Question

Inertia problem

Answer 1

Can we discuss this problem before we solve? We have not touched on this subject yet in class. We are going to start talking about it today.

Answer 2

Absolutely...we can discuss tonight.

Answer 3

Okay sounds good

Answer 4

In preparation, please look up:
a) moment of inertia of a rectangle
b) Parallel axis theorme

Answer 5

You will probably have in your textbook a location which list moments of inertia for common geometries (rectangles, circles etc.)

Answer 6

The two concepts you mentioned were actually a few things we discussed last class period. We never got around doing any examples in class though. I learn best with examples and then referring back to the concept.

Answer 7

Agreed...engineering is all about learning how to solve problems.

Answer 8

What is inertia? You can't see it. Is it when a beam or a metal rod is being put under strain? Like an object being warped?

Answer 9

We will clearly differentiate tonight between inertia (a la Newtons first law) and moment of inertia.

Answer 10

Gotta run.

Answer 11

ok

Answer 12

Let me describe how to approach the problem. If you refer to my markup you will see this is composite section (made of from several basic shapes....in this case three rectangles) each having a moment of inertia of 1/12*base*height^3. In order to get the overall moment of inertia about the x axis we have:

I total about x = I1 - I2- I3

now:

rectangle 1 is the overall rectangle without the cutouts. You can determine the base and height and therefore can get its moment of inertia through its centroidal axis which corresponds to the x axis of the overall section. Therefore there is no need to use the parallel axis theorem for this piece.

rectangle 2 is the cut out on top and you can also obtain its moment of inertia about its centroidal axis as indicated in my mark up. Once you have this value, you must however use the parallel axis theorem to get the moment of inertia about the x axis.

rectangle 3 is the cut out on the bottom and you follow the same approach as for rectangle 2 which means you must also use the parallel axis theorem.

Parallel axis theorem says:

Ix = Icentroidal + Area*(distance to the x-axis)^2....see my mark up

So for rectangles 2 and 3, you will have calculated Icentroidal for each using base and height values. Now you will have to calculate the additional term for each which is the area*(distance to the x axis)^2. Adding them will then give you the moments of inertia about x.

The radius of gyration about x = (Ix/Total area)^1/2