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Engineering Mechanics: Statics Appendix A: Area Moments of Inertia Moment of Inertia Hydrostatic pressure Bending moment in beam When forces are distributed continuously over an area, it is often necessary to calculate moment of these forces about some axis (in or perpendicular to the plane of area) Frequently, intensity of the distributed force is proportional to the distance of the line of action from the moment axis, p = ky dM = y(pdA) = ky2dA M k y2dA Moment of inertia of area/ Second moment of area (I ) Torsion in shaft I is a function of geometry only! Definitions Rectangular moment of inertia Ix y2dA Iy x2dA -- Moment of inertia about x-axis Polar moment of inertia Iz r 2dA Ix Iy • Notice that Ix, Iy, Iz involve the square of the distance from the inertia axis -- always positive! • dimensions = L4 (ex. m4 or mm4) Sample Problem A/1 Determine the moments of inertia of the rectangular area about the centroidal x0- and y0-axes, the centroidal polar axis z0 through C, the xaxis, and the polar axis z through O. 3 3 -- Must remember!: for a rectangular area, Ix bh , Iy hb 12 12 4 : for a circular area, Ix Iy r - see sample problem A/3 4 Radius of Gyration For an area A with moment of inertia Ix and Iy Visualize it as concentrated into a long narrow strip of area A a distance kx from the x-axis. The moment of inertia about x-axis is Ix. Therefore, kx2A Ix The distance kx = radius of gyration of the area about x-axis kx Ix A Radius of Gyration Similarly, ky Iy A kz Iz A Do not confused with centroid C! kz2 kx2 ky2 Transfer of Axes Moment of inertia of an area about a noncentroidal axis dIx y2dA (y0 dx )2 dA Ix dIx y02dA 2dx y0dA dx2 dA Ix Ix Adx 2 Ix Ix Adx2 Iy Iy Ady 2 Ay0 and y0 0 with the centroid on x0-axis Parallel-axis theorems Iz Iz Ad2 The axis between which the transfer is made must be parallel One of the axes must pass through the centroid of the area Composite Areas Centroid of composite areas: x Ai xi y Ai Ai yi Ai 400 mm 100 mm Part Area, A Sum SA x y Ax Ay S Ax S Ay 400 mm 100 mm Composite Areas The moment of inertia of a composite area about a particular axis is the sum of the moments of inertia of its component parts about the same axis. I = SI + SAd2 o The radius of gyration for the composite area cannot be added, k = I/A Part Sum Area, A SA dx dy Adx2 Ady2 Ix SAdx2 SAdy2 S Ix Iy S Iy Example A/7 Calculate the moment of inertia and radius of gyration about the x-axis for the shaded area shown Table D/3 Products of Inertia Unsymmetrical cross section Ixy = xydA may be positive, negative or zero Ixy = 0 when either the reference axes is an axis of symmetry because x(-y)dA cancel x(+y)dA Transfer of Axes Ixy = (x0+dy)(y0+dx)dA Ixy Ixy Adxdy Sample Problem A/8 & A/10 Determine the product of inertia of the area shown with respect to the x-y axes. Rotation of Axes To calculate the moment of inertia of an area about an inclined axes Ix’ = y’2 dA = (ycos q – xsin q )2 dA Iy’ = x’2 dA = (ysin q – xcos q )2 dA -- expand & substitute Ix ' Iy ' Ix Iy 2 Ix Iy Ix ' y ' 2 Ix Iy 2 Ix Iy 2 Ix Iy 2 sin2q = (1- cos 2q)/2 cos2q = (1+ cos 2q)/2 cos 2q Ixy sin2q cos 2q Ixy sin2q sin2q Ixy cos 2q 2 Ix Iy I x' 2 I x 'y ' 2 I x Iy 4 2 Ixy 2 Mohr’s Circle of Inertia R Imin S 1. Draw x-axis as I and y-axis as Ixy 2. Plot point A at (Ix, Ixy) and B at (Iy, -Ixy) 3. Find the center of the circle at O 4. Radius of the circle is OA or OB R OS2 AS2 Imax 5. Angle 2 is found from AS and OS as tan2 6. AS OS Imax = O + R and Imin = O - R Rotation of Axes The critical angle : tan2 2Ixy Iy Ix This equation gives two value of 2 [tan 2 = tan (2+) ] obtain two values for (differ by /2) axis of minimum moment of inertia axis of maximum moment of inertia Imax Iave R Imin Iave R called “Principal Axes of Inertia” Sample Problem A/11 Determine the orientation of the principal axes of inertia through the centroid of the angle section and determine the corresponding maximum and minimum moments of inertia.