Question

Moment of inertia problem

Answer 1

Surry99 said---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

Answer 2

My solution to the first moment of inertia problem

Answer 3

ok let see...

Answer 4

d is suppose to be the distance to the center of the ref to the x axes

Answer 5

yes

Answer 6

You broke it up into a lot more pieces than you needed to.
Let me attach my solution and we can discuss.

Answer 7

Why can't I go about it this way?

Answer 8

you can it just takes a lot longer and on exams you usually don't have a lot of time

Answer 9

At least, I didn't give my students a lot of time.

Answer 10

yeah tell me about it... Well in my class we have weekly quizzes with one problem. We have a week and half left before finals. He will have four problems on the final, one of which will be the inertia problem and another will be the friction problem. The other two not sure.

Answer 11

Take a look at my solution method, this is what I tried to describe.

Answer 12

Let me eat something and I will study your solution a bit more... I have no clue how you are approaching this problem. Looks like a short cut.

Answer 13

ok, it is no shortcut though. While you are going over the solution, read my description on how to solve it...hopefully I explained it clearly enough that you can follow it.
If not, just ask.

Answer 14

The key is first seeing you are dealing with three rectangles.

Answer 15

I'll be back in an hour.

Answer 16

why did you make a axis for 2 and 3?

Answer 17

good question, when you use 1/12 b*h^3 for any rectangle, by definition you are calculating the moment of inertia about the centroidal (center of mass) axis of that section.
But I need their values about the x axis which is why I need the parallel axis theorem for rectangles 2 and 3.
That is also why you see y2 and y3 for those sections ....the distances from the centroid of the sections to the parallel axis.

Answer 18

How do you know this? (made of from several basic shapes....in this case three rectangles) each having a moment of inertia of 1/12*base*height^3.

Answer 19

each having a moment of inertia 1/12*base*height^3

Answer 20

Good question lets do it one at a time:
1) do you know where the the 1/12*base*height^3 comes from?

Answer 21

no

Answer 22

ah...do you have a table of moments of inertia for simple geometric shapes in your book?

Answer 23

I see.. Yes

Answer 24

ok good...now look at the section again...can you visualize it as follows:
one big rectangle, with two smaller rectangles cut out of it?

Answer 25

why do we use the triangle?

Answer 26

why not the rectangle?

Answer 27

oops....

Answer 28

I see where the rectangle is, proceed

Answer 29

ok. good. so do you see the big rectangel with two smaller ones cut out?

Answer 30

The rectangle as a whole? I see that you made three smaller rectangles with their own axis. Actually you gave rectangle 3 an axis and rectangle 2 an axis. Those two rectangles you have to aplly the theorem? Because?

Answer 31

Try sketching it on a piece of paper ....sketch a big rectangle, then cut out the top rectange , then cut out the bottom and you will have our section.

Answer 32

yeah that is what I am doing right now. Let me look at it some more.. I think i've almost got the idea.

Answer 33

looking at rectangle 2 with the axis drawn in the middle/center. what is the distance? From the axis you drew to the center at o?

Answer 34

The big rectangle is:
5 inches wide by 6 inches high

The cut at the top is:
4 inches wide x 2 inches high
the top of the cut lines up with the top of the big rectangle

The cut at the bottom is:
4 inches wide x 1 inch high
the top of this cut is 1 inch down form the x axis

Answer 35

It is 1.5 inches

Answer 36

oops...rectangle 2 ..it is 2 inches

Answer 37

rectangle 3 is 1.5 inches from the x axis

Answer 38

Ok I see. Let me work on it some more

Answer 39

ok, if you can now see it as three rectangles, my solution should make sense.

Answer 40

Okay your solution does make sense... How do you know when you can draw a new x axis?

Answer 41

Or axis in general?

Answer 42

dont even answer that lets do the same problem but find the moment of inertia and the radius of gyration of the shaded area with respect to the y axis

Answer 43

the key is knowing that whatever simple geometric section you have the moment of inertia
equation for, it is always relative to the location of the center of mass of the section.

Answer 44

Look carefully at the page in your textbook with the moments of inertia, every section shows the center of mass (centroid).
I promised to help another student but will get to the y axis problem tomorrow.
Try to see if you can calculate it please.

Answer 45

Yeah I will. Thanks

Answer 46

np.

Answer 47

Got it figured out solving it the way I showed you.

Answer 48

Great...whatever method works best for you is all that matters.

Answer 49

I got started on the other question where I had to find the moment of interia respect to the y axis. When I find d i am looking for the distance from the y axis to the center of the rectangles?

Answer 50

^solution to last problem

Answer 51

yes that is correct

Answer 52

Remember that since you are now calculating the moment of inertia about y, the "base" of each rectangle is now parallel to the y axis. When you did the moment of inertial about the x axis, the base was parallel to the x axis.