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

Question 3: Coffin–Manson fit and 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 tests, and a two-block service history to be assessed against them:

Given data — strain cycling tests and service blocks
Plastic strain range $\Delta\varepsilon$	Cycles to failure $N$
0.0370	260
0.0211	970
0.0140	2 800
0.0080	12 000
Service: 200 cycles at $\Delta\varepsilon = 0.017$, then the balance of life at $\Delta\varepsilon = 0.010$

Find. The best-fit constants $C$ and $\alpha$ of the Coffin–Manson law, and the total number of cycles the component survives under the two-block history.

Figure 3.1 — the four test points and the least-squares Coffin–Manson line on log–log axes. The fit is essentially exact over one and a half decades of life.

Approach. Take logarithms to turn the power law into a straight line, fit it by least squares, then convert each service strain range into an allowable life and sum the damage fractions to unity.

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 $\ln C$ on log–log axes. The four data pairs become $(\ln N,\ \ln \Delta\varepsilon) = (5.561,\,-3.297)$, $(6.877,\,-3.858)$, $(7.937,\,-4.269)$ and $(9.393,\,-4.828)$.
Fit by least squares (part a). Applying the standard normal equations to those four points,$$\alpha = \frac{n\sum xy - \sum x \sum y}{n\sum x^{2} - (\sum x)^{2}} = -0.3985, \qquad \ln C = \bar{y} - \alpha\bar{x} = -1.0975,$$so $C = e^{-1.0975} = 0.3337$ and the fitted law is$$\boxed{\Delta\varepsilon = 0.3337\,N^{-0.3985}} .$$The fit is excellent — the coefficient of determination is $R^{2} = 0.9988$ — and the exponent is of the order expected for a plastic-strain (Coffin–Manson) law — a little shallower than the −0.5 to −0.7 usually quoted for metals, but far steeper than the −0.05 to −0.12 of the elastic (Basquin) term — which is a useful sanity check that the data are plastic-strain controlled rather than elastic.
Convert each service block to an allowable life. Inverting the fitted law, $N = (\Delta\varepsilon/C)^{1/\alpha}$. For the first block,$$N_1 = \left(\frac{0.017}{0.3337}\right)^{1/(-0.3985)} = 1\,756\ \text{cycles},$$and for the second,$$N_2 = \left(\frac{0.010}{0.3337}\right)^{1/(-0.3985)} = 6\,650\ \text{cycles}.$$Both interpolate inside the tested range, so no extrapolation is involved.
Apply Miner’s rule (part b). Failure occurs when the accumulated damage reaches unity:$$\frac{n_1}{N_1} + \frac{n_2}{N_2} = 1 \;\Longrightarrow\; \frac{200}{1756} + \frac{n_2}{6650} = 1 .$$The first block consumes $0.1139$, or 11.4 % of the life, leaving $0.8861$ for the second. Hence$$n_2 = 6650(0.8861) = 5\,892\ \text{cycles},$$and the total life is$$\boxed{N_{\text{total}} = 200 + 5\,892 = 6\,092\ \text{cycles}} .$$It is worth seeing how disproportionate the first block is: 200 cycles at the higher strain — only 3 % of the total count — eat 11 % of the available life, because life varies as roughly the 2.5th power of the inverse strain range.

Final results — Question 3
Quantity	Value
Fatigue ductility exponent $\alpha$	−0.3985
Coefficient $C$	0.3337
Quality of fit $R^{2}$	0.9988
$N_1$ at $\Delta\varepsilon = 0.017$	1 756 cycles
$N_2$ at $\Delta\varepsilon = 0.010$	6 650 cycles
Damage used by the first block	0.1139
Cycles survived in the second block	5 892
Total life	6 092 cycles

Topic. Low-cycle (strain-life) fatigue: the Coffin–Manson relation and linear cumulative damage.

Key relations. $\Delta\varepsilon_p = C N^{\alpha}$ (equivalently $\Delta\varepsilon_p/2 = \varepsilon_f'(2N_f)^{c}$); least squares on $\ln \Delta\varepsilon$ against $\ln N$; and the Palmgren–Miner sum $\sum n_i/N_i = 1$.

Why this works. When cycling is plastic-strain controlled the damage per cycle is governed by the irreversible hysteresis loop, and experiment shows that plotting strain range against life on log–log axes gives a straight line over several decades. Fitting in log space rather than by eye is what makes the exponent trustworthy, because it weights all four decades of life equally instead of letting the two high-strain points dominate. Miner’s rule then treats each block as consuming a fixed fraction of the total life, independent of the order in which the blocks are applied.

Common pitfall. Fitting the raw $(\Delta\varepsilon, N)$ pairs by ordinary least squares instead of the logarithms; that minimises the wrong residual and typically shifts $\alpha$ by several per cent. The second trap is arithmetic: $1/\alpha = -2.509$, so a sign slip when inverting the power law turns 1 756 cycles into 0.00057. Miner’s rule itself is also an idealisation — it ignores load-sequence effects, and a high-low sequence like this one in reality fails a little earlier than the rule predicts.

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 cumulative damage

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