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Determine the distance to the centroid (ȳ) and calculate Ix & Ix’

To determine the distance to the centroid (\(\bar{y}\)) and calculate the moments of inertia (\(I_x\) and \(I_{x'}\)) for this composite shape, we can follow these steps:

### 1. Break the shape into simple geometric parts:
Looking at the diagram, the structure can be divided into three distinct rectangles:
- **Rectangle 1**: The top horizontal section.
- **Rectangles 2 and 3**: The two vertical sections below the top.

### 2. Calculate the area and centroid of each part:

#### **Rectangle 1** (Top section):
- **Width**: 6 in
- **Height**: 2 in
- **Area**: \( A_1 = 6 \times 2 = 12 \text{ in}^2 \)
- **Centroid (y-coordinate)**: Since the top is 2 in high, the centroid is at \( y_1 = 2 + \frac{2}{2} = 3 \) in from the base.

#### **Rectangle 2 and 3** (Two vertical sections):
- **Width**: 1 in each (since there are two sections, both have the same dimensions)
- **Height**: 4 in
- **Area of each**: \( A_2 = A_3 = 1 \times 4 = 4 \text{ in}^2 \)
- **Centroid (y-coordinate)**: The centroid is halfway through their height, \( y_2 = y_3 = \frac{4}{2} = 2 \) in from the base.

### 3. Find the total area of the composite shape:
- Total area \( A_{\text{total}} = A_1 + A_2 + A_3 = 12 + 4 + 4 = 20 \text{ in}^2 \)

### 4. Calculate the centroid location (\(\bar{y}\)):
Using the formula for the centroid in the y-direction for composite areas:
\[
\bar{y} = \frac{\sum A_i y_i}{\sum A_i} = \frac{(A_1 y_1) + (A_2 y_2) + (A_3 y_3)}{A_{\text{total}}}
\]
Substituting the values:
\[
\bar{y} = \frac{(12 \times 3) + (4 \times 2) + (4 \times 2)}{20} = \frac{36 + 8 + 8}{20} = \frac{52}{20} = 2.6 \text{ in}
\]

Thus, the centroid is located \( \bar{y} = 2.6 \) inches from the base.

### 5. Calculate the moment of inertia (\( I_x \)):
The moment of inertia about the x-axis is calculated using the parallel axis theorem:

\[
I_x = I_{\text{centroid}} + A d^2
\]
where \( I_{\text{centroid}} \) is the moment of inertia of each part around its centroid, and \( d \) is the distance from the centroid of each part to the total centroid.

#### For each part:

- **Rectangle 1**:
  \[
  I_{x1} = \frac{1}{12} \times 6 \times 2^3 + 12 \times (3 - 2.6)^2 = 4 + 12 \times 0.16 = 5.92 \text{ in}^4
  \]

- **Rectangles 2 and 3**:
  \[
  I_{x2} = I_{x3} = \frac{1}{12} \times 1 \times 4^3 + 4 \times (2.6 - 2)^2 = 5.33 + 4 \times 0.36 = 6.77 \text{ in}^4
  \]

Thus, the total moment of inertia about the x-axis is:
\[
I_x = 5.92 + 6.77 + 6.77 = 19.46 \text{ in}^4
\]

### 6. Moment of inertia about \( x' \)-axis (\( I_{x'} \)):
Since the \( x' \)-axis passes through the centroid, the moment of inertia about the centroidal axis is the sum of the individual moments of inertia:
\[
I_{x'} = I_{x1} + I_{x2} + I_{x3}
\]
So:
\[
I_{x'} = 5.92 + 6.77 + 6.77 = 19.46 \text{ in}^4
\]

### Summary:
- Centroid location (\(\bar{y}\)): **2.6 in**
- Moment of inertia about \(x\)-axis (\(I_x\)): **19.46 in\(^4\)**
- Moment of inertia about centroidal \(x'\)-axis (\(I_{x'}\)): **19.46 in\(^4\)**

Would you like more details or have any further questions?

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Here are some related questions for deeper understanding:
1. How does the parallel axis theorem affect moment of inertia calculations?
2. What role does the centroid play in structural analysis?
3. How would the moment of inertia change if the vertical sections were wider?
4. How can we calculate the moment of inertia for more complex shapes?
5. How does the distance between components impact the total moment of inertia?

**Tip:** In structural engineering, minimizing the moment of inertia in beams can lead to increased deflection, making them less stable.

Learn how to calculate the centroid (ȳ) and moment of inertia (Ix & Ix’) for a composite shape using the Parallel Axis Theorem. This step-by-step solution includes geometric division, centroid location, and moment of inertia calculations. Suitable for advanced high school or college-level students.

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