3" 1" 2" Find Ix 2" 2"
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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![### Calculation of Moment of Inertia (Ix)
**Objective:**
Find the moment of inertia (Ix) of the provided shape.
**Diagram Explanation:**
The given diagram is a composite shape consisting of different rectangles and a right triangle. Here are the key dimensions:
- Bottom horizontal rectangle:
- Width = 6 inches (sum of three 2-inch segments)
- Height = 1 inch
- Left vertical rectangle:
- Width = 2 inches
- Height = 3 inches (combining the height of 1 inch from the bottom rectangle and the height of 2 inches for the vertical part)
- Top right triangle:
- Base = 2 inches
- Height = 2 inches (same height as the vertical rectangle above the bottom horizontal rectangle)
The composite figure can be divided into simpler shapes:
1. A bottom horizontal rectangle (6" x 1")
2. A left vertical rectangle (2" x 3")
3. A right triangle (base = 2", height = 2")
### Steps to Calculate Ix
1. **Identify individual shapes**: Divide the composite shape into basic geometric shapes (rectangles and a triangle).
2. **Calculate individual moments of inertia (I) about their centroids**:
- **Rectangle 1 (bottom horizontal part)**:
\[
I_1 = \frac{1}{12} b h^3 = \frac{1}{12} (6 \, \text{in}) (1 \, \text{in})^3 = 0.5 \, \text{in}^4
\]
- **Rectangle 2 (left vertical part)**:
\[
I_2 = \frac{1}{12} b h^3 = \frac{1}{12} (2 \, \text{in}) (3 \, \text{in})^3 = 4.5 \, \text{in}^4
\]
- **Triangle**:
\[
I_3 = \frac{1}{36} b h^3 = \frac{1}{36} (2 \, \text{in}) (2 \, \text{in})^3 = 0.444 \, \text{in}^4
\]
3. **Apply the parallel axis theorem** to shift the moments of inertia](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6abdb772-2674-4ccf-b97c-7a87bba16f16%2Fc1bc533d-fd5b-4243-9825-bfcc274e5216%2Fv7dc4oe_processed.png&w=3840&q=75)
Transcribed Image Text:### Calculation of Moment of Inertia (Ix)
**Objective:**
Find the moment of inertia (Ix) of the provided shape.
**Diagram Explanation:**
The given diagram is a composite shape consisting of different rectangles and a right triangle. Here are the key dimensions:
- Bottom horizontal rectangle:
- Width = 6 inches (sum of three 2-inch segments)
- Height = 1 inch
- Left vertical rectangle:
- Width = 2 inches
- Height = 3 inches (combining the height of 1 inch from the bottom rectangle and the height of 2 inches for the vertical part)
- Top right triangle:
- Base = 2 inches
- Height = 2 inches (same height as the vertical rectangle above the bottom horizontal rectangle)
The composite figure can be divided into simpler shapes:
1. A bottom horizontal rectangle (6" x 1")
2. A left vertical rectangle (2" x 3")
3. A right triangle (base = 2", height = 2")
### Steps to Calculate Ix
1. **Identify individual shapes**: Divide the composite shape into basic geometric shapes (rectangles and a triangle).
2. **Calculate individual moments of inertia (I) about their centroids**:
- **Rectangle 1 (bottom horizontal part)**:
\[
I_1 = \frac{1}{12} b h^3 = \frac{1}{12} (6 \, \text{in}) (1 \, \text{in})^3 = 0.5 \, \text{in}^4
\]
- **Rectangle 2 (left vertical part)**:
\[
I_2 = \frac{1}{12} b h^3 = \frac{1}{12} (2 \, \text{in}) (3 \, \text{in})^3 = 4.5 \, \text{in}^4
\]
- **Triangle**:
\[
I_3 = \frac{1}{36} b h^3 = \frac{1}{36} (2 \, \text{in}) (2 \, \text{in})^3 = 0.444 \, \text{in}^4
\]
3. **Apply the parallel axis theorem** to shift the moments of inertia
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