Coffin–Manson constants and Miner’s cumulative damage: 22-Mec-B9 Advanced Engineering Structures, Undated paper, Q1

Question 1: Coffin–Manson constants and Miner’s cumulative damage (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Four constant-amplitude low-cycle-fatigue results for an aircraft metallic alloy, and a three-block variable-amplitude service spectrum.

Plastic strain range Δε_p	Cycles to failure N
0.0390	210
0.0215	890
0.0125	3850
0.0075	13000

Service spectrum: Δε_p = 0.011 for the first 500 cycles, then Δε_p = 0.02 for the next 300 cycles, then Δε_p = 0.009 for the remainder of the life.

Find. (a) the coefficient $C$ and exponent $\alpha$ of $\Delta\epsilon_p = C N^{\alpha}$ that best fit the four points; (b) the total number of cycles the component survives under the three-block spectrum, using Miner’s linear cumulative-damage law.

Figure 1.1 — the four test results on log–log axes with the least-squares Coffin–Manson line. A straight line on these axes is exactly the statement Δε_p = C N^α.

Approach. Take base-10 logarithms to turn the power law into a straight line, fit that line by least squares, then invert it to get an allowable life at each block strain and sum the damage fractions to unity.

Part (a) — linearise the power law. Taking logarithms of $\Delta\epsilon_p = C N^{\alpha}$ gives a straight line in the transformed variables $X = \log_{10} N$ and $Y = \log_{10}\Delta\epsilon_p$: $$Y = \log_{10} C + \alpha X.$$ The slope of that line is the exponent and its intercept is $\log_{10} C$, so an ordinary least-squares fit on the transformed data is all that is required.
Form the least-squares sums. With $n = 4$ points, $$\sum X = 12.97101,\quad \sum Y = -7.10453,\quad \sum X^2 = 43.87166,\quad \sum XY = -23.75548.$$
Solve for the slope and intercept. The normal equations give $$\alpha = \frac{n\sum XY - \sum X \sum Y}{n \sum X^2 - (\sum X)^2} = -0.3963,$$ and the intercept follows as $\log_{10} C = (\sum Y - \alpha \sum X)/n = -0.4910$, so that $$\boxed{\;C = 0.3228,\qquad \alpha = -0.3963\;}$$ The coefficient of determination is $R^2 = 0.99945$, i.e. the four points lie on the fitted line to within a fraction of a percent — this alloy obeys the Coffin–Manson form over the whole range tested.
Part (b) — allowable life at each block strain. Inverting the fitted law, $N = (\Delta\epsilon_p / C)^{1/\alpha}$. Substituting each of the three block strains, $$N_1(0.011) = 5049,\qquad N_2(0.02) = 1117,\qquad N_3(0.009) = 8377 \text{ cycles.}$$ Note that the largest strain gives the shortest life, as it must.
Accumulate the damage of the two completed blocks. Miner’s rule assigns each block a damage fraction $n_i/N_i$ and treats the fractions as additive: $$D_{1+2} = \frac{500}{5049} + \frac{300}{1117} = 0.0990 + 0.2686 = 0.3676.$$ Just over a third of the fatigue life is used up in the first 800 cycles, most of it by the short 0.02-strain block.
Spend the remaining damage in the third block. Failure occurs when the total damage reaches unity, so the third block may contribute $1 - D_{1+2} = 0.6324$: $$n_3 = (1 - D_{1+2})\,N_3 = 0.6324 \times 8377 = 5298 \text{ cycles.}$$
Total life. Adding the three blocks, $$\boxed{\;N_{\text{total}} = 500 + 300 + 5298 = 6098 \text{ cycles}\;}$$ which is about 73% of the life the component would have had if it had seen the mild 0.009 strain range alone.

Quantity	Symbol	Value
Coffin–Manson coefficient	C	0.3228
Coffin–Manson exponent	α	-0.3963
Quality of fit	R²	0.99945
Life at Δε_p = 0.011	N₁	5049 cycles
Life at Δε_p = 0.02	N₂	1117 cycles
Life at Δε_p = 0.009	N₃	8377 cycles
Damage after the first two blocks	D₁₊₂	0.3676
Cycles survived in the third block	n₃	5298 cycles
Total life	N_total	6098 cycles

Topic. Low-cycle fatigue: the Coffin–Manson relation and Miner’s linear damage summation.

Key relations. $\Delta\epsilon_p = C N^{\alpha}$ (Coffin–Manson, with $\alpha$ typically between $-0.5$ and $-0.4$ for metals); $D = \sum n_i/N_i = 1$ at failure (Palmgren–Miner).

Why this works. Above about 10⁴ cycles the strain amplitude that matters is elastic and Basquin’s stress-life law governs; below it the plastic strain dominates and the number of reversals a metal survives is set almost entirely by how much it is plastically worked each cycle. Coffin and Manson found empirically that plotting plastic strain range against life on log–log axes gives a straight line, which is why the fit is done in log space rather than on the raw numbers — a least-squares fit on the untransformed data would let the single high-strain point dominate. Miner’s rule then assumes damage accumulates linearly and independently of the order in which the blocks are applied, which makes the arithmetic trivial and is accurate to roughly a factor of two in practice.

Common pitfall. Fitting the line by eye through the two end points, or fitting $N$ on $\Delta\epsilon_p$ instead of the other way round, both shift $\alpha$ by several percent — and because $\alpha$ sits in an exponent, a small error in it becomes a much larger error in predicted life. The second common slip is to add the block cycles rather than the damage fractions.

Source. Dowling, Mechanical Behavior of Materials, 4th ed., §14.4–14.6 (strain-life) and §9.7 (Miner); Megson, Aircraft Structures, 6th ed., §12.2.

Question 1 of 8: Coffin–Manson constants and Miner’s cumulative damage

Question 1: Coffin–Manson constants and Miner’s cumulative damage (20 marks)