Sure! Let's break down the problem step by step. --- # **Given Data** - **Bending moment, \( M \) = 10,000 lb·in** - **Cross section: I-beam** - Flange width = 3 in - Flange thickness = 0.25 in - Web height = 3 in (overall height) - Web thickness = 0.25 in We'll: 1. **Find the moment of inertia (\( I \))** 2. **Find the distance from the neutral axis to the extreme fiber (\( c \))** 3. **Calculate max tensile and compressive stress using the bending equation** 4. **Find the stress at a point 1 inch from the bottom** --- ## **1. Calculate the Moment of Inertia (\( I \))** The **I-beam** consists of two flanges and a web. ### **Step 1: Find the Neutral Axis (Centroid) Location** Let’s place the bottom of the beam at \( y = 0 \). - **Flanges:** Each 3 in × 0.25 in, at top and bottom - **Web:** 0.25 in × (3 - 0.5) in = 0.25 in × 2.5 in (subtracting flange thickness from total height for web height) #### **Areas and Centroids:** - **Bottom Flange:** \( A_1 = 3 \times 0.25 = 0.75 \) in², centroid at \( y_1 = 0.125 \) in - **Top Flange:** \( A_2 = 3 \times 0.25 = 0.75 \) in², centroid at \( y_2 = 3 - 0.125 = 2.875 \) in - **Web:** \( A_3 = 0.25 \times 2.5 = 0.625 \) in², centroid at \( y_3 = 0.25 + 2.5/2 = 1.5 \) in #### **Total Area \( A \):** \[ A = 0.75 + 0.75 + 0.625 = 2.125 \text{ in}^2 \] #### **Centroid (\( \bar{y} \)):** \[ \bar{y} = \frac{(A_1 y_1 + A_2 y_2 + A_3 y_3)}{A} \] \[ = \frac{(0.75 \times 0.125) + (0.75 \times 2.875) + (0.625 \times 1.5)}{2.125} \] \[ = \frac{0.09375 + 2.15625 + 0.9375}{2.125} = \frac{3.1875}{2.125} = 1.5 \text{ in} \] So the neutral axis is at **1.5 in from the bottom**. --- ### **Step 2: Find Moment of Inertia (\( I \)) about the Neutral Axis** #### **Use Parallel Axis Theorem:** \[ I = \sum (I_{\text{local}} + Ad^2) \] #### **Bottom Flange:** - \( I_{\text{local}} = \frac{1}{12} b h^3 = \frac{1}{12} \times 3 \times (0.25)^3 = 0.0039 \) in⁴ - \( d_1 = 1.5 - 0.125 = 1.375 \) in - \( A_1 d_1^2 = 0.75 \times (1.375)^2 = 1.417 \) in⁴ - \( I_1 = 0.0039 + 1.417 = 1.421 \) in⁴ #### **Top Flange:** - \( I_{\text{local}} = 0.0039 \) in⁴ (same as bottom) - \( d_2 = 2.875 - 1.5 = 1.375 \) in - \( A_2 d_2^2 = 1.417 \) in⁴ - \( I_2 = 1.421 \) in⁴ #### **Web:** - \( I_{\text{local}} = \frac{1}{12} b h^3 = \frac{1}{12} \times 0.25 \times (2.5)^3 = 0.3255 \) in⁴ - \( d_3 = 1.5 - 1.5 = 0 \) - \( I_3 = 0.3255 \) in⁴ #### **Total \( I \):** \[ I = I_1 + I_2 + I_3 = 1.421 + 1.421 + 0.3255 = 3.1675 \text{ in}^4 \] --- ## **2. Maximum Tensile and Compressive Bending Stress** Bending stress at a distance \( y \) from the neutral axis: \[ \sigma = \frac{M y}{I} \] - Distance from neutral axis to top/bottom = \( c = 1.5 \) in ### **Maximum Stress:** \[ \sigma_{\text{max}} = \frac{M c}{I} \] \[ = \frac{10,000 \text{ lb·in} \times 1.5 \text{ in}}{3.1675 \text{ in}^4} = \frac{15,000}{3.1675} = 4,737 \text{ psi} = 4.74 \text{ ksi} \] - **Top fiber**: compressive (if moment causes top to be in compression) - **Bottom fiber**: tensile --- ## **3. Bending Stress 1 inch from the Bottom** - Distance from neutral axis: \( y = 1.5 - 1 = 0.5 \) in \[ \sigma_{y=0.5} = \frac{M y}{I} = \frac{10,000 \times 0.5}{3.1675} = \frac{5,000}{3.1675} = 1,578 \text{ psi} = 1.58 \text{ ksi} \] --- ## **4. Stress Distribution Diagram** ``` ^ | σ = 4.74 ksi (compression, top) |---------------------- | | | | <-- linear stress gradient | | |---------------------- | σ = 0 (neutral axis, mid-height) |---------------------- | | | | <-- linear stress gradient | | |---------------------- | σ = 4.74 ksi (tension, bottom) v ``` At 1 inch from the bottom, σ = 1.58 ksi (tension). --- ## **Summary Table** | Location | Distance from NA (in) | Stress (ksi) | Type | |-----------------------|----------------------|--------------|--------------| | Top (compression) | 1.5 | 4.74 | Compressive | | 1 in from bottom | 0.5 | 1.58 | Tensile | | Bottom (tension) | 1.5 | 4.74 | Tensile | --- ## **Key Formulas Used** - **Centroid:** \( \bar{y} = \frac{\sum (A_i y_i)}{\sum A_i} \) - **Moment of Inertia:** \( I = \sum (I_{\text{local}} + Ad^2) \) - **Bending Stress:** \( \sigma = \frac{M y}{I} \) --- ### **Let me know if you'd like a hand-drawn stress distribution or any step explained further!**