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HW11_Solution

# HW11_Solution - ‘imtgﬁk-AW 32mm man W...

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Unformatted text preview: . ‘imtgﬁk-AW. 32mm, man W ~ - "WWW'u'iﬁmz‘mm'ﬂgf‘ From Problem 9.5% the mass and stiffness matrices are. m 2 1 ma 6 l 2 Ii22E; :4 *5] ; L3 —5 2 nus '3 2—2}. —5—/?. k-mzmzﬂ—seﬁt 14—21] (a) where 3 _ iii; 0): (b) ibstrtuurw Eq (:1) in Eq {\$0.12 6) crwes the frequency station A” — 3411 + 1 = 0 'he solution of this equation gives: 2.1 = O. 0718 and =.i3 928. The corresponding natural frequencies are El E1 2 0.928 #:3- 502 = £2.59 mL ML] Foliowing the procedure used in Example 10.1 the nature? :3des are determined from Eq {10._ ” .5}: , t _ 1 pl:{27321i \$2 m[—0.7321} (d) First mode Second mode (/r/f w‘ = 0.923 «,IEI/mrf‘ w: x 12.923 «fEI/mLi LJ'I Part I; The free vibration response of the system is computed using Erg. (10.8.6): ‘ q (0) u(t)=2¢n[qn(0)coswnr+ Z)" sing)”: (e) 11:1 where -r 0 mu{0) . ¢ TmL'i(0) O = "I ._ Q = W q" ) 95mm q" ) 93mm {3 Substituting are) = (0 1)’ and um} = (o of gives: £10 2.7321)[1 3H3} 6‘1 (0)=w~————————~—-~«--=0.2382 Lg“ 1732i)? IHZJIBQI} m 1 D E(i~c3.73:1.1)[ 1H1} ——0'>837(a7) ‘jﬁmzwpﬂ'T—“T— 51(1 —o.7321)[ H } 6 1 —o.7321 42(0) = O (M {2-1) 41(0) = Substituting Eqs, (g). (h) and (d) in Eq. (e) gives the displacements: u[(f) _ m3 0.31iOsinmlr—OﬂZZSsinwzt “3m _ £1 esgwsrnwlr+0.9154srnm2: .Probiem 10.11 From Problem 9.7, l m = m i 05 2 —-1 0 9 - = :3?!» —1 2 -1 h 0 ,1 1 Part a: Natural frequencies and modes. 9 2 - ﬂ. ~1 0 “mam-eff! —1 2&1 —1 (a) ‘ 0 —1 1 ~— 058 where . 111113 3 =3 Cr) 24E] Substituting Eq. (:1) in Eq. (10.2.6) gives the frequency 1quation: 213—681+92.-2=0 The solution gives 17.) = 2 — «E = 0.2679, xi: = 2 end x13 : 2 + J3 a 3. 7321. The corresponding natured g frequencies are 2.53 Elw_=619282 J13- £03 =9.4641 «5% 11113311111 ml: {5) iollowing the procedure used in Example 101, the mode shapes are determined from Eq. (10.2.5): il 05 — 1 06 0.866 82-: 0 \$3 = - 0866 (c) 1 1 1} (.151: Part b: Very”); modal orthogonality. 1 —1 6311163 = 111(05 0.866 1) 1 0 a 0 0.5 1 05 636163 = 111(05 0.866 1) 1 ~0.866 m 0 0.5 1 1 05 ¢§m¢3 = 111(—1 0 1) 1 -—0.866 :0 05 1 (e) 2—10—1 7 6m»; =l§§1(05 0.866 1) —1 2 ~1 0 =0 0 —1 1 1 451 2 —1 0 05 {1163 =l~m(05 0.866 1) —1 2 —1 «0.866 =0 113 0 ~1 1 1 7451’ 2 —1 0 05 1183:?(81 0 1) —1 2 —1 —0.866 =0 0 —1 1 1 (0 Thus the computed modes satisfy the orthogonaiity properties. Part c: Normalize mode: so that M” = 1 M1 = Qfmqjl l 05 2111(05 0.866 1) 1 0.866 05 1 = 1511: M2 = 105111133 = 1.5m M3 = 6§m63 = 1.5111 Divide 9; by 41.5112, 9'3 by «11.5111 and {1‘13 by 41.5111 to obtain the normalized modes: .- —'\ 2 0‘3 I2 {H1 2 031 1;} a — 0366 02 r: »— 0 1-?— - —~«~» —{).Séo } . i J 1 { 3m ‘ Yam = -m 1. 1 1 11 (g) These modes are scalar moltipi eso fthe modes in (al; 111:" shapes 0:" the U 0 6e: 3 U1 modes ar e the some il imbiem 10.23 l'ﬂff a ’ From Problem 9&3, the mass and stiffness matrices of ‘te system are 5 m=m l I 28 6—6 k— m 6 7 3 -'i1eo . 28—577. _6 —6 k—w2m=3ﬂ3 6 7—71 3 (a) 10’3- 6 3 7~;t 7h6re 3 A = lOmL a): (b) 351 fiebstituting Eq. (3) in Eq. (10.2.6) gives the frequency quﬂtion: “3—9333+52071~400=0 “he roots of this equation are all i1 = 8.6780 and 2.3 "equencies are 7. = 0.3259 35—"; 777. = 1.6135 £73 673 = 1.7321 El. _‘ ml. ' mL mJLJ (C) ’oilowing the procedure used in Example 10. i, the natural iodes are determined from Eq. (10.2.5): r» 0.9219, = 10. The corresponding natural i] J 1 0 l; ..—:. —l.9492[ {92 = 1.2826 £7.53 = i (d) 1 Third mode Part I) The vectors of initial displacements and velocities are l 0 um) 2 0 77(6): 0 (e) O 0 Substituting m, k and Eqs. (d)—{e) in Eq. 80.8.5) gives 'ITmMO) 91{0)=iIT—‘—~ Q1 mél T g? mu(0) . qgt0)='—=~——= 0.6031 qztm=o (6 20.3969 «(ope (l3 (0)“0 The free vibration response is given by Eq. (10.8.6) as 3 . 0 11(7):: 7,3,,[qn (G)coswnt+-g—:;(—)sinwnt_i (g) 71:1 71 Substituting for {3'77- qn(0) and cjn(0) gives 771(7)): 0.3969] [ 0.6031 7737(7) 0—.77 36E-cos 676+) O7736’CDSﬁ92! (6) 773ml i 0.7 736E {—0.773l The third mode does not contribute to the freewibration response because the initial conditions do not contain a component in this mode. ...
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