Calculation of Moment of Inertia of Triangular Section

The moment of inertia of a triangular section with respect to an axis through its centroid can be calculated using the following formula:

I = (bh^3)/36

Where,
b = base width of triangular section
h = height of triangular section

Explanation

The moment of inertia is a measure of an object's resistance to changes in its rotational motion. It depends on the shape of the object and the axis of rotation. In this case, we are calculating the moment of inertia of a triangular section with respect to an axis passing through its centroid.

To calculate the moment of inertia, we need to consider the shape of the triangular section and the distribution of its mass. The centroid is the point where the mass of the triangular section is concentrated. It is located one-third of the way from the base to the top of the triangle.

We can divide the triangular section into small vertical strips of width dx, each having a height y. The moment of inertia of each strip with respect to the axis passing through the centroid is given by the formula:

dI = (1/12)*dx*y^3

The total moment of inertia of the triangular section can be obtained by integrating this expression over the entire height of the triangle:

I = ∫[0,h] (1/12)*x*(h-x)^3 dx

This integral can be simplified using integration by substitution. Let u = h-x, then du = -dx. The integral becomes:

I = (1/12)*∫[0,h] u^3 (h-u)^3 du

This integral can be further simplified using the binomial expansion:

I = (1/12)*∫[0,h] (u^6 - 3h*u^5 + 3h^2*u^4 - h^3*u^3) du

I = (1/12)*[(1/7)*h^7 - (3/6)*h^6 + (3/5)*h^5 - (1/4)*h^4]

I = (bh^3)/36

Therefore, the moment of inertia of a triangular section with respect to an axis passing through its centroid is (bh^3)/36.