moment of inertia of composite sections

8/13/2019 Moment of Inertia of Composite Sections

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52 7.154

Copy Right: Rai University

ENGINEERINGMEC

HANICS

LESSON 15:

MOMENT OF INERTIA OF COMPOSITE SECTIONS

IntroductionWe have derived moment of inertia of certain standard sections

in our last session. Sections of beams and columns commonly

used in real structures are combination of the standardsections. So, we need to have a method by which moment of

inertia of such composite sections can be determined.

Moment of inertia of composite sections about an axis can be

found by the following steps:

1. Divide the given figure into a number of simple figures of

known cross sectional properties..

2. Locate the centroid of each simple figure by inspection or

using standard expression.

3. Find the moment of inertia of each figure about its

centroidal axis. Add the term 2Ay where A is the area ofthe simple figure and y is the distance of the centroid of the

simple figure from the reference axis. This gives moment ofinertia of the simple figure about the reference axis.

4. Sum up the moment of inertia of all simple figures to get

the moment of inertia of the composite section.

The procedure given above is illustrated below. Referring to the

Fig. 15.1, it is required to find out the moment of inertia of the

section about axis A B.

Fig. 15.1

1. The section in the figure is divided into a rectangle, a triangle

and a semicircle. The areas of the simple figures

321 , AandAA are calculated.

2. The centroids of the rectangle ( )1g , triangle ( )2g andsemicircle ( )3g are located. The distances 21,yy and

3y are found from the axis AB..3. The moment of inertia of the rectangle about its centroid

( )1gI is calculated using standard expression. To this, theterm 2

11yA is added to get the moment of inertia aboutthe axis AB

2

1111 yAII g +=

Similarly, the moment of inertia of the triangle

( )22222

yAII g +=

and of semicircle

( )23333 yAII g += about axis AB are calculated.

4. Moment of inertia of the composite section about AB is

given by:

321 IIIIAB ++=

2

333

2

222

2

111 yAIyAIyAI ggg +++++= 15.1

In most engineering problems, moment of inertia about the

centroidal axis is required. In such cases, first locate the

centroidal axis and then find the moment of inertia about this

axis.

Fig. 15.2

Referring to Fig.15.2, first the moment of area about any

reference axis, say AB, is taken and is divided by the total area of

the section to locate centroidal axis x-x. Then the distances of

centroid of individual figures 321 , ccc yandyy from the axisx-x are determined. The moment of inertia of the composite

section about the centroidal axis x-x is calculate using the

expression:2

333

2

222

2

11 1 cgcgcgxx yAIyAIyAII +++++= 15.2

Sometimes, the moment of inertia is found about a convenient

axis and then using parallel axis theorem, the moment of

inertia about centroidal axis is found. In the above example, the

moment of inertiaABI

is found andy

, the distance of CG

from axis AB is calculated. Then from parallel axis theorem,

2yAII xxAB +=

2yAII ABXX =

Example 1

Determine the moment of inertia of the symmetric I-section

shown in Fig.15.3 about its centroidal axes x-x and y-y.


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Fig. 15.3

Also, determine moment of inertia of the section about a

centroidal axis perpendicular to x-x axis and y-y axis. The section

is divided into three rectangles .,, 321 AAA

Area 21 18009200 mmA ==

( ) 22 4.15547.629250 mmA ==

2

3 18009200 mmA ==

Total Area

24.5154 mmA =

The section is symmetrical about both x-x and y-y axes.

Therefore, its centroid will coincide with the centroid of

rectangle .2AWith respect to the centroidal axes x-x and y-y, the centroid of

rectangle 1A is 1g (0.0,120.5), that of 2A is 2g (0.0,0.0) andthat of

3A is 3g (0.0,120.5).

=xxI Moment of inertia of +1A Moment of inertia of

+2A Moment of inertia of 3A about x-x

( )233

2

3

5.120180012

92000

12

2327.65.1201800

12

9200+

++

++

=xxI

469202,92,5 mm=Similarly,

12

2009

12

7.6232

12

2009 333 +

+

=xxI

405815,20,1 mm=

Moment of inertia of the section about a centroidal axisperpendicular to x-x and y-y axis is nothing but polar moment

of inertia, and is given by:

yyzz IIxxI += = 05815,20,169202,92,5 +

475017,12,7 mmIzz = Ans.

Example 2

Compute the second moment of area of the channel section

shown in Fig.15.4 about centroidal axes x-x and y-y.

Fig. 15.4

The section is divided into three rectangles21 ,AA and .3A

Area, 21 13505.13100 mmA == ,

( ) 22 3.30211.827400 mmA ==2

3 00.13505.13100 mmA ==

Total Area

23.5721 mmA =The given section is symmetric about horizontal axis passing

through the centroid2g of the rectangle .2A A reference axis

(1)-(1) is chosen as shown in Fig. 15.4.

The distance of the centroid of the section from (1)-(1)

3.5721

5013502

1.83.3021501350 ++

= = 25.73 mm

With reference to the centroidal axes x-x and y-y, the centroid of

the rectangle1A is ( ),25.19,27.241g that of 2A is

2g (21.68,0.0) and that of 3A is 3g (24.27,193.25).

Moment of inertia of 21 ,AA and 3A about x-x

332

3

25.193135012

5.13100

12

3731.825.1931350

12

5.13100+

+

++

=

4810359.1 mmIxx = Ans.

Example 3

Determine the moment of inertia of the built-up section

shown in Fig. 15.5 about its centroidal axes x-x and y-y.

The given composite section may be divided into simple

rectangles and triangles as shown in the Fig. 15.5.

Fig. 15.5


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54 7.154


ENGINEER

INGMECHANICS

Area1A =

2300030100 mm= ,

2A =2250025100 mm=

3A =2400020200 mm=

4A =2

875205.872

1mm=

5A =2875205.87

2

1mm=

Total Area

A= 211250mm

Due to symmetry, centroid lies on the axis y-y. A reference axis

(1)-(1) is choosen as shown in the figure. The distance of the

centroidal axis from (1)-(1)

areaTotal

aboutareasofmomentsofSum 1(1( =

areaTotal

aboutareasofmomentsofSum )1()1( =

= 59.26 mm

With reference to the centroidal axes x-x and y-y, the centroid of

the rectangle1A is 1g (0.0,75.74), that of 2A is 2g

(0.0,10.74), that of 3A is 3g (0.0,49.26), the centroid oftriangle

4A is 4g (41.66,32.59) and that of 5A is

5g (41.66,32.59).

12

2020074.102500

12

1002574.753000

12

30100 323

23

++

++

=xxI

23

23

2 59.3287536

205.8759.32875

36

205.8726.494000 +

++

++

443447,15,3 mmIxx = Ans.

2

3333

66.4187536

5.8720

12

20020

12

25100

12

10030+

+

+

+

=yyI

23

66.4187536

5.8721+

+ ,

445122,97,1 mmIyy = Ans.

Problems

1. The C/ S of a gantry girder is shown in fig . 15.6. It is made

up of an I sec of depth 450 mm, flange width 200 mm and

a channel of size 400 mm x 150 mm. Thickness of all

members is 10 mm. Calculate the moment of inertia abouthorizontal centroidal axis.(Ans. I

xx= 4.2198 x 108mm 4)

Fig. 15.6

Fig. 15.7

2. A plate girder is made up of a web plate of size 400mm x

100mm, 4 angles of size 100mm x 100mm x 10mm and

covered plates of size 300mm x 10 mm as shown in fig.

15.7. Calculate the moment of inertia about horizontal and

vertical centroidal axes.

[Ans. Ixx = 5.35786 x 10 8 mm 4 , Iyy = 60850667 mm 4 ]

3. The c/ s of a plain concrete culvert is as shown in fig. 15.8.

Calculate the moment of inertia about horizontal centroidalaxis.(Ans. I

xx= 5.45865 x 1010mm 4)

Fig. 15.8


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Fig. 15.9

4. Determine the centroid of built up section shown in fig.15.9

and calculate the moment of inertia and radius of gyration

about horizontal centroidal axis.

[Ans. Ixx= 1267942mm4 ; Kxx = 18.55a mm]

Question

1. How would you find out the moment of inertia of a plane

area?

2. What is Rouths rule for finding out the moment of inertiaof an area?

3. Derive an equation for moment of inertia of the following

sections about centroidal axis:

a. a rectangular section

b. a hollow rectangular section

c. a circular section , and

d. a hollow circular section

4. State and prove the theorem of perpendicular axis, as applied

to moment of inertia.

5. Prove the parallel axis theorem, in the determination of

moment of inertia of areas with the help of a neat sketch..6. Describe the method of finding out the moment of inertia

of a composite section.

Notes

moment of inertia of composite sections

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