ENB312-T1-S2-2010

ENB312-T1-S2-2010 - Surname Given Name/s III-I..-...

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Unformatted text preview: Surname Given Name/s III-I..- Examination Paper SEMESTER: SECOND SEMESTER EXAMINATIONS 2010 UNIT: ENB312 DYNAMICS OF MACHINERY - THEORY 1 DURATION OF EXAMINATION: PERUSAL: 10 MINUTES WORKING: 2 HOURS EXAMINATION MATERIAL SUPPLIED BY THE UNIVERSITY: EXAMINATION BOOKLETS - TWO (2) PER STUDENT EQUATION SHEETS - THREE (3) PAGES ATTACHED GRAPH PAPER MM - THREE (3) PER STUDENT EXAMINATION MATERIAL SUPPLIED BY THE STUDENT: WRITING IMPLEMENTS GRAPH TOOLS (COMPASSES, RULER, PROTRACTOR) CALCULATORS - NON PROGRAMMABLE INSTRUCTIONS TO STUDENTS: Students are prohibited from having mobile phones or any other device capable of communicating information (either verbal 0r written) in their possessions during the examination NOTES MAY BE MADE QNLX ON THE EXAMINATION PAPER DURING PERUSAL TIME SECTION A - TWO (2) QUESTIONS ONLY ARE TO BE ATTEMPTED m - TWO (2) QUESTIONS ONLY ARE TO BE ATTEMPTED ATTEMPT EACH SECTION IN A SEPARATE EXAMINATION BOOKLET ALL QUESTIONS ARE OF EQUAL VALUE GUT Gardens Point SECTION A QUESTION 1 (a) For the free vibrations of a system, and the following given information; determine (i) the natural frequency, (ii) damping ratio and (iii) the damping constant of the system. (10 marks) Spring constant 10 kN/m Mass 1 2 kg Amplitude of the first cycle 61 mm Amplitude of the second cycle 48 mm Amplitude of the third cycle 36 mm Amplitude of the fourth cycle 27 mm (b) An instrument package installed in the nose of rocket is cushioned against vibration. The rocket is fired vertically from rest with a constant acceleration ii = c, where c is a constant. Derive the equation of motion for the instrument package if it has a mass, m. Find the expressions for the relative displacement of the package with respect to the rocket. Assume initial conditions 2(0) : 0, and 2(0) = 0, where z = x- u . (10 marks) QUESTION 2 For the rotor system shown in the figure, determine the fundamental frequency and mode shape of the system in torsional Vibration. Neglect the inertia of the shaft. (20 marks) N- N-m _ 5 fl k,‘=1.2>< 105:31 k,2=8.7X105 rad k,3—5X10 rad ENB312T1.102 Cont .... .. UESTION 3 (a) Define principal mode. (4 marks) (b) Tension spring packaging is a means of supporting extremely fragile packages for shipping. To simplify the problem, idealised it as two masses and three springs. What are the two natural frequencies and mode shapes? (16 marks) ENB312T1.102 Cont .... .. SECTION B QUESTION 4 For a general fourbar linkage shown in the figure below, Link 2 is the input link, and Link 4 is the output link. Link lengths [cm], angle [degree], and input angular velocity [rad./sec.] for this mechanism are given in the table below. Onlflie open circuit is considered. B Figure (Figure is not to scale) Link 1 Link 2 Link 3 Link 4 62 mg 8 cm 3.5cm 10 cm 9 cm 15° 36.7rad/s (a) Accurately draw the linkage for the open circuit to scale on the graph paper provided. (4 marks) (b) Graphically find the solutions for 63 and 04. (4 marks) (0) Calculate the transmission angle. (2 marks) ((1) Calculate angular velocity ratio, my. (5 marks) (6) Determine (you can use either the graphical method or the analytical method) angular velocity (03 for link 3; (5 marks) ENB312T1.102 Cont .... .. QUESTION 5 The figure below depicts a slider crank mechanism. Link lengths [cm], angles [degrees] , input angular velocity [rad. /sec.] for this mechanism are given in the table below. ‘ NB: Graphical solutions to this problem are NOT permitted. c (Offset) Coordinates of B: (x3, ya) Figure (Figure is not to scale) Offset 62 a); c -2 cm 50.0(degree) 52.4 rad/sec. (constant) (a) Analytically calculate, for both open and crossed circuits, solutions for: (i) Position (x3, y3) of the slider B. (ii) Angular velocity, (03, for Link 3, (10 marks) (b) For the open circuit, determine the velocity (magnitude and direction) of the slider B. (5 marks) (c) For the open circuit, determine the maximum and minimum magnitude of x3 for the position of slider. (5 marks) ENB312T1.102 Cont .... .. QUESTION 6 The system, as shown in the figures (i) and (ii), has the following data: m1 = 1.2 kg, R1 = 1.135m, 61 = 113.40 and m2 = 1.8 kg, R2 = 0.822m, 92 = 48.8° Ifa) = 40rad/sec Moving global mass center - unbalanced (i)Unbalanced link (ii) Dynamic balanced model To balance this system, an additional counterweight mass, mb with the CG radius Rb, is designed, as shown in the figure (ii) of Dynamic balanced model . (a) Find mass-radius product and its angular location needed to statically balance the system. (12 marks) (b) If the mass required for this counterweight design is mb = 2.98kg, calculate the CG radius, Rb , of the counterweight. (4 marks) (c) If Rb = 1.0m, calculate the counterweight mass, mb. (4 marks) END OF PAPER ENB312T1.102 (i) ENB312 Part I--Vibrati0ns: Equation Sheet Forced Vibration mx' + kx = F(t) Solution of equation . F sinwt x=Acoscont+Bsmwnt+—lz-—"2 m(wn —a) ) . Fem)! x = Acos cant+Bsmwnt+12—2 "7(a)" ‘60 ) Steady state displacement X l EM_F% Steady state displacement for unbalanced force _ moeco2 x_ 2 2 m(a),, —(o ) Magnification ratio MR : mX moe Transmission ratio 1 TR 2 2 1 _ a) 2 wn Rectangular step forcing function , F x: Acoscont+Bsmcont+—° Linearly increasing forcing function . Ct x: Acoscont+Bsmwnt+—k— Exponentially decaying forcing fimction —aI Foe x = Acoscont+Bsin wnt+ 2 ma +k Combination of forcing functions I F e7!!! x = Acos cunt + Bsm wnt+—°[l — k 1+% (2),, for initial conditions x(0):0 and x(0)=0 then F0 F0 0 a - rat x:__— —— —cosa)nt+smwnt +6 k k(1—;—:) a)" a)" ENB312T1.102 Damping and damped forced Vibrations ccr=2mw é’=— wd=wn l—§2 x— _kJa—:pz+a;$)2 Amplitude ratio _ X E/k Phase angle ca) 240% tan(0) = ——2-T = ‘—wZ-" m(a)n —a)) (1— ml Transmission ratio 1+ 2 i 2 TR:£ k2+c2w2= (an) 1 0-%Y+QC%Y Forced vibration (Impulse! x = —F——exp(—4’a)nt) sin(,/1 — 4’2 cont) ma)" 1—4’2 Forced vibration gStep! x=51_wcos( 1‘42wn,_¢) k 1_;2 where tan(0)= 4’ 1—42 Vibration absorbers 2 )clk1 _ 1—(wi22) " k 2 2 k F0 [1+fi-(fg)][1-(wfi)]—f where k 2 6011 = —L wzzz =£ m1 m2 2 _ m2 k2 _ (022 #1 — —‘ — — ‘T m] k1 0)“ (ii) ENB312 Part II--Dynamics of Mechanisms: Equation Sheet Gruebler’s equation M=3 (L- 1)-2J1-J2 where: M: DOF or mobility L: number of links J1: Number of l DOF(full) joint J2: Number of 2 DOF (halt) joint Grashof Condition S+LSP+Q where: S: Length of shortest link L: Length of longest link P: Length of one remaining link Q: Length of other remaining link Vector osition anal sis ora ourbar linka e Euler identity for a vector :19 e . = c056 i j sin 6 where j denotes imaginary unit Position vector loop equation for afourbar linkage R2 +R3 —R4—Rl =0 If 6. for grounded link 1 is 0°. The two scalar equations are bcos 63 = —acos«92 + ccostél4 +d bsinH3 = —asin62 + csin 04 where a, b, c and d are used to denote lengths for Links 2, 3,4 and 1 respectively. The solution for the above equations is: [ [—BiJBz—4AC] 04” =2 arctan —-—————— 2A where A = c0562 —K1—K2 c05492 +K3 B = —25in62 C = K1 —(K2 +1)cost92 +K3 d. . 2_ 2 z 2 K1=_, Kzzi, K =M a C 3 2ac Position of a Point on a link RA = ae’g2 = a(cost9Z +jsin02) Rm 2 pe’(6‘+5~‘) = p[cos(6l3 +53)+jsin(63 +60] RP = RA + R“ l Vector osition anal sis or slider-crank Position vector loop equation Rz—R3 —R4—Rl :0 If 61 =0°, the two scalar equations are acosa2 —bcos«93 —ccos€4 —d = 0 l asinH2 —bsint93 —csin64 = 0 ENB312Tl.102 the solution for the above equations is: . asin0 —c 03 = arcsm d = acostél2 —bcos€3 anal sis ora ourbar linka e Angular Velocity Ratio woutput (04 mV : = — winput wZ Mechanical Advantage a)mpul llinput W2 [input m A = = — - woutput routput Q4 roulput Velocity vector loop equation a 5(R2+R3 —R4—R1)=0 If 61 for grounded link 1 is 0‘? The two scalar equations are —an2 sin 02 — bro3 sin (93 + 0604 sin 64 = 0 {an2 cos 62 + ba)3 cos 6'3 — ca)4 cos 64 = 0 The solution for the above equations is: _ awz sin(6l4 —02) _ 7' sin(t93 494) _ aa)2 sin(6l2 —03) w c Isin(t94 -63) ‘03 ‘04 Vector velociQ analysis [or slider-crank Velocity vector loop equation a 5(R2—R3—R4—R1)=0 If 61 =00, the two scalar equations are {—awz sin 92 +ba)3 sin 63 — cl = 0 aa)2 cos 62 — ba)3 cos 03 = 0 The solution for the above equations is: aa)2 cos 492 [7 cl = —aw2 sm 62 + bat3 5m 03 60} cos 193 Vector acceleration analysis for a [ourbar linka e Acceleration vector loop equation 62 EARZ +R3 —R4 —R1)= 0 If 61 for grounded link 1 is 0°. The two scalar equations are —aa2 sin 02 — awz2 cos 62 — b0:3 sin 63 —ba)32 cos 63 +cot4 sin 04 + cwf cos 04 = 0 aoz2 cos 62 — acuz2 sin 62 + ba3 cos 63 —ba)32 sin 63 — cot4 cos 494 + co): sin 64 = 0 The solution for the above equations is: _CD—AF 3 AE—BD _CE—BF AE — BD where A = c sin 64 B = [3 sin 63 _ ~ 2 2 2 C — aa2 sm 62 + aw2 cos 02 + bu)3 cos 03 — 0% cos 04 (2 a4 D = ccosél4 E=bcos¢93 _ 2 . 2 . 2 . i‘F — aa2 cos 62 — awz sm 02 — ba)3 sm 6’3 + ca)4 sm 64 Vector acceleration ana_lysis [or slider-crank Velocity vector loop equation 62 Elez—RS—R4—RJ=0 If 61 =00, the two scalar equations are . 2 a —aa2 sm 02 ~ aa)2 cos 62 + ba3 s1n 03 +bco32 00563 — a’ = 0 2 . aoc2 cos 62 — aa)2 s1n 62 — b0:3 cos 03 +boo32 sin 63 = 0 The solution for the above equations is: aa2 cos 62 — aw: sin 62 + [76032 sin 63 a 2—— 3 b cos 6'3 .. _ _ 2 . d — —aot2 5m 62 — aa)2 cos 02 + bot3 sm 03 + bag2 cos 63 ENB3 l2Tl . 102 (iii) Force analysis or single link in Pure Rotation Vector equation 2F = FP +1712 = "7236 ET = T12 +(R12 XF12)+(RP XFP) = [Ga Scalar equations FPX +Fle z mZGGx 1 FPy +Fl2y = mZaGy T12 +(R12xFlZy ‘ R12y‘F12x)+(RP"FPy _ RPyFPX) : [Ga Matrixform l 0 0 Fizx "72an ‘pr 0 l 0 Fwy = mzaGy —F1,y _R12y R121: 1 T12 [Ga—RP):pr +RPyFPX J Balancing a rotating system Static balance: Vector equation: 7"th = _m1Rl ‘msz j Scalar equations mb be = _(mlRlx + mszx) mthy = —(mlRly + mZRZy) Dynamic balance: Vector equation: mARA +mBRB = —mIR1 —m2R2 —m3R3 mBRBlB = —mIR1[1—m2R212 —m3R3l3 Scalar equations mARAx + mBRBx = '(miRix +"'2R2x + m3R3x) mARAy +mBRBy = —(m]R1y +m2R2y +m3R3y) _ m1 Rlxll + mszxlz + m3R3xl3 mBRBx — —_ " ls mlRlyl1 + m2R2y12 + m3R3yl3 mBRBy = — L n ...
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This note was uploaded on 11/03/2011 for the course EN 40 taught by Professor Mcgregor during the Three '10 term at Queensland Tech.

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ENB312-T1-S2-2010 - Surname Given Name/s III-I..-...

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