Let's break down the problem step by step: --- ## **Given Data** - **Semi-circular flaw radius (a₀)**: .5 mm = .0005 m - **Fatigue crack growth law:** \[ \frac{da}{dN} = 2 \times 10^{-10} (\Delta K)^3 \] where \( da/dN \) is in m/cycle, \( \Delta K \) in MPa\(\sqrt{\text{m}}\) --- ### **Step 1: Define Fatigue Crack Growth Law** \[ \frac{da}{dN} = C (\Delta K)^n \] Here, \( C = 2 \times 10^{-10} \), \( n = 3 \) --- ### **Step 2: Stress Intensity Factor for a Surface Crack** For a semi-circular surface crack in a plate, \[ \Delta K = Y \Delta \sigma \sqrt{\pi a} \] Assume \( Y \approx 1 \) (common for semi-circular surface cracks). --- ### **Step 3: Set Up the Integral** Rewriting the crack growth law: \[ \frac{da}{dN} = 2 \times 10^{-10} \left( \Delta \sigma \sqrt{\pi a} \right)^3 \] \[ \frac{da}{dN} = 2 \times 10^{-10} (\Delta \sigma)^3 (\pi a)^{3/2} \] Rearrange for \( dN \): \[ dN = \frac{da}{2 \times 10^{-10} (\Delta \sigma)^3 (\pi)^{3/2} a^{3/2}} \] --- ### **Step 4: Integrate from Initial Crack Size to Critical Size** Let \( a_ = .0005 \) m (initial), \( a_c \) = critical flaw size before burst. \[ N = \int_{a_}^{a_c} \frac{da}{2 \times 10^{-10} (\Delta \sigma)^3 (\pi)^{3/2} a^{3/2}} \] \[ N = \frac{1}{2 \times 10^{-10} (\Delta \sigma)^3 (\pi)^{3/2}} \int_{a_}^{a_c} a^{-3/2} da \] \[ \int a^{-3/2} da = -2 a^{-1/2} \] \[ N = \frac{1}{2 \times 10^{-10} (\Delta \sigma)^3 (\pi)^{3/2}} \cdot \left[-2 a^{-1/2} \right]_{a_}^{a_c} \] \[ N = \frac{-2}{2 \times 10^{-10} (\Delta \sigma)^3 (\pi)^{3/2}} \left[a_c^{-1/2} - a_^{-1/2}\right] \] \[ N = \frac{1}{10^{-10} (\Delta \sigma)^3 (\pi)^{3/2}} \left[a_^{-1/2} - a_c^{-1/2}\right] \] --- ### **Step 5: Find the Critical Crack Size (\( a_c \))** For a thin-walled cylinder, the hoop stress (assuming internal pressure at burst): \[ \sigma = \frac{P r}{t} \] Bursting occurs when \( K_c = Y \sigma \sqrt{\pi a_c} \) Let \( K_c \) be the fracture toughness (not given, but you could use a typical value for aluminum, e.g., 30 MPa\(\sqrt{\text{m}}\)). If you have the burst pressure, you could use that to get \( \sigma \). --- ### **Step 6: Plug in Values** Let's **assume**: - \( \Delta \sigma = 100 \) MPa (typical fatigue range for such problems; you can adjust if you have more info) - \( K_c = 30 \) MPa\(\sqrt{\text{m}}\) - \( Y = 1 \) - \( a_ = .0005 \) m **Find \( a_c \):** \[ K_c = \Delta \sigma \sqrt{\pi a_c} \] \[ a_c = \frac{K_c^2}{\pi (\Delta \sigma)^2} \] \[ a_c = \frac{30^2}{\pi \times 100^2} = \frac{900}{31415.93} \approx .0286 \text{ m} \] --- ### **Step 7: Calculate Number of Cycles** \[ N = \frac{1}{10^{-10} \times (100)^3 \times (\pi)^{3/2}} \left[.0005^{-1/2} - .0286^{-1/2}\right] \] Calculate each term: - \( (100)^3 = 1 \times 10^6 \) - \( (\pi)^{3/2} \approx 5.568 \) - \( .0005^{-1/2} = \frac{1}{\sqrt{.0005}} \approx 44.72 \) - \( .0286^{-1/2} = \frac{1}{\sqrt{.0286}} \approx 5.91 \) \[ N = \frac{1}{10^{-10} \times 1 \times 10^6 \times 5.568} (44.72 - 5.91) \] \[ N = \frac{1}{5.568 \times 10^{-4}} \times 38.81 \] \[ \frac{1}{5.568 \times 10^{-4}} \approx 1796.3 \] \[ N = 1796.3 \times 38.81 \approx 69693 \text{ cycles} \] --- ## **Final Answer** **The cylinder could withstand approximately \( \boxed{69,\!700} \) pressure cycles before bursting.** --- **Note:** - If you know the real value of \( \Delta \sigma \) or \( K_c \), substitute those instead. - For more accurate results, use the exact stress range and fracture toughness for your specific aluminum alloy and geometry.

# Step-by-Step Solution for Fatigue Crack Growth in a Cylinder ## Given Data - **Semi-circular flaw radius (\(a_0\))**: \(0.5 \, \text{mm} = 0.0005 \, \text{m}\) - **Fatigue crack growth law**: \[ \frac{da}{dN} = 2 \times 10^{-10} (\Delta K)^3 \] where \( \frac{da}{dN} \) is in m/cycle and \( \Delta K \) in MPa\(\sqrt{\text{m}}\) - **Cylinder diameter**: \(2.6 \, \text{m}\) ## Step 1: Define Fatigue Crack Growth Law The general form of the crack growth law is: \[ \frac{da}{dN} = C (\Delta K)^n \] Where: - \( C = 2 \times 10^{-10} \) - \( n = 3 \) ## Step 2: Stress Intensity Factor for a Surface Crack For a semi-circular surface crack: \[ \Delta K = Y \Delta \sigma \sqrt{\pi a} \] Assuming \( Y \approx 1 \). ## Step 3: Set Up the Integral Rewriting the crack growth law: \[ \frac{da}{dN} = 2 \times 10^{-10} (\Delta \sigma \sqrt{\pi a})^3 \] Rearranging gives: \[ dN = \frac{da}{2 \times 10^{-10} (\Delta \sigma)^3 (\pi)^{3/2} a^{3/2}} \] ## Step 4: Integrate from Initial Crack Size to Critical Size Let \( a_0 = 0.0005 \, \text{m} \) (initial), \( a_c \) = critical flaw size. \[ N = \int_{a_0}^{a_c} \frac{da}{2 \times 10^{-10} (\Delta \sigma)^3 (\pi)^{3/2} a^{3/2}} \] Calculating the integral: \[ N = \frac{1}{2 \times 10^{-10} (\Delta \sigma)^3 (\pi)^{3/2}} \left[-2 a^{-1/2} \right]_{a_0}^{a_c} \] \[ N = \frac{1}{10^{-10} (\Delta \sigma)^3 (\pi)^{3/2}} \left[a_0^{-1/2} - a_c^{-1/2}\right] \] ## Step 5: Find the Critical Crack Size (\(a_c\)) Using the relationship for a thin-walled cylinder: \[ K_c = Y \sigma \sqrt{\pi a_c} \] Assuming typical values: - \( K_c = 30 \, \text{MPa}\sqrt{\text{m}} \) - \( Y = 1 \) Calculating \( a_c \): \[ K_c = \Delta \sigma \sqrt{\pi a_c} \Rightarrow a_c = \frac{K_c^2}{\pi (\Delta \sigma)^2} \] ## Step 6: Plug in Values Assuming: - \( \Delta \sigma = 100 \, \text{MPa} \) Calculating \( a_c \): \[ a_c = \frac{30^2}{\pi \times 100^2} = \frac{900}{31415.93} \approx 0.0286 \, \text{m} \] ## Step 7: Calculate Number of Cycles Substituting into the equation for \( N \): \[ N = \frac{1}{10^{-10} \times (100)^3 \times (\pi)^{3/2}} \left[0.0005^{-1/2} - 0.0286^{-1/2}\right] \] Calculating each term: - \( (100)^3 = 1 \times 10^6 \) - \( (\pi)^{3/2} \approx 5.568 \) - \( 0.0005^{-1/2} \approx 44.72 \) - \( 0.0286^{-1/2} \approx 5.91 \) Calculating \( N \): \[ N = \frac{1}{5.568 \times 10^{-4}} (44.72 - 5.91) \approx \frac{1}{5.568 \times 10^{-4}} \cdot 38.81 \] \[ N \approx 1796.3 \times 38.81 \approx 69693 \text{ cycles} \] ## Final Answer The cylinder could withstand approximately \( \boxed{69,700} \) pressure cycles before bursting.