Notes_08_04 1 of 20

Two-mass Equivalent Link

total mass 3C3B3 mmm =+

centroid location BC

BGmm

BCCG

mm 33C3

33B3 ==

check approximate mass moment ( ) ( )2

3C32

3B3APP_3G3G CGmBGmJJ +=≈

(for slender rod ACTUAL_3GAPP_3G J3J = )

B C G3

m3 JG3

B C

G3 m3B

m3C

Notes_08_04 2 of 20

Shaking Force for Slider Crank

assume crank is statically balanced G2 = A constant crank speed 0=θ split link 3 into m3B and m3C assume φ is small Case I - static force analysis for only d’Alembert inertial force P with no friction

( ) θ=φ=µθ+θθ−=+−= sinRsinL02coscosRss)mm(P LR2

4C3

φ−==φ=−=φ

φ+θ= tanPF0FtanPFPF

cos)sin(RPT Y

4on1X

4on1Y

2on1X

2on12on1

Case II - static force analysis for only inertial force 2

B3 Rm ω

θω−=θω−== sinRmFcosRmF0T 2

B3Y

2on12

B3X

2on12on1 Case III - place additional counterbalance mBAL on the crank at radius R and 180º from pin B Superposition = Case I + Case II + Case III

)3sin2sinsin(R)mm(T L2R3

L2R22

4C321

2on1 θ−θ−θθ+−=

( ) ( )smmcosRmmF 4C32

BALB3x

1on2

+−θθ−=

( ) ( ) φ++θθ−= tansmmsinRmmF 4C32

BALB3y

1on2

( ) φ+−= tansmmF 4C3y

1on4

1 1

2

3

4 A=G2

B

φ θ P

T12 G3

C

m3B

m3C+m4

mBAL

Notes_08_04 3 of 20

{ } { } { } ( ) ( )( )( )

θ−θ+θ++θ−

θ=+=sinmm

2coscosmmcosmmRFFF

BALB3

LR

4C3BALB321on41on2SHAKING

Notes_08_04 4 of 20

Model 1 – no additional balancing mass mBAL = 0 Model 2 – unidirectional in-line shaking force

{ } ( ) ( )( )( )

θ−θ+θ++θ−

θ=sinmm

2coscosmmcosmmRF

BALB3

LR

4C3BALB32SHAKING

{ } ( )( )

θ+θ+

θ==0

2coscosmmRFmm L

R4C32

SHAKINGB3BAL

Model 3 – minimize the maximum magnitude of shaking force (approximate)

{ } ( ) ( )( )( )

θ−θ+θ++θ−

θ=sinmm

2coscosmmcosmmRF

BALB3

LR

4C3BALB32SHAKING

{ } ( )( )

θ−−θ++θ

θ=+=sinmmm

2cosmmcosmRFmmm

4C3B3

LR

4C3B32SHAKING4C3BAL

Model 4 – minimize in-line shaking force (approximate)

{ } ( ) ( )( )( )

θ−θ+θ++θ−

θ=sinmm

2coscosmmcosmmRF

BALB3

LR

4C3BALB32SHAKING

{ } ( )

θ−θ

+θ=

++=+=

sin2cos

mmRF

mmmmmm

LR

4C32

SHAKING

4C3B343BAL

Notes_08_04 5 of 20

-40 -20 0 20 40-40

-20

0

20

40mBAL = 0

-40 -20 0 20 40-40

-20

0

20

40mBAL = m3B

-40 -20 0 20 40-40

-20

0

20

40mBAL = m3C + m4

-40 -20 0 20 40-40

-20

0

20

40mBAL = m3 + m4

0 0.2 0.4 0.6 0.8 1 1.2 1.40

5

10

15

20

25

30

35

40

45

Balancing mass [lbm]

Max

imum

sha

king

forc

e [lb

f]

Notes_08_04 6 of 20

% shake.m - single cylinder shaking force Notes_08_04 % HJSIII, 13.03.14 clear % constants d2r = pi / 180; % geometry [inches] R = 0.985; L = 4.33; % crank speed [rad/sec] w = 104.719; % masses [lbm] m3B = 0.351; m3C = 0.111; m4 = 0.781; % crank angle th_deg = (0:360)'; th = th_deg * d2r; % piston acceleration [inch/s/s] sdd = -R*w*w*(cos(th) +R*cos(2*th)/L); % plot four models figure( 1 ) clf res = []; %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Model 1 - mBAL = 0 subplot( 2,2,1 ) mBAL = 0; % shaking force [lbm.in/s/s] * [lbf*s*s / 32.174 lbm.ft] * [ ft / 12 in ] Fx = ( (m3B-mBAL)*R*w*w*cos(th) - (m3C+m4)*sdd ) / 386; % [lbf] Fy = ( (m3B-mBAL)*R*w*w*sin(th) ) / 386; Fs = sqrt( Fx.*Fx + Fy.*Fy ); Fs_max = max( Fs ); res = [ res ; mBAL Fs_max ]; % plot shaking force curve plot( Fx, Fy ) axis square axis( [ -40 40 -40 40 ] ) title( 'mBAL = 0' ) %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Model 2 - mBAL = m3B subplot( 2,2,2 ) mBAL = m3B; % shaking force [lbm.in/s/s] * [lbf*s*s / 32.174 lbm.ft] * [ ft / 12 in ] Fx = ( (m3B-mBAL)*R*w*w*cos(th) - (m3C+m4)*sdd ) / 386; % [lbf] Fy = ( (m3B-mBAL)*R*w*w*sin(th) ) / 386; Fs = sqrt( Fx.*Fx + Fy.*Fy ); Fs_max = max( Fs ); res = [ res ; mBAL Fs_max ]; % plot shaking force curve plot( Fx, Fy ) axis square axis( [ -40 40 -40 40 ] ) title( 'mBAL = m3B' )

Notes_08_04 7 of 20

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Model 3 - mBAL = m3C + m4 subplot( 2,2,3 ) mBAL = m3C + m4; % shaking force [lbm.in/s/s] * [lbf*s*s / 32.174 lbm.ft] * [ ft / 12 in ] Fx = ( (m3B-mBAL)*R*w*w*cos(th) - (m3C+m4)*sdd ) / 386; % [lbf] Fy = ( (m3B-mBAL)*R*w*w*sin(th) ) / 386; Fs = sqrt( Fx.*Fx + Fy.*Fy ); Fs_max = max( Fs ); res = [ res ; mBAL Fs_max ]; % plot shaking force curve plot( Fx, Fy ) axis square axis( [ -40 40 -40 40 ] ) title( 'mBAL = m3C + m4' ) %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Model 4 - mBAL = m3 + m4 subplot( 2,2,4 ) mBAL = m3B + m3C + m4; % shaking force [lbm.in/s/s] * [lbf*s*s / 32.174 lbm.ft] * [ ft / 12 in ] Fx = ( (m3B-mBAL)*R*w*w*cos(th) - (m3C+m4)*sdd ) / 386; % [lbf] Fy = ( (m3B-mBAL)*R*w*w*sin(th) ) / 386; Fs = sqrt( Fx.*Fx + Fy.*Fy ); Fs_max = max( Fs ); res = [ res ; mBAL Fs_max ]; % plot shaking force curve plot( Fx, Fy ) axis square axis( [ -40 40 -40 40 ] ) title( 'mBAL = m3 + m4' ) %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % plot maximum shaking force versus balancing mass keep = []; for mBAL = 0 : 0.01 : (m3B+m3C+m4), % shaking force [lbm.in/s/s] * [lbf*s*s / 32.174 lbm.ft] * [ ft / 12 in ] Fx = ( (m3B-mBAL)*R*w*w*cos(th) - (m3C+m4)*sdd ) / 386; % [lbf] Fy = ( (m3B-mBAL)*R*w*w*sin(th) ) / 386; Fs = sqrt( Fx.*Fx + Fy.*Fy ); Fs_max = max( Fs ); keep = [ keep ; mBAL Fs_max ]; end % plot results figure( 2 ) clf plot( keep(:,1),keep(:,2),'b', res(:,1),res(:,2),'ro' ) axis( [ 0 1.4 0 45 ] ) xlabel( 'Balancing mass [lbm]' ) ylabel( 'Maximum shaking force [lbf]' ) % bottom of shake

Notes_08_04 8 of 20

Model 5 – minimize the maximum magnitude of shaking force (exact) does not work ?????

( ) ( )( ) ( )( ) ( ) )2coscosmm

2coscoscosmmmm2mm(RF2

LR2

4C3

LR

4C3BALB32

BALB3422

SHAKING

θ+θ++

θ+θθ+−+−θ=

( ) ( ) ( ) ( ) 0)2coscoscosmm2mm2(R

mF

LR

4C3BALB342

BAL

2SHAKING =θ+θθ+−−−θ=

∂∂

( )( )θθ+θ++= 2coscoscosmmmm L

R24C3B3BAL

( )( )θθ−θ+θ++= 2

LR3

LR2

4C3B3BAL sincoscoscosmmmm

( )( )θ−θ+θ++= coscoscos2mmmm LR23

LR

4C3B3BAL

( ) 0sinsincos2sincos6coscoscos2 LR2

LR

LR23

LR =θ+θθ−θθ−=θ∂θ−θ+θ∂

( )( )L

R4C3B3BAL 1mmmm0,0sin +++=°=θ=θ

( )( )L

R4C3B3BAL 1mmmm180,0sin −++=°=θ=θ

( ) ( )( )θ−θ+θ++=+±−

=θ coscoscos2mmmm6

611cos L

R23LR

4C3B3BALLR

2LR

% model5.m - min/max shaking force for slider crank % does not work ??????????????? % HJSIII, 13.03.14 clear % geometry [inches] R = 0.985; L = 4.33; % masses [lbm] m3B = 0.351; m3C = 0.111; m4 = 0.781; % cosine solution rho = R / L; c1 = (-1 + sqrt( 1 + 6*rho*rho ) ) /6 /rho; c2 = (-1 - sqrt( 1 + 6*rho*rho ) ) /6 /rho; % values v1 = 2*rho*c1*c1*c1 + c1*c1 - rho*c1; v2 = 2*rho*c2*c2*c2 + c2*c2 - rho*c2;

Notes_08_04 9 of 20

% balancing masses mBAL1 = m3B + (m3C+m4)*v1 mBAL2 = m3B + (m3C+m4)*v2

Notes_08_04 10 of 20

Shaking Force for In-line Two Cylinder Air Compressor

only ABC { }

( )( )

( )

( )

θ−

θ++θ++θ−

θ=

sinmm

2cosmmcosmmcosmm

RF

BALB3

LR

4C3

4C3

BALB3

2SHAKING

both { }

( ) ( )( )( ) ( )( )

( ) ( )( )

( ) ( )( )

°+θ+θ−

°+θ+θ++°+θ+θ++°+θ+θ−

θ=

180sinsinmm

3602cos2cosmm180coscosmm180coscosmm

RF

BALB3

LR

4C3

4C3

BALB3

2SHAKING

( )

( )( ) 0180sincos180cossinsin180sinsin

2cos23602cos2cos0180sinsin180coscoscos180coscos

=°θ+°θ+θ=°+θ+θθ=°+θ+θ

=°θ−°θ+θ=°+θ+θ

both { } ( )

θ+

θ=0

2cosmm2RF 4C3L

R2

SHAKING

shaking moments ( ) X

1on2YSHAKING

Y1on4

Y1on2

XSHAKING FaMFFaM +=+−=

same crank, connecting rods

and pistons

A = G2

B

C

D

E

y in z

x

a

x

B

θΒ = θ θD = θ + 180°

C

y

D

E

A = G2

z out

Notes_08_04 11 of 20

In-line Four Cylinder Engine

Notes_08_04 12 of 20

Shaking Force for In-line Four Cylinder Engine

°+θ=θ=θθ=θ=θ

18032

41

same as two two-cylinder compressors mirrored end to end

{ } ( )

θ+

θ=0

2cosmm4RF 4C3L

R2

SHAKING

shaking moments 0M0M Y

SHAKINGXSHAKING ==

same crank, connecting rods

and pistons

2

x

1 4 3

y in

z

x

1,4

C

y

2,3

E

A = G2

z out

Notes_08_04 13 of 20

In-line Six Cylinder Engine

Notes_08_04 14 of 20

Shaking Force for In-line Six Cylinder Engine

°+θ=θ=θ°+θ=θ=θ

θ=θ=θ

240120

43

52

61

( ) ( )

( ) ( )( ) ( ) 0240sin120sinsin

04802cos2402cos2cos0240cos120coscos

=°+θ+°+θ+θ=°+θ+°+θ+θ

=°+θ+°+θ+θ

{ } 0M0M00

F YSHAKING

XSHAKINGSHAKING ==

=

same crank, connecting rods

and pistons

x

1,6

y

A = G2

z out

2,5

3,4

y in

z

x

1 2 3 4 5 6

Notes_08_04 15 of 20

V8 separate cranks

V8 dual cranks

Notes_08_04 16 of 20

V8 split cranks

V8 with pistons

Notes_08_04 17 of 20

Shaking Force for Four Bar

assume crank is statically balanced G2 = A split link 3 into m3B and m3C assume 3θ is small link 3 becomes two-force member static force analysis with d’Alembert inertial forces

Notes_08_04 18 of 20

ΣM on 4 about D CCW + ( ) ( ) ( )( )2

C32

444G44on3 CDmDGmJCDsinF ++θ=γ

( ) ( )( ) ( )34

2C3

2444G44on3 sinCD/CDmDGmJF θ−θ++θ=

( ) ( )( ) ( ) ( )( ) 34on344C3444

24C344

x1on4 cosFsinCDmDGmcosCDmDGmF θ−θθ++θθ+=

( ) ( )( ) ( ) ( )( ) 34on344C3444

24C344

y1on4 sinFcosCDmDGmsinCDmDGmF θ−θθ+−θθ+=

Notes_08_04 19 of 20

ΣM on 2 about A CCW + ( )( ) 0ABsinFT 322on32_on_1 =θ−θ−

( ) ( )( ) ( )( )34

322C3

2444G42on1 sinCD

sinABCDmDGmJT

θ−θθ−θ

++θ=

( ) ( )( )2

42C3

2444G42on1 CDmDGmJT

θθ

++θ=

from Notes_03_02

44 and θθ from Notes_03_02

( )( ) 32on32

22BALB3

x1on2 cosFcosABmmF θ+θθ−=

( )( ) 32on32

22BALB3

y1on2 sinFsinABmmF θ+θθ−=

Notes_08_04 20 of 20

{ }

++

= y1on4

y1on2

x1on4

x1on2

SHAKING FFFF

F

{ }( )( )

( ) ( )( )( ) ( )( )

θ+θ+

θ−

θθ−θθθθ

=24C344

4C344

22BALB3

442

442SHAKING

CDmDGmCDmDGm

ABmm

sincossincossincos

F