Great tutorial. You are much more helpful and clear than my college lecturer. Now I understand everything. I cannot thank you enough. Thank you so much.
Thx a lot Mr. Mike , I’ve learnt a lot from ur channel, pls don’t stop making these very useful videos . Specially because there’s is no one on RUclips who explains the manual design to Ec2 .
Thanks a lot Mike. Your reply clarifies my query. I hope you will have time to provide more tutorials because they are simply great. You are doing a fantastic job :-)
Omg you're a life saver! My newly hired lecturer didn't even teach this! He just expects us to know how to calculate by only teaching theories! Thank you very much sir! Salute!
well I haven't watched this video; just downloading it. But from comments by others I think this video is a good one. Plz Tutor continue witht the good work. Thanks for that.
Too many incompetant structural engineers these days who act as if what you has shown is some form of black art, which it certainly is not. Your tutorial is great
Excellent video - you explain everything exceptionally. This sums up a month of lectures in my RC design course. Thank you so much, all the best from Singapore!
Thank you for making these videos, much appreciated. Just one question, the tributary area you used had a 5m width. Due to continuity of the slab, should the width not have been 1.25 x 5m = 6.25m? The 1.25 factor is used for a continuous 2 span system?
Hi Name, Last Name, thanks for your comment. Yes, you are correct. If the two span slab is continuous over the central beam support, the reaction should be greater there due to elastic shear. I have squeezed in an awful lot into this rather short video (which I was trying to keep under 15 minutes) and did not feel that I had time to explain this. I hope that I have not misled anyone by taking such a simple approach to loading, Mike
@@mikebather Understood, the main focus of your video was how to design a RC beam, mentioning continuity or non continuity of the slab may have added unneccesary confusion. Loading would need to be covered in a separate video I guess. Thanks Mike, the time and knowledge you share is much appreciated.
Great tutorial thank you very much. I have one doubt. In our Engineering studies we are not studying about the concealed beam. So i think if i put any concealed beam inside a slab,its also taking any load or not.
Hi Jeyaganesh, Thanks for your comment. This example is rather simplified to focus on the design steps for reinforced concrete.In reality the beam in the example would act as a T beam. By 'concealed beam' I am guessing that this is a heavily reinforced section within a slab. This section could act in a very similar way to a normal beam. Because it is stiffer (due to the heavier reinforcement) than the adjacent slabs, it will attract load and begin to function in a similar way to a beam. I hope that this helps.
If the beams are continuous the inner supports would experience hogging, so the tension reinforcement at the support would be in upper side of the beam (or possibly in the slab since it is monolithic), in that case how you would calculate the required tension reinforcement please?
Can we provide 25mm for the hanger bars, I mean the main bars are 32 dia and can we reduce the dia of the hanger bars as it is just a support, the self weight will be reduced? Thank you sir
Good video! But you should perhaps perfect the end by doing further calcs, showing that your obtained tensile reinforcement area is within the "acceptable" range? I am referering to As(min) = 0.26 Fctm/Fyk * bd ; As (max) = 0.04Ac, etc etc...
Hello Mike, Thanks for the tutorial. Could you please elaborate on the k value? I see that when calculating the ultimate moment capacity it is good practice to take 0.168 as this will ensure ductile failure. Is this because you are ensuring the steel yields prior to concrete crushing? If so it is strange to me as the Mu = 0.168*fcu*b*d2 has no mention of the bar size or steel strength. It seems that as a larger k value will give a shorter lever arm and therefore a smaller moment capacity. I am unsure what this value represents? Please help! Homer
Hi Homer, Thanks for the comment and I can comment a little on the k value. It would be better with some diagrams to help really. The balanced section of the RC beam is the case when the steel reinforcement will yield in a ductile way just before the concrete fails in compression in a brittle way and the two materials reach their ultimate strains at the same time. As more and more tension steel is used in a beam, the depth of the compression block in the concrete gets deeper and deeper (as does the depth to the neutral axis x). It has been shown by experiment that the steel will not reach its yield point when x/d is greater than around 0.5. EC2 limits x/d to 0.45 to ensure ductile and not brittle failures. Now,the ultimate moment capacity of the RC beam is the force in the steel x the lever arm distance to the centre of the compression block in the concrete (i.e. Mu = Fs x z). This is exactly the same as the force in the concrete compressive stress block x the lever arm distance to the centre of the steel reinforcement bars (i.e. Mu = Fc x z). This is because the horizontal tension force in the steel is always the same as the compressive force in the concrete. Knowing this firstly, means that we can define the bending strength of an RC beam (even without reference to the amount of steel there) and secondly, allows us to make an equation for Mu (the ultimate moment capacity of the beam) including the size of the compressive stress block of the concrete, the strength of the concrete and the lever arm distance to the steel. When this equation is tidied up a little, it equates to what you have written in your comment: Mu = 0.168 x fcu x b x d x d. I hope that this helps.
Hi, Do you happen to know the equation for the line on the graph which is used to find z/d or where to find it? Lesson was really useful by the way thanks alot
hi mike, your tutorial is great & elaborate.kindly clarify my one query that how we can select the dimension of beam i.e LxBxD and you chose half of one part of slab i.e 2.5 m while calculating live load which transfers a load on the beam.Kindly revert on the same. Thanks
Sir, i have a question here. for the selfweight of the slab, in the previous video, you calculated in per metre^2 ( as in 1x1xthicknessof slab). however in this video you used the area of the slab in the selfweight formula ( bxthickness x length). i am kind of confused here sir. which one to use Sir?
Hi, is the previous video on slab design? If so, in the slab design video, I calculate the weight of one square metre of slab say W kN/sqm. then I consider a one metre wide slab spanning a certain distance. The UDL (self weight of the slab) becomes W kN/m for the slab as it is only one metre wide. However for the beam in this video, the beam supports some slab on both sides. It so happens that each one metre length of beam supports around 5.0m width of slab in total. So, for the beam, its UDL (from the self weight of the slab) is 5W. Not sure if this is what you asking about. Hope it helps.
Hi Omar Caruna, There is no particular reason why I chose C32/40 as the strength class for my concrete. Perhaps this is a little stronger than a designer may normally use in a situation like this. Normally, the structural engineer chooses the strength class of the concrete right at the start of a job after considering things like exposure and durability. I hope that this helps.
Hi Arman Chua, the purpose of the two bars in the compression area in this example is simply to support the links and to help form a reinforcement cage that is reasonably stable for when the concrete is poured into the mould. So their size is a practical design issue, not really an analysis one. Hope this helps.
Mike Bather Thank you very much for your response. I figured that as well. Just wanted to reconfirm. Another thing...what if we do not decide on the size of the bars in the start and instead, work out the As,req before determining the size and number of the rebars? I was taught a different way to your example. Please advice.
Arman Chua Hi Arman Chua, using the method of design in the example, the size of the main bars is estimated at the start, and it is possible that, at the end of the calculation, you may prefer to use a different size of main bar. In this case, you need to go back and amend your calculations. Not very efficient. However, an experienced designer will normally predict bar size correctly. Also, using spreadsheets from The Concrete Centre (www.concretecentre.com/online_services/design_tools_and_software/rc_spreadsheets_v4b.aspx) or other places makes revising calculations easy.
Hi Mike. Yours tutorials are excellent. Do you have one covering Two way spanning slabs on downstand beams. I have to calculate the edge beams and internal sizes, steel arrangement. Check for shear and bending. i would be grateful of any help. The bays are 7.5m wide (2 bay) and 6.75m long 7 bay)
Hi Gaelle Mabassa, for a singly reinforced RC element, the size of the hanger bars does not matter. If you changed the design to be doubly reinforced, i.e. using the rebar in the compression zone to carry some of the internal compression forces, then, yes, the size of the rebar matters - as this affects the lever arm between it and the tension steel. I hope that this helps, Mike
Hi Hasabo, Mu is the ultimate bending moment that can be carried by the beam without compression reinforcement and while remaining ductile. Simply, it is the maximum moment that the beam can safely carry.
Very very helpful! please please please could you upload a worked example of shear reinforcement and deflection asap as I have an assignment due in next week but have no idea where to start! thanks!
Hi Ethan, Thanks for the comments. I am in the process of filming something on shear design and will upload asap - probably not in time to help you with your assignment - sorry. The IStructE Manual for the design on reinforced concrete buildings is pretty easy to follow, otherwise the Concrete Centre also have some good free guidance - even easier to follow.
Hi Nicholas, there are a few different ways of finding z (the lever arm) after you have calculated the factor k. I like to use a chart as it shows an alternative to using equations. Mosley, Bungey and Hulse use z = d[0.5 + sqrt(0.25-k/1.134)] and the Concrete Centre use z = 0.5d[1+sqrt(1-3.53k)]. The various formulae are found from solving a rather clunky quadratic equation and then presenting the result in a convenient way. So there is more than one way to skin a cat, Mike
Hello Mike, thanks for this video, I wonder if you could have similar video on the topic of deep beams or may be you know where I can find it in eurocode or in others regulatory documentation? I would like to find step by step instructions, can you help me somehow?
Hi Настасья Пономарёва, I am sorry to say that I am no help to you on this. If I were in your shoes, I would contact the Concrete Centre. They provide superb technical advice for structural engineers in the UK, Mike
Hi dear Sir, thank you so much for your useful videos. I have a question if you have time plz answer it. What if both slabs on both side of beam have different spans e.g. first 5m and second 4 meter?
Hi 786Hosai, If the slabs on either side of the beam have different spans then your design is very similar. Assume that half of each slab is loading your beam (this may not be the case in reality but should be OK for most student exercises). So if one slab is 5m (take a load width of 2.5m for this) and the other slab is 4m (take a load width of 2.0m for this). I hope that this helps, Mike
Hi 786Hosai, Apologies for the delayed response. Approximately half of the load of each slab would be supported by the beam. So the total load width would be 2.5m plus 2m, making 4.5m loading the beam. I hope that this helps.
Hi Hla Soe Naing, Design of compression steel is not needed in this example. The bars in the top of a beam are called hanger bars and their main purpose is to stiffen up the reinforcement cage to prevent its distortion when pouring and vibrating the concrete. Mike
Hi abhisek sukla, I am not sure why the beam would be considered to be over-reinforced. We have checked that Mu is greater than the applied moment and in doing this, we have made sure that at ultimate limit state, the gradual ductile failure of the steel will occur rather than the brittle failure of the concrete. Mike
Hi uwalakapaul7970, at around 1.35 there is an explanation of how the slab loads the beam (both VA and PA). Sorry but this is only very brief as I have tried to keep the video as short as possible. If you find this difficult, then you need to look elsewhere for a better explanation. Best wishes, Mike
Hi, Table 5.5 is from the "Manual for the design of concrete building structures to Eurocode 2" by the IStructE. If your company or university subscribes to CIS then you may be able to find it there. Don't worry if you cannot find the book, as there is another way to find z using this formula: z = d [ 0.5 + SQRT( 0.25 - k/1.134) ] SQRT just means the square root of everything in the (brackets). Hope this helps.
Hi The Sage, w l squared over 8 applies to a simply supported beam or slab supporting a udl over its full length. It is possibly the only equation that all practicing engineers can remember. It is worth going back to your structural analysis notes and revising this. I hope that this helps, Mike
Hi Arjonnel Lim, Thanks for the message. Yes different text books use different figures. This is fine. Please remember that civil engineering is a very approximate discipline, after all we use load factors that increase our best estimate of loads by up to 50%. If you were one of my students I would not mind which figure you used.
Hi, this is the method given by the Institution of Structural Engineers from the UK in accordance with the Euro-codes from Europe as a whole. I hope that that helps, Mike
Great video Mike, very clear. Love the use of the table from IStructE.
Great tutorial. You are much more helpful and clear than my college lecturer. Now I understand everything. I cannot thank you enough. Thank you so much.
Absolutely Fantastic Mike! Thank you a great deal! You've made things so clear, so simply!
Thx a lot Mr. Mike , I’ve learnt a lot from ur channel, pls don’t stop making these very useful videos . Specially because there’s is no one on RUclips who explains the manual design to Ec2 .
Thanks a lot Mike. Your reply clarifies my query. I hope you will have time to provide more tutorials because they are simply great. You are doing a fantastic job :-)
This literally simple and easy way for the design of reinforcement, thank you so much and sir you explained it very well
thank you from an engineer to an engineer. i wish i could draw lines as straight as you!
Omg you're a life saver! My newly hired lecturer didn't even teach this! He just expects us to know how to calculate by only teaching theories! Thank you very much sir! Salute!
I have these lecturers now 😢
@@naveedladha4211 word of advice, quit civil while you can
@@spidermancrawlingtheweb why? I can see it's been 9 years since you commented, you didn't continue civil as a career?
@@spidermancrawlingtheweb like i would need more explanation on that pls
@@spidermancrawlingtheweb 🤣🤣🤣🤣
A very clear presentation...well done!! Keep making these videos...they are excellent.
Merci, beaucoup Mr pour ton disponibilité de nous faire cet exemple de calcul.......
Thank you so much, Mike. very good video, and much more easier to follow than my uni lecturer.
well I haven't watched this video; just downloading it. But from comments by others I think this video is a good one. Plz Tutor continue witht the good work. Thanks for that.
Thankyou very much
Too many incompetant structural engineers these days who act as if what you has shown is some form of black art, which it certainly is not. Your tutorial is great
Thank you so much Mike. I love your videos i wish you were our University lecturer
Excellent video - you explain everything exceptionally. This sums up a month of lectures in my RC design course. Thank you so much, all the best from Singapore!
This is wonderful.. Pls make more videos on Ec2
Great Tutorial Mike..Do you have other similar tutorials?? Pls share link
Thank you for making these videos, much appreciated. Just one question, the tributary area you used had a 5m width. Due to continuity of the slab, should the width not have been 1.25 x 5m = 6.25m? The 1.25 factor is used for a continuous 2 span system?
Hi Name, Last Name, thanks for your comment. Yes, you are correct. If the two span slab is continuous over the central beam support, the reaction should be greater there due to elastic shear. I have squeezed in an awful lot into this rather short video (which I was trying to keep under 15 minutes) and did not feel that I had time to explain this. I hope that I have not misled anyone by taking such a simple approach to loading, Mike
@@mikebather Understood, the main focus of your video was how to design a RC beam, mentioning continuity or non continuity of the slab may have added unneccesary confusion. Loading would need to be covered in a separate video I guess. Thanks Mike, the time and knowledge you share is much appreciated.
Great tutorial thank you very much. I have one doubt. In our Engineering studies we are not studying about the concealed beam. So i think if i put any concealed beam inside a slab,its also taking any load or not.
Hi Jeyaganesh, Thanks for your comment. This example is rather simplified to focus on the design steps for reinforced concrete.In reality the beam in the example would act as a T beam. By 'concealed beam' I am guessing that this is a heavily reinforced section within a slab. This section could act in a very similar way to a normal beam. Because it is stiffer (due to the heavier reinforcement) than the adjacent slabs, it will attract load and begin to function in a similar way to a beam. I hope that this helps.
If the beams are continuous the inner supports would experience hogging, so the tension reinforcement at the support would be in upper side of the beam (or possibly in the slab since it is monolithic), in that case how you would calculate the required tension reinforcement please?
Can we provide 25mm for the hanger bars, I mean the main bars are 32 dia and can we reduce the dia of the hanger bars as it is just a support, the self weight will be reduced? Thank you sir
Thanks For the video
very easy and simple solutions
Good video! But you should perhaps perfect the end by doing further calcs, showing that your obtained tensile reinforcement area is within the "acceptable" range? I am referering to As(min) = 0.26 Fctm/Fyk * bd ; As (max) = 0.04Ac, etc etc...
ur a life saver
Hello Mike,
Thanks for the tutorial. Could you please elaborate on the k value?
I see that when calculating the ultimate moment capacity it is good practice to take 0.168 as this will ensure ductile failure. Is this because you are ensuring the steel yields prior to concrete crushing? If so it is strange to me as the Mu = 0.168*fcu*b*d2 has no mention of the bar size or steel strength.
It seems that as a larger k value will give a shorter lever arm and therefore a smaller moment capacity. I am unsure what this value represents?
Please help!
Homer
Hi Homer, Thanks for the comment and I can comment a little on the k value. It would be better with some diagrams to help really. The balanced section of the RC beam is the case when the steel reinforcement will yield in a ductile way just before the concrete fails in compression in a brittle way and the two materials reach their ultimate strains at the same time. As more and more tension steel is used in a beam, the depth of the compression block in the concrete gets deeper and deeper (as does the depth to the neutral axis x). It has been shown by experiment that the steel will not reach its yield point when x/d is greater than around 0.5. EC2 limits x/d to 0.45 to ensure ductile and not brittle failures.
Now,the ultimate moment capacity of the RC beam is the force in the steel x the lever arm distance to the centre of the compression block in the concrete (i.e. Mu = Fs x z). This is exactly the same as the force in the concrete compressive stress block x the lever arm distance to the centre of the steel reinforcement bars (i.e. Mu = Fc x z). This is because the horizontal tension force in the steel is always the same as the compressive force in the concrete.
Knowing this firstly, means that we can define the bending strength of an RC beam (even without reference to the amount of steel there) and secondly, allows us to make an equation for Mu (the ultimate moment capacity of the beam) including the size of the compressive stress block of the concrete, the strength of the concrete and the lever arm distance to the steel. When this equation is tidied up a little, it equates to what you have written in your comment: Mu = 0.168 x fcu x b x d x d.
I hope that this helps.
Hi,
Do you happen to know the equation for the line on the graph which is used to find z/d or where to find it?
Lesson was really useful by the way thanks alot
hi mike, your tutorial is great & elaborate.kindly clarify my one query that how we can select the dimension of beam i.e LxBxD and you chose half of one part of slab i.e 2.5 m while calculating live load which transfers a load on the beam.Kindly revert on the same.
Thanks
Sir, i have a question here. for the selfweight of the slab, in the previous video, you calculated in per metre^2 ( as in 1x1xthicknessof slab). however in this video you used the area of the slab in the selfweight formula ( bxthickness x length). i am kind of confused here sir. which one to use Sir?
Hi, is the previous video on slab design? If so, in the slab design video, I calculate the weight of one square metre of slab say W kN/sqm. then I consider a one metre wide slab spanning a certain distance. The UDL (self weight of the slab) becomes W kN/m for the slab as it is only one metre wide.
However for the beam in this video, the beam supports some slab on both sides. It so happens that each one metre length of beam supports around 5.0m width of slab in total. So, for the beam, its UDL (from the self weight of the slab) is 5W.
Not sure if this is what you asking about. Hope it helps.
Great job!
More videos please
Great tutorials :) One small question, why are you using fck=32Kn/m2 instead of 30? Thanks.
Hi Omar Caruna, There is no particular reason why I chose C32/40 as the strength class for my concrete. Perhaps this is a little stronger than a designer may normally use in a situation like this. Normally, the structural engineer chooses the strength class of the concrete right at the start of a job after considering things like exposure and durability. I hope that this helps.
Great tutorial sir. May I ask about the 2 bars in the compression area. How do we decide on the size and number?
Hi Arman Chua, the purpose of the two bars in the compression area in this example is simply to support the links and to help form a reinforcement cage that is reasonably stable for when the concrete is poured into the mould. So their size is a practical design issue, not really an analysis one. Hope this helps.
Mike Bather Thank you very much for your response. I figured that as well. Just wanted to reconfirm. Another thing...what if we do not decide on the size of the bars in the start and instead, work out the As,req before determining the size and number of the rebars? I was taught a different way to your example. Please advice.
Arman Chua Hi Arman Chua, using the method of design in the example, the size of the main bars is estimated at the start, and it is possible that, at the end of the calculation, you may prefer to use a different size of main bar. In this case, you need to go back and amend your calculations. Not very efficient. However, an experienced designer will normally predict bar size correctly. Also, using spreadsheets from The Concrete Centre (www.concretecentre.com/online_services/design_tools_and_software/rc_spreadsheets_v4b.aspx) or other places makes revising calculations easy.
wow! awesome tutorial! keep them coming for concrete design :)
Hi Mike. Yours tutorials are excellent. Do you have one covering Two way spanning slabs on downstand beams. I have to calculate the edge beams and internal sizes, steel arrangement. Check for shear and bending. i would be grateful of any help.
The bays are 7.5m wide (2 bay) and 6.75m long 7 bay)
thank you soooo much just wondering does it matter what size of reinforcement i put in the compression zone
Hi Gaelle Mabassa, for a singly reinforced RC element, the size of the hanger bars does not matter. If you changed the design to be doubly reinforced, i.e. using the rebar in the compression zone to carry some of the internal compression forces, then, yes, the size of the rebar matters - as this affects the lever arm between it and the tension steel. I hope that this helps, Mike
what about the spacing for the main bar? If it is 2 main bar, what is the spacing for crack control?
Well explaining thanks alot .. but would i know what is the purpose of Mu ?
Hi Hasabo, Mu is the ultimate bending moment that can be carried by the beam without compression reinforcement and while remaining ductile. Simply, it is the maximum moment that the beam can safely carry.
Very good videos! Thank you!
Great tutorial thank you very much
Very very helpful! please please please could you upload a worked example of shear reinforcement and deflection asap as I have an assignment due in next week but have no idea where to start! thanks!
Hi Ethan, Thanks for the comments. I am in the process of filming something on shear design and will upload asap - probably not in time to help you with your assignment - sorry. The IStructE Manual for the design on reinforced concrete buildings is pretty easy to follow, otherwise the Concrete Centre also have some good free guidance - even easier to follow.
refreshed me. thank you
Brilliant Video!!!
Thanks Mike
thank you very much. Really helpful.
wow it is so amaizing tutorial keep it on
it is nice and easy to follow. i remember pt
Hi Mike, fantastic video on this. I just have a query for the z value. Is it not z= d (0.5*sq root 0.25 - k / 1.134
Hi Nicholas, there are a few different ways of finding z (the lever arm) after you have calculated the factor k. I like to use a chart as it shows an alternative to using equations. Mosley, Bungey and Hulse use z = d[0.5 + sqrt(0.25-k/1.134)] and the Concrete Centre use z = 0.5d[1+sqrt(1-3.53k)]. The various formulae are found from solving a rather clunky quadratic equation and then presenting the result in a convenient way. So there is more than one way to skin a cat, Mike
Mr Mike Bather
Thanks for the RCC video
I want to know do have such videos related to Steel design of EC3
Hello Mike, thanks for this video, I wonder if you could have similar video on the topic of deep beams or may be you know where I can find it in eurocode or in others regulatory documentation? I would like to find step by step instructions, can you help me somehow?
Hi Настасья Пономарёва, I am sorry to say that I am no help to you on this. If I were in your shoes, I would contact the Concrete Centre. They provide superb technical advice for structural engineers in the UK, Mike
Thank mate. This is great.
Hi dear Sir, thank you so much for your useful videos. I have a question if you have time plz answer it. What if both slabs on both side of beam have different spans e.g. first 5m and second 4 meter?
Hi 786Hosai,
If the slabs on either side of the beam have different spans then your design is very similar. Assume that half of each slab is loading your beam (this may not be the case in reality but should be OK for most student exercises). So if one slab is 5m (take a load width of 2.5m for this) and the other slab is 4m (take a load width of 2.0m for this). I hope that this helps,
Mike
+Mike Bather
Thank you sir so much for your reply. Your lectures are amazing. You are so generous :)
+Mike Bather
Thank you sir. I summed half of both slabs and it worked :)
Hi 786Hosai, Apologies for the delayed response. Approximately half of the load of each slab would be supported by the beam. So the total load width would be 2.5m plus 2m, making 4.5m loading the beam. I hope that this helps.
where can I find the z/d vs K factor graph?
Dear Mike,
I cannot find next video for top reinforcements of this beam. Where can I see it, please?
Hi Hla Soe Naing, Design of compression steel is not needed in this example. The bars in the top of a beam are called hanger bars and their main purpose is to stiffen up the reinforcement cage to prevent its distortion when pouring and vibrating the concrete. Mike
I see, Mike... Thanks a lot for your reply.
Hi Hla Soe Naing, I have not done a video for the top reinforcement of an RC beam yet, Mike
I will wait for next video to learn thr shear and deflection checks
sir is not this beam become over reinforced section, which we always should avoid
Hi abhisek sukla, I am not sure why the beam would be considered to be over-reinforced. We have checked that Mu is greater than the applied moment and in doing this, we have made sure that at ultimate limit state, the gradual ductile failure of the steel will occur rather than the brittle failure of the concrete. Mike
sir, please what of the load transfer from the slab to the beam
Hi uwalakapaul7970, at around 1.35 there is an explanation of how the slab loads the beam (both VA and PA). Sorry but this is only very brief as I have tried to keep the video as short as possible. If you find this difficult, then you need to look elsewhere for a better explanation. Best wishes, Mike
from where we can get the table 5.5 for finding value of with z/d value????
Hi, Table 5.5 is from the "Manual for the design of concrete building structures to Eurocode 2" by the IStructE. If your company or university subscribes to CIS then you may be able to find it there. Don't worry if you cannot find the book, as there is another way to find z using this formula:
z = d [ 0.5 + SQRT( 0.25 - k/1.134) ]
SQRT just means the square root of everything in the (brackets). Hope this helps.
Actually sir euro code is a part of my thesis.. And yeah this is very helpful... Thanks a lot ...
Q: doesnt wl^2/8 apply to one support fixed and the other pinned ?
Hi The Sage, w l squared over 8 applies to a simply supported beam or slab supporting a udl over its full length. It is possibly the only equation that all practicing engineers can remember. It is worth going back to your structural analysis notes and revising this. I hope that this helps, Mike
Can you explain why Mu = 0.168fckbd2 not 0.167fckbd2? It's because I've been taught that it is 0.167fckbd2
Hi Arjonnel Lim, Thanks for the message. Yes different text books use different figures. This is fine. Please remember that civil engineering is a very approximate discipline, after all we use load factors that increase our best estimate of loads by up to 50%. If you were one of my students I would not mind which figure you used.
Where is your new vedios
this method you are using is British system or indian system Eng
Hi, this is the method given by the Institution of Structural Engineers from the UK in accordance with the Euro-codes from Europe as a whole. I hope that that helps, Mike
ok eng can you give me pdf book to refer in this course if you have my address is ciidyoonis@gmil.com
alas he is my goat
can i say it again? very well explained....thanks Sir.
k means na constant of balanced section totaly depends on grade of steel fe500=.253
thanks man
I love you man
Artistic engineer. Yea you easily designed.
thx
Design sapn1 -5m and span2 -3m
Hi amazing Engineering
I'd like to make a tutorial os DESIGN OF FLAT SLAB.
I wish all the best for you.
MIGUEL FEIJÓ
ANGOLA
I have a question which am struggling to solve is it possible that you send me your email and you do solve it step by step please