Question

A specimen containing maximum initial surface crack of size 1.5 mm is subjected to cyclic loading with \( \sigma_{max} = 300 \) MPa and \( \sigma_{min} = 0 \) MPa. Assuming specimen geometric factor of 1, and referring to the given figure, the crack growth rate in \(\mu\)m cycle\(^{-1}\) is ................... (round off to nearest integer). 

Hint:

Pay close attention to units when calculating \(\Delta K\). The standard units are MPa m\(^{0.5}\). Ensure your stress is in MPa and your crack length is in meters. Also, be careful reading logarithmic scales on graphs; the intervals are not linear.

Answer 1

Step 1: Understanding the Concept:
This problem requires using the principles of fatigue crack growth. We need to calculate the stress intensity factor range (\(\Delta K\)) for the given loading conditions and initial crack size. Then, we use the provided experimental plot of crack growth rate (da/dN) versus \(\Delta K\) to find the corresponding rate.
Step 2: Key Formula or Approach:
1. Stress Intensity Factor Range (\(\Delta K\)): For a crack of length \(a\) under a cyclic stress range \(\Delta \sigma\), it is given by:
\[ \Delta K = Y \Delta \sigma \sqrt{\pi a} \] 2. Stress Range (\(\Delta \sigma\)): \( \Delta \sigma = \sigma_{max} - \sigma_{min} \).
3. Stress Ratio (R): \( R = \frac{\sigma_{min}}{\sigma_{max}} \). This is needed to select the correct curve on the plot.
Step 3: Detailed Calculation:
1. Calculate Stress Ratio (R): \[ R = \frac{0 \text{ MPa}}{300 \text{ MPa}} = 0 \] We must use the curve labeled "R = 0". 2. Calculate Stress Range (\(\Delta \sigma\)): \[ \Delta \sigma = 300 \text{ MPa} - 0 \text{ MPa} = 300 \text{ MPa} \] 3. Calculate \(\Delta K\): First, convert crack size to meters: \( a = 1.5 \text{ mm} = 0.0015 \text{ m} \). The geometric factor \(Y = 1\). \[ \Delta K = 1 \times (300 \text{ MPa}) \sqrt{\pi \times (0.0015 \text{ m})} \] \[ \Delta K = 300 \sqrt{0.004712} \] \[ \Delta K \approx 300 \times 0.06865 \approx 20.595 \text{ MPa m}^{0.5} \] 4. Read the Crack Growth Rate from the Figure: - Locate \(\Delta K \approx 20.6\) on the horizontal axis. This is just to the right of the "20" mark.
- Move vertically up to intersect the "R = 0" curve.
- From the intersection point, move horizontally to the left to read the value on the vertical (da/dN) axis.
- The intersection point is slightly above the horizontal line for da/dN = 10 \(\mu\)m/cycle.
- By visual interpolation between \(\Delta K = 20\) (where da/dN \(\approx\) 10) and \(\Delta K = 30\) (where da/dN \(\approx\) 85), the value at 20.6 will be slightly higher than 10. A reasonable estimate is around 12 \(\mu\)m/cycle.
5. Round to the nearest integer: The estimated value is 12 \(\mu\)m/cycle.
Step 4: Final Answer:
The crack growth rate is 12 \(\mu\)m cycle\(^{-1}\).