Interactive Lecture
Multi-Degree of Freedom Systems, Modal Analysis, and Mode Shapes
by Dr Arnaldo Delli Carri
Multi-degree of freedom (MDOF) systems are mechanical systems that require more than one coordinate to describe their motion. Unlike SDOF systems, MDOF systems exhibit coupled behaviour where motion in one coordinate affects motion in others.
A 3-DOF system requires three independent coordinates to describe its configuration. Each mass can move independently, but the spring and damper couplings link them — producing multiple natural frequencies and mode shapes.
For a 3-DOF undamped system with masses \(m_1, m_2, m_3\) and springs \(k_1, k_2, k_3, k_4\), the equations of motion are:
\[m_1\ddot{x}_1 + (k_1 + k_2)x_1 - k_2x_2 = 0\]
\[m_2\ddot{x}_2 - k_2x_1 + (k_2 + k_3)x_2 - k_3x_3 = 0\]
\[m_3\ddot{x}_3 - k_3x_2 + (k_3 + k_4)x_3 = 0\]
In matrix form:
\[\mathbf{M}\ddot{\mathbf{x}} + \mathbf{K}\mathbf{x} = \mathbf{0}\]
where the mass matrix \(\mathbf{M}\) and stiffness matrix \(\mathbf{K}\) are:
\[\mathbf{M} = \begin{bmatrix} m_1 & 0 & 0 \\ 0 & m_2 & 0 \\ 0 & 0 & m_3 \end{bmatrix}, \quad \mathbf{K} = \begin{bmatrix} k_1+k_2 & -k_2 & 0 \\ -k_2 & k_2+k_3 & -k_3 \\ 0 & -k_3 & k_3+k_4 \end{bmatrix}\]
Modal analysis decomposes the coupled motion of an MDOF system into independent modal coordinates.
Eigenvalue Problem:
\[(\mathbf{K} - \omega_i^2\mathbf{M})\mathbf{\Phi}_i = \mathbf{0}\]
Solving this gives:
Each mode shape represents a particular pattern of vibration where all masses oscillate at the same frequency with fixed amplitude ratios:
Mode shapes are orthogonal with respect to both the mass and stiffness matrices. This orthogonality lets you decouple the system equations and solve each mode independently.
Orthogonality Conditions:
For any two distinct mode shapes \(\mathbf{\Phi}_i\) and \(\mathbf{\Phi}_j\) where \(i \neq j\):
\[\mathbf{\Phi}_i^T \mathbf{M} \mathbf{\Phi}_j = 0 \quad \text{(Mass orthogonality)}\]
\[\mathbf{\Phi}_i^T \mathbf{K} \mathbf{\Phi}_j = 0 \quad \text{(Stiffness orthogonality)}\]
These conditions are often normalised such that:
\[\mathbf{\Phi}_i^T \mathbf{M} \mathbf{\Phi}_i = 1 \quad \text{and} \quad \mathbf{\Phi}_i^T \mathbf{K} \mathbf{\Phi}_i = \omega_i^2\]
The orthogonality conditions transform the coupled system \(\mathbf{M}\ddot{\mathbf{x}} + \mathbf{K}\mathbf{x} = \mathbf{0}\) into \(n\) independent SDOF equations in modal coordinates. This is the essence of modal decomposition and makes solving MDOF systems tractable.
Proof of Orthogonality: Consider two mode shapes satisfying the eigenvalue equations:
\[\mathbf{K}\mathbf{\Phi}_i = \omega_i^2\mathbf{M}\mathbf{\Phi}_i\]
\[\mathbf{K}\mathbf{\Phi}_j = \omega_j^2\mathbf{M}\mathbf{\Phi}_j\]
Pre-multiplying the first equation by \(\mathbf{\Phi}_j^T\) and the second by \(\mathbf{\Phi}_i^T\), then using the symmetry of \(\mathbf{M}\) and \(\mathbf{K}\), we get:
\[(\omega_i^2 - \omega_j^2)\mathbf{\Phi}_j^T\mathbf{M}\mathbf{\Phi}_i = 0\]
For distinct natural frequencies (\(\omega_i \neq \omega_j\)), this implies \(\mathbf{\Phi}_j^T\mathbf{M}\mathbf{\Phi}_i = 0\).
Physical coordinates relate to modal coordinates by:
\[\mathbf{x}(t) = \mathbf{\Phi}\mathbf{q}(t) = \sum_{i=1}^{3} \mathbf{\Phi}_i q_i(t)\]
For the undamped case, each modal equation becomes:
\[\ddot{q}_i + \omega_i^2 q_i = 0\]
This is identical to an undamped SDOF system!
In real systems, damping is always present. However, general damping matrices make modal decomposition ineffective because the modal equations remain coupled through the damping terms.
A special case that preserves the advantage of modal decomposition is proportional damping (also called Rayleigh damping), where the damping matrix is a linear combination of the mass and stiffness matrices:
\[\mathbf{C} = \alpha \mathbf{M} + \beta \mathbf{K}\]
where \(\alpha\) and \(\beta\) are Rayleigh damping coefficients.
With proportional damping, the mode shapes remain the same as for the undamped system, and the modal equations are decoupled. Each modal damping ratio becomes:
\[\zeta_i = \frac{\alpha}{2\omega_i} + \frac{\beta\omega_i}{2}\]
Determining Rayleigh Coefficients: If we know the damping ratios \(\zeta_i\) and \(\zeta_j\) at two different frequencies \(\omega_i\) and \(\omega_j\), we can solve for \(\alpha\) and \(\beta\):
\[\begin{bmatrix} \frac{1}{2\omega_i} & \frac{\omega_i}{2} \\ \frac{1}{2\omega_j} & \frac{\omega_j}{2} \end{bmatrix} \begin{bmatrix} \alpha \\ \beta \end{bmatrix} = \begin{bmatrix} \zeta_i \\ \zeta_j \end{bmatrix}\]
Solving this gives:
\[\alpha = \frac{2\omega_i\omega_j(\zeta_i\omega_j - \zeta_j\omega_i)}{\omega_j^2 - \omega_i^2}\]
\[\beta = \frac{2(\zeta_j\omega_j - \zeta_i\omega_i)}{\omega_j^2 - \omega_i^2}\]
Damped Modal Equations: With proportional damping, the system equations in matrix form are:
\[\mathbf{M}\ddot{\mathbf{x}} + \mathbf{C}\dot{\mathbf{x}} + \mathbf{K}\mathbf{x} = \mathbf{0}\]
After modal transformation, each mode satisfies:
\[\ddot{q}_i + 2\zeta_i\omega_i\dot{q}_i + \omega_i^2 q_i = 0\]
This is identical to a damped SDOF system with solution:
\[q_i(t) = e^{-\zeta_i\omega_i t}\left(A_i\cos(\omega_{di}t) + B_i\sin(\omega_{di}t)\right)\]
where \(\omega_{di} = \omega_i\sqrt{1-\zeta_i^2}\) is the damped natural frequency.
When external forces are applied to an MDOF system, the equations of motion become:
\[\mathbf{M}\ddot{\mathbf{x}} + \mathbf{C}\dot{\mathbf{x}} + \mathbf{K}\mathbf{x} = \mathbf{f}(t)\]
where \(\mathbf{f}(t) = [f_1(t), f_2(t), f_3(t)]^T\) is the vector of external forces applied to each mass.
Using the modal transformation \(\mathbf{x}(t) = \mathbf{\Phi}\mathbf{q}(t)\) and pre-multiplying by \(\mathbf{\Phi}^T\), we obtain decoupled modal equations:
\[\ddot{q}_i + 2\zeta_i\omega_i\dot{q}_i + \omega_i^2 q_i = Q_i(t)\]
where the modal force is:
\[Q_i(t) = \mathbf{\Phi}_i^T\mathbf{f}(t)\]
For harmonic excitation \(\mathbf{f}(t) = \mathbf{F}_0\sin(\Omega t)\), where \(\Omega\) is the forcing frequency, each modal equation becomes:
\[\ddot{q}_i + 2\zeta_i\omega_i\dot{q}_i + \omega_i^2 q_i = Q_{i0}\sin(\Omega t)\]
where \(Q_{i0} = \mathbf{\Phi}_i^T\mathbf{F}_0\).
The steady-state solution for each modal coordinate is:
\[q_i(t) = \frac{Q_{i0}/\omega_i^2}{\sqrt{(1-r_i^2)^2 + (2\zeta_i r_i)^2}}\sin(\Omega t - \phi_i)\]
where \(r_i = \Omega/\omega_i\) is the frequency ratio and \(\phi_i = \tan^{-1}\left(\frac{2\zeta_i r_i}{1-r_i^2}\right)\) is the phase angle — lowercase \(\phi_i\) here, distinct from the bold mode-shape matrix \(\mathbf{\Phi}_i\).
Resonance occurs when the forcing frequency \(\Omega\) approaches one of the natural frequencies \(\omega_i\). At resonance, the amplitude of that particular mode becomes very large (theoretically infinite for undamped systems), and that mode dominates the response. Machinery running near a natural frequency sees that mode's amplitude climb sharply, driving fatigue and failure.
The total physical response is obtained by summing all modal contributions:
\[\mathbf{x}(t) = \sum_{i=1}^{3} \mathbf{\Phi}_i q_i(t)\]
Each mode contributes based on:
General Forcing: For arbitrary time-varying forces \(\mathbf{f}(t)\), the Duhamel integral (convolution) can be used for each modal equation:
\[q_i(t) = \frac{1}{\omega_{di}}\int_0^t Q_i(\tau)e^{-\zeta_i\omega_i(t-\tau)}\sin(\omega_{di}(t-\tau))d\tau\]
where \(\omega_{di} = \omega_i\sqrt{1-\zeta_i^2}\) is the damped natural frequency.
Explore how changing masses, spring stiffnesses, and initial conditions affects the vibrational behaviour. Observe both modal and physical responses.
Three degree of freedom modal analysis appears throughout structural and mechanical engineering:
| Concept | Equation | Description |
|---|---|---|
| Equations of motion (undamped) | \(\mathbf{M}\ddot{\mathbf{x}} + \mathbf{K}\mathbf{x} = \mathbf{0}\) | \(\mathbf{M}\) = mass matrix; \(\mathbf{K}\) = stiffness matrix; \(\mathbf{x}\) = displacement vector |
| Eigenvalue problem | \((\mathbf{K} - \omega_i^2\mathbf{M})\mathbf{\Phi}_i = \mathbf{0}\) | \(\omega_i\) = natural frequency; \(\mathbf{\Phi}_i\) = mode shape (eigenvector) |
| Mass orthogonality condition | \(\mathbf{\Phi}_i^T\mathbf{M}\mathbf{\Phi}_j = 0\) for \(i \neq j\) | Mode shapes are orthogonal with respect to mass matrix |
| Stiffness orthogonality condition | \(\mathbf{\Phi}_i^T\mathbf{K}\mathbf{\Phi}_j = 0\) for \(i \neq j\) | Mode shapes are orthogonal with respect to stiffness matrix |
| Modal transformation | \(\mathbf{x}(t) = \mathbf{\Phi}\mathbf{q}(t)\) | \(\mathbf{q}(t)\) = modal coordinates; \(\mathbf{\Phi}\) = modal matrix |
| Uncoupled modal equation | \(\ddot{q}_i + \omega_i^2 q_i = 0\) | Each mode behaves as independent SDOF system |
| Rayleigh damping matrix | \(\mathbf{C} = \alpha\mathbf{M} + \beta\mathbf{K}\) | \(\alpha, \beta\) = Rayleigh damping coefficients |
| Modal damping ratio | \(\zeta_i = \frac{\alpha}{2\omega_i} + \frac{\beta\omega_i}{2}\) | \(\zeta_i\) = damping ratio for mode \(i\) |
| Damped modal equation | \(\ddot{q}_i + 2\zeta_i\omega_i\dot{q}_i + \omega_i^2 q_i = 0\) | Each damped mode as damped SDOF system |
| Modal force (forced vibration) | \(Q_i(t) = \mathbf{\Phi}_i^T\mathbf{f}(t)\) | \(\mathbf{f}(t)\) = external force vector; \(Q_i(t)\) = modal force |
| Modal superposition | \(\mathbf{x}(t) = \sum_{i=1}^{3} \mathbf{\Phi}_i q_i(t)\) | Total response is sum of all modal contributions |
A 3-DOF system has equal masses m = 2 kg and spring stiffnesses k₁ = k₂ = k₃ = k₄ = 1000 N/m. Calculate the three natural frequencies.
Show Solution
Solution:
Mass matrix: \(\mathbf{M} = m\mathbf{I} = 2\mathbf{I}\)
Stiffness matrix:
\[\mathbf{K} = \begin{bmatrix} 2k & -k & 0 \\ -k & 2k & -k \\ 0 & -k & 2k \end{bmatrix} = 1000\begin{bmatrix} 2 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 2 \end{bmatrix}\]
Eigenvalue problem: \(\det(\mathbf{K} - \omega^2\mathbf{M}) = 0\)
This gives three natural frequencies:
\(\omega_1 = \sqrt{\frac{k}{m}(2 - \sqrt{2})} = \sqrt{\frac{1000}{2}(2 - \sqrt{2})} = \sqrt{500 \times 0.5858} = \sqrt{292.9} = 17.1\) rad/s
\(\omega_2 = \sqrt{\frac{2k}{m}} = \sqrt{\frac{2000}{2}} = \sqrt{1000} = 31.6\) rad/s
\(\omega_3 = \sqrt{\frac{k}{m}(2 + \sqrt{2})} = \sqrt{\frac{1000}{2}(2 + \sqrt{2})} = \sqrt{500 \times 3.4142} = \sqrt{1707} = 41.3\) rad/s
Answer: \(\omega_1 = 17.1\) rad/s, \(\omega_2 = 31.6\) rad/s, \(\omega_3 = 41.3\) rad/s.
Explain why the mode shapes of a 3-DOF system are orthogonal with respect to the mass matrix.
Show Solution
Solution:
Mode shapes \(\mathbf{\Phi}_i\) and \(\mathbf{\Phi}_j\) satisfy the eigenvalue equations:
\(\mathbf{K}\mathbf{\Phi}_i = \omega_i^2\mathbf{M}\mathbf{\Phi}_i\)
\(\mathbf{K}\mathbf{\Phi}_j = \omega_j^2\mathbf{M}\mathbf{\Phi}_j\)
Pre-multiplying the first by \(\mathbf{\Phi}_j^T\) and the second by \(\mathbf{\Phi}_i^T\):
\(\mathbf{\Phi}_j^T\mathbf{K}\mathbf{\Phi}_i = \omega_i^2\mathbf{\Phi}_j^T\mathbf{M}\mathbf{\Phi}_i\)
\(\mathbf{\Phi}_i^T\mathbf{K}\mathbf{\Phi}_j = \omega_j^2\mathbf{\Phi}_i^T\mathbf{M}\mathbf{\Phi}_j\)
Since \(\mathbf{K}\) and \(\mathbf{M}\) are symmetric:
\((\omega_i^2 - \omega_j^2)\mathbf{\Phi}_j^T\mathbf{M}\mathbf{\Phi}_i = 0\)
For \(\omega_i \neq \omega_j\), we must have:
\(\mathbf{\Phi}_j^T\mathbf{M}\mathbf{\Phi}_i = 0\) (orthogonality condition)
Answer: Because \(\mathbf{K}\) and \(\mathbf{M}\) are symmetric, distinct natural frequencies force \(\mathbf{\Phi}_j^T\mathbf{M}\mathbf{\Phi}_i = 0\) — the mode shapes are mass-orthogonal.
A 3-DOF system has natural frequencies \(\omega_1 = 10\) rad/s and \(\omega_3 = 50\) rad/s. If we want damping ratios of \(\zeta_1 = 0.02\) and \(\zeta_3 = 0.05\), determine the Rayleigh damping coefficients \(\alpha\) and \(\beta\), and calculate the damping ratio for the second mode at \(\omega_2 = 30\) rad/s.
Show Solution
Solution:
Step 1: Use the Rayleigh damping formula \(\zeta_i = \frac{\alpha}{2\omega_i} + \frac{\beta\omega_i}{2}\)
For the first and third modes:
\[0.02 = \frac{\alpha}{2(10)} + \frac{\beta(10)}{2}\]
\[0.05 = \frac{\alpha}{2(50)} + \frac{\beta(50)}{2}\]
Step 2: Simplify:
\[0.02 = \frac{\alpha}{20} + 5\beta\]
\[0.05 = \frac{\alpha}{100} + 25\beta\]
Step 3: Solve for \(\beta\) from the system of equations:
From equation 1: \(\alpha = 0.4 - 100\beta\)
Substituting into equation 2: \(0.05 = \frac{0.4 - 100\beta}{100} + 25\beta\)
\[0.05 = 0.004 - \beta + 25\beta\]
\[0.046 = 24\beta\]
\[\beta = 0.00192 \text{ s}\]
Step 4: Find \(\alpha\):
\[\alpha = 0.4 - 100(0.00192) = 0.208 \text{ rad/s}\]
Step 5: Calculate \(\zeta_2\) at \(\omega_2 = 30\) rad/s:
\[\zeta_2 = \frac{0.208}{2(30)} + \frac{0.00192(30)}{2} = 0.00347 + 0.0288 = 0.0323\]
Answer: \(\alpha = 0.208\) rad/s, \(\beta = 0.00192\) s, and \(\zeta_2 = 0.0323\) (3.23%)
A 3-DOF system with natural frequencies \(\omega_1 = 15\) rad/s, \(\omega_2 = 30\) rad/s, and \(\omega_3 = 45\) rad/s is subjected to a harmonic force \(f_1(t) = 100\sin(32t)\) N on the first mass only. If all modes have damping ratio \(\zeta_i = 0.05\) and the modal participation factor for the second mode is \(Q_{20} = 50\) N, find the amplitude of the second modal coordinate.
Show Solution
Solution:
Given: \(\omega_2 = 30\) rad/s, \(\Omega = 32\) rad/s, \(\zeta_2 = 0.05\), \(Q_{20} = 50\) N
This problem treats \(Q_{20}\) in N and \(q_2\) in m directly, as a simplified convention — the mass-normalised eigenvectors used earlier in the lecture (\(\mathbf{\Phi}^T\mathbf{M}\mathbf{\Phi} = \mathbf{I}\)) would carry \(Q_i\) in N·kg\(^{-1/2}\) and \(q_i\) in kg\(^{1/2}\)·m instead.
Step 1: Calculate the frequency ratio:
\[r_2 = \frac{\Omega}{\omega_2} = \frac{32}{30} = 1.067\]
Step 2: Calculate the static deflection:
\[\delta_{st} = \frac{Q_{20}}{\omega_2^2} = \frac{50}{30^2} = \frac{50}{900} = 0.0556 \text{ m}\]
Step 3: Calculate the dynamic magnification factor:
\[D_2 = \frac{1}{\sqrt{(1-r_2^2)^2 + (2\zeta_2 r_2)^2}}\]
\[D_2 = \frac{1}{\sqrt{(1-1.067^2)^2 + (2 \times 0.05 \times 1.067)^2}}\]
\[D_2 = \frac{1}{\sqrt{(-0.138)^2 + (0.1067)^2}} = \frac{1}{\sqrt{0.0190 + 0.0114}} = \frac{1}{0.174} = 5.74\]
Step 4: Calculate the amplitude of the second modal coordinate:
\[|q_2| = \delta_{st} \times D_2 = 0.0556 \times 5.74 = 0.319 \text{ m}\]
Answer: The amplitude is 0.319 m or 319 mm. Note that this is a significant amplitude because the forcing frequency (32 rad/s) is very close to the second natural frequency (30 rad/s), creating a near-resonance condition.