We were discussing “The difference between centroid and centre of gravity”, “Relationship between young modulus of rigidity and Poisson's ratio” and “Elongation of uniformly tapering rectangular rod” and
we have also seen the “Basic principle of complementary shear
stresses” with the help of our previous posts.

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Now we are going further to start our discussion to
understand the basic concept of area moment of inertia with the help of this
post.

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**Let
us first see here the basic concept of area moment of inertia**

Area moment of inertia is basically defined as the sum
of second moment of area of individual sections about an 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

_{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 area moment of inertia for this 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
Total area of the lamina = Sum of all small elemental
areas

A = a

_{1 }+ a_{2 }+ a_{3 }+ a_{4 }+ ----------
Moment of area about the OY axis will be determined
by multiplying the area A with the perpendicular distance between the C.G and
axis OY i.e. X.

Moment of area about the OY axis = A .X

Let us determine the moments of all small elemental areas
about the OY axis and we will have following
equation.

Moments of all small elemental areas about the OY
axis = a

_{1}.x_{1 }+ a_{2}.x_{2}+ a_{3}.x_{3}+ a_{4}.x_{4 }+…….
Above equation will be termed as the first moment of
area about the OY axis and it is used to determine the centroid or centre of
gravity of the area of lamina.

If the first moment of area will again multiplied
with the vertical distance between the C.G and axis OY, then we will have
second moment of area i.e. A.X

^{2}.
By taking the definition of area moment of inertia
in to consideration, we can write the equation for the sum of second moment of area
of all small elemental areas about the OY axis and we will have following
equation.

Second moments of all small elemental areas 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}
Similarly, we will determine the sum of second
moment of area of all small elemental areas about the OX axis and we will have following
equation.

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}
Therefore, we can say here that area moment of
inertia about an axis could be calculated by taking the product of area and the
square of the perpendicular distance between the centre of gravity of the area and
that axis.

Therefore, we have following formula for determining
the area moment of inertia around the X axis and Y axis and it is as mentioned
here.

####
*I*_{xx}
= Σ a.y^{2}

*I*

_{xx}= Σ a.y^{2}####
*I*_{yy}
= Σ a.x^{2}

*I*

_{yy}= Σ a.x^{2}
Do you have any suggestions or any amendment
required in this post? 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.
Basic concept of radius of gyration in the category of strength of material
in our next post.

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