Dynamics of Machines - Part II - IFS.pdf

Demanding (λ 2 + 2ξωnλ + ω 2 n) = 0, because Ce λt = 0 in eq.(28), one gets two values of λ, or
two roots for the equation (λ 2 + 2ξωnλ + ω 2 n) = 0:
λ1 = −ξωn − ωn 1 − ξ2 · i
λ2 = −ξωn + ωn 1 − ξ2 · i
It is important to highlight that all analyzes carried out here will be concentrated
in cases where ξ < 1 (sub-critically damped system). Cases where ξ = 1 or ξ > 1
are called critically or super-critically damped systems respectively. In these cases
no problem related to amplification of vibration amplitudes can be found while
crossing resonances. Using the roots λ1 and λ2 the homogenous solution is:
yh(t) = C1e λ1t + C2e λ2t
where C1 and C2 are defined as a function of the initial condition of the movement when t = 0:
• initial displacement y(0) = yini
• initial movement ˙y(0) = vini
It is important to point out that these constants have to be calculated by using the general
solution, which will be presented later.
Permanent Solution and Steady-State Analysis – The permanent solution of a linear
differential equation is also called steady-state solution. The complete differential equation is
achieved when the right side of the equation is completed with the excitation (see eq.(30)), or
in other words, when static or dynamic forces act on the system. All analyzes and conclusions
obtained from the permanent solution are called steady-state analyzes, and provide information
about the dynamical behavior of the system while excitation forces are introduced into the
system. Let us introduce an excitation force which value oscillates in time with frequency ω
[rad/s]:
¨y(t) + 2ξωn ˙y(t) + ω 2 ny(t) = f
m eiωt
(29)
(30)
yh(t) = Ae iωt , (assumption) (31)
˙yh(t) = jωAe iωt
¨yh(t) = (jω) 2 Ae iωt = −(ω) 2 Ae iωt
The assumption (31) and its derivatives are introduced into the differential equation (30), leading
to:
− (ω) 2 Ae iωt + 2ξωn(jωAe jωt ) + ω 2 nAe iωt = f
m eiωt
⇕
(−ω 2 + ω 2 n + i2ξωnω)A = f
m
11
(32)