Moment of inertia of triangular section having base 80 mm and hei
Moment of inertia of triangular section having base 80 mm and height 60 mm about axis passing through CG and parallel to base is
A. 15 × 10<sup>6</sup> mm<sup>4</sup>
B. 20 × 10<span style="position: relative; line-height: 0; vertical-align: baseline; top: -0.5em;">6</span> mm<sup>4</sup>
C. 480 × 10<sup>3</sup> mm<sup>4</sup>
D. 1440 × 10<span style="position: relative; line-height: 0; vertical-align: baseline; top: -0.5em;">3</span> mm<sup>4</sup>
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Right Answer is: C
SOLUTION
Concept:
The moment of Inertia of the triangular section about an axis through its Centroid and parallel to the base is
\(\Rightarrow {I_G} = {I_{x'x'}} = \frac{{b{h^3}}}{{36}}\)
Calculation:
Given:
base (b) = 80 mm, height (h) = 60 mm
The moment of Inertia of the triangular section about an axis through its Centroid and parallel to the base is
\(\Rightarrow {I_G} = {I_{x'x'}} = \frac{{b{h^3}}}{{36}}\)
\(I_G = \frac{{80 \times {{60}^3}}}{{36}} = 480 \times {10^3}\;m{m^4}\)
