Question

Consider a beam with a square box cross-section as shown in the figure. The outer square has a length of 10 mm. The thickness of the section is 1 mm. The area moment of inertia about the x-axis is ........... mm\(^4\) (in integer). 

Hint:

The area moment of inertia for a hollow section (like a square box) can be calculated by subtracting the moment of inertia of the inner section from the outer section.

Answer 1

Step 1: The area moment of inertia for a rectangular section about an axis is given by: \[ I_x = \frac{1}{12} b h^3 \] where \( b \) is the base (width) and \( h \) is the height. 
Step 2: The given section is a square box. The outer square has a side length of 10 mm, and the thickness of the section is 1 mm. This means the inner square has a side length of \( 10 - 2 \times 1 = 8 \) mm. The area moment of inertia for the square box is the difference between the moment of inertia of the outer square and the inner square. So, we calculate the area moment of inertia for both squares. 
For the outer square (side = 10 mm): \[ I_{{outer}} = \frac{1}{12} \times 10 \times (10)^3 = \frac{1}{12} \times 10 \times 1000 = 833.33 \, {mm}^4 \]
For the inner square (side = 8 mm): \[ I_{{inner}} = \frac{1}{12} \times 8 \times (8)^3 = \frac{1}{12} \times 8 \times 512 = 341.33 \, {mm}^4 \] Step 3: The area moment of inertia for the box section is the difference: \[ I_x = I_{{outer}} - I_{{inner}} = 833.33 - 341.33 = 492 \, {mm}^4 \] Step 4: Therefore, the area moment of inertia is approximately 492 mm\(^4\), which lies between 490 and 494, as per the given range.