centroid (aĞirlik merkezİ ) · 2014-06-02 · moment of inertia is not a physical quantity such...

CENTROID (AĞIRLIK MERKEZİ )

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. 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)

Area-a flat plate with constant

thickness (Alan)

Volume-a sphere or a cone

(Hacim)

Composite Composite Composite

dl

xdlx

dl

ydly

dl

zdlz

i

ii

i

ii

i

ii

l

lzz

l

lyy

l

lxx

dA

zdAz

dA

ydAy

dA

xdAx

i

ii

i

ii

i

ii

A

Azz

A

Ayy

A

Axx

dV

zdVz

dV

ydAy

dV

xdVx

i

ii

i

ii

i

ii

V

Vzz

V

Vyy

V

Vxx

Thin Rod in the Shape of a Quarter Circle (Çeyrek Çember Şeklinde İnce Çubuk)

r=radius

dA

ydAy

4

rA

2

dA=rdqdr

3π

4Ryx

3π

4R

πR

4

3

R

3

R 1-0-

3

R cosθ-

3

ρ

dsin3

ρddsinρρsinθρdρdθydA

2

3

33π/2

0

R

0

3

π/2

0

R

0

3R

0

π/2

0

2

qqqrq

x

dA

y=rsin

q

x=rcosq

R

r

q

dq

dr

y

G

x

y

0y

3

r4x

x

G

x

y

3

hy

3

bx

x

y

h

b

h

b

w=b-x

y

x

b

x

w=b-x

b/2

y

dy

dx

dy

y

dx

yupper

ylower

xright

xleft

x

y

623

2 2

0

2

0

23

0

212121

2121

axxadxxxadAA

dxxxadxyyldxdA

dA

xdAx

aa

/a

///

//lowerupper

--

--

5

2

6

15

1535

2

2

3

3

0

3

0

25

0

212121 a

a

a

xax

xadxxxaxxdA

aa

/a

/// --

dx

l

y2=ax y=a1/2x1/2

yupper

ylower

xright

xleft

x

y

2

6

12

12434

1

3

2

3

333

0

4

0

3

0

2

a

a

a

y

aaay

a

ydy

a

yyydyxxyywdyydA

aaa

leftright

--

--

dy

w

dA

ydAy

6

2aA

y2=ax x=y2/a

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 (pressure or stress) are proportional to distance of the line action

of the force from the moment axis. The elemental force acting on an

element of area, then is proportional to distance times differential area,

and the elemental moment is proportional to distance squared times

differential area.

Thus, the total moment: dM=M=d2dA.

This integral is named as “Area Moment of Inertia” or

“Second Moment of Area”.

Elemantary moment is proportional to distance2 × differential area:

dM=d2dA

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.

Ix=y2dA Inertia moment of area A with respect to x-axis

Iy=x2dA Inertia moment of area A with respect to y-axis

Rectangular Moments of Inertia

Polar Moments of Inertia

Io=Iz=r2dA r2=x2+y2

Io=Iz= Ix+ Iy

Product of Inertia

(Çarpım Alan Atalet Momenti)

Ixy=xydA

In certain problems involving unsymmetrical cross sections and in calculation of

moments of inertia about rotated axes, an expression dIxy=xydA occurs, which

has the integrated form

Properties of moments of inertia :

1. Area moments of inertia Io, Ix , Iy are always positive .

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

taken for length (L4).

2. Ixy may be (-), (+) or zero whenever either of the

reference axes is an axis of symmetry, such as the x axis

in the figure.

Properties of moments of inertia :

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 increases 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.

Jirasyon (Atalet – Eylemsizlik) Yarıçapı

Consider an area A, which has rectangular moment of inertia Ix. 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

Ix=kx2A

The distance kx is called the “radius of gyration” of the area about the x

axis.

A

Ik x

x kx

y

x

A

O

y

O

A

z

x

Radii of gyration about the y-and z axes are obtained in the same manner.

ky y

x

A

O

y

x O

kz

A

AkI A kI2

zz

2

yy

A

Ik

y

y A

Ikk o

oz

2y

2x

2zzyx kkk III Also since,

G

x

y

O

d

e

A

r

x

y

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.

AdeII

ArII

AeII

AdII

xyxy

2zz

2yy

2xx

2

2

AeII

AdII

yy

xx

The sum of these two equations gives

2ArII zz zyxzyx IIIIII ,and since

G

x

y

O

d

e

A

r

AdeII xyxy

For product of inertia

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

The Parallel-axis theorems also hold for radii of gyration as:

dekk 22

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

1) RECTANGLE

b

G

x

y b/2

h/2

h

b

y

x

y

y

x

dy

dA=bdy

b

h

dy b

3

bh

3

bydyybbdyydAyI

3h

0

3h

0

222x

12

bh

2

hbh

3

bhAdII

2

hdAdII

3232

xx

2

xx

--

1) RECTANGLE

dx

G

x

y b/2

h/2

h

b

y

x

y

x

x

dA=hdx

h

h

3

hb

3

hbdxxhhdxxdAxI

3h

0

3b

0

222y

12

hb

2

bbh

3

hbAeII

2

beAeII

3232

yy

2

yy

--

dx

2. TRIANGLE

G x

y b/3

h/3

h

b

y

x

h

b

y

x

dy h-y

y

n

y

h

b

yh

n

-From similarity of the triangles,

12

bh

h

b

4

y

3

bydyyh

h

byndyyAdyI

3h

0

4h

0

3h

0

222

x --

36

bh

3

h

2

bh

12

bhAdII

3232

xx

--

yhh

bn -

2. TRIANGLE

G x

y b/3

h/3

h

b

y

x

h

b

y

x dx

m x

12

hbdAxI

32

y

In a similar manner,

3. SOLID CIRCLE

y

G x

y

R

y

x

G x

x

z

y

z

r dr

2

πR

4

R2

4

r2πI

drr2π rdr2rI rdr 2dA dArI

44R

0

4

z

R

0

32

z

2

z

yxzo IIII

Due to symmetry; 4

πRII

4

yx

4. SEMI CIRCLE

G x

y

O x

4R/3

8

R

2

4

R

I4

4

x

-

-

-

9

8

8R

3

R4

2

R

8

RAdII 4

2242

xx

8

R

2

4

R

I4

4

y

5. QUARTER CIRCLE

y

x

x

y

G

4R/3

4R/3 16

R

4

4

R

II4

4

yx

-

--

9

4

163

4

416

4224

2 RRRR

AdII xx

-

--

9

4

163

4

416

4224

2 RRRR

AeII yy

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.

Given dAxyI,dAxI,dAyI xyyx22

qq

qq

qq

2cos2sin2

2sin2cos22

2sin2cos22

xyyx

yx

xyyxyx

y

xyyxyx

x

III

I

IIIII

I

IIIII

I

-

-

-

--

qq

qq

sincos

sincos

xyy

yxx

-

Note:

We wish to determine moments and product of inertia with respect to new

axes x and y.

22

minmax 4

2

1

2xyyx

yxIII

III -

yx

xym

II

I

--

22tan q

Imax and Imin are the principal moments of inertia.

The product of inertia is zero for the principle axes of inertia.

The equation for qm defines two angles 90o apart which correspond to the principal

axes of the area about O.

Sample Problem: Determine the orientation of the

principal axes of the inertia through the centroid the

angle section and determine the corresponding maximum

and minimum moments of inertia.

50

mm

40 mm

10 mm

10 mm

7.5 mm G

2.5 mm

The product of inertia about the x-y axes for part I is;

40 mm

10 mm

10 mm

7.5 mm

G

2.5 mm

I

II

G1

G2

44

1

1

1075.3

)1040(5.2102055.70

111

mmI

AyxII

xy

Add

xyxy

yx

-

---

The product of inertia about the x-y axes for part II is;

44

2

2

105.7

)1040(5.255.7200

222

mmI

AyxII

xy

Add

xyxy

yx

-

--

dy1

dx1

dy2

dx2 50

mm

x

y

x

y

For the complete section;

Ixy=-7.5.104 mm4

The moments of inertia about the x and y axes for part I is;

40 mm

10 mm

10 mm

7.5 mm

G

2.5 mm

I

II

G1

G2

44232

1

442321

1058.75.7400401012

1

1058.65.12400104012

1

mmAeII

mmAdII

yy

xx

dy1

dx1

dx2

dy2 50

mm

x

y

The moments of inertia about the x and y axes for part II is;

44232

2

442322

1058.25.7400104012

1

1058.115.12400401012

1

mmAeII

mmAdII

yy

xx

For the complete section;

Ixy=-7.5.104 mm4

Ix=18.17.104 mm4 Iy=10.17.104 mm4

om

yx

xy

mII

I31

17.1817.10

)5.7(222tan

-

--

-- qq

44min

44max

22

minmax

1067.5

107.2242

1

2

mmI

mmIIIIII

I xyyx

yx

-