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IIII.. Examination Paper SEMESTER: SECOND SEMESTER EXAMINATIONS 2009 UNIT: ENBS12 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
GRAPH PAPER MM  TWO (2) PER STUDENT
EQUATION SHEETS  THREE (3) PAGES ATTACHED EXAMINATION MATERIAL SUPPLIED BY THE STUDENT: WRITING IMPLEMENTS
GRAPH TOOLS (COMPASSES, RULER, PROTRACTOR) CALCULATOR  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 SE TI N A AND B  FOUR (4) QUESTIONS ONLY ARE TO BE ATTEMPTED INCLUDING NO MORE
THAN TWO (2) QUESTIONS FROM EACH SECTION ATTEMPT EACH SECTION IN A SEPARATE EXAMINATION BOOKLET ALL QUESTIONS ARE OF EQUAL VALUE Queensland University of Technology GUT Kelvin Grove SECTION A QUESTION 1 A general fourbar linkage is shown in Figure. The Link 2 is the input link, and the Link 4 is
the output link. Link lengths [cm], angles [degrees], and input angular velocity [rad/sec] are
given for this mechanism in the table below. Only the open circuit is considered. B Figure (Figure is not to scale) Link4
81° 00
O
B
U3
LII
0
EN
3E:
0%
Boa (a) Prove the degree of freedom of this mechanism is one. (2 marks)
(b) Calculate the transmission angle. (2 marks)
(0) Check the Grashof condition. (2 marks)
((1) Calculate angular velocity ratio, mv. (4 marks) (6) If input force is applied to point A (perpendicular to link 2) and the output force is applied to point B (perpendicular to link 4), determine the Mechanical Advantage, m A.
(2 marks) (f) Determine (you can use either the graphical method or the analytical method): (i) angular velocity (03 for link 3; (3 marks) (ii) velocities (magnitude and direction) of the pin joint A and the pin joint B. (5 marks) ENB312T1.092 Cont ...... QUESTION 2 A general fourbar linkage conﬁguration and terminology are shown in Figure. Link lengths and the input angle 62 are given as:
Linklzd=6cm Link22a=2cm
Link32b=7cm Link4zc=9cm 62=30° \ g /\
3/ 4 Crossed b Figure (Figure is not to scale) (a) Accurately draw the linkage (for both open and crossed circuits) to scale on the graph
paper provided. (6 marks) (b) Graphically ﬁnd all possible solutions (both open and crossed circuits) for 93 and 04.
(4 marks) (0) Analytically determine 63 and 64 for the linkage (for open circuit only). (10 marks) ENB312T1.092 Cont ...... QUESTION 3 The SI units are used for all following parameters. A single rotating link with the length: L2
= 10, and mass: m2 =1. Its CG is on the line of centres at the 5 point (the middle point of the link). Its mass moment of inertia about its CG is I G = 2. An external force, F p, of 20 at 150°
is applied at point P. Its kinematic data are: link angle: 62 = 30° ; angular acceleration: a = 1
(clockwise); and the acceleration at CG: aG = 100 at 270°. Figure (Figure is not to scale) (a) Develop the Freebody Diagram (4 marks)
(b) For this link, ﬁnd force F12 at pin joint 02 and the driving torque T12 (12 marks) (C) Find the shaking force and shaking torque (moment) acting on the ground ﬁxed pivot
(4 marks) ENB312T1.092 Cont ...... SECTION B QUESTION 4
(20 marks) Two springs with different stiffness support a rigid rod and two identical masses. Determine
the two natural frequencies of motion for the twomass system. QUESTION 5
(20 marks) For the rotor system shown in the ﬁgure, determine the fundamental frequency and mode
shape of the system in tensional vibration. Neglect the inertia of the shaft. 11 = 60 1(ng ,12 =12 kg.m2, 13 = 9 1(ng and 14 = 80 1(ng K1 = 4000 KN.m/rad, K2 = 2650 KN.m/rad and K3 = 2500 KN.m/rad I
1
 fllllllllll m 1;. ENB312T1.092 Cont ...... QUESTION 6
(20 marks) (a) Deﬁne static and dynamic coupling. (b) The double pendulum system shown in the ﬁgure is constrained to oscillate in one plane.
Determine the two natural frequencies and mode shapes. Given that m1 is equal to m 2. END OF PAPER ENB312T1.092 ENB312 Part IDynamics of Mechanisms: Equation Sheet Gruebler’s equation
M=3 (LI )—2J 1J2 where: M: DOF or mobility
L: number of links
J1: Number of 1 DOF(full) joint
J2: Number of 2 DOF (half) 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 e Euler identity for a vector
eijg = c036: jsin0 where j denotes imaginary unit Position vector loop equation for a fourbar linkage
R2 +R3 —R4 —R1 =0 If 61 for grounded link 1 is 0°. The two scalar equations
are bcosol3 = —acos€2 +ccost94 +d
bsinél3 =—asin6l2 +csint94 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: _ + 2 _
a“ z {WAMH
.2 2A where
A = c0502 —K1 —K2 c03192 +K3
B = —2 sin 02
C = K1 «(K2 +1)c056l2 +K3 2 _ 2 2 2
Kl =1; K2 =1; K3 =ujf.+_d a 0 Zac Position of a Point on a link
RA = aej”2 = a(cost92 +jsin 02)
RPA = Pejw’wl) = P[COS(03 + 63) + jsin(493 + 63 )1
RP = RA + RM losition anal sis or slidercrank Position vector loop equation
R2 —R3 —R4 —R1 = 0
If 61 =0°, the two scalar equations are acosé’2 —bcost93 —ccost94 —d =0
asinH2 —bsinl93 —csin6?4 =0 ENB312T1.092 the solution for the above equations is: 03 = arcsin [M] d = acosa2 —bcos«93 e
Angular Velocity Ratio woutput (04 Cl) input 02 Mechanical Advantage
60 r input input m A = 
woutput routput Velocity vector loop equation a
a(112+113—R4—121)=0 If 01 for grounded link 1 is 0 ‘f The two scalar equations
are —an2 sin 62 —ba)3 sin 63 + ca)4 sin 64 = 0
{an}2 cos 62 + ba)3 cos 63 — ca)4 cos 64 = 0
The solution for the above equations is:
a) = aa)2 _sin(04 —t92) 3 b sin(l93 _ o4) _ aa)2 'sin(t92 —63)
c anal sis or slidercrank Velocity vector loop equation
6
6—t(R2 —R3 —R4 —R1)= 0 If 01 =0°, the two scalar equations are
—aco2 sin 02 + bag sin 93 — d = o { aa)2 cos (92 —ba)3 cos 63 = 0 The solution for the above equations is: a) _ aa)2 c05192
3 __.
cosé’3 d = —aco2 Sln 62 + bco3 Sln 03 Vector acceleration anal sis or a ourbar linka_
Acceleration vector loop equation 62 57(R2 +R3 —R4 —R1)= 0
If 01 for grounded link 1 is 0°. The two scalar equations
are e —aot2 sin 02 — acoz2 cos 62 — bat3 sin (93
—ba)32 cos 63 + 00:4 sin (94 + ca): cos 64 = O act2 cos 62 —— aw; sin 02 + bat3 cos 63 2  2 . _
—ba)3 sm (93 —coz4 cos 04 +ca)4 $11104 — 0 The solution for the above equations is:
CD — AF a3 = —
AE — BD _ CE — BF AE — BD
where
A = csin 04 B=bsin03 _ ' 2 2 2
C — act2 Sln 02 + aa)2 cos (92 + ba)3 cos 63 — ca)4 cos 04 a4 D = ccost94
E =bc0503 _ 2  2  2 
F — act2 cos ()2 — aa)2 5m (92 —ba)3 $11103 + ca)4 51n64 Vector acceleration anal sis or slidercrank Velocity vector loop equation at2
If 01 =0°, the two scalar equations are —aoc2 sin 62 — 616022 cos 62 + ba3 sin 63
+bao32 cos 63 — d = 0 2  ..
aoc2 cos 62 — aa)2 811102 — bot3 cos 63 +ba)32 sin 63 = 0 The solution for the above equations is: a : a0;2 cos 62 — acoz2 sin (92 + bu): sin 03
3 b cos 03 —a0t2 sin 62 — acuz2 cos 92 + bot3 sin (93 + bcof cos 03 ENB312T1.092 (ii) Force anal sis or sin_ le link in Pure Rotation Vector equation
ZF = FP +F12 = "1230 ZT=T12 +(R12 xF12)+(RP xFP)= [Ga
Scalar equations
FPx' + 52:: =. ”’2an FPy + FlZy = mZaGy 712 +(R12xF12y “R12yF12x)+(RPxFPy ‘RPyFPx) = [Ga Matrixform
1 0 0 Fle "12an — FPx
O 1 0 Fi2y = m2 610), — FPy
_R12y R12x 1 712 [Ga—RPXFPy +RPyFPX Balancin a rotatin s stem Static balance:
Vector equation:
mbRb = —mlR1—m2R2 Scalar equations
”112be = _(m1R1x + m2R2x)
mbRby = _(mlR1y + mZRZy) Dynamic balance:
Vector equation: mARA + mBRB = —m]R1—m2R2 m3R3 mBRBlB = —m]Rll1 — m2R212 —m3R3l3
Scalar equations mARAx +mBRBx = _(mlRlx “"2sz +m3R3x) mARAy + mBRBy = —(m1R]y +m2R2y + m3R3y)
m1 Rlxll + mszxlz + m3 R3x13 mBRBx = — l
B mlRlyll + mszylz + m3R3yl3 B (iii) ENB312 Part IIVibrations: Equation Sheet Forced vibration
mié + kx = F (t) Solution of eguation . F sin a) t
x = Acos cont + Bsm wnt + ——1—,—"—2
”1(a); — a) )
. F em
x = Acos w"! + Bsm cant + —'2——2—
"1(0),.  w )
Steady state displacement
X _ 1
Fl / k l  ~3—
Steady state displacement. for unbalanced force
. __ moea)2
A _ 2 2
112(0),, — a) ) Magniﬁcation ratio mX MR = —— moe Transmission ratio
1 TR = ,
1 _ g; 012 Rectangular step forcing function . F
x= Acoswnt+ Bsm w,,r+—]:”— Linearly increasing forcing function . Ct
x = Acoswnt +Bsmwnt+7 Exponentially decaying forcing function ”I"
Foe ., x= Acoswnt +Bsin w,,r+
ma’+k Combination of forcing functions ' F 6.1"
x=Acoswnt+Bsmwut+—" l— I
k 1+;— for initial conditions 1(0)=o and i(0)=0 then k —k(l_:,_:) wn w" ENB312T1.092 F0 F0 (1 a . _,,
x—— _. — —cosa)nt+sma),,t +e 'l Dam in and dam ed forced vibrations cc, =2ma)" (=5. (0,, =60 1—{2 F] x:——————————————
k1/(1—gi)2+(2§:—,)2 Amplitude ratio _ X
F./k
we
ca) 24’ 7.3";
tan(0) = ——2—2— = (1)2
171(0)" — a) ) (1' 0.?)
Tran smission ratio
1+ (24' :3): TR = £de +c'2w2 =
l (1—95? + (2%)] Forced vibration (Impulse) —F—Texp(—§wnt)sin(‘ll  {2 (out) ma)" 1—; x: Forted vibration (Step) x—— ——
k /]_{2
where '
tan(0)= 4' 1—;2
Vibration absorbers _ _w_ 3 xlkl _ 1  (“’22 _~Fo—_[1+:—f(w%)2][1(%n)2]%] where 2 _  2 _ k2
mu“— 22‘
mI m2
ﬁzz"; Lawn);
m. k. wf. ...
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