MENG 371, Chapter 8
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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=IgT12
+(R12F12)+(RPFP)
• Breaking down into
components:
SFx=m2ax=FPx+F12x
SFy=m2ay=FPy+F12y
ST=IgT12
(R12xF12y-R12yF12x)
+(RPxFPy-RPyFPx)
Single Link in Pure Rotation
SFx=m2ax=FPx+F12x
SFy=m2ay=FPy+F12y
ST=IgT12(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