Moment Inertia.pdf

Chapter 9: Distributed Forces: Moments of Inertia 최해진 [email protected]

School of Mechanical Engineering

Contents Introduction Moments of Inertia of an Area Moment of Inertia of an Area by Integration Polar Moment of Inertia Radius of Gyration of an Area Sample Problem 9.1 Sample Problem 9.2 Parallel Axis Theorem Moments of Inertia of Composite Areas Sample Problem 9.4 Sample Problem 9.5 Product of Inertia Principal Axes and Principal Moments of Inertia

Sample Problem 9.6 Sample Problem 9.7 Mohr’s Circle for Moments and Products of Inertia Sample Problem 9.8 Moment of Inertia of a Mass Parallel Axis Theorem Moment of Inertia of Thin Plates Moment of Inertia of a 3D Body by Integration Moment of Inertia of Common Geometric Shapes Sample Problem 9.12 Moment of Inertia With Respect to an Arbitrary Axis Ellipsoid of Inertia. Principle Axes of Axes of Inertia of a Mass School of Mechanical Engineering

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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. School of Mechanical Engineering 9- 3

Moment of Inertia of an Area r • Consider distributed forces DF whose magnitudes are proportional to the elemental areas DA on which they act and also vary linearly with the distance of DA 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. r DF = kyDA R = k ò y dA = 0 M = k ò y 2 dA

ò y dA = Qx = first moment 2 ò y dA = second moment

• Example: Consider the net hydrostatic force on a submerged circular gate. DF = pDA = gyDA R = g ò y dA M x = g ò y 2 dA

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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, I x = ò y 2 dA

I y = ò x 2 dA

• 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, 2

h

I x = ò y dA = ò y 2bdy = 13 bh 3 0

• The formula for rectangular areas may also be applied to strips parallel to the axes, dI x = 13 y 3 dx

dI y = x 2 dA = x 2 y dx School of Mechanical Engineering 9- 5

Polar Moment of Inertia • The polar moment of inertia is an important parameter in problems involving torsion of cylindrical shafts and rotations of slabs. J 0 = ò r 2 dA

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

(

)

J 0 = ò r 2 dA = ò x 2 + y 2 dA = ò x 2 dA + ò y 2 dA = I y + Ix

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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. I I x = k x2 A kx = x A kx = radius of gyration with respect to the x axis • Similarly, Iy =

k y2 A

ky =

J O = kO2 A kO =

Iy A JO A

kO2 = k x2 + k y2 School of Mechanical Engineering 9- 7

Sample Problem 9.1 SOLUTION: • A differential strip parallel to the x axis is chosen for dA. dI x = y 2 dA dA = l dy • For similar triangles,

Determine the moment of inertia of a triangle with respect to its base.

l h- y = b h

l =b

h- y h

dA = b

h- y dy h

• Integrating dIx from y = 0 to y = h, h- y bh 2 I x = ò y dA = ò y b dy = ò hy - y 3 dy h h0 0 2

h

(

2

h

b é y3 y 4 ù = êh - ú hë 3 4û 0

)

bh3 I x= 12 School of Mechanical Engineering 9- 8

Sample Problem 9.2 SOLUTION: • An annular differential area element is chosen, dJ O = u 2 dA

dA = 2p u du r

r

2

J O = ò dJ O = ò u (2p u du ) = 2p ò u 3du 0

0

JO =

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.

p 2

r4

• From symmetry, Ix = Iy, JO = I x + I y = 2I x

p 2

r 4 = 2I x I diameter = I x =

p 4

r4

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Parallel Axis Theorem • Consider moment of inertia I of an area A with respect to the axis AA’ I = ò y 2 dA

• The axis BB’ passes through the area centroid and is called a centroidal axis. I = ò y 2 dA = ò ( y ¢ + d )2 dA = ò y ¢ 2 dA + 2d ò y ¢dA + d 2 ò dA I = I + Ad 2

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,

( )

I T = I + Ad 2 = 14 p r 4 + p r 2 r 2 = 54 p r 4

• Moment of inertia of a triangle with respect to a centroidal axis, I AA¢ = I BB¢ + Ad 2

( )2

1 bh 3 - 1 bh 1 h I BB¢ = I AA¢ - Ad 2 = 12 2 3 1 bh 3 = 36

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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 9.4 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 The strength of a W200 x 46.1 w W360 x 44 rolled steel beam is increased by attaching composite section centroidal axis. 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.

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

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Sample Problem 9.4 SOLUTION: • Determine location of the centroid of composite section with respect to a coordinate system with origin at the centroid of the beam section.

Y å A = å yA

Y =

.0008184 = 0.08 m .01013

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Sample Problem 9.4 • Apply the parallel axis theorem to determine moments of inertia of beam section and plate with respect to composite section centroidal axis.

I x¢,beam section = I x + AY 2 = 122 + (0.00573)(0.08) 2 = 2158.67 ´ 10 6 m 4 I x¢,plate = I x + Ad 2 = 121 (0.22) (0.02) 3 + (0.0044) (0.186 - 0.08) 2 = 4.95 ´ 10 -5 m 4 I x¢ = I x¢,beam section + I x¢,plate = 2158.67 ´ 10 6 + 4.95 ´ 10 -5 I x¢ = 208.17 ´ 10 -6 m 4 • Calculate the radius of gyration from the moment of inertia of the composite section. -6

I x¢ 208.17 ´ 10 m 4 = k x¢ = A .00573

k x¢ = 0.143 m School of Mechanical Engineering 9 - 16

Sample Problem 9.5 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. Determine the moment of inertia of the shaded area with respect to the x axis.

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Sample Problem 9.5 SOLUTION: • Compute the moments of inertia of the bounding rectangle and half-circle with respect to the x axis. Rectangle: I x = 13 bh3 = 13 (240 )(120 ) = 138.2 ´ 106 mm 4

Half-circle: moment of inertia with respect to AA’, I AA¢ = 18 pr 4 = 18 p (90 )4 = 25.76 ´ 106 mm 4

moment of inertia with respect to x’, 4r (4 )(90 ) a= = = 38.2 mm 3p 3p b = 120 - a = 81.8 mm 2

2

A = 12 pr = 12 p (90 )

= 12.72 ´ 103 mm 2

(

)(

I x¢ = I AA¢ - Aa 2 = 25.76 ´ 106 12.72 ´ 103

)

= 7.20 ´ 106 mm 4

moment of inertia with respect to x,

(

)

I x = I x¢ + Ab 2 = 7.20 ´ 106 + 12.72 ´ 103 (81.8)2 = 92.3 ´ 106 mm 4 School of Mechanical Engineering 9 - 18

Sample Problem 9.5 • 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.

Ix

=

138.2 ´ 106 mm 4

-

92.3 ´ 106 mm 4

I x = 45.9 ´ 106 mm 4

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Product of Inertia • Product of Inertia: I xy = ò xy dA

• 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: I xy = I xy + x yA

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Principal Axes and Principal Moments of Inertia • The change of axes yields Ix + I y Ix - I y I x¢ = + cos 2q - I xy sin 2q 2 2 Ix + I y Ix - I y I y¢ = cos 2q + I xy sin 2q 2 2 Ix - I y I x¢y¢ = sin 2q + I xy cos 2q 2 Given I x = ò y 2 dA I y = ò x 2 dA I xy = ò xy dA

we wish to determine moments and product of inertia with respect to new axes x’ and y’. Note: x¢ = x cosq + y sin q y ¢ = y cosq - x sin q

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

(I x¢ - I ave )2 + I x2¢y¢ = R 2 Ix + I y I ave = 2

æ Ix - I y ö 2 ÷÷ + I xy R = çç è 2 ø

• The equations for Iy’ and Ix’y’ lead to the same circle. School of Mechanical Engineering

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Principal Axes and Principal Moments of Inertia • At the points A and B, Ix’y’ = 0 and Ix’ is a maximum and minimum, respectively. I max,min = I ave ± R tan 2q m = -

2 I xy Ix - I y

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

(I x¢ - I ave )2 + I x2¢y¢ = R 2 Ix + I y I ave = 2

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

æ Ix - I y ö 2 ÷÷ + I xy R = çç è 2 ø School of Mechanical Engineering 9 - 22

Sample Problem 9.6 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.

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. School of Mechanical Engineering 9 - 23

Sample Problem 9.6 SOLUTION: • Determine the product of inertia using direct integration with the parallel axis theorem on vertical differential area strips æ xö æ xö y = hç1 - ÷ dA = y dx = hç1 - ÷dx è bø è bø æ xö xel = x yel = 12 y = 12 hç1 - ÷ è bø Integrating dIx from x = 0 to x = b, b

I xy = ò dI xy = ò xel yel dA = ò x 0 3

2æ

b

2

é x 2 x3 x 4 ù x ö÷ 2 çx x =h ò + dx = h ê - + 2 ú ç 2 b 2b 2 ÷ 0è ø ë 4 3b 8b û 0 bæ

2

2

xö h ç1 - ÷ dx è bø

() 1 2

1 b2h2 24 School of Mechanical Engineering 9 - 24

I xy =

Sample Problem 9.6 • Apply the parallel axis theorem to evaluate the product of inertia with respect to the centroidal axes.

x = 13 b

y = 13 h

With the results from part a, I xy = I x¢¢y ¢¢ + x yA I x¢¢y ¢¢ =

1 b2h2 24

( )(13 h)(12 bh)

- 13 b

1 b2h2 I x¢¢y ¢¢ = - 72

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Sample Problem 9.7 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). For the section shown, the moments of inertia with respect to the x and y axes are Ix = 4.32 x 106 m4 and Iy = 2.901 x 10-6 m4. 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. School of Mechanical Engineering 9 - 26

Sample Problem 9.7 SOLUTION: • Compute the product of inertia with respect to the xy axes by dividing the section into three rectangles. Apply the parallel axis theorem to each rectangle,

(

I xy = å I x¢y ¢ + x yA

)

Note that the product of inertia with respect to centroidal axes parallel to the xy axes is zero for each rectangle.

Rectangle Area, cm 2 x , cm y , cm I 1.5 - 1.25 + 1.75 II 1.5 0 0 III 1.5 + 1.25 - 1.75

x yA, cm 4 - 3.28 0 - 3.28 å x yA = -6.56

I xy = å x yA = -2.723 x 10 -6 m 4 School of Mechanical Engineering 9 - 27

Sample Problem 9.7 • Determine the orientation of the principal axes (Eq. 9.25) and the principal moments of inertia (Eq. 9. 27).

q m = 37.7° and q m = 127.7° I max,min =

Ix + Iy 2

æ Ix - Iy ± çç è 2

2

ö ÷÷ + I xy2 ø

I a = I max = 6.4244 x 10 -6 m 4 I b = I min = 10.7966 x 10 -6 m 4 School of Mechanical Engineering 9 - 28

Mohr’s Circle for Moments and Products of Inertia • The moments and product of inertia for an area are plotted as shown and used to construct Mohr’s circle, I ave =

Ix + I y 2

æ Ix - I y ö 2 ÷÷ + I xy R = çç è 2 ø

• 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 9.8 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. 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

• Based on the circle, evaluate the moments and product of inertia with respect to the x’y’ axes.

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Sample Problem 9.8 SOLUTION: • Plot the points (Ix , Ixy) and (Iy ,-Ixy). Construct Mohr’s circle based on the circle diameter between the points.

(

)

OC = I ave = 12 I x + I y = 4.925 ´ 106 mm 4

(

)

CD = 12 I x - I y = 2.315 ´ 106 mm 4 R=

I x = 7.24 ´ 106 mm 4

I xy = -2.54 ´ 10 mm

= 3.437 ´ 106 mm 4

• Based on the circle, determine the orientation of the principal axes and the principal moments of inertia.

I y = 2.61 ´ 106 mm 4 6

(CD )2 + (DX )2

4

tan 2q m =

DX = 1.097 2q m = 47.6° CD

q m = 23.8°

I max = OA = I ave + R

I max = 8.36 ´ 106 mm 4

I min = OB = I ave - R

I min = 1.49 ´ 106 mm 4 School of Mechanical Engineering 9 - 31

Sample Problem 9.8 • 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 Q = 2(60o) = 120o. The angle that CX’ forms with the x’ axes is f = 120o - 47.6o = 72.4o. I x ' = OF = OC + CX ¢ cos j = I ave + R cos 72.4o I x¢ = 5.96 ´ 106 mm 4

I y ' = OG = OC - CY ¢ cos j = I ave - R cos 72.4o I y ¢ = 3.89 ´ 106 mm 4

I x¢y ' = FX ¢ = CY ¢ sin j = R sin 72.4o OC = I ave = 4.925 ´ 106 mm 4 R = 3.437 ´ 106 mm 4

I x¢y ¢ = 3.28 ´ 106 mm 4 School of Mechanical Engineering 9 - 32

Moment of Inertia of a Mass • Angular acceleration about the axis AA’ of the small mass Dm due to the application of a couple is proportional to r2Dm. r2Dm = moment of inertia of the mass Dm with respect to the axis AA’ • For a body of mass m the resistance to rotation about the axis AA’ is I = r12 Dm + r22 Dm + r32 Dm + L = ò r 2 dm = mass moment of inertia

• The radius of gyration for a concentrated mass with equivalent mass moment of inertia is I = k 2m

k=

I m School of Mechanical Engineering 9 - 33

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

(

)

I y = ò r 2 dm = ò z 2 + x 2 dm

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

( ) I z = ò (x 2 + y 2 )dm I x = ò y 2 + z 2 dm

• In SI units,

(

I = ò r 2 dm = kg × m 2

)

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

( ) [ ] = ò (y ¢ 2 + z ¢ 2 )dm + 2 y ò y ¢dm + 2 z ò z ¢dm + (y 2 + z 2 )ò dm I x = I x ¢ + m (y 2 + z 2 ) I y = I y¢ + m(z 2 + x 2 ) I z = I z¢ + m(x 2 + y 2 ) I x = ò y 2 + z 2 dm = ò ( y ¢ + y )2 + ( z ¢ + z )2 dm

• Generalizing for any axis AA’ and a parallel centroidal axis, I = I + md 2

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Moments of Inertia of Thin Plates • For a thin plate of uniform thickness t and homogeneous material of density r, the mass moment of inertia with respect to axis AA’ contained in the plate is I AA¢ = ò r 2 dm = rt ò r 2 dA = r t I AA¢,area

• Similarly, for perpendicular axis BB’ which is also contained in the plate, I BB¢ = r t I BB¢,area

• For the axis CC’ which is perpendicular to the plate, I CC ¢ = r t J C ,area = r t (I AA¢,area + I BB¢,area ) = I AA¢ + I BB¢ School of Mechanical Engineering 9 - 36

Moments of Inertia of Thin Plates • For the principal centroidal axes on a rectangular plate,

(121 a 3b) = 121 ma 2 1 ab 3 ) = 1 mb 2 I BB¢ = r t I BB¢,area = r t (12 12 1 m(a 2 + b 2 ) I CC ¢ = I AA¢,mass + I BB¢,mass = 12 I AA¢ = r t I AA¢,area = r t

• For centroidal axes on a circular plate,

(

)

I AA¢ = I BB¢ = r t I AA¢,area = r t 14 p r 4 = 14 mr 2 I CC ¢ = I AA¢ + I BB¢ = 12 mr 2 School of Mechanical Engineering 9 - 37

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 I = r ò r 2 dV

• 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. School of Mechanical Engineering 9 - 38

Moments of Inertia of Common Geometric Shapes

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Sample Problem 9.12 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. Determine the moments of inertia of the steel forging with respect to the xyz coordinate axes, knowing that the specific weight of steel is 7896 kg/m3 10,000 N/m3 . School of Mechanical Engineering 9 - 40

Sample Problem 9.12 SOLUTION: • Compute the moments of inertia of each component with respect to the xyz axes.

cylinders

(a = .025 , L = .076m, x = .06m, y = .05m ) :

I x = 12 ma 2 + my 2 2 2 = 12 (.15kg )(.025) + (.15 kg )(.05 m ) = 42.1875 x 10 -5 kgm 2

[

]

I y = 121 m 3a 2 + L2 + mx 2

[

2

2

]

= 121 (.15kg ) 3(.025 m ) + (.076 m ) + (.15 kg )(.06 m )

2

= 63.56375 x 10 -5 kgm 2

[

] [

I y = 121 m 3a 2 + L2 + m x 2 + y 2 each cylinder : m = gV = (10000) p x 0.025 2 x .076

(

m = 0.15 kg

[

]

2

2

]

= 121 (.15 kg ) 3 x (.025m ) + (.076 m ) +

)

(.15 kg )([ .06 m )2 + (.076m )2 ] = 150.20375 x 10 -5 kgm 2 School of Mechanical Engineering 9 - 41

Sample Problem 9.12 prism (a = 0.5m, b = 0.15m, c = 0.05 m):

[

]

[

]

[

2

I x = I z = 121 m b 2 + c 2 = 121 x .38 (0.15 m ) + (0.5 m ) = 7.9167 x 10 - 4 kgm 2

[

2

I y = 121 m c 2 + a 2 = 121 (0.38 kg )(0.05) + (0.05) = 1.5833 x 10 -4 kgm 2

2

2

]

]

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

prism : m = gV = 10,000 kg/m 3 (.05 m) (.05 m) (.05 m) = 3.75 N

(

)

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Moment of Inertia With Respect to an Arbitrary Axis • IOL = moment of inertia with respect to axis OL r r2 2 I OL = ò p dm = ò l ´ r dm r

r • Expressing l and r in terms of the vector components and expanding yields

I OL = I x l2x + I y l2y + I z l2z - 2 I xy l x l y - 2 I yz l y l z - 2 I zx l z l x

• The definition of the mass products of inertia of a mass is an extension of the definition of product of inertia of an area I xy = ò xy dm = I x¢y¢ + mx y I yz = ò yz dm = I y¢z¢ + myz I zx = ò zx dm = I z¢x¢ + mz x School of Mechanical Engineering 9 - 43

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 OQ = 1 I OL • 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.

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