me 101: engineering mechanics · respect to a tangent to the circle, ( ) 4 4 5 ... for problems...

ME 101: Engineering MechanicsRajib Kumar Bhattacharjya

Department of Civil Engineering

Indian Institute of Technology Guwahati

M Block : Room No 005 : Tel: 2428

www.iitg.ernet.in/rkbc

Area Moments of Inertia

Parallel Axis Theorem• Consider moment of inertia I of an area A

with respect to the axis AA’

∫= dAyI2

• The axis BB’ passes through the area centroid

and is called a centroidal axis.

• Second term = 0 since centroid lies on BB’

(∫y’dA = ycA, and yc = 0

( )

∫∫∫

∫∫

+′+′=

+′==

dAddAyddAy

dAdydAyI

22

22

2

2AdII += Parallel Axis theorem

Parallel Axis theorem:

MI @ any axis =

MI @ centroidal axis + Ad2

The two axes should be

parallel to each other.

Area Moments of InertiaParallel Axis Theorem

• Moment of inertia IT of a circular area with

respect to a tangent to the circle,

( )4

45

224

412

r

rrrAdIIT

π

ππ

=

+=+=

• Moment of inertia of a triangle with respect to a

centroidal axis,

( )3

361

2

31

213

1212

2

bh

hbhbhAdII

AdII

AABB

BBAA

=

−=−=

+=

′′

′′

• The moment of inertia of a composite area A about a given axis is obtained by adding the

moments of inertia of the component areas A1, A2, A3, ... , with respect to the same axis.

Area Moments of Inertia: Standard MIs

Moment of inertia about �-axis �� =1

3�ℎ

� =1

3�ℎ

Answer

Moment of inertia about �-axis

Moment of inertia about ��-axis �� =1

12�ℎ

� =1

12�ℎMoment of inertia about ��-axis

Moment of inertia about �-axis passing through C �� =1

12�ℎ �� + ℎ�


12�ℎ


36�ℎ


4��

Moment of inertia about �-axis passing through O �� =1

2��

Moment of inertia about ��-axis �� = � =1

8��


4��

Moment of inertia about ��-axis �� = � =1

16��


8��


4��


4�� + ��

Moment of inertia about �-axis � =1

4��

Area Moments of InertiaExample:

Determine the moment of inertia of the shaded area with respect to the x axis.

SOLUTION:

• Compute the moments of inertia of the

bounding rectangle and half-circle with

respect to the x axis.

• The moment of inertia of the shaded area is

obtained by subtracting the moment of

inertia of the half-circle from the moment

of inertia of the rectangle.

Area Moments of InertiaExample: Solution

SOLUTION:

• Compute the moments of inertia of the bounding

rectangle and half-circle with respect to the x axis.

Rectangle:

( )( ) 463

3

13

3

1 102.138120240 mmbhI x ×===

Half-circle:

moment of inertia with respect to AA’,

( ) 464

814

81 mm1076.2590 ×===′ ππrI AA

( )( )

( )2

mm

rA

mm 81.8a-120b

mm r

a

3

2

2

12

2

1

1072.12

90

2.383

904

3

4

×=

==

==

===

ππ

ππ

Moment of inertia with respect to x’,

( ) ( )( )4

AAx

mm

AaII

6

2362

1020.7

2.381072.121076.25

×=

×−×=−= ′′

moment of inertia with respect to x,

( )( )4

xx

mm

AbII

6

2362

103.92

8.811072.121020.7

×=

×+×=+= ′


Example: Solution

• The moment of inertia of the shaded area is obtained by

subtracting the moment of inertia of the half-circle from

the moment of inertia of the rectangle.

46 mm109.45 ×=xI

xI = 46 mm102.138 × − 46 mm103.92 ×

Consider area (1)

�� =1

3�ℎ =

1

3× 80 × 60 = 5.76 × 10 ""�

Consider area (2)

�� =1

4

��

4=

�

1630 � = 0.1590 × 10 ""�

��$ = 0.1590 × 10 −

�

430 � × 12.73 � = 0.0445 × 10 ""�

�� = 0.0445 × 10 +�

430 � 60 − 12.73 = 1.624 × 10 ""�

Consider area (3)

�� =1

12�ℎ =

1

12× 40 × 30 = 0.09 × 10 ""�

�� = 5.76 × 10 − 1.624 × 10 − 0.09 × 10 = &. '( × )'* ++&

, = 60 × 80 −1

4� 30 � −

1

240 × 30 = 3490""�

-� =��

,=

4.05 × 10

3490= .&. ''++

Determine the moment of inertia and the radius of gyration of

the area shown in the fig.

�1� �1

12�/ �

1

12� 24 � 6 � 432 ""�

�2� �1

12�/ �

1

12� 8 � 48 � 73728 ""�

�3� �1

12�/ �

1

12� 48 � 6 � 864 ""�

�1� � �1� � ,�� 432 � 24 � 6 � 24 � 3 � � 105408 ""�

�3� � �3� � ,�� 864 � 48 � 6 � 24 � 3 � � 210816 ""�

�� 105408 � 73728 � 210816 � 390 � 10""�

-� ��

,�

390 � 10

816� 0). 1 ++


Products of Inertia: for problems involving unsymmetrical cross-sections

and in calculation of MI about rotated axes. It may be +ve, -ve, or zero

• Product of Inertia of area A w.r.t. x-y axes:

x and y are the coordinates of the element of area dA=xy

∫= dAxyI xy

• When the x axis, the y axis, or both are an

axis of symmetry, the product of inertia is

zero.

• Parallel axis theorem for products of inertia:

AyxII xyxy +=

+ Ixy

+ Ixy - Ixy

- Ixy

Quadrants

Area Moments of InertiaRotation of Axes Product of inertia is useful in calculating MI @ inclined axes.� Determination of axes about which the MI is a maximum and a minimum

∫

∫∫=

==

dAxyI

dAxIdAyI

xy

yx22

Moments and product of inertia

w.r.t. new axes x’ and y’ ?

θθ

θθ

sincos

sincos

xyy

yxx

−=′

+=′Note:

( )

( )

( )( )dAxyyxdAyxI

dAyxdAxI

dAxydAyI

yx

y

x

∫∫

∫∫

∫∫

−+==

+==

−==

θθθθ

θθ

θθ

sincossincos''

sincos'

sincos'

''

22

'

22

'

θθθθθθ

θθ

θθ

2cossincos2sin2/1cossin

2

2cos1cos

2

2cos1sin

22

22

=−=

+=

−=

θθ

θθ

θθ

2cos2sin2

2sin2cos22

2sin2cos22

' xy

yx

yx

xy

yxyx

y

xy

yxyx

x

III

I

IIIII

I

IIIII

I

+−

=

+−

−+

=

−−

++

=

′

′

′


Rotation of Axes Adding first two eqns:

Ix’ + Iy’ = Ix + Iy = Iz � The Polar MI @ O

θθ

θθ

θθ

2cos2sin2

2sin2cos22

2sin2cos22

' xyyx

yx

xyyxyx

y

xyyxyx

x

III

I

IIIII

I

IIIII

I

+−

=

+−

−+

=

−−

++

=

′

′

′

Angle which makes Ix’ and Iy’ either max or

min can be found by setting the derivative

of either Ix’ or Iy’ w.r.t. θ equal to zero:

Denoting this critical angle by α

( ) 02cos22sin' =−−= θθ

θxyxy

x IIId

dI

xy

xy

II

I

−=2

2tan α

� two values of 2α which differ by π since tan2α = tan(2α+π)� two solutions for α will differ by π/2� one value of α will define the axis of maximum MI and the other defines the

axis of minimum MI

�These two rectangular axes are called the principal axes of inertia


Rotation of Axes

θθ

θθ

θθ

2cos2sin2

2sin2cos22

2sin2cos22

' xyyx

yx

xyyxyx

y

xyyxyx

x

III

I

IIIII

I

IIIII

I

+−

=

+−

−+

=

−−

++

=

′

′

′

xy

xy

xy

xy

II

I

II

I

−=⇒

−=

22cos2sin

22tan ααα

Substituting in the third eqn for critical value

of 2θ: Ix’y’ = 0

� Product of Inertia Ix’y’ is zero for the

Principal Axes of inertia

Substituting sin2α and cos2α in first two eqns

for Principal Moments of Inertia:

( )

( )0

42

1

2

42

1

2

@

22

min

22

max

=

+−−+

=

+−++

=

αxy

xyyxyx

xyyxyx

I

IIIII

I

IIIII

I

�� =�� + �

2+

�� − �

223425 − ��46725

�� =�� − �

246725 + ��23425

�� −�� + �

2=

�� − �

223425 − ��46725

Squaring both the equation and adding

�� −�� + �

2

�

+ ��

=�� − �

223425 − ��46725

�

+�� − �

246725 + ��23425

�

�� −89:8;

�

�

+ ��

=89<8;

�

�

234�25 − 289<8;

�� 2342546725 + ��

�467�25

+89<8;

�

�

467�25 + 289<8;

�� 2342546725 + ��

�467�25

�� −�� + �

2

�

+ ��

=�� − �

2

�

+ ��

�

�� −�� + �

2

�

+ ��

=�� − �

2

�

+ ��

�

Defining �� + �

2= �=>?

And @ =�� − �

2

�

+ ��

�

�� − �=>?

�+ ��

�= @� Which is a equation of circle with center �=>? , 0 and radius @

�

��

�=>?

@

25

Area Moments of InertiaMohr’s Circle of Inertia: Following relations can be represented

graphically by a diagram called Mohr’s CircleFor given values of Ix, Iy, & Ixy, corresponding values of Ix’, Iy’, & Ix’y’ may be

determined from the diagram for any desired angle θ.

θθ

θθ

θθ

2cos2sin2

2sin2cos22

2sin2cos22

' xyyx

yx

xyyxyx

y

xyyxyx

x

III

I

IIIII

I

IIIII

I

+−

=

+−

−+

=

−−

++

=

′

′

′

xy

xy

II

I

−=2

2tan α

( )

( )0

42

1

2

42

1

2

@

22

min

22

max

=

+−−+

=

+−++

=

αxy

xyyxyx

xyyxyx

I

IIIII

I

IIIII

I

( )

2

2

222

22xy

yxyxave

yxavex

III

RII

I

RIII

+

−=

+=

=+− ′′′ • At the points A and B, Ix’y’ = 0 and Ix’ is

a maximum and minimum, respectively.

RII ave ±=minmax,

I

Ixy

Area Moments of InertiaMohr’s Circle of Inertia: Construction

θθ

θθ

θθ

2cos2sin2

2sin2cos22

2sin2cos22

' xyyx

yx

xyyxyx

y

xyyxyx

x

III

I

IIIII

I

IIIII

I

+−

=

+−

−+

=

−−

++

=

′

′

′

xy

xy

II

I

−=2

2tan α

( )

( )0

42

1

2

42

1

2

@

22

min

22

max

=

+−−+

=

+−++

=

αxy

xyyxyx

xyyxyx

I

IIIII

I

IIIII

I

Choose horz axis � MI

Choose vert axis � PIPoint A – known {Ix, Ixy}

Point B – known {Iy, -Ixy} Circle with dia AB

Angle α for Area� Angle 2α to horz (same

sense) � Imax, Imin

Angle x to x’ = θ

� Angle OA to OC = 2θ� Same sense

Point C � Ix’, Ix’y’

Point D � Iy’


Example: Product of Inertia

Determine the product of inertia of

the right triangle (a) with respect

to the x and y axes and

(b) with respect to centroidal axes

parallel to the x and y axes.

SOLUTION:

• Determine the product of inertia using

direct integration with the parallel axis

theorem on vertical differential area strips

• Apply the parallel axis theorem to

evaluate the product of inertia with respect

to the centroidal axes.

Area Moments of InertiaExamples

SOLUTION:

• Determine the product of inertia using direct integration

with the parallel axis theorem on vertical differential

area strips

−===

−==

−=

b

xhyyxx

dxb

xhdxydA

b

xhy

elel 1

11

21

21

Integrating dIx from x = 0 to x = b,

( )

bb

b

elelxyxy

b

x

b

xxhdx

b

x

b

xxh

dxb

xhxdAyxdII

02

4322

02

322

0

22

21

83422

1

+−=

+−=

−===

∫

∫∫∫

22

241 hbIxy =


Examples

• Apply the parallel axis theorem to evaluate the

product of inertia with respect to the centroidal axes.

hybx31

31 ==

With the results from part a,

( )( )( )bhhbhbI

AyxII

yx

yxxy

21

31

3122

241 −=

+=

′′′′

′′′′

22

721 hbI yx −=′′′′

SOLUTION


Example: Mohr’s Circle of Inertia

The moments and product of inertia

with respect to the x and y axes are Ix =

7.24x106 mm4, Iy = 2.61x106 mm4, and

Ixy = -2.54x106 mm4.

Using Mohr’s circle, determine (a) the

principal axes about O, (b) the values of

the principal moments about O, and (c)

the values of the moments and product

of inertia about the x’ and y’ axes

SOLUTION:

• Plot the points (Ix , Ixy) and (Iy ,-Ixy).

Construct Mohr’s circle based on the

circle diameter between the points.

• Based on the circle, determine the

orientation of the principal axes and the

principal moments of inertia.

• Based on the circle, evaluate the

moments and product of inertia with

respect to the x’y’ axes.


Example: Mohr’s Circle of Inertia

46

46

46

mm1054.2

mm1061.2

mm1024.7

×−=

×=

×=

xy

y

x

I

I

I

SOLUTION:

• Plot the points (Ix , Ixy) and (Iy ,-Ixy). Construct Mohr’s

circle based on the circle diameter between the points.

( )

( )

( ) ( ) 4622

46

21

46

21

mm10437.3

mm10315.2

mm10925.4

×=+=

×=−=

×=+==

DXCDR

IICD

IIIOC

yx

yxave

• Based on the circle, determine the orientation of the

principal axes and the principal moments of inertia.

°=== 6.472097.12tan mmCD

DXθθ °= 8.23mθ

Area Moments of InertiaExample: Mohr’s Circle of Inertia

46

46

mm10437.3

mm10925.4

×=

×==

R

IOC ave

• Based on the circle, evaluate the moments and product

of inertia with respect to the x’y’ axes.

The points X’ and Y’ corresponding to the x’ and y’ axes

are obtained by rotating CX and CY counterclockwise

through an angle θ= 2(60o) = 120o. The angle that CX’

forms with the horz is φ = 120o - 47.6o = 72.4o.

oavey RIYCOCOGI 4.72coscos' −=′−== ϕ

46mm1089.3 ×=′yI

oavex RIXCOCOFI 4.72coscos' +=′+== ϕ

46mm1096.5 ×=′xI

oyx RYCXFI 4.72sinsin' =′=′=′ ϕ

46mm1028.3 ×=′′yxI

Determine the product of the inertia of the shaded area shown below about the x-y axes.

Solution:

Parallel axis theorem: Ixy = Īxy +dx dy A

Both areas (1) and (2) are symmetric @ their

centroidal

Total Ixy = Ixy1 + Ixy2 = 18.40×106 mm 4

Similarly, for Area (2): Ixy2 = dx2 dy2 A2

Ixy2 = 60×20×130×30 = 4.68×106 mm 4

Therefore, for Area (1): Ixy1 = dx1 dy1 A1

Ixy1 = 20×140×70×70 = 13.72×106 mm 4

Axis � Īxy = 0 for both area.

Ix < Iy , Ixy (+)

Ix >Iy , Ixy (+)

Ix > Iy , Ixy (-)

Ix < Iy , Ixy (-)

Area Moments of InertiaMass Moments of Inertia (I): Important in Rigid Body Dynamics- I is a measure of distribution of mass of a rigid body w.r.t. the

axis in question (constant property for that axis).- Units are (mass)(length)2 � kg.m2

Consider a three dimensional body of mass m

Mass moment of inertia of this body about axis O-O:

I = ∫ r2 dm

Integration is over the entire body.

r = perpendicular distance of the mass element dm

from the axis O-O

Area Moments of InertiaMoments of Inertia of Thin Plates

• For a thin plate of uniform thickness t and homogeneous

material of density ρ, the mass moment of inertia with

respect to axis AA’ contained in the plate is

areaAA

AA

It

dArtdmrI

,

22

′

′

=

== ∫∫ρ

ρ

• Similarly, for perpendicular axis BB’

which is also contained in the plate,

areaBBBB ItI ,′′ = ρ

• For the axis CC’ which is perpendicular to the plate,

( )

BBAA

areaBBareaAAareaCCC

II

IItJtI

′′

′′′

+=

+== ,,, ρρ

Area Moments of InertiaMoments of Inertia of Thin Plates

• For the principal centroidal axes on a rectangular plate,

( ) 2

1213

121

, mabatItI areaAAAA === ′′ ρρ

( ) 2

1213

121

, mbabtItI areaBBBB === ′′ ρρ

( )22121

,, bamIII massBBmassAACC +=+= ′′′

• For centroidal axes on a circular plate,

( ) 2

414

41

, mrrtItII areaAABBAA ==== ′′′ πρρ

Area Moments of InertiaMoments of Inertia of a 3D Body by Integration

• Moment of inertia of a homogeneous body

is obtained from double or triple

integrations of the form

∫= dVrI2ρ

• For bodies with two planes of symmetry,

the moment of inertia may be obtained

from a single integration by choosing thin

slabs perpendicular to the planes of

symmetry for dm.

• The moment of inertia with respect to a

particular axis for a composite body may

be obtained by adding the moments of

inertia with respect to the same axis of the

components.

Q. No. Determine the moment of inertia of a slender rod of

length L and mass m with respect to an axis which is

perpendicular to the rod and passes through one end of the rod.

Solution

Q. No. For the homogeneous rectangular prism shown,

determine the moment of inertia with respect to z axis

Solution

Area Moments of InertiaMI of some common geometric shapes

Q. Determine the angle α which locates the principal axes of inertia through point O for the

rectangular area (Figure 5). Construct the Mohr’s circle of inertia and specify the

corresponding values of Imax and Imin.

x

y

y’

x’

O

2b

b

α

Ix = B

�(2�) �

E

��

Iy = B

2�(�) �

�

��

Ixy = 2��(F

�)(�) � ��

With this data, plot the Mohr’s circle, and using trigonometry

calculate the angle 2 α

2 α =tan-1(b4/b4) = 45o

Therefore, α =22.5o (clockwise w.r.t. x)

From Mohr’s Circle:

Imax = G

�� + �� 2 � 3.08��

Imin = G

�� − �� 2 � 0.252��

Q. No. Determine the moments and product of inertia of the area of the square with respect to the x'- y' axes.

Ix =B

� � =

B

��; Iy =

B

��

Ixy = 0 + (F

�×

F

�× ��) =

B

��

With θ = 30o, using the equation of moment of inertia about any inclined axes

��

= �� + �

2+

�� − �

2H34 25 − �� I67 25

��

= �� + �

2−

�� − �

2H34 25 + �� I67 25

��

= �� − �

2I67 25 + �� H34 25

we get,

�Ix' =FJ

+ 0 −

B

��I67 60° = (

B

−

E)�� = 0.1168��

�Iy' =FJ

+ 0 +

B

��I67 60° = (

B

+

E)�� = 0.5498��

�Ixy' = 0 + (B

�� ×

B

�) = (

B

E× ��) = 0.1250��

Alternatively, Mohr’s Circle may be

used to determine the three

quantities.

me 101: engineering mechanics · respect to a tangent to the circle, ( ) 4 4 5 ... for problems...

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