Low-cycle fatigue: strain-life fit and cumulative damage: 22-Mec-B9 Advanced Engineering Structures, May 2013, Q5

Question 5: Low-cycle fatigue: strain-life 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 constant-amplitude strain-cycling results, then a two-block service history.

Test point	Value
Plastic strain range $\Delta\varepsilon$	Cycles to failure $N$
0.0400	200
0.0211	1 000
0.0160	2 000
0.0084	10 000
First service block	$\Delta\varepsilon = 0.015$ for 500 cycles
Second service block	$\Delta\varepsilon = 0.010$ to failure

Find. (a) the constants $C$ and $\alpha$ that fit the four test points, and (b) the total number of cycles the two-block history survives.

The four test points on logarithmic axes. They lie on a straight line to within the width of the markers, which is what demonstrates the power law; the slope of that line is $\alpha$ and its intercept at $N = 1$ is $C$.

Approach. Take logarithms to turn the power law into a straight line, fit it by least squares, confirm the fit reproduces every data point, then use the fitted law to convert each service strain range into a life and sum the damage fractions to unity.

Linearise the proposed law. Taking natural logarithms of $\Delta\varepsilon = CN^{\alpha}$ gives$$\ln\Delta\varepsilon = \ln C + \alpha \ln N$$so the proposition to be demonstrated is simply that the four points are collinear on log–log axes.
Fit the line by least squares. Using all four points,$$\alpha = \frac{4\sum xy - \sum x\sum y}{4\sum x^{2} - \left(\sum x\right)^{2}} = -0.3989, \qquad C = \exp\left(\bar{y} - \alpha\bar{x}\right) = 0.3316$$with $x = \ln N$ and $y = \ln\Delta\varepsilon$, so $\boxed{\Delta\varepsilon = 0.3316\,N^{-0.3989}}$. The exponent is within one per cent of $-0.4$ and $C$ is close to one third, the values a candidate would quote from a two-point calculation using the extreme data.
Demonstrate that the fit represents the data. Substituting each test life back into the fitted law returns 0.04005, 0.02107, 0.01598, 0.00841 against the measured 0.0400, 0.0211, 0.0160, 0.0084. The largest discrepancy is under half a per cent and the coefficient of determination is $R^{2} = 1.00000$, so the power law is not merely plausible but essentially exact over this range — which is the answer part (a) asks for.
Invert the law to obtain lives (part b). Rearranging, $N = \left(\Delta\varepsilon/C\right)^{1/\alpha}$. For the two service blocks,$$N_{1} = \left(\frac{0.015}{0.3316}\right)^{1/-0.3989} = 2\,345, \qquad N_{2} = \left(\frac{0.010}{0.3316}\right)^{1/-0.3989} = 6\,479\ \text{cycles}$$Note that reducing the strain range by only one third nearly triples the life; the reciprocal exponent of $-2.5$ is what makes low-cycle fatigue life so sensitive to strain amplitude.
Apply Miner's rule. Failure occurs when the damage fractions sum to unity:$$\frac{500}{2\,345} + \frac{n_{2}}{6\,479} = 1\;\Longrightarrow\; n_{2} = \left(1 - 0.2133\right)6\,479 = 5\,097\ \text{cycles}$$The first block consumes 21.3 per cent of the life in only 500 cycles.
Report the total life.$$N_{\text{total}} = 500 + 5\,097 = \boxed{5\,597\ \text{cycles}}$$

Quantity	Result
Fatigue ductility exponent $\alpha$	-0.3989
Fatigue ductility coefficient $C$	0.3316
Quality of fit $R^{2}$	1.00000
Life at $\Delta\varepsilon = 0.015$	2 345 cycles
Life at $\Delta\varepsilon = 0.010$	6 479 cycles
Damage used by the first block	0.213 (21.3 per cent)
Cycles at the second strain range	5 097 cycles
Total life	5 597 cycles

Check

A least-squares fit is used because it employs all four data points; taking only the first and last points gives $\alpha = -0.399$ and $C = 0.333$ and shifts the final answer by under one per cent, so either route earns full marks. Miner's rule is assumed by the question; it takes no account of the order of the blocks, although in practice a high-strain block applied first is more damaging than the linear sum predicts.

Concept & Theory

Topic. The Coffin–Manson plastic strain-life law and linear cumulative damage.

Key relations. $\Delta\varepsilon_{p} = CN^{\alpha}$ (equivalently $\Delta\varepsilon_{p}/2 = \varepsilon_{f}'(2N_{f})^{c}$); $N = \left(\Delta\varepsilon/C\right)^{1/\alpha}$; Miner: $\sum n_{i}/N_{i} = 1$ at failure.

Why this works. In low-cycle fatigue the plastic strain range, not the stress range, is the controlling variable, because above the yield point stress is a very insensitive measure of how hard the material is being worked. Coffin and Manson found independently that plotting plastic strain range against life on logarithmic axes gives a straight line of slope close to $-0.5$ for most metals, with the intercept related to the monotonic fracture ductility. Miner's rule then assumes damage accumulates linearly and independently of sequence, so each block of loading consumes a fraction of life equal to its cycles divided by the life it would have on its own.

Common pitfall. Two errors dominate. The first is fitting the line on linear rather than logarithmic axes, which weights the high-strain point enormously and gives a badly wrong exponent. The second is inverting the exponent incorrectly: the life follows $N \propto \Delta\varepsilon^{1/\alpha}$ with $1/\alpha = -2.5$, and using $\alpha$ rather than $1/\alpha$ in the inversion produces lives that are wrong by orders of magnitude.

Source. Dowling, Mechanical Behavior of Materials, 4th ed., §14.3 (strain-based approach, Coffin–Manson) and §9.7–9.8 (Palmgren–Miner rule); Megson, Aircraft Structures, 6th ed., §10.5.

Question 5 of 8: Low-cycle fatigue: strain-life fit and cumulative damage

Question 5: Low-cycle fatigue: strain-life fit and cumulative damage (20 marks)