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Machine Elements · Chapter 8 of 10 · Intermediate

Mechanical Springs

A helical spring is a bar in torsion, coiled up. That insight gives both the stress in the wire and the deflection per unit load, the two numbers a spring is designed around.

Readiness check

This chapter treats a coiled wire as a torsion member. Tick only what you can do closed-notes.

Recall torsional shear stress τ = Tr/J.
Use the shear modulus G.
Relate force, deflection, and a spring rate.
Evaluate a power law for material strength.
Keep track of wire and coil diameters separately.

0 or 1 weak itemsContinue with this chapter.

2 weak itemsRevisit the spring rate in Chapter 3.

3 or more weak itemsReview torsion in Mechanics of Materials, Chapter 3.

The core idea

Loading a helical spring twists its wire. The shear stress, raised by curvature through the Wahl factor, and the spring rate both follow from the coil and wire diameters.

C = D/dτ = K_W · 8FD/(πd³)k = d⁴G/(8D³N_a)

An axial force F on a coil of mean diameter D applies a torque FD/2 to the wire of diameter d, so the wire sees a torsional shear stress. The tight curvature of the inside of the coil raises that stress, captured by the Wahl correction factor K_W, which depends on the spring index C = D/d. The same torsional flexibility, summed over N_a active coils, gives the spring rate k. Stress sets whether the wire survives; the rate sets how the spring feels.

The skill works when: you treat the wire as a torsion bar, apply the Wahl factor, and compute both stress and rate.

The skill breaks down when: the Wahl factor is dropped, or coil and wire diameters are confused.

The concept. An axial load twists the spring wire. The coil diameter D sets the torque arm; the wire diameter d carries it as a torsional shear stress, amplified at the inner fibre by curvature.

The skills, taught in order

Five skills define spring geometry, the corrected shear stress, the spring rate, the deflection limits, and the wire strength.

8.1 Spring terminology

A helical compression spring is described by its wire diameter d, mean coil diameter D, the spring index C = D/d (usually 4 to 12), and the number of active coils N_a. The index measures how tightly the wire is wound; a small index means sharp curvature and high stress.

8.2 Shear stress and the Wahl factor

The wire carries a torsional shear stress τ = K_W · 8FD/πd³. The Wahl factor K_W = (4C − 1)/(4C − 4) + 0.615/C corrects for both the direct shear and the curvature that crowds stress onto the inside of the coil. It is always greater than one and grows as the index shrinks.

8.3 The spring rate

Summing the wire's torsional twist over the active coils gives the rate k = d⁴G/8D³N_a. The wire diameter enters to the fourth power, so it is by far the strongest lever: a small change in d transforms the stiffness.

Spring wire	ASTM	G (GPa)
Music wire	A228	81.0
Hard-drawn	A227	79.3
Oil-tempered	A229	77.2
Chrome-vanadium	A232	77.2

Shear modulus by wire type (Shigley, Table 10-5). Wire strength rises as diameter falls: S_ut = A/d^m.

8.4 Deflection and solid length

The deflection is δ = F/k, and at solid length all coils touch and the spring bottoms out. A working spring is designed to stay below the solid deflection with margin, and its free length, solid length, and pitch all follow from the coil count.

8.5 Wire strength

Spring wire is strongest in small diameters: S_ut = A/d^m, with A and m from the wire type. The allowable shear stress is a fraction of S_ut (often about 0.45 for static use), so the corrected τ must stay below it with a factor of safety.

Engineering connection: when a spring cycles, the Wahl-corrected stress feeds the fatigue and Goodman methods of Chapter 5.

Worked example 1: shear stress in a helical spring

A helical compression spring has wire diameter d = 3 mm and mean coil diameter D = 24 mm. It carries an axial load of 150 N. Find the spring index, the Wahl factor, and the corrected shear stress in the wire.

Figure 1. The load twists the wire; the spring index sets the curvature correction, and the Wahl factor raises the nominal torsional stress to the true peak at the inner fibre.

ProblemFind C, K_W, and τ for the spring in Figure 1.
Given / findd = 3 mm, D = 24 mm, F = 150 N. Find C, K_W, τ.
AssumptionsThe wire acts as a torsion bar; the Wahl factor captures direct shear and curvature.
ModelCompute the index C = D/d, the Wahl factor, then τ = K_W · 8FD/πd³.
EquationsC = D/dK_W = (4C − 1)/(4C − 4) + 0.615/Cτ = K_W · 8FD/(πd³)
SolveC = 24/3 = 8. K_W = (31/28) + 0.615/8 = 1.107 + 0.077 = 1.18. τ = 1.18 × 8(150)(24)/(π·3³) = 1.18 × 28 800/84.82 = 1.18 × 339.5 = 402 MPa.
CheckFor music wire of d = 3 mm, S_ut = 2211/3^0.145 ≈ 1885 MPa, and an allowable shear near 0.45 S_ut ≈ 848 MPa. The 402 MPa stress gives a factor of safety near 2, so the wire is comfortable.
ConclusionThe Wahl factor raises the stress about 18 percent here; ignoring it would under-predict the peak and overstate the spring's life.

Result. C = 8, K_W = 1.18, and τ = 402 MPa.

Worked example 2: spring rate and deflection

The same spring (d = 3 mm, D = 24 mm) is wound from music wire (G = 81.0 GPa) with 8 active coils. Find the spring rate and the deflection under the 150 N load.

Figure 2. Summing the wire's torsional flexibility over the active coils gives the rate. The wire diameter enters to the fourth power, dominating the result.

ProblemFind the spring rate and deflection for the spring in Figure 2.
Given / findd = 3 mm, D = 24 mm, G = 81 000 MPa, N_a = 8, F = 150 N. Find k and δ.
AssumptionsLinear spring; only the active coils contribute; ends do not add stiffness.
ModelUse k = d⁴G/8D³N_a, then δ = F/k.
Equationsk = d⁴G/(8D³N_a)δ = F/k
Solvek = (3⁴ × 81 000)/(8 × 24³ × 8) = (81 × 81 000)/(8 × 13 824 × 8) = 6 561 000/884 736 = 7.42 N/mm. δ = 150/7.42 = 20.2 mm.
CheckIf the wire grew to 4 mm, the rate would jump by (4/3)⁴ = 3.16 times, to about 23 N/mm. The fourth-power dependence on d is the dominant design lever, far more than coil count.
ConclusionStress and rate come from the same geometry: wire and coil diameters set both. Designing a spring is balancing a safe stress against a target rate.

Result. k = 7.42 N/mm and δ = 20.2 mm under 150 N.

Misconceptions and diagnostics

Mistake	Symptom	Diagnostic question	Correction
Dropping the Wahl factor	Peak stress under-predicted	"Did I correct for curvature?"	Multiply by K_W, always greater than one.
Confusing d and D	Stress or rate off by large factors	"Is this the wire or the coil diameter?"	Wire d carries the stress; coil D sets the torque arm.
Using total instead of active coils	Rate too low	"Are end coils active?"	Use only the active coils N_a in the rate.
Constant wire strength	Allowable stress wrong for the size	"Did I use S_ut = A/d^m?"	Thin wire is stronger; compute strength from the diameter.

Practice ladder

Level 1 · Direct skill

A spring has d = 4 mm and D = 32 mm. Find its spring index and Wahl factor.

Show answer

C = 32/4 = 8. K_W = (31/28) + 0.615/8 = 1.18, the same as the worked example, because the index is the same.

Level 2 · Mixed concept

The Worked Example 2 spring is rewound with 12 active coils instead of 8. What is the new rate?

Show answer

The rate scales with 1/N_a: k = 7.42 × (8/12) = 4.95 N/mm. More coils make a softer spring.

Level 3 · Independent problem

A spring with d = 5 mm, D = 40 mm carries 300 N. Find the Wahl-corrected shear stress.

Show answer

C = 8, K_W = 1.18. τ = 1.18 × 8(300)(40)/(π·5³) = 1.18 × 96 000/392.7 = 1.18 × 244.5 = 288 MPa.

Level 4 · Transfer to real engineering

Find a real compression spring (a pen, a valve, a suspension). Estimate its index and explain whether it is designed for a target rate or a stress limit.

What good work looks like

A reasonable index from measured diameters, and a judgement about whether the spring's job is a specific feel (rate-driven) or carrying a high load safely (stress-driven).

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that I used the wire diameter for stress and applied the Wahl factor."

"Give me three spring geometries; I will find the index and rate for each."

"What is the spring stress?" Treating the wire as a torsion bar yourself is the skill.

"Design this spring." Balancing stress against rate is the point.

Portfolio task

Analyse a real helical spring: measure d and D, find the index and Wahl factor, compute the shear stress at a working load, and find the spring rate.

Must include: the spring index, the Wahl-corrected stress, and the spring rate with deflection.

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. Define the spring index.

C = D/d, the ratio of coil diameter to wire diameter.

2. What does the Wahl factor correct?

Direct shear plus the curvature that crowds stress onto the inner fibre; it is always greater than one.

3. Write the spring rate.

k = d⁴G/8D³N_a, with the wire diameter to the fourth power.

4. Why is wire diameter so influential?

It enters the rate to the fourth power and the stress to the third, so small changes have large effects.

5. How does wire strength vary with diameter?

S_ut = A/d^m: thinner wire is stronger.

TodayFinish this quiz and Levels 1 and 2 of the ladder.

+1 dayRe-derive the shear stress and spring rate from a blank page.

+3 daysDesign a spring to a target rate.

+7 daysCarry the spring into bearings, Chapter 9.

+30 daysRun the Wahl-corrected stress through a fatigue check.

Textbook mapping

Item	Mapping
Primary source	Budynas and Nisbett, Shigley's Mechanical Engineering Design, Chapter 10 (Mechanical Springs)
Cross-reference	Norton, Ch. 14 · Mechanics of Materials, Ch. 3
Core topics	8.1 Spring terminology · 8.2 Shear stress and Wahl factor · 8.3 Spring rate · 8.4 Deflection and solid length · 8.5 Wire strength
Engineering connection	The Wahl-corrected stress feeds the spring fatigue check.
Read next	Chapter 9: Rolling-Contact Bearings.