Sure! Let's break this down step by step: --- # **Problem Overview** You have a cantilever (overhung) diving board subjected to a concentrated load \( P = 780 \) N at its free end. You need to: 1. **Find the largest principal (bending) stress** in the board. 2. **Find the maximum deflection** of the board. The board is: - Length: \( L = 2.7 \) m (from wall to free end) - Width: \( b = 305 \) mm = .305 m - Thickness: \( h = 32 \) mm = .032 m - **Young's modulus:** \( E = 10.3 \) GPa = \( 10.3 \times 10^9 \) Pa --- ## **A. Largest Principal Stress \((\sigma_1)\)** For a cantilever beam with a point load at the free end, the maximum bending moment occurs at the fixed support: \[ M_{max} = P \cdot L \] The maximum bending stress is at the top/bottom fiber: \[ \sigma_{max} = \frac{M_{max} \cdot c}{I} \] Where: - \( c = \frac{h}{2} \) (distance from neutral axis to outer fiber) - \( I = \frac{1}{12} b h^3 \) (second moment of area for a rectangle) **Step-by-step:** 1. **Calculate \( M_{max} \):** \[ M_{max} = 780\,\text{N} \times 2.7\,\text{m} = 2106\,\text{N}\cdot\text{m} \] 2. **Calculate \( c \):** \[ c = \frac{h}{2} = \frac{.032}{2} = .016\,\text{m} \] 3. **Calculate \( I \):** \[ I = \frac{1}{12} \times .305 \times (.032)^3 = \frac{1}{12} \times .305 \times 3.2768 \times 10^{-5} \] \[ I \approx \frac{1}{12} \times .305 \times 3.2768 \times 10^{-5} \approx 8.33 \times 10^{-7} \,\text{m}^4 \] 4. **Calculate \( \sigma_{max} \):** \[ \sigma_{max} = \frac{2106 \times .016}{8.33 \times 10^{-7}} \] \[ \sigma_{max} \approx \frac{33.696}{8.33 \times 10^{-7}} \approx 4.05 \times 10^{7} \,\text{Pa} = 40.5 \,\text{MPa} \] --- ## **B. Maximum Deflection \((y_{max})\)** For a cantilever with a point load at the end: \[ \delta_{max} = \frac{P L^3}{3 E I} \] **Step-by-step:** 1. **Plug in the values:** \[ \delta_{max} = \frac{780 \times (2.7)^3}{3 \times 10.3 \times 10^9 \times 8.33 \times 10^{-7}} \] 2. **Calculate numerator:** \[ 780 \times (2.7)^3 = 780 \times 19.683 = 15,351 \] 3. **Calculate denominator:** \[ 3 \times 10.3 \times 10^9 \times 8.33 \times 10^{-7} = 3 \times 10.3 \times 8.33 \times 10^2 = 257.139 \times 10^2 = 25,713.9 \] 4. **Calculate \( \delta_{max} \):** \[ \delta_{max} = \frac{15,351}{25,713.9} \approx .597 \text{ m} \] --- ## **Summary of Formulas** - **Maximum Bending Stress:** \[ \sigma_{max} = \frac{P L \cdot (h/2)}{\frac{1}{12} b h^3} \] - **Maximum Deflection:** \[ \delta_{max} = \frac{P L^3}{3 E I} \] --- ## **Tips** - Always check units! - Use the outer fiber (top/bottom) for maximum bending stress. - For deflection, positive = up, negative = down (here, down due to gravity). --- Would you like to see step-by-step calculations with all numbers plugged in?

Sure! Here are the final answers for the problem: --- # **Final Answers** ## **A. Largest Principal Stress \((\sigma_1)\)** - **Value:** \( \sigma_{max} \approx 40.5 \, \text{MPa} \) ## **B. Maximum Deflection \((\delta_{max})\)** - **Value:** \( \delta_{max} \approx -0.597 \, \text{m} \) (negative indicates downward deflection) --- ## **Summary** - Largest Principal Stress: \( 40.5 \, \text{MPa} \) - Maximum Deflection: \( -0.597 \, \text{m} \) This indicates that the diving board bends downwards by approximately 0.597 meters when a 780 N person stands at the free end.