# Calculating the Centroid of the C-Channel Beam Let's walk through the steps to calculate the centroid **C** of the C-shaped (channel) section. --- ## **Step 1: Break the Section Into Simple Shapes** The C-channel can be divided into three rectangles: - **Top flange** (horizontal part at the top) - **Web** (vertical part) - **Bottom flange** (horizontal part at the bottom) --- ## **Step 2: Assign Dimensions** Given values: - **b** = 50.7 mm (width of flanges) - **h** = 70. mm (height of web) - **t** = 6.24 mm (thickness of all parts) --- ## **Step 3: Area of Each Part** 1. **Top flange:** \( A_1 = b \times t \) 2. **Web:** \( A_2 = (h - 2t) \times t \) 3. **Bottom flange:** \( A_3 = b \times t \) --- ## **Step 4: Locate Centroid of Each Part** Let's use the **bottom left corner as the origin**. ### **y-coordinates (vertical):** - Centroid of **bottom flange**: \( y_1 = \frac{t}{2} \) - Centroid of **web**: \( y_2 = t + \frac{h - 2t}{2} = t + \frac{h}{2} - t = \frac{h}{2} \) - Centroid of **top flange**: \( y_3 = h - \frac{t}{2} \) ### **x-coordinates (horizontal):** Since all parts are symmetric in x (flange widths are the same), centroids will be at: - \( x_1 = x_2 = x_3 = \frac{t}{2} \) (for the web) - \( x_1 = x_3 = \frac{b}{2} \) (for the flanges) But in a typical C-channel, the web is offset to one side. So, let's clarify: - **Bottom flange:** Centered at \( x = \frac{b}{2} \) - **Web:** Centered at \( x = t/2 \) from the left - **Top flange:** Centered at \( x = \frac{b}{2} \) --- ## **Step 5: Calculate Centroid Coordinates** ### **Areas:** - \( A_1 = b \times t \) - \( A_2 = (h - 2t) \times t \) - \( A_3 = b \times t \) ### **x-coordinates:** - \( x_1 = x_3 = \frac{b}{2} \) - \( x_2 = \frac{t}{2} \) ### **y-coordinates:** - \( y_1 = \frac{t}{2} \) - \( y_2 = \frac{h}{2} \) - \( y_3 = h - \frac{t}{2} \) --- ### **Centroid Formula** \[ \bar{x} = \frac{\sum (A_i x_i)}{\sum A_i} \] \[ \bar{y} = \frac{\sum (A_i y_i)}{\sum A_i} \] --- ### **Calculate Each Term** #### **Areas:** - \( A_1 = 50.7 \times 6.24 = 316.368 \) mm² - \( A_2 = (70. - 2 \times 6.24) \times 6.24 = (70. - 12.48) \times 6.24 = 57.52 \times 6.24 = 359.9488 \) mm² - \( A_3 = 316.368 \) mm² #### **x-coordinates:** - \( x_1 = x_3 = \frac{50.7}{2} = 25.35 \) mm - \( x_2 = \frac{6.24}{2} = 3.12 \) mm #### **y-coordinates:** - \( y_1 = \frac{6.24}{2} = 3.12 \) mm - \( y_2 = \frac{70.}{2} = 35. \) mm - \( y_3 = 70. - \frac{6.24}{2} = 70. - 3.12 = 66.88 \) mm --- ## **Step 6: Plug Into the Formula** ### **Total Area:** \[ A_{total} = A_1 + A_2 + A_3 = 316.368 + 359.9488 + 316.368 = 992.6848 \text{ mm}^2 \] ### **\(\bar{x}\):** \[ \bar{x} = \frac{A_1 x_1 + A_2 x_2 + A_3 x_3}{A_{total}} = \frac{316.368 \times 25.35 + 359.9488 \times 3.12 + 316.368 \times 25.35}{992.6848} \] \[ = \frac{8024.224 + 1123.023 + 8024.224}{992.6848} \] \[ = \frac{17171.471}{992.6848} = 17.30 \text{ mm} \] ### **\(\bar{y}\):** \[ \bar{y} = \frac{A_1 y_1 + A_2 y_2 + A_3 y_3}{A_{total}} = \frac{316.368 \times 3.12 + 359.9488 \times 35. + 316.368 \times 66.88}{992.6848} \] \[ = \frac{987.133 + 12598.208 + 21161.150}{992.6848} \] \[ = \frac{34746.491}{992.6848} = 35.01 \text{ mm} \] --- ## **Final Answer** ### **Centroid location \((\bar{x}, \bar{y})\) from the bottom left corner:** \[ \boxed{ (\bar{x}, \bar{y}) = (17.3 \text{ mm}, 35. \text{ mm}) } \] --- ### **Summary Table** | Part | Area (\(mm^2\)) | \(x_i\) (mm) | \(y_i\) (mm) | |-----------------|-----------------|--------------|--------------| | Top flange | 316.368 | 25.35 | 66.88 | | Web | 359.9488 | 3.12 | 35.00 | | Bottom flange | 316.368 | 25.35 | 3.12 | --- ### **Key Steps** 1. Divide the cross-section into rectangles. 2. Find area and centroid of each. 3. Use centroid formulas for composite sections. 4. Plug in values and compute. Let me know if you need a step-by-step drawing or have any questions on each step!