Interactive Lecture
S-N Curves, Stress Concentration, and Fatigue Life Prediction
by Dr Arnaldo Delli Carri
Fatigue is a failure mechanism that occurs in materials subjected to cyclic loading, even when the stresses are well below the material's ultimate tensile strength. By common estimates it drives roughly 80–90% of all in-service mechanical failures.
Key aspects that differentiate fatigue from static failure:
The tools below work through fatigue in four stages: characterising a stress cycle, reading life off an S-N curve, correcting for mean stress, and summing damage under variable-amplitude loading.
In fatigue analysis, we characterise cyclic loading using several key parameters:
\[\sigma_{max} = \text{Maximum stress in cycle}\]
\[\sigma_{min} = \text{Minimum stress in cycle}\]
\[\sigma_a = \frac{\sigma_{max} - \sigma_{min}}{2} = \text{Stress amplitude}\]
\[\sigma_m = \frac{\sigma_{max} + \sigma_{min}}{2} = \text{Mean stress}\]
\[R = \frac{\sigma_{min}}{\sigma_{max}} = \text{Stress ratio}\]
August Wöhler conducted the first systematic investigation of fatigue failure in the 1850s whilst studying railway axle failures. His experiments involved applying cyclic loads to test specimens until failure and plotting the results.
The relationship between stress amplitude and fatigue life:
\[\sigma_a = \sigma'_f (2N_f)^b\]
Where:
On log-log scale: Appears as a straight line with slope = b
On semi-log scale: Appears as a curved (power law) line
Log-log: Both axes logarithmic - shows power law as straight line
Semi-log: Only N-axis logarithmic - shows power law as curve
Understanding Scale Types:
In practical applications, many components experience cyclic loading with a non-zero mean stress. Tensile mean stress generally reduces fatigue life, whilst compressive mean stress tends to improve it.
Goodman: \(\frac{\sigma_a}{S_e} + \frac{\sigma_m}{S_{ut}} = 1\)
Gerber: \(\frac{\sigma_a}{S_e} + \left(\frac{\sigma_m}{S_{ut}}\right)^2 = 1\)
Soderberg: \(\frac{\sigma_a}{S_e} + \frac{\sigma_m}{S_y} = 1\)
In real applications, components often experience variable amplitude loading. The Palmgren-Miner rule provides a way to assess cumulative damage.
\[D = \sum_{i=1}^{k} \frac{n_i}{N_i}\]
Where:
Cycles to failure \(N_i\) for each block come from Basquin's equation for Steel 1045 (\(\sigma'_f = 1120\) MPa, \(b = -0.092\)), matching the Steel 1045 curve in the S-N section above.
Limitations of the Palmgren-Miner Rule:
Fatigue analysis applies wherever components experience cyclic loading:
| Concept | Equation | Description |
|---|---|---|
| Stress Amplitude | \(\sigma_a = \frac{\sigma_{max} - \sigma_{min}}{2}\) | Half the stress range in cyclic loading |
| Mean Stress | \(\sigma_m = \frac{\sigma_{max} + \sigma_{min}}{2}\) | Average stress during cycle |
| Basquin's Equation | \(\sigma_a = \sigma'_f (2N_f)^b\) | S-N curve relationship for fatigue life |
| Goodman Criterion | \(\frac{\sigma_a}{S_e} + \frac{\sigma_m}{S_{ut}} = \frac{1}{SF}\) | Mean stress correction for fatigue design |
| Palmgren-Miner Rule | \(D = \sum_{i=1}^{k} \frac{n_i}{N_i}\) | Cumulative damage assessment (failure at D ≥ 1) |
A steel component (σ'f = 1000 MPa, b = -0.09) experiences fully reversed loading with a stress amplitude of 300 MPa. Calculate the expected fatigue life.
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Solution:
Using Basquin's equation: \(\sigma_a = \sigma'_f (2N_f)^b\)
\(300 = 1000 \times (2N_f)^{-0.09}\)
\(0.3 = (2N_f)^{-0.09}\)
Taking logarithms: \(\log(0.3) = -0.09 \times \log(2N_f)\)
\(\log(2N_f) = \dfrac{\log(0.3)}{-0.09} = \dfrac{-0.5229}{-0.09} = 5.810\)
\(2N_f = 10^{5.810} = 6.46 \times 10^5\)
Answer: \(N_f \approx 3.23 \times 10^5\) cycles
A component experiences σₐ = 200 MPa and σₘ = 150 MPa. If Se = 400 MPa and Sut = 800 MPa, calculate the safety factor using the Goodman criterion.
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Solution:
Goodman equation: \(\dfrac{\sigma_a}{S_e} + \dfrac{\sigma_m}{S_{ut}} = \dfrac{1}{SF}\)
\(\dfrac{200}{400} + \dfrac{150}{800} = \dfrac{1}{SF}\)
\(0.5 + 0.1875 = \dfrac{1}{SF}\)
\(0.6875 = \dfrac{1}{SF}\)
Answer: \(SF = 1.45\)
A component experiences three stress blocks: Block 1: 400 MPa for 1000 cycles (Nf = 10,000), Block 2: 300 MPa for 20,000 cycles (Nf = 100,000), Block 3: 200 MPa for 500,000 cycles (Nf = 10^6). Calculate the cumulative damage.
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Solution:
Using the Palmgren-Miner rule: \(D = \sum \dfrac{n_i}{N_i}\)
\(D_1 = \dfrac{1000}{10{,}000} = 0.1\)
\(D_2 = \dfrac{20{,}000}{100{,}000} = 0.2\)
\(D_3 = \dfrac{500{,}000}{1{,}000{,}000} = 0.5\)
Total \(D = 0.1 + 0.2 + 0.5 = 0.8\)
Component is safe (\(D < 1\)) with 20% remaining life.
Answer: \(D = 0.8\) (safe; 20% life remaining)