hi morning everyone welcome again to and you read you about reinforced concrete design and I am dr. sherry fisherman and today I'm going to give you a video about this line or reinforce it concrete beams this will include part one and will be introduction about the design of beams and all requirements and chicks that you need to follow to design reinforced concrete beams first of all let's talk about different types of beams that you may see you may see simply supported beams when you have only one span it could be like simply supported like this one or sometimes it would be also fix it from ends and this is easy to find the bending moment and design for reinforcement and this type so the structure analysis of the same way support is easier one then you may also see a continuous beam of two three or more expands as you can see in this case usually this type of structure is indeterminate structures and the structure analysis is somehow more difficult than the previous case when you have a simply supported but usually different codes are giving some factors that you can use to help you to find bending moments feel horses and then you can use them to make your design of course you also can use any computer program to make the structure analysis for you the third type it could be I can cleaver you may have a cantilever beam when it is this is a free end here and also this one is similar to the first one which is simply supported both of them are determinate and would be easy to find the bending moment which is W in the square over 2 or PL and you can find also this year and you can make you design in the cantilever here don't forgive that the reinforcement should be of the main reinforcement should be at the top because the tension will be at the top and the simply supported the main reinforcement will be at the bottom for the continuous you have some reinforcement allows us of course will be top reinforcement and in the span it will be bottom reinforcement also you may classify beams into shallow beams and deep in shadow beams when you have the span to depth ratio is greater than 2.5 this is the span divided by the debt ratio is greater than 2.5 in this case we call it shallow beans and the design is based mainly on bending moments another type also you can see which is called the beams in this type the span over this ratio is less than or equal to 0.5 so you have big dips here compared to the span in this case the design is based mainly on shear forces also you may see different types of beam like we call drop beams when you have the beam is here and supporting the slab which is the slab in this case is above the beam okay so the slab is supported on the beam here we have the web is under the slab and this is the famous pipes that you may see it in my most cases also you may see something called the inverted beams what is the inverted payment it mean when you have the slab is at the bottom of the beam still the beam is supporting the stuff but the slab in this case is at the bottom of the beam you may see this by which we call it inverted beam here at roof plan when you have rules and you want the roof to be smooth from the bottom so sometimes you invert the beam and make it at the top so we call this type inverted beam also a third private call hidden beams hidden beams you may use a beam with the same thickness of the slab you have a slab thickness here and you have a beam with the same thickness or like little bit bigger sickness so in this case we call it hidden beam because you cannot see once you makes a concrete casting you will not be able to see usually this type of hidden beams could be wide beam and requires a lot of reinforcement because of the small thickness sector so you have drop beams inverted beams and also hidden beams okay let's see what is the difference between rectangular rectangular and the flanged sections okay when can I design a six to be a rectangular section and when I should be signed the section to the flanged section let's see here an example of a simply supported beam aborted in two columns as you can see here so under loads you would see that at the top you will have a completion force at the bottom you will have a tension force so if we took a section here at this point okay so this is showing the cross section this is we call it the flange or the slab and this is hole the beam the beam it has this part to pull the rip of the beam and also this part in the slab is a part of the beam so this is the beam and the flange in this section you will see that there is a neutral axis here above the neutral axis you will have a compression force under the neutral axis you will have a tension force as you can you know most of the concrete design codes says that concrete will not resist any tensile forces once you have a tension force on the concrete they assume that will be cracked and will not carry anything side forces so all the party here under the neutral axis okay we assume that it is already cracked so once it is cracked only the steering enforcement will be carrying is a tension force and this part above the neutral axis where you have a compression also will be carried by the concrete so in this is the flange is carrying compression holes the flange is carrying compression holes so we have to consider the flange in this case and will be designed in this case as a flanged section when you have a compression on the flange or on the slab this is you will consider it and will be designed as a flanged section so as a conclusion for that if the flange is in compression so we design this section as a flanged section okay on the other case if you have a beam or this is part of the beam supported in a column here and extend we just took only part above the column in this case under load you will have the opposite you will have tensile stresses at the top and you have compressive stresses at the bottom so if we took a section at this point here so this is a neutral axis but opposite to the previous one you will have a tension force at the top and compression at the bottom and as we said here any concrete under tension will be assumed as crack and will be neglected so in this case anything above the neutral axis here it will be removed only I will have the steel reinforced means that will carry the tension force so in this case you can see that all the flange here is not carrying any load because it is under tension and we assume that the concrete will not resist tension so only the part here under the neutral axis which is a rectangular part as you can see here is taking the load the flange is not taking anything in this case so we cannot consider the flange so in this case when we design we design this friction to be as a rectangular section because the flange is already cracked so as a conclusion also it is a flange is the intention we design as a rectangular section okay so in this case the flange is the intention so we design this section as a rectangular section however in the simply supported here as dispense the flange is in compression so we design it as a flanged section let's see here if you have a continuous beam so using the same concept in the previous slide we can see that okay in this beam all these spans at the spans okay this tension will be down and the top will be under compression so all of this will be designed as a flange flanged section okay however add the supports or above the supports here you have a tensile force in the flange so you will neglect the flange and will be designed as thank love fiction just be careful if you have an inverted beam because if you have an inverted beam the slab will be at the bottom and in this case it will be opposite to what we have here okay but in most cases you have the slab is a regular slab the slab is above the beam and in this case above the support will be designed as a rectangular section and the spans will be designed as flange it section okay so it will come now the question okay if I want to design a section as a flanged section what will be the width of the flange okay what will be this B flange for the section we have B web this is hole is a beam here the web and the top here we call it B flange how much it will be this D flange according to the bridge standard 40 beams okay when you have transplanted orange from both sides the B flange equals b1 okay plus something called LZ over five we will see what is Elizabeth in a fuse in few seconds in a kiss if you have a flange section but it is we call the L beam when you have a flange only from one side so the B flange in this case equals B web Plus a lizard over 10 so though the difference between these t beams and L beams here it is a lizard over five and here it is LZ over ten so in case of t beam you have B flange is greater than the B flange in the case of le beam so what is this Elizab l is it is the distance between points of zero moments if you have a beam let's say simply supported beam so the bending moment L is it will be equal to the L because this all positive woman from the support to the support so they elicit in this case will be equals to the spans L in case of continuous beams the L is it will be less because you have part is under negative moment part is positive moment so the distance between the zero moments will be less so in if you have a section in continuous beam a lizard will be taken as point 7 L if you have assembly supported beams L is it equals to L just you put the L is it here or this in this equation you will be able to find the B flash okay so how to transfer loads from slabs to bins if you want to design a beam you need to find the load in this beam and then you make some structural analysis to finds a bending moment and shear forces then you go for the design steps so the first thing is to find the loads okay you cannot find the bending moment and shear forces without findings at all so to find the loads usually beams are supporting different types of loads loads from the slabs loads from walls self-weight of the beam itself okay so first thing is that note from slabs okay we have two different types of slabs as we know one-way slabs when the l1 the long side divided by the short side is greater than 2 in this case it is easy to transfer the load because for the one-way we assume that the load will be transferred only in the short direction so how to do that we take make a line here at the middle of the short direction and then this part of the load about this line here will go to the long beam here this part here is a bottom part will go to Salonga team in this case the short beams are not supporting any road from the slab okay I have to do that we take this area here and this is the beam that we are designing now this is the beam okay you have a support here a support here and this will be the area carried on this pin area one here it will be the same areas that you have it here so how much is the height here is a height it will be equal to the LX over to the shorter span divided by two because we always divides the short span okay so if you want to find the load here okay it is easy if you have a load on the beam N or W equals any value kilonewton per meter square this is the load on the slab total load so how to find the load on the beam kilonewton per meter so it equals this lens this height here multiplied by n so the W equals M kilo Newton per meter square finds lengths per meter so it will be only kilonewton per meter so this will give you the load once you have the road here kilonewton per meter you will be able to get the concentrated load if you want to get the resultant you can get the reaction you can process yet entered in the moment in this case so in one way is elapsed is easy all the load will be going to the long beams because the loads goes in the short span of the slab so it will go to the long beam ears along with him here this beam and this beam will not take any load from Destler how about two-way slabs into his lapses you have the NY over L X is less than or equals to so how to transfer the load in this case first of all you have to make inclined lines at 45 degrees from all the corners and then they will need to intersect at one point then you connect these two points you can see here that you will have a part will be like Travis oil the loads and you have triangle loads as you can see for the Tommy say the load this part at the bottom will go to the bottom beam the bottom long beam here the top Robbie so it will go to the top long beam as you can see and then you have a triangle part will go to the short beam how much is this height here it is always LX over to the same here because this is 45 degrees the angle so this distance equal to this distance so the height here it is LX over to the height here it legs over to how much is this distance between the this point and this point it equals l y- l X minus L X so it is L y -2 and X in this case so pull the bottom beam here or the top beam the load will be this area one will come and will be supported in the bottom beam and as I told you the height will be an x over two to give the load here as kilonewton per meter so again it will be n times l x over two to something kilo Newton per meter once you have this one you will be able to get the bending moment reactions and Cheer horses okay then pours a short beam it will take always a triangular load as you can see and this could be area tool and the road again it will be the same load here because both of them they have the same height in the x over 2 and le x over 2 this is showing to you how to transfer the load in case of one-way slab to the beam and for two-way slabs called one-way slabs against the road will be transferred only to the long beam short beams are not taking any route from the slaw it may take a load from a wall the self weight is still there but load from the slab it is neglected and all the load will be transferred to the long beam in cases in case of you if you have two-way slabs you know that all beams will carry loads because the loads in the slabs will go in both directions and the short direction of the slab and in the long direction of the slideshow all beams will take loads the long beam will take attributes Odin load and the short beam will take a triangular load as you can see if you have here a square slab everything will be triangles and in this case you'll not see gravity that note okay in continuous beam say bridge standard is giving a table table 3.5 that can be used to design for 0 to find the ultimate bending moment and shear forces but under some conditions okay we cannot use table 3.5 unless we satisfy the following three conditions first the live load on the beam should be less than or equal to the dead load live load should be less than or equal to the dead load the second point loads should be uniformly distributed over three spans or not so we have to have uniformly distributed load and the span should be three spans or more you cannot use this world table 3.5 when you have only two strands okay because the minimum it really spans all more okay you cannot use table 3.5 if you have a concentrated load on the beam equals saying uniformly distributed load also the live load should be less than or equal to the dead load what will be the third condition variations in the span length shouldn't exceed 15% of the longest span so the span should be somehow close to each other five meters 5.5 4.9 okay is the difference between the span should be less than or equal to 15% of the longest span because if you have big difference this table 3.5 is not going to give you accurate results and in this case your design may be wrong and is not satisfying the requirement so in this case if to use table 3.5 you should satisfy the three conditions what is this table 3.5 let's see it together it is similar to the table that we saw when we were designing continuous one-way slabs but with a little bit difference in the values so you have the first row here is called the bending moment the bottom hole is for the shear and you can see here adds outer supports as soon as 0 equals assume the right Allison be supported at the end or the connection between the beam to the column will be as a simply supported needs a middle of end span it is post positive moment point 0 9 F capital times L at first the interior supported will be negative minus point 11 F capital L middle of the interior span will be point seven point zero 7 FL and then at material spend again it will be negative and this is showing that shear forces of course the shear will be at the supports that shear adds that spans or near middle of the spans will be always minimum so it is similar to the table 312 in the bridge standard that we use to analyze simply supported one-way slabs and also similar conditions were bleak about there so what is this capital F here capital F is the total design ultimate load ok the resultant of the rope not the uniform load it equal 1.4 G capital K plus 1 point 6 Q capital K capital and capital here here means there is all controls alone not the uniform so if you have a uniform note we should multiplies this uniform load by the span to get it as a resultant force is a capital F okay and of course is the effective span and as we did for the slabs if you have at philistine period support if you have a span on the left and the spans all right it is not similar so you have to take the average you hear from both values so it will be minus buoyant 11 minus F 2 1 L 1 plus F 2 a little divided by 2 you get the average from both sides okay let's see it here more clear if you have a span in the one span in the tools when a literary or whatever so you have to Harris to get the resultant of the road as F capital one or the priestess pan F P capital 2 in the second span and then you will be able to apply table three point five so for the bending moment at the beginning assume that is zero he will have a positive moment was the point zero nine f1 l1 then add the second span again it will be able to hear point zero seven f2 a little at first the Imperial support it will be minus point 11 and you get the average from both sides f1 l1 plus two a little all divided by two and then you continue if this continuous you continue as a second here if this is continuous here it would be the interior support and so on then for the shear also in use a table 3.5 the bottom row at this point four five f1 naught multiplied by L in this case minus 0.06 f1 then 0.55 f2 and so on if you have an n disband here like this one again you will use the same values that you used at the first expand because there is this platform lift or press the span from right they are the same this point called 5 F point 6 F assuming that we have here f1 is similar to the exponents that we have it here okay so this is showing how to apply the 3.5 here and the British standard to find the bending moment and shear horses in continuous beam satisfying the four d3 conditions okay so now we will be we are ready to go to the design steps of being this in Section three point four in the BS eight one one zero first thing we have to make initiate proportion english' proportioning it means i need to find the dimensions okay what will be the total height of the beam what will be the thickness of the beam okay we need to make initial proportion because everything in the design will be depending on the depth there be web that we have it and also it will affect the self weight of the beam okay so we have to estimate the effective depth okay then again we use table three point nine and page 215 and the BS but in this is we use here usually we are designing flanged beams so usually we are using this one be web over b is less than or equal to point two three so in this case if you have a simply supported this will be the factor sixteen continuous this will be the factor but in this case we don't divide by modification factor we take the value here so to find the depth basic depth or the minimum depth it will equal to the stand divided by this fact okay it will give you an initial value but now rounded up always rounded up to be in the safe five so the same table here but don't use any modification factor that we used in the case of slabs okay so this is the first step to get the d okay then we have to estimate cover okay we have to estimate the cover to the steel it will be the larger from two values durability table three three okay this is exactly similar to what we did in case of slabs this table three three it will give you the nominal cover two or reinforcement including links to meet the ordinary requirements and it depends on the exposure condition mild exposure moderate severe severe exposure very severe and and so on and also to be expected by the complete compressive strength so if you have let's say a severe environment you can see that in this case we cannot use concrete less than here - here it means you cannot use complete sir and concrete 35 so in a case of severe environment you have to use a better quality concrete to be c40 and from here you can find that this will be the cover let's say you have a mild exposure and in this case you can use if you have a concrete 30 mega Pascal the cover will be 25 if you are using concrete 35 of course the cover will be less and will be 20 in this case so the using this table is exactly similar to what we did in case of slabs and if you want more you can go and check my videos about one-way slabs and two-way slabs this is for durability now the second trick for the cover it will be about fire resistance we have table three point four this is the same table here and at the first column you have set fire resistance per hour half an hour one hour one and a half two or three and four hours and here it will give you the cover that you need based on which structure element you are designing in cloves you have to use a one with float simply supported or continuous in a key so we are designing a beam as we are doing in this video so we have to use the values here in a case of simply supported beam it will be these values in a case of continuous beams it will be these values we can see here the difference is almost no different that from zero point half half an hour to 1.5 hours after that you can see there is a difference between an assembly supported and a continuous so once you calculated the cover from durability is a cover from the fire resistance you will take the larger from Basel 10 this will be the cover that you are going to use then to find the total edge h equals d the effective depth that you calculated from table 3 point 9 plus cover the larger from these two values plus 1 plus Phi bar divided by 2 half diameter plus Phi link in the beams we have an additional link which will where was not there when we were designing for slabs so for beams we increase here Phi of the link so as a guide poor the high of the link assume it as ten millimeter sometimes 20 millimeter it's okay it's up to you I own a ten millimeter and for the the emitter of the bar in this case we assume it 20 millimeter because usually the size of the partisans the beam is greater than the size of the bar in the slack since the slabs we use two assumes a high bar ten millimeter however for the beams here just assume a 22 all substitute these values and to here you can get the inch okay usually this edge from table here it depends on D and this is like little bit smaller value so always round edge up okay rounded up to the nearest 50 millimeter okay and keep in mind that the edge minimum would be 300 millimeter so if you have the H is less than 300 millimeter take it as 300 millimeter okay if this is greater than 300 millimeter it is fine but always rounded to the nearest 50 millimeter so if you have 300 the second one will be 350 then 400 450 and so on so always it will be better for the workers in the side to go with 50 millimeter nearest 50 millimeter 5 centimeter higher for the beam so the edge will be 300 this will be the minimum then you will go 350 400 450 500 and so on okay so this will be the edge now you need to find though it's how much it will be the width ropes a beam normally the widths will be ranging between H over 3 to H over 2 K from H over 3 to H over 2 and round it to the nearest 25 millimeter ok we're on to the nearest 25 millimeter also you may take an another edge which which is not similar to this because another be sorry because this width of the beam is usually is you have a wall under the beam so sometimes we keep alright it is preferable to use to keep the same width of the beam as the same weight so loser wall under the beam okay so if you have a wall of 25 millimeter width okay this is the thickness of the wall so it is better use the same thickness of the beam okay don't use a smaller thickness or a smaller be a smaller B of the beam than the world under the beam but we may use bigger B than the world if you have a wall of 10 centimeter only okay 100 millimeter okay in this case you may use a 20 millimeter width beam or 25 to be able to put your reinforcement because making a beam within 7 centimeters will be very small sometimes will be difficult to put your reinforcement inside okay so don't use a beam with a be less than the wall under the beam but you can use it bigger than the wall thickness okay and always round the to near is 25 Mille meter once you calculate the edge okay and because we round it so we have to recalculate the D again the D it will be using the same equation here you can use it if the different other side could be equal to H - cover pass filing plus Phi bar over 2 by doing that so you finish the initial proportioning you know their edge and also you know the beam so you are ready now to go to the second step about final proportioning and always final proportion is starting by calculating nodes so we start by the grid loads in this case did loads we have different types of loads usually we have it on the beams the first thing is the self weight of the beam okay the self weight of the beam self weight of the beam equal H minus HF times B find the gamma times L okay what are these values it will be the cross-section of the beam okay we need to get the volume multiplied by Dynasty okay by the total length of the beam to get it as a kilonewton so the cross-section which would be H minus HF because this HF is a segment of the slab it is already considered in the slab no need to repeat it okay so H is the total height of the beam minus H of the flange okay x ZB multiplied by gamma this will give you this part here will give you per meter if you want to get it as for the total length or is it even want to drive by the lens it will give you as kilonewton okay this will be the self weight of the beam then we have did lowered from the slab did love the load from the slab equals did load multiplied by the area supported okay if you have an area of the slab supported unibeam so you'll have this area it is meter square multiplied by the load kilonewton per meter square it will give you as a load in kilonewton this will visited load from the slab if you have wounds on the slab if you have walls not partitions like real world on the beam supported on the beam so again you need to consider that because usually it will be like easy walls to do that it will be similar to the self-weight but in this case we get the self weight of the wall it will be the width of the world finds height of the wall times gamma finding the span again it will be kilonewton so it is repeated of this one okay the height of the wall it is similar to the height of the beam h minus HF which solves the word it is the similar to the air B and the case of a self weight of the beam it is B multiplied by gamma multiplied by L but in this case the gamma of the wall should be taken into consideration and doneski observe what it depends on the five of the words that you have you have Allah blocks if you have solid blocks what is the type of the material that you are using light weight or heavy weight and so on so based on top of the wall you have to know how much it could be the gamma of the wall however for the concrete we use the game of course of concrete which is usually 24 kilo Newton per meter cube so if you calculated the self weight of the beam kilo Newton dead load from the slab in kilo Newton world load as in kilo Newton to get the total debt load you get the submission of the three values here self weight slab load and more load it will be as kilo Newton okay now you need to add the live load live load from only live load that comes from the slab so it will be the live load multiplied by the area supported area or slab supported one is a beam it is similar to they'd load but in this case we instead of using that it load we use a wide load okay so usually live load is kilonewton per meter squared could be to killing it and three kilonewton per meter square multiplied by the area supported and in this case you will get the road as kilo Newton once you have the total dead load and the total live load in this case it will be able you will be able to get that total load ultimate load in this case will be one point four times dead load plus one point point six point five load as a kilo Newton this is still kilo Newton okay so this is the resultant of the road if you want to give the road as kilonewton per meter as a uniform load you need to divide this total load here where the span of the beam okay the end so if you divide the total load kilo Newton divided by the span of the beam it will be as kilonewton per meter as i only form the word on this beam and in this case you will be able to use this uniform load to make your design okay usually we you do this in a case of continuous beams in a case of simply supported easier to do that with the trapezoid load or the triangle load as we saw together okay once you have the loads on the beams you go to the step of the structure analysis structural analysis for continuous beams under uniformly distributed load we can use table three point five is that we saw a few minutes ago and this case you will be able to find the moment and to find the shear once you have the bending moment you will calculate for the reinforcement your as huge when section three point four or always start by calculating Z K equals M ultimate divided by f CU b square concrete the zip you calculate the is required and then you will be able to get the area of the steel you can do this manually and the number of bars and the diameter or you can use this table here table number one that help you to to get that number and diameter directly so let's say you have an area of steering mentor 900 millimeter square okay 900 millimeter square so you can choose let's say if you choose 16 millimeter okay and this is the number of bars so 1 bar it will give you 200 and 1/2 bar 3 bars and so on so 800 we said if you have 900 millimeter square so it will be between 4 & 5 in this case you use 565 bars over the mcat 16 or if you want to use diameter 20 so it will be 3 bar than 20 or - bartender 25 and so on okay for beams don't use a diameter less than 10 millimeters and in practice we don't go less than 12 millimeters for beeps okay and practice we don't go less than 12 millimeters for beams for the slabs usually we go we don't go less than 8 millimeters but also in practice it is better to use 10 millimeters and other okay so this is showing you how to get that a number of bars and then come over bars and once you did that you would know how much is they supervise it you finish the design but you still need to make some each it's okay so we have to check the collections using the largest span moment the maximum post the running from table 3 10 which is the same table that we used for checking deflections in slabs okay the same here you have to check your own dedication factor it depends on something called the FS the stress in the steel reinforcement and M over B B Square FS you can get it from this equations here 2 over 3 F field times s require divided by issue provided and beat Abby always music as one then M in this part here always M that we use it is the maximum positive moment maximum span moment means maximum positive moment b and b is b web and B is the dips of that beam so this is the same equation and the same modification factor that we use to treat deflection and beams and slaps okay exam deflection is safe then you have to design for shear okay not only check for shear and slabs we check for shear but usually you don't use steroids or shear reinforcement in solid slabs but for beams you have to use links or stirrups so we have to design for shear left and right of each support you will have different value so you have to click that and then you design based on table three point seven and table three point eight is to go and review together table three point seven and there'll be a school it tells you if the shear stress is greater than point five VC and less than VC plus point for where DC is a shear carried by shear stress carried by the country in this case you use minimum links minimum links you can get the area of the minimum links from this equation here is the minimum if you have the shear stress is greater than VC plus point four and listen the maximum shear which is the minimum of point eight square root FC u or five in this case if this between these two values you have to design for shear reinforcement and in this case you design using this equation the difference between this equation and this one here we use only point four but here we use V minus VC which is greater than point four if you have a case that the shear is greater than the V Max in this case you cannot design for shear is the only choice that you have in this case to increase that by mention of the beam you have to increase the cross section dimension can increase the B you can increase the dips in this case but you cannot design because could be unsafe even if you put a lot of shear reinforcement this is Table three point seven and also three table three point eight it tells you how to get the VC okay how much is the shear carried by the concrete we use it from this equation point seven nine times 100 is / bv d to our one over three times four hundred over v2 or 1 over or divided all by gamma M and gamma M in this case is one point two five and keep in mind that 100 is over bv x d should be taken as greater than shouldn't be taken as greater than 3 so the maximum for this value will be 3 and 400 over d should be greater than 1 if it is less than 1 you take as 1 if you have a concrete strength of greater than 25 mega Pascal all the values here or this value should not be multiplied by FCC divided by 25 to power 1 over 3 this is again repetition all what you have taken and cheer it shakes for slabs it is the same equation applied for snaps and applied for concrete from this table we get that VC from the previous table we get we design for the shear reinforcement so table 37 and table 3 8 for each span design the required shear reinforcement and when I am talking required she reinforcement means you need to know the diameter of the link and the spacing between links okay then once you did that you have check to check cracking from table 325 and section 312 11 to from table 325 it will give you the minimum percentage of reinforcement okay you have here for planchet beams when the web intention for flange it being doing the flange is intention okay and you have different values here based on your the steel that you are using usually we are using a steel or high yield steel so these are the values you have different values here it depends on which type of beam you are designing it's a flange flange it beams and the web is in tension okay it depends on the be web divided by B if this is some point for a grill greater than or equal to 0.4 so you have a value of point 18% 1.15 percent for flange it beams and the flange intention like it means like a rectangular flange it being but the flange is the transpose will be designed as a rectangular section so if t beam and le beam you have here also some veg okay let's go back how about this section section 312 11 - let's see together review also the spacing of reinforcement minimum distance between bars okay so the minimum distance between bars in a case of beams it is equals Z bar soil winds up oxides exceeds H aggregate plus 5a spacing less than the part size or equivalent part should be avoided okay what does this mean here he likes a conclusion on this part it tells you that the spacing between parts should be taking greater than or equal to H aggregate plus 5 millimeter if sagres is the aggregate size that you are using plus 5 millimeter to allow you to have a enough space for the aggregate to war between the steel bars usually we take this as a minimum of 25 millimeters H aggregate plus 5 millimeter okay also if the bond size is greater than H agric plus 5 millimeter in this case it will govern the spacing and in this case take it greater than the bar side so it is the spacing will be between the bars the minimum spacing will be the maximum of two values attract this aggregate +5 or the fie of the bar okay which one is greater you take it also in the vertical distance between the bar if you are using more than one layer also the vertical distance between bars shouldn't be less than 2 H aggregate over 3 a 2 HR it over three and usually we take it also similar to the 5i a meter or 25 millimeters okay how about the maximum spacing also we have a maximum spacing between bars the maximum spacing in a case of 460 if you assume no redistribution so the value will be 155 millimeter so we have a range of the spacing the minimum will be H aggregate plus 5 or 5 diameter which one is greater and use this is about 25 millimeters okay and the maximum is spacing between bars shouldn't exceed 155 millimeter in a case of 460 still high in steel you are using okay so between 25 and 155 millimetres then once you did that you have to make your drawing and detailing of the beam don't piddle section and the cross section at mid span at at support and we have figure 324 in the code that shows how to do this let's see here together like modification of this drawing or this drawing to make it easier for you to understand in a case of simply supported beam the code is allowing you to cut 60% of the bars and extended 50% of the bar so if you have here 100% at the middle because you have the maximum moment here so it allows you to cut 50% of the bar and what will be the distance here the distance from the center line of the column to this cut it is point zero eight L where the L is the span from centerline to centerline so this in case of simply supported beams in case of continuous beams also here it allows you to extend 30% of the world and you can cut 70% and the distance here will be from the end support in doverton or point 1 L and from the middle support will be point 15 and usually we cut fifty fifty percent fifty percent but the code is allowing you to cut 70 percent and extend 30 percent the top reinforcement here okay above the support you will have the required reinforcement because this will be a negative woman here and you have a high tensile stresses so it allows you to cut the part of the board here sixty percent here will be extended to a distance of span over four point two five of the span but from the face of the beam then if you want to cut another part of the bar or is a minimum distance here it will be point fifteen and fifteen percent of the span again from the face so for the top reinforcements the distance is always measured from the face of the beam for the bottom reinforcement the distance is always measured from the centerline of the beam how about this reinforcement here which is 20% this is hangers we call them hangers theoretically you don't need reinforcement here because you don't have a tinfoil forces at this part or this part of the beam but you need some reinforcement minimum reinforcement for shrinkage for temperature changes and also for hanging the links because you have stirrups and you know this why we call these types of bars hangers and this part is you always take in 20% of the bottom steel reinforcement take it as a minimum 20% of the bottom misty reinforcement so if you have here let's say 10 bars or a diameter you take 2 bars as minimum okay you can increase this you can increase than 20% of course because 20% is that minimum for second Cleaver also you can extend 50% will be with the Vorlon so they can cleaver another 50% for short will be shorter where this distance is the greater of L over 2 the span over 2 or 45 times bar diameter so this is showing you how to make curtailment of the steering foresman how to cut your is here reinforcement according to the bridge standard also at the end the anchorage at simply supported here at the end here you have this steel bar is under tension because hazard inside stresses at the middle of the span so how to extend what will be this distance from the center line to the end it is beyond 12 x bar diameter if you have about the inter ha here it's fine okay so this distance from the center line to the end of the bar it is 12 x bar damper so what if this distance is not enough so you have to bend the bar we have to use some hook here 90 degree hook and we bend the bar to extend and satisfy this requirement by the code also beyond the fish'll to support this distance from the face here to the end it is should be greater than D over 2 plus 12 times or them okay so you have to satisfy these two values I usually this will be enough and usually this distance is not you will not be able to satisfy this requirement so we use hook a 90-degree hook and we extended this part a little bit up to satisfy this requirement okay additional reinforcement in the beam so the good also is have some requirements about transverse reinforcement in flanged beams this transverse reinforcement it means the reinforcement above the beam this reinforcement in the slab it is telling us that this reinforcement as a minimum should be taken as point 15% of the cross section of the flange point 15% of the cross section of the flange if you did that point 15% point 15 divided by 100 times 1000 where if you assume like one meter width of the flange we take one meter because it's a slab times the edge so this will give us a value 1 point 5 hf okay if you still remember the minimum reinforcement ratio in the slabs it was point 13 percent okay here we have a special requirement above the beams the minimum reinforcement in the slab above the beam shouldn't be less than point 15 percent which is little bit greater than the minimum of the slab so you should be careful in this part and increase little bit the reinforcement to satisfy this requirement okay s should be greater than or equal to 1.5 HF or point 15 percent of the area of the slab also side reinforcement in the beam if you have a beam here with H is greater than 750 millimeter if this H is greater than 750 millimeter so the code is requiring to both something called side bars these are side parts it's not allowing to have only top reinforcement and bottom reinforcement we should have some side policy here this is only in case if H is greater than 750 millimeter so in case of H is less than 700 millimetre is not required to both the soil part but if this greater you should put sidewalks okay how much is this side Barnes okay it tells us the spacing between this side parts shouldn't be less than 250 should be less than or equal 250 me limit okay so if you are using side part the spacing s be okay should be less than or equal to 150 millimeter and we'll cover two over three of the two that I chose the beam 2 over 3 H we have to use side ones here spacing should be less than or equals 250 and the diameter of these side bars here should be greater than the square root of B times s B divided by a field a field is usually 460 B is the B web and s CB is the spacing so based on the spacing you can choose the diameter or based on the diameter you will be able to get the spacing between this side part okay by doing this we reach at the end of the first part of that presentation we will have two other videos one video about design example of simply supported beam another video which will be part three it will be a design of continuous beam this will include the examples will include six for shear chips for cracking also shear design so it will be full design examples or all details that you need thank you and see you in the coming video ok goodbye