CHAPTER 1 UNSYMMETRICAL BENDING Summary The second moments of area of a section are given by I=ydA and Iy=dA The product second moment of area of a section is defined as ln =xydA which reduces to Ixy =Ahk for a rectangle of area A and centroid distance h and k from the X and Y axes. The principal second moments of area are the maximum and minimum values for a section and they occur about the principal axes.Product second moments of area about principal axes are zero. With a knowledge of I,Iyy and Ixy for a given section,the principal values may be determined using either Mohr's or Land's circle construction. The following relationships apply between the second moments of area about different axes: Iu=(lx +Iyy)+(Ixx -Iyy)sec 20 =(Ixx+lyy)(lx-Iyy)sec20 where 6 is the angle between the U and X axes,and is given by 2Ixy tan29=1y-1x） Then Iu+Iv=Ia+lyy The second moment of area about the neutral axis is given by IN.A.=(u+)+(u-I)cos2au where o is the angle between the neutral axis(N.A.)and the U axis. Also Ix Iu Cos20+I sin20 Iyy =Iv cos20+Iu sin20 1xw=l。-lu)sin29 Ix-Iyy=(Iu-I)cos 20 1
CHAPTER 1 UNSYMMETRICAL BENDING Summary The second moments of area of a section are given by I, = 1 y2 dA and I,, = 1 x2 dA The product second moment of area of a section is defined as I,, = xydA which reduces to I,, = Ahk for a rectangle of area A and centroid distance h and k from the X and Y axes. The principal second moments of area are the maximum and minimum values for a section and they occur about the principal axes. Product second moments of area about principal axes are zero. With a knowledge of I,, I,, and I,, for a given section, the principal values may be determined using either Mohr’s or Land’s circle construction. The following relationships apply between the second moments of area about different axes: s I, = ;(I,, +I,,) + ;(I= - 1,,)sec28 I, = ;(I,, + I,,) - ;(I= - I,,)sec20 where 0 is the angle between the U and X axes, and is given by Then I, + I, = I.r, + I,, The second moment of area about the neutral axis is given by IN.^,. = ;(I, + I,) + 4 (I, - I,) COS 2a, where u, is the angle between the neutral axis (N.A.) and the U axis. Also I, = I, cos2 8 + I, sin2 8 I,, = I, cos2 8 + I, sin2 0 I,, = ;(I~ - 1,)sin20 I, - I,, = (I, - I,>) cos 28 1
2 Mechanics of Materials 2 Stress determination For skew loading and other forms of bending about principal axes a=Mi Mou where M and M,are the components of the applied moment about the U and V axes. Alternatively,with a Px+Oy Mxt PIxy +Qlx Myy=-Plyy -Qlxy Then the inclination of the N.A.to the X axis is given by P tana = 0 As a further alternative, M'n 0三 IN.A. where M'is the component of the applied moment about the N.A.,IN.A.is determined either from the momental ellipse or from the Mohr or Land constructions,and n is the perpendicular distance from the point in question to the N.A. Deflections of unsymmetrical members are found by applying standard deflection formulae to bending about either the principal axes or the N.A.taking care to use the correct component of load and the correct second moment of area value. Introduction It has been shown in Chapter 4 of Mechanics of Materials I that the simple bending theory applies when bending takes place about an axis which is perpendicular to a plane of symmetry.If such an axis is drawn through the centroid of a section,and another mutually perpendicular to it also through the centroid,then these axes are principal axes.Thus a plane of symmetry is automatically a principal axis.Second moments of area of a cross-section about its principal axes are found to be maximum and minimum values,while the product second moment of area,fxy dA,is found to be zero.All plane sections,whether they have an axis of symmetry or not,have two perpendicular axes about which the product second moment of area is zero.Principal axes are thus defined as the axes about which the product second moment of area is zero.Simple bending can then be taken as bending which takes place about a principal axis,moments being applied in a plane parallel to one such axis. In general,however,moments are applied about a convenient axis in the cross-section; the plane containing the applied moment may not then be parallel to a principal axis.Such cases are termed "unsymmetrical"or"asymmetrical"bending. The most simple type of unsymmetrical bending problem is that of"skew"loading of sections containing at least one axis of symmetry,as in Fig.1.1.This axis and the axis EJ.Hearn,Mechanics of Materials 1.Butterworth-Heinemann.1997
2 Mechanics of Materials 2 Stress determination For skew loading and other forms of bending about principal axes M,v M,u c=-+- 1, 1, where Mu and M, are the components of the applied moment about the U and V axes. Alternatively, with 0 = Px + Qy M, = PI,, + QIM Myy = -Plyy - QIxy Then the inclination of the N.A. to the X axis is given by P tana! = -- Q As a further alternative, M’n 1N.A. o=- where M’ is the component of the applied moment about the N.A., IN.A. is determined either from the momenta1 ellipse or from the Mohr or Land constructions, and n is the perpendicular distance from the point in question to the N.A. Deflections of unsymmetrical members are found by applying standard deflection formulae to bending about either the principal axes or the N.A. taking care to use the correct component of load and the correct second moment of area value. Introduction It has been shown in Chapter 4 of Mechanics of Materials 1 that the simple bending theory applies when bending takes place about an axis which is perpendicular to a plane of symmetry. If such an axis is drawn through the centroid of a section, and another mutually perpendicular to it also through the centroid, then these axes are principal axes. Thus a plane of symmetry is automatically a principal axis. Second moments of area of a cross-section about its principal axes are found to be maximum and minimum values, while the product second moment of area, JxydA, is found to be zero. All plane sections, whether they have an axis of symmetry or not, have two perpendicular axes about which the product second moment of area is zero. Principal axes are thus de$ned as the axes about which the product second moment of area is Zero. Simple bending can then be taken as bending which takes place about a principal axis, moments being applied in a plane parallel to one such axis. In general, however, moments are applied about a convenient axis in the cross-section; the plane containing the applied moment may not then be parallel to a principal axis. Such cases are termed “unsymmetrical” or “asymmetrical” bending. The most simple type of unsymmetrical bending problem is that of “skew” loading of sections containing at least one axis of symmetry, as in Fig. 1.1. This axis and the axis EJ. Hearn, Mechanics of Murerids I, Buttenvorth-Heinemann, 1997
§1.1 Unsymmetrical Bending 3 (a)Rectangular (b)I-section (c)Channei (d)T-section section section Fig.1.1.Skew loading of sections containing one axis of symmetry. perpendicular to it are then principal axes and the term skew loading implies load applied at some angle to these principal axes.The method of solution in this case is to resolve the applied moment MA about some axis A into its components about the principal axes. Bending is then assumed to take place simultaneously about the two principal axes,the total stress being given by Mv Mou 0= With at least one of the principal axes being an axis of symmetry the second moments of area about the principal axes I and I can easily be determined. With unsymmetrical sections (e.g.angle-sections,Z-sections,etc.)the principal axes are not easily recognized and the second moments of area about the principal axes are not easily found except by the use of special techniques to be introduced in $s1.3 and 1.4.In such cases an easier solution is obtained as will be shown in $1.8.Before proceeding with the various methods of solution of unsymmetrical bending problems,however,it is advisable to consider in some detail the concept of principal and product second moments of area. 1.1.Product second moment of area Consider a small element of area in a plane surface with a centroid having coordinates (x,y)relative to the X and Y axes (Fig.1.2).The second moments of area of the surface about the X and Y axes are defined as Ie=y'dA and In =xda (1.1) Similarly,the product second moment of area of the section is defined as follows: (1.2) Since the cross-section of most structural members used in bending applications consists of a combination of rectangles the value of the product second moment of area for such sections is determined by the addition of the /x value for each rectangle(Fig.1.3), ie. Iry Ahk (1.3)
$1.1 Unsymmetrical Bending 3 V V V (c) Rectangular (b) I-sectam (c) Channel (d) T-sectton section section Fig. 1 .I. Skew loading of sections containing one axis of symmetry. perpendicular to it are then principal axes and the term skew loading implies load applied at some angle to these principal axes. The method of solution in this case is to resolve the applied moment MA about some axis A into its components about the principal axes. Bending is then assumed to take place simultaneously about the two principal axes, the total stress being given by M,v M,u a=-+- 1, I, With at least one of the principal axes being an axis of symmetry the second moments of area about the principal axes I, and I, can easily be determined. With unsymmetrical sections (e.g. angle-sections, Z-sections, etc.) the principal axes are not easily recognized and the second moments of area about the principal axes are not easily found except by the use of special techniques to be introduced in $3 1.3 and 1.4. In such cases an easier solution is obtained as will be shown in 51.8. Before proceeding with the various methods of solution of unsymmetrical bending problems, however, it is advisable to consider in some detail the concept of principal and product second moments of area. 1.1. Product second moment of area Consider a small element of area in a plane surface with a centroid having coordinates (x, y) relative to the X and Y axes (Fig. 1.2). The second moments of area of the surface about the X and Y axes are defined as zXx = Jy’d~ and zYy = /x’& (1.1) Similarly, the product second moment of area of the section is defined as follows: zXy = Jxy (1.2) Since the cross-section of most structural members used in bending applications consists of a combination of rectangles the value of the product second moment of area for such sections is determined by the addition of the I,, value for each rectangle (Fig. 1.3), i.e. Zxy = Ahk (1.3)
4 Mechanics of Materials 2 12 Fig.1.2. where h and k are the distances of the centroid of each rectangle from the X and Y axes respectively(taking account of the normal sign convention for x and y)and A is the area of the rectangle. k- h+ h+ X k- k+ h- h- X Fig.1.3. 1.2.Principal second moments of area The principal axes of a section have been defined in the introduction to this chapter. Second moments of area about these axes are then termed principal values and these may be related to the standard values about the conventional X and Y axes as follows. Consider Fig.1.4 in which GX and GY are any two mutually perpendicular axes inclined at to the principal axes GV and GU.A small element of area A will then have coordinates (u,v)to the principal axes and (x,y)referred to the axes GX and GY.The area will thus have a product second moment of area about the principal axes given by uudA. .'total product second moment of area of a cross-section uvdA (x cos+y sin)(y cos-x sin)dA
4 Mechanics of Materials 2 51.2 Y t Fig. 1.2. where h and k are the distances of the centroid of each rectangle from the X and Y axes respectively (taking account of the normal sign convention for x and y) and A is the area of the rectangle. k- kt h- I hFig. 1.3. 1.2. Principal second moments of area The principal axes of a section have been defined in the introduction to this chapter. Second moments of area about these axes are then termed principal values and these may be related to the standard values about the conventional X and Y axes as follows. Consider Fig. 1.4 in which GX and GY are any two mutually perpendicular axes inclined at 8 to the principal axes GV and GU. A small element of area A will then have coordinates (u, v) to the principal axes and (x, y) referred to the axes GX and GY. The area will thus have a product second moment of area about the principal axes given by uvdA. :. total product second moment of area of a cross-section I,, = /"uvdA = S(xcosO+ysin8)(ycos8-xsine)~A
§1.2 Unsymmetrical Bending 5 (ycosysincoscossin-xysin0dA =aosg-sm2o∫yda+sn6ecos[/ah-/ran Ixy cos 20+(lx-Iyy)sin 20 dA Principal axis Fig.14. Now for principal axes the product second moment of area is zero. 0=Ixy cos20+(lxx-Iyy)sin 20 tan 20= -2= 2Ixy (Lx -Iyy)(Iyy -Ixx) (1.4) This equation,therefore,gives the direction of the principal axes. To determine the second moments of area about these axes, 1.da=(ycos0-xsina da =cos0ydA+sin20dA-2cossin xydA =Lxx cos20+lyy sin20-Ixy sin 26 (1.5) Substituting for Ixy from eqn.(1.4), 1.=1+cos201w+1-cos291n-1c0s200n-1a) =0+m2a+0-om20y[A62un-1a =(1 cos20)xx+(1-cos 20)Iyy-i sec 20(lyy -Ixx)+cos 20(lys -Ixx) =(l+Iyy)+(Ixx -Iyy)cos 20-(Iyy-Lx)sec 20+(Iyy -Ixx)cos 20
91.2 Unsymmetrical Bending 5 = /(x y cos2 8 + y2 sin 8 cos 8 - x2 cos 8 sin 8 - xy sin2 8) dA = (cos2 8 - sin2 8) /xy dA + sin 8 cos 8 [/” y2 dA - /x2 dA] Y Principal axis Fig. 1.4. Now for principal axes the product second moment of area is zero. .. o = I,, COS 28 + 4 (I, - zYy) sin 28 This equation, therefore, gives the direction of the principal axes. To determine the second moments of area about these axes, I, = v2 dA = (y cos 8 - x sin dA ss = cos2 8 y2 dA + sin2 8 /x2 dA - 2cos8 sin 8 xydA I = I, cos2 8 + I,, sin2 8 - I,,~ sin 28 Substituting for I,, from eqn. (1.4), sin228 I,= ;(1+cos28)Ixx+;(1-cos28)z,,--- 2 cos28 (I,r - 1,) (1.4) = ;(I + cos 2011, + ; (1 - cos 28)1,, - sec 213(1,, - I,) + ; cos 20(1,,. .. - I,) I - (I, + I,,) + (I, - I,,) cos 28 - (I,, - I,) sec 28 + (Ivv - I,) cos 213
6 Mechanics of Materials 2 §1.3 i.e. Iu =(Ixx +Iyy)+(Ix-Iyy)sec 20 (1.6) Similarly, 1.=da=coso+ysineydA =(Ixx +Iy)-(Itx -Iyy)sec 20 (1. 7 N.B.-Adding the above expressions, Iu+1=Ix+lyy Also from eqn.(1.5), Iu=Ixt cos20+Iyy sin20-Ixy sin 20 =(1 cos 20)/xx+(1-cos 20)lyy-Ixy sin 20 Iu=(Ixx Iyy)+(Ixx -Iyy)cos 20-Ixy sin 20 (1.8) Similarly, I=(Ixx Iy)-(Itz -Iyp)cos 20+Igy sin 20 (1.9) These equations are then identical in form with the complex-stress eqns.(13.8)and(13.9) with I lyy,and Iy replacing ox,oy and txy and Mohr's circle can be drawn to represent I values in exactly the same way as Mohr's stress circle represents stress values. 1.3.Mohr's circle of second moments of area The construction is as follows (Fig.1.5): (1)Set up axes for second moments of area (horizontal)and product second moments of area (vertical). (2)Plot the points A and B represented by (,I)and (Iyy,-I). (3)Join AB and construct a circle with this as diameter.This is then the Mohr's circle (4)Since the principal moments of area are those about the axes with a zero product second moment of area they are given by the points where the circle cuts the horizontal axis. Thus OC and OD are the principal second moments of area I and I. The point A represents values on the X axis and B those for the Y axis.Thus,in order to determine the second moment of area about some other axis,e.g.the N.A.,at some angle a counterclockwise to the X axis,construct a line from G at an angle 2a counterclockwise to GA on the Mohr construction to cut the circle in point N.The horizontal coordinate of N then gives the value of IN.A. EJ.Hearn,Mechanics of Materials 1,Butterworth-Heinemann,1997
6 Mechanics of Materials 2 i.e. 1 1, = TUxx +zYy> + Similarly, - 1,,)sec20 $1.3 (1.6) I, = u2dA = (xcos8+ysin8)2dA JJ 1 = z(zxx + zyy) - ;(L - zyy) sec 28 N.B .-Adding the above expressions, I, +I, = I,, + I,, Also from eqn. (1 S), I, = I, cos2 8 + I,, sin2 8 - I,, sin 20 = (1 + cos B)I, + (1 - cos 20)1,, - I,, sin 28 Z, = ;(z~ +I,,)+ ;(zxx -Z,.~)COS~O-Z~S~~~~ (1.8) Similarly, I,, = ;(zXx + zYy) - ;(zX, - zYy) cos 28 + z, sin 28 (1.9) These equations are then identical in form with the complex-stress eqns. (1 3 .S) and (1 3.9)t with I,, I,,, and I,, replacing a,, oy and txy and Mohr’s circle can be drawn to represent I values in exactly the same way as Mohr’s stress circle represents stress values. 13. Mohr’s circle of second moments of area The construction is as follows (Fig. 1.5): (1) Set up axes for second moments of area (horizontal) and product second moments of (2) Plot the points A and B represented by (I,, I,,) and (I,,, -Ixy). (3) Join AB and construct a circle with this as diameter. This is then the Mohr’s circle. (4) Since the principal moments of area are those about the axes with a zero product second area (vertical). moment of area they are given by the points where the circle cuts the horizontal axis. Thus OC and OD are the principal second moments of area I, and I,. The point A represents values on the X axis and B those for the Y axis. Thus, in order to determine the second moment of area about some other axis, e.g. the N.A., at some angle a! counterclockwise to the X axis, construct a line from G at an angle 2a! counterclockwise to GA on the Mohr construction to cut the circle in point N. The horizontal coordinate of N then gives the value of IN.A. t E.J. Hem, Mechanics ofMuteriuls I, Butterworth-Heinemann, 1997
§1.4 Unsymmetrical Bending 7 2a D Fig.1.5.Mohr's circle of second moments of area. The procedure is therefore identical to that for determining the direct stress on some plane inclined at a to the plane on which ox acts in Mohr's stress circle construction,i.e.angles are DOUBLED on Mohr's circle. 1.4.Land's circle of second moments of area An alternative graphical solution to the Mohr procedure has been developed by Land as follows (Fig.1.6): 0 Fig.1.6.Land's circle of second moments of area. (1)From O as origin of the given XY axes mark off lengths OA =I and AB =lyy on the vertical axis
$1.4 Unsymmetrical Bending YV 7 Fig. 1.5. Mohr's circle of second moments of area. The procedure is therefore identical to that for determining the direct stress on some plane inclined at CY to the plane on which uX acts in Mohr's stress circle construction, i.e. angles are DOUBLED on Mohr's circle. 1.4. Land's circle of second moments of area An alternative graphical solution to the Mohr procedure has been developed by Land as follows (Fig. 1.6): Y t V Fig. 1.6. Land's circle of second moments of area. (1) From 0 as origin of the given XY axes mark off lengths OA = I, and AB = I,, on the vertical axis
8 Mechanics of Materials 2 §15 (2)Draw a circle with OB as diameter and centre C.This is then Land's circle of second moment of area. (3)From point A mark off AD=Ixy parallel with the X axis. (4)Join the centre of the circle C to D,and produce,to cut the circle in E and F.Then ED=I and DF=I are the principal moments of area about the principal axes OV and OU the positions of which are found by joining OE and OF.The principal axes are thus inclined at an angle 0 to the OX and OY axes. 1.5.Rotation of axes:determination of moments of area in terms of the principal values Figure 1.7 shows any plane section having coordinate axes XX and YY and principal axes UU and VV,each passing through the centroid O.Any element of area dA will then have coordinates (x,y)and (u,v),respectively,for the two sets of axes. X Momental ellipse Fig.1.7.The momental ellipse. Now y dsin dA -cosdA+2usinecosdA+sinodA But UU and VV are the principal axes so thatI=fuvdA is zero. Ixx Iu cos20+Iv sin20 (1.10)
8 Mechanics of Materials 2 $1.5 (2) Draw a circle with OB as diameter and centre C. This is then Land's circle of second (3) From point A mark off AD = I,, parallel with the X axis. (4) Join the centre of the circle C to D, and produce, to cut the circle in E and F. Then ED = I, and DF = I, are the principal moments of area about the principal axes OV and OU the positions of which are found by joining OE and OF. The principal axes are thus inclined at an angle 8 to the OX and OY axes. moment of area. 15. Rotation of axes: determination of moments of area in terms of the principal values Figure 1.7 shows any plane section having coordinate axes XX and Y Y and principal axes UU and VV, each passing through the centroid 0. Any element of area dA will then have coordinates (x, y) and (u, v), respectively, for the two sets of axes. I" Y Fig. I .7. The momental ellipse. Now 1, = /y2dA = /(vcos8+usin8)2dA = /u2cos28dA+ J~uvsin~cos~dA+ s u2sin28dA But UU and VV are the principal axes so that I,, = SuvdA is zero. .. zXx = I, cos2 8 + Z, sin' e (1.10)
§1.6 Unsymmetrical Bending 9 Similarly, Idn=f(ucoso-vsinoydn cdA-sincdA+sindA and with∫uvdA=0 Iyy Iv cos2+In sin20 (1.11) Also I,=∫xydA=∫-sins8+usin))dA =/[uu(cos20-sin20)+(u2)sin0cos 0ldA =Iuv cos 20+(-I)sin 20 and Iie =0 Ixy=(I-Iu)sin 20 (1.12 From eqns.(1.10)and (1.11) Ixx -Iyy Iu cos20+1 sin20-I cos20-Iu sin20 =(Iu-1,)cos20-(Iw-1)sin20 Ixx -Iyy (Iu -I)cos 26 (1.13) Combining eqns.(1.12)and (1.13)gives 2Ix tan20=Iyy -Ixx (1.14) and combining eqns.(1.10)and (1.11)gives Ix +Iyy=lu+lv (1.15) Substitution into egns.(1.10)and (1.11)then yields Iu =(Ix +I)+(Ixz -Iy)sec 20] (1.16)as(1.6) I=[(I +lyy)-(Itx-Iyy)sec20] (1.17)as(1.7) 1.6.The ellipse of second moments of area The above relationships can be used as the basis for construction of the moment of area ellipse proceeding as follows: (1)Plot the values of I and I.on two mutually perpendicular axes and draw concentric circles with centres at the origin,and radii equal to I and I(Fig.1.8). (2)Plot the point with coordinatesx =l cos and y =I sin,the value of 0 being given by eqn.(1.14)
$1.6 Unsymmetrical Bending 9 Similarly, I,, = /x2dA = /(ucos6 - wsinQ2dA = /u2cos26dA - 2uvsin6cos6dA+ and with S uvdA = 0 zYy = I,, cos2 e + I, sin2 e Also I,, = /xydA = /(ucos8- wsin8)(vcos6+usin6)dA (1.11) = J [uw(cos2 8 - sin2 6) + (u2 - w2) sin 6 cos 61 dA (1.12) = I,, cos 26 + :(I, - I,) sin 26 and I,, = 0 .. Zxy = z(Z. 1 - Z,)sin28 From eqns. (1.10) and (1.11) I,, - I,, = I, cos2 6 + I, sin2 8 - I, cos2 6 - I, sin2 6 = (I, - 1,) cos2 0 - (I, - I,) sin2 8 z, - iyy = (I, - 1.1 COS 28 (1.13) Combining eqns. ( 1 .12) and (1 .13) gives (1.14) and combining eqns. (1 .lo) and (1.1 1) gives I, + I,, = I, + I, (1.15) Substitution into eqns. (1.10) and (1.11) then yields 1, = [(zXx + + (zXx - zYy) sec 281 (1.16) as (1.6) 1. = $ [(L + zYy) - (zXx - zYy ) sec 281 (1.17) as (1.7) 1.6. The ellipse of second moments of area The above relationships can be used as the basis for construction of the moment of area ellipse proceeding as follows: (1) Plot the values of I, and I, on two mutually perpendicular axes and draw concentric (2) Plot the point with coordinates x = I, cos 6 and y = I,, sin 6, the value of 6 being given circles with centres at the origin, and radii equal to I, and I, (Fig. 1.8). by eqn. (1.14)
10 Mechanics of Materials 2 §1.6 0 R 8 Fig.1.8.The ellipse of second moments of area. It then follows that t2 y2 (=1 This equation is the locus of the point P and represents the equation of an ellipse-the ellipse of second moments of area. (3)Draw O0 at an angle 0 to the I axis,cutting the circle through I in point S and join SP which is then parallel to the I axis.Construct a perpendicular to through P to meet O0 in R. Then OR=00-RO =Iu-(lu sin-I sin0)sin0 =Iu-(Iu-Iv)sin20 =Iu Cos20+1 sin20 =Ix Similarly,repeating the process with Oo1 perpendicular to O0 gives the result OR=Iyx Further, PR=POcos0 =(Iu sin0-I sin0)cos0 =(lu-1v)sin 20=Ixy Thus the construction shown in Fig.1.8 can be used to determine the second moments of area and the product second moment of area about any set of perpendicular axis at a known orientation to the principal axes
10 Mechanics of Materials 2 $1.6 (3) Fig. 1.8. The ellipse of second moments of area. It then follows that X2 Y2 -+-=l (Id2 (I,>2 This equation is the locus of the point P and represents the equation of an ellipse - the ellipse of second moments of area. Draw OQ at an angle 8 to the I, axis, cutting the circle through I, in point S and join SP which is then parallel to the I, axis. Construct a perpendicular to OQ through P to meet OQ in R. Then OR = OQ - RQ = I,-(I,sine-I,sine)sine = I, - (I, - I,) sin2 e = I, cos2 e + I, sin2 e = I, Similarly, repeating the process with OQ1 perpendicular to OQ gives the result OR, = I,, Further, PR = PQcose = (I, sin 8 - I, sin 8) cos 0 = (I, - I,) sin 28 = I,, I Thus the construction shown in Fig. 1.8 can be used to determine the second moments of area and the product second moment of area about any set of perpendicular axis at a known orientation to the principal axes
