© 2004 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics SeventhEdition 1 - 1 CE 102 Statics Chapter 8 Distributed

© 2004 The McGraw-Hill Companies, Inc. All rights reserved.

Vector Mechanics for Engineers: Statics

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CE 102 Statics

Chapter 8

Distributed Forces: Moments of Inertia



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ContentsIntroductionMoments of Inertia of an AreaMoment of Inertia of an Area by

IntegrationPolar Moment of InertiaRadius of Gyration of an AreaSample Problem 8.1Sample Problem 8.2Parallel Axis TheoremMoments of Inertia of Composite

AreasSample Problem 8.3Sample Problem 8.4Product of InertiaPrincipal Axes and Principal Moments

of Inertia

Sample Problem 8.5Sample Problem 8.6Mohr’s Circle for Moments and Products

of InertiaSample Problem 8.7Moment of Inertia of a MassParallel Axis TheoremMoment of Inertia of Thin PlatesMoment of Inertia of a 3D Body by

IntegrationMoment of Inertia of Common Geometric

ShapesSample Problem 8.89Moment of Inertia With Respect to an

Arbitrary AxisEllipsoid of Inertia. Principle Axes of

Axes of Inertia of a Mass



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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 moment of the force distribution with respect to the axis.

- The point of application of the resultant depends on the second moment of the distribution with respect to the axis.

• Current chapter will present methods for computing the moments and products of inertia for areas and masses.



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Moment 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.

moment second

momentfirst 022

dAydAykM

QdAydAykR

AkyF

x

• Example: Consider the net hydrostatic force on a submerged circular gate.

dAyM

dAyR

AyApF

x2



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Moment 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 dto 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



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Polar Moment of Inertia

• The polar moment of inertia is an important parameter in problems involving torsion of cylindrical shafts and rotations of slabs.

dArJ 20

• The polar moment of inertia is related to the rectangular moments of inertia,

xy II

dAydAxdAyxdArJ

222220



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Radius of Gyration of an Area• Consider area A with moment of inertia

Ix. Imagine that the area is concentrated in a thin strip parallel to the x axis with equivalent Ix.

A

IkAkI x

xxx 2

kx = radius of gyration with respect to the x axis

• Similarly,

A

JkAkJ

A

IkAkI

OOOO

yyyy

2

2

222yxO kkk



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Sample Problem 8.1

Determine the moment 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



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Sample Problem 8.2

a) Determine the centroidal polar moment of inertia of a circular area by direct integration.

b) Using the result of part a, determine the moment of inertia of a circular area with respect to a diameter.

SOLUTION:

• An annular differential area element is chosen,

rr

OO

O

duuduuudJJ

duudAdAudJ

0

3

0

2

2

22

2

4

2rJO

• From symmetry, Ix = Iy,

xxyxO IrIIIJ 22

2 4

4

4rII xdiameter



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Parallel Axis Theorem

• Consider moment of inertia I of an area A with respect to the axis AA’

dAyI 2

• The axis BB’ passes through the area centroid and is called a centroidal axis.

dAddAyddAy

dAdydAyI

22

22

2

2AdII parallel axis theorem



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Parallel Axis Theorem

• Moment of inertia IT of a circular area with respect to a tangent to the circle,

4

45

224412

r

rrrAdIIT

• Moment of inertia of a triangle with respect to a centroidal axis,

3

361

2

31

213

1212

2

bh

hbhbhAdII

AdII

AABB

BBAA



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Moments of Inertia of Composite Areas• 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.



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Moments of Inertia of Composite Areas



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Sample Problem 8.3

The strength of a W14x38 rolled steel beam is increased by attaching a plate to its upper flange.

Determine the moment of inertia and radius of gyration with respect to an axis which is parallel to the plate and passes through the centroid of the section.

SOLUTION:

• Determine location of the centroid of composite section with respect to a coordinate system with origin at the centroid of the beam section.

• Apply the parallel axis theorem to determine moments of inertia of beam section and plate with respect to composite section centroidal axis.

• Calculate the radius of gyration from the moment of inertia of the composite section.



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Sample Problem 8.3SOLUTION:

• Determine location of the centroid of composite section with respect to a coordinate system with origin at the centroid of the beam section.

12.5095.17

0011.20Section Beam

12.50425.76.75Plate

in ,in. ,in ,Section 32

AyA

AyyA

in. 792.2in 17.95

in 12.502

3

A

AyYAyAY



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Sample Problem 8.3• Apply the parallel axis theorem to determine moments of

inertia of beam section and plate with respect to composite section centroidal axis.

4

23

43

1212

plate,

4

22sectionbeam,

in 2.145

792.2425.775.69

in3.472

792.220.11385

AdII

YAII

xx

xx

• Calculate the radius of gyration from the moment of inertia of the composite section.

2

4

in 17.95

in 5.617

A

Ik x

x in. 87.5xk

2.145 3.472plate,section beam, xxx III

4in 618xI



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Sample Problem 8.4

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.



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Sample Problem 8.4SOLUTION:• Compute the moments of inertia of the bounding

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

Rectangle:

46313

31 mm102.138120240 bhIx

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

464814

81 mm1076.2590 rI AA

23

2212

21

mm1072.12

90

mm 81.8a-120b

mm 2.383

904

3

4

rA

ra

moment of inertia with respect to x’,

46

362

mm1020.7

1072.121076.25

AaII AAx

moment of inertia with respect to x,

46

2362

mm103.92

8.811072.121020.7

AbII xx



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Sample Problem 8.4• 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.

46mm109.45 xI

xI 46mm102.138 46mm103.92



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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



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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’.

2cos2sin2

2sin2cos22

2sin2cos22

xyyx

yx

xyyxyx

y

xyyxyx

x

III

I

IIIII

I

IIIII

I

• The change of axes yields

• The equations for Ix’ and Ix’y’ are the parametric equations for a circle,

2

222

22 xyyxyx

ave

yxavex

III

RII

I

RIII

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

sincos

sincos

xyy

yxx

Note:



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Principal Axes and Principal Moments of Inertia

2

222

22 xyyxyx

ave

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,

yx

xym II

I

22tan

• Imax and Imin are the principal moments of inertia of the area about O.

• The equation for m defines two angles, 90o apart which correspond to the principal axes of the area about O.



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Sample Problem 8.5

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.



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Sample Problem 8.5

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

22241 hbIxy



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Sample Problem 8.5• 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

22721 hbI yx



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Sample Problem 8.6

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 (Eq. 9.25) and the principal moments of inertia (Eq. 9. 27).



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Sample Problem 8.6SOLUTION:

• 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



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Sample Problem 8.6• Determine the orientation of the principal axes (Eq. 9.25)

and the principal moments of inertia (Eq. 9. 27).

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

yxyx IIIII

I

4min

4max

in 897.1

in 45.15

II

II

b

a



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Mohr’s Circle for Moments and Products of Inertia

2

22 xyyxyx

ave III

RII

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 and principal moments and products of inertia.



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Sample Problem 8.7

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.



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Sample Problem 8.7

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

4621

4621

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 mm CD

DX 8.23m

RIOAI ave max46

max mm1036.8 I

RIOBI ave min46

min mm1049.1 I



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Sample Problem 8.7

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 = 120o - 47.6o = 72.4o.

oavey RIYCOCOGI 4.72coscos'

46 mm1089.3 yI

oavex RIXCOCOFI 4.72coscos'

46 mm1096.5 xI

oyx RYCXFI 4.72sinsin'

46 mm1028.3 yxI



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Moment of Inertia of a Mass• Angular acceleration about the axis AA’ of the

small mass m due to the application of a couple is proportional to r2m.

r2m = moment of inertia of the mass m with respect to the axis AA’

• For a body of mass m the resistance to rotation about the axis AA’ is

inertiaofmomentmassdmr

mrmrmrI

2

23

22

21

• The radius of gyration for a concentrated mass with equivalent mass moment of inertia is

m

IkmkI 2



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Moment of Inertia of a Mass

• Moment of inertia with respect to the y coordinate axis is

dmxzdmrI y222

• Similarly, for the moment of inertia with respect to the x and z axes,

dmyxI

dmzyI

z

x

22

22

• In SI units,

22 mkg dmrI

In U.S. customary units,

222

2 sftlbftft

slbftslugI



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Parallel Axis Theorem• For the rectangular axes with origin at O and parallel

centroidal axes,

dmzydmzzdmyydmzy

dmzzyydmzyI x

2222

2222

22

22 zymII xx

22

22

yxmII

xzmII

zz

yy

• Generalizing for any axis AA’ and a parallel centroidal axis,

2mdII



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Moments of Inertia of Thin Plates• For a thin plate of uniform thickness t and homogeneous

material of density , the mass moment of inertia with respect to axis AA’ contained in the plate is

areaAA

AA

It

dArtdmrI

,

22

• Similarly, for perpendicular axis BB’ which is also contained in the plate,

areaBBBB ItI ,

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

BBAA

areaBBareaAAareaCCC

II

IItJtI

,,,



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Moments of Inertia of Thin Plates

• For the principal centroidal axes on a rectangular plate,

21213

121

, mabatItI areaAAAA

21213

121

, mbabtItI areaBBBB

22121

,, bamIII massBBmassAACC

• For centroidal axes on a circular plate,

2414

41

, mrrtItII areaAABBAA

221 mrIII BBAACC



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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

dVrI 2

• 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.



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Moments of Inertia of Common Geometric Shapes



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Sample Problem 8.8

Determine the moments of inertia of the steel forging with respect to the xyz coordinate axes, knowing that the specific weight of steel is 490 lb/ft3.

SOLUTION:

• With the forging divided into a prism and two cylinders, compute the mass and moments of inertia of each component with respect to the xyz axes using the parallel axis theorem.

• Add the moments of inertia from the components to determine the total moments of inertia for the forging.



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Sample Problem 8.8

ftslb0829.0

sft2.32ftin1728

in31lb/ft490

:cylindereach

2

233

323

m

g

Vm

23

2

125.22

1232

121

121

222121

sftlb1017.4

0829.030829.0

3

xmLamI y

23

2

1222

125.22

1232

121

121

2222121

sftlb1048.6

0829.030829.0

3

yxmLamI y

23

2

1222

121

21

2221

sftlb1059.2

0829.00829.0

ymmaIx

cylinders :in.2.,in5.2.,in3,.in1 yxLaSOLUTION:• Compute the moments of inertia

of each component with respect to the xyz axes.



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Sample Problem 8.8

ftslb211.0

sft2.32ftin1728

in622lb/ft490

:prism

2

233

33

m

g

Vm

prism (a = 2 in., b = 6 in., c = 2 in.):

23

2

1222

126

12122

121

sftlb 1088.4

211.0

cbmII zx

23

2

1222

122

12122

121

sftlb 10977.0

211.0

acmI y

• Add the moments of inertia from the components to determine the total moments of inertia. 33 1059.221088.4 xI

23 sftlb1006.10 xI

33 1017.4210977.0 yI

23 sftlb1032.9 yI

33 1048.621088.4 zI23 sftlb1084.17

zI



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Moment of Inertia With Respect to an Arbitrary Axis• IOL = moment of inertia with respect to axis OL

dmrdmpIOL22

• Expressing in terms of the vector components and expanding yields

r

and

xzzxzyyzyxxy

zzyyxxOL

III

IIII

222

222

• The definition of the mass products of inertia of a mass is an extension of the definition of product of inertia of an area

xzmIdmzxI

zymIdmyzI

yxmIdmxyI

xzzx

zyyz

yxxy



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Ellipsoid of Inertia. Principal Axes of Inertia of a Mass• Assume the moment of inertia of a body has been

computed for a large number of axes OL and that point Q is plotted on each axis at a distance OLIOQ 1

• The locus of points Q forms a surface known as the ellipsoid of inertia which defines the moment of inertia of the body for any axis through O.

• x’,y’,z’ axes may be chosen which are the principal axes of inertia for which the products of inertia are zero and the moments of inertia are the principal moments of inertia.

4545

Problem 8.9

b

y2 = mx

ax

y1 = kx2

y

Determine by direct integrationthe moments of inertia of theshaded area with respect tothe x and y axes.

4646

b

y2 = mx

ax

y1 = kx2

y

1. Calculate the moments of inertia Ix and Iy. These moments ofinertia are defined by:

Ix = y2 dA and Iy = x2 dA

Where dA is a differential element of area dx dy.

1a. To compute Ix choose dA to be a thin strip parallel to the xaxis. All the the points of the strip are at the same distance y fromthe x axis. The moment of inertia dIx of the strip is given by y2 dA.

Solving Problems on Your Own


Problem 8.9

4747

b

y2 = mx

ax

y1 = kx2

y

1b. To compute Iy choose dA to be a thin strip parallel to the yaxis. All the the points of the strip are at the same distance x fromthe y axis. The moment of inertia dIy of the strip is given by x2 dA.

1c. Integrate dIx and dIy over the whole area.



Problem 8.9

4848

Problem 8.9 Solution

b

y2 = mx

ax

y1 = kx2

yDetermine m and k:

At x = a, y2 = b: b = m a

m =

y1 = k x2: b = k a2

k =

ba

ba2Express x in terms of y1 and y2:

y1 = k x2 then: x1 = y1

y2 = mx then x2 = y2

1k1/2

1m

1/21

4949

To compute Iy choose dA tobe a thin strip parallel to the yaxis.

b

a

x

y

y1

dx

y2

x

dA = ( y2 - y1 ) dx

= ( m x _ k x2 ) dx

Iy = x2 dA = x2 ( m x _ k x2 ) dx

= ( m x3 _ k x4 ) dx = [ m x4 _ k x5 ]

= m a4 _ k a5 = b a3

0

a

a

0

a

0

14

120

14

15

15

Substituting k = , m =ba2

ba


Iy = b a3/ 20

5050

To compute Ix choose dA tobe a thin strip parallel to the xaxis.

b

a

x

y

y

dy

x2

x1

dA = ( x1 - x2 ) dy

= ( y _ y ) dy1k1/2

1m

1/2

Ix = y2 dA = y2 ( y _ y ) dy

= ( y _ y3 ) dy = [ y _ y4 ]

= b _ b4 = a b3

1k1/2 1/2

1m

1k1/2

5/2 1m

0

b

b

0

27

1k1/2

7/2 1m

14

b

0

27

1k1/2

1m

14

7/2 128

Substituting k = , m =ba2

ba


Ix = a b3/ 28

5151

Problem 8.10

For the 5 x 3 x -in. angle crosssection shown, use Mohr’s circleto determine (a) the moments ofinertia and the product of inertiawith respect to new centroidalaxes obtained by rotating the xand y axes 30o clockwise, (b) theorientation of the principal axesthrough the centroid and thecorresponding values of themoments of inertia.

12

3 in

5 in

0.5 in

1.75 in

x

0.5 in0.75 in

y

L5x3x 12

5252

Solving Problems on Your Owny

3 in

5 in

0.5 in

1.75 in

x

0.5 in0.75 in

L5x3x 12

For the 5 x 3 x -in. angle cross sectionshown, use Mohr’s circle to determine(a) the moments of inertia and theproduct of inertia with respect to newcentroidal axes obtained by rotatingthe x and y axes 30o clockwise, (b) theorientation of the principal axes throughthe centroid and the correspondingvalues of the moments of inertia.

12

1. Draw Mohr’s circle. Mohr’s circle is completely defined by thequantities R and IAVE which represent, respectively, the radius ofthe circle and the distance from the origin O to the center C of thecircle. These quantities can be obtained if the moments andproduct of inertia are known for a given orientation of the axes.

Problem 8.10

5353

2. Use the Mohr’s circle to determine moments of inertia ofrotated axes. As the coordinate axes x-y arerotated through an angle , the associatedrotation of the diameter of Mohr’s circle isequal to 2 in the same sense(clockwise or counterclockwise).

x

y

2

Ix , Iy

Ixy

x’

y’



12

Problem 8.10

3 in

5 in

0.5 in

1.75 in

x

0.5 in0.75 in

L5x3x 12

5454

2. Use the Mohr’s circle to determine the orientation of principalaxes, and the principal moments of inertia. Points A and B wherethe circle intersects the horizontal axisrepresent the principal moments of inertia.The orientation of the principal axes isdetermined by the angle 2m.

x

yA

B

2m

Ix, Iy

Ixy



12

Problem 8.10

3 in

5 in

0.5 in

1.75 in

x

0.5 in0.75 in

L5x3x 12

5555


From Fig. 9.13A:

Ix = 9.45 in4, Iy = 2.58 in4

Product of inertia Ixy:

Ixy = (Ixy)1 + (Ixy)2

For each rectangle Ixy = Ix’y’ + x y A

and Ix’y’ = 0 (symmetry)

Thus: Ixy = x y A

A, in2 x, in y in x y A, in4

1 1.5 0.75 -1.5 -1.6875

2 2.25 -0.5 1.0 -1.125

Ixy = x y A = -2.81

y

1.5 inx

0.5 in

1 in

0.75 in1

2

3 in

5 in

0.5 in

1.75 in

x

0.5 in0.75 in

L5x3x 12

5656

Problem 8.10 SolutionDraw Mohr’s circle.y

x

The Mohr’s circle is defined by thediameter XY where X(9.45, -2.81)and Y(2.58, 2.81).The coordinate of the center C andthe radius R are calculated by:

OC = Iave = (Ix + Iy)

OC = (9.45 + 2.58) = 6.02 in4

R = [ (Ix - Iy)]2 + Ixy

R = [ (9.45 - 2.58)]2 + (-2.82)2

R = 4.44 in4

12

12

X (9.45, -2.81)

Ix, Iy (in4)

Ixy (in4)

Y (2.58, +2.81)

CO

2m

F tan 2m = = FXCF

2.819.45-6.02

m = 19.6o

1212

5757


Use the Mohr’s circle to determinemoments of inertia of rotated axes.

y

x

x’

y’

(a) The coordinates of X’ and Y’ give themoments and product of inertia withrespect to the x’y’ axes.

Ix’ = OE = OC _ CE = 6.02 _ 4.44 cos 80.7o

Ix’ = 5.30 in4

Iy’ = OD = OC + CD = 6.02 + 4.44 cos 80.7o

Iy’ = 6.73 in4

X

Ix, Iy (in4)

Ixy (in4)

Y

X’

Y’

2 = 60o

O

6.02 in4

R = 4.44 in4

80.7o Ix’y’ = EX’ = _ 4.44 sin 80.7o

Ix’y’ = _ 4.38 in439.3o

CDE

30o

5858

Problem 8.10 Solutiony

x

Use the Mohr’s circle to determine the orientation of principal axes, andthe principal moments of inertia.

(b) The principal axes are obtained byrotating the xy axes through angle m.

tan 2m = = FXCF

2.819.45-6.02 m = 19.6o

The corresponding moments of inertiaare Imax and Imin :

X (9.45, -2.81)

Ix, Iy (in4)

Ixy (in4)

Y (2.58, +2.81)

CO

2m

F ImaxImin

Imax, min = OC + R = 6.02 + 4.44

R = 4.44 in4

Imax = 10.46 in4

Imin = 1.574 in4

a

b

m = 19.6o

a axis corresponds to Imax

b axis corresponds to Imin

__

5959

x

y

z

120 mm

150 mm

150 mm

120 mm

150 mm150 mm

Problem 8.11

A section of sheet steel 2 mmthick is cut and bent into themachine component shown.Knowing that the density ofsteel is 7850 kg/m3,determine the mass momentof inertia of the componentwith respect to (a) the x axis,(b) the y axis, (c) the z axis.

6060


x

y

z

120 mm

150 mm

150 mm

120 mm

150 mm150 mm

1. Compute the mass moments of inertia of a composite body withrespect to a given axis.

1a. Divide the body into sections. The sections should have asimple shape for which the centroid and moments of inertia canbe easily determined (e.g. from Fig. 9.28 in the book).


Problem 8.11

6161


x

y

z

120 mm

150 mm

150 mm

120 mm

150 mm150 mm


1b. Compute the mass moment of inertia of each section. Themoment of inertia of a section with respect to the given axis isdetermined by using the parallel-axis theorem:

I = I + m d2

Where I is the moment of inertia of the section about its owncentroidal axis, I is the moment of inertia of the section about thegiven axis, d is the distance between the two axes, and m is thesection’s mass.

Problem 8.11

6262


x

y

z

120 mm

150 mm

150 mm

120 mm

150 mm150 mm


1c. Compute the mass moment of inertia of the whole body. Themoment of inertia of the whole body is determined by adding themoments of inertia of all the sections.

Problem 8.11

6363


Divide the body into sections.x

y

z

120 mm

150 mm

150 mm

120 mm

150 mm150 mm

x

y

z

x’

y’

z’

y’’

x’’

z’’

2

1

3

6464

x

y

z

120 mm

150 mm

150 mm

120 mm

150 mm150 mm

x

y

z

x’

y’

z’

y’’

x’’

z’’

2

1

3

Computation of Masses:

Section 1: m1 = V1 = (7850 kg/m3)(0.002 m)(0.300 m)2 = 1.413 kg

Section 2:

m2 = V2 = (7850 kg/m3)(0.002 m)(0.150 m)(0.120 m) = 0.2826 kg

Section 3: m3 = m2 = 0.2826 kg


6565

x

y

z

120 mm

150 mm

150 mm

120 mm

150 mm150 mm

x

y

z

x’

y’

z’

y’’

x’’

z’’

2

1

3Compute the moment of inertia of each section.

(a) Mass moment of inertia with respect to the x axis.

Section 1: (Ix)1 = (1.413) (0.30)2 = 1.06 x 10-2 kg . m21

12

Section 2: (Ix)2 = (Ix’)2 + m d2

(Ix)2 = (0.2826) (0.120) 2 + (0.2826)(0.152 + 0.062)

(Ix)2 = 7.71 x 10-3 kg . m2

112

Section 3: (Ix)3 = (Ix)2 = 7.71 x 10-3 kg . m2


6666

x

y

z

120 mm

150 mm

150 mm

120 mm

150 mm150 mm

x

y

z

x’

y’

z’

y’’

x’’

z’’

2

1

3

Compute the moment of inertia of the whole area.

For the whole body:

Ix = (Ix)1 + (Ix)2 + (Ix)3

Ix = 1.06 x 10-2 + 7.71 x 10-3 + 7.71 x 10-3 = 2.60 x 10-2 kg . m2


Ix = 26.0 x 10-3 kg . m2

6767

x

y

z

120 mm

150 mm

150 mm

120 mm

150 mm150 mm

x

y

z

x’

y’

z’

y’’

x’’

z’’

2

1


(b) Mass moment of inertia with respect to the y axis.

Section 1: (Iy)1 = (1.413) (0.302 + 0.302) = 2.12 x 10-2 kg . m21

12

Section 2: (Ix)2 = (Ix’)2 + m d2

(Iy)2 = (0.2826) (0.150)2 + (0.2826)(0.152 + 0.0752)

(Iy)2 = 8.48 x 10-3 kg . m2

112

Section 3: (Iy)3 = (Iy)2 = 8.48 x 10-3 kg . m2


6868

x

y

z

120 mm

150 mm

150 mm

120 mm

150 mm150 mm

x

y

z

x’

y’

z’

y’’

x’’

z’’

2

1

3


For the whole body:

Iy = (Iy)1 + (Iy)2 + (Iy)3

Iy = 2.12 x 10-2 + 8.48 x 10-3 + 8.48 x 10-3 = 3.82 x 10-2 kg . m2


Iy = 38.2 x 10-3 kg . m2

6969

x

y

z

120 mm

150 mm

150 mm

120 mm

150 mm150 mm

x

y

z

x’

y’

z’

y’’

x’’

z’’

2

1


(c) Mass moment of inertia with respect to the z axis.

Section 2: (Iz)2 = (Iz’)2 + m d2

(Iz)2 = (0.2826) (0.152 + 0.122) + (0.2826)(0.0602 + 0.0752)

(Iz)2 = 3.48 x 10-3 kg . m2

112

Section 3: (Iz)3 = (Iz)2 = 3.48 x 10-3 kg . m2

Section 1: (Iz)1 = (1.413) (0.30)2 = 1.059 x 10-2 kg . m21

12


7070

x

y

z

120 mm

150 mm

150 mm

120 mm

150 mm150 mm

x

y

z

x’

y’

z’

y’’

x’’

z’’

2

1

3


For the whole body:

Iz = (Iz)1 + (Iz)2 + (Iz)3

Iz = 1.06 x 10-2 + 3.48 x 10-3 + 3.48 x 10-3 = 1.755 x 10-2 kg . m2


Iz = 17.55 x 10-3 kg . m2

7171

Problem 8.12

Determine the moment ofinertia and the radius ofgyration of the steel machineelement shown with respect tothe x axis. (The density ofsteel is 7850 kg/m3.)

15 mm15 mm

x

y

z

40 mm50 mm

60 mm

45 mm

45 mm

45 mm

38 mm

7272

1. Compute the mass moments of inertia of a composite body withrespect to a given axis.

1a. Divide the body into sections. The sections should have asimple shape for which the centroid and moments of inertia canbe easily determined (e.g. from Fig. 9.28 in the book).

15 mm15 mm

x

y

z

40 mm50 mm

60 mm

45 mm

45 mm

45 mm

38 mm



Problem 8.12

7373

1b. Compute the mass moment of inertia of each section. Themoment of inertia of a section with respect to the given axis isdetermined by using the parallel-axis theorem:

I = I + m d2

Where I is the moment of inertia of the section about its owncentroidal axis, I is the moment of inertia of the section about thegiven axis, d is the distance between the two axes, and m is thesection’s mass.

15 mm15 mm

x

y

z

40 mm50 mm

60 mm

45 mm

45 mm

45 mm

38 mm



Problem 8.12

7474

1c. Compute the mass moment of inertia of the whole body. Themoment of inertia of the whole body is determined by adding themoments of inertia of all the sections.

2. Compute the radius of gyration. The radius of gyration k of abody is defined by:

k = Im



Problem 8.12

15 mm15 mm

x

y

z

40 mm50 mm

60 mm

45 mm

45 mm

45 mm

38 mm

7575

Divide the body into sections.

x

y

z

15 mm15 mm

x

y

z

40 mm50 mm

60 mm

45 mm

45 mm

45 mm

38 mm

xz

y

x

y

z


7676


Compute the mass momentof inertia of each section.

x

y

z

Rectangular prism:

m = VV = (0.15 m)(0.09 m)(0.03 m)V = 4.05 x 10-4 m3

m = (7850 kg/m3)(4.05x10-4 m3)m = 3.18 kg

x’75 mm

(Ix) = ( Ix’) + m d2

(Ix) = (3.18 kg)[(0.15 m)2+(0.03 m)2]

+ (3.18 kg) (0.075 m)2

(Ix) = 2.408 x 10-2 kg . m2

112

15 mm15 mm

x

y

z

40 mm50 mm

60 mm

45 mm

45 mm

45 mm

38 mm

7777


Semicircular cylinder:

m = VV = (0.045 m)2 (0.04 m)V = 1.27 x 10-4 m3

m = (7850 kg/m3)(1.27x10-4 m3)m = 1.0 kg

1215 mm

15 mm

x

y

z

40 mm50 mm

60 mm

45 mm

45 mm

45 mm

38 mm

x

y

z

x’130 mm

15 mm

x’’

4r3 4.453 = 19.1 mm=

(centroidal axis) C

xx’x’’

40 mm

7878


x

y

z

x’130 mm

15 mm

x’’

4r3 4.453 = 19.1 mm=

(centroidal axis) C

xx’x’’

Ix’ = Ix’’ + m d2

(1.0 kg) [3 (0.045 m)2 + (0.04 m)2] = Ix’’ + (1.0 kg)(0.0191 m)2

Ix’’ = 27.477 x 10-5 kg . m2

Ix = ( Ix’’) + m d2 = 27.477 x 10-5 kg . m2

+ (1.0 kg)[(0.13 m)2 + (0.015 m + 0.0191 m)2]

Ix = 18.34 x 10-3 kg . m2

112

40 mm

7979

15 mm15 mm

x

y

z

40 mm50 mm

60 mm

45 mm

45 mm

45 mm

38 mm

Circular cylindrical hole:

m = VV = (0.038 m)2 (0.03 m)V = 1.361 x 10-4 m3

m = (7850 kg/m3)(1.36x10-4 m3)m = 1.068 kg

(Ix) = ( Ix’) + m d2

(Ix) = (1.07 kg) [ 3 (0.038 m)2

+ (0.03 m)2 ]

+ (1.07 kg) (0.06 m)2

(Ix) = 4.31 x 10-3 kg . m2

112

xz

y

x’60 mm


8080

15 mm15 mm

x

y

z

40 mm50 mm

60 mm

45 mm

45 mm

45 mm

38 mm

Compute the mass momentof inertia of the whole body.

For the entire body, adding thevalues obtained:

Ix = 2.408 x 10-2 + 1.834 x 10-2 - 4.31 x 10-3

Ix = 3.81 x 10-2 kg . m2

Compute the radius of gyration.The mass of the entire body:

m = 3.18 + 1.0 - 1.07 = 3.11 kg

k = =

k = 0.1107 m

Im 3.81 x 10-2 kg . m2

3.11 kg

Radius of gyration:


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© 2004 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics SeventhEdition 1 - 1 CE 102 Statics Chapter 8 Distributed