APPENblX B

OF I ~ E R T I A ~ OF A P L A E

B.l MOMENTS OF INERTIA OF A PLANE AREA ~~~~~ ~

The derivation of load-stress formulas for torsion members and beams may require solu- tions of one or more of the following integrals:

where dA is an element of the plane area A lying in the (x, y) plane in Figure B. 1. Area A represents the cross-sectional area of a member subjected to bending and/or torsional loads.

The integrals in Eqs. B.l-B.3 are commonly called moments of inertia of the area A because of the similarity with integrals that define the mass moment of inertia of bodies in the field of dynamics. Since an area cannot have an inertia, moment of inertia of an area is a misnomer. We use the term because of common usage.

The integral represented by Eq. B.4 is called the product of inertia. Its sign can be neg- ative. The moment of inertia and product of inertia are given the symbol I if the axes about which the moments are taken lie in the plane of the area (see Eqs. B.l , B.2, and B.4). When the axis about which the moment is taken is perpendicular to the area (see Eq. B.3), the moment of inertia is given the symbol J and is called the polar moment of inertia of the area.

2

0

FIGURE 8.1 Plane area A in the (x, yl plane.

660

8.2 PARALLEL AXIS THEOREM 661

B.2 PARALLEL AXIS THEOREM

In the application of Eqs. B.l-B.4 to engineering problems, it is convenient to know these integrals for coordinate axes at the centroid of area A. The values of the integrals for a few basic cross sections are listed in Table B.l. Often, practical members have cross sections that are composed of two or more simple cross sections (Table B.l). Moments of inertia for composite areas are obtained by application of the parallel axis theorem.

TABLE B.l Moments of Inertia of Common Plane Areas

Rectangle

Right Triangle

Circle

E I I i pse

2

bh3 I = - 12

bh3 + hb3 Jo = 7 I, = 0

bh3 I = - 36

bh3 + hb3 Jo = 36

I, = 0

zbh3 4

I , = -

zhb3 I = - y 4

zbh(h2 + b2) 4

J o =

I, = 0 (continues)

662 APPENDIX B SECOND MOMENT (MOMENT OF INERTIA) OF A PLANE AREA

TABLE B.1 Moments of Inertia of Common Plane Areas (continued)

I , = 0

Semiellipse

Trapezoid

h

I , = nbh3( f - 2) nbh3 I = -

y 8

J o = nbh[ h2 - 9$ 8h2+b2 .) I , = 0

h3 a + 4 a b + b I’ = [ ’36(u+b) ‘1 I = -[. h 4 +b4+2ab(a 2 +b2)

Y 36(a + b ) 3 2 2 3

- C ( U + 3 ~ b - 3 ~ b - b )

I + c2(a2 + 4ab + b2)

J , = l , + I y

I = -[b(3a h2 2 -3ab-b2) x y 72(a + b )

+ a3 - c(2a2 + 8ab + 2b2)]

Let it be required to obtain moments of inertia for area A in Figure B.2 for coordinate axes (x’, y’, z’). AreaA lies in the (x’, y’) plane. First, locate coordinate axes (x, y, z) with axes parallel, respectively, to the (x’, y’, z’) axes and with the origin 0 at the centroid of A. Let the

distances of the centroid 0 from the axes (x’, y’) be (K, J ) . Then, 7 = Jr2 x + y isthedis- tance between the z’axis and the z axis. Using Eqs. B.l-B.4, we obtain

8.2 PARALLEL AXIS THEOREM 663

FIGURE B.2 Plane area A in the (x', f ) plane.

EXAMPLE B.l Moments of

Inertia for a Z-bar

Solution

(B.6)

(B.7)

(B.8)

where integrals fy dA and fx dA are zero since the first moment of an area with respect to an axis through the centroid of the area vanishes. Equations B.5-B.8 represent parallel axes formulas for moments of inertia of an area. They may be employed to obtain the moments of inertia of composite areas.

I, , = I ( x + X)2dA = ] , + A X 2

Jot = j [ ( x + X ) 2 + ( y + L)2]dA = J , + A T 2

Ix,yt = I ( x + K)(y + 7 ) d A = I , + AX 7

A Z-bar has the cross section shown in Figure EB. 1. Determine I,, I,,, and Ixr for the centroidal axes (x, y) shown.

1

FIGURE EB.l

The area is divided into three rectangular areasAl,A2, and A, (Figure EB.l). Using Eqs. B.5, B.6, and B.8 and Table B.l, we obtain

60( 1 0 ) ~ 2 io (120)~ I , = - + 60( lo)(%) + - + 120( 10)(0)2 12 12

+- 60(10)3 + 60( = 5.08 x lo6 mm4 12

664 APPENDIX B SECOND MOMENT (MOMENT OF INERTIA) OF A PLANE AREA

I =- 2 120(10)~ 2 10(60)3 + 60( lo)(-35) + - + 120( 10)(0) Y 12 12

10(60)3 2 6 4 + - + 60(10)(35) = 1.84 x 10 mm 12

I , = 60( 10)(-35)(55) + 120( 10)(0)(0) + 60( 10)(35)(-55) 6 4 = -2.31 X 10 ITIIII

B.3 TRANSFORMATION EQUATIONS FOR MOMENTS AND PRODUCTS OF INERTIA

Let Zx, Zy, and Zv be known moments and product of inertia for area A (Figure B.3) for (x , y) rectangular axes that lie in the plane of the area. Consider the (X, Y ) coordinate axes that have the same origin and same plane as the (x, y) axes. We wish to derive transforma- tion equations by which Zx, I,, and Zm are obtained in terms of I,, Zy, Zv, and 6, the angle through which the (x , y) axes must be rotated to coincide with the (X, Y) axes; 8 is positive in the counterclockwise sense. Consider an element of area dA at X and Y coordinates given by the relations

Substitution of Eqs. B.9 into Eqs. B.l, B.2, and B.4 gives

= (Zx-Zy)sinO cosO+Zxy(cos 2 0-sin 2 0)

FIGURE 8.3 Transformation from (x, v) axes to (X, Y ) axes.

8.3 TRANSFORMATION EQUATIONS FOR MOMENTS AND PRODUCTS OF INERTIA 665

EXAMPLE B.2 principal Axes for

a 2-Bar

Solution

With double angle identities, Eqs. B.10 can be written in the form

Locate the principal axes and detennine the principal moments of inertia I, and I , for the Z-bar whose dimensions are specified in Figure EB. 1.

Since I, = 5.08 x lo6 mm4 I = 1.84 x lo6 mm4, and Ixy = -2.31 x lo6 mm4, the principal values for the moments of inertia are given by Eqs. B.13: ’ .y

Ix+Iy zx-zy

zx+zy zx-zy I , = -+-cos2e-z sin28

I , = -- - cos 28 + I, sin 28

2 2 XY

2 2 (B. 1 1)

I,, = Ix -sin2e+ - I , I COS 28 2 XY

Note the similarity between the transformation equations for moments and products of inertia and the transformation equations of stress given by Eq. 2.3 1. Like stress compo- nents and strain components, moments and products of inertia transform according to the rule for second-order symmetric tensors.

B.3.1 Principal Axes of Inertia

There are two values of 8 for which I, = 0. To determine these values, let I, = 0. Then, the third of Eqs. B.11 yields

(B.12)

The two values of 8 given by Eq. B.12 locate two positions of axes (X, Y) that represent the principal axes of inertia for a given cross-sectional area. In the discussion that follows, we assume for definitiveness that I , is greater than Z y . Then, the maximum moment of iner- tia, which we take to be I , and is associated with the X axis, will occur for the X axis located at the smallest of the two values of 8 from the x axis; the direction is counterclock- wise for a positive value of 8 and clockwise for a negative value of 8. Since I , - Zy > 0, if we substitute the value of 8 given by Eq. B.12 into the first and second of Eqs. B.11, we find

2 I ,+I 2 1, = zx-zY +I,

(B.13)

as the principal moments of inertia for the cross-sectional area A.

666 APPENDIX B SECOND MOMEM (MOMEM OF INERTIA) OF A PLANE AREA

6 4 = 0.639 x 10 mm I

= 6.281 x lo6 mm4

~ The location of the X axis is given by Eq. B. 12. Thus,

I , = 5.08 x lo6 + 1.84 x lo6 5.08 x lo6 - 1.84 x lo6 2

= 1.4259 tan28 = - 2(-2.31 x lo6) 5.08 x lo6 - 1.84 x lo6

8 = 0.4796 rad

Hence, the X axis is located at 0.4796 rad, measured counterclockwise from the x axis, as shown in Figure EB.2.

FIGURE EB.2

PROBLEMS

Section B.2 B.l. Derive the expressions for 1, and Iv for the right triangle in Table B.l. B.2. Derive the expression for I, for the semiellipse in Table B.l.

B3. Determine I,, Zy, and Zv for the centroidal axes for the cross-sectional area shown in Figure PB.3. B.4. Determine I,, Zy, and Zq for the centroidal axes for the cross-sectional area shown in Figure PB.4.

PROBLEMS 667

Section 8.3 B.5. Locate principal axes (X, Y ) and determine Ix and Z, for the cross-sectional area in Problem PB.3. B.6. Locate the principal axes (X, Y) and determine Zx and I , for the cross-sectional area in Figure PB.4.

FIGURE PB.3

FIGURE PB.4