How to find the moment of inertia for composite shapes 

Glad I could help!! =)

+1

Wow I'm glad I can help man, I hope you crush this course with a good grade!

Your teacher wants to know your location

@@Engineer4Free You posted this video 4 years ago, and people, like me, are still watching it. It just proves how good of a teacher you are. Thank you for saving several students. ❤️

@@ahmedelfarra7674 Glad I can still help!! =)

@@Engineer4Free Sorry, this might be a dumb question, but I'm studying for the fe. How do you know when to use Ixc, Iyc, Ix, IY, or J for area moment of Inertia?

Hey Yitzchak, thanks for the feedback, glad you're finding them helpful!! You can find the full list of software and hardware that I use at engineer4free.com/tools Using an external pen tablet like I do is probably too cumbersome for taking to class, you might be better off using a tablet or laptop that you can draw directly on the screen, but still check out the set up that I use!

Might be a little late, but if you are not. One Note is really good for taking notes (especcially if you have some sort of drawing thing to connect with your pc)

Are u major in phy

Hahah it happens to everyone! Glad I could help, Sami!

Thanks!!

Good luck! How did it go?

It went really good. The questions were a little bit complicated but I managed to solve them. Thanks again!

No worries, glad I could help!

Amazing, thanks for sharing!! You're GPA is better than mine was, but I'm really glad my videos have helped you do so well!! So cool to hear. Keep it up, and good luck getting into the masters =)

Yes this is the moment of inertia about the composite shape's centroidal x-axis. If you wanted to find the moment of inertia about a different x-axis that is given to you, then skip the first part of the problem that is written in black. That whole part is done to locate where the centroid is, because it begins as an unknown in this problem and we need to find it so we will know what "d" is for each section in the next part. If you are given an axis to calculate moment of inertia about, then you will be able to easily measure the "d" for each composite part from the diagram. If for example you wanted to know the moment of inertia of the composite shape in this video about an axis that was 10mm below the very bottom of the overall shape, then dA would be 90mm and dB would be 40mm and you would only need to do the equation that is written in blue from the last half of the video. That would be a pretty random example though, as we are almost always concerned with the centroidal moment of inertia for mechanics of materials and structural analysis problems. I guess every once in a while a question like that might show up. Some common shortcuts are even printed in the moment of inertia tables in textbooks, like for example the moment of inertia of a rectangle about an axis that is right on it's lower edge is (1/3)bh^3. It comes from the simplification of (1/12)bh^3+ad^2 = (1/12)bh^3+(bh)(h/2)^2 = (1/12)bh^3+bh^3(1/4) = (1/12)bh^3+(3/12)(h^3) = (4/12)bh^3 = (1/3)bh^3. Hope that clears it all up for you!

Great explanation! Quick question, did you mean that dA would be 80mm - instead of 90mm? Since the centroid on A is at 70mm above the bottom, and the axis of rotation is 10mm below the bottom the total distance for that MoI component is 80mm, correct?

Hey Lewis, it is not a mistake. See this i.imgur.com/ot98den.png for the unit conversion 👍

@@Engineer4Free thanks for that. I guess my maths needs brushing up too!

@@LewisBeckman No worries! It's a common mistake to forget that all 4 mm's need to be converted into 4 m's

That would be most appreciated. Good luck!!!

Hope it went well!!

Ah yes, such was my life too when I first took it. Glad the video helps, do check out the rest of the videos I made for this course: engineer4free.com/mechanics-of-materials cheers :)

Yeah X bar is irrelevant to the problem. We only need to know how far away the y component of the centroid is from the neutral axis. Often you will find that these members are symmetrical from left to right anyways, which would put Xbar in the middle. More context in videos 22-28 here: engineer4free.com/mechanics-of-materials

Thanks Diego 👍

You're welcome Kay!! 🤜🤛

So am I. I can’t work out what’s going on here.

So nice of youThanks Mehraj =)

Thanks!! 👍

Hi Regdor, this is not a standalone video, it is #25 of 56 videos that cover a full course on Mechanics of Materials. I recommend watching the previous videos in the series, and any other pre-requisite course as needed (ie, statics, calculus) to become familiar with the variables and lingo, as I do not re-introduce it all in every video. And it is not intended to resemble drafting, rather hand calculations, as students would perform in class.

Welcome

Hey I don’t usually solve specific questions for people... I just make the general tutorials that everyone can access all the time. You can check all the related vidz I made at engineer4free.com/mechanics-of-materials ✌️

We’re doing it by parts, and 80 x 20 is referring to the area of section A only

Mmmm, you’re looking “down the barrel” of a beam in this video that is subject to positive bending, which means it’s basically sagging down in the middle, or looks like a smile from the side. That smile forms a radius of curvature about an axis in the x direction (would be left to right) as viewed down the barrel on the screen. So we really are calculating bending moment about an x axis, not a y axis. The deflection of the beam parts are in the y axis rather.

Thanks n0m!! 🐐

Hi Seth. There is not Xbar in this video. At 5:26 I label the centroidal axis as "the x axis." Section A and Section B both have their own unique Ybars, but because it is written in table form, using a subscript would be unnecessary, as A and B already define the rows of the table. See any edition of Beer and Johnston Mechanics of Materials for further reference on this method.

For beam bending problems with a section that is symmetrical from left to right, it's not necessary. I recommend watching videos 22 - 28 here: engineer4free.com/mechanics-of-materaisl

Thanks for watching! Tons more mechanics of materials videos linked in the playlist in video description 🙌🙏

You're wel ome, thanks for watching! Check out engineer4free.com/mechanics-of-materials for the full playlist 😊

♥️

You need to find the centroid of a triangle. Check out videos 58, 59, 63 for some related videos I did on triangle centroids: engineer4free.com/statics

haha, just me sent me ;) Thanks!!

Thanks Ahmed!!!!

🙌🙌

Yes, applies to any type of simple or composite shape. Technically it's the centre of area, but if you assume a uniform mass distribution of the cross section then it essentially becomes the same thing. Take a look at videos 58-60 here: engineer4free.com/statics for some insight on centroids

Luv ya too

Yes

You're very welcome!!! =)

Glad you liked it! You should check out the others I did at engineer4free.com/mechanics-of-materials =)

Thanks bruddah 🙌

Thanks Rodolfo!

I got this program for casio 9860gii the answers match

You're welcome 😊

❤️

You're welcome!! More pure bending videos @ engineer4free.com/mechanics-of-materials =)

You’re welcome!!

Same idea, you will just have three sections to consider instead of two

Thanks =)

Should be the same

All the hardware and software that I use to make my videos is listed here: engineer4free.com/tools

haha nice! you should check these links, as there is crossover between all of them: engineer4free.com/statics engineer4free.com/mechanics-of-materials engineer4free.com/structural-analysis 👍

And the figure too is it one shape or 2 different shapes

Thanks yo, means a lot.

🙌

You're welcome Iebe! =)

Thanks!!! =)

You pick the axis you want to measure from which will just be the bottom of the cross section, and then you measure the vertical distance from that axis to the centroid of each individual shape. Given that the shapes are always just rectangles, the centroid will be 1/2 the height. If your beam is like this one, you might have to add 1/2 the height of the top section plus all of the height of the bottom one to find how far the top one's centroid is from the axis that touches the bottom of the bottom one.

You're welcome, hope it went well!

Oh sorry, the way I labelled the pink arrow on the right is a little misleading. Look at the arrows on the right side, section B is 60mm tall, and A is 20mm tall. 60 + 20/2 = 60 + 10 = 70mm. I see how reading the left side would lead you to that conclusion. Sorry for that, but it is 70mm!

it will not be the same