Let's break down the problem and solve it step by step. ### **Given Data** - Width, \( w = 100 \) mm - Length, \( L = 200 \) mm - Thickness, \( t = 18 \) mm - Crack length, \( a = 20 \) mm - Fracture toughness, \( K_{IC} = 24 \) MPa·m\(^{1/2}\) - Factor of safety, \( n = 2.2 \) - Material: 7075-T7351 aluminum alloy ### **Step 1: Formula for Maximum Load (Fracture Mechanics)** The critical stress intensity factor for a plate with an edge crack is: \[ K_{I} = Y \sigma \sqrt{\pi a} \] Where: - \( K_{I} \): Stress intensity factor (should not exceed \( K_{IC}/n \)) - \( Y \): Geometry factor (\( Y \approx 1.12 \) for an edge crack) - \( \sigma \): Applied stress - \( a \): Crack length ### **Step 2: Allowable Stress** \[ K_{IC,allowable} = \frac{K_{IC}}{n} = \frac{24 \text{ MPa} \cdot \sqrt{\text{m}}}{2.2} = 10.91 \text{ MPa} \cdot \sqrt{\text{m}} \] \[ K_{IC,allowable} = Y \sigma_{max} \sqrt{\pi a} \] \[ \sigma_{max} = \frac{K_{IC,allowable}}{Y \sqrt{\pi a}} \] ### **Step 3: Insert Values** - \( a = 20 \) mm \( = 0.02 \) m - \( Y = 1.12 \) - \( \pi \approx 3.1416 \) \[ \sigma_{max} = \frac{10.91}{1.12 \times \sqrt{3.1416 \times 0.02}} \] Calculate denominator: - \( 3.1416 \times 0.02 = 0.062832 \) - \( \sqrt{0.062832} = 0.2507 \) - \( 1.12 \times 0.2507 = 0.2808 \) \[ \sigma_{max} = \frac{10.91}{0.2808} = 38.85 \text{ MPa} \] ### **Step 4: Maximum Load** The maximum load \( P \): \[ P = \sigma_{max} \times (\text{cross-sectional area}) \] \[ \text{Area} = w \times t = 100 \text{ mm} \times 18 \text{ mm} = 1800 \text{ mm}^2 = 1.8 \times 10^{-3} \text{ m}^2 \] \[ P = 38.85 \text{ MPa} \times 1.8 \times 10^{-3} \text{ m}^2 \] \[ 1 \text{ MPa} = 10^6 \text{ N/m}^2 \] \[ P = 38.85 \times 10^6 \times 1.8 \times 10^{-3} \text{ N} \] \[ P = 38.85 \times 1.8 \times 10^3 \text{ N} \] \[ P = 69.93 \times 10^3 \text{ N} \] \[ P \approx \boxed{69,930 \text{ N}} \] --- ## **Final Answer** ### **The maximum load \( P \) that can be applied is:** \[ \boxed{69,930 \text{ N}} \]