The document discusses the topic of moment of inertia. It will cover moment of inertia concepts including introduction, parallel axis theorem, perpendicular axis theorem, and moment of inertia calculations for rectangles, triangles, and circles. Examples are provided for calculating the moment of inertia of a rectangle about its centroidal axis and for a triangular section about its base and centroidal axis. The last example calculates the second moment of inertia of a circular lamina about its centroidal axis using the polar moment of inertia method.

1. Department of
Mechanical
Engineering.
Prof. Malay Badodariya
+91 9429 158833
FundamentalOfMachineDesign (01ME0504)
Centroid and Moment of
inertia
(Part 2 - MI)

2. Topic will be cover:
1. Moment of Inertia:
a) Introduction
b) Parallel Axis Theorem
c) Perpendicular Axis Theorem
d) Moment Of Inertia of Rectangle
e) Polar Moment of Inertia
f) Moment of Inertia of Triangle
g) Moment Of Inertia of Circle
Department of Mechanical
Engineering.
Prof. Malay Badodariya
malay.badodariya@marwadie
ducation.edu.in
Unit no.: 5
Subject Name: FMD
01ME0305

3. Topic will be cover:
1. Moment of Inertia:
h) Formula of Moment of Inertia
i) Steps of problem solving
j) Example (1 to 8 With Solution)
k) Example (Home Work)
Department of Mechanical
Engineering.
Prof. Malay Badodariya
malay.badodariya@marwadie
ducation.edu.in
Unit no.: 5
Subject Name: FMD
01ME0305

4. Topic will be cover:
1. Moment of Inertia:
a) Introduction
b) Parallel Axis Theorem
c) Perpendicular Axis Theorem
d) Moment Of Inertia of Rectangle
e) Polar Moment of Inertia
f) Moment of Inertia of Triangle
g) Moment Of Inertia of Circle
Department of Mechanical
Engineering.
Prof. Malay Badodariya
malay.badodariya@marwadie
ducation.edu.in
Unit no.: 5
Subject Name: FMD
01ME0305

5. a)
Introduction
 It is defined as the algebraic sum of product of area and the
square of distance from the particular axis.
 It is denoted by 𝐼 𝑋𝑋 or 𝐼 𝑌𝑌
 It is given by I =Aℎ2
 Unit - 𝑚𝑚4
 It is second moment of area which is measure of résistance
to bending & forms basic to strength of material.
 Mass MI is the résistance to rotation & forms basic to
dynamics to rigid bodies.

6.  Consider a lamina of area A shown in fig.
 Let, this lamina is split up in to an infinite number of small
elements each of area da.
 Let, x1, x2, x3 ... are the distances of small elements from
OY axis.
 y1 , y2, y3 ... are the distances of small elements from OX
axis.
 Taking second moment of all the small elements about OX
axis,

7.  Taking second moment of all the small elements about OY
axis,

8. Topic will be cover:
1. Moment of Inertia:
a) Introduction
b) Parallel Axis Theorem
c) Perpendicular Axis Theorem
d) Moment Of Inertia of Rectangle
e) Polar Moment of Inertia
f) Moment of Inertia of Triangle
g) Moment Of Inertia of Circle
Department of Mechanical
Engineering.
Prof. Malay Badodariya
malay.badodariya@marwadie
ducation.edu.in
Unit no.: 5
Subject Name: FMD
01ME0305

9. b)
ParallelAxis
Theorem
 M.I of plane section about any axis is equal to MI of plane
section about the parallel axis parallel axis passing through
its CG plus the product of area and square of distance
between two axes.
 𝑰 𝑨𝑩 = 𝑰 𝒈 + a𝒉 𝟐
Where,
𝐼𝐴𝐵 = M.I. of area about AB
𝐼𝑔 = M.I. of area about C.G.
a = Area of the section
h = distance between c.g.
of section and axis AB.

10. Proof:
Condition 1: Two axis must be parallel.
Condition 2: One axis should pass through CG.
Condition 3: h is perpendicular distance between two axis.
 Y = distance of strip from CG of section.

11.  M.I. of strip about CG of section,
 I = da.𝑦2
 M.I. of whole section about an axis passing through CG,
 𝐼𝑔 = 𝑑𝑎. 𝑦2
 M.I. of whole section about an axis AB,
 𝐼𝐴𝐵 = 𝑑𝑎. ℎ + 𝑦 2
= 𝑑𝑎. (ℎ2 + 𝑦2 + 2ℎ𝑦)
= 𝑑𝑎. ℎ2 + 𝑑𝑎. 𝑦2+ 𝑑𝑎. 2ℎ𝑦
= aℎ2+ 𝐼𝑔 + 0
 𝐼𝐴𝐵 = 𝐼𝑔 + aℎ2

12. Topic will be cover:
1. Moment of Inertia:
a) Introduction
b) Parallel Axis Theorem
c) Perpendicular Axis Theorem
d) Moment Of Inertia of Rectangle
e) Polar Moment of Inertia
f) Moment of Inertia of Triangle
g) Moment Of Inertia of Circle
Department of Mechanical
Engineering.
Prof. Malay Badodariya
malay.badodariya@marwadie
ducation.edu.in
Unit no.: 5
Subject Name: FMD
01ME0305

13. c)
Perpendicular
AxisTheorem
 The moment of inertia of polar axis is equal to algebraic
sum of MI of two axis about which axis is perpendicular.
 𝑰 𝒛𝒛 = 𝑰 𝒙𝒙 + 𝑰 𝒚𝒚
Proof:
Condition 1: Only for 2D Body (Laminar Plate)
Condition 2: X,Y & Z axis mutually perpendicular
Condition 3: X &Y axis should be in the plane of the body
Condition 4: Z should be perpendicular to the plane of body
2D Body

14.  Consider a small area da on lamina plate.
 X andY are co-ordinate of P along OX and OY axis.
 OP = r, (distance between P and z axis)
 Now use Pythagoras theorem,
 𝑟2 = 𝑋2 + 𝑌2
 M.I. of P about X – X axis,
 𝐼 𝑋𝑋 = 𝑑𝑎. 𝑦2
 M.I. of P about Y –Y axis,
 𝐼 𝑌𝑌 = 𝑑𝑎. 𝑥2
2D Body
 M.I. of P about Z – Z axis,
 𝐼zz = 𝑑𝑎. r2

15.  M.I. of P about Z – Z axis,
 𝐼zz = 𝑑𝑎. r2 = da (𝑋2 + 𝑌2) = 𝑑𝑎. 𝑥2 + 𝑑𝑎. 𝑦2
 𝐼zz = 𝐼 𝑋𝑋 + 𝐼 𝑌𝑌
2D Body

16. Topic will be cover:
1. Moment of Inertia:
a) Introduction
b) Parallel Axis Theorem
c) Perpendicular Axis Theorem
d) Moment Of Inertia of Rectangle
e) Polar Moment of Inertia
f) Moment of Inertia of Triangle
g) Moment Of Inertia of Circle
Department of Mechanical
Engineering.
Prof. Malay Badodariya
malay.badodariya@marwadie
ducation.edu.in
Unit no.: 5
Subject Name: FMD
01ME0305

17. Example on Second Moment of Inertia
Example 1: Determine the second moment of area of a rectangle about an axis through the centroid.
Solution:
Given Data,
 Consider a rectangle of width ‘b’ and depth ‘d’.
 Also consider a small strip of thickness ‘dy’ at distance ‘y’ from centroidal axis
 We know that second moment of area of a strip about xx axis = da.𝑦2
Where, da = area of small strip
 For M.I. of whole area, integrate between limits
−𝑑
2
𝑎𝑛𝑑
𝑑
2
∴ 𝐼 𝑋𝑋 =
−𝑑
2
𝑑
2
𝑑𝑎. 𝑦2
∴ 𝐼 𝑋𝑋 =
−𝑑
2
𝑑
2
(𝑏. 𝑑 𝑦). 𝑦2

18. Example on Second Moment of Inertia
Example 1: Determine the second moment of area of a rectangle about an axis through the centroid.
Solution:
Given Data,
∴ 𝐼 𝑋𝑋 =
−𝑑
2
𝑑
2
𝑑𝑎. 𝑦2
∴ 𝐼 𝑋𝑋 =
−𝑑
2
𝑑
2
(𝑏. 𝑑 𝑦). 𝑦2
∴ 𝐼 𝑋𝑋 = 𝑏
−𝑑
2
𝑑
2
(𝑑 𝑦). 𝑦2
∴ 𝐼 𝑋𝑋 = 𝑏
𝑦3
3 −𝑑
2
𝑑
2
∴ 𝐼 𝑋𝑋 = b
𝑑3
24
−
−𝑑3
24
=
𝑏.𝑑3
12
𝐼 𝑌𝑌 =
𝑑.𝑏3
12

19. Topic will be cover:
1. Moment of Inertia:
a) Introduction
b) Parallel Axis Theorem
c) Perpendicular Axis Theorem
d) Moment Of Inertia of Rectangle
e) Polar Moment of Inertia
f) Moment of Inertia of Triangle
g) Moment Of Inertia of Circle
Department of Mechanical
Engineering.
Prof. Malay Badodariya
malay.badodariya@marwadie
ducation.edu.in
Unit no.: 5
Subject Name: FMD
01ME0305

20. d)
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 Fig. is called polar moment of
inertia at point O, and is given by,
𝐼𝑧𝑧 = 𝑟2
da

21. Topic will be cover:
1. Moment of Inertia:
a) Introduction
b) Parallel Axis Theorem
c) Perpendicular Axis Theorem
d) Moment Of Inertia of Rectangle
e) Polar Moment of Inertia
f) Moment of Inertia of Triangle
g) Moment Of Inertia of Circle
Department of Mechanical
Engineering.
Prof. Malay Badodariya
malay.badodariya@marwadie
ducation.edu.in
Unit no.: 5
Subject Name: FMD
01ME0305

22. Example on Second Moment of Inertia
Example 2: Calculate moment of inertia of a triangular section with base b and height h, about base and about axis
passing through centroid.
Solution:
Given Data,  Consider a triangular section as shown in figure,
 let, base = b height = h
 Consider a small strip PQ of thickness dx at distance x from A.
∴
𝑃𝑄
𝐵𝐶
=
𝑥
ℎ
∴ 𝑃𝑄 = 𝐵𝐶.
𝑥
ℎ
 But, BC = base = b,
∴ 𝑃𝑄 =
𝑏. 𝑥
ℎ
 Now area of strip PQ =
𝑏.𝑥
ℎ
. dx

23. Example on Second Moment of Inertia
Example 2: Calculate moment of inertia of a triangular section with base b and height h, about base and about axis
passing through centroid.
Solution:
Given Data, ∴ 𝑀. 𝐼. 𝑜𝑓 𝑠𝑡𝑟𝑖𝑝 𝑎𝑏𝑜𝑢𝑡 𝑏𝑎𝑠𝑒 𝐵𝐶, = 𝐴𝑟𝑒𝑎 𝑋 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 2 =
𝑏.𝑥
ℎ
. dx ℎ − 𝑥 2
 M.I. of whole triangle section may be found out by integrating above equation
between limits 0 to h.
∴ 𝐼 𝐵𝐶 = 0
ℎ 𝑏.𝑥
ℎ
ℎ − 𝑥 2
. dx =
𝑏
ℎ 0
ℎ
𝑥 ℎ2
+ 𝑥2
− 2. ℎ. 𝑥 . 𝑑𝑥
∴ 𝐼 𝐵𝐶 =
𝑏
ℎ 0
ℎ
𝑥ℎ2
+ 𝑥3
− 2. ℎ. 𝑥2
𝑑𝑥 =
𝑏
ℎ
ℎ2
.
𝑥2
2
+
𝑥4
4
−
2.ℎ.𝑥3
3 0
ℎ
∴ 𝐼 𝐵𝐶 =
𝑏
ℎ
ℎ4
2
+
ℎ4
4
−
2.ℎ4
3
=
𝑏
ℎ
6.ℎ4+3.ℎ4−8ℎ4
12
=
𝑏
ℎ
ℎ4
12
=
𝑏.ℎ3
12
________M.I. about
Base BC.

24. Example on Second Moment of Inertia
Example 2: Calculate moment of inertia of a triangular section with base b and height h, about base and about axis
passing through centroid.
Solution:
Given Data,  M.I. about c.g. of section can be found out by parallel axis theorem.
 𝐼 𝐵𝐶 = 𝐼 𝑥𝑥𝐺 + 𝐴𝑑2
 𝐼 𝑥𝑥𝑔 = 𝐼 𝐵𝐶 − 𝐴𝑑2
=
𝑏ℎ3
12
−
1
2
𝑏ℎ
ℎ
3
2
=
𝑏.ℎ3
12
−
𝑏.ℎ3
18
=
3𝑏.ℎ3−2𝑏.ℎ3
36
 𝐼 𝑥𝑥𝑔 =
𝑏.ℎ3
36
___________________ M.I. about c.g. and parallel to base.

25. Topic will be cover:
1. Moment of Inertia:
a) Introduction
b) Parallel Axis Theorem
c) Perpendicular Axis Theorem
d) Moment Of Inertia of Rectangle
e) Polar Moment of Inertia
f) Moment of Inertia of Triangle
g) Moment Of Inertia of Circle
Department of Mechanical
Engineering.
Prof. Malay Badodariya
malay.badodariya@marwadie
ducation.edu.in
Unit no.: 5
Subject Name: FMD
01ME0305

26. Example on Second Moment of Inertia
Example 3: Find second moment of inertia of a circular lamina, about its centroidal axis.
Solution:
Given Data,
Using method of polar M.I.
Consider a circular lamina of centre ‘O’ and radius ‘r’.
Consider an elementary ring of radius ‘x’ and thickness ‘dx’.
Area of elementary ring : da = 2.π.x.dx
M.I. of ring about z-z axis = Area X 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 2
= 2.π.x.dx X 𝑥 2
= 2.π. 𝑋3
dx
Now M.I. of whole section about centroidal axis can be
found out by integrating the above equation from limits ‘O’
to ‘r’.

27. Example on Second Moment of Inertia
Example 3: Find second moment of inertia of a circular lamina, about its centroidal axis.
Solution:
Given Data,
∴ 𝐼𝑧𝑧 =
0
𝑟
2. 𝜋. 𝑋3
. 𝑑𝑥
𝐼𝑧𝑧= 2.𝜋.
𝑋4
4 0
𝑟
= 2.π.
𝑟4
4
=
𝜋.𝑟4
2
∴ 𝐼𝑧𝑧 =
𝜋
𝑑
2
4
2
∴ 𝐼 𝑍𝑍 =
𝜋
32
𝑑4

28. Example on Second Moment of Inertia
Example 3: Find second moment of inertia of a circular lamina, about its centroidal axis.
Solution:
Given Data,
 As per perpendicular axis theorem, 𝐼 𝑋𝑋 = 𝐼 𝑌𝑌
 Now, 𝐼 𝑍𝑍 = 𝐼 𝑋𝑋 + 𝐼 𝑌𝑌 = 2 𝐼 𝑋𝑋
∴
𝜋
32
𝑑4 = 2 𝐼 𝑋𝑋
∴ 𝐼 𝑋𝑋 =
𝜋
64
𝑑4
∴ 𝑆𝑖𝑚𝑖𝑙𝑎𝑟𝑙𝑦, 𝐼 𝑦𝑦 =
𝜋
64
𝑑4

29. Topic will be cover:
1. Moment of Inertia:
h) Formula of Moment of Inertia
i) Steps of problem solving
j) Example (1 to 8 With Solution)
k) Example (Home Work)
Department of Mechanical
Engineering.
Prof. Malay Badodariya
malay.badodariya@marwadie
ducation.edu.in
Unit no.: 5
Subject Name: FMD
01ME0305

30. Formula of Moment of Inertia
Sr. No. Geometrical Shape Area 𝑰 𝒙𝒙 𝑰 𝒚𝒚
1
b X d 𝐼 𝑥𝑥 =
𝑏𝑑3
12
𝐼 𝑦𝑦 =
𝑑𝑏3
12
2
𝜋
4
𝑑2 𝐼 𝑥𝑥 =
𝜋
64
𝑑4
𝐼 𝑦𝑦 =
𝜋
64
𝑑4

31. Formula of Moment of Inertia
Sr. No. Geometrical Shape Area 𝑰 𝒙𝒙 𝑰 𝒚𝒚
3
1
2
𝑏.h
𝐼 𝑥𝑥 =
𝑏. ℎ3
36
𝐼 𝑦𝑦 =
ℎ. 𝑏3
48
4
1
2
𝑏. h 𝐼 𝑥𝑥 =
𝑏. ℎ3
36
𝐼 𝑦𝑦 =
ℎ. 𝑏3
36

32. Topic will be cover:
1. Moment of Inertia:
h) Formula of Moment of Inertia
i) Steps of problem solving
j) Example (1 to 8 With Solution)
k) Example (Home Work)
Department of Mechanical
Engineering.
Prof. Malay Badodariya
malay.badodariya@marwadie
ducation.edu.in
Unit no.: 5
Subject Name: FMD
01ME0305

33. h)
Steps of
solving
problem of
M.I.
Number Discription
Step 1: Split figure from unknown geometry to known geometry and draw X
andY Axis.
Like – Square, Rectangle,Triangle, Circle etc.
Step 2: Find Centroid of Individual geometry (X andY Component)
Like - 𝑋1, 𝑌1, 𝑋2, 𝑌2 (Use Centroid Formula)
Step 3: Find centroid of whole geometry.
Like - 𝑋 & 𝑌
Step 4: Find Moment about x axis of Individual geometry
(i.e.: 𝐼 𝑋𝑋1, 𝐼 𝑋𝑋2, 𝐼 𝑋𝑋3, 𝐼 𝑋𝑋4)
(Use parallel axis theorem, 𝐼 𝑋𝑋1 = 𝐼𝑔 + 𝑎ℎ2 )
Where, 𝐼𝑔 = 𝑀𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑖𝑛𝑒𝑟𝑡𝑖𝑎 𝑜𝑓 𝑖𝑛𝑑𝑖𝑣𝑖𝑑𝑢𝑎𝑙 𝑔𝑒𝑜𝑚𝑒𝑡𝑟𝑦 𝐼 𝑋𝑋
a = Area of geometry
h = 𝑌1 - 𝑌
Step 5: Find Moment aboutY axis of Individual geometry
(i.e.: 𝐼 𝑌𝑌1, 𝐼 𝑌𝑌2, 𝐼 𝑌𝑌3, 𝐼 𝑌𝑌4)
(Use parallel axis theorem, 𝐼 𝑌𝑌1 = 𝐼𝑔 + 𝑎ℎ2
)
Where, 𝐼𝑔 = 𝑀𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑖𝑛𝑒𝑟𝑡𝑖𝑎 𝑜𝑓 𝑖𝑛𝑑𝑖𝑣𝑖𝑑𝑢𝑎𝑙 𝑔𝑒𝑜𝑚𝑒𝑡𝑟𝑦 𝐼 𝑌𝑌
a = Area of geometry
h = 𝑋1 - 𝑋

34. Number Discription
Step 6: Find Total M.I. of geometry (X andY Component)
Like - 𝐼 𝑋𝑋 = 𝐼 𝑋𝑋1 + 𝐼 𝑋𝑋2 + 𝐼 𝑋𝑋3 + 𝐼 𝑋𝑋4 + ________
Like - 𝐼 𝑌𝑌 = 𝐼 𝑌𝑌1 + 𝐼 𝑌𝑌2 + 𝐼 𝑌𝑌3 + 𝐼 𝑌𝑌4 + ________

35. Topic will be cover:
1. Moment of Inertia:
h) Formula of Moment of Inertia
i) Steps of problem solving
j) Example (1 to 8 With Solution)
k) Example (Home Work)
Department of Mechanical
Engineering.
Prof. Malay Badodariya
malay.badodariya@marwadie
ducation.edu.in
Unit no.: 5
Subject Name: FMD
01ME0305

36. Example on Moment of Inertia
Example 1: Calculate 𝑰 𝑿𝑿 𝒂𝒏𝒅 𝑰 𝒀𝒀 for a ‘T’ section as per figure.
Solution:
Given Data,
20 cm
2 cm
30 cm
2 cm
Step 1:Split figure from unknown geometry to known
geometry and draw X andY Axis.
20 cm
2 cm
30 cm
2 cm

37. Example on Composite Area (2D)
Example 1: Calculate 𝑰 𝑿𝑿 𝒂𝒏𝒅 𝑰 𝒀𝒀 for a ‘T’ section as per figure.
Solution:
Given Data,
20 cm
2 cm
30 cm
2 cm
Now Consider Element 1
Area of Element 1 = 2 x 20 = 40 𝐶𝑀2
𝑋1=
𝐿
2
=
20
2
= 10 cm
Y1 = 30 +
𝑊
2
= 30 +
2
2
= 30 + 1 = 31 cm
Note:
𝑋 = Distance between Centroid of element toY axis,
𝑌 = Distance between Centroid of element to X axis,
X1
Y1
Step 2 :Find Centroid of Individual geometry (X andY Component)
Like - 𝑿 𝟏, 𝒀 𝟏, 𝑿 𝟐, 𝒀 𝟐 (Use Centroid Formula)

38. Example on Composite Area (2D)
20 cm
2 cm
30 cm
2 cm
Now Consider Element 2
Area of Element 2 = 2 x 30 = 60 𝑐𝑚2
𝑋2= 10 cm
Y2 =
𝑊
2
=
30
2
= 15 cm
Note:
𝑋 = Distance between Centroid of element toY axis,
𝑌 = Distance between Centroid of element to X axis,
X2 Y2
Example 1: Calculate 𝑰 𝑿𝑿 𝒂𝒏𝒅 𝑰 𝒀𝒀 for a ‘T’ section as per figure.
Solution:
Given Data,

39. Example on Composite Area (2D)
20 cm
2 cm
30 cm
2 cm
Note:
𝑋 = Distance between Centroid of element toY axis,
𝑌 = Distance between Centroid of element to X axis,
𝑿
𝒀
Element Area (𝒄𝒎 𝟐 ) Xi Yi
1 40 X1=10 cm Y1=31 cm
2 60 X2=10 cm Y2=15 cm
𝑋 = Distance between Centroid of element toY axis,
𝑋 =
𝐴1 𝑋1+ 𝐴2 𝑋2
𝐴1+𝐴2
=
400+600
100
= 10 cm
𝑌 = Distance between Centroid of element to X axis,
𝑌 =
𝐴1 𝑌1+ 𝐴2 𝑌2
𝐴1+𝐴2
=
1240+900
100
= 21.40 cm
G
Step 3: Find centroid of whole geometry.
Like - 𝑿 & 𝒀
Example 1: Calculate 𝑰 𝑿𝑿 𝒂𝒏𝒅 𝑰 𝒀𝒀 for a ‘T’ section as per figure.
Solution:
Given Data,

40. Example on Composite Area (2D)
Example 1: Calculate 𝑰 𝑿𝑿 𝒂𝒏𝒅 𝑰 𝒀𝒀 for a ‘T’ section as per figure.
Solution:
Given Data,
20 cm
2 cm
30 cm
2 cm
𝑿
𝒀
G
Step 4: Find Moment about x axis of Individual geometry
Like - 𝑰 𝑿𝑿𝟏, 𝑰 𝑿𝑿𝟐
𝑰 𝑿𝑿𝟏 = 𝑰 𝒈𝟏 + 𝒂𝒉 𝟐
𝐼𝑔 = M.I. of area about C.G.
Let’s Find 𝐼𝑔1 =
𝑏𝑑3
12
=
(20)(2)3
12
= 13.33 𝑐𝑚4
Area, a = b X d = 20 X 2 = 40 𝑐𝑚2
h = 𝑌1 - 𝑌 = 31 – 21.40 = 9.6 cm
𝑰 𝑿𝑿𝟏 = 𝑰 𝒈𝟏 + 𝒂𝒉 𝟐 = 13.33 + (40 X 9.62)
= 3699.73 𝑐𝑚4
X1
Y1h

41. Example on Composite Area (2D)
Example 1: Calculate 𝑰 𝑿𝑿 𝒂𝒏𝒅 𝑰 𝒀𝒀 for a ‘T’ section as per figure.
Solution:
Given Data,
20 cm
2 cm
30 cm
2 cm
𝒀
G
Step 4: Find Moment about x axis of Individual geometry
Like - 𝑰 𝑿𝑿𝟏, 𝑰 𝑿𝑿𝟐
𝑰 𝑿𝑿𝟐 = 𝑰 𝒈 + 𝒂𝒉 𝟐
𝐼𝑔 = M.I. of area about C.G.
Let’s Find 𝐼𝑔2 =
𝑏𝑑3
12
=
(2)(30)3
12
= 4500 𝑐𝑚4
Area, a = b X d = 2 X 30 = 60 𝑐𝑚2
h = 𝑌2 - 𝑌 = 15 – 21.40 = (-6.4) cm
𝑰 𝑿𝑿𝟐 = 𝑰 𝒈𝟐 + 𝒂𝒉 𝟐 = 4500 + (60 X 6.42)
= 6957.6 𝑐𝑚4
h
X2 Y2

42. Example on Composite Area (2D)
Example 1: Calculate 𝑰 𝑿𝑿 𝒂𝒏𝒅 𝑰 𝒀𝒀 for a ‘T’ section as per figure.
Solution:
Given Data,
20 cm
2 cm
30 cm
2 cm
𝑿
𝒀
G
Step 5: Find Moment aboutY axis of Individual geometry
Like - 𝑰 𝒀𝒀𝟏, 𝑰 𝒀𝒀𝟐
𝑰 𝒀𝒀𝟏 = 𝑰 𝒈𝟏 + 𝒂𝒉 𝟐
𝐼𝑔 = M.I. of area about C.G.
Let’s Find 𝐼𝑔1 =
𝑑𝑏3
12
=
(2)(20)3
12
= 1333.33 𝑐𝑚4
Area, a = b X d = 20 X 2 = 40 𝑐𝑚2
h = 𝑋1 - 𝑋 = 10 – 10 = 0 cm
𝑰 𝒀𝒀𝟏 = 𝑰 𝒈𝟏 + 𝒂𝒉 𝟐 = 1333.33 + (40 X0)
= 1333.33 𝑐𝑚4
X1
Y1

43. Example on Composite Area (2D)
Example 1: Calculate 𝑰 𝑿𝑿 𝒂𝒏𝒅 𝑰 𝒀𝒀 for a ‘T’ section as per figure.
Solution:
Given Data,
20 cm
2 cm
30 cm
2 cm
𝒀
G
Step 5: Find Moment about x axis of Individual geometry
Like - 𝑰 𝒀𝒀𝟏, 𝑰 𝒀𝒀𝟐
𝑰 𝒀𝒀𝟐 = 𝑰 𝒈 + 𝒂𝒉 𝟐
𝐼𝑔 = M.I. of area about C.G.
Let’s Find 𝐼𝑔2 =
𝑑𝑏3
12
=
(30)(2)3
12
= 20 𝑐𝑚4
Area, a = b X d = 2 X 30 = 60 𝑐𝑚2
h = 𝑋2 - 𝑋 = 0 – 0 = 0 cm
𝑰 𝒀𝒀𝟐 = 𝑰 𝒈𝟐 + 𝒂𝒉 𝟐 = 20 + (60 X0)
= 20 𝑐𝑚4
X2 Y2
𝑿

44. Example on Composite Area (2D)
Example 1: Calculate 𝑰 𝑿𝑿 𝒂𝒏𝒅 𝑰 𝒀𝒀 for a ‘T’ section as per figure.
Solution:
Given Data,
20 cm
2 cm
30 cm
2 cm
𝒀
G
Step 6: Find Total M.I. of geometry (X andY Component)
Like - 𝑰 𝑿𝑿, 𝑰 𝒀𝒀
𝐼 𝑋𝑋 = 𝐼 𝑋𝑋1 + 𝐼 𝑋𝑋2
𝐼 𝑋𝑋 = 3699.7 + 6957.6
𝐼 𝑋𝑋 = 10657.3 𝑐𝑚4
𝐼 𝑌𝑌 = 𝐼 𝑌𝑌1 + 𝐼 𝑌𝑌2
𝐼 𝑌𝑌 = 1333.33 + 20
𝐼 𝑌𝑌 = 1353.33 𝑐𝑚4
X2
𝑿

45. Example on Composite Area (2D)
10 cm
2.5 cm
10 cm
2.5 cm
2.5
cm
10 cm
2.5 cm
10 cm
2.5 cm
2.5
cm
Example 2: Calculate 𝑰 𝑿𝑿 𝒂𝒏𝒅 𝑰 𝒀𝒀 for as per figure.
Solution:
Given Data, Step 1:Split figure from unknown geometry to known
geometry and draw X andY Axis.

46. Example on Composite Area (2D)
10 cm
2.5 cm
10 cm
2.5 cm
2.5
cm
Note:
𝑋 = Distance between Centroid of element toY axis,
𝑌 = Distance between Centroid of element to X axis,
Now Consider Part 1
Area of Element 1 = 2.5 x 10 = 25 𝑐𝑚2
𝑋1=
𝐿
2
=
10
2
= 5 cm
Y1 = 2.5 + 10 +
𝑊
2
= 2.5 + 10 +
2.5
2
= 13.75 cm
X1
Y1
Example 2: Calculate 𝑰 𝑿𝑿 𝒂𝒏𝒅 𝑰 𝒀𝒀 for as per figure.
Solution:
Given Data, Step 2 :Find Centroid of Individual geometry (X andY Component)
Like - 𝑿 𝟏, 𝒀 𝟏, 𝑿 𝟐, 𝒀 𝟐 (Use Centroid Formula)

47. Example on Composite Area (2D)
10 cm
2.5 cm
10 cm
2.5 cm
Note:
𝑋 = Distance between Centroid of element toY axis,
𝑌 = Distance between Centroid of element to X axis,
Now Consider Part 2
Area of Element 1 = 2.5 x 10 = 25 𝑐𝑚2
𝑋2=
𝐿
2
=
2.5
2
= 1.25 cm
Y2 = 2.5 +
𝑊
2
= 2.5 +
10
2
= 7.5 cm
X2
Y2
Example 2: Calculate 𝑰 𝑿𝑿 𝒂𝒏𝒅 𝑰 𝒀𝒀 for as per figure.
Solution:
Given Data,

48. Example on Composite Area (2D)
10 cm
2.5 cm
10 cm
2.5 cm
2.5
cm
Note:
𝑋 = Distance between Centroid of element toY axis,
𝑌 = Distance between Centroid of element to X axis,
Now Consider part 3
Area of Element 3 = 2.5 x 10 = 25 𝑐𝑚2
𝑋3=
𝐿
2
=
10
2
= 5 cm
Y3 =
𝑊
2
=
2.5
2
= 1.25 cm
X3
Y3
Example 2: Calculate 𝑰 𝑿𝑿 𝒂𝒏𝒅 𝑰 𝒀𝒀 for as per figure.
Solution:
Given Data,

49. Example on Composite Area (2D)
10 cm
2.5 cm
10 cm
2.5 cm
Note:
𝑋 = Distance between Centroid of element toY axis,
𝑌 = Distance between Centroid of element to X axis,
Part Area (𝒄𝒎 𝟐 ) Xi Yi
1 25 X1=5 cm Y1=13.75 cm
2 25 X2=1.25 cm Y2=7.5 cm
3 25 X3=5 cm Y3=1.25 cm
𝑋 = Distance between Centroid of element toY axis,
𝑋 =
𝐴1 𝑋1+ 𝐴2 𝑋2+𝐴3 𝑋3
𝐴1+𝐴2+𝐴3
=
125+31.25+125
75
= 3.75 cm
𝑌 = Distance between Centroid of element to X axis,
𝑌 =
𝐴1 𝑌1+ 𝐴2 𝑌2+𝐴3 𝑌3
𝐴1+𝐴2+𝐴3
=
343.75+187.5+31.25
75
= 7.5 cm
𝑿
𝒀
G
Example 2: Calculate 𝑰 𝑿𝑿 𝒂𝒏𝒅 𝑰 𝒀𝒀 for as per figure.
Solution:
Given Data,
Step 3: Find centroid of whole geometry.
Like - 𝑿 & 𝒀

50. Example on Composite Area (2D)
10 cm
2.5 cm
10 cm
2.5 cm
Note:
𝑋 = Distance between Centroid of element toY axis,
𝑌 = Distance between Centroid of element to X axis,
𝑿
𝒀
G
Example 2: Calculate 𝑰 𝑿𝑿 𝒂𝒏𝒅 𝑰 𝒀𝒀 for as per figure.
Solution:
Given Data,
Step 4: Find Moment about x axis of Individual geometry
Like - 𝑰 𝑿𝑿𝟏, 𝑰 𝑿𝑿𝟐
𝑰 𝑿𝑿𝟏 = 𝑰 𝒈𝟏 + 𝒂𝒉 𝟐
𝐼𝑔 = M.I. of area about C.G.
Let’s Find 𝐼𝑔1 =
𝑏𝑑3
12
=
(10)(2.5)3
12
= 13.02 𝑐𝑚4
Area, a = b X d = 10 X 2.5 = 25 𝑐𝑚2
h = 𝑌1 - 𝑌 = 13.75 – 7.5 = 6.25 cm
𝑰 𝑿𝑿𝟏 = 𝑰 𝒈𝟏 + 𝒂𝒉 𝟐 = 13.02 + (25 X 6.252)
= 989.58 𝑐𝑚4
X1
Y1

51. Example on Composite Area (2D)
10 cm
2.5 cm
10 cm
2.5 cm
Note:
𝑋 = Distance between Centroid of element toY axis,
𝑌 = Distance between Centroid of element to X axis,
𝑿
𝒀
G
Example 2: Calculate 𝑰 𝑿𝑿 𝒂𝒏𝒅 𝑰 𝒀𝒀 for as per figure.
Solution:
Given Data,
𝑰 𝑿𝑿𝟐 = 𝑰 𝒈𝟐 + 𝒂𝒉 𝟐
𝐼𝑔 = M.I. of area about C.G.
Let’s Find 𝐼𝑔2 =
𝑏𝑑3
12
=
(2.5)(10)3
12
= 208.33 𝑐𝑚4
Area, a = b X d = 10 X 2.5 = 25 𝑐𝑚2
h = 𝑌2 - 𝑌 = 7.5 – 7.5 = 0 cm
𝑰 𝑿𝑿𝟐 = 𝑰 𝒈𝟐 + 𝒂𝒉 𝟐 = 208.33 + (25 X0)
= 208.33 𝑐𝑚4
X2
Y2

52. Example on Composite Area (2D)
10 cm
2.5 cm
10 cm
2.5 cm
𝑿
𝒀
G
Example 2: Calculate 𝑰 𝑿𝑿 𝒂𝒏𝒅 𝑰 𝒀𝒀 for as per figure.
Solution:
Given Data,
𝑰 𝑿𝑿𝟑 = 𝑰 𝒈𝟑 + 𝒂𝒉 𝟐
𝐼𝑔 = M.I. of area about C.G.
Let’s Find 𝐼𝑔3 =
𝑏𝑑3
12
=
(10)(2.5)3
12
= 13.02 𝑐𝑚4
Area, a = b X d = 2.5 X 10 = 25 𝑐𝑚2
h = 𝑌3 - 𝑌 = 1.25 – 7.5 = - 6.25 cm
𝑰 𝑿𝑿𝟑 = 𝑰 𝒈𝟑 + 𝒂𝒉 𝟐 = 13.02 + (25 X6.252)
= 989.58 𝑐𝑚4
X3
Y3

53. Example on Composite Area (2D)
10 cm
2.5 cm
10 cm
2.5 cm
Note:
𝑋 = Distance between Centroid of element toY axis,
𝑌 = Distance between Centroid of element to X axis,
𝑿
𝒀
G
Example 2: Calculate 𝑰 𝑿𝑿 𝒂𝒏𝒅 𝑰 𝒀𝒀 for as per figure.
Solution:
Given Data,
Step 5: Find Moment aboutY axis of Individual geometry
Like - 𝑰 𝒀𝒀𝟏, 𝑰 𝒀𝒀𝟐
𝑰 𝒀𝒀𝟏 = 𝑰 𝒈𝟏 + 𝒂𝒉 𝟐
𝐼𝑔 = M.I. of area about C.G.
Let’s Find 𝐼𝑔1 =
𝑑𝑏3
12
=
(2.5)(10)3
12
= 208.33 𝑐𝑚4
Area, a = b X d = 10 X 2.5 = 25 𝑐𝑚2
h = 𝑋1 - 𝑋 = 5 – 3.75 = 1.25 cm
𝑰 𝒀𝒀𝟏 = 𝑰 𝒈𝟏 + 𝒂𝒉 𝟐 = 208.33 + (25 X 1.252)
= 247.39 𝑐𝑚4
X1
Y1

54. Example on Composite Area (2D)
10 cm
2.5 cm
10 cm
2.5 cm
Note:
𝑋 = Distance between Centroid of element toY axis,
𝑌 = Distance between Centroid of element to X axis,
𝑿
𝒀
G
Example 2: Calculate 𝑰 𝑿𝑿 𝒂𝒏𝒅 𝑰 𝒀𝒀 for as per figure.
Solution:
Given Data,
𝑰 𝒀𝒀𝟐 = 𝑰 𝒈𝟐 + 𝒂𝒉 𝟐
𝐼𝑔 = M.I. of area about C.G.
Let’s Find 𝐼𝑔2 =
𝑑𝑏3
12
=
(10)(2.5)3
12
= 13.02 𝑐𝑚4
Area, a = b X d = 10 X 2.5 = 25 𝑐𝑚2
h = 𝑋2 - 𝑋 = 1.25 – 3.75 = -2.5 cm
𝑰 𝒀𝒀𝟐 = 𝑰 𝒈𝟐 + 𝒂𝒉 𝟐 = 13.02 + (25 X 2.52)
= 169.27 𝑐𝑚4
X2
Y2

55. Example on Composite Area (2D)
10 cm
2.5 cm
10 cm
2.5 cm
Note:
𝑋 = Distance between Centroid of element toY axis,
𝑌 = Distance between Centroid of element to X axis,
𝑿
𝒀
G
Example 2: Calculate 𝑰 𝑿𝑿 𝒂𝒏𝒅 𝑰 𝒀𝒀 for as per figure.
Solution:
Given Data,
𝑰 𝒀𝒀𝟑 = 𝑰 𝒈𝟑 + 𝒂𝒉 𝟐
𝐼𝑔 = M.I. of area about C.G.
Let’s Find 𝐼𝑔3 =
𝑑𝑏3
12
=
(2.5)(10)3
12
= 208.33 𝑐𝑚4
Area, a = b X d = 10 X 2.5 = 25 𝑐𝑚2
h = 𝑋3 - 𝑋 = 5 – 3.75 = 1.25 cm
𝑰 𝒀𝒀𝟑 = 𝑰 𝒈𝟑 + 𝒂𝒉 𝟐 = 208.33 + (25 X 1.252)
= 247.39 𝑐𝑚4
X3
Y3

56. Example on Composite Area (2D)
10 cm
2.5 cm
10 cm
2.5 cm
Note:
𝑋 = Distance between Centroid of element toY axis,
𝑌 = Distance between Centroid of element to X axis,
𝑿
𝒀
G
Example 2: Calculate 𝑰 𝑿𝑿 𝒂𝒏𝒅 𝑰 𝒀𝒀 for as per figure.
Solution:
Given Data,
Step 6: Find Total M.I. of geometry (X andY Component)
Like - 𝑰 𝑿𝑿, 𝑰 𝒀𝒀
𝐼 𝑋𝑋 = 𝐼 𝑋𝑋1 + 𝐼 𝑋𝑋2 + 𝐼 𝑋𝑋3
𝐼 𝑋𝑋 = 989.58 + 208.33 + 989.58
𝐼 𝑋𝑋 = 2187.49 𝑐𝑚4
𝐼 𝑌𝑌 = 𝐼 𝑌𝑌1 + 𝐼 𝑌𝑌2 + 𝐼 𝑌𝑌3
𝐼 𝑌𝑌 = 247.39 + 169.27 +247.39
𝐼 𝑌𝑌 = 664.05 𝑐𝑚4

57. Example on Composite Area (2D)
Example 3: Calculate 𝑰 𝑿𝑿 𝒂𝒏𝒅 𝑰 𝒀𝒀 for a given figure.
Solution:
Given Data,
2 cm
2 cm
2 cm
8 cm
16 cm
Part Area (𝒄𝒎 𝟐 ) Xi Yi
1 16 X1=4 cm Y1=13 cm
2 20 X2=7 cm Y2=7 cm
3 32 X3=14 cm Y3=1 cm
𝑋 = Distance between Centroid of element toY axis,
𝑋 =
𝐴1 𝑋1+ 𝐴2 𝑋2+𝐴3 𝑋3
𝐴1+𝐴2+𝐴3
=
64+140+448
68
= 9.58 cm
𝑌 = Distance between Centroid of element to X axis,
𝑌 =
𝐴1 𝑌1+ 𝐴2 𝑌2+𝐴3 𝑌3
𝐴1+𝐴2+𝐴3
=
208+140+32
68
= 5.58 cm
𝑿 𝒀
G
12 cm
Y1=13 cm,Y2=7 cm,Y3=1 cm, 𝒀 = 5.58 cm

58. Example on Composite Area (2D)
Example 3: Calculate 𝑰 𝑿𝑿 𝒂𝒏𝒅 𝑰 𝒀𝒀 for a given figure.
Solution:
Given Data,
2 cm
2 cm
2 cm
8 cm
16 cm
𝑿 𝒀
12 cm
𝐼 𝑋𝑋 = 𝐼 𝑋𝑋1 + 𝐼 𝑋𝑋2 + 𝐼 𝑋𝑋3
𝐼 𝑋𝑋 = 886.23+206.99+681.99
𝐼 𝑋𝑋 = 1775.21 𝑐𝑚4
1
2
3
𝑏. 𝑑3
12
= 5.33
𝑏. 𝑑3
12
= 166.67
𝑏. 𝑑3
12
= 10.66
16
20
32
7.42
1.42
-4.58
𝑰 𝑿𝑿𝟏 = 886.23
𝑰 𝑿𝑿𝟐 = 206.99
𝑰 𝑿𝑿𝟑 = 681.90
Y1=13 cm,Y2=7 cm,Y3=1 cm, 𝒀 = 5.58 cm

59. Example on Composite Area (2D)
Example 3: Calculate 𝑰 𝑿𝑿 𝒂𝒏𝒅 𝑰 𝒀𝒀 for a given figure.
Solution:
Given Data,
2 cm
2 cm
2 cm
8 cm
16 cm
𝑿 𝒀
G
12 cm
𝐼 𝑌𝑌 = 𝐼 𝑌𝑌1 + 𝐼 𝑌𝑌2 + 𝐼 𝑌𝑌3
𝐼 𝑌𝑌 = 583.51+139.80+1307.83
𝐼 𝑌𝑌 = 2031.14 𝑐𝑚4
1
2
3
𝑑. 𝑏3
12
= 85.33
𝑑. 𝑏3
12
= 6.67
𝑑. 𝑏3
12
= 682.67
16
20
32
-5.58
-2.58
4.42
𝑰 𝒀𝒀𝟏 = 583.51
𝑰 𝒀𝒀𝟐 = 139.80
𝑰 𝒀𝒀𝟑 = 1307.83
X1=4 cm, X2=7 cm, X3=14 cm, 𝑿 = 9.58 cm

60. Example on Composite Area (2D)
Example 4: Calculate 𝑰 𝑿𝑿 𝒂𝒏𝒅 𝑰 𝒀𝒀 for a given figure.
Solution:
Given Data,
9cm
3 cm
6 cm
𝑿
𝒀
Element Area (𝒄𝒎 𝟐 ) Xi Yi
1 27 X1=1.5 cm Y1=4.5 cm
2 13.5 X2=4 cm Y2=3 cm
𝑋 = Distance between Centroid of element toY axis,
𝑋 =
𝐴1 𝑋1+ 𝐴2 𝑋2
𝐴1+𝐴2
= 2.33 cm
𝑌 = Distance between Centroid of element to X axis,
𝑌 =
𝐴1 𝑌1+ 𝐴2 𝑌2
𝐴1+𝐴2
= 4 cm

61. Example on Composite Area (2D)
Example 4: Calculate 𝑰 𝑿𝑿 𝒂𝒏𝒅 𝑰 𝒀𝒀 for a given figure.
Solution:
Given Data,
9cm
3 cm
6 cm
𝑿
𝒀
1
2
𝑏. 𝑑3
12
= 182.25
𝑏.ℎ3
36
= 60.75
27
13.5
Y1=4.5 cm,Y2=3 cm, 𝒀 = 4 cm
0.5
- 1
189
74.25
𝐼 𝑋𝑋 = 𝐼 𝑋𝑋1 + 𝐼 𝑋𝑋2
𝐼 𝑋𝑋 = 189+74.25
𝐼 𝑋𝑋 = 263.25 𝑐𝑚4

62. Example on Composite Area (2D)
Example 4: Calculate 𝑰 𝑿𝑿 𝒂𝒏𝒅 𝑰 𝒀𝒀 for a given figure.
Solution:
Given Data,
9cm
3 cm
6 cm
𝑿
𝒀
1
2
𝑑. 𝑏3
12
= 20.25
ℎ.𝑏3
36
= 6.75
27
13.5
-0.83
1.67
38.85
44.40
𝐼 𝑌𝑌 = 𝐼 𝑌𝑌1 + 𝐼 𝑌𝑌2
𝐼 𝑌𝑌 = 38.85+44.40
𝐼 𝑌𝑌 = 83.25 𝑐𝑚4
X1=1.5 cm, X2=4 cm, 𝑿 = 2.33 cm

63. Example on Composite Area (2D)
Example 5: A rectangular section is 400 mm wide and 800 mm deep. Two circular holes of 200 mm diameter each are
cut on Y-Y axis at distance 200 mm and 600 mm from the top edge. Calculate M.I. about xx and yy axis.
Solution:
Given Data,
800
400
200200200200
1
2
3
A1 = 320000
A2 = 31415.92
A3=31415.92
X1= 0
X2= 0
X3= 0
Y1= 0 mm
Y2= 200 mm
Y3= -200 mm
𝑋 = Distance between Centroid of element toY axis,
𝑋 =
𝐴1 𝑋1− 𝐴2 𝑋2−𝐴3 𝑋3
𝐴1+𝐴2+𝐴3
= 0 mm
𝑌 = Distance between Centroid of element to X axis,
𝑌 =
𝐴1 𝑌1− 𝐴2 𝑌2−𝐴3 𝑌3
𝐴1−𝐴2−𝐴3
= 0 mm

64. Example on Composite Area (2D)
Example 5: A rectangular section is 400 mm wide and 800 mm deep. Two circular holes of 200 mm diameter each are
cut on Y-Y axis at distance 200 mm and 600 mm from the top edge. Calculate M.I. about xx and yy axis.
Solution:
Given Data,
800
400
200200200200
1
2
3
𝜋
64
𝑑4 = 7.85𝑋107
𝜋
64
𝑑4 = 7.85𝑋107
𝑏. 𝑑3
12
= 1.7𝑋1010 320000
31415.92
31415.92
0
200
200
1.7𝑋1010
1.3𝑋109
1.3𝑋109
𝐼 𝑋𝑋 = 𝐼 𝑋𝑋1 − 𝐼 𝑋𝑋2 − 𝐼 𝑋𝑋3
𝐼 𝑋𝑋 = (1.7𝑋1010) − 1.3𝑋109 − 1.3𝑋109
𝐼 𝑋𝑋 = 1.44 𝑋 1010 𝑚𝑚4
Y1=0 mm,Y2=200 mm,Y3=-200 mm, 𝒀 = 0 mm

65. Example on Composite Area (2D)
Example 5: A rectangular section is 400 mm wide and 800 mm deep. Two circular holes of 200 mm diameter each are
cut on Y-Y axis at distance 200 mm and 600 mm from the top edge. Calculate M.I. about xx and yy axis.
Solution:
Given Data,
800
400
200200200200
1
2
3
𝑑. 𝑏3
12
= 4.27𝑋109
𝜋
64
𝑑4
= 7.85𝑋107
𝜋
64
𝑑4 = 7.85𝑋107
320000
31415.92
31415.92
0
0
0
4.27𝑋109
7.85𝑋107
7.85𝑋107
𝐼 𝑌𝑌 = 𝐼 𝑌𝑌1 − 𝐼 𝑌𝑌2 − 𝐼 𝑌𝑌3
IYY = (4.27X109) –(7.85X107) − (7.85X107)
𝐼 𝑌𝑌 = 4.1 𝑋109 𝑚𝑚4
X1=0 mm, X2=0 mm, X3=0 mm, 𝑿 = 0 mm

66. Example 6: Calculate 𝑰 𝑿𝑿 𝒂𝒏𝒅 𝑰 𝒀𝒀 for a given figure.
Solution:
Given Data,
Example on Composite Area (2D)
1
2
3
A1 = 30X100=3000
A2 = 25X100=2500
X1= 100 Y1= 20+20+80+
30
2
= 135
Y2= 20+
100
2
= 70
Y4= 20+
20
3
= 26.664
5 A5 =
1
2
𝑋87.5𝑋20=875
A4 =
1
2
𝑋87.5𝑋20=875
A3 = 20X200=4000
X2= 100
X3= 100
X4=
2
3
𝑋87.5 = 58.33
X5=
87.5+25+
87.5
3
=141.67
Y3=
20
2
= 10
Y5= 20+
20
3
= 26.66
𝑋 = Distance between Centroid of element toY axis,
𝑋 =
𝐴1 𝑋1+ 𝐴2 𝑋2+𝐴3 𝑋3+𝐴4 𝑋4+𝐴5 𝑋5
𝐴1+𝐴2+𝐴3+𝐴4+𝐴5
= 100 mm
𝑌 = Distance between Centroid of element to X axis,
𝑌 =
𝐴1 𝑌1+ 𝐴2 𝑌2+𝐴3 𝑌3+𝐴4 𝑌4+𝐴5 𝑌5
𝐴1+𝐴2+𝐴3+𝐴4+𝐴5
= 59.26 mm

67. Example 6: Calculate 𝑰 𝑿𝑿 𝒂𝒏𝒅 𝑰 𝒀𝒀 for a given figure.
Solution:
Given Data,
Example on Composite Area (2D)
1
2
3
4
5
𝑏. 𝑑3
12
= 2.25𝑋105
𝑏. 𝑑3
12
= 2.08𝑋106
𝑏. 𝑑3
12
= 1.33𝑋105
𝑏. ℎ3
36
= 1.94𝑋104
𝑏. ℎ3
36
= 1.94𝑋104
3000
2500
4000
875
875
Y1=135,Y2=70,Y3=10,Y4=26.66,Y5=26.66, 𝒀 = 69.91 mm
65.09
0.09
-59.91
-43.25
-43.25
1.29𝑋107
2.08𝑋106
1.45𝑋107
1.66𝑋106
1.66𝑋106
𝐼 𝑋𝑋 = 𝐼 𝑋𝑋1 + 𝐼 𝑋𝑋2 + 𝐼 𝑋𝑋3 + 𝐼 𝑋𝑋4 + 𝐼 𝑋𝑋5
𝑰 𝑿𝑿 = 3.28 X 𝟏𝟎 𝟕 𝒎𝒎 𝟒

68. Example 6: Calculate 𝑰 𝑿𝑿 𝒂𝒏𝒅 𝑰 𝒀𝒀 for a given figure.
Solution:
Given Data,
Example on Composite Area (2D)
1
2
3
4
5
𝑑. 𝑏3
12
= 2.5𝑋106
𝑑. 𝑏3
12
= 1.30𝑋105
𝑑. 𝑏3
12
= 1.33𝑋107
ℎ.𝑏3
36
= 3.72𝑋105
ℎ.𝑏3
36
= 3.72𝑋105
3000
2500
4000
875
875
X1=X2= X3=100, X4=58.33, X5=141.67, 𝑿 = 100 mm
0
0
0
-41.67
41.67
2.5𝑋106
1.30𝑋105
1.33𝑋107
1.89𝑋106
1.89𝑋106
𝐼 𝑌𝑌 = 𝐼 𝑌𝑌1 + 𝐼 𝑌𝑌2 + 𝐼 𝑌𝑌3 + 𝐼 𝑌𝑌4 + 𝐼 𝑌𝑌5
𝑰 𝒀𝒀 = 1.97 X 𝟏𝟎 𝟕 𝒎𝒎 𝟒

69. Example on Composite Area (2D)
Example 7: Determine location of centroid, 𝑰 𝑿𝑿 𝒂𝒏𝒅 𝑰 𝒀𝒀 of a lamina shown in figure. (All Dimension in mm)
Solution:
Given Data, Part Area (𝒎𝒎 𝟐 ) Xi (mm) Yi (mm)
1 1,20,000 X1=200 Y1= 150
2 30,000 X2=400+
200
3
= 466.67 Y2=
300
3
= 100
3 𝜋.𝑟2
2
=35342.91 X3= 100 + 150 =250 Y3=
4.𝑟
3.𝜋
= 63.66
𝑋 = Distance between Centroid of element toY axis,
𝑋 =
𝐴1 𝑋1+ 𝐴2 𝑋2−𝐴3 𝑋3
𝐴1+𝐴2−𝐴3
= 254.36 mm
𝑌 = Distance between Centroid of element to X axis,
𝑌 =
𝐴1 𝑌1+ 𝐴2 𝑌2−𝐴3 𝑌3
𝐴1+𝐴2−𝐴3
= 163.53 mm
100 300 200
300
1
2
3

70. Example on Composite Area (2D)
Example 7: Determine location of centroid, 𝑰 𝑿𝑿 𝒂𝒏𝒅 𝑰 𝒀𝒀 of a lamina shown in figure. (All Dimension in mm)
Solution:
Given Data,
100 300 200
300
1
2
3
1
2
3
𝑏. 𝑑3
12
= 9𝑋108
0.11.𝑟4
= 5.57𝑋107
𝑏. ℎ3
36
= 1.5𝑋108
1,20,000
30,000
35342.91
Y1=150,Y2=100,Y3=63.66, 𝒀 = 163.53 mm -13.53
-63.53
-99.87
9.21𝑋108
2.71𝑋108
4.08𝑋108
𝐼 𝑋𝑋 = 𝐼 𝑋𝑋1 + 𝐼 𝑋𝑋2 − 𝐼 𝑋𝑋3
𝑰 𝑿𝑿 = 7.84 X 𝟏𝟎 𝟖
𝒎𝒎 𝟒

71. Example on Composite Area (2D)
Example 7: Determine location of centroid, 𝑰 𝑿𝑿 𝒂𝒏𝒅 𝑰 𝒀𝒀 of a lamina shown in figure. (All Dimension in mm)
Solution:
Given Data,
100 300 200
300
1
2
3
1
2
3
𝑑. 𝑏3
12
= 1.6𝑋109
ℎ.𝑏3
36
= 6.67𝑋107
𝜋.𝑟4
8
= 1.99𝑋108
1,20,000
30,000
35342.91
X1=200, X2=466.67, X3=250, 𝑿 = 254.36 mm
-54.36
212.31
-4.36
1.95𝑋109
1.41𝑋109
1.99𝑋108
𝐼 𝑌𝑌 = 𝐼 𝑌𝑌1 + 𝐼 𝑌𝑌2 − 𝐼 𝑌𝑌3
𝑰 𝒀𝒀 = 3.16 X 𝟏𝟎 𝟗
𝒎𝒎 𝟒

72. Example on Composite Area (2D)
Example 8: Determine location of centroid, 𝑰 𝑿𝑿 𝒂𝒏𝒅 𝑰 𝒀𝒀 of a lamina shown in figure. (All Dimension in mm)
Solution:
Given Data, Part Area
(𝒎𝒎 𝟐
)
Xi (mm) Yi (mm)
1 12X9=108 X1=
12
2
= 6 Y1=
9
2
= 4.5
2 27 X2= 6+
2
3
(6) = 10 Y2=
1
3
(9) = 3
3 9 X3=
1
3
(6) = 2 Y3=6+
2
3
(3) = 8
6 6
66
6
3
1 2
3
𝑋 = Distance between Centroid of element toY axis,
𝑋 =
𝐴1 𝑋1− 𝐴2 𝑋2−𝐴3 𝑋3
𝐴1−𝐴2−𝐴3
= 5 mm
𝑌 = Distance between Centroid of element to X axis,
𝑌 =
𝐴1 𝑌1+ 𝐴2 𝑌2−𝐴3 𝑌3
𝐴1+𝐴2−𝐴3
= 4.63 mm

73. Example on Composite Area (2D)
Example 8: Determine location of centroid, 𝑰 𝑿𝑿 𝒂𝒏𝒅 𝑰 𝒀𝒀 of a lamina shown in figure. (All Dimension in mm)
Solution:
Given Data,
6 6
66
6
3
1 2
3
1
2
3
𝑏. 𝑑3
12
= 729
𝑏. ℎ3
36
= 121.5
𝑏. ℎ3
36
= 4.5
108
27
9
Y1=4.5,Y2=3,Y3=8, 𝒀 = 4.63 mm -0.13
-1.63
3.37
730.82
193.24
106.71
𝐼 𝑋𝑋 = 𝐼 𝑋𝑋1 − 𝐼 𝑋𝑋2 − 𝐼 𝑋𝑋3
𝑰 𝑿𝑿 = 𝟒𝟑𝟎. 𝟖𝟕 𝒎𝒎 𝟒

74. Example on Composite Area (2D)
Example 8: Determine location of centroid, 𝑰 𝑿𝑿 𝒂𝒏𝒅 𝑰 𝒀𝒀 of a lamina shown in figure. (All Dimension in mm)
Solution:
Given Data,
6 6
66
6
3
1 2
3
1
2
3
108
27
9
1
5
-3
1404
729
99
𝐼 𝑌𝑌 = 𝐼 𝑌𝑌1 − 𝐼 𝑌𝑌2 − 𝐼 𝑌𝑌3
𝑰 𝒀𝒀 = 𝟓𝟕𝟔 𝒎𝒎 𝟒
𝑑. 𝑏3
12
= 1296
ℎ.𝑏3
36
= 54
ℎ.𝑏3
36
= 18
X1=6, X2=10, X3=2, 𝑿 = 5 mm

75. Topic will be cover:
1. Moment of Inertia:
h) Formula of Moment of Inertia
i) Steps of problem solving
j) Example (1 to 8 With Solution)
k) Example (Home Work)
Department of Mechanical
Engineering.
Prof. Malay Badodariya
malay.badodariya@marwadie
ducation.edu.in
Unit no.: 5
Subject Name: FMD
01ME0305

76. Example on Composite Area (2D)
Example 9: Determine location of centroid, 𝑰 𝑿𝑿 𝒂𝒏𝒅 𝑰 𝒀𝒀 of a lamina shown in figure. (All Dimension in mm)
Solution:
Given Data,
3
6
3
3 6
3
3
6
4

77. Department of Mechanical
Engineering.
Prof. Malay Badodariya
malay.badodariya@marwadie
ducation.edu.in