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Unit -3
Proprieties of areas and solids.
Introduction:-The basic computations in engineering and
mathematics relate to determination of length, area, volume, mass
etc. we may define all these parameters as physical properties of
objects.
Centroid and moment of inertia are two important properties of a
section, which are required frequently in the analysis of many
engineering problems.
Wewill discussfollowing parameters in detail.
i) Centroid, ii) Center of mass,
iii) Center of gravity, iv) moment of area,
v) Mass moment of inertia vi) Radius of gravity ration etc.
Centroid and Center of Gravity
The physical object may be one-dimensional, two-
dimensional or three-dimensional. In case of a plane geometric
Figure or a plane area of we can identify a point where the entire

2. amount of the area can stay concentrated, that point is designated
as centroid of area. The term centroid is purely related to any
geometric figure which does not involve the property of mass or
weight.
When we take effect of mass density, the term center of
mass appears. Center of mass is applicable to any physical body
like a rod, plate or solid.
Centroid:Thus centroid can be defined as the point where the
whole area of the body is concentrated.
Center of Gravity:Center of gravity may be defined as that point
through which the whole weight of a body is assumed to act
Difference between center of gravity and centroid:-
1. The term center of gravity applies to bodies with mass and
weight, and centroid applies to plane areas.
2. Centre of gravity of a body is a point through which the
resultant gravitational force (weight) acts for any orientation of
the body whereas centroid is a point in a plans area such that
the moment of area about any axis through that point is zero.
Centroids Determined by Integration
We recall that integration is the process of summing up
infinitesimal quantities. Except for a change in symbols and
procedure, integration is equivalent to a finite summation. In the
preceding section.
Example:-If the area of an element had been expressed as the
differential dA (i.e., a small part of the total area A), the equations
for determining the centroid of an area would have become
 xdAxA and  ydAyA
And for determining the centroid of a line, we could have used
 xdLxL and  ydLyL
When we determine the centroid by integration, the figure is
divided into differential elements so that:
1. All points of the element are located the same distance from
the axis of moment, or
2. The position of the centroid of the element, its known s that
the moment of the element about the axis of moments is the

3. product of the element and the distance of its centroid from
the axis.
Use of Axis of Symmetry
Centroid of an area line on the axis of symmetry if it exists. This is
useful theorem to locate the centroid of an area. This theorem can
be proved as follows:
Consider the area shown in Fig.9.3. In this figure y-axis is the axis
of symmetry. From Eqn. 9.6.the distance of centroid from this axis
is given by:
A
xΣA ii
Consider the two elemental areas shown in Fig.9.3 which are equal
in size and are equidistant from the axis, but on either side. Now
the sum of moments of these areas cancel each other since the
areas and distance are the same, but signs of distances are
opposite. Similarly, we can go on considering one area on one side
of symmetric axis and corresponding image area on the other side,
and prove that total moment of area (Aixi) about the symmetric
axis is zero. Hence the distance of centroid from the symmetric
axis is zero, i.e., centroid line on symmetric axis.
Fig No. 9.3 page no. 228 bhavikatti
Making use of the symmetry we can conclude that:
1) Centroid of a circle is its centre (Fig. 9.4)
2) Centroid of a rectangle of sides b and d is a distance
2
b
and
2
d
from any corner (Fig. 9.5).
Determination of centroid of simple figures from first principle
For simple figures like triangle and semicircle, we can write
general expression for the elemental area and its distance from an
axis. Then equations 9.5 and 9.6 reduce to:
A
ydA
y
 …… (9.7)
A
xdA
X
 …… (9.8)

4. The location of the centroid using the above equations may
be considered as finding centroid from first principles. Now, let us
find centroid of some standard figures from first principles.
Centroid of a Triangle:-Consider the triangle ABC of base width
‘b’ and height ‘h’ as shown in Fig. 9.6. Let us locate the distance of
centroid from the base. Let b be the with of elemental strip of
thickness dy at a distance y from the base. Since ABC are similar
triangle, we can write;
Fig. No. 9.6
h
yh
b
b1 

b
h
y
1b
h
yh
b1 










 

 Area of the element = dA = b1dy
dyb
h
y
1 






Area of the triangle bh
2
1
A 
 From Eqn. (9.7)
A
ydA
areaTotal
areaofMoment
y

Now,    












h
0
h
0
2
dyb
h
y
ydyb
h
y
1yydA
6
bh
3h
y
2
y
b
2h
0
32







  
bh
2
1
1
6
bh
A
ydA
y
2

3
h
y 

5. Thus the centroid of a triangle is at a distance
3
h
from the base (or
3
2h
from the apex) of the triangle where h is the height of the
triangle where ‘h’ is height of the triangle.
Cenrtroid of a semicircle – consider the semicircle of radius
R as shown in Fig. 9.7. Due to symmetry centroid must lie on y
axis. Let its distance from diametral axis be y . To find y , consider
an element at a distance r from the centre O of the semicircle,
radial width dr and bound by radii at  and  + d.
Fig No. 9.7
Area of element = r d dr.
Its moment about diametral axis x is given by:
rddr r sin  = r2 sin dr d
 Total moment of area about diametral axis
 






π
0
R
0
3
R
0
2
π
0
dθθsin
3
r
dθdrθsinr
3
2R
1][1
3
R
cos][
3
R 33
π
0
3

Area of semicircle 2
R
2
1
A 

 3
4R
R
2
1
3
2R
Totalarea
eaMomentofar
y
2
3

Thus, the centroid of the semicircle is at a distance
3
4R
from the
diametral axis.
Centroid of sector of a circle consider the sector of a circle of
angle 2 as shown in Fig. 9.8. Due to symmetry, centroid lies on x
axis. To find its distance from the centre O, consider the elemental
area show.
Area of the element = r ddr
Its moment about y axis

6. Fig No. 9.8
= rddr r cos = r2cosdr d
 Total moment of area about y axis
 



dθdrθcosr
R
0
2
 






 θsin
3
r
R
0
3
2sin
3
R3

Total area of the sector
 



dθdrr
R
0
2
 







-
R
0
2
dθ
2
r
  

2
2
Rθ
2
R

 The distance of centroid from centre O
figuretheofArea
axisyaaboutareofMoment




sin
3
2R
R
sin
R
2R
2
3

Centroid of parabolic spandrel- consider the parabolic spandrel
shown in Fig 9.9 Height of the element at a distance x from O is y =
kx2
Width of element = dx
 Area of the element = kx2 dx
 Total area of spandrel 
a
0
2
dxkx

7. 3
ka
3
kx 3a
0
3






Fig No. 9.9
Moment of area about y axis
 
a
0
3
a
0
2
dxkxxdxkx
4
ka4

Moment of area about x axis
 
a
0
52422
a
0
2
10
ak
dx
2
xk
2
kx
dxkx

4
3a
3
ka
4
ka
x
34

2
352
ka
10
3
3
ka
10
ak
y 
From the Fig 9.9, at x = a, y = h
 2
2
a
h
korkah 

10
3h
a
a
h
10
3
y 2
2

Thus, centroid of spandrel is located at 





10
3h
,
4
3a
Centroids of some common figures are shown in Table 9.1
6. Moment Of Inertia
Consider the area shown in figure 9.18(a).dA is an elemental area
with co-ordinates as x and y. the term y2dA is called moment of
inertia of the area about x axis and is denoted as Ixx. Similarly,
the moment of inertia about axis is Iyy=  x2dA

8. In general, if r is the distance of element area dA from the axis AB
[fig.9.18 (b)], the sum of the terms r2dA to cover the entire area is
called moment of inertia of the area about the axis AB. If r and dA
can be expressed in general terms, for any element then the sum
becomes an integral. Thus,
IAB =  r2dA =  r2dA
FIGURE bhavikattipg 240 fig no. 9.18
The term rdA may be called as moment of area, similar to moment
of a force, and hence r2dA may be called as moment of moment of
area of the second moment of area. Thus, the moment of inertia of
a plane figure is nothing but second moment of area. In fact, the
term ‘second moment of area’ appears to correctly signify the
meaning of the expression  r2dA.
Though moment of inertia of plane area is a purely mathematical
term, it is one of the important properties of areas. The strength of
members subject to bending depends on the moment of inertia of
its cross-sectional area. Students will find this property of area
very useful when they study subjects like strength of materials, the
moment of inertia is a fourth dimensional term since it is a trm
obtained by multiplying area by the distance. Hence, in SI units, if
meter (m) is the unit for linear measurements used then m4 is the
unit of moment of inertia.
7. Polar Moment of Inertia
Moment of inertia about an axis perpendicular to the plane of an
area is known as polar moment of inertia. It may be denoted as
J or Izz. Thus, the moment of inertia about an axis perpendicular to
the plane of the area at o in figure 9.19 is called polar moment of
inertia at point o, and given by
Figure 4m bhavilattipg no. 241 giu no. 9.19
Izz =  r2dA
= x2 + y2dA
= x2dA+ y2dA
Izz = Ix + Iy
8. Radius of Gyration
The term radius of gyration is used to describe another
mathematical expression and appears most frequency in column

9. formulas. Radius of gyration is usually denoted by the symbol k
(sometimes by r) and is defined by the relation
2
AkIor
A
I
k 
Where I is the moment of inertia and A the cross-sectional area
and k is radius of gyration.
In view of this discussion, the radius of gyration is frequently
considered to be the uniform distance from the reference axis at
which the entire area may be assumed to be distributed. For an
area whose dimensions perpendicular to a reference axis are
negligibly small compared with its distance from that axis, radius
of gyration is practically equivalent to the centroidal location of the
area.
9. Axis Transfer Theorem
There are two theorems which are very useful when we know the
moment of inertia about some axis and which to calculate the
same about some other axis.
(1) Perpendicular axis theorem, and
(2) Parallel axis theorem.
Perpendicular Axis Theorem
The moment of inertia of an area about an axis perpendicular to its
plane (polar moment of inertia) at any point o is equal to the sum
of moments of inertia about any two mutually perpendicular axis
through the same point o and lying in the plane of the area.
FIGURE f4 bhavikattipgno. 242 fig no 9.21
Referring to figure 9.21, if z – z is the axis normal to the plane of
paper passing through point o, as per this theorem,
Izz = Ixx+ Iyy
The above theorem can be easily proved. Let us consider an
elemental area dA at a distance r from o. let the co-ordinates of
dAbe x and y. them from definition:
Izz=  r2dA = (x2 + y2) dA
=  x2dA +  y2dA
Izz = Ixx+ Iyy
Parallel Axis Theorems

10. Moment of inertia about any axis in the plane of an area is equal to
the sum of moment of inertia about a parallel centroidal axis and
the product of area and square of the distance between the two
parallel axis. Referring to figure 9.22, the above theorem means:
IAB = IGG + A yc2
Where IAB – moment of inertia about the axis AB
IGG – moment of inertia about centroidal axis GG
parallel to AB.
A – The area of the plane figure given and
Yc – the distance between the axis AB and the parallel
centroidal axis GG.
Proof: consider an elemental parallel strip dA at a distance y from
the centroidal axis (fig. 9.22.)
Then, IAB = (y + yc)2dA =  (y2 + 2yyc + yc2)dA
=  y2dA +  2yycdA +  yc2dA
FIGURE 4m bhavikattipg no. 242 fig no. 9.22
y2dA = Moment of inertia about the axis GG
= IGG
=  2 yycdA + 2 ycy dA
=
A
ydA
A2yc

In the above term 2yc A is constant and
A
ydA is the distance of
centroid from the reference axis GG. Since GG is passing through
the centroid it self
A
ydA
is zero and hence the term 2yycdA is zero.
Now, the third term,
 yc2dA = yc2dA = Ayc2
 IAB = IGG + Ayc2
Note: The above equation cannot be applied to any two parallel
axes (GG) must be centroidal axis only.
10. Table for moment of inertia
4m bhavikattipg no. 247 to 249 table no. 9.2

11. 11. Moment of inertia of standard sections
Rectangle – referring to figure 9.26,
(a)
12
bd
I
3
xx as derived in Art. 9.12.
(b)
12
bd
I
3
yy  can be derived on the same lines.
(c) About the base AB, from parallel axis theorem,
IAB = Ixx+ A yc2
,
2
d
bd
12
bd
23






 




2
d
yc
4
bd
12
bd 33

3
bd 3

Hollow rectangular section – referring to figure. 9.27, Moment of
inertia Ixx = Moment of inertia of larger rectangle – moment of inertia of
hollow portion. That is,
12
bd
12
BD 33

 33
bdBD
12
1

Triangle – Referring to figure. 9.28,
(a) about the base:
As found in Art 9.12,
12
bh
I
3
AB 
(b) About centroidal axis x – x parallel to bas:
From parallel axis theorem,
IAB = Ixx+ A yc2
Now, yc, the distance between the non-centroidal axis AB and
centroidal axis x – x, is equal to
3
h

12. 
18
bh
I
3
h
bhI
12
bh 3
xx
2
xx
3








36
bh
18
bh
-
12
bh
I
333
xx 
Moment of inertia a circle about any diametral axis
64
d4
π

Moment of inertia of a hollow circle – referring to figure 9.29
64
πd
64
πD 44

 44
dD
64
d

Moment of inertia of a semicircle – (a) About Diametral Axis:
If the limit of a integration is put as 0 to  instead of 0
to 2 in the derivation for the moment of inertia of a circle a about
diametral axis (ref. art. 9.12), the moment of inertia of semicircle is
obtained. It can be observed that the moment of inertia of a
semicircle (figure 9.30) about the diametral axis AB:
128
πd
64
πd
2
1 44

(b) About centroidal axis s – x: Now, the distance of centroidal
axis yc from the diametral axis is given by:
3π
2d
3π
4R
yc 
Area
8
πd
4
πd
2
1
A
44

From parallel axes theorem,
IAB = Ixx+ A yc2
22
xx
2
3π
2d
8
πd
I
128
πd







Moment of Inertia of Composite Section:

13. Beams and columns having composite sections are commonly
used in structures. Moment of inertia of these sections about an
axis can be found by the following steps:
(1) Divide the given figure into a number of simple figures.
(2) Locate the centroid of each simple figure by inspection or
using standard expressions.
(3) Find the moment of inertia of each simple figure about its
centroidal axis. Add theterra Ay2 where A is the area of the simple
figure and is the distance of the centroid
of the simple figure from the reference axis. This gives moment of
inertia of thesimple figure about the reference axis.
(4) Sum up moments of inertia of all simple figures to get the
moment of inertia of thecomposite section!"
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