Blue Book Steel Design - Laterally Restrained Steel Beams
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- Опубліковано 27 вер 2024
- This is a short video tutorial to show the design of a laterally restrained steel beam to EC3, using the Blue Book by the SCI. This video has been made to help students in their design project work. It is in no way endorsed by the SCI, Tata Steel or the BCSA. It is recommended to watch the other videos relating to this topic to best understand the concepts discussed and the calculations carried out. I hope that you find it helpful.
Thank you very much. I am a structural engineer but I have worked in other fields since 15 years. I needed a refresher and man was your explanation easy to follow, clear and very useful.
very helpful video for studying for exam in design of structural elements, thanks
Hi krishshah63, good question, this was given to me by a steel fabricator a long time ago but is still valid and you can check this yourself. Set the deflection to span/360 and substitute this into the usual deflection formula for a simply supported beam with a constant UDL, you can also add in the E value for steel and the span length. So, the only unknown would be w, the UDL. Similarly, you can take the general formula for the mid-span moment of a beam with a constant UDL. You know the span length L. You can rearrange the two formula to show that I required = ML/56. You need to take care of your units and remember that your answer is cm to the power 4. I hope that this helps, Mike
Great video and great explanation, excited for new videos!!
SO SO GOOD, Absolute gold!!!!!!!!!!!!!!!
Your videos are really helpful, thanks
Thanks sir,really good explanation and very easy to understand
Grate video, can't wait to more.
Thank you Mike, I can't wait to see more of your videos
Regards:)
Hi Mr Bather: Thank you for all your U Tube videos and appreciated. Could you please do one video how to design steel column using EC3 and carry moment due to (lateral load wind load + eccentricity) + Axial load. I am sure all your followers do appreciate !!! Thank You
Please do put videos on moment resistant connections sir.....,,Thank you sir for your clear cut explanation.....,,,👍
Legend
Please, very interesting and educative video. Please, can you do a video on timber design, please??
Thanks for the video. Would it be possible to explain how you came to the 'short-cut formula', I = ML/56 to find the second moment of area? Thanks
Hi Ryan, apologies for such a slow response (I thought that I had replied earlier). ML/56 is found by rearranging the deflection formula that appears at 11.30 in the video. This is a standard formula for a UDL on a simply supported beam. The deflection is given in terms of span/360 and the moment is given by the usual w x L x L / 8. This is algebraic manipulation to make I the subject of the equation. You have to be very careful of the units and always be aware of the assumptions for this shortcut. I hope that this belatedly helps, Mike
@@mikebather thank you very much Mike!
@mikebather
I was wondering what is the ML/56 formula you used for the 2nd moment calculation...
How did you derive it?
A quick response would be appreciated. Thanks.
I'm confused: what's the differendfe between the BM in blue and the one in red at around 4min 30 secs? And why does it suffice to work out required second moment using only the (lower) red BM? Brilliant video by the way
Hi A A, at around 4 mins 30 secs, the blue BM relates to factored loads and the red BM is solely unfactored live loads. So the blue is ULS and the red is SLS. I hope that this helps, Mike
How do you know whether to find a higher required plastic modulus Wpl,y or larger 2nd moment of area for our trial section?
Hi Mil, this is a good question - do you design for strength or deflection. A rather vague answer is that for shorter spans, strength is likely to limit and for longer spans it is deflection. The problem is that most beams are somewhere in between. This is something that experience will help, but until then, you need to plump for one or the other - for typical floor beams, I generally go for strength first and then check deflection. I hope that this helps, Mike
@@mikebather Hi Mike, thank you for your response. It does make sense now.
Also, when establishing Wpl,y required we are automatically assuming the section is plastic right? Then is the cross-section classification a verification step to confirm the section is Class 1 or Class 2 plastic?
@@mil9102 Hi Mil, yes it is a check. Take a look at Draycott and Bullman's textbook where is shows that ALL UB sections are Class 1 or 2 in bending and nearly all UC sections too. But, you may wish to design for other sections and so, for these, you need to check classification, hope this helps, Mike
Sir u should upload more vids. U have great works
Hi Mike, why are you using Iy, why not Ix? I'm guessing that live load will load the beam in its major axis. So shouldn't just we check Ix value? TIA
Hi Gagan Kataria, thanks for the question - it is something that I did not make clear in the video. In the UK the strong axis (vertical bending for a vertically aligned I beam) used to be called (BS5950 and BS449) the Ix axis and the weak axis (side to side bending) was the Iy axis. EC3 has renamed these axes so that the strong axis is now the Iy axis and the weak axis is now the Iz axis. I hope that this helps, Mike
Cheers Mike, that makes it very clear. Iy is basically the strong/major axis, what you're checking against.
I tried to see if I could use blue book for Australian beam sizes but I figured names and some properties were different. Therefore didn't proceed.
However I do have Fionna Cobbs structural pocket book, which I refer to sometimes.
А что с расчетом на устойчивость?
Hi Andrejs Jakovlevs, using google translate, I understand that you are asking: what about stability? These short videos only really cover element design, overall lateral stability is another matter. Is it the lateral stability of the beam that you are asking about? A laterally restrained beam has its top flange connected to a structure that is able to restrain it from lateral buckling - this is considered to be so when the adjacent structure is able to resist a force of 2.5% of the compression force in the top flange. I hope that this helps. If not, please could you reply again in English, thanks.