Let's break down the problem from the image: --- **Question:** Calculate the rate of crack growth, when a mild steel plate is subjected to a cyclic stress with stress in tension as 500 MPa and stress in compression as 60 MPa. Consider the geometric factor as √πa and the plain strain fracture toughness of mild steel as 50 MPa√m. --- Let's extract and solve step by step: ### 1. **Given Values** - Maximum stress, \(\sigma_{max}\) = 500 MPa - Minimum stress, \(\sigma_{min}\) = 60 MPa - Fracture toughness, \(K_{IC}\) = 50 MPa√m - Geometric factor is \(\sqrt{\pi a}\) - We are to calculate the **crack length, a** at which fast fracture occurs (since options are in meters). ### 2. **Critical Stress Intensity Factor (\(K_{IC}\))** The stress intensity factor for a crack of length \(a\) is: \[ K = Y \sigma \sqrt{\pi a} \] Where \(Y\) is the geometric factor (here, \(Y = 1\)), and \(\sigma\) is the applied stress. At the point of fracture: \[ K_{IC} = \sigma_{max} \sqrt{\pi a} \] ### 3. **Solve for \(a\)** \[ K_{IC} = \sigma_{max} \sqrt{\pi a} \implies \frac{K_{IC}}{\sigma_{max}} = \sqrt{\pi a} \implies \left(\frac{K_{IC}}{\sigma_{max}}\right)^2 = \pi a \implies a = \frac{1}{\pi}\left(\frac{K_{IC}}{\sigma_{max}}\right)^2 \] Plug in the values: - \(K_{IC} = 50 \text{ MPa} \sqrt{\text{m}}\) - \(\sigma_{max} = 500 \text{ MPa}\) \[ a = \frac{1}{\pi} \left(\frac{50}{500}\right)^2 = \frac{1}{\pi} \left(.1\right)^2 = \frac{1}{\pi} (.01) = \frac{.01}{\pi} \approx \frac{.01}{3.1416} \approx .00318 \text{ m} \] ### 4. **Choose Correct Option** From the provided options: - .00102 m - .0026 m - .0081 m - .0042 m The closest value is **.0042 m**. --- ## **Final Answer** ### **The correct option is:** **.0042 m** --- Let me know if you need a more detailed explanation or further calculations!