LN 10 - Rotating unbalance PDF

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Rotating unbalance Small irregularities in the distribution of the mass in the rotating component of a machine can produce substantial vibration. A rotating unbalance can be represented as a mass m0 rotating with an angular velocity ω r at a distance e from the center of rotation (Fig. 1). Therefore the x – component of the motion of the mass m0 is: xr = esin ω r t (1) Thus the x-component of the acceleration of the mass m0 is: ax = −eω 2r sin ωr t (2) Therefore the x-component of the force imparted by the structure is: Fx = −m0eω2r sin ωr t In Fig. 1, m is the total mass of the machine and includes the unbalance mass m0.

Fig. 1 Schematic of a rotating balance The equilibrium of forces on mass m yields: m x = −kx − cx − Fx or m x + cx + kx = m0eω 2r sin ω r t The steady-state solution (particular solution) of the equation of motion, Eq. (3) is: x p (t) = X sin(ω r t − θ )

where the amplitude is: X = the phase is: tanθ =

r2 mo e m (1 − r 2 )2 + ( 2ζ r)2

2ζ r 1− r 2

ωr . ωn The amplitude can be no-dimensionalized as:

and the frequency ratio is: r =

(3)

mX mo e

=

r2 2 2

( )

(1 − r ) + 2ζ r

2

.

A plot of the dimensionless amplitude versus the frequency ratio r is presented in Fig. 2.

Fig. 2 Dimensionless amplitude vs. frequency ratio From Fig. 2 it can be seen that: • The maximum deflection is less than or equal to 1 for any ζ >1 • For ζ...