Transcript MENG 371, Chapter 8

MENG 372
Chapter 11
Dynamic Force Analysis
All figures taken from Design of Machinery, 3rd ed. Robert Norton 2003
Solution using Newton’s Law
• Newton’s Law:
 F  ma
G
M
G
 IG
• For planar motion we have:
F
x
 max
F
y
 may
M
G
 I g
Center of Percussion (from Ch.10)
• The center of percussion (P) is a point on a body
which, when struck with a force, will have associated
with it another point called the center of rotation (R)
at which there will be a zero reaction force
F  maG Fl p  IG
aG  F / m
  Fl p / I G
ai / G  ri
ai  aG  ai / G
ai/G
R
ai
aG
P
Single Link in Pure Rotation
• From free body diagram:
SF=m2a=FP+F12
ST=IgT12
+(R12F12)+(RPFP)
• Breaking down into
components:
SFx=m2ax=FPx+F12x
SFy=m2ay=FPy+F12y
ST=IgT12
(R12xF12y-R12yF12x)
+(RPxFPy-RPyFPx)
Single Link in Pure Rotation
SFx=m2ax=FPx+F12x
SFy=m2ay=FPy+F12y
ST=IgT12(R12xF12y-R12yF12x)
+(RPxFPy-RPyFPx)
• Putting into a matrix format
 1

 0
 R
 12 y
0
1
R12x
0  F12x  
m2 aGx  FPx



0  F12 y   
m2 aGy  FPy
1  T12   I G  RPx FPy  RPy FPx


(






Force Analysis of a Fourbar Linkage
• Free Body Diagrams
Links 2 and 3
Link 2
F12 x  F32 x  m2 aG2 x
F12 y  F32 y  m2 aG2 y
(

T12  R12 x F12 y  R12 y F12 x
(

 R32 x F32 y  R32 y F32 x  I G2 2
Link 3 (F23=-F32)
F43x  F32 x  FPx  m3aG3 x
F43 y  F32 y  FPy  m3aG3 y
(R F  R F 
 (R F  R F 
 (R F  R F   I 
43x 43 y
23x 32 y
Px Py
43 y 43x
23 y 32 x
Py Px
G3 3
Link 4
• F34=-F43
F14 x  F43x  m4 aG4 x
F14 y  F43 y  m4 aG4 y
(
T4  R14 x F14 y  R14 y F14 x
(


 R43x F43 y  R43 y F43x  I G4 4
In One Matrix Equation
• We have 9 equations and 9 unknowns
 1
 0

 R12 y

 0
 0

 0
 0

 0
 0

0
1
0
0
0
0
0
1
0
1
0
0
0
0
R12 x
 R32 y
R32 x
0
0
0
0
0
1
0
0
0
0
1
0
0
1
0
1
0
0
0
R23 y
 R23x
 R43 y
0
0
0
0
0
1
R43x
0
1
0
0
0
0
0
1
0
1
0
0
0
R34 y
 R43x
 R14 y
R14 x
0  F12  
m2 aG2 x

x


0  F12  
m
a
2
G
2y

 y 
1  F32  

I G2 2
x



0  F32  
m
a

F
3 G3 x
Px

 y

0  F43   
m3aG3 y  FPy
x


 
0  F43   I G3 3  RPx FPy  RPy FPx 
y



0 F14  
m4 aG4 x
 x  

0  F14  
m4 aG4 y

y

0  T12  
I


T
G
4
4
4


Crank Slider
Free Body Diagrams:
Crank Slider
• For Link 4:
F14x  F43x  FPx  m4 aG4 x
F14 y  F43 y  FPy  0
F14 x   F14 y
• 8 equations, 8 unknowns
In One Matrix Equation
• We have 8 equations and 8 unknowns
 1
 0

 R12 y

 0
 0

 0
 0

 0
0
1
1
0
0
1
0
0
0
0
R12 x
0
0
0
0
0
 R32 y
1
0
R23 y
0
0
R32 x
0
1
 R23x
0
0
0
1
0
 R43 y
1
0
0
0
1
R43x
0
1
0  F12 x   m2 aG2 x 


0  F12 y   m2 aG2 y 
0 1  F32 x   I G2 2 

  ma

0 0  F32 y  
3 G3 x


m3aG3 y 
0 0  F43x 

 

0 0  F43 y   I G3 3 
  0  F14 y  m4 aG4 x  FPx 

 

 FPy

1 0  T12  
0
0
Inverted Crank Slider
(error in the book)
Free Body Diagrams:
T43
T34=-T43
T43
Links 3 and 4
• Link 3 (F23=-F32)
F43x  F32 x  m3aG3 x
F43 y  F32 y  m3aG3 y
(R F
 (R F
43x 43 y
 R43 y F43x
23x 32 y


 R23 y F32 x  T43  I G3 3
• Link 4(F34=-F43)
F14 x  F43x  m4 aG4 x
F14 y  F43 y  m4 aG4 y
(
 T43  R14 x F14 y  R14 y F14 x
(


 R43x F43 y  R43 y F43x  I G4 4
T34=-T43
Other equations for F43
• We know the direction of F43n
F43t   F43 n
F43 x   F43 n sin  3  F43 n cos 3  F43 n ( sin  3   cos 3 
F43 y  F43 n cos 3  F43 n sin  3  F43 n (cos 3   sin  3 
F43n
F43t
3
T43
Matrix equation with no friction
• 9 equations, 9 unknowns:
 1
 0

  R12 y

 0
 0

 0

 0
 0

 0

0
1
R12x
1
0
 R32 y
0
1
R32x
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
R23y
0
1
 R23x
 sin 3
0
R43
0
cos 3
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
sin 3
0
 sin 3 R34 y  cos 3 R43x
0
 cos 3
1
1
0
 R14 y
0
1
R14 x
F43x   F43n sin 3  F43n (  sin 3 
F43y  F43n cos 3  F43n ( cos 3 
ma
0   F12x   2 G2 x 


0   F12 y   m2 aG2 y 


1   F32   I G2  2 
 x  
m3 aG3 x 
0   F32 y 


 

0  F    m3aG3 y 
 43n


0  T   I  
  43   G3 3 
0   F14x  m4 aG
4x 

 

0  F  m a 
 14 y
 4 G4 y 


0 
  T12   I G  4 
 4 
Shaking Forces and Shaking Torque
• Shaking Force: sum of forces acting on the
ground frame
FS=F21+F41
• Shaking Torque (Ts): reaction torque felt by the
ground.
Ts=T21=-T12
T21