Modal Analysis

Abstract

Chapter 2 describes how to use modal analysis, or the study of a system’s dynamic properties in the frequency domain, to describe the tool point dynamics for tool-holder combinations. It first reviews the fundamentals of single and two degree of freedom free and forced vibrations to establish notation conventions for a description of modal analysis. The text then discusses frequency response functions, details a modal fitting technique for extracting modal parameters, and describes the experimental procedures and equipment used to measure tool point frequency response functions.

All these primary impulses, not easily described in words, are the springs of man’s actions.

—Albert Einstein

Appendices

Exercises

• 1. A harmonic motion has an amplitude of 0.2 cm and a period of 15 s.

  • (a) Determine the maximum velocity (m/s) and maximum acceleration (m/s²) of the periodic motion.

  • (b) Assume that the motion expresses the free vibration of an undamped single degree of freedom system and that the motion was initiated with an initial displacement and no initial velocity. Express the motion (in units of meters) in each of the following four forms:

    • 1. A cos (ω_nt + Φ_c).

    • 2. A sin (ω_nt + Φ_s).

    • 3. B cos (ω_nt) + C sin (ω_nt).

    • 4. $D{\mathrm{e}}^{i\left({\omega}_{\mathrm{n}}t\right)}+E{\mathrm{e}}^{-i\left({\omega}_{\mathrm{n}}t\right)}$.

• 2. A single degree of freedom lumped parameter system under free vibration can be modeled with the following mass, stiffness, and damping values: m = 1 kg, k = 4 × 10⁴ N/m, and c = 10 N s/m.

  • (a) Determine the natural frequency, f_n (Hz), and the damping ratio, ζ.

  • (b) Given an initial displacement of 5 mm and an initial velocity of 0 mm/s, find an expression for the time response of the damped free vibration, x(t), of the form:

    $$ x(t)={\mathrm{e}}^{-{\zeta \omega}_{\mathrm{n}}t}\left(A\sin \left({\omega}_{\mathrm{d}}t\right)+B\cos \left({\omega}_{\mathrm{d}}t\right)\right), $$

    where x(t) is expressed in meters. Plot the first 25 cycles of motion.

• 3. A single degree of freedom lumped parameter system has mass, stiffness, and damping values of 1.2 kg, 1 × 10⁷ N/m, and 364.4 N s/m, respectively. Generate the following plots of the frequency response function:

  • (a) Magnitude (m/N) versus frequency (Hz) and phase (deg) versus frequency (Hz).

  • (b) Real part (m/N) versus frequency (Hz) and imaginary part (m/N) versus frequency (Hz).

  • (c) Argand diagram, real part (m/N) versus imaginary part (m/N).

• 4. For the two degree of freedom, damped lumped parameter system shown in Fig. 2.37, complete parts (a) through (f).

  ka = 2 × 105 N/m	kb = 5.5 × 104 N/m
  ca = 60 N s/m	cb = 16.5 N s/m
  ma = 2.5 kg	mb = 1.2 kg
  x1(0) = 1 mm	x2(0) = 0 mm
  ${\dot{x}}_1(0)=0$ mm/s	${\dot{x}}_2(0)=0$ mm/s

  Fig. 2.37

  

  Two degree of freedom, damped lumped parameter system

  • (a) Obtain the equations of motion in matrix form and transform them into modal coordinates q₁ and q₂. Normalize your eigenvectors to coordinate x₂. Verify that proportional damping exists.

  • (b) Determine the time responses q₁(t) and q₂(t) in mm. Express your solutions in the form: ${q}_{1,2}(t)={\mathrm{e}}^{-{\zeta}_{q1,2}{\omega}_{\mathrm{n}1,2}t}\left(A\cos \left({\omega}_{\mathrm{d}1,2}t\right)+B\sin \left({\omega}_{\mathrm{d}1,2}t\right)\right)$.

  • (c) Transform the modal coordinate solutions, q₁(t) and q₂(t), back into local coordinates, x₁(t) and x₂(t).

  • (d) Plot x₁(t) and x₂(t) (in mm) versus time (in seconds).

  • (e) Determine the time responses x₁(t) and x₂(t) (in mm) if the initial velocities are zero and the initial displacements are x₁(0) = 0.312 mm and x₂(0) = 1 mm.

• 5. Assume that a harmonic force, f₂ = f₀e^iωt, is applied to the lower mass (at coordinate x₂) in Fig. 2.37. Obtain the FRFs $\frac{Q_1}{R_1}$, $\frac{Q_2}{R_2}$, and $\frac{X_2}{F_2}$. Express them in equation form and then plot the real and imaginary parts (in m/N) versus frequency (in Hz).

• 6. For a single degree of freedom spring-mass-damper system under free vibration, determine the values for the mass, m (kg), viscous damping coefficient, c (N s/m), and spring constant, k (N/m), given the following information:

  • The damping ratio is 0.1.

  • The undamped natural frequency is 100 Hz.

  • The initial displacement is 1 mm.

  • The initial velocity is 5 mm/s.

  • If the system was critically damped, the value of the damping coefficient would be 586.1 N s/m.

• 7. For a single degree of freedom spring-mass-damper system with m = 2.5 kg, k = 6 × 10⁶ N/m, and c = 180 N s/m, complete the following for the case of forced harmonic vibration:

  • (a) Calculate the undamped natural frequency (in rad/s) and damping ratio.

  • (b) Sketch the imaginary part of the system FRF versus frequency. Identify the frequency (in Hz) and amplitude (in m/N) of the key features.

  • (c) Determine the value of the imaginary part of the vibration (in mm) for this system at a forcing frequency of 1500 rad/s if the harmonic force magnitude is 250 N.

• 8. Given the eigenvalues and eigenvectors for the two degree of freedom system shown in Fig. 2.38, determine the modal matrices m_q (kg), c_q (N s/m), and k_q (N/m).

  $$ {\displaystyle \begin{array}{l}{s_1}^2=-1\times 1{0}^6\kern0.5em \mathrm{rad}/{\mathrm{s}}^2\\ {}{s_2}^2=-7\times 1{0}^6\kern0.5em \mathrm{rad}/{\mathrm{s}}^2\end{array}} $$
  $$ {\psi}_1=\left\{\begin{array}{c}0.5\\ {}1\end{array}\right\}\kern1em {\psi}_2=\left\{\begin{array}{c}-2.5\\ {}1\end{array}\right\} $$

• 9. Determine the mass, damping, and stiffness matrices in local coordinates for the model shown in Fig. 2.39.

• 10. For a single degree of freedom spring-mass-damper system subject to forced harmonic vibration, the measured FRF is displayed in Figs. 2.40 and 2.41. Using the peak picking method, determine m (in kg), k (in N/m), and c (in N s/m).

Fig. 2.38

Two degree of freedom spring-mass-damper system

Fig. 2.39

Three degree of freedom spring-mass-damper model

Fig. 2.40

Measured FRF

Fig. 2.41

Measured FRF (smaller frequency scale)

Appendix: Orthogonality of Eigenvectors

As described in Chap. 2, the orthogonality of eigenvectors with respect to the system mass and stiffness matrices is the basis for modal analysis. In general, we can say that two vectors are perpendicular if their scalar, or dot, product is zero. Consider the two vectors:

$$ \left[U\right]=\left\{\begin{array}{c}{u}_{11}\\ {}{u}_{21}\end{array}\right\}\kern1em \mathrm{and}\kern1em \left[V\right]=\left\{\begin{array}{c}{v}_{11}\\ {}{v}_{21}\end{array}\right\}. $$

(2.77)

Their dot product is:

$$ \left[U\right]\bullet \left[V\right]={\left[U\right]}^{\mathrm{T}}\left[V\right]=\left\{{u}_{11}\kern0.5em {u}_{21}\right\}\left\{\begin{array}{c}{v}_{11}\\ {}{v}_{21}\end{array}\right\}={u}_{11}\cdot {v}_{11}+{u}_{21}\cdot {v}_{21}. $$

(2.78)

This product is zero if the vectors are perpendicular. Orthogonality can be considered a generalization of the concept of perpendicularity.

From Chap. 2, we have seen that we can write the matrix form of the system equations of motion ([M]s² + [K]){X}e^st = {0} if we assume harmonic vibration. We used the characteristic equation, |[M]s² + [K]| = 0, to find the eigenvalues, s₁² and s₂². We then substituted the eigenvalues into either of the linearly dependent equations of motion to find the eigenvectors, or mode shapes. Using s₁² =  − ω_n1², we can write:

$$ \left(-\left[M\right]{\omega_{\mathrm{n}1}}^2+\left[K\right]\right)\left\{{\psi}_1\right\}=\left\{0\right\}, $$

(2.79)

where ψ₁ is the corresponding mode shape. Equation 2.79 can be expanded to:

$$ -{\omega_{\mathrm{n}1}}^2\left[M\right]\left\{{\psi}_1\right\}+\left[K\right]\left\{{\psi}_1\right\}=\left\{0\right\}. $$

(2.80)

Premultiplying Eq. 2.80 by the transpose of the second mode shape ψ₂, which corresponds to vibration at ω_n2, yields:

$$ -{\omega_{\mathrm{n}1}}^2{\left\{{\psi}_2\right\}}^{\mathrm{T}}\left[M\right]\left\{{\psi}_1\right\}+{\left\{{\psi}_2\right\}}^{\mathrm{T}}\left[K\right]\left\{{\psi}_1\right\}=0. $$

(2.81)

Performing the transpose operation on Eq. 2.81 gives:

$$ -{\omega_{\mathrm{n}1}}^2{\left\{{\psi}_1\right\}}^{\mathrm{T}}\left[M\right]\left\{{\psi}_2\right\}+{\left\{{\psi}_1\right\}}^{\mathrm{T}}\left[K\right]\left\{{\psi}_2\right\}=0, $$

(2.82)

where the transpose properties ([A][B])^T = [B]^T[A]^T and ([A]^T)^T = [A] (using matrices of appropriate dimensions) have been applied.

Completing the same operations using s₂² =  − ω_n2² gives:

$$ -{\omega_{\mathrm{n}2}}^2{\left\{{\psi}_1\right\}}^{\mathrm{T}}\left[M\right]\left\{{\psi}_2\right\}+{\left\{{\psi}_1\right\}}^{\mathrm{T}}\left[K\right]\left\{{\psi}_2\right\}=0. $$

(2.83)

Taking the difference of Eqs. 2.82 and 2.83 yields:

$$ \left({\omega_{\mathrm{n}2}}^2-{\omega_{\mathrm{n}1}}^2\right){\left\{{\psi}_1\right\}}^{\mathrm{T}}\left[M\right]\left\{{\psi}_2\right\}=0. $$

(2.84)

Provided ω_n2² ≠ ω_n1², then {ψ₁}^T[M]{ψ₂} = 0. Substituting this result into either Eq. 2.82 or Eq. 2.83 gives {ψ₁}^T[K]{ψ₂} = 0. Collecting these results, we obtain the orthogonality conditions shown in Eqs. 2.85 through 2.88.

$$ {\displaystyle \begin{array}{l}{\left\{{\psi}_1\right\}}^{\mathrm{T}}\left[M\right]\left\{{\psi}_2\right\}=0\\ {}{\left\{{\psi}_2\right\}}^{\mathrm{T}}\left[M\right]\left\{{\psi}_1\right\}=0\end{array}} $$

(2.85)

$$ {\displaystyle \begin{array}{c}{\left\{{\psi}_1\right\}}^{\mathrm{T}}\left[M\right]\left\{{\psi}_1\right\}={m}_{q1}\\ {}{\left\{{\psi}_2\right\}}^{\mathrm{T}}\left[M\right]\left\{{\psi}_2\right\}={m}_{q2}\end{array}}\kern1em \left(\mathrm{These}\ \mathrm{products}\ \mathrm{are}\ \mathrm{not}\ \mathrm{necessarily}\ \mathrm{zero}.\right) $$

(2.86)

$$ {\displaystyle \begin{array}{l}{\left\{{\psi}_1\right\}}^{\mathrm{T}}\left[K\right]\left\{{\psi}_2\right\}=0\\ {}{\left\{{\psi}_2\right\}}^{\mathrm{T}}\left[K\right]\left\{{\psi}_1\right\}=0\end{array}} $$

(2.87)

$$ {\displaystyle \begin{array}{c}{\left\{{\psi}_1\right\}}^{\mathrm{T}}\left[K\right]\left\{{\psi}_1\right\}={k}_{q1}\\ {}{\left\{{\psi}_2\right\}}^{\mathrm{T}}\left[K\right]\left\{{\psi}_2\right\}={k}_{q2}\end{array}}\kern1em \left(\mathrm{These}\ \mathrm{products}\ \mathrm{are}\ \mathrm{not}\ \mathrm{necessarily}\ \mathrm{zero}.\right) $$

(2.88)

Using the modal matrix, $\left[P\right]=\left[{\psi}_1\kern0.5em {\psi}_2\right]$, and the orthogonality conditions we obtain the diagonalized modal mass and stiffness matrices:

$$ {\left[P\right]}^{\mathrm{T}}\left[M\right]\left[P\right]=\left[\begin{array}{cc}{\left\{{\psi}_1\right\}}^{\mathrm{T}}\left[M\right]\left\{{\psi}_1\right\}& {\left\{{\psi}_1\right\}}^{\mathrm{T}}\left[M\right]\left\{{\psi}_2\right\}\\ {}{\left\{{\psi}_2\right\}}^{\mathrm{T}}\left[M\right]\left\{{\psi}_1\right\}& {\left\{{\psi}_2\right\}}^{\mathrm{T}}\left[M\right]\left\{{\psi}_2\right\}\end{array}\right]=\left[\begin{array}{cc}{m}_{q1}& 0\\ {}0& {m}_{q2}\end{array}\right]=\left[{M}_q\right]\kern1em \mathrm{and} $$

(2.89)

$$ {\left[P\right]}^{\mathrm{T}}\left[K\right]\left[P\right]=\left[\begin{array}{cc}{\left\{{\psi}_1\right\}}^{\mathrm{T}}\left[K\right]\left\{{\psi}_1\right\}& {\left\{{\psi}_1\right\}}^{\mathrm{T}}\left[K\right]\left\{{\psi}_2\right\}\\ {}{\left\{{\psi}_2\right\}}^{\mathrm{T}}\left[K\right]\left\{{\psi}_1\right\}& {\left\{{\psi}_2\right\}}^{\mathrm{T}}\left[K\right]\left\{{\psi}_2\right\}\end{array}\right]=\left[\begin{array}{cc}{k}_{q1}& 0\\ {}0& {k}_{q2}\end{array}\right]=\left[{K}_q\right]. $$

(2.90)

These diagonal modal mass and stiffness matrices uncouple the equations of motion and enable the solution of independent single degree of freedom systems in modal coordinates. The individual modal contributions can then be transformed back into local (physical) coordinates as discussed in Chap. 2.