Question 10 of 10

Question 8 (Part B, 20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Bed+patient of mass $M_t$ and centroidal mass moment of inertia $I_G=0.2M_tL^2$, supported on two identical compressive springs (stiffness $k$ each) a distance $L/2=1\,\text{m}$ either side of $G$; disturbance frequencies at $\omega$ (10–20 Hz) and $4\omega$ (40–80 Hz).

Find. A spring stiffness $k$ (equal springs, symmetric placement, so bounce and pitch modes are decoupled) that isolates the bed from both disturbance bands, and the resulting static deflection.

Approach. With equal springs placed symmetrically about $G$, the bounce and pitch equations of motion decouple, giving two independent natural frequencies; size $k$ so that both natural frequencies sit well below the lowest disturbance frequency (10 Hz) — the standard vibration-isolation design rule — then verify the transmissibility at every disturbance frequency and report the resulting static deflection.

1. Decoupled natural frequencies (equal springs $k$, symmetric at $\pm L/2$): $$\omega_{n,\text{bounce}} = \sqrt{\dfrac{2k}{M_t}}, \qquad \omega_{n,\text{pitch}} = \sqrt{\dfrac{2k(L/2)^2}{I_G}} = \sqrt{\dfrac{kL^2}{2\,I_G}} = \sqrt{\dfrac{k}{0.4\,M_t}} = \sqrt{2.5\,\dfrac{k}{M_t}}$$ so for this specific $I_G=0.2M_tL^2$, $\omega_{n,\text{pitch}}/\omega_{n,\text{bounce}} = \sqrt{2.5/2} = 1.118$ regardless of $k$ — the two modes always track together, 11.8% apart.
2. Design target. Standard isolation practice keeps $\omega_n \le \omega_{\min}/3$ so that transmissibility stays comfortably below 1 across the whole operating band; with $\omega_{\min}=2\pi(10)=62.83\,\text{rad/s}$: $$\boxed{\omega_{n,\text{bounce}} \le 62.83/3 = 20.94\ \text{rad/s}\ (3.33\ \text{Hz})}$$
3. Solve for the required stiffness per spring (per unit $M_t$): $$k = \dfrac{M_t\,\omega_{n,\text{bounce}}^2}{2} = \boxed{219.3\,M_t\ \ \text{N/m (each spring, }M_t\text{ in kg)}}$$ giving $\omega_{n,\text{bounce}}=20.94\,\text{rad/s}=3.33\,\text{Hz}$ and, from step 1, $\omega_{n,\text{pitch}}=23.42\,\text{rad/s}=3.73\,\text{Hz}$.
4. Static deflection check (the requested design consideration): $\delta_{st}=M_tg/(2k)=g/\omega_{n,\text{bounce}}^2$: $$\boxed{\delta_{st} = 9.81/20.94^2 = 22.4\ \text{mm per spring}}$$ — a practical, achievable deflection for a compression spring mount (not so soft that the bed sways excessively under normal handling).
5. Verify isolation (transmissibility, undamped) across the full disturbance range $TR=1/|1-(\omega/\omega_n)^2|$:

   Frequency	$TR$ (bounce, $\omega_n{=}3.33$ Hz)	$TR$ (pitch, $\omega_n{=}3.73$ Hz)
   10 Hz ($\omega_{\min}$)	0.125	0.161
   20 Hz ($\omega_{\max}$)	0.029	0.036
   40 Hz ($4\omega_{\min}$)	0.007	0.009
   80 Hz ($4\omega_{\max}$)	0.002	0.002

   Every combination of disturbance and mode gives $TR<0.16$ (at most 16% of the disturbance force is transmitted, worst case at $\omega_{\min}=10\,\text{Hz}$) — both bounce and pitch are safely isolated with no resonance risk anywhere in the 10–80 Hz range.