Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Engineering 36 Chp10: Moment of Inertia Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu

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

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

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-of-Materials Analysis Shows that forces vary linearly with distance from the NEUTRAL AXIS which passes through the section centroid. In This Case 2nd moment inplies the [Lever-Arm]2 times some intensity

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 = pA Locate Point of Resultant Application, yP, by Equating Moments about the x-Axis The FORCE is equal to the sp-wt times the distance to GEOMETRIC Centroid of the total are in question as dF is proportional to y*dA. The incremental moment, dM, about the x-axis is proportional to the SQUARE of the distance to dA. The 2nd power implies a squaring of the lever arm; hence the SECOND Moment designation More on this Later

Second Moment The Areal Moments of Inertia Take these forms In both the Integral and the Sum Observe that since u is a distance, so the expressions are of the form Thus “moments of Inertia” are the SECOND Moment of Area or Mass

Moment of Inertia by Integration SECOND MOMENTS or MOMENTS OF INERTIA of an area with respect to the x and y axes: Evaluation of the integrals is simplified by choosing dA to be a thin strip parallel to one of the coordinate axes.

Base Case → Rectangle Find Moment of Inertia for a Rectangular Area 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

Use BaseLine Case for Iy

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 Relate JO to Ix & Iy Using The Pythagorean Theorem

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 Similarly in the Polar Case the Radius of Gyration

Example 1 SOLUTION: By Similar Triangles See that l in linear in y Choose dA as a differential strip parallel to the x axis By Similar Triangles See that l in linear in y Integrating dIx from y = 0 to y = h l in LINEAR in y. l = b => y = 0. l = 0 => y = h. So slope m = del-l/del-y = (0-b)/(h-0). Intercept => l = b, when y = 0. So l = (-b/h)y +b Determine the moment of inertia of a triangle with respect to its base.

Example 2 SOLUTION: By symmetry, Ix = Iy, so Choose dA as a differential annulus of width du By symmetry, Ix = Iy, so Determine the centroidal POLAR moment of inertia of a circular area by direct integration. Using the result of part a, determine the moment of inertia of a circular area with respect to a diameter. Polar moments are taken about the origin

Parallel Axis Theorem With Respect to The OffSet Axis AA’ Now Consider the Middle Integral 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 But this is a 1st Moment about an Axis Thru the Centroid, Which By Definition = 0

Parallel Axis Theorem cont. Thus The Formal Statement of the Parallel Axis Theorem Next Consider the Last Integral Note: This Theorem Applies ONLY to a CENTROIDAL Axis and an Axis PARALLEL to it The Relation for I w.r.t. the Centroidal Axis and a parallel Axis:

Parallel Axis Theorem Exmpls Find the Moment of inertia IT of a circular area with respect to a tangent to the circle Find the Moment of inertia of a triangle with respect to a centroidal axis

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

Standard Shapes are Tabulated

Example 3 SOLUTION PLAN 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 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.

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 Calc the Centroid

Example 3 cont.2 Apply the parallel axis theorem to determine moments of inertia of beam section and plate with respect to COMPOSITE SECTION centroidal axis. Then the Composite Moment Of Inertia About the Composite Centroidal Axis, at x’

Example 3 cont.3 Finally Calculate the Radius of Gyration for the Composite From Quantities Previously Determined So Finally

Let’s Work This Nice Problem WhiteBoard Work Let’s Work This Nice Problem H13e P10-53 => ENGR36_H13_Tutorial_Areal_Moment_of_Inertia_P10_53_1207.pptx Find the Area Moment of Inertia of the Parallelogram about its Centroidal axis y’

Registered Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Engineering 36 Appendix Bruce Mayer, PE Registered Electrical & Mechanical Engineer BMayer@ChabotCollege.edu