ME 101: Engineering Mechanics Rajib Kumar Bhattacharjya Department - - PowerPoint PPT Presentation

ME 101: Engineering Mechanics

Rajib 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’

• =

dA y I

2

• 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

( )

• +

′ + ′ = + ′ = = dA d dA y d dA y dA d y dA y I

2 2 2 2

2

2

Ad I I + =

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 Inertia

Parallel Axis Theorem

• Moment of inertia IT of a circular area with

respect to a tangent to the circle,

( )

4 4 5 2 2 4 4 1 2

r r r r Ad I IT π π π = + = + =

• Moment of inertia of a triangle with respect to a

centroidal axis,

( )

3 36 1 2 3 1 2 1 3 12 1 2 2

bh h bh bh Ad I I Ad I I

A A B B B B A A

= − = − = + =

′ ′ ′ ′

• 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 ℎ Moment of inertia about -axis Moment of inertia about -axis passing through C = 1 12 ℎ + ℎ

Moment of inertia about -axis = 1 12 ℎ Moment of inertia about -axis = 1 36 ℎ

Moment of inertia about -axis = 1 4 Moment of inertia about -axis passing through O = 1 2

Moment of inertia about -axis = = 1 8 Moment of inertia about -axis passing through O = 1 4

Moment of inertia about -axis = = 1 16 Moment of inertia about -axis passing through O = 1 8

Moment of inertia about -axis = 1 4 Moment of inertia about -axis passing through O = 1 4 + Moment of inertia about -axis = 1 4

Area Moments of Inertia

Example:

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
• btained by subtracting the moment of

inertia of the half-circle from the moment

• f inertia of the rectangle.

Area Moments of Inertia

Example: Solution

SOLUTION:

• Compute the moments of inertia of the bounding

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

( )( )

4 6 3 3 1 3 3 1

10 2 . 138 120 240 mm bh I x × = = =

Half-circle: moment of inertia with respect to AA’,

( )

4 6 4 8 1 4 8 1

mm 10 76 . 25 90 × = = =

′

π πr I A

A

( )( ) ( )

2

mm r A mm 81.8 a

• 120

b mm r a

3 2 2 1 2 2 1

10 72 . 12 90 2 . 38 3 90 4 3 4 × = = = = = = = = π π π π

Moment of inertia with respect to x’,

( ) ( )(

)

4 A A x

mm Aa I I

6 2 3 6 2

10 20 . 7 2 . 38 10 72 . 12 10 76 . 25 × = × − × = − =

′ ′

moment of inertia with respect to x,

( )(

)

4 x x

mm Ab I I

6 2 3 6 2

10 3 . 92 8 . 81 10 72 . 12 10 20 . 7 × = × + × = + =

′

Area Moments of Inertia

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.

4 6mm

10 9 . 45 × =

x

I

x

I =

4 6mm

10 2 . 138 × −

4 6mm

10 3 . 92 ×

Consider area (1)

= 1 3 ℎ = 1 3 × 80 × 60 = 5.76 × 10 ""

Consider area (2)

= 1 4

• 4

= 16 30 = 0.1590 × 10 ""

• $ = 0.1590 × 10 −

4 30 × 12.73 = 0.0445 × 10 "" = 0.0445 × 10 + 4 30 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 2 40 × 30 = 3490""

• =
• , =

4.05 × 10 3490 = .&. ''++

Determine the moment of inertia and the radius of gyration of the area shown in the fig.

/ !"" / !"" / !""

, "" , "" # ""

• ,

#

• 0) 1!++

Area Moments of Inertia

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

• =

dA xy I 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:

A y x I I

xy xy

+ =

+ Ixy + Ixy

• Ixy
• Ixy

Quadrants

Area Moments of Inertia

Rotation 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

• =

= = dA xy I dA x I dA y I

xy y x 2 2

Moments and product of inertia w.r.t. new axes x’ and y’ ? θ θ θ θ sin cos sin cos x y y y x x − = ′ + = ′ Note:

( ) ( ) ( )( )dA

x y y x dA y x I dA y x dA x I dA x y dA y I

y x y x

• −

+ = = + = = − = = θ θ θ θ θ θ θ θ sin cos sin cos ' ' sin cos ' sin cos '

' ' 2 2 ' 2 2 '

θ θ θ θ θ θ θ θ θ θ 2 cos sin cos 2 sin 2 / 1 cos sin 2 2 cos 1 cos 2 2 cos 1 sin

2 2 2 2

= − = + = − =

θ θ θ θ θ θ 2 cos 2 sin 2 2 sin 2 cos 2 2 2 sin 2 cos 2 2

' xy y x y x xy y x y x y xy y x y x x

I I I I I I I I I I I I I I I I + − = + − − + = − − + + =

′ ′ ′

Area Moments of Inertia

Rotation of Axes

Adding first two eqns: Ix’ + Iy’ = Ix + Iy = Iz The Polar MI @ O

θ θ θ θ θ θ

• +

− = + − − + = − − + + =

′ ′ ′

Angle which makes Ix’ and Iy’ either max or min can be found by setting the derivative

• f either Ix’ or Iy’ w.r.t. equal to zero:

Denoting this critical angle by

( )

• =

− − = θ θ θ

• −

=

• α

two values of 2 which differ by since tan2 = tan(2+) two solutions for will differ by /2

• ne value of 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

Area Moments of Inertia

Rotation of Axes

θ θ θ θ θ θ

• +

− = + − − + = − − + + =

′ ′ ′

• −

=

• −

=

• α

α α

Substituting in the third eqn for critical value

• f 2: Ix’y’ = 0

Product of Inertia Ix’y’ is zero for the Principal Axes of inertia Substituting sin2 and cos2 in first two eqns for Principal Moments of Inertia:

( ) ( )

• =

+ − − + = + − + + =

α

= + 2 + − 2 23425 − 46725 = − 2 46725 + 23425 − + 2 = − 2 23425 − 46725

Squaring both the equation and adding

− + 2

• +

=

− 2 23425 − 46725

• + −

2 46725 + 23425

• −

89:8;

• +

= 89<8;

• 23425 − 2

89<8;

• 2342546725 +

46725

+

89<8;

• 46725 + 2

89<8;

• 2342546725 +

46725

− + 2

• +

=

− 2

• +

− + 2

• +

=

− 2

• +
• Defining

+ 2 = =>?

And

@ = − 2

• +
• − =>?

+ = @

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

• =>?

@

25

Area Moments of Inertia

Mohr’s Circle of Inertia: Following relations can be represented

graphically by a diagram called Mohr’s Circle For given values of Ix, Iy, & Ixy, corresponding values of Ix’, Iy’, & Ix’y’ may be determined from the diagram for any desired angle .

θ θ θ θ θ θ

• +

− = + − − + = − − + + =

′ ′ ′

• −

=

• α

( ) ( )

• =

+ − − + = + − + + =

α

• (

)

• +
• −

= + = = + −

′ ′ ′

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

a maximum and minimum, respectively. R I I

ave ±

=

min max,

I Ixy

Area Moments of Inertia

Mohr’s Circle of Inertia: Construction

θ θ θ θ θ θ

• +

− = + − − + = − − + + =

′ ′ ′

• −

=

• α

( ) ( )

• =

+ − − + = + − + + =

α

• Choose horz axis MI

Choose vert axis PI Point A – known {Ix, Ixy} Point B – known {Iy, -Ixy} Circle with dia AB Angle for Area Angle 2 to horz (same sense) Imax, Imin Angle x to x’ = Angle OA to OC = 2 Same sense Point C Ix’, Ix’y’ Point D Iy’

Area Moments of Inertia

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 Inertia

Examples

SOLUTION:

• Determine the product of inertia using direct integration

with the parallel axis theorem on vertical differential area strips

• −

= = =

• −

= =

• −

= b x h y y x x dx b x h dx y dA b x h y

el el

1 1 1

2 1 2 1

Integrating dIx from x = 0 to x = b,

( )

b b b el el xy xy

b x b x x h dx b x b x x h dx b x h x dA y x dI I

2 4 3 2 2 2 3 2 2 2 2 2 1

8 3 4 2 2 1

• +

− =

• +

− =

• −

= = =

• 2

2 24 1

h b Ixy =

Area Moments of Inertia

Examples

• Apply the parallel axis theorem to evaluate the

product of inertia with respect to the centroidal axes. h y b x

3 1 3 1

= = With the results from part a,

( )( )( )

bh h b h b I A y x I I

y x y x xy 2 1 3 1 3 1 2 2 24 1

− = + =

′ ′ ′ ′ ′ ′ ′ ′ 2 2 72 1

h b I

y x

− =

′ ′ ′ ′

SOLUTION

Area Moments of Inertia

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

• f 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
• rientation 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.

Area Moments of Inertia

Example: Mohr’s Circle of Inertia

4 6 4 6 4 6

mm 10 54 . 2 mm 10 61 . 2 mm 10 24 . 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.

( ) ( )

( ) ( )

4 6 2 2 4 6 2 1 4 6 2 1

mm 10 437 . 3 mm 10 315 . 2 mm 10 925 . 4 × = + = × = − = × = + = = DX CD R I I CD I I I OC

y x y x ave

• Based on the circle, determine the orientation of the

principal axes and the principal moments of inertia. ° = = = 6 . 47 2 097 . 1 2 tan

m m

CD DX θ θ ° = 8 . 23

m

θ

Area Moments of Inertia

Example: Mohr’s Circle of Inertia

4 6 4 6

mm 10 437 . 3 mm 10 925 . 4 × = × = = R I OC

ave

• Based on the circle, evaluate the moments and product
• f 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 = 2(60o) = 120o. The angle that CX’ forms with the horz is φ = 120o - 47.6o = 72.4o.

• ave

y

R I Y C OC OG I 4 . 72 cos cos

'

− = ′ − = = ϕ

4 6mm

10 89 . 3 × =

′ y

I

• ave

x

R I X C OC OF I 4 . 72 cos cos

'

+ = ′ + = = ϕ

4 6mm

10 96 . 5 × =

′ x

I

• y

x

R Y C X F I 4 . 72 sin sin

'

= ′ = ′ =

′

ϕ

4 6mm

10 28 . 3 × =

′ ′y x

I

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

Solution: Parallel axis theorem: Ixy = xy +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 xy = 0 for both area.

Ix < Iy , Ixy (+) Ix >Iy , Ixy (+) Ix > Iy , Ixy (-) Ix < Iy , Ixy (-)

Area Moments of Inertia

Mass 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 Inertia

Moments 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

area A A A A

I t dA r t dm r I

, 2 2 ′ ′

= = =

• ρ

ρ

• Similarly, for perpendicular axis BB’

which is also contained in the plate,

area B B B B

I t I

, ′ ′ = ρ

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

( )

B B A A area B B area A A area C C C

I I I I t J t I

′ ′ ′ ′ ′

+ = + = =

, , ,

ρ ρ

Area Moments of Inertia

Moments of Inertia of Thin Plates

• For the principal centroidal axes on a rectangular plate,

( )

2 12 1 3 12 1 ,

ma b a t I t I

area A A A A

= = =

′ ′

ρ ρ

( )

2 12 1 3 12 1 ,

mb ab t I t I

area B B B B

= = =

′ ′

ρ ρ

( )

2 2 12 1 , ,

b a m I I I

mass B B mass A A C C

+ = + =

′ ′ ′

• For centroidal axes on a circular plate,

( )

2 4 1 4 4 1 ,

mr r t I t I I

area A A B B A A

= = = =

′ ′ ′

π ρ ρ

Area Moments of Inertia

Moments 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

• =

dV r I

2

ρ

• 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 Inertia

MI of some common geometric shapes

• Q. Determine the angle 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

Ix =

B () E

Iy =

B ()

• Ixy = (

F )()

With this data, plot the Mohr’s circle, and using trigonometry calculate the angle 2 2 =tan-1(b4/b4) = 45o Therefore, =22.5o (clockwise w.r.t. x) From Mohr’s Circle: Imax =

G + 3.08

Imin =

G − 0.5

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

• = +
• + −
• H34 5 − I67 5
• = +
• − −
• H34 5 + I67 5
• = −
• I67 5 + H34 5

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 × ) = 0.150

Alternatively, Mohr’s Circle may be used to determine the three quantities.