Area Moments of Inertia by Integration

1 Area Moments of Inertia by Integration Second momentsor Moments of inertiaof an area with respect to the xand yaxes, dAxIdAyIyx22 Evaluation of the integrals is simplified by choosing dAto be a thin strip parallel to one of the coordinate axes1ME101 -Division IIIK austubh DasguptaArea Moments of InertiaProducts of Inertia : for problems involving unsymmetrical cross-sections and in calculation of MI about rotated axes. It may be +ve, -ve, or zero Product of Inertia of area A axes:x andy are the coordinates of the element of area dA=xy dAxyIxy When the xaxis, the yaxis, or both are an axis of symmetry, the product of Inertia is zero. Parallel axis theorem for products of Inertia :AyxIIxyxy + Ixy+ Ixy-Ixy-IxyQuadrants2ME101 -Division IIIK austubh DasguptaArea Moments of InertiaRotation of Axes Product of Inertia is useful in calculating MI @ inclined axes. Determination of axes about which the MI is a maximum and a minimum dAxyIdAxIdAyIxyyx22 Moments and product of Inertia new axes x and y ?

2 Sincossincosxyyyxx Note: dAxyyxdAyxIdAyxdAxIdAxydAyIyxyx sincossincos''sincos'sincos'''22'22' 2cossincos2sin2/1cossin22cos1cos22cos1si n2222 2cos2sin22sin2cos222sin2cos22'xyyxyxxyyx yxyxyyxyxxIIIIIIIIIIIIIIII 3ME101 -Division IIIK austubh DasguptaArea Moments of InertiaRotation of Axes Adding first two eqns:Ix + Iy = Ix + Iy = Iz The Polar MI @ O 2cos2sin22sin2cos222sin2cos22'xyyxyxxyyx yxyxyyxyxxIIIIIIIIIIIIIIII Angle which makes Ix and Iy either max or min can be found by setting the derivative of either Ix or Iy equal to zero:Denoting this critical angle by 02cos22sin' xyxyxIIIddIxyxyIII 22tan two values of 2 which differ by since tan2 = tan(2 + ) two solutions for will differ by /2 one value of will define the axis of maximum MIand the otherdefines the axis of minimum MI These two rectangular axes are called the principal axes of inertia4ME101 -Division IIIK austubh DasguptaArea Moments of InertiaRotation of Axes 2cos2sin22sin2cos222sin2cos22'xyyxyxxyyx yxyxyyxyxxIIIIIIIIIIIIIIII xyxyxyxyIIIIII 22cos2sin22tan Substituting in the third eqn for critical value of 2 : Ix y = 0 Product of Inertia Ix y is zero for the Principal Axes of inertiaSubstituting sin2 and cos2 in first two eqns for Principal Moments of Inertia : 042124212@22min22max xyxyyxyxxyyxyxIIIIIIIIIIIII5ME101 -Division IIIK austubh DasguptaArea Moments of InertiaMohr s Circle of Inertia .

3 Graphical representation of the MI equations-For given values of Ix, Iy,& Ixy, corresponding values of Ix , Iy , & Ix y may be determined from the diagram for any desired angle . 2cos2sin22sin2cos222sin2cos22'xyyxyxxyyx yxyxyyxyxxIIIIIIIIIIIIIIII xyxyIII 22tan 042124212@22min22max xyxyyxyxxyyxyxIIIIIIIIIIIII 2222222xyyxyxaveyxavexIIIRIIIRIII At the pointsAandB, Ix y = 0 andIx takes the maximum and minimum valuesRIIave minmax,IIxy6ME101 -Division IIIK austubh DasguptaArea Moments of InertiaMohr s Circle of Inertia : Construction 2cos2sin22sin2cos222sin2cos22'xyyxyxxyyx yxyxyyxyxxIIIIIIIIIIIIIIII xyxyIII 22tan 042124212@22min22max xyxyyxyxxyyxyxIIIIIIIIIIIIIC hoose horzaxis MIChoose vertaxis PIPoint A known {Ix, Ixy} Point B known {Iy, -Ixy} Circle with diaABAngle for Area Angle 2 to horz(same sense) Imax, IminAngle xto x = Angle OA to OC = 2 Same sensePoint C Ix , Ix y Point D Iy 7ME101 -Division IIIK austubh DasguptaArea Moments of InertiaExample: Product of InertiaDetermine the product of Inertia of the right triangle (a)with respect to the xandyaxes and (b)with respect to centroidal axes parallel to the.

4 Determine the product of Inertia using direct Integration with the parallel axis theorem on vertical differential area strips Apply the parallel axis theorem to evaluate the product of Inertia with respect to the centroidal -Division IIIK austubh DasguptaArea Moments of InertiaExamplesSOLUTION: Determine the product of Inertia using direct Integration with the parallel axis theorem on vertical differential area strips bxhyyxxdxbxhdxydAbxhyelel1112121 Integrating dIxfrom x= 0 to x= b, bbbelelxyxybxbxxhdxbxbxxhdxbxhxdAyxdII02 43220232202221834221 22241hbIxy 9ME101 -Division IIIK austubh DasguptaArea Moments of InertiaExamples Apply the parallel axis theorem to evaluate the product of Inertia with respect to the centroidal With the results from part a, bhhbhbIAyxIIyxyxxy21313122241 22721hbIyx SOLUTION10ME101 -Division IIIK austubh DasguptaArea Moments of InertiaExample: Mohr s Circle of InertiaThe Moments and product of Inertia with respect to the xand yaxes are Ix= mm4, Iy= mm4, and Ixy= Mohr s circle, determine (a)the principal axes about O, (b)the values of the principal Moments about O, and (c) the values of the Moments and product of Inertia about the x and y axesSOLUTION: Plot the points (Ix , Ixy) and (Iy ,-Ixy).

5 Construct Mohr s circle based on the circle diameter between the points. Based on the circle, determine the orientation of the principal axes and the principal Moments of Inertia . Based on the circle, evaluate the Moments and product of Inertia with respect to the x y -Division IIIK austubh DasguptaArea Moments of InertiaExample: Mohr s Circle of xyyxIIISOLUTION: Plot the points (Ix , Ixy) and (Iy ,-Ixy). Construct Mohr s circle based on the circle diameter between the points. DXCDRIICDIIIOC yxyxave Based on the circle, determine the orientation of the principal axes and the principal Moments of Inertia . 12ME101 -Division IIIK austubh DasguptaArea Moments of InertiaExample: Mohr s Circle of RIOCave Based on the circle, evaluate the Moments and product of Inertia with respect to the x y points X and Y corresponding to the x and y axes are obtained by rotating CXand CYcounterclockwise through an angle 2(60o) = 120o. The angle that CX forms with the horz is f= 120o ' ' ' yxI13ME101 -Division IIIK austubh DasguptaMass moment of Inertia Application in rigid body dynamics-Measure of distribution of mass of a rigid body the axis (constant property for that axis)I = r2dmr= perpendicular distance of the mass element dmfrom the axis O-Or2 m.

6 Measure of the Inertia of the system14ME101 -Division IIIK austubh DasguptaMass moment of Inertia About individual coordinate axes15ME101 -Division IIIK austubh DasguptaMass moment of Inertia Parallel Axis Theorem16ME101 -Division IIIK austubh DasguptaMass moment of InertiaMoments of Inertia of Thin Plates For a thin plate of uniform thickness tand homogeneous material of density r, the mass moment of Inertia with respect to axis AA contained in the plate isareaAAAAItdArtdmrI,22 rr Similarly, for perpendicular axis BB which is also contained in the plate,areaBBBBItI, r For the axis CC which is perpendicular to the plate, BBAA areaBBareaAAareaCCCIIIItJtI ,,,rr17ME101 -Division IIIK austubh DasguptaMass moment of InertiaMoments of Inertia of Thin Plates For the principal centroidal axes on a rectangular plate, 21213121,mabatItIareaAAAA rr 21213121,mbabtItIareaBBBB rr 22121,,bamIIImassBBmassAACC For centroidal axes on a circular plate, 241441,mrrtItIIareaAABBAA rr18ME101 -Division IIIK austubh DasguptaMass moment of InertiaMoments of Inertia of a 3D Body by Integration moment of Inertia of a homogeneous body is obtained from double or triple integrations of the form dVrI2r For bodies with two planes of symmetry, the moment of Inertia may be obtained from a single Integration by choosing thin slabs perpendicular to the planes of symmetry for dm.

7 The moment of Inertia with respect to a particular axis for a composite body may be obtained by adding the Moments of Inertia with respect to the same axis of the -Division IIIK austubh DasguptaMass moment of InertiaMI of some common geometric shapes20ME101 -Division IIIK austubh Dasgupta