Page 59 - Singiresu S. Rao-Mechanical Vibrations in SI Units, Global Edition-Pearson (2017)

56 Chapter 1 Fundamentals oF Vibration
For small values of x, the higher-order derivative terms can be neglected to obtain
dF
F + F = F1x*2 + 2 1 x2 (1.5)
dx x*
Since F = F1x*2, we can express F as
F = k x (1.6)

where k is the linearized spring constant at x* given by
dF
k = 2 (1.7)
dx x*
We may use Eq. (1.6) for simplicity, but sometimes the error involved in the approximation
may be very large.

equivalent linearized spring Constant
example 1.2
A precision milling machine, weighing 5000 N, is supported on a rubber mount. The force-deflection
relationship of the rubber mount is given by
F = 375x + 0.1x 3 (E.1)
where the force (F) and the deflection (x) are measured in newtons and millimeters, respectively.
Determine the equivalent linearized spring constant of the rubber mount at its static equilibrium position.

Solution: The static equilibrium position of the rubber mount 1x*2, under the weight of the milling
machine, can be determined from Eq. (E.1):
3
5000 = 375x* + 0.11x*2
or
3
0.11x*2 + 375x* - 5000 = 0 (E.2)
The roots of the cubic equation, (E.2), can be found (e.g., using the function roots in MATLAB) as
x* = 12.77, -6.38 + 62.2i, and -6.38 - 62.2i
The static equilibrium position of the rubber mount is given by the real root of Eq. (E.2):
x* = 12.77 mm. The equivalent linear spring constant of the rubber mount at its static equilibrium
position can be determined using Eq. (1.7):
dF
2
2
k eq = 2 = 375 + 0.31x*2 = 375 + 0.3112.772 = 423.9 N>mm
dx x*
Note: The equivalent linear spring constant, k eq = 423.9 N>mm, predicts the static deflection of the
milling machine as
F 5000
x = = = 11.79 mm
k eq 423.9
which is slightly different from the true value of 12.77 mm. The error is due to the truncation of the
higher-order derivative terms in Eq. (1.4).
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