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

2. 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

3. 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

4. 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

5. 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

6. 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

7. 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.

8. 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

9. Use BaseLine Case for Iy

11. 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

12. 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

13. 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.

14. 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

15. 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

16. 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:

17. 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

18. 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

19. Standard Shapes are Tabulated

20. 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.

22. 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

23. 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’

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

25. 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’

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