We were
discussing “Area moment of inertia” and “The
difference between centroid and centre of gravity”,” and also we have seen
the “Basic
principle of complementary shear stresses” with the help of our previous
posts.

Now we are going further to start our discussion to understand the basic concept of radius of gyration with the help of this post.

Now we are going further to start our discussion to understand the basic concept of radius of gyration with the help of this post.

###
**Let
us first see here the basic concept of ****radius of gyration**

Radius of
gyration of a body or a given lamina is basically defined as the distance from
the given axis up to a point at which the entire area of the lamina will be
considered to be concentrated.

We can
also explain the radius of gyration about an axis as a distance that if square
of distance will be multiplied with the area of lamina then we will have area
moment of inertia of lamina about that given axis.

Let us see the following figure which indicates one lamina with area A. Let us assume that lamina, displayed here, is made with number of small elemental areas a

Let us see the following figure which indicates one lamina with area A. Let us assume that lamina, displayed here, is made with number of small elemental areas a

_{1}, a_{2}, a_{3}, a_{4}…etc. As we have considered above that the total area of the lamina is A, now we need to determine here the radius of gyration about an axis for the given lamina.
Where,

x

_{1}= Distance of the C.G of the area a_{1}from OY axis
x

_{2}= Distance of the C.G of the area a_{2}from OY axis
x

_{3}= Distance of the C.G of the area a_{3}from OY axis
x

_{4}= Distance of the C.G of the area a_{4}from OY axis
Similarly,
we will have

y

_{1}= Distance of the C.G of the area a_{1}from OX axis
y

_{2}= Distance of the C.G of the area a_{2}from OX axis
y

_{3}= Distance of the C.G of the area a_{3}from OX axis
y

_{4}= Distance of the C.G of the area a_{4}from OX axis
Area
moment of inertia about the OY axis = a

_{1}.x_{1}^{2}_{ }+ a_{2}.x_{2}^{2}+ a_{3}.x_{3}^{2}+ a_{4}.x_{4}^{2}_{ }+…….
I

_{yy}= a_{1}.x_{1}^{2}_{ }+ a_{2}.x_{2}^{2}+ a_{3}.x_{3}^{2}+ a_{4}.x_{4}^{2}_{ }+…….
I

_{yy}= Σ a.x^{2}
Let us
assume that entire area of the given lamina is concentrated at a distance k
from the reference axis i.e. OY axis and therefore area moment of inertia for
the entire area about the reference axis i.e. OY axis will be A.k

^{2}.
I

_{yy}= A.k^{2}
Similarly,
Area moment of inertia about the OX axis

I

_{xx}= a_{1}.y_{1}^{2}_{ }+ a_{2}.y_{2}^{2}+ a_{3}.y_{3}^{2}+ a_{4}.y_{4}^{2}_{ }+…….
I

_{xx}= Σ a.y^{2}
Let us
assume that entire area of the given lamina is concentrated at a distance k
from the reference axis i.e. OX axis and therefore area moment of inertia for
the entire area about the reference axis i.e. OX axis will be A.k

^{2}.
I

_{xx}= A.k^{2}
Do you
have any suggestions or any amendment required in this post? Please write in
comment box.

###
**Reference:**

Strength
of material, By R. K. Bansal

Image Courtesy: Google

Image Courtesy: Google

We will see another important topic i.e. state the perpendicular axis theorem with proof, in the category of
strength of material, in our next post.

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