Harmonic Balance Flutter

Since blade flutter is a matter of vibration, it can be expressed in terms of sine functions and therefore can be addressed in terms of Fourier series.

Real and Imaginary Displacements

The Fourier series that are the basis for harmonic balance equations can be expressed as trigonometric functions or, equivalently, as exponential functions of complex numbers. In the latter form, the expression includes both real and imaginary displacements.

If $v_{r}$ and $v_{i}$ are the real and imaginary displacements that you specify, then the physical displacements $d x$ at every time level $k$ are computed as:

Figure 1. EQUATION_DISPLAY

\begin{matrix} d x (t_{k}) & = & (v_{r}, v_{i}) exp (\frac{i (2 π k)}{N}) + (v_{r}, {- v}_{i}) exp (\frac{- i (2 π k)}{N}) \\ = & 2 v_{r} (\cos \frac{(2 π k)}{N}) - 2 v_{i} (\sin \frac{(2 π k)}{N}) \end{matrix}

(5058)

where $t_{k}$ represents the physical time at time level $k$ (where $k$ varies from 0 to $N - 1$ ), $N$ is the number of time levels, and $(v_{r}, v_{i})$ represents a complex number with $v_{r}$ and $v_{i}$ being the real and imaginary parts of the complex number.

Fluttering Motion of a Rotor

In the rotor frame of reference, an inter-blade phase angle $σ$ , and a flutter frequency $ω$ describe the motion. For example,

Figure 2. EQUATION_DISPLAY

h_{i} = h_{0} e^{j ω t} e^{(j σ θ_{i}) / G_{θ}}

(5059)

where $G_{θ}$ is the blade-to-blade gap (in radians). Written in terms of nodal diameters, Eqn. (5059) becomes

Figure 3. EQUATION_DISPLAY

h_{i} = h_{0} e^{j ω t} e^{j N θ_{i}}

(5060)

where $N = σ / G_{θ}$ . This motion represents a traveling wave, which is like the motion of a rotor. In this case, we have an $N$ nodal diameter wave moving with a rotational rate of

Figure 4. EQUATION_DISPLAY

Ω_{r e l} = - ω / N = - (ω G_{θ}) / σ

(5061)

This rotor rate is relative to the frame of reference of the fluttering rotor. Therefore, the rotation rate in the absolute reference frame is

Figure 5. EQUATION_DISPLAY

Ω_{a b s} = Ω_{r o t o r} - ω / N

(5062)

To compute the frequencies and nodal diameters, another row is added to the mode table. For example, consider a stator with 36 blades and a rotor with 42 blades, rotating at 366 rad/s. Assume that the rotor blade is fluttering with an inter-blade phase angle of 180 $°$ with a frequency of 2000 rad/s. To construct the mode table:

Compute the nodal diameter $N$ for the flutter motion:
$N = σ / G_{θ} = (180 °) / ((360 °) / 42) = 21$
Compute the rate of disturbance $Ω_{a b s}$ :
$Ω_{a b s} = Ω_{r o t o r} - ω / N = 366 rad/s - (2000 rad/s) / 21 = 270.76 rad/s$
Construct the mode table with three rows:
$(B_{0}, B_{1}, B_{2}) = (21, 36, 42)$

$(Ω_{0}, Ω_{1}, Ω_{2}) = (270.76, 0.0, 366.0) rad/s$