Ch10 - Chapter 10 Vibration Measurement and Applications...

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Unformatted text preview: Chapter 10 Vibration Measurement and Applications Volfage sensitivity: 1) = @038 yoLt—meter/Ncwton thicxntss: {7: 2mm: 7' m . ‘ Iooo ou’crm‘: voltage: 7.2.0 voLts . Wes—WU“: “H’h‘d: f1: '2 2. 2 vi: 1; _=;. .220 = (°'°?8)(looo> 19x 1’“: l-I’Z-Z4 Xxo‘ “/1722 H mzo-S kg, 1:: (0000 N/m, C2 0 10.2 _ 4 amrltfude =,Y= [000 m “fo‘t'al di‘splacemcn‘t of mass: x: ”/1000 m re‘a'tivg dI'SP’RoCSYnCfl‘t = 3,: 1—3.: SAOOO Th 2. Z r; r Y i-e» _...8...... = (””009 r2. :2} 71:2/3 1-!" loco ’_ r7“; _ - as __ ‘ r- w“ - o 8165 ‘ (.3: rash: o-SIGS #9553? :: “5.4.705 mA/sec 1213-3777 Hz Z Z = Y7 Y 2. Wkere ‘5': Z . : TL— Q*r‘)2+ (Mr) ‘“ 2' TZY . r2 Y \/G-r‘)z+ (‘5 02 1+ r“ ‘ Speeol range: -' 500 rpm 252-36 ra—J/sec -‘ 600 rpm: [57-08 raJ/Sec :: ; T1 Y x X £05 (at z (E!) -/(1~r1)1 + (2T‘r)2 93502: Le): T: o E8-(E1)»9t'ves) for 2% error, '2.— ——-———-—-————-—-r1 —- [.02 9, _ w .. Y lr;_'l " Y—TBY, _- 7'I4I4 H L19 __ Q3 _ 500 or {500 “ «mam ' 741m» 749m 2701“” M 2'0'0428 r'bm :: l-lGé‘i or 3.5007 Hz 2 7533!? or 21-9957 rad/Sec Q9”: {-1469 HZ 2435.052; Let r: 0.5 53- (E) grves Z yz Y W _ E = "‘32 => 0-0404. r4— 0-5826 r1+a-o4o4 = o ’ ' mm. aid/es r: z-zssz, 3:054-6 Since H18 guantify E attains maximum at 1 r=~—-——-—; = 1-3898 J3or ‘5'=o-6 JI—zr we use r: 2-2562- .‘. Maximum G3": 03/]. = goo/:ZQJSGZXQO) == 3°6935 HZ Error factor for v£brometer {s r2 ' 21-0 ’21) +0302 Maximumoj: E occurs ad: * r = ———"-‘ $-2y” T7" FOT “SH-=0: E = and T*=1 Ilvr" S.'nc,e the I‘D-“38 "5 4- f r 4 00) ’We use. r: 4 for WhXt-ml-‘m error. 1 + E] = 1 — l ose‘l r= [2-4 I Percent error : (E—I)l00 = 5‘77. r2 Error factor = E : “Mr—'- ‘ . x/G-r‘) + (arr): E attains maxiraum at r*: I/l,__2TZI For 3:20-67, rdr: 3.31M anal 5/ i =1 1.00153 r=r .‘;=o-(.7 1 Farcer‘t error: 0.53% Selecf the vibromcter on Hue basis of lowed {resuming [Being measured. (a) w = o = 7. From 53.00.19), 3. - 1.03- __r___ 2. M; .. - — r : __ :3 -3333 Y {vi—q ’ w; + cs r: :9— : 27r(-;oo) * .95. 5 " so can " 5 ’9 L4% = 3-5359 moi/sec = 14222 Hz Cb) T= 0-6 > _ 2 From Ez- 00-19), .3.— = 1.03 z r z E - (i-TLDL-f‘ (171,-)?- (han' {1+ r4c-2rz + 4r‘T‘} : H” .'.e,, 0-057404 r4 — 0.55 r1 + 1 = o 6-9.» v1 = 2.3535 , 74am be” r = 1.53411} 2-7206 Ry ‘selecb‘ng r= 1.7205, 03" .___ 1000 7!" so (2-7206) = 19.2458 “4/8“ + The gum-154:3 E affains' maximum of ' 1 xix-1‘s“ for T: o-Q. Hence we luau-re to tan r = 2.7206 ‘t'o avoid fear , of E. 854: = $20 In , (.9: 4000 rpm : 418-88 rad/sec. an: woo 5/834: 3 9‘3|( m ) := 3:. 2209 read /sec T= ”/60“ = «viii/3:42” : 13-3733 Leif: ‘S‘: o : Error facfidr 2: r JQ‘ “‘3” (2 Tr)‘ = hoot-36 r: :: 1'8993 z '2. Z V“, jt—r‘l [1*i3-37381 (3) Mameum cla'slbfcuement : Y: Z 1-0056 : %aoo‘3€ = 0.9944— mm (ii) Maxu‘mum velocity: 05 Y : (4;g.3g)(0,5344§ = 446- 5473 "fin/SEC (Eli) Maxfmum accelerafx‘on : wzy = (418-8x)2(a.9944)= 17442345 Tm"/sm‘. 3:——-£L——~— _-~_ Y V'LL—r‘)1+(zrr>2 (El) . 1 . Maximum of 3— occurs when r = r” 2 —-—-——-—— (see 58‘ (3.34)) Y J1 - 2 Ti ' : . * .._1_ 1 For T 05’ r. = m 214442 , (r‘) :2 2 _z_} = -—————e-‘—-——————————- : 3-;— 21.1547 Y r* “(1—2.)“ + (zxo-S x «4142.)1 when error is one percenf, Z!— =1.o| or‘ 3%.: 0.9901 ‘1' E3.(Ei) Can Be rewritten as: , 131‘: — (1431 + 49erI Z ” {‘4‘ _"r*(1'!§-!2) mm) = o «22> For .21: 0.3901 axial 7-:— 0-5', (E1)’Be¢ome$ _ 2 0.0197 r“-— r1 +1 =0 t/Y’ r“: (.0203, 49-741! . _ g..- . r— can“ [OIOI, 7-0527 Lowest freauency , for> ‘ one Percent «Lowrac; '= 7.0527 (5) = 35.2635 Hz FreguenCy, range > :00 Hz , maximum error = 27; «:4000 N/TTI: C20¢r201 3713-? For vibrome‘tEr wifh ‘g':o ’ ______[E________._ .2..- Y - /Q-r‘)z+(2rr)z [3:0 T1 = 7:757- : 1-02 7. 0V Y/(!-—r") = ~3'02 sfnce r must- be, greater than one for LigI-yer freguencres z 0" Y“ = S'I E'e') 1‘]: 74434 ml'm‘mum impressed acreguencif : ca: 10 Hz St'nce r= 0763,. : loo/Q9" = 74414 , (.311 = 14 002‘? Hz — 37 9927 2:5 9; J4: m‘ ,3 m: k/wnz : 4ooc/(37-9927f: 0-6167 kg ® :49": to Hz, (‘34: 8 Hz: (.3 1—; oJ‘—y1 =2» ‘3": 0-6 Let ‘H'xc lamest grezuencj : COO-:- r0: 63%;" Error 3 1% in fine range r2. 1/Q--r’")2'+ (2x 0-6 at)" => o~o404 r4— 0-5826 r2 + 1-0404 :0 rosr<oa Erro-r: E: 1'0?- : —_-> r: 2-2562, 3-0546 Let r0: 2-1562: (do: rO a9“: 2.2552 (Io): 22-562 Hz Let r0: 3.0544: use: rawIn = 3.0546(M) : 30.546 Hz Lowest- frezuancj' : 22'562 HZ: l4l-7616 rad/53‘ Error factor for aficeiewmetcr: E:——-—-—-—-—-——1—-——————- ' ' A JG'V’YHZ‘FOZ Mam‘rnum of E occurs aJl' ‘r*= ,f‘- 23-1 when 3:0, 7*:i 1 [l—r‘l Since Hue range [5‘ of r < 0'65 , we. use. r: 0:65 3 ._.___.1.______ EL: 0‘5 - 11~ was", ; “73:6 'Percen-k error :: CE-1)Ioo :: 73-16% 1 Error 5041’”? = E = E ./(1—r*)7- + (“01 Value of r a}; what. E affains' maxamum .‘s r“: \[1—23’1 When T: 0-75 , 1 SCnce Hue range is 05 r_<_ 0-6, we _______1________________ E! = {——~———————-——-—~ = 0.9055 r20‘6 (1_ O 3‘):- + 2 25 (o 6)?" _ Per-Cent error = (5-1) 100 = — 9 45/ c.1161 E :: use r: 0-6 m: 0.05 1:? , max error = 379 over freguency range 'Of 0 1‘2: (00 Hz Find ’9: and C. 1 For accelerémetcr, error fax/afar : E =——-—-—;-—-——*— , . V ‘ JQ-r‘) +0301 E aired-{n5 mmxn'mum ad: 1- = * = ‘I‘ _ 2 T1 T 23"“..3" (I) Considermi‘fon of maximum error: ‘ “I“? upon reamramaemeni‘J this (and; to ~~§4 ~T2 +023 555.. 0 or T: 0-964, 0-7274. 11} Error: 6:: E—1: o~03 Yr 0* Y20'6‘64‘ :1 \ll— 2Co.€|‘4)1 : 0.4-9 (U) Consiaerwtabn of mimmum error: 1 1/(1— M)2 + (arr)z *l (5,) with ‘5': o 6164; E; (E1) Cam be simplified M ' r4 -—a-4802 "'r —o.oezs :0 =} r7“: a-SQ'H, - O'IOG'? or r: 0-7662 M 8Yr0r=e: E-‘ : ‘0'032 At Hm: maximum :Frcguency, a}: 27W") = 622.31 rad/sex; a“: ”/r = 523.32 0.7ch :2 820-0470 raJ/sce 'k: m as}: 0-05‘(920.047)z = 3-3 623-854! N/m cc = 2"“ C43,, '7: 737' 7»? :2: am can? 2 2(o-o§)(920-047) (OI-6'64) I : 50-54 77 N-J/m m=o-||<3, :IocooN/m, c=o‘=>‘5'=o A ® can: \zk/m- 2JIOODO/ou: : 3‘6-2273 Tad/53C Engine and :: L0: loco rpm = 104.72 rad/sec ”(a/w" = m4.7z/3;g.27_73 = 0-3312 rack—to—feak tram/cl of may = (0 mm Fo'rwl: Y, (DY, ale. We 4mm, £mm (95.00.17), 2 z z r 0- 33:2 .232 .__.- 2 —-~-———- : -——————--—-—-—-—-—_——_ o. Y g=o [t—rlf !1_ 0.33121! Sine: Pmk— fo— Peak trowel of mag: =. Io mm; 2 2 5- TN“ V Y= 2/0-1232 = 5/04232 = 40-5844 mm ma-Y Jr‘sfa{¢Cemenf 03C feunciafu‘on : Y: 4-0-5944 mm Max vealed-{=3 of foundmh‘onzas‘r = 4243-3334. rum/3a Max Man of founol¢fion G9,"! = 445053-3131 mm/secz Maximum speed: 3000 rpm: so Hz ";r=._.é;’ ~.-. —:—° 2 0.5 o H , For acceleromefer, ___.._..__...3:____.__._._.. , 2. . 2 =1+error= 0-9 (ED “(t—r") + (ZTT‘) Here c: 20 N-J/‘m ; :_ C ' 2° __ o-OIS‘HS .. 7.111th 27:» (nooxzvr)_ m For r=o-5, E3.CE:> gives r :o. 9193- C: = ‘g‘ = 2? 0-8199 = 24‘3“," N‘J/m c . ; m: c = 2.4 3?62 = (”0.941 kg = (9.44 grams 209,, 2(aooxzvr) l = mwt: o-ol94l(ooo::17r)z - 7é22-79G‘7 N/m __ 2: 5 n= m“ — 4m _ 4 . 0%,, — s, wa/ww .2. ¢; = fan—l (2TH) l— r: —i #1: n (2 (0'23)4 = -- 3-4-9340 t—IS ¢z= tan-Yum“; = -4. «75° 1—64 ‘ 5‘3: hh"'(2xo-28 1.12.): _ 2.4.3050 l- I44 7. 4;...“ 3:.2. 2 x20: 2""‘”4 Q' rl ) + (2}?,) z J64?)z+ (1T9) r32 (i—T31)2 + (2Tr3)z Rewrd indicahsd by viLromefcr is given by 50:): Zf-O‘??<t sin (47“: + 9-4934°) + {0433! St'n (errh— 4-oc75') + 5.0294 sl'n (lzrt + 2-6905”) mm x 5 ': 5.o294 ‘ x8): 20 sin Sat + 5' sin 1504: mm . (5,) I. . L ‘ 1— ‘ g t m/se 2 1U?) = -zo (in) sun sot .— s (use) Sun I o m c , : ~50 000 Sin sot — Ileoo sin 1501‘: ”mm/secz (Ea) wn=WIoo rad/Sec , (at, = a)“ ,_ 3.2. = 80 z? T: O 6 0’: So (.5 (-9“ too ' 2 can —--——mo (5 _ 2T r _, . . . ¢,~— an—'( I): tam (2x06x0§)= 33-6‘5‘7? 1— r! - 1- 0-2.5 952: {15"(7'1— E): fin‘l(2)<o.éxl-5>= _g5.2122_° {" r1 5'0 coo .. 2 == 51°57' 9206 ‘/Q—¥11) + (1T7: )2 H2. 500 G * ’2‘)“ GT r2)“ Gui-Put of the accglerometer is given by : 5| 335- 622.7 31(19): —-52 057-9209 sin (50%: -— 33.6923") _ 51-335. 4229 Sin (wot + 55.11291") «um/m2 . (53) It Can Be Seen Hard: 52‘ (ES) is substantially different from 53. (£2). ® For ya'veh beam-n . IL A: 1015‘) : 0-0625 Inz 1;} I: TLOHTLG): 20-35 x7136 5'14 I I Viv/J 1‘ f: 2” to lo" For a. cdntzleVEr Lem , 59. 8.15 9431:: can: W >‘(—f—'—-)" fAX“ where (fi,l)z:@.975|o4 )1: 3-SléoIS (151! )z 2 6“ H4070" _. zz-os+4? 0 (($33)2 = (7.854757); = chm-121 0342)“ ~.— @4955“? = 12.040192 For sprina stceL, E = 30 x406 Psi , o 283 M/ma (_______ Efli'i: {5°ch (20 39x10 ‘)}'/-’-: 3&32-0 a 283 4 ’ J’A (396-4) (0-0615) ,Q '03 = fl 1),. £53.19. \ n ( h ( £1. ) . Tl»: fa'rsi.’ four frezuenu'es are given below: "-"r--*f-1 ----------- : ---------- :‘"'""“""‘"fl ------------- -_--4mu{__4-_-ffi_.---+____%%;«___L__--E€i‘---_i--_-EE-.-_u_- 1:2”: 4. ,' 32:0 .020 .zou'l- 4894 :56319 5527 {3103834530 —----|—-—-—-—-u—-~---—---—~—'--------——-r-——-----——--|------------— 1=n&fi loo : :29 404? 3 904 6996 : 2253 :32: ; 44:5 33:: _..__|_-.._.._-..___._...___- ————....._..-...L....----.._.___.__...-—__----___...- Hence {he range a; freauenues fine-t cam be meaSureJ 15' given 53 c.) > [23 404? rad/Sec. However, for fIrSf mode 011‘}; (whid. is easiest {to excffc) the range of frcguencfcs .‘s 123. 4049 33;; .<_ 09< 3210 M {ii . 5w cc ‘ I k XR‘ = 1 - r2 — }: (assume) F0 (1 - r2); + (2 5‘ 1-)2 D dN dD i—kh =D3”N§:_0n DEE—Nfll—o ' dr F0 D2 ° " dr ‘ dr {(142)? +(2gr)2}(——2r)—(1—r2){2(1—r*)(—2r)+2(2cr)(2s~)}=° This equation can be simplified as r4—2fi+{1—48)=0 and its solution is given by =li2§ or r=VV1+2§ ; V142; Since ka I =— 1 F0 T— 1+2§ 4§U+0 and kXRl 1 F0 r-m—m (2) we note that r = R1 = V 1 ~25” corresponds to a maximum and r =R2 = V 1 +23‘ corresponds to a minimum of X3. . 1(XI: —-2§r ® ‘ F0 (1——r2)2 +45‘2 1'2 dN dD 1)...- '__. :1 “‘1 =_____d_r_____N__dx;_=o (1) dl‘ F0 1)2 dN . dB 2 “=_ ___—_—._. __ 8 where dr 2g and dr 2(1 r)(2 r)+ (2r By setting the numerator of Eq. (1) equal to zero, we obtain ' {1+r‘1 —-2r2 +4 8 r2}(—2§)—(—-2§r){—4r+4r3 +851 r}=0 which can be simplified to obtain 3:4'+(4g3——2)r2—1=o _ (2) The roots of Eq. (2) are given by 2: 1—2g2—:2'Vg4——;2 +1 _ 1~23‘+2'Vg4—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 5‘ = 0.1, for example, Eq. (3) gives 1' r2 = —o.0030 ; 0.6583 This shows that 1 x/T— z ' ,._. 1-2§2+2 §—§’+1 . (4) 3 corresponds to the minimum of XI. For small values of 5', (2 << 1 and Eq. (4) gives m, (5) Thus X1 attains its minimum value close to r = 1. _........ Response of a single d.o.f. system with hysteretic damping is given by Eq. (3.106): ' _X_ = ________L_______ F0 . k—mu12+ik,5 2 X 1 Re —— + [In -— = ——-—-—————-—-—~———— 1 Fe Fe] {k—mwz)2+(kfi)2 M It can be verified that Eq. (1) can be rewritten as 2 ‘ 2 X X 1 1 _ —— = , 2 Re F0] +1111 Fo]+2flk [Zflk] () Eq. (2) shows that the locus of E:- as (U moreasw from zero 13 part of a. crrcle, Wlth 0 , 1 1 0, _ . 1 . . . center [ 2 k )6] and radlus 2 k 3 as shown :11 the followmg figure 10 23 The peak of Bode diagram is equal to 2 "213111 the present case, peak-to-peak value is plotted; hence X '21 0—;i mil = 0.225 mil. X 0.225 3; ‘ 0.05 ‘4'5 ‘3? or 5‘ = 0.1111. Reduction in amplitude from 6.8 in/sec to 0.8 in/sec in 7 cycles or 22 milliseconds. Eq. (2.92) gives: 1 X1 ’ 1 6.8 , -— ln —- =- = — ———- = 7 [X3 1 6 or 6 7 In (0.8 ) 0.3057 Hence g‘ = ~215— = 0.04866. 7r __...._._. Typical Bode plot of phase angle , is shown in the figure, LU (9.) ¢ = 90° at r = Z;— z 1. Hence the value of wn can be determined from the value of n r corresponding to €15 = 90°. (b) Since . 2 g r c w = ta. "'1 — = t ‘1 — —--—-- ¢ 11 [ 1—1-2] an [ k—mwz] we find c ‘ c k "‘ In. W] k ‘— m 012 where cal and 0J2 correspond to the half power points. Hence, by finding the-values of w corresponding to‘gb = — 45° and ql = — 135°, we obtain w] and (.02. From these values, the damping ratio can be found using the relation: gzfl 2wn W 0.09576 Dominant frequency of vibration Inner race 67 7 9.4 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 cyclea/ 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) (343.25 Hz) v ‘ Cage 326.3 cycles/ min 437 .3 cycles/min 678.2 cycles/min defect ' (51.93 Hz) 1 (69.61 Hz) ‘ (107.93 Hz) . 1 . f(x)=:;1$x$5 5 5 __ 1 x2 x=meanva1ueofx=ff(x)xdx=—4: «2— =3 1 1 1 5 1 °° a 5 1 k=kurtosis=— x~x4fxdx=m ~34 ~—-d a4 _jm( () 1611.0: )(4) X Let = x— 3 so thatd =2 dx. This ives 2 64 y y 5 ._____. if = mean Value of x = 2 f(x;) x1 i _;2m+§fl+%m+gfl+7m+fim+%m=4 02 = (standard deviation)2 =’ 2‘ (X1 — if f(xi) i =(1‘4)2—-+(2- 4)2-—+(3- 4)2——+(4—— 4‘—“)z 166 +(5——_4)2 —1—(-5—+(6—-4)2 5‘932—+(7-—4)2 ———-=——=1.6875 2m ~14 f(xi)=<1—4)4 31—2-+(‘2— 43y H+<3~4>4 ' 6 3 1 135 4—44—-— 5-—4 6—44—. 7—4 4 =—-—=8.4375 +( )16+( )16+( ) +( 93—2— 16 , . 1 ‘ 8.4375 k = kurtosis = -—— x- - 4 f x- = = 2.9630 0“ :4: ( ‘ j ( J 1.68752 Range of O9: ‘61-832 to 31446 maul/sec 2:600 éo 3000 rpm Max mlekafion level. = {03 -: ‘F'i’d m/secz Max YWCIE'jkt‘ of Specimen : (o N Max vibroita‘on amplféude :: 0-0025 m V Frezuencj range: Vaxiaaufi dread Czectr‘fc mofor cam be used to obtain {:h: fray-«Erica range (for a. mecham‘ml sflaxcr). Vibration amph'tuo‘e: If fit): A Sin (at, _ (2,) accc‘crab‘orx = , A031- M: (43:3ch Yuri/sec, amplifiuo'e needed! 1'50 a—C‘I"e—VE H“: maximum Malerafcbn 55‘: amrlf‘tuolc (A) : meal: Ruffian/:52 At 09:: 62- 8’37. rad/sac , am‘b’ni‘qdez needed to ackueve {he mammum amleragflbn i5 - (amplitude (A) : accelerat'on/wz = W'Vfiz‘mz): = 0-02495 7" (amplitude 1's 1'?» high; heme cured: arrll‘caftbn 5}: aft) {5‘ not Permitted), Mechanical Ahavkcr of the type Show: in Fla. 30.16 can be usecl. Electrodynamfc Shaker of the tjlpe shah/h in Fig. io-l7k(cv) com also be used. Electromanih‘siheser: __...—.~- —— -. '- Max force (Emu) available depends on: (Ca) magnetic field strength Lb) number of turns (,2) coil: diameter (A) current §iowm3 Limitations are: (0..) material. Serength, (A) cool-'n3 provided. Max welerafc‘on .—_ Fmax > Where m3 : me.“ of S‘Pecfmen and "mt. = mass of shaker lie—U6. Speed range: 300 - 600 rpm ‘ - Frequency range: 31.416 1 62.832 rad / sec Number of reeds = 12 p ‘ Uniform spacing of frequencies give the reed frequencies__as: {21, ..., nu = 31.416, 34.272, 37.128, 39.984, 42.340, 45.696, 43.552, 51.408, 54.264, 57.120, 59.976, 62.832 rad/sec Let each reed be: considered as a cantilever beam of cross section axb 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 __ _ I _ ml ='(51€)2{pi164}2 =(1.87512) {pir }2 ’ 1 . l .1 = 3,515 {my = 69.1142 (104) {XE-[7} 2 ‘ (1) _ I _ By equating 011 given by Eq. (1) to 01 , ..., (212, in turn, the proper value of {A (4} needed for different reeds can be computed. By selecting a common value of 6’ for all reeds, thle cross section of any reed can then be found to achieve the required value of I? Af“ Iterative process is to be used. I l , p - 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: m=%pA€ . ' m k = stiffness of a cantilever beam: 3 E I k = ,3 (2) 3. Equations of motion: mi+k(x—y)==0 or mi+kz=-m§ . (3) Where 2 2 relative displacement of end mass. .4. Since 'jmu = 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 2(t). 5. From the known 2m“ value, compute the maximum stress (Umn) induced in the beam. If am“ 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. r'“: I I -.....____ -...._----_.4 l” 7‘; }M H 30.58 Motion, }- <-—-‘> ...
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