Fatigue and Damage Tolerance Questions and Answers

Answer: C

Both statements are true and the second is the reason for the first. Lap joints and bolted fittings are the usual locations, and the remedy is to reduce the slip through clamping or to protect the interface with a coating or shim.

Answer: D

The assertion is false and the reason is true. Local plasticity and the steep stress gradient at a notch root mean the material does not feel the full elastic peak, so the fatigue factor is smaller, related to the elastic one through the notch sensitivity.

Answer: A

Both are true and the second explains the first. Investment in eddy current or ultrasonic inspection therefore pays back directly in reduced maintenance frequency, which is one of the clearer economic arguments in structural management.

Answer: B

Correct answer: 1 and 2 only

Statements 1 and 2 are correct: ferrous materials show a genuine endurance limit, typically near half the ultimate, below which the life is effectively unlimited, whereas aluminium alloys do not - their curve continues to fall, so a fatigue strength must be quoted at a stated number of cycles. Statement 3 is wrong: the cycle count spans several decades and is always plotted logarithmically.

Answer: C

Correct answer: 1 and 2 only

Statements 1 and 2 are correct: a tensile mean stress shortens the life at a given amplitude, and the Goodman relation is the standard correction for it. Statement 3 has the definition inverted - \( R = -1 \) is fully reversed loading, equal tension and compression. Wholly tensile loading gives a positive \( R \).

Answer: D

Correct answer: 1 and 2 only

Statements 1 and 2 state the rule correctly: the damage at each level is \( n_i/N_i \), and failure is predicted when the sum reaches unity. Statement 3 is the rule’s known weakness - it is linear and order-independent, so it cannot represent the retardation that follows an overload or the acceleration that can follow a compressive one.

Answer: A

The first two are correct. The third confuses it with fracture toughness: the stress intensity factor describes the loading of a particular cracked body, while the toughness is the critical value the material can sustain.

Answer: B

Correct answer: 1 and 2 only

Statements 1 and 2 are correct: \( da/dN = C(\Delta K)^m \), with \( m \) typically three to four for metals. Statement 3 overstates its range. The law describes the straight middle portion of the growth curve only; near the threshold the growth falls away far more steeply, and near the toughness it accelerates far more sharply than the power law predicts.

Answer: C

The first two are correct. The third inverts the philosophy entirely: damage tolerance depends absolutely on inspection, since finding the crack before it becomes critical is the whole mechanism by which safety is maintained.

Answer: D

The first two describe the two techniques correctly. The third misattributes the mechanism: neither changes the material's properties, and both work by introducing a residual stress that applied tensile cycles must overcome.