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To solve this problem, we need to calculate the moments of inertia of the combined section (C250 × 22.8 channel welded to W460 × 113 wide-flange section) with respect to its centroidal axes \(x\) and \(y\).

### Step-by-step process:

#### 1. **Given Data**:
   From the table:
   - **C250 × 22.8**:
     - Area (\(A_C\)) = 2900 mm²
     - Moment of inertia about its own centroid:
       - \(I_{xC}\) = \(0.945 \times 10^6\) mm⁴
       - \(I_{yC}\) = \(28 \times 10^6\) mm⁴
     - Distance from its own centroid to the centroid of the combined section (\(\bar{y}_C\)) = 512.9 mm

   - **W460 × 113**:
     - Area (\(A_W\)) = 14400 mm²
     - Moment of inertia about its own centroid:
       - \(I_{xW}\) = \(554 \times 10^6\) mm⁴
       - \(I_{yW}\) = \(63.3 \times 10^6\) mm⁴
     - Distance from its own centroid to the centroid of the combined section (\(\bar{y}_W\)) = 211.5 mm

#### 2. **Determine the Centroid of the Combined Section**:
   The centroid \(y_{cg}\) can be found using the following formula:
   \[
   y_{cg} = \frac{A_C \cdot \bar{y}_C + A_W \cdot \bar{y}_W}{A_C + A_W}
   \]

   Plugging in the values:
   \[
   y_{cg} = \frac{(2900 \, \text{mm}^2 \cdot 512.9 \, \text{mm}) + (14400 \, \text{mm}^2 \cdot 211.5 \, \text{mm})}{2900 \, \text{mm}^2 + 14400 \, \text{mm}^2}
   \]

#### 3. **Apply the Parallel Axis Theorem**:
   Once the centroid is known, we can calculate the moments of inertia of the individual sections about the centroidal axes of the combined section using the parallel axis theorem:

   \[
   I_{x_{\text{total}}} = I_{xC} + A_C \cdot d_{xC}^2 + I_{xW} + A_W \cdot d_{xW}^2
   \]
   where \(d_{xC}\) and \(d_{xW}\) are the distances from the individual centroids of the channel and wide-flange to the combined centroid.

   Similarly for the \(y\)-axis:
   \[
   I_{y_{\text{total}}} = I_{yC} + A_C \cdot d_{yC}^2 + I_{yW} + A_W \cdot d_{yW}^2
   \]

#### 4. **Final Calculation**:
   I will now calculate these values step by step.The combined centroid \( y_{cg} \) of the section is approximately \( 262.02 \, \text{mm} \) from the reference axis.

Next, I'll calculate the moments of inertia of the combined section using the parallel axis theorem. Let's proceed with the calculation of the distances \( d_{xC} \) and \( d_{xW} \) for both sections and then compute the total moments of inertia about the \( x \)- and \( y \)-axes.The moments of inertia of the combined section are as follows:

- Moment of inertia about the \( x \)-axis: \( 774.23 \times 10^6 \, \text{mm}^4 \)
- Moment of inertia about the \( y \)-axis: \( 264.3 \times 10^6 \, \text{mm}^4 \)

Would you like further details or have any questions? Here are some related questions you might want to explore:

1. How would the position of the centroid change if the channel were placed on the bottom instead of the top?
2. How does the parallel axis theorem affect the final moments of inertia?
3. What is the significance of calculating the moments of inertia for structural analysis?
4. How would the design change if the sections were not welded together?
5. Could we calculate the radius of gyration from these moments of inertia?

**Tip:** The moments of inertia are crucial for determining the resistance of a structural member to bending stresses.

Determine the moments of inertia of the combined section (C250 × 22.8 channel welded to W460 × 113 wide-flange section) with respect to its centroidal axes x and y.

This tutorial explains how to calculate the moments of inertia for a combined C250 × 22.8 channel and W460 × 113 wide-flange section. Learn how to find the centroid of the composite section and apply the parallel axis theorem for both x and y axes.

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