Vibrations_Rao_4thSI_ch10

Vibrations_Rao_4thSI_ch10 - Chapter 10 Vibration...

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Unformatted text preview: Chapter 10 Vibration Measurement and Applications VoLfaage sensitivity : V = 0'038 VoL‘t-meter/Ncwton thl'CKhCS'S: t: 2mm: 7- m taco cuffed: voL‘Eage: zzo VoLts . Pressure applied: fl: 2, 5= W 1; => 22° = (0-098)(.oio) 1"x ' 1;: 142.24. xw‘ u/mz m=O-5 R3, 1t: (0000 N/m, C2 0 amrlituale = Y: 4/0“ m 'i’o‘tal displacement: of mass: 2:: “A,” m relafwe displacement = 5,: 1-7,: 3/,000 m . L Z: 1...!- ;.e., 8 = (“AMNDI'Z =5 r":_2./3 1-r" loco t_ r1. (4., r=r a: = o-8l65 69: rca = ~9l65 "0—00—0- n O 0.5 “5-4-705 rad/sec =t8-3777 Hz r2 Y c Z =W walker: 3': Cc“. = J—E = r" Y _ r2 Y I-r") +6]? 1‘): - 1+ r4' speed range: 500 rpm : 52'36 moi/sec - woo rpm: [57-08 raJ/sec x: X cos (at ; 2 VI Y CASE (i): Le’c *5: o : —— (E, [(1—1‘1)‘ + (23%)" ) Eg.(E|) gives, for 27, error, E. _"—z (.5 Y - lrz-Il zl'oz => rz—n=7"4’4 628 as“: Ca 500 or I500 __ 7-ILH4 74414 749“, - 70'ot43 or z:o-o#zs_ rpm = I-l669 or 3.5007 Hz = 7-33“? or 2:4757 mJ/sec (an: (466‘? Hz seEEJJELa Let r: 0.6 0-0404. r4 _ 0-5826 r" + 10404 = o gives :5 r : 2- 2562, 3-054-6 Since the guantu'tg E affains maximum ad: 1 r— —-—-—z = #8898 Jzor no.6 \ll-Z‘E we use 1': 2-2562- Maxfmum C13": C‘s/r = goo/(2-2562x60> = 3-6935 Hz @ Error factor for vibrometer is E _ ._Ef_____ _. ’(‘_r1)2+(2’yr)z * 1 Maximumof E occurs at r :— .h—zr“ For E: r1 m and r* = 1 Since the range is 4- E r < 0° » we use r: 4 for maximum error. L E - —-L-—- - I 0667 |r=+ [1-4‘l .- Percerrb error: (E-QIOO = 6477. Error factor = E = -—L-— \IQ‘VLYW- (LTr)z E obtains “10.x:mum ad: r* _ 1 m1 For 3:0-67, far: 3418' and = l-0053 Farcent error = 0.532 Elf: I", ‘S'=o.67 629 @ Selecl: the vibrometer on Hue basis of lowest freguency being mecuured. (0.)) T: O 1 7. From EZ-(IOL’9), 3— - 1-03- r r"— "03 -3 '3- - .. , - _ - 333 - Y [fl-1| 0.03 + r: '3’ = m6”) — 5-9595 n 60 (an .- wn = 8-9359 .raJ/sec = P4222. Hz (by: 0.; = r2 From 63- (“349), = 1.03 = -———— = E (1_r1)1+ (zTr)1 (1'03)2{1+F4-2r1 + 4-r‘r‘} = r” .'.e., 0-05-7404 r4” -— 0-“ r2 + 1. foe.) = O = 1-534-1T, 2-7206 By selecting r: 1-7206, 1r (an ___ 1000 so (1.7206) r1 -.- 2.3535 , 7.4m: f'eu Y = 19-1458 mJ/m 1- The guru-154:3 E a.'H:a.n‘n$ maximum or r: -—1'-—— = 138998 41-11'1 for T: 0-6. Hence we Anne b i'aJu. r: 2.720; {-0 cwau'a‘ rear of E. 8s{: = 472;, 'm ’ (.9: 4000 rpm = 443.33 rad/sec IOOO an: W = 9'“( .0 ) = 3I-3209 rad/sec r: c"/40" = 4‘3'33/31-3203 = I3-3738 Lat ‘5': o = z z 2 Error factor = ___.r____ = r ___ '3'3738 JQ—r‘)‘+(zrr)" T=° [1-H] !1_13.31381 =I-0056 630 (i) Maximum oln'sflacemeh‘l: = Y: Z [:0056 = %.0055 = 0-9344 .mm (ii) Maximum velocity: c.9Y = (4|8-88)(o-9944§ = 4.15.5473 rum/sec (iii) Maximum accelerah'on : cazY = (4&8-88)2(a-9944)= l74-483-35 "m/secl. ' l: __Yi___ 1 Maximum of % occurs when r=r*='J-1=2.T_? (see. E5.(3o84)) For T= 0-5 . r* -_- fi = 1-4142 , (r*)2 = 2 _z_i = —i————— =—Z—= 14547 Y r* {3- When error is one percenf, .2— :l-ol or %— = 0.9901 E5.(E1) can be rewritéen as: ‘1“ = £1451 + 4T‘r‘ . 2 r4' 0 ——-- (52) “(pg—‘2) —1"' (2-412) + 1 For g = 0.9901 and ‘r= 'o-S. (E1) becomes c.0197 r4" — r2 + 1 = O lz/Yl r = l°0203, 49-74-14 - E= l-ouol. 7-0527 Lowest freguency for one percent mm} .—. 7-0527 (5) = 35-2535 Hz Freguency range > 100 Hz , maximum error: 2 . k: 4.000 N/m. c=o=> ‘s-zo , "1:2. For Vibrometer war}: ‘5': o , Z —L-——- \2 Y Q r‘)z+(zrr)z 3:0 r2. 7 — [brat = 102 or Y (l—r‘) = -l°02 since. r must be greater Han—n one for {atgher fregucnc-‘es 0' Y2: 5| 5-6» r: 74414 631 Minimum :‘mpreSSea‘ freguencj = (.9: Io Hz Since r: “ya,” : loo/09h rad 7.54:4 , can = I4-ooz‘? Hz : 87-9917 — rec : "4: m m: k/whz = 4ooc/(37.9927)2: 0.5«57 k3 ® a": '° “" “’4: 8 “== «Wu—q: = mm? => 3—: 0.6 Let the lowest grezuemy = (do? r'zfiso/én O a . Error = 2/, m the range r0 5 ‘- < co r2 Err0r= E = l-OZ - - ‘IQ—r‘)z+ (Zxo-G r)‘ z} c.0404 r4 .. c.5826 r1 + n-o4o4. =0 e» r: 2-2562, 3-0545 Let r,= 2-2562: (.30: r9 a9“ = 2.2;“ (to): 22.552 Hz Let r. = 3.0546: «9,: 136,, = 3.0545 (to) = 30-54; Hz Lowest fugue”, = 22-562 Hz _ l4l-76l6 rad/sec Error factor for accelerometer: E=--—1——— [(I—r‘)"+(z‘s-r)‘ Mam‘mum of E occan aJ.‘ r*= "_ 2T2! ll-r‘l Since. the range is 05 r 5 0.65, we use r: 0.55 : E =_1— _ L”.- 0.65 I," 0.651, - I-‘73l6 Percen‘l: error = (2-0100 = 73462 Error 54d" = E = ——1————— J(1—r*)7' + (21%)" Value of r at which E attains maximum is r“: 41—2 1" 1 -—-—— l- r ' E- and T = 1-1-125 =ima.3;na.rj Since Hue range is 05 ff 0.5, we use r: 0,6 1 E, = 7—— : 0-9055 T=O-6 (1— 0.3‘)l+ 2.25 (O_6)Z Percent error = (5—1.) .00 = _9.4sz Much T: 0-75, 632 m: 0.05 1:, , max error = 3 7, over frezuency range of o *0 ma Hz Find 4< and C. For accelerometer, error factor = E =—1__ \ié'rz)z+(zyr)'* E affains maximum a): 1-: *= "l_2Tz “a ENE]: E'rg. —— FIJI-3‘ Considerafion of mo.ximurn error: M el’I'Ol"=’— e: E-‘z 0.03: ——_1 “Po” "arr‘ufiemenf’ “"5 13“” t" Y4 - T2 + 0.23 56 5 = a or T = o'-6|64 , 037874. ' r* of 7:0-m54 =\/1— 29-064)" = 0-4-9 1"- afi' Y=on7874 = ’1- 1(o.7g74)7- = imogI'nO-Ty (ii) Consideration of minimum error: 111' error=e= E-‘ _._ _O.°3= 1 (1- r"-)z+(2.‘3'v)z .' (6") with T: o-6l64; (51) Can Le “my-{ml M r4 - 0-4802 Tz—oquZS =0 9 1-7- : o-SE‘H. — 0-4069 0" r= 0'7552 At Hue maximum :Frczuency, w= W000) = 628-37. rad/sec Un= c'-"/1r = “Viz/Luca 3 320-0470 rad/sec “= T" “n1 = 0-05 (920.047); : 33623-354: N/m _ _ _°_ C-c - 2m GS." _ .3, =) c: 2". can? ._. 2(o.o;)(9zo-o47) (o-él64) = 50- 54 77 N—z/m - m=o~| K7, 1:: loooo N/m, c=o=> ‘S’=O mn=¢1ym =‘/|OOOO/o.. : SIG-2278 rad/sec Engine dyes-J = b): loco rrm = “34-72 Moi/sec mos/w“ = 104-72/3l6-zz78 = 0-33” Peak— to-fecu: {travel a; man = 10 TM" Find: Y; WY; 631‘!- 633 We 41mm, gram Ea. (Io-I7), z r" a-BZIzz —— ='_——' = _= 0-1232 Y ‘$=° l1‘r‘l lt- o-aalzzl Since Peak- to— peak trmvd of mass = lo mm, 2 = 5- mm Y= z/O-l232 = 5/0-(232 = 40-58447 ‘mm Max Jn‘anAemenf a} found¢flbn _.__ Y: 40.5944 mm Max ve‘ocH-‘j of foundmfc‘on=69‘t = 4249.9384 rum/sec M” W of founalo-fron = 0927 = 445053-823! mun/ml ' — _ . U - 5° .. Mammum sfeul- 3000 rpm _ so Hz .579: _ m _. 0.5 For muelcromE’cer; _______1 (5.) z = (+errar= 0.9 Jo—r*>‘+ (m) Here c: 20 N-l/‘m 3 .S._ C = 2° = o-olE‘HS 7.me 21-h (aooxz'fl’) m For r:o.5, Ez.(E1) QI-VCS r =0- 9|?8 Ct '-' {7'2 ZO/o.3|93 = 24-3962- N-J/rn c:c m: = 2—4496?" _—_ mom“ #3 = (9-H era-ms 109" 2000x210 = m 69: = o-ol‘Ml (Ioowz'fl’)a = 7621-7967 N/m @ as": z1r (o-s) : 1r moi/sec ; as; = as" I- T" 1— r‘ = w/an = °'4%.s => r = °'7-9 "I = “'fl’n = 4-7r/1r = 4 ; r2: 631/43,, = a ; r,= cM’s/1.9," = I: ¢: = em." (2TH) I. l— r; A = fi;'(z(o.zs)4 = _ 8.4934) l-IC #2: fan-.<M = -4. oG7s° 1—64 ¢3= hm" = _ 2- 6305' 1- I44 1 TI ——__1 :20 = Zl- 0H4 Q' rut) +03“); 2. ——r’-——— xlo = Io-l33l («_T1;)1+(1Tr1)2' 634 r,“ 1/(1— r3")z + (2113)" Record «Nubia—{red L: vl'Lrome‘Eer FS‘ given by x5 = 5-0294 5(1‘.) 2 ZI-o‘??q Sn'n (41H: + s.4.934°) + loqaal Su'n(81'l"t+ 4-0675") + 5-0294 sin (IZTTt + 2.5905?) mm at“): 20 sin 501‘: + 5' Sc'n "so-t mm 2 (E') 260:) = —7.0 (“)1 sin 50%: -— § 050)" Sin IS'ot mun/sec . l = -50 00° sin Sat _. uzsoo Sm 150* "m/Sec (52) an: 100 ra-J/Sec , w; = a)“ I—3": so =y ‘5'=o-6 __ U: 5’0 _ _ 6’2. ISO r" 0n=loo “(’5’ r1- wn‘EJ" _ -' 2 l‘ -l o 95,- am ( 3' '>= bu (1"°"‘°‘)= 38 6579 I—r,‘ 1-o-zs 5‘ _ +4".a(zT r1) = fan-'(zxonIS): _;5.zzzz° 5'0 000 JQ—n‘) +(zrn)‘ H7— 500‘ - 5| 335- 622? do ' raj->22? (1T r1); Ou'l'Put of the accelerometer r is given by 3,(+:) = -52 057420; sin (Set — 39-5593“) — 5| 335- can sin (1501': + '55- zzzz") run/sec” (Ea) It can be seen Hard: 5;. (Es) .‘s Substantiallfl different from 53. (Ea). ' 635 For given beam, A = (1.6 X 25) mm2 = 40 x 10‘6 m2, 1 I = E (0.025) (0.0016)3 = 8.53333 x 10‘12 m4, 6 = 0.05 m to 0.25 m. 1 E I 2 For a cantilever beam, Fig. 8.15 gives can = (Bn 6)2 [ ] where (Bl ,)2 = (1.875104)2 = 3.516015 ([32 (3)2 = (4.694091)2 2 22.03449 (33 {)2 = (7.854757)2 = 61.69721 ([34 02 = (10.995541)2 = 120.90192 For spring steel, E = 200 x 109 Pa, p = 7800 kg/m3 1 .L [ EI )2 Z {(200 x 109) (8.53333 x 10—12 2 = 2.33882 p A 44 (7800) (40 x 10-6) 74 72 2.33882 wn:(Bn€)2[ [2 \J The first four frequencies are given below: _______ €=0.05m I 0.0025 I 3289.33 I 20613.88 I 57719.47 I 113107.13 - _ _ - - - - __I___—-_'I-_‘—___T-__-_'___l'----_-_l____-----_ €=0.25m - 0.0625 n 131.573 - 824.6996 - 2308.78 - 4524.29 _ _ _ _ _ _ _ __|______J-____-—L---______L_______L_________ Hence, the range of frequencies that can be measured is given by (1) > 131.573 rad/sec. However, for first mode only (which is easiest to excite), the range of frequencies is 131.573 rad/sec 5 (0 S 3289.33 rad/sec. 636 ka =;’z__=§_(assume) Fo (1—x-2)=+(2gr)2 D dN NdD d Min 3" E . dN ".dD dr F0 — D2 —0 1.8., or {(1-r2)2+(2gr)2}(-2r)—(1—r2){2(1—r2)(—2r)+2(2§r)(2§)}=0 This equation can be simplified as r4—2r’+(1—4g2)=o and its solution is given by r2=1:!:2§ or r=V1+2 ;V1-—2§ Since kXR 1 _ =_.__ 1 F0 '1' 1+2; and kXR 1 -V _ =_ 2 F0 I! 1 25 we note that r=R1 = V1—25’ corresponds to a. maximum and r=R2 = V1 +2; corrponds to a. minimum of XR. 637 kx1_ —2§r Fo (I-r’)”+4’c’r2 dN ('11) i m =DdT-‘NT;=0 (1) dr Fo D2 dN__ fill=_ _2 where E— 2; and dr 2(1 r)(2r)+8§zr By setting the numerator of Eq. (1) equal to zero, we obtain {1+r4—2r2+452r2}(—2§)—(—2gr){—4r+4r3+8§2r}=0 which can be simplified to obtain _ 3r‘+(4g’—2)r2—1=o (2) The roots of Eq. (2) are given by 2:1-28-2 §‘—c’+1_1—28+2‘Vg‘—g2+1 3 3 , (3) Since it is difficult to determine, from Eq. (3), the correct value of r that corresponds to the minimum of X1, we use a numerical computation. For g = 0.1, for example, Eq. (3) gives ‘ 1' r2 = —o.ooso ; 0.6583 This shows that 1 r={1—28+2m}7 (4) 3 corresponds to the minimum of X]. For small values of g, {2 << 1 and Eq. (4) gives rzl (5) Thus X1 attains its minimum value close to r = 1. 638 Response of a single d.o.f. system with hysteretic damping is given by Eq. (3.106): X 1 X _ X _ ° (It—ma?) +(kfi)2 o (k—mwz)2+(kfl)2 2 2 X 1 R — 1m — = —————— e F0 + F0 (1: — m J)’ + (k 5)2 (1) It can be verified that Eq. (1) can be rewritten as, 2 2 2 X 1 1 Re [F0 + 1111 F0 + m — (2) Eq. (2) shows that the locus of T] as w increases from zero is part of a circle, with o 1 center [0, — —] and radius 1 2kfl , 3 shown in the following figure. 2kfi 639 1 The peak of Bode diagram is equal to z In the present case, peak-to-peak value is 2 '5’ 11.43 plotted; hence X z T = 5.715 11 m. X _= 5.715 :45: i 235t 1.27 2C or c: 0.1111. ° (.0: GS“ ® Reduction in amplitude from 173 mm/s to 20 mm/s in 7 cycles or 22 ms. Eq. 2.92 gives: 1 x1 1 173 — ln — = 5 or 8 = - 1n = 0.308223 7 X8 7 2O 5 Hence C = -—- = 0.049055 2 11 Typical Bode plot of phase angle is shown in the figure. (a) 45 = 90° at r = fi % 1. Hence the value of can can be determined from the value of r corresponding to (15 = 90° (b) Since 2 g r c w = tan-1 — = t ‘1 — —— we find c c k — In (.01 k — In (.02 where cal and wz correspond to the half power points. Hence, by finding the values of to corresponding to = — 45° and d) = — 135°, we obtain cal and wz. From these values, the damping ratio can be found using the relation: g = fl 2 wn 640 Characteristic Problem 10.26 Problem 10.27 Problem 10.28 16 18 18 1500 rpm 10 mm 80 mm 40° _ o a 0.09576 D c 3 Dominant frequency of vibration Inner race 6779.4 cycles/min 10127.7 cycles/ min 14792.8 cycles/ min defect (1078.97 Hz) (1611.87 Hz) (2354.33 Hz) Outer race 5220.6 cycles/min 7872.3 cycles/min 12207.2 cycles/min defect (830.88 Hz) (1252.91 Hz) (1942.84 Hz) Ball or roller 1214.6 cycles/min 1761.7 cycles/min 2188.1 cycles/min defect (193.31 Hz) (280.39 Hz) (348.25 Hz) Cage 326.3 cycles/min 437.3 cycles/min 678.2 cycles/min defect (51.93 Hz) (69.61 Hz) (107.93 Hz) f(x)=%;1$x$5 5 2 5 i=meanvalueofx=ff(x)xdx=—:-[%—] =3 1 1 s a2 = (standard deviation)2 = f (X — fl2 11") dx 1 5 5 1 2 1 x3 4 =— -3 dx=— —-3 2 9 =— 4((x ) 4[3 x + XL 3 . 1 °° 9 5 1 k=k t =_ _ 4: dx=_ _34 _ dx ur 0813 0A _f°°(x E (x) 16{(11 ) (4) Lety=x- 350 that. dy = dx. This gives 9 2 k=fi I 1'4dy=i y=-2 5 641 x = mean value ofx = 2 xi '1 3 1 3 —-:E-.(1)+§(2)+16 “H116(4)+%(5)+%(6)+3%(7)=4 (72 = (standard deviation)2 = (Xi - —)2 f(xi) _ _ _ 2L _ 2_3_ _ 21 _ 2_6_ _(1 4) 32+(2 4) +(3 4) 16 +(4 ) 16 2 3 2 3 2 1 27 — — — — — —=—=.67 +(5 4) 16 +(6 4) (7 4) 32 16 1 85 2(2-—i)‘r(xi)=(1—4)‘—L+(2—4)‘1+(a—4ri i ' 32 32 16 4 3 4 3 4 3 4 1 135 _. ___ —.4 ___ —— —__ —- .—_.== ————-== _ +(4 4) 16+(5 )16+(s 4).32 +(7 )32 16 s 1 8.4375 k=kurtosis=— x--x4fx- = =2.9630 a4 2):“ j (‘) 1.68752 2000 100' Magn-‘éude (4") 10 IOOO 'nso lFref/iJency (Hz) 186" e O 3 . A 9. "8° @ (9 ® @ FIB m- 46 £750 M 0 mt reSonamaa, Hae magniiude will have m skarFPEak. . Al: TESOnant—e, Phase will be 90° and! the Phase changes by I80" 015' Erezuency crosses the naJ'uraJ freauency. Usu‘na these rules, we {Aenb'fy four remnant frezuencu'es in FI'3.Io-46. Recena-hi: §reizuencz Damlainfl raJn'o 235. 3948— 2:7.8276 09,: 223- 4433 Hz 3": 2(Ms-(mu) =o-033962 «91: 474-1377 Hz {2: 482-7586—458.52o7 2 (474.|379) =aooz'5459 ,7 ' (‘53: mm, 5.72 Hz ‘3: Izzs-Hqsz- Ins-2753 z(|7-'5-5I72) = o-oIZlHI (.94 .—. mas-7.7579 Hz 3": WEB-Hwa— uni-0345 20%84159) = 0-007738 Raoul” o§ circle 1 = l-ZS— 4.5 l ‘5: _— = 0.; 4-025) FIGURE 10.47 643 Range of 09: 62-832. to 3l4-u6 rut-J/sec=coo £0 3000 -rPIn Max acceleration level = (03 = SW-I Irv/Sec: weight of Specimen : (ON Mas: vibraft‘oh amplrfude = 0-0025 m Frezuenctj range: Varid-“e- Areas! electric mo‘kor 'cmn Be used to obtain the freauencj range (30" a. mechanical shaker). Vibra‘bfon amplitude: If fit): A srn of, (En) o-cce‘crwl-t'oh = A951. ' A1: w=3M-l6 RAJ/Sec, amr'n'lzuale needed to acmeve the maximum underahbn .‘s: mmrfl'budc (A) : “Lela rafub'n/wz = 98'V(314o16)'— = 0-9939 “6'3 m At 05: 62'832 Tad/Sec 2 amrla'i'ucle needed to a_c;“.eve {he ma-xfmum CLCCLICraA'n'On is: amplihale (A) = meleraxh'on/wz = 78‘V(621931)z = 0.02495 Tn (amph'hulc I'S ’l'bo ‘13"; Anne: Aired: mrHI‘cafl'an Of 3(t) is not rermfffed u'Vled'N-‘IJWI'COrt “ta—kc" 053 Ha: {are shown in 53- I048 can Le used. Elecfrodjn¢m£c shaker of the fa": Shown in F03. m. .9 (a) can aJ$0 be used. 644 E"°*’°m°i“fi*."£-.$.":eee:: Max force (Fun) available defends on: (at) magnetic field strength LL) numLer of turns (c) coil diameter (A) Current HOW-'7': lelfa'tlm's are: C“) m"+e"i°'l strenfl‘tlu (L) cool-"n3 provided. Max ucelerafubn = Fmax > Where Wu: mass of specimen and me: mass of sinker fable. Speed range: 300 - 600 rpm Frequency range: 31.416 - 62.832 rad/ sec Number of reeds = 12 Uniform spacing of frequencies give the reed frequencies as: (ll, ..., 012 = 31.416, 34.272, 37.128, 39.984, 42.840, 45.696, 48.552, 51.408, 54.264, 57.120, 59.976, 62.832 rad/sec Let each reed be considered as a cantilever beam of cross section ax b inches. Let lengths of all reeds be same and the material be aluminum for light weight. The fundamental natural frequency of a reed is given by (Fig. 8.15): 1 1 E1 2 E1 7 _ 2 = . 2 1 i .1. (10 ) I 2 4 I 2 = 3.516 ——-—— = 69.1142 10 1 {(0.1/386.4) A t" l ) A 4" ( ) 1 I _ By equating wl given by Eq. (1) to n, , ..., flu, in turn, the proper value of { A (A }2 needed for different can be computed. By selecting a common value of 't’ for all reeds, the cross section of any reed can then be found to achieve the required value of 1 {i7}?- ______—_—_—__________—____—__— 645 Iterative process is to be used. 1. Select trial values of the design parameters (material of the beam and its dimensions). 2. Model the beam as a spring—mass system with: m = end mass = 50% of mass of beam: 111 = i p A e (1) k = stifiness of a cantilever beam: 3 E I k = — 2 e. ( ) 3. Equations of motion: mii+k(x—y)=0 or m'z'+kz=-m'y (3) where z = relative displacement of end mass. 4. Since in,“ = 0.2 g, assume a constant force of - m (0.2 g) on the right hand side of Eq. (3) and solve the equation to find z(t). 5. From the known 2m; value, compute the maximum stress (Gnu) induced in the beam. If on“ is less than the yield stress of the material, the design is complete. Otherwise, go to step 1 and change one or more dign parameters and repeat the procedure until a. satisfactory design is found. x“) We) Base motion , 3c. 646 ...
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This note was uploaded on 07/07/2011 for the course MAE 315 taught by Professor Wu during the Spring '08 term at N.C. State.

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Vibrations_Rao_4thSI_ch10 - Chapter 10 Vibration...

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