Beam(part 02)/shear force and bending moment diagram for ssb with udl/strength of material/in tamil.
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- Опубліковано 29 лип 2020
- In this video you can easily understand How to draw the Shear force and Bending moment diagram for simply supported beam with udl and point load.
Before watching this video, I suggested to watch Beam (part 01) video to understand what is beam?, types of beam, types of load and concept of shear force and bending moment.
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/ @learnmechtamizh
Please watch the video fully to understand the concept....
Mohr's Circle Method:
• Introduction to princi...
Beam Part 01:
• Beam(part 01)/types of...
Beam Part 02:
• Beam(part 01)/types of...
Moment of Inertia
• Moment of inertia/unde...
Shear stress distribution
• Shear stress distribut...
Hashtags:
#learnmechtamizh #shearforce #bendingmoment #beam #strengthofmaterials #mech #mechsubjects
Sir you posted the vdo 3yrs ago.but now it is useful for me .thank you so much .good explanation
Nalla vilanguthu sir....
❤❤❤❤ thank you so much sir..
Sir, you clearly explained the concept. It will be helpful for gate student.please put more mechanical core subjects 👍
Thank you pa ..will do it.. keep supporting
Easy to understand sir...please do put more videos on the subjects of DME...it will be useful for part time students....as we are working and studying its easy to follow your lessons while travelling also....
Simple and clear cut explanation sir awesome👌👌
I'm from Cochin, could understand Tamil, ur teaching skill looks perfect, Well Done.,
Thank You
I am part time student of DME.. Really I understand well by ur teach... I request u pls do continue SOM syllabus torque subject in 3rd sem for us may helpfull.. thank u very much sir...
ok sure..
மிகவும் எளிமையான வழிகாட்டல்.நன்றி ஐயா
I clearly understand the concept sir really good
Thank you.. Keep Watching..
Very good teaching sir...👌👌
Thanks and welcome
Good explanation sir, and my request is please put the video solve the problems
Very good explanation.. Regards Veeraraghavan Brunei
Thank you...
Easy to understand.... Way of explanation was good... Try to put sf and bm for shaft with pully and gear...
Hmm ok Guna will make it soon
sir very clear explanation please continue these things...
Ya sure, Thank you.. keep supporting.. and share to your friends
Very clear explanation sir tq ❤❤
Thank you keep watching
Good lecture sir. Regards, Prof. P. Sakthivel.
Thank you so much... keep watching
@@LearnMechTamizh Please look at my channel and need your comments on two of my courses. 1. Engineering Mechanics and 2. Vehicle Dynamics. Thanks and regards, Dr. P. Sakthivel.
@@sakthivelp5894 your channel name sir
@@LearnMechTamizh click on the logo 'S'
Superb 🎉
One of the best video , thank you so much sir
Thank you so much ...
Sir, Your teaching is awesome....
Keep watching Thank you for your comments
Thanks brother. Very nicely explained and nice MS. Blue shirt.
Thank you brother..
Simple and clear explanation sir
Thank you.. Keep watching...
Superb sir your explanation vera level sir thanks a lot
Thank you so much.. share to your friends too...
Sema sir super ah solitharinga 💥👍🏻
Thank you...
Supera concept purinchikittan sir💛
Thank you pa
Sir super class.... thanks 🙏🏾
Thank you keep watching
Thank you for the video sir...❤
Thank you so much.. keep watching
Thank you sir .my mind is broken for 4 days this type of problem.
It's very useful to my exam
Thank you sir. Very helpful
Thank you.. Keep watching...
Really a gud explanation
Sir final BM at C, RA should be negative right?...or I understood wrongly? Kindly clarify me sir.
Useful sir💪👏
Vera level explanation sir
Thank you keep watching
well done sir ...Thankyou
Thank you.. Keep watching...
👍
Thankyou sir😊😊😊
A simple supported beam of length 10 meters carries a uniformly distributed load of 50kn/m for its entire length and point load of 40kn at distance of 4 meters from the left end support draw the S. F AND BMD video podunga sir pls
Hello Sir i would like to learn design of steel structure for warehouse in manual method. Plz advice how can i approach to you. You have done a good job.
Mail me @ learnmechtamizh@gmail.com
super jii
Thank you Keep watching...
Vera level thala ❤
Thank you.. Keep watching...
Nice explanation bro tq so much
Thank you for comment
Excellent teaching
Thank You
Please make a tutorial with an example and staad pro
Thank you very much sir. I received this comment clearly from you. Can you take Mechanical gate course, it will definitely help the aspirants.
Thank you so much
Awesome
thank you so much....keep watching
Thanku for the class sir🙏🏻
Thank you so much... keep watching
Super explaining sir
Thanks and welcome
Superb sir
Thank-you keep watching
Your videos are nice sir
Super sir 👍
Thank you
Thank you sir..
thank you so much....keep watching
Thank you sir🙏
❤️
thank you so much....keep watching
nice explanation sir
Thank you
Nice explanation
Thanks and welcome
Sir,
Very clear explanation about the concept.
Is the point X is point of contraflexure
Thank you.. Keep Watching..
@@LearnMechTamizh
Sir,
Pl do videos about influence lines
👍👍
Thank you.. Keep watching...
good work brother
Thank you bro
Sir SF Negative agum pothu Beam Break akiruma Sir...... 🤔
I am going to design one enclosure, i have plan to use 2 mm aluminum sheet but i dont know load carrying capacity of sheet,
Could you please explain how i calculate the load capacity of 2mm sheet ,
Very good
Super sir
Nice sir 🔥
Thanks ✌️
Sir moment diagram la parabolic curve A to X varaya illa A to point C varaya sir?
Sir can you please solve a problem with thickness of beam or plate.
Or suggest me how to find SF and BM for 2mm thikness sheet
Super
Sir udl vara edathula uvl vandha andha load epdi point loada cinvert pantradhu sollunga
Thank you so much
Welcome
Great job!!! Clearly explained
Thank-you keep watching
Sir roller support ah erundha vertical reaction mattu thana varum....hinged ah erundha 1vertical and 1horizontal force thana varum....neenga hinged support thana potrukinga appo 1vertical and 1 horizontal varum thana sir
There is no horizontal external force acting on beam. So no reaction in horizontal reaction
Beammmmmmmmmmmm
Sir last aah oru curve draw pannunathu en endu welangala sir. X and C kku etaiala
That is for maximum bending moment find pandrathuku
Practicala epdi apply panrom ethukaka entha diagram
Sir explain the impact load
Nice
Thank you
Even the HOD of my college haven't taught me this trick, If he would have done it then I would have loved the subject (SOM)😔
Thank you
In shear force diagram, why -60 KN act on 6 m not in cenyre 3 m, As same like point load -40 KN
Sir i want sfd and bmd for fixed beam
Nandri guru😍😍
Thank you.. Keep Watching..
😅
why you didn't considered udl at moment calculation
Yes considered .. for finding max bending moment
Formulae irukuma sir
Long day doubt ah clear pannitingo
Thank you ...
@@LearnMechTamizh more videos
Hh
Thank you Keep watching...
Thanks for your video.
How to provide rebar in that max bm location.
Ungala like panna kudadhu koil katti kumbudalam
Thank you for your comment.. keep watching and support
Hi,
Bending moment RB*2=40 OR 100
100 🤷🏻♂️
Sir Ra+RB=--100 sir
Ana enga +100
Ethu pureyalasir
Ra+RB-100=0
RA+RB=100
Thank you very much sir
Thank you