https://file.aiquickdraw.com/mathbot/file/172670966945286q4aalp.jpg

To compute the moment of inertia about both the X-axis and Y-axis for the given I-beam, we can use the parallel axis theorem and standard formulas for calculating the moment of inertia of rectangular shapes.

The beam in the image consists of three rectangular sections: 
1. The top flange
2. The bottom flange
3. The web (vertical section between the two flanges)

### Step 1: Dimensions of each section

- **Top flange**: Width = 200 mm, Height = 15 mm
- **Bottom flange**: Width = 200 mm, Height = 15 mm
- **Web**: Width = 15 mm, Height = 300 mm

### Step 2: Moment of inertia about the X-axis (I\(_x\))

For the X-axis (horizontal axis passing through the centroid), we need to compute the moment of inertia for each part and then use the parallel axis theorem for the flanges, since their centroid is not at the overall centroid of the I-beam.

1. **Web**:
   - Moment of inertia (I\(_x_{\text{web}}\)) for a rectangle about its own centroid:
     \[
     I_x = \frac{b \cdot h^3}{12} = \frac{15 \cdot 300^3}{12}
     \]
   
2. **Flanges**:
   For the flanges, their centroids are not aligned with the overall centroid of the section, so we use the parallel axis theorem.
   
   - Moment of inertia for each flange (I\(_x_{\text{flange}}\)) about its own centroid:
     \[
     I_x = \frac{b \cdot h^3}{12} = \frac{200 \cdot 15^3}{12}
     \]
   - Parallel axis theorem: Add the contribution from the distance \( d \) between the flange centroid and the overall beam centroid.
     \[
     I_x = I_x + A \cdot d^2
     \]
     where \( d \) is half the distance between the centroids (150 mm from the web center to the flange center).

### Step 3: Moment of inertia about the Y-axis (I\(_y\))

For the Y-axis (vertical axis passing through the centroid):
1. **Web**:
   - Moment of inertia (I\(_y_{\text{web}}\)) for the web about its own centroid:
     \[
     I_y = \frac{h \cdot b^3}{12} = \frac{300 \cdot 15^3}{12}
     \]
2. **Flanges**:
   - Moment of inertia (I\(_y_{\text{flange}}\)) for each flange about its own centroid:
     \[
     I_y = \frac{h \cdot b^3}{12} = \frac{15 \cdot 200^3}{12}
     \]

### Final Computation

Let me calculate both I\(_x\) and I\(_y\).The computed moments of inertia for the given I-beam are:

- Moment of inertia about the X-axis (I\(_x\)): \( 168,862,500 \, \text{mm}^4 \)
- Moment of inertia about the Y-axis (I\(_y\)): \( 20,084,375 \, \text{mm}^4 \)

Would you like further details or have any questions on this calculation? Here are some related questions for deeper understanding:

1. How does the parallel axis theorem affect the moment of inertia for offset shapes?
2. What is the significance of the centroid location in these calculations?
3. How would the moment of inertia change if the beam's dimensions were doubled?
4. Can the moment of inertia be directly used to calculate the beam's bending stress?
5. How do different cross-sectional shapes affect the structural properties?

**Tip**: The moment of inertia is crucial in determining how resistant a shape is to bending under load; the higher it is, the less the structure will bend.

Computation of Moment of Inertia about the X-axis and Y-axis for an I-beam with dimensions as shown: top and bottom flange width = 200 mm, height = 15 mm; web height = 300 mm, width = 15 mm.

Calculate the moment of inertia of an I-beam with a top flange width of 200 mm, a bottom flange height of 15 mm, and a web height of 300 mm. Learn how the Parallel Axis Theorem is applied to compute the moment of inertia about both the X and Y axes for structural analysis.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation