dom ppt 01
TRANSCRIPT
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Dynamics of Machinery IIME 3015
B.Tech (First Year)Second Semester
Daw Thida Oo Lecturer
Mechanical Engineering Department Yangon Technological University
Lectured by
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CHAPTERS
• Balancing• Forces• Cam Profile• [References:1.Dynamics of Machinery written by
U Kyaw Sein. 2.Theory of Machine (vol. 1) written by N.C Pandya & C.S Shah. 3.Theory of Machines and Mechanism written by Joseph Edward Shigley & John Joseph Uicker, JR]
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BALANCING• The technique of correcting or eliminating
unwanted inertia forces and moments in rotating or reciprocating masses.
• The two equations to determine the amount and location of the correction
F
CCW
M
Reference plane
Σ F = 0 and Σ M = 0
• Upwards force is positive &• Counter clockwise couple is
positive.
Assumptions
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Resultant Effects of Engine• 1. Σ F = 0 & Σ M = 0
– Complete balanced• 2. Σ F = 0 & Σ M ≠ 0
– Unbalanced being due to a couple.
• 3. Σ F ≠ 0 & Σ M = 0– Unbalanced being due to a single resultant force in the
reference plane.
Ref; PlaneF
F
5
Continued
• 4. Σ F ≠ 0 & Σ M ≠ 0– Unbalanced being due to a single resultant force which
locates at a distance z from the reference plane and
FRef; Plane
FRef; Plane z
FMz
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Balancing of Multi-cylinder In-line Engines (Analytical Method)
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Continued• In-line Engine Mechanism
1
n
2
θΦ2
Φn
ω
F1
F2
F3
x x
Reference
Plane
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Continued
nn
nn
1
02cos
nn
1n
nn
1n
nn
1n
nn
1nnnnn
2 2sin2sinlr2cos2cos
lrsinsincoscosr
gWF
• The resultant inertia force,
(Primary Force) (Secondary Force)
• For Primary force Balance,
nn
nn
1
0cos 0sin1
n
nn
n
and
02sin1
n
nn
n
• For Secondary force Balance,
and
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Continued• The resultant moment,
nn
n
nn
n
nn
n
nn
nnnnn x
lrx
lrxxr
gWM
1 1 1 1
2 2sin2sin2cos2cossinsincoscos
(Primary Moment) (Secondary Moment)
• For Primary Moment Balance,
• For Secondary Moment Balance,
nn
nnx
1
0sin
nn
nnx
1
02cos
nn
nnx
1
02sin
nn
nnx
1
0cos and
and
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Example (1)• 2 stroke, 2 cylinder, In-line Engine
– Firing interval = 1802
360
θ
Φ2
1
2 x
ωrad/s
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ContinuedΦ cosΦ sinΦ 2Φ cos2Φ sin2Φ x xcosΦ xsinΦ xcos2Φ xsin2Φ
Φ1=0˚
1 0 0˚ 1 0 0 0 0 0 0
Φ2 =180˚
-1 0 360˚ 1 0 x -x 0 x 0
0 0PrimaryForces
Balanced
2 0Secondary
ForcesUnbalanced
-x 0Primary
MomentsUnbalanced
x 0SecondaryMoments
Unbalanced
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Continued• Secondary Unbalanced Force, 2cos2
22
lr
gW
(Upwards)
• Primary Unbalanced Moment,
cos2rxgW
(CW)
• Secondary Unbalanced Moment, 2cos2
2 xlr
gW
(CCW)
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Arrangement to balance the secondary force (Upwards)
2cos22cos)2(22
22
lr
gWr
gW
bb For balance,
2θ
ω
2ω2θ
2ω
2cos)2( 2b
b rgW
2sin)2( 2b
b rgW
2cos)2( 2b
b rgW
2sin)2( 2b
b rgW
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Arrangement to balance the primary moment (C.W)
coscos 22 rxgWxr
gW
bbb For balance,
cos2b
b rgW
sin2b
b rgW cos2
bb rgW
xb
ω
θ ωω
θ
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Arrangement to balance the secondary moment (C.C.W)
2cos2cos)2(2
22 xlr
gWxr
gW
bbb For balance,
2cos)2( 2b
b rgW
2sin)2( 2b
b rgW 2sin)2( 2
bb rgW
2cos)2( 2b
b rgW
2ω
2ω2θ
2θ
2ω
2ω
ω
xb
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Balancing of multi-cylinder V-Engine
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Continued
• V-Engine Mechanism
Odd no cylinders (1,3,5,..)
2
3
I
II
III
2 FAFB
Even no cylinders (2,4,6,..)
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Continued• The resultant vertical inertia force,
sincossincoscos[cosrgW2F 222V
]2sin2coscos2sinlr2cos2coscos2cos
lr
• The resultant horizontal inertia force,
sinsincoscossin[sinrgW2F 222H
]2sin2sinsin2coslr2cos2sinsin2sin
lr
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Assumptions
+FV
+FH
+MH
C.W
+MV
C.C.W
Reference Plane
x
y
z
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Example-4• 8-cylinder, 4-stroke,V-engine
– Firing Order 1-5-4-2-6-8-7-3– Firing Interval = 90˚– V angle = 90˚
I(1,2)
III(5,6)
IV(7,8)
II(3,4)
IIIV
III
90
8
720
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ContinuedΦ cosΦ sinΦ 2Φ cos2Φ sin2Φ x xcosΦ xsinΦ xcos2Φ xsin2Φ
ΦI=0˚
1 0 0 1 0 0 0 0 0 0
ΦII=90˚
0 1 180 -1 0 x 0 X -x 0
ΦIII=270˚
0 -1 540 -1 0 2x 0 -2x -2x 0
ΦIV=180˚
-1 0 360 1 0 3x -3x 0 3x 0
0 0PrimaryForces
Balanced
0 0 Secondary
ForcesBalanced
-3x -xPrimary
MomentsUnbalanced
0 0SecondaryMomentsBalanced
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Continued
)57.71sin(rxgW10 2
• Primary Unbalanced Horizontal Moment =
• Primary Unbalanced Vertical Moment =(CW)
)43.18sin(10 2 rxgW
)57.71cos(10 2 rxgW
(CCW)
Note- sinα=cos(α - 90˚)
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Continued• Arrangement for
Balancing 57.71sin2b
b rgW
bb2b xr
gW
57.71cos2b
b rgW
For Balance,
71.57
71.57
I(1,2)II(3,4)
III(5,6)IV(7,8)
xb
rxgW 210
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FORCES• To study forces acting on machine members
Statically and Dynamically.• To determine the magnitudes, directions and
locations of forces.• Assumptions.
– A member of a machine composed of all external forces and inertia forces is equilibrium.
– The forces acting on machines having plane motion are for the most part situated in parallel plane.
– The friction is disregarded.– The system will be applied Newton’s Law.
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Static Force Analysis
• Static forces exits if the system is in equilibrium among them, when not running.– For equilibrium, Σ F = 0 and Σ M = 0.– Two forces in equilibrium, (Two force member)
F1
F2
• F1 = - F2 Σ F = 0
• Shared same line of action Σ M = 0
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Continued– Three forces in equilibrium
F2
F1
F3
F1
F2F3
• F1 + F2 + F3 = 0, Σ F = 0
• Have a common point of application, Σ M = 0
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Continued• To find a line of action of unknown force in
three force member having a known force, known line of action of another force but not magnitude and known point acting the last force.
F2
F1
F3
Line of action of F3
OF
F3
F2
F1
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Dynamics Force Analysis
• m = total mass of body concentrated at the centroid, C.G, of body.
• AG = absolute acceleration of the centre of mass of the body.
• IG = mass moment of inertia.• α = angular acceleration of the body.
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Inertia Forces and D’Alembert’s Principle
• ΣF = F1+F2+F3 , the resultant force will not be through the mass centre, and results the unbalanced force system.
• The effect of this unbalanced system is to produce an acceleration, AG, of the centre of mass of the body. ΣF = mAG (1)
• Taking moment about centre of mass of the body results the unbalanced moment system. It causes angular acceleration, α, of the body. ΣM = IG α (2)
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Continued• (a) An unbalanced set of forces on a rigid
body.• (b) The acceleration which result from the
unbalanced forces.
O
y
x
G
∑MG=Iα
∑F=mAG
(b)
F2
∑F
O
F3
F1
y
x
h
Gm
(a)
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Conitued
• From (1) and (2),• ΣF +(- mAG) = 0 and ΣM +(- IG α) = 0• (- mAG) is called inertia force which has the same
line of action as the absolute acceleration AG but is opposite in sense.
• (- IG α) is called inertia torque which is opposite in sense to the angular acceleration α.
• The equations above are known as D’Alembert’s principle.
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Continued
• To describe graphically,
G
G
AmI
h3
3The distance between the forces and couple,
AG
F23
F43+F23
α
3
G
F43
(a)
G
3
h
-mAG
-mAG
-Iα3
+mAG
(b)
3
G
F23
-mAG
h
F43
(c)
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Example - 1
CWsradABA BAt
,/856 23
B
A
GFA
2
3
4O x
y A nBA =533ft/s 2A
t BA=713ft/s
2
AB=888ft/s2 AG=444ft/s2Oa
b G
inAm
IhG
G 35.13
3
-m3AG=30.33lb
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Continued
A
FA
F12
F32
F23
F14
-m3AG
Free body diagrams
h
-mAG
F14
F23
OForce polygon of 3 & 4
F12
F32
FA
Force polygon of 2
Ans, FA=27 lb
Line of action of F23
-m3AG
F14
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