Moment of Inertia (MI) of a triangle of base 'b' and height 'h' about its base is expressed as

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  1. \(\frac{b^2h}{12}\)
  2. \(\frac{bh^3}{12}\)
  3. \(\frac{bh^3}{36}\)
  4. \(\frac{b^3h}{36}\)

Answer (Detailed Solution Below)

Option 2 : \(\frac{bh^3}{12}\)
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Concept:

Moment of inertia of a triangle with base b and height h

\({I_{base}} = \frac{{b{h^3}}}{{12}}\)

Parallel axis theorem

  • Parallel axis theorem is used to find a moment of inertia about an axis which is at some distance from the centroidal axis and parallel to centroidal.
  • Suppose an axis is at distance d from the centroidal axis and parallel to the centroidal axis.

According to the parallel axis theorem, the moment of inertia about that axis is given by,

\({I_{about~axis~parallel~to~centroida~laxis}} = {I_{about~centroidal~axis}} + A{d^2}\)

Moment of inertia about the centroidal axis of the triangle

F1 Ashik Madhu 14.08.20 D15

Applying parallel axis theorem we get

\({I_{xx}} = {I_{cc}} + A{\left( {\frac{h}{3}} \right)^2}\)

\({I_{cc}} = {I_{xx}} - A{\left( {\frac{h}{3}} \right)^2}\)

\(I_{cc}=\frac{{b{h^3}}}{{12}} - \frac{{bh}}{2} \times \frac{{{h^2}}}{9} = \frac{{b{h^3}}}{{36}}\)

\({I_{centroid}} = \frac{{b{h^3}}}{{36}}\)

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