘9 w + W1 [m % zigkt +(;~§) it; (a?) j 6% ‘ e a 1" gg[u§méé‘t+(§'5) M136} 1 ._ C [fig-é» “ggktj .mém g g W—y Ufiégflbmcfim Qfigfiifiy A 699:3ng H H g LURKEURSLE gig -—-——-—-> m: wag m M‘! Q QWE fig/é) m,WAWMA gm}; ; , E 22%; ii; ‘ Lima vmémw m . . 5‘ x t H _ “"2 m: 7M WW: 2 3m Maw 3w; R “WM Ma mm {a g b 8 { i r" i" ‘13 w; : 92% 233553? is: Eiaémifi Eéié’ii} + 8%; fl 2 2% 3 EVALUME {11(8) A» Mg} 5% ‘EZtm {At g“ £9?“ ti} 1 * 2' .. 2 .W.\\ a a, biLth): 21%;”? {43+ '5” A}; Egg-é. gig—— _flw{;§§>~ £2... "’5 g; l ’ . H {fig .‘I /".V W I if > . ‘ ‘ uéyfigfjs uh“: 1A,; ‘L U; *3 §* «if H‘; l ’24; At g 3... fails: “flié; “32%;?” “32%) ~ ’ EB < “‘1‘”: 3 § § 2 a S % NéwHkKK Lmékfi Mfiéwfikfi§>g Wa’gmfigwfi ng-ggg; 2;»; 4; %) At I LVAKEF’EQQN 5m MébJHRKK Lméséssz kigéixfiézfifiééxé; “fiéflw $5 Mg 394$: “Agggng EXQmMQ gigggm VAKVE % L m gagiw . L E I E /,7<’“m% g A“ “’Wfibkx “Wu,an Q Mew m {fix/,kmw Mil/“M” a» “m” .mw‘w‘vww WM m4 Example: Determine the response of the tower subject to a half sine impluse load. Assume the structure is initially at rest. ' m = 2.533 kip-é2/in.. k z 100 kips/in k f 100 can —- \/;-2- —- 6.283 rad/s .5 = 10% c -— 2§wnm = 3.183 kip—s/in 1. Central Difference Method am I 269.21%-+1 = pg+405.6u,- “2373925-, ' [$2 * it 11M = 5(uj+1 *uj-i) we - Wxi*l"ud"‘3/Mt 2 21M 2 100(uj+1—2uj+uji1)emwmwwzufld-O/M z: . 2. Constant-Acceleration Method . ‘ Uj+1 Uj + 0.11% + 0.005% 11.34.] = 11.3” + 0.12% X _ ‘ u 1 o a I (g «01.2%034 \2 ‘2 Uj+1 '3 M(Pj+1 —" ~ '— .3.” ‘ j ” A ' r pair/kg: 4 v u (theoretical) £1244”"‘“"”‘7‘ 0.0 0.000 0.1 0.0323 W704» I 0.2 0.2254 (D 00. 0;, .= 0.3 0.6240 57) U.“ gimigzm 3mm 0.4 1.0961 _Q§w)\3 K? 0.5 1.4251 , (mg “aigwg 0.6 1.3772 . ' ‘2 'v "t? f". 0.9 05974WW ‘40" 3‘1 A “J r r _ -10073 . 3. Average-Acceleration Method 1176.915“ = pj+1 + 1076.925 + 104.57% + 2.53311]- w M 71:41 = 20(uj+1 “ uj) ‘“ 730‘ I .2 am = 400(uj+1 —— uj) -- 402'... —- 0]- fla-V’17W(pw—cug..—é<uga 1 . my ~ - ,f:1;::;l._ww,. 0-0323 .3. mm- 6;,” 0.2 . 21:21—65. 100.0 0.2254 ‘“ “O‘EMW 0.3 10.8151> 86.6 0.624000 02.... 0.4 «11.779 1.0961 @fiwfifi“) My” 0.5 -35.806 1.4251 Rfigmkmx¢t+ 1,33%le 0.6 ~49.613 1.3772 ‘1 0.7 —28.066 0.8683 1' a {222; $2322} 0 8 0.686 0.1105 26.089 ' 39.871 —0.5974@ Ki", 4’11 1233511420 '31.} aid? 08H '7“ _fig At _ (5532.1: 4. Linear-Acceleration Method . 1715.315.H “H1 pj+1 + 1615.32” + 158.35%- + 5.22552:- 30(’Uj+1 — Uj) - 2113‘ - 1.1.74.1 = 600(Uj+1 '— Uj) - —- 5. Wilson-0 Method (0 = 1.5) . 839.13Uj+g = pj+g + 739.13Uj +107.6921j + 5.305513“ .11149 = 266.7(Uj+9 - Uj) *- — 274+} = 0.667fij+9 + 11m = 111' + 0-05032' + 13.241) uj+1 = W + 0.1%- + 0,0033%- + 0.0016711]-+1 Mm» M _ fl 4 30 _ “ LU” Wielxti‘ "’ 93g *3“ gave ‘ , if m Qaygii‘viyg’Mi/Mfi fii- f)“: {:3 ‘9‘» {73.55 5V1: q a,” g‘ ‘ . : a} a. u- »» ¥ 4,‘ t . at: £9; 14- 1 i Spectral radius, plA) r“ O p 00 T. p m l 0.4 L- 0.27- 0.01 __1_ Central difference method NA) Houboult's method principal roots Houboult’s method spurious root \ Wilson — 8 method, 8 = Md spurious root _L_l__l_0|i L_ _l_ Linear acceleration method Average acceleration method Vlfilson — 6, 9 = 1.4 principal roots ‘ Wilson - 8, 6 = 213 principal roots Vlfilson — 8, e = 2.0 spurious root ...
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This note was uploaded on 05/09/2008 for the course CIVIL ENGI CE 511 taught by Professor Lynch during the Spring '06 term at Michigan State University.

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Class13 - é‘iéWMwQ—fi afggmmaé ', {VIEW}...

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