Hi Mr. Techy......Thanks for appreciating my work…and all the best for your examI am sure this channel is helping you learn…You can support my channel by donating and do watch this video: ua-cam.com/video/WfyUEhxhjZUH/v-deo.htmlave a nice day.
Sir ...I am very thankful to you.....last semester I studied the whole engineering graphics from ur videos.....I suggested it to my frnds........they really helped us......I scored good marks too.......this semester ur engineering mechanics lectures are really helpful ......thanks a lot sir
Sir you have literally helped me to infinity cz I was not knowing anything i.e. how to calculate it as I missed my mechanics lacture But got a great social tutor.
For semi circle and quarter circle you have taken radius instead of diameter in your formula. Please correct it or else people get confuse. remaining good explanation
Here we are asked to calculate moi abt AB so......but if we asked to calculate moi abt centroidal axus then we have to calculate it. Correct me if m wrong
how do we understand whether we have to take the x distance or y distance from the given line? like in the previous video we got the x component and the y component, we dont have to do that in this?
Sir, in your previous videos (eg. I beam), the "y" value was (value of centroid "y" - y1/y2/y3) but in this video you directly took y1/y2/y3 value. I couldn't understand.
Hi Sriram... thats just the aplication of formula in the right sense... Welcome to the "The Library of Manas Patnaik" Topic-wise Playlist of Engineering Drawing / Engineering Graphics in English 1. Orthographic Projection: ua-cam.com/video/WG6H2pISUzQ/v-deo.html 2. Projection of Point: ua-cam.com/video/fK4h5gM73w8/v-deo.html 3. Projection of Line: OLD: ua-cam.com/play/PLIhUrsYr8yHz_FkG5tGWXaNbIxVcibQvV.html NEW: ua-cam.com/video/pzE0D-Ijxd4/v-deo.html 4. Projection of Plane: ua-cam.com/play/PLIhUrsYr8yHx7TVB51jN3HZVyW3R6RiBg.html 5. Projection of Solid: ua-cam.com/play/PLIhUrsYr8yHxARPzEFz1nXgt8j6xF_tEm.html 6. Section of Solid: ua-cam.com/play/PLIhUrsYr8yHwAbiCATZUbzd_CpF0EHF3v.html 7. Development of Surface: ua-cam.com/play/PLIhUrsYr8yHwdB96ft6c0Uwc4SDCLuG1v.html 8. Auxiliary Plane: ua-cam.com/play/PLIhUrsYr8yHxHWk2TCg1SrphF1vKpHXoB.html 9. Interpenetration of Solids: ua-cam.com/play/PLIhUrsYr8yHyTGceE1TGhCSxxfHyFpnNM.html 10. Isometric Drawings: ua-cam.com/play/PLIhUrsYr8yHxVky7bfrnbRcdXcHjT_K83.html 11. Conic Sections: ua-cam.com/play/PLIhUrsYr8yHx0TuqsY0FUZ7Z24BbKxJoQ.html 12. Cycloidal Curves: ua-cam.com/play/PLIhUrsYr8yHwwA8eBea78pJW1lw9XAv6z.html 13. Involute: ua-cam.com/play/PLIhUrsYr8yHzu8pqYvGwQoJKVpuS1Jusa.html 14. Plain Scale and Diagonal Scale: ua-cam.com/play/PLIhUrsYr8yHwTgo_zu_ELqOC1ypcp5Gr9.html 15. Complete Engineering Drawing in one single Playlist: ua-cam.com/play/PLIhUrsYr8yHwDUrVYmUNYkEeZgZTvoIfS.html Topic-wise Playlist of Engineering Drawing / Engineering Graphics in Hindi 1. Orthographic Projection in Hindi: ua-cam.com/play/PLIhUrsYr8yHy3chHVzYA4BMbY9DbhyYSY.html 2. Projection of Point in Hindi: ua-cam.com/play/PLIhUrsYr8yHwinNUxVA1NW0xxLRlDfINH.html 3. Projection of Line in Hindi: ua-cam.com/play/PLIhUrsYr8yHyvWb87YtG3A3jDEnChgzir.html 4. Projection of Plane in Hindi: ua-cam.com/play/PLIhUrsYr8yHxj9N5e1oZA0x_k5ZVssf3U.html 5. Projection of Solid in Hindi: ua-cam.com/play/PLIhUrsYr8yHylYPjrulRGm8DSPiimxP3P.html 6. Scales Hindi-Urdu: ua-cam.com/play/PLIhUrsYr8yHzJA426HjlEmiRqUkJTkozB.html Topic-wise Playlist of Engineering Mechanics / Applied Mechanics / Fundamentals of Mechanical Engineering in English 1. Forces: ua-cam.com/play/PLIhUrsYr8yHzoyUutEK57JzWy1n-vxIS0.html 2. Friction: ua-cam.com/play/PLIhUrsYr8yHwY1TDraJpyYkU8Zug-kM4m.html 3. Shear Force and Bending Moment: ua-cam.com/play/PLIhUrsYr8yHy1lc0hEhbpEAGsGwNmUWt7.html 4. Trusses: ua-cam.com/play/PLIhUrsYr8yHwIXEt9cqjEsEO8l83xcIf7.html 5. Centroid and Centre of Gravity: ua-cam.com/play/PLIhUrsYr8yHzZ52dVA3wQ_0IQVl7CKL7p.html 6. Moment of Inertia: ua-cam.com/play/PLIhUrsYr8yHw8qWxAVOx-5Y2XfGPOmcDy.html 7. Dynamics: ua-cam.com/play/PLIhUrsYr8yHwOZxmY63A_lURu6c-jNk-z.html 8. Complete Engineering Mechanics / Applied Mechanics in one single Playlist: ua-cam.com/play/PLIhUrsYr8yHxfOmcGIeUThHEr9lQ3IXRS.html Complete Playlist of ENGINEERING SERVICES EXAMINATION: ua-cam.com/video/ca5IwhrCNbQ/v-deo.html Complete Playlist of Machine Drawing: ua-cam.com/video/iFCnFUR9Gdc/v-deo.html Complete Playlist of Theory of Machines / Kinematics of Machines / Dynamics of Machines: ua-cam.com/video/Co4YlavCpeQ/v-deo.html
I will check it John in just a moment..... If there is an error...... I will make a comment and pin it at the top of comment section. Because I can't make alterations to the video..... I am sure that this video has helped you gain some insight into calculating moment of inertias......
Please, I have a couple of questions, isn't Ix of a semicircle meant to be equal to Iy? what of quarter circles, if we were to find Ixx and Iyy when parrallel and perpendicular to a quarter circle, what values do we use? Also, for semicircles, I know how they obtained that 0.393r^4 =(pi.r^4/8). But how did you get 0.11r^4?
I would like to correct that this is not a right triangle but actually an iscoceles triangle. indicating it as a "Right Angled Triangle" is totally wrong because an isosceles triangle has a summation of 180° inside. Also take note that a right triangle has 90° only. Anyway, the solution still correct.
Yes it is but it is about its own centroidal axis and then you can apply the parallel axis theorem to calculate the moment of inertia with respect to any axis
lol. The video is correct. look at the location of the two 8/3. both of them located at the 1/3 position of C.G. take note that the 2/3 is located from the vertex of the triangle to the C.G. the vertices I'm saying base to the triangle in the video is located at the top and the left of the triangle. The distance he indicated there is located at the 1/3. He didn't use 2/3. If you still didn't get it then try to focus your eyes at the distances at triangle from B to X and B to Y. If still didn't get then you really an idiot. I'd better suggest you to quit learning Inertia. LOL
I could have never passed my sem exam(bem)without this knowledge ,thank u for this wonderful lecture
Hi Mr. Techy......Thanks for appreciating my work…and all the best for your examI am sure this channel is helping you learn…You can support my channel by donating and do watch this video: ua-cam.com/video/WfyUEhxhjZUH/v-deo.htmlave a nice day.
for semi circle we can use (pi/8 x r^4) where r = 4 directly ... also for circle we can use(pi/4 x r^4) where r = 2 directly
Sir ...I am very thankful to you.....last semester I studied the whole engineering graphics from ur videos.....I suggested it to my frnds........they really helped us......I scored good marks too.......this semester ur engineering mechanics lectures are really helpful ......thanks a lot sir
I wish u all the very best in life....
For the semi circle. It’s not meant to be 0.11*(4)^4. It should be pi over 8 then multiply be 4^4.
A very clear and good teaching! Thanks a lot for this sir
Sir you have literally helped me to infinity cz I was not knowing anything i.e. how to calculate it as I missed my mechanics lacture
But got a great social tutor.
Smart way of teaching sir ..❤
It cleared my dbt
How did you get the semicircle MOI and Circle MOI
That semi circle
= (3.1415)4^4 /8 +ad^2
for that Circle
MOcm=pi*r^4/4+0=201.061
For semi circle and quarter circle you have taken radius instead of diameter in your formula. Please correct it or else people get confuse. remaining good explanation
Why didn't we calculate centroid for this question ?as we do for all until now
Here we are asked to calculate moi abt AB so......but if we asked to calculate moi abt centroidal axus then we have to calculate it. Correct me if m wrong
Please tell me more about Moi of composite sections (about semicircle problems)
WHY IS IT THAT INDIAN GUYS ON UA-cam ARE BETTER TEACHERS/EXPLAINERS THAN ANY OF MY PROFS WITH PHD'S AT A TOP SCHOOL IN CANADA
Thanks a lot sir this explanation helped me a lot in my CA.
how do we understand whether we have to take the x distance or y distance from the given line? like in the previous video we got the x component and the y component, we dont have to do that in this?
i have the same concern...
you are a life savior, holy macaroni
Hello, usually in a moment of inertia problem do I have to determine the centroid first or it will be given?
Sir, for the same figure how to find MOI about its centroidal x-axis?
sir the semicircle is in 3 4 quadarant and in 3 4 quadarant the y values is negative ,then sir why you should not represent the y distance in (-) form
Thanku for ur wonderful lectures . I am watching both ed and mechanics ur videos
Brilliant explanation of and hour long concept in 10 min
Sir, in your previous videos (eg. I beam), the "y" value was (value of centroid "y" - y1/y2/y3) but in this video you directly took y1/y2/y3 value. I couldn't understand.
Same question
Sir, how did u get the 0.11 in IAB2? How to calculate? Plss help!
The MI through its C.G
= MI through its base - A×y^2
=0.393r^4 - A×y^2, (y= 4r/3 pi) & ( A= pi× r^2/2) . calculate this you will get 0.11r^4.
Do you also know how to solve the polar and radius of gyration? And where is the y axis. Thank you
Thank you for that helpful explaination. But why did you come up to that formula of an area?
Sir , for semicircle how did you get that 0.11*(4)^4
Hi Sriram...
thats just the aplication of formula in the right sense...
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I was asking the exact same question.
It’s not meant to be 0.11*(4)^4. It should be pi over 8 then multiply be 4^4
Very nice video and very helpful , with Explanation
Hindi peoples are too good in teaching👍👍😊😊☺️
Which book are you using? Tell me the name please.
Some time y2 is (y-ybar)...
Some time y2 is directly substitute....
Please tell the answer sir...
Same question
Sir,Cetroid of right angled triangle is (2*8÷3 , 8÷3)
correct but look at the distance marked in video its from opposite side so the distance from opposite side is (8-(2*8)/3)=8/3
I will check it John in just a moment.....
If there is an error...... I will make a comment and pin it at the top of comment section.
Because I can't make alterations to the video.....
I am sure that this video has helped you gain some insight into calculating moment of inertias......
@@ManasPatnaikofficial no the video is correct but you dint mention the origin direction thats why some people have this confusion
Sir, how did u get the 0.11 in IAB2? How to calculate?
It's a formula for a semicircle
@@abhishekprajwal5691 can you please type how it was obtained? that's my major problem right now.
Fabulous one sir....love you so much sir....
All the best.
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Moi of the triangle is this correct formula ? sir
can we take MOI of semi circle about the base formula in this problem ie pi * r^4/8 directly
really awesome
Please, I have a couple of questions, isn't Ix of a semicircle meant to be equal to Iy? what of quarter circles, if we were to find Ixx and Iyy when parrallel and perpendicular to a quarter circle, what values do we use?
Also, for semicircles, I know how they obtained that 0.393r^4 =(pi.r^4/8). But how did you get 0.11r^4?
Sir same problem in vertical centroidal axis, for circle x distance how much sir
thank u so much sir it helped me a lot tomorrow is my exam😀
All the best.....
Rock it tomorrow
What will be the solutuon if we want to find moi about centroidal axis parallel to AB?
Moi of semi circle is not mentioned and how did we get that 0.11*4^4
Pls, I want to know how this is going to help us future engineers??
is the line not passing throough the base not center to make the moment of inertia bh3/12 not bh3/36
Sir u are simply awesome!!!
Please sir, is there differences between Engineers and Engineering Technologist. Please sir help me out in the distinction.
Sir, why did u take MOI of triangle as bh 3/36 instead of bh3/12 ? In some other videos they have taken bh3/12. Totally confused .
the MOI of Triangle is bh^3/36 & MOI of Rectangle is bh^3/12
Thank you for neat explanation
why you did NOT CALCULATE X AND Y BAR PLEASE TELL DONT IGNORE
Because we don't have to calculate moi about centroidal axis
Bro it's wrt to AB
Sir is it possible to use direct formula about it's base instead of parallel axis..
Absolutely Rakesh
U teach very well, sir
I would like to correct that this is not a right triangle but actually an iscoceles triangle. indicating it as a "Right Angled Triangle" is totally wrong because an isosceles triangle has a summation of 180° inside. Also take note that a right triangle has 90° only. Anyway, the solution still correct.
Sir Iab2 =0.11×4 to the power of 4..why
Sir hum isme kisi axis ke respect m hi nikalte h either x axis or y axis
Why is the distance of the centroid in respect to the y axis divided by 3
Why you have not applied parallel axis theorem to using y-axis...?
Sorry , I Got it
Why you keep using b/3 instead of 2/3*b?
Gd evening sir.can I know the ix and iy values of a triangle other than right angle triangle
you wrote value of r of triangle 8/3 how? can you explain plz
Sir plz send finding moment of inertia of circle , semicircle and quarter circle video
Ok MAdhu...that will be uploaded soon
Thank you sir
Mai apka abhari hu 😄😄😄😄😄
sir same problem how to find centroid
SIR please make a video on moment of inertIa of builtup sections
Ixx=(bh^3)/36
Where is 36 come from?
Thank u!! Helped a lot 🧡
Thank a lot.
For right angle triangle formula if bh^3/12 .. Why u have used 36 instead of 12 ?
36 is from the Centroidal axis & 12 is from the XX axis(directly)
in same question about centroidal axes?
area of circle is A=πr2 then wht divided by 2
Ufffff tq u sir
When u took the moi of circle why did u use radius 4 instead 2?
Bro, that's ''Diameter'' not the ''Radius''. Sir said that the formula is pi/64 X D^4.
Can you please upload mass moment of inertia,it's our need to know them.Please upload as soon as possible......🙏
It's in the planning already.
Ok sir👍
Thank you very much, man, you have saved me.
I thought that the moment of inertia of right triangle is bh^3/12
And why are you using diameters if you can use the radii.
Yes it is but it is about its own centroidal axis and then you can apply the parallel axis theorem to calculate the moment of inertia with respect to any axis
Manas Patnaik okay. I got it now. Thanks lmao good luck for my exam later.
thank you very much sir ,, ur videos r clearly understable
Why don't we calculate directly I.e.., bd^3/12+pid^4/128_pid^4/64
Thank you soo much sirr
Thank you sir ❤️
How ur taking ,only Inertia of XX axis
Please add the archimedian spiral
Sir in case of semicircle the radius is 2 cm but you write it as 4.How?
he took the formula in diameter form.
Sir Iyy also have to calculate
sir may I get more composite figures
Thank you very much sir
Sir, whats the MOI of a quarter circle?
4r/3pi
@@kmlfunny817 what you wrote is centroid and not moment of inertia.
Like Igxx is we need to calculate Igyy
Thank you for speaking in English
All the best
How to find y value?
Sir why did u take 8/3 for triangle
CENTROID OF A TRIANGLE IS B/3
This sum is slightly different y u not find out y square
Can you assist me with a question am really struggling and I was using your method
Sir couple ke video please.......
thanks borthers
It is 12 rather than 36
Air i have a question
Then write it down.....🙂
Sir for the triangle moment of inertia is bh^3/12 about base ab but y did u consider bh^3/36
G fad diya exam ne toh
How take 0.11 value in IAB
That is the formula
🙂
Wrong formula of MOI of triangle
S
It is not b \3 it is 2b/3
lol. The video is correct. look at the location of the two 8/3. both of them located at the 1/3 position of C.G. take note that the 2/3 is located from the vertex of the triangle to the C.G. the vertices I'm saying base to the triangle in the video is located at the top and the left of the triangle. The distance he indicated there is located at the 1/3. He didn't use 2/3. If you still didn't get it then try to focus your eyes at the distances at triangle from B to X and B to Y. If still didn't get then you really an idiot. I'd better suggest you to quit learning Inertia. LOL
Its a shortcut dude.Incomplete tbh
sir rolling and collision ke video kB banoge sir btao naa pls
Abhishek...that will take sometime..
I cannot promise anything as of now......
Manas Patnaik thanks for information..I will wait for your videos because your videos are highly understandable and conceptual...