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Engineering Mechanics: Statics

Engineering Mechanics: Statics. Appendix A: Area Moments of Inertia. Moment of Inertia. 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)

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Engineering Mechanics: Statics

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  1. Engineering Mechanics: Statics Appendix A: Area Moments of Inertia

  2. Moment of Inertia 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 I is a function of geometry only! Hydrostatic pressure Bending moment in beam Moment of inertia of area/ Second moment of area(I ) Torsion in shaft

  3. Rectangular moment of inertia Polar moment of inertia Definitions -- Moment of inertia about x-axis • Notice that Ix, Iy, Iz involve the square of the distance from the inertia axis • -- always positive! • dimensions = L4 (ex. m4 or mm4)

  4. 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 x-axis, and the polar axis z through O. -- Must remember!: for a rectangular area, : for a circular area, - see sample problem A/3

  5. 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 kxfrom the x-axis. The moment of inertia about x-axis is Ix. Therefore, The distance kx = radius of gyration of the area about x-axis Radius of Gyration

  6. Similarly, Do not confused with centroid C! Radius of Gyration

  7. Transfer of Axes Moment of inertia of an area about a noncentroidal axis The axis between which the transfer is made must be parallel One of the axes must pass through the centroid of the area and with the centroid on x0-axis Parallel-axis theorems

  8. Composite Areas Centroid of composite areas: 400 mm Part Area, A 100 mm 400 mm Sum SA S S 100 mm

  9. 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 The radius of gyration for the composite area cannot be added, k =I/A Part Area, Adxdy Adx2Ady2 Sum SA SAdx2SAdy2S S

  10. Example A/7 Calculate the moment of inertia and radius of gyration about the x-axis for the shaded area shown

  11. 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 + Adx dy

  12. Sample Problem A/8 & A/10 Determine the product of inertia of the area shown with respect to the x-y axes. 

  13. Rotation of Axes To calculate the moment of inertia of an area about an inclined axes Ix’ = y’2 dA = (ycosq– xsinq)2 dA Iy’ = x’2 dA = (ysinq– xcosq)2 dA -- expand & substitute sin2q = (1- cos 2q)/2 cos2q = (1+ cos 2q)/2

  14. Rotation of Axes The angle which makes Ix’ and Iy’ either max or min dIx’/dq = (Iy - Ix)sin 2q - 2Ixycos 2q = 0 The critical angle a: • This equation gives two value of 2a [tan 2a = tan (2a+p) ] • obtain two values for a (differ by p/2) • axis of minimum moment of inertia • axis of maximum moment of inertia called “Principal Axes of Inertia”

  15. Mohr’s Circle of Inertia R Imin S Imax 5. Angle 2a is found from AS and OS as 6. Imax = O + R and Imin = O - R • Draw x-axis as I and y-axis as Ixy • Plot point A at (Ix, Ixy) and B at (Iy, -Ixy) • Find the center of the circle at O 4. Radius of the circle is OA or OB

  16. Sample Problem A/11 Determine the orientation of the principle axes of inertia through the centroid of the angle section and determine the corresponding maximum and minimum moments of inertia.

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