ME 321 Kinematics and Dynamics of Machines Steve

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ME 321 Kinematics and Dynamics of Machines Steve Lambert Mechanical Engineering, U of Waterloo

ME 321 Kinematics and Dynamics of Machines Steve Lambert Mechanical Engineering, U of Waterloo 9/17/2020

Rotating Unbalance r t e m 0 x m c 9/17/2020 k m 0

Rotating Unbalance r t e m 0 x m c 9/17/2020 k m 0 Fr xr x

Rotating Unbalance Equations of Motion: For the unbalanced mass: For the net mass: 9/17/2020

Rotating Unbalance Equations of Motion: For the unbalanced mass: For the net mass: 9/17/2020

Rotating Unbalance Assume a particular solution of the form: By analogy to earlier solutions:

Rotating Unbalance Assume a particular solution of the form: By analogy to earlier solutions: and for r = r/ n 9/17/2020

Rotating Unbalance 9/17/2020

Rotating Unbalance 9/17/2020

Rotating Unbalance Example 6. 5: A machine has a rotating unbalance, which results in

Rotating Unbalance Example 6. 5: A machine has a rotating unbalance, which results in a maximum steady-state deflection of 1 cm at resonance. Based on a measurement of the free decay of the system, it is estimated that the damping ratio is = 0. 1. The total mass of the system is 100 kg, and it is estimated, from manufacturing considerations, that the magnitude of the mass unbalance is 5 kg. Estimate the effective radius, e of the unbalance, and the amount of mass that would have to be added to the machine to reduce the vibration to 1 mm. 9/17/2020

Base Excitation x(t) m c k y(t) 9/17/2020

Base Excitation x(t) m c k y(t) 9/17/2020

Base Excitation Assume a base excitation of the form: This gives the following governing

Base Excitation Assume a base excitation of the form: This gives the following governing differential equation: Or, in normalized form: 9/17/2020

Base Excitation There are two particular solutions. One due to the force of the

Base Excitation There are two particular solutions. One due to the force of the damper: And one due to the force of the spring: 9/17/2020

Base Excitation These two solutions have the same frequency, and can be combined as

Base Excitation These two solutions have the same frequency, and can be combined as follows: 9/17/2020

Base Excitation Displacement transmissibility: 9/17/2020

Base Excitation Displacement transmissibility: 9/17/2020

Base Excitation The force acting on the mass through the damper and spring is:

Base Excitation The force acting on the mass through the damper and spring is: or: 9/17/2020

Base Excitation The transmitted force can be rewritten as: with: or, in normalized form:

Base Excitation The transmitted force can be rewritten as: with: or, in normalized form: 9/17/2020

Base Excitation Force transmissibility: 9/17/2020

Base Excitation Force transmissibility: 9/17/2020

Base Excitation Force transmissibility (dashed line) and displacement transmissibility (solid line) for = 0.

Base Excitation Force transmissibility (dashed line) and displacement transmissibility (solid line) for = 0. 05 9/17/2020

Base Excitation Example 6. 6: Consider a single degree-of-freedom model of an automobile suspension

Base Excitation Example 6. 6: Consider a single degree-of-freedom model of an automobile suspension travelling over a rough road. The road is modeled as providing a base excitation, in m, of The equivalent stiffness of the suspension is k = 4 105 N/m, a damping coefficient, c = 40 103 kg/s, and a mass of 1000 kg. Determine the steady-state amplitude and displacement of the automobile mass. 9/17/2020