Engineering Mechanics

Prof. Manoj Harbola

Indian Institute of Technology, Kanpur
Module - 03
Lecture - 03
Properties of Surfaces - III
In the previous lecture, we have been talking about the first moment of a plane area and a
centroid. Continuing on that in this lecture we define some more mathematical quantities
and just work out some examples with them. The utility of defining such quantities
would be clear later, when you do rotation dynamics and so on. But in this lecture we are
just going to restrict to their definition and working them out. The quantities I will be
talking about would be 1.
(Refer Slide Time: 00:54)

Second moment of a plane area 2 transfer theorems transfer, while transfer theorems we
mean, how if we know the second moment of a plane area and one particular set of axis,
how do we transfer them to another set of axis, 3. We will be talking about relation
between second moment and product of area in particular.
We will be focusing on how second moments and products of area, which we will later
in the lecture. Change when we go from one set of axis to another set, which allotted
with respect to the other. And then we will be talking about polar moment of an area. So,
to begin with let us define what is second moment and product of an area and product.

(Refer Slide Time: 02:23)

Suppose, I am given an area, a plane area in x y plane, like this is the x axis, this is the y
axis. Then, the second moment I x x about the y axis is defined as I take a small area. Let
me make it in blue delta a multiply this by the perpendicular distance from the x axis
square. So, I x x is defined as take area delta A i multiply by its perpendicular distance
from the x axis and add it up. This is the second moment of this area with respect to the x
axis and this; obviously, goes to the integration y square d A.
Similarly, I y y that is the second moment about the y axis, is defined as the distance of
the area from the y axis is chosen and then we write this as summation x i square delta Ai
or limit of integration this becomes x square d A. These are just mathematical definitions
and then the product of this area is defined as integral x y d A and I am going to call this
I x y equals this. So, we have defined the second moment of a plane area and product of
a plane area. Let us now work out some examples of how to calculate these.

(Refer Slide Time: 04:41)

As a first example I take a square of side a, and calculate for it the moment of moment of
area about the x axis, and about the y axis and its product of area I x x is equal to by
definition y square d A. To calculate d A, I choose a strip at height y because for this
entire strip the moment of the area is going to be the same. So, this becomes integral y
square a d y and y changes from minus a by 2 to a by 2. And therefore, I x x is going to
be a times one third y cube minus a by 2 to a by 2 or this comes out to be a raise to 4
over 12.
Similarly, to calculate I y y this is the square. I choose a strip like this and calculate I y y
as integral x square d A, with this case would become integral, this is at distance x. So,
integral x square a d x from minus a by 2 to a by 2, and this also in this case by
symmetry would come out to be a raise to 4 over 12 third for the product of area.

(Refer Slide Time: 06:42)

You see that I x y is equal to integral x y d A minus a by 2 to a by 2, and if I choose a

small area d A here. This will be equal to integral minus a by 2 to a by 2 x d x integral
minus a by 2 to a by 2 y d y and by and the symmetry of the wave of the function x and y
this goes to 0. So, for a square of side a I x x equals I y y equals a raise to 4 over 12 and I
x y equals 0.
(Refer Slide Time: 07:51)

As a second example, let me take a rectangle of length a, and width b placed

symmetrically about the origin. So, this side is precisely a by 2. So, is this is not made to

scale please understand, this is a by 2 a by 2 divided on both sides, and I want to

calculate I x x for that again I choose a strip here of width d y because for this entire strip
y square is the same and calculate integral y square a d y is that small area d A and y
changes from minus b by 2 to b by 2. And this comes out to be a over three y cubed
minus b by 2 to b by 2 or a b cubed over 12.
Similarly, when I calculate I y y, for that I choose a strip parallel to y axis and calculate
integral x square d A, which now becomes x square b d x x varying from minus a by 2 to
a by 2, and this comes out to be a cubed b over 12, how about I x y? Again you will see
by symmetry because the area is equally distributed on the negative side and the positive
side of the y and x axis comes out to be 0. So, you have already also calculated the
moment second moment and product of area for a rectangle. As the third example will
make it slightly more complicated and I wish to calculate.
(Refer Slide Time: 10:09)

Second moment and product of area for an ellipse. The quarter of an ellipse here with
semi major axis a, and semi minor axis b. The equation for the ellipse is x square over a
square plus y square over b square is equal to 1. To calculate I x x, which is integral y
square d A, I choose a strip parallel to the x axis of width d y and write this quantity as y
square. The length of the strip, let us call it delta x d y.
My job is to calculate delta x and y varies from 0 to b. From the equation x lower is 0
that is this point and I also know from the equation of the ellipse that x is equal to a over

b square root of b square minus y square. Therefore, I x x is going to be 0 to b a over b y

square a square root of b square minus y square d y. This is what, to what we got to
integrate to do this, we substitute y equals b sin of theta by doing.
(Refer Slide Time: 12:03)

So, I get I x x, which is a over b integral y square the square root of b square minus y
square d y 0 to b as a over b integral 0 to pi by 2 y square is b square sin square theta b
square minus y square root, is going to be b cosine theta and d y is again b cosine theta d
theta. And this gives me this b cancels, and I get a b cube integral sin square theta cosine
square theta d theta 0 to pi by 2. This can be written as a b cubed divided by 4 0 to pi by
2 sin square 2 theta d theta, which is a b cube over 4 integral 0 to pi by 2 1 minus cosine
2 theta over 2 d theta. Doing this integral, when we do cosine 2 theta integral that gives
me 0 and the first integral is going to be pi by 4 and therefore, I get for this shape where
this is a.

(Refer Slide Time: 13:36)

This is b I x x equals pi a b cube over 16. Similarly, I can calculate I y y, which is

integral x square d A, where now I am going to choose a strip parallel to the y axis. The
integral is very similar to what we just now did, and this will come out to be pi a cube b
over 16, how about I x y? To calculate I x y I have to calculate the integral x y d A,
where d A I take 2 be a small area at point x and y.
This can therefore, written as x d x y d y x square is from 0 to a. For any given x, the
given x here y varies from 0 to a y upper, where y upper is given by the equation x
square by a square plus y square over b square equals 1, and that gives for any given x,
that y is going to be b over a the square root of a square minus x square.

(Refer Slide Time: 15:28)

And therefore, I x y for this is going to be integral x d x 0 to a integral 0 to b over a is

square root of a square minus x square y d y, which is nothing but one half 0 to a x d x b
square over a square times a square minus x square. This is a standard integral. So, this
gives me the answer a square b square divided by 8. Therefore, what we have determined
is that for the quarter of an ellipse with semi major axis a and semi minor axis b, I x x is
pi over 16 a b cubed, and I y y is pi over 16 a cube b, and I x y is a square b square over
8.
I will now leave it for you as an easy exercise to calculate I x x and I y y. That is the
second moment of an area with respect to x and y axis for the entire ellipse, and also
show that I x y for the entire ellipse is going to be 0, because of the symmetry between x
and y axis. Using the second moment of area, we can also define, when we call the
radius of gyration.

(Refer Slide Time: 17:08)

By gyration you can already see that these quantities are going to be useful, when we
describe rotation. The radius of gyration of k x of an area A, about the x axis is defined
through the relationship A k x square equals I x x equals integral y square d A. And
radius of gyration for about the y axis is defined as A k y square equals I y y equals
integral x square d A. Thus for example, when we look at a rectangle of length a, and
width b, we have already calculated that I x x is equal to a b cubed over 12. And
therefore, k x square is going to be a b cubed over 12 a b or equals b square over 12.
Similarly, k y square is going to be a square over 12, and from these area of gyration
about the x axis and the y axis can be calculated. Having defined these quantities the
second moment of inertia and the product of inertia. We now describe relationship
between the second moments of an area about a set of axis passing through the centroid
of the body, and another set x y axis, which are parallel to those passing through the
centroid.

(Refer Slide Time: 19:10)

Thus suppose, I have an area the centroid with respect to a given set of axis x and y with
origin at O such that the coordinates of a centroid are x 0 and y 0. Let us chose another
set of parallel axis x prime and y prime passing through the centroid, X prime is parallel
to x and y prime is parallel to y. And we want to now calculate show that I x x and I x
prime x prime are related by a very simple theorem. So, are the other moments? Thus I x
x is equal to y square d A.
So, suppose I have a small area d A here. This is y, which is also equal to y is equal to y
prime, where y prime is the y coordinate of the same area with respect to x prime and y
prime axis plus y 0 square d A. This is equal to y prime square d A plus y 0 square d A
plus 2 y 0 integral y prime d A. Notice that y prime square d A is the I x prime, x prime.
That is the second moment of inertia about the x axis passing through the centroid, and y
prime d A is nothing but area times the y y coordinate of the centroid, in the centroid
frame.
So, this is going to be 0, if I calculate the centroid with the origin of the centroid. The
coordinates of centroid are going to come to 0. And therefore, I see that I x x is equal to
y prime square d A. That is I x prime, x prime plus y 0 square times entire area. This
way, if I know the second moment of inertia of a body about an axis passing through a
centroid. I can easily calculate the second moment of inertia on the same body with
respect to an axis, which is parallel to the first axis were displace by amount y 0.

(Refer Slide Time: 22:14)

In the same manner, we can now calculate I y y which is equal to integral x square d A,
which I can write as x prime plus x 0 square d A. Let me for convenience show in the
picture, again what we are talking about? This is the body; this is the centroid set of axis
x prime y prime. This is x y 0, this is x 0, this is y 0. If, I choose an area here, this is x
and this is x prime, x is x prime plus x naught. So, this can be written as equal to integral
x prime square d A plus x 0 square integral d A plus 2 x 0 x prime d A.
This again has the same logic as we applied earlier that this is the x coordinate of the
centroid in the coordinate system, which has this origin at the centroid itself. So, this is 0.
So, I get I y y is equal to I y prime y prime plus x naught square A. So, it is as if the
entire area is concentrated at the centroid plus whatever the moment second moment of
area is allowed the axis passing through the centroid.

(Refer Slide Time: 24:02)

Next let us calculate the product of area in making these axis x prime y prime x y 0, and I
take an area here. The product is I x y, which is equal to integral x y d A, and I substitute
for x and y as x prime plus x 0 y prime plus y 0 d A, which comes out to be x prime y
prime d A plus x 0 integral y prime d A plus y 0 integral x prime d A plus x 0 y 0 d A.
Again by the arguments that we have used earlier these 2 terms drop to 0.
And therefore, I x y is equal to I x prime y prime plus x naught y naught times per area.
So, what we have learnt is, if we know the second moment of inertia and the product of
inertia about a set of axis passing through the centroid. I can calculate about any other set
of axis, which are parallel to those passing through the centroid. Let us summarize these.

(Refer Slide Time: 25:35)

So, for an area whose second moment and product of area are known about the axis
passing through the centroid. I have in general I x x equals I x prime x prime plus y 0
square times the area, where y 0 is the coordinate of the centroid. Y coordinate of the
centroid I y y is equal to I y prime, y prime plus x 0 square times the full area, and I x y
equals I x prime, y prime plus x 0 y 0 times entire area. These are known as transfer
theorems, using these I can transfer the moment of area or the product of area from one
coordinate system to another. As an example of the application of transfer theorem.
(Refer Slide Time: 26:46)

Let us take case of an ellipse with its length being 2 a, and this being 2 b with its centroid
at point x 0 y 0. And calculate its moment of area second moment of area and product of
area with respect to the x y axis shown here. So, by transfer theorems I have I x x equals
I x prime x prime plus y naught square times the area of the ellipse I y y. Similarly is I y
prime y prime plus x naught square times the area of ellipse and I x y is equal to I x
prime y prime plus x 0 y 0 times the area of the ellipse, where I x x prime, x prime is the
second moment of area with respect to the x prime axis parallel to the x axis passing
through the centroid. I y prime y prime is the second moment of area with respect to the
y prime axis parallel to the y axis and passing through the centroid.
Previously we have calculated I x x as pi over 16 times a b cube for quarter of an ellipse
like this. So, for the full ellipse and this I left as an exercise for you there this is going to
be 4 times as much. So, this is going to be pi over 16 a b cubed, which is pi over 4 a b
cubed.
(Refer Slide Time: 28:56)

Similarly, I y y prime for the ellipse is going to be 4 times pi over 16 a cubed b, which is
pi over 4 a cubed b and I x prime y prime is 0. Therefore, for this ellipse we will have I x
x as pi by 4 a b cubed, which is the second moment of area about the x prime axis
passing through the centroid plus y 0 square times pi a b, where pi a b is the area of the
ellipse.

I y y is going to be pi over 4 a cubed b plus x 0 square pi a b, and I x y is going to be 0,

which is the I x prime y prime by symmetry is the 0 for axis passing through the centroid
plus x 0 y 0 pi a b. So, using transfer theorems we could calculate the second moment of
area and the product of area, when it was given about the centroid, so far what we
considered in the transfer theorem is the product.
(Refer Slide Time: 30:28)

And second moment of area, when the centroid is displaced with respect to the origin of
a given system. Now, we want to look at another transformation, where given an area,
and its second moment of inertia and product of inertia about in set of axis x y. We wish
to calculate it about another set x prime y prime, which is rotated with respect to the first
strip by an angle theta. Let us see, what happens in this case?
So, if I want to calculate I x prime y x prime, x prime in the rotated set. This is going to
be equal to integral y prime square d A, I y prime y prime is going to be equal to x prime
square d A. And I x prime y prime in the second frame is going to be x prime y prime d
A, where we chose a small area d A, whose coordinates in the origin system are x and y.
In the new system x prime y prime, we can find out the relationship of I x x prime, with
those similar quantities in the unrotated frame by a simple transformation laws of x and y
coordinates. So, let us do that now.

(Refer Slide Time: 32:18)

So, what we given is an area and we used to calculate its second moment of area and
product of area with respect to a set of axis x prime y prime, when they are given in x
and y. We know from our previous lectures that x prime for a given point is equal to x
cosine of theta plus y sin of theta. Similarly, y prime is equal to minus x sin of theta plus
y cosine of theta using these let us find what I x prime x prime is from the previous slide
we know this is equal to y prime square d A, where d A is a small area.
Chosen Y prime square is going to be equal to integral minus x sin theta plus y cosine of
theta square d A, which I can write as x square d A integral sin square theta plus integral
y square d A cosine square theta minus 2 x y d A sin theta cosine of theta. But x square d
A is nothing but I x x y square I y y. Sorry, I y y y square d A is nothing but I x x, and x
y d A is nothing but I x y. Therefore, I can write this quantity as let us go to next page.

(Refer Slide Time: 34:31)

I x prime x prime as I y y sin square theta plus I x x cosine square theta minus I x y 2 sin
theta cosine theta, which can be written as I y y divided by 21 minus cosine of 2 theta
plus I x x divided by 2 1 plus cosine of 2 theta minus I x y sin of 2 theta, which is
nothing but I x x plus I y y divided by 2 plus I x x minus I y y divided by 2 cosine of 2
theta minus I x y sin of 2 theta. Thus, if I know the second moment of area and product
of area in one frame I can calculate it in the rotated frame. Let us do the same exercise
for I y y prime.
(Refer Slide Time: 36:00)

Y prime y prime, which is going to be equal to x prime square d A, but I know x prime is
equal to x cosine theta plus y sin of theta. And therefore, I can write this as integral x
square d A cosine square theta plus integral y square d A sin square theta plus 2 integral
x y d A sin theta cosine of theta, which is this is nothing but I y y. This is nothing but I x
x, and this is nothing but I x y. So, this whole thing can be written as I y y over 2 1 plus
cosine 2 theta plus I x x over 2 1 minus cosine 2 theta plus I x y sin of 2 theta., which is
nothing but I x x plus I y y divided by 2 minus I x x minus I y y divided by 2 cosine 2
theta plus I x y sin of 2 theta.
(Refer Slide Time: 37:36)

First we calculate I x prime y prime, which is nothing but x prime y prime d A, which is
equal to integral x cosine theta plus y sin of theta times minus x sin theta plus y cosine
theta d A. It comes out to be integral minus x square sin theta cosine theta plus x y cosine
square theta minus x y sin square theta d A plus y square sin theta cosine theta d A. This
is also d A x square d A is nothing but minus I y y.
This can be written as sin 2 theta divided by 2 plus y square d A is I x x sin 2 theta
divided by 2 and x y d A is nothing but I x y cosine square theta minus sin square theta is
cosine 2 theta. So, I x prime y prime is nothing but I x x minus I y y divided by 2 sin 2
theta plus I x y cosine of 2 theta. Let us summarize what we are looking for, is if we
know for a body.

(Refer Slide Time: 39:27)

The products and second moments of inertia in one particular frame, how about its
values in the rotated frame. In the rotated frame, let me now write it in blue I x prime x
prime is nothing but I x x plus I y y divided by 2 plus I x x minus I y y divided by 2
cosine 2 theta minus I x y sin of 2 theta. Similarly, I y prime y prime is going to be I x x
plus I y y divided by 2 minus I x x minus I y y divided by 2 cosine 2 theta plus I x y sin
of 2 theta.
And I x prime y prime is going to be equal to I x x minus I y y divided by 2 sin of 2 theta
plus I x y cosine of 2 theta, what these transformation laws give me? It is if I am given
the second moment and product of area about a set of axis. I can calculate about any
other set of axis, which is rotated with respect to the first set of axis. Let me just illustrate
this thing by couple of examples, which are very interesting.

(Refer Slide Time: 41:21)

Suppose, I take a circle circular area for a circular area, no matter how I chose my rotated
set of axis. Let us take third one, like this the circle always looks the same. And
therefore, I x x and I y y should always come out to be the same no matter what cosine
theta or sin theta is, and I x y should always come out to be 0. Let us see, if that happens.
So, I x x we saw already is I x prime x prime is I x x plus I y y divided by 2 plus I x x
minus I x y divided by 2 cosine of 2 theta minus I x y sin of 2 theta.
Now, for a circular area I x x is equal to I y y and I x y is 0. And therefore, I x prime x
prime is going to be equal to I x x, and I y prime y prime is also going to be equal to I x x
equals I y y. And I x prime y prime, which is equal to I x x minus I y y divided by 2 sin
of 2 theta plus I x y cosine of 2 theta is also going to be 0. This is expected for a circle,
what is very interesting that is the same thing comes out to be true for a square. Let us
look at that case.

(Refer Slide Time: 43:19)

So, if I take a square of side a, we have already calculated that I x x for such a square is a
raise to 4 over 12. So, is I y y where these are the x and y axis, and I x y is 0. The fact
that I x x and I y y are equal, and I x y are 0 makes these quantities the same no matter,
which other frame we look at. So, let me write it in red I x prime x prime, which is equal
to I x x plus I y y divided by 2 plus I x x minus I y y divided by 2 cosine 2 theta plus I x
y sin of 2 theta. This is not plus this is minus is going to be equal to I x x again.
Similarly, I y prime y prime is going to be equal to I y y and I x prime y prime is equal to
I x x minus I y y divided by 2 sin 2 theta plus I x y cosine 2 theta. This is always going to
come out to be 0, no matter how much you rotate the axis y. So, for a square about any
set of axis I x prime x prime is always equal to a raise to 4 divided by 12 I y prime y
prime is always equal to a raise to 4 divided by 12, and an I x prime y prime is always 0.
Having given these two examples, I use these transformations to define something called
the principal.

(Refer Slide Time: 45:20)

Set of axes. So, given an area I look for those set of axes, let me call them x and y. So,
that I x y to 0 or rather given them a special name I x bar y bar. So, that I x bar y bar is 0,
how do we accomplish that since we already known that I x bar y bar is going to be equal
to I x x minus I y y divided by 2 sin of 2 theta plus I x y cosine of 2 theta. This implies
that if I choose rotate the new set of axes thus a tangent 2 theta is equal to 2 I x y divided
by I y y minus I x x.
I will get new I x bar y bar is equal to 0. Such a set of axis where the product of area
vanishes is known as the principles of set of axes. And you can see from the construction
that you can always find one set of axes because tangent 2 theta is varies from minus
infinity to plus infinity, when always find a set of axes where the product of area would
be 0. An interesting fact about the principle set of axis is that about.

(Refer Slide Time: 47:06)

The principle set of axes the one product of area vanishes and 2 the moment or second
moment of area is maximum about one axis. Say, the x axis and minimum about the
other one, if it is maximum about the x axis, the other one is going to be y axis. Let us
see, how does that come about?
So, let us look at I x bar x bar, which is I x x plus I y y divided by 2 plus I x x minus I y y
divided by 2 cosine 2 theta minus I x y sin 2 theta. And asked for a new frame such that I
x bar x bar is a maximum. So, for that I got to do I x bar x bar over d theta is equal to 0,
and when I do that. Here, this implies that I x x minus I y y divided by 2 times minus 2
sin 2 theta minus 2 I x y cosine 2 theta is equal to 0, and that immediately gives me.

(Refer Slide Time: 49:14)

The tangent of 2 theta is equal to I x y times 2 divided by I y y minus I x x. So, when I

accomplished by this rotation the fact that the product of area vanishes at the same time
it maximizes or minimizes the moment of area, this equation has 2 solutions. Suppose,
one of the solutions is theta equals alpha, then theta equals alpha plus pi by 2 is also a
solution. So, by rotating it by angle alpha I maximize or minimize the moment of inertia
about that particular axis. You can show that about the axis at alpha plus pi by 2. It will
be the other way, if it maximizes at alpha at alpha plus pi by 2, it will minimize and the
and vice versa.
So, we found a set of axis principle such that not only the product of inertia vanishes the
second moment of area is also either maximum or minimum, if it is maximum about the
x axis about the other y axis it becomes a minimum. If, it is minimum about the x axis it
becomes maximum about the other axis the y axis. Have been made this point, this point
let me now define something for you, which is known as the polar moment of area.

(Refer Slide Time: 50:48)

This is quite usually written as J, which is nothing but I x x plus I y y. And therefore, is
equal to integral x square plus y square d A or r square d A. Given any area r square for a
small area chosen is independent of which set of axes, we are talking about. So, this is
independent of a set of the set of axes chosen.
And through this discussion you also see that for a square the any set of axes is the
principle set of axes because as we have seen earlier the principle set of axes gives
product of area 0, and second moment of area maximum or minimum for a square, any
set of axes gives you product of area 0. So, therefore, any set of axes chosen for a square
or a circle is the principle axes, what we have covered in this lecture. So, far is the
second moment and product of an area a related quantity, which we will talk about in
later lectures. We will discuss dynamics of rigid bodies would be the moment of inertia
and product of inertia, and we will be using it, then in describing the rotational motion of
a rigid body.