Exercise set

Material Fatigue and Fracture Exercises

Worked materials engineering exercises for fatigue and fracture covering stress amplitude, stress ratio, Goodman screening, Miner damage, notch fatigue, critical crack size, crack-growth inspection intervals, corrosion-fatigue, residual strength, and validation evidence.

These exercises practise fatigue and fracture as engineering evidence problems. They cover cyclic stress quantities, mean-stress screening, cumulative damage, notch effects, critical crack size, crack-growth inspection intervals, corrosion-fatigue, residual strength, and validation records.

The goal is not only to calculate a fatigue number. The goal is to decide whether the material, surface condition, geometry, load spectrum, environment, and inspection plan can prevent crack initiation, unstable crack growth, or unacceptable deformation over the intended service life.

Assume simplified screening models unless an exercise states otherwise. Real fatigue and fracture assessments should also check local stress definition, product form, heat treatment, surface finish, residual stress, weld quality, corrosion environment, load sequence, inspection probability of detection, and consequence of failure.

How to Use These Exercises

For each calculation, define:

  1. the local detail or crack-like flaw being assessed;
  2. the stress history, stress ratio, and cycle count;
  3. the material condition, surface condition, and environment;
  4. the inspection method and detectable flaw size where relevant;
  5. the engineering action if the fatigue or fracture margin is weak.

The common mistake is treating fatigue strength as a single material property. Fatigue and fracture are system properties of load, geometry, defects, surface, environment, inspection, and time.

Use the exercises as integrity gates: accept or reject a mean-stress screen, identify a damaging load block, redesign a notch, set a crack-growth inspection interval, challenge a probability-of-detection assumption, reduce corrosion-fatigue exposure, or hold validation until the test spectrum and failure criteria match service.

Exercise 1: Cyclic Stress Quantities

A component sees maximum stress:

S_{max}=180\ \text{MPa}

and minimum stress:

S_{min}=20\ \text{MPa}

Calculate stress range, stress amplitude, mean stress, and stress ratio:

\Delta S=S_{max}-S_{min}
\displaystyle S_a=\frac{S_{max}-S_{min}}{2}
\displaystyle S_m=\frac{S_{max}+S_{min}}{2}
\displaystyle R=\frac{S_{min}}{S_{max}}

Solution

Stress range:

\Delta S=180-20=160\ \text{MPa}

Stress amplitude:

\displaystyle S_a=\frac{160}{2}=80\ \text{MPa}

Mean stress:

\displaystyle S_m=\frac{180+20}{2}=100\ \text{MPa}

Stress ratio:

\displaystyle R=\frac{20}{180}=0.111

Engineering Comment

The cycle is not fully reversed because mean stress is tensile and R is positive. That matters because tensile mean stress usually reduces fatigue life compared with fully reversed loading at the same amplitude.

Fatigue evidence should always state the stress convention used.

Exercise 2: Modified Goodman Screening

A steel detail has:

S_a=95\ \text{MPa}
S_m=120\ \text{MPa}

Corrected endurance limit:

S_e=210\ \text{MPa}

Ultimate tensile strength:

S_{UTS}=620\ \text{MPa}

Use the Goodman utilization:

\displaystyle U=\frac{S_a}{S_e}+\frac{S_m}{S_{UTS}}

Check whether the screen passes for a required design factor of:

N=1.5

where the acceptance condition is:

\displaystyle U\leq\frac{1}{N}

Solution

Goodman utilization:

\displaystyle U=\frac{95}{210}+\frac{120}{620}=0.452+0.194=0.646

Allowable utilization:

\displaystyle \frac{1}{N}=\frac{1}{1.5}=0.667

Comparison:

0.646<0.667

The screen passes, but with a narrow margin.

Engineering Comment

The Goodman screen barely passes. The result is sensitive to endurance-limit correction, surface condition, notch effect, residual stress, and whether the local stress is represented correctly.

A narrow pass should trigger a check of measurement uncertainty, load spectrum, and manufacturing variation.

Exercise 3: Miner Damage for a Two-Level Spectrum

A welded bracket experiences two cyclic stress blocks per year:

Stress blockApplied cyclesCycles to failure
Normal operation180,0008,000,000
Startup transient12,000300,000

Use Miner’s rule:

\displaystyle D=\sum\frac{n_i}{N_i}

Calculate annual damage and estimated repeated-spectrum life.

Solution

Annual damage:

\displaystyle D=\frac{180000}{8000000}+\frac{12000}{300000}
D=0.0225+0.0400=0.0625

Repeated-spectrum life:

\displaystyle Life=\frac{1}{D}=\frac{1}{0.0625}=16\ \text{years}

Engineering Comment

The startup transient contributes more damage than normal operation despite fewer cycles. This is exactly why load spectra should not be summarized only by total cycles.

Miner life is a screening estimate. It does not include sequence effects, weld residual stress, corrosion, crack closure, inspection, repair, or scatter.

Exercise 4: Notch Fatigue Stress

A nominal alternating stress is:

S_{a,nom}=55\ \text{MPa}

The theoretical stress concentration factor is:

K_t=2.4

The notch sensitivity is:

q=0.65

Use:

K_f=1+q(K_t-1)

and:

S_{a,local}=K_fS_{a,nom}

Solution

Fatigue stress concentration factor:

K_f=1+0.65(2.4-1)=1+0.91=1.91

Local alternating stress:

S_{a,local}=1.91(55)=105.1\ \text{MPa}

Engineering Comment

The notch nearly doubles the alternating stress. Fatigue often initiates at local details, not at smooth nominal sections.

Design improvement should consider radius increase, surface finish, residual compressive treatment, load-path change, or inspection of the notch region.

Exercise 5: Critical Crack Size

A plate has fracture toughness:

K_c=42\ \text{MPa}\sqrt{\text{m}}

The maximum nominal tensile stress is:

\sigma=180\ \text{MPa}

Geometry factor:

Y=1.12

Estimate critical crack size:

\displaystyle a_c=\frac{1}{\pi}\left(\frac{K_c}{Y\sigma}\right)^2

Solution

Critical crack size:

\displaystyle a_c=\frac{1}{\pi}\left(\frac{42}{1.12(180)}\right)^2
\displaystyle a_c=\frac{1}{\pi}(0.2083)^2=0.0138\ \text{m}

In millimeters:

a_c=13.8\ \text{mm}

Engineering Comment

The simplified critical crack size is 13.8 mm. This is not a permission to wait until cracks are large. The inspection interval should be based on detectable flaw size, crack-growth rate, stress spectrum, and consequence of failure.

Fracture toughness calculations are strongest when tied to inspection capability.

Exercise 6: Residual Strength at a Detected Crack

An inspection finds a crack:

a=4.0\ \text{mm}

The applied maximum stress is:

\sigma=150\ \text{MPa}

The geometry factor is:

Y=1.1

Estimate stress intensity:

K=Y\sigma\sqrt{\pi a}

Use meters for crack size.

Solution

Convert crack size:

a=4.0\ \text{mm}=0.0040\ \text{m}

Stress intensity:

K=1.1(150)\sqrt{\pi(0.0040)}
K=165(0.1121)=18.5\ \text{MPa}\sqrt{\text{m}}

Engineering Comment

The detected crack has estimated stress intensity of 18.5 MPa square-root meters. Residual strength assessment should compare this with the relevant toughness and include uncertainty in crack size, stress, geometry factor, temperature, and material condition.

Crack disposition should be controlled by engineering review, not by visual judgement alone.

Exercise 7: Crack-Growth Inspection Interval

A damage-tolerance review estimates stable crack-growth life from detectable crack size to critical crack size as:

N_{growth}=240{,}000\ \text{cycles}

The inspection policy uses inspection factor:

F=4

Use:

\displaystyle N_{inspect}\leq\frac{N_{growth}}{F}

Solution

Inspection interval:

\displaystyle N_{inspect}\leq\frac{240{,}000}{4}=60{,}000\ \text{cycles}

Engineering Comment

The inspection interval should not exceed 60,000 cycles under the assumptions. That interval is only valid if the inspection method can reliably detect the assumed starting crack size and if the load spectrum remains representative.

Inspection planning is part of fracture control, not an administrative afterthought.

Exercise 8: Corrosion-Fatigue Damage Trigger

A rotating detail has corrected endurance limit in clean air:

S_{e,air}=180\ \text{MPa}

Service exposure is expected to reduce the endurance limit by 35 percent:

Reduction=35\%

The local alternating stress is:

S_a=128\ \text{MPa}

Estimate exposed endurance limit and compare.

Solution

Exposed endurance limit:

S_{e,exposed}=180(1-0.35)=117\ \text{MPa}

Comparison:

128>117

The alternating stress exceeds the exposed endurance-limit screen.

Engineering Comment

The detail may have passed in clean-air assumptions but fails the corrosion-fatigue screen. The engineering response may include improved surface protection, material change, stress reduction, drainage, inspection interval reduction, or environmental control.

Environment can turn an acceptable fatigue design into an unacceptable one.

Exercise 9: Probability-of-Detection Gap

A fracture-control plan assumes that inspection detects cracks of:

a=2.0\ \text{mm}

with high reliability. The qualified ultrasonic procedure has high-confidence detection only above:

a=3.5\ \text{mm}

Find the detection gap and state the implication.

Solution

Detection gap:

3.5-2.0=1.5\ \text{mm}

The inspection method is not qualified to support the assumed detectable crack size.

Engineering Comment

This is a serious evidence gap. The fracture-control plan assumes detection earlier than the inspection method can support. The team must revise the crack-growth calculation, shorten the interval, improve inspection, change access, reduce stress, or use a different inspection technology.

Probability of detection is part of the engineering model.

Exercise 10: Validation Margin for Fatigue Test Duration

A component is required to survive:

N_{service}=1.5\times10^6\ \text{cycles}

The validation plan tests:

N_{test}=3.0\times10^6\ \text{cycles}

Calculate cycle factor:

\displaystyle F_N=\frac{N_{test}}{N_{service}}

Solution

Cycle factor:

\displaystyle F_N=\frac{3.0\times10^6}{1.5\times10^6}=2.0

Engineering Comment

The test has a factor of 2.0 on cycle count. That may be useful, but validation strength also depends on stress level, spectrum, mean stress, environment, sample size, product form, manufacturing route, surface finish, and failure criteria.

A long test at the wrong load spectrum can still be weak evidence.

Exercise 11: Fatigue Test Failure Rate

A validation batch tests 24 specimens. Three fail before the required cycle count. Calculate observed failure rate and pass fraction.

Solution

Observed failure rate:

\displaystyle F=\frac{3}{24}\times100=12.5\%

Pass fraction:

\displaystyle P=\frac{21}{24}\times100=87.5\%

Engineering Comment

An 87.5 percent pass fraction may be unacceptable for a fatigue-critical component. The failed specimens matter more than the average result. The team should investigate fracture origin, surface condition, defect population, heat treatment, fixture alignment, load spectrum, and whether the failures represent production risk.

Fatigue validation should produce design learning, not only pass/fail counts.

Review Checklist

When reviewing fatigue and fracture evidence, ask:

  • Are stresses local, nominal, hot-spot, principal, equivalent, or notch stresses?
  • Are stress ratio, mean stress, cycle count, and load spectrum explicit?
  • Do S-N data and endurance limits match material condition, surface condition, and environment?
  • Are notches, weld toes, pits, scratches, threads, pores, and residual stress represented?
  • Does fracture toughness analysis match crack geometry, temperature, thickness, and product form?
  • Is the inspection method qualified for the flaw size assumed in the analysis?
  • Does crack-growth life define a defensible inspection interval?
  • Does validation reproduce the relevant load spectrum, environment, geometry, and manufacturing state?
  • Are scatter, uncertainty, sample size, and failed-specimen evidence treated as design inputs rather than as statistical inconvenience?
  • Is every life-extension or continued-service decision tied to updated loads, detected flaws, inspection capability, repair history, and residual risk?

Fatigue and fracture engineering is credible when stress analysis, material evidence, defects, environment, inspection, and validation all describe the same component reality.

REF

See also