Coffin–Manson fit and Miner’s cumulative damage rule: 22-Mec-B9 Advanced Engineering Structures, May 2018, Q4

Question 4: Coffin–Manson fit and Miner’s cumulative damage rule (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 strain-cycling test points for an aircraft component, and a two-block service history.

Range of plastic strain Δε	Cycles to failure N
0.0500	150
0.0211	1100
0.0160	2200
0.0084	12000

In service the component sees $\Delta\varepsilon = 0.015$ for the first 500 cycles and $\Delta\varepsilon = 0.009$ for the remainder of its life.

Find. (a) the constants C and α in $\Delta\varepsilon = C N^{\alpha}$, and (b) the total number of cycles to failure under the two-block history, using Miner’s rule.

Approach. The proposed law is a straight line in log–log axes, so fit it by least squares on the logarithms; then invert the law to get the allowable life at each service strain and sum the damage fractions to unity.

Linearise the Coffin–Manson form. Taking natural logarithms of $\Delta\varepsilon = CN^{\alpha}$, $$\ln \Delta\varepsilon = \ln C + \alpha \ln N$$ so a least-squares straight line through the four $(\ln N, \ln \Delta\varepsilon)$ points delivers both constants at once.
Least-squares slope and intercept. With $x = \ln N$ and $y = \ln \Delta\varepsilon$, $$\alpha = \frac{n\sum xy - \sum x \sum y}{n \sum x^2 - (\sum x)^2} = -0.4076, \qquad \ln C = \frac{\sum y - \alpha \sum x}{n} \;\Rightarrow\; C = 0.3766$$ so the fitted law is $$\boxed{\Delta\varepsilon = 0.3766\,N^{-0.4076}}$$ with a coefficient of determination of 0.9985 on the logarithms — the four points really are collinear in log–log axes.
Allowable life at each service strain (part b). Inverting the fitted law, $N_f = (\Delta\varepsilon/C)^{1/\alpha}$: $$N_1 = \left(\frac{0.015}{0.3766}\right)^{1/(-0.4076)} = 2717\ \text{cycles}, \qquad N_2 = \left(\frac{0.009}{0.3766}\right)^{1/(-0.4076)} = 9514\ \text{cycles}$$
Damage used by the first block. Miner’s rule charges each block its fraction of the life it consumes: $$D_1 = \frac{n_1}{N_1} = \frac{500}{2717} = 0.184$$ so 18.4% of the component is spent in the first 500 cycles.
Cycles available in the second block. Failure arrives when the damage sum reaches unity: $$\frac{n_2}{N_2} = 1 - D_1 \;\Rightarrow\; n_2 = (1-0.184)(9514) = 7763\ \text{cycles}$$
Total life. $$N_{\text{total}} = n_1 + n_2 = 500 + 7763 = \boxed{8263\ \text{cycles}}$$ Checking the closure, $500/2717 + 7763/9514 = 1.000$.

Question 4 — the four test points and the fitted Coffin–Manson line on log–log axes. The two service strains, 0.015 and 0.009, both lie inside the tested range, so the life estimates are interpolations rather than extrapolations.

It is worth noticing how lopsided the answer is. The high-strain block is short but expensive, and the low-strain block is long but cheap: 500 cycles at 0.015 cost as much life as 1750 cycles at 0.009 would. That is the practical value of a cumulative-damage calculation — it converts a mixed history into a single number that can be compared with an inspection or retirement interval.

Quantity	Value
(a) Coefficient C	0.3766
(a) Exponent α	−0.4076
Quality of the fit (r² on logarithms)	0.9985
Life at Δε = 0.015	2717 cycles
Life at Δε = 0.009	9514 cycles
Damage used in the first block	0.184
Cycles endured in the second block	7763 cycles
(b) Total life	8263 cycles

Topic. Low-cycle fatigue: the Coffin–Manson law and Palmgren–Miner cumulative damage.

Key relations. $\Delta\varepsilon_p = C N^{\alpha}$ (a straight line in log–log axes); least squares on $\ln N$, $\ln \Delta\varepsilon$; Miner’s rule $\sum n_i/N_{f,i} = 1$ at failure.

Why this works. In low-cycle fatigue the plastic strain range, not the stress range, is the controlling variable, and over several decades of life it follows a power law with a fatigue-ductility exponent near −0.4 to −0.6 for most metals — the fitted −0.408 sits squarely in that band. Miner’s rule then assumes damage accumulates linearly and independently of order, so each block spends the fraction of the life it uses, and failure is the moment the fractions total one. Linear summation ignores load-sequence effects, but it is the standard first estimate and is what the question asks for.

Common pitfall. Fitting the law to the raw numbers instead of their logarithms, which lets the single high-strain point dominate the regression; and rounding α too early — it sits in an exponent $1/\alpha \approx -2.45$, so an error in the third decimal moves the predicted life by several per cent.

Source. Dowling, Mechanical Behavior of Materials, 4th ed., ch. 14 (strain-based fatigue) and §9.9 (Palmgren–Miner); Megson, 6th ed., §15.4 (cumulative damage in airframes).

Question 4 of 8: Coffin–Manson fit and Miner’s cumulative damage rule

Question 4: Coffin–Manson fit and Miner’s cumulative damage rule (20 marks)