basics of moment of inertia

7/29/2019 Basics of Moment of Inertia

1/44

EngineeringMechanics

Continued(6)

National Institute of Technology Calicut

Mohammed Ameen, Ph.D

Professor of Civil Engineering


2/44

Rolling Resistance

P

W

r

N

a


,Wat the centre, moving without slipping

along a horizontal surface.

A force Pis needed to maintain uniformmotion.

This can be explained by considering

the deformation of the surface.

The reaction Nis thus oriented at anyangle .


3/44

As is small,

Coulomb suggested that a depends onthe materials, irrespective of Wand r.

cosNW = sinNP =

W

P

= tan

r

a sintan

r

a

W

P=

r

WaPor =


There are other opinions too in this

regard.

Coef. of Rolling Resistance (a mm)

Steel on steel 0.18 0.38

Steel on wood 1.52 2.54

Tyre on smooth road 0.50 0.76

Tyre on mud road 1.00 1.50

Hardened steel on h.s 0.005 0.01


4/44

Example: What is the rolling resistanceof a railway coach weighing 1500 kN?The wheels are of 750 mm diameter,

and the coefficient of rolling resistancebetween the wheel and the rail is 0.025mm.

*

If it were a truck with the same weight,

N50

kN05.0750

025.01500

=

=

==r

WaP


w a s e va ue o e ro ngresistance? The diameter of the tyres

are 1.2 m, and a= 0.62 mm.

(* the number of wheels has noinfluence; we divide by nand thenmultiply by it again).

N77510001200

62.01500=

==

r

WaP


5/44

Example: A block Cweighing 10 kN isbeing moved on rollers A and B, eachweighing 1 kN. What force Pis neededto maintain steady motion? Thecoefficient of rolling resistance betweenthe rollers and the ground is 0.6 mm,

A B

C

300

P


between block Cand the rollers is 0.4mm.

10 kN

P

N22

N22

0.6

0.4

N1

N2


6/44

Properties of Surfaces

A variety of quantitative descriptions of

surfaces are necessary in engineeringwork.

First Moment of Area and theCentroid

y

dAx


The first moment of a coplanar surfaceof area A about the x-axis is defined as

Similarly, the first moment of the areaabout y-axis is

=A

x dAyM

=A

y dAxM

x

y


7/44

These two quantities Mxand Myconveya certain idea about the shape, size andorientation of the area which is useful inmechanics.

We can notice the similarity of this withthe case of a distributed parallel forcesystem.

In that case, we could replace the forcesystem by a single resultant forcelocated at a articular oint .


Likewise, we can imagine the entire

area to be concentrated at a single pointcalled the centroidwith the coordinates(xc, yc).

To compute these coordinates, we

equate the moments of the distributedarea with that of the concentrated areaabout both the axes.


8/44

Thus,

Therefore,

x

y

centroid

yc

xc

= Ac dAyyA

dAy


Similarly,

and

AAy xAc ==

=A

c dAxxA

A

M

A

dAxx

yAc ==


9/44

The location of centroid of an area isindependent of the location of the

reference axes.That is, the centroid is a property only ofthe area itself.

centroid

yc


x

x

b

yc

bybM

A

dAbyy c

xAc +=+=

+

= )(

'


10/44

All axes passing through the centroidare called centroidal axes.

The first moment of an area about anyof its centroidal axes is zero.

Examples: Determine the centroid of thefollowing areas:

h

(a)

h

(b)


b b

(c)

b

h

x

y (d)

x

y

R


11/44

(c)

Atx = b,y = h.

Hence, C = h/bn

b

hx

y

y = C xn

xdx

100

+=== nbhdxxbhdxyA

b

nn

b

2bb


For a rectangle: n= 0

For a triangle: n= 1

For a parabola: n= 2

)2(0

1

0

+===

+

ndxx

bdxyxxA n

n

)2(

)1(

+

+=

n

nbx


12/44

(d)

x

y

R

dr

d r

2R

R


20 0

rrr

==

= =

3

2)sin(

3

0 0

RrddrryA

R

r

== = =

3

4Ry =


13/44

Area with One Axis of Symmetry

x

y

x dA

x


(as for every +x dA, there exists a x dA)

Hence, the centroid must lie on the axis of

symmetry.

0== Ac dAxxA


14/44

Composite Areas

Example: Determine the centroid of the

following area:

(a)

30 mm

30 mm 20 mm

A B

C


Area A = 302 + 3020/2 = 1200 mm2

Taking moment about AB,

Similarly, taking moments about AC

mm75.131200500,16

)30()2030(1530303

1

2

1

==

+=

y

yA

mm4167.201200500,24

)]20(30[)2030(1530303

1

2

1

==

++=

x

xA


15/44

Example: Determine the centroid of thefollowing area (all dimensions shownare in mm): 50

305

12


Area,

Taking moments of area about thebottom edge

Similarly,

22

mm46.14214

10

3050==

A

mm1658.1546.1421522.557,21

124

10153050

2

==

=

y

yA

mm7183.2546.1421522.557,36 ==x


16/44

Exercise: Determine the centroid of thefollowing areas:

50 mm

10 mm

20 mm

A B

C

8 mm

(a)


10

20

(b)

All dimensions

are in mm


17/44

Second Moments andthe Product of Area

Second moments of an area A about xand ycoordinates are defined as

=A

xx dAyI2

=A

yy dAxI2


x yy ,

contrast to the first moments.

Similar to the concept of centroid, theentire area is assumed to be

concentrated at (kx, ky) such that

yyy

xxx

IkA

IkA

=

=

2

2

A

Ik

A

Ik

yy

y

xxx

=

=Or


18/44

The distances kx and ky are called radii

of gyration.

They depend on both the shape of thearea and the position of the x, yaxes(unlike centroid).

The product of area is defined as

= Axy dAxyI


xy

x

y

x dA

x

If the area has anaxis of symmetry,the product of areafor this axis andany axis

orthogonal to thisaxis is zero.


19/44

Transfer Theorems

x

dA

y

x

d

y

y y

c

c


In the above figure,x andy are the

centroidal axis.

2

''

22

22

'2)'(

)'(

dAI

dAdAyddAy

dAdydAyI

xx

AA

AA

xx

+=

++=

+==


20/44

Iabout any axis =Iabout any parallel

axis through centroid +A d2

AdcI

dAcddAxddAycdAyx

dAdycxdAxyI

yx

AAAA

AA

xy

+=

+++=

++==

''

''''

)')('(


Ixy about any axis =Ixy about any

parallel axis through centroid + c d A

Important Note: The distances c and d

are measured from thex andy axes tothe centroid.


21/44

Example: FindIxx,Iyy andIxy of a

rectangle of size band dabout the xand yaxes shown in figure.

x

d

b

y

x

d

b

y

y

dyb


3

3

0

22 bdbdyydAyI

d

Axx

===

3

3

0

22 dbdxdxdAxI

b

A

yy ===

4

22

0 0

dbdydxxydAxyI

b d

A

xy ===


22/44

Example: FindIxx,Iyy andIxy of the

rectangle of size band dabout thecentroidal xand yaxes shown in figure.

xd

b

y

y

dyb

32/

22 bdd


Similarly, we get

and

which is due to the lines of symmetry.

122/dA

xx

12

32/

2/

22 dbdxdxdAxI

b

bA

yy ===

02/

2/

2/

2/

===

b

b

d

dA

xy dydxxydAxyI


23/44

From the above example, we can seethe validity of transfer theorems. Thus,we have the second moment Ixxabout

the base is

x

d

y

y

dyb


34122)(

12

33323

bdbdbddbdbdIxx =+=

+=

440

22)(

16

222222dbdbdb

bddb

Ixy =+=

+=

and


24/44

Example: FindIxx of a circle about oneof its diameters.

x

y

d/2

dr

d r

d 2/ 2

Ixy = 0 asthere are aninfinitenumber oflines of

symmetry!


yy

A

xx

Id

ddrrrdAyI

==

==

64

)()sin(

40 0

22

x

sin2x

2

1


25/44

Example: DetermineIxx,Iyy andIxy of the

section shown in figure about itscentroidal axes.

x50

50

y

10

10

A1

A2

Dividing thearea into two

A1 andA2we

get:

A1 = 500 mm2

A2 = 400 mm2

= 2


The centroidal distances xcand ycare

obtained by taking moments of thearea about the bottom and left sideedges. Thus, we obtain

mm111.16

255005400900

=

+=

c

c

y

y

Due to symmetry,xc =yc


26/44

Redraw the figure, now marking thecentroidal distances too. Thus

x

16.111

y

10

10

A1

A2

33.889

16.111

33.889


yy

xx

I

I

==

++

+=

45

23

12

1

23

12

1

mm1096389.1

)5111.16)(400()10)(40()25889.33)(500()50)(10(

45mm1066667.0

)400)(111.11)(111.1620()500)(889.8)(111.11(

=

+

=xyI


27/44

Example: DetermineIxx,Iyy andIxy of the

section shown in figure about the xandyaxes shown. Also determine the

centroid and calculate the secondmoments and the product of the areaabout the centroidal axes. Given tisvery small when compared to R.

y

txxtdRRI = )sin(

2/

2


The remaining part is homework.

x

RyyItR

== 4

3

0

2

)cos)(sin(32/

0

tRtdRRRIxy ==


28/44

Coordinate TransformationConcept of Cartesian Tensor

A. Scalar (Zeroth order tensor)A scalar quantity, say the temperature Tat a point, remains invariant when thecoordinate axes are rotated.

B. Vector (First order tensor)

Next, let us consider a vector, say aforce vector F.


Although the vector as such remains thesame, its components keep varying as

the coordinate axes are rotated.

The question is this: If Fxand Fycomponents of F are known withrespect to xand ycoordinates,

determine the components Fxand Fywith respect to xand ycoordinates,which are obtained by rotating xand yby an angle .


29/44

From the above, it is easy to verify that

x

y

F

Fx

x

y

Fy

Fx

Fy

Fx

Fx


and

which can be written using matrixnotation as

sincos yxx FFF +=

cossin yxy FFF +=

=

y

x

y

x

F

F

F

F

cossin

sincos


30/44

The above transformation can bewritten more concisely as

where

is called the rotation transformationmatrix. It is an orthogonal matrix (which

means thatR1 =RTasRRT=I).

{ } { }FF ][' R=

=

cossin

sincos][R

A


Eq. [A] represents the transformation

law for vectors (in other words, all thequantities that transform according toEq. [A] are called vectors).

The position vector at a point alsotransforms according to the same law

which can be written as

=

y

x

y

x

cossin

sincos

'

'


31/44

B. Second Order TensorDyadic

The second moment of area and the

product of area put together for a givenarea at a point can be written as follows:

which is a symmetric matrix. Now the

=

yyxy

xyxx

x II

III ][


Given the above components of the

second moment area tensor withrespect to xand ycoordinates,

determine the components (Ixx,Iyy and

Ixy) with respect to xand y

coordinates, which are obtained byrotating xand yby an angle .

We proceed as follows.


32/44

x

y

P

x

x

y

y

x

y

)'(2

''xx dAyI =


which is written as

cossin2cossin

)cossin(

22

2

xyxxyy

A

III

dAyx

+=

+=

2sin2cos22

'' xy

yyxxyyxx

xx IIIII

I

++

=


33/44

Iyy is obtained from the above by

replacing by +/2. Thus, we obtain

Similarly, we obtainIxy as

The above three results can be

2sin2cos22

'' xyyyxxyyxx

yy IIIIII ++=

2cos2sin2'' xy

yyxx

yx III

I +

=


represented using matrix notation as

Or

=

cossin

sincos

cossin

sincos

''''

''''

yyxy

xyxx

yyyx

yxxx

II

II

II

II

T

xx RIRI ]][][[][ ' =


34/44

Principal Axes

Thus, we have seen that as the angle changes, we start getting differentcomponents for the area tensor.

Now the question arises: Is there any

value of at whichIxx takes on a

maximum (or minimum value)?

Ixx is maximum when

'' III


or

where corresponds to an extreme

value ofIxx.

There are two possible values of which are /2 radians apart.

coss n2

==

xy

xxyy

xy

II

I

=

22tan


35/44

The axes corresponding to these anglesare called the principal axes.

The second moments of area aboutthese axes are called the principalmoments of areaone being the majorprincipal momentand the other theminor principal moment.

Next, let us determine the product ofarea about the principal axes.


Dividing the above by cos 2we get

Thus, we see that the product of area iszeroabout the principal axes.

2cos2sin2

'' xy

yyxx

yx II +=

02tan22cos

''=+

= xy

yyxxyxI

III


36/44

Moreover, we see that

is a constant.Now, for any orthogonal set of axes wehave

yyxxyyxx IIII +=+ ''''

y

r

dA

y

x


Ixx+Iyy is called the polar moment ofareadenoted byJorIP and isindependent of the orientation of theaxes.

x

=+=+AA

yyxx dArdAxyII222

)(


37/44

SinceIxx+Iyy = a constant, it is termed

as an invariant. We can also show that

IxxIyyIxy2 is also an invariant under

rotation of axes.

Example: Determine the principalmoments of area of the plane areashown below about the centroidal axes.

y10We have seenearlier that thearea, the


x50

50

10

A1

A2

mm111.16

mm9002

==

=

cc yx

A

centroidaldistances andthe momentarea tensorare

45mm1096389.1 == yyxx II

45mm1066667.0 =xyI


38/44

The principal axes are given by

or, 2= /2.

That is, 1 = /4 and 2 = 3/4.

The principal moments of area are:

==

=0

222tan

xy

xxyy

xy I

II

I

'' 2sin2cos22

++

= xyyyxxyyxx

xx IIIII

I


and

45

5

5

mm106306.2

)4/(2sin)106667.0(

01096389.1

=

+=

45

55

''

mm102972.1

106667.01096389.1

2sin2cos22

=

=

+

+

= xyyyxxyyxx

yy I

IIII

I


39/44

It may be verified that

45

'''' mm109278.3 =+=+ yyxxyyxx IIII

02cos2sin2

'' =+

= xyyyxx

yx III

I

=

''''

''''

yyyx

yxxx

II

II


45cos45sin45cos45sin yyxy

xyxx

II

IxxIyyIxy2 can also be verified to remain

unchanged during the transformation.

x50

50

y

10

10

A1

A2

xy


40/44

Principal Moments of Area as anEigenvalue Problem

Find the direction cosines land mof theprincipal axis (the axis which

corresponds to an extreme value forIxx.

Let

Then, the equation forIxx is given by

sinandcos == ml

mlImIlII xyyyxxxx 222

'' +=


constraint that

This can be done using the Lagrangemultiplier technique: Thus, maximise

which leads to

122

=+ ml

)1(22222

++= mlmlImIlIF xyyyxx

0222 ==

lmIlI

l

Fxyxx


41/44

That is

Similarly, equating derivative of Fwithrespect to mto zero and putting themtogether, we get

which is a matrix eigenvalue problem ofthe form

As the matrix I is s mmetric the

lmIlI xyxx 22 =

=

m

l

m

l

II

II

yyxy

xyxx

}{}]{[ XXA =


eigenvalues are always real (and not

complex).Hence, the principal moments of areaare always real.

Exercise: Determine the principal axes

and the principal moments of area ofthe following plane area at the point A:


42/44

50

305

15

10x

y

A

23)15)(3050()30)(50(

12

1+=xxI


46224

mm10441655.0)10(

4

)10(

64

)10(=

46224

23

mm1023184.1)15(4

)10(

64

)10(

)25)(3050()50)(30(12

1

=

+=

yyI

46

2

mm10550719.0

)10)(15(4

)10()15)(25)(3050(

=

=

xyI


43/44

Thus the area tensor is given by:

Solving the eigenvalue problem, we get

which corresponds to the characteristicequation given by

610

23184.155072.0

55072.044166.0][

=xI

01023184.155072.0

55072.044166.06

=

2=


where

and

The solution of equation [A] gives theeigenvalues.

46

61

mm106735.1

10)23184.144166.0()(

=

+== xItrI

46

122

2

mm10240762.0

10)55072.023184.144166.0()det(

=

== xII

6

2

6

1 10158968.0and105145.1 ==


44/44

Determine the principal axes and theprincipal moments of areas for thefollowing sections at A.

x

10

y

10

10

8

80

10

10 10

8

80

10

10

40

8

80

basics of moment of inertia

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