Notes_08_02
1 of 4
Matrix Dynamic Analysis for Four Bar
The four bar linkage shown below operates in a vertical plane. Each link is a uniform bar with 2
cm by 2 cm square cross-section stainless steel. Assume that the masses of the bearings and the
effects of friction are negligible. Do not neglect the effects of gravity.
2 = 45 deg
3 = 20 deg
4 = 117.4 deg
2 = 20 rad/s CW
3 = 12.82 rad/s CCW
4 = 6.20 rad/s CW
2 = 100 rad/s/s CCW
3 = 39.6 rad/s/s CW
4 = 482.5 rad/s/s CCW
m2 = 0.248 kg
m3 = 0.372 kg
m4 = 0.341 kg
JG2’ = 1.405 kg.cm2
JG3’ = 4.588 kg.cm2
JG4’ = 3.552 kg.cm2
 = 7.75 g/cm3
AG2 = -1414.2 - j 848.5 cps2
AG3 = -3672.5 - j 2260.9 cps2
AG4 = -2262.4 - j 1412.4 cps2
VG2 = 56.57 - j 56.57 cps
VG3 = 86.83 - j 40.86 cps
VG4 = 30.27 - j 15.71 cps
C
Y
AD = 22 cm
AB = 8 cm
BC = 12 cm
CD = 11 cm

B
T


D
A
X
F34y
F34x
F43
F32x
RB2/G2
T12
G2
x
F12
A
W2
F43x
RC3/G3
B
RA2/G2
C
y
F32y
RB3/G3
C
F23x
RC4/G4
G4
G3
B
F23y
W3
RD4/G4
W4
F14x
F12y
D
F14y
Notes_08_02
2 of 4
F on 2 right +
F12x + F32x = m2 AG2x
F on 2 up +
F12y + F32y + W2 = m2 AG2y
- F12x rA2/G2y + F12y rA2/G2x - F32x rB2/G2y + F32y rB2/G2x + T12 = JG2’ 2
M on 2 about G2 CCW +
F23x + F43x = m3 AG3x
F23y + F43y + W3 = m2 AG3y
- F23x rB3/G3y + F23y rB3/G3x - F43x rC3/G3y + F43y rC3/G3x = JG3’ 3
F on 3 right +
F on 3 up +
M on 3 about G3 CCW +
F on 4 up +
F34x + F14x = m4 AG4x
F34y + F14y + W4 = m4 AG4y
M on 4 about G4 CCW +
- F34x rC4/G4y + F34y rC4/G4x - F14x rD4/G4y + F14y rD4/G4x = JG4’ 4
F on 4 right +
 1
 0

 rAy 2 / G 2

 0
 0

 0
 0

 0
 0

0
1
0
0
0
0
0
1
0
1
0
0
0
0
 rAx 2 / G 2
rBy2 / G 2
 rBx2 / G 2
0
0
0
0
0
1
0
1
0
0
0
1
0
1
0
0
x
B3 / G 3
y
C3 / G 3
0
0
0
0
0
r
y
B3 / G 3
r
r
r
x
C3 / G 3
0
0
0
1
0
1
0
0
0
0
0
1
0
1
0
0
0
 rCy4 / G 4
rCx4 / G 4
 rDy 4 / G 4
rDx 4 / G 4
W2 = -j 2.433 N
W3 = -j 3.649 N
W4 = -j 3.345 N
m2 AG2 = -3.507 - j 2.104 N
m3 AG3 = -13.662 - j 8.411 N
m4 AG4 = -7.715 - j 4.816 N
0  F12x   m 2 A Gx 2 

  
0  F12y  m 2 A Gy 2  W2 

1 F23x   J 'G 2  2

 y  
x
0 F23   m 3 A G 3 
  

0 F34x    m 3 A Gy 3  W3 


0 F34y   J 'G 3  3
  

0  F14x   m 4 A Gx 4 
  

0  F14y  m 4 A Gy 4  W4 

0 T12   J 'G 4  4
JG2’ 2 = 1.405 N.cm
JG3’ 3 = -1.817 N.cm
JG4’ 4 = 17.138 N.cm
0
1
0
0
0
0
0
 1
 0
1
0
1
0
0
0
0

2.828  2.828 2.828  2.828
0
0
0
0

0
1
0
1
0
0
0
 0
 0
0
0
1
0
1
0
0

0
2.052  5.638 2.052  5.638
0
0
 0
 0
0
0
0
1
0
1
0

0
0
0
0
1
0
1
 0
 0
0
0
0
 4.883  2.531 4.883 2.531

0  F12x    3.507 
 
0  F12y   0.329 
1 F23x   1.405 
  

0 F23y   13.662
  

0 F34x     4.762 

0 F34x    1.817 
  

0  F14x    7.715 
  

0  F14y    1.471 
0 T12   17.138 
Notes_08_02
 F12x   - 22.236 N 
 y 

 F12   - 6.221 N 
F23x   - 18.729 N 
 y 

F23   - 6.550 N 
 x 

F34    - 5.067 N 
F x   - 1.788 N 
 34  

 F14x   - 2.648 N 
 y 

 F14    0.317 N 
T12   81.135 N.cm
3 of 4
Notes_08_02
4 of 4
Matrix Dynamic Analysis for Slider Crank
B
T12
3
2
A
G3
C


4
P
1
1
F32y
T12 B
G2 = A

F23x
F32x
2
+ direction for
VC to right
F23y
F34y

B

F43y
G3
3
C
F12y
F on 2 up +
M on 2 about A CCW +
F34x
F14x

P
F14y
F12x + F32x = m2 AG2x = 0
F12y + F32y = m2 AG2y = 0
- (F32x sin) AB + (F32y cos) AB + T12 = JG2 2
F23x + F43x = m3 AG3x
F23y + F43y = m3 AG3y
F on 3 right +
F on 3 up +
M on 3 about G3 CCW +
- (F23x sin) BG3 - (F23y cos) BG3
+ (F43x sin) CG3 + (F43y cos) CG3 = JG3 3
F on 4 right +
F14x + F34x + P = m4 AG4x
F14y + F34y = m4 AG4y = 0
F14x = -  abs(F14y) sign(VC)
F on 4 up +
friction
1
0

0

0
0

0
0

0
0

4
F43x
F12x
F on 2 right +
C
0
1
0
0
0
0
1
0
1
0
0
0
0
 AB sin 
 AB cos 
0
0
0
0
1
0
1
0
0
0
0
1
0
1
0
0  BG 3 sin   BG 3 cos   CG 3 sin   CG 3 cos  0
0
0
0
1
0
1
0
0
0
0
1
0
0
0
0
0
0
1
0  F12x  
0

 y 


0
0  F12 
0


0
1 F23x   J G 2  2 
  

0
0 F23y   m 3 A Gx 3 
  

0
0 F34x    m 3 A Gy 3 

0
0 F34y   J G 3  3 
  

0
0  F14x  m 4 A Gx 4  P 
  

1
0  F14y  
0


  sign (VC ) 0 T12  
0
0