CHAPTER VECTOR MECHANICS FOR ENGINEERS: 9 STATICS

Seventh EditionCHAPTER9VECTOR MECHANICS FOR ENGINEERS:STATICSFerdinand P. BeerE. Russell Johnston, Jr.Lecture Notes:J. Walt OlerTexas Tech UniversityDistributed Forces:Moments of Inertia 2003 The McGraw-Hill Companies, Inc. All rights reserved.

SeventhEditionVector Mechanics for Engineers: DynamicsContentsIntroductionMoments of Inertia of an AreaMoment of Inertia of an Area byIntegrationPolar Moment of InertiaRadius of Gyration of an AreaSample Problem 9.1Sample Problem 9.2Parallel Axis TheoremMoments of Inertia of CompositeAreasSample Problem 9.4Sample Problem 9.5Product of InertiaPrincipal Axes and Principal Momentsof InertiaSample Problem 9.6Sample Problem 9.7Mohr’s Circle for Moments and Productsof InertiaSample Problem 9.8Moment of Inertia of a MassParallel Axis TheoremMoment of Inertia of Thin PlatesMoment of Inertia of a 3D Body byIntegrationMoment of Inertia of Common GeometricShapesSample Problem 9.12Moment of Inertia With Respect to anArbitrary AxisEllipsoid of Inertia. Principle Axes ofAxes of Inertia of a Mass 2003 The McGraw-Hill Companies, Inc. All rights reserved.9- 2

SeventhEditionVector Mechanics for Engineers: DynamicsIntroduction Previously considered distributed forces which were proportional to thearea or volume over which they act.- The resultant was obtained by summing or integrating over theareas or volumes.- The moment of the resultant about any axis was determined bycomputing the first moments of the areas or volumes about thataxis. Will now consider forces which are proportional to the area or volumeover which they act but also vary linearly with distance from a given axis.- It will be shown that the magnitude of the resultant depends on thefirst moment of the force distribution with respect to the axis.- The point of application of the resultant depends on the secondmoment of the distribution with respect to the axis. Current chapter will present methods for computing the moments andproducts of inertia for areas and masses. 2003 The McGraw-Hill Companies, Inc. All rights reserved.9- 3

SeventhEditionVector Mechanics for Engineers: DynamicsMoment of Inertia of an Area Consider distributed forces F whose magnitudes areproportional to the elemental areas A on which theyact and also vary linearly with the distance of Afrom a given axis. Example: Consider a beam subjected to pure bending.Internal forces vary linearly with distance from theneutral axis which passes through the section centroid. F ky AR k y dA 0M k y 2 dA y dA Qx first moment2 y dA second moment Example: Consider the net hydrostatic force on asubmerged circular gate. F p A y AR y dAM x y 2 dA 2003 The McGraw-Hill Companies, Inc. All rights reserved.9- 4

SeventhEditionVector Mechanics for Engineers: DynamicsMoment of Inertia of an Area by Integration Second moments or moments of inertia ofan area with respect to the x and y axes,I x y 2 dAI y x 2 dA Evaluation of the integrals is simplified bychoosing d to be a thin strip parallel toone of the coordinate axes. For a rectangular area,hI x y dA y 2bdy 13 bh 320 The formula for rectangular areas may alsobe applied to strips parallel to the axes,dI x 13 y 3 dx 2003 The McGraw-Hill Companies, Inc. All rights reserved.dI y x 2 dA x 2 y dx9- 5

SeventhEditionVector Mechanics for Engineers: DynamicsPolar Moment of Inertia The polar moment of inertia is an importantparameter in problems involving torsion ofcylindrical shafts and rotations of slabs.J 0 r 2 dA The polar moment of inertia is related to therectangular moments of inertia, J 0 r 2 dA x 2 y 2 dA x 2 dA y 2 dA Iy Ix 2003 The McGraw-Hill Companies, Inc. All rights reserved.9- 6

SeventhEditionVector Mechanics for Engineers: DynamicsRadius of Gyration of an Area Consider area A with moment of inertiaIx. Imagine that the area isconcentrated in a thin strip parallel tothe x axis with equivalent Ix.II x k x2 Akx xAkx radius of gyration with respectto the x axis Similarly,Iy k y2 Aky J O kO2 A kO IyAJOAkO2 k x2 k y2 2003 The McGraw-Hill Companies, Inc. All rights reserved.9- 7

SeventhEditionVector Mechanics for Engineers: DynamicsSample Problem 9.1SOLUTION: A differential strip parallel to the x axis is chosen fordA.dI x y 2 dAdA l dy For similar triangles,Determine the moment ofinertia of a triangle with respectto its base.l h y hbl bh yhdA bh ydyh Integrating dIx from y 0 to y h, h ybh 2I x y dA y bdy hy y 3 dyhh002h2hb y3 y 4 h h 34 0 2003 The McGraw-Hill Companies, Inc. All rights reserved.bh3I x 129- 8

SeventhEditionVector Mechanics for Engineers: DynamicsSample Problem 9.2SOLUTION: An annular differential area element is chosen,dA 2 u dudJ O u 2 dArrJ O dJ O u 2 u du 2 u 3du200JO a) Determine the centroidal polarmoment of inertia of a circulararea by direct integration. 2r4 From symmetry, Ix Iy,JO I x I y 2I xb) Using the result of part a,determine the moment of inertiaof a circular area with respect to adiameter. 2003 The McGraw-Hill Companies, Inc. All rights reserved. 2r 4 2I xI diameter I x 4r49- 9

SeventhEditionVector Mechanics for Engineers: DynamicsParallel Axis Theorem Consider moment of inertia I of an area Awith respect to the axis AA’I y 2 dA The axis BB’ passes through the area centroidand is called a centroidal axis.I y 2 dA y d 2 dA y 2 dA 2d y dA d 2 dAI I Ad 2 2003 The McGraw-Hill Companies, Inc. All rights reserved.parallel axis theorem9 - 10

SeventhEditionVector Mechanics for Engineers: DynamicsParallel Axis Theorem Moment of inertia IT of a circular area withrespect to a tangent to the circle, I T I Ad 2 14 r 4 r 2 r 2 54 r 4 Moment of inertia of a triangle with respect to acentroidal axis,I AA I BB Ad 2I BB I AA Ad21 bh 3 12 211 2 bh 3 h1 bh 3 36 2003 The McGraw-Hill Companies, Inc. All rights reserved.9 - 11

SeventhEditionVector Mechanics for Engineers: DynamicsMoments of Inertia of Composite Areas The moment of inertia of a composite area A about a given axis isobtained by adding the moments of inertia of the component areasA1, A2, A3, . , with respect to the same axis. 2003 The McGraw-Hill Companies, Inc. All rights reserved.9 - 12

SeventhEditionVector Mechanics for Engineers: DynamicsMoments of Inertia of Composite Areas 2003 The McGraw-Hill Companies, Inc. All rights reserved.9 - 13

SeventhEditionVector Mechanics for Engineers: DynamicsSample Problem 9.4SOLUTION: Determine location of the centroid ofcomposite section with respect to acoordinate system with origin at thecentroid of the beam section.The strength of a W14x38 rolled steelbeam is increased by attaching a plateto its upper flange.Determine the moment of inertia andradius of gyration with respect to anaxis which is parallel to the plate andpasses through the centroid of thesection. Apply the parallel axis theorem todetermine moments of inertia of beamsection and plate with respect tocomposite section centroidal axis. Calculate the radius of gyration from themoment of inertia of the compositesection. 2003 The McGraw-Hill Companies, Inc. All rights reserved.9 - 14

SeventhEditionVector Mechanics for Engineers: DynamicsSample Problem 9.4SOLUTION: Determine location of the centroid of composite sectionwith respect to a coordinate system with origin at thecentroid of the beam section.SectionA, in 2y , in. yA, in 3Plate6.757.425 50.12Beam Section 11.2000 A 17.95 yA 50.12Y A yA 2003 The McGraw-Hill Companies, Inc. All rights reserved.yA 50.12 in 3 Y 2.792 in.2A17.95 in 9 - 15

SeventhEditionVector Mechanics for Engineers: DynamicsSample Problem 9.4 Apply the parallel axis theorem to determine moments ofinertia of beam section and plate with respect to compositesection centroidal axis.I x , beam section I x AY 2 385 11.20 2.792 2 472.3 in 4 31 9 3 6.75 7.425 2.792 2I x , plate I x Ad 2 124 145.2 in 4I x I x , beam section I x , plate 472.3 145.2I x 618 in 4 Calculate the radius of gyration from the moment of inertiaof the composite section.k x I x 617.5 in 4 A 17.95 in 2 2003 The McGraw-Hill Companies, Inc. All rights reserved.k x 5.87 in.9 - 16

SeventhEditionVector Mechanics for Engineers: DynamicsSample Problem 9.5SOLUTION: Compute the moments of inertia of thebounding rectangle and half-circle withrespect to the x axis. The moment of inertia of the shaded area isobtained by subtracting the moment ofinertia of the half-circle from the momentof inertia of the rectangle.Determine the moment of inertiaof the shaded area with respect tothe x axis. 2003 The McGraw-Hill Companies, Inc. All rights reserved.9 - 17

SeventhEditionVector Mechanics for Engineers: DynamicsSample Problem 9.5SOLUTION: Compute the moments of inertia of the boundingrectangle and half-circle with respect to the x axis.Rectangle:I x 13 bh3 13 240 120 138.2 106 mm 4Half-circle:moment of inertia with respect to AA’,I AA 18 r 4 18 90 4 25.76 106 mm 4moment of inertia with respect to x’,4r 4 90 38.2 mma 3 3 b 120 - a 81.8 mmA 12 r 12 90 22 12.72 103 mm 2 I x I AA Aa 2 25.76 106 12.72 103 7.20 106 mm 4moment of inertia with respect to x, I x I x Ab 2 7.20 106 12.72 103 81.8 2 92.3 106 mm 4 2003 The McGraw-Hill Companies, Inc. All rights reserved.9 - 18

SeventhEditionVector Mechanics for Engineers: DynamicsSample Problem 9.5 The moment of inertia of the shaded area is obtained bysubtracting the moment of inertia of the half-circle fromthe moment of inertia of the rectangle.Ix 138.2 106 mm 4 92.3 106 mm 4I x 45.9 106 mm 4 2003 The McGraw-Hill Companies, Inc. All rights reserved.9 - 19

SeventhEditionVector Mechanics for Engineers: DynamicsProduct of Inertia Product of Inertia:I xy xy dA When the x axis, the y axis, or both are anaxis of symmetry, the product of inertia iszero. Parallel axis theorem for products of inertia:I xy I xy x yA 2003 The McGraw-Hill Companies, Inc. All rights reserved.9 - 20

SeventhEditionVector Mechanics for Engineers: DynamicsPrincipal Axes and Principal Moments of Inertia The change of axes yieldsIx Iy Ix IyI x cos 2 I xy sin 2 22Ix I y Ix I y I y cos 2 I xy sin 2 22Ix IyI x y sin 2 I xy cos 2 2Given I x y 2 dA I y x 2 dAI xy xy dAwe wish to determine momentsand product of inertia withrespect to new axes x’ and y’.Note: x x cos y sin y y cos x sin The equations for Ix’ and Ix’y’ are theparametric equations for a circle, I x I ave 2 I x2 y R 2Ix IyI ave 2 Ix Iy 2 I xyR 2 The equations for Iy’ and Ix’y’ lead to thesame circle. 2003 The McGraw-Hill Companies, Inc. All rights reserved.9 - 21

SeventhEditionVector Mechanics for Engineers: DynamicsPrincipal Axes and Principal Moments of Inertia At the points A and B, Ix’y’ 0 and Ix’ isa maximum and minimum, respectively.I max,min I ave Rtan 2 m 2 I xyIx I y The equation for m defines twoangles, 90o apart which correspond tothe principal axes of the area about O. I x I ave 2 I x2 y R 2Ix IyI ave 2 Ix Iy 2 I xyR 2 Imax and Imin are the principal momentsof inertia of the area about O. 2003 The McGraw-Hill Companies, Inc. All rights reserved.9 - 22

SeventhEditionVector Mechanics for Engineers: DynamicsSample Problem 9.6SOLUTION: Determine the product of inertia usingdirect integration with the parallel axistheorem on vertical differential area strips Apply the parallel axis theorem toevaluate the product of inertia with respectto the centroidal axes.Determine the product of inertia ofthe right triangle (a) with respectto the x and y axes and(b) with respect to centroidal axesparallel to the x and y axes. 2003 The McGraw-Hill Companies, Inc. All rights reserved.9 - 23

SeventhEditionVector Mechanics for Engineers: DynamicsSample Problem 9.6SOLUTION: Determine the product of inertia using direct integrationwith the parallel axis theorem on vertical differentialarea strips x x y h 1 dA y dx h 1 dx b b x xel xyel 12 y 12 h 1 b Integrating dIx from x 0 to x b,bI xy dI xy xel yel dA x0b 12 2 2x h 1 dx b b2 x 2 x3 x 4 x 2 x x h dx h 2 2 b 2b 2 0 4 3b 8b 023I xy 2003 The McGraw-Hill Companies, Inc. All rights reserved.1 b2h2249 - 24