Tutorial 7 for Mechanical Vibrations, Exercises of Engineering

Download Tutorial 7 for Mechanical Vibrations and more Exercises Engineering in PDF only on Docsity! 1. Consider the different systems shown in Figure 1 and 2. Find the relevant Equations of Motion, linearised and in matrix-vector form, for each of these systems. Leave the EOMS in symbolic form. You can assume any hinges, pivots or rollers are frictionless. my m4 “Es |" me oa cm f= ky 2 3 ™1 bw] 73 LW kg me (b)D (a) C ib) B Figure 1; Some saulli DOF systeius “| | cy ky Cae) Efe: M1, 30, - k%,- Cz, -~ k,Cx,-x,) - C,Cz,-%,) = Me, M+ Cc + bark Cx.-x,) + CC%- %,) = oO M+ Cx, ~- CX + C.m, * Kix, —bet24 Kee. = O MX. + CG Ce) % - Cote + Chye ke) 21 - be%2 = O ZF : oh xX -G Cz,-%) - k2Cx2-x,) = Me Sa MX. + C22, Cx, + bed - eX =O Max. ~ Cor + Coke — bax + kere = O Mm, Oo {% + Cl, -C, x k+ke -ke a: e - . + = O M2 lee “C, C, Xe ke ke xX O  (b) B 2M5- Loe (-b, 2,)(@ ces ©) - - [kx (x- =)1C @) we)= le Ce. a5i0 Ca case) - [kz ((art)sia o-- a) \as)cos e) = FMLe SAA (-tae)(a)-k.C@wo-s) @) = 3mv'o mL @ + kee O + baer) (@8) Om) = 0 Zml O+hae + b:ew'O -kz@n) oc =O ame t(aos Kr @O')O - ken) x =O 2Fe om,x ~ kya - kzCx-x2) = Mex Mex + ke Cx-22) + bgoxc = O Mix 4+ keer - Kote 4+ kg 2 O Mex - k2rLO +(ketks)x =O mgt o] § 8 , feta Cay - one {2 = {3} o mil - ki (a+b) berks | C% ° 2. Calculate the modal frequencies and shapes as follows: (a) For Fig. la, if the damping is ignored, calculate the modal frequencies and shapes for the following constants: ky =500Nm7!, ko = 150Nm™!, my = 5kg and ma = 3kg. ie leis eh SES EHS UES 3 det (K-amM)= 6 680 - 25 -150 -150 1S0-A3 =O (6s0 - 5a) (150-32) - 150° = 0 91 500 - 1 450.” — 7502 +187 -130" =O ISA*-2700A% + 75 000 =O A, = 34, 32 Da= 145,678 Kew W, = J 34.32 W, s 5,36 cud |S Wi: I 14S, 673 Wr = 12,07 cad/S As = 34, 32a 680-25 -1S0 x = D2= 145,678 -150 is0-A3]} 1X25 [Lo 6S0 -5 (34.52) -1S0 x 4 ot -150 1S0- 3( 34.32) Xz ° 418.4 -180 M7 fet -150 7.0% x.J Lo 473.4 ~150 ; -1$0 &7-0% @, > ish e £2 —> *B Re 0 0 X, - 0.3136¥2 = 6 0.3136 Xz =X, Ka = 3.188%, def X= I Xa = 3.1% X= « (sues) l -0.3136 | - O,3136 N: A, = 34, 3a 650-25 -150 %, = N2= 145,678 -150 is0-A3]} 1X25 [Lo 6S0 -5 (145.679) -1S0 % t - | ot -1$0 1S0- 3 (luS-679) Xz Oo “BL -1SO am 4 ot “150 _as7q,) [X25 LO “76-4 -150 | 0 -'50 -287.oi,| O 1 @,- < 78. eb Ba —> te Re l 1.4136 | 0 | 1.9136 | 0 X, +1-4136¥%2 = 6 X%, == 1.436 XR Yr = = 1.9136 deh X42 1 x= pC) (b) For Fig. 1b,calculate the modal frequencies and shapes for the following constants? ky = 800Nm—?,_ ky = 300Nm™, ky = 500Nm~!, m, = 9kg, my = 4kg, a= 0.5m, b= 0.3m and L=1.0m. [ime olf ets [rarer Rewlfer fe) [2 oifh[e eH deb Ce-xm) = 0 | 22-78 -240 | =0 - 440 B00 -Ale (a2- 32) (800 - WA) - - 240. =0 313 600 -1 S68 2- - 2 UOoA + 12% - 240° =0 256 000 - 3 3% +17R'=0 12% -3 468A + 256 C00 = O Dd, = 87.36 Ar= 242.3 ew W, 37.96 W) = 4.373 cadls WW: [242.4 We = 15-53 cad/S ql Ms 392-23 - 240 xX, oO Me ne - ALO go0-2u} 1 X20 ot metas (™ -3 (97.36) - Ried X ) - Ano 300 -4-(87.86) x 13h 126.42 =n | Viet 4 ot ~ AuO 443.56) | Xz “Lo ["* be 240 Jo | a40 443.56 @, aa g, Pe oe l - 1.864 | 9 | - 1.64) 0 X, - 1.969%, = 0 Xr = 1-869 F2 Lh X21 Xa = 0.5351 x = « (oss) * D = $7.36 392-23 = - aud \( X, = A= 242.3 - AnD B00 -AL Xz oO 392 - 32928) - 240 { oat - { al - Ano $00 -U- (222-8) Xz ° 336. — 240 ) { am -| ot ~ Avo -171-2 Xz oO -336.4 -240 |o 0 £2 —> aS Re ~ 240 -111.2 | 0.7133 | O | 0.71331 0 X, +0.713 Xz = 6 x: = -0.13%2 1 C, 336, bs heb %= 1 X, = - 9-133 X= p Cr) (c) For Fig. 2a, calculate the modal frequencies and shapes for the following constants: |ky = 820Nm-~ 1,» Pe hth [A US 18h [2 ehh [ee teh th deb Ce-2m) = 0 $09.25-22 -'20 | =0 -120 BUD -Ale (soa.25 -22.) (80 -u2) - 120° = oO 427 7110 - 2 037A - 16802 +BR —I20 = 0 Sar - 3 7170 +427 770 -—1A0°= O $2 -3 7112 + 413 370 =0 D> 184.3766 Ar> 280.248 Kew W, = J 164.3 166 Ws 13.6 cadls Wi 2 | 280. 2ug Wr = 16.74) cad/S ‘Dv = 134.377 M: $01.25-22 -120 x = Ar= 280.248 -120 gup-a~u } | Xz oO 504.25- 2 (184.877) -1290 x - | ot 120 340-4437) | Xe oO : Wb ~10 | \f am -| ot 102.492) | Xz o 140. mae -120 J02.442 oO 0 i monk g 1 fa © “iz0 Fe l -0.954(| O | -0.354\| 0 X, - 0.35uI%2 = O X, = O-B4S1 Xe def % =] Xa = 1:1832% X, = (ess) ‘: > ‘Dv = 184.377 $04.25-22 -'20 x = Ar* 280.248 -120 BUD -Ale Xz Oo 504.2S- 2. (280.206) -120 x “42 ot -120 B40 -b- (280. zug) Xz -51.2ub =u | \f am -| ot -120 ~ag0.0q) | Xz Oo -S1-2ub -1z0 Jo -120 — — 280.99 > = R, : > mize Fe l 32-5416 | 0 | 2-3u16 | 0 X, +2306 %2 =O Xi = -2-3Hl6 Xa deb X21 X= -a-3ub X.- B (-*""") (d) For Fig. 2b, calculate the modal frequencies and shapes for the following constants: kj) =100N m7’, mM 0 Ol) (&% ki+ke 9 -ke mH © om O Ar + ° kgrky - ks Ze 23 - ke -k3 Korky x3 3 0 O a 300 Oo -200 | |~ oO o 6 O Ab +] 0 yoo 300 |4% 0 u 3 -200 -300 seo | (73 © deb C k- am) =O B00 -3” Oo -2c0 ° 100 ~ 6” -300 =0O -2100 -3009 S00 -uA (300-3) | Geo-sn) (soo 4A) - 00"| - 200 [- C200) (700-62)| =o (300-3) | Qeo-n) (seo-na) - 00 +200 [ C200) (700-6)| =0 (300-32) [ @e0)(s00) -SB00% -306 | + 200 Fino 000 + 1Z00 al =0 (300-3) [.Ge0}(so0) - 5 800 0 -306 | + A00 [-1:0000 + 12000 )= 0 -7ax? + Ae 600° - Z 230 0008/2 + 50 600000 =O A,= 31.361 w, = 9.6 Az = 108.004 Wr = 10.4 Az = A0l.796 Wz = 144 Ms A,= 300-32 - wo X, oO) Aez 108.009 ° 100 ~ 6” -300 X24=40 Az = Aol.796 - Leo -300 $00-4A k Oo 300 -3(31.81) © -tco Xx oO ° 700 ~ 6(31.561) -300 X2 b= ° - 100 -300 S00 - uw CBlsél) By; 20.417 Oo - 290 X oO oO 508-34 - 300 X2z4=4 0 - 200 -300 372-556 A; o Aly. 47 O - 290 oO Oo S08. S34 - 300 0 - 200 - 390 372.556| 9 @, = 0.978370, +h AO. 47 O - 290 oO oO S08-%34 - 300 0 oO -300 176.878 | 9 -300%2 + 176.378 X3 = 0 Zu .ui7 K, — Zeoxe = (Z Book, = 176.873 } 2Ote- wI7 ¥, = 200 © Uh toe1 ¥, = 0.973834 X, = 0.584 In: 300-37 oO - 200 X, Oo ° 700 ~ 6A -300 Xz4=4 0 - 100 -300 S00-uA RK Oo B00 -3(i08.001) 0 -%e0 % Oo ° 100 ~ 6 (103.0¢4) -300 Xe 7 = $ -100 -300 Soo- uw (108.004) As ~2l.027 O - 290 X oO oO S\.446 - 300 X2 = °o -200 -300 67.964 hg 9 A,= 31.26l Az = 108.004 Az = Al. 796 ~Aur.027 O - 7290 oO oO 1.946 -300 | O -200 -300 67.96% | 9 @, = $.3a390,-@3 ~Alr.027 oO -7290 oO oO S1.446 -300 | O oO 300 -1732.15 | 9 gooke —1732-1817k3 = O = u.0z7 X, — Leoys = () Books = 1732-15714 X3 ~Aboz1¥%, = AOOKS %, = - 8.32396 heh 43 21 X. = 5.7788 X 3 p [= = | 1 % fim” x MY, =(s.32096 5.716 1) 30 Of 3229 o 6 0 $.1758 o Oo | \ =(-8.32346 $.71S8 1) “aH aTe 3. 6S43 ue re [ Coz316x-24072 ) ¥ (8-788 rv. 65ug) + Cuxidl = Lgl. 1858 | 6, = fxemy, % -8. 32396 == G $.1758 ) - 0.410049 o - | o.2¢us5¢ t 0.04426 -0.65449 = ~ 0.3% X= i 73 1 - = X O,- [ximy, *3 X™MY = Co.csq -0.5873% 1) 3 0 O}f ~o.6se4 3 3 ¢ su 0 6 Oo 0.58734 oo 4 \ =(-0.0549 - 0.8734 1) H-I647 -3.524 tt z [ Go.tsuax - 1.9667) + (-0.58734n-3.524) + Cuxid] 2 738646 \ eS CX 6, 7 | 1 MY; ; =0.6844 | q ~ Tastee |( ~ 58734 \ -O.2u1uS @ = 0.216842 3 0.36 8613 ©. 326498 0.41007 -O-2u1wS o= 0.196845 -0.264859g -0.216SKR 0. 33h2 ~0-04926 0.368615 3. Consider the system shown in Figure 3. Where F(t) = Fosinwt, m = 1.0kg, k = 1000N m~!, Fy = 5N, w= 10rads7!. (a) Determine the equations of motion for the system. (b) Determine the modal frequencies and the mode shapes. Q) a Bes M= 2m T= Ma = Ms am gs (#:-%3) a: bn F, Ge.) + = { bebers) iG G;-x,) m Ts Ta ii dal ing gia ZFas Mz, ~- hx - Fe Cx -22) ~ kp Ce x3) = Mx, + Fix, + brCx-2x2) + £3 Cau.g) =6 Mx, a Kroc, 4 bexr- kixX, + b32t, —k3%> = } MX, + (erate ths) x - Vv. x2 -k; 2 = 3 ZF. = m,2, — Kefts- 2) + Ft) = Mex, Maxie + keCe2-x,) = FC) Mexe -heDi+ kere = Fc) ZFi= Mx, — kz C#,-=.) = = M35 M3 dg + Ky Gr -™) = Oo Msc, — kg x, + k32c3 = 0 m oOo 0 1|% (Fs +¥.+ 5) -Ee ks 0 om O =x,| + o ) = | Fee) O 0 m3)\l xs ~k3 oO ks oO A, = 219,223 3000-A2n -J000 -1000 xX, Az = 1000 -1080 1000 -” ° he: O Az; = 2.280.776 -(600 1000 -” ae -1000 1000 -joe0 o 2 -l000 wr O ts -j000 1900 x ° 1000 - \oo0 S 1000 -)000 —_-1000 % -1060 o ° Yi: O -loo0o oO oO BS 1900 ~)000 — -1000 °o -1060 ° © 0 -l000 oO O Oo looo%, - \0OO%2 - loooy; = O Le 4524 lo00%, -lW0ofe - loo0 =0 1000 Ch -1) = 100072 X,-l 2%. 8 Ke X21 xp (3) Lb #e=-) A, = 219,223 3000-A2n -J000 -1000 xX, Az = 1900 -1000 ooo-A 8 We: O Az; = 2.280.776 - 000 1000-7 B 3000 - 2 (ase.m) - 1000 -1000 % ee 1000 - 2290-776 o hie: O o 1000 - 230.776 B “1 S61.8% -)000 = -1000 xX, -10800 -1200716 9° Xi: O -l000 6 -12¢0-776 BS ~1S6].552 -)000 —-1000 9° -1080 -1280.776 6 o -l000 oO -\750.716 | O - 1090 KX - 1%80.776 $2 = -1o0e X, = 1280.776 Xz -looo f, -178077642= O -12%6.776 f2 = 1900 KF, hed fy 2 | x = (eet Y Gr 30776) ¥, = - 1.280776 2 ~ [-1t8e-77% Fra = 1 1.28677 x; = St \ \ Q,= [yr eny, : MYX, = (0. 78077 ! nm) 2 0 0 0.73077 o 1 Oo ' Oo o 1 | 2 (0.7%77 1 1) 1.S61SH ) ) > (0.18077 x 1.56sh) +2 = 3.2192 __\ o.-fimr Ky 1 0.73077 0, 2 [3-292 \ l 0.4352 gb =| 0.5573 0.8573 4. Consider the damped two degree of freedom system shown below in Figure 4. Where: my = m2 =m =1kg; ky = kg =k =500Nm7!: c) =e =c= 200Nsm"!; Py =5N andw=Irads7!. Derive the equations of motion of the system and find the natural frequencies and mode shapes for the ° (“ kx, Gs, ky Lda (t) = Po sin wt o= Ta 9 be(e,-*,) oy CeCe) 4 FGa-x) CzGe,-*,) Figure 4: Damped Spring Multi Mass System Xr Xun Zple mx, = eae, = Cady — KeCxr2te) - C20, - Sn) th = mx, Mi2ii +t Kirt Cede kee Carne) + CC, -%2) = P, Ce) Mixer + Cx, 4C, %, - GxXz +d + bea, - Kexz c P,Ct) (3, (Cale) & — Cote +(eith) x, ~ exes RCH) system. X. Zfye Mx, = kz Cx 2-1) - CrCes -x,) = MX, M2%2. 4 Cre. -X) + k2Cx2-,) 20 Mrx, - Gz, + C2X2 — fey + [222 = 0 5s Gra) -Q1(3 gake tz] [Pee mM, 0 et fee ella? oO mM, -Q Cr | (xe “kt ke ° oe Xe det C r-2m)=0 hoo- 7” - 200 -209 200-2 (00-7) (200 -n) - G0 900 - L90A - AOD +2"- tov = 0 n - 60027 + 40 000 =O =O d, = 7h .393 A, = $a3. 607 A boo- Xn - 200 ee | nt. a -Aavo Leoo - 16.393 -Z00 -720 two - 76-393 | © (323.007 -200 © ) O ~Aoo 123.607 323.607 -200 - 200 123.607 © 0 1 — 0.618 033 F ) - ©. 61033 |9 Let Fre) Arz 1.61%0 X,= ( mn) (er slid hoo - 923-61 -Z00 -200 Loo - $236| 9° © 0 oO oO -123.607 -200 -A0o0 — 32.667 \ 1.61903 1 161363 K h 5. A vibration system has the equation of motion 2 0 0] (& 10 10 —10] (a 20-10 —10] (a, 0 0 10 O} 2%$4]-10 15 -10}¢a$+ ]-10 20 -10) 4a,$= 40 0 0 20} [a5 -10 -10 10| [as -10 -10 20] [a3 0 (a) Find the ffimdamental frequencies and mass normalised mode shapes associated with the system. (b) Using diagrams, describe the mode shapes. (c) What does the first modal frequency suggest? Q) deb Ce-am)= 0 w-2zor = =- 10 -10 -10 20-10” -10 = 06 -lo -10 90-70” 3 = t; ~@, wW-aor = - 10 -10 -10 20-10 -10 = 0 O ~30+10a___- J9- 2OA CG; = Cs - Cz w-aor = -10 oO -10 20-0A = -gorlon | = O O ~30+10A 60 -30”0 (z0- goa) [ (20-102) (0-30n) ~- (-3 2110] t o[-wcte-24)| =0 4.000 2° + 1 b0002* - 1S000A = 0 AZO Wiz O cadls Aw Ligid Body mode nm z wr {2 cols w-aor -lo -10 -to 20-107 -10. -1o -10 J0-20”A w -10 -10 © -'o 20 “10 | © -lo -10 40 oO a s hr ake g; = @, ta C3 w -10 -10 o ° 30 -30 | o oO -30 30 | 0 -30%, + 30%, =0 he = X3 he a2 Zork, -10 -1lo =O x, = | X= K | ) a> Plu niu O nlu O A, 1-202 -10 “10 ° At -10 20-10% -10 ° m= -1o -10 J0-20”A Oo A; = -10 -loO -10 a -\o 5 -10 oO “lo -10 -l0 | O g; = ¢, - ZL, -lO -10 -10 oO -10 Ss -10 Oo 0 o o 10 -10¥, -10¥, -10¥3; =O “l0O¥i+S42 -lo%s =O -lo (-f2-1) + Shr.-jo = 0 Lb X32) 10%, 410 +S%2-Jjo = 0 %=90 ~-)O%, -JO%2. -1O = O ~J]O%, -10 = 10fz Ai2- -10 (441) = 10% X, +) = -X2 A, = -¥2-1 Xa “a — Cc -ol nN” Plu  Xe: ( . = x 2 x Riga © wnt = Oe alfa) (4) | 220 (+) oO Oo to =( 0 ) = 40 % 1 33 05° HIM 4, Ys 1M Y,2( “+ 1) 20 9 a =(1 yt {2 20 + (u+u0) +20 AoO woul 0.° Tema, ¥; y iF 2 |- % Cco™M~N ' ~3- N.Y