Strengths of Materials - 3 - Moments of an Area

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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.



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 .



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:




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




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,




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:




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 =

−=⋅=⋅=⋅== ∫∫∫ .




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:




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.




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




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




68

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:




69

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.




70

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.




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).




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




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:




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

Strengths of Materials - 3 - Moments of an Area

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