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Unformatted text preview: GUT surname GivenNamers
IIII.. Examination Paper SEMESTER: SECOND SEMESTER EXAMINATIONS 2008 UNIT: ENBB‘IZ DYNAMICS OF MACHINERY  THEORY 1 DURATION OF EXAMINATION: PERUSAL: 10 MINUTES
WORKING: 2HOURS EXAMINATION MATERIAL SUPPLIED BY THE UNIVERSITY: EXAMINATION BOOKLETS  TWO (2) PER STUDENT
EQUATIONS SHEETS  THREE (3) PAGES ATTACHED
GRAPH PAPER MM  FOUR (4) PER STUDENT EXAMINATION MATERIAL SUPPLIED BY THE STUDENT: WRITING IMPLEMENTS
CALCULATORS  NON PROGRAMMABLE INSTRUCTIONS TO STUDENTS: Students are prohibited from having mobile phones or any other device capable of communicating information
(either verbal or written) in their possession during the examination NOTES MAY BE MADE QNLX ON THE EXAMINATION PAPER DURING PERUSAL TIME SEC IQNS A AND B  FOUR (4) QUESTIONS ONLY ARE TO BE ATTEMPTED
NO MORE THAN TWO (2) QUESTIONS FROM EACH SECTION ATTEMPT EACH SECTION IN A SEPARATE EXAMINATION BOOKLET MARKS FOR EACH QUESTION ARE AS INDICATED Queensland University of Technology GUT GUT GUT Gardens Point Kelvin Grove Carseldine SECTION A QUESTION 1 A general fourbar linkage is shown in Figure Q1. The Link 2 is the input link, and the Link 4
is the output link. Link lengths [cm], angles [degrees], input angular velocity [rad/see] are
given for this mechanism in the table below. Only the open cirCuit is considered. Figure Q1(Figure is not to scale) 8 cm 3.5 cm 10 cm 9 cm 15 53 81 36.7 25 l8.7 0 908
deg deg deg r/s r/s r/s r/s2 r/s2
(a) Prove the degree of freedom of this mechanism is one. (2 marks) (b) Calculate the transmission angle. (3 marks) (0) If input force is applied to point A and the output forCe is applied to point B, determine the
Mechanical Advantage, mA. (2 marks) (d) Analytically determine velocities (magnitude and direction) of the pin joint A and the pin
 joint B. (4 marks) (e) Determine magnitude and direction of acceleration for pin joint B. You can use either the
graphical method or the analytical method. (9 marks) ENB312T1.082 Cont .... .. ' QUESTION 2 Design a fourbar mechanism to give the three positions (C1D1, C2D2, C3D3) shown in the
following Figure Q2(a). All units use Standard International (SI) system of units. NB: The solutions to this question must be obtained graphically 3.744 —>i Figure Q2(a) (Figure is not to scale) (a) Accurately draw the linkage to scale on the graph paper provided. (7 marks)
(b) Obtain lengths for all links. (3 marks)
(c) Check the Grashof condition. (2 marks) ((1) Based on the results obtained in above a) and b), for the fourbar linkage at the Position #1
(CD1), as shown in Figure Q2(b), the link 0201 is the input link and the angular velocity (02:10. The link 04D. is the output link. i) Graphically determine the angular velocity, (04, of link O4D1. (5 marks)
ii) Calculate the angular velocity ratio, mV. (3 marks)
D1
C1
C02
04
02 , ' % J) Figure Q2(b) (Figure is not to scale) ENB312T1.082 Cont .... .. QUESTION 3 Figure Q3 depicts a slider crank mechanism. Link lengths [cm], angles [degrees] , input
angular velocity [rad. /sec.] are given for this mechanism in the table below. NB: Graphical solutions to this problem are NOT permitted. c (Offset) Coordinates of B:
(x3: yB) Figure Q3 (Figure is not to scale) ' ‘
0A AB c
(constant) rad/sec. (a) Analytically calculate for both open and crossed, solutions for: Lmk 2 Link 3 i) Angle (95 ii) Position (x3, y3) of the slider.
(8 marks) (b) For the open circuit, determine the velocities (magnitude and direction) of the pin joint A and the slider B.
(7 marks) (0) For the open circuit, determine the maximum and minimum magnitude of x3 for the
position of slider. (5 marks) ENB312T1.082 Cont . . . . .. SECTION B QUESTION 4 An instrument package installed in the nose of rocket is cushioned against vibration. The
rocket is ﬁred vertically from rest with an acceleration which increases linearly with time ii : bt, where b 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 .
(20 marks) QUESTiON 5 Determine the two natural frequencies and mode shapes of the two~mass systems. Given that k1:2k,k2 =k3=kandm1=mandm2=2m
(20 marks) ENB312T1.082 Cont .... .. QﬂﬁﬁTIQN 6 A gas turbine rotor is arranged schematically as show in the ﬁgure below. Neglect the inertial
of the shafts and determine the fundamental frequency and mode shape of the system. 1(compressor) = 15 kgIl'l2 K1 2 3.0 X 106 Nm/rad
[(turbine rotor) = 10 kgmz K2 ; 1.5 X 106 Nm/rad
{generator} = 8 kgm2 (20 marks) Generator Compressor Turbine END OF PAPER ENB312T1.082 (i) ENB312 Part IDynarnics of Mechanisms: Equation Sheet Gruebler’s equation M=3(L1)—2J,J2 where: M: DOF or mobility
L: number of links
J1: Number of 1 DOF(full) joint
J3: 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 3 Euler identity for a vector
oils = cos 9 i j sing where j denotes imaginary unit Position vector loop equation for a fourbar linkage
R2+R3 —R4 Rl :0 If 91 for grounded link 1 is On. The two scalar equations
are bcosﬁ3 : —acos I92 +ccost94 +d
bsin 93 = —a sin 62 + c sin 94
where a, l). c and d are used to denote lengths for Links 2, 3,4 and 1 respectively.
The solution for the above equations is: “W 2/1 where
A = (:0st —K1K2 cost?2 +K3
.l3=—325int5l2
C=Kl —(K2 +1)cose2 +K3
d d _a2—b2+c2+d2 K,=—; K2=—; K}—
a c 2ac Position ofa Point on a link RA = new” = a(cosﬂ2 +jsin61) Rm = pew“ = p[cos(6]+5_1)+jsin(93+c5'3)]
RP=RA+RN Vector osition anal sis or slider—crank
Position vector loop equation
R2 —R3 —R4 —R1 = 0 If .91 =0°, the two scalar equations are
{acost —.bcost93 ~ccost94 —cl = 0 asintﬁl2 mbsin63 —csin94 =0 ENB312T1.082 the solution for the above equations is:
. asin 6 we
63 = arcs1n[—l———] d: acostﬁl2 wbcoség e
Angular Velocity Ratio moutput _ g.
a’ ‘02 input
Mechanical A dvantage m: V woulput romp at Velocity vector loop equation a
g(112+1i3—124—ni)=0 Ij'é‘, for grounded link 1 is 0‘! The two scalar equations
are —aa)Z sin 92 — butJ sin 93 + ca)4 sin 34 = 0
{coal cos 62 +th3 cos 63 —caJ4 cos 64 = O
The solution for the above equations is: _ act)2 sin(t94 — 92)
m’ ‘T' sin(93 —a,) = 3312“ sin(82 —63) C anal sis or slider«crank Velocity vector loop equation a
5(R2”R3 _R4_Rl)=0 If 6] =0: the two scalar equations are
—aa)2 sine2 +bcq, sin63 ma: 0 { aa)2 c03192 “bag cos 63 = 0 The solution for the above equations is: b ‘cost93
cl = —aa11 sin 62 +1560j sin 93 If 61 for grounded link 1 is 0°. The two scalar equations
are (ii) Force anal sis or sin lo link in Pure Rotation
Vector equation —o'o:2 sin 02  an): cos 62 — but3 sin I9j
who)? cos6_1 +cot4 sin 64+co): cost?4 =0
1’ a I: I" 1  n E! 2 P+ I ‘r 3 G S‘TT” +(R.3 :<l'3:I+\:l{p xl’P)m.'Go. .—J aaz cos 92  an): sin 92 +bo:3 cos 03
4m): sin 03 “c054 cos 94 +80): sin 6'4 = 0
The solution for the above equations is: a _ CD—AF Scalar equations 3 — AE—BD F‘Px + Flzlt . minor
a =M pr +1713“ nmzac.‘ 4 AE—BD ' ' .
where 2 3.5:)  RI:__F12_V)+{RE‘FF{F Rp)_pr] lot!
A = csin 64 Matrix form
B =bsin6l '3 2 l 0 0 Fizx "12%, '5;
C = (30:2 3111 61 + no): cos 92 +601? cos 63 —c«ro4 cos 94
~ 1 0 F12 i= mzaG —FP
= y
D ccos 94 y y
E=bcosﬂ3 Tu [Ga—RHF%+RPyFB F : aw2 cosfil2 —aa)22 sin 92 —b(o§ sin 93 +caJ42 sin 64 Vector acceleration anal sis or slidercrank Velocity vector loop equation If 91 =05, the two scalar equations are
—oro:2 sin 92 — an): cos 62 + bot3 sin 63 +bwfcos€3 —cii=0 aa'2 cos 92 — an): sin 62 — bot3 cos 63 +bto32 sin 63 = 0
The solution for the above equations is:
aaz cos 02 — an): sin 92 + barf sin 63
1) cos 6] d = —aa2 sin 62 — an): cos 62 + bot!3 sin (93 a3: + [)an2 cos 03 ENB312T1.082 Forced vibration (iii) Equations in Vibrations mk' + for z ("(1)
Solution of eguation x = Acoswnt+ Bsin cont+ — x: AcosmnH 831nm": + m( File“)! Steady state displacement
X E/k Steady state displacement for unbalanced force
m(J e (02 x__ Magnification ratio Transmission ratio TR* _ m2
1— x = Acoswnt + Bsin canr + I”— MRm Rectan rular ste forcin v m(w,'f 1
}_ a’JZ 1
a"! _. 02") mX mac l a) function F Linearly increasing forcing function x = Acos can: + £381an + « Cr I» Exgonentiaily decaying forcing function x = A cos war + Bsin (Dut' + "(H
Fuel '1 ma'+k Combination of forcing functions .1: : Acosta”: + Bsin cant + J ¢zr (f F in 1+ 51‘; 01,] l FI sin a)”:
'2 "f
a)“ _ a) ) m(wf m (at) for initial conditions mud) and limbo then k(1— :12) X3— k ENB312T1.082 l (I w” l a .
—cos a)"; + Sm a)“:
a)” l” Dam in and dam ed forced vibrations C“ : 2mm" 6‘ 3 (0d a," E‘ {2
CRT
5 in X0 . 22v: m, _c_. a Amplitude ratio m X
Fl [k
Phase angle
ca) 2? all tan(_6) m ~——2————~2 = mm)" 60 Transmission ratio
1+ {29" as): Tszggwlk1 +8691 2 1 (1—3;): + (26—533)? Forced vibration (Impulse)
F
q ex.p(—§a)nt)sin(1ll— faint) ma)” 1—? .X: Forced vibration Ste 3 F __
m—r‘l IWEE—Emgwcoﬂ 1— §2wﬂt w?)
k I] _ 9’2
where
tan(6) =— ‘5
1w {'2
Vibration absorbers
4 2
xlkl _ 1_ i _ 5'9, .. 2 m. 2 .. F0 [1 + (0,“) in. ) 1 M]
where a it" a: = 11
m1 m2
W # = ﬂ E w p 0:2
ml kl (012i ...
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 Three '10
 MCGREGOR
 Acceleration, Force, Velocity, Displacement, Sin

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