centroid & moment of inertia

3UNIT

Centroid & Moment ofInertia

Learning ObjectivesAfter studying this unit, the student will be able to

• Know what is centre of gravity and centroid

• Calculate centroid of geometric sections

Centre of Gravity

Centre of Gravity (or) mass centre of a point in the body where entire mass weight – is assumed to be concentrated. In other words, it is a point in thebody, through which the resultant of the weights of different parts of the body isassumed to be acting. It is generally written as C.G.

Centroid:

The plane figure like triangle, rectangle circle etc have only areas and massis negligible. The centre of area of such plane figures is called ‘Centroid’ (or)“Centre of Area”. It is generally denoted by “G”

Centroidal Axis

The axis which passes through centre of gravity (or)centroid is known as “Centroidal Axis” XX1, YY1, ZZ1

are called Centroidal Axis

Fig 3.1

Construction Technology340

Axis of Symmetry

Axis of Symmetry is the line dividing the figure into two equal parts likemirror images the centroid always lies on the axis of symmetry.

Fig 3.2

A figure may contain one (or) more axis of symmetry. If there are moreaxis of symmetry the cntroid lies at the intersection of axis of symmetry

Fig 3.3

Position of centroids for Standard Geometric Sections.

S. No Name Shape of figure Position of centroid

1 Rectangle

2 Triangle

At int er sec tions ofDiagonals

Hy31A BH2


Lx2By2

A L X B

Paper - III Engineering Mechanics 341

3 Parallelogram

4 Circle

5 Semicircle

6 Trapezium (slopingon both sides)

7 Trapezium (One sideis vertical andother side is sloping)


Lx2By2

2a ba b

2a b1 a b

hy ( )3hy ( )3hA (a b)2

2


Dx Radius2Dy Radius2

DA2

2 2a ab bx3 (a b)

hA (a b)2


Centroid of Composite sections

A composite section is a combination of simple regular shapes as rectangle,triangle circle, semi circle etc. For determining the centroid of composite sections,the entire area is divided into two (or) more regular simple shapes, Then theprinciple of moments is applied to determine the centroid.

Centoid of plane figure having hollow Portion

The Centroid of plane figure having hallow portion is determined similar tothe composite sections by applying principle of moments, However the negativesign is taken into consideration of hollow positions which are enclosed in aregular shape.

Sections Symmetrical about both X and Y axes

Sections Symmetrical about – horizontal axis (XX)

Sections Symmetrical about the vertical axes (YY)

Fig 3.4

Fig 3.5

Fig 3.6


Sections un symmetric about – the both axes (X-X,Y-Y)

Methods of determination of centroid

The following three methods are available to locate the cntroid of an area.

1. Analytical method

2. Graphical method

3. Experimental method

Analytical method for location of the centroid

Principle: The sum of the moments of a system of a coplanar forces aboutany point in the plane is equal to the moment of their resultant about the samepoint.

Fig 3.8

Consider a lamina in area “A” divided into number of elementary areas A1,A2, A3, ….etc as shown in fig. 3.8.Let the centroids of these elementary areasbe at a distance of x1, x2, x3….. etc from vertical axis and y1 , y2, y3 from thehorizontal axis.

Let the centroids of the total area “A” is at a distance of x and y fromvertical and horizontal axis respectively. As per the principle of moments, thesum of moments of all the elementary areas about horizontal axis OX is equal tothe moment of the total area about the same horizontal axis i.e OX.

Fig 3.7


1 1A yy

A

where A=A1+A2 +A3 +….

Similarly taking moments of areas about vertical axis i.e. OY

1 1A xx

A

The terms 1 1 1 1A y & A x are know as First movement of areaabout y-axis and x-axis respectively

First moment of area: The First moment of area about a line is the productof area and the perpendicular distance of its centroid from the given line.

Important – Note

1. If the axis passer through the centroid, the moments of areas on oneside of the axis will be equal to the moments of areas on the other side of theaxis.

Example 3.2

Locate the centroids if the trapegezium as shown in figure 3.09

Fig 3.09

Solution

Dived the trapezium into rectangle of size a x h and triangle of base (b-a)and height – “h”

Area of rectangle (1) A1 = a.h

Area of rectangle (2) A2 = 12 (b-a)h

Total area of trapezium hA a b2

Let the centroid of the trapezium be at a distance y above base and xfrom last vertical side. Centroidal distance of rectangle from A i. e ,

1ax2


Centroid distance of triangle from A i. e ; 2b ax a

3

3a b a 2a b3 3

Similarly

Example 3.3

Locate the position of centroid of lamina in fig 3.10

Solution

Y.Y Axis as symmetry centroid lies in this axis divide the section in to asquare and a triangle A1 = 100x100 = 10,000mm

1 1 2 2

1 2

22 2 2

2 2

a 1 2a bah b a hA x A x 2 2 3x hA A a b

2b a 2a bh a

2 3 3a 2ab b 2a abh 3 a ba b2

a ab b3 a b

1 1 2 21 2

1 2

2 2

A y A y h hy y , yA A 2 3

h h h h hah b a ah b a2 2 3 2 3y h ha b a b

2 23ah bh ah 2ah bh h 2a b

3 a b 3 a h 3 a b

a ab b h 2a bcentroid x , y3 a b 3 a b


y1 = 100 50mm2

22

1A 100 60 3000mm2

260Y 100

0

3

120mmabove base

Total area = A1 +A2 = 10,000+3000 = 13000 mm2

Let be centroid distance from base

The centroid of the lamina is 66.15 mm above the base.

Example 3.4

A trapezoidal lamina has uniform batter on both sides. Its top width is 200mm. bottom width is 300mm and height is 600 mm. determine position of centroidfrom base.

Solution

Top width a = 200 mm

Bottom width b = 30 mm

Height h = 600 m

= Position of centroid from the basey

Fig 3.10

y

66.15 mm

500000 + 3,60,000 13,000

8,60,000 13,000

10000(50) + 3000(120) 13,000

A1y1+A2y2 A1+A2

= =

=

=

=

Fig 3.11


Example 3.5

Find the centroid of the following T – section

Solution

Fig 3.12 shows T- Sections Y-Y axis as symmetrical axis. In this axis only.

Taking base X-X axis as reference line.

Dividing the “T” section in to two rectangles areas. (flange +web)

Area of rectangles flange A1 = 100x10 = 1000 mm2

from the bottom ofthe base

Area of rectangle web (2) A2 =140x10 = 1400 mm2

from the bottom of theweb X.X axis

distance of centroid from bottom of web XX i. e.,

y h3

2a+b a+b( )= = 2x200 + 300

200 + 300( )600 3

400 + 300 500

200 ( )=

280 mm=

Fig 3.12

1 1 2 2

1 2

A y A y 1000 145 1400 70yA A 1000 1400

98000 145000 101.25mm2400

y2 = 140 2 = 70 mm

y1 = 10 2

= 145 mm140 +


Example 3.6(imp)

Find the position of the centroid of an I section given.

Top angle : 60 x 20 mm

Web : 20 x 100 mm

Bottom angle : 100x20 mm

Solution

Fig 3.13 shows given I sectrim Y-Y axis as axis of symmetry so centroidlien in this axis only.

we can find

X-X axis is base of the bottom taken as reference line. Dividing I section into three rectangles.

Area of rectangle (1) A1 = 60 x 20 =1200 mm2

from the base of thebottom flange

Area of rectangle (2) A2 = 100 x 20 = 2000 mm2

from the base of the bottom flange


Y3 = = 10 mm from the base of the botttom of the flange

y1 = 20 + 100 + 20 2

130 mm=

y2 = 20 + 100 2

70 mm=

20 2

y


Example 3.7(imp)

A masonry dam of the trapezoidal section with one face is vertical. Topwidth of dam is 3m, bottom width of dam is 6m and height is 6m. Find theposition of centroid.

Solution

(i) Applying for multa

Top width a = 3 mt

Bottom width b = 6 mt

Height of the dam = 6 mt

∴one face is vertical.

Let centroid of the dam be at a distance about baseand turn use vertical face centroid.

IInd method

Trapezium OBCD is divided in to tow simple areas

1. Rectangle OLCD 2. Triangle CLB

x

Fig 3.14

x =a2 + ab + b2

3(a+b)

32 + 3 x 6 + 62

3(3+6)=

9+18+36 27=

= 2.33 mt

y =h3

2a + b a+b( )

63

2x3+6 3+6( )=

12 92 ( )=

= 2.67 mt

Y


There is no axis as axis of symmetry line.

∴ We can find both x and yFinding Vertical face OD as resume line.

For Rectangle OLCD area. A1 = 3x 6 = 18 m2

Xl = = 1.5. from vertical face OD.

For triangle CLB

Area

m

form vertical face.

Finding y Base of the dam OB taken as reference line.

Area of rectangle (1) A1 = 18m2

from the base of the dam

Area of triangle (2) A2 = 9m2

from the baseof the dam

Example 3.8 (imp)

Determine the centroid of the channel section 200 x 100 x 10 mm as shownis fig 3.15

Solution

Fig 3.15 shown the given channel section. X-X axis as axis of symmetryline is this only centroid lies it.

x

16y 3mt2

21A2

3 6 3 2

2

9m

3x 3 4m3

1 1 2 2

1 2

A y A y 18 3 9 2y 2.67mA A 18 9

1 1 2 2

1 2

A x A x 18 1.5 9 4x 2.33A A 18 9


Finding only taking vertical outerface AB as reference line dividing thesection in to three rectangles

Rectangle (1) AreaA1 = 100 10 = 1000 mm2

1100x 50mm

2 from vertical face AB

Area of Rectangle(2) Area A2 = 180 10 = 1800 mm2

Area of rectangle (3) A3 =100 10 = 100mm2

from verticalface AB

From the vertical face AB

Example 3.9

Find the position of centroid for an angle of section from base as shown infig. 3.16

Solution

Fig 3.16 shows given angle section there is no X-X and Y-Y axis are axisof symmetry.

Fig 3.15

23

100x 50mm2

1 1 2 2 3 3

1 2 3

A x A x A xxA A A

1000 50 1800 5 1000 501000 1800 1000

50000 9000 500003800

28.68mm

x2 = = 5mm from vertical face AB10 2


We have to find both x and y

Finding

Vertical face AC as reference line.

Dividing the angle section as two rectangularareas.

Areas of Rectangle 1 A1 = 120 x 20 =2400 mm2

120x 10mm2

from vertical face AC

Area of Rectangle 2 A2 = 100 x 20 = 2000 mm2

from vertical from AC

= 37.27mm from vertical face AC.

Finding

Bottom AB as axis of reference.

1120y 60mm

2 from bottom base AB

from the bottom base AB

= 37.27 from the bottom base AB.

Example 3.10

Determine the centroids of the selection shown in figure 3.17

y

x

1 1 2 2

1 2

A x A x 2400 10 2000 70 24000 140000xA A 2400 2000 4400

2100x 20 70mm

2

220y 10mm2

1 1 2 2

1 2

A y A y 2400 60 2000 10yA A 2400 2000

Fi.g 3.16

120

120

20

20


Solution

Figure 3.17 shows selection has

no X-X and Y-Y axis as axis of

symmetry. So both and can be

determined.

Finding

CD line vertical face taken as axis of reference line.

Dividing given Z selection in to three rectangular areas.

Area of rectangular A1 = 80 x 20 = 1600mm2

180x 40mm2

(from vertical face CD)

Area of rectangle (2) A2 = 220x20 = 4400 mm2

from vertical face CD


from the vertical face CD

= 92.02 mm from the vertical face CD

Finding yBottom base AB as axis of reference

120y 200 210mm2

from bottom base AB

2200y 110mm


from bottom base of AB

x

1 1 2 2 3 3

1 2 3

A x A x A x 1600 40 4400 90 1920 140xA A A 1600 4400 1920

64000 396000 2688007920

380x 100 140mm2

32y 12mm2

x y

220x 80 90mm2

100

200

80

80

20

10020

2 0

Fig. 3.17


= 106.44 mm from bottom base AB

Example 3.11

Find the cenrodidal distance for the built up section shown in figure 3.18

Solution

Figure 3.18 shown Y-Y axis of symmetry

centroid lies in it we can find

Finding

Bottom most layer AB line a as axis of reference.

Built up section has divided in 5 rectangular areas.

Rectangular (1) A1= 100 x 10 = 1000mm2

110y 10 20 150 20 205mm2

Rectangular (2) A2 = 100 x 20 = 2000mm2

220y 10 20 150 190mm2

Rectangular (3) A3 = 150 x 20 = 3000 mm2

3150y 10 20 105mm

2

Rectangular (4) A4 = 20 x 200 = 4000mm2

420y 10 20mm2

Rectangle (5) A5 = 10 x 200 = 2000mm2

510y 5mm2

1 1 2 2 3 3

1 2 3

A y A y A y 1600 210 4400 110 1920 12yA A A 1600 4400 1920

336000 484000 230407920

8430407920

Fig 3.18

y

y


= 82.5mm from bottom base AB

Review QuestionsShort Answer Type Questions

1. Define centre of gravity

2. Determine the center of gravity and centroid

3. Locate the position of centroid of the following figures with a neat sketch

1) rectangle 2) triangle 3) circle 4) Semi circle

4. Find the centroid of triangle of base 80 mm and height 120 mm from thebase and the apex

Essay Answer Type Questions1. A masonry dam is trapezoidal in section with one face vertical. Top

width is 3m and bottom width is 10 m height is 10 m. Find the position ofcentroid axis

Ans. x 3.564m and y = 4.260 m

2. Determine the centre of gravity of I section having the following dimensions

Bottom flange = 300x100mm

Top flange = 150x50mm

Web = 50x400mm

Ans. 198.9mm from bottom flange


3. Find out the centroid of an un equal angle section 100mm x 80mm x20mm

Ans = 25 mm from left face

= 35mm from bottom face

4. Find the centre of gravity of channel section 100 x 50 x 15 mm

Ans = 17.8 mm from outer face of web

5. Find the centroid of the given “T” section

Top flange of 250mmx50mm

Web 50mmx200mm

Ans: y = 169.44mm from bottom of the web.

6. Find the centroid of the section shown in figure

Ans: x = 95.56mm from the left edge

y = 85.55mm from the bottom edge

Fig 3.19

xy

x


Moment of Inertia (M.I)Definition

The product area (A) and perpendicular distance (x) between the point isknow as “the first moment of area (Ax) about the point. If this moment is againmultipled by the distance ‘x’ i.e. Ax.x = Ax2 is called moment of moment of area(or) the second moment of area or simply moment of inertia. Its unit in SI systemis mm4.

Moment of inertia for some regular geometrical sections

Position of centroids for Standard Geometric Sections.

S. No Name Shape of figure MI about MI about XX(Ixx) yy(Iyy)

1 Rectangle

2 Hollow

Rectangle

3 Solid Circular

section

4 4D D64 64

4 Hollow Circular

section

5 Triangle

3 32 2BD DB12 12

db3

12DB3

12-bd3

12BD3

12-

about cg about baseBC

3 3b h b h3 6 1 2

4 4 4 4(D d ) (D d )64 64


Parallel Axis Theorem

It states that if the moment of inertia of a plane area about an axis throughits centre of gravity (IGG), is shown in fig 3.20. Then the moment of inertia of thearea about an axis AB parallel to IGG at a distance of “h” from centre of gravityis given by IAB = IGG + Ah2

Fig 3.20

Where IAB = M.I of the area about an axis AB

IGG = M. I of the area about its C.G

A = Area of the section

h = distance between C.G of the section and the axis AB.

Radius of Gyration

Radius of gyration about a given axis is defined as the effective distancefrom the given axis at which the whole are may be considered to be located withrespect to axis of rotation. It is denoted by “k” or “r”

I = Ak2 (or) Ar2

Where I = moment of inertia

K(or) r = radius of gyration k (or) r =

A= area of cross section

Units for k or r in S.I system is mm

Perpendicular axis theorem

It states that if IXX and IYY be the moment of inertia of plane section abouttwo perpendicular axes meeting at “o” shown in figure 3.21 then, the moment ofinertia IZZ about the axis Z Z which is perpendicular to both XX and YY axises,is given by

IZZ = IXX + IYY

IA


For symmetrical section like circular IXX = IYY

IZZ = IXX + Ixx

IZZ = 2IXX

J = 2 I where “J” is known as polar moment of Inertia

Polar moment of inertia.

Definition :- The moment of inertia of an area( IZZ ) about an axisperpendicular to its plane is called “polar moment of Inertia”. It is denoted by“J”

Solved ProblemsProblem 3.12

Find the moment of inertia of a rectangular

section 400mm wide and 800mm deep about

its base.

Solution

Breadth of bearn B = 400mm

Depth of beam D = 800 mm

M.I . about C.G i.e.

M. I about its base 2ABI I Ah

IAB = 1.706 x 1010 + (400x 800) (400)2

= 1.706 x 1010 + 5.12 x 1010

= 6.826 x 1010 mm4

4 3310 2

XX

400 800BDI 1.706 10 mm12 12

Fig 3.21

Fig 3.22


Problem 3.13

Find the M.I of hollow circular sections whose external diameter is 60mmand internal diameter is 50mm about Centroidal axis

Solutions

External dia D = 60 mm

Internal dia d = 50mm

4 4XX YY

4 4

4

I I D d64

60 5064329.2mm

Moment of inertia about Centroidal axis is = 329.2 mm4

Problem 3.14

Find the moment of inertia of a rectangle 60mm wide and 120mm deepabout Centroidal axis. Find also least radius of gyration.

Solutions

B = 60mm

D = 120mm

M. I about Centroidal axis

Area of rectangle A = BD = 60 x 120 = 7200mm2

Least radius of gyrations k (or) r =

Fig 3.23

IcG A

8.64 x 106

7200=

Fig 3.24


∴ Least radius of gyration = 34.64mm

Problem 3.15

Find the radius of gyration of hollow circular sectors of external diameter300mm and internal dia 200mm.

Solution

External dia D = 300mm

d = 200mm

Area 2 2D d4

2 2 4 2300 200 3.927 10 mm4

=

Radius of gyration

Alternate method

= 90.14mm

Problem 3.16

Find the radius of gyration of a triangle whose base is 40mm and height is60mm about an axis passing through C.G and parallel to base.

Base b = 40mm

H = 60mm

Area = 1 bh2

M.I of triangle about Centroidal axis

8

4

I 3.191 10K 90.14mmA 3.927 10

4 42 2 2 2

2 2

D d D d 300 20064K4 4D d

64

Fig 3.25

Fig 3.26


∴ Radius of gyration

3bh 2 hK36 bh 18

60K 14.14mm18

Problem 3.17

Find the moment of inertia about Centroidal axis of hollow rectangularsections shown in fig 3.27

Solution

B = 200mm

D = 400mm

b = 100mm

d = 200mm

M.I about XX axis for hollow rectangular sections.

= 1000x106 mm4

M. I about Y Y Axis for a hollow rectangular section

= 250 x 106mm4

Problem 3.18

Determine the position of centroid and calculate the moment of inertia aboutits horizontal centroidal axis of a T – beam shown in figure 3.28

Solution

Ixx A

K =

Ixx = 112

[ 200 x 4003 - 100 x 2003]

XXbhI36

Fig 3.27

3 33 3

YYDB db 1I 400 200 200 10012 12 2


Finding Centroid

YY axis is axis of symmetry centroid lies on it.

To Finding y

Take AB line axis of reference.

Dividing T section into two rectangular areas

Area of rectangle (1) A 1 = 300 x 100 = 30000mm2

1100y 200 250mm


Area of rectangle (2) A2 = 200 x 100 = 2000mm2

from bottom base AB

Centroidal distance y from bottom 1 1 2 2

1 2

A y A yA A

30000 250 20000 10030000 20000

= 190mm from bottom base AB.

M. I of a rectangle 1 about centroidal axis

IXX at (1) = IG + Ah12

= 2.5 x 107x108 x 106 = 25 x 106+108 x 106

= 133 x 106 mm4

Fig 3.28

2200y 100mm

2

32

271 1

7 2

300 100 300 100 y y12

2.5 10 300 100 250 190 h y y

2.5 10 300 100 60


M. I of a rectangle 2 of about Centroidal axis

Moment inertia of T – beam about its Centroidal axis

Problem 3.19

An un symmetrical I section has top flange 100x20mm web 100 x 120mmand bottom flange 80x20 mm over all depth is 160mm.

Calculate centroid

Solution

Figure 3.30 Shows given I section.

YY-axis is axis of symmetric line

so centroid lies on it.

Finding yTake line AB, passing through the bottom edge as axis of reference

Divide the section into three rectangular areas.

Area of rectangle (1) A 1 = 100 x 20 = 2000 mm2

120y 20 120 150mm2

from base

2G 2I @ 2 I Ah

32100 200 100 200 90

12

6 6

6 4

I at (1)d I at(2)

133 10 228.7 10361.70 10 mm

7 6

6 6

6 4

6.67 10 162 1066.7 10 162 10228.7 10 mm

since h2 = y- y2

h2 = 190-100 = 90mm



2120y 20

60

280mm from base


320y 10mm2

from the base

1 1 2 2 3 3

1 2 3

A y A y A y 2000 150 1200 80 1600 10yA A A 2000 1200 1600

300000 96000 16000 85.83mm

4800

from the base AB.

Finding M.I of “I” section about X-X axis about centroid

M.I of rectangular (1) about X-X axis

h1 = 150-85.83 = 64.17

M. I of rectangle (2) about a X-axis

h2 = y - y2 = 85.83-50 = 5.83mm

M. I of rectangle (3) about X-X axis

h3 = - y3 = 85.83 - 10 = 75.83 mm

Ixx

1

2xx G 1 1 1 1

3

6 4

I @I I A h h y y

100 20 100 20 64.1712

8.3 10 mm

2

2xx G 2 2

32

6 4

I at 2 I A h

10 120 10 120 5.8312

1.48 10 mm

3

2xx G 3

32

6 4

I at 3 I A h

80 20 80 20 75.8312

9.25 10 mm


Moment of inertia of given I section about X axis

Problem 3.20

Determine the moment of inertia of the un equal angle section of size 150mmx 100mm x 25mm about Centroidal axis.

Solution

Finding centroid

FindingVertical face CD has axis of reference, dividing L section has two

rectangular areas.

Area of rectangle 1 A1= 125 x 25 = 3125 mm2

125x 12.5mm2


Area of rectangle 2 A2 = 100 x 25 = 2500mm2


= 29.17mm from vertical face CD

Finding yBottom base AB has taken as axis of reference

1125y 25 87.5mm

2 from base

225y 12.52

from base

y

xx xx xx6 6 6

6 4

I at1 I at 2 I at 3

8.3 10 1.48 10 9.25 1019.03 10 mm

x

2100x 50mm

2

1 1 2 2

1 2

A x A x 3125 12.5 2500 50xA A 3125 2500

39062.5 1250005625

164062.55625


= 54.17mm from the base AB.

Finding xxI

M. I of rectangle (1) about x-x axis

h1 = y - y1 =54.17-37.5 = 33.33mm

= 7.54 106 mm4

M I of rectangle (2) about x-x axis

h2 = y - y2 =54.17-12.5

= 41.67mm

= 0.13 x 106 + 4.34 x 106

= 4.47 x 106 mm4

Moment Inertia of given angular section about X-X axis

Finding YYI M I of rectangle (1) about Y-Y axis

YYI at 1 = IG1 + AA1h12

= 29.17 - 12.5

y

1

322

xx G 1 1

6 6

25 125I at1 I A h 25 125 33.3312

4.07 10 3.47 10

2

32

xx G 2 2100 25I @.2 I A h 100 25 41.67

12

xx xx6 6

6 4

I at I I at2

7.54 10 4.47 1012.01 10 mm

273437.5 + 31250 5625

=

304687.5 5625

=

y = A1y1 + A2y2 A1 + A2

3125 x 87.5 + 2500 x 12.5 31.25 + 2500=

h1 = x - x1


M.I Rectangular (2) about Y-Y axis

M. I of a given angular section about Y-Y axis

Review QuestionsShort Answer Type Questions

1. Explain a) Parallel axis Theorem

b) Perpendicular axis theorem

2. Define the termsa) Moment of inertia

b) Radius of gylation

3. Find the radius of gyration of circle having diameter “d”

Ans: d4

4. Find the radius of gyration of hollow circular plate of 60mm inner diameterand 100mm outer diameter

(Ans:29.15mm)

2

2YY G 2 2

32

6 6

6 4

I at 2 I A h

25 100 2500 20.8312

2.08 10 1.08 103.16 10 mm

YY YY6 6

6 4

I at I I at2

1.063 10 3.16 104.223 10 mm

32

1 1

32

6 6

6 4

DB A x x12

150 25 3125 29.17 12.512

0.195 10 0.868 101.063 10 mm

h2 = x2 - x = 50 - 29.17 = 20.83


5. Find M.I of a rectangular section 200mm width and 400mm depth aboutthe base

(Ans. 4.267 x 109mm4)

Essay Answer Type Questions1. Find the moment of Inertia of a T Section having flange150mm x 50mm

and web 50 x 150mm about xx and yy-axis through the C. G of thesection.

[ Ans: Ixx = 53.125 x 106 mm4

IYY =15.625 x 106mm4]

2. Determine the moment of Inertia of an unequal angle section of size100mm x 80mm x 20mm about Centroidal axis

[ Ans: Ixx = 2.907 x 106 mm4

IYY =1.627 x 106mm4]

3. Determine the moment of inertia of an I section about XX axis given thattop flange 100mm x 10mm web = 200mm x 10mm different flange160mm x 10mm

[Ans: Ixx = 34.38 x 106 mm4]

4. A built up section is formed by an I section and to flange plates of size280 x 20mm are an each flange find the moment of inertia aboutcentrodial X-X axis as shown in below figure

[Ans: Ixx = 188.22 x 106 mm4]

Key Concepts1. The C.G of a body is the fixed point at which its weight is assumed to be

concentrated.

2. The centroid of a surface is the fixed point at which the area of thesurface is assumed to be concentrated.


3. The centroid of a surface is determined from the equations:

4. The centroid of a composite area is treated by the principle of moments,dividing it into regular simple figures.

5. The M.I of an area about a given axis is the sum of the values of “ax2”where “a” is the area of each element and “x” is the distance of thecentroid of the element from the given axis

2I ax 6. Radius of gylation (Kxx) of an area about given axis is the distance from

the axis at which the area may be assumed concentrated to given the M.I of the area about the given axis

IKA

7. Parallel axis theorem :- if “XX” is an axis is parallel to the centrodal axisC.G of surface of area A and if “d” is the distance between the twoparallel axis.

2CGI I Ad

8. Perpendicular axis theorem: If XX and YY are two perpendicular axisis the plane of the area and ZZ is an axis perpendicular to both of themthrough their intersection.

zz xx YYI I I

9. The M.I about an axis perpendicular to its plane is known as its polarM.I

10. M.I of a built up section = Sum of M.I of all elements of the sectionabout the same axis.

11.M.I of a rectangle bxd about axis through centroid parallel to

side

12. M. I of a triangle ‘bxh’ about axis through centroid parallel to base3bh

36

13. M. I of circle of dia ‘d’ about any diameter 4d

64

1 1 1 1A x A yx and yA A

3bdb12


14. M.I of hollow circular section of diameters “D” and ‘d’ about anydia

15. Polar M.I of a solid shaft of dia ‘d’ about axis 4d32

16. Polar M.I of hollow shaft of dia of diameter ‘D’ and ‘d’

=

4 4D d64

4 4D d32

centroid & moment of inertia

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