ME 302 DYNAMICS OF MACHINERY

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Presentation transcript:

ME 302 DYNAMICS OF MACHINERY Dynamic Force Analysis II Dr. Sadettin KAPUCU © 2007 Sadettin Kapucu

Example The slender bar of mass m is released form rest in the horizontal position as shown. At that instant, determine the force exerted on the bar by the support A. Freebody Diagram Equations Of Motion

Example Freebody Diagram Equations Of Motion The slender bar of mass m is released form rest in the horizontal position as shown. At that instant, determine the force exerted on the bar by the support A. Freebody Diagram Equations Of Motion Kinematics of the slender rod

At the instant bar is released, its angular velocity Example The slender bar of mass m is released form rest in the horizontal position as shown. At that instant, determine the force exerted on the bar by the support A. Kinematics of the slender rod r At the instant bar is released, its angular velocity

Example Equations Of Motion Freebody Diagram The slender bar of mass m is released form rest in the horizontal position as shown. At that instant, determine the force exerted on the bar by the support A. Equations Of Motion Freebody Diagram

D’Alembert’s Principle D’Alembert’s principle permits the reduction of a problem in dynamics to one in statics. This is accomplished by introducing a fictitious force equal in magnitude to the product of the mass of the body and its acceleration, and directed opposite to the acceleration. The result is a condition of kinetic equilibrium. m, I a SF CG a CG SF m, I a -Ia fictitious force and torque -ma The meaning of the equation; i.e. indication of a dynamic case still holds true, but equation, having zero on right hand side becomes very easy to solve, like that in a “static force analysis” problem.

Solution of a Dynamic Problem Using D’Alembert’s Principle Do an acceleration analysis and calculate the linear acceleration of the mass centers of each moving link. Also calculate the angular acceleration of each moving link. Masses and centroidal inertias of each moving link must be known beforehand. Add one fictitious force on each moving body equal to the mass of that body times the acceleration of its mass center, direction opposite to its acceleration, applied directly onto the center of gravity, apart from the already existing real forces. Add fictitious torque on each moving body equal to the centroidal inertia of that body times its angular acceleration, direction or sense opposite to that of acceleration apart from the already existing real torques. Solve statically.

Example 1 In the figure, a double- slider mechanism working in horizontal plane is shown. The slider at B is moving rightward with a constant velocity of 1 m/sec. Calculate the amount of force on this mechanism in the given kinematic state. AB=10 cm, AG3=BG3=5 cm, q=60o m2=m4=0.5 kg, m3=0.8 kg, I3=0.01 kg.m2

Example 1 cont In the figure, a double- slider mechanism working in horizontal plane is shown. The slider at B is moving rightward with a constant velocity of 1 m/sec. Calculate the amount of force on this mechanism in the given kinematic state.

Example 1 cont G3

Example 1 cont G3 D’Alembert forces and moments

Example 1 cont

Example 1 cont y + x

Example 1 cont h h

Example 1 cont h h

Example 1 cont y + x h

Example 2 Crank AB of the mechanism shown is balanced such that the mass center is at A. Mass center of the link CD is at its mid point. At the given instant, link 4 is translating rightward with constant velocity of 5 m/sec. Calculate the amount of motor torque required on crank AB to keep at the given kinematics state. AC=10 cm, CB=BD=4 cm, AF=2 cm CG3=4 cm, m2=m3=m4=5 kg, I2=I3=I4=0.05 kg-m2

Example 2 cont. Crank AB of the mechanism shown is balanced such that the mass center is at A. Mass center of the link CD is at its mid point. At the given instant, link 4 is translating rightward with constant velocity of 5 m/sec. Calculate the amount of motor torque required on crank AB to keep at the given kinematics state. AC=10 cm, CB=BD=4 cm, AF=2 cm CG3=4 cm, m2=m3=m4=5 kg, I2=I3=I4=0.05 kg-m2 5 m/s

Example 2 cont. Crank AB of the mechanism shown is balanced such that the mass center is at A. Mass center of the link CD is at its mid point. At the given instant, link 4 is translating rightward with constant velocity of 5 m/sec. Calculate the amount of motor torque required on crank AB to keep at the given kinematics state. AC=10 cm, CB=BD=4 cm, AF=2 cm CG3=5 cm, m2=m3=m4=5 kg, I2=I3=I4=0.05 kg-m2

Example 2 cont. Crank AB of the mechanism shown is balanced such that the mass center is at A. Mass center of the link CD is at its mid point. At the given instant, link 4 is translating rightward with constant velocity of 5 m/sec. Calculate the amount of motor torque required on crank AB to keep at the given kinematics state. AC=10 cm, CB=BD=4 cm, AF=2 cm CG3=5 cm, m2=m3=m4=5 kg, I2=I3=I4=0.05 kg-m2 a3 a2 B

Example 2 cont t D’Alembert forces and moments -m3aB -I3a3 -I2a2 B m2=m3=m4=5 kg, I2=I3=I4=0.05 kg-m2 -m3aB -I3a3 t -I2a2 B

Example 2 cont -m3aB -I3a3 t -I2a2 4000 N t 1000 Nm 504 Nm tA4

Example 2 cont 4000 N t 1000 Nm 504 Nm tA4

Example 2 cont 4000 N t 1000 Nm 504 Nm tA4

Example 2 cont 4000 N t 1000 Nm 504 Nm tA4

504 Nm t 4000 N 1000 Nm tA4 Example 2 cont