centroids moment of inertia: introduction to the concept

MODULE – IV CENTROIDS AND MOMENTS OF INERTIA OF ENGINEERING

SECTIONS

G. Ravindra Kumar, Associate Prof, Govt Engineering College, Chamarajanagar of 46

Centroids: Introduction to the concept, centroid of and area, centroid of basic

geometrical figures, computing centroid for T, L, I, Z and full / quadrant circular

sections and their built up sections, Numerical problems.

Moment of Inertia: Introduction to the concept, radius of gyration, Parallel axis

theorem, perpendicular axis theorem, Moment of inertia of basic planar figures,

computing Moment of inertia for T, L, I, Z and full / quadrant circular sections and

their built up sections, Numerical problems.

Centroid: The point at which the total area of a plane figure (like rectangle, square,

triangle, quadrilateral, circle etc.) is assumed to be concentrated, is known as the

centroid of that area.

Centre of Gravity: Centre of gravity of a body is the point through which the

whole weight of the body acts. A body is having only one centre of gravity for all

positions of the body. It is represented by C.G, or simple G.

Difference between Centre of Gravity and Centroid :

1. The terms centre of gravity applied to bodies with weight, and centroid applied to

lines, plane areas and volumes.

2. Centre of gravity of a body is a point through which resultant gravitational force

(weight) acts for any orientation of the body whereas centroid is a point in a

line/plain area/volume such that the moment of area about any axis through

that point is zero.

Reference Axis: The C.G. of a body is expressed in terms of the distances

measured from some axis of reference. In the plain figures the axis of reference is

generally the lowest line of the figure for calculating y and the left line of the figure

for calculating x . 1 – 1 and 2 – 2 axis are reference axis.

1

2

1

2

40mm

80mm

40mm

1

2

2

1


SECTIONS


Axis of symmetry:

In plain figures such as circle, square, rectangle etc., the section is

symmetrical about X –X and Y – Y axis. The area (section) is said to lie on axis of

symmetry. X –X and Y – Y axes are called symmetrical axis.

Centroid of regular plates/ Laminas / figures:

Rectangle: The c.g of a rectangle is at a point where its

diagonal meet each other.

Square: The c.g of a square is at a point where its

diagonal meet each other.

Circle: The c.g of a circle lies at its centre.

Triangle: The c.g of a triangle lies at a point where

the three medians of a triangle meet.

Y

Y

X X

Circle

Y

Y

X X

Rectangle

Y

Y

X X

Square

Y

X X

Y

cg

Y

Y

X X

Y

Y

X X

Square

cg

X

Y

X

Y

cg


SECTIONS


Centroid or centre of gravity of figures by the method of moments:

:X of ionDeterminat

Resultant R = a1 + a2 + a3 + a4 + ….

Applying principle of moments

...... xa xa xa xa X R 44332211

...... a a a a

...... xa xa xa xa X

4321

44332211

a

ax X

:Y of ionDeterminat

Resultant R = a1 + a2 + a3 + a4 + ….

Applying principle of moments

...... ya ya ya ya Y R 44332211

...... a a a a

...... ya ya ya ya y

4321

44332211

a

ay y

Centroid of an Equilateral Triangle:

Consider a triangle ABC with base

width „b‟ and height „h‟.

Consider a elemental strip „DE‟ of

width „b1‟ and thickness „dy‟ at a distance

„y‟ from the reference axis 1 – 1.

From the similar triangles „ABC‟

and „ADE‟.

h

y-h

b

b1

b h

y1b

h

y-h b1

Area opf elemental strip da = b1 x dy

dyb h

y1 ad

Moment of this elemental area about 1 – 1 axis = da x y =

y

b

A

h

h - y

dy

B C

1 1

b1

E D

O

Y

Area a1

X

Area a2 Area a3

Area a4

X1

X2

X3

X4

X

CG

Y4

O

Y

Area a1 Area a2

Area a3 Area a4

Y1

Y2

Y3 Y

X

CG


SECTIONS


bdy

h

yyydyb

h

y1

2

Moment of whole triangle about 1 – 1 axis h

0

32h

0

2

h3

y

2

ybbdy

h

yyay

6

bh

h6

h2h3b

h3

h

2

hbay

23332

2

bhhb

2

1a triangle whole of Area

2

bh6

bh

a

ay y base from centroid of Distance

2

base from 3

h y

The triangle is symmetrical about vertical axis (Y – Y )

axis.

Centroid of an Right angled Triangle:

Consider a triangle ABC with base width „b‟ and

height „h‟.

The Procedure of calculating cg from base is same as

above.

base from 3

h y i.e.,

The Procedure of calculating cg from vertical

edge AB is as below.

Consider a elemental strip „DE‟ of depth „h1‟ and

thickness „dx‟ at a distance „x‟ from the reference axis 1

– 1.

From the similar triangles „ABC‟ and „DEC‟.

b

x-b

h

h1

C b

A

h

h/3

B

X X

2h/3

Y

cg

Y

B x

b

A

h

b - x dx C

1

1

E

h1

D


SECTIONS


h b

x1h

b

x-b h1

Area of elemental strip da = h1 x dx

dxh b

x1 ad

Moment of this elemental area about 1 – 1 axis = da x x =

hdx

b

xxxdxh

b

x1

2

Moment of whole triangle about 1 – 1 axis b

0

32h

0

2

b3

h

2

xhhdx

b

xxax

6

hb

b6

b2b3h

b3

b

2

bhax

23332

2

bhhb

2

1a triangle whole of Area

2

bh6

hb

a

ax x edge verticalfrom centroid of Distance

2

edge verticalfrom 3

b X

Semicircle:

Consider semi - circular section of radius „R‟ with „O‟ as centre.

Consider an elemental radial area OPP‟. The angle POP‟ being dϴ, Area of

triangle = d2

r rdr

2

1

2

The distance of centroid of this area = insr3

2

Moment of the elemental area about AB is

dins3

r insr

3

2d

2

r

32

Moment of the whole area about AB is dins23

r dins

3

r

2

0

3

0

3

Y

C b

A

h

h/3

B

X X

2h/3

b/3 2b/3 Y

cg

O

A B

Y

r r

3

2

Y


SECTIONS


3

2r 0ins

2ins2

3

r

33

2

r semicirle the of Area

2

3

r4

2r

3r2

a

ayY

2

3


SECTIONS


Geometrical Properties:

Shape Figure X Y Area

Rectangle

Hallow rectangle

Square

a2

Circle

4

d2

d

Y

X Y

X X cg

1 1

2

2

d /2 d /2

a

Y

X Y

X X

cg

1 1

2

2

Y

a a /2 a /2

B /2 D/2 B x D

cg

X

1

B

d Y

Y

X X

1

2

2

Y

D

b

cg

Y

Y

X X

B

Y

1

2

2

D

X

1

B /2 D/2 B x D – bxd

d


SECTIONS


Hollow circular

4

dd 2i

2e

Triangle

2

b

3

h

2

bh

Right angled triangle

3

b

3

h

2

bh

Triangle

3

aL

3

h h

2

1 L

Semicircle

D/2or R

3

4R

2

2R

Or

4

D

2

1 2 4R/3π

c

g

Y

Y

X X

D

X

Y R

Y Y

X X

h/3

b

cg

h

X

Y

h/3

b

cg

h

Y

Y

X X

X

Y

X

h/3

b

cg

h

a

L

Y

Y

X X

Y

Y

de/2

Y

X Y

X X cg

1 1

2

2

de

di de/2


SECTIONS


Quarter Circle

3

4R

3

4R

4

2R

OR

4

D

4

1 2

Note:

i) The axis about which moments of areas are taken, is known as axis of

reference. 1-1 and 2-2 axis are called axis of reference.

ii) The axis of reference, of plane figures, are generally taken as the lowest line

of the fig for determining Y , and left line of the fig for calculating X .

iii) If the given section is symmetrical about X –X axis or Y – Y axis, then the cg

of the section will lie on the axis of symmetry.

Problem:

Find the centre of gravity of the L – section shown in fig.

Solution:

Divide the given section into 2 components as shown

in fig.

Consider reference axis 1-1 at the left

extreme edge of component 1 and reference

axis 2-2 at the bottom most edge of component

2.

Sl No

Component

Area – (a cm2)

C.D from 1 -1 axis

(x) - cm

C.D from 2 -2 axis

(y) - cm

Moment of the area about 1 -1

axis (ax – cm3)


axis (ay –

cm3)

1 Rectangle –

1

2 x 9 = 18 2/2 = 1 3 + 9/2 =

7.5

18 135

2 Rectangle – 2

8 x 3 = 24 8/2 = 4 3 /2 = 1.5

96 36

∑a = 42 ∑ax =114 ∑ay =171

X

R

4R/3π

cg

Y

Y

X X

Y

2 cm

12 cm

8 cm

3 cm

2 cm

9 cm

3 cm

8 cm

1

1 2 2

1

2


SECTIONS


cm 714.242

114

a

ax x

1- 1 axis reference from centroid of Distance

cm 07.442

171

a

ay y


Problem:

Find the centre of gravity of the T – section shown in fig.

Solution:

Divide the given section into 2 components as

shown in fig.

The section (fig) is symmetrical about vertical axis

(y – y). y can be calculated either from bottom or from

top edge. Let us consider reference axis 1-1 at the top most

edge of component 1.

cm 82.356

214

a

ay y


Sl No

Component

Area – (a cm2)

C.D from 1 -1 axis

(y) - cm


axis (ay – cm3)

1 Rectangle – 1

12 x 3 = 36

3/2 = 1.5 54

2 Rectangle –

2

2 x 10 =

20

3 + 10/2

= 8

160

∑a = 56 ∑ay =214

2 cm

12 cm

8 cm

3 cm

Y

cg

cm 07.4 y

cm .7142 X

Y

X X

2 cm

12 cm

3 cm

1 1

1

2

10 cm

2 cm

12 cm

3 cm

10 cm

cg

Y

cm .823 y

Y

X X

2 cm

12 cm

3 cm

1 1

1

2

10 cm


SECTIONS


Problem:

Find the cg of Z section shown in

the figure

Solution:

Divide the given section into

3 components as shown in fig.

Consider reference axis 1-1

at the left extreme edge of

component 3. The distance of

component 1 and 2 from axis 1- 1

= 200 – 20 = 180 mm.

Consider reference axis 2-2

at the top most edge of component

1.

Sl

No

Compone

nt

Area – (a

mm2)

C.D from 1 -

1 axis (x) – mm

C.D from 2

-2 axis (y) - mm

Moment of the

area about 1 -1 axis (ax –

mm3)

Moment of the

area about 2 -2 axis (ay –

mm3)

1 Rectangle –

1

300 x 30 =

9000

180 + 300/2

= 230

30/2 = 15 2.07 x 106 135 x 103

2 Rectangle – 2

20 x 250 = 5000

180 + 20/2 = 190

30 + 250 /2 = 155

950 x 103 775 x 103

3 Rectangle –

3

200 x 20 =

4000

300/2 = 150 30 + 250 +

20/2 = 290

600 x 103 1.16 x 106

∑a = 18000 ∑ax = 3.62 x 106 ∑ay = 2.07 x 106

mm 11.20118000

1062.3

a

ax x


6

mm 11518000

1007.2

a

ay y


6

300 mm

300 mm

30 mm

20 mm

20 mm

200 mm

300 mm

250 mm

30 mm

20 mm

20 mm

200 mm

1

1

2 2

1

2

3

180 mm

cg

300 mm

250 mm

30 mm

20 mm

20 mm

200 mm

1

1

2 2

1

2

3

180 mm

Y

mm 151 y

mm 01.112 X

Y

X X


SECTIONS


Problem: Dec 2015/ Jan 2016 – 8 marks

Determine centroidal of the area shown in fig.

Solution:

Let the quadrant to the right of OY axis be

component (1), square to the left of OY axis be

component (2), and the quadrant to the left of OY

axis be component (3).

„x‟ distance measured to the right of Y – axis is taken +ve and to the left of`

Y – axis –ve. „y‟ distance measured upward of X – axis is taken +ve.

Sl

No

Compon

ent

Area – (a

mm2)

C.D from 1 -1

axis (x) – mm

C.D from 2 -2

axis (y) - mm

Moment

of the area

about 1 -1 axis (ax –

mm3)

Moment

of the area

about 2 -2 axis

(ay – mm3)

1 Quadrant

– 1

86.706

4

230

4

2r

732.12

3

304

3

r4

732.12

3

304

3

r4

9000 9000

2 square –

2

30 x 30

= 900

-30/2 = -15 + 30 /2 = +15 - 13500 +13500

3 Deduct

Quadrant – 3

86.706

4

230

4

2r

268.17

3

30430

3

r4r

268.17

3

30430

3

r4r

12206.06 -12206.06

∑a = 900 ∑ax =

7700.06

∑ay =

10293.94

y

x 30 mm

30 mm

y

x 30 mm

30 mm 1

3

2


SECTIONS


mm 56.8900

06.7700

a

ax x

axis -Y from centroid of Distance

mm 44.11900

94.10293

a

ay y

axis - X from centroid of Distance

Problem:

Find the position of the centroid of the shaded area with

respect to the axis as shown in figure.

Solution:

The given section is divided into 4 components.

Let the quadrant to the right of 1 - 1 axis be

component (1), rectangle below 2-2 axis be component

(2), the square to the left of 1-1 axis be (3) and the

quadrant to the left of 1-1 axis be component (4).

„x‟ distance measured to the right of 1-1 –

axis is taken +ve and to the left of`1-1 – axis –ve.

„y‟ distance measured above 2-2 – axis is taken +ve

and „y‟ distance measured belowe 2-2 – axis is

taken -ve.

Sl

No

Compon

ent

Area – (a

mm2)

C.D from 1 -1

axis (x) – mm

C.D from 2 -2

axis (y) - mm

Moment

of the area

about 1 -1

axis (ax – mm3)

Moment of

the area about 2 -2

axis (ay

– mm3)

1 Quadrant

– 1

98.7853

4

2100

4

2r

4.42

3

1004

3

r4

4.42

3

1004

3

r4

333 x 103 333 x 103

2 Rectangl

e –2

200 x 20 =

4000 0

- 20 /2 = -10

0 - 40 x 103

3 Square –

3

100 x 100 =

10,000

-100/2 = -50 + 100 /2 = +50 - 500 x 103 +500 x 103

y

30 mm

30 mm

CG

mm 44.11 Y

mm 56.8 X

20 mm

1

2 100 mm

1

2

100 mm

20 mm

1

2 100 mm

1

2

100 mm

1

2

4

3


SECTIONS


4 Deduct

Quadrant – 3

98.7853

4

2100

4

2r

6.57

3

1004100

3

r4r

6.57

3

1004100

3

r4r

452.40 x

103

- 452.40 x

103

∑a = 14000 ∑ax =

285.1 x

103

∑ay =

340.6 x 103

mm 36.2014000

101.285

a

ax x

axis - 1-1 from centroid of Distance 3

mm 33.2414000

106.340

a

ay y



Locate the centroid of the shaded area as shown in fig

Solution:


Let the rectangle to the right of O - Y axis be

component (1), triangle be component (2) and the semicircle

be component (3).

„x‟ distance measured to the right of O – Y axis is

taken +ve, „y‟ distance measured above O - X axis is taken

+ve and „y‟ distance measured below O - X axis is taken

-ve.

mm 33.24 Y

20 mm

1

2

1

2

100 mm

x x

Y

Y

mm 36.20 X

CG

4 cm

R = 4cm

X

15 cm

10 cm Y

O

4 cm

R = 4cm

X

15 cm

10 cm Y

1

2

3

O


SECTIONS


Sl

No

Compone

nt

Area –

(a cm2)

C.D from O - Y

axis (x) – cm

C.D from O - X

axis (y) - cm

Moment

of the area

about O -

Y axis (ax – cm3)

Moment

of the area

about O -

X axis (ay –

cm3)

1 Rectangle

–1

10 x 15 =

150 10/2 = 5

15 /2 = 7.5

750 1125

2 Triangular componen

t - 2

12

4102

bh2

1

1

33.33

10

3

b

33.1

3

4

3

h

39.96 - 15.96

3 Deduct Semi –

circular componen

t – 3

13.25

24

2r

2

2

30.8

3

4410

3

r410

4

- 208.60 - 100.52

∑a = 136.87

∑ax = 581.36

∑ay = 1008.52

mc 25.487.136

36.581

a

ax x

axisY -O from centroid of Distance

mc 37.787.136

52.1008

a

ay y

axisX -O from centroid of Distance

Problem:

Locate the centroid of the plane section as shown in fig.

Solution:

mc 25.4 X

mc 37.7 Y

4 cm

R = 4cm

X

15 cm

10 cm Y

O

CG

30 mm

60 m

m

30 mm

30 m

m

O

30 mm

60 m

m

30 mm

30 m

m

O

1

1

2 2

1

2 3

4


SECTIONS


Solution:


Let the triangle to the right of 1-1 axis be component (1), Square be component (2),

another triangle be component (3) and the quadrant be component (4).

Sl

No

Compone

nt

Area –

(a mm2)

C.D from 1 - 1

axis (x) – mm

C.D from 2 - 2

axis (y) - mm

Moment

of the area

about 1 -1

axis (ax – mm3)

Moment

of the area

about 2 -

2 axis (xy –

mm3)

1 Triangular

component

-1

900

60302

bh2

1

1

103

30

3

b

50

330

3

h30

60

9000 45000

2 Square component

-2 a2 = 30 x 30

= 900

a/2 = 30/2

=15

a/2 = 30/2 =15

13500 13500

3 Triangular

component - 3

450

30302

bh2

1

1

40

30

30

3

30

3

b

20

3

302

3

h2

18000 9000

4 Deduct

Quadrantal component

– 4

86.706

230

2r

4

4

73.12

3

304

3

r4

73.12

3

304

3

r4

- 8998.33 - 8998.33

∑a =

1543.14

∑ax =

31501.67

∑ay =

58501.67

mm 41.2014.1543

67.31501

a

ax x

axis 1-1 from centroid of Distance

mm 91.3714.1543

67.58501

a

ay y

axis 2-2 from centroid of Distance

30 mm

60 m

m

30 mm

30 m

m

O

1

1

2 2

mm 41.20 X

mm 91.37 Y

CG X

Y

Y

X


SECTIONS


Problem: June/ July 2016 – 10 marks

Locate the centroid of area shown in fig with

respect to the Cartesian coordinate system

shown.

Solution:

The given section is divided into 4

components.

Let the triangle to the right of 0 -Y axis be

component (1), Rectangle be component (2),

another triangle be component (3) and the

rectangle be component (4).

Sl No

Component Area – (a m2)

C.D from O - Y axis (x) – m

C.D from O - X axis (y) - m

Moment of the area about OY axis (ax – m3)

Moment of the area about OX axis (xy –m3)

1 Triangular

component -1

6622

bh2

1

1

33.1

223

2

3

b

2

33

h 6

7.98 12

2 Rectangula

r component

-2

bd= 2 x 7.5 = 15 2+b/2 = 2+2/2 =3.0 d/2 = 7.5/2 =3.75 45 56.25

3 Triangular component

- 3

5

522

bh2

1

1

67.4

4

22

3

2

3

b

67.2

3

51

3

h1

23.35 13.35

4 Rectangula

r component

– 4

bd = 3 x 1 =3

2+2+3/2=5.5 1/2 = 0.5

16.5 1.5

∑a = 29 ∑ax = 92.83

∑ay = 83.1

x

2m y 2m 2m 1m

1m

6 m

1.5 m

2m

y

2m 2m 1m

1m

5m

7.5 m

1

2

3

4 x

6m


SECTIONS


mm 20.329

83.92

a

ax x

axisY -O from centroid of Distance

mm 87.229

1.83

a

ay y

axisX -O from centroid of Distance

Problem:

Determine the cg of the fig shown

Solution:


components as shown in fig.

Consider reference axis 1-1 at the

left extreme edge of the section.


bottom most edge of the section.

R = 4m

R =

4m

2 m 4 m 8 m

8 m

4 m

4 m 4 m 6 m

R = 4m

6 m

4m

1

2 2

R = 4m

R =

4m

2 m 4 m 8 m

8 m

4 m

4 m 4 m 6 m

R = 4m

6 m

4m

1

2 3

4 5

x

2m y 2m 2m 1m

1m

6 m

1.5 m

m2.3 X

m87.2 Y

Y

Y

X X cg


SECTIONS


Sl

No

Compone

nt

Area – (a

m2)

C.D from 1 -1

axis (x) – m

C.D from 2 -

2 axis (y) - m

Moment of

the area about 1 -1

axis (ax –

m3)

Moment of

the area about 2 -2

axis (ay

- m3)

1 Rectangular component – 1

14 x 12

=168 7

2

14

6

2

12

2

h

1176 1008

2 Triangular

component - 2

12

462

bh2

1

1

1641

482

3

6

3

b

33.1

h

3

4

3

192 15.96

Deduct

3 Semi – circular

component

– 3

13.25

24

2r

2

2

6

2

82

70.1

3

44

3

r4

- 150.78 - 42.72

4 Square component – 4

- 4 x 4 = - 16

2

2

4

102

48

- 32 - 160

5 Quadrant component – 5

57.12

4

24

4

2r

30.12

3

4414

3

r414

30.10

3

4412

3

r412

-154.61 -129.47

∑a = 126.3 ∑ax =1030.61

∑ay = 691.77

m 16.83.126

61.1030

a

ax x

axis - 1-1 from centroid of Distance

m 48.53.126

77.691

a

ay y

axis 2 -2 from centroid of Distance


SECTIONS


Problem:

Locate the centroid of area shown in fig. Solution:


shown in fig.

Consider reference axis 1-1 at the left extreme

end of the triangle.

Consider reference axis 2-2 at the bottom most

edge of the semi-circle.

Sl

No

Compone

nt

Area – (a

mm2)

C.D from 1 -1

axis (x) – mm

C.D from 2 -

2 axis (y) – mm

Moment of

the area about 1 -1

axis (ax –

mm3)

Moment of

the area about 2 -2

axis (ay –

mm3)

1 Triangle

component – 1

3200

8080

hb

2

2

33.53

803

2

67.66

340

340

80

h

170.66 x 103 213.344 x

103

2 Semi – circular

component

– 2

27.2513

240

2r

2

2

40

04.23

3

40440

3

r4r

100.53 x 103 57.91 x 103

3 circular

component – 3

63.1256

4

240

4

2d

40

40

- 50.27 x 103 - 50.27 x 103

∑a =

4456.64

∑ax =

220.92 x 103

∑ay =

220.98x 103

mm57.4964.4456

1092.220

a

ax x


40mm

80mm

40mm

2

1

1

2

1

2

3

mm 57.49 X

40mm

80mm

40mm

2

1

1

2

Y

Y

X X cg

mm 60.49 Y

mm 60.4964.4456

1098.220

a

ay y

axis 2 -2 from centroid of Distance 3


SECTIONS



Locate the centroid of the plane area shown in

fig.

Solution:

Let us divide the given section into 5


Component No 1 – semicircle, Component

No 2 – rectangle, Component No 3 – Left side

Triangle, Component No 4 – right side

triangle and Component No 5 – circle,

Consider reference axis 1-1 at the left

extreme end of the section.


bottom most edge.

Sl N

o

Component Area – (a mm2)

C.D from 1 -1 axis (x) – mm

C.D from 2 -2 axis (y) –

mm

Moment of the area

about 1 -1 axis (ax –

mm3)

Moment of the area

about 2 -2 axis (ay –

mm3)

1 Semi – circular component –

1

3927

250

2r

2

2

45 + 50 = 95

2. 101

3

50480

3

r480

373.07 x 103 397.41 x 103

2 Rectangular component –

2 100 x 80 =

8000

45 + 100 /2 = 95 80 /2 = 40

760 x 103 320 x 103

3 Triangle

component – 3

1800

8045

hb

2

2

30

453

2

b3

2

67.26

3

80

3

h

54 x 103 48.01 x 103

4 Triangle component –

4

3200

8080

hb

2

2

67.171

803

145

b3

10045

1

1

67.26

3

80

3

h

549.34 x 103 85.34 x 103

45 mm

225 mm

80mm

50 mm

80mm

30 mm

80mm 80 mm

2

1

1

2

45 mm

225 mm

80mm

50 mm

80mm

30 mm

80mm

80 mm

1

5

2 3 4


SECTIONS


5 Deduct :

Circular component –

5

43.2827

4

260

4

2d

45 + 50 = 95

80

- 268.61 x 103 - 226.20 x 103

∑a = 14.10

x 103

∑ax = 1.47 x

106

∑ay = 624.56

x 103

mm 26.1041010.14

1047.1

a

ax x

axis - 1-1 from centroid of Distance

3

6

mm 30.441010.14

1056.624

a

ay y

axis 2 -2 from centroid of Distance

3

3

2

45 mm

225 mm

80mm

50 mm

80mm

30 mm

80mm

80 mm

Y

mm 26.104 X

mm 30.44 Y

X X

2

1

1

cg

Y


SECTIONS


Moment of Inertia: Introduction to the concept, radius of gyration, Parallel axis

theorem, perpendicular axis theorem, Moment of inertia of basic planar figures,

computing Moment of inertia for T, L, I, Z and full / quadrant circular sections and

their built up sections, Numerical problems.

Moment of inertia or second moment of area

(I) :

Consider a thin lamina of area (A) as shown

in fig.

Let X = distance of the cg of area A from OY axis.

Let Y = distance of the cg of area A from OX axis.

Moment of area about OY axis = Area x

perpendicular distance of cg of area from axis OY

= A X

The above equation is known as first moment of area about the axis OY.

If the moment of the area is again multiplied by the perpendicular distance between

the cg of area and axis OY (i.e., distance X ),

Then the quantity (A X ) x X = 2 X A is known as moment of the area or

second moment of area or moment of inertia about OY axis.

Similarly, 2 Y A is known as moment of the area or second moment of area

or moment of inertia about OX axis.

Hence, the product of the area and the square of the distance of the centre

of gravity of the area from an axis is known as moment of inertia of the

area about that axis.

Theorem:

i) Parallel axis theorem

ii) Perpendicular axis theorem

i) Parallel axis theorem :

Statement “Moment of inertia of a plane lamina about any axis parallel to the

centroidal axis is equal sum of the M.I of the lamina about its centroidal axis and the

product of the area and square of the perpendicular distance between them”.

I1-1 = Ig + Ah2

Consider an elemental strip of area „da‟ at

a distance „y‟ from centroidal axis (X – X).

Moment of inertia of the elemental strip

about the reference axis 1 – 1

= da x (h +y) x h +y)

y

1

Elemental Area da

X

h

X

1

Lamina of

Area (A)

O

Y

X

X

CG

Y


SECTIONS


= da x (h +y)2 = da x (h2 +y2 + 2hy)

MI of the entire lamina about 1 – 1 axis = Ah2 + Ay2 + 2Ahy

But, Ay2 = IXX = MI of the lamina about its axis

2Ahy = Moment of area about centroidal axis = 0.

I1 - 1 = Ah2 + IXX = Ig + Ah2

Similarly,

I2 - 2 = Ax2 + Iyy = Ig + Ax2

XA I XA II 2g

2YY22

ii) Perpendicular axis theorem :

The moment of inertia of plane lamina about the centroidal axis perpendicular to the

plane of the lamina is equal to the sum of its moment of inertia about two mutually

perpendicular axis that lie in the plane of the lamina.

i.e., Ixx and Iyy be the moment of inertia of a plane section about two mutually

perpendicular axis XX and YY in the plane of the section, then the moment of inertia

of the section Izz about the axis ZZ, perpendicular to the plane and passing through

the intersection of XX and YY axis is given by

IZZ = Ixx + Iyy

Proof:

A plane section of area A and lying in the plane x – y

is shown in fig. let OX and OY be the two mutually

perpendicular axes, and OZ be the perpendicular

axis.

Consider a small area „da‟.

Let, x = distance of da from the axis OY

y = distance of da from the axis OX

r = distance of da from the axis OZ

then, r2 = x2 + y2

Moment of Inertia of da about x – axis

= da x (distance of da from x - axis)2

Moment of Inertia of total area A about x – axis

Ixx = ∑day2

Similarly, Moment of Inertia of total area A about y – axis

Iyy = ∑dax2

Moment of Inertia of total area A about z – axis

Izz = ∑dar2 = ∑da (x2 + y2) = ∑da x2 + ∑da y2) = Ixx + Iyy

Izz = Ixx + Iyy

2

Y

Y

2

cg

X

Y

da

Z

O X

X

y r


SECTIONS


Radius of gyration:

Radius of gyration of a body (or a given

lamina) is defined as the distance from an axis of

reference where the whole mass (or area) of a

body is assumed to be concentrated so as not to

alter the moment of inertia about the given axis.

Consider a plane area which is split up into

small areas a1, a2, a3 ……etc.

Let the moment of inertia of the plane area about the given axis is given by

...... ra ra ra ra I 244

233

222

211

2ar I

let the whole mass (or area) of a body is concentrated at

a distance „k‟ from the axis of reference, then the

moment of inertia of the whole area about the given axis

will be equal to Ak2.

Ak2 = I, then „k‟ is known as radius of gyration about

the given axis.

A

I k

Moment of inertia of a rectangular section about centroidal X – X axis:

Consider a rectangle having width = b and

depth = d.

Let X - X be the horizontal axis passing through

the C.G of the rectangular section.

Consider a rectangular elemental strip of

thickness „dy‟ at a distance „y‟ from the X - X axis.

Area of elemental strip da = b x dy

Moment of inertia of the elemental area about X-X

= da x y2 = bdy x y2 = b y2 dy

Moment of inertia of the whole section will be obtained by integrating the above

equation between the limits – d/2 to + d/2

O

Y

Area a1

X

Area a2 Area a3

Area a4

r1

r2

r3

r4

O

Y

Area a1

Area a2

Area a3

Area a4

k

Y

Y

O X

Area a1 Area a2 Area a3 Area a4

k

X X

dy

b

x

d/2

y d

d/2

x


SECTIONS


332

d

2d

32d

2d

2XX

2

d

2

d

3

b

3

ybdyybI

8

dd

3

b

8

d

8

d

3

b

8

d

8

d

3

bI

333333

XX

12

bdI

3

XX

Similarly, 12

dbI

3

YY

Moment of inertia of a rectangular section about

base:

Consider a rectangle having width = b and

depth = d.

Consider a rectangular elemental strip of thickness

„dy‟ at a distance „y‟ from the base 1-1.

Area of elemental strip da = b x dy

Moment of inertia of the elemental area about base 1-1.

= da x y2 = bdy x y2 = b y2 dy

Moment of inertia of the whole section will be obtained by

integrating the above equation between the limits 0 to d.

3

db

3

ybdyybI

3d

0

3d

0

2XX

3

bdI

3

XX

Similarly, 3

dbI

3

YY

Moment of inertia of a Hollow rectangular section:

Moment of inertia of bigger section about X – X axis

12

BDI

3

XX

Moment of inertia of cut section about X – X axis

12

bdI

3

XX

IXX of the hollow section = IXX of bigger section – IXX of cut section

IXX of the hollow section = 12

bd

12

BD

33

dy

b

1 y

d

1

b

x x

d

B

D

Y

Y


SECTIONS


Moment of inertia of a Circular section:

Consider circular section of radius „R‟ with „O‟ as centre.

Consider an elementary circular ring of radius „r‟

and thickness „dr‟.

Area of elementary circular ring = 2r x dr

Moment of inertia is calculated with respect to Z –

Z axis and then MI about X – X axis and Y – Y axis is

obtained by applying perpendicular axis theorem.

IZZ = Area of ring x (radius of ring from O)2.

IZZ = (2r x dr) x r2

IZZ = 2 r3 dr

Moment of inertia of the whole circular section is obtained by integrating above

equation between the limits 0 to R.

2

r

4

rR2

4

r2drr2drr2I

44R

0

4R

0

3R

0

3ZZ

sectioncircular the ofdiameter the is D 2

D RBut

32

D

2

D

2I

44

ZZ

From the theorem of perpendicular axis

Izz = Ixx + Iyy

But due to symmetry, Ixx = Iyy

Ixx = Iyy = Izz / 2

32

D

2

1I

4

XX

YY

4

XX I64

DI

4

RII

4

YYXX

Moment of inertia of a Hollow Circular section:

Let D = diameter of outer circle and d = diameter of

cut out circle

64

DI

axis X -about X circleouter of inertia ofMoment 4

XX

64

dI

axis X -about X circleout cut of inertia ofMoment 4

XX

O

X X

Y

Y

R

dr

r

O

X X

Y

Y

D

d


SECTIONS


64

d

64

DI

axis X -about X sectioncircular hollow of inertia ofMoment 44

XX

44XX dD

64I

44YY dD

64I Similarly,

Moment of inertia of a Semi - Circular section:

Consider semi - circular section of radius „R‟ with „O‟ as

centre.

Consider an elementary semi - circular ring of

radius „r‟ and thickness „dr‟.

Area of elementary circular ring = r x dr

Moment of inertia is calculated with respect to Z –

Z axis and then MI about X – X axis and Y – Y axis is

obtained by applying perpendicular axis theorem.

IZZ = Area of ring x (radius of ring from O)2.

IZZ = (r x dr) x r2

IZZ = r3 dr

Moment of inertia of the whole circular section is obtained by integrating above

equation between the limits 0 to R.

4

R

4

R

4

rdrrdrrI

44R

0

4R

0

3R

0

3ZZ

From the theorem of perpendicular axis

Izz = IAB + Iyy

But due to symmetry, IAB = Iyy

IAB = Iyy = Izz / 2

8

R

4

R

2

1I

44

AB

Centroidal axis X – X acts at a distance ABbase from 3

4R y

Applying parallel axis theorem:

Moment of inertia about base AB = Moment of inertia about cg + Area x (Distance

between X – X and AB)2. 2

gAB Y A II Ig = Ixx

44

2242

ABXX R283.0R393.03

R4

2

R

8

RY A II

4XX R11.0I

O

A B

Y

Y

R

dr

r

x x


SECTIONS


Moment of inertia of a Quadrant section:

Moment of inertia of a Equilateral Triangle: Consider a triangle ABC with base

width „b‟ and height „h‟.

Consider a elemental strip „DE‟ of

width „b1‟ and thickness „dy‟ at a distance

„y‟ from the reference axis 1 – 1 or

(base BC) .

From the similar triangles „ABC‟

and „ADE‟.

h

y-h

b

b1

b h

y1b

h

y-h b1

Area of elemental strip da = b1 x dy

y

b

A

h

h - y

dy

B C

1 1

b1

E D


SECTIONS


dyb h

y1 ad

Moment of inertia of the elemental area about base BC = da x y2 =

bdy

h

yyydyb

h

y1

322

Moment of inertia of the whole triangle about the base BC h

0

43h

0

32

BCh4

y

3

ybdy

h

yybI

h12

bh

h12

h3h4b

h4

h

3

hbI

44443

BC

12

bhI

3

BC

Moment of inertia of the Triangular section about an axis (centroidal) passing through the cg and parallel to the base: Centroidal axis X – X acts at a distance

base from 3

h y

Applying parallel axis theorem:

Moment of inertia about base BC = Moment of inertia

about cg + Area x (Distance between X – X and BC)2. 2

gBC Y A II Ig = Ixx

36

bh2bh3

18

bh

12

bh

9

h

2

bh

12

bh

3

h A II

3333232

BCXX

36

bhI

3

XX

Moment of inertia of the Triangular section about an symmetrical Y – Y axis passing through the cg:

The triangle is symmetrical about vertical axis (Y – Y ) axis.

48

hb

122

bh2I

33

YY

C b

A

h

h/3

B

X X

2h/3

Y

cg

Y


SECTIONS


Area’s and Moment of Inertia’s about X and Y axis:

Shape Figure Area M.I about X – X axis

M.I about Y – Y axis

Rectangle

bd 12

bdI

3

xx

3

bdI

3

base

12

dbI

3

YY

3

dbI

3

leftedge

Hollow

Rectangle

bd - BD 12

bd

12

BDI

33

xx

33xx bdBD

12

1I

12

db

12

DBI

33

YY

33YY dbDB

12

1I

Triangle

2

bh

36

bhI

3

xx

12

bhI

3

base 48

hbI

3

YY

Right

angled triangle

2

bh

36

bhI

3

xx

12

bhI

3

base

36

hbI

3

YY

12

hbI

3

leftedge

Triangle

h2

1 L

36

LhI

3

xx

To be determined by considering two separate triangles

Y Y

X X

h/3

b

cg

h

X

Y

h/3

b

cg

h

Y

Y

X X

X

Y

X

h/3

b

cg

h

a

L

Y

Y

X X

Y

b

x x d

b

x x

d

B

D


SECTIONS


Circular

4

D2

or

2R

64

DI

4

xx

4

RI

4

XX

64

DI

4

YY

4

RI

4

YY

Hollow circular

64

d

64

DI

44

xx

44xx dD

64I

64

d

64

DI

44

YY

44YY dD

64I

Semicircle

2

2R

Or

4

D

2

1 2

4xx R11.0I

8

RI

4

Base

8

RI

4

YY

Quarter Circle

4

2R

OR

4

D

4

1 2

4xx R055.0I

16

RII

4

2211

4YY R055.0I

4R/3π

c

g

Y

Y

X X

D

X

Y R

X

R

4R/3π

cg

Y

Y

X X

Y

D


SECTIONS


Problem:

Find the moment of inertia about centroidal X – X axis and Y – Y axis of the angle

section with measurements 100 x 80 x 20 mm

Solution:


shown in fig.

Consider reference axis 1-1 at the left extreme

edge of component 2 and reference axis 2-2 at the

bottom most edge of component 1.

mm 353200

10112

a

ax x


3

mm 253200

1080

a

ay y


3

46662gy 1-1 mm1083.61012.51071.1ax I I

46262

1-1yy

2

yy 1-1

mm1091.23532001083.6xaII

xa I I

46632gx 2-2 mm1063.3102.31067.426ay I I

46262

2-2yy

2

xx2-2

mm1063.12532001063.3yaII

ya I I

20 mm

80 mm

100 mm

20 mm

20 mm

60 mm

20 mm

100 mm

1

1 2 2

2

1

MODULE – IV CENTROIDS AND MOMENTS OF INERTIA OF ENGINEERING SECTIONS


Sl

No

Compone

nt

Area – (a

mm2)

C.D from

1 -1 axis (x) - mm

C.D from

2 -2 axis (y) – mm

Moment of the

area about 1 -1 axis (ax –

mm3)

Moment of

the moment of area

about 1 -1 axis (ax2 –

mm4)

Moment of

the area about 2 -2

axis (ay – mm3)

Moment of

the moment of area

about 2 -2 axis (aY2 –

mm4)

MI about its

centroidal axis - Igx -

mm4

MI about its

centroidal axis - IgY -

mm4

1 Rectangle – 1

100 x 20bgtbggg

.++ = 2000

100/2 = 50 20/2 = 10 100 x 103 5 x 106 20 x 103 200 x 103

31067.66

12

302100

12

3bd

gxI

61067.1

12

310002

12

3db

gyI

2 Rectangle – 2

20 x 60 = 1200

20/2 = 10 20 + 60 /2 = 50

12 x 103 120 x 103 60 x 103 3 x 106

310360

12

30620

12

3bd

gxI

31040

12

32006

12

3db

gyI

∑a = 3200 ∑ax = 112 x 103

∑ax2 =5.12 x 106

∑ay = 80 x 103

∑ay2 = 3.2 x 106

∑Igx = 426.67x 103

∑Igy = 1.71 x 106


SECTIONS


Problem:

Find the moment of inertia of a hollow section shown

in fig about an axis passing through its centre of

gravity and parallel to X – X axis

Solution:


in fig.

Consider reference axis 1-1 at the bottom

most edge of component 1.

Sl

No

Component Area –

(a mm2)

C.D

from

1-1 axis (y) –

mm

Moment of

the area

about 1 -1 axis

(ay – mm3)

Moment of

the moment

of area about 1-1

axis (aY2 –

mm4)

MI about its

centroidal

axis - Igx - mm4

1 Rectangle component

– 1

200 x 300 = 60,000

300/2 = 150

9 x 106 1.35 x 109

610450

12

3003200

12

3bd

gxI

2 Deduct

circle

component – 2

31068.17

4

150

d

2

2

4

200 3.53 x 106 706.86 x 106

61085.24

64

150

d

4

4

64

∑a =77.68

x 103

∑ay =

12.53 x

103

∑Igx =

474.85 x

106

300 m

m

100 mm

200 mm

150 mm

300 m

m

100 mm

200 mm

150 mm

1

2

1 1


SECTIONS


Problem: June/ July 2016 – 14 marks 14 CIV

13/23

Determine the second moment of area about

the horizontal centroidal axis as shown in fig. Also

find radius of gyration.

Solution:



Consider reference axis 1-1 at the bottom most edge

of component 1.

Sl No


C.D from 1-1 axis (y)

– mm

Moment of the

area

about 1 -1 axis

(ay – mm3)

Moment of the

moment

of area about 1-1

axis (aY2 – mm4)

MI about its centroidal axis -

Igx - mm4

1 Rectangular

component - 1

80 x 40 = 3200 20

2

40

64 x 103 1.28 x 106

310426.67

34080

3bd

gxI

12

12

2 Triangular

component - 2

1200

30802

bh2

1

1

5040

h40

3

30

3

1

60 x 103 3 x 106

31060

30380

36

3bh

gxI

36

3 Deduct: Semi-circular

component - 3

32.628

302

2r

2

2

0.424 x r = 0.424 x 20

= 8.48

- 5328.15 - 45.20 x 103

- 0.11 x r4 = - 0.11 x 204

= - 17600

∑a =3771.68

∑ay = 118.67 x

103

∑ay2 = 4.23 x 106

∑Igx = 469.07 x 103

30 mm 50 mm

30 mm

20

mm

20

mm 20

mm

40 mm

30 mm 50 mm

30 mm

20

mm

20

mm 20

mm

40 mm 1

2

3

1 1


SECTIONS



mm 46.3169.3771

1067.118

a

ay y

3

46632gx 1-1 mm1070.41023.41007.469ay I I

4326

2

1-1xx

2

xx1-1

mm1004.96746.3169.37711070.4yaII

ya I I

:GYRATION OF RADIUS

mm1669.3771

1004.967

a

I k gyration of Radius

3xx

Problem: June/ July 2016 – 10 marks 15

CIV 13/23

Find the moment of inertia of the region in

fig about horizontal axis 1-1 and also find the

radius of gyration about the same axis.

Solution:



Component No 1 – Triangle,

Component No 2 – Square and Component

No 3 – Quadrant.

Consider reference axis 1-1 at the bottom

most edge.

Sl

No

Compone

nt

Area –

(a mm2)

C.D from

1-1 axis (y) – mm

Moment

of the area

about 1 -1 axis

(ay –

mm3)

Moment

of the moment

of area about 1-

1 axis

(aY2 – mm4)

MI about its

centroidal axis - Igx - mm4

100 mm

150 mm

1 1

100 mm

100 mm

150 mm

1 1

100 mm

1

2

3


SECTIONS


1 Triangular

component - 1

7500

1001502

bh2

1

1

33.33

h

3

100

3

1

249.98 x

103

8.33x 106

31017.4

36

3001150

36

3bh

gxI

2 Square

component

– 2

100 x 100

= 10 x 103

100/2 = 50 500 x 103 25 x 106

61033.8

12

3001100

12

3bd

gxI

3 Deduct Quadrant

component – 3

98.7853

4

100

r4

1

2

2

0.424 r = 0.424 x 100

= 42.4

- 333 x 103

- 14.12 x 106

6105.5

4100055.0

4r055.0

∑a = 9646.02

∑ay = 416.97 x

103

∑ay2 = 19.21 x

106

∑Igx = 2.83 x 106


mm 32.4302.9626

1097.416

a

ay y

3

46662gx 1-1 mm1004.221021.191083.2ay I I

4626

2

1-1xx

2

xx1-1

mm1094.332.4302.96461004.22yaII

ya I I

:GYRATION OF RADIUS

mm 23.2002.9626

1094.3

a


6xx


SECTIONS



Determine the moment of inertia and radii of

gyration of the area shown in fig about the base AB

and centroidal axis parallel to AB.

Solution:

Sl

No

Component Area –

(a mm2)

C.D from 1-1

axis (y) – mm

Moment

of the area

about 1 -1 axis

(ay –

mm3)

Moment

of the moment

of area about 1-

1 axis

(aY2 – mm4)

MI about its


mm4


4500

900012

bh2

1

1

30

90

h

3

1

3

1

135 x 103 4.05 x 106

610025.2

36

390100

36

3bh

gxI

2 Deduct:

Rectangular component

20 x 30

= - 600

30 + 30/2

= 45

- 27 x 103 -1.215 x

106

31045

12

30320

12

3bd

gxI

∑a =3900

∑ay =

108 x

103

∑ay2 =

2.835 x

106

∑Igx = 1.98

x 106

mm 69.273900

10108

a

ay y

B. - Abase from centroid of Distance 3

46662gx B-A mm10815.410835.21098.1ay I I

4626

2

B -A xx

2

xxB -A

mm10825.169.27390010815.4yaII

ya I I

30 mm

40 mm

30 mm

30 mm

20 mm

40 mm


SECTIONS


:GYRATION OF RADIUS

mm 14.353900

10815.4

a

I k ABbaseabout gyration of Radius

6AB

AB

mm 63.213900

10825.1

a

I k X - X axis Centroidalabout gyration of Radius

6XX

XX

Problem:

Determine the moment of inertia of the area shown in

fig about the base AB.

Solution:

Sl No


C.D from 1-1 axis (y) – mm

Moment of the

area

about 1 -1 axis

(ay – mm3)

Moment of the

moment

of area about 1-

1 axis (aY2 –

mm4)

MI about its centroidal

axis - Igx -

mm4


4050

90902

bh2

1

1

30

90

h

3

1

3

1

121.5 x 103

3.65 x 106

61082.1

36

39090

36

3bh

gxI

2 Deduct circle

component

– 2

86.706

4

30

d

2

2

4

45

302

30

- 31.81 x 103

- 1.43 x 106

31076.39

64

30

d

4

4

64

∑a = 3343.14

∑ay =

89.69 x

103

∑ay2 =

2.22 x

106

∑Igx = 1.78

x 106

mm 83.2614.3343

1069.89

a

ay y

B. - Abase from centroid of Distance 3

46662gx B-A mm100.41022.21078.1ay I I

30 mm

90 mm

30 mm

30 mm

A B


SECTIONS


Problem:

Determine the moment of inertia of the shaded area

shown in fig about the X - X axis.

Solution:

Problem:

Determine the second moment of the area about the

horizontal centroidal axis as shown in fig. Also determine

radius of gyration.

Solution: Divide the given section into 3 components as

shown in fig.

Consider reference axis 1-1 at the bottom most

edge of the semi-circle.

Sl No

Component MI about its centroidal axis – Ixx

- mm4


61041.3

38080

3bh

xxI

12

12

2 Semi –

circular component

61001.1

40

rxxI

8

84

4

3 Deduct:

Circular

component

31066.125

40

dxxI

64

644

4

∑Ixx =4.29 x 106

X X

40mm

80mm

40mm

X X

40mm

80mm

40mm

1 1

1

2

3


SECTIONS


Sl

No

Component Area –

(a mm2)

C.D from 1-1

axis (y) – mm

Moment

of the area

about 1

-1 axis (ay –

mm3)

Moment

of the moment

of area

about 1-1 axis (aY2 –

mm4)

MI about its

centroidal axis - Igx - mm4

1 Triangular

component

– 1

3200

8080

hb

2

2

67.66

340

340

80

h

213.34 x

103

14.22 x 106

61041.3

38080

3bh

xxI

12

12

2 Semi – circular

component – 2

27.2513

240

2r

2

2

04.23

3

40440

3

r4r

57.91 x 103

1.33 x 106

61001.1

40

rxxI

8

84

4

3 Deduct:

Circular component

– 3

63.1256

4

240

4

2d

40

- 50.27 x

103

2.01 x 106

31066.125

40

dxxI

64

644

4

∑a = 4456.64

∑ay = 220.98 x

103

∑ay2 = 17.56 x

106

∑Igx =4.29 x 106

mm 60.4964.4456

1098.220

a

ay y


46662gx 1-1 mm1085.211056.171029.4ay I I

4626

2

1-1xx

2

xx1-1

mm1089.1060.4964.44561085.21yaII

ya I I

:GYRATION OF RADIUS

mm43.4964.4456

1089.10

a


6xx


SECTIONS


Problem: June / July 2015 – 12 marks

Find the moment of inertia of the lamina shown in

fig about its horizontal (xx) centroidal axis

(unshaded area)

Solution:



Component No 1 – Triangle, Component No

2 – semicircle, Component No 3 – rectangle.

Consider reference axis 1-1 at the bottom most edge.

Sl

No

Component Area –

(a mm2)

C.D from 1-1

axis (y) – mm

Moment

of the area

about 1 -1 axis

(ay – mm3)

Moment

of the moment

of area about 1-

1 axis (aY2 –

mm4)

MI about its


mm4


– 1

5000

1000012

bh2

1

1

33.33

100

h

3

1

3

1

166.67 x 103

5.56 x 106

61078.2

36

3100100

36

3bh

gxI

2 Semi –

circular component

– 2

3927

250

2r

2

2

= 50

196.35 x

103

9.82x 106

61045.2

50

rxxI

8

84

4

3 Deduct:

Rectangular

component – 3

20 x 40 = - 800

= 50 - 40 x 103

-2 x 106

31067.106

12

30420

12

3bd

gxI

∑a =8127

∑ay = 323.02

x 103

∑ay2 = 13.38 x

106

∑Igx = 5.12 x 106

Y

X

40 mm 100 mm

20 mm

100 mm

1

2

1

Y

1

40 mm 100 mm

20 mm

100 mm

3


SECTIONS


mm 75.398127

1002.323

a

ay y


46662gx 1-1 mm105.181038.131012.5ay I I

4626

2

1-1xx

2

xx1-1

mm1066.575.398127105.18yaII

ya I I

Problem: Dec 07 / Jan 08 - 06 Civ 13 / 23

Determine the second moment of the area about the horizontal centroidal axis as

shown in fig. Also determine radius of gyration.


SECTIONS



For the cross section shown in fig, calculate the MI

about the centroidal axis parallel to top edge. Also

determine the radius of gyration.

Solution:


in fig.

Consider reference axis 1-1 at the bottom most edge of the

section.

Sl No


C.D from 1-1 axis (y)

– mm

Moment of the area

about 1 -1

axis (ay – mm3)

Moment of the

moment

of area about 1-1

axis (aY2 – mm4)

MI about its centroidal axis

- Igx - mm4

1 Square

component – 1

25 x 25

= 625

5.12

2

25

2

d

7812.5 97.66 x 103

31055.32

32552

3bd

xxI

12

12

2 Rectangular

component – 2

15 x 35

= 525

5.42

2

3525

225

d

22.31 x 103 948.28 x

103

31060.53

33551

3bd

xxI

12

12

3 Rectangular

component

– 3

100 x 10 = 1000

65

2

1060

23525

d

65 x 103 4.23 x 106

33.8333

310001

3bd

xxI

12

12

∑a = 2150 ∑ay =

95.12 x

103

∑ay2 = 5.28 x 106

∑Igx = 94.48 x 103

25 mm

35 mm

10mm

15 mm

80mm

25 mm

80mm

100 mm

1

2

3

25 mm

35 mm

10mm

15 mm

80mm

25 mm

80mm

100 mm

1 1


SECTIONS


mm 24.442150

1012.95

a

ay y


46632gx 1-1 mm1037.51028.51048.94ay I I

4626

2

1-1xx

2

xx1-1

mm1016.124.4421501037.5yaII

ya I I

:GYRATION OF RADIUS

mm23.232150

1016.1

a


6xx

centroids moment of inertia: introduction to the concept

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