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:

For hollow symmetric sections, simply use the difference formula: \[ I = \frac{a^4 - b^4}{12} \] Always remember to subtract the inner cut-out contribution from the outer solid shape.

Answer 1

Correct Answer: 490

Step 1: Recall the formula.
For a solid square section of side \(a\), the centroidal moment of inertia about the x-axis is: \[ I_x = \frac{a^4}{12} \] For a hollow square (box section), the net inertia is obtained by subtracting the inertia of the inner (cut-out) square from that of the outer square: \[ I_{net} = \frac{a^4 - b^4}{12} \] where \(a\) = outer side length, \(b\) = inner side length.

Step 2: Dimensions.
Outer side length: \[ a = 10 \, \text{mm} \] Thickness = 1 mm, so the inner square side = \(a - 2t\): \[ b = 10 - 2(1) = 8 \, \text{mm} \]

Step 3: Compute the fourth powers.
\[ a^4 = 10^4 = 10000 \] \[ b^4 = 8^4 = 4096 \]

Step 4: Net moment of inertia.
\[ I_{net} = \frac{a^4 - b^4}{12} = \frac{10000 - 4096}{12} \] \[ I_{net} = \frac{5904}{12} = 492 \, mm^4 \] Final Answer:
\[ \boxed{492 \, mm^4} \]