8.2 / Moments of Inertia of a Plane Area 529
8.1 Introduction
In this chapter we study the inertia properties of plane areas. One reason for studying this topic in statics is that these properties arise in the formu las for locating the resultant of hydrostatic pressure forces on a sub merged body (which we shall xamine in Section 9.2). A more important reason for this study is that it is sometimes considered a prerequisite for courses in strength of materials (or deformable bodies), which follow statics. In these later courses the student will find that stresses in a transversely loaded beam are, under special but important circum stance , inversely proportional to a moment of inertia of the cross sectional area of the beam. Deflections of the beam will likewise be inversely proportional to this moment ofinertia, which forms part of the resistance to bending of the b am. Similarly, th "polar moment of iner tia" is a factor in the resistance of a shaft to torsion, or twisting. The four sections 8.2-8.5 of this chapter can be read by a student who is familiar with no more than single integration. These are the sections normally needed in the first course in the mechanics of deform able solids. The last three sections, however, utilize double integrals when dealing with products of inertia, Moments and products of inertia of rna s are needed in dynamics; we hall cover this related topic in our sec nd volume at the pint where the subject arises naturally,
8.2 Moments of Inertia of a Plane Area
y For the plane area shown in Figure 8.1, the moments of inertia" with respect to the x and y axes are defined to be x dA Ix = L y2 dA (8.1) x and lure 8.1 ly = L x 2 dA (8.2) These definitions illustrate why a moment of inertia is sometimes called a "second moment" - because of the square of the distance from the x axis for Ix (and from the y axis for I y )' We have s en "first moments" in Chapter 7 in relation to the concept of centroid. Because a moment of inertia is made up of areas multiplied by squares of distances, it has the dimension (length)4.
• Many prefer to use the term "second moments of area," feeling that the word "iner tia" suggests mass and sh uld be reserved for similar integrals that reflect the mass dis tribution of a body. Nevertheless, the terms "area moment of inertia" and "moment of inertia of area" are widely used in text on the mechanics af deformable solids, 530 CHAPTI!R 8 / INERTIA PROPERTIES OF PLANE AREAS
Equations (8.1) and (8.2) also tell us that a moment of inertia is always positive and is a measure of "how much area is located how far" from a line. If we wish to be specific about the origin of the x and y coordinates, we may write, for example, Ixcif the origin is the centroid, or Ix. if the origin is some other point P. We now proceed to use the above definitions to find moments of inertia of several common shapes in the following examples.
EXAMPLE 8.1 y Find the moments of inertia of the rectangular area of Figure E8.1a about the centroidal x and y axes. Solution
To find Ixc ' we need the integral f y2 dA. Using the horizontal strip shown in h Figure EB.lb for our diHerential area dA, and noting that the y coordinate is the c x same for all parts of the strip, we get
h/2 I = y2 dA = y2(b dy) xc j b J -h/2 Fiyurtl E8.11 = by3lh/2 3 -h/2 bh3 Yj dA = bdy /' I ~L 12 IT ( dy A similar integration with dA = h dx as shown in Figure E8.1c gives Iyc : It y b/2 X3!bf2 2 h I = x (h dx) = h c x yc lx--b/2 '--.r-J 3 -b/2 dA hb3 b 12 Figure E8. Ib
D.uestion 8.1 Was the lyC calculation really necessary? That is, could y the answer for lyC have been deduced from the result for I xc obtained first?
dA = h dx --1-nl...--/' For those familiar with double integrals, we use them below to reproduce the result for I xc: hi c~---t h/2 lb/2 I = y2 ax dy Xc jy--h/2 dx %--b/2 '--.r-J dA b Figure E8. It
AnSWlf 8 1 No, it was not necessary, because if we change the names of x and y, and of band h, then the same integral as before yields lyC = hb3 / 12 without having to inte grate again. 8.2 / Mom nts of Inertia of a Plane Area 531
J '1 / 2 Ib/2 fh /2 yJ Ih /2 bh = X y2 dy = y2 b dy = b - = jy--hI2 -b/2 -hl2 3 -h/2 12 ote that the first integration (on x) produces the strip b dy used earlier.
Question 8,2 CouJd the strip 11 dx have been used for "dA" in the 1:<,0 calculation?
ADsWIll' 8.2 0, because then y is not the same [or every element of the differential strip.
EXAMPLE 8.2 y Find the moment of inertia of a circuJar area (see Figure E8.2a) about any diameter. " Solution 2 Since x + y2 = R2 on the boundary, our dA in Figur E8.2b is given by x dA = 2 JR2 - y2 dy so that Figure E8.2a [Xc = Jy2 d = iRR 2y2JR2 - y2 dy R = 2l 2y2JR2 - y2 dy
Substituting sin efor y/R, and noting that d(y / R) = cos 0 dO,
1rc=4R·l~/2 sin2 0c sO(cos dO)
Figure E8.2b where for the integral limits, y / R = 0 when 8 = 0, and y/ R = 1 when () = n/2. Continuing,
n/ 2 sin2 20 Ire = 4R· i0 --4- de ~/2 1 - cos 40 =R2 dO 1 2 v / = R· (e _sin 48) In 2 = nR· 24 0 4 For the reader acquainted with double integrals, we can obtain the above result with somewhat less effort using polar co rdinates ( e Figure E8.2c) as figul E8.2c follows: 532 CHAPTER 8 / INERTIA PROPIlRTIES OF PLANE AR£AS
I: _ (2. r4IR 2 Jo 4" 0 5in 0 dO 2 -_ R412. (1 - cos 20) dO-_R4 [ 0--sin 20] 1 " 4 0 2 8 2 0 1 = nR4 .te 4 This is, of course, also lYe or 1about any other diameter of the circle. EXAMPLE 8.3 -, b Find the moment of inertia of the triangular area in Figure E8.3 about the y axis. Solution I For our differential area, we shall use the shaded strip in the figure; thus, using Y h to locate the lowest (boundary) point of the strip, dA = (h - Y) dx y 211 . .. But Y = b x for the sIde of the triangle In the first quadrant, so that x Figure E8.3 dA = ( h - 2:X) dx Therefore, 2 (b/2 2 ( 2h) I.~ = Jx dA = 2 Jo x h b x dx 3 3 4 3 b/2 2 ) [X x Ib/2 hb = 211 (2xx -- dx = 211 --- J = lo b 3 2b 0 48 PROBLEMS • Section 8.2 8.1 In Figure P8,1: b a. Find the moment of inertia of the shaded area about the x axis. b. Tell why I Y is identical to lYe of Example 8.1. h Yif\ 8.2 Find 1% for the shaded area in Figure P8.2. 8.3 In Problem 8.2 find ly for the same area. x Figure P8.1 534 CHAPTER 8 / INERTIA PROPERTIE OF PLANE AREAS 8.3 The Polar Moment of Inertia 01 a Plane Area In the study of deformable solids, the "torsion problem" refers to what happens to a shaft when it is twisted. In the same way that the moment of inertia forms part of the resistance of a beam to bending, the polar moment of inertia forms part of its resistance to twisting. For this reason, then, we shall discuss the polar moment of inertia in this section. The polar moment of inertia of an area about a point P is defined to be (see Figure 8.2). 2 ]p = J(x + y2) dA '-1 ~- dA where the axes (x, y) have origin at P. Since the polar coordinate ris given 2 t , /1 by· x + y2 = r2, this simplifies to y 2 // ]p = J r dA 7 P~x x If we recall that both I % and I yare the sums (in integration) of (differential areas) times (squares of their distances from the axis), then we see that]p fIgure 8.2 is the same type of quantity, this time with respect to the z axis (normal to the area) through a point P. Also, 2 Jp = Jy2 dA + Jx dA = I%p + Iyp (8.3) so that for an area moment of inertia, the x and y (in-plane) inertias add up to the z (out-of-plane) inertia. y EXAMPLE 8.4 Find the polar moment of inertia of the circular cross section of Figure E8.4a with respect to point C. CR .' Solution \ By Ie = I xe + lYe' we obtain immediately from the results of Example 8.2: ~ nR4 nR4 nR4 Figure E8.4a lc=-4-+4=-2 Alternatively, the shaded "circular strip" in Figure E8.4b has the feature that all its elements are the same distance r from the z axis. Thus with dA = 21[T dr, R Ie = Jr 2 dA = i r2(2nT dr) 4 = 2n~ III = nR x 4 0 2 And if double integrals are familiar to the reader, we obtain the same result a Figure E8.4b • Hence the name "polar moment of inertia." 8.3 / The Polar Moment of Inertia of a Plane Area 535 third time using polar coordinates: 2 2 2 Ie = Jr dA = 1" fR r (r dr de) 8-0 ,-0 2 R4 1 " ltR4 =2" -r4\R de=-o = 10-0 4 0 4 0 2 EXAMPLE y 8.5 Find the polar moment of inertia ofthe rectangular cross secti n of Figure E8.5, at point C. Solution h Because Ie = I xc + I Yc' we se previous results and get c x 3 3 2 bh hb bhW + h ) le=12+12= 12 b or Ftgur E8 PROBLEM Section 8.3 In Problems 8.13-8.19 find the polar moment of inertia 8.15 with respect to the origin for ach of th shaded areas. 8 13 b h h y 0 x 0 x Figure PH.1S Figul1l 8 13 8, I y(m) 8.16 y x = 4em 4 x o 2 x(m) o I Figu P8 14 Figure PB.16