Let's solve the problem step by step. --- ### **Step 1: Given Data** - **Timber:** - \( E_{timber} = 3 \times 10^3 \) ksi - \( \sigma_{all}^{timber} = 1 \) ksi - **Steel:** - \( E_{steel} = 30 \times 10^3 \) ksi - \( \sigma_{all}^{steel} = 10 \) ksi - **Section Dimensions:** - Timber cross-section: \( 6 \) in wide \( \times \) \( 12 \) in high - Steel plates: \( 2 \) in \( \times \) \( 2 \) in (two plates at the bottom) --- ### **Step 2: Modular Ratio (n)** \[ n = \frac{E_{steel}}{E_{timber}} = \frac{30,000}{3,000} = 10 \] --- ### **Step 3: Transformed Section** #### **Steel Area (Transformed to Timber):** - Area of one steel plate = \( 2 \) in \( \times \) \( 2 \) in = \( 4 \) in² - Total steel area = \( 2 \times 4 = 8 \) in² - Transformed steel area to timber equivalent: \[ A_{steel,eff} = n \times A_{steel} = 10 \times 8 = 80 \text{ in}^2 \] #### **Total Transformed Area:** - Area of timber section (minus steel cutout at bottom): - Timber area above steel = \( 6 \) in × \( (12 - 2) \) in = \( 6 \times 10 = 60 \) in² - Timber area at bottom (excluding steel) = \( (6 - 4) \) in × \( 2 \) in = \( 2 \times 2 = 4 \) in² Total timber area = \( 60 + 4 = 64 \) in² - Total transformed area = Timber area + Transformed steel area \[ A_{total,trans} = 64 + 80 = 144 \text{ in}^2 \] --- ### **Step 4: Location of the Neutral Axis (\( y_{NA} \))** Let bottom of section be \( y = 0 \). - Timber area at bottom: \( A_1 = 4 \) in², centroid \( y_1 = 1 \) in - Timber area above steel: \( A_2 = 60 \) in², centroid \( y_2 = 7 \) in - Transformed steel area: \( A_3 = 80 \) in², centroid \( y_3 = 1 \) in \[ y_{NA} = \frac{A_1 y_1 + A_2 y_2 + A_3 y_3}{A_1 + A_2 + A_3} = \frac{(4 \times 1) + (60 \times 7) + (80 \times 1)}{4 + 60 + 80} = \frac{4 + 420 + 80}{144} = \frac{504}{144} = 3.5 \text{ in} \] --- ### **Step 5: Moment of Inertia about the Neutral Axis (\( I_{NA} \))** For each area, \( I = I_{own} + Ad^2 \): #### **A. Timber Bottom (A1)** - \( A_1 = 4 \) in², centroid at \( y_1 = 1 \) in - \( d_1 = 3.5 - 1 = 2.5 \) in - \( I_{1, own} = \frac{1}{12} b h^3 = \frac{1}{12} \times 2 \times 2^3 = \frac{1}{12} \times 16 = 1.333 \) in⁴ - \( I_1 = 1.333 + 4 \times (2.5)^2 = 1.333 + 25 = 26.333 \) in⁴ #### **B. Timber Above Steel (A2)** - \( A_2 = 60 \) in², centroid at \( y_2 = 7 \) in - \( d_2 = 7 - 3.5 = 3.5 \) in - \( I_{2, own} = \frac{1}{12} b h^3 = \frac{1}{12} \times 6 \times 10^3 = \frac{1}{12} \times 6000 = 500 \) in⁴ - \( I_2 = 500 + 60 \times (3.5)^2 = 500 + 735 = 1235 \) in⁴ #### **C. Transformed Steel (A3)** - \( A_3 = 80 \) in², centroid at \( y_3 = 1 \) in - \( d_3 = 3.5 - 1 = 2.5 \) in - \( I_{3, own} = 2 \) plates × \( \frac{1}{12} b h^3 = 2 \times \frac{1}{12} \times 2 \times 2^3 = 2 \times 1.333 = 2.667 \) in⁴ - \( I_3 = 2.667 + 80 \times (2.5)^2 = 2.667 + 500 = 502.667 \) in⁴ #### **Total Moment of Inertia** \[ I_{total} = I_1 + I_2 + I_3 = 26.333 + 1235 + 502.667 = 1764 \text{ in}^4 \] --- ### **Step 6: Maximum Moment (M) Calculation** **Maximum compressive stress in timber:** - Distance from NA to top: \( y_{top} = 12 - 3.5 = 8.5 \) in - \( \sigma_{timber, max} = 1 \) ksi **Maximum tensile stress in steel:** - Distance from NA to bottom: \( y_{steel} = 3.5 - 1 = 2.5 \) in - \( \sigma_{steel, max} = 10 \) ksi (but in timber transformed section, allowable timber stress is \( 10/10 = 1 \) ksi in timber units) #### **For Timber (top fiber):** \[ \sigma_{top} = \frac{M \cdot y_{top}}{I} \] \[ 1 = \frac{M \cdot 8.5}{1764} \] \[ M_{timber} = \frac{1764}{8.5} = 207.53 \text{ kip-in} \] #### **For Steel (bottom fiber, transformed):** \[ \sigma_{bottom} = \frac{M \cdot y_{steel}}{I} \] \[ 1 = \frac{M \cdot 2.5}{1764} \] \[ M_{steel} = \frac{1764}{2.5} = 705.6 \text{ kip-in} \] --- ### **Step 7: **Maximum Moment That Can Be Applied** The maximum moment is governed by the smaller value: \[ \boxed{M_{max} = 207.5 \;\text{kip-in}} \] --- #### **Summary Table** | Limiting Material | Max Moment (kip-in) | |-------------------|--------------------| | Timber (top) | 207.5 | | Steel (bottom) | 705.6 | --- ### **Final Answer** \[ \boxed{M_{max} = 207.5 \;\text{kip-in}} \] This is the maximum horizontal bending moment that can be applied on the cross-section. If you need the answer in ft-lb or another unit, let me know!

Let's revisit and verify the solution considering the correction that the width of the transformed section should be \( 2n = 20 \) inches. --- ### **Given Data Revisited:** - **Timber:** - \( E_{timber} = 3 \times 10^3 \text{ ksi} \) - \( \sigma_{all}^{timber} = 1 \text{ ksi} \) - **Steel:** - \( E_{steel} = 30 \times 10^3 \text{ ksi} \) - \( \sigma_{all}^{steel} = 10 \text{ ksi} \) - **Section Dimensions:** - Timber cross-section: 6 in wide, 12 in high - Steel plates: 2 in × 2 in (two plates at the bottom) --- ### **Step 1: Modular Ratio** \[ n = \frac{E_{steel}}{E_{timber}} = \frac{30,000}{3,000} = 10 \] ### **Step 2: Steel Area** - Area of one steel plate: \[ A_{steel} = 2 \times 2 = 4 \text{ in}^2 \] - Total steel area: \[ A_{steel, total} = 2 \times 4 = 8 \text{ in}^2 \] ### **Step 3: Transformed Steel Area** \[ A_{steel,eff} = n \times A_{steel} = 10 \times 8 = 80 \text{ in}^2 \] --- ### **Step 4: Transformed Section Width** The **corrected assumption** is that the **total transformed section width** is: \[ b_{transformed} = 2n = 20 \text{ in} \] This implies that the **section is modeled as a homogeneous section** with width 20 in, representing both timber and steel transformed to the timber's modulus. --- ### **Step 5: Cross-Section Area and Centroids** **Original timber section:** - Width = 6 in - Height = 12 in - Steel plates at bottom, each 2 in × 2 in, located at the bottom **Transformed section:** - Width = 20 in - The steel plates are transformed to equivalent timber areas. --- ### **Step 6: Neutral Axis Location** Assuming the **steel plates are at the bottom**: - **Steel centroid:** 1 in from bottom (since plates are 2 in high at bottom) - **Timber above steel:** from top to steel plates Calculations: - **Area of timber (excluding steel):** - Top timber: 6 in wide, 10 in high - Area: \(6 \times 10 = 60 \text{ in}^2\) - **Steel area (transformed):** 80 in² - **Centroids:** - Timber top: at \( 1 + 8 = 9 \) in from bottom (since total height is 12 in, steel at bottom 2 in, timber above: 10 in high) - Steel centroid: at 1 in from bottom - Timber bottom: at 1 in from bottom (steel plates) Calculating the **overall neutral axis**: \[ y_{NA} = \frac{A_{timber} \times y_{timber} + A_{steel} \times y_{steel}}{A_{timber} + A_{steel}} \] Where: - \(A_{timber} = 60 \text{ in}^2\), centroid approx at \( y_{timber} = 8 \text{ in} \) (from bottom) - \(A_{steel} = 80 \text{ in}^2\), centroid at \( y_{steel} = 1 \text{ in} \) \[ y_{NA} = \frac{60 \times 8 + 80 \times 1}{60 + 80} = \frac{480 + 80}{140} = \frac{560}{140} = 4 \text{ in} \] --- ### **Step 7: Moment of Inertia** Calculate \( I_{total} \) about the neutral axis at \( y_{NA} = 4 \text{ in} \). - **For timber (top part):** - \( A_{timber} = 60 \text{ in}^2 \) - Own moment of inertia (about centroid): \[ I_{own, timber} = \frac{1}{12} \times b \times h^3 = \frac{1}{12} \times 6 \times 10^3 = 500 \text{ in}^4 \] - Distance from centroid to NA: \( 8 - 4 = 4 \text{ in} \) - Using parallel axis theorem: \[ I_{timber} = 500 + 60 \times 4^2 = 500 + 960 = 1460 \text{ in}^4 \] - **For steel plates (transformed):** - \( A_{steel} = 80 \text{ in}^2 \) - Own moment of inertia (about centroid at \( y=1 \)): \[ I_{own, steel} = 2 \times \frac{1}{12} \times 2 \times 2^3 = 2 \times 1.333 = 2.667 \text{ in}^4 \] - Distance from steel centroid to NA: \[ | 1 - 4 | = 3 \text{ in} \] - Parallel axis theorem: \[ I_{steel} = 2.667 + 80 \times 3^2 = 2.667 + 80 \times 9 = 2.667 + 720 = 722.667 \text{ in}^4 \] - **Total Moment of Inertia:** \[ I_{total} = 1460 + 722.667 = 2182.667 \text{ in}^4 \] --- ### **Step 8: Maximum Bending Moment Calculation** - **Maximum tensile stress in timber:** \[ \sigma_{max} = \frac{M \times c}{I} \] Where \( c \) is the distance from NA to extreme fiber. - **Top fiber (in timber):** \[ c_{top} = 12 - 4 = 8 \text{ in} \] - **Bottom fiber (steel plates):** \[ c_{bottom} = 4 - 1 = 3 \text{ in} \] The maximum stress (allowable) in timber is 1 ksi, and in steel (transformed to timber units) is 10 ksi. --- ### **In Timber (top fiber):** \[ \sigma_{timber} = \frac{M \times 8}{2182.667} \leq 1 \text{ ksi} \] \[ M \leq \frac{2182.667}{8} = 272.83 \text{ kip-in} \] --- ### **In Steel (bottom fiber):** \[ \sigma_{steel} = \frac{M \times 3}{2182.667} \leq 10 \text{ ksi} \] \[ M \leq \frac{10 \times 2182.667}{3} = 7275.56 \text{ kip-in} \] --- ### **Final Maximum Moment:** The limiting value is **from the timber (top fiber)**: \[ \boxed{ M_{max} \approx 273 \text{ kip-in} } \] --- ### **Summary:** - The **corrected total transformed section width** is 20 in. - The **neutral axis** is approximately 4 in from the bottom. - The **moment of inertia** about this axis is approximately 2183 in\(^4\). - **Maximum permissible moment** based on timber stress is approximately **273 kip-in**. --- **Conclusion:** **The maximum horizontal bending moment that can be applied is approximately \(\boxed{273 \text{ kip-in}}\).** --- If you need this converted to other units or further details, please let me know!