write the moment of inertia of the triangle section about centroidal Axis
Answers
The moment of inertia of a triangle with respect to an axis passing through its centroid, parallel to its base, is given by the following expression:
I=bh^2/36
where b is the base width, and specifically the triangle side parallel to the axis, and h is the triangle height (perpendicular to the axis and the base).
The moment of inertia of a triangle with respect to an axis passing through its base, is given by the following expression:
I=bh^2/12
This can be proved by application of the Parallel Axes Theorem (see below) considering that triangle centroid is located at a distance equal to h/3 from base.
The moment of inertia of a triangle with respect to an axis perpendicular to its base, can be found, considering that axis y'-y' in the figure below, divides the original triangle into two right ones, A and B. These triangles, have common base equal to h, and heights b1 and b2 respectively. Thus their combined moment of inertia is:
Iy′=hb1^2/12+hb2^2/12
Taking into account that b2=b−b1 and that centroidal parallel axis y-y is at a distance 23(b2−b1) from y'-y' makes possible to find the moment of inertia Iy, using the Parallel Axes Theorem (see below). After algebraic manipulation the final expression is:
Iy=hb/36(b^2−b1b+b1^2)