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

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External ForcingBase Excitation Types of Forcing: 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”

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3 Rotating unbalance Rotor excitation

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

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Rotor Excitation M is total mass INCLUDING eccentric mass Non-rotating part of the machine has position x, Rotating piece has position: Equations of motion for the two masses (vertical direction only): -kx-cv -F Mm FmMFmM R1R1 R2R2 FBDs for the two masses: Standard Form but with k c =t=t m M e x(t )

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

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

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Force Transmission to Base 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) What is the force transmitted? Position, velocity are always 90 o out of phase, so transmitted force is Draw FBD for base structure: Since max of over all is we have Maximum transmitted force: k c =t=t m m e x(t) k c kxkx cvcv

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

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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,

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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

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12 Summary: =e R=e (eccentricity) Force transmitted to the base:

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