A component cycles between \( 20 \) and \( 100 \) MPa. Its stress ratio is
Answer: C
Correct answer: 0.2
\[ R = \frac{20}{100} = 0.2. \] Both stresses are tensile, so the cycle never crosses zero and the ratio is positive. A positive stress ratio means a tensile mean stress, which is the less favourable condition.
Please log in or register to report this question.
For that same cycle, the stress amplitude and mean stress are
Answer: D
Correct answer: 40 and 60 MPa
\[ \sigma_a = \frac{100-20}{2} = 40\ \mathrm{MPa}, \qquad \sigma_m = \frac{100+20}{2} = 60\ \mathrm{MPa}. \] The amplitude is half the range and the mean is the average of the extremes. Together with the range and the ratio, any two of these four quantities fix the cycle completely.
Please log in or register to report this question.
A detail has \( K_t = 3.0 \) and a notch sensitivity of \( 0.8 \). Its fatigue notch factor is
Answer: A
Correct answer: 2.6
\[ K_f = 1 + q(K_t - 1) = 1 + 0.8 \times 2.0 = 2.6. \] The notch sensitivity \( q \) lies between zero and one and measures how nearly the material feels the full elastic concentration. At \( q = 0 \) the notch has no fatigue effect at all; at \( q = 1 \) the fatigue factor equals the elastic one. Coarse-grained and ductile materials tend to have the lower sensitivities.
Please log in or register to report this question.
A crack takes \( 12\,000 \) flights to grow from detectable to critical. With two inspections required in that period, the interval is
Answer: B
Correct answer: 6000 flights
If two inspections must fall within the interval during which a crack grows from detectable to critical, the interval can be no more than half that period: \[ \frac{12{,}000}{2} = 6{,}000\ \mathrm{flights}. \] The requirement for two rather than one is deliberate redundancy - a single missed detection must not lose the aircraft.
Please log in or register to report this question.
A structure demonstrates \( 120\,000 \) cycles in test. With a scatter factor of \( 3 \), the certified life is
Answer: C
Correct answer: 40,000 cycles
\[ \frac{120{,}000}{3} = 40{,}000\ \mathrm{cycles}. \] The scatter factor accounts for the spread of fatigue life between nominally identical specimens, which is wide. A single test demonstrates one sample from that distribution, and dividing by three brings the certified figure down to something the whole fleet can be expected to achieve.
Please log in or register to report this question.
A fuselage skin at \( 68.8 \) MPa hoop stress cycles once per flight from zero. Over \( 40\,000 \) flights it accumulates
Answer: D
Each flight applies one full pressurisation cycle from zero to the differential and back, so the count equals the number of flights and the stress ratio is zero. It is the single largest identifiable contributor to fuselage fatigue damage.
Please log in or register to report this question.
Improving an inspection method so the detectable crack size halves, with growth from detection to critical then taking \( 20\,000 \) rather than \( 12\,000 \) flights, changes the interval from 6000 to
Answer: A
Halving the interval requirement over the longer growth period gives 10,000 flights. Investing in a better inspection method therefore buys directly reduced maintenance cost, which is why eddy current and ultrasonic techniques displaced visual inspection for critical details.
Please log in or register to report this question.
A crack grows at \( 2 \times 10^{-5} \) mm per cycle. To extend by \( 4 \) mm it needs
Answer: B
Correct answer: 200,000 cycles
\[ N = \frac{4}{2 \times 10^{-5}} = 2 \times 10^{5}\ \mathrm{cycles}. \] This treats the growth rate as constant, which it is not - the Paris law makes it rise as the crack lengthens, so the real figure is smaller. The calculation is a first estimate, and a real analysis integrates the growth rate over the crack length.
Please log in or register to report this question.
A cycle of amplitude \( 40 \) MPa at zero mean has a life of \( 10^6 \) cycles. Adding a tensile mean stress will make the life
Answer: C
Correct answer: Shorter than a million cycles
Adding a tensile mean stress to a cycle of the same amplitude always shortens the life, because the crack spends more of the cycle open. The Goodman relation gives the size of the effect: the permissible amplitude falls roughly linearly as the mean stress approaches the ultimate strength.
Please log in or register to report this question.
Two details have theoretical stress concentration factors of \( 2.5 \) and \( 3.5 \). At equal nominal stress, the fatigue-critical one is
Answer: D
The higher factor produces a higher local stress at the notch root and therefore initiates a crack sooner. Reducing stress concentration, by radiusing corners and easing transitions, is the most direct fatigue improvement available.
Please log in or register to report this question.
