# Beam Design for Deflection and Bending ## 1. **Given Data** - **Span, \( L \):** 30 ft - **Live load, \( w_1 \):** 160 lb/ft - **Dead load, \( w_2 \):** 185 lb/ft - **Assumed beam size:** 10 in × 16 in (actual: 9.25 in × 15.25 in typical) - **Material:** Douglas Fir-Larch No.1 & Better - **Density:** 30 lb/ft³ ## 2. **Total Uniform Load** Sum the live and dead loads: \[ w = w_1 + w_2 = 160\, \text{lb/ft} + 185\, \text{lb/ft} = 345\, \text{lb/ft} \] ## 3. **Maximum Moment Calculation** For a simply supported beam with uniform load: \[ M_{max} = \frac{wL^2}{8} \] Plug in the values: \[ M_{max} = \frac{345 \times (30)^2}{8} = \frac{345 \times 900}{8} = \frac{310,500}{8} = 38,812.5\, \text{lb-ft} \] Convert to in-lb: \[ M_{max} = 38,812.5 \times 12 = 465,750\, \text{in-lb} \] ## 4. **Section Properties** Assume actual dimensions: **9.25 in (b) × 15.25 in (d)** - **Section modulus, \( S \):** \[ S = \frac{b d^2}{6} = \frac{9.25 \times (15.25)^2}{6} \approx \frac{9.25 \times 232.56}{6} \approx \frac{2,150.18}{6} \approx 358.36\, \text{in}^3 \] - **Moment of inertia, \( I \):** \[ I = \frac{b d^3}{12} = \frac{9.25 \times (15.25)^3}{12} \approx \frac{9.25 \times 3548.64}{12} \approx \frac{32,817.92}{12} \approx 2,734.83\, \text{in}^4 \] ## 5. **Bending Stress Check** - **Allowable bending stress, \( F_b \):** (from NDS supplement for DF-Larch No.1&Btr) - \( F_b = 1,500\, \text{psi} \) (typical, verify for code year) \[ f_b = \frac{M_{max}}{S} = \frac{465,750}{358.36} \approx 1,299\, \text{psi} \] - **Check:** \( 1,299 < 1,500 \) psi → **OK** ## 6. **Deflection Check** - **Allowable deflection:** \( L/360 \) for live load \( \delta_{allow} = \frac{30 \times 12}{360} = 1.\, \text{in} \) - **Deflection due to total load:** For a simply supported beam: \[ \delta_{max} = \frac{5wL^4}{384EI} \] - Modulus of elasticity, \( E \), for DF-Larch No.1&Btr: \( 1,600,000\, \text{psi} \) - Convert w to lb/in: \[ w = 345\, \text{lb/ft} \div 12 = 28.75\, \text{lb/in} \] - \( L = 30\, \text{ft} = 360\, \text{in} \) \[ \delta_{max} = \frac{5 \times 28.75 \times (360)^4}{384 \times 1,600,000 \times 2,734.83} \] \[ (360)^4 = 16,796,160,000 \] \[ Numerator: 5 \times 28.75 \times 16,796,160,000 = 2,414,515,200,000 \] \[ Denominator: 384 \times 1,600,000 \times 2,734.83 = 1,680,000,000 \] \[ \delta_{max} = \frac{2,414,515,200,000}{1,680,000,000} \approx 1.438\, \text{in} \] - **Check:** \( 1.438 > 1.\, \text{in} \) → **Deflection NOT OK** ## 7. **Summary & Adjustment** - **Bending:** OK - **Deflection:** **NOT OK** (exceeds allowable) ### **Possible Solutions** - Use a deeper beam (increase \( d \)) - Use higher modulus material - Add supports (reduce span) - Use double or triple beams --- ## **Final Answers** - **Maximum moment:** \( 38,813\, \text{lb-ft} \) - **Required section modulus:** \( \geq 310.5\, \text{in}^3 \) - **Actual section modulus:** \( 358.4\, \text{in}^3 \) (**OK**) - **Maximum bending stress:** \( 1,299\, \text{psi} \) (**OK**) - **Maximum deflection:** \( 1.44\, \text{in} \) (**NOT OK**, should be \(\leq 1.\, \text{in}\)) --- ### **Conclusion** **A 10"×16" (actual 9.25"×15.25") Douglas Fir-Larch No.1 beam is OK for bending but fails deflection. Use a deeper beam or reduce span/load for deflection compliance.**