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Unformatted text preview: - 1.06∠ − 179.16° mm -or- 1.06∠ − 3.075rad mm
Therefore, the time domain response of the system to the given forcing function, f (t ) , is expressed by*: f (t ) = 20cos(20π t ) N x(t ) = 1.06cos(20π t − 3.075)mm * TODO: Need to show equivalence of two sided FRF solution. -4406/16/06 12:25 PM Lecture Notes Mechanical Vibrations I #9 - Rotating Unbalance
For many situations, the source of excitation is an unbalance in a rotating machine part. This internal (self) excitation is a function of the operation of the mechanism. However, the techniques developed are applicable to this condition, as well. By defining the position of the unbalance mass ( me ) as x(t ) + e(t ) and the mass of the system moving in translation only as m − me , we can write the equation of motion for this representative single degree-of-freedom system ω e me as: x(t) me ( x(t ) + e (t )) + ( m − me ) x(t ) + cx(t ) + k x(t ) = 0 Rearranging terms yields: mx(t ) + cx(t ) + k x(t ) = − mee (t ) m k c Comparing this solution with the previously developed equation of motion shows that the expression mee (t ) is equivalent in form to the original external forcing function f (t ) . As such we can use the same harmonic function assumed solution form to convert this to a steady-state solution. By using both, x(t ) = Xe st , x(t ) = sXe st & x(t ) = s 2 Xe st and e(t ) = Ee st , e(t ) = sEe st & e (t ) = s 2 Ee st where both X & E are complex scalars. Substituting, the above solution forms into the differential equation of motion produces a solution that must be valid for every value of s. This results in the following Laplace Domain solution. ms 2 Xe st + csXe st + kXe st = − me s 2 Ee st By collecting common terms, the expression reduces to: ( ms 2 + cs + k ) Xe st = − me Es 2e st Again, comparison with the Laplace Domain steady-state solution shows that F ≡ − me Es 2 . Calculation of the effective Transfer Function yields: Lecture Notes -45- 06/16/06 12:25 PM Mechanical Vibrations I H (s) = X −s2 ( s) = 2 me E ms + cs + k Or expressed in the Frequency Domain as:
X ω2 H (ω ) = (ω ) = −ω 2 m + jω c + k me E In the Frequency Domain, the effective forcing function is F ≡ me Eω 2 . As can be seen from the form, the magnitude of the force is linearly proportional to both the unbalance mass and eccentricity, however, it is proportional to the square of the rotational frequency ω . Because of this, the unbalance is normally expressed in mass times length units ( mee ) and hence the above frequency response form (displacement / unbalance).
M agnitude 6 M agnitude
5 0 -5 Om ega [rad/s ec] -10 -1 -0.5 S igm a [rad/s ec ] 0.5 0 1 5 4 3 2 1 0 -10 4 2 0 10 -8 -6 -4 -2 0 2 Om ega [rad/sec] 4 6 8 10 4 3
2 P hase [rad] 1
0 P hase [rad]
5 0 -5 Om ega [rad/s ec] -10 -1 -0.5 S igm a [rad/s ec ] 0.5 0 1 0 -2 -1 -4 10 -2 -3 -4 -10 -8 -6 -4 -2 0 2 Om ega [rad/sec ] 4 6 8 10 Lecture Notes -46- 06/16/06 12:25 PM Mechanical Vibrations I Rota...
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This note was uploaded on 09/29/2013 for the course MECHANICAL ME taught by Professor Regalla during the Fall '11 term at Birla Institute of Technology & Science, Pilani - Hyderabad.
- Fall '11