Let's solve the problem step by step. --- ## **Step 1: Understand the Problem** We have a composite beam (wood + steel strips) subjected to bending. The maximum bending moment is limited by the allowable stresses in wood (\(\sigma_{wood}\)) and steel (\(\sigma_{steel}\)). **Given:** - \(\sigma_{wood} = 1.6\) ksi - \(\sigma_{steel} = 35.7\) ksi - \(L_1 = 10.7\) in (height of wood) - \(L_2 = 1.6\) in (height of steel strip) - \(L_3 = 4.1\) in (width of beam) --- ## **Step 2: Find the Modular Ratio** The modular ratio (\(n\)) compares the modulus of elasticity of steel (\(E_s\)) to that of wood (\(E_w\)). Typically: - \(E_{steel} \approx 29,000\) ksi - \(E_{wood} \approx 1,200\) ksi (for pine) \[ n = \frac{E_{steel}}{E_{wood}} = \frac{29,000}{1,200} \approx 24.17 \] --- ## **Step 3: Transform the Section** Convert the steel strip areas to an equivalent wood area using the modular ratio. - **Width of the beam:** \(b = L_3 = 4.1\) in - **Area of one steel strip:** \(A_{steel} = b \times L_2 = 4.1 \times 1.6 = 6.56 \, \text{in}^2\) Equivalent area in wood: \[ A_{steel,eq} = n \times A_{steel} = 24.17 \times 6.56 \approx 158.6 \, \text{in}^2 \] --- ## **Step 4: Locate the Neutral Axis** Let’s place the bottom of the beam at \(y = \). - **Wood:** From \(y = L_2\) to \(y = L_2 + L_1\) - **Top steel:** From \(y = L_2 + L_1\) to \(y = 2L_2 + L_1\) - **Bottom steel:** From \(y = \) to \(y = L_2\) Calculate centroid (\(\bar{y}\)) of the equivalent section: \[ \text{Total Area} = A_{wood} + 2A_{steel,eq} \] \[ A_{wood} = b \times L_1 = 4.1 \times 10.7 = 43.87 \, \text{in}^2 \] \[ \text{Total Area} = 43.87 + 2 \times 158.6 = 361.07 \, \text{in}^2 \] Calculate the first moment of area about bottom: \[ Q = A_{wood} \cdot y_{wood} + 2 \cdot A_{steel,eq} \cdot y_{steel} \] Where: - \(y_{wood}\) = distance from bottom to centroid of wood = \(L_2 + \frac{L_1}{2} = 1.6 + 5.35 = 6.95\) in - \(y_{steel, bottom}\) = centroid of bottom steel = \(.8\) in - \(y_{steel, top}\) = centroid of top steel = \(L_2 + L_1 + .8 = 1.6 + 10.7 + .8 = 13.1\) in \[ Q = (43.87 \times 6.95) + (158.6 \times .8) + (158.6 \times 13.1) \] \[ Q = 305.89 + 126.88 + 2077.66 = 251.43 \] \[ \bar{y} = \frac{Q}{\text{Total Area}} = \frac{251.43}{361.07} \approx 6.95 \, \text{in} \] --- ## **Step 5: Calculate Moment of Inertia (\(I_{trans}\))** Moment of inertia for each part (about neutral axis): 1. **Wood:** \[ I_{wood} = \frac{b L_1^3}{12} + A_{wood} (y_{wood} - \bar{y})^2 \] \[ = \frac{4.1 \times 10.7^3}{12} + 43.87 \times (6.95 - 6.95)^2 \] \[ = \frac{4.1 \times 1224.3}{12} = \frac{502}{12} = 418.3 \, \text{in}^4 \] 2. **Steel (bottom):** \[ I_{steel, bottom} = \frac{b L_2^3}{12} + A_{steel,eq} (y_{steel, bottom} - \bar{y})^2 \] \[ = \frac{4.1 \times 1.6^3}{12} + 158.6 \times (.8 - 6.95)^2 \] \[ = \frac{4.1 \times 4.096}{12} = 1.4 + 158.6 \times 37.82 = 1.4 + 6001.85 = 6003.25 \, \text{in}^4 \] 3. **Steel (top):** \[ I_{steel, top} = \frac{4.1 \times 1.6^3}{12} + 158.6 \times (13.1 - 6.95)^2 \] \[ = 1.4 + 158.6 \times 37.82 = 1.4 + 6001.85 = 6003.25 \, \text{in}^4 \] \[ I_{trans} = 418.3 + 6003.25 + 6003.25 = 12424.8 \, \text{in}^4 \] --- ## **Step 6: Find Maximum Moment Based on Allowable Stresses** The maximum moment is found by checking both materials: \[ M = \frac{\sigma I}{c} \] where \(c\) is the distance from the neutral axis to the extreme fiber. - **For wood:** \(c_{wood} = \bar{y} - (L_2) = 6.95 - 1.6 = 5.35\) in - **For steel (top):** \(c_{steel} = (L_2 + L_1 + L_2) - \bar{y} = (1.6 + 10.7 + 1.6) - 6.95 = 13.9 - 6.95 = 6.95\) in ### **A. Wood Controls** \[ M_{wood} = \frac{\sigma_{wood} I_{trans}}{c_{wood}} \] \[ M_{wood} = \frac{1.6 \times 12424.8}{5.35} = \frac{19879.68}{5.35} = 3718.6 \, \text{kip-in} \] ### **B. Steel Controls** *Remember, the steel area was multiplied by \(n\), so we must use the actual steel stress:* \[ M_{steel} = \frac{\sigma_{steel} I_{trans}}{n \cdot c_{steel}} \] \[ M_{steel} = \frac{35.7 \times 12424.8}{24.17 \times 6.95} \] \[ = \frac{444,367.36}{168.92} = 2631.1 \, \text{kip-in} \] --- ## **Step 7: Final Answer** The maximum allowable moment is the **lower** of the two: \[ \boxed{M_{max} = 263 \ \text{kip-in}} \] (Rounded to 3 significant digits.) --- ### **Summary Table** | Material | Allowable Moment (kip-in) | |----------|--------------------------| | Wood | 3719 | | Steel | 263 | **Final Answer:** \[ \boxed{M_{max} = 263 \ \text{kip-in}} \] Let me know if you want any step clarified further!