Surge and natural frequency of valve spring

(a) Surge

Surge is resonance of the spring coils in longitudinal vibration. When the frequency of the applied load (for example, valve motion from a cam, including its harmonics) equals the natural frequency of the spring, a compression wave travels to and fro along the coils. The stress rises sharply, the spring may fail by fatigue, and the valve may bounce.

Reduction:

• Make the natural frequency high: increase $d$, reduce $D$ and the number of active turns, and use a material with a high $G/\rho$.
• Use dual (concentric) springs with different natural frequencies; friction between them also damps surge.
• Use springs with unequal (variable) pitch, or fit friction dampers.

(b) Numerical

Spring mass $m = \rho\cdot\dfrac{\pi d^2}{4}\cdot\pi D n$ and stiffness $k = \dfrac{Gd^4}{8D^3n}$. Fundamental frequency with both ends fixed:

$$f = \frac{1}{2}\sqrt{\frac{k}{m}} = \frac{d}{2\pi D^2 n}\sqrt{\frac{G}{2\rho}}$$

Substituting $d = 0.004$ m, $D = 0.032$ m, $n = 7$:

$$\sqrt{\frac{G}{2\rho}} = \sqrt{\frac{79.3\times10^9}{2\times7850}} = 2247.6\ \text{m/s}$$ $$f = \frac{0.004}{2\pi\,(0.032)^2\,7}\times2247.6 = 0.08883\times2247.6 = 199.6\ \text{Hz}$$

Cam frequency at 3000 rpm $= 50$ Hz. The ratio $f/f_{cam} = 199.6/50 = 3.99 \approx 4$, so the 4th harmonic of the cam excites the spring at maximum speed (resonance at about 2994 rpm, almost the top speed). Higher harmonics (5th, 6th, ...) resonate at lower cam speeds within the operating range.

Cam speed at which the 13th harmonic resonates:

$$N = \frac{60 f}{13} = \frac{60\times199.6}{13} = 921\ \text{rpm}$$

(The fundamental would resonate at $60f = 11\,976$ rpm.)

Acceptance: the required natural frequency is $\ge 13\times50 = 650$ Hz, but $f = 199.6$ Hz. The spring is not acceptable; it needs a stiffer, lighter design (larger $d/D^2$, fewer turns) or dual springs.

Answer: $f = 199.6$ Hz; the 4th harmonic resonates near maximum cam speed; the 13th harmonic resonates at 921 rpm; the spring is unsafe against surge.