Home / Courses / Machine Elements / Fatigue

Machine Elements · Chapter 5 of 10 · Advanced

Fatigue Failure from Variable Loading

Most machine parts break under loads far below yield, simply because the load repeats. Fatigue is the single most important failure mode in machine design, and it has its own strength and its own rules.

Readiness check

This chapter introduces a new strength and new criteria. Tick only what you can do closed-notes.

Recall ultimate tensile strength S_ut.
Read a log-log graph.
Evaluate a power law such as a·x^b.
Split a fluctuating signal into mean and amplitude.
Form a factor of safety from a linear criterion.

0 or 1 weak itemsContinue with this chapter.

2 weak itemsReview fatigue basics in Materials, Chapter 7.

3 or more weak itemsRevisit stress and strength before continuing.

The core idea

Under repeated load a part has a fatigue strength, the endurance limit, well below its static strength. We estimate it from the ultimate strength, correct it for reality, and check the stress against a mean-stress line.

S'_e = 0.5 S_ut (steel)S_e = k_a k_b k_c S'_eσ_a/S_e + σ_m/S_ut = 1/n

A polished test specimen of steel has an endurance limit of about half its ultimate strength; below it, the part survives indefinitely. A real part is rougher, larger, and differently loaded, so Marin factors k_a, k_b, and others reduce that ideal value to a usable S_e. A fluctuating load is split into a steady mean stress σ_m and an alternating amplitude σ_a; the Goodman line combines them against S_e and S_ut to give a factor of safety.

The skill works when: you estimate S'_e, apply the Marin factors, and check σ_a and σ_m on the Goodman line.

The skill breaks down when: a static check is trusted for a cyclic load, or the modifying factors are skipped.

The concept. The S-N curve: fatigue strength falls as the number of cycles rises, then flattens at the endurance limit for steel near a million cycles. Below that line the part has effectively infinite life.

The skills, taught in order

Five skills introduce cyclic loading, the endurance limit, the modifying factors, the mean-alternating split, and the failure criteria.

5.1 Cyclic loading and the S-N curve

Repeated loading drives cracks that grow until the part breaks, often with no warning. Plotting fatigue strength against the number of cycles gives the S-N curve, which falls through the low-cycle and high-cycle regions and, for steel, flattens at the endurance limit near 10⁶ cycles.

5.2 The endurance limit

For steels, the endurance limit of a polished specimen is estimated as S'_e = 0.5 S_ut (capped near 700 MPa). It is the alternating stress the ideal specimen can survive forever. Nonferrous metals lack a true limit and are rated at a chosen life instead.

5.3 Marin modifying factors

A real part differs from the test specimen, so S_e = k_a k_b k_c k_d k_e S'_e. Surface finish k_a = a S_ut^b and size k_b are the two that move the number most; load type, temperature, and reliability supply the rest.

Factor	What it corrects	Form
k_a	surface finish	a S_ut^b (machined: 4.51, −0.265)
k_b	size (bending, torsion)	1.24 d^−0.107, 2.79 ≤ d ≤ 51 mm
k_c	load type	1 bending, 0.85 axial, 0.59 torsion
k_d, k_e	temperature, reliability	from tables

Surface and size parameters from Shigley, Eqs. 6-19 and 6-20 (SI). Machined surface, room temperature, 50 percent reliability give k_a from S_ut and k_b from diameter.

5.4 Mean and alternating stress

A fluctuating stress has a mean σ_m = (σ_max + σ_min)/2 and an amplitude σ_a = (σ_max − σ_min)/2. A fully reversed stress has σ_m = 0; a steady offset raises σ_m and shortens life. Both components must be tracked.

5.5 Fatigue criteria

The modified Goodman line, σ_a/S_e + σ_m/S_ut = 1/n, is the common, slightly conservative choice for the safe combination of mean and alternating stress. Soderberg (using S_y) is stricter; Gerber (a parabola) fits data better but is less conservative.

Engineering connection: the endurance limit and Goodman line drive the shaft sizing of Chapter 6 and the fatigue check of springs and bolts.

Worked example 1: estimating the endurance limit

A machined rotating shaft of AISI 1050 cold-drawn steel (S_ut = 690 MPa) has a critical diameter of 10 mm and sees fully reversed bending. Estimate its corrected endurance limit using the surface and size factors.

Figure 1. The ideal endurance limit, half the ultimate strength, is knocked down by the surface and size factors to the value the real shaft can sustain.

ProblemFind the corrected endurance limit S_e for the shaft in Figure 1.
Given / findS_ut = 690 MPa, machined surface (a = 4.51, b = −0.265), d = 10 mm, bending. Find S_e.
AssumptionsSteel below the 700 MPa cap, so S'_e = 0.5 S_ut; bending load (k_c = 1); room temperature and 50 percent reliability (k_d = k_e = 1).
ModelEstimate S'_e, then apply k_a = a S_ut^b and k_b = 1.24 d^−0.107.
EquationsS'_e = 0.5 S_utk_a = 4.51 S_ut^−0.265, k_b = 1.24 d^−0.107S_e = k_a k_b S'_e
SolveS'_e = 0.5 × 690 = 345 MPa. k_a = 4.51 × 690^−0.265 = 0.80. k_b = 1.24 × 10^−0.107 = 0.97. S_e = 0.80 × 0.97 × 345 = 267 MPa.
CheckThe factors pull the ideal 345 MPa down to 267 MPa, a 23 percent reduction that is typical for a small machined part. Both factors are below one, as they must be.
ConclusionThe corrected endurance limit, not half the ultimate strength, is the fatigue strength to design against. It feeds the Goodman check next.

Result. S'_e = 345 MPa, k_a = 0.80, k_b = 0.97, so S_e = 267 MPa.

Worked example 2: the Goodman fatigue check

The same shaft (S_e = 267 MPa, S_ut = 690 MPa) carries a fluctuating stress with alternating amplitude σ_a = 120 MPa and mean σ_m = 80 MPa. Find the fatigue factor of safety by the modified Goodman criterion.

Figure 2. The Goodman line runs from the endurance limit on the alternating axis to the ultimate strength on the mean axis. The load point sits inside it, and the factor of safety is how far it can scale outward.

ProblemFind the fatigue factor of safety for the load point in Figure 2.
Given / findS_e = 267 MPa, S_ut = 690 MPa, σ_a = 120 MPa, σ_m = 80 MPa. Find n.
AssumptionsConstant-amplitude loading; the modified Goodman line; proportional scaling of σ_a and σ_m.
ModelApply the Goodman relation σ_a/S_e + σ_m/S_ut = 1/n and solve for n.
Equations1/n = σ_a/S_e + σ_m/S_ut
Solve1/n = 120/267 + 80/690 = 0.449 + 0.116 = 0.566, so n = 1.77.
CheckThe alternating term dominates, as it should, since fatigue is driven by the cyclic part. A factor of 1.77 means the load could grow about 77 percent before reaching the Goodman line.
ConclusionThe Goodman check combines mean and alternating stress into one factor of safety against fatigue. This is the calculation a shaft must pass in Chapter 6.

Result. 1/n = 0.566, so n = 1.77 against fatigue.

Misconceptions and diagnostics

Mistake	Symptom	Diagnostic question	Correction
Static check on a cyclic load	Part sized for yield, fails early	"Does this load repeat?"	Use S_e and a fatigue criterion when the load cycles.
Skipping the Marin factors	Endurance limit far too high	"Did I correct S'_e for surface and size?"	Apply k_a, k_b, and the rest before designing.
Ignoring mean stress	Life over-predicted for an offset load	"Is there a steady component σ_m?"	Carry σ_m into the Goodman line.
Endurance limit for nonferrous	Aluminium assumed to have a flat limit	"Is this steel?"	Nonferrous metals have no true limit; design to a stated life.

Practice ladder

Level 1 · Direct skill

Estimate the ideal endurance limit of a steel with S_ut = 600 MPa.

Show answer

S'_e = 0.5 × 600 = 300 MPa, below the 700 MPa cap, so the half rule applies directly.

Level 2 · Mixed concept

For the Worked Example 1 material (S_ut = 690 MPa) but a 30 mm diameter, find the new size factor and S_e (keep k_a = 0.80).

Show answer

k_b = 1.24 × 30^−0.107 = 0.86. S_e = 0.80 × 0.86 × 345 = 237 MPa. Larger parts have a lower endurance limit.

Level 3 · Independent problem

A part has S_e = 220 MPa, S_ut = 600 MPa, σ_a = 90 MPa, σ_m = 120 MPa. Find the Goodman factor of safety.

Show answer

1/n = 90/220 + 120/600 = 0.409 + 0.200 = 0.609, so n = 1.64. The mean stress takes a larger share here than in the worked example.

Level 4 · Transfer to real engineering

Find a part that failed by fatigue (a bracket, an axle, a spring). Explain where the crack started and why a static check would have missed it.

What good work looks like

A crack origin at a surface or stress concentration, a recognition that the peak stress was below yield, and the point that only a cyclic, S_e-based check captures the failure.

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that I applied the surface and size factors before designing."

"Give me four load histories; I will find σ_m and σ_a for each."

"What is the endurance limit?" Estimating and correcting it yourself is the skill.

"Is it safe in fatigue?" Placing the point on the Goodman line is the point.

Portfolio task

Take a cyclically loaded part, estimate its corrected endurance limit, split its stress into mean and alternating, and report a Goodman factor of safety with the modifying factors listed.

Must include: S'_e, the Marin factors, σ_m and σ_a, and a Goodman factor of safety.

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. Estimate the endurance limit of a steel.

S'_e = 0.5 S_ut for the polished specimen, capped near 700 MPa.

2. Which Marin factors move S_e the most?

Surface finish k_a and size k_b.

3. Define mean and alternating stress.

σ_m = (σ_max + σ_min)/2 and σ_a = (σ_max − σ_min)/2.

4. Write the modified Goodman criterion.

σ_a/S_e + σ_m/S_ut = 1/n.

5. Do aluminium alloys have an endurance limit?

No true one; they are designed to a chosen finite life.

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

+1 dayRe-derive S_e and the Goodman factor from a blank page.

+3 daysRun two new Goodman checks.

+7 daysCarry fatigue into shaft design, Chapter 6.

+30 daysReuse the endurance limit for springs and bolted joints.

Textbook mapping

Item	Mapping
Primary source	Budynas and Nisbett, Shigley's Mechanical Engineering Design, Chapter 6 (Fatigue Failure Resulting from Variable Loading)
Cross-reference	Norton, Ch. 6 · Materials, Ch. 7
Core topics	5.1 S-N curve · 5.2 Endurance limit · 5.3 Marin factors · 5.4 Mean and alternating stress · 5.5 Fatigue criteria
Engineering connection	The endurance limit and Goodman line size shafts, springs, and bolts.
Read next	Chapter 6: Shafts and Shaft Components.