MENG 371, Chapter 8 - The American University in Cairo

MENG 371, Chapter 8 - The American University in Cairo

MENG 372 Chapter 11 Dynamic Force Analysis All figures taken from Design of Machinery, 3rd ed. Robert Norton 2003 Solution using Newtons Law Newtons Law: M G I G F maG For planar motion we have: F x max F y ma y 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 I G 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: F=m2a=FP+F12 T=IgT12 +(R12F12)+(RPFP) Breaking down into components: Fx=m2ax=FPx+F12x Fy=m2ay=FPy+F12y T=Ig R12xF12y-R12yF12x) +RPxFPy-RPyFPx) Single Link in Pure Rotation Fx=m2ax=FPx+F12x Fy=m2ay=FPy+F12y T=IgR12xF12y-R12yF12x) +RPxFPy-RPyFPx) Putting into a matrix format 1 0 0 F12 x m2 aGx FPx

1 0 F12 y m2 aG y FPy 0 R T I R F R F R 1 12 x Px Py Py Px 12 y 12 G 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 R12 x 0 0 0 0 0 0 1 0 R32 y

1 0 R23 y 0 0 0 0 1 R32 x 0 1 R23x 0 0 0 0 0 0 1 0 R43 y 1 0 R34 y 0 0 0 0 1 R43x 0 1 R43x 0 0 0 0 0

0 1 0 R14 y 0 0 0 0 0 0 0 1 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: F14 x 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 1 0 0 0 0 R12 y 0 1 R12 x 0 0 R32 y

1 1 R32 x 0 0 0 1 0 0 0 0 0 0 0 0 R23 y 1 R23x 0 R43 y 1 R43x 0 0 0 0 0 0 0

0 1 0 0 1 0 F12 x m2 aG2 x 0 0 F12 y m2 aG2 y I G2 2 0 1 F32 x ma 0 0 F32 y 3 G3 x m3aG3 y 0 0 F43x I G3 3 0 0 F43 y 0 F14 y m4 aG4 x FPx FPy 1 0 T12 0 Inverted Crank Slider (error in the book) Free Body Diagrams: T43

T34=-T43 Links 3 and 4 T43 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 F43x 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 0

0 0 0 0 1 R12 x 0 R32 y 1 R32 x 0 0 0 0 0 0 0 0 0 1 0 sin 3 0 0

0 0 0 1 0 cos 3 0 0 0 R23y R23x R43 1 0 0 0 0 0 sin 3 0 1

0 0 0 0 0 cos 3 0 1 0 0 0 sin 3 R34 y cos 3 R43x 1 R14 y R14x F43x F43n sin 3 F43n sin 3 F43 y 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 m3 aG3 y 43n 0 T I 43 G3 3 0 F14x m4 aG4 x 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

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