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Let's break down the problem and solve it step by step. The column is constructed from four plates, each with dimensions 300 mm × 16 mm, made of A36 steel. Here's how we'll approach each part:

### Given:
- Four plates: each 300 mm (height) × 16 mm (thickness)
- A36 Steel
- Symmetrical cross-section as shown in the diagram.

### Required:
1. **Moment of Inertia along X and Y axes**
2. **Polar Moment of Inertia (about the center)**
3. **Section Modulus along X and Y axes**
4. **Radius of Gyration**

#### Step 1: **Moment of Inertia (I\_x and I\_y)**

For each of the plates, we can calculate the moment of inertia about its own centroid and then use the parallel axis theorem to find the total moment of inertia of the whole section.

1. **For vertical plates (about the X-axis):**
   - \( I_{x} \text{(vertical)} = \frac{1}{12}bh^3 = \frac{1}{12}(16)(300^3) \) mm\(^4\)
   - There are two vertical plates placed symmetrically.

2. **For horizontal plates (about the Y-axis):**
   - \( I_{y} \text{(horizontal)} = \frac{1}{12}bh^3 = \frac{1}{12}(16)(300^3) \) mm\(^4\)
   - These also have two plates placed symmetrically.

Since the cross-section is symmetrical, these results need to be summed.

#### Step 2: **Polar Moment of Inertia (J)**

The polar moment of inertia is given by:

\[
J = I_x + I_y
\]

#### Step 3: **Section Modulus (S\_x and S\_y)**

Section modulus is defined as:

\[
S = \frac{I}{c}
\]

Where \( c \) is the distance from the neutral axis to the outermost fiber.

#### Step 4: **Radius of Gyration (r\_x and r\_y)**

The radius of gyration is given by:

\[
r = \sqrt{\frac{I}{A}}
\]

Where \( A \) is the area of the cross-section.

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Do you want me to continue with the detailed calculations for each part? Also, feel free to ask any clarifications or more details about the next steps. 

### Additional Questions:
1. How does the moment of inertia impact the design of structural columns?
2. What is the significance of the polar moment of inertia in this context?
3. How do section modulus values relate to beam bending capacity?
4. What is the role of the radius of gyration in buckling analysis?
5. How does the material (A36 steel) influence these properties?

**Tip**: Always verify the symmetry and placement of components when applying the parallel axis theorem for moments of inertia.

The column shown in the figure is made from four 300 mm x 16 mm A36 Steel plates. Determine the following: a. Moment of Inertia along X and Y axes. b. Polar Moment of Inertia with respect to X and Y axes. c. Section Modulus, S_x and S_y. d. Radius of Gyration, r_x and r_y.

Learn how to calculate the moment of inertia, polar moment of inertia, section modulus, and radius of gyration for a steel column section made of four 300 mm x 16 mm A36 steel plates. Detailed steps and formulas are provided for structural engineering students.

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