22-Mec-B9 Advanced Engineering Structures · May 2014
Question 3 of 8: Coffin-Manson constants from strain cycling data and cumulative damage life
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, May 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 and fracture 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. For every thin-walled question a right-handed set is used: the span axis runs along the member, $Y$ is vertical (upward) in the cross-section, and $Z$ is horizontal. A vertical shear $S_y$ acting on a section whose product of inertia vanishes produces the open-section shear flow $q_s = -\dfrac{S_y}{I_{zz}}\displaystyle\int_0^s y\,t\,\mathrm{d}s$, measured positive in the direction of increasing $s$. Closed cells add the constant redundant flow $q_{s,0}$ fixed either by zero twist (shear-centre problems) or by moment equilibrium (a load of known line of action). All shear flows below are quoted with respect to a stated walking direction, and each question closes with the two statical checks — the resolved vertical force must return $S_y$ and the moment of the flows about a chosen origin must return the moment of the applied load.
Question 3: Coffin-Manson constants from strain cycling data and cumulative damage life (20 marks)
Given. Four low-cycle fatigue test points and a two-block service history.
Quantity
Symbol
Value
Test point 1
$(\Delta\epsilon,\,N)$
0.0380, 250
Test point 2
$(\Delta\epsilon,\,N)$
0.0221, 900
Test point 3
$(\Delta\epsilon,\,N)$
0.0150, 2200
Test point 4
$(\Delta\epsilon,\,N)$
0.0080, 12000
Service block 1
$\Delta\epsilon_1$, $n_1$
0.018 for 300 cycles
Service block 2
$\Delta\epsilon_2$
0.010 for the remainder
Find. (a) the Coffin-Manson constants $C$ and $\alpha$; (b) the total number of cycles to failure under the two-block history.
The four test points and the fitted power law on logarithmic axes. A power law plots as a straight line, whose slope is the exponent and whose intercept at N = 1 is the coefficient.
Approach. Taking logarithms turns the power law into a straight line, so a least-squares fit in log-log space returns the exponent as the slope and the coefficient as the intercept; the fitted law then converts each service strain range into an allowable life, and Miner’s rule sums the damage fractions to unity.
Part (a) — linearise the power law. Taking natural logarithms of $\Delta\epsilon = C N^{\alpha}$ gives $$\ln(\Delta\epsilon) = \ln C + \alpha \ln N$$ which is a straight line of slope $\alpha$ and intercept $\ln C$ in the variables $X = \ln N$, $Y = \ln(\Delta\epsilon)$. This is why low-cycle fatigue data are always plotted on log-log axes.
Form the least-squares sums. With the four points, $\overline{X} = 7.3532$ and $\overline{Y} = -4.0276$, and $$\begin{aligned}S_{XY} &= \sum (X_i-\overline{X})(Y_i-\overline{Y}) = -3.1981, \\ S_{XX} &= \sum (X_i-\overline{X})^{2} = 7.9357\end{aligned}$$ These two sums are all that a linear regression needs.
Slope and intercept give the two constants. $$\alpha = \frac{S_{XY}}{S_{XX}} = \frac{-3.1981}{7.9357} = \boxed{-0.4030}$$$$\ln C = \overline{Y} - \alpha\overline{X} = -1.0642 \;\Rightarrow\; C = \boxed{0.3450}$$ so the fitted law is $\Delta\epsilon = 0.3450\,N^{-0.4030}$. The coefficient of determination is $R^{2} = 0.99844$, and the fitted strain ranges (0.03728, 0.02225, 0.01552, 0.00783) reproduce the measured values to better than four per cent at every point.
Part (b) — allowable life at each service strain range. Inverting the fitted law, $N = (\Delta\epsilon/C)^{1/\alpha}$ with $1/\alpha = -2.4814$: $$\begin{aligned}N_1 &= \left(\frac{0.018}{0.3450}\right)^{-2.4814} = 1522\ \text{cycles}, \\ N_2 &= \left(\frac{0.010}{0.3450}\right)^{-2.4814} = 6544\ \text{cycles}\end{aligned}$$ Dropping the strain range from 0.018 to 0.010 multiplies the allowable life by 4.30, which is the practical meaning of an exponent near $-0.4$.
Apply Miner’s linear damage summation. Failure is predicted when the accumulated damage reaches unity: $$\frac{n_1}{N_1} + \frac{n_2}{N_2} = 1 \;\Rightarrow\; \frac{300}{1522} + \frac{n_2}{6544} = 1$$$$0.1971 + \frac{n_2}{6544} = 1 \;\Rightarrow\; n_2 = 6544(0.8029) = 5255\ \text{cycles}$$ The first block consumes just under a fifth of the life even though it lasts only 300 cycles, because the strain range is nearly twice as large.
Total life. Adding the two blocks, $$N_{\text{total}} = n_1 + n_2 = 300 + 5255 = \boxed{5\,555\ \text{cycles}}$$ of which 5255 are at the lower strain range. For comparison, the component would survive 6544 cycles if the 0.018 block never occurred, so the short severe block costs about 1290 cycles of remaining life.
