Question 1 of 8: Coffin–Manson fit and Miner cumulative damage

Question 1: 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 constant-plastic-strain-range fatigue tests on one alloy, then a three-block service history applied to a component made from it.

Service history: 500 cycles at $\Delta\epsilon = 0.011$, then 300 cycles at $\Delta\epsilon = 0.020$, then the balance of life at $\Delta\epsilon = 0.009$.

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

Approach. Take logarithms to turn the power law into a straight line, fit it by least squares to recover $\alpha$ and $C$, invert the law to get an allowable life for each service block, and close the Miner damage sum on unity.

1. Part (a) — linearise the Coffin–Manson law. Taking natural logarithms of $\Delta\epsilon = CN^{\alpha}$ gives $$\ln \Delta\epsilon = \ln C + \alpha \ln N,$$ which is a straight line of slope $\alpha$ and intercept $\ln C$ in the variables $(\ln N,\ \ln \Delta\epsilon)$. The four test points therefore over-determine two unknowns, and least squares is the right way to use all four.
2. Form the regression sums. With $x_i = \ln N_i$ and $y_i = \ln \Delta\epsilon_i$ over the four tests, $$\alpha = \frac{n\sum x_i y_i - \sum x_i \sum y_i}{n\sum x_i^2 - \left(\sum x_i\right)^2}.$$ Substituting $n = 4$, $\sum x_i = 30.0647$, $\sum y_i = -16.4807$, $\sum x_i^2 = 235.3635$ and $\sum x_i y_i = -127.7590$ gives the slope.
3. Recover the exponent and the coefficient. The regression returns $$\boxed{\alpha = -0.4139}, \qquad \ln C = \frac{\sum y_i - \alpha \sum x_i}{n} = -1.0095 \ \Rightarrow\ \boxed{C = 0.3644}.$$ The coefficient of determination is $R^2 = 0.9998$, so the four points lie essentially on one line and the fit can be used with confidence over this strain range.
4. Part (b) — invert the law to get an allowable life per block. Rearranging, $N = (\Delta\epsilon / C)^{1/\alpha}$. Applying it to the three service strain ranges: $$N_1 = \left(\frac{0.011}{0.3644}\right)^{-1/0.4139} = 4711 \ \text{cycles},$$ and in the same way $N_2 = 1111$ cycles at $\Delta\epsilon = 0.020$ and $N_3 = 7651$ cycles at $\Delta\epsilon = 0.009$. Note how steeply life falls with strain range — not quite doubling the strain from 0.011 to 0.020 cuts the life by a factor of more than four.
5. Accumulate the damage already spent. Miner's rule adds the fractions of life consumed in each block, $$D = \sum \frac{n_i}{N_i} = \frac{500}{4711} + \frac{300}{1111} = 0.1061 + 0.2700 = 0.3761.$$ The first two blocks therefore use just under 38 % of the fatigue life, leaving $1 - D = 0.6239$ available for the third block.
6. Convert the remaining damage into cycles at the final strain range. The third block runs until the damage sum reaches unity, so $$n_3 = (1 - D)\,N_3 = 0.6239 \times 7651 = 4773 \ \text{cycles},$$ and the total life is $$\boxed{N_{\text{total}} = 500 + 300 + 4773 = 5573 \ \text{cycles}}.$$ Substituting back, $500/4711 + 300/1111 + 4773/7651 = 1.000$, which is the closure check that the damage sum was solved consistently.

The answer is dominated by the third block simply because it is the mildest: at $\Delta\epsilon = 0.009$ the alloy tolerates roughly seven times as many cycles as it does at 0.020, so the last block contributes 86 % of the cycles but only 62 % of the damage.