1.9.2 System Classification
Systems of the type (1.129) can be classified according to the forces involved:
1. Conservative system: All forces are conservative
1:1. Gyroscopic conservative system: Conservative system with gyroscopic
forces
1:2. Non-gyroscopic conservative system: Conservative system without gyroscopic forces
2. Non-conservative system:
2:1. [Purely] Dissipative system: [Only conservative and] Dissipative forces
2:2. [Purely] Circulatory system: [Only conservative and] Circulatory forces
2:3. [Purely] Instationary system: [Only conservative and] Instationary forces
Some of the descriptors may combine, so that a system can be classified as, for
example, ‘purely and completely dissipative’, or ‘circulatory and dissipative’, or
‘gyroscopic conservative with non-circulatory forces’.
System classification may point to certain types of dynamic behavior to expect,
in particular to the types of instabilities that may occur, and – as described next – to
possible ways of examining this.
1.9.3 Stability Assessment
Without external forcing (f = 0) Eq. (1.129) has a static equilibrium solution x ¼
_
x ¼ 0: This equilibrium can be stable or unstable, or marginally stable (i.e. at the
border separating stable from unstable). Image the system starting at the equilibrium at
rest, i.e. xð0Þ ¼ _
xð0Þ ¼ 0; and then subjected to a small disturbance, e.g. _
xð0 þ Þ 6 ¼ 0:
As time t increases the state x(t) may reapproach the equilibrium asymptotically so that
x!0 for t!∞; in that case the equilibrium x = 0 is stable. Or x may diverge exponentially away from 0 so that |x|!∞, in which case x = 0 is unstable. Or the state may
stay close to the equilibrium, and x = 0 is marginally unstable.
Note that it is the (equilibrium) state of the system that can be stable or unstable,
not ‘the system’ itself. Think of the unforced mathematical pendulum, having two
equilibria – a down-pointing/stable, and an up-pointing/unstable (in the absence of
damping the down-pointing equilibrium is only marginally stable). It makes no
sense to address stability of ‘the pendulum’, but the pendulum state has two static
equilibria, one stable (down) and the other unstable (up).
There are several ways to calculate equilibrium stability (more on this in
Chaps. 3–5.). Ziegler (1968) provides a number of theorems for assessing how stability of the zero solution of (1.129) (with f = 0) can be assessed in a possibly simple
manner, dependent on the system class. It is useful to be able to identify such cases,
which can save efforts as compared to the more elaborate methods in Chaps. 3–5.
44
1 Vibration Basics
Systems of the type (1.129) can be classified according to the forces involved:
1. Conservative system: All forces are conservative
1:1. Gyroscopic conservative system: Conservative system with gyroscopic
forces
1:2. Non-gyroscopic conservative system: Conservative system without gyroscopic forces
2. Non-conservative system:
2:1. [Purely] Dissipative system: [Only conservative and] Dissipative forces
2:2. [Purely] Circulatory system: [Only conservative and] Circulatory forces
2:3. [Purely] Instationary system: [Only conservative and] Instationary forces
Some of the descriptors may combine, so that a system can be classified as, for
example, ‘purely and completely dissipative’, or ‘circulatory and dissipative’, or
‘gyroscopic conservative with non-circulatory forces’.
System classification may point to certain types of dynamic behavior to expect,
in particular to the types of instabilities that may occur, and – as described next – to
possible ways of examining this.
1.9.3 Stability Assessment
Without external forcing (f = 0) Eq. (1.129) has a static equilibrium solution x ¼
_
x ¼ 0: This equilibrium can be stable or unstable, or marginally stable (i.e. at the
border separating stable from unstable). Image the system starting at the equilibrium at
rest, i.e. xð0Þ ¼ _
xð0Þ ¼ 0; and then subjected to a small disturbance, e.g. _
xð0 þ Þ 6 ¼ 0:
As time t increases the state x(t) may reapproach the equilibrium asymptotically so that
x!0 for t!∞; in that case the equilibrium x = 0 is stable. Or x may diverge exponentially away from 0 so that |x|!∞, in which case x = 0 is unstable. Or the state may
stay close to the equilibrium, and x = 0 is marginally unstable.
Note that it is the (equilibrium) state of the system that can be stable or unstable,
not ‘the system’ itself. Think of the unforced mathematical pendulum, having two
equilibria – a down-pointing/stable, and an up-pointing/unstable (in the absence of
damping the down-pointing equilibrium is only marginally stable). It makes no
sense to address stability of ‘the pendulum’, but the pendulum state has two static
equilibria, one stable (down) and the other unstable (up).
There are several ways to calculate equilibrium stability (more on this in
Chaps. 3–5.). Ziegler (1968) provides a number of theorems for assessing how stability of the zero solution of (1.129) (with f = 0) can be assessed in a possibly simple
manner, dependent on the system class. It is useful to be able to identify such cases,
which can save efforts as compared to the more elaborate methods in Chaps. 3–5.
44
1 Vibration Basics