Talk:Shear Force and Bending Moment Diagrams

Latest comment: 6 years ago by Taltastic in topic Units

A few comments

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This is really an superb coverage of SFD & BMD. All the complicated topics are covered in lucid language & its really very worth for all engineers. I had some minor doubts mentioned as below -

1. Will the concept of SFD & BMD differ for various types of support (such as rollers, or cantilever)

2. Thanks a lot for the valuable techniques. It was really fun filled to learn SFD and BMD from your page. It will be great if you include topics on Macaulay's method, Double integration method etc for bending of beams.


--41.133.21.67 09:37, 14 October 2012 (UTC)This was absolutely brilliant. Thank you, as an architect trying to teach tructures to architecture students, it was simply put and concisely explained. Graphic quality to the article is fantastic. Architects understand pictures! Many Thanks Ken Stucke. South Africa.Reply


I found this far more useful than the textbook and the lecture notes I have. My thanks to you.

Thanks a lot sir, for such an interesting and useful material.--Rajeev


Some very usefull information. would be nice if you wrote some about Torsional moments as well.
-Thanks. sorry, I never did anything on torsional moments.


--Mack paris 10:33, 14 October 2010 (UTC)could you please add something about moments and couple along with the diagram.--Mack paris 10:33, 14 October 2010 (UTC)Reply

-Sure. Hope what I added helps. Taltastic 23:24, 17 October 2010 (UTC)Reply


The explanation provided here with simple and easy to understand figures are absolutely brilliant. Looking forward to your continuation, sir. Please do go ahead sir. Your article has conveyed the ideas way easily than when I tried even harder to understand what my sir was trying to explain, by complicating it in class.

By the way sir, just 1 doubt. " Taking moments about A (clockwise is positive): 40·2 - 20 + 6·R2 = 0 "

This is the line I've picked from your example about point moments. I believe the equation should be: 40*2 - 20 - 6*R2 = 0

Thanks again for this brilliant article. Looking forward..

- Deepak.ail 10:49, 18 October 2010 (UTC)Reply

--Thanks a lot for the positive feedback, and thanks for calling me Sir :) And yes, you were right about my mistake, thanks for noticing. Taltastic 21:33, 19 October 2010 (UTC)Reply

Fantastic article, really helped me with pre exam study.

--== Thanks! ==

Really Interesting! I am going to university this year to study civil engineering and find this article very helpful. Is this a writing error?

"If you slice the cube in half halfway up"

BTW This article is written very well. All you need to know really is some basic mathematical concepts (e.g. how to draw a diagram) and the concept of moments. Please continue to make more articles. I personally (considering my career) would prefer any article related to civil engineering.

Thanks once again! :)

--92.15.215.59 18:00, 14 January 2011 (UTC)Reply

Thanks for the nice message! nope, not an error. I may continue writing articles one day, right now I have to find a job... Taltastic 14:34, 15 January 2011 (UTC)Reply

Doubt.

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Hello Sir,

I have a doubt to draw Shear force Diagram for a beam when a force is acting at the point. For instance, in the example, Why do you have to consider both E and E' while drawing SFD??

- So let me first check something, you understand how the method I have taught works. i.e. you know which situations to take the shear force before and after, and you know which situations to it just at the point. However you don't understand WHY you are doing this. I'm afraid I can't think of a good way of explaining why, other than to say that at C the point force of 40N up, causes the shear force to increase by 40N (i.e. go from -10N to 30N. so from c to c')Taltastic

Thanks!

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You nailed it!! Thank you for the effort..

This is very very helpful!

Thanks a ton.....

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This is worth reading and remembering......All that what looked so complicated in the school back then, you have nailed it here so effectively.....that it gets registered easily in the users' minds. Thanks for your help.

Thanks

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Thank You so much, It helped me recall the basics I learnt at school.

i found this extremely easy to understand in comparison to most websites and textbooks (pdfs included) THANKS =)

Thanks an Ocean...

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I like this article. I understand shear and bending moments diagrams clearly now.

Please improve this article if you can. Thanks

Shear Force at Point

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In the example under the heading "Basic Shear Diagram" you have a point force of 20N acting down. I understand the need to sum the forces before and then after the 20N force which gives you a straight vertical line on the diagram. My confusion is involving what is the shear exactly at that point of the 20N acting down? Since according to the diagram which displays the strait vertical line starting at 10N at the top and -10N at the bottom, would the Shear force exactly on the 20N point be 20N, 10N (as it is before the force has acted), -10N (as it is after the force has acted) or 0N (Since the line crosses 0)? Thank You! --Bmullin (discusscontribs) 01:54, 12 February 2013 (UTC)Reply


I'll be honest, I'm not an expert on the topic. However, I suspect that on the 20N point, you would say that there is a shear force transition from -10N, to 10N. (So the shear force isn't a single value, like -10N,0N,10N or 20N). Taltastic (discusscontribs) 01:48, 14 February 2013 (UTC)Reply

What is shear force?

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As stated in the example here, it is understandable that when the force is applied to an object it is redistributed all over. However the distribution cannot be linear as suggested in diagrams, having an instant rise or fall at a definite point, in an elastic case, it is very hard to visualize. The only explanation of this phenomena could be that we consider the object totally inelastic for consideration of the Shear Force. Hence rendering it to be proofing against inelastic failure. Please elaborate if this is the case or if there is another explanation.

Thank You!


You make an interesting observation...To quickly answer your question - no, you will still get a uniform shear force distribution even when the material is elastic! I think I gave too brief of an explanation on page on what shear force is, so I will write a better one (with pictures). Give me a week or 2 to do it :) Taltastic (discusscontribs) 15:13, 20 May 2013 (UTC)Reply


why are the right portions covered with paper? I mean why the loads towards the right hand side neglected while considering the moments?

Moments and bending moments are 2 different things. When you work out moments, you look at both sides. But when you work out bending moments you only look at one side. I've chosen the left side. Let me know if that doesnt answer your question/you have more questions Taltastic (discusscontribs) 20:19, 12 November 2013 (UTC)Reply

Hello Sir! This article was simply amazing, helped me a lot. But i have doubt,

Y do we have to calculate shear force before and after at a particular point? And What does maximum bending moment indicate?

The shear force before a point is different to the shear force after a point, sometimes. Which is why you have to calculate both, and show them on the diagram. To try to answer your next question: In all the examples I've shown in this document, you can imagine that the beam will bend under the forces/moments its experiencing. This is because the forces are producing a bending moment at a given point on the beam. It is that bending moment which causes the beam to bend at that point. The more bending moment there is at that particular point, the more the beam will bend at that particular point.Taltastic (discusscontribs) 20:54, 27 November 2013 (UTC)Reply

IT IS PERFECT,thanks a lot

Thank you for the topic on SFD and BMD, was finding it difficult to recollect after my Engineering days. Found your topic very easy to follow. Please continue with such basics of Engineering.

Regards

Can I get some similar examples of SFD and BFD for roller support and Cantilevers.

Thanks

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This helped me :) Thanks a lot

THIS IS SUPERB! Please create more tutorials on....

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  1. Thin-walled pressure vessels
  2. Axial loading
  3. Stress and Strain
  4. Piles, pad footings and retaining walls
  5. Fluid Statics
  6. Centre of Gravity, Centroid, and Moments of Inertia

Your way of teaching is really easy to understand. Thank you very much.

Thanks

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Super and crystal clear explanation, I have read many books and articles and was totally confused about SFD and BMD. I am very thankful to the writer about clearing my doubts on sfd and BMD in a simpler way.

Left on Resource page at [1] by User:Umashankarshastry, 24 November 2016

I question the validity of this statement in Uniformly Distributed Load (UDL)

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Are these really equivalent?

I think an IP editor made a valid point about the diagram shown. It seems to be only an approximation. The IP editor wrote:

Oh no, this is wrong, the placement of the force along the x-axis will influence the deflection of the beam. With your logic (whoever wrote this), we could spread out the load over the entire beam and it would be equivalent, but everyone knows a force applied directly above or near the supports will not bend the beam anywhere near as much as if you apply the same force at the center. 30N distributed over an area of the beam will not deflect it as much as 30N concentrated at the center...

See this edit. Could somebody look into this? --Guy vandegrift (discusscontribs) 03:15, 20 August 2017 (UTC)Reply

Those loads are not equivalent for bending moment and shear force under the uniformly distributed load. Reaction forces will be the same, Shear force and bending moments will be equivalent between the uniformly distributed load and the supports. They will differ over the centre section where the loads are different. IP editor is correct in their reasoning. Pbsouthwood (discusscontribs) 12:09, 19 October 2017 (UTC)Reply


Thank you both for looking into this. I wrote this article. And yes you are right, these situations are not equivalent for the reasons you've mentioned. Diagram B is a simlification of Diagram A, and should be done in order to help with plotting shear force and bending force diagrams. Hopefully I've that more clear now. Please edit if you don't believe that I have made it clear.Taltastic (discusscontribs) 08:38, 21 October 2017 (UTC)Reply

thanks for the article, I was totally confused by all the lectures & assignment. Yet, it is very easily understandable in succinct language here. Really appreciate. --JY 14:16, 17 November 2017 (UTC)

Units

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Shouldn't the units of Bending Moments be Nm, not N? Your BMD's vertical axes all show units of Newtons, not Newton-meters. by Turbocentric | 21 March 2018‎

@Guy vandegrift: Please review. -- Dave Braunschweig (discusscontribs) 00:00, 22 March 2018 (UTC)Reply
Yes, I agree that there is a problem. The person might mean the contribution of bending-moment (per meter). If these calculations are correct (a BIF "if"), then this needs to be clarified.--Guy vandegrift (discusscontribs) 13:25, 22 March 2018 (UTC)Reply
Thank you for spotting, and yes it is a mistake I made. I made a note in the document to make this clear to readers. Taltastic (discusscontribs) 21:10, 13 October 2018 (UTC)Reply
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