Moments and Product of Inertia

Contents

• Introduction（绪论）

• Moments of Inertia of an Area（平面图形的惯性矩）

• Moments of Inertia of an Area by Integration（积分法求惯性矩）

• Polar Moments of Inertia（极惯性矩）

• Radius of Gyration of an Area（惯性半径）

• Parallel Axis Theorem（平行移轴定理）

• Moments of Inertia of Common Shapes of Areas（常见平面图形的惯性矩）

• Product of Inertia（惯性积）

• Principal Axes and Principal Moments of Inertia（主惯性轴与主惯性矩）

• Mohr’s Circle for Moments of Inertia（惯性矩和惯性积莫尔圆）

• Principal Points（主惯性点）2

Introduction

• Previously considered distributed forces which were proportional to the

area or volume over which they act.

- The resultant was obtained by summing or integrating over the

areas or volumes.

- The moment of the resultant about any axis was determined by

computing the first moments of the areas or volumes about that

axis.

• Will now consider forces which are proportional to the area or volume

over which they act but also vary linearly with distance from a given axis.

- It will be shown that the magnitude of the resultant depends on the

first moments of the force distribution with respect to the axis.

- The point of application of the resultant depends on the second

moments of the distribution with respect to the axis.

• Current chapter will present methods for computing the moments and

products of inertia for areas.

3

Moments of Inertia of an Area

• Consider distributed forces whose magnitudes are

proportional to the elemental areas on which they

act and also vary linearly with the distance of

from a given axis.

F

A

A

• Example: Consider a beam subjected to pure bending.

Internal forces vary linearly with distance from the

neutral axis which passes through the section centroid.

2 2

0 first moment

second moment

x

F ky A

R k y dA y dA S

M k y dA y dA

• Example: Consider the net hydrostatic force on a

submerged circular gate.

2

x

F p A gy A

R g y dA

M g y dA

4

Moments of Inertia of an Area by Integration

• Second moments or moments of inertia of

an area with respect to the x and y axes,

dAxIdAyI yx22

• Evaluation of the integrals is simplified by

choosing dA to be a thin strip parallel to

one of the coordinate axes.

• For a rectangular area,

331

0

22 bhbdyydAyIh

x

• The formula for rectangular areas may also

be applied to strips parallel to the axes,

dxyxdAxdIdxydI yx223

31

5

Polar Moments of Inertia

• The polar moments of inertia is an important

parameter in problems involving torsion of

cylindrical shafts and rotations of slabs.

2

pI r dA

• The polar moments of inertia is related to the

rectangular moments of inertia,

2 2 2 2 2

p

y x

I r dA x y dA x dA y dA

I I

6

Radius of Gyration of an Area

• Consider area A with moments of

inertia Ix. Imagine that the area is

concentrated in a thin strip parallel to

the x axis with equivalent Ix.

2 xx x x

II i A i

A

ix = radius of gyration with respect

to the x axis

• Similarly,

2

2

y

y y y

p

p p p

II i A i

A

II i A i

A

2 2 2

p x yi i i

7

pi

yixi

Sample Problem

Determine the moments of

inertia of a triangle with respect

to its base.

SOLUTION:

• A differential strip parallel to the x axis is chosen for

dA.

dyldAdAydIx 2

• For similar triangles,

dyh

yhbdA

h

yhbl

h

yh

b

l

• Integrating dIx from y = 0 to y = h,

h

hh

x

yyh

h

b

dyyhyh

bdy

h

yhbydAyI

0

43

0

32

0

22

43

12

3bhI x

8

Sample Problem

a) Determine the centroidal polar

moments of inertia of a circular

area by direct integration.

b) Using the result of part a,

determine the moments of inertia

of a circular area with respect to a

diameter.

SOLUTION:

• An annular differential area element is chosen,

2

2 3

0 0

2

2 2

p

r r

p p

dI u dA dA u du

I dI u u du u du

4

2pI r

• From symmetry, Ix = Iy,

42 22

p x y x xI I I I r I

diamete

4

r4

x II r

9

Parallel Axis Theorem

• Consider moments of inertia I of an area A

with respect to the axis AA’

2

AAI y dA

• The axis BB’ passes through the area centroid

and is called a centroidal axis.

22

2 2

AAI y dA y d dA

y dA d y dA

2d dA

2 2

AA BBI I Ad I Ad

10

• For a group of parallel axes, the moment of inertia reaches the minimum value

when the reference axis is the centroid axis.

Parallel Axis Theorem

• Moments of inertia IT of a circular area with

respect to a tangent to the circle,

4

45

224412

r

rrrAdIIT

• Moments of inertia of a triangle with respect to a

centroidal axis,

3

361

2

31

213

1212

2

bh

hbhbhAdII

AdII

AABB

BBAA

11

Moments of Inertia of Common Shapes of Areas

• The moments 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.

12

pI

pI

pI

pI

pI

Sample Problem

Determine the moments 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 moments of inertia of the shaded area

is obtained by subtracting the moments of

inertia of the half-circle from the moments

of inertia of the rectangle.

13

SOLUTION:

• Compute the moments of inertia of the bounding

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

Rectangle:

46

313

31 mm102.138120240 bhIx

Half-circle:

moments of inertia with respect to AA’,

464

814

81 mm1076.2590 rI AA

23

2

212

21

mm1072.12

90

mm 81.8a-120b

mm 2.383

904

3

4

rA

ra

moments of inertia with respect to x’,

2 6 3 2

6 4

25.76 10 12.72 10 38.2

7.20 10 mm

x AAI I Aa

moments of inertia with respect to x,

46

2362

mm103.92

8.811072.121020.7

AbII xx

14

• The moments of inertia of the shaded area is obtained

by subtracting the moments of inertia of the half-circle

from the moments of inertia of the rectangle.

46mm109.45 xI

xI 46mm102.138 46mm103.92

15

Product of Inertia

• Product of Inertia:

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

16

Sample Problem

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.

17

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 dIxy 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

18

• 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

19

Principal Axes and Principal Moments of Inertia

Given

dAxyI

dAxIdAyI

xy

yx22

We wish to determine moments

and product of inertia with

respect to new axes x’ and y’.

cos 2 sin 22 2

si

cos 2 sin 2

n 22

2

cos

2

2

x y x y

x xy

x y

x y x

x y x y

y

y xy

I I

I I I II I

I II I

I II I

• The change of axes yields

• The equations for Ix’ and Ix’y’ are the

parametric equations for a circle,

2 2 2

2

2

2 2

x ave x y

x y x y

ave xy

I I I R

I I I II R I

• The equations for Iy’ and Ix’y’ lead to the

same circle.

sincos

sincos

xyy

yxx

Note:

20

Principal Axes and Principal Moments of Inertia

2 2 2

2

2

2 2

x ave x y

x y x y

ave xy

I I I R

I I I II R I

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

a maximum and minimum, respectively.

RII ave minmax,

tan 2 2m xy x yI I I

• Alternatively, the principal moments of

inertia may be determined by

• The equation for m defines two angles,

90o apart which correspond to the

principal axes of the area about O.

21

• One method to determine the principal

moments of inertia of the area about O

is to substitute these m back into the

equation for Ix’.

• The advantage is that we know which of

the two principal angles corresponds to

each principal moment of inertia.

Sample Problem

For the section shown, the moments of

inertia with respect to the x and y axes

are Ix = 10.38 in4 and Iy = 6.97 in4.

Determine (a) the orientation of the

principal axes of the section about O,

and (b) the values of the principal

moments of inertia about O.

SOLUTION:

• Compute the product of inertia with

respect to the xy axes by dividing the

section into three rectangles and applying

the parallel axis theorem to each.

• Determine the orientation of the

principal axes and the principal

moments of inertia.

22

SOLUTION:

• Compute the product of inertia with respect to the xy axes

by dividing the section into three rectangles.

56.6

28.375.125.15.1

0005.1

28.375.125.15.1

in,in. ,in. ,in Area,Rectangle 42

Ayx

III

II

I

Ayxyx

Apply the parallel axis theorem to each rectangle,

AyxII yxxy

Note that the product of inertia with respect to centroidal

axes parallel to the xy axes is zero for each rectangle.

4in 56.6 AyxIxy

23

• Determine the orientation of the principal axes and the

principal moments of inertia.

4

4

4

in 56.6

in 97.6

in 38.10

xy

y

x

I

I

I

255.4 and 4.752

85.397.638.10

56.6222tan

m

yx

xym

II

I

7.127 and 7.37 mm

22

22

minmax,

56.62

97.638.10

2

97.638.10

22

xy

yxyxI

IIIII

4min

4max

in 897.1

in 45.15

II

II

b

a

24

Mohr’s Circle for Moments of Inertia

2

2

2 2

x y x y

ave xy

I I I II R I

• The moments and product of inertia for an area

are plotted as shown and used to construct Mohr’s

circle,

• Mohr’s circle may be used to graphically or

analytically determine the moments and product

of inertia for any other rectangular axes

including the principal axes.

25

Sample Problem

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.

26

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

DX 8.23m

RIOAI ave max46

max mm1036.8 I

RIOBI ave min46

min mm1049.1 I

27

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 2(60o) = 120o. The angle that CX’

forms with the x’ axes is f = 120o - 47.6o = 72.4o.

' cos cos72.4o

y aveI OG OC CY I Rf

46mm1089.3 yI

' cos cos72.4o

x aveI OF OC CX I Rf

46mm1096.5 xI

' sin sin 72.4o

x yI FX CY Rf

46mm1028.3 yxI

28

Principal Points

• Consider a pair of principal axes with origin at a given point O.

• If there exists a different pair of principal axes through that same point, then every

pair of axes through that point is a set of principal axes and all moments of inertia

are the same.

cos 2 sin 22 2

sin 2 cos 22

x y x y

x xy

x y

x y xy

I I I II I

I II I

• A point so located that every axis through the point is a principal axis, and hence

the moments of inertia are the same for all axes through the point, is called a

principal point.

• In general, every plane area has two principal points. These points lie equidistant

from the centroid on the principal centroidal axis having the larger principal

moment of inertia.

29

• Apply the concepts described above to axes through the centroid of an area.

• If an area has three or more axes of symmetry, the centroid is a principal point and

every axis through the centroid is a principal axis and has the same moment of

inertia.

• These conditions are fulfilled for a circle, for all regular polygons (equilateral

triangle, square, regular pentagon, regular hexagon, and so on), and for many other

symmetric shapes.

• When the two principal centroidal moments of inertia are equal; then the two

principal points merge at the centroid, which becomes the sole principal point.

Principal Points

30