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Thomas Dehaeze
2024-12-17 16:38:55 +01:00
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content/phdthesis/li01_simul_fault_vibrat_isolat_point.md
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@@ -131,7 +131,7 @@ Define a new input and a new output:
u\_1 = J^T f\_m, \quad y = J^{-1} (l - l\_r)
\end{equation}
Equation <eq:hexapod_eq_motion> can be rewritten as:
Equation \eqref{eq:hexapod\_eq\_motion} can be rewritten as:
\begin{equation} \label{eq:hexapod\_eq\_motion\_decoup\_1}
\begin{split}

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content/phdthesis/rankers98_machin.md
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@@ -335,17 +335,17 @@ These eigenvectors have the following orthogonality properties, or can always be
\phi\_i^T M \phi\_j = 0 \quad (i \neq j)
\end{equation}
For \\(i=j\\) the result of the multiplication according to equation <eq:eigenvector_orthogonality_mass> yields a non-zero result, which is normally indicated as **modal mass** \\(\mathit{m}\_i\\):
For \\(i=j\\) the result of the multiplication according to equation \eqref{eq:eigenvector\_orthogonality\_mass} yields a non-zero result, which is normally indicated as **modal mass** \\(\mathit{m}\_i\\):
\begin{equation} \label{eq:modal\_mass}
\phi\_i^T M \phi\_i = \mathit{m}\_i
\end{equation}
Because only the direction but not the length of an eigenvector is defined, several scaling methods are used, all based on equation <eq:modal_mass>:
Because only the direction but not the length of an eigenvector is defined, several scaling methods are used, all based on equation \eqref{eq:modal\_mass:}
- \\(|\phi\_i| = 1\\). Each eigenvector \\(\phi\_i\\) is scaled such that its length is equal to \\(1\\). The modal mass are then calculated from equation <eq:modal_mass>.
- \\(\max(\phi\_i) = 1\\). Each eigenvector \\(\phi\_i\\) is scaled such that its largest element is equation to \\(1\\). The modal mass is then calculated from equation <eq:modal_mass>.
- \\(m\_i = 1\\). The modal mass \\(\mathit{m}\_i\\) is set to \\(1\\). The scaling of the mode vector \\(\phi\_i\\) follows from equation <eq:modal_mass>.
- \\(|\phi\_i| = 1\\). Each eigenvector \\(\phi\_i\\) is scaled such that its length is equal to \\(1\\). The modal mass are then calculated from equation \eqref{eq:modal\_mass}.
- \\(\max(\phi\_i) = 1\\). Each eigenvector \\(\phi\_i\\) is scaled such that its largest element is equation to \\(1\\). The modal mass is then calculated from equation \eqref{eq:modal\_mass}.
- \\(m\_i = 1\\). The modal mass \\(\mathit{m}\_i\\) is set to \\(1\\). The scaling of the mode vector \\(\phi\_i\\) follows from equation \eqref{eq:modal\_mass}.
The orthogonality properties also apply to the stiffness matrix \\(K\\):
@@ -450,7 +450,7 @@ In such an analysis one is typically interested in the transfer function between
Applying the principle of modal decomposition, any transfer function can be derived by first calculating the behavior of the individual modes, and then **summing all modal contributions**.
The contribution of one single mode \\(i\\) to the transfer function \\(x\_l/f\_k\\) can be derived by first considering the response of the modal DoF \\(q\_i\\) to a force vector \\(f\\) with only one non-zero component \\(f\_k\\).
In that case, equation <eq:eoq_modal_i> is reduced to:
In that case, equation \eqref{eq:eoq\_modal\_i} is reduced to:
\begin{equation}
m\_i \ddot{q}\_i(t) + k\_i q\_i(t) = \phi\_{ik} f\_k(t)
@@ -481,7 +481,7 @@ The overall transfer function can be found by summation of the individual modal
### Graphical Representation {#graphical-representation}
Due to the equivalence with the differential equations of a single mass spring system, equation <eq:eoq_modal_i> is often represented by a single mass spring system on which a force \\(f^\prime = \phi\_i^T f\\) acts.
Due to the equivalence with the differential equations of a single mass spring system, equation \eqref{eq:eoq\_modal\_i} is often represented by a single mass spring system on which a force \\(f^\prime = \phi\_i^T f\\) acts.
However, this representation implies an important loss of information because it neglects all information about the mode-shape vector.
Consider the system in [Figure 8](#figure--fig:rankers98-mode-trad-representation) for which the three mode shapes are depicted in the traditional graphical representation.
@@ -516,7 +516,7 @@ The resulting moment of inertia \\(J\_i\\) of the i-th modal lever then is:
J\_i = \sum\_{k=1}^n m\_k \phi\_{ik}^2
\end{equation}
This result is identical to the modal mass \\(m\_i\\) found with Equation <eq:modal_mass>, because the mass matrix \\(M\\) is a diagonal matrix of physical masses \\(m\_k\\), and consequently the expression for the modal mass \\(m\_i\\) yields:
This result is identical to the modal mass \\(m\_i\\) found with Equation \eqref{eq:modal\_mass}, because the mass matrix \\(M\\) is a diagonal matrix of physical masses \\(m\_k\\), and consequently the expression for the modal mass \\(m\_i\\) yields:
\begin{equation}
m\_i = \phi\_j^T M \phi\_j = \sum\_{k=1}^n m\_k \phi\_{ik}^2
@@ -637,7 +637,7 @@ whereas the modal stiffnesses follow from \\(k\_i = \omega\_i^2 m\_i\\).
{{< figure src="/ox-hugo/rankers98_example_2dof_modal.png" caption="<span class=\"figure-number\">Figure 15: </span>Graphical representation of modes and modal parameters of the two mass spring system" >}}
From these results, the effective modal parameters for each mode, and for each individual DoF can be defined using equations <eq:m_modal_eff> and <eq:k_modal_eff>.
From these results, the effective modal parameters for each mode, and for each individual DoF can be defined using equations \eqref{eq:m\_modal\_eff} and \eqref{eq:k\_modal\_eff}.
The results are summarized in [Table 2](#table--tab:2dof-example-modal-params-eff).
<a id="table--tab:2dof-example-modal-params-eff"></a>
@@ -716,7 +716,7 @@ f\_{\text{new},i}(\Delta k) &= \frac{1}{2\pi}\sqrt{\frac{k\_{\text{eff},i} + \De
Let's use the two mass spring system in [Figure 14](#figure--fig:rankers98-example-2dof) as an example.
In order to analyze the effect of an extra mass at \\(x\_2\\), the effective modal mass at that DoF needs to be known for both modes (see [Table 2](#table--tab:2dof-example-modal-params-eff)).
Then using equation <eq:sensitivity_add_m>, one can estimate the effect of an extra mass \\(\Delta m = 1 kg\\) added to \\(m\_2\\).
Then using equation \eqref{eq:sensitivity\_add\_m}, one can estimate the effect of an extra mass \\(\Delta m = 1 kg\\) added to \\(m\_2\\).
To estimate the influence of extra stiffness between the two DoF, one needs to calculate the effective modal stiffness that corresponds to the relative motion between \\(x\_2\\) and \\(x\_1\\).
This can be graphically done as shown in [Figure 19](#figure--fig:rankers98-example-sensitivity-2dof):
@@ -726,7 +726,7 @@ k\_{\text{eff},1,(2-1)} &= 0.46 \cdot 10^7 / 0.07^2 = 93.9 \cdot 10^7 N/m \\\\
k\_{\text{eff},2,(2-1)} &= 1.23 \cdot 10^7 / 1.1^2 = 1.0 \cdot 10^7 N/m
\end{align}
And using equation <eq:sensitivity_add_m>, the effect of additional stiffness on the frequency of the two modes can be computed.
And using equation \eqref{eq:sensitivity\_add\_m}, the effect of additional stiffness on the frequency of the two modes can be computed.
The results are summarized in [Table 3](#table--tab:example-sensitivity-2dof-results).
@@ -860,13 +860,13 @@ Let's introduce a variable \\(\alpha\\), which **relates the high-frequency cont
\alpha = \frac{\frac{\phi\_{i,\text{servo}} \phi\_{i,\text{force}}}{m\_i}}{\frac{1}{m}}
\end{equation}
which simplifies equation <eq:effect_one_mode> to:
which simplifies equation \eqref{eq:effect\_one\_mode} to:
\begin{equation} \label{eq:effect\_one\_mode\_simplified}
\boxed{\frac{x\_{\text{servo}}}{F\_{\text{servo}}} = \frac{1}{ms^2} + \frac{\alpha}{m s^2 + m \omega\_i^2}}
\end{equation}
Equation <eq:effect_one_mode_simplified> will be the basis for the discussion of the various patterns that can be observe in the frequency response functions and the effect of resonances on servo stability.
Equation \eqref{eq:effect\_one\_mode\_simplified} will be the basis for the discussion of the various patterns that can be observe in the frequency response functions and the effect of resonances on servo stability.
Three different types of intersection pattern can be found in the amplitude plot as shown in [Figure 25](#figure--fig:rankers98-frf-effect-alpha).
Depending on the absolute value of \\(\alpha\\) one can observe: