Question

The moment of inertia of a triangular section of base (b) and height (h) about an axis passing through its vertex and parallel to the base is ____________ times as that passing through its centre of gravity and parallel to the base.

• A. Twelve
• B. Nine
• C. Six
• D. Four

Hint:

First find the moment of inertia about the base, then shift it again to the vertex using the parallel axis theorem, being careful about which distance you use each time.

Answer 1

The Correct Option is B

Step 1: For a triangular section, the standard moment of inertia about the centroidal axis parallel to the base is \(I_{cg} = \dfrac{bh^3}{36}\).

Step 2: The centroid of a triangle lies at a distance h/3 from the base, so the distance from the centroid to the vertex, measured along the height, is \(h - \dfrac{h}{3} = \dfrac{2h}{3}\).

Step 3: To get the moment of inertia about the axis through the vertex, apply the parallel axis theorem using this distance. The area of the triangle is \(A = \dfrac{1}{2}bh\), so \(I_{vertex} = I_{cg} + A\left(\dfrac{2h}{3}\right)^2 = \dfrac{bh^3}{36} + \dfrac{1}{2}bh \times \dfrac{4h^2}{9} = \dfrac{bh^3}{36} + \dfrac{2bh^3}{9}\).

Step 4: Bring both terms to a denominator of 36, \(\dfrac{2bh^3}{9} = \dfrac{8bh^3}{36}\), so \(I_{vertex} = \dfrac{bh^3}{36} + \dfrac{8bh^3}{36} = \dfrac{9bh^3}{36} = \dfrac{bh^3}{4}\).

Step 5: The ratio asked for is \(\dfrac{I_{vertex}}{I_{cg}} = \dfrac{bh^3/4}{bh^3/36} = \dfrac{36}{4} = 9\).

Step 6: So the moment of inertia through the vertex is nine times that through the centroid, parallel to the base. Twelve, Six and Four come from using the wrong distance in the parallel axis step, such as h/3 or h instead of 2h/3.