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CHAPTER 4: HARMONIC RESPONSE WITH A SINGLE DEGREE OF FREEDOM

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CHAPTER 4: HARMONIC RESPONSE WITH A SINGLE DEGREE OF FREEDOM
Mechanical Vibration by Palm CHAPTER 4: HARMONIC RESPONSE WITH A SINGLE DEGREE OF FREEDOM Instructor: Dr Simin Nasseri, SPSU © Copyright, 2010 Based on a lecture from Brown university (Division of engineering)

Types of Forcing: External Forcing Base Excitation Rotor Excitation
All of these situations are of practical interest. Some subtle but important distinctions to consider, so we will look at each. But the strategy is simple: derive Equation of Motion and put into the “Standard Form” 2

Rotating unbalance Rotor excitation 04_04_01 04_04_01.jpg

Unbalanced. More mass one side than other.
Rotor Excitation Rotating machinery not always perfectly balanced e.g. your car’s wheels; Imperfectly machined rotating disks; Turbine engine with cracked turbine blades w Unbalanced. More mass one side than other. Leads to a sinusoidally-varying force k c q=wt m M w e x(t) Engineering Model: Effective unbalanced mass m Effective eccentricity e In a machine of total mass M 4

Rotor Excitation q=wt w  -kx -cv -FMm FmM R1 R2
x(t) M is total mass INCLUDING eccentric mass Non-rotating part of the machine has position x, Rotating piece has position: FBDs for the two masses: -kx -cv -FMm FmM R1 R2 Equations of motion for the two masses (vertical direction only): Standard Form but with 5

Steady State Solution (same as before)
Standard Form but with Steady State Solution (same as before) Same form as relative base excitation…….

High frequencies: imbalance force grows at same rate as magnification factor decreases, leading to amplitude ratio =1 (not usually a problem) Low frequencies: not moving fast enough to generate a force large enough to drive the mass

Force Transmission to Base
k c q=wt m w e x(t) With vibrating machinery, forces exerted on the supporting structures can become large near resonance. Equipment is thus constructed on isolating mounts (springs and dashpots – dashpots suppress resonance) k c kx cv What is the force transmitted? Draw FBD for base structure: Position, velocity are always 90o out of phase, so transmitted force is Since max of over all q is we have Maximum transmitted force: 8

And we found earlier Force required to stretch the spring by a distance me/M

i.e. m = 10 kg Here, “Total mass of the device” = 10 kg
Example: “Total mass of the device” = 10 kg i.e. m = 10 kg Determine the two possible values of the equivalent spring stiffness k for the mounting to permit the amplitude of the force transmitted to the fixed mounting due to the imbalance to be 1500 N at a speed of 1800 rpm Here,

i.e. we can choose a stiff spring and run the machine BELOW the natural freq
or we can choose a soft spring and run the machine ABOVE the natural freq

04_04_01tbl Summary: =e R=e (eccentricity)
04_04_01tbl.jpg Force transmitted to the base:

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