22-Mec-B9 Advanced Engineering Structures · December 2014
Question 3 of 8: Coffin–Manson fit and cumulative fatigue damage
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, December 2014 — 07-Mec-B9 Advanced Engineering Structures. Three hours, open book, any non-communicating calculator permitted. Eight questions of equal total value (20 marks each); the rubric states that any five constitute a complete paper. All eight are solved here, because the complete set is the study resource.
Reference texts.
T. H. G. Megson, Aircraft Structures for Engineering Students, 6th ed. — open and closed thin-walled sections, shear flow, shear centre, structural idealisation (Ch. 16, 17, 18, 20, 21, 23).
A. C. Ugural and S. K. Fenster, Advanced Strength and Applied Elasticity, 5th ed. — yield criteria, torsion of thin-walled members, statically indeterminate axial members (Ch. 2, 4, 6).
A. P. Boresi and R. J. Schmidt, Advanced Mechanics of Materials, 6th ed. — torsion of open and closed sections, multiply connected cells (Ch. 6).
N. E. Dowling, Mechanical Behavior of Materials, 4th ed. — linear elastic fracture mechanics, the Paris law, strain-life and Miner’s rule (Ch. 8, 11, 14).
R. C. Hibbeler, Mechanics of Materials, 10th ed. — thermal stress in restrained members and combined loading (Ch. 4, 8).
Axes and sign convention used throughout. A right-handed set is used for every thin-walled question: the span axis runs along the member, $Y$ is vertical (upward) in the cross-section and $Z$ is horizontal. Section constants are written $I_{ZZ}=\int Z^{2}\,dA$, $I_{YY}=\int Y^{2}\,dA$ and $I_{YZ}=\int ZY\,dA$ about the centroid, and the bending stress is carried as the linear field $\sigma = aZ + bY$. Shear flows are counted positive in the direction of the walk stated with each figure; a negative answer simply means the flow runs the other way round the circuit.
Question 3: Coffin–Manson fit and cumulative fatigue damage (20 marks)
Given. Four low-cycle fatigue test points and a two-block service history.
Plastic strain range
Cycles to failure
Role
0.0400
120
test point 1
0.0211
800
test point 2
0.0160
1900
test point 3
0.0084
8000
test point 4
0.015
to be found
service block 1, 500 cycles applied
0.010
to be found
service block 2, run to failure
Find. (a) the constants $C$ and $\alpha$ of the Coffin–Manson power law together with evidence that the law actually fits, and (b) the total number of cycles the component survives under the two-block history.
Figure Q3 — the four test points fall on a straight line when plotted on logarithmic axes, which is the visual proof that a power law applies. The blue squares are the two service strain ranges read off the fitted line.
Approach. Take logarithms to turn the power law into a straight line, fit it by least squares and report the coefficient of determination as the evidence for part (a); then invert the fitted law at each service strain range and close Miner’s sum at unity for part (b).
Part (a) — linearise the proposed law. Taking base-ten logarithms of $\Delta\epsilon = C N^{\alpha}$ gives $$\log_{10}\Delta\epsilon = \log_{10}C + \alpha\log_{10}N$$so if the law holds, a plot of $\log\Delta\epsilon$ against $\log N$ must be a straight line of slope $\alpha$ and intercept $\log_{10}C$. Figure Q3 shows that it is, which is the qualitative half of the answer.
Least-squares fit. With four points $(\log N_i,\ \log\Delta\epsilon_i)$ the standard normal equations give slope and intercept directly. Evaluating the sums ($\sum x = 12.9836$, $\sum y = -7.1174$, $\sum x^{2} = 43.3946$, $\sum xy = -22.4680$) yields $$\boxed{\alpha = -0.3673,\qquad C = 0.2403}$$so the material law is $\Delta\epsilon = 0.2403\,N^{-0.3673}$.
How well it fits. The coefficient of determination on the logarithmic pair is $R^{2} = 0.9932$, and back-substituting the four measured strain ranges reproduces lives of 132, 752, 1 597 and 9 226 cycles against the measured 120, 800, 1 900 and 8 000. The worst discrepancy is about 16 per cent in life — entirely normal scatter for low-cycle fatigue data, where a factor of two is often accepted — so the proposed form is confirmed.
Part (b) — life at each service strain range. Inverting the fitted law, $N = (\Delta\epsilon/C)^{1/\alpha}$. At the first block, $N_1 = (0.015/0.2403)^{-1/0.3673} = 1\,903$ cycles; at the second, $N_2 = (0.010/0.2403)^{-1/0.3673} = 5\,739$ cycles. Both are pure interpolation inside the tested range, which is why the fit can be trusted here.
Miner’s cumulative damage. The first block consumes a damage fraction $500/1\,903 = 0.2627$, leaving $0.7373$ of the life available at the lower strain range. Failure occurs when $$\sum \frac{n_i}{N_i} = \frac{500}{1\,903} + \frac{n_2}{5\,739} = 1$$so $n_2 = 0.7373 \times 5\,739 = 4\,232$ cycles at $\Delta\epsilon = 0.010$.
Total life. Adding the two blocks, $$\boxed{N_{\text{total}} = 500 + 4\,232 = 4\,732\ \text{cycles}}$$Note how expensive the first block is: 500 cycles at the higher strain range, only about 10 per cent of the total count, eat 26 per cent of the life.
Check: Miner’s rule is sequence-independent by construction, so the same 4 732 cycles would be predicted if the blocks were reversed. Real materials are not sequence-independent — a high-strain block first is generally more damaging than the rule predicts — so the answer should be read as the nominal design estimate the question asks for, not a guaranteed life.
