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 \( Y = 1 \) and the plain strain fracture toughness of mild steel as 50 MPa\(\sqrt{m}\). **Given:** - Maximum stress in tension, \( \sigma_{max} = 500 \) MPa - Minimum stress in compression, \( \sigma_{min} = 60 \) MPa - Geometric factor, \( Y = 1 \) - Fracture toughness, \( K_{IC} = 50 \) MPa\(\sqrt{m}\) **We need to find the critical crack length for failure.** ### Step 1: Calculate the stress range (\( \Delta\sigma \)) \[ \Delta\sigma = \sigma_{max} - \sigma_{min} = 500 - (-60) = 560 \text{ MPa} \] *Note: If compression is negative, the difference is 560 MPa. If both are positive, then 500 - 60 = 440 MPa. But typically, compression is negative for stress range calculation.* ### Step 2: Formula for critical crack length (\( a_c \)) The fracture toughness is given by: \[ K_{IC} = Y \cdot \sigma \cdot \sqrt{\pi a_c} \] Given \( K_{IC} = 50 \) MPa\(\sqrt{m}\), \( Y = 1 \), and \(\sigma = 500 \) MPa, Rearrange for \( a_c \): \[ a_c = \frac{(K_{IC})^2}{\pi Y^2 \sigma^2} \] ### Step 3: Substitute the values \[ a_c = \frac{50^2}{\pi \times 1^2 \times 500^2} \] \[ a_c = \frac{2500}{\pi \times 250000} \] \[ a_c = \frac{2500}{785398} \] \[ a_c \approx 0.0032 \text{ m} = 3.2 \text{ mm} \] ### Step 4: Find the closest value from the options - 0.0025 m - 0.0081 m - 0.0002 m - 0.0042 m The closest value is **0.0042 m**. --- **Final Answer:** The rate of crack growth (critical crack length for failure) is **0.0042 m**. Let me know if you need detailed steps on calculating the rate of crack growth (da/dN) using Paris' Law!