Report

Engineering 36 Chp10: Moment of Inertia Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected] Engineering-36: Engineering Mechanics - Statics 1 Bruce Mayer, PE [email protected] • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx Introduction Previously we considered distributed forces which were proportional to the area or volume over which they act; i.e., Centroids • The resultant Force was obtained by summing or integrating over the areas or volumes • The moment of the resultant about any axis was determined by Calculating the FIRST MOMENTS of the areas or volumes about that axis – FIRST MOMENT = An Area/Volume/Mass INCREMENT (or INTENSITY) times its LEVER ARM (kind of like F•d for Torque) Lever Arm = the ┴ Distance to Axis of Interest Engineering-36: Engineering Mechanics - Statics 2 Bruce Mayer, PE [email protected] • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx Introduction cont. Next consider forces which are proportional to the Area or Volume over 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 the FIRST MOMENT of the force distribution with respect to the axis of interest • The Equivalent point of application of the resultant depends on the SECOND moment of the Force distribution with respect to the given axis Engineering-36: Engineering Mechanics - Statics 3 Bruce Mayer, PE [email protected] • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx Area Moment of Inertia Consider distributed forces, F , whose magnitudes are proportional to the elemental areas, A, on which they act, and also vary LINEARLY with the Distance of A from a given axis. Example: Consider a beam subjected to pure bending. A Mechanics-ofMaterials Analysis Shows that forces vary linearly with distance from the F kyA M yF NEUTRAL AXIS which passes through k Matl/GeomCONST the section centroid. In This Case R k y dA 0 y dA Qy first moment(ResultantForce) M k y 2 dA 2 y dA secondmoment(ResultantMoment) Engineering-36: Engineering Mechanics - Statics 4 Bruce Mayer, PE [email protected] • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx Example → 2nd Moment of Area Find the Resultant hydrostatic force on a submerged CIRCULAR gate, and its Point of Application • Recall That the gage Pressure, p, is Proportional to the Depth, y, by p = γy • Consider an Increment of Area, A • Then The Associated Force Increment F = pA • Locate Point of Resultant Application, yP, by Equating Moments about the x-Axis F pA yA Let A dA dF pdA ydA taking dF R y dA y A also dM x ydF y ydA A 2 dM M y R y x x P dA Engineering-36: Engineering Mechanics - Statics 5 y 2 dA I x k x2 yP y dA y A y Bruce Mayer, PE [email protected] • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx More on this Later Second Moment The Areal Moments of Inertia Take 2 2 I u dA I u these forms v i Ai v In both the Integral and the Sum Observe that since u is a distance, so the expressions are of the form 2 I v LeverArm Intensity Thus “moments of Inertia” are the SECOND Moment of Area or Mass Engineering-36: Engineering Mechanics - Statics 6 Bruce Mayer, PE [email protected] • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx Moment of Inertia by Integration SECOND MOMENTS or MOMENTS OF INERTIA of an area with respect to the x and y axes: 2 2 I x y dA I y x dA Evaluation of the integrals is simplified by choosing dA to be a thin strip parallel to one of the coordinate axes. I x y dA y a x dy 2 2 I y x 2 dA x 2 ydx Engineering-36: Engineering Mechanics - Statics 7 Bruce Mayer, PE [email protected] • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx Base Case → Rectangle Find Moment of Inertia for a Rectangular h Area 1 3 2 2 I x y dA y bdy bh 3 0 Apply This Basic formula to strip-like rectangular areas That are Most conveniently Parallel to the Axes Consider a Horizontal Strip • dIx: b∙y → [a-x]∙y • dIy → Subtract strips: b+ → a; b− → x dIx y 2 dA y 2 a x dy dI y dI y Strip of Lengtha dI y Strip of Length x 1 3 1 3 1 3 3 dI y a dy x dy a x dy 3 3 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics 3 8 [email protected] • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx Use BaseLine Case for Iy Engineering-36: Engineering Mechanics - Statics 9 Bruce Mayer, PE [email protected] • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx Engineering-36: Engineering Mechanics - Statics 10 Bruce Mayer, PE [email protected] • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx Polar Moment of Inertia The polar moment of inertia is an important parameter in problems involving torsion of cylindrical shafts, Torsion in Welded Joints, and the rotation of slabs In Torsion Problems, Define a Moment of Inertia Relative to the Pivot-Point, or “Pole”, at O J O r dA 2 Relate JO to Ix & Iy Using The Pythagorean Theorem J O r 2 dA x 2 y 2 dA x 2 dA y 2 dA Iy Ix Engineering-36: Engineering Mechanics - Statics 11 Bruce Mayer, PE [email protected] • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx Radius of Gyration (k↔r) Consider area A with moment of inertia Ix. Imagine that the area is concentrated in a thin strip parallel to the x axis with equivalent Ix, Then Define the RADIUS OF GYRATION, r I x r A rx 2 x rx ry rO Engineering-36: Engineering Mechanics - Statics 12 Ix A I y r A ry 2 y Similarly in the Polar Case the Radius of Gyration JO 2 J O rO A rO A rO2 rx2 ry2 Bruce Mayer, PE [email protected] • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx Iy A Example 1 SOLUTION: • Choose dA as a differential strip parallel to the x axis dIx y 2dA dA l dy By Similar Triangles See that l in linear in y l h y b h l b h y h dA b h y dy h Integrating dIx from y = 0 to y = h h y b I x y 2 dA y 2b dy hy2 y 3 dy h h0 0 h Determine the moment of inertia of a triangle with respect to its base. Engineering-36: Engineering Mechanics - Statics 13 h h b y3 y4 h h 3 4 0 bh3 I x 12 Bruce Mayer, PE [email protected] • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx Example 2 SOLUTION: • Choose dA as a differential annulus of width du dA 2 u du dJO u 2 dA r r J O dJO u 2 u du 2 u 3 du 2 0 JO 2 0 r4 By symmetry, Ix = Iy, so a) Determine the centroidal POLAR moment of inertia 4 of a circular area by direct J I I 2 I r O x y x integration. 2 b) Using the result of part a, 4 determine the moment of I diameter I x r inertia of a circular area 4 with respect to a diameter. Engineering-36: Engineering Mechanics - Statics 14 Bruce Mayer, PE [email protected] • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx 2I x Parallel Axis Theorem With Respect to The OffSet Axis AA’ I AA' y 2 dA y d dA 2 y2 dA 2d ydA d 2 dA Consider the moment of inertia I of an area A with respect to the axis AA’ Next Consider Axis BB’ That Passes thru The Area Centriod Engineering-36: Engineering Mechanics - Statics 15 Now Consider the Middle Integral 2d ydA 2d 1st Moment WRT BB' But this is a 1st Moment about an Axis Thru the Centroid, Which By Definition = 0 Bruce Mayer, PE [email protected] • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx Parallel Axis Theorem cont. I AA' y2 dA 2d ydA d 2 dA I AA' I BB ' 0 d 2 A but BB' is a Cent roidalAxis, so I BB' I Next Consider the Last Integral d 2 dA d 2 A The Relation for I w.r.t. the Centroidal Axis and a parallel Axis: Engineering-36: Engineering Mechanics - Statics 16 Thus The Formal Statement of the Parallel Axis Theorem I I d A 2 • Note: This Theorem Applies ONLY to a CENTROIDAL Axis and an Axis PARALLEL to it Bruce Mayer, PE [email protected] • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx Parallel Axis Theorem Exmpls Find the Moment of inertia IT of a circular area with respect to a tangent to the circle IT I Ad 2 14 r 4 r 2 r 2 54 r 4 Find the Moment of inertia of a triangle with respect to a centroidal axis I AA I BB Ad I Ad 2 I I AA Ad 2 1 12 2 bh bh h 3 1 2 I 361 bh3 Engineering-36: Engineering Mechanics - Statics 17 Bruce Mayer, PE [email protected] • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx 1 3 2 Composite Moments of Inertia The moment of inertia of a composite area, A, about a given axis is obtained by ADDING the moments of inertia of the COMPONENT areas A1, A2, A3, ... , with respect to the same axis. • This Analogous to Finding the Centroid of a Composite Engineering-36: Engineering Mechanics - Statics 18 Bruce Mayer, PE [email protected] • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx Standard Shapes are Tabulated 7230 mm 2 11.20 in 2 Engineering-36: Engineering Mechanics - Statics 19 160.2 106 mm 4 385 in 4 Bruce Mayer, PE [email protected] • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx Example 3 SOLUTION PLAN Abeam 11.20in 2 I beam 385in 4 The strength of a W14x38 rolled steel beam is increased by welding a ¾” plate to its upper flange. Determine the Moment Of Inertia and Radius Of Gyration with respect to an axis which is parallel to the plate and passes through the centroid of the NEW section. Engineering-36: Engineering Mechanics - Statics 20 • Determine location of the Centroid for the Composite section with respect to a coordinate system with origin at the centroid of the BEAM section. • Apply the PARALLEL AXIS THEOREM to determine moments of inertia of the beam section and plate with respect to COMPOSITE SECTION Centroidal Axis. • Calculate the Radius of Gyration from the Moment of Inertia of the Composite Section Bruce Mayer, PE [email protected] • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx Engineering-36: Engineering Mechanics - Statics 21 Bruce Mayer, PE [email protected] • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx Example 3 cont. Determine location of the CENTROID of the COMPOSITE SECTION with respect to a coordinate system with origin at the CENTROID of the BEAM section. The First Moment Tabulation Section A, in 2 y , in. P late 6.75 7.425 50.12 Beam Section 11.20 0 A 17.95 2.792 in. 0 yA 50.12 Calc the Centroid Y A yA Engineering-36: Engineering Mechanics - Statics 22 yA, in 3 yA 50.12 in 3 Y 2.792 in. 2 A 17.95 in Bruce Mayer, PE [email protected] • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx Example 3 cont.2 2.792 in. Apply the parallel axis theorem to determine moments of inertia of beam section and plate with respect to COMPOSITE SECTION centroidal axis. I x,beam section I x AY 2 385 11.202.792 2 472.3 in 4 I x,plate I x Ad 2 121 9 34 6.757.425 2.792 3 2 145.2 in 4 Then the Composite Moment Of Inertia About the Composite Centroidal Axis, at x’ I x I x,beamsection I x,plate 472.3 145.2 I x 617.5 in 4 Engineering-36: Engineering Mechanics - Statics 23 Bruce Mayer, PE [email protected] • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx Example 3 cont.3 Finally Calculate the Radius of Gyration for the Composite From Quantities Previously Determined 4 I x 617.5 in k x A 17.95in 2 So Finally k x 5.87 in. 2.792 in. Engineering-36: Engineering Mechanics - Statics 24 Bruce Mayer, PE [email protected] • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx WhiteBoard Work Let’s Work This Nice Problem Find the Area Moment of Inertia of the Parallelogram about its Centroidal axis y’ Engineering-36: Engineering Mechanics - Statics 25 Bruce Mayer, PE [email protected] • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx Engineering 36 Appendix dy µx µs sinh dx T0 T0 Bruce Mayer, PE Registered Electrical & Mechanical Engineer [email protected] Engineering-36: Engineering Mechanics - Statics 26 Bruce Mayer, PE [email protected] • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx Engineering-36: Engineering Mechanics - Statics 27 Bruce Mayer, PE [email protected] • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx Engineering-36: Engineering Mechanics - Statics 28 Bruce Mayer, PE [email protected] • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx