Moment of Inertia of a Triangle

MOI of a triangle can be stated in various frames of discussion. These techniques are of three types, such as:

1. The passage of an axis via the centroid.

2. The passage of the line through the base.

3. The axis which is perpendicular to the base.

We will look at each expression below.

1. The Passage of an Axis via the Centroid

The picture is showing a triangle and a line that is passing through the centroid. 

This line is parallel to the base of the triangle. 

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The moment of inertia is expressed as:

I = bh3 / 36

Where, 

b = base width

h = height

2. The Passage of the Line through the Base

If the passage of the line is through the base, then the moment of inertia of a triangle about its base is:

I = bh3 / 12

Do you know about the Parallel Axis Theorem?

We can calculate the moment of inertia of any figure. 

The theorem of parallel axis states that the moment of inertia of a body about an axis parallel to an axis passing through the center of mass is equal to the sum about an axis passing through the center of mass, and product of mass and square of the distance between the two axes.  

We can also write the parallel axis theorem, which is given below:

I' = I + Ad²

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Here,

I' = the moment of inertia about an arbitrary axis

I = the moment of inertia about a centroidal axis which is parallel to the first one

d =  distance within the two parallel axes and the area of the shape 

Where, area = bh/2 in case of a triangle

3. The Axis Perpendicular to its Base

The method to calculate the moment of inertia of a triangle at the time of its axis is perpendicular to its base is mentioned below.

Firstly, we require that the line y’y’ in the triangle is used in dividing the whole triangle into two right triangles, respectively A and B. 

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Here in the picture, the colors are different for both triangles, and b1 and b2 are the base, and ‘h’ is the common height of the triangles.

However, for both, the moment of inertia will be;

Iy’ = hb13 / 12 + hb13 / 12

We need to consider that b2 = b – b1 

Here, the parallel line (axis) yy/ through the centroid 

Distance from y’-y’ = ⅔ (b2 / 2 – b1) 

By doing so, we can simply manipulate the moment of inertia value ly’. 

Here is the application of the parallel axis theorem to calculate ly’. 

So we just need to memorize the formula of the MOI of ly’

Iy’ = hb / 36 (b2 – b1 b + b12)

How to Find the Moment of Inertia of a Triangle?

Let’s find the moment of inertia of the triangle with proper derivation.

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dA = x dy

From the similar triangles,

 \[\frac{x}{b}\] =  \[\frac{h-y}{h}\]

x = \[\frac{b}{h}\](h-y)

We can now calculate the moment of inertia of a triangle w.r.t. its base.

 \[I_x\] =  \[ \int y^2dA\]

= \[\int_{0}^{h}y^2xdy\]….(1) 

By placing the value of x in the eq (1), we get:

= \[\int_{0}^{h}y^2\frac{b}{u}(h-y)dy\]

= \[\frac{b}{h}\int_{0}^{h}(y^2h -y^3)dy\]

= \[\frac{b}{h}|y^3\frac{h}{3} - \frac{y^4}{4}|_0^h\]

=  \[\frac{b}{h}(\frac{h^4}{3} - \frac{h^4}{4})\]

= \[\frac{bh^4}{h}(\frac{1}{3} - \frac{1}{4})\]

So, after solving it, we can conclude that,

Ix = bh3 / 36

Area Moment of Inertia of a Triangle

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Enlarging the selected area in the triangle we will get:

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 \[I_n\] = \[\frac{1}{12}bh^3 + bhd^2\]

y(x) = \[\frac{b}{a}x\]

In = \[\frac{1}{12}(\frac{b}{a}x)^3dx + (\frac{b}{a}x)(\frac{b}{a}\frac{x}{2} - \frac{b}{3})^2dx\]

Itotal = \[\int_{0}^{a}\frac{1}{12}(\frac{b}{a}x)^3dx + (\frac{b}{a}x)(\frac{b}{a}\frac{x}{2} - \frac{b}{3})^2dx\]

I = \[\frac{1}{36}ab^3\]

Moment of Inertia of a Right-angled Triangle

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Y = mx + b

m = slope = - h/a

b = y intercept = 2h/3

y = \[-\frac{hx}{a}+ \frac{2}{3}h\]

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dA = x dy

Ix = \[\int y^2dA\] = \[\int y^2xdy\] 

y = \[-\frac{h}{a}x+ \frac{2}{3}h\]

So, x = \[\frac{2}{3}a - \frac{a}{h}y\]

Ix =  \[\int_{\frac{-h}{3}}^{\frac{2h}{y}2}(\frac{2}{3}a - \frac{a}{h}y)dy\]

= \[\frac{ah^3}{36}\]

So, IG =  \[\frac{ah^3}{36}\]

Moment of Inertia of a Triangle about its Vertex

We all know that the moment of inertia of a triangle about its center of gravity can be written as

IG =  \[\frac{ah^3}{36}\]

Let’s consider a triangle ABC

Here, Base = b, Height = h

Consider an elemental strip PQ at a distance of ‘x’ from vertex ‘A’

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Moment of inertia of the PQ about base BC will be

(h – x)2 da……(1)

da = PQ * dx…..(2)

From geometrically similar triangles of ABC and APQ

We have derived the equation as

PQ / BC = x / h

Here, BC can be written as

PQ / b = x / h

PQ = bx / h

By substituting PQ in equation (2)

da = PQ * dx

da = (bx / h) dx

Now, substituting da in equation (1)

Moment of inertia of the strip PQ about base BC = (h – x)2 da

= (h – x)2 da

= (h – x)2 (bx / h) dx

So, MOI for the whole triangle can be obtained by applying limits “0 to h”

IBC =  \[\int_{0}^{h}(h-x)^2\frac{bx}{h}dx\]

= \[ \frac{b}{h}\int_{0}^{h}(h^2 + x^2 -2hx)xdx\]

= \[\frac{b}{h}[\frac{h^2x^2}{2}+\frac{x^4}{4}-\frac{2hx^3}{3}]_0^h\]

= \[\frac{b}{h}[\frac{h^4}{2}+\frac{h^4}{4}-\frac{2h^4}{3}]\]

= \[\frac{bh^4}{h}[\frac{1}{2}+\frac{1}{4}-\frac{2}{3}]\]

By solving the above, we will get;

IBC = bh3 / 12

From parallel axes theorem, MOI about the parallel axis can be written as

IBC = IG + Ad2

IG = IAB – Ad2

=  \[\frac{bh^3}{12} - \frac{bh}{2}[\frac{h}{3}]^2\]

=  \[\frac{bh^3}{12} - \frac{bh^3}{18}\]

  IG =  \[\frac{bh^3}{36}\]