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3.FIRST MOMENTS AND SECOND MOMENTS OF AN
AREA
Many engineering formulas and applications such as those relating to strength
of beams, columns, shafts, etc., involve the use of different mathematical expressions
which describe, from the mathematical point of view, the shape and dimensions of
the cross sections. These mathematical expressions are called: geometrical
characteristics.For members under axial loading (tension or compression) the single involved
geometrical characteristic is represented by the cross-sectional area A of the member.
A higher value of the cross-sectional area A of the member means a higher strength of the member under axial loading (Fig.3.1).
For structural elements in bending,
torsion etc. there are also other
geometrical characteristics involved
within the strength calculus: the static
moments (first moments of an area) and
the moments of inertia (second moments
of an area). Fig. 3.1
3.1 FIRST MOMENTS OF AN AREA. CENTROID OF AN
AREA
Consider an area A located in the zOy plane (Fig.3.2). Denoting by z and y the
coordinates of an element of area d A, we define the first moment of area A with
respect to z axis as the integral:
∫= A
z A yS d . (3.1)
Similarly, the first moment of area A with respect to the y axis is defined as the
integral:
∫=
A
y A zS d . (3.2)
We note that each of these integrals may be positive, negative or zero,
depending upon the position of the coordinate axes. The first moments of area, S z andS y, are expressed in mm3, cm
3, m
3, etc.
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Strength of Materials
58
Fig. 3.2
Since in almost all cases the area A of Fig.3.2 is assimilated to the cross-
sectional area of a beam, a shaft etc., it is
presented as being located in the zOy
plane, the Ox axis being directed along
the beam, shaft etc.The centroid of area A is defined as
the point G of coordinates zG and yG
(Fig.3.2), which satisfy the relations:
⋅=⇒==
⋅=⇒==
∫
∫
.
d
;
d
G z z A
G
G y
y AG
y AS AS
A
A y
y
z AS A
S
A
A z
z(3.3)
Comparing (3.1) and (3.2) with (3.3), we note that the first moments of area A may be
expressed as the products of the area and the coordinates of its centroid:
;G y z AS ⋅= .G z y AS ⋅= (3.4)
Fig. 3.3
When an area possesses an axis of
symmetry, the first moment of the areawith respect to that axis is zero. Indeed,
considering the area A of Fig.3.3, which
is symmetric with respect to the O y axis,
we observe that to every element of area
d A of abscissa z corresponds an element
of area d A’ of abscissa - z. It follows that
the integral in (3.2) is zero and, thus,
S y=0.
It does also follow from the first of the relations (3.3) that zG = 0. Thus, if an area A
possesses an axis of symmetry, its centroid G is located on that axis. If an area
possesses two axes of symmetry (Fig.3.4) the centroid G coincides with its geometric
center.The coordinate axes passing through the centroid of a given area are called
centroidal (or central) axes.
It is to be observed that the integrals involved in relations (3.1) and (3.2) are
actually double integrals, which have to be calculated with the well known
mathematical methods (Fig. 3.5).
;ddd
)(∫ ∫∫==
A D z
y z y A yS
∫ ∫∫==
A D
y y z z A zS
)(
ddd .
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First moments and second moments of an area
59
Fig. 3.4 Fig. 3.5
3.2 SECOND MOMENTS OF AN AREA
Consider again an area A located in the zOy plane (Fig. 3.6) and an element of area d A of coordinates z and y.
Fig. 3.6
The second moment , or moment of
inertia, of area A with respect to the O z
axis, and the second moment, or the
moment of inertia, of area A with respect
to the O y axis are defined, respectively,as:
;d2
∫=
A
z A y I .d2
∫=
A
y A z I (3.5)
While each of the above integrals is
actually a double integral, it is possible in
many applications to select elements of
area d A
in the shape of thin horizontal or vertical strips, and thus reduce the computation to
simple integration. This will be illustrated later.
We now define the centrifugal moment of inertia (or the product of inertia) of area A with respect to Oz and Oy axes (Fig. 3.6) as the integral:
.d∫=
A
zy A zy I (3.6)
Relations (3.5) show that the moments of inertia of an area are positive
quantities and are expressed in mm4, cm
4, m
4etc. On the other hand, relation (3.6)
shows that the centrifugal moment of inertia may be positive, negative or zero,
depending upon the locations of the area relative to the involved axes. It is positive if
the area lies principally in the first or third quadrants and negative if the area lies
principally in the second or fourth quadrants.We define the polar moment of inertia of area A with respect to point O (Fig.
3.6) as the integral:
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Strength of Materials
60
,d2
∫=
A
p Ar I (3.7)
where r is the distance from O to the element d A. While this integral is again a double
integral, it is possible in the case of a circular area to select elements of area d A in the
shape of thin circular rings, and thus reduce the computation of I p to a simple
integration. This will be illustrated later. It is to be noted that the polar moment of
inertia is also a positive quantity, being expressed in mm4, cm
4, m
4etc.
An important relation may be established between the polar moment of inertia
I p of a given area and the moments of inertia I z and I y of the same area. Noting thatr
2= z
2+ y
2(Fig. 3.6) we write:
∫+∫=∫
+=∫=
A A A A
p A y A z A y z Ar I dddd 22222
or
y z p I I I += . (3.8)
Fig. 3.7
If an area has an axis of symmetry,this axis together with any axis
perpendicular to it will form a set of axes
for which the centrifugal moment of
inertia is zero. Consideration of the
symmetrical section shown in Fig. 3.7
will disclose that, for any differential aread A, there is a symmetrically placed equal
differential area d A’. With respect to theOy axis of symmetry, the z coordinates of
d A and d A’ are equal but of opposite sign,
whereas their y coordinates are equal and
of the same sign regardless of the
position of the Oz axis.
Hence the sum of the products zy⋅d A for each such pair of symmetrically
placed elements d A and d A’ will be zero. It follows, therefore, that the value of
∫ A
A zy d for the entire area will be zero if either or both reference axes are axes of
symmetry.
3.3 PARALLEL - AXIS THEOREM (STEINER’S RELATIONS)
Consider the moments of inertia I z and I y and the centrifugal moment of inertia I zy of an area A with respect to two arbitrary perpendicular axes Oz and Oy (Fig. 3.8).We assume to know the quantities I z, I y and I zy, where
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First moments and second moments of an area
61
;2
∫=
A
z A y I d ;2
∫=
A
y A z I d
.
∫=
A
zy A zy I d
Let us now consider another
coordinate system z1O1 y1, translated with
quantities a and b with respect to the axes
Oy and Oz of the first coordinate system.
The problem which arises consists in
determining the quantities1
z I ,1
y I andFig. 3.8
11 y z I of the same area A but with respect to the axes of the new coordinate system.
We write:
( ) ( ) =+−=−== ∫∫∫ A A A
z Abby y Ab y A y I ddd2222
1 21
;22222 AbbS I Ab A yb A y z z
A A A
+−=+−= ∫∫∫ ddd
( ) =+−=−== ∫∫∫ A A A
y Aaaz z Aa z A z I ddd2222
1 21
;22222 AaaS I Aa A za A z y y
A A A
+−=+−= ∫∫∫ ddd
( )( ) ( ) =+−−=−−== ∫∫∫ A A A
y z Aabay zb zy Aa zb y A y z I ddd1111
.abAaS bS I Aab A ya A zb A zy z y zy
A A A A
+−−=+−−= ∫∫∫∫ dddd
Thus, the mathematical connection between the moments of inertia z I , y I and zy I of an area A and the same quantities
1 z I ,
1 y I and
11 y z I calculated with respect to
the translated coordinate system z1O1 y1, is described by the following relations:
+−−=
+−=
+−=
,
2
2
11
1
1
2
2
abAaS bS I I
AaaS I I
AbbS I I
z y zy y z
y y y
z z z
(3.9)
wherea is the distance between axes Oy and O1 y1,
b is the distance between axes Oz and O1 z1,
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Strength of Materials
62
S z, S y are the static moments (first moments) of area A with respect to axes Oz and Oy.
If the point O is the centroid of area A, it follows from relations (3.4) that S z =
S y = 0 and we may write:
+=
+=
+=
.
;
;
11
1
1
2
2
abA I I
Aa I I
Ab I I
zy y z
y y
z z
(3.10)
Relations (3.10) are known as Steiner’s formulas.
For example, the first relation of (3.10) expresses that the moment of inertia
1 z I of an area with respect to an arbitrary Oz1 axis is equal to the moment of inertia I z
of the same area with respect to the centroidal Oz axis parallel to the Oz1, plus theproduct b2 A of area A and the square of the distance b between the two axes. This
result is also known as the parallel-axis theorem. It makes it possible to determine themoment of inertia of an area with respect to a given axis, when its moment of inertia
with respect to a centroidal axis of the same direction is known. Conversely, it makes
it possible to determine the moment of inertia I z of an area A with respect to a
centroidal axis Oz, when the moment of inertia1
z I of A with respect to a parallel axis
is known, by subtracting from1
z I the product b2 A. We should note that the parallel-
axis theorem may be used only if one of the two axes involved is a centroidal axis.
3.4 MOMENTS OF INERTIA OF SIMPLE SURFACES
a) Rectangular area
For the rectangular area A shown in
Fig. 3.9, determine the moments of
inertia I z , I y and I zy with respect to the
centroidal Oz and Oy axes.
As mentioned before:
∫=
A
z A y I d2
.
We select as an element of area (d A)
a horizontal strip of length b and
thickness d y (Fig.3.9). We write:
d A = b⋅d y.
Fig. 3.9
It follows that:
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First moments and second moments of an area
63
.1224
2
883
1
3dd
3333
2
232
2
22 bhhbhhb
by yb y A y I
h
hh
h A
z =⋅
=
−−====
−−
∫∫
Thus, the moment of inertia I z of a rectangular area with respect to the centroidal Oz
axis is:
12
3bh I z = . (3.11)
In the same manner, it follows that :
12
3hb I
y= . (3.12)
Since Oz and Oy are axes of symmetry, we have:
.0= zy I
b) Circular area
For the circular area shown in Fig. 3.10 determine the polar moment of inertia I p
and the moments of inertia I z , I y and I zy.
Fig. 3.10 Fig. 3.11
We select as an element of area (d A) a ring of radius r and thickness dr ,
(Fig.3.11). The polar moment of inertia of area A is:
32
0
1624
2d2d2d44
0
242
0
32
0
22 d d r r r r r r Ar I
d d d
A
p
π π π π π =
−=⋅=⋅=⋅== ∫∫∫ .
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Strength of Materials
64
Thus
32
4d
I pπ
= . (3.13)
Due to the symmetry of the circular area, we have I z = I y.Recalling (3.8), we write:
32
4d
2I 2I I I I y z y z p
π ===+=
and, thus
64
4d
I I y z
π == . (3.14)
c) Triangular area
Determine the moments of inertia for a triangle of base b and altitude h with
respect to an axis coinciding with its base and a centroidal axis parallel to its base.
Select the differential element as shown in Fig. 3.12.
From similar triangles, we have
( )h
yhb m
−= . The moment of inertia with respect
to z1 axis is obtained from:
( )=
−⋅=⋅== ∫∫∫
hh
A
z yh
yhb y ym y A y I
0
2
0
22ddd
1
.124343
33343bhbhbhh
h
bbh=−=⋅−=
We have thus obtained:Fig. 3.12
12
3
1
bh I z = . (3.15)
To determine the centroidal moment of inertia z I , we transfer the known value of
1 z I , from the base axis z1 to the parallel axis z. Since the transfer distance is
3
has
shown in Fig. 3.12, we write:
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First moments and second moments of an area
65
23
2
1
bhh I I z z ⋅
+= .
It follows that
36181223
3332
1
bhbhbhbhh I I z z =−=⋅
−= . (3.16)
3.5 MOMENTS OF INERTIA OF COMPLEX SURFACES
(COMPOSITE AREAS)
To determine of the moments for inertia of a complex surface the following steps
have to be covered:- the complex surface (area) A has to be divided into several component parts
of areas A1, A2 ...;
- determination of the centroidal point G of the complex area;
- since the integral representing the moment of inertia of area A may be
subdivided into integrals extending over A1, A2 ..., the moment of inertia of A with respect to a given axis will be obtained by adding the moments of
inertia of areas A1, A2 ... , with respect to the same axis. Before adding the
moments of inertia of the component areas, however, the parallel-axis
theorem should be used to transfer each moment of inertia to the desiredaxis. This is shown in the following example.
Determine the moments of inertia I z , I y and I zy of area A shown in Fig. 3.13, with respect to
the centroidal axes.
Fig. 3.13 Fig. 3.14
We first divide the complex area A into the two rectangular areas A1 and A2 (Fig 3.14) and
denote their centroids and their own centroidal axes by G1 ,G2 , z1 , y1 , z2, y2 respectively.
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Strength of Materials
66
We may now determine the coordinates zG and yG of the centroid G of the composite area A,
using, for example, the coordinate system z1 G1 y1 as follows:
;5,15,144
45,13
2a
aaa
aaa
A
A y
ii
iii
G y =
⋅+
⋅⋅=
∑
∑
= (i = 1,2)
aaaa
aaa
a
A
A z
z
ii
iii
G 75,05,144
5,142
5,12
2−=
⋅+
⋅⋅−−
=
∑
∑
=
.
Recalling the formulas (3.11) and (3.12) and using the parallel-axis theorem we may write
the moments of inertia of the composite area A as follows:
( ) ( ) ( ) ( ) 42322333,235,145,15,2
12
45,145,1
12
4aaaaa
aaaa
aa I z =⋅⋅−+
⋅+⋅+
⋅= ;
( )( )
( )
;2,10
5,1475,02
5,12
12
5,14475,0
12
4
4
232
3
a
aaaa
aaa
aaaaa
I y
=
⋅⋅−−+⋅
+⋅⋅+⋅
=
( ) ( )
.43,7
5,14652,0
2
5,12565,15,204652,0565,10
4
2
a
aaaa
aaaaaa I zy
−=
⋅⋅−−⋅−−++⋅⋅−+=
3.6 MOMENTS OF INERTIA WITH RESPECT TO INCLINED
AXES
Fig. 3.15
In some cases, it is necessary to
determine the moments of inertia withrespect to axes that are inclined to the
usual axes. The moments of inertia insuch cases can be obtained by formal
integration, but a general formula is
usually easier to use.
The problem may be stated as
follows: assuming the values I z, I y and I zy with respect to the Oz and Oy axes
to be known, determine the values of
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First moments and second moments of an area
67
1 z I ,1 y I and
11 y z I with respect to the Oz1 and Oy1 axes inclined at an angle α with
Oz and Oy axes, as shown in Fig. 3.15.
The coordinates for a typical differential area d A are given by z and y with
respect to the y and z axes, and by y1 and z1 relative to the z1 and y1 axes. The relation
between these coordinates can be obtained by projecting the coordinates z and y on
the z1 and y1 axes. This gives (Fig. 3.15):
−=
+=
.sincos
;sincos
1
1
α α
α α
z y y
y z z (3.17)
By definition, the values of 1
z I and1
y I are:
∫=
A z A y I d
2
11 ; ∫=
A y A z I d
2
11 ; ∫=
A y z A y z I d1111 .
Replacing the values of z1 and y1 from (3.17) we have:
( ) ( ) =+−=−== ∫∫∫ A A A
z A z yz y A z y A y I dsinsincos2cosdsincosd222222
11
α α α α α α
α α α 22 sin2sincos y zy z I I I +−= ;
( ) ( )∫∫∫ =++=+== A A A
y A y zy z A y z A z I dsinsincos2cosdsincosd222222
11 α α α α α α
α α α 2sinsincos 22 zy z y I I I ++= ;
( )( ) =−+== ∫∫ A A
y z A z y y z A y z I dsincossincosd1111
α α α α
=−+−= ∫ A
A yz y z zy dsincossincossincos2222α α α α α α
( ) α α α α 22 sincoscossin −+−= zy y z I I I .
We, thus, obtain:
( ) ( )
−+−=
++=
−+=
.sincoscossin
;cossin2cossin
;cossin2sincos
22
22
22
11
1
1
α α α α
α α α α
α α α α
zy y z y z
zy y z y
zy y z z
I I I I
I I I I
I I I I
(3.18)
If the relations
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Strength of Materials
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2
2cos1sin
2 α α
−= ,
2
2cos1cos
2α
+=
are substituted in (3.18), we may write:
+⋅−
=
++
⋅+−
⋅=
−−
⋅++
⋅=
α α
α α α
α α α
2cos2sin2
;2sin2
2cos1
2
2cos1
;2sin2
2cos1
2
2cos1
11
1
1
zy y z
y z
zy y z y
zy y z z
I I I
I
I I I I
I I I I
or
⋅+⋅−
=
⋅+⋅−
−+
=
⋅−⋅−
++
=
.2cos2sin2
;2sin2cos22
;2sin2cos22
11
1
1
α α
α α
α α
zy y z
y z
zy y z y z
y
zy y z y z
z
I I I
I
I I I I I
I
I I I I I
I
(3.19)
When the values of I z, I y and I zy are known, relations (3.19) permit the values of
1111and, y z y z I I I with respect to the Oz1 and Oy1 axes, inclined at an angle α to the
Oz and Oy axes, to be determined without further integration. In a sense, these
relations do for inclined axes what the Steiner’s formula does for parallel axes.
A simple analysis of relations (3.19) tells us that1111
and, y z y z I I I are
functions of angle α . One could ask: which are the values of angle α that make these
quantities (1111
and, y z y z I I I ) maximum or minimum? The angles defining the
maximum and the minimum moments of inertia, also called the principal moments of
inertia, may be found by differentiating (3.19) with respect toα
and setting thederivative equal to zero:
==⋅+−
⋅=
=−=⋅−−
⋅−=
.022cos22
2sin2d
d
;022cos22
2sin2d
d
11
1
11
1
y z zy y z y
y z zy y z z
I I I I I
I I I I I
α α
α
α α
α (3.20)
We find that:
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First moments and second moments of an area
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y z
zy
I I
I tg
−−=
22α . (3.21)
Equation (3.21) gives us two values of α (
2
and 121π
α α α += ) for which1
z I and
1 y I have extreme values. This is why the equation (3.21) is always written as:
y z
zy
I I
I tg
−−=
22 2,1α . (3.22)
The extreme conditions for1
z I ,1
y I – (3.20 ) - mean in fact that the product of
inertia11
y z I equals zero. In the same time, a second differentiation of (3.20) shows
that:
2
2
2
2
d
d
d
d11
α α
y z I I −= , (3.23)
which means that a maximum value of 1
z I implies a minimum value of 1
y I and
vice versa.
Substituting for α from equation (3.21) into (3.19) we obtain the extreme
values of quantities1
z I and1
y I , called the principal moments of inertia:
( ) 222,1 4
2
1
2 zy y z
y z I I I
I I I +−±
+= (3.24)
with respect to the axes Oz1 and Oy1, rotated with angle α 1.
In this way we found two
perpendicular directions given by α1
and2
12
π α α += for which the moments
of inertia of the original area have
extreme values (a maximum value I 1 withrespect to one of these directions and a
minimum value I 2 with respect to the
other direction). Usually, the axis of
maximum is denoted by 1 while the axis
of minimum by 2.Fig. 3.16
It is important to be mentioned again that the product of inertia of the originalarea with respect to the coordinate system 1O2 (Fig. 3.16) is zero.
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Axes 1 and 2 are called principal axes.One could demonstrate that if I zy < 0 the axis of maximum is placed in the first
quadrant while, if I zy > 0, the axis of maximum is placed in the second quadrant.
3.7 RADIUS OF GYRATION. ELLIPSE OF INERTIA
The term radius of gyration is used to describe another mathematical
expression and occurs most frequently in column formulas. Radius of gyration is
usually denoted by the symbol i and is defined as:
A
I i = , (3.24)
where I is the moment of inertia and A the area.
Thus, we have:
A
I i z z = ;
A
I i
y y = ;
A
I i 11 = ;
A
I i 22 = .
(3.25)
The following is a geometric interpretation of this relation. Assume the area of Fig.
3.2 to be squeezed into a long narrow strip as shown in Fig. 3.17.
Fig. 3.17
Each differential element of area
d A will then have the same distance i z
from the Oz axis. The moment of inertia
is given by:
=== ∫∫ A
z
A
z Ai A y I dd22
.d 22 Ai Ai z
A
z ⋅== ∫
The strip may be placed on either side of the reference axis, since if i z is negative,
squaring it will automatically make it plus. Also, part of the strip may be at a distancei z from one side of the reference axis and the remainder of the strip at equal distance i z
from the other side of the axis.
In view of this discussion, the radius of gyration is frequently considered to be
the uniform distance from the reference axis at which the entire area may be assumed
to be distributed. For an area whose dimensions perpendicular to a reference axis arenegligibly small compared with its distance from that axis, the radius of gyration is
practically equivalent to the centroidal location of the area.
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First moments and second moments of an area
71
The ellipse of equation
0121
2
22
2
=−+
i
y
i
z(3.26)
represents the centroidal principal ellipse of inertia with respect to a certain area A.
Sample problems
1. For the area shown in Fig. 3.18 determine: (a) the centroidal point G of the area A; (b) the
moments of inertia I z , I y and I zy with respect to the centroidal reference system zGy; (c) the
principal axes of inertia 1 and 2; (d) the principal moments of inertia I 1 and I 2; (e) the
principal radii of inertia i1 and i2 and (f) draw the ellipse of inertia.
Solution
We first divide the area A of the whole
surface into three rectangular areas with centroidal
points G1, G2 and G3 (Fig. 3.18). We observe that
the centroid of the second rectangular area G2
coincides with the centroid G of the whole area A.
Thus, axes z2 and y2 coincide with the centroidal
axes G z and G y of the whole area A.
Recalling (3.11) and (3.12) and using the
parallel-axis theorem we may write the moments of
inertia of the composite area A as follows:
=⋅
+⋅⋅⋅+⋅
=
12
24030216530135
12
301653
23
z I
;mm48
102,1573 ⋅= Fig. 3.18
483
23
109038,012
302402301655,82
12
16530mm I y ⋅=
⋅+⋅⋅⋅+
⋅=
;
( ) 48101026,12165305,82135 mm I zy ⋅=⋅⋅⋅⋅= .
The principal directions of inertia are given by
.81,592
;19,309038,01573,2
1026,1222 1212,1
oo
=+=−=⇒−
⋅−=
−
−=π
α α α α
y z
zy
I I
I tg
Since I zy > 0, the principal axis of maximum is placed in the second quadrant (Fig. 3.18).
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Strength of Materials
72
The principal moments of inertia are:
( ) =+−±
+
=22
2,1 42
1
2 zy y z
y z I I I
I I I
( ) ( )2828888101026,14109038,0101573,2
2
1
2
109038,0101573,2⋅+⋅−⋅±
⋅+⋅= .
We finally have:
⋅=
⋅=
.10262,0
;10799,248
2
481
mm I
mm I
The principal centroidal radii of inertia are:
;94,12724030216530
10799,28
11 mm A
I i =
⋅+⋅⋅
⋅
==
.14,3924030216530
10262,08
22 mm
A
I i =
⋅+⋅⋅
⋅==
The principal centroidal ellipse of inertia has been represented in Fig. 3.18.
2. For the composite area of Fig. 3.19, composed of two U-shaped profiles, determine the same
quantities like in the previous example.
Fig 3.19
In these tables we may also find the location
of the centroidal point with respect to the U -
shaped section (22,3 mm - Fig. 3.19).
Using the coordinate system z1 G1 y1 , we
can compute now the position of the centroidal
point G of the entire area as follows:
SolutionWe first divide the area A of the whole surface
into two areas 1 and 2 (the two U shapes) having
the centroidal points at G1 and G2 (Fig. 3.19).
From the appropriate tables containing the
geometrical characteristics of rolled-steel shapes
(APPENDIX III) we may get all the necessary
data:
⋅=
⋅=
⋅=
;103,42
;103600
;10248
:1
22
1
44
1
44
1
mm A
mm I
mm I
y
z
⋅=
⋅=
⋅=
.103,42
;10248
;103600
:2
22
2
44
2
44
2
mm A
mm I
mm I
y
z
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First moments and second moments of an area
73
( );85,48
103,422
103,423,22120
2
2
mm A
A y y
i
iiG =
⋅⋅
⋅⋅−=
∑
∑=
( ).15,71
103,422
103,423,22120
2
2
mm
A
A z z
i
iiG =
⋅⋅
⋅⋅+=
∑
∑=
We have thus located the centroidal reference coordinate system of the composite area: zGy
(Fig.3.19).
Recalling again the formulas (3.11) and (3.12) and using the parallel –axis theorem we may
write the moments of inertia of the composite area as follows:
( ) =⋅⋅−−+⋅+⋅⋅+⋅=224224
103,4285,483,22120103600103,4285,4810248 z I
;10668,5846
mm⋅=
( ) =⋅⋅−−+⋅+⋅⋅+⋅=224224
103,423,2215,7112010248103,4215,71103600 y
I
;1030,8146
mm⋅=
( )( ) ( ) ( ) ⋅+−⋅−−++⋅⋅−−+= 3,2215,7112085,483,221200103,4215,7185,4802
zy I
.104,29103,42
462
mm⋅=⋅⋅
The principal directions of inertia are given by
( )⇒
−
⋅⋅−=
−
−=6
6
2,11030,81668,58
104,29222
y z
zy
I I
I tg α
=
=
.47,124
;47,34
2
1o
o
α
α
Since I zy > 0, the principal axis of maximum is placed in the second quadrant (Fig 3.19). The
principal moments of inertia are
( ) 222,1 4
2
1
2 zy y z
y z I I I
I I I +−±
+
= .
We finally have
⋅=
⋅=
.1048,38
;1048,10146
2
461
mm I
mm I
The principal centroidal radii of inertia are:
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Strength of Materials
74
;52,109103,422
1048,101
2
61
1 mm A
I i =
⋅⋅
⋅== .44,67
103,422
1048,38
2
62
2 mm A
I i =
⋅⋅
⋅==
The principal centroidal ellipse of inertia is shown in Fig. 3.19.
PROBLEMS TO BE ASSIGNED
P.3
For the areas shown in the figures below, locate the centroids of the areas and then determine:
• the second moments of the involved areas ( I z , I y and I zy ) with respect to the centroidal axes;
• the principal axes of inertia 1 and 2 ;
• the principal moments of inertia I 1 and I 2;
• the principal radii of inertia i1 and i2 , and, finally, draw the ellipse of inertia.
Fig. P.3.1 Fig. P.3.2 Fig. P.3.3
Fig. P.3.4 Fig. P.3.5 Fig. P.3.6
Fig. P.3.7 Fig. P.3.8 Fig. P.3.9 Fig. P3.10