Uniformly Distributed Load (UDL): Shear Force and Bending Moment Diagram [SFD BMD Problem 4]

In this video we are Going to Learn about How to solve problems on Shear Force diagram [SFD] and Bending Moment Diagram [BMD] for overhanging beam with three Loads. In this problem three loads is acting downward because of that the reaction offered by the supports will be in upward direction. So at point A there will be Ra i.e. reaction at point A
and at point B there is reaction force Rb.
Timecodes
0:00 - Problems on Shear force and Bending Moment Diagram [SFD and BMD] for Uniformly Distributed Load
0:06 - How to Convert Uniformly Distributed Load into Point Load
0:29 - Calculations of Reaction forces for Uniformly Distributed Load
2:16 - Shear force Calculations for Uniformly Distributed Load
6:36 - Bending Moment Calculations for Uniformly Distributed Load
This problem we are going to solve it, in 3 simple steps.
1. Calculate the values of support Reaction.
2. Calculation of Shear force
3. Calculation of Bending Moment
Note: In shear force diagram, whatever the Portion drawn above the reference line, I will show this by positive sign. And the Portion drawn below the reference line, I will show this by negative sign. So here I have completed shear force diagram
Note: In bending moment diagram, whatever the Portion drawn above the reference line, I will show this by positive sign. And the Portion drawn below the reference line, I will show this by negative sign.
You can also watch separate videos for Shear Force and Bending Moment Problems on 4 types of beams -
Cantilever Beam - • Cantilever Beam: Shear...
Simply Supported Beam - • Simply Supported Beam:...
Overhanging Beam - • Overhanging Beam: Shea...
Simply Supported Beam with Uniformly Distributed Load - • Uniformly Distributed ...
You can also watch Playlist on Shear Force and Bending Moment Solved Problem
Shear force and Bending Moment Diagram Solved Problems - • Shear force and Bendin...
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Faculty - Shubham Kola
Email ID - [e-mail]
From - Maharashtra ( India )
UA-cam Channel ( Shubham Kola ) - / @ifinfotech
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video is for educational purpose only. Copyright Disclaimer Under Section 107 of the Copyright Act 1976, allowance is made for "fair use" for purposes such as criticism, comment, news reporting, teaching, scholarship, and research. Fair use is a use permitted by copyright statute that might otherwise be infringing. Non-profit, educational or personal use tips the balance in favour of fair use. The content in this video is collected from books, media, internet space etc. This is strictly for educational and information purposes and is not intended to be advice or recommendation of any kind whatsoever. The faculty is not a subject expert. The faculty is sharing his knowledge in the form of videos. The faculty giving small basics information on mechanical engineering topics. We endeavour to keep the information up to date and correct, we make no representations or warranties of any kind, express or implied, about the completeness, accuracy, reliability, suitability or availability with respect to the video or the information, related graphics contained in the video for any purpose. any reliance you place on such information is therefore strictly at your own risk. In no event will we be liable for any loss or damage without limitation, indirect or consequential loss or damage, or any loss or damage whatsoever arising from loss of data or profits arise out of, or in connection with, the use of this video.

[@md.hasanuzzamanjihad4623]

Great and easiest explanation... Thanks bro. Pls upload more videos on SFD and BMD problem solving

[@FilmonAron-qx4gg]

Thank you so much for the clear explanation!

[@ifinfotech]

Glad it was helpful!

[@AnanyaAugustine]

Thank you for the detailed explanation

[@ifinfotech]

You are welcome!

[@user-sc2zw8tl6z]

Thank you so much

[@kenken578]

Can you also do for moving loads, and triangular loads example? Thank you.

[@baiklahtu]

Sorry sir, SFdr for SF diagram should be - 4.14 or - 5.86?

[@gabrielatem1121]

Much appreciated your explanation❤

[@user-eq1ih2oc9p]

Great video, have being trying to understand udl and your video helped me a lot, much love from Kenya 🇰🇪

[@charukadilshan4599]

Great Teaching

[@gluton07]

while taking moment at E its length is 3.57, doesnt the point load 8 kN will be counted as Moment at E = 7.14*3.57-2*3.57²/2-8*1.57 ??

[@satazer5824]

Perfect method. Thank you.

[@ifinfotech]

You're welcome!

[@strdpra0z]

Sir when you find out moment of udl that time 8×4×(4÷2) +5×5=Rb×6 hona chahiye

[@olivergrennan309]

your the goat

[@prashanthakumar4573]

how did you get Rb value 5.86kn

[@mahendratripura800]

41/7

[@user-zw2qq8gx1r]

你超猛的,聽不懂你的語言我卻看得懂整個分布圖,

[@gabrielatem1121]

Could you clarify how length x=3.57 become force at maximum point E?

[@aqsanazir4819]

Same question

[@zyler548]

Look back 10 Standard Maths Similarity of Triangles

[@patilgouda]

How Rb =5.86kn

[@swasthik.B]

41/7

[@kerinekylie]

There's something wrong here

[@richardrichard886]

💩