--- title: "Surge and natural frequency of valve spring" subject: "Machine Design II" chapter: 6 key: "semester_7/machine_design_ii:spring-surge-natural-frequency" source: https://www.bctengineer.com/q/semester_7/machine_design_ii/spring-surge-natural-frequency site: BCT Engineering --- # Surge and natural frequency of valve spring Machine Design II · Chapter 6: Spring design - Asked in: year not given, 3+5 marks - Video: [Natural frequency of the spring](https://www.youtube.com/watch?v=ITDosu34YMU) — Yerbolat Magazov - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/machine_design_ii/spring-surge-natural-frequency - Whole chapter: https://www.bctengineer.com/md/semester_7/machine_design_ii/chapter-6 ## Question (a) What is surging of a helical spring and how can it be reduced? (b) A valve spring is made of steel wire of 4 mm diameter, mean coil diameter 32 mm and 7 active turns, with both ends fixed. G = 79.3 GPa and density = 7850 kg/m³. Find its fundamental natural frequency. If the camshaft can run up to 3000 rpm, find which harmonic of the cam frequency can resonate with the spring, and the cam speed at which the 13th harmonic would excite it. State whether the spring is acceptable if the natural frequency should be at least 13 times the cam frequency. ## Short answer Surging is longitudinal resonance when load frequency equals spring natural frequency, causing high stress. It can be reduced by raising natural frequency (larger wire, smaller diameter, fewer coils) or using dual/variable‑pitch springs. For the given spring: f = 199.6 Hz. The 4th harmonic can resonate near 3000 rpm, the 13th harmonic at 921 rpm. Not acceptable (required ≥ 650 Hz). ## Answer #### (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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is surge in a helical spring? A. Permanent set after overload B. Buckling of a long spring C. Resonance of the coils in longitudinal vibration when load frequency matches the spring natural frequency D. Increase of stiffness with temperature Answer: C. When the load frequency (including cam harmonics) equals the natural frequency, a compression wave travels along the coils and stress rises sharply. (Key point: Meaning of surge) 2. Which is a way to reduce surge? A. Use a spring with more active turns and larger D B. Use thicker coils spaced equally in a heavy wire C. Use dual concentric springs or variable pitch springs D. Lower the natural frequency of the spring Answer: C. Raising the natural frequency (larger d, smaller D, fewer turns), concentric springs whose friction damps, or unequal pitch reduce surge. (Key point: Reducing surge) 3. What is the fundamental frequency formula used for both ends fixed? A. f = (d / (2 pi D^2 n)) sqrt(G / (2 rho)) B. f = (D / (2 pi d^2 n)) sqrt(G / rho) C. f = 2 pi sqrt(k/m) D. f = (d / (pi D n)) sqrt(G / (2 rho)) Answer: A. f = (1/2) sqrt(k/m) with k = G d^4/(8 D^3 n) and m = rho (pi d^2/4)(pi D n) gives this expression, 199.6 Hz here. (Key point: Natural frequency formula) 4. With f = 199.6 Hz and a cam frequency of 50 Hz, what is the conclusion? A. Spring resonates with the 4th harmonic and fails the requirement of 650 Hz B. Spring resonates with the 13th harmonic and is acceptable C. Spring is acceptable because 199.6 > 50 D. Spring resonates with the fundamental at 3000 rpm Answer: A. f/f_cam is about 4, and the required f is 13 x 50 = 650 Hz, so the spring is not acceptable. (Key point: Harmonic and acceptance) ## Related in Machine Design II - [Design of a helical compression spring](https://www.bctengineer.com/question.md?key=semester_7%2Fmachine_design_ii%3Acompression-spring-design-static) — Machine Design II, chapter 6 - [Helical spring under fluctuating load](https://www.bctengineer.com/question.md?key=semester_7%2Fmachine_design_ii%3Aspring-fatigue-soderberg-design) — Machine Design II, chapter 6 - [Stress and deflection of helical spring](https://www.bctengineer.com/question.md?key=semester_7%2Fmachine_design_ii%3Ahelical-spring-stress-deflection-derive) — Machine Design II, chapter 6 - [Extension springs, ends and spring combinations](https://www.bctengineer.com/question.md?key=semester_7%2Fmachine_design_ii%3Aextension-spring-ends-series-parallel) — Machine Design II, chapter 6 ## Related in other subjects - [Euler's formula both ends fixed, with limitations](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aeuler-both-ends-fixed-limitation) — Strength of Materials, chapter 8 - [Bar fixed at both ends: load shares, displacements](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Acircular-bar-fixed-both-ends-50kn) — Strength of Materials, chapter 3 - [Euler's formula: both ends fixed](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aeuler-both-ends-fixed) — Strength of Materials, chapter 8, asked 2× - [Water hammer in 20 m steel pipeline with valve](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Awater-hammer-20m-pipeline-pressure-time-curves) — Hydraulics, chapter 4