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Calculate the second moment of area of the cross section of the element INA

To solve this problem, we need to calculate the second moment of area (also known as the moment of inertia) for the given T-section.

### Step-by-Step Solution:

1. **Determine the centroid (neutral axis) of the T-section:**

   The centroid \( y_c \) of the T-section can be found using the formula for the centroid of composite areas:
   \[
   y_c = \frac{\sum (A_i y_i)}{\sum A_i}
   \]
   where \( A_i \) is the area of each section, and \( y_i \) is the distance from the reference axis to the centroid of each section.

   The T-section consists of two rectangles:
   - Top flange: \( 200 \, \text{mm} \times 80 \, \text{mm} \)
   - Web: \( 60 \, \text{mm} \times 100 \, \text{mm} \)

   **Areas:**
   \[
   A_1 = 200 \times 80 = 16000 \, \text{mm}^2
   \]
   \[
   A_2 = 60 \times 100 = 6000 \, \text{mm}^2
   \]

   **Centroids:**
   - For the top flange:
     \[
     y_1 = \frac{80}{2} = 40 \, \text{mm}
     \]
   - For the web:
     \[
     y_2 = 80 + \frac{100}{2} = 130 \, \text{mm}
     \]

   **Calculate \( y_c \):**
   \[
   y_c = \frac{16000 \times 40 + 6000 \times 130}{16000 + 6000} = \frac{640000 + 780000}{22000} = \frac{1420000}{22000} = 64.55 \, \text{mm}
   \]

2. **Calculate the second moment of area (moment of inertia) about the centroidal axis \( I_{NA} \):**

   The second moment of area for each rectangle is calculated using the parallel axis theorem:
   \[
   I_{NA} = I_{c} + A \cdot d^2
   \]
   where \( I_c \) is the moment of inertia about the centroidal axis of the shape, \( A \) is the area, and \( d \) is the distance from the centroid of the shape to the neutral axis.

   **Top flange:**
   \[
   I_{c1} = \frac{1}{12} \times 200 \times 80^3 = \frac{1}{12} \times 200 \times 512000 = 8533333.33 \, \text{mm}^4
   \]
   \[
   d_1 = 64.55 - 40 = 24.55 \, \text{mm}
   \]
   \[
   I_1 = 8533333.33 + 16000 \times 24.55^2 = 8533333.33 + 16000 \times 602.70 = 8533333.33 + 9643200 = 18176533.33 \, \text{mm}^4
   \]

   **Web:**
   \[
   I_{c2} = \frac{1}{12} \times 60 \times 100^3 = \frac{1}{12} \times 60 \times 1000000 = 5000000 \, \text{mm}^4
   \]
   \[
   d_2 = 130 - 64.55 = 65.45 \, \text{mm}
   \]
   \[
   I_2 = 5000000 + 6000 \times 65.45^2 = 5000000 + 6000 \times 4283.20 = 5000000 + 25699200 = 30699200 \, \text{mm}^4
   \]

   **Total second moment of area \( I_{NA} \):**
   \[
   I_{NA} = I_1 + I_2 = 18176533.33 + 30699200 = 48875733.33 \, \text{mm}^4
   \]

Thus, the second moment of area of the cross-section \( I_{NA} \) is approximately \( 48.88 \times 10^6 \, \text{mm}^4 \).

Would you like more details on this solution, or do you have any specific questions?

### Additional Questions for Understanding:
1. How does the centroid location affect the second moment of area calculation?
2. What is the significance of the parallel axis theorem in calculating the second moment of area?
3. Can you explain why the second moment of area is important in structural engineering?
4. How would you calculate the second moment of area if the section was not symmetric?
5. What are the units of the second moment of area, and why is it important to keep track of units in these calculations?

### Tip:
Always double-check your calculations, especially when dealing with complex shapes and multiple components, to ensure accuracy in your final results.

A structural element, as shown in Figure, is used as a simply supported beam for a span of 10 m. It carries a distributed load of 120 kN/m and a concentrated load of 80 kN at mid-span. (a) Calculate the second moment of area of the cross-section of the element (INA). (b) The location and magnitude of the maximum bending moment (kNm).

Solve a beam mechanics problem involving a simply supported beam with distributed and concentrated loads. Calculate the second moment of area (moment of inertia) and determine the location and magnitude of the maximum bending moment. Suitable for undergraduate engineering students.

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