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Fatigue and Damage Tolerance Questions and Answers

71.

A student concludes that because a structure passed its ultimate load test, it will not crack in service. The conclusion is

Answer: C

A structure can carry ultimate load comfortably and still crack after twenty thousand flights at a small fraction of that load. Fatigue is demonstrated by a separate cyclic test lasting several times the intended service life.

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Answer: D

The aluminium S-N curve continues to fall rather than flattening, so no stress is safe indefinitely. This is why aluminium airframes are managed by finite life and inspection rather than declared immune below some threshold.

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Answer: A

Correct answer: It accounts for load sequence

Miner’s rule is a linear summation and is entirely order-independent: it gives the same answer whether the large cycles come first or last, when experiment shows they do not. The other three statements are all true, including the last - the rule remains the standard first estimate despite this defect, because nothing simpler works better.

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74.

Which of the following would NOT be expected to increase the critical crack length?

Answer: B

Correct answer: Selecting a lower-toughness alloy

Since \( a_c = (K_{IC}/Y\sigma)^2/\pi \), the critical crack length rises with toughness and falls with stress and geometry factor. A lower-toughness alloy therefore reduces it - which is exactly why a strong but brittle alloy can be the wrong choice for a damage-tolerant structure, even though it carries more static load.

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75.

An engineer proposes a test spectrum containing frequent large overloads to be conservative. The proposal is

Answer: C

Each overload leaves compressive residual stress at the crack tip that slows subsequent growth, so a spectrum rich in them demonstrates a longer life than the real usage would deliver. Spectrum development takes considerable care over this.

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Answer: D

Correct answer: Increasing the applied stress in the member

A longer interval requires the crack growth period to be longer, so the useful levers are a smaller detectable crack, a larger critical crack, or slower growth. Raising the stress does the opposite on two counts: it shortens the critical length as the inverse square, and it raises \( \Delta K \) and so speeds the growth.

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Answer: A

As details of the same design reach similar ages they crack together, and the ligaments between them can fail progressively. This is precisely what fail-safe design does not anticipate, and it is central to ageing aircraft programmes.

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78.

A component shows a margin of seven per cent between its current stress intensity and the fracture toughness. The correct action is

Answer: B

Correct answer: Ground the aircraft; the crack is close to critical

Seven per cent is not an engineering margin. The stress intensity grows with the square root of the crack length, so a small further extension exhausts it, and the structure is one overload away from unstable fracture. The aircraft is withdrawn and the member repaired or replaced before further flight.

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79.

The de Havilland Comet accidents of 1954 were ultimately attributed to

Answer: C

Repeated pressurisation cycles initiated cracks at cut-out corners, and the investigation included a full-scale fuselage tested in a water tank. It established pressurisation fatigue as a first-order design consideration and reshaped how transport fuselages are certified.

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Answer: D

A large section of upper fuselage separated after many small cracks at a lap joint linked together. The aircraft had accumulated an unusually high cycle count on short sectors, and the accident launched the ageing aircraft programmes.

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