 I am Dr. Patil Sunil Kumarish, Professor and Head Surgeoning Department at Valdeshen Institute at Akhmer Soulapur. So today, let us discuss on numerical example on design of multi-span RCC continuous beam. Learning outcomes. At the end of this session, the viewers will be able to design a multi-span RCC continuous beam. This is an example design a continuous rectangular beam of span 7 meter each to carry a dead load of 12 kilonewton per meter and a live load of 16 kilonewton per meter. The beam is continuous over a more than 3 spans and is supported by columns use M20 concrete and FE 415 steel. So it is given span 7 meter WD 12 kilonewton per meter WL that is live load 16 kilonewton per meter FCK that is characteristic strength 20 Newton per Newton square and yield strength FY is 415 Newton per Newton square. First step. So we have to assume the cross-sectional dimensions of the beam based on the serviceability criteria that is span to depth ratio. So effective depth is taken usually 115th to 120th of the span. Span is here 7000 mm. So therefore D will be 450 mm. Let us consider D equal to effective depth equal to 450 mm. The overall depth will be equal to let us say 500 mm by considering 50 mm effective cover. So assume breadth as 250 mm. Design moment M U and design shear force V U are to be determined. So for this we have to calculate the dead load because now we have chosen the section. So therefore we have to calculate the dead load which consists of self weight of the beam based on the cross-section chosen then dead load from slab and dead load from floor finish. So self weight of the beam it is 0.25 into 0.5 into 25. 25 is density of concrete 25 kilonewton per meter. So therefore this is 3.125 kilonewton per meter. So dead load from slab is 12 kilonewton per meter floor finish load we have assumed 0.875 kilonewton per meter. So total dead load is 16 kilonewton per meter. So the live load is already given it is 16 kilonewton per meter. So W we are supposed to find out design loads. So design load is the partial safety factor into dead load that is 1.5 into 16 and this is again 1.5 into 16 that is the live load. So these are design loads. So one has to refer the IS4562000 table number 12 for bending moment coefficients and table number 13 for shear force coefficients. So this is table number 12 and this is table number 13 for the shear force coefficients. So here we get the first dead load and for live load the coefficients are given for different conditions. So these are span moments and here we get support moments. So again span moment at two sections. So near middle of the end span and at the middle of the interior span. Again for the span moments we are having at the span next to end support and at the support next to end support and at interior supports. So now can you guess what are the coefficient bending moment coefficients as per the table number 12 of IS4562000 for dead load and imposed load to get a highest span moment and this is the question for highest shear force that is dead load and the imposed load coefficients to get a maximum shear force. Can you guess please? So this is the answer for bending moment coefficient that is the minus 1 by 10 for WD and minus 1 by 9 for WL and for shear it is 0.6 for dead load and 0.6 for live load. Now the moment is maximum at the support next to end support. The design moment is 1.5 times the maximum bending moment. Therefore we have multiplied by 1.5 everywhere. So MU support moment and the moment at mid span. So we are having two types of moment. This is support moment is hogging and mid span moment is sagging. Now the shear force is maximum at the support next to end support that is on the outer side. The design shear is 1.5 times the maximum shear force. Again it is multiplied by 1.5 into 16 into 0.6 into 7. So we get 201.6 kilo Newton. Now the next is we are step 3 that is design of longitudinal reinforcement. Since IS 456 is Fe 415 is used, steel is used, XU limit is equal to 0.48 D as per IS. So we get 216 MU limit is 1 by 0.36 FCKB into XU limit into D minus 0.42 XU limit. So this is G.1.1C. So accordingly we are getting 139.688 into 10 to the power of 6 Newton mm. So MU is greater than MU limit. So hence a doubly reinforced section is to be designed. So using equilibrium equation of a singly reinforced section find first the steel for the singly reinforced section that is EST1. So this is the equation 0.87 FYST into D minus 0.42 XU limit is equal to the this is your MU limit which we have calculated for singly reinforced section. We get EST1 is equal to 1077 mm square. Now equating the moment of resistance offered by force in compression steel and additional steel EST2 we get that is the force offered by EST2 for this is required for equilibrium. So therefore 0.87 FYST2 into D minus data this is lever arm is equal to MU minus MU limit. So this is moment. So we get EST2 is equal to 752 mm square. So EST2 so total steel to be provided EST is EST1 plus EST2. Then since strain varies linearly we get this equation that is epsilon SC is equal to 216 minus 50 divided by 216 into 0.0035. This is 0.0035 is maximum strain at the extreme fiber. So we get the strain then by referring IS456 figure number 23 we can find out what is the stress in compression steel. So it works out to be 0.83 FY. So equating compression force in steel to tensile force in additional steel which we are providing we get FSC into ESC is equal to 0.87 FY into EST2. So from this we can find out ESC that is the compression steel it is 788 mm square. Provide 6 bars of 20 mm bars in tensile zone and 3 20 mm bars in compression zone. So that is which will be sufficient to have the area will be more than this. At mid span again MU is different this is sagging moment. So MU again at mid span is also greater than MU limit. So therefore EST1 is the same we have done for the support we get EST1 is equal to 1077 mm square and EST2 here it is 525 mm square because here moment is little less than the support moment. So next FSC again determined from figure 23 of IS456 we get EST is equal to 550 mm square and EST that is at bottom here it is sagging therefore EST is the tensile steel at bottom it is 15 EST1 plus EST2 it is 150 to mm square. So provide 6 bars of 20 mm diameter in tension zone and 2 bars of 20 mm diameter in the compression zone. Now for shear reinforcement so VU is equal to we have already calculated maximum shear force that is 201.6 kilo Newton. So tau V that is nominal shear stress is equal to V upon BD it is 1.792 Newton per Newton square. So we have to calculate percentage steel to find out tau C that is the shear strength of concrete from table 19 of IS456. So percentage steel is 1.676 and by referring table 19 we can get what is strength of concrete. So shear strength that can be taken by concrete tau C it is 0.74 Newton per Newton square. So VUS is balance shear force which is to be taken by the stirrups it is VU minus tau CBD. So we have to calculate VUS and as per clause number 40.4.4A of IS456 we get VUS is equal to 0.87 FYSV into D divided by SV from this we can find out SV is the spacing of stirrups ASV is the area of stirrup. So since 2 legged is used 8 mm 2 legged 8 mm stirrups are used here it is 2 into area of pi by 4 into 8 square. So we will find out what is the spacing it is 137.2 mm. So provide 2 legged stirrups at 130 mm center to center. Now check for deflection so we have to calculate L by D max. So basic value as per IS456 clause number 23.2.1A and it is 26 for continuous. So IS from figure 4 of IS456 for PT is equal to that is percentage still is equal to 1.676 we will get modification factor F1 is equal to 0.9. So FS so you have modification factor is determined from figure 4. So FS we have to calculate for that then percentage still is this much. So from since we are having a compression still we have to refer figure number 5 for finding the modification factor F2 that was 1.23. So from figure number 6 so because it is again a T beam here for the mid span we will be designing it as T beam therefore F3 again we get. So if you consider it as rectangle beam then it is 1. If you consider it as T beam then it will give you some value. So therefore we have to complete the L by D max so L by D max work out to be 15.55. So L by D provided is 15.55 which is less than L by D max which is 28.78. Hence deflection control is satisfactory. Now this is the reinforcement details of continuous beam here you will find the steel is maximum at bottom here and here it is at top. So over the support we get maximum steel so because we are having end moment that is this is only one span we have shown here only one span we have shown out of three spans. So for the center we get six bars at bottom and at the end we have six bars at top. So therefore this is the section at mid span and this is the section at support. So these are the references which we have used for this particular presentation thank you very much.