How to find the moment of inertia for composite shapes
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- เผยแพร่เมื่อ 2 มิ.ย. 2017
- This mechanics of materials tutorial shows how to find the moment of inertia for composite shapes.
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Thanks for watching, I hope it helps!
Wasteful past 6 months of Online University STATICS class, I've learned more from your channel than those 6 months.
Glad I could help!! =)
+1
Watched a lot of videos to learn all sorts of stuff Uni fails to teach. Ive never ever been more impressed with an explanation before, truely, this is awesome. beyond words.
You're the reason I'm passing class
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?
Really shouldn't have missed my lecture on second moment of area. Thanks a lot you saved me!
Hahah it happens to everyone! Glad I could help, Sami!
Yikes this is some of the stuff that used to mess me up in college. Thanks for all that you do for the community man!!
Are u major in phy
I'm a civie student in college and even though I'm a pretty good student I don't know if I'd be in as good a situation without you. At the moment I'm at a 3.7 cumulative GPA as a junior in my undergrad. Your videos have been invaluable to me and I don't think I would be able to choose the masters programs I am applying for without your help, so thank you.
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 =)
I can't thank you enough for this video. You have no idea how helpful this was.
Excellent step by step process. Thanks a lot man.
I have my mechanics of solids exam tomorrow, and you really helped me in this issue. Hope that I will pass the exam. Wish me luck.
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!
just start from introduction of centriod in the statics part and learn the very basic concepts, then i fully understand what is going on when i come back to this video. Now i am confident about my physics
Thanks bro clear explanation, I'll definitely dive into your other videos.
Thanks for taking time to clear this complicated information up. In class it is not explained very 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 :)
Thank you so much!! Your videos are so helpful
This channel is super helpful 👍
Isnt the moment of inertia in kg*m^2? And also, could you use the PAT but with mass instead of Area (in case of same density)?
Legend tbh there isn’t a better Channel that explains this well
Thanks yo, means a lot.
I originally found this video when I took statics, now I am rewatching it when I am taking mechanics and materials, so I am now subscribed.
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
👍
Do you have anything like this for egg shapes. I would like to be able to determine centroids and moments of inertia for egg brick sewers
you are the savior my friend i love you !!
Luv ya too
ty sir i understood finally.
When calculated stress, is the c value at 10:09 27mm or 53mm (or both)?
"Orray gaes welkombek tu dis video.."
I like your style of teaching 👍
Thanks!!
this is very helpful. thank you so much!
You're very welcome!!! =)
Hi! First of all, I'm using your videos to study, and I've got to say, they're absolutely amazing!
Just a question, what software do you use to write these notes? I have been looking for a good digital note-taking software for a while now, and what you are using seems pretty good.
Edit:
For all those future engineers out there, just know that although it's difficult, you can do whatever you put your mind to. I failed every math and science exam in high school, but I'm doing exceptionally well in college, acing my mathematics and engineering courses! You just need to push yourself. Engineering won't be easy, but I guarantee it will be worth it :)
Push yourself, no matter how hard it may seem, and only then will you see results that make you realize it was all worth it
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)
i want to bolt a short lenght of of 100 x 100 mm T section steel to a horizontal timber beam to support a 900 x 100 x 100 newel post with a slot at the bottom. 1. What gauge steel do I use? 2. How tall does the center of the T need to be so the newel post supports the weight of a person leaning against it? Is 100 mm sufficient?
Thank you for saving my statics grade 🙏
Thank you so much for this video!! 😭😭😭
You're welcome Kay!! 🤜🤛
thank you so much!
does that table work for every example like that? I tried it for problem 10-29 in engineering mechanics statics 14th edition by R.C. Hibbeler and it didn't work
This was really helpful.Thank you
You’re welcome! Thanks for watching 🙂🙂
Would the moment of inertia about the y-axis just be the moment of inertia of shape A + shape B? Since dx would be 0 since the centroid of x is directly in the middle
Excellent video 👍
you're a life saver
❤️
Sir, You are amazing god bless you
So nice of youThanks Mehraj =)
Helpfull....thanks so much
You're welcome Iebe! =)
This is great. Thanks.
Thanks for watching and commenting :). Make sure to check out the whole free course at engineer4free.com/mechanics-of-materials if you haven't already!
Good video.best teaching effort
Thanks!! 👍
Great video, thank you
Glad you liked it! You should check out the others I did at engineer4free.com/mechanics-of-materials =)
Just to clarify, this is the moment of inertia about the centroidal x-axis? If we wanted to find it about the general x-axis could we use the same method but 'd' would be Ybar???
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?
When it comes to labelling a section is they a law I need to follow or not ..because if the shape was any (i) it either I can make my top part 1 or the bottom part and the calculations will not be the same when it comes to my y axis
Thank you!
Hello, quick question pls. When taking the individual Y bars are we using the bottom as the reference axis ?
Thank you so much. This helped me a lot.
Great!! Glad I could help 🙂
Thank you SO MUCH
You’re WELCOME!!! 😋
im having an exam later, Wish me luck dude, thanks forball of your videos, it helps me alot
Hope it went well!!
Mark my words, I am a broke college student now, but if I graduate I will make sure I buy your patrion.
That would be most appreciated. Good luck!!!
Thanks 🌹🌹🌹
You are the best 👑
Thanks!!! =)
Need more people so spread this video more
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You are amazing
Thank you ❤
You're welcome!! More pure bending videos @ engineer4free.com/mechanics-of-materials =)
Thank you so so much 🙏🙏🙏🙏
You're welcome!!! 🙌
Wow this physics I haven’t took this kind of maths heard from my lecturer is primary school maths. Not like calculus. But have to understand the concept and learn how to read shapes.
Thx man😃
This is the centroidal moment of inertia, right? I'm doing a similar practice problem now for the MOI about the x-axis and it says dy for the top rectangle is (70 - 0). Shouldn't it be 70 - 10, because the y-centroid of this rectangle if it were sitting atop the x axis would be 10 mm, not 0. I didn't take statics in undergrad and this review book is killing me T_T
You're awesome!
Thanks =)
How do you include mass into this?
Thankyou
Welcome
nice work 👍👍
Thank you 👍
Thanks man
You’re welcome!!
you may have many children, thank you man!!!!
Thanks Diego 👍
Best video ever
Thanks Rodolfo!
I got this program for casio 9860gii the answers match
hey. thank you for your explanation.
I have a question. if we take different coordinate system. so the result will be same or not ? for example if i take (y) from top not from bottom.
Should be the same
do we need to calculate Iy as well?
thx bro
Thanks for watching! Tons more mechanics of materials videos linked in the playlist in video description 🙌🙏
Is centre of mass also known as centroid??...coz I have learnt that centroid is the point of intersection of medians of a triangle.....
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
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Thanks Ahmed!!!!
I'm one of those who put a thumbs UP, those who put thumbs down are doing business they dont belong here. Good job bro
Thanks bruddah 🙌
Thanks 🌹🌹❤️❤️
You're welcome 😊
i would like to ask you that isn't this inertia is in y direction rather than x direction. i mean it should be Iy not Ix? is that right or not
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.
Why do you only include Y bar in the table when calculating the centroid? Is including X bar incorrect?
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
Is Neutral Axis and Centroid thesame thing?
whats ix for triangles is it the same or divided by 36?
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
thank u so much for tomorrow exam
You're welcome, hope it went well!
What application do you use to do your videos?
I keep a list at engineer4free.com/tools that is current with all of my hardware and software. Cheers!
can you also do for finding Iy?
why at the end there (the Ix, u times 80x20 for area A? why not 80x80?
We’re doing it by parts, and 80 x 20 is referring to the area of section A only
Hello, what if the material is rotated? Will it be the same or not?
it will not be the same
Love videos that go straight to the point. Quick question. The final units at the end are millimeters to the power of 4. What does that even mean? Fourth dimension quantity? Asking just as a first-year taking mechanics.
I don’t have a quick answer for that. It’s just how it works out mathematically. When you use this quantity in other expressions that it’s involved with, the other variables also get their correct units. In mechanics of materials you end up getting a lot of weird units, but it all cancels out and works at the end. Here’s a very quick read with some further vocab in there to search: en.wikipedia.org/wiki/Second_moment_of_area
Hi i have some question about some example please can you solve for me?
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 ✌️
Can you do more videos on this concept.
Hey Philip, check out videos #22-34 here: engineer4free.com/mechanics-of-materials cheers!
surely not the real chancellor of the exchequer is watching this
Why isnt the chancellor allowed to learn about new concepts in his spare time?.......May you explain further!
Isn't the second moment of area of a rectangle 1/3bh^3?
God sent you to us right?. keep it bro.
haha, just me sent me ;) Thanks!!
Why do we not find the centroid of the X axis as well?
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
program you use please??
All the hardware and software that I use to make my videos is listed here: engineer4free.com/tools
Been following you for a while.! Your videos are amazing!
Never the less, when/how do we chose "C" for "Sigma=MC/I" ... I see you lebeled "c" two times... one "from Y-bar to the top" and the other "from Y-bar to the bottom"...
Hey that's awesome glad to hear it! The equation σ_m=Mc/I can give us the max stress either above or below the neutral axis (NA). In the case of basic pure bending where we have compression above the NA and tension below the NA, we apply that formula to each side. Stress will be maximum at the furthest point away from the NA. c represents the distance from the NA to the furthest fibre of the member in the section (above/below or compressive/tensile) that we're currently considering. So for each case, you might have a different c value if the NA is not the same distance from each end, vertically.
@@Engineer4Free Thanks a lot
What if we have to solve for a i section
Same idea, you will just have three sections to consider instead of two
ur the goat
Thanks n0m!! 🐐
how can this be apply to parabola shape ?
Hey thanks for watching. I've got a few videos that discuss finding centroids of parabolas in the "centroids and distributed loads" section of engineer4free.com/statics, specifically recommend watching video 62 and 65, but 59 and 60 might be good too to give some context. Cheers
Thank you. Great video by the way, I will support your channel and site when I can :D
what is the answer for Ac?
I still dont get how to find y bar for each shape? can someone just give me a formula or something?
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.
super
duper
shoukd the Ix be to the power 10^-6 not 10^-9
Please is it solved with respect to the centroid
And the figure too is it one shape or 2 different shapes
I think that in the table Ybar for shape A should be 100 not 70 given that we measure from the bottom. @2:41
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!
I am getting 15.11*10^5 to be the answer (or 1511*10^3)
So am I. I can’t work out what’s going on here.
god bless
🙌
I don’t know how you figure 1511533mm^4 is 1511x10^-9m^4 when it’s actually just 1511m^4. Nobody else is picking this up?
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
You made the concept clear but can you please write it more clearly. It is too difficult to understand sometimes
Sir one doubt sir
When you write on the boarded some one can not see the detail work out please ignore that 🙏🙏