Dynamics of Machines - Part II - IFS.pdf

and
˙y(0) = ˙y0
˙y1(0)
˙y2(0)
=
v1ini
v2ini
and the equation of motion eq.(52), which has to be solved, one can calculate the acceleration,
when t = t0 = 0:
¨y0 = −M −1 D ˙y0 + Ky0 − ¯ jωt0 fe
The first predicted values of displacement, velocity and acceleration in time t1 = ∆t , using the
approximation given by eq.(71), are:
t1 = ∆t
˙y1 = ˙y0 + ¨y0∆t
y1 = y0 + ˙y1∆t
¨y1 = −M −1 D ˙y1 + Ky1 − ¯ jωt1 fe
The second predicted values of displacement, velocity and acceleration in time t2 = t1 + ∆t ,
using the approximation given by eq.(71), are:
t2 = 2∆t
˙y2 = ˙y1 + ¨y1∆t
y2 = y1 + ˙y2∆t
¨y2 = −M −1 D ˙y2 + Ky2 − ¯ jωt2 fe
The N-th predicted values of displacement, velocity and acceleration in time tN = tN−1 + ∆t ,
using the approximation given by eq.(74), are:
tN = N∆t
˙yN = ˙yN−1 + ¨yN−1∆t
yN = yN−1 + ˙yN∆t
¨yN = −M −1 D ˙yN + KyN − ¯ jωtN fe
Plotting the points [y1,y2,y3, ...,yN] versus [t1, t2, t3, ..., tN], one can observe the numerical
solution of the differential equation, which describes the displacements of the mass-dampingspring
system in time domain. Plotting the points [ ˙y1, ˙y2, ˙y3, ..., ˙yN] versus [t1, t2, t3, ..., tN] or
[¨y1, ¨y2, ¨y3, ..., ¨yN] versus [t1, t2, t3, ..., tN] one can also observe the velocity and acceleration of
the mass-damping-spring system in time domain. The analytical and numerical solutions of the
second order differential equation, eq.(52), are illustrated using a Matlab code.
36
(74)
(73)