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.

Today we will see here the method to determine the moment
of inertia for the triangular section about a line passing through 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 one line is passing through
the base of the triangular section and let us consider this line as line BC and
we will determine the moment of inertia for the triangular section about this
line BC.

b = Width of the triangular section ABC

h = Depth or height of the triangular section ABC

I

_{BC}= Moment of inertia of the triangular section about the BC line###
*Now
we will determine the value or expression for the moment of inertia of the ***triangular*** section about the
line BC*

*Now we will determine the value or expression for the moment of inertia of the*

*section about the line BC*

Let us consider one small elementary strip with
thickness dy and at a distance y from the vertex of triangular
section ABC as displayed in above figure.

*Let us determine first the area and moment of inertia of the*

*small*

*elementary strip of*

*triangular*

*section about the line BC*

Area of triangular
elementary strip, dA = Width of strip (DE) x Thickness of strip (dy)

Area of the triangular
elementary strip, dA = DE x dy

Let us find the value of DE by considering the two
triangular sections i.e. ABC and ADE and we can easily conclude that

DE/BC = y/h

DE/b = y/h

DE = b.y/h

Therefore we can write here area of triangular elementary strip, dA = (b.y/h) dy

Moment of inertia of the triangular
elementary strip about the line BC = dA. (h - y)

^{ 2}
Moment of inertia of the triangular
elementary strip about the line BC = (b.y/h) (h - y)

^{ 2}dy
Now in order to secure the moment of inertia of the triangular section ABC about the line BC, we will
have to integrate the above equation from 0 to h and therefore we can write
here the moment of inertia of triangular section about the base line and we
will have as mentioned here.

Moment of inertia of the triangular section about
the BC line.

Do you have any suggestions? Please write in comment
box

###
**Reference:**

Strength of material, By R. K. Bansal

Image Courtesy: Google

We will see another important topic i.e.Moment of inertia for the triangular section about a line passing through the center of gravity and parallel to the base line of the triangular section in the category of strength of material.

###

We will see another important topic i.e.Moment of inertia for the triangular section about a line passing through the center of gravity and parallel to the base line of the triangular section in the category of strength of material.

## No comments:

## Post a comment