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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
Çeyrek Çember Şeklinde İnce Çubuk
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
0y
dddAR
A 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
bhdyyh
hb
wdydAyhhb
wyh
whb
h
dyy)(hh
by)(h
h
by
h
axdA
h
02
3
bax
h
a
y
x
wxx
2
dA
xdAx
dA
xdAx
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
x2Ad
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
x
y
x’
Y’
z
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
4
24
22224
0
4
0
322 πR
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
2
4πRI z
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
27212132
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
xydAI xy
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
πR
-R
πR
πRπR
- R
AdeII
R
--R
θR
I
xyxy
π/
xy
94
834
34
48
811
162cos
21
84424
442
0
4
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