Coffin–Manson fit and Miner cumulative damage: 22-Mec-B9 Advanced Engineering Structures, May 2016, Q3

Question 3: Coffin–Manson fit and Miner 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 strain-controlled fatigue results and a two-block service history to be assessed against them.

Given data — Question 3
Plastic strain range $\Delta\varepsilon$	Cycles to failure $N$
0.0380	230
0.0221	910
0.0150	2 300
0.0080	12 500
Service: $\Delta\varepsilon = 0.019$ for 350 cycles, then $\Delta\varepsilon = 0.011$ to failure

Find. (a) the constants $C$ and $\alpha$ of the Coffin–Manson law $\Delta\varepsilon = CN^{\alpha}$ that best fit the four points, and (b) the total life under the two-block history using Miner’s rule.

Linearise the law. Taking base-ten logarithms of $\Delta\varepsilon = CN^{\alpha}$ turns it into a straight line, $$\log_{10}\Delta\varepsilon = \log_{10}C + \alpha\log_{10}N$$ so a least-squares fit in log–log coordinates gives the exponent as the slope and the coefficient as the intercept. This is the only fit that treats the four decades of life evenly; fitting the raw numbers would let the 12 500-cycle point dominate.
Form the regression sums. With $X = \log_{10}N$ and $Y = \log_{10}\Delta\varepsilon$ over the four points, $$\begin{aligned} n &= 4, \\ \sum X &= 12.77941, \\ \sum Y &= -6.99664, \\ \sum XY &= -22.97551, \\ \sum X^{2} &= 42.41957 \end{aligned}$$
Solve for the slope and intercept. The standard least-squares formulae give $$\alpha = \frac{n\sum XY - \sum X\sum Y}{n\sum X^{2}-\left(\sum X\right)^{2}} = \boxed{-0.3911}$$ $$\log_{10}C = \bar{Y}-\alpha\bar{X} = -1.74916 - (-0.3911)(3.19485) = -0.49978 \;\Rightarrow\; C = \boxed{0.3164}$$ so the fitted law is $\Delta\varepsilon = 0.3164\,N^{-0.3911}$.
Check the quality of the fit. Substituting the four test lives back gives predicted strain ranges of 0.03773, 0.02203, 0.01533 and 0.00791, within 2.3 per cent of every measurement, and the coefficient of determination is $R^{2} = 0.9995$. An exponent near $-0.4$ and a coefficient of order 0.3 are also the textbook range for a ductile aerospace alloy, so nothing here looks like a transcription error.

Figure 3.1 — the four test points and the fitted Coffin–Manson line, with the two service strain ranges marked on it.

Part (b) uses the fitted law twice, once for each block of the service history.

Invert the law to get the allowable life at each strain range. Rearranging $\Delta\varepsilon = CN^{\alpha}$ gives $N = \left(\Delta\varepsilon/C\right)^{1/\alpha}$, so $$N_{1} = \left(\frac{0.019}{0.3164}\right)^{1/(-0.3911)} = \boxed{1329\ \text{cycles}}$$ $$N_{2} = \left(\frac{0.011}{0.3164}\right)^{1/(-0.3911)} = \boxed{5376\ \text{cycles}}$$ Halving the strain range multiplies the life by about six, which is what an exponent of $-0.39$ implies.
Accumulate damage on Miner’s rule. Miner adds the cycle ratios and calls failure when they reach unity: $$\frac{n_{1}}{N_{1}} + \frac{n_{2}}{N_{2}} = 1 \;\Rightarrow\; \frac{350}{1329} + \frac{n_{2}}{5376} = 1$$ The first block therefore consumes $350/1329 = 0.2634$, or just over a quarter of the total damage capacity, in only 350 cycles.
Solve for the length of the second block. Rearranging, $$n_{2} = \left(1-0.2634\right)\times 5376 = \boxed{3960\ \text{cycles}}$$
Add the two blocks for the total life. $$N_{\text{total}} = n_{1} + n_{2} = 350 + 3960 = \boxed{4310\ \text{cycles}}$$ This sits between the two single-block lives, much closer to the milder one, because only a quarter of the damage was spent on the severe block. Had the whole life been run at $\Delta\varepsilon = 0.011$ it would have reached 5376 cycles, so the 350 severe cycles cost about 1070 mild ones — a penalty of roughly three to one per cycle.

Question 3 — results
Quantity	Result
(a) Coffin–Manson exponent	$\alpha = -0.3911$
(a) Coffin–Manson coefficient	$C = 0.3164$
(a) Fitted law	$\Delta\varepsilon = 0.3164\,N^{-0.3911}$
(a) Coefficient of determination	$R^{2} = 0.9995$
(b) Allowable life at $\Delta\varepsilon = 0.019$	1 329 cycles
(b) Allowable life at $\Delta\varepsilon = 0.011$	5 376 cycles
(b) Damage from the first block	0.2634
(b) Cycles available in the second block	3 960 cycles
(b) Total life	4 310 cycles

Concept & Theory

Topic. Low-cycle fatigue: fitting the Coffin–Manson plastic strain-life law and combining variable-amplitude blocks with Miner’s linear damage rule.

Key relations. $\Delta\varepsilon_{p} = C N^{\alpha}$; its logarithmic form $\log\Delta\varepsilon = \log C + \alpha\log N$; least squares for the slope and intercept; and $\sum n_{i}/N_{i} = 1$ at failure.

Why this works. Low-cycle fatigue is governed by cyclic plastic strain rather than by stress, and over the plastic range the strain-life relation is a straight line on log–log axes, so a two-parameter power law captures decades of data with only a handful of tests. Miner’s rule then treats damage as a fraction of life consumed at each amplitude, independent of the order in which the blocks are applied. That order-independence is an approximation, not a physical law — high-then-low sequences typically fail at sums below unity because the severe block leaves residual damage that the mild block exploits — but it is the standard assumption and it is what the question asks for.

Common pitfall. Fitting the raw $\Delta\varepsilon$ and $N$ values rather than their logarithms; the resulting exponent can be off by ten per cent or more because the fit is then dominated by the longest-life point. The second trap in part (b) is adding lives rather than damage fractions, which would give a much longer and quite wrong answer.

Source. Dowling, Mechanical Behavior of Materials, 4th ed., Ch. 14 (strain-based fatigue) and Ch. 9 (Palmgren–Miner rule).

Question 3 of 8: Coffin–Manson fit and Miner cumulative damage

Question 3: Coffin–Manson fit and Miner cumulative damage (20 marks)