Question

For the section having width \( b \) and depth \( d \), the second moment of the area about an axis \( d/4 \) distance above the bottom of the area is ............

• A. \( \frac{bd^3}{48} \)
• B. \( \frac{bd^3}{12} \)
• C. \( \frac{7bd^3}{48} \)
• D. \( \frac{bd^3}{3} \)

Hint:

When calculating the second moment of area for a rectangular section about an axis not passing through the centroid, apply the parallel axis theorem to adjust for the distance of the axis from the centroid.

Answer 1

The Correct Option is C

The second moment of area (also known as the area moment of inertia) for a rectangular section about an axis a distance \( y \) from the centroid of the section is given by the formula: \[ I = \frac{bd^3}{12} - A y^2 \] where:
- \( b \) is the width of the section,
- \( d \) is the depth of the section,
- \( A \) is the area of the section.
For the given question, the axis is at a distance \( d/4 \) from the bottom of the area. The area of the section is \( A = b \cdot d \). The distance from the centroid (which is at \( d/2 \) from the bottom) to the axis \( d/4 \) is: \[ y = \frac{d}{2} - \frac{d}{4} = \frac{d}{4} \] Substitute the values into the formula: \[ I = \frac{bd^3}{12} - b \cdot d \cdot \left( \frac{d}{4} \right)^2 \] \[ I = \frac{bd^3}{12} - b \cdot d \cdot \frac{d^2}{16} = \frac{bd^3}{12} - \frac{bd^3}{16} \] Now, finding a common denominator: \[ I = \frac{4bd^3}{48} - \frac{3bd^3}{48} = \frac{7bd^3}{48} \] Thus, the second moment of area is \( \frac{7bd^3}{48} \).