Question: A 12-ft-long overhanging timber beam AC with an 8-ft span AB is to be designed to support the distributed and concentrated loads shown. Knowing that timber of 4-in. nominal width (3.5-in. actual width) and a 17.5-ksi allowable stress is to be used, what will be the (a)Maximum absolute value of the bending moment (b) Minimum allowable section modulus (c) Minimum required depth (h) of the beam? 400 lb/ft 4.5 kips 3.5 in. h 8 ft 4 ft A: left support at the bottom-left end of the beam; B: roller support near the center with 8 ft between A and B; C: free right end 4 ft beyond B with a downward point load of 4.5 kips at C; a uniformly distributed load of 400 lb/ft acts downward over span AB from A to B; to the right of the beam is a rectangular cross-section sketch labeled 3.5 in. width and depth h.

1. **Problem Statement:** We have a 12-ft beam AC with an 8-ft span AB and a 4-ft overhang BC. Loads: - Uniformly distributed load (UDL) $w = 400$ lb/ft on AB (8 ft) - Concentrated load $P = 4.5$ kips = 4500 lb at C (free end) Material: - Width $b = 3.5$ in - Allowable stress $F_b = 17.5$ ksi = 17500 psi Find: (a) Maximum absolute bending moment $M_{max}$ (b) Minimum allowable section modulus $S_{min}$ (c) Minimum required depth $h$ of the beam cross-section --- 2. **Calculate reactions at supports A and B:** Let $R_A$ and $R_B$ be the vertical reactions at A and B. Sum of vertical forces: $$R_A + R_B = w \times 8 + P = 400 \times 8 + 4500 = 3200 + 4500 = 7700 \text{ lb}$$ Taking moments about A (counterclockwise positive): $$R_B \times 8 = w \times 8 \times \frac{8}{2} + P \times 12$$ $$R_B \times 8 = 400 \times 8 \times 4 + 4500 \times 12$$ $$R_B \times 8 = 12800 + 54000 = 66800$$ $$R_B = \frac{66800}{8} = 8350 \text{ lb}$$ Then, $$R_A = 7700 - 8350 = -650 \text{ lb}$$ Negative $R_A$ means the support at A is pulling down (uplift). --- 3. **Calculate bending moments at key points:** - At A ($x=0$): $M_A = 0$ - At B ($x=8$ ft): Moment due to $R_A$ and UDL: $$M_B = R_A \times 8 - w \times 8 \times \frac{8}{2} = (-650) \times 8 - 400 \times 8 \times 4 = -5200 - 12800 = -18000 \text{ lb-ft}$$ - At C ($x=12$ ft): Moment at C is zero (free end). - At point between B and C (overhang), maximum moment usually occurs at B or between B and C. Calculate moment at B from right side: Moment at B due to $P$ at C: $$M_B = -P \times 4 = -4500 \times 4 = -18000 \text{ lb-ft}$$ This matches the previous calculation. --- 4. **Maximum absolute bending moment:** Maximum absolute value is $|M_B| = 18000$ lb-ft. Convert to lb-in: $$M_{max} = 18000 \times 12 = 216000 \text{ lb-in}$$ --- 5. **Minimum allowable section modulus $S_{min}$:** Using bending stress formula: $$F_b = \frac{M_{max}}{S_{min}} \Rightarrow S_{min} = \frac{M_{max}}{F_b}$$ Substitute values: $$S_{min} = \frac{216000}{17500} = 12.34 \text{ in}^3$$ --- 6. **Minimum required depth $h$ of the beam:** Section modulus for rectangular section: $$S = \frac{b h^2}{6}$$ Solve for $h$: $$h = \sqrt{\frac{6 S_{min}}{b}} = \sqrt{\frac{6 \times 12.34}{3.5}} = \sqrt{21.14} = 4.60 \text{ in}$$ --- **Final answers:** (a) Maximum absolute bending moment: $$\boxed{18000 \text{ lb-ft}}$$ (b) Minimum allowable section modulus: $$\boxed{12.34 \text{ in}^3}$$ (c) Minimum required depth of beam: $$\boxed{4.60 \text{ in}}$$