We were discussing “Moment of inertia for rectangular section”,
“Moment of inertia for the hollow
rectangular section” and similarly we have also seen “Moment of inertia for circular section”
and “Moment of inertia for the hollow circular
section" in our previous posts.

Let us consider one triangular section ABC as displayed in following figure. Let us assume that axis X-X is passing through the center of gravity and parallel to the base of the triangular section.

We have already derived the moment of inertia of the triangular section about its base and it will be as mentioned here.

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Today we will see here the method to determine the
moment of inertia for the triangular section about a line passing through the center
of gravity and parallel to the base of the triangular section with the help of
this post.

Let us consider one triangular section ABC as displayed in following figure. Let us assume that axis X-X is passing through the center of gravity and parallel to the base of the triangular section.

We will have
to determine the moment of inertia for the triangular section about axis XX
which is passing through the center of gravity and parallel to the base of the
triangular section.

b = Width of the triangular section ABC

h = Depth or height of the triangular section ABC

I

_{XX}= Moment of inertia of the triangular section about the axis XX which is passing through the center of gravity and parallel to the base of the triangular section
I

_{BC}= Moment of inertia of the triangular section about its base i.e. BC lineWe have already derived the moment of inertia of the triangular section about its base and it will be as mentioned here.

I

_{BC}= b.h^{3}/12
Area of triangle, A = b. h/2

Distance between C.G and Base of the triangular
section = h/3

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*Now we will determine the value or
expression for the moment of inertia of the **triangular section about the axis XX which is passing
through the center of gravity and parallel to the base of the triangular
section*

We will use here the concept of parallel axis
theorem in order to secure the value
or expression for the moment of inertia of the triangular section about the axis XX. Let us
recall the theorem of parallel axis which is related with the determination of
moment of inertia.

I

_{BC}= I_{G}+ A. (h/3)^{2}
I

_{G}= I_{BC}- A. (h/3)^{2}
I

_{G}= b.h^{3}/12 – (b. h/2). (h/3)^{2}
I

_{G}= b.h^{3}/12 – b. h^{3}/18###
**I**_{G}
= b.h^{3}/36

_{G}= b.h

^{3}/36

Do you have any suggestions? Please write in comment
box

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**Reference:**

Strength of material, By R. K. Bansal

Image Courtesy: Google

We will see another important topic i.e. Moment of
inertia of a uniform thin rod in the category of strength of material.

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