CENTROIDS AND

MOMENTS OF INERTIA

CENTROIDS

(AĞIRLIK MERKEZLERİ)

A centroid is a geometrical

concept arising from parallel

forces. Thus, only parallel

forces possess a centroid.

Centroid is thought of as the

point where the whole weight

of a physical body or system

of particles is lumped.

CENTROIDS (AĞIRLIK MERKEZLERİ)

If proper geometrical bodies possess an axis of symmetry,

the centroid will lie on this axis. If the body possesses two

or three symmetry axes, then the centroid will be located

at the intersection of these axes.

If one, two or three dimensional

bodies are defined as analytical

functions, the locations of their

centroids can be calculated using

integrals.

A composite body is one which is comprised of the

combination of several simple bodies. In such bodies, the

centroid is calculated as follows:

Line – a thin rod (Çizgi – ince çubuk)

Area - a flat plate with constant

thickness (Alan – sabit kalınlıklı

düz plaka)

Volume – a sphere or a cone

(Hacim – küre ya da koni)

Composite Composite Composite

dl

xdlx

dl

ydly

dl

zdlz

i

ii

l

lxx

i

ii

l

lyy

i

ii

l

lzz

dA

xdAx

dA

ydAy

dA

zdAz

i

ii

A

Axx

i

ii

A

Ayy

i

ii

A

Azz

dV

xdVx

dV

ydVy

dV

zdVz

i

ii

V

Vxx

i

ii

V

Vyy

i

ii

V

Vzz

CENTROIDS OF SOME

GEOMETRIC SHAPES

CENTROIDS OF SOME GEOMETRIC SHAPES

Thin Rod in the Shape of a Quarter Circle

dl

xdlx

dl

ydly

sin

cos

ryy

rxx

drdl

el

el

dl

d

r = radius

x

y

xxel

yyel

ry

r

r

rd

rdr

dl

ydly

rx

r

r

rd

rdr

dl

xdlx

2cossin

2sincos

2/

0

2/

0

2

2/

0

2/

0

2/

0

2/

0

2

2/

0

2/

0

r

Circular Arc

r=radius

Solid Quarter Circle

dddAR

AdA

ydAy

4

2

a a

x

y

G 0sin

yr

xa

a

3

4

3

44

3

310

3cos

3

sin3

sinsin

2

3

332

0

0

3

2

00

3

0

2

0

2

π

Ryx

π

R

πR

R

R -

R - θ-

ρ

θdθρ

θdρdθρθρdρdθρydA

π/

R

π/RR π/

x

dA

y=sin

x=cos

R

d d

y

İçi Dolu Daire Yayı

İçi Dolu Çeyrek Daire

Solid Half Circle

Üçgen Triangle

G x

y İçi Dolu Yarım Daire

dA

xdAx

0y

dddA

RA

2

2

2/

2/0

3

0

2/

2/

2 cos3

coscos

dddddxdA

RR

3

4Rx

h

dy

y

x

w

x

y

b

x

a

dywdAh

yx 3

?

2

0

bhdyy

h

bbwdydAyh

h

bw

yh

w

h

bh

dyy)(hh

by)(h

h

by

h

axdA

h

02

3

bax

h

a

y

x

wxx

2

Parabola

dy

dx

b/2

y

x

w=b-x y

x

b

Parabol

b

xy

2

2

dA

xdAx

dA

ydAy

62

2

0

2 bdx

b

xydxAdA

b

82

3

0

2 bdx

b

xxxydxxdA

b

40

2(3

2/12/12/1

2/

0

bdybybydyxbyywdyydA

b

4

3bx

20

3by

b

xy

2

2

Area Between a Line and a Curve

Bir Doğru ile Bir Eğri Arasında Kalan Alan

dy

dx

y=x

xleft

w

y

x

xright

ybottom

ytop l

(a, a)

axy 2

dA

xdAx

623

2

22

3

2

3

2

2

2

0

2

0

2321

0

2121

2121

aaaA

xxadxxxadA

dxxxadxyyldxdA

)()(

aa//a

//

//

bottomtop

a

y x , xaya xax , xax , y

xy x ,y

//2

212122

22

dA

ydAy

2

6

1243ydA

2

3

0

4

0

0

32 ay ,

a

a

a

yy)dy

a

yy(ydyxxy a

a

a

leftright

5

26

151535

2

32

5 2

33

5

3

3

3

0

3

0

0

25212121 a

a

ax ,

aaaxxax)dxxx(axdA

)()(

a

a

a//

//

dA

dF

d

AREA MOMENT OF INERTIA

ALAN ATALET (EYLEMSİZLİK) MOMENTİ

It is often necessary to calculate the

moments of uniformly distributed loads

about an axis lying within the plane they

are applied to or perpendicular to this

plane. Generally, the magnitudes of these

forces per unit area (named as pressure or

stress) are directly proportional with the

distance of their lines of action from the

moment axis. This way, an elementary

force acting in an elementary area will be

proportional to,

distance x differential area

dA

dF

d

AREA MOMENT OF INERTIA ALAN ATALET (EYLEMSİZLİK) MOMENTİ

Thus, the total moment:

This integral is named as “area moment of inertia” or

“second moment of area”.

dAddFddA

dF

dA

dFdPddP , , ,

Elementary moment is proportional to

distance2 x differential area: dM=d2dA

dAdMdAddM 22 ,

AREA MOMENT OF INERTIA ALAN ATALET (EYLEMSİZLİK) MOMENTİ

Moment of inertia is not a physical quantity such as

velocity, acceleration or force, but it enables ease of

calculation; it is a function of the geometry of the area.

Since in Dynamics there is no such concept as the inertia of

an area, the moment of inertia has no physical meaning.

But in mechanics, moment of inertia is used in the

calculation of bending of a bar, torsion of a shaft and

determination of the stresses in any cross section of a

machine element or an engineering structure.

Rectangular (Cartesian) and Polar Area Moments of Inertia and Product of Inertia

(Kartezyen ve Kutupsal Alan Atalet Momentleri ve Çarpım Alan Atalet Momenti)

By definition, the moments of inertia of area dA with respect to x and y axes are

z

dAxdI

dAydI

y

x

2

2

The moments of inertia of the total area A

with respect to x and y axes are

dAxI

dAyI

y

x

2

2 The moment of inertia of area A with respect to x axis

The moment of inertia of area A with respect to y axis

These moments of inertia are named as “Rectangular (Cartesian) moments of inertia”.

The moment of inertia of area dA with

respect to z axis or pole O is by definition

The moment of inertia of total area A

with respect to z axis or pole O is z

dArordIordI Oz

2J

dArI z

2The moment of inertia of area A with respect to z axis

Since the z axis is perpendicular to the plane of the area and cuts

the plane at pole O, the moment of inertia is named “polar moment

of inertia”.

222 yxr Therefore, yxz III

In certain problems involving unsymmetrical cross sections and in the

calculation of moments of inertia about rotated axes, an expression

xydAdI xy

occurs, which has the integrated form

xydAI xy

where x and y are the coordinates of the

element of area dA.

z

Ixy is named as the “product of inertia” of the area A with respect to

the xy axes.

Properties of moments of inertia:

1) Area moments of inertia Ix, Iy, Iz are always positive (+).

2) Ixy can be positive (+), negative (-) or

zero whenever either of the

reference axes is an axis of

symmetry, such as the x axis in the

figure.

3) The unit for all area moments of inertia is the 4. power of that

taken for length (L4).

4) The smallest value of an area moment of inertia that an area can have

is realized with respect to an axis that passes from the centroid of

this area. The area moment of inertia of an area increaes as the area

goes further from this axis.

The area moment of inertia will get smaller when the distribution of an

area gets closer to the axis as possible.

Properties of Moments of Inertia:

A A

Ix=103.13 L4

y

x

Ix=108 L4

y

x

A A

Ix=243 L4 Ix=752 L4

y

x

y

x A=36 L2

Radius of Gyration

Jirasyon (Atalet – Eylemsizlik) Yarıçapı

Consider an area A, which has rectangular moments of inertia Ix and Iy

and a polar moment of inerta Iz about O. We now visualize this area as

concentrated into a long narrow strip of area A a distance kx from the x

axis. By definition, the moment of inertia of the strip about the x axis will

be the same as that of the original area if

xx IAk 2

The distance kx is called the

“radius of gyration” of the

area about the x axis.

z

A similar relation for the y axis is written by considering the area as

concentrated into a narrow strip parallel to the y axis as seen in the

figure. Also, if we visualize the area as concentrated into a narrow ring of

radius kz, we may express the polar moment of inertia as .

In summary, zz IAk

2

AkIAkIAkI zzyyxx

222

A

Ik

A

Ik

A

Ik z

z

y

yx

x

Also since, 222

zyxzyx kkkIII

Transfer of Axes

Paralel Eksenler (Steiner) Teoremi

The moment of inertia of an area about a

noncentroidal axis may be easily expressed in

terms of the moment of inertia about a

parallel centroidal axis.

In the figure, axes pass through the centroid G of the area. Let

us now determine the moments of inertia of the area about the parallel

xy axes. By definition, the moment of inertia of the element dA about

the x axis is

yx

A2ddydI Ox

Expanding to the whole area

dAddAyddAyI OOx

222

We see that the first integral is by definition,

the moment of inertia about the centroidal

axis. The second integral is zero, since

and is automatically zero

with the centroid on the axis. The third

term is simply . Thus, the expression for

Ix and the similar expression for Iy become

x

OO yAdAy Oy

x2

xAd

xI

2

2

AeII

AdII

yy

xx

The sum of these two equations give

2ArII zz zyxzyx IIIIII ,and since

dAddAyddAyI OOx

222

AdeII xyxy

For product of inertia

The parallel axes theorems also hold for

radii of gyration as:

222rkk zz

where is the radius of gyration about a centroidal axis parallel

to the axis about which k applies and r is the perpendicular

distance between the two axes. For product of inertia:

dekk xyxy 22

k

Two points that should be noted in particular about the transfer of axes

are:

The two transfer axes must be parallel to each other

One of the axes must pass through the centroid of the area

If a transfer is desired between two parallel axes neither of which

passes through the centroid, it is first necessary to transfer from one

axis to the parallel centroidal axis and then to transfer from the

centroidal axis to the second axis. So, the parallel axis theorem must be

used twice.

Rotation of Axes, Principle Axes of Inertia,

Principle Moments of Inertia

Eksenlerin Döndürülmesi, Asal Atalet Eksenleri, Asal Atalet

Momentleri

In Mechanics it is often necessary to calculate the moments of inertia

about rotated axes.

The product of inertia is useful when we need to calculate the moment of

inertia of an area about inclined axes. This consideration leads directly

to the important problem of determining the axes about which the

moment of inertia is a maximum and a minimum.

From the figure, the moments

of inertia of the area about

the x' and y' axes are:

dAxydAxI

dAxydAyI

y

x

22

22

cossin

sincos

Expanding and substituting the

trigonometric identities:

2

2cos1os 1cos22cos

2

2cos1sin sin212cos

22

22

c

2sin2cos22

2sin2cos22

xy

yxyx

y

xy

yxyx

x

IIIII

I

IIIII

I

Defining the relations in terms of Ix, Iy, Ixy give

In a similar manner we write the product

of inertia about the inclines axes as:

dAxyxydAyxI yx cossinsincos

Expanding and substituting the trigonometric identities:

2sin2

1cossin , cossin22in and sincos2cos 22 s

2cos2sin2

xy

yx

yx III

I

Defining the relations for Ix, Iy, Ixy give

zyxyx IIIII

Adding, gives

The angle which makes and either maximum or minimum may be

determined by setting the derivative of either or with respect

to equal to zero. Thus,

xI

yI xI

yI

02cos22

2sin2

xy

yxx III

d

dI

Denoting this critical angle by a gives

xy

xy

II

I

22tan a

The equation gives two values for 2a which differ by , since tan2a

= tan (2a + ). Consequently, the two solutions for a will differ by

/2.

One value defines the axis of maximum moment of inertia and the

other value defines the axis of minimum moment of inertia. These

two rectangular axes are called “principle axes of inertia”.

When the critical value of 2 is substituted into the equations, it is

seen that the product of inertia is zero for the principle axes of

inertia. Substitution of sin2a and cos2a into the equations, gives the

expressions for the principle moments of inertia as:

2

2

minmax

22xy

yxyxI

IIIII

Area Moments of Inertia of

Basic Geometric Shapes

Area Moments of Inertia of Basic Geometric Shapes 1) Rectangle

b dy

h

dx

33

3

0

3

0

222 bhbydyybbdyydAyI

hh

x

33

3

0

3

0

222 hbhbdxxhhdxxdAxI

hb

y

12232

32322 bhh

bhbh

AdIIh

dAdII xxxx

hb

I bh

I hb

I bh

I

hbbbh

hbAeII

beAeII

yxyx

yyyy

121233

12232

3333

32322

Area Moments of Inertia of Basic Geometric Shapes

2) Square

Since b=h=a

G

a/2

a/2

a

a

y

x a

I

a

I

a

I

a

I

y

x

y

x

12

12

3

3

4

4

4

4

Area Moments of Inertia of Basic Geometric Shapes 3) Triangle

From similar triangles

y

G x

y

b/3

h/3

h

b

y

x

h

b

x dx

dy h-y

y

n

m

dyndAdAyI x 2

h

b

yh

n

yh

h

bn

12

3

04

4

03

3

0

22 bh

xI

h

h

byh

byhdyyh

h

byndyy

xI

363212

3232 bh

Ihbhbh

AdII xxx

12

32 hbdAxI y

36

32 hb

AeII yy

Area Moments of Inertia of Basic Geometric Shapes 4) Circle

Due to symmetry

24

2

42222

44

0

4

0

322 πRπR

rπIdrrπ πrdrrI πrdr dA dArI

R

z

R

zz

yxzo IIII

4

4πRII yx

πR

I I πR

RπRπR

AdII yxxx4

5

4

5

4

4422

42

G x

y , y

O x

4R/3

Area Moments of Inertia of Basic Geometric Shapes 5) Semicircle

6) Quarter circle

82

44

4

RI

R

I xx

9

8

8,

3

4

28

4

2242 RI

RRRAdII xxx

82

44

4

RII

R

II yyyy

164

44

4

RR

II yx

y

x

x

y

G

4R/3

4R/3

Applications of Area Moment of Inertia

dy

dx

b/2

y

x

w=b-x

y

x

b

* Determine the area moments of inertia of the area under the curve with respect to x and y axes.

? I? I yx b

xy

2

2

1052

1

22

168282427

22

23

27

2

32

4

0

5

0

222

22

4

6

4

7

4

27

272121

3

3

2

0

21212132

0

21212122

bx

b I dx

b

xxdAxdx

b

xydxdA dAxI

b

b

- b

bb

bb

I

yb-

bydyybby dA y I

b

y

b

y

/

///

x

b////b/

///

x

* Determine the area moments of inertia of the area between a curve and a line with respect to x and y axes.

dy

dx

y=x

xleft

w

y

x

xright

ybottom

ytop l

(a, a)

axy 2

2054

54

2847

2

42

7

4

4

4

5

4

00

54222

4

7

4

4

4

0

42721

0

212122

21212

aa

- a

I

a

yydy

a

yy-ydAyI

aa

- a

I

xxa dx-xxaxdAx

dx-xxadA , dAxI

x

aa

x

y

a//a

//

//

y

The product of inertia is zero whenever either of the centroidal axes is an axis of symmetry.

* Determine the product of inertias of the following areas.

G x

y

b/2

h/2

h

b

y

x

dx

dy

0224

422

22

22

0

2

0

2

0 0

hb

-bhhb

- Ade IIAde , II

hb

y

xxydxdyI

dxdy dAxydA I? I

xyxyxyxy

hbb h

xy

xyxy

G x

y e

d

y

x

G x

y e

d

y

x

AdeII

I

xyxy

xy

0

0

AdeII

I

xyxy

xy

0

0

24834

43

2

2243

2

222

2

2

22

0

22

3

22

8

22

6

22

444

2

2

0

4322

2

2

0

22

2

2

22

2

22

2

bhbhbh

-bh

bbb

b

hxbxxb

b

hdxxbxb

b

hxIdI

xbxbb

hy

b

b-xhy

b

h

b-x

y dx

yx

yxydx dI

dAde I d dI I d ydx , dA

bb

xyxy

xy

xyxyxy

for this area

writing in terms of x,

G x

y

b/3

h/3

h

b

y

x

* Determine the product of inertia of the triangular area.

h

b

y

x dx

y

x

y/2

dA

b-x

?I

? I

xy

xy

and

Considering only the differential rectangular area

For the triangle

72182433224

22

4

22

3

2222 bh-

bh-

bh

bhbh-

bh-AdeII xyxy

Products of inertia for various configurations

G x

y

2b/3

h/3

h

b

y

x

G

x

y

b/3

2h/3

h

b

y

x

728

2222 bhI

bhI xyxy

Products of inertia for various configurations

x

dA

y=sin

x=cos

R

d d

y

?I ? I xyxy

cos1

sin2sin2

1

4cossin

4

cossin22sinsincos

2

0

42

00

4

0

2

0

axa

axdx θdθ R

θdθθρ

θθθ θρdθdρθρρI

ρdθdρ dA , xydAI

π/π/R

R π/

xy

xy

π

R-

R

π

R

π

RπR-

RAdeII

R

--R

θR

I

xyxy

π/

xy

9

4

83

4

3

4

48

811

162cos

2

1

8

4444

442

0

4