Chapter 4 Linear Differential Operators

146 CHAPTER 4. LINEAR DIFFERENTIAL OPERATORS
The kinetic energy of the whirling shaft is
T = 1
L
ρ{ ˙x
2 −L
2 + ˙y 2 }dz,
and the strain energy due to bending is
V [x, y] = 1
L
γ{(x
2 −L
′′ ) 2 + (y ′′ ) 2 } dz.
a) Write down the Lagrangian, and from it obtain the equations of motion
for the shaft.
b) Seek whirling-mode solutions of the equations of motion in the form
x(z, t) = ψ(z) cos ωt,
y(z, t) = ψ(z) sin ωt.
Show that this quest requires the solution of the eigenvalue problem
γ d
ρ
4ψ dz4 = ω2 nψ, ψ′ (−L) = ψ(−L) = ψ ′ (L) = ψ(L) = 0.
c) Show that the critical frequencies are given in terms of the solutions ξn
to the transcendental equation
as
tanh ξn = ± tan ξn, (⋆)
ωn =
γ
ρ
2 ξn
,
L
Show that the plus sign in ⋆ applies to odd parity modes, where ψ(z) =
−ψ(−z), and the minus sign to even parity modes where ψ(z) = ψ(−z).
Whirling, we conclude, occurs at the frequencies of the natural transverse
vibration modes of the elastic shaft. These modes are excited by slight imbalances
that have negligeable effect except when the shaft is being rotated at
the resonant frequency.
Insight into adjoint boundary conditions for an ODE can be obtained by
thinking about how we would impose these boundary conditions in a numerical
solution. The next exercise problem this.