Moments of Inertia 10 Engineering Mechanics: Statics in SI Units, 12e Copyright © 2010 Pearson...

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Moments of Inertia1010

Engineering Mechanics: Statics in SI Units, 12e


Chapter Objectives

• Method for determining the moment of inertia for an area

• Introduce product of inertia and show determine the maximum and minimum moments of inertia for an area

• Discuss the mass moment of inertia


Chapter Outline

1. Definitions of Moments of Inertia for Areas

2. Parallel-Axis Theorem for an Area

3. Radius of Gyration of an Area

4. Moments of Inertia for Composite Areas

5. Product of Inertia for an Area

6. Moments of Inertia for an Area about Inclined Axes

7. Mohr’s Circle for Moments of Inertia

8. Mass Moment of Inertia


10.1 Definition of Moments of Inertia for Areas

• Centroid for an area is determined by the first moment of an area about an axis

• Second moment of an area is referred as the moment of inertia

• Moment of inertia of an area originates whenever one relates the normal stress σ or force per unit area



Moment of Inertia • Consider area A lying in the x-y plane• Be definition, moments of inertia of the differential

plane area dA about the x and y axes

• For entire area, moments of

inertia are given by

Ay

Ax

yx

dAxI

dAyI

dAxdIdAydI

2

2

22



Moment of Inertia • Formulate the second moment of dA about the pole O

or z axis• This is known as the polar axis

where r is perpendicular from the pole (z axis) to the element dA

• Polar moment of inertia for entire area,

dArdJO2

yxAO IIdArJ 2


10.2 Parallel Axis Theorem for an Area

• For moment of inertia of an area known about an axis passing through its centroid, determine the moment of inertia of area about a corresponding parallel axis using the parallel axis theorem

• Consider moment of inertia of the shaded area• A differential element dA is

located at an arbitrary distance y’ from the centroidal x’ axis



• The fixed distance between the parallel x and x’ axes is defined as dy

• For moment of inertia of dA about x axis

• For entire area

• First integral represent the moment of inertia of the area about the centroidal axis

AyAyA

A yx

yx

dAddAyddAy

dAdyI

dAdydI

22

2

2

'2'

'

'



• Second integral = 0 since x’ passes through the area’s centroid C

• Third integral represents the total area A

• Similarly

• For polar moment of inertia about an axis perpendicular to the x-y plane and passing through pole O (z axis)

2

2

2

0;0'

AdJJ

AdII

AdII

ydAydAy

CO

xyy

yxx


10.3 Radius of Gyration of an Area

• Radius of gyration of a planar area has units of length and is a quantity used in the design of columns in structural mechanics

• For radii of gyration

• Similar to finding moment of inertia of a differential area about an axis

dAydIAkI

AJ

kA

Ik

AI

k

xxx

Oz

yy

xx

22


Example 10.1

Determine the moment of inertia for the rectangular area with respect to (a) the centroidal x’ axis, (b) the axis xb passing through the base of the rectangular, and (c) the pole or z’ axis perpendicular to the x’-y’ plane and passing through the centroid C.


Solution

Part (a)

Differential element chosen, distance y’ from x’ axis.

Since dA = b dy’,

Part (b)

By applying parallel axis theorem,

32/

2/

22/

2/

22

12

1')'('' bhdyybdyydAyI

h

h

h

hAx

32

32

3

1

212

1bh

hbhbhAdII xxb


Solution

Part (c)

For polar moment of inertia about point C,

)(12

112

1

22'

3'

bhbhIIJ

hbI

yxC

y


10.4 Moments of Inertia for Composite Areas

• Composite area consist of a series of connected simpler parts or shapes

• Moment of inertia of the composite area = algebraic sum of the moments of inertia of all its parts

Procedure for AnalysisComposite Parts• Divide area into its composite parts and indicate the

centroid of each part to the reference axisParallel Axis Theorem• Moment of inertia of each part is determined about its

centroidal axis


10.4 Moments of Inertia for Composite Areas

Procedure for Analysis

Parallel Axis Theorem• When centroidal axis does not coincide with the

reference axis, the parallel axis theorem is used

Summation• Moment of inertia of the entire area about the

reference axis is determined by summing the results of its composite parts


Example 10.4

Compute the moment of inertia of the composite area about the x axis.


Solution

Composite Parts

Composite area obtained by subtracting the circle form the rectangle.

Centroid of each area is located in the figure below.


Solution

Parallel Axis Theorem

Circle

Rectangle

46224

2'

104.117525254

1mm

AdII yxx

4623

2'

105.1127515010015010012

1mm

AdII yxx


Solution

Summation

For moment of inertia for the composite area,

46

66

10101

105.112104.11

mm

I x


10.5 Product of Inertia for an Area

• Moment of inertia for an area is different for every axis about which it is computed

• First, compute the product of the inertia for the area as well as its moments of inertia for given x, y axes

• Product of inertia for an element of area dA located at a point (x, y) is defined as

dIxy = xydA

• Thus for product of inertia,

Axy xydAI


10.5 Product of Inertia for an Area

Parallel Axis Theorem• For the product of inertia of dA with respect to the x

and y axes

• For the entire area,

• Forth integral represent the total area A,

dAdydxdIA yxxy ''

AyxAyA Ax

A yxxy

dAdddAxddAyddAyx

dAdydxdI

''''

''

yxyxxy dAdII ''


Example 10.6

Determine the product Ixy of the triangle.


Solution

Differential element has thickness dx and area dA = y dx

Using parallel axis theorem,

locates centroid of the element or origin of x’, y’ axes

yxdAIddI xyxy~~

yx ~,~


Solution

Due to symmetry,

Integrating we have

2/~,~0 yyxxId xy

82

22

0

32

2 hbdxx

b

hI

b

xy

dxxb

hxb

hxxdx

b

hyxydxdI xy

32

2

222)(0


Solution

Differential element has thickness dy and area dA = (b - x) dy.

For centroid,

For product of inertia of element

yyxbxbxx ~,2/)(2/)(~

dyy

h

bbyy

yhbbdyy

h

bb

yxb

dyxbyxdAIddI xyxy

22

22

2

1

2

/

2)(0~~~


10.6 Moments of Inertia for an Area about Inclined Axes

• In structural and mechanical design, necessary to calculate Iu, Iv and Iuv for an area with respect to a set of inclined u and v axes when the values of θ, Ix, Iy and Ixy are known

• Use transformation equations which relate the x, y and u, v coordinates

dAxyyxuvdAdI

dAyxdAudI

dAxydAvdI

xyv

yxu

uv

v

u

)sincos)(sincos(

)sincos(

)sincos(

sincos

sincos

22

22



• Integrating,

• Simplifying using trigonometric identities,

)sin(cos2cossincossin

cossin2cossin

cossin2sincos

22

22

22

xyyxuv

xyyxv

xyyxu

IIII

IIII

IIII

22 sincos2cos

cossin22sin



• We can simplify to

• Polar moment of inertia about the z axis passing through point O is,

2cos22sin2

2sin2cos22

2sin2cos22

xyyx

uv

xyyxyx

v

xyyxyx

u

III

I

IIIII

I

IIIII

I

yxvuO IIIIJ



Principal Moments of Inertia

• Iu, Iv and Iuv depend on the angle of inclination θ of the u, v axes

• The angle θ = θp defines the orientation of the principal axes for the area

2/2tan

02cos22sin2

2

yx

xyp

p

xyyxu

II

I

III

d

dI



Principal Moments of Inertia• Substituting each of the sine and cosine ratios, we

have

• Result can gives the max or min moment of inertia for the area

• It can be shown that Iuv = 0, that is, the product of inertia with respect to the principal axes is zero

• Any symmetric axis represent a principal axis of inertia for the area

2

2

maxmin 22 xy

yxyx IIIII

I


Example 10.8

Determine the principal moments of inertia for the beam’s cross-sectional area with respect to an axis passing through the centroid.


Solution

Moment and product of inertia of the cross-sectional area,

Using the angles of inclination of principal axes u and v,

494949 1000.31060.51090.2 mmImmImmI zyx

1.57,9.32

2.1142,8.652

22.22/1060.51090.2

1000.3

2/2tan

21

21

99

9

pp

pp

yx

xyp II

I


Solution

For principal of inertia with respect to the u and v axes

49

min49

max

99maxmin

29

299

99

2

2

maxmin

10960.0,1054.7

1029.31025.4

1000.32

1060.51090.2

2

1060.51090.2

22

mmImmI

I

IIIII

I xyyxyx


10.7 Mohr’s Circle for Moments of Inertia

• It is found that

• In a given problem, Iu and Iv are variables and Ix, Iy and Ixy are known constants

• When this equation is plotted on a set of axes that represent the respective moment of inertia and the product of inertia, the resulting graph represents a circle

222

2

2

2

2

22

RIaI

III

III

I

uvu

xyyx

uvyx

u



• The circle constructed is known as a Mohr’s circle with radius

and center at (a, 0) where

2

2

2 xyyx III

R

2/yx IIa




Determine Ix, Iy and Ixy

• Establish the x, y axes for the area, with the origin located at point P of interest and determine Ix, Iy and Ixy

Construct the Circle• Construct a rectangular coordinate system such that

the abscissa represents the moment of inertia I and the ordinate represent the product of inertia Ixy



Construct the Circle• Determine center of the circle O, which is located at a

distance (Ix + Iy)/2 from the origin, and plot the reference point a having coordinates (Ix, Ixy)

• By definition, Ix is always positive, whereas Ixy will either be positive or negative

• Connect the reference point A with the center of the circle and determine distance OA (radius of the circle) by trigonometry

• Draw the circle



Principal of Moments of Inertia• Points where the circle intersects the abscissa give the

values of the principle moments of inertia Imin and Imax

• Product of inertia will be zero at these points

Principle Axes• This angle represent twice the angle from the x axis to

the area in question to the axis of maximum moment of inertia Imax

• The axis for the minimum moment of inertia Imin is perpendicular to the axis for Imax


Example 10.9

Using Mohr’s circle, determine the principle moments of the beam’s cross-sectional area with respect to an axis

passing through the centroid.


Solution

Determine Ix, Iy and Ixy

Moments of inertia and the product of inertia have been determined in previous examples

Construct the Circle

Center of circle, O, lies from the origin, at a distance

494949 1000.3 1060.51090.2 mmImmImmI xyyx

25.42/)60.590.2(2/ yx II


Solution

Construct the Circle

With reference point A (2.90, -3.00) connected to point O, radius OA is determined using Pythagorean theorem

Principal Moments of Inertia

Circle intersects I axis at points (7.54, 0) and (0.960, 0)

29.300.335.1 22 OA

49

min

49max

10960.0

1054.7

mmI

mmI


Solution

Principal Axes

Angle 2θp1 is determined from the circle by measuring CCW from OA to the direction of the positive I axis

The principal axis for Imax = 7.54(109) mm4 is therefore orientated at an angle θp1 = 57.1°, measured CCW from the positive x axisto the positive u axis.

v axis is perpendicular to this axis.

2.11429.3

00.3sin180sin1802 11

1

OA

BAp


10.8 Mass Moment of Inertia

• Mass moment of inertia is defined as the integral of the second moment about an axis of all the elements of mass dm which compose the body

• For body’s moment of inertia about the z axis,

• The axis that is generally chosen for analysis, passes through the body’s mass center G

mdmrI 2



• If the body consists of material having a variable density ρ = ρ(x, y, z), the element mass dm of the body may be expressed as dm = ρ dV

• Using volume element for integration,

• When ρ being a constant,

V

dVrI 2

VdVrI 2




Shell Element• For a shell element having height z, radius y and

thickness dy, volume dV = (2πy)(z)dy

Disk Element• For disk element having radius y, thickness dz, volume

dV = (πy2) dz


Example 10.10

Determine the mass moment of inertia of the cylinder about the z axis. The density of the material is constant.


Solution

Shell Element

For volume of the element,

For mass,

Since the entire element lies at the same distance r from the z axis, for the moment of inertia of the element,

drhrdmrdI

drrhdVdm

drhrdV

z32 2

2

2


Solution

Integrating over entire region of the cylinder,

For the mass of the cylinder

So that

hRdrrhdmrIR

mz4

0

32

22

2

02 hRrdrhdmm

R

m

2

2

1mRI z



Parallel Axis Theorem• If the moment of inertia of the body about an axis

passing through the body’s mass center is known, the moment of inertia about any other parallel axis may be determined by using parallel axis theorem

• Using Pythagorean theorem,

r 2 = (d + x’)2 + y’2

• For moment of inertia of body about the z axis,

mmm

mm

dmddmxddmyx

dmyxddmrI

222

222

'2''

''



Parallel Axis Theorem• For moment of inertia about the z axis,

I = IG + md2

Radius of Gyration• For moment of inertia expressed using k, radius of

gyration,

m

IkormkI 2


Example 10.12

If the plate has a density of 8000kg/m3 and a thickness of 10mm, determine its mass moment of inertia about an axis perpendicular to the page and passing through point O.


Solution

The plate consists of 2 composite parts, the 250mm radius disk minus the 125mm radius disk.

Disk

For moment of inertia of a disk,

Mass center of the disk is located 0.25m from point O

2

2

1mrIG

22222

2

.473.125.071.1525.071.152

1

2

1

71.1501.025.08000

mkgdmrmI

kgVm

ddddO

ddd


Solution

Hole

For moment of inertia of plate about point O,

222

22

2

.276.025.093.3125.093.32

12

1

93.301.0125.08000

mkg

dmrmI

kgVm

hhhhO

hhh

2.20.1276.0473.1 mkg

III hOdOO


QUIZ

1. The definition of the Moment of Inertia for an area involves an integral of the form

A) x dA. B) x2 dA.

C) x2 dm. D) m dA.

2. Select the SI units for the Moment of Inertia for an area.

A) m3

B) m4

C) kg·m2

D) kg·m3


QUIZ

3. A pipe is subjected to a bending moment as shown. Which property of the pipe will result in lower stress (assuming a constant cross-sectional area)?

A) Smaller Ix B) Smaller Iy

C) Larger Ix D) Larger Iy

4. In the figure to the right, what is the differential moment of inertia of the element with respect to the y-axis (dIy)?

A) x2 ydx B) (1/12)x3dy

C) y2 x dy D) (1/3)ydy

My

M

Pipe section

x

x

y=x3

x,y

y


QUIZ

5. The parallel-axis theorem for an area is applied betweenA) An axis passing through its centroid and any corresponding parallel axis.B) Any two parallel axis.C) Two horizontal axes only.D) Two vertical axes only.

6. The moment of inertia of a composite area equals the ____ of the MoI of all of its parts.A) Vector sumB) Algebraic sum (addition or subtraction)C) Addition D) Product


QUIZ

7. For the area A, we know the centroid’s (C) location, area, distances between the four parallel axes, and the MoI about axis 1. We can determine the MoI about axis 2 by applying the parallel axis theorem ___ .

A) Directly between the axes 1 and 2.

B) Between axes 1 and 3 and then between the axes 3 and 2.

C) Between axes 1 and 4 and then axes 4 and 2.

D) None of the above.

d3

d2

d1

432

1

A

•C

Axis


QUIZ

8. For the same case, consider the MoI about each of the four axes. About which axis will the MoI be the smallest number?

A) Axis 1

B) Axis 2

C) Axis 3

D) Axis 4

E) Can not tell.

d3

d2

d1

432

1

A

•C

Axis


QUIZ

9. For the given area, the moment of inertia about axis 1 is 200 cm4 . What is the MoI about axis 3?

A) 90 cm4 B) 110 cm4

C) 60 cm4 D) 40 cm4

10. The moment of inertia of the rectangle about the x-axis equals

A) 8 cm4. B) 56 cm4 .

C) 24 cm4 . D) 26 cm4 .

A=10 cm2

•Cd2

d1

32

1

•C

d1 = d2 = 2 cm

2cm

2cm

3cm

x


QUIZ

11. The formula definition of the mass moment of inertia about an axis is ___________ .

A) r dm B) r2 dm

C) m dr D) m dr

12. The parallel-axis theorem can be applied to determine ________ .

A) Only the MoI B) Only the MMI

C) Both the MoI and MMI D) None of the above.

Note: MoI is the moment of inertia of an area and MMI is the mass moment inertia of a body


QUIZ

13. Consider a particle of mass 1 kg located at point P, whose coordinates are given in meters. Determine the MMIof that particle about the z axis.A) 9 kg·m2 B) 16 kg·m2

C) 25 kg·m2 D) 36 kg·m2

14. Consider a rectangular frame made of 4 slender bars with four axes perpendicular to the screen and passing through P, Q, R, and S respectively. About which of the four axes will the MMI of the frame be the largest?A) zP B) zQ C) zRD) zS E) Not possible to determine

z

xy

·P(3,4,6)

P •

S •

Q • • R


QUIZ

15. A particle of mass 2 kg is located 1 m down the y-axis. What are the MMI of the particle about the x, y, and z axes, respectively?

A) (2, 0, 2) B) (0, 2, 2)

C) (0, 2, 2) D) (2, 2, 0)•

1 m

x y

z

Moments of Inertia 10 Engineering Mechanics: Statics in SI Units, 12e Copyright © 2010 Pearson...

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