Design for Manufacture

6. Dynamics of Machine

Design for Manufacture

Grass cutting

Scissoring - Using fixed and moving blades, a cylinder mower cuts grass like a pair of scissors.

Impact cutting - A rotary mower rotates about a vertical axis with the blade spinning at high speed relying on impact to cutwith the blade spinning at high speed relying on impact to cut the grass, resulting in a rougher cut.

High speed blade tip approaches the grass first.

Cylinder lawn mower

Rotary lawnmower

Rotary lawnmower

VV=ω r

Impactp

Grass and air flow

Grass and air flow

Grass exit

Torque

Fconstant

process Polytropicnpv =

x

0

0 vv

pp

n

x

x⎟⎟⎠

⎞⎜⎜⎝

⎛=

β0

0

Force

pvvpthen

n

xx ⎟⎟

⎠

⎞⎜⎜⎝

⎛=

l0

0 ApvvApF

Forcen

xx ⎟⎟

⎠

⎞⎜⎜⎝

⎛==

θr

θθ sinsin:

ArpFrTTorque

x==

Fθsin

..

00 ArvpT

eip

n

x

⎟⎟⎠

⎞⎜⎜⎝

⎛=

Design for Manufacture

0 vp

x⎟⎠

⎜⎝

Torque

Fθβ )cos1()1(cos rlx −+−=

x

( )θβ )cos1()1(cos00 rlAvAxvvx −+−−=−=

βθβθβ

θβ

sin1cos,sinsin

sinsin2

lr

lr

rl

⎟⎠⎞

⎜⎝⎛−==

=

l

0 AvF

Forcell

n

⎟⎟⎞

⎜⎜⎛

⎠⎝

θr

( )θβ )cos1()1(cos 00

0

T

AprlAv

F ⎟⎟⎠

⎜⎜⎝ −+−−

=

F( ) θθβ

sin)cos1()1(cos 0

0

0 ArprlAv

vT

Torquen

⎟⎟⎠

⎞⎜⎜⎝

⎛−+−−

= ( )θβ )cos1()1(cos0 rlAv ⎠⎝ +

Design for Manufacture

Forces on piston and bearing

gmxmmApF srprpx +++⎟⎠⎞

⎜⎝⎛ ++= &&

31

forces 3 includemay bearingscrank on the acting Force

Fx

⎠⎝

( ) ( )+++−=

+−=

−+−=

θθθθββββ

θθββ

θβ

22 sincossincossinsin

)cos1()1(cos

rlxrlx

rlx

&&&&&&&&

&&&

β

( ) ( )

=

θ

θ

ββββ

sinsin

since

lrβ

l

====

=

θωβθβωθωθ

θβθβ

.0,coscos,,

coscos

lrt

lr

&&&&

&&

θr

⎟⎞

⎜⎛

⎟⎟⎞

⎜⎜⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛ −+=

βθωωθθβω

θβ

βθβθβθθβθβ

β

2

2

sincoscossincos

cossincossincos

coscos

r

lr &

&&&&&&

F⎞⎛

⎟⎟⎟⎟⎟

⎠⎜⎜⎜⎜⎜

⎝

⎟⎟⎠

⎜⎜⎝

−=

β

βθωωβ

θβω

2

2cos

sincoscos

sincosl

lr

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

ββθ

βθω 3

22

cossincos

cossin

lr

lr

Design for Manufacture

Vibration Source

Combustion| Exhaust Induction |Compression

revolution 1 revolution 2revolution 1 revolution 2

10

Vibration SourceRotating masses are balanced

Reciprocating masses cannot be balanced

Piston inertia force:

⎞⎛

massrodconnectingmasspistonml

rrmmaf cc

3/1

2coscos22

2

+−

⎟⎟⎠

⎞⎜⎜⎝

⎛+==

θωθω

massrodconnectingmasspistonm ___3/1_ +−

θωθωθ 0t cc === &&&

θθθθθθ

θθαα

θα

coscoscoscoscos

sinsin

rrrrl

rl=

=

&&&

&&

θθθ

θ

θθαθθα

sincos

sin1

coscoscos

2222

2 rlr

lrl

rl

r−

=

⎟⎠⎞

⎜⎝⎛−

==&

αθ coscos)( lrlrx −−+=⎠⎝

θωθωθ 0t cc === &&&

θαθα sinsinsinsinlrrl =→=

θθθ

αθθαθθαα

222 sincos

coscoscoscos

rlr

lrrl ==→=

&&&&&

αθ

θα

coscos)(

sincos

lrlrx

rll

−−+=

−

θθθθθθ

θθθθθααθθ

222

2

222 sin22sinsinsin

sincossinsinsin

rlrr

lr

rlrlrlrx

−+=

−+=+=

&&

&&&&&

θθ sin2sin rllrl −−

( )rl

rrlrrrx sin2

cossin22sin2cos22sinsincossin

222

22222

22 −

−−+−

++= θθθθθθθθθθθ

θθθθ

&&&&&

&&&&&

( )r

rlrrx

2sin

sin2cossin

222

222 −++=

θθ

θθθθθ

&& ( )

rlrl

rrlrr

sinsin22sinsin2cos2

2cos 222

222

22222

2

−−

+−+=

θθ

θθθθθθθ

&

&

( )( )

( )rl

r

rl

rrsin4

2sin

sin

2coscos 2/32/1222

224

222

222

−+

−+=

θ

θθ

θ

θθθθ&&

&

( ) ( )

ei

rlrl

..

sin4sin

⎤⎡

θθ

( ) ( )( )rln

tn

t

tn

ttrxc

c

c

cc /

sin8

4cos1

sin

2coscos 2/32/12222

2 =⎥⎥⎦

⎤

⎢⎢⎣

⎡

−

−+

−+=

ω

ω

ω

ωωω&&

Piston first and second inertia exciting forces

tcωθ =

tn

rmtrmf cc

cc

c

ωωωω 2coscos2

2 +=

forceinertiaorderFirst :___trm cc ωω cos2

trmforceinertiaorderSecond

c ωω 2cos

:___2

tn cω2cos

Noise sources

• Exhaust noise is the main noise source.Exhaust noise is the main noise source.

• Structural borne noise: vibrating frequency is greater than 100hz.

If main shaft speed 3600 rpm = 60hz its basic harmonic does not generate audible structural borne noise noise but 2nd and higher harmonics g gdo.

Flywheel

A flywheel is a rotating mechanical device that is used to store rotational energy.

Flywheels have a significant moment of inertia and thus resist changes in rotational speedresist changes in rotational speed.

Flywheel

:inertiaofMoment

gyrationofRadius:

2

kmkI =

gyrationofRadius:k

Quantum Power Engine flywheel

4 stroke engine torque

Design for Manufacture 19

Work done per cycle

π4720o

ππ

4Tcycleper donework4720

mean ⋅==

Design for Manufacture 20

Green area = blue area

Energy fluctuation

JoulIω )(21Energy 2=

WattTωθ )(TPower

2

== &

:speedofn fluctuatiooft coefficien

)(

meanωωωφ

speedmean speedin variation

p

minmax −== mean

:energyofn fluctuatiooft coefficien

p

( )III ωωωωβ 2

121

gy

2min

2max

2min

2max −

=−

=

Design for Manufacture 21

meanTππβ

8T4 mean

Energy fluctuation

( )2min

2maxI ωωβ −

=

22

88

mean

mean

TI

T

ωωβπ

πβ

=

=

minmax

Since

ωω −

( )( ) 2minmaxminmax

2min

2max 22 meanmeanmean φωφωωωωωωωω =≈−+=−

2

4Therefore

meanTIφβπ

= 2meanφω

Design for Manufacture 22

Example of flywheel

A 3.5 horse power, running at 2500rpm mean speed, 4 stoke single cylinder IC engine requires a flywheel with a fluctuation of speed ±2% and fl t ti f 5% It di f ti i 100 Fi d th ffluctuation of energy 5%. Its radius of gyration is 100mm. Find the mass of the flywheel. Power: 3 5 ×745=2600 (Watts)Power: 3.5 ×745 2600 (Watts)Mean speed: 2500rpm/60×2π=261.8rad/s Mean Torque:2600/261.8=9.93(Nm)

kgmTImean

mean )(0023.08.26104.0

93.905.044 222 =

××××

==π

φωβπ

( )kgI 2300023.0m

:flywheel of mass

( )kgk

23.01.0

m 22 ===

Design for Manufacture 23