moment of inertia

Moments of Inertia

CHAPTER OBJECTIVES• To develop a method for determining the moment of inertia for

an area.

• To introduce the product of inertia and show how to determine themaximum and minimum moments of inertia for an area.

• To discuss the mass moment of inertia.

10.1 Definition of Moments of Inertia for Areas

Whenever a distributed loading acts perpendicular to an area and itsintensity varies linearly, the computation of the moment of the loadingdistribution about an axis will involve a quantity called the moment ofinertia of the area. For example, consider the plate in Fig. 10–1, which issubjected to a fluid pressure p. As discussed in Sec. 9.5, this pressure pvaries linearly with depth, such that , where is the specificweight of the fluid. Thus, the force acting on the differential area ofthe plate is . The moment of this force about the

axis is therefore , and so integrating over theentire area of the plate yields .The integral is calledthe moment of inertia of the area about the axis. Integrals of thisform often arise in formulas used in fluid mechanics, mechanics ofmaterials, structural mechanics, and mechanical design, and so theengineer needs to be familiar with the methods used for theircomputation.

xIx

1y2dAM = g1y2dAdMdM = y dF = gy2dAx

dF = p dA = (g y)dAdA

gp = gy

10

y

x

z

y

dF

dAp ! gy

Fig. 10–1

512 CH A P T E R 10 MO M E N T S O F IN E RT I A

10

Moment of Inertia. By definition, the moments of inertia of adifferential area dA about the x and y axes are and

respectively, Fig. 10–2. For the entire area A the momentsof inertia are determined by integration; i.e.,

(10–1)

We can also formulate this quantity for dA about the “pole” O orz axis, Fig. 10–2. This is referred to as the polar moment of inertia. It isdefined as where r is the perpendicular distance from thepole (z axis) to the element dA. For the entire area the polar moment ofinertia is

(10–2)

This relation between and is possible since Fig. 10–2.

From the above formulations it is seen that and will alwaysbe positive since they involve the product of distance squared and area.Furthermore, the units for moment of inertia involve length raised to thefourth power, e.g., or

10.2 Parallel-Axis Theorem for an Area

The parallel-axis theorem can be used to find the moment of inertia of anarea about any axis that is parallel to an axis passing through the centroidand about which the moment of inertia is known.To develop this theorem,we will consider finding the moment of inertia of the shaded area shownin Fig. 10–3 about the x axis.To start, we choose a differential element dAlocated at an arbitrary distance from the centroidal axis. If thedistance between the parallel x and axes is then the moment ofinertia of dA about the x axis is . For the entire area,

= LAy¿2 dA + 2dyLA

y¿ dA + dy2LA

dA

Ix = LA1y¿ + dy22 dA

dIx = 1y¿ + dy22 dAdy,x¿

x¿y¿

in4.ft4,mm4,m4,

JOIy,Ix,

r2 = x2 + y2,IyIx,JO

JO = LAr2 dA = Ix + Iy

dJO = r2 dA,

Ix = LAy2 dA

Iy = LAx2 dA

dIy = x2 dA,dIx = y2 dA

Ox

y

y

x

r

dA

A

Fig. 10–2

Ox

y

d

dx

dy

x¿

y"

x¿

y¿dA

C

Fig. 10–3

10.3 RADIUS OF GYRATION OF AN AREA 513

10

The first integral represents the moment of inertia of the area about thecentroidal axis, The second integral is zero since the axis passesthrough the area’s centroid C; i.e., sinceSince the third integral represents the total area A, the final result istherefore

(10–3)

A similar expression can be written for i.e.,

(10–4)

And finally, for the polar moment of inertia, since and, we have

(10–5)

The form of each of these three equations states that the moment ofinertia for an area about an axis is equal to its moment of inertia about aparallel axis passing through the area’s centroid plus the product of thearea and the square of the perpendicular distance between the axes.

10.3 Radius of Gyration of an Area

The radius of gyration of an area about an axis has units of length and isa quantity that is often used for the design of columns in structuralmechanics. Provided the areas and moments of inertia are known, the radiiof gyration are determined from the formulas

(10–6)

The form of these equations is easily remembered since it is similar tothat for finding the moment of inertia for a differential area about an axis. For example, whereas for a differential area,dIx = y2 dA.

Ix = kx2A;

kO = DJO

A

ky = DIy

A

kx = DIx

A

JO = JC + Ad2

d2 = d2x + d2

y

JC = Ix¿ + Iy¿

Iy = Iy¿ + Adx2

Iy;

Ix = Ix¿ + Ady2

y¿ = 0.1y¿ dA = y¿1dA = 0x¿Ix¿.

In order to predict the strength anddeflection of this beam, it is necessary tocalculate the moment of inertia of thebeam’s cross-sectional area.


10

y

(a)

y

x

dy

x

(x, y)

y ! f(x)

dA

x

(b)

y

x

y

dx

(x, y)

dA

y ! f(x)

Procedure for Analysis

In most cases the moment of inertia can be determined using asingle integration. The following procedure shows two ways inwhich this can be done.• If the curve defining the boundary of the area is expressed as

, then select a rectangular differential element such thatit has a finite length and differential width.

• The element should be located so that it intersects the curve atthe arbitrary point (x, y).

Case 1• Orient the element so that its length is parallel to the axis about

which the moment of inertia is computed.This situation occurs whenthe rectangular element shown in Fig. 10–4a is used to determine for the area.Here the entire element is at a distance y from the x axissince it has a thickness .Thus .To find , the elementis oriented as shown in Fig. 10–4b. This element lies at the samedistance x from the y axis so that .

Case 2• The length of the element can be oriented perpendicular to the axis

about which the moment of inertia is computed; however, Eq. 10–1does not apply since all points on the element will not lie at the samemoment-arm distance from the axis. For example, if the rectangularelement in Fig. 10–4a is used to determine , it will first benecessary to calculate the moment of inertia of the element aboutan axis parallel to the y axis that passes through the element’scentroid, and then determine the moment of inertia of the elementabout the y axis using the parallel-axis theorem. Integration of thisresult will yield . See Examples 10.2 and 10.3.Iy

Iy

Iy = 1x2dA

IyIx = 1y2dAdy

Ix

y = f(x)

Fig. 10–4


10

EXAMPLE 10.1

Determine the moment of inertia for the rectangular area shown inFig. 10–5 with respect to (a) the centroidal axis, (b) the axis passing through the base of the rectangle, and (c) the pole or axisperpendicular to the plane and passing through the centroid C.

SOLUTION (CASE 1)

Part (a). The differential element shown in Fig. 10–5 is chosen forintegration. Because of its location and orientation, the entire elementis at a distance from the axis. Here it is necessary to integratefrom to Since then

Ans.

Part (b). The moment of inertia about an axis passing through thebase of the rectangle can be obtained by using the above result of part(a) and applying the parallel-axis theorem, Eq. 10–3.

Ans.

Part (c). To obtain the polar moment of inertia about point C, wemust first obtain which may be found by interchanging thedimensions b and h in the result of part (a), i.e.,

Using Eq. 10–2, the polar moment of inertia about C is therefore

Ans.JC = Ix¿ + Iy¿ = 112

bh1h2 + b22

Iy¿ = 112

hb3

Iy¿,

= 112

bh3 + bhah2b2

= 13

bh3

Ixb= Ix¿ + Ady

2

Ix¿ = 112

bh3

Ix¿ = LAyœ2 dA = L

h>2-h>2yœ21b dy¿2 = bL

h>2-h>2yœ2 dyœ

dA = b dy¿,y¿ = h>2.y¿ = -h>2 x¿y¿

x¿ –y¿z¿

xbx¿

x¿

y¿

y¿

xb

C

dy¿

b2

b2

h2

h2

Fig. 10–5


10

x

y

200 mm

100 mm

y

xdy

y2 ! 400x

(a)

(100 – x)

x

y

200 mm

x

y

100 mm

dx

x¿

y2 ! 400x

(b)

y !~ y––2

Fig. 10–6

Determine the moment of inertia for the shaded area shown in Fig. 10–6a about the x axis.

SOLUTION I (CASE 1)A differential element of area that is parallel to the x axis, as shown inFig. 10–6a, is chosen for integration. Since this element has a thicknessdy and intersects the curve at the arbitrary point (x, y), its area is

Furthermore, the element lies at the samedistance y from the x axis. Hence, integrating with respect to y, from

to yieldsy = 200 mm,y = 0

dA = 1100 - x2 dy.

EXAMPLE 10.2

Ans.= 10711062 mm4

= L200 mm

0y2a100 -

y2

400b dy = L

200 mm

0a100y2 -

y4

400bdy

Ix = LAy2 dA = L

200 mm

0y21100 - x2 dy

SOLUTION II (CASE 2)A differential element parallel to the y axis, as shown in Fig. 10–6b, ischosen for integration. It intersects the curve at the arbitrary point (x, y).In this case, all points of the element do not lie at the same distancefrom the x axis, and therefore the parallel-axis theorem must be usedto determine the moment of inertia of the element with respect to thisaxis. For a rectangle having a base b and height h, the moment ofinertia about its centroidal axis has been determined in part (a) ofExample 10.1.There it was found that For the differentialelement shown in Fig. 10–6b, and and thus

Since the centroid of the element is from thex axis, the moment of inertia of the element about this axis is

(This result can also be concluded from part (b) of Example 10.1.)Integrating with respect to x, from to yields

Ans.= 10711062 mm4

Ix = LdIx = L100 mm

0

13

y3 dx = L100 mm

0

131400x23>2 dx

x = 100 mm,x = 0

dIx = dIx¿ + dA y'2 = 1

12dx y3 + y dxay

2b2

= 13

y3 dx

y' = y>2dIx¿ = 1

12 dx y3.h = y,b = dx

Ix¿ = 112 bh3.


10

EXAMPLE 10.3

Determine the moment of inertia with respect to the x axis for thecircular area shown in Fig. 10–7a.

SOLUTION I (CASE 1)Using the differential element shown in Fig. 10–7a, sincewe have

Ans.

SOLUTION II (CASE 2)When the differential element shown in Fig. 10–7b is chosen, thecentroid for the element happens to lie on the x axis, and since

for a rectangle, we have

Integrating with respect to x yields

Ans.

NOTE: By comparison, Solution I requires much less computation.Therefore, if an integral using a particular element appears difficult toevaluate, try solving the problem using an element oriented in theother direction.

Ix = La

-a

231a2 - x223>2 dx = pa4

4

= 23

y3 dx

dIx = 112

dx12y23Ix¿ = 112 bh3

= La

-ay2 A22a2 - y2 B dy = pa4

4

Ix = LAy2 dA = LA

y212x2 dy

dA = 2x dy,

x

y

y

x#x

dy

(#x, y)(x, y)

x2 $ y2 ! a2

(a)

O

a

O x

y

a

(x, y)

(x, #y)dx

y

#y

(b)

(x, y)~ ~

x2 $ y2 ! a2

Fig. 10–7


10

y

x

1 m

1 m

y3 ! x2

y

x

1 m

1 m

y3 ! x2

F10–2

y

x

1 m

1 m

y3 ! x2

y

x

1 m

1 m

y3 ! x2

F10–4

F10–3. Determine the moment of inertia of the shadedarea about the y axis.

F10–2. Determine the moment of inertia of the shadedarea about the x axis.

F10–1. Determine the moment of inertia of the shadedarea about the x axis.

F10–4. Determine the moment of inertia of the shadedarea about the y axis.

FUNDAMENTAL PROBLEMS

F10–1 F10–3


10

y

x

2 m

2 m

y ! 0.25 x3

Probs. 10–1/2

•10–5. Determine the moment of inertia of the area aboutthe axis.

10–6. Determine the moment of inertia of the area aboutthe axis.y

x

10–3. Determine the moment of inertia of the area aboutthe axis.

*10–4. Determine the moment of inertia of the area aboutthe axis.y

x

•10–1. Determine the moment of inertia of the area aboutthe axis.

10–2. Determine the moment of inertia of the area aboutthe axis.y

x

10–7. Determine the moment of inertia of the area aboutthe axis.

*10–8. Determine the moment of inertia of the area aboutthe axis.

•10–9. Determine the polar moment of inertia of the areaabout the axis passing through point .Oz

y

x

PROBLEMS

y

x

y2 ! x3 1 m

1 m

Probs. 10–3/4

y

x

y2 ! 2x

2 m

2 m

Probs. 10–5/6

y

xO

y ! 2x4 2 m

1 m

Probs. 10–7/8/9


10

y

x

2 in.

8 in.

y ! x3

Probs. 10–10/11

10–14. Determine the moment of inertia of the area aboutthe x axis. Solve the problem in two ways, using rectangulardifferential elements: (a) having a thickness of dx, and (b) having a thickness of dy.

10–15. Determine the moment of inertia of the area aboutthe y axis. Solve the problem in two ways, using rectangulardifferential elements: (a) having a thickness of dx, and (b) having a thickness of dy.

*10–12. Determine the moment of inertia of the areaabout the x axis.

•10–13. Determine the moment of inertia of the areaabout the y axis.

10–10. Determine the moment of inertia of the area aboutthe x axis.

10–11. Determine the moment of inertia of the area aboutthe y axis.

x

y

1 in.

2 in.y ! 2 – 2 x 3

Probs. 10–12/13

1 in. 1 in.

4 in.

y ! 4 – 4x2

x

y

Probs. 10–14/15 k

*10–16. Determine the moment of inertia of the triangulararea about the x axis.•10–17. Determine the moment of inertia of the triangulararea about the y axis.

y ! (b # x)h––b

y

x

b

h

Probs. 10–16/17


10



*10–20. Determine the moment of inertia of the areaabout the x axis.

•10–21. Determine the moment of inertia of the areaabout the y axis.



*10–24. Determine the moment of inertia of the areaabout the axis.

•10–25. Determine the moment of inertia of the areaabout the axis.

10–26. Determine the polar moment of inertia of the areaabout the axis passing through point O.z

y

x

x

y

b

h

y ! x2 h—b2

Probs. 10–18/19

y

x

y3 ! x2 in.

8 in.

Probs. 10–20/21

y

x

y ! 2 cos ( x)––8

2 in.

4 in.4 in.

π

Probs. 10–22/23

y

x

x2 $ y2 ! r2

r0

0

Probs. 10–24/25/26

For design or analysis of this Tee beam,engineers must be able to locate thecentroid of its cross-sectional area, andthen find the moment of inertia of thisarea about the centroidal axis.


10

10.4 Moments of Inertia for Composite Areas

A composite area consists of a series of connected “simpler” parts orshapes, such as rectangles, triangles, and circles. Provided the moment ofinertia of each of these parts is known or can be determined about acommon axis, then the moment of inertia for the composite area aboutthis axis equals the algebraic sum of the moments of inertia of all its parts.

Procedure for Analysis

The moment of inertia for a composite area about a reference axiscan be determined using the following procedure.

Composite Parts.• Using a sketch, divide the area into its composite parts and

indicate the perpendicular distance from the centroid of eachpart to the reference axis.

Parallel-Axis Theorem.• If the centroidal axis for each part does not coincide with the

reference axis, the parallel-axis theorem, should beused to determine the moment of inertia of the part about thereference axis. For the calculation of , use the table on the insideback cover.

Summation.• The moment of inertia of the entire area about the reference axis

is determined by summing the results of its composite partsabout this axis.

• If a composite part has a “hole,” its moment of inertia is found by “subtracting” the moment of inertia of the hole from themoment of inertia of the entire part including the hole.

I

I = I + Ad2,

10.4 MOMENTS OF INERTIA FOR COMPOSITE AREAS 523

10

x

100 mm

75 mm

75 mm

25 mm

(a)x

100 mm

75 mm

75 mm

25 mm

–

(b)

Fig. 10–8

EXAMPLE 10.4

Determine the moment of inertia of the area shown in Fig. 10–8aabout the x axis.

SOLUTION

Composite Parts. The area can be obtained by subtracting thecircle from the rectangle shown in Fig. 10–8b. The centroid of eacharea is located in the figure.

Parallel-Axis Theorem. The moments of inertia about the x axisare determined using the parallel-axis theorem and the data in thetable on the inside back cover.

Circle

Rectangle

Summation. The moment of inertia for the area is therefore

Ans.= 10111062 mm4

Ix = -11.411062 + 112.511062= 1

1211002115023 + 110021150217522 = 112.511062 mm4

Ix = Ixœ + Ady2

= 14p12524 + p1252217522 = 11.411062 mm4

Ix = Ixœ + Ady2


10

Determine the moments of inertia for the cross-sectional area of themember shown in Fig. 10–9a about the x and y centroidal axes.

SOLUTION

Composite Parts. The cross section can be subdivided into the threerectangular areas A, B, and D shown in Fig. 10–9b. For the calculation,the centroid of each of these rectangles is located in the figure.

Parallel-Axis Theorem. From the table on the inside back cover, orExample 10.1, the moment of inertia of a rectangle about itscentroidal axis is Hence, using the parallel-axis theoremfor rectangles A and D, the calculations are as follows:

Rectangles A and D

Rectangle B

Summation. The moments of inertia for the entire cross section are thus

Ans.

Ans.= 5.6011092 mm4

Iy = 2[1.9011092] + 1.8011092= 2.9011092 mm4

Ix = 2[1.42511092] + 0.0511092

Iy = 11211002160023 = 1.8011092 mm4

Ix = 11216002110023 = 0.0511092 mm4

= 1.9011092 mm4

Iy = Iy¿ + Adx2 = 1

1213002110023 + 1100213002125022= 1.42511092 mm4

Ix = Ix¿ + Ady2 = 1

1211002130023 + 1100213002120022

I = 112 bh3.

EXAMPLE 10.5

100 mm

400 mm

100 mm

100 mm600 mm

400 mm

x

y

(a)

C

100 mm

100 mm

x

y

300 mm

300 mm200 mm

250 mm

200 mm

(b)

A

B

D

250 mm

Fig. 10–9

10.4 MOMENTS OF INERTIA FOR COMPOSITE AREAS 525

10

FUNDAMENTAL PROBLEMS

F10–7. Determine the moment of inertia of the cross-sectional area of the channel with respect to the y axis.

F10–6. Determine the moment of inertia of the beam’scross-sectional area about the centroidal x and y axes.

F10–5. Determine the moment of inertia of the beam’scross-sectional area about the centroidal x and y axes.

F10–8. Determine the moment of inertia of the cross-sectional area of the T-beam with respect to the axispassing through the centroid of the cross section.

x¿

300 mm

200 mm

30 mm 30 mm

30 mm

30 mm

x

y

F10–6

x

y

50 mm

50 mm

300 mm

50 mm

200 mm

F10–7

30 mm

150 mm

150 mm

30 mm

y

x¿

F10–8

F10–5

200 mm

150 mm 150 mm

200 mm

50 mm

50 mm

x

y

moment of inertia

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