12.2 Vibration Isolation, Multi-DOF Systems & Transmissibility

12.2 Vibration Isolation, Multi-DOF Systems & Transmissibility

Vibration isolation prevents the transmission of dynamic forces from operating machinery into supporting structures or, conversely, protects sensitive instrumentation from floor vibration. Achieving effective isolation requires proper matching of mount stiffness, operating speeds, and damping ratios. This section examines force and motion transmissibility, the fundamental isolation threshold $r > \sqrt{2}$, isolator selection, and multi-degree-of-freedom (MDOF) modal dynamics including tuned mass dampers.

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Principles of Vibration Isolation: Source vs. Receiver

Vibration isolation addresses two complementary engineering problems:

1. Source Isolation (Direct Force Excitation): A vibrating machine generates internal unbalanced dynamic forces $F(t) = F_0 \cos(\omega t)$. The objective is to minimize the dynamic force $F_T$ transmitted through the mounts into the supporting floor or foundation.
2. Receiver Isolation (Base Motion Excitation): Ground or foundation displacement $y(t) = Y_0 \cos(\omega t)$ (from seismic activity, traffic, or nearby forging presses) vibrates the mounts. The objective is to minimize the absolute displacement response $X$ transmitted to a sensitive instrument or payload.

          Source Isolation                            Receiver Isolation
    +---------------------------+              +---------------------------+
    | Machine: F(t)=F_0 cos(wt) |              |  Sensitive Metrology CMM  |
    +-------------+-------------+              +-------------+-------------+
           |             |                            |             |
          [k]           [c]                          [k]           [c]
           |             |                            |             |
    =======+=============+=======              =======+=============+=======
    Foundation: Transmitted Force F_T          Moving Base: y(t)=Y_0 cos(wt)

Force Transmitted to the Foundation

In source isolation, the machine mass moves with displacement $x(t) = X \cos(\omega t - \phi)$. The dynamic force transmitted to the foundation is the vector sum of spring force and viscous damping force:

$$f_T(t) = k x(t) + c \dot{x}(t) = k X \cos(\omega t - \phi) - c \omega X \sin(\omega t - \phi)$$

Because the spring force and damping force are $90^\circ$ out of phase, their resultant peak amplitude is found via vector (phasor) addition:

$$F_T = \sqrt{(k X)^2 + (c \omega X)^2} = X \sqrt{k^2 + c^2 \omega^2} = k X \sqrt{1 + \left(\frac{c\omega}{k}\right)^2}$$

Recalling that $\frac{c}{k} = \frac{2\zeta \omega_n}{\omega_n^2} = \frac{2\zeta}{\omega_n}$ and $r = \frac{\omega}{\omega_n}$:

$$F_T = k X \sqrt{1 + (2\zeta r)^2}$$

Substituting the harmonic displacement amplitude $X = \frac{F_0 / k}{\sqrt{(1 - r^2)^2 + (2\zeta r)^2}}$ into the force equation gives:

$$F_T = F_0 \left[\frac{\sqrt{1 + (2\zeta r)^2}}{\sqrt{(1 - r^2)^2 + (2\zeta r)^2}}\right]$$

Force and Motion Transmissibility Formula

The Transmissibility ($TR$) is defined as the ratio of transmitted force amplitude to applied excitation force amplitude:

$$TR = \frac{F_T}{F_0} = \sqrt{\frac{1 + (2\zeta r)^2}{(1 - r^2)^2 + (2\zeta r)^2}}$$

For base excitation (receiver isolation) where base displacement is $y(t) = Y_0 \cos(\omega t)$, Newton's equation yields the identical mathematical ratio for motion transmissibility:

$$TR_{\text{motion}} = \frac{X}{Y_0} = \sqrt{\frac{1 + (2\zeta r)^2}{(1 - r^2)^2 + (2\zeta r)^2}}$$

Force transmissibility and motion transmissibility are mathematically equivalent.

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The Isolation Criteria: The $r > \sqrt{2}$ Rule

To achieve true vibration isolation, the transmitted force must be strictly less than the excitation force ($TR < 1.0$). Setting $TR = 1.0$ in the transmissibility equation reveals the critical operating boundary:

$$\frac{1 + (2\zeta r)^2}{(1 - r^2)^2 + (2\zeta r)^2} = 1.0 \implies (1 - r^2)^2 = 1.0$$

Taking the square root:

$$1 - r^2 = -1 \implies r^2 = 2 \implies r = \sqrt{2} \approx 1.414$$

                  Transmissibility TR vs Frequency Ratio r
      TR ^
         |          /|\  Light Damping (Resonance Peak)
     4.0 +         / | \
         |        /  |  \
     2.0 +       /   |   \
         |      /    |    \
     1.0 +-----+-----+-----\----+-------------------------> TR = 1.0
         |    /      |      \    \____ Higher damping has HIGHER TR here!
     0.0 +---+-------+-------+----+----------------------->
         0  0.5     r=1.0   r=sqrt(2)                     r
             Amplification   Crossover     ISOLATION REGION
                Region        Boundary        (TR < 1.0)

Frequency Regimes of Transmissibility

1. $0 < r < 1.0$ (Stiffness Zone): $TR > 1.0$. Transmitted force exceeds applied force. At $r = 0$, $TR = 1.0$.
2. $r = 1.0$ (Resonance Zone): $TR$ reaches its maximum value, governed by damping: $TR_{r=1} = \frac{\sqrt{1 + 4\zeta^2}}{2\zeta} \approx \frac{1}{2\zeta}$.
3. $1.0 < r < \sqrt{2}$ (Amplification Zone): $TR > 1.0$. The mounts amplify force transmitted to the foundation.
4. $r = \sqrt{2}$ (Crossover Point): $TR = 1.0$ regardless of the value of damping ratio $\zeta$. All curves pass through $(r = \sqrt{2}, TR = 1.0)$.
5. $r > \sqrt{2}$ (Isolation Zone): $TR < 1.0$. True vibration isolation occurs. Effective isolation requires operating well into this region, typically $r \ge 2.5$ to $4.0$.

Isolation Efficiency

The effectiveness of an isolator is expressed as Isolation Efficiency $\eta_I$:

$$\eta_I = (1 - TR) \times 100\%$$

For an undamped or lightly damped isolator ($\zeta \ll 1$) operating in the isolation region ($r > \sqrt{2}$), transmissibility simplifies to:

$$TR \approx \frac{1}{r^2 - 1}$$ $$\eta_I \approx \left(1 - \frac{1}{r^2 - 1}\right) = \frac{r^2 - 2}{r^2 - 1}$$

To achieve $90\%$ isolation efficiency ($\eta_I = 0.90$, meaning $TR = 0.10$):

$$0.10 \approx \frac{1}{r^2 - 1} \implies r^2 - 1 = 10 \implies r = \sqrt{11} \approx 3.317$$

The operating speed must be at least $3.32$ times the natural frequency of the mounting system.

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The Damping Paradox in Vibration Isolation

Examining the transmissibility plot reveals an important engineering trade-off known as the Damping Paradox:

• In the resonant region ($r < \sqrt{2}$): Damping is essential to limit excessive vibration amplitudes and structural stresses as machinery passes through resonance during startup and shutdown.
• In the isolation region ($r > \sqrt{2}$): Higher damping increases transmissibility ($TR$) and reduces isolation efficiency! The dashpot creates a direct dynamic path between the machine and foundation. At very high frequencies ($r \gg 1$), the damper force dominates the spring force, causing $TR$ to decay as $2\zeta / r$ rather than $1 / r^2$.

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Isolator Media Selection & Properties

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Multi-Degree-of-Freedom (MDOF) Systems

Real mechanical installations possess multiple independent displacement coordinates. For an $N$-degree-of-freedom system, the generalized coordinate vector is $\mathbf{x}(t) = [x_1(t), x_2(t), \dots, x_N(t)]^T$.

Matrix Equations of Motion

For undamped free vibration, dynamic equilibrium yields the matrix system:

$$[M] \mathbf{\ddot{x}} + [K] \mathbf{x} = \mathbf{0}$$

where $[M]$ is the $N \times N$ symmetric, positive-definite mass matrix, and $[K]$ is the $N \times N$ symmetric stiffness matrix.

               Two-Degree-of-Freedom (2-DOF) System Model
         +------------------------------------------------+
         |                   Fixed Wall                   |
         +-------+--------------------------------+-------+
                 |                                |
                [k_1]                            [k_3] (Optional)
                 |                                |
                 v                                v
           +-----------+                    +-----------+
           | Mass (m_1)| <--- [k_2] ------->| Mass (m_2)|
           +-----------+                    +-----------+
              x_1(t)                           x_2(t)

Assuming synchronous harmonic motion at circular frequency $\omega$:

$$\mathbf{x}(t) = \mathbf{u} e^{i \omega t}$$

Substituting into the matrix equation produces the algebraic generalized eigenvalue problem:

$$([K] - \omega^2 [M]) \mathbf{u} = \mathbf{0}$$

The Characteristic Equation & Eigenvalues

For non-trivial eigenvector solutions ($\mathbf{u} \ne \mathbf{0}$), the characteristic determinant must equal zero:

$$\det([K] - \omega^2 [M]) = 0$$

For a two-degree-of-freedom (2-DOF) system:

$$\det \begin{bmatrix} k_{11} - \omega^2 m_{11} & k_{12} - \omega^2 m_{12} \\ k_{21} - \omega^2 m_{21} & k_{22} - \omega^2 m_{22} \end{bmatrix} = 0$$

Expanding this determinant yields a quadratic equation in $\lambda = \omega^2$ (the characteristic polynomial). Solving gives two roots: $\omega_1^2$ and $\omega_2^2$:

• $\omega_1$: First (fundamental) circular natural frequency.
• $\omega_2$: Second circular natural frequency ($\omega_2 > \omega_1$).

Mode Shapes (Eigenvectors) and Orthogonality

Substituting each natural frequency $\omega_r$ back into $([K] - \omega_r^2 [M]) \mathbf{u}^{(r)} = \mathbf{0}$ yields the corresponding mode shape vector $\mathbf{u}^{(r)}$:

• Mode 1 ($\,\omega_1\,$): Masses vibrate in-phase with one another (no internal node between masses).
• Mode 2 ($\,\omega_2\,$): Masses vibrate out-of-phase with one another (a stationary nodal point exists between masses).

Mode shapes satisfy mass-orthogonality and stiffness-orthogonality:

$$\mathbf{u}^{(1)T} [M] \mathbf{u}^{(2)} = 0, \qquad \mathbf{u}^{(1)T} [K] \mathbf{u}^{(2)} = 0$$

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The Dynamic Vibration Absorber (Tuned Mass Damper - TMD)

When a primary machine structure of mass $M$ and stiffness $K$ is excited by a harmonic force $F(t) = F_0 \sin(\Omega t)$ operating at or near its resonance frequency $\omega_n = \sqrt{K/M}$, excessive vibration occurs. A dynamic vibration absorber attaches an auxiliary mass $m_a$ and spring $k_a$ directly to the primary mass.

                 Tuned Mass Damper (TMD) Architecture
            +--------------------------------------------+
            | Primary Machine: Mass M, Stiffness K       | ---> F_0 sin(Omega*t)
            +---------------------+----------------------+
                                  |
                                 [k_a] (Auxiliary Spring)
                                  |
                                  v
                            +-----------+
                            | Mass(m_a) | (Auxiliary Absorber Mass)
                            +-----------+

The undamped coupled equations of motion are:

$$M \ddot{x}_1 + (K + k_a) x_1 - k_a x_2 = F_0 \sin(\Omega t)$$ $$m_a \ddot{x}_2 + k_a (x_2 - x_1) = 0$$

Solving for the steady-state harmonic amplitudes $x_1(t) = X_1 \sin(\Omega t)$ and $x_2(t) = X_2 \sin(\Omega t)$:

$$X_1 = \frac{F_0 (k_a - m_a \Omega^2)}{(K + k_a - M \Omega^2)(k_a - m_a \Omega^2) - k_a^2}$$ $$X_2 = \frac{F_0 k_a}{(K + k_a - M \Omega^2)(k_a - m_a \Omega^2) - k_a^2}$$

The Tuning Condition

Examining the numerator of $X_1$, the motion of the primary mass becomes identically zero ($X_1 = 0$) when the numerator vanishes:

$$k_a - m_a \Omega^2 = 0 \implies \omega_a = \sqrt{\frac{k_a}{m_a}} = \Omega$$

When the absorber's natural frequency $\omega_a$ is tuned exactly to the excitation frequency $\Omega$:

• The primary mass $M$ is completely stationary ($X_1 = 0$).
• The auxiliary spring exerts an oscillating force $-k_a X_2$ that is exactly equal in magnitude and opposite in phase to the applied external force $F_0$:

$$X_2 = -\frac{F_0}{k_a}$$

Operational Limitation

Attaching the absorber converts the original 1-DOF system into a 2-DOF system. While it neutralizes vibration at $\Omega$, it introduces two new resonant frequencies: one below $\Omega$ and one above $\Omega$. Therefore, an undamped TMD is suitable only for constant-speed machinery. If excitation frequency drifts, the machine operates on one of the new resonance peaks unless damping is added to the absorber (creating a damped Tuned Mass Damper).

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Step-by-Step Worked Vibration Isolation Design Problem

Problem Statement

A rooftop industrial HVAC compressor package has a total operating mass $M = 1,\!200\,\text{kg}$. The compressor operates at a constant running speed $N = 1,\!800\,\text{rpm}$. Structural criteria require that the vibration isolation system achieve a minimum isolation efficiency of $88\%$ (transmissibility $TR \le 0.12$) transmitted to the building concrete floor slab. The selected isolators provide a damping ratio $\zeta = 0.05$.

Calculate:

1. The operating circular frequency $\omega$.
2. The minimum required frequency ratio $r$ to satisfy $TR = 0.12$.
3. The maximum permissible system natural frequency $\omega_n$ and cyclic natural frequency $f_n$.
4. The required static deflection $\delta_{st}$ of the mounts under the machine's static weight.
5. The total required mounting stiffness $K_{\text{tot}}$, and the stiffness $k_i$ for each of four identical corner isolator mounts.
6. The appropriate isolator medium selection.

Step-by-Step Solution

Step 1: Compressor Operating Frequency

Convert running speed from rpm to radians per second:

$$\omega = \frac{2\pi N}{60} = \frac{2\pi \times 1,\!800}{60} = 60\pi \approx 188.50\,\text{rad/s}$$ $$f_{\text{operating}} = \frac{1,\!800}{60} = 30.0\,\text{Hz}$$

Step 2: Determine Required Frequency Ratio ($r$)

The transmissibility equation is:

$$TR = \sqrt{\frac{1 + (2\zeta r)^2}{(1 - r^2)^2 + (2\zeta r)^2}}$$

Squaring both sides with $TR = 0.12$ and $\zeta = 0.05$:

$$(0.12)^2 = \frac{1 + [2(0.05) r]^2}{(1 - r^2)^2 + [2(0.05) r]^2} \implies 0.0144 = \frac{1 + 0.01 r^2}{(1 - r^2)^2 + 0.01 r^2}$$

Letting $u = r^2$:

$$0.0144 \left[(1 - u)^2 + 0.01 u\right] = 1 + 0.01 u$$ $$0.0144 \left[u^2 - 2u + 1 + 0.01 u\right] = 1 + 0.01 u$$ $$0.0144 \left[u^2 - 1.99 u + 1\right] = 1 + 0.01 u$$ $$0.0144 u^2 - 0.028656 u + 0.0144 = 1 + 0.01 u$$ $$0.0144 u^2 - 0.038656 u - 0.9856 = 0$$

Dividing through by $0.0144$:

$$u^2 - 2.6844 u - 68.444 = 0$$

Applying the quadratic formula:

$$u = \frac{2.6844 + \sqrt{(-2.6844)^2 - 4(1)(-68.444)}}{2} = \frac{2.6844 + \sqrt{7.206 + 273.778}}{2} = \frac{2.6844 + 16.763}{2} = 9.724$$

Since $u = r^2$:

$$r = \sqrt{9.724} \approx 3.118$$

(Note: Using the undamped approximation $TR \approx 1/(r^2 - 1)$ gives $r^2 - 1 = 1/0.12 = 8.333 \implies r = \sqrt{9.333} = 3.055$. Exact solution with damping yields $r = 3.118$.)

Step 3: Maximum Allowable Natural Frequency

To maintain $r \ge 3.118$:

$$\omega_n = \frac{\omega}{r} = \frac{188.50\,\text{rad/s}}{3.118} = 60.455\,\text{rad/s}$$ $$f_n = \frac{\omega_n}{2\pi} = \frac{60.455}{2\pi} \approx 9.62\,\text{Hz}$$

The mounting system natural frequency must not exceed $9.62\,\text{Hz}$.

Step 4: Required Static Deflection

Using the gravitational static deflection relation ($g = 9.81\,\text{m/s}^2$):

$$\delta_{st} = \frac{g}{\omega_n^2} = \frac{9.81\,\text{m/s}^2}{(60.455\,\text{rad/s})^2} = \frac{9.81}{3,\!654.8} = 0.002684\,\text{m} = 2.68\,\text{mm} \quad (\approx 0.106\,\text{in})$$

Step 5: Required Mounting Stiffness

The total system stiffness is:

$$K_{\text{tot}} = M \omega_n^2 = 1,\!200\,\text{kg} \times 3,\!654.8\,\text{rad}^2/\text{s}^2 = 4,\!385,\!800\,\text{N/m} = 4.386\,\text{MN/m}$$

For four identical corner mounts supporting the load equally in parallel:

$$k_i = \frac{K_{\text{tot}}}{4} = \frac{4,\!385,\!800\,\text{N/m}}{4} = 1,\!096,\!450\,\text{N/m} \approx 1,\!096\,\text{kN/m} = 1,\!096\,\text{N/mm}$$

Step 6: Isolator Selection Analysis

The required static deflection is $\delta_{st} = 2.68\,\text{mm}$ with a natural frequency $f_n = 9.62\,\text{Hz}$. Consulting the isolator selection table:

• Elastomeric (neoprene) shear mounts easily support deflections up to $6 - 12\,\text{mm}$ and natural frequencies in the $7 - 12\,\text{Hz}$ range.
• Neoprene isolators provide the required internal damping $\zeta \approx 0.05$ without acoustic ringing.
• Steel springs are unnecessary because the required deflection does not demand large spring travels (> $25\,\text{mm}$).

Final Engineering Specification: Install four commercial elastomeric (neoprene-in-shear) vibration isolators, each rated for $k_i = 1,\!096\,\text{kN/m}$ ($6,\!260\,\text{lbf/in}$) with a rated static load of $300\,\text{kg}$ ($2,\!943\,\text{N}$) each.