The document defines several terms: 1. Centroid - The point at which the total area of a plane is concentrated. It is where the average position of the total weight of the plane would balance. 2. Radius of gyration - The distance from the axis of rotation to the centroid. It is calculated as the square root of the moment of inertia divided by the total area. 3. Area moment of inertia - The product of the plane area and the square of the perpendicular distance to the axis of reference. It is a measure of the resistance offered by a plane figure to bending or twisting about an axis.

1. Define following terms
1.Centroid
2. Radius of gyration
3.Area moment of Inertia
JNTU – Dec2006
1.Centroid
centroid
The point at which the total area of plane to be considered is known as
centroid , the centroid is represented by C.G The centroid and centre of
gravity are at the same point Where centre of gravity consider to be
whole mass of an object act at a point
C.G

2. x axis
Y
y axis
X
Let consider dA is small area on the plane
We know that moment of area is = area multiplied by
Perpendicular distance to axis of reference
i.e. - area moment = XdA
Then moment of moment or second moment
is = X2dA with respect to x axis
Area moment of Inertia
dA
= Y2dA with respect to y axis

3. Area moment of Inertia
x axis
Y
y axis
centroid
C.G
X
Ixx = Moment of Inertia about x axis
Iyy = Moment of Inertia about y axis
𝑿 𝟐 dAIxx =
𝒀 𝟐 dAIyy =
The area moment about a point is
the product of the plane area and
the normal distance of the point
from the reference line area. If this
is multiplied by the perpendicular
distance is called second moment
of area Let consider dA is small area on the plane

4. Radius of gyration
K
A= total area of
lamina concentrated
at this point
A
Radius of gyration =
axis
I = Ak2
I = moment of inertia
K = I
A

5. Parallel axis Theorem
X X
h
X axis
A B
Plane
centroid
Moment of Inertia about AB axis IAB = IXX + Ah2
Area of plane
Moment Inertia about
centroid
Moment of Inertia about AB axis IAB = moment of Inertia
about centroid + area Multiplied by square
of distance between
centroid and
axis parallel to centroid

6. O
b
b/2
h/2
h
Centroid of
rectangle is known to
us by looking at
component
O
Regular rectangle

7. Radius = r
Diameter = D = 2r
O O = centroid of
circle is known to
us by looking at
the component
Regular circle

8. A hole on a rectangle
whose diameter = 10mm
80mm
160mm
120mm
100mm
Non Regular component of hole on a rectangle
Centroid of the component to be calculated

9. 80mm
160mm
100mm
160mm
X2=80mm
Y2=120mm
100mm
X1=50mm
Y1=80mm
10mm
0,0 0,0
Y axis
X axis
Y axis
X axis
Non regular component regular component of rectangle regular component of circle
A1= 100X160=16000 mm2 A2= π r2 = π102 = 314 mm2

10. Y
X X=
A1X1 - A2X2
A1 - A2
Y=
A1Y1 - A2Y2
A1 - A2
=
=
A1= 100X160=16000 mm2
A2= π r2 = π102 = 314 mm2
X1=50mm Y1=80mm
X2=80mm Y2=120mm
Area X cordinate Y cordinate
16000X50 - 314X80
16000 - 314
16000X80 - 314X120
16000 - 314
800000 - 25120
=
=
15686
1280000 - 37680
15686
= 49.3mm
=79.2mm
X = 49.3mm
Y
= 79.2mm

11. 50 mm
80mm
49.3mm
79.2
mm
Centroid of rectangle before making hole Centroid of rectangle after making hole
Pl note that shift of centroid

12. b
h/3
b/3
h
For triangle note the
centroid
For triangle note centroid distance

13. centroid
Radius = rRadius = r
4r/3π
Radius = r
For semicircle note centroid distance

14. centroid
Radius = rRadius = r
4r/3π
Radius = r
4r/3π
4r/3π
For quarter circle note the centroid

15. R= 2m
6m 3m
4m
Determination of the centroid of the area of
composite section is explained in different
method developed by me

16. R= 2m
6m 3m
4m
Draw the axis as shown

17. R= 2m
Consider semi circle by splitting the composite section and mark
the centroid with respect to axis drawn considering + and –
(positive and negative with respect to axis)
- 4r/3π = (4x2)/3π = - 0.849
Area = A1= π22 /2= 6.28 m2
X1= - 0.849
Y1= 2.0
Y1= 2.0
- Ve sign is with respect to axis
Centroid of semi circle

18. 6m
4m
X2=3m
Y2=2m
Area = A2= 6X4 24 m2
X2=3
Y2=2
Centroid of rectangle

19. 6m
3m
4m
3/3=1
Y3=4/3=1.333
A3=(1/2)x 3x 4=6 m2
X3= 7
Y3= 1.333
X3
Centroid of triangle

20. Area X Coordinate Y Coordinate
A1=6.28 X1= - 0.849 Y1= 2.0
A2=24 X2=3 Y2=2
A3=6 X3= 7 Y3= 1.333
X =
A1X1 + A2X2 + A3X3
A1+ A2 + A3
=
(6.28 X ( - 0.849)) +( 24X3) +(6X7)
(6.28+ 24 + 6)
= 2.995 mm
Y =
A1Y1 + A2Y2 + A3Y3
A1+ A2 + A3
=
(6.28 X ( 2.0) +( 24X2) +(6X1.333)
(6.28+ 24 + 6)
= 1.890mm

21. R= 2m
6m
3m
4m
X = 2.995 mm
Y = 1.890 mm
Centroid of composite bar

22. Moment of Inertia of standard and regular shape component
h/2
h
b/2
b
xx
y
y
X axis
y axis
Centroid
I xx =
bh3
12
I yy =
hb3
12
Where I xx = moment of Inertia about xx axis
Where I yy = moment of Inertia about yy axis

23. x
X axis
x
Y
Y
Y1
BA
Moment of Inertia of
rectangle about the axis AB
parallel to axis xx is equal
to
IAB= Ixx + A(Y1) 2
Where Ixx moment of Inertia about xx axis
A area of rectangle and Y1 is distance
between the axis AB and x axis Pl note that
AB axis is parallel to x axis
IAB =
bh3
12
+{( b X h) (Y1)2
b
h
}

24. XX
Y
Y
X axis
Y axis
d = Diameter of circle
Circle Ixx =
πd4
64
centroid
Circle Iyy =
πd4
64
Plolar moment of Inertia Izz = Ixx+Iyy =
πd4
64
+
πd4
64
=
πd4
32

25. centroid
Radius = rRadius = r
4r/3π
Radius = r
For semicircle note centroid distance
Moment of Inertia of semicircle
For circle moment of Inertia about this
axis is known that is πd4/64 and for semi
circle it is( πd4/64)X ½ = πd4/128 pl note
that we do not have moment f Inertia of
semi circle about centroid axis which is to
be calculated applying parallel axis
theorem
x x
BA
moment of Inertia
for semi circle it is
( πd4/64)X ½ = πd4/128
Now IAB= Ixx +(AY1)2
Y1 =
Ixx =0.0068598d4
Ixx =0.0068598d4
Where IAB= πd4/128
A=πr2 or πd4/4
Ixx = IAB - (AY1)2
Y1
4r/3πY1 =

26. Let us know the moment of inertia of Triangle
A
B
moment of inertia of
Triangle about this axis
or base of triangle
is=
b
Y1=h/3
Ixx= moment of Inertia
Of triangle about
Centroid of triangle is
moment of inertia of Triangle about centroid= Ixx =IAB-A(y1)2 =
h
A = area of triangle
IAB = moment of inertia triangle about centroid
bh3
12
bh3
36
bh3
36
Applying parallel axis theorem

27. A hole on a rectangle
whose diameter = 10mm
80mm
160mm
120mm
100mm
Non Regular component of hole on a rectangle
Centroid of the component to be calculated

28. 80mm
160mm
100mm
160mm
X2=80mm
Y2=120mm
100mm
X1=50mm
Y1=80mm
10mm
0,0 0,0
Y axis
X axis
Y axis
X axis
Non regular component regular component of rectangle regular component of circle
A1= 100X160=16000 mm2 A2= π r2 = π102 = 314 mm2

29. Y
X X=
A1X1 - A2X2
A1 - A2
Y=
A1Y1 - A2Y2
A1 - A2
=
=
A1= 100X160=16000 mm2
A2= π r2 = π102 = 314 mm2
X1=50mm Y1=80mm
X2=80mm Y2=120mm
Area X cordinate Y cordinate
16000X50 - 314X80
16000 - 314
16000X80 - 314X120
16000 - 314
800000 - 25120
=
=
15686
1280000 - 37680
15686
= 49.3mm
=79.2mm
X = 49.3mm
Y
= 79.2mm

30. 50 mm
80mm
49.3mm
79.2
mm
Centroid of rectangle before making hole Centroid of rectangle after making hole
Pl note that shift of centroid

31. A hole on a rectangle
whose diameter = 10mm
80mm
160mm
120mm
100mm
Non Regular component of hole on a rectangle
Centroid of the component to be calculated

32. 80mm
160mm
100mm
160mm
X2=80mm
Y2=120mm
100mm
X1=50mm
Y1=80mm
10mm
0,0 0,0
Y axis
X axis
Y axis
X axis
Non regular component regular component of rectangle regular component of circle
A1= 100X160=16000 mm2 A2= π r2 = π102 = 314 mm2

33. Y
X X=
A1X1 - A2X2
A1 - A2
Y=
A1Y1 - A2Y2
A1 - A2
=
=
A1= 100X160=16000 mm2
A2= π r2 = π102 = 314 mm2
X1=50mm Y1=80mm
X2=80mm Y2=120mm
Area X cordinate Y cordinate
16000X50 - 314X80
16000 - 314
16000X80 - 314X120
16000 - 314
800000 - 25120
=
=
15686
1280000 - 37680
15686
= 49.3mm
=79.2mm
X = 49.3mm
Y
= 79.2mm

34. 50 mm
80mm
49.3mm
79.2
mm
Centroid of rectangle before making hole Centroid of rectangle after making hole
Pl note that shift of centroid

35. 50 mm
80mm
49.3mm
79.2mm
Centroid of rectangle before making hole
Centroid of rectangle after making hole
Pl note that shift of
centroid and we
need to calculate
moment Inertia of
Composite section
about the shifted
axis let us do it
Ixx =
bh3
12
Ixx of circle = πd4
64 d=10 mm
0.8 mm
A B
IA B = to be calculated by applying parallel axis theorem
IA B of rectangle Ixx of rectangle about its centroid + area of rectangle X square axis shift in y axis
160mm
100 mm
=
IA B =
bh3
12
+ bxh (y1)2
=
100x803
12
+ 100x80x (0.8)2 = 4266666 +5120=4271787mm4
40.8mm
80mm
80mm
40mm

36. IA B of Circle =
πd4
64
+ π r2 (y1)2
=
πx104
64
+ πx52x (40.8)2 =490.625+130674.2=131164.8 mm4
Moment of Inertia of composite section = IA B of rectangle - IA B of Circle
4271787 -= 131164.8 =4140622mm4

37. To find the centroids of given object
Let h= 50mm ,D=50mm and R=10mm
h
D
R
A B
Hole

38. A1 = (1/2)50*50 = 1250mm2
Y1=50/3=16.667mm
A1Y1=20833.75 mm3
A1
Y1= =
h = 50mm
A1 area of triangle
centroid
Y1 = centroid distance from base
h
3
50
3
X axis
Y axis

39. r=10mm
A2
Y2=0 since centroid of
Circle coincides and with axis
i.e. it lies in x axis
A2 = πr2 = π*102= 314mm2
A2Y2=0
centroid
A2 area of circle
Y axis
X axis
A

40. X axis
Y axis
Y3 = - (4r/3π) =-(4*25)/3π= - 10.617mm
Pl note –ve sign is due to the centroid
of Semi circle fall below the x axis
centroid
Radius of circle r=25mm
A3y3= - 10417.9
A3 = (1/2)πr2 = 0.5*π*252=981.25mm2
A3 = area of semi circle
A

41. A1 y1-A2Y2 +A3Y3
Y =
A1 -A2 +A3
=
20833.75 - 0 – 10417.9
1250 - 314 +981.25
= 5.4327mm
area Y= centroid distance Area X centroid distance = AXY
A1 Y1 A1Y1
A2 Y2 = 0 A2Y2 = 0
A3 Y3 A3Y3
=1250mm2
=16.667mm =20833.75 mm3
=314mm2
=981.25mm2 =10.617mm = - 10417.9 mm3

42. h=50mm
D= 50mm
r= 10mm
A B
5.4327mm
Centroid of composite section
Calculated

43. Finding moment of inertia of object about the axis AB
h=50
D=b=50mm
R=D/2=25mm
r=d/2= 10mm
A B

44. Moment of inertia of Triangle about AB=(1/12)bh3 =
h=50 mm
B
50 mm
A
1
12
50 x 503
=520833 mm4
Axis about which moment
of inertia
To be taken

45. r=d/2= 10mm
A B
Moment of inertia of circle about AB=(πd4)/64 =
π 204
64
Centroid
=7850mm4
Axis about which moment
of inertia
To be taken

46. Moment of inertia of semi circle about AB=(1/128)*(πD4) =
π 504
128
D=b=50mm
A B
Centroid
=153320mm4
Axis about which moment
of inertia
To be taken

47. Moment of inertia of object about AB =
Moment of inertia of Triangle about AB
+
Moment of inertia of semi circle about AB(
-
Moment of inertia of circle about AB{(πd4)/64}
1
12
50 x 503= +
π 504
128
-
π 204
64
=x 520833 + 153320 - 7850 = 666303mm4
h=50
D=b=50mm
R=D/2=25mm
r=d/2= 10mm
A B
π 504
128
(1/12)bh3
(

48. To find the moment of Inertia about centroids of given object
Let h= 50mm ,D=50mm and R=10mm
h
D
R
A B
Hole

49. h=50mm
D= 50mm
r= 10mm
A B
5.4327mm
Centroid of composite section
Calculated

50. Moment of inertia of object about AB =
Moment of inertia of Triangle about AB
+
Moment of inertia of semi circle about AB(
-
Moment of inertia of circle about AB{(πd4)/64}
1
12
50 x 503= +
π 504
128
-
π 204
64
=x 520833 + 153320 - 7850 = 666303mm4
h=50
D=b=50mm
R=D/2=25mm
r=d/2= 10mm
A B
π 504
128
(1/12)bh3
(

51. To find the moment of Inertia about centroid of given object
Let h= 50mm ,D=50mm and R=10mm
h
D
R
A B
Hole
5.4327mm
Centroid of composite section
Calculated
Moment of inertia of
object about AB = 666303mm4
moment of Inertia
about centroid is to be
calculated

52. Moment of Inertia about centroid calculation
IAB = IXX+ Ah2 Moment of Inertia about an axis= IAB= 666303mm4
Moment of Inertia about centroid to be calculated
A = Triangle area + semicircle area with radius 25mm – circle area with radius 10mm
h = distance between axis to centroid = 5.4327mm
1250= + 981.25 – 314 =1917.25mm2
IXX = IAB Ah2
= 1917.25 x666303 –
–
5.43272 = 610395.99mm4