1 ME 302 DYNAMICS OF MACHINERY Dynamic Force Analysis Dr. Sadettin KAPUCU © 2007 Sadettin Kapucu.
Dynamic Force Analysis of Machinery
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Transcript of Dynamic Force Analysis of Machinery
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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; Plane
F
F
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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
F
Mz
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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 2sin2sinl
r2cos2cos
l
rsinsincoscosr
g
WF
•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
l
rx
l
rxxr
g
WM
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 0
Primary
Forces
Balanced
2 0
Secondary
Forces
Unbalanced
-x 0
Primary
Moments
Unbalanced
x 0
Secondary
Moments
Unbalanced
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Continued• Secondary Unbalanced Force, 2cos2
22
l
r
g
W (Upwards)
• Primary Unbalanced Moment, cos2rxg
W (CW)
• Secondary Unbalanced Moment, 2cos2
2 xl
r
g
W (CCW)
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Arrangement to balance the secondary force (Upwards)
2cos22cos)2(22
22
l
r
g
Wr
g
Wb
b For balance,
2θ
ω
2ω
2θ
2ω
2cos)2( 2b
b rg
W
2sin)2( 2b
b rg
W
2cos)2( 2b
b rg
W
2sin)2( 2b
b rg
W
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Arrangement to balance the primary moment (C.W)
coscos 22 rxg
Wxr
g
Wbb
b For balance,
cos2b
b rg
W sin2
bb rg
W cos2b
b rg
W
xb
ω
θ ωω
θ
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Arrangement to balance the secondary moment (C.C.W)
2cos2cos)2(2
22 xl
r
g
Wxr
g
Wbb
b For balance,
2cos)2( 2b
b rg
W 2sin)2( 2
bb rg
W 2sin)2( 2b
b rg
W
2cos)2( 2b
b rg
W
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[cosrg
W2F 222V
]2sin2coscos2sinl
r2cos2coscos2cos
l
r
•The resultant horizontal inertia force,
sinsincoscossin[sinrg
W2F 222H
]2sin2sinsin2cosl
r2cos2sinsin2sin
l
r
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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 0
Primary
Forces
Balanced
0 0
Secondary
Forces
Balanced
-3x -x
Primary
Moments
Unbalanced
0 0
Secondary
Moments
Balanced
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Continued
)57.71sin(rxg
W10 2
•Primary Unbalanced Horizontal Moment =
• Primary Unbalanced Vertical Moment =
(CW)
)43.18sin(10 2 rxg
W
)57.71cos(10 2 rxg
W
(CCW)
Note- sinα=cos(α - 90˚)
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Continued
• Arrangement for Balancing 57.71sin2
bb rg
W
bb2b xr
g
W
57.71cos2b
b rg
W
For Balance,
71.57
71.57
I(1,2)II(3,4)
III(5,6)IV(7,8)
xb
rxg
W 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
Am
Ih
3
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
CWsradAB
A BAt
,/856 23
B
A
GFA
2
3
4O x
y A nBA =533ft/s 2A
tBA=713ft/s
2
AB=888ft/s2 AG=444ft/s2Oa
b G
inAm
Ih
G
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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