Transmissibility and isolator design

(a) Transmissibility

Transmissibility $T_R$ is the ratio of the amplitude of the force transmitted to the foundation to the amplitude of the force applied on the machine (force transmissibility). The same ratio applies to motion (displacement transmissibility) for base excitation.

For $m\ddot{x} + c\dot{x} + kx = F_0\sin\omega t$, the steady-state amplitude is

$$X = \frac{F_0/k}{\sqrt{(1-r^2)^2 + (2\zeta r)^2}}, \qquad r = \frac{\omega}{\omega_n}$$

The force reaching the foundation is the sum of spring and damper forces (in quadrature):

$$F_T = X\sqrt{k^2 + (c\omega)^2} = kX\sqrt{1 + (2\zeta r)^2}$$ $$T_R = \frac{F_T}{F_0} = \sqrt{\frac{1 + (2\zeta r)^2}{(1-r^2)^2 + (2\zeta r)^2}}$$

Key points:

• $T_R > 1$ for $r < \sqrt2$ (the foundation sees more force than applied).
• $T_R = 1$ at $r = \sqrt2$ for every damping ratio.
• Isolation ($T_R < 1$) begins only for $r > \sqrt2$. Above this, more damping increases transmissibility, so light damping is best, while some damping is needed to limit the peak when passing resonance.

(b) Isolator design (undamped)

For $\zeta = 0$:

$$T_R = \frac{1}{r^2 - 1} \;(r > \sqrt{2})$$ $$0.2 = \frac{1}{r^2 - 1} \Rightarrow r^2 = 6 \Rightarrow r = 2.4495$$ $$\omega = \frac{2\pi \times 1200}{60} = 125.66\ \text{rad/s}, \qquad \omega_n = \frac{\omega}{r} = 51.30\ \text{rad/s}$$ $$k = m\omega_n^2 = 400 \times 51.30^2 = 1052758\ \text{N/m} = 1052.8\ \text{kN/m}$$ $$\delta_{st} = \frac{mg}{k} = \frac{g}{\omega_n^2} = 0.00373\ \text{m} = 3.73\ \text{mm}$$

Isolation efficiency $= 1 - T_R = 80\%$.

Answer: Spring stiffness $\approx 1052.8$ kN/m (total, all springs); static deflection $\approx 3.73$ mm.