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Vector Mechanics for Engineers: Statics

CE 102 Statics

Chapter 8

Distributed Forces: Moments of Inertia

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

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.

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.

• 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

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

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,

22 bhbdyydAyIh

• The formula for rectangular areas may also be applied to strips parallel to the axes,

dxyxdAxdIdxydI yx223

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,

dAydAxdAyxdArJ

222220

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.

IkAkI x

kx = radius of gyration with respect to the x axis

• Similarly,

222yxO kkk

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,

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

dyyhyh

yhbydAyI

3bhI x

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,

duuduuudJJ

duudAdAudJ

• From symmetry, Ix = Iy,

xxyxO IrIIIJ 22

4rII xdiameter

1 - 10

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

2AdII parallel axis theorem

1 - 11

Parallel Axis Theorem

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

224412

rrrAdIIT

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

hbhbhAdII

1 - 12

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.

1 - 13

Moments of Inertia of Composite Areas

1 - 14

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.

1 - 15

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

in. 792.2in 17.95

in 12.502

AyYAyAY

1 - 16

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.

plate,

22sectionbeam,

in 2.145

792.2425.775.69

in3.472

792.220.11385

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

in 17.95

in 5.617

x in. 87.5xk

2.145 3.472plate,section beam, xxx III

4in 618xI

1 - 17

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.

1 - 18

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

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

Rectangle:

31 mm102.138120240 bhIx

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

464814

81 mm1076.2590 rI AA

mm1072.12

mm 81.8a-120b

mm 2.383

moment of inertia with respect to x’,

mm1020.7

1072.121076.25

AaII AAx

moment of inertia with respect to x,

mm103.92

8.811072.121020.7

AbII xx

1 - 19

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

1 - 20

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

1 - 21

Principal Axes and Principal Moments of Inertia

dAxIdAyI

we wish to determine moments and product of inertia with respect to new axes x’ and y’.

2cos2sin2

2sin2cos22

xyyxyx

• The change of axes yields

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

22 xyyxyx

yxavex

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

sincos

1 - 22

Principal Axes and Principal Moments of Inertia

22 xyyxyx

yxavex

• At the points A and B, Ix’y’ = 0 and Ix’ is a maximum and minimum, respectively. RII ave minmax,

xym II

• 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.

1 - 23

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.

1 - 24

Sample Problem 8.5

SOLUTION:

• Determine the product of inertia using direct integration with the parallel axis theorem on vertical differential area strips

xhyyxx

xhdxydA

elel 1

Integrating dIx from x = 0 to x = b,

elelxyxy

xhxdAyxdII

22241 hbIxy

1 - 25

Sample Problem 8.5• Apply the parallel axis theorem to evaluate the

product of inertia with respect to the centroidal axes.

hybx31

With the results from part a,

bhhbhbI

22721 hbI yx

1 - 26

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).

1 - 27

Sample Problem 8.6SOLUTION:

• Compute the product of inertia with respect to the xy axes by dividing the section into three rectangles.

28.375.125.15.1

0005.1

28.375.125.15.1

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

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

1 - 28

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

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

in 56.6

in 97.6

in 38.10

255.4 and 4.752

85.397.638.10

56.6222tan

xym II

7.127 and 7.37 mm

minmax,

97.638.10

yxyx IIIII

in 897.1

in 45.15

1 - 29

Mohr’s Circle for Moments and Products of Inertia

22 xyyxyx

ave III

• 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.

1 - 30

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.

1 - 31

Sample Problem 8.7

mm1054.2

mm1061.2

mm1024.7

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

circle based on the circle diameter between the points.

mm10437.3

mm10315.2

mm10925.4

• 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

1 - 32

Sample Problem 8.7

mm10437.3

mm10925.4

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

1 - 33

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

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

IkmkI 2

1 - 34

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,

• In SI units,

22 mkg dmrI

In U.S. customary units,

2 sftlbftft

slbftslugI

1 - 35

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

centroidal axes,

dmzydmzzdmyydmzy

dmzzyydmzyI x

22 zymII xx

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

1 - 36

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

dArtdmrI

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

areaBBBB ItI ,

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

areaBBareaAAareaCCC

IItJtI

1 - 37

Moments of Inertia of Thin Plates

• For the principal centroidal axes on a rectangular plate,

, mabatItI areaAAAA

, mbabtItI areaBBBB

,, bamIII massBBmassAACC

• For centroidal axes on a circular plate,

, mrrtItII areaAABBAA

221 mrIII BBAACC

1 - 38

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.

1 - 39

Moments of Inertia of Common Geometric Shapes

1 - 40

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.

1 - 41

Sample Problem 8.8

ftslb0829.0

sft2.32ftin1728

in31lb/ft490

:cylindereach

125.22

222121

sftlb1017.4

0829.030829.0

xmLamI y

125.22

2222121

sftlb1048.6

0829.030829.0

yxmLamI y

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.

1 - 42

Sample Problem 8.8

ftslb211.0

sft2.32ftin1728

in622lb/ft490

:prism

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

sftlb 1088.4

cbmII zx

sftlb 10977.0

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

1 - 43

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

xzzxzyyzyxxy

zzyyxxOL

• 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

1 - 44

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.

Problem 8.9

y2 = mx

y1 = kx2

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

y2 = mx

y1 = kx2

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

y2 = mx

y1 = kx2

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

Problem 8.9 Solution

y2 = mx

y1 = kx2

yDetermine m and k:

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

y1 = k x2: b = k a2

ba2Express x in terms of y1 and y2:

y1 = k x2 then: x1 = y1

y2 = mx then x2 = y2

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

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

Substituting k = , m =ba2

Iy = b a3/ 20

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

dA = ( x1 - x2 ) dy

= ( y _ y ) dy1k1/2

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

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

= b _ b4 = a b3

1k1/2 1/2

5/2 1m

7/2 1m

7/2 128

Substituting k = , m =ba2

Ix = a b3/ 28

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.

0.5 in

1.75 in

0.5 in0.75 in

L5x3x 12

Solving Problems on Your Owny

0.5 in

1.75 in

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.

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

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).

Ix , Iy

Problem 8.10

0.5 in

1.75 in

0.5 in0.75 in

L5x3x 12

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.

Ix, Iy

Problem 8.10

0.5 in

1.75 in

0.5 in0.75 in

L5x3x 12

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

1.5 inx

0.5 in

0.75 in1

0.5 in

1.75 in

0.5 in0.75 in

L5x3x 12

Problem 8.10 SolutionDraw Mohr’s circle.y

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

X (9.45, -2.81)

Ix, Iy (in4)

Ixy (in4)

Y (2.58, +2.81)

F tan 2m = = FXCF

2.819.45-6.02

m = 19.6o

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

(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

Ix, Iy (in4)

Ixy (in4)

2 = 60o

6.02 in4

R = 4.44 in4

80.7o Ix’y’ = EX’ = _ 4.44 sin 80.7o

Ix’y’ = _ 4.38 in439.3o

Problem 8.10 Solutiony

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)

F ImaxImin

Imax, min = OC + R = 6.02 + 4.44

R = 4.44 in4

Imax = 10.46 in4

Imin = 1.574 in4

m = 19.6o

a axis corresponds to Imax

b axis corresponds to Imin

120 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.

120 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

120 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

120 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

Divide the body into sections.x

120 mm

150 mm

120 mm

150 mm150 mm

y’’

x’’

z’’

120 mm

150 mm

120 mm

150 mm150 mm

y’’

x’’

z’’

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

120 mm

150 mm

120 mm

150 mm150 mm

y’’

x’’

z’’

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

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

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

120 mm

150 mm

120 mm

150 mm150 mm

y’’

x’’

z’’

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

120 mm

150 mm

120 mm

150 mm150 mm

y’’

x’’

z’’

(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

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

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

120 mm

150 mm

120 mm

150 mm150 mm

y’’

x’’

z’’

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

120 mm

150 mm

120 mm

150 mm150 mm

y’’

x’’

z’’

(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

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

120 mm

150 mm

120 mm

150 mm150 mm

y’’

x’’

z’’

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

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

40 mm50 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).

15 mm15 mm

40 mm50 mm

Problem 8.12

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

40 mm50 mm

Problem 8.12

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

40 mm50 mm

Divide the body into sections.

15 mm15 mm

40 mm50 mm

Compute the mass momentof inertia of each section.

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

15 mm15 mm

40 mm50 mm

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

40 mm50 mm

x’130 mm

x’’

4r3 4.453 = 19.1 mm=

(centroidal axis) C

xx’x’’

x’130 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

15 mm15 mm

40 mm50 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

x’60 mm

15 mm15 mm

40 mm50 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 = 0.1107 m

Im 3.81 x 10-2 kg . m2

3.11 kg

Radius of gyration:

© 2004 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics...

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