Let's break down how to **find the centroidal moments of inertia** (\(\bar{I}_x\) and \(\bar{I}_y\)) for composite areas step by step. I’ll focus on the process and provide formulas you can apply to **each shape**. If you want a full solution for a specific region, let me know which one--- # **Step-by-Step Solution Guide** ## **1. General Approach** 1. **Divide the Area**: Break each shape into basic geometric parts (rectangles, semicircles, etc.). 2. **Find Area & Centroid**: Calculate the area and centroid location (\(\bar{x}, \bar{y}\)) for each part. 3. **Find Moment of Inertia About Own Centroid**: Use standard formulas for each part. 4. **Parallel Axis Theorem**: If needed, transfer the moment of inertia to the centroid of the composite area: \[ I = I_{\text{centroid}} + Ad^2 \] where \(d\) = distance between the centroid of the part and the centroid of the composite section. 5. **Sum or Subtract**: Add (or subtract, for holes) the moments of inertia for all parts. --- ## **2. Standard Area Moment of Inertia Formulas** | Shape | \(I_{x,\text{centroid}}\) | \(I_{y,\text{centroid}}\) | |-----------------|---------------------------------------------|---------------------------------------------| | Rectangle (\(b \times h\)) | \(\frac{1}{12} b h^3\) | \(\frac{1}{12} h b^3\) | | Semicircle (\(r\)) | \(\frac{1}{8} \pi r^4 - \frac{8r^4}{9\pi}\) | \(\frac{1}{8} \pi r^4\) | | Triangle (\(b \times h\)) | \(\frac{1}{36} b h^3\) | \(\frac{1}{48} b^3 h\) | --- ## **3. Example: Top Right (I-beam)** ### **Step 1: Divide the shape** - Top flange (rectangle) - Web (rectangle) - Bottom flange (rectangle) ### **Step 2: Calculate centroids** #### Find the **centroidal axis** (\(\bar{y}\)): \[ \bar{y} = \frac{\sum (A_i y_i)}{\sum A_i} \] where \(A_i\) is the area of each part and \(y_i\) is the centroid location from the base. #### Areas: - Top flange: \(A_1 = 90 \times 20\) - Web: \(A_2 = 10 \times 70\) - Bottom flange: \(A_3 = 50 \times 20\) #### Centroid positions (\(y\) from bottom): - Top flange: \(y_1 = 110 - 10 = 100\) mm (center of top flange) - Web: \(y_2 = 20 + 35 = 55\) mm (center of web) - Bottom flange: \(y_3 = 10\) mm (center of bottom flange) Calculate \(\bar{y}\) using these. ### **Step 3: Find \(I_{x,\text{part centroid}}\) for each** \[ I_{x} = \frac{1}{12} b h^3 \] ### **Step 4: Use Parallel Axis Theorem** \[ I_{x, \text{total}} = \sum \left[ I_{x,\text{centroid}} + A_i (d_i)^2 \right] \] where \(d_i\) = distance from each part's centroid to the composite centroid. ### **Step 5: \(I_y\) Calculation** \[ I_{y} = \sum \left[ I_{y,\text{centroid}} \right] \] For symmetric sections about the \(y\)-axis and vertical axis through centroid, the parallel axis theorem may not be needed as centroids of parts align. --- # **Apply This to Each Shape** - **For regions with holes:** Subtract the area and moments of inertia of the hole (negative sign). - **For composite shapes:** Add or subtract as needed. --- ## **Summary Table for Your Steps** | Step | What to Do | Formula/Note | |------|------------------------|-----------------------------------------------| | 1 | Divide into shapes | Use rectangles, triangles, circles, etc. | | 2 | Area and centroid | \(A_i, (\bar{x}_i, \bar{y}_i)\) | | 3 | Centroid of full shape | \(\bar{x} = \frac{\sum A_i x_i}{\sum A_i}\) | | 4 | \(I_{x}, I_{y}\) of each| Standard formulas above | | 5 | Parallel Axis Theorem | \(I = I_{\text{centroid}} + A d^2\) | | 6 | Total | Add/subtract all \(I\) values | --- # **If you want a full, worked-out example for a specific shape, tell me which one!**

To find the centroidal moments of inertia (\(I_x\) and \(I_y\)) for the regions shown in the provided image, we need to follow these steps: ## **Step-by-Step Solution** ### **1. Identify Shapes and Dimensions** From the image, identify the shapes and their dimensions. It appears that you have rectangles and possibly composite shapes. #### **Dimensions (assumed from the image)** - Rectangles: - Top flange: Width = 70 mm, Height = 10 mm - Bottom flange: Width = 70 mm, Height = 20 mm - Web: Width = 10 mm, Height = 70 mm - Additional regions may need similar measurements. ### **2. Calculate Area and Centroid for Each Shape** Using the dimensions, calculate the area \(A\) and the centroid position for each rectangle. #### **Areas** - For the **top flange**: \[ A_1 = 70 \times 10 = 700 \, \text{mm}^2 \] - For the **bottom flange**: \[ A_2 = 70 \times 20 = 1400 \, \text{mm}^2 \] - For the **web**: \[ A_3 = 10 \times 70 = 700 \, \text{mm}^2 \] ### **3. Calculate Centroid Locations** Calculate the centroid positions \(y_i\) for each part, measured from a reference point (usually the bottom of the entire shape). Assuming the following positions: - Top flange: \(y_1 = 10 + 5 = 15 \, \text{mm}\) (half of height from the bottom) - Web: \(y_2 = 10 + 70/2 = 45 \, \text{mm}\) (half of height from bottom) - Bottom flange: \(y_3 = 10 + 20/2 = 20 \, \text{mm}\) (half of height from bottom) ### **4. Calculate Composite Centroid \( \bar{y} \)** Using the formula: \[ \bar{y} = \frac{\sum (A_i y_i)}{\sum A_i} \] Substituting the areas and centroids: \[ \bar{y} = \frac{(700 \cdot 15) + (1400 \cdot 20) + (700 \cdot 45)}{700 + 1400 + 700} \] \[ \bar{y} = \frac{10500 + 28000 + 31500}{2800} = \frac{70000}{2800} = 25 \, \text{mm} \] ### **5. Calculate Moment of Inertia for Each Shape** Using the standard formulas for \(I_x\) and \(I_y\): #### **For the Top Flange** \[ I_{x1} = \frac{1}{12} b h^3 = \frac{1}{12} (70)(10^3) = \frac{1}{12} (70)(1000) = 5833.33 \, \text{mm}^4 \] #### **For the Bottom Flange** \[ I_{x2} = \frac{1}{12} b h^3 = \frac{1}{12} (70)(20^3) = \frac{1}{12} (70)(8000) = 46666.67 \, \text{mm}^4 \] #### **For the Web** \[ I_{x3} = \frac{1}{12} b h^3 = \frac{1}{12} (10)(70^3) = \frac{1}{12} (10)(343000) = 285833.33 \, \text{mm}^4 \] ### **6. Apply the Parallel Axis Theorem** For each part, calculate \(I_{x,\text{total}}\): \[ I_{x,\text{total}} = \sum \left[ I_{x,\text{centroid}} + A_i (d_i)^2 \right] \] Where \(d_i\) is the distance from the centroid of each part to the centroid of the composite shape (\( \bar{y} = 25 \, \text{mm} \)). - **Top flange**: \[ d_1 = 25 - 15 = 10 \, \text{mm} \] \[ I_{1} = 5833.33 + 700(10^2) = 5833.33 + 70000 = 75833.33 \, \text{mm}^4 \] - **Bottom flange**: \[ d_2 = 25 - 20 = 5 \, \text{mm} \] \[ I_{2} = 46666.67 + 1400(5^2) = 46666.67 + 35000 = 81666.67 \, \text{mm}^4 \] - **Web**: \[ d_3 = 25 - 45 = -20 \, \text{mm} \] \[ I_{3} = 285833.33 + 700(-20^2) = 285833.33 + 280000 = 565833.33 \, \text{mm}^4 \] ### **7. Total Moment of Inertia \(I_x\)** Finally, add all contributions: \[ I_{x,\text{total}} = 75833.33 + 81666.67 + 565833.33 = 732333.33 \, \text{mm}^4 \] ### **8. Calculate \(I_y\)** For \(I_y\) (similar calculations using the standard formula): - **Top Flange**: \[ I_{y1} = \frac{1}{12} h b^3 = \frac{1}{12}(10)(70^3) = \frac{1}{12}(10)(343000) = 285833.33 \, \text{mm}^4 \] - **Bottom Flange**: \[ I_{y2} = \frac{1}{12}(20)(70^3) = \frac{1}{12}(20)(343000) = 571388.89 \, \text{mm}^4 \] - **Web**: \[ I_{y3} = \frac{1}{12}(70)(10^3) = 5833.33 \, \text{mm}^4 \] ### **Total Moment of Inertia \(I_y\)** \[ I_{y,\text{total}} = 285833.33 + 571388.89 + 5833.33 = 859055.55 \, \text{mm}^4 \] ## **Final Results** - \(I_x = 732333.33 \, \text{mm}^4\) - \(I_y = 859055.55 \, \text{mm}^4\) If you need further assistance or a breakdown of any specific region, please specify!