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Saturday, 4 March 2017

MOMENT OF INERTIA OF TRIANGULAR SECTION ABOUT CENTROIDAL AXIS

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 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
IXX = 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

IBC = Moment of inertia of the triangular section about its base i.e. BC line
 
We have already derived the moment of inertia of the triangular section about its base and it will be as mentioned here.
IBC = b.h3/12

Area of triangle, A = b. h/2
Distance between C.G and Base of the triangular section = h/3  

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.

IBC = IG + A. (h/3)2
IG = IBC - A. (h/3)2
IG = b.h3/12 – (b. h/2). (h/3)2
IG = b.h3/12 – b. h3/18

IG = b.h3/36

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 of a uniform thin rod in the category of strength of material.

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