RC Slab Design EC2 - Worked example - Bending reinforcement
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- Опубліковано 28 вер 2024
- A short tutorial showing how the main reinforcement in a simply supported slab is designed using EC2. A second video shows how the shear resistance of the slab and its deflection are checked.
I wish i had you as a lecturer while completing my degree..we struggled unnecessarily with structures
Thank you very much for such an explicit and detailed explanation Mr Bather.
very clear and practical example. Thank you! What cool explanation!
Hi Mike, very helpful thank you. I appreciate you explaining all the terms and going back to basics.
Henry
Great videos! This has made everything so much clearer for me. Would it be possible for you to post a link for your tables? I've found most of them however struggling to find the sectional areas of bars with spacings. Found a few but all really low quality images.
Thank you! It was very helpful! I would love to see more your videos!
Thank you very much for this explanations. 😃
Brilliant video, I know that it will be taking it to very basics, but for educational purpose the guys should try to use their units - that will help them to keep track of what they do and verify results. Nevertheless a great explanation with lots of tutorial value.
Great explanation and pace, thank you !
i have no words to thank you enough. thank you thank you so much .
Thank you for this amazing explanation and excellent instructions of solving the EC beams. I just subscribed to your channel Mr. Bather.
Really you are doing a great job. I really appreciate your effort
Great Video, however I would like to know how we can get the distribution steel....
Hi keithkagiri1604, good question. You do need distribution steel and I would look at the minimum cross sectional area of steel required for a slab. You would also need to check minimum bar sizes and spacings in relation to cracking. All in all, this really would need a separate video - the alternative would be to speak to an experienced RC detailer who is likely to know this better than the design engineer. Sorry for not directly answering your question - Mike
I must say a very big thank you for making me understand this topic, but to calculate the area of the steel you used moment/ 0.87 X Fyk X z. The question is in some textbooks, 0.95 and also 0.87 are used. Are these constants, or what determines when to use 0.87 or 0.95 to multiply with the Fyk and Z
Hi Afolabi Olaosebikan, thanks for your comment. In my equation I multiply the yield strength of the steel rebar by 0.87. In the Eurocodes the design strength is found by dividing the yield strength by the material factor of safety of 1.15. The reciprocal of 1.15 is 0.87 and this explains how it comes to be in the equation for the area of rebar.
Now, the lever arm (z) is a different matter and this is the distance between the centre of the compression stress block in the concrete and the centroid of the steel rebar in tension. For a few good reasons, the lever arm must not be too large and a distance of 0.95d is given as its limit. d is the distance from the top of the slab to the centroid of the steel rebar. I hope that this helps, Mike
@@mikebather thank you so much, Mike. I didn't understand at first but now I do get what you explained. Well done and keep up the good job. Thank you
Good explanation, let's say I have a slab with one edge contineous and at the contineous edge I have hogging moments of 35KNm and a sagging moment of 26kNm at midspan, how do I get area of reinforcement required? do I just consider the highest moment or what? pls help
well explained thank you very much, jst for anchorage and curtailment how did you calculate it
Anchorage was rule of thumb based on old BS. For curtailment he used a specific formula as shown in the video.
learned a lot from this lecture.. can u also do an example for a slab which also needed to provide compression steel bars
Hi Engr. Nikko, I will make a note of the slab - but am really busy with my students at the moment. Mike
Can you solve an example about designing two way slab
Hi Mike, thank you for sharing the knowledge. One thing I am not clear when you used 12 x diameter of bar for the anchorage beyond the centre length of the support. Shouldn't this be at least 40 dia of the tension bar ?
Hi Qasim Muhammad, the anchorage at the support already includes a bob on the bars and only needs to transmit reduced tension forces in the bars. The 12 x D is a rule of thumb in this situation. I hope this helps.
Thanks so much that's appreciated but I wonder if you can use fixed end support ??
Great explanation . . . is that someone snoring in the background? Big distraction!
Hi Joeaby Vasslo, thanks for your comments and yes it may be snoring in the background, but not a person, it is probably my dog Dale!
Haha cute! Can you recommend me any publications or links online where by I can find the tables used in EuroCode Design procedures?
Joeaby Vassallo Hi again, I have just searched the internet for some publications that the Concrete Centre in the UK used to offer as free downloads. It seems that these have been withdrawn. They do however, have a version of their spreadsheets for the design of RC structures which I understand is free for educational use. The spreadsheets are really excellent and a very quick way to design RC elements.
great thanks a lot!
hi Mike, I'm assuming the reason for no steel on the top side of the slab at the support is because it was designed as a simply supported and not fixed end connection? when would we design it as a fixed end connection?
Hi Josh D, Yes, you are correct. Continuous slabs develop tension forces at supports and so are designed with rebar in their tops - this is similar to design for a fixed end connection. Most RC slabs are continuous and so are designed like this. Slabs fixing into walls and very large beams could be considered fixed to some degree at their supports, Mike
Excellent presentation with neat and comprehensive list of calculations. Hope to see more videos soon.
Hello..Great explanation. But why we are using always 1m strip, in fact we've to consider full slab. and also there is no consideration of the another dimension of slab expect thickness and span.
Hi Nipurn Khatri, thanks for your comment. At the end of the calculations we specify the amount of reinforcement needed for 1.0m width of slab. So if you have a 2.0m wide slab this needs double the steel over double the length and so it goes for a slab 8.6m wide then 8.6x the amount of steel. Considering a unit width of slab is a useful way of allowing our calculations to be used flexibly over a variety of widths. I hope that this helps, Mike
Hi Prof Mike. Why did you not account for negative moments in your calculation? Why did you not provide top reinforcement to counter act hogging at the support ends?
Hi szymon starosta, good question. I can only cover so much in a 15 minute long video and so I chose a slab that is genuinely simply supported with no fixity at its ends. Yes, this is not a common condition for slabs but it does mean that there are no moments at the ends of the slabs due to fixity. I just wanted to focus on the main sagging bending reinforcement. There are a few other issues that I do not cover also. Best wishes, Mike
@@mikebather thanks. Please do a video on calculating the cracking moment of a slab. You have no idea what a great help these vids are!
This is great. Do you have example for two way slab design to eurocode? Thanks
thanks for the video ,can u make a video too on "two way slab"?
What if there is a hole in the slab, for instance a manhole cover or stairwell? Does the same apply?
Hi dougaldog89, Thanks for the question. The quick answer is yes. There is a good book used in design offices in the UK (Reynolds's Reinforced Concrete Designer's Handbook) which gives really clear advice on matters like this. Broadly, if you remove some of the slab and if you increase the loading on the remaining slab then you need to add extra reinforcement to cope with this. So, if you remove half the slab and don't increase the load, then you will need to roughly double the reinforcement. I hope that this helps, Mike
how would I design a floor slab that spans 10m by 7.5m
easy to understand,i gonna subscribe.
Great Video! How do you determine if one or two layers of reinforcement?
What is covered in the video is only a part of the overall design process. In the design office, the engineer must also consider: fire, durability, crack width control, detailing, etc. When all this is done, the final arrangement of rebar can be completed. As a start, it is convenient to keep to a single layer of tension steel and, if apropriate, to make use of mesh reinforcement. Sorry not to be able to fully answer your question.
helow mike. thanks for the tutorial. I am just wondering how did you decide 12mm rebar for the reinforcement of slab. thanks
Hi Dondie Amoroto, this is a good question. Really, it is the designer's choice at the start of the design. Large diameter bars will have wide spacings which may be a cause of cracking. Small diameter bars will have narrow spacings which causes extra work on site and may be problematic at junctions and when pouring concrete. This is something for experience. I have just picked a bar size that will not give problems with cracking (I checked this separately) - in the design office, an experienced RC detailer will be able to give advice to young designers. I hope that this helps.
hi...first, thanks for this very good tutorials, it helped me a lot. but I have one question about the area of steel part, where did the .87 value came from? sorry for this ignorance of mine. thanks again thou.
+Peter Pandesal 0.87 comes from the safety factor of steel which is fyk/1.15. Converting 1/1.15 becomes 0.87.
+Goh khen hui , oh thanks! now I get it. Appreciate your kindness for answering my question.
Plz sir can you answer my question since i have a project
I have continuos
And i design for simply ?
Its ok?
it's great video apart from that snoring!
Oh sir on the example you made why did you calculate "Mu" for?
Hi nyakallo letsie, Thanks for your question. Mu lets us know the greatest moment that the slab can support without compression reinforcement in the top of the slab, i.e. without extra steel rebar that helps the concrete to carry the compressive forces at the top of the element. If your design moment is greater than Mu then you need to design some extra compression reinforcement. That is it really, Mike
thank you very much, nice and clear
any chance of shear reinforcement calculation?
Hi Umar, Thanks for the comments.There is now a video covering a shear reinforcement check.
Hi! I have a question, No more positive and negative moments for the slab?
Hi Neil Ryan Lim, In this example, I have used a simply supported slab. And so, there are only sagging moments in the mid-span of the slab. This is a condition that almost never occurs in real life, but that is fine as the point of the video is to introduce the design of reinforcement in a simple situation, so as to not overcomplicate things. I hope that this helps.
Oh thank you!
Hi sir. Again another great video. May I ask how do we use BRC instead of the bar?
Hi Arman Chua, thanks for your comment. I am not sure what BRC is. In the UK there is a company that makes steel reinforcement called BRC but I am not sure if this is what you mean?
Mike Bather Hi Mike, some countries called it BRC. In UK, we called it mesh or mesh reinforcement.
FANTASTIC! ur camera going in and out of focus a lil though. but thanx for ur help!
thanks helpful
Is that snoring in the background?
Plz may have the figure that you calculate d/z from it i think its (5.5) because i search for it but no result
thanks .
Hi Amjad Sharba, I am afraid that I don't know how to give you the diagram. There is however another way to find z. You can use the following equation which give just the same answer:
z = d x [0.5 + square root of (0.25 - K/1.134)]
If you are unsure how to make use of the equation, you could substitute the numbers from the UA-cam example into it to check your understanding. I hope that this helps.
Hi, i would like to ask, how to differentiate continuous slab and simply supported slab? In a building design eg shop lots, my slab should be in continuous right?
Hi Zian Yang Chiew, a simply supported slab will have just two supports. It may be supported by beams or walls. A continuous slab will have more that two supports (say three supports) and its reinforcement will be continuous across the middle support(s). It is important to differentiate between the two, as continuous slabs typically hog over their middle supports (which means that tension is in the top of the slab - and so this is where reinforcement must be placed). Simply supported slabs just sag in the middle of their single span (and here, the tension is in the bottom of the slab and this is where the reinforcement is placed). I hope that this is helpful.
Hey thanks for the explanation. So it is up to me to design it as a continuous slab or a simply supported slab? Lets say i want a slab to be design as simply supported slab, meaning that sagging moment will be larger compare to continuous slab (bigger Med), hence, reinforcement will be more at the mid span (to provide bigger Mrd) while since there is no sagging moment at the support, no top reinforcement is needed for simply supported slab. Does this sound correct?
Zian Yang Chiew Hi again, you seem to understand the idea behind where to put your main reinforcement very well. With reinforced concrete you must also consider placing additional reinforcement (as well as the main reinforcement) for example for shear, or to limit cracking. So even when you do not need main reinforcement to cater for bending, you may need other reinforcement for other reasons. Really, this is beyond the scope of the short video that I posted.
hello,
can I get help with a two way simply supported slab on all directions ?
regards,
Hi Razan, I am sorry but I don't have time to help - I am too busy with my own students. However, there are some great resources. Do you have access to the CIS / IHE database? If so, you can download the IStructE Manual for the design of concrete building structures to Eurocode 2. Also, the Concrete Centre has some good web pages for students: www.concretecentre.com/Publications-Software/Design-tools-and-software/CALcrete.aspx. I hope that this helps, Mike
Hi Mike, how would you design a two-way supporting slab ?
Hi Athtech Designs, two-way spanning slabs can have linear supports along their edges (say beams) or just be supported on columns (in which case they are generally called flat slabs). Either way, for a manual design (i.e. making hand written calculations), I would make use of design advice given in the IStructE Manual for the design of concrete building structures to Eurocode 2. This gives simple advice allowing a similar design approach to that outlined above to be used. For a linear two-way slab square shaped on plan, with length and breadth the same, then the applied load is carried equally by the slab spanning from one side to the other and also from one end to the other. So the reinforcement in each direction will be equal. For a linear two-way slab with length different to breadth, then more of the load is carried by the shorter span and less by the longer span. The manual noted above gives coefficients for bending moments and shear forces for these situations. I hope that this helps, Mike.PS I am sure that other design guidance is available elsewhere
Life saver
Thanks
Reinforcement tables are available at:www.brc-reinforcement.co.uk/pdfs/catalogue.pdf
Thanks.
hello please this link doesnt work anymore, can you please send another link and also can you include the link for the lever arm graph thanks
Where do you find that klin=0,168 factor ? at t=6:35
It is often recommended in the UK that Klim should be limited to 0.168 to ensure ductile failure.
what is klim?
can some one explain this to me please: concrete self weight = 0.29 m^3 x 25 KN/m^3, that equals 7.25 KN, then he says 7.25 KN/m^2. Does he multiply it by 1 m^2 after?
7.25 KN is the self weight derived for a volume of 1 x 1 x 0.29 m of concrete. However on plan view, 7.25KN is the weight for 1 x 1 m area of concrete. Hence so. I hope this answers your question.
Excellent - couldn't have put it better myself!
@@kirati thank you for your answer i confuse that the video is .29 i do not see 0.29 make me wondering
@@14_sreyniruth_m34 you are very welcome.
@@mikebather my pleasure.
thanks
Hello sir,
Can you please share that site where I can find free spread sheets, Also please deliver some lectures on flat slabs designing and flat Plate designing
Sir can u pls share the Graph for the value of Z
Hi Waheed, I cannot upload the graph for z but here is a formula that many engineers use (especially if you have excel):
z = d [1 + SQRT(1 - 3.529K)]/2
In this formula K is found from:
K = M / (fck x b x d x d)
I hope that this helps, Mike
Didn't know EC is used in the UK. Would've thought you guys used some sort of "Her Majesty the Queen of England Elizabeth II"- standard for reinforced concrete...
not two way bending.
Amazing tutorial
Good explanation, let's say I have a slab with one edge contineous and at the contineous edge I have hogging moments of 35KNm and a sagging moment of 26kNm at midspan, how do I get area of reinforcement required? do I just consider the highest moment or what? pls help
+Herta Mbadhi I think that you need to design them separately, as the reinforcements for hogging moment and sagging moment are placed differently.
well explained thank you very much, jst for anchorage and curtailment how did you calculate it
Hi Hasabo, Thanks for the message. I will not be covering anchorage or curtailment in detail with my students this year. A general rule in design offices in the UK has been to ensure 40 x diameter of the bar for anchorage. The Eurocode is far more complicated than this. Best wishes.
thanks.
Nicely Explained Mr. Bather
well explained. Thank you sir :) please do upload more videos on reinforced concrete beams.... thank you.
dan
nice presentation ,but little loud for other language speakers
you changed my life
may I know K=m/fckbd2 what is the m mean?
m=moment
my goat
Hi,thanks for good explanation ,know i understand there is a differentiate between simply supported and continuous
plz when i calculate is there a different in length span .For example, in simply supported we take L eff but for Continuous we take Ln is the clear span from column to column ? is that right ?
and for the moment in simply support we take W L2/8 but for Continuous it will be W L2/12 ?
Hello sir, very good explanation. I have few questions sir, could you please help me.
Hi. This video is very useful. What is fck and fyk?
Hi Regina Tasya, thanks for the comment. These terms refer to the characteristic compressive strength of the concrete (fck) and the characteristic strength of the reinforcing steel (fyk). I hope this helps, Mike
The link below no longer works and so I had a quick Google for 'reinforcement area tables' and this link seems to be OK at the moment
www.sarcea.co.za/mass&area.php
I don't have a link to the lever arm graphs and now tend to use a formula in place of the graphs. You can use the following equation which gives just the same answer:
z = d x [0.5 + square root of (0.25 - K/1.134)]
If you are unsure how to make use of the equation, you could substitute the numbers from the UA-cam example into it to check your understanding. I hope that this helps, Mike
Hi. How about a concrete flat slab design with and without drops?
U are simply the greatest!!! Keep doing what you do. I would love to see your take on a moments connection of a UB to UC to eurocode
You are beautiful teacher
Thank you I love you
eng.kapi I agree with the lectures
This is absolutely a great explanation, but there is something it does not convince me at all !
You jus said that the dead load is 7.25 KN and that is totally right , but how come that the live load is 5 KN/m2 !
I mean the difference in their units ! the live load is on Kn/m2 and on the other side the dead load is only KN !
abdulaziz alhajeri The dead load will be divided by the area, in this case 1m by 1m hence you can say that the 7.25KN will be divided by 1m2 to get the units right - 7.25KN/1m2 = 7.25KN/m2
Sir, please can you do for Beam!!
Thank you you are the best !
Purushoottamnishadgomal
So Great.. thank you so much
Hi Mike
Excellent tutorial. I have a question to add to this. How will you go about calculating the total factored moment with an additional point load (DL and LL) occurring @ 1/3 of the span? Is it wl^2/8 + Pab/L for total moment?
lets say a dead load of 10kn and live load of 15kn acting @1/3 span in addition to the UDL.
Hi XXYellowFlashXX1, what you are proposing seems reasonable. It is not quite correct as the point load moment occurs at the third-point and the UDL moment occurs at the mid-point, but heh, bearing in mind the enormous safety factors already applied, this is conservative and near enough for a small job. If there is a lot of repetition or large areas of floor, then really you should use basic structural analysis and actually find the moments along the length of the slab. This second approach only takes a couple of minutes, as long as you are comfortable with the analysis. I hope that this helps, Mike
Sorry I could not understand the solution. For the udl the moment calculation is wl^2 and the point load will be the load multiplied by the distance to support. We calculate the moment separately for both udl then point load then sum the moments together. Then apply the safety factors to get the total factored moment? Can you verify the above please? If otherwise, please explain. Much appreciated.
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