Fatigue and Damage Tolerance Questions and Answers

Answer: A

Correct answer: Below the static ultimate, under repetition

This is the whole point of fatigue and the reason it took so long to be recognised. A member loaded once to a stress well within its static capability may fail after enough repetitions of that same stress, because each cycle does a small amount of irreversible damage at a slip band or an inclusion. A static calculation cannot detect the danger.

Answer: B

The curve is obtained from constant-amplitude testing and falls steeply at first before flattening. It is the basic input to any stress-life fatigue calculation, and the number of cycles is conventionally plotted on a logarithmic scale.

Answer: C

A steel S-N curve flattens to a horizontal asymptote below which failure does not occur, but an aluminium curve continues to fall. Aluminium components are therefore quoted with a fatigue strength at a stated number of cycles rather than an endurance limit.

Answer: D

Correct answer: Minimum stress divided by maximum stress

\[ R = \frac{\sigma_{min}}{\sigma_{max}}. \] Fully reversed loading gives \( R = -1 \), loading from zero to a peak gives \( R = 0 \), and a wholly tensile cycle gives a positive \( R \) less than one. It is the standard way of describing where a cycle sits relative to zero, which matters because the mean stress affects the life.

Answer: A

Correct answer: Reduces fatigue life

A tensile mean stress holds the crack open through more of the cycle, so a larger part of the cycle contributes to growth. The Goodman and Gerber relations quantify the effect, converting a cycle at some mean stress into an equivalent fully reversed one so that ordinary S-N data may be used.

Answer: B

It draws a straight line between the fatigue strength at zero mean stress and the ultimate tensile strength at zero amplitude. Gerber uses a parabola through the same end points and is less conservative for ductile metals.

Answer: C

It is a purely elastic, geometric quantity, being the ratio of peak local stress to nominal stress. The fatigue notch factor is smaller, because real materials redistribute stress locally, and the difference is captured by the notch sensitivity.

Answer: D

Local plasticity and the stress gradient at the notch root mean the material does not feel the full elastic peak. Notch sensitivity, which runs from zero to one, relates the two, and it rises with material strength and with notch root radius.

Answer: A

Each block of cycles contributes damage equal to the cycles applied divided by the cycles to failure at that level, and failure is predicted when the accumulated damage reaches unity. It ignores sequence effects entirely, which is its main weakness.

Answer: B

A large cycle followed by small ones behaves differently from the reverse order, because an overload leaves a compressive residual stress at the crack tip that retards subsequent growth. Miner's linear summation cannot represent that.