Moment of inertia

Dungarwal Ankita Sanjay

1. Name of Faculty : Ms. A. S. Dungarwal
Lecturer in Civil Engg. Department,
SHHJB Polytechnic, Chandwad

2. Unit Title:
Moment of Inertia (M.I.)
Course Outcome (CO) :
Articulate Practical
Application Of Moment Of Inertia Of
Symmetrical And Unsymmetrical Structural
Section
Dungarwal A.S.

3. CONTENT:
Center of Gravity
Definition, Radius Of Gyration, Section Modulus, parallel
& Perpendicular Axes Theorems, M.I Of Basic
Geometrical Sections.
M.I. of Symmetrical & Unsymmetrical Sections- I,C,T,L-
sections, Hollow Sections ,Built up Section.
Polar M. I. of Solid Circular section.
Dungarwal A.S.

4. CENTER OF GRAVITY (C.G.)
The C.G. Refers Only To The Vertical Direction On
Which Gravity Acts.
The Center Of Mass Does Not Depends On Vertical
Orientation W.R.T. Gravity.
At Uniform Gravitational Field Center Of Mass & Center
Of Gravity Are Same.
X =
??????
1
�
1
:??????
2
�
2
:??????
3
�
3
: ……………….:????????????�??????
??????
1
:??????
2
:??????
3
: …………..:????????????


Y =
??????
1
�
1
:??????
2
�
2
:??????
3
�
3
: ……………….:????????????�??????
??????
1
:??????
2
:??????
3
: …………..:????????????

Unit- m ,cm ,mm , etc.


Centroidal Axis X & Y
Dungarwal A.S.

5. EXAMPLES:
C.G.
X
X -AXIS
Y -AXIS

Y

b
d
X =
??????
2

Y =
??????
2

X -
AXIS

D
X = Y =
??????
2
= R
X
WHERE,

b = BREADTH
d = DEPTH
D = DIAMETER OF
CIRCLE
Dungarwal A.S.

6. ①
X1

X2

Y1
Y2
Y
X
Draw The Section in x-y coordinate.

Split section into basic geometric shapes.

Find C.G. of each section by diagonal
crossing method.

Draw X & Y axis for each section from its
C.G.

Calculate Area (A1,A2,….An) of each section
and distance of C.G.(X1,X2,….., Xn &
Y1,Y2,….., Yn)

Put this value in Formula

Find distance of centroidal axis X & Y



X
Y
X1
X2
Y2
Y1
STEPS OF FINDING C.G. :
x
Y
Dungarwal A.S.

7. First Moment of Area
The moment of area about any axis is the product of area
of the distance of its centroid from that axis is called as first
Moment of Area.
C.G. A
Y
X X
M1=A X Y
Dungarwal A.S.

8. Second Moment Of Area
The moment of first moment area about the same axis
is called its second moment of area.

This is also called as Moment of Inertia of section.

It is written as M.I. or I

Unit – mm
4
,cm
4
,m
4 ,
etc.

M.I. =M1 x Y =(A X Y )X Y = A X Y
2
Dungarwal A.S.

9. DEFINITION OF Moment of Inertia (M.I.)
M.I. of body about any axis is equal to the product of
the area of the body and the square of the distance of
its centroid from that axis.
i.e M.I. = A x Y
2
Generally , M.I. = A x h
2
Where, A= area of section,
h= distance of centroid of area of a section from the axis
considered.
I= M.I. about the axis considered.

Unit – m
4
, cm
4
,mm
4
,etc.
Dungarwal A.S.

10. Dungarwal A.S.

11. Radius of Gyration
The distance at which the area ‘A’ is supposed to be
concentrated to give the same Moment of Inertia is called
as Radius of Gyration.








Unit – mm, cm, m ,etc.




A1
A2
A3
C.G.
An
Y1
Y2
Y Yn
Y3
Ixx = ????????????????????????
2

=A1Y1
2
+A2Y2
2
+…..



k
Ixx = Ak
2

Kxx =??????��/??????
Kyy= ??????��/??????
A
where, k= radius of gyration about
the axis considered.
x
x x x
Dungarwal A.S.

12. Section Modulus
The ratio of M.I. and distance from Neutral axis (N.A.)
to the extreme fiber is called as Section Modulus.

It is denoted by ‘Z’ or ‘S’.

Unit – mm
3
,cm
3
,m
3
, etc

Formula: Section modulus = Z=
??????
�

Dungarwal A.S.

13. Theorem of parallel axis :
It states that, “The M.I. of a plane section about any
axis (PQ) parallel to the centroidal axis is equal to the
M.I. of section about the centroidal axis (XX) plus (+)
the product (x) of the area (A) of the section &square
of the distance (h
2
)between the two axis”.
A x x
P Q
h1
Y
Y
A
B
A
h2
I
PQ = I
XX + Ah1
2
I
AB= I
YY +Ah2
2
Dungarwal A.S.

14. Theorem of Perpendicular Axis (Polar axis
theorem):
The third axis ZZ is perpendicular to XX &YY axis is
called as Polar Axis.
It’s State that, “ If I
XX & I
YY be the M.I. of a Plane
section About two mutually perpendicular axis passing
through origin ‘O’ , then moment of inertia of polar
axis (I
ZZ) denoted by I
P is given by I
XX +I
YY
I
P= I
ZZ= I
XX + I
YY
Y
Polar axis
X
Z
Dungarwal A.S.

15. Dungarwal A.S.

16. Dungarwal A.S.

17. Dungarwal A.S.

18. Dungarwal A.S.

19. Numerical
Problems
Dungarwal A.S.

20. Problems
Find Ixx and Iyy and also kxx and kyy
200 mm
450
mm
Solution:
Given- b=200mm ,A=b x d=200 x450=90x10
3
mm2
d=450mm
Find- Ixx , Iyy, Kxx, Kyy
Formula- Ixx=
????????????
3
12
, Iyy=
????????????
3
12
, kxx=
??????��
??????

Soln-
Ixx=
????????????
3
12
Iyy=
????????????
3
12

Ixx=
200 � 4503
12
Iyy=
450 � 2003
12

s
Ixx=1.518x10
9
mm
4
Iyy=300x10
6
mm
4

Kxx= kyy=
Dungarwal A.S.

21. Find M.I of circular lamina about any tangent .
P
X
Q
X
D/2
Sol:
I
PQ = Ixx + Ah
2
=
π
64
D
4
+ [
π
4
D
2
x (
D
2
)
2
]
=
π
64
x 100
4
+ [
π
4
x 100
2
x (
100
2
)
2
]
I
PQ = 24.54 x 10
6
mm
4
Ans:
Given:
D= 100mm
Find:
MI @ tangent PQ i.e. I
PQ
Formula:
I
PQ = Ixx + Ah2

Dungarwal A.S.

22. A T-section has a flange 100mm x 10mm and web 10mm x 120mm. Total
depth of section is 130 mm. Calculate MI about Xx, yy.
Y1 X1= X2
Y2
� = 50mm due symmetry about
Y axis
� =89.55mm
�
�
Sol:
Ixx = I
XX1 + I
XX2
= I
X1X1+ A1h1
2
+ I
X2X2+A2h2
2

= [
??????1??????1
3
12
+(b1xd1 x h1
2
)]
+[
??????2??????2
3
12
+(??????2xd2 x h2
2
)]
=[
100�10
3
12
+(100x10 x 35.45
2
)]
+[
10�120
3
12
+(10x120 x 29.55
2
)]
Ixx =3.752 x10
6
mm
4
h2
h1
Dungarwal A.S.

23. Iyy = I
yy1 + I
yy2
= I
y1y1+ A1h1
2
+ I
y2y2+A2h2
2

= [
??????1??????1
3
12
+(b1xd1 x h1
2
)] +[
??????2??????2
3
12
+(??????2xd2 x h2
2
)]
Due to symmetry about Y-Y axis Value of h1 and h2 is 0
=[
10�100
3
12
+(100x10 x 0

)] +[
120�10
3
12
+(10x120 x 0

)]
Iyy =843.33x10
3
mm
4
Dungarwal A.S.

24. Sol:
Ixx = I
XX1 + I
XX2
= I
X1X1+ A1h1
2
+ I
X1X1+A1h1
2

= [
??????1??????1
3
12
+(b1xd1 x h1
2
)]
+[
??????2??????2
3
12
+(??????2xd2 x h2
2
)]
=[
50�10
3
12
+(50x10 x 29.5
2
)]
+[
8�90
3
12
+(8x90 x 20.5
2
)]
Ixx =1.277 x10
6
mm
4
� = 12.6mm
� =34.5mm
Y1
X2
Y2
�
�
h2
X1
h1
X1
A L-section has a flange 50mm x 10mm and web 8mm x 90mm.
Total depth of section is 100 mm. Calculate MI about Xx, yy.
Dungarwal A.S.

25. Y1
Y2
�
�
h1 h2
X2
X1
Iyy = I
yy1 + I
yy2
= I
y1y1+ A1h1
2
+ I
y2y2+A2h2
2

= [
??????1??????1
3
12
+(b1xd1 x h1
2
)]
+[
??????2??????2
3
12
+(??????2xd2 x h2
2
)]
=[
10�50
3
12
+(50x10 x 8.6
2
)]
+[
90�8
3
12
+(90x8 x 12.4
2
)]
Iyy =238.137x10
3
mm
4
Dungarwal A.S.

26. Y1
X1= X2=100mm
Y2 �
h2
A Inverted T-section has a flange 200mm x 9mm and web 7.8. Total
depth of section is 130 mm. Calculate MI about Xx, yy.
Sol:
Ixx = I
XX1 + I
XX2
= I
X1X1+ A1h1
2
+ I
X2X2+A2h2
2

= [
??????1??????1
3
12
+(b1xd1 x h1
2
)]
+[
??????2??????2
3
12
+(??????2xd2 x h2
2
)]
=[
200�9
3
12
+(200x9x 14.14
2
)]
+[
7.8�121
3
12
+(7.8x121 x 35.86
2
)]
Ixx =1.77x10
6
mm
4
h1
� =200/2 =100mm
� =18.64mm
Dungarwal A.S.

27. Iyy = I
yy1 + I
yy2
= I
y1y1+ A1h1
2
+ I
y2y2+A2h2
2

= [
??????1??????1
3
12
+(b1xd1 x h1
2
)]
+[
??????2??????2
3
12
+(??????2xd2 x h2
2
)]
=[
9�200
3
12
+(9x200x 0
2
)]
+[
121�7.8
3
12
+(121x7.8 x 0
2
)]
Iyy =6.00x10
6
mm
4
Y1
X1= X2=100mm
Y2 �
�
Dungarwal A.S.

28. Y1
Y2
�
�
h1
h2
X2
X1
� =14.70mm
� =14.70mm

Sol:
Ixx = I
XX1 + I
XX2
= I
X1X1+ A1h1
2
+ I
X1X1+A1h1
2

= [
??????1??????1
3
12
+(b1xd1 x h1
2
)]
+[
??????2??????2
3
12
+(??????2xd2 x h2
2
)]
=[
50�6
3
12
+(50x6 x 11.70
2
)]
+[
6�44
3
12
+(6x44 x 13.3
2
)]
Ixx =131.468 x10
3
mm
4
Find MI of Angle Section of 50x50x6mm about X-X and Y-Y
axis.
x1 x2 y1 y2
50/2 6/2 6/2 (44/
2)+6
Dungarwal A.S.

29. Iyy = I
yy1 + I
yy2
= I
y1y1+ A1h1
2
+ I
y2y2+A2h2
2

= [
??????1??????1
3
12
+(b1xd1 x h1
2
)]
+[
??????2??????2
3
12
+(??????2xd2 x h2
2
)]
=[
6�50
3
12
+(50x6 x 10.3
2
)]
+[
44�6
3
12
+(44x6 x 11.7
2
)]
Iyy =131.257x10
3
mm
4
Y1
Y2
�
�
X2
X1
h1 h2
Dungarwal A.S.

30. Y1
Y2
�
�
h1
h2
X1=X2
X3
h3
Sol:
Ixx = I
XX1 + I
XX2 +I
XX3
= I
X1X1+ A1h1
2
+ I
X2X2+A2h2
2
+ I
X3X3+A3h3
2

= [
??????1??????1
3
12
+(b1xd1 x h1
2
)]+[
??????2??????2
3
12
+(??????2xd2 x h2
2
)]
+[
??????3??????3
3
12
+(??????3xd3 x h2
2
)]
=[
170�20
3
12
+(170x20 x 126.10
2
)]
+[
20�220
3
12
+(20x220 x 6.10
2
)]
+[
200�20
3
12
+(200x20 x 113.9
2
)]
Ixx =124.113 x10
6
mm
4
Y3
An I section is made up of 3 rectangle top flange 200x20mm, web
220x20mm, bottom flange 170x20mm find Ixx and Iyy
� =100mm ,� =136.10mm
Dungarwal A.S.

31. Sol:
Ixx = I
yy1 + I
yy2 +I
yy3
= I
y1y1+ A1h1
2
+ I
y2y2+A2h2
2
+ I
y3y3+A3h3
2

= [
??????1??????1
3
12
+(b1xd1 x h1
2
)]+[
??????2??????2
3
12
+
(??????2xd2 x h2
2
)] +[
??????3??????3
3
12
+(??????3xd3 x h2
2
)]
Due to symmetry about Y-Y axis Value of h1 , h2
and h3 are 0
=[
20�170
3
12
] +[
220�20
3
12
] +[
20�200
3
12
]
Ixx =124.113 x10
6
mm
4
Y2
�
�
X1
X1=X2
Y3
Dungarwal A.S.

32. 200mm
150mm
100mm
150mm
� =100mm
,� =129.125mm

�
�
Sol:
Ixx = I
XX1 + I
XX2
= I
X1X1+ A1h1
2
- I
X1X1+A1h1
2

= [
??????1??????1
3
12
+(b1xd1 x h1
2
)]
-[ +( x h2
2
)]
=[
200�300
3
12
+(200x300x 20.875
2
)]
- [
??????
64
x 150
4
+(
??????
4
x150
2
x 70.875
2
)]
Ixx =362.54 x10
6
mm
4
??????
4
x D
2
??????
64
x D
4
Dungarwal A.S.

33. Sol:
Ixx = I
yy1 + I
yy2
= I
y1y1+ A1h1
2
– I
y2y2+A2h2
2

= [
??????1??????13
12
+(b1xd1 x h1
2
)]
-[
??????
64
x D
4
+(
??????
4
xD
2
x h2
2
)]
Due to symmetry h1 and h2 are 0.
=[
300�200
3
12
] - [
??????
64
x 150
4
]
Iyy =175.149 x10
6
mm
4
200mm
150mm
�
�
150mm
100mm
Dungarwal A.S.