---
title: "Stress and deflection of helical spring"
subject: "Machine Design II"
chapter: 6
key: "semester_7/machine_design_ii:helical-spring-stress-deflection-derive"
source: https://www.bctengineer.com/q/semester_7/machine_design_ii/helical-spring-stress-deflection-derive
site: BCT Engineering
---

# Stress and deflection of helical spring

Machine Design II · Chapter 6: Spring design

- Asked in: year not given, 8 marks
- Video: [Mechanical Springs - Stress, Deflection, and Spring Constant in Just Over 10 MINUTES!](https://www.youtube.com/watch?v=5Vv1NBvpG4Y) — Less Boring Lectures
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/machine_design_ii/helical-spring-stress-deflection-derive
- Whole chapter: https://www.bctengineer.com/md/semester_7/machine_design_ii/chapter-6

## Question

Derive expressions for the shear stress and the deflection of a close-coiled helical compression spring of circular wire. Explain the Wahl stress correction factor and state its expression.

## Short answer

tau = 8FD/(pi d^3) times Ks with Ks=1+0.5/C. Wahl factor K = (4C-1)/(4C-4)+0.615/C includes curvature effect; tau_max = K * 8FD/(pi d^3). Deflection delta = 8FD^3 n/(G d^4).

## Answer

Notation: $F$ = axial load, $D$ = mean coil diameter, $d$ = wire diameter, $C = D/d$ (spring index), $n$ = number of active coils, $G$ = shear modulus.

#### Stress
A section of the wire is subjected to a **direct shear force** $F$ and a **torque** $T = FD/2$.

```text
        F
        |
    ---(  )---   wire section: shear F + torque T = F D/2
       D/2
```

Torsional shear stress: $\tau_t = \dfrac{16T}{\pi d^3} = \dfrac{8FD}{\pi d^3}$

Direct shear stress: $\tau_d = \dfrac{F}{\pi d^2/4} = \dfrac{4F}{\pi d^2}$

Maximum stress (inner fibre) by simple addition:

$$
\tau = \frac{8FD}{\pi d^3}\left(1 + \frac{d}{2D}\right) = K_s\,\frac{8FD}{\pi d^3}, \quad K_s = 1 + \frac{1}{2C}
$$

#### Wahl correction factor
Because the wire is curved, the stress is higher at the inner side of the coil, and the direct shear is not uniform. **A. M. Wahl** included the curvature effect and direct shear in one factor:

$$
K = \frac{4C-1}{4C-4} + \frac{0.615}{C}
$$

$$
\tau_{max} = K\,\frac{8FD}{\pi d^3} = K\,\frac{8FC}{\pi d^2}
$$

For $C = 6$: $K = 1.2525$. For small $C$ (tight coils) $K$ is large; for static loads of ductile wire $K_s$ may be used, but for fatigue $K$ is used.

#### Deflection (strain energy method)
Strain energy due to torsion of the wire (length $l = \pi D n$, polar moment $J = \pi d^4/32$):

$$
U = \frac{T^2 l}{2GJ} = \frac{(FD/2)^2(\pi D n)}{2G(\pi d^4/32)} = \frac{4F^2D^3 n}{G d^4}
$$

Work done by load $= \tfrac{1}{2}F\delta$. Equating, $\tfrac12 F\delta = U$:

$$
\delta = \frac{8FD^3 n}{G d^4} = \frac{8FC^3 n}{G d}
$$

#### Stiffness
$$
k = \frac{F}{\delta} = \frac{G d^4}{8D^3 n} = \frac{G d}{8C^3 n}
$$

Stiffness increases with the fourth power of the wire diameter and falls with the cube of coil diameter.

## Check yourself

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

1. What does a section of wire in a close-coiled helical spring carry under axial load F?
   A. Only a bending moment F D/2
   B. A direct shear force F and a torque T = F D/2
   C. A direct shear force F and a bending moment F D
   D. Only a torque T = F D
   Answer: B. Because the load acts along the axis, offset D/2 from the wire, the section sees direct shear F plus a torque FD/2. (Key point: Loads on wire section)

2. Why is the Wahl correction factor K used instead of simple K_s = 1 + 1/(2C)?
   A. It includes wire curvature (higher stress at the inner side) as well as direct shear
   B. It accounts for buckling of long springs
   C. It corrects for the number of active coils
   D. It corrects only for the non-uniform direct shear
   Answer: A. Wahl's factor includes both the curvature effect and direct shear, whereas K_s covers only direct shear. (Key point: Wahl factor purpose)

3. How is the deflection of the spring obtained in the answer?
   A. By equating (1/2)F*delta to the torsional strain energy of the wire
   B. By equating F*delta to the bending strain energy
   C. By using beam deflection formula for the wire
   D. By equating (1/2)F*delta to the direct-shear energy only
   Answer: A. Work done (1/2)F*delta equals the torsional strain energy U = T^2 l/(2GJ), giving delta = 8FD^3 n/(G d^4). (Key point: Deflection by strain energy)

4. According to k = G d^4 / (8 D^3 n), how does stiffness change?
   A. Increases with D^3 and falls with d^4
   B. Increases with d^4 and falls with D^3
   C. Increases with n and with d
   D. Is independent of the wire diameter
   Answer: B. Stiffness rises with the fourth power of wire diameter and falls with the cube of coil diameter (and with n). (Key point: Stiffness dependence)

## 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

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