Fatigue and Damage Tolerance Questions and Answers

Answer: A

Bombarding the surface plastically deforms a thin layer that the bulk then holds in compression. Since fatigue cracks almost always start at the surface, that compression must be overcome before any tensile cycle is felt there.

Answer: B

Micro-slip at a joint interface damages the surface and creates initiation sites far earlier than plain fatigue would. Lap joints, bolted fittings and spline couplings are the usual locations, and it can cut fatigue strength by half or more.

Answer: C

A hole roughly triples the local stress, and the drilling process itself leaves a damaged edge. That is why hole quality, cold working and interference-fit fasteners receive so much attention in fatigue-critical structure.

Answer: D

A corrosion pit is a stress concentration and a ready-made initiation site, so corrosion fatigue produces far shorter lives than either mechanism alone. It is a major driver of structural maintenance on ageing aircraft, particularly in humid operating environments.

Answer: A

A scatter factor of two or more on demonstrated life covers material variability and the difference between the test spectrum and real usage. The test also reveals where cracks appear first, which drives the inspection programme.

Answer: B

Correct answer: 14.0 MPa root m

\[ K = Y\sigma\sqrt{\pi a} = 1.12 \times 100 \times \sqrt{\pi \times 0.005} = 1.12 \times 100 \times 0.1253 = 14.0\ \mathrm{MPa}\sqrt{\mathrm{m}}. \] Note the crack length must be in metres for the answer to come out in the usual units - using millimetres here is the commonest arithmetic slip in the subject.

Answer: C

Correct answer: 10 mm

Rearranging \( K_{IC} = Y\sigma\sqrt{\pi a_c} \), \[ a_c = \frac{1}{\pi}\left(\frac{K_{IC}}{Y\sigma}\right)^2 = \frac{1}{\pi}\left(\frac{30}{1.12 \times 150}\right)^2 = \frac{0.0319}{\pi} = 0.0102\ \mathrm{m}, \] that is about 10 mm. Since the critical length varies as the inverse square of the stress, halving the working stress quadruples the crack the structure can tolerate.

Answer: D

Stress intensity varies with the square root of crack length, so a fourfold growth doubles it. This square root is why growth accelerates so sharply near the end of life, since the driving force keeps climbing as the crack extends.

Answer: A

The Paris law gives the growth rate as a constant times the range raised to the exponent, so \( 2^3 = 8 \). A modest increase in stress therefore has a dramatic effect on crack growth life, far more than on static margin.

Answer: B

Correct answer: 0.25

Miner’s rule sums the damage fractions and predicts failure when they reach unity: \[ 0.30 + 0.45 = 0.75, \qquad 1 - 0.75 = 0.25. \] A quarter of the life remains. The rule is linear and takes no account of the order in which the levels are applied, which is its principal limitation.