Forced vibration: isolation starts at root two

24 pushes a second

• A small unbalance pushes the pump 24 times a second. Whether the floor feels more or less than that push depends only on the frequency ratio $r = \omega/\omega_n$ and the damping.
• The steady amplitude is $X = (F_0/k)/\sqrt{(1 - r^2)^2 + (2\zeta r)^2}$. At resonance, $r = 1$, only damping limits it.

Transmissibility

• The transmissibility, force to the floor over force applied, is $T = \sqrt{[1 + (2\zeta r)^2]/[(1 - r^2)^2 + (2\zeta r)^2]}$.
• $T = 1$ at $r = \sqrt2$ for every $\zeta$: below it mounts amplify, above it they isolate. Vibration isolation wants a low natural frequency and low damping.

The present mounts

Designing for 90% isolation

• $T \le 0.1$ at $\zeta = 0.076$ needs $r \ge 3.40$, so $f_n \le 7.06$ Hz. Choose 6 Hz: $k = m(2\pi \times 6)^2 = \mathbf{170.5}$ kN/m, a static deflection of 6.9 mm.
• Then $r = 4.0$ and $T = \mathbf{0.078}$. The cost: softer mounts move more when a pipe pulls, so flexible pipe connections join the design.

Isolation from a spring's sag

• For a spring mount, $f_n = \tfrac{1}{2\pi}\sqrt{g/\delta_{st}}$, which is $15.76/\sqrt{\delta_{st}}$ Hz with $\delta_{st}$ in mm. Suppliers sell mounts by deflection for that reason.
• The 6 Hz design needs $\delta_{st} = 6.9$ mm; a 25 mm deflection spring gives $15.76/5 = \mathbf{3.15}$ Hz, for even better isolation and a much softer installation.

Worked example: an inertia block

• Bolt the pump to a 200 kg concrete block and size the springs for the same 6 Hz: $k = 320 \times (2\pi \times 6)^2 = 455$ kN/m.
• $T$ is unchanged at $r = 4$, but the motion is $X = F_0/[k(r^2 - 1)] = 27.3/(455\,000 \times 15) = \mathbf{4.0}$ µm instead of 10.7. The block steadies the pump for the pipes, not the floor.

Three isolation slips

Recap