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12. Moments of Inertia 12.1. Moment of Inertia of an Area The moment of inertia I of an element of area about an axis in its plane is the product of the area of the element and the square of its distance from the axis. The moment of inertia[1] is also called the second moment of the area. In Fig. 12-1, the moments of inertia of the differential area element are

(12.1) The moment of inertia of an area is the sum of the moments of inertia of its elements:

(12.2) The radius of gyration k of an area with respect to an axis is defined by

(12.3)

12.2. Polar Moment of Inertia of an Area The polar moment of inertia J of an element about an axis perpendicular to its plane is the product of the area of the element and the square of its distance from the axis. This can also be thought of as the second moment about the z-axis. In Fig. 12-1, the polar moment of inertia of the area element is

(12.4)

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Figure 12-1 An area element used in the definition of the second moment of an area.

The polar moment of inertia of an area is the sum of the polar moments of inertia of its elements:

(12.5)

12.3. Product of Inertia of an Area The product of inertia of an element of area in the figure is defined as

(12.6) The product of inertia of an area is the sum of the products of inertia of its elements:

(12.7)

12.4. Parallel Axis Theorem The parallel axis theorem states that the axial or polar moment of inertia of an area about any axis equals the moment of inertia Ī of the area about a parallel axis through the centroid of the area plus the product of the area and the square of the distance between the two parallel axes. In Fig. 12-2, x and y are any axes through O, while x′ and y′ are coplanar parallel axes through the centroid C. The parallel axis theorem then states that

(12.8)

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Figure 12-2 The axes used in the parallel axis theorems.

The product of inertia of an area with respect to any two axes equals the product of inertia about two parallel centroidal axes plus the product of the area and the distances from respective axes:

(12.9) where b and d are the coordinates of C relative to the (x, y) axes through O, or the coordinates of O relative to the (x′, y′) axes through C. In the first case, d and b are positive, whereas in the second case, they are negative. In either case, their product is positive. See Fig. 12-2.

12.5. Composite Area The axial or polar moment of inertia, or product of inertia, of a composite area is the sum of the axial or polar moments of inertia, or products of inertia, of the component areas making up the whole. Units of any of the foregoing are the fourth power of length. In SI, the units are m4 or mm4 . In the U.S. Customary system, the units are usually in 4 . In using mm 4 , it is convenient to use 10 6 mm4 (particularly in tables listing properties in SI units). Moments of inertia of a number of common areas are listed in Table 12.1. They can be used in certain problems.

12.6. Rotated Set of Axes The moment of inertia of any area with respect to a rotated set of axes (x′, y′) may be expressed in terms of the moments and product of inertia with respect to the (x, y) axes as follows:

(12.10) where

I x , I y = moments of inertia with respect to the (x, y) axes I x′ , I y′ = moments of inertia with respect to the (x′, y′) axes, which have the same origin as the (x, y) axes but are rotated through an angle θ

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I xy = product of inertia with respect to the (x, y) axes I x′y′ = product of inertia with respect to the (x′, y′) axes For a proof of this, refer to Problem 12.21. Maximum moments of inertia of any area occur with respect to principal axes: a special set of (x′, y′) axes for which 2θ′ = tan [−2I xy/(I x − I y )]:

−1

(12.11) For details, refer to Problem 12.22.

12.7. Mohr’s Circle Mohr’s circle is a device that makes it unnecessary to memorize the formulas connected with rotation of axes. Refer to Problems 12.23 and 12.24.

12.8. Moment of Inertia of a Mass The moment of inertia of an element of mass is the product of the mass of the element and the square of the distance of the element from the axis. The moment of inertia of a mass is the sum of the axial moments of all its elements. Thus, for a mass of which dm is one element with coordinates (x, y, z), the following definitions hold (see Fig. 12-3):

(12.12) where I x, I y, I z = axial moments of inertia (with respect to the x-, y-, and z-axes, respectively). For a thin plate essentially in the xy-plane, the following relations hold:

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(12.13) where I x, I y = axial moments of inertia about the x- and y-axes, respectively, and J o = axial moment of inertia about the z-axis. Figure 12-3 The mass element used for the moment of inertia of a mass

Figure 12-4 The mass element used for the product of inertia of a mass

The radius of gyration k of a body with respect to an axis is

(12.14) that is, the square root of the quotient of its moment of inertia divided by the mass.

12.9. Product of Inertia of a Mass The product of inertia of a mass is the sum of the products of inertia of its elements (seeFig. 12-4):

(12.15)

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The product of inertia is zero if one, or both, of the reference axes is an axis of symmetry.

12.10. Parallel Axis Theorem for a Mass The parallel axis theorem states that the moment of inertia of a body about an axis is equal to the moment of inertia Ī about a parallel axis through the center of gravity of the body plus the product of the mass of the body and the square of the distance between the two parallel axes (refer to Fig. 12-2):

(12.16)

12.11. Composite Mass The moments of inertia, or product of inertia, of a composite mass, with respect to an axis, is the sum of the moments of inertia, or products of inertia, of the component masses, with respect to the same axes. Units of all the foregoing moments involve those of mass and square of length. In SI, the units are kg · m2 . In the U.S. Customary system, the units are lb-sec 2 -ft. Centroidal moments of inertia of masses for some common shapes are listed in Table 12-2 with reference to problems in which these results are derived.

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Table 12.1 Moments of Inertia of Areas FIGURE

AREA AND MOMENT

FIGURE

RECTANGLE

A = bh

TRIANGLE

HOLLOW RECTANGLE

A = BH − bh

ELLIPSE

CIRCLE

A = πr 2

SEMIELLIPSE

AREA AND MOMENT

A = πab

I

SEMICIRCLE

x = 0.11ab

3

I x = 0.055r

4

QUARTER CIRCLE I x = 0.11r

4

Table 12.2 Centroidal Moments of Inertia of Masses PROBLEM

FIGURE

NAME

Ix

Iy

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Iz

12.26

Slender bar

12.28

Rectangular parallelepiped

12.29

Thin circular disk

12.32

Right circular cylinder

—

12.33

12.34

Sphere

12.36

Right circular cone

12.12. SOLVED PROBLEMS © McGraw-Hill Education. All rights reserved. Any use is subject to the Terms of Use, Privacy Notice and copyright information.

12.1. Determine the moment of inertia of the rectangle with base b and altitude h about a centroidal axis parallel to the base. Refer to Fig. 12-5.SolutionChoose an element of area dA parallel to the base and at a distance y from the centroidal x-axis as in Fig. 12-6:

Figure 12-5

Figure 12-6

12.2. Determine the moment of inertia of the rectangle in Problem 12.1 with respect to the base. Refer to Fig. 12-7.Solution

Figure 12-7

12.3. Determine the moment of inertia of a rectangle with respect to its base by means of the parallel axis theorem. See Fig. © McGraw-Hill Education. All rights reserved. Any use is subject to the Terms of Use, Privacy Notice and copyright information.

12-8. Assume that the results of Problem 12.1 are known.Solution

Figure 12-8

12.4. Determine the moment of inertia for a triangle of base b and height h about a centroidal axis parallel to the base. Refer to Fig. 12-9 and Table 12.1.Solution

since by similar triangles,

Figure 12-9

12.5. Determine the moment of inertia for a triangle of base b and height h about the base. Refer to Fig. 12-10.Solution

since, by similar triangles,

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Figure 12-10

12.6. Knowing the results of Problem 12.5 (a comparatively simple integration problem), find the moment of inertia of the triangle about a centroidal axis parallel to the base. See Fig. 12-11.SolutionThis is an application of the parallel axis theorem (12.8) in reverse:

Figure 12-11

12.7. Determine the moment of inertia for a circle of radius r about a diameter. Refer to Fig. 12-12.Solution

Note that this integral could have been evaluated from −r to r instead of as twice the integral from 0 to r. This is permissible since the moment of inertia of the two halves of the area is equal to the moment of inertia of the whole.

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Figure 12-12

12.8. Determine the moment of inertia of a circle of radius r about a diameter, using the differential area dA shown in Fig. 1213.Solution

where y = ρ sin θ and dA = ρ dρ dθ. Then,

This problem illustrates the calculation of the moment of inertia with a different choice of the area element. Figure 12-13

12.9. Determine the polar moment of inertia for a circle of radius r about an axis through its center and perpendicular to the plane of the circle. Refer to Fig. 12-14.SolutionChoose as the differential area an annular ring of radius ρ and thickness dρ. Hence, dA equals the circumference 2πρ times the thickness dρ:

At this point, it is possible to derive the value of the axial moment of inertia about a diameter. Since

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Figure 12-14

12.10. Determine the moments of inertia I x , I y , and

where x 2/ a 2 + y 2 /b 2 = 1 or

for the ellipse shown in Fig. 12-15.Solution

. Then (a table of integrals may be helpful)

A similar integration would yield

Then,

Figure 12-15

12.11. Determine the moment of inertia of the T-section shown in Fig. 12-16 about the centroidal axis parallel to the base.SolutionThe first step is to locate the centroid C using the two rectangles shown. Hence,

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Next, determine I for each rectangle about its own centroidal axis parallel to the base, using

:

The final step is to transfer from each subdivision’s centroidal axis to the axis through C to determine Ī for the entire area. By the parallel axis theorem (d 1 = 36 mm; d 2 = 54 mm),

Figure 12-16

12.12. Determine the moment of inertia about a centroidal axis parallel to the base of the composite area shown in Fig. 1217.SolutionThe first step is the location of the centroid of the composite area. Let T represent the top rectangular area, B the bottom rectangular area, and C the circular area. Using the base as the reference line, we have

The distance d T from the centroid of the top area to the common centroid is

Similarly,

The values of I for each component area about that area’s centroidal axis parallel to the base of the composite area are as follows:

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Finally, using the parallel axis theorem,

Figure 12-17

12.13. Determine the moment of inertia for the channel shown in Fig. 12-18 about a centroidal axis parallel to the base b.SolutionIn this case, the centroidal axis is, by symmetry, at half the height above the base. Consider the channel to be made up of a rectangle of base b and altitude h from which have been deleted two triangles of altitudet and base a and a rectangle of base a and height 2d. Refer to Fig. 12-19. To determine I x for the channel, subtract I x of the triangles and the smaller rectangle from I x for the large rectangle. That is,

(1) Then, using Table 12.1 and the parallel axis theorem,

Finally, equation (1) provides

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Of course, this result will take on a numerical value when numbers are assigned to the dimensions. Figure 12-18

Figure 12-19

12.14. A column is built up of 40-mm planks, as shown in Fig. 12-20. Determine the moment of inertia about a centroidal axis parallel to a side.SolutionThe centroid is located by inspection at the midpoint. To determine Ī, it is only necessary to double the summation of the axial moments of areas 1 and 2 about a line through the midpoint.

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Another easier technique is to subtract I for the inner square from I for the outer square, both about an axis through the midpoint and parallel to a side:

Figure 12-20

12.15. Determine the moment of inertia about the horizontal centroidal axis of the Z-section shown in Fig. 12-21. What is the radius of gyration?SolutionThe centroidal axis is an axis of symmetry in this case. The moment of inertia Ī is then equal to the sum of the second moments of area 2 and two areas 1:

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Figure 12-21

12.16. What is the product of inertia about two adjacent sides of a rectangle of baseb and altitude h?SolutionAs shown in Fig. 12-22(a), the x- and y-axes are along two adjacent sides. Denote the product of inertia as I xy . Then

Next, choose the y-axis as the right side of the rectangle [see Fig. 12-22(b)]. Then

This indicates that the product of inertia may be positive or negative, depending on the location of the area relative to the axes. Figure 12-22

12.17. Determine the product of inertia about two centroidal axes parallel to the sides of a rectangle of baseb and height h.SolutionIn Fig. 12-23, it is seen that the x′-limits of integration are from −b/2 to b/2. The y′-limits of integration are from −h/2 to h/2. Hence,

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This could also be deduced from the result of Problem 12.16 using the parallel axis theorem for product of inertia:

where

and

are the perpendicular distances between the x, y- and x′, y′-axes:

Figure 12-23

12.18. Determine the product of inertia with respect to the base and altitude of a right triangle. Refer to Fig. 12-24.SolutionBy definition,

The upper limit of the x-integration depends on y. Therefore, it must be evaluated from the equation of the sloping line, which is

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Figure 12-24

12.19. Determine the product of inertia with respect to the bounding radii of a quadrant of a circle of radius r. Use ( a) the element of the area shown in Fig. 12-25, and ( b) the element of the area shown in Fig. 12-26.Solution (a) By definition,

Here the double integration means a summation first with respect to the variable x, which depends on y according to the equation x 2 + y 2 = r 2 . Substituting,

(b) As before,

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Figure 12-25

Figure 12-26

12.20. In Problem 12.19, determine the product of inertia for a quadrant of a circle of radius r equal to 50 mm.Solution

12.21. Show that the moments of inertia of an area with respect to the rotated set of axes (x′, y′) may be expressed as follows:

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I x , I y = moments of inertia with respect to (x, y) axes

where

I x′ , I y′ = moments of inertia with respect to (x′, y′) axes I xy = product of inertia with respect to (x, y) axes I x′y′ = product of inertia with respect to (x′, y′) axes Solution Figure 12-27 indicates an element dA of the area. By definition,

(1) But from the figure, x′ = x cos θ + y sin θ and y′ = −x sin θ + y cos θ. Squaring,

Also,

Substituting into equation (1),

Since the integration over the area is independent of θ, the above equations may be written (using I x = ∫ y2 dA, Iy = ∫ x2 dA, and I xy = ∫ xy dA) as

Using

and

, we finally obtain

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Figure 12-27

12.22. Determine the values of I x′ and I y′ in Problem 12.21 with respect to the principal axes, the axes that yield maximum or minimum values of I.SolutionTo determine the value of θ that will make I x′ a maximum, take the derivative of I x′ with respect to θ and equate the resulting expression to zero. Thus,

This is the value θ′ of θ that will make I x ′ a maximum (or minimum). It is necessary to evaluate sin 2θ′ and cos 2θ′; this is most easily done from Fig. 12-28. Thus,

Substituting these values and simplifying, we obtain

By taking the derivative of I y′ with respect to θ, it can be seen that the same value θ′ makes I y ′ a maximum (or minimum). Substituting the values of sin 2θ′ and cos 2θ′ yields

Since I y ′ has a negative s...