Notes_08_04
1 of 20
Two-mass Equivalent Link
m3 JG3
B
G3
C
m3B
G3
m3C
C
B
total mass
centroid location
m 3B  m 3C  m 3
m 3B  m 3
check approximate mass moment
CG 3
BC
m 3C  m 3
BG 3
BC
J G 3  J G 3 _ APP  m 3B BG 3   m 3C CG 3 
(for slender rod J G 3 _ APP  3 J G 3 _ ACTUAL )
2
2
Notes_08_04
2 of 20
Shaking Force for Slider Crank
B m3B
T12
2
A=G2
3
G3


C m3B+m4
4 P
1
1
mBAL
assume crank is statically balanced G2 = A
split link 3 into m3B and m3C
constant crank speed   0
assume  is small
Case I - static force analysis for only d’Alembert inertial force P with no friction
P  (m 3C  m 4 )s
T1on 2 
s  R 2 cos   RL cos 2
P R sin(   )
cos 
F1Xon 2  P
F1Yon 2  P tan 
0
F1Xon 4  0
Case II - static force analysis for only inertial force m 3B R
T1on 2  0
F1Xon 2  m 3B R 2 cos 
L sin   R sin 
F1Yon 4  P tan 
2
F1Yon 2  m 3B R 2 sin 
Case III - place additional counterbalance mBAL on the crank at radius R and 180º from pin B
Superposition = Case I + Case II + Case III
T1on 2   12 (m 3C  m 4 )R 2  2 ( 2RL sin   sin 2  32RL sin 3)
F2xon1  m 3B  m BAL R 2 cos   m 3C  m 4 s
F2yon1  m 3B  m BAL R 2 sin   m 3C  m 4 s tan 
F4yon1  m 3C  m 4 s tan 
Notes_08_04
m  m BAL  cos   m 3C  m 4 cos   RL cos 2
FSHAKING   F2on1   F4on1   R 2  3B

m 3B  m BAL sin 


3 of 20
Notes_08_04
4 of 20
Model 1 – no additional balancing mass
mBAL = 0
Model 2 – unidirectional in-line shaking force
m  m BAL  cos   m 3C  m 4 cos   RL cos 2
FSHAKING   R 2  3B

m 3B  m BAL sin 


m BAL  m 3B
m  m 4 cos   RL cos 2
FSHAKING   R 2  3C


0

Model 3 – minimize the maximum magnitude of shaking force (approximate)
m  m BAL  cos   m 3C  m 4 cos   RL cos 2
FSHAKING   R 2  3B

m 3B  m BAL sin 


m BAL  m 3C  m 4
m 3B cos   m 3C  m 4  RL cos 2
m 3B  m 3C  m 4 sin  

FSHAKING   R 2 
Model 4 – minimize in-line shaking force (approximate)
m  m BAL  cos   m 3C  m 4 cos   RL cos 2
FSHAKING   R 2  3B

m 3B  m BAL sin 


m BAL  m 3  m 4  m 3B  m 3C  m 4
 R cos 2

  sin  
FSHAKING   R 2 m 3C  m 4  L
Notes_08_04
5 of 20
mBAL = 0
mBAL = m3B
40
40
20
20
0
0
-20
-20
-40
-40
-20
0
20
40
mBAL = m3C + m4
40
-40
-40
20
0
0
-20
-20
-40
-40
-20
0
20
40
0
20
40
mBAL = m3 + m4
40
20
-20
-40
-40
-20
0
20
40
45
40
Maximum shaking force [lbf]
35
30
25
20
15
10
5
0
0
0.2
0.4
0.6
0.8
Balancing mass [lbm]
1
1.2
1.4
Notes_08_04
% 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' )
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Model 3 - mBAL = m3C + m4
subplot( 2,2,3 )
6 of 20
Notes_08_04
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
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Notes_08_04
8 of 20
Model 5 – minimize the maximum magnitude of shaking force (exact) does not work ?????
2
FSHAKING
 R 2  4 ( m 3B  m BAL   2m 3B  m BAL m 3C  m 4  cos cos   RL cos 2
2
 m 3C  m 4  cos   RL cos 2 )
2
2


2
 FSHAKING
 R 2  4 (  2m 3B  m BAL   2m 3C  m 4  cos cos   RL cos 2 )  0
m BAL

m BAL  m 3B  m 3C  m 4  cos 2   RL cos  cos 2


m BAL  m 3B  m 3C  m 4  cos 2   RL cos 3   RL cos  sin 2 

m BAL  m 3B  m 3C  m 4  2 RL cos 3   cos 2   RL cos 




 2 RL cos 3   cos 2   RL cos     6 RL cos 2  sin   2 cos  sin   RL sin   0
m BAL  m 3B  m 3C  m 4 1  RL 
sin   0,   0
m BAL  m 3B  m 3C  m 4 1  RL 
sin   0,   180
 1  1  6 RL 
2
cos  
6
R
L
% 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;
% balancing masses
mBAL1 = m3B + (m3C+m4)*v1

m BAL  m 3B  m 3C  m 4  2 RL cos 3   cos 2   RL cos 

Notes_08_04
mBAL2 = m3B + (m3C+m4)*v2
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Notes_08_04
10 of 20
Shaking Force for In-line Two Cylinder Air Compressor
x
x
C
C
same crank,
connecting rods
and pistons
a
E
E
B
B
y in
z out
z
y
  
D   + 180°
only ABC
both
A = G2
D
A = G2
D
 m 3B  m BAL  cos  
  m  m  cos  
3C
4


2

FSHAKING   R  m 3C  m 4  RL cos 2




 m 3B  m BAL sin  
 m 3B  m BAL cos   cos  180 
  m  m cos   cos  180 
3C
4


2
R

FSHAKING   R  m 3C  m 4  L cos 2  cos2  360




 m 3B  m BAL sin   sin   180 
cos   cos  180  cos   cos  cos180  sin  sin 180  0
cos 2  cos2  360  2 cos 2
sin   sin   180  sin   sin  cos180  cos  sin 180  0
both
2 RL m 3C  m 4  cos 2

0


FSHAKING   R 2 
shaking moments

X
M SHAKING
 a F2Yon1  F4Yon1

Y
M SHAKING
 a F2Xon1
Notes_08_04
In-line Four Cylinder Engine
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Notes_08_04
12 of 20
Shaking Force for In-line Four Cylinder Engine
x
x
y in
C
same crank,
connecting rods
and pistons
E
1,4
z out
y
A = G2
z
2,3
1
1   4  
 2   3    180
same as two two-cylinder compressors mirrored end to end
4 RL m 3C  m 4  cos 2

0


FSHAKING   R 2 
shaking moments
X
M SHAKING
0
Y
M SHAKING
0
2
3
4
Notes_08_04
In-line Six Cylinder Engine
13 of 20
Notes_08_04
14 of 20
Shaking Force for In-line Six Cylinder Engine
x
x
y in
same crank,
connecting rods
and pistons
z out
1,6
z
y
3,4
A = G2
2,5
1
1   6  
 2   5    120
 3   4    240
cos   cos  120  cos  240  0
cos 2  cos2  240  cos2  480  0
sin   sin   120  sin   240  0
FSHAKING  
0

0
X
M SHAKING
0
Y
M SHAKING
0
2
3
4
5
6
Notes_08_04
V8 separate cranks
V8 dual cranks
15 of 20
Notes_08_04
V8 split cranks
V8 with pistons
16 of 20
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  3 is small
static force analysis with d’Alembert inertial forces
link 3 becomes two-force member
Notes_08_04
M on 4 about D CCW +

F3on4 sin  CD  4 J G4  m4 DG 4 2  m3C CD2 

F3on4   4 J G 4  m 4 DG 4   m3C CD / CD sin 4  3 
2
2
F4xon1  m 4 DG 4   m 3C CD 24 cos  4  m 4 DG 4   m 3C CD 4 sin  4  F3on 4 cos  3
F4yon1  m 4 DG 4   m 3C CD 24 sin  4  m 4 DG 4   m 3C CD 4 cos  4  F3on 4 sin  3
18 of 20
Notes_08_04
M on 2 about A CCW +

T1 _ on _ 2  F3on 2 sin 2  3 AB  0
T1on 2   4 J G 4  m 4 DG 4   m 3C CD

2
2
T1on 2  4 J G 4  m 4 DG 4   m3C CD
2
2
AB sin 
 CD
sin 
 

4
 3 
4  3 
2
from Notes_03_02
2
 4 and 4 from Notes_03_02
F2xon1  m 3B  m BAL AB 22 cos  2  F3on 2 cos  3
F2yon1  m 3B  m BAL AB 22 sin  2  F3on 2 sin  3
19 of 20
Notes_08_04
F2xon1  F4xon1 

y
y
F2on1  F4on1 
FSHAKING   
cos  2
 sin  2
FSHAKING   
sin  4
 cos  4
 m 3B  m BAL AB  22 
cos  4  

m 4 DG 4   m 3C CD 4 

sin  4  
2
m 4 DG 4   m 3C CD 4 
20 of 20