Spring design for a helicopter‑suspended patient bed (bounce & pitch): 22-Mec-A2 Kinematics and Dynamics of Machines, December 2019, Q8

Question 8 — Spring design for a helicopter‑suspended patient bed (bounce & pitch) (20 marks — Part B)

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 / Find. Bed length $L=2$ m, four corner springs at $\pm a=\pm L/2=\pm1$ m from the C.G., excitation band $10$–$20$ Hz. Design the spring rate so that both the bounce (vertical) and pitch (rotational) natural frequencies lie safely below the excitation band, and check the static deflection.

Check — missing/corrupted data. The paper gives no patient‑bed mass $M$ and no units for $I_G$ (printed “$0.2$”). Taken literally, $I_G=0.2$ kg·m$^2$ makes the pitch frequency $\sim100$ Hz for a 2‑m bed — physically impossible — so the value is incomplete. The design below uses a representative $M=120$ kg (bed + patient) and the physical uniform‑bed inertia $I_G=ML^2/12=40$ kg·m$^2$; the method and criterion are exact, only the numbers are illustrative.

Approach. Model the bed as a 2‑DOF rigid body: vertical bounce $y$ and pitch $\theta$ about the C.G. Vibration isolation requires the natural frequencies well below the forcing band (transmissibility $T_R\lt1$ needs frequency ratio $r=f/f_n\gt\sqrt2$), so target $f_n\le f_{\min}/\sqrt2\approx7$ Hz; a softer $f_n\approx4$ Hz gives good isolation. Size the springs from that target and the static‑deflection relation.

Natural frequencies (2‑DOF, symmetric mounting). With four equal springs (total rate $k_{\text{tot}}=4k$) symmetric about the C.G., bounce and pitch decouple: $$f_{\text{bounce}}=\frac{1}{2\pi}\sqrt{\frac{k_{\text{tot}}}{M}},\qquad f_{\text{pitch}}=\frac{1}{2\pi}\sqrt{\frac{K_\theta}{I_G}},\quad K_\theta=k_{\text{tot}}\,a^2.$$
Static‑deflection design (bounce). Target $f_{\text{bounce}}=4$ Hz. Since $f_n=\tfrac1{2\pi}\sqrt{g/\delta_{st}}$, $$\delta_{st}=\frac{g}{(2\pi f_n)^2}=\frac{9.81}{(2\pi\cdot4)^2}=\boxed{15.5\ \text{mm}},$$ a sensible static sag. The total and per‑spring rates: $$k_{\text{tot}}=\frac{Mg}{\delta_{st}}=\frac{120(9.81)}{0.0155}\approx75.8\ \text{kN/m},\qquad k=\frac{k_{\text{tot}}}{4}\approx\boxed{18.9\ \text{kN/m per spring}}.$$
Pitch check. $K_\theta=k_{\text{tot}}a^2=75.8(1)^2=75.8$ kN·m/rad, so $$f_{\text{pitch}}=\frac{1}{2\pi}\sqrt{\frac{75\,800}{40}}=\boxed{6.9\ \text{Hz}}.$$ Both $f_{\text{bounce}}=4.0$ Hz and $f_{\text{pitch}}=6.9$ Hz are below the $10$ Hz lower excitation, so neither mode resonates in the $10$–$20$ Hz band.
Isolation performance. At the worst (lowest) forcing $f=10$ Hz, the bounce frequency ratio is $r=10/4=2.5$, giving transmissibility $$T_R=\frac{1}{r^2-1}=\frac{1}{2.5^2-1}=\boxed{0.19},$$ i.e. only $\sim19\%$ of the blade excitation reaches the patient in bounce (less at higher forcing) — “disastrous bouncing” is avoided. Add the dampers (each spring‑damper element) sized for $\zeta\approx0.1$–$0.2$ to control the pass through resonance on start‑up/shut‑down.

Quantity	Value (representative, flagged)
Target bounce frequency	4.0 Hz ($\le f_{\min}/\sqrt2\approx7$ Hz)
Static deflection $\delta_{st}$	15.5 mm
Spring rate per corner	18.9 kN/m ($k_{\text{tot}}=75.8$ kN/m)
Pitch frequency	6.9 Hz (below band)
Transmissibility at 10 Hz	0.19

Topic: 2‑DOF rigid‑body vibration isolation (bounce + pitch); static‑deflection spring sizing.

Key relations: $f_{\text{bounce}}=\tfrac1{2\pi}\sqrt{k_{\text{tot}}/M}=\tfrac1{2\pi}\sqrt{g/\delta_{st}}$; $f_{\text{pitch}}=\tfrac1{2\pi}\sqrt{k_{\text{tot}}a^2/I_G}$; isolation needs $r=f/f_n>\sqrt2$; transmissibility $T_R=1/(r^2-1)$ (undamped).

Why this works: Symmetric spring placement decouples heave and pitch, so each is a single‑DOF isolator. Placing both natural frequencies well below the forcing band puts the system in the isolation region ($T_R<1$); the static deflection then fixes the required softness independent of the (uncertain) mass, since $f_n$ depends only on $\delta_{st}$.

Common pitfall: Designing $f_n$ inside or just below the band (resonance / poor isolation); forgetting the pitch mode, which is stiffer ($K_\theta=k_{\text{tot}}a^2$) and can fall in‑band even when bounce is safe.

Source: Rao, Mechanical Vibrations 6e, Ch. 9 (vibration isolation, transmissibility) and Ch. 5–6 (multi‑DOF systems).

Question 8 of 8: Spring design for a helicopter‑suspended patient bed (bounce & pitch)

Question 8 — Spring design for a helicopter‑suspended patient bed (bounce & pitch) (20 marks — Part B)