---
title: "Design of a helical compression spring"
subject: "Machine Design II"
chapter: 6
key: "semester_7/machine_design_ii:compression-spring-design-static"
source: https://www.bctengineer.com/q/semester_7/machine_design_ii/compression-spring-design-static
site: BCT Engineering
---

# Design of a helical compression spring

Machine Design II · Chapter 6: Spring design

- Asked in: year not given, 8 marks
- Video: [Design of Helical Compression Springs - Numerical 3 Part 1 - Machine Design I](https://www.youtube.com/watch?v=W__j79G4sGw) — Ekeeda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/machine_design_ii/compression-spring-design-static
- Whole chapter: https://www.bctengineer.com/md/semester_7/machine_design_ii/chapter-6

## Question

Design a helical compression spring to carry a maximum axial load of 800 N with a deflection of 30 mm under this load. Take the spring index as 6, allowable shear stress 450 MPa and modulus of rigidity 81.4 GPa. The spring has squared and ground ends. Find the wire diameter (use Wahl factor), mean coil diameter, number of active and total turns, solid length and free length (allow 15% of the maximum deflection as clearance).

## Short answer

Wahl factor K=1.2525; d = sqrt(8*1.2525*800*6/(pi*450)) = 5.83 mm, select 6 mm. D = 36 mm. Deflection formula gives n = (G d delta)/(8 F C^3) = 10.6, take 11 active turns; total turns 13. Solid length 78 mm; with 15% clearance, free length = 114 mm.

## Answer

#### Data
$F = 800$ N, $\delta = 30$ mm, $C = 6$, $\tau = 450$ MPa, $G = 81.4$ GPa $= 81\,400$ N/mm².

#### Wire diameter
Wahl factor:

$$
K = \frac{4C-1}{4C-4} + \frac{0.615}{C} = \frac{23}{20} + 0.1025 = 1.2525
$$

From $\tau = K\dfrac{8FC}{\pi d^2}$:

$$
d^2 = \frac{8KFC}{\pi\tau} = \frac{8\times1.2525\times800\times6}{\pi\times450} = 34.02
\;\Rightarrow\; d = 5.83\ \text{mm}
$$

Select standard wire $d = 6$ mm. Check: $\tau = 1.2525\times\dfrac{8\times800\times6}{\pi\times 36} = 425\ \text{MPa} < 450$ MPa (safe).

#### Mean diameter
$$
D = C d = 6\times6 = 36\ \text{mm}, \qquad D_o = D + d = 42\ \text{mm}
$$

#### Active turns
From $\delta = \dfrac{8FC^3 n}{Gd}$:

$$
n = \frac{G d\,\delta}{8FC^3} = \frac{81400\times6\times30}{8\times800\times216} = 10.6
$$

Take $n = 11$ active turns. The actual stiffness is $k = \dfrac{Gd}{8C^3n} = \dfrac{81400\times6}{8\times216\times11} = 25.7$ N/mm, so the actual deflection is $800/25.7 = 31.1$ mm.

#### Total turns and lengths
Squared and ground ends add 2 inactive turns: $n_t = 11 + 2 = 13$.

- Solid length: $L_s = n_t d = 13\times6 = 78$ mm
- Clearance: $0.15\times31.1 \approx 4.7$ mm
- Free length: $L_f = L_s + \delta + \text{clearance} = 78 + 31.1 + 4.7 = 113.8$ mm

Pitch $p \approx \dfrac{L_f - 2d}{n} = \dfrac{113.8-12}{11} = 9.26$ mm.

Check for buckling: $L_f/D = 113.8/36 = 3.16$, which is below the usual limit of about 4, so the spring is stable without a guide. Squared and ground ends seated on parallel plates make buckling even less likely.

**Answer:** $d = 6$ mm, $D = 36$ mm, $n = 11$, total 13 turns, $L_s = 78$ mm, $L_f \approx 114$ mm.

## Check yourself

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

1. Which relation is used to find the wire diameter in this design?
   A. d^2 = 8 K F C / (pi tau)
   B. d^3 = 16 F D / (pi tau)
   C. d^2 = 4F / (pi tau)
   D. d^4 = 8 F D^3 n / (G delta)
   Answer: A. From tau = K 8FC/(pi d^2), solving for d^2 gives d^2 = 8KFC/(pi tau), giving 5.83 mm, rounded to 6 mm. (Key point: Wire diameter from Wahl stress)

2. How are the active turns n obtained?
   A. From the solid length of the spring
   B. From delta = 8 F C^3 n / (G d), giving n = G d delta / (8 F C^3)
   C. From the allowable shear stress directly
   D. By dividing free length by wire diameter
   Answer: B. The deflection formula is rearranged for n, giving 10.6, taken as 11. (Key point: Active turns)

3. For squared and ground ends with 11 active turns, what are the total turns and solid length (d = 6 mm)?
   A. 11 turns, 66 mm
   B. 12 turns, 72 mm
   C. 13 turns, 78 mm
   D. 15 turns, 90 mm
   Answer: C. Squared and ground ends add 2 inactive turns, so n_t = 13 and L_s = 13 x 6 = 78 mm. (Key point: Total turns and solid length)

4. How is the free length found?
   A. Solid length + deflection + clearance (15% of deflection)
   B. Solid length + deflection only
   C. Number of active turns x pitch only
   D. Solid length x 1.15
   Answer: A. L_f = L_s + delta + clearance = 78 + 31.1 + 4.7 = 113.8 mm. (Key point: Free length)

## Related in Machine Design II

- [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

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