Dynamics of Machines - Part II - IFS.pdf

zini = U c + A ⇒ c = U −1 {(zini − A)} (68)
Summarizing, below is the analytical solution of a second order differential equation, which is
responsible for describing the displacements and velocities of the coordinates y1(t) and y2(t) in
time domain, as a function of the excitation force and the initial condition of displacement and
velocity:
z(t) = C1u1e λ1t + C2u2e λ2t + C3u3e λ3t + C4u4e λ4t + Ae iωt
where
λ1 = −ξ1ωn1 − ωn1 1 − ξ2 1 · i
and u1
λ2 = −ξ1ωn1 + ωn1 1 − ξ2 1 · i
and u2
λ3 = −ξ2ωn2 − ωn2 1 − ξ2 2 · i
and u3
λ4 = −ξ2ωn2 + ωn2 1 − ξ2 2 · i and u4
⎧
⎪⎨
⎪⎩
C1
C2
C3
C4
A = [jωA + B] −1 f
⎫
⎪⎬
= [ u1 u2 u3 u4 ]
⎪⎭
−1 { zini − A}
Numerical Solution – The numerical solution of the system of differential equations can be
found by using the approximation according to Taylor’s expansion. Thus, one equation can be
approximated by:
f(t) ⋍ f(t0) + df
dt
t=t0
(t − t0) + d2 f
dt 2
t=t0
(t − t0)... + dn f
dt n
t=t0
(69)
(t − t0) (70)
Assuming a very small time step t −t0 ≪ 1, the higher order terms of eq.(71) can be neglected.
It turns:
f(t) ⋍ f(t0) + df
dt
t=t0
(t − t0) (71)
Knowing the initial conditions of the movement, when t = t0 = 0,
y(0) = y0
y1(0)
y2(0)
=
y1ini
y1ini
35
(72)