Machinery's Handbook, 31st Edition
MOMENTS OF INERTIA AND OTHER FORMULAS 251 Polar Area Moment of Inertia and Section Modulus.— The polar moment of inertia J of a cross section with respect to a polar axis, that is, an axis at right angles to the plane of the cross section, is defined as the moment of inertia of the cross section with respect to the point of intersection of the axis and the plane. The polar moment of inertia may be found by taking the sum of the moments of inertia about two perpendicular axes lying in the plane of the cross section and passing through this point. Thus, for example, the polar mo- ment of inertia of a circular or a square area with respect to a polar axis through the center of gravity is equal to twice the moment of inertia with respect to an axis lying in the plane of the cross section and passing through the center of gravity. The polar moment of inertia with respect to a polar axis through the center of gravity is required for problems involving the torsional strength of shafts since this axis is usually the axis about which twisting of the shaft takes place. The polar section modulus (also called section modulus of torsion) Z p for circular sec tions may be found by dividing the polar moment of inertia J by the distance c from the center of gravity to the most remote fiber. This method may be used to find the approxi mate value of the polar section modulus of sections that are nearly round. For other than circular cross sections, however, the polar section modulus does not equal the polar mo- ment of inertia divided by the distance c . The accompanying table Polar Moment of Inertia and Polar Section Modulus on page 252 gives formulas for the polar section modulus for several different cross sections. The polar section modulus multiplied by the allowable torsional shearing stress gives the allowable twisting moment to which a shaft may be subjected; see Formula (7) on page 296 . Polar Mass Moment of Inertia.— Starting on page 253 , formulas for mass moment of inertia * J M of various solids are given in a series of tables. The example that follows illus- trates the derivaion of J M for one of the bodies given on page 253 . Example, Polar Mass Moment of Inertia of a Hollow Circular Section: Referring to the figure Hollow Cylinder on page 253 , consider a strip of width dr on a hollow circular section whose inner radius is r and outer radius is R . The mass of the strip = 2 π rdr ρ , where ρ is the density of material. In order to get the mass of an individual section, integrate the mass of the strip from r to R . M 2 πρ rdr r R ∫ 2 πρ rdr r R ∫ 2 πρ r 2 2 -- r R = = = 2 πρ R 2 2 --- r 2 2 – -- = πρ R 2 r 2 – ( ) = The second moment of the strip about the AA axis = 2 π r dr ρ r 2 . To find the polar mo- ment of inertia about the AA axis, integrate the 2nd moment from r to R . J r dr r r dr r R r R r R r R r R r M R r 2 2 2 4 2 4 4 2 2 2 M r R r R r R 2 3 4 4 4 2 2 2 2 2 2 2 2 2 2 π ρ πρ πρ πρ πρ πρ = = = = − = − + = − + = + ^ a ^ ^ ^ ^ ^ ^ h h k h h h h h : D ∫ ∫
*In some books the symbol I denotes the polar moment of inertia of masses; J M is used in this Handbook to avoid confusion with moments of inertia of plane areas.
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