Question
Download Solution PDFMoment of Inertia (MI) of a triangle of base 'b' and height 'h' about its base is expressed as
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
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
Applying parallel axis theorem we get
\({I_{xx}} = {I_{cc}} + A{\left( {\frac{h}{3}} \right)^2}\)
\({I_{cc}} = {I_{xx}} - A{\left( {\frac{h}{3}} \right)^2}\)
\(I_{cc}=\frac{{b{h^3}}}{{12}} - \frac{{bh}}{2} \times \frac{{{h^2}}}{9} = \frac{{b{h^3}}}{{36}}\)
\({I_{centroid}} = \frac{{b{h^3}}}{{36}}\)
Last updated on Jun 24, 2025
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