Interactive Lecture
Base Excitation, Transmissibility, and Vibration Isolation
by Dr Arnaldo Delli Carri
Base excitation is a fundamental concept in vibration engineering where a system is excited through motion of its support or foundation rather than by directly applying a force to the mass. This type of vibration is commonly encountered in:
Base excitation occurs when the input to a mechanical system is the motion of its base or foundation, rather than a force applied directly to the mass.
A Single Degree of Freedom (SDOF) system with base excitation consists of:
To derive the equation of motion for a base-excited SDOF system, we need to identify all forces acting on the mass and apply Newton's Second Law.
For a mass m with absolute displacement x(t) and base with displacement y(t):
Spring force: $F_s = k(x - y)$
Damping force: $F_d = c(\dot{x} - \dot{y})$
Using Newton's Second Law: $m\ddot{x} + c(\dot{x} - \dot{y}) + k(x - y) = 0$
Rearranging: $m\ddot{x} + c\dot{x} + kx = c\dot{y} + ky$
If we define the relative displacement between the mass and the base as $z = x - y$, then:
$m\ddot{z} + c\dot{z} + kz = -m\ddot{y}$
Where $-m\ddot{y}$ is the effective force applied to the mass due to base acceleration.
In base excitation the effective forcing term is the base acceleration times the mass, \(-m\ddot{y}\) — not a force applied directly to the mass, as it would be in direct force excitation.
When the base motion is harmonic (sinusoidal), we can express it as:
$y(t) = Y\sin(\omega t)$
Where $Y$ is the amplitude of base motion and $\omega$ is the angular frequency.
The acceleration of the base is: $\ddot{y}(t) = -Y\omega^2\sin(\omega t)$
Substituting into the equation of motion:
$m\ddot{z} + c\dot{z} + kz = m Y \omega^2 \sin(\omega t)$
The steady-state solution for the relative displacement is:
$z(t) = Z\sin(\omega t - \varphi)$
Where $Z$ is the amplitude of relative motion and $\varphi$ is the phase angle between the base motion and the mass response.
Displacement transmissibility is the ratio of the amplitude of mass displacement to the amplitude of base displacement:
$T_d = \frac{|X|}{|Y|} = \sqrt{\frac{1+(2\zeta \rho)^2}{(1-\rho^2)^2+(2\zeta \rho)^2}}$
Where:
Force transmissibility is the ratio of the force transmitted to the foundation to the force generated by the base motion:
$T_f = \frac{F_{transmitted}}{F_{base}} = \sqrt{\frac{1+(2\zeta \rho)^2}{(1-\rho^2)^2+(2\zeta \rho)^2}}$
This has the same mathematical form as displacement transmissibility.
The frequency ratio decides whether base motion is amplified or isolated.
Many real-world vibrations are random rather than perfectly harmonic. Random vibration is characterised by:
The PSD describes how the power of a signal is distributed across different frequencies:
For base excitation with input displacement PSD $S_y(f)$, the response PSD is:
$S_x(f) = |H(f)|^2 S_y(f)$
Here $H(f)$ is the displacement transmissibility — the FRF relating base displacement to mass displacement:
$H(f) = \frac{1 + j\,2\zeta\rho}{(1-\rho^2) + j\,2\zeta\rho}, \qquad \rho = \frac{f}{f_n}$
This is dimensionless, so $S_x$ and $S_y$ carry the same units (m²/Hz). The force-input receptance $1/(-m\omega^2 + j\omega c + k)$ has units m/N and is the wrong transfer function for a displacement-driven base.
The root-mean-square (RMS) response can be calculated from the PSD:
$x_{RMS} = \sqrt{\int_{0}^{\infty} S_x(f) df}$
For a broadband random excitation with constant PSD $S_0$ and lightly damped system:
$x_{RMS} \approx \sqrt{\frac{\pi f_n S_0}{4\zeta}}$
Where $f_n$ is the natural frequency in Hz.
| Concept | Equation | Description |
|---|---|---|
| Equation of Motion | $m\ddot{z} + c\dot{z} + kz = -m\ddot{y}$ | Base excitation equation |
| Natural Frequency | $\omega_n = \sqrt{\frac{k}{m}}$ | Undamped natural frequency |
| Damping Ratio | $\zeta = \frac{c}{2\sqrt{km}}$ | Damping ratio |
| Frequency Ratio | \(\rho = \frac{\omega}{\omega_n}\) | Excitation to natural frequency ratio |
| Transmissibility | \(T_d = \frac{|X|}{|Y|} = \frac{\sqrt{1+(2\zeta \rho)^2}}{\sqrt{(1-\rho^2)^2+(2\zeta \rho)^2}}\) | Mass-to-base displacement ratio (same form as force transmissibility \(T_f\)) |
| Absolute Response | \(X = Y \cdot T_d\) | Mass displacement amplitude |
| Relative Response | \(Z = Y\cdot\frac{\rho^2}{\sqrt{(1-\rho^2)^2+(2\zeta \rho)^2}}\) | Displacement relative to base |
| Isolation Criterion | \(T_d < 1\) when \(\rho > \sqrt{2}\) | Condition for vibration isolation |
| Response PSD | $S_x(f) = |H(f)|^2 S_y(f)$ | Response power spectral density |
| RMS from PSD | $x_{RMS} = \sqrt{\int_{0}^{\infty} S_x(f) df}$ | Root-mean-square response |
A sensitive electronic device with mass 5 kg needs to be isolated from a base that vibrates with a frequency of 50 Hz. What should be the stiffness of the isolator to ensure the transmissibility is less than 0.1?
Show Solution
For transmissibility to be less than 0.1:
$T_d = \frac{1}{\rho^2} < 0.1$ (assuming high frequency, where $\rho \gg 1$)
$\rho^2 > 10$
$\rho > 3.16$
Since $\rho = \frac{\omega}{\omega_n} = \frac{2\pi \times 50}{\sqrt{k/m}}$, we need:
$2\pi \times 50 > 3.16 \times \sqrt{k/5}$
$\sqrt{k/5} < \frac{2\pi \times 50}{3.16} = 99.4 \text{ rad/s} \Rightarrow k/5 < 9884 \text{ N/m/kg}$
$k < 9884 \times 5 \approx 49{,}400 \text{ N/m}$
To ensure transmissibility less than 0.1, the stiffness should be less than approximately 49,400 N/m.
Answer: \(k < 49{,}400\ \text{N/m}\) (about 49.4 kN/m).
A machine with mass 100 kg is mounted on a platform using springs with total stiffness 40,000 N/m and dampers with damping coefficient 1,000 Ns/m. If the platform undergoes harmonic motion with amplitude 5 mm and frequency 10 Hz, calculate:
Show Solution
(a) Natural frequency:
$\omega_n = \sqrt{\frac{k}{m}} = \sqrt{\frac{40{,}000}{100}} = 20 \text{ rad/s}, \qquad f_n = \frac{\omega_n}{2\pi} = \frac{20}{2\pi} = 3.18 \text{ Hz}$
The frequency and damping ratios feed into parts (b) and (c):
$\rho = \frac{\omega}{\omega_n} = \frac{2\pi \times 10}{20} = 3.14, \qquad \zeta = \frac{c}{2\sqrt{km}} = \frac{1{,}000}{2\sqrt{40{,}000 \times 100}} = 0.25$
(b) Steady-state relative amplitude:
$Z = Y\cdot\frac{\rho^2}{\sqrt{(1-\rho^2)^2+(2\zeta \rho)^2}} = 0.005 \cdot \frac{3.14^2}{\sqrt{(1-3.14^2)^2+(2 \times 0.25 \times 3.14)^2}}$
$(1-\rho^2)^2 + (2\zeta\rho)^2 = (1-9.87)^2 + (1.571)^2 = 78.67 + 2.467 = 81.14$
$Z = 0.005 \cdot \frac{9.87}{\sqrt{81.14}} = 0.005 \cdot \frac{9.87}{9.008} = 0.00548 \text{ m} = 5.48 \text{ mm}$
(c) Transmissibility:
$T_d = \sqrt{\frac{1+(2\zeta \rho)^2}{(1-\rho^2)^2+(2\zeta \rho)^2}} = \sqrt{\frac{1+(2 \times 0.25 \times 3.14)^2}{(1-3.14^2)^2+(2 \times 0.25 \times 3.14)^2}}$
$T_d = \sqrt{\frac{1+2.467}{81.14}} = \sqrt{\frac{3.467}{81.14}} = 0.207$
The motion is attenuated to about 20.7% of the base motion.
Answer: \(\omega_n = 20\ \text{rad/s}\ (f_n = 3.18\ \text{Hz})\); relative amplitude \(Z = 5.48\ \text{mm}\); transmissibility \(T_d = 0.207\).
A building with mass 200,000 kg is equipped with a seismic isolation system. During an earthquake, the ground experiences harmonic motion with amplitude 100 mm and frequency 2 Hz. The isolation system has:
Calculate:
Show Solution
(a) Natural frequency and frequency ratio:
\[\omega_n = \sqrt{\frac{k}{m}} = \sqrt{\frac{8 \times 10^6}{200,000}} = \sqrt{40} = 6.32 \text{ rad/s}\]
\[f_n = \frac{\omega_n}{2\pi} = \frac{6.32}{2\pi} = 1.01 \text{ Hz}\]
Excitation frequency: \(\omega = 2\pi f = 2\pi \times 2 = 12.57 \text{ rad/s}\)
\[\rho = \frac{\omega}{\omega_n} = \frac{12.57}{6.32} = 1.99\]
(b) Damping ratio:
\[\zeta = \frac{c}{2\sqrt{km}} = \frac{400,000}{2\sqrt{8 \times 10^6 \times 200,000}} = \frac{400,000}{2 \times 1,264,911} = 0.158\]
(c) Transmissibility:
\[T_d = \sqrt{\frac{1 + (2\zeta \rho)^2}{(1 - \rho^2)^2 + (2\zeta \rho)^2}}\]
\[T_d = \sqrt{\frac{1 + (2 \times 0.158 \times 1.99)^2}{(1 - 1.99^2)^2 + (2 \times 0.158 \times 1.99)^2}}\]
\[T_d = \sqrt{\frac{1 + 0.396}{(1 - 3.96)^2 + 0.396}} = \sqrt{\frac{1.396}{8.76 + 0.396}} = \sqrt{\frac{1.396}{9.16}} = 0.390\]
(d) Absolute displacement amplitude of building:
\[X = Y \times T_d = 100 \text{ mm} \times 0.390 = 39.0 \text{ mm}\]
(e) Effectiveness assessment:
The isolation system reduces ground motion to 39% of its original amplitude (61% reduction). Since \(\rho = 1.99 \approx 2\), the system operates near the resonance region boundary. For \(T_d < 1\) (isolation), we need \(\rho > \sqrt{2} = 1.414\), which is satisfied. The system is effective, reducing building motion by 61% compared to the ground motion. However, for better isolation (e.g., \(T_d < 0.2\)), a softer spring system or higher frequency ratio would be needed.
Answer: \(\omega_n = 6.32\ \text{rad/s}\), \(\rho = 1.99\), \(\zeta = 0.158\), \(T_d = 0.390\), \(X = 39.0\ \text{mm}\); isolation is effective (\(\rho > \sqrt{2}\)).