Coffin–Manson constants and Miner cumulative damage: 22-Mec-B9 Advanced Engineering Structures, December 2017, Q3

Question 3: Coffin–Manson constants 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 constant-amplitude low-cycle fatigue tests on one alloy, followed by a three-block service history applied to a component made from it.

Given data — strain-cycling test results
Range of plastic strain $\Delta\varepsilon$	Cycles to failure $N$
0.0360	260
0.0211	990
0.0120	2700
0.0070	15 500

Service history: 390 cycles at $\Delta\varepsilon = 0.019$, then 100 cycles at $\Delta\varepsilon = 0.025$, then the balance of the life at $\Delta\varepsilon = 0.010$.

Find. (a) the best-fit constants $C$ and $\alpha$ in $\Delta\varepsilon = CN^{\alpha}$; (b) the total number of cycles the component survives under the three-block history, using the linear Palmgren–Miner damage rule.

The four test points and the least-squares Coffin–Manson line. A power law plots as a straight line on logarithmic axes, so fitting it is an ordinary linear regression of $\ln \Delta\varepsilon$ on $\ln N$.

Approach. Linearise the power law by taking logarithms, fit by least squares to obtain $\alpha$ and $C$, invert the fitted law to get an allowable life at each of the three service strain ranges, and close the Miner sum at unity to solve for the unknown third block.

Part (a) — linearise the power law. Taking natural logarithms of $\Delta\varepsilon = CN^{\alpha}$ gives $\ln\Delta\varepsilon = \ln C + \alpha \ln N$, a straight line of slope $\alpha$ and intercept $\ln C$ in the transformed variables $x = \ln N$, $y = \ln\Delta\varepsilon$.
Fit the line by least squares. With the four transformed points, the normal equations give $$\alpha = \frac{n\sum xy-\sum x\sum y}{n\sum x^{2}-(\sum x)^{2}} = -0.4079, \qquad \ln C = \frac{\sum y - \alpha \sum x}{n} = -1.0819 .$$ Exponentiating the intercept, $$\boxed{\Delta\varepsilon = 0.3390\,N^{-0.4079}}$$ with a coefficient of determination $R^{2} = 0.987$, confirming that a single power law represents all four decades of the data well.
Sanity-check the exponent. A Coffin–Manson exponent near $-0.4$ to $-0.6$ is the expected range for structural metals, and the fitted $-0.408$ sits comfortably inside it, so neither a transcription error nor a units slip has crept into the data.
Part (b) — invert the law to get allowable lives. Rearranging, $N = \left(\Delta\varepsilon / C\right)^{1/\alpha}$, so for the three service blocks $$N_1 = 1169\ \text{cycles at } 0.019, \qquad N_2 = 597\ \text{cycles at } 0.025, \qquad N_3 = 5641\ \text{cycles at } 0.010 .$$ The larger strain range gives the shorter life, as it must.
Accumulate the damage from the two completed blocks. Miner's rule adds the fractional lives consumed: $$D_{1,2} = \frac{n_1}{N_1}+\frac{n_2}{N_2} = \frac{390}{1169}+\frac{100}{597} = 0.3335 + 0.1676 = 0.5011 .$$ Just over half the fatigue life is used up in the first 490 cycles.
Solve for the third block. Failure occurs when the total damage reaches unity, so the number of further cycles the component tolerates at $\Delta\varepsilon = 0.010$ is $$n_3 = \left(1 - D_{1,2}\right) N_3 = 0.4989 \times 5641 = 2815\ \text{cycles}.$$
Add the blocks for the total life. $$\boxed{N_{\text{total}} = 390 + 100 + 2815 = 3305 \ \text{cycles}}$$ Rounding to the precision the data justify, the component survives approximately $3.3\times10^{3}$ cycles.
Verify the damage closes on unity. Re-evaluating $390/1169 + 100/597 + 2815/5641$ returns $1.0000$ to ten decimal places, which is the arithmetic check that the third block was solved rather than assumed.

Final results — Question 3
Quantity	Value
(a) Exponent	$\alpha = -0.4079$
(a) Coefficient	$C = 0.3390$
(a) Goodness of fit	$R^{2} = 0.987$
Allowable life at $\Delta\varepsilon = 0.019$	1169 cycles
Allowable life at $\Delta\varepsilon = 0.025$	597 cycles
Allowable life at $\Delta\varepsilon = 0.010$	5641 cycles
Damage after the first two blocks	$D = 0.501$
Cycles endured in the third block	2815
(b) Total life	3305 cycles

Topic. Strain-life (low-cycle) fatigue and linear cumulative damage.

Key relations. Coffin–Manson: $\Delta\varepsilon_p = C N^{\alpha}$ (equivalently $\Delta\varepsilon_p/2 = \varepsilon_f'(2N)^{c}$). Palmgren–Miner: $\sum n_i/N_i = 1$ at failure. Least-squares slope $\alpha = [n\sum xy - \sum x \sum y]/[n\sum x^{2}-(\sum x)^{2}]$ on $x=\ln N$, $y = \ln \Delta\varepsilon$.

Why this works. Low-cycle fatigue is controlled by plastic strain, not by stress, because once the material yields each cycle the stress amplitude barely changes while the plastic strain range does all the damage. Plotting the data logarithmically turns the empirical power law into a straight line, so all four points contribute to one slope and one intercept instead of being paired off. Miner's rule then assumes damage accumulates linearly and independently of the order in which the blocks are applied, which lets the unknown third block be recovered from a single equation.

Common pitfall. Fitting the line to $\Delta\varepsilon$ against $N$ rather than to their logarithms — that weights the largest-strain point almost exclusively and typically shifts $\alpha$ by 10 % or more. A second pitfall is forgetting to add the completed blocks back in at the end: $n_3 = 2815$ is not the answer, $390+100+2815$ is.

Source. Dowling, Mechanical Behavior of Materials, 4th ed., §14.4 (strain-life curves and the Coffin–Manson relation) and §9.7 (the Palmgren–Miner rule).

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

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