good morning everyone this is dr. Sharif el-gamal and today we are going to have a lecture about important grid design in this lecture I'm going to talk to you about solid slabs part 1 that would be introduction to slabs and we'll concentrate more on one way solid slab supported on beams as you know reinforced concrete slabs are usually used in floors and roofs of buildings to cover big areas they carry distributed loads and primary by bending moments or there are some shear stresses differs from one pile slack to another type of slab there are different types of slabs you will see the first type is solid this lab supported on beams straight on the slick shot when this type of slaps and the solid slap supported on beams you can see that you have a slab then the load will be transferred from the slab to beams and the beams will transfer the load to the columns okay so you have a slab supported on beams and the beams will transfer the load to the columns another type of slabs that you may also see flat slabs and flat plates in flat slabs and flat plates you will have a slab supported directly on the column without beams so the difference between the flat slab and flat plates here in the flat slabs and flat plates you don't have beams loads would be transferred directly from the slab to the column and in this case the shear will be effective and will be important to here you have to take care about the ship because you will have a high shear stresses at the column area this is why sometimes they make some drop panels around the columns and sometimes we also use column heads the third type over slabs hollow block slabs and rip the slabs hollow block slabs and rip the slabs in the hollow block they use some blocks it could be lightweight blocks or any type of boom or any other material to reduce the self-weight of the slab and they arrange these blocks to have a spacing between them we call them ribs and in this case you will have a small slab thickness at the top this would be 5 centimeter to 7 centimeters usually supported on trips here and the spacing between reps depends on the size of the loop that you are using here and the Hama blocks you use some blocks or like food or any other material in the red the slabs you have only the top slab and the rich was how it looks here the last type of slab set called waffle slabs the slabs have some similarities to the red this laughs but two-way so it is similar to the rib the slab to where it the slabs and add the columns because of the high shear stresses so you have some solid part this will be able to resist the higher shear stresses act around the column area there are different classifications like the spans you may have simply supported slab or continuous this is valid for slabs or for beams or for any structure you may have simply supported or one span only or mini spans you'll recall them continuous also for the suppose sometimes could be fixed supports or pin supports and in some cases you have some steps with free end like balconies you will have a free end or sometimes you have a slab supported from two or three sides and the last side of the slab is free for the solidus slabs supported on beams which is a topic of our lecture today there are two main types one-way solid slabs and two-way solid slabs what is the difference between one-way and two-way solid slabs it depends on the geometry so let's say this is a slab the long side of the slab here we call it long span and the short will be the shortest span so if this long disband divided by the short span is greater than two we classify this as one way solid islam if the longest ban divided by the shorter span is less than or equals to even if it is equal to yes if this less than or equals to win all this two-way solid slab okay let's see more details about one-way solid slabs as we said the longest band divided by the shorter span should be greater than two in this case most of the roads will be transferred in the short direction of the slab it means it will go to the long column and in this case we will assume if we classified if the longest band divided by the shorter span greater than two we will assume that all the load would be transferred only in the short direction and no loads will be going in the long direction of the slap so in this case if you want to design this slab you will design only in the short direction of the slab and will not be designing for the long direction so it means to design for the slab you take a strip of one meter and you get the load on this one meter of course the slab here will be supported on this beam will be like a support another long beam here will be another support and then we'll design this one as a rectangular section of B equals one meter 1000 millimeter and the H equals H f4 is the h of the slab so the reinforcement that will we get in this case it will be the reinforcement in the short direction or we call it main reinforcement okay how about long direction do we need to have reinforcement here of course we need to have reinforcement for shrinkage for temperature it changes but we are not going to design for this one we just use the minimum according to the code so we designed only for the short direction and for the long direction we just take the minimum according to the code okay in that two way solid slabs the difference between longus van and the shortest ban is not that big as in the case of one-way the difference if this long is ban divided by the short span less than or equals to so in this case the road will be transferred in both directions most of the loads will be going to the or more loads will be going to the short direction and also some loads will be going to the long direction so in this case if you want to design this you have to design it in both directions so you have to take a strip in that short direction get the road get the moment and design for that one to get the reinforcement in the short direction then you have to repeat again and take a strip in the long direction of 1 meter and get that load on that one gets a moment and then you design and give the reinforcement in the lone direction so we have to design for the short reinforcement we have to design for the long reinforcement what are the design steps as you know in this course we are full wings a BS code 8110 and we are going to follow section three point five point one the design steps the first thing is we design the slab as a beam as I told you we take a strip of one meter so the B will be assumed as 1000 millimeter so the design will be as a rectangular section of high it equals that it shows the slab and B width of the section equals 1000 millimeter the first thing after you take one meter strip we start by something called initial proportioning because when you design you don't know how much is the thickness of the slab what is the H slab what is HF so we need to assume it okay how to make this initial proportioning we have to start by estimating the effective depth what is that tips how to get the depth so we get it from Table three point nine and this is page 215 and the B is 8 1 1 0 and let's go and see what is this table three point nine this table three point nine it gives you something called basic span to effective depth ratio for rectangular or flange it beams and we are going to use this also for slabs but by adding something called modification factor what is this one for the slabs as you know the section is a rectangular section so we have three different types of rectangular section or slabs it will be simply supported and the basic span to depth ratio in this case will be equal to 20 if it is a continuous slab it will be 20 if it is a cantilever it would be assumed this span to depth ratio is seven okay how to use this one okay to find that depth of the slab the minimum depth of the slab equals this van divided by this basic span-to-depth ratio from that table here multiplied by something called notification factor and as a guide you can assume the certification factor as 1.3 but there is something important here what she's been I'm going to use because for each slab you have to spend you have long expand you have short the span always the span here that will govern our design and our thickness it will be the shortest and not the long span so here it would be the shortest path as spam divided by basic span to depth ratio even 7/8 case of cantilever 20 in case of simply supported when we slab or 26 in case of continuous and then this modification factor is assumed as 1 point 2 3 just to clarify this let's see some examples if you have a simply supported slab of 3.3 times 7 meter so the dimension of the slab 3 meters short this vamp I'm 7 meter long span at it is simply supported so how much it will be the effective dips in in this case the effective depth it was the shortest man K 3000 divided by basic span-to-depth ratio this is simply supported soul and simply supported the basic span to depth ratio in the table is 20 so it will be 3000 divided by 20 times the modification factor always we'll assume it 1.3 so it gives us an effective depth of 1 1 5 point 4 millimeter another example if you have a continuous slab of 5 5 7 meters ok so again how much it will be the dips or the effective depth okay again we are going to apply this equation the ID equals the shortest ban in this case the short the span is 5,000 divided by basic span to depth ratio in this case because it is continuous so from the table it is 26 times of difficult actor always take it as 1 point 2 3 so it gives us 147 and again if you have a cantilever of 1.3 times 7 meter and the span is one point three meter so again the depth Z equals the span but you put it in millimeter to get the tip seven millimeter divided by base expand to Dipsy ratio in this case for a cantilever is seven times the beautification factor and you have a value of one hundred forty two point nine millimeter okay this is the first step to estimate the effective depth why it is important because it will help us to get the total edge total thickness because it will affect on the self weight and the dead load of the structure so once we got that depth okay again I just want to concentrate that the short span cover the effective depth okay now we need to estimate the cover to steel reinforcement the cover to the steel reinforcement it depends on the the larger of two values the first value about durability we have a table three point three page 206 to go and see this table and this table in the BS code it will give you the nominal cover to all reinforcements okay and the bins on the exposure mild exposure moderate severe and so on and also it depends on the on the concrete compressive strains that you are using starting from grade 30 it means 30 mega Pascal going to 35 and 250 mega Pascal so it means how to use this one okay if we assume that we have a moderate exposure your structure in an area of moderate condition in this case you will go take models and go horizontal you will find that you cannot use concrete strength of 30 mega Pascal in a moderate exposure so the minimum concrete according to CBS and a moderate exposure equals 35 so in this case how much it will be the cover it will be thirty five-millimeter cover as given here if you have let's say severe exposure and you have a concrete strength of 45 mega Pascal so you will go severe exposure with a 45 you will find that the cover is 30 min limit okay so based on the exposure and the concrete strains you are using you will be able to get the minimum cover according to table 3.3 and called durability because it will the cover helps to protect the steel against corrosion problems okay how about the second missile to get the cover we get it also for fire resistant because as you know the function of the cover the cover has two main functions first one to protect steel reinforcement against environmental conditions we get it from table three point three and also the second one is to protect the reinforcement against fire and we get it from table three point four as we are going to see now this table three point four it gives us cover for the reinforcement but four different fire resistance per hour so in the first column here you have the fire resistance you need your building to resist fire for one hour for one and a half or two and till four hours and here you have the cover based on the type of the structure limit you have you have four beams you have four floors four ribs and four columns so if we are talking about floor slabs you have case of simply supported and case of continuous okay so let's say you have to resist the fire for two hours and you have a simply supported slab so how much it will be the cover at this case you will go for two hours go horizontal since the slabs floors you have simply supported will be 35 millimeter if you have a fire resistance of one hour and you have a column you need to design a column to resist a fire for one hour you will go here one hour and you will go for columns you will find it is 20 millimeter with some additional conditions can be seen here you can do it yourself so for the cover we get the larger from the two values Table three point three it will give you a value they would three point four it give you and give you another value and then you get the maximum the larger of the two vendors then in this case you can give the total dips or total height of the slab equals what we call it edge the H equals the effective dip switches from the compression side and tells a centerline of the tension steel you need to add the cover to the steel that we just calculated from Table three three and pivot three point four and then you have to add the diameter divided by two half of the diameter of the bar to get the total edge of the slab the total height of the slab so is the total height which is this one is H equals D you get it from Table three point nine plus cover the maximum of three point three and three point four plus five over two and like for Phi we can assume the Phi as ten millimeter for our calculation and you don't need to recalculate this one if you design and you find a Phi which is like 12 volts in millimeter no need to go and repeat this so this is how to get the H and by getting the H or the total height we finish the first part about initial proportioning then we can move to a second the third step which is the final proportioning why not proportioning it will include different paths the first one to get the loads loading on the slab the main two loads on the slabs are dead loads and live loads did loads it will include the self weight of the slab and in this case for self weight you need to get the edge which we got it on the first the second step in the previous slide initial proportioning if you have finishes weight of partitions weight of ceiling and services all of this also will be considered as dead load in addition to the dead load we have also live load or sometimes we call it in post we take it from the hood and depends on the function of your building okay once you have the loads you have to go to the structure and let's step to find the bending moment and shear forces on the slabs after you get the bending moment and the shear force we have to design for reinforcement calculation for reinforcement we use section three point four point four four and the V is code you design for reinforcement in the short direction only if you are dealing with one way so it is loud or you will design for both directions if you are designing for two way solidus laps once we design for reinforcement this is not the end you need to make some it checks to be sure that your structure is safe and can resist the loads and can satisfy the ultimate limit state and serviceability limit state the first check here is a check for deflection and checking for deflection we use stable 3.10 as you can see here and it will come later after two slides we will talk in details about how to use this one to check our deflections so once you check your deflection and the point is safe also we have to check for shear and checking for shear it is similar to that shakes the shear shakes that we did for beams but you have to keep in mind that no shear reinforcement and slabs should be provided if the H is less than or equals 200 millimeter so if the height of the slab the thickness of the slab is less than or equals 400 millimeter you cannot actually reinforcement so what can you do if the shear is not safe the only choice in this case to increase the H to use more edge because if the H is less than 200 you cannot it is not allowed by the code to actually reinforcement so what is Table three point eight let's go and see this table it is similar to that they will use for beams the main equations that we have it here for the VC the shear karatbar is a concrete equal point seven nine times between brackets 100 a s / v bv x d - power 1 over 3 + x 400 / did - power 1 / 4 divided by gamma M gamma M which is a material safety factor for shear it is always taken as 1.25 according to CBS code it s is the area of the tension steel reinforcement BV is the width of one millimeter that we have it for slabs V is the tips of the step okay but you have to keep in mind that this value here 100 a s divided by B V times D shouldn't be taken as greater than 3 so if it is less than 3 it is fine okay you take it as if it is greater than 3 take it as 3 okay so this value here 100 s / bv x d the maximum value is 3 as you can see here from this table the maximum is 3 also for 400 over d shouldn't be taken as less than 1 and any slabs of course usually this one will be greater than 1 so don't worry about this part only you have to check about this one that shouldn't be greater than 3 and you should also keep in mind that this VC the value of BC here is calculated for concrete strength of 25 mega Pascal or less so if you have a concrete strength of 30 mega Pascal or 40 mega Pascal you have to multiply the values that you can get from this table from this equation by this factor here FC u divided by 25 to power 1 over 3 so you multiply the values that you get from this table or from this equation by Fe C u divided by 25 to power or 1 over 3 okay this is about the shear and it will be more clear when we go and solve an example together ok so we can create reinforcement shift for deflection checked for shear and it is safe we have also to check for cracking we have to be sure that the cracking will satisfy the serviceability requirements the BS code satisfies the cracking procedure by controlling the area of the steel reinforcement and the spacing between the steering forcement so it is not telling you you have to check the crack widths and you calculate the crack width some like difficult calculations though what is easier in the BS code and mini codes are using the same method now so you have section 312 5 3 it gives you the minimum percentage of reinforcement so your reinforcement ratio should be greater or equal to the minimum reinforcement ratio so this table 325 it will give you the minimum area of steel reinforcement for different cases we are going to use this for beams for slabs but for the slabs here rectangular number seed rectangular section and solidus laps this minimum should be equal to this value you have two values here point two four percent and point thirteen percent depends on which type of steel you are using if you are using high yield steel 460 mega Pascal so the minimum reinforcement ratio is point 13 percent if you are using mild steel this minimum reinforcement ratio is 0.4 percent okay so you have to control the crack widths by keeping your reinforcement greater than or equal to the minimum given by the code in table 325 how about the spacing also again you have to check about the spacing and this Scrolls three twelve eleven to seven about slabs it says in no case should the clear spacing between bars exceeds the lesser of three times effective depth or 750 millimeter so the maximum spacing allowed by the code is 3 times the effective depth so it's three times D or 703 in millimeter when you choose the spacing between your reinforcement should be sure that it will be less than 3d and also less than 750 millimeter so if you are satisfying the minimum area with steel reinforcement and the spacing between the reinforcement you will be sure that your cracks will not exceed the crack widths allowed by the code and it will you have no problem in this cracking after you finished all of this three you finished all the design but you have to show your design by drawing and detailing of the straps as you know as in an engineer the drawing is the only way to communicate between engineers so the engineer in society would have a drawing this drawing should have all details so you can use this drool to build the structure easily with all information given on that drawing so for detailing we can see here figure 325 in the code just showing where to cut the steel reinforcement to be steer important and bottom steel reinforcement maybe it is this one is not clear but it's just to show you here a second one here for the bottom steel reinforcement in case and to be steel in case of continuous slabs can and continue verse but I will give you this drawing here it makes more clear in a case of assembly supported a slab one is van here the codes it fills you at the middle of the slab I need hundred percent off very important because you know at the middle of the slab you have a maximum positive moment so you need more reinforcement but close to the edge the moment will be less so you don't need to use all the reinforcements that you have it here so what this drawing tells us it tells us that the code like asking you or pin you to you should use at least forty percent of the steel reinforcement extent from the beginning to the end and 60 is the remaining 60 percent you can cut them shorter and this will give you the lens here from this point to that point this distance here similar to this case is this point 1l about the stand divided by 10 so in a case of simply supports it tells you 40% will continue from the support to support the remaining 60% you can cut them so if you have let's say you have thin bars per meter so you can make four bars long and four bars shorter usually we take 50 50 50 percent longer 50 percent shorter in a case of continuous as you know for continuous you have some positive moment at the middle of the spans or close to the middle and add the supports above the beam you will have negative so in this case it will requires both on the steel reinforcement and double steel reinforcement for the bottom is still again forty percent will continue and you can cut 60 percent for this 60 percent from the first support it will be L over 10 and from the middle support or interior support that will be about 2.2 l or the span over five from the centerline of the support how about top steel reinforcement Dobie steel reinforced it tells you like 50 percent should extend from the edge of the beam not from the centerline as in the case of bottom reinforcement no from the edge of the beam it extends to point III span 30% of the span which is this is the span from centerline to centerline so the top reinforcement 50% at least should extend to 30% another 50% of the reinforcement should extend at least point 15 percent of the L and should be also greater than 40 times per diameter at the end support here because there are some fixation or some continuity between the slab and the last beam here so it will result in some smaller amount of bending moments here negative bending moment so you should have also some reinforcement usually this reinforcement will be taken as 50 percent of the main reinforcement if you have here let's say 10 bars per meter you have to take here at least 5 parts per meter and it should be greater than the minimum allowed by the code which is 0.7 percent of the area of the country again the distance here will be equal to C point 15 or 15 percent of the span and not less than 45 times a bar diameter here about the Anchorage or how to extend this bottom reinforcement here you just stop it add the support at the center line or you extend so is a code is saying no this one should be extended if the V is less than 0.5 VC this is about the shear the shear stress on the slab less than 50% of the shield carried by the concrete this distance should be greater of BS over 3 or 30 millimeter what is this vs is the thickness of the support the width of the beam here this is BS so you divide this by 3 and this will be the extension here from the center line it will extend to the greater of these two values okay this is about detailing of reinforcement now how to make the structural analysis of if you have a slab of continuous spans okay for continuous bands we have to use this table 312 for table 312 it tells you ultimate bending moment and shear forces in one way spanning slabs okay so you have values for shear you have also values for moments is this will be more clear by seeing this drawing okay okay what do we have here we have the first row is for the bending moment okay these values give you the bending moment values at different points of the slab so as a first out of support here in a case if you have assembly support at the beginning here so the bending moment will be equal to 0 so this will be 0 near the middle of the Indus band this is the end this fan so near the middle of the Indus bands a bending moment will be point zero eight six FL okay and we'll see what is this FL late okay I will keep it later but let's go for at the first interior support first interior support it which is this one how much is the bending moment the bending moment will be a negative moment here and the value will be minus because negative moment bowing 0 8 6 FL then middle interior span again it is positive value less than the first a span and the value will be point 0 63 FL okay how about the case if you have the end support here have some continuity between the slab and the beam at this point so it means I would have some negative moment here and this is why we have this part in case of continuous so you will have some negative moment at the beginning it equals minus point zero 4 FL and also to effect the positive moment it would be read use because you took some of the moment negative so it will reducible step moment from point zero eight six two point zero seven five FL but what is this F and what is this L L is this man this is van and this is parent air disband and so on but how about F V capital F the capital is the total new Zion ultimate load it means the ultimate load equals one point four dead load plus one point six live load but this is chemical it so it means that resultant to observe so if you have a road here which is a uniform load so this uniform load you get it at the ultimate multiply animal it means x one point four and one point six live load and the dead load and live load and you should multiply to give the resultant Multan by multiplying this by the span so you get a concentrated load here as will be a capital S then you will use this F it will be multiplied by F L a times this sector to give the bending moment - okay then the second part of this table it gives us shear okay so to get the shear we get it from the second row and again to show this this is showing the stance so add the first support here you have a shear of point 4f no L anymore here because this F here equals a uniform load multiplied by the span as a concentrated load so it is point four times concentrated load point four F then at the first interior support at this point six F then later on it will be point five F in case again if you have a continuity here at the beginning we have to use this table here the value here it will be just to change it or the red one from point four it would be point four six or seven this is only difference so how to use this if you have a problem or you have a slab continuous one-way slab and you have different loads here so if you have a slab with s ban and one and a span in the tool the first disband is loaded by I only form load all W one kilo Newton per meter the second span has a load of W 2 equal also kilonewton per meter so what you should do you should good the resultant of the first expand the result all of the road and the second span so we call it f1 equals W 1 times n1 this is the F physics we use it here in this equation then you will repeat this for the second span so you will have FP 2 2 equals W 2 x LD 2 it will give you a force here as kilo Newton here as in a Newton here okay then we have to use this F to get the bending moment values so from the table the value here will be zero at near the middle of the first span will be point zero eight six F 1 times L 1 then here F 2 and here also let's write this one okay good here to add this n2 okay so point 0 63 ft 2 x + 2 and how about the middle support here because in the middle support you cannot say point 0 8 6 FL ok because you have F 1 and F 2 2 and L 1 and L 2 so here we get the average so it will be point zero eight six times F 1 and 1 plus F 2 le 2 divided by 2 so you get the average from both and this will be the ones that we take it you take it to get the negative moment at the interior support here okay then for this year we are going to do the same you will use the values at the bottom here so it will be point 4 F 1 then here point 6 F 1 then it will be point 5 F 2 and point 5 52 and so on ok so in a case if you have a real problem with different bands with different loads the important thing here at the middle one you get the average of F 1 L 1 plus F 2 and the 2 divided by 2 multiplied by this factor here which is point zero eight six about checking deflection how we check the deflection okay as I told you we have to use a table in the cold the deflections are important to be checked because if you have existence deflection it may cause damage of ceiling finishes or other architectural details to avoid this we have to use the minimum effective dips which is the span divided by basic span-to-depth ratio from the table 3.9 x the modification factor let's remove this one okay then to do that or to check the deflection we have to get a value called modification factor is this modification factor we have to check at the beginning we assume a duplication factor 1.3 but now we need to check how much is this real modification factor after we make our calculation and we good a dips and total it shows the step to do this really smooth this we get this modification factor from they would 3.10 which is this table and this table you have a modification factor equal this value 0.55 + 4 7 7 minus FS divided by 120 times 0.9 + m / BD square and this should be less than or equals to 2 okay we have some values here we need to know what is the FS this is the stress in the tension steel the stress in the bottom tension steel how much is this is stressed FS also what is M M is the maximum positive moment not negative moment because deflection will be at the span not at the support so this M here is the maximum negative maximum positive moment B as you know is 1000 millimeter for slabs and D is the depth of the slab okay so how to use this one to use this one we have to get the value of M / BD square and we have to get the FS how much is this a fest we get it from this equation here according to the code it equals 2 over 3 F field times its required / s provided it is required by design and the read areas that you provided times 1 divided by beta B and beta B this is something for redistribution and we use it as 1 in this case so to use this one you need to get the FS once you get the FPS you put it here into this equation then get M / BD square and you put it there you will substitute into the equation get the modification factor okay let's go back and see so again this is the end from the same equation that we just saw together okay as I just explained M here is maximum positive moment and the short span because usually you have more moment in the short disband or always you would have more moment in the short disband then in the wrong dis pan because takes more amount of loads then to get the FS created from this equation as I just explained it is required this is the area of reinforcement click well at mid-span it is su provided area or reinforcement provide provided at the mid span and beta B is the ratio of mid span moment after and before redistribution and usually the code is not allowing to make redistribution if you are using the coefficients from the table to get the bending moment and the shear okay once we got em so we have to calculate something called limiting span to depth ratio okay limiting we span to depth ratio it equals basic span to depth ratio from the code the values that you get from the hold which was 20 in cases simply supported or 26 in case of continuous multiply it by M so the M is that you calculated from the previous equation in the previous slide here okay so you multiply this M by the basic ratios that you used at the beginning so M multiplied by 20 or M multiplied by 26 then we calculate the actual span to depth ratio which is the shortest ban divided by the actual depth and then we compare this one to that one the first one to that limiting one to the actual one if the limiting we spent to depth ratio the first value here greater than the actual span to depth ratio it means okay the deflection is says you can't continue and it will not have any problem in the serviceability regarding deflection if the limiting the span to depth ratio was less than the actual span to depth ratio it means it is not safe okay is unsafe and the solution in this case to increase the depth and in some cases also if you cannot increase the depths you can increase the reinforcement because it will will help the tool makes it flex and more safe because it will affect this value of FS okay this is the end of our present station today the coming one we were women to talk about real example and how to solve a problem with simply supported one-way slab and then we will have a second example about how to solve a continuous one-way slab thank you and see you in a minute video