Dynamics of Machines - Part II - IFS.pdf
Dynamics of Machines - Part II - IFS.pdf
Dynamics of Machines - Part II - IFS.pdf
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Permanent Solution and Steady-State Analysis – The permanent solution takes into<br />
account the right side <strong>of</strong> the differential equation (see eq.(60)), or in other words, the effect <strong>of</strong><br />
the force on the system. In case <strong>of</strong> harmonic excitation, one can write:<br />
A˙zp(t) + Bzp(t) = fe jωt<br />
(60)<br />
zp(t) = Ae jωt , (assumption!) (61)<br />
˙zp(t) = jωAe jωt<br />
The assumption adopted in eq.(61) and its derivative are introduced into the differential equation<br />
(60), leading to:<br />
[jωA + B]Ae jωt = fe jωt ⇒ A = [jωA + B] −1 f (62)<br />
The permanent solution <strong>of</strong> the equation <strong>of</strong> motion is given by:<br />
zp(t) = Ae jωt ⇒ zp(t) = [jωA + B] −1 fe jωt<br />
General Solution – The general solution <strong>of</strong> a linear differential equation is achieved by adding<br />
the homogenous and the permanent solutions, and by sequentially defining the initial conditions<br />
<strong>of</strong> the movement. This solution will provide information about the transient and steady-state<br />
response <strong>of</strong> the mechanical model. Considering that the order <strong>of</strong> the mechanical model is correct<br />
(in this case, the two degree-<strong>of</strong>-freedom system), the solution <strong>of</strong> the linear differential equation<br />
will be useful for predicting the dynamical behavior <strong>of</strong> the physical system, if the coefficients<br />
<strong>of</strong> the differential equation (M, D and K, or A and B) are properly chosen, either by using<br />
theoretical or experimental information or both. The general solution <strong>of</strong> the differential equation<br />
<strong>of</strong> motion is given by:<br />
z(t) = C1u1e λ1t + C2u2e λ2t + C3u3e λ3t + C4u4e λ4t + Ae iωt<br />
where z(t) gives information about the displacement and velocity <strong>of</strong> the coordinates y1 and y2.<br />
Introducing the initial conditions <strong>of</strong> displacement and velocity<br />
zini = { v1ini<br />
v2ini y1ini y2ini }T<br />
into eq.(64), when t = 0, one obtains<br />
z(0) = zini = C1u1e λ10 + C2u2e λ20 + C3u3e λ30 + C4u4e λ40 + Ae iω0<br />
or<br />
⎧<br />
⎪⎨<br />
zini = C1u1 + C2u2 + C3u3 + C4u4 + A = [ u1 u2 u3 u4 ]<br />
⎪⎩<br />
C1<br />
C2<br />
C3<br />
C4<br />
(63)<br />
(64)<br />
(65)<br />
(66)<br />
⎫<br />
⎪⎬<br />
+ A (67)<br />
⎪⎭<br />
Solving the linear system by inverting the modal matrix U = [ u1 u2 u3 u4 ] one achieves the<br />
vector c = { C1 C2 C3 C4 } T :<br />
34