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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 looo cuffed: voL‘Eage: zzo VoLts . Pressure applied: fl: 2, 5= ”’5 1; => 22° = (o-o?8)(.oio) 1"x ' 1;: 142.24. xw‘ u/mz m=O-5 R3, 1t: (0000 N/m, C2 0 amrlitude = Y: 4/0“ m finial displacement of mass: at: 'z/iooo m relative displacement = 5.: 1-}: 8/,000 m ' L z = _r___‘[_ i.e., _3_ = { (“ALMMD r1} =5 1-1: 2/3 _ 2. 1 r loco t— r" —_0-l_— . r—rwh—08l65 69= ”an: 0-9'65 ,lef’? = “Sui-705 rad/sec =l8-3777 Hz r2 Y c Z =_—— walker: 3’: = '— Q—r*)z + (n r)‘ Cert V? _ r" Y r‘ Y - _ 2. 3' 7- - 4' Q" ) + (V77 V) 1+ r Speed range: 500 rpm : 52'36 rad/sec - woo rpm: [57-08 ra-J/sec x:Xcasc.3f; TZY =—————— E Z ’(1—1‘1)‘ + (23‘1”)2 (J cAs_E(_i)__: Le’c ‘y=o E’s-.(Ep) gives, for 27, error, 2' r1 7=W=hoz => Y‘=%=7'l4l4 628 Q“: Ca 500 or I500 __ 7-ILH4 744:4 7-0919 - 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 Y : r——————I(l—r1)2+ (z:r)z : E = (-02. 0-0404. r4 _ 0-5826 r" + 10404 = o Wh-I'Ch gives :5 r : 2- 2562, 3-054-6 Since the guantu'tg E affains maximum ad: 1 r— ..———z = [.3399 :For r=o.6 JI-Z‘S' we use 1': 2-2562- Maxfmum C13": C‘s/r = goo/(2-ZSGZx60) = 3-6935 Hz Error factor for vibrometer u's- ._Ef_____ E: 1 2 z ’(I-r) +(sz) * 1 Maximumof E occurs ail: r :— .h—zr“ For :30, E: '1 m am! r*=1 Since the range is 4- E r < 0° » \Ne use r: 4. f3: wuthnuwl error. 2. E - —-L-—- - I 0667 |r=+ “-4."! .- Percerrb error: (E-QIOO = 6477. Error factor = E = -—L-— \[Q-r‘YW- (L‘s'r)z E attains THQX:I‘!IU~W\ a_£ r*_ 1/J—-——z1 1—2.}- 3' = l-0053 Farcent error: 0.53% I" , 3-:0'67 629 @ Selecl: the vibrometer on Hue basis of lowest freguency being measured. (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- = ”6”) — 5-9595 n 60 (an .- wn = 8-9359 .raJ/sec = Mrzzz Hz (by: 0.; = r2 From 63- (“349), 3;!— = 1.03 = -———— = E (1_r1)1+ (lTT)1 (1:03):{1+r“’—2rz + 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 03r- ___ 1000 60 (1.7206) r1 -.- 2.3535 , 7.4m: f'eu Y = 19-1458 mJ/m 1- The guru-154:3 E a.‘H:a.-‘n$ maximum at 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 = 9'3|( .0 ) = 31-320? rad/sec r: c"/40" = 4‘3'33/31-3203 = I3-3738 Lei” ‘5': o = z 2. 2 Error factor = ____r____ = r ___ '3'3738 JQ—r‘)‘+(zrr)" T=° [1-H] !1_13.313s7' =I-0056 630 (i) Maxfirnum oln'sflacemeh‘l: = Y: Z [:0056 = %.0056 = 0-9344 .mm (ii) Maximum veloaisz c.9Y = (4|8-88)(o-9944§ = 4.15.5473 run/sec (iii) Maximum accelerah'on : cazY = (4&8-88)2(a-9944)= l74-483-35 "m/secl. ' 2 __L‘___ —— = _ _ __ E Y «——-— (m2 <1? Maximum of % occurs when r=r*='J-1___;2.T_? (see. E5.(3o84)) For T= 0-5 , r* -_- fi = 1-4442 , (r*)2 = —2— : 14547 _-L*= ‘/—_—_———(1—2)12 + (2x0 5 x I- 4442.) When error is one percenf __Z._ _ --l -ol or %— = 0.9901 E5.(E1) can be rewritéen as: ‘1“ — (1451 + 4T‘r‘ _ r+ o ——-- (52) +(1-,;_.2>(2-uz)+1 For g = 0.9901 and ‘r= 'o-S. (52.) becomes 0:0197 r4' — r2 + 1 = o lz/Yl r1 = l°0203 , 49-74- H _ (.9 - 33:: l-Olol, 7-0527 Lowest freguency for one percent mm} .—. 7-0527 (5‘) = 35-2535 Hz Freguency range > 100 Hz , maximum error: 2?; 1<= 4-000 N/m. c:o=> r=o , m: 2. For Vibrometer war}: ‘5': o , Z —L-——- Y Q rz)z+(2TY')z 3:0 r2. 7 — [brat = 102 or Y (l—r‘) = -l°02 since. r must be greater Han—n one for {atgher freguchI'eS 0' Y2: 5| 5-6» r: 74414 631 Minimum :‘mpreSSea‘ freguencj = (.9: lo Hz Sl'nce r: cyan : loo/09h = 7.54”, , on = [4.0029 Hz : 87-9917 E4. rec : "4: m m: k/whz = 4ooc/(37.9927)2: 0.5157 «3 ® a": '° “" ”A: 8 H== «Wu—q: = wiry? => 3—: 0.6 Let the lowest §reguency = 63°? r,="‘9°/a',n O O . Error = 2/, m the range r0 5 ‘- < oo r2 Error= 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 (n) = 30-54; Hz Lowest fugue”, = 22-562 Hz _ l4l-76l6 rad/sec Error factor for accelerometer: E=--—1——— ’(1-r‘)‘+(2‘gr)‘ Mam‘mum of E occur: aJ.‘ r*= "_ 2T2! ll-r‘l Since. Hue range I'S' 05 r 5 0-65, we use r: 0.55 : E =_1— _ L”.- 0.65 I," 0.651, - I-‘73l6 Percen‘l: error = CE-i)loo = 73462 Error 54d" = E = ——1————— J(1—r*)7' + (z‘rr)" Value of r at which E attains maximum is r“: 41—2 1" 1 -—-—— l- r ' E- W and T = 1-1-125 =ima.3;na.rj Since Hue range is 05 rs 0.5, we use r= 0-6 1 E, = 7—— : 0-9055 T=O-6 (1— 0.3‘)l+ 215(04):" 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-: *= "I‘ZTZ and Ema: = E' g. = —_ (I) Considerafion of mo.ximurn error: M el’I'Ol"=’— e: E-‘z 0.03: ——_1 “Po" "arm“?menf’ “"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=017874 = ’1- 10.7974)‘ = imogI'nO-Ty (ii) Consideration of minimum error: 111' error=e= E-l _._ -o-03= 1 (1- r"-)z+(2.‘3'v)z .' (6") with T: o-6l64; E3- (51) Can Le s‘imri-‘fa'CJ M r4 - 0-4802 Tz—o:o£;18 =0 9 1-7- : o-SE‘H. — 0-4069 0" r: 0'7552 At Hue maximum :Frczuency, as: W000) = 628-37. rad/sec 0n= 09/, = “Viz/Luca 3 320-0470 rad/sec “= T" ”n1 = 0-05 (920.047); : 33623-354: N/m _ _ _°_ C-c - 2m GS." _ T =) c: 2m can? ._. z(o-o$)(920-047) (o-él64) = so- 54 77 N—z/m - m=o~| K7, 1:: loooo N/m, c=o=> ‘S’=O mn=¢1ym =‘/|OOOO/o.. : SIG-2278 rad/sec Engine dyes-J = b): loco rrm = lo4-72 Moi/sec mos/w“ = 104-72/3l6-zz78 = 0.33.2 reak— to-fecu: travel a; man = 10 TM" Find: Y; ”Y; 631‘!- 633 We 41mm, gram Ea. (Io-I7), z r" a-BZIzz —— ='_——' = _= 0-1232 Y ‘$=° l1‘r‘l lt- o-aalzzl Since Peak- to— peak trawd of mass = lo mm, 2 = 5- mm Y= z/O-l232 = 5/0-(232 = 40-58447 ‘mm Max Jn‘anAemenf a} foundah‘on _.__ Y: 40.5944 mm Max ve‘oc-"L’j of founda-fc‘on=69‘t = 4249'9384 mm/Sc: M” W Of founalo-fron = 0927 = 445053-823! mun/ml ' — _ . U - 5° .. Mammum sfesal- 3000 rpm _ so Hz .579: _ R _. 0.5 For muelcromE’cer; _______1 (ED 2 = !+errar= 0.9 Jo—rtfi (m) Here c: 20 N-l/‘m 3 3“ 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: zo/O-3l98 = 24-3962. N-J/m c:c m: = 2—4496?" _—_ o-OI94I kg = (9-H era-ms 109" 2000x210 = m 69: = o-ol‘Ml (Ioowz'fl’)a = 7621-7967 N/m @ as": z1r(o-5) : 1r moi/sec ; ”a = as" I-T‘ A = fi;'(z(o.zs)4 = _ 8.4934. me #2: tan-.(1XO'23X8 = —4'O‘_’5° 1—64 753: hn"(1"_ 1‘23}. '71) = _ 2- 6305' 1- :44 1. TI ——T__— :20 = z:-a??4 Q' rut) +03%); 2. __r,_____ xlo = lO-l33l («_T1;)1+(1Tr1)2' 634 r,“ 1/(1— r3")z + (2113)" Record «Nubia—fed L: viLrome‘lzcr 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 (Iz‘n't + 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 . 2. = -50 000 sin Sat _. ”7.500 5m 150* 7"""‘/5¢'-¢= (52) an: 100 ra-J/sec , w; = a)“ I—3": so =y ‘5'=o-6 .. “v '50 _ _ 6’2. _ :50 r... 0n=loo “(’5’ r2- wn’fia‘" _ -' 2 l‘ -l o 95,- am ( 3' '>= bu (1"°"‘°‘)= 38 6579 I—r,‘ 1-o-zs 5‘ _ +4".a(zT r1) = fan-'(zxosxls)= _;5.zzzz° 5'0 006 JQ—n‘) +(zrn)‘ H7. 500' - 5| 335- 622? JO ' raj->22? 0'“ r1); Ou'l'Put of the accelerometer , is given by 3,(+:) = -52 057420; sin (sot — 39-5593“) — 5| 335- can s-‘n (1501': + '55- zzzz") warn/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 (on = (Bn 6)2 [ ] where (Bl ,)2 = (1.875104)2 = 3.516015 ([32 (3)2 = (4.694091)2 2 22.03449 (B3 {)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 = {(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: _________ 68889 €=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 . 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=—__1—r2 =y—(assume) Fo (1—x-2)=+(2gr)2 D dN NdD d Min 3" E dN dD dr F0 " D2 "°‘°’D$‘NE‘° 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; 4§(1+§) () and kXR 1 I _ _ =_ 2 F0 I! 1 25 4§(1_§) ' () we note that r=R1 = V1—25’ corresponds to a. maximum and r=R2 = V1 +2; corresponds to a. minimum of XR. 637 kX1_ —2§r Fo (1-r’)”+4’c’r2 dN (1D i m =DdT-‘NT;=0 (1) dr Fo D2 dN__ fill=_ _2 where F_ 2; and dr 2(1 r)(2r)+8§zr By setting the numerator of Eq. (1) equal to zero, we obtain {1+r4—2r2+48r2}(—2g)—(—2gr){—4r+4r3+ngr}=o 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 _ Re F— = k m 012 ’ In: L = - k [3 0 (k—mu/Z)2+(kfi)2 F0 (k—mwz)2+(kfl)2 2 2 X X 1 R — 1m — = —————— 1 e F0 + F0 (k—mwz)2+(kfi)2 H It can be verified that Eq. (1) can be rewritten as, 2 2 2 X 1 1 Re [F0 + 1111 F0 + m — m] (2) X . . Eq. (2) shows that the locus of T] as w increases from zero 1s part of a cu'cle, with o 1 1 0 _ __ . . . . center [ , 2 k fl] and radms 2 k 3, 3 shown 1n the followmg figure 639 1 The peak of Bode diagram is equal to x 2—C' In the present case, peak-to-peak value is 2 '5’ 11.43 plotted; hence X z T = 5.715 11 In. X _= 5.715 :45: i 55: 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 = & z 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 w 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 1Problem 10. 26 18Problem 10. 27 8Problem 10. 28 —cosa D Dominant frequency of vibration Inner race defect Outer race defect Ball or roller defect Cage f(x)=% ; 1 6779.4 cycles/min (1078.97 Hz) 5220.6 cycles / min (830.88 Hz) 1214.6 cycles / min (193.31 Hz) 326.3 cycles/min (51.93 Hz) 10127.7 cycles / min (1611.87 Hz) 7872.3 cycles/min (1252.91 Hz) 1761.7 cycles/min (280.39 Hz) 437.3 cycles/min (69.61 Hz) 5 1 2 i=meanvalueofx=ff(x)xdx=2_ 1%. 1 0.2 = (standard deviation)2 5 ={(,; 3 5 1 2 1 x =— -—3 dx_.—-_ __. 2 4“£71: ) 4[3 31: +9 00 k= kurtosis= 111.7 f( (x —- 3‘ f(x) dx = -m Lety= x - 3 so that dy = dx. This gives 641 — a” f(x) dx 1500 rpm 10 mm 80 mm 40° 0.09576 14792.8 cycles/min (2354.33 Hz) 12207.2 cycles/min (1942.84 Hz) 2188.1 cycles / min (348.25 Hz) 678.2 cycles / min (107.93 Hz) 5 —: 1 3 5 i_f((x—3)4 1)dx 16 1(4) 2 9 64y=_2 5 x = mean value ofx = 2 f(x,) 1:; i 3 =%(I)+:—2(2)+—(3)+f46(4)+lis(s)+%<s)+.313(7)=4 16 (72 = (standard deviation)2 = 2 (Xi - —)2 f(xi) _ 2 1 2 3 2 3 2 5 = _ _ _ _ .. _ _4 __ (1 4) 32 +(2 4) 32 +(3 4) 16 +(4 ) 16 2 3 z 3 2 1 27 _ _ 6— __ —-4 —=—= .6875 +(5 4) 16+( 4) 32+(7 ,) 32 1 2(2-—i)‘r(xi)=(1—4)‘—1—+(2—4)‘1+(a—4r—3— i ' 32 32 16 4 5 4 3 4 3 4 1 _ _ _4 _. _ _ _ _ +(4 4) 16+(5 ) 16+(6 4).32 +(7 4) 32 1 8.4375 k=kurtosxs=— x--x4fx- = =2.9630 a4 i (' j (‘) 1.68752 2000 100' Magniéude (db) '10 50b P150 ! looo I l I I I iFref/im racy (Hz) hsoo £86" ID 0 3 A —| ‘“ 8° (9 © @ [3.3 10- 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. Recona-hi: §reizuencz Damlainfl raJn'o 235. 3948— 2:7.8276 09,: 223- 4433 Hz 3": 2(Ms-Imam) =o-033962 «91: 474-1377 Hz {2: 482-7586—458.52o7 2 (474.|379) =ao025459 ,7 ' (‘53: ”,5, 5.72 Hz ‘3: Izzs-Hqsz- Ins-2753 z(|7-'5-5I72) = o-oIZlHI (.94 .—. mas-7.7579 Hz 3": WEB-Hwa— uni-0345 201:“.2159) = 0-007738 .Raolfius 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 . z ‘ Max accelerator! level = (03 = SW-I Inn/sec Max weight of Specimen : (ON Mas: vibraft‘oh amplrfude = 0-0025 m Freauencu range: Varid-“e- Areas! electric mofor 'cmn Be and to obtain the freauermj range (30" a. mechanical shaker). Vibration amplitude: If fit): A srn of, (En) o-cce‘crwl-t'oh = A951. ' A1: w=3M-l6 RAJ/Sec, amr'n'lzuale needed to achieve, {the maximum underah'on .‘s: mmrfl'budc (A) : “Lela rafub'n/wz = 98'V(314o 16)‘ = O-9739x153 m At (a: 62 832 rad/Sec , amrla‘l'ucle needed to a.cHeVe. {Lg mufmum Wlermh'on I'S' - amplihale (A) = meleraxh'on/wz = 79‘V(62. .931)2 = O' .0249; 7" (amrll'fhulc I3. ’l'bo 41-3’13A¢nce Jul-ed: mrHICG-i'lan °f 3(1') «‘5' not Pet‘mrH-‘ecl ). -"'|eckr-I.mce«t dad-Ref of ‘Hme {are shown in Fig lo- i8 can Le USCJ' E'cc‘trodjnamlc Shaker of He {are shown in F03. I0- .99...) can aJ$o be used 644 E"°*’°m°i“fi*."£-.$."2§%t: Max force (Fun) available defends on: (at) magnetic field strength Lb) numLer of turns (c) coil diameter (A) Current HOW-'7': Limitations are: 6“) m"+e"i°'l Strength (L) cool-"n3 provided. Max ucelerahbn = Fmax > Where Wu: mass of specimen and me: mass of Shaker 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 WW} wH 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 reeds 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 design parameters and repeat the procedure until a satisfactory design is found. x“) “I a“ We) Base motion , 3c. 646 ...
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    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

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    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

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    Dana University of Pennsylvania ‘17, Course Hero Intern

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    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

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    Jill Tulane University ‘16, Course Hero Intern