---
title: "Helical spring under fluctuating load"
subject: "Machine Design II"
chapter: 6
key: "semester_7/machine_design_ii:spring-fatigue-soderberg-design"
source: https://www.bctengineer.com/q/semester_7/machine_design_ii/spring-fatigue-soderberg-design
site: BCT Engineering
---

# Helical spring under fluctuating load

Machine Design II · Chapter 6: Spring design

- Asked in: year not given, 8 marks
- Video: [L 38 Design of Helical Spring Against Fluctuating Load  | Design of machine Element | Mechanical](https://www.youtube.com/watch?v=94HJjGvF2Wk) — MECHANICAL & AUTOMOBILE ENGINEERING_LJIET
- 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-fatigue-soderberg-design
- Whole chapter: https://www.bctengineer.com/md/semester_7/machine_design_ii/chapter-6

## Question

A helical compression spring is subjected to a load that varies from 150 N to 450 N. The spring index is 6 and the factor of safety is 1.4 on the Soderberg line. The wire material has a torsional yield strength of 680 MPa and a torsional endurance limit (completely reversed) of 300 MPa. Find the wire diameter, and, if the deflection between the two loads is to be 12 mm, the number of active turns (G = 79.3 GPa). Use the Wahl factor for the alternating stress and the direct shear factor K_s = 1 + 0.5/C for the mean stress.

## Short answer

Fm=300 N, Fa=150 N; Soderberg criterion d^2 = (8*C*n/pi)(Kw Fa/Sse+Ks Fm/Ssy), d=4.86 mm -> 5 mm (D=30 mm). Stiffness k=25 N/mm gives active turns n = G d/(8 C^3 k) = 9.18 -> 9 turns.

## Answer

#### Loads
$$
F_m = \frac{450+150}{2} = 300\ \text{N}, \qquad F_a = \frac{450-150}{2} = 150\ \text{N}
$$

Spring index $C = 6$:

$$
K_w = \frac{23}{20}+\frac{0.615}{6} = 1.2525, \qquad K_s = 1 + \frac{0.5}{6} = 1.0833
$$

#### Stresses (with $D = Cd$)
$$
\tau_a = K_w\frac{8F_aC}{\pi d^2}, \qquad \tau_m = K_s\frac{8F_mC}{\pi d^2}
$$

#### Soderberg line
$$
\frac{1}{n} = \frac{\tau_a}{S_{se}} + \frac{\tau_m}{S_{sy}}
$$

Substitute and solve for $d^2$:

$$
d^2 = \frac{8C\,n}{\pi}\left(\frac{K_wF_a}{S_{se}} + \frac{K_sF_m}{S_{sy}}\right)
$$

$$
d^2 = \frac{8\times6\times1.4}{\pi}\left(\frac{1.2525\times150}{300} + \frac{1.0833\times300}{680}\right) = 21.39\,(0.6263 + 0.4779) = 23.62
$$

$$
d = 4.86\ \text{mm} \;\Rightarrow\; \text{use } d = 5\ \text{mm}
$$

#### Check with $d = 5$ mm
$D = 30$ mm.

$$
\tau_a = 1.2525\times\frac{8\times150\times30}{\pi\times125} = 114.8\ \text{MPa}, \quad \tau_m = 1.0833\times\frac{8\times300\times30}{\pi\times125} = 198.6\ \text{MPa}
$$

$$
\frac{1}{n} = \frac{114.8}{300} + \frac{198.6}{680} = 0.3827 + 0.2921 = 0.6748 \;\Rightarrow\; n = 1.48 \ (>1.4)
$$

Maximum stress $= 1.2525\times\dfrac{8\times450\times30}{\pi\times125} = 344$ MPa, well below the yield 680 MPa.

#### Number of active turns
Stiffness required: $k = \dfrac{450-150}{12} = 25$ N/mm.

$$
n = \frac{G d}{8C^3 k} = \frac{79300\times5}{8\times216\times25} = 9.18 \;\Rightarrow\; 9\ \text{turns}
$$

With $n = 9$: $k = 25.5$ N/mm and the deflection between loads is $300/25.5 = 11.8$ mm.

**Answer:** $d = 5$ mm ($D = 30$ mm), safety factor 1.48, 9 active turns.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For a load varying from 150 N to 450 N, what are F_m and F_a?
   A. F_m = 450, F_a = 150
   B. F_m = 300, F_a = 300
   C. F_m = 150, F_a = 450
   D. F_m = 300, F_a = 150
   Answer: D. F_m = (450+150)/2 = 300 N and F_a = (450-150)/2 = 150 N. (Key point: Mean and alternating loads)

2. Which stress factors are applied to the alternating and mean stresses?
   A. K_s for alternating, K_w for mean
   B. K_w for both
   C. K_w for alternating, K_s for mean
   D. No factor for either
   Answer: C. As stated in the problem, Wahl factor K_w multiplies the alternating stress and K_s = 1+0.5/C the mean stress. (Key point: Stress factors used)

3. What is the Soderberg design equation used?
   A. 1/n = tau_a/S_sy + tau_m/S_se
   B. 1/n = tau_a/S_se + tau_m/S_sy
   C. n = tau_a/S_se + tau_m/S_sy
   D. 1/n = tau_a/S_su + tau_m/S_se
   Answer: B. The alternating stress is divided by the endurance limit and the mean stress by the yield strength. (Key point: Soderberg criterion)

4. How is the number of active turns found?
   A. From stiffness k = (450-150)/12 = 25 N/mm and n = G d / (8 C^3 k)
   B. From k = 450/12 and n = G d / (8 C^3 k)
   C. From the Soderberg line directly
   D. From n = d^2 / D
   Answer: A. The change in load over the change in deflection gives k = 25 N/mm, then n = 79300x5/(8x216x25) = 9.18, taken as 9. (Key point: Turns from required stiffness)

## 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
- [Spring materials and yield strength estimation](https://www.bctengineer.com/question.md?key=semester_7%2Fmachine_design_ii%3Aspring-materials-yield-strength) — 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

## Related in other subjects

- [Max shear for 150 MPa stress limit; principal stresses](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Adirect-stresses-100t-60c-shear-limit) — Strength of Materials, chapter 4
- [Soderberg design with reversed bending and steady torque](https://www.bctengineer.com/question.md?key=semester_6%2Fmachine_design_i%3Asoderberg-bending-torsion-shaft) — Machine Design I, chapter 5
- [True length of 680 m line, 20 m tape](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atape-20m-680m-line) — Surveying I, chapter 2, asked 3×
- [Population from spring yield of 5 L/s](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Apop-s71) — Water Supply Engineering, chapter 3