Let's break down and solve the problem step by step: --- ## **Step 1: Restate the Problem** **Given:** - Span of wood beam \( L = 16' = 16 \times 12 = 192 \) in - Beam material: Douglas Fir - Allowable bending stress, \( F_{b,allow} = 120 \) psi - Find the most economical (smallest safe) beam cross-section. **Assumption:** No load (w) is given. For residential bedroom floors, a common live load is 40 psf and dead load is about 10 psf, totaling **50 psf**. --- ## **Step 2: Calculate Maximum Moment** For a **simply supported beam** with **uniformly distributed load**: \[ M_{max} = \frac{w L^2}{8} \] Where: - \( w \) = load per unit length (lb/in) - \( L \) = span (in) ### **Calculate total load per foot:** If the beam is spaced at 16" o.c. (common for floor joists), tributary width = 16" = 1.33 ft. \[ w = (50 \text{ psf}) \times (1.33 \text{ ft}) = 66.7 \text{ plf} \] Convert to pounds per inch: \[ w = \frac{66.7}{12} = 5.56 \text{ lb/in} \] --- ## **Step 3: Maximum Bending Moment Calculation** \[ M_{max} = \frac{w L^2}{8} \] \[ M_{max} = \frac{5.56 \times (192)^2}{8} \] \[ M_{max} = \frac{5.56 \times 36,864}{8} \] \[ M_{max} = \frac{204,926.4}{8} \] \[ M_{max} = 25,616 \text{ in-lb} \] --- ## **Step 4: Section Modulus Needed** The **allowable bending stress** relationship: \[ F_{b,allow} = \frac{M_{max}}{S} \implies S = \frac{M_{max}}{F_{b,allow}} \] \[ S = \frac{25,616}{120} = 21.35 \text{ in}^3 \] --- ## **Step 5: Select Economical Section** For a **rectangular section** (width \( b \), depth \( h \)): \[ S = \frac{b h^2}{6} \] We want the smallest size, so we usually pick a standard width (say, 2" nominal = 1.5" actual for dimensional lumber) and solve for \( h \): \[ S = \frac{1.5 \times h^2}{6} \] Set \( S = 21.35 \): \[ 21.35 = \frac{1.5 \times h^2}{6} \] \[ 21.35 \times 6 = 1.5 h^2 \] \[ 128.1 = 1.5 h^2 \] \[ h^2 = \frac{128.1}{1.5} = 85.4 \] \[ h = \sqrt{85.4} = 9.24 \text{ in} \] --- ## **Step 6: Choose Standard Lumber Size** The next standard size up is a **2x10** (actual: 1.5" x 9.25"). - Section modulus for 2x10: \[ S = \frac{1.5 \times (9.25)^2}{6} = \frac{1.5 \times 85.56}{6} = \frac{128.34}{6} = 21.39 \text{ in}^3 \] This meets the requirement. --- ## **Final Answer** ### **The most economical (smallest) cross-section is:** \[ \boxed{\textbf{2" x 10" (actual: 1.5" x 9.25") Douglas Fir beam}} \] This size safely supports the assumed floor load for a 16' span using Douglas Fir with an allowable stress of 120 psi. If load spacing is different, adjust tributary width accordingly.