dom ppt 01

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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

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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

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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