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7/29/2019 Basics of Moment of Inertia
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EngineeringMechanics
Continued(6)
National Institute of Technology Calicut
Mohammed Ameen, Ph.D
Professor of Civil Engineering
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Rolling Resistance
P
W
r
N
a
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,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 .
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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 =
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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
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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
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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
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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
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between block Cand the rollers is 0.4mm.
10 kN
P
N22
N22
0.6
0.4
N1
N2
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Properties of Surfaces
A variety of quantitative descriptions of
surfaces are necessary in engineeringwork.
First Moment of Area and theCentroid
y
dAx
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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
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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 .
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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.
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Thus,
Therefore,
x
y
centroid
yc
xc
= Ac dAyyA
dAy
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Similarly,
and
AAy xAc ==
=A
c dAxxA
A
M
A
dAxx
yAc ==
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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
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x
x
b
yc
bybM
A
dAbyy c
xAc +=+=
+
= )(
'
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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)
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b b
(c)
b
h
x
y (d)
x
y
R
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(c)
Atx = b,y = h.
Hence, C = h/bn
b
hx
y
y = C xn
xdx
100
+=== nbhdxxbhdxyA
b
nn
b
2bb
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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
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(d)
x
y
R
dr
d r
2R
R
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20 0
rrr
==
= =
3
2)sin(
3
0 0
RrddrryA
R
r
== = =
3
4Ry =
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Area with One Axis of Symmetry
x
y
x dA
x
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(as for every +x dA, there exists a x dA)
Hence, the centroid must lie on the axis of
symmetry.
0== Ac dAxxA
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Composite Areas
Example: Determine the centroid of the
following area:
(a)
30 mm
30 mm 20 mm
A B
C
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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
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Example: Determine the centroid of thefollowing area (all dimensions shownare in mm): 50
305
12
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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
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Exercise: Determine the centroid of thefollowing areas:
50 mm
10 mm
20 mm
A B
C
8 mm
(a)
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10
20
(b)
All dimensions
are in mm
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Second Moments andthe Product of Area
Second moments of an area A about xand ycoordinates are defined as
=A
xx dAyI2
=A
yy dAxI2
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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
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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
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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.
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Transfer Theorems
x
dA
y
x
d
y
y y
c
c
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In the above figure,x andy are the
centroidal axis.
2
''
22
22
'2)'(
)'(
dAI
dAdAyddAy
dAdydAyI
xx
AA
AA
xx
+=
++=
+==
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Iabout any axis =Iabout any parallel
axis through centroid +A d2
AdcI
dAcddAxddAycdAyx
dAdycxdAxyI
yx
AAAA
AA
xy
+=
+++=
++==
''
''''
)')('(
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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.
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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
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3
3
0
22 bdbdyydAyI
d
Axx
===
3
3
0
22 dbdxdxdAxI
b
A
yy ===
4
22
0 0
dbdydxxydAxyI
b d
A
xy ===
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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
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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
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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
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34122)(
12
33323
bdbdbddbdbdIxx =+=
+=
440
22)(
16
222222dbdbdb
bddb
Ixy =+=
+=
and
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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!
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yy
A
xx
Id
ddrrrdAyI
==
==
64
)()sin(
40 0
22
x
sin2x
2
1
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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
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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
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Redraw the figure, now marking thecentroidal distances too. Thus
x
16.111
y
10
10
A1
A2
33.889
16.111
33.889
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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
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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
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The remaining part is homework.
x
RyyItR
== 4
3
0
2
)cos)(sin(32/
0
tRtdRRRIxy ==
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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.
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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 .
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From the above, it is easy to verify that
x
y
F
Fx
x
y
Fy
Fx
Fy
Fx
Fx
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and
which can be written using matrixnotation as
sincos yxx FFF +=
cossin yxy FFF +=
=
y
x
y
x
F
F
F
F
cossin
sincos
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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
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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
'
'
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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 ][
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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.
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x
y
P
x
x
y
y
x
y
)'(2
''xx dAyI =
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which is written as
cossin2cossin
)cossin(
22
2
xyxxyy
A
III
dAyx
+=
+=
2sin2cos22
'' xy
yyxxyyxx
xx IIIII
I
++
=
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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 +
=
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represented using matrix notation as
Or
=
cossin
sincos
cossin
sincos
''''
''''
yyxy
xyxx
yyyx
yxxx
II
II
II
II
T
xx RIRI ]][][[][ ' =
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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
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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
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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.
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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
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Moreover, we see that
is a constant.Now, for any orthogonal set of axes wehave
yyxxyyxx IIII +=+ ''''
y
r
dA
y
x
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Ixx+Iyy is called the polar moment ofareadenoted byJorIP and isindependent of the orientation of theaxes.
x
=+=+AA
yyxx dArdAxyII222
)(
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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
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x50
50
10
A1
A2
mm111.16
mm9002
==
=
cc yx
A
centroidaldistances andthe momentarea tensorare
45mm1096389.1 == yyxx II
45mm1066667.0 =xyI
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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
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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
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It may be verified that
45
'''' mm109278.3 =+=+ yyxxyyxx IIII
02cos2sin2
'' =+
= xyyyxx
yx III
I
=
''''
''''
yyyx
yxxx
II
II
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45cos45sin45cos45sin yyxy
xyxx
II
IxxIyyIxy2 can also be verified to remain
unchanged during the transformation.
x50
50
y
10
10
A1
A2
xy
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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
'' +=
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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
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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 =
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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:
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50
305
15
10x
y
A
23)15)(3050()30)(50(
12
1+=xxI
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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
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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=
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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 ==
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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