 Dr. Patil Sunil Kumar S, Professor in Head Swinging Department, Balchan Institute at Konosolapur. So I am going to discuss about analysis and design of multi span RCC continuous beam. Learning outcomes. At the end of this session, the viewers will be able to explain the steps of analysis and design for a multi span RCC continuous beam. First step, so one has to assume a cross section of the beam. So effective depth D is taken as 115 to 120th of span, where span is centre to centre distance between the columns. Oral depth D is equal to effective depth plus half the diameter of the bar plus clear cover that will give us oral depth and width of beam B that is minimum is 200 mm or it may again be in the multiple of or it may be 230 to 50 or in the multiple of 50 mm for that. Step two, determine nation of design moment EMU and design shear force VU. So for this, we have to find out what is the total dead load. It consists of self-heat of the beam that is width into depth of the overall depth of the beam into density of concrete. And dead load from slab usually it is given in the example and dead load of floor finish we have to assume if it is given it is alright otherwise you can assume what is the dead load of floor finish depending upon flooring type. So live load usually it is given in the example. Then we have to refer the winning and coefficients as per IS456-2000. If you see the coefficients here, so we are having first column dead load and live load and this is for span moments and this is for support moments. So when we want to find out the EMU and VU we are supposed to find out the highest winning moment. So if you see the coefficients here for span moments it is lesser, denominator is more. So therefore coefficient is less and for support moment this is having lesser denominator therefore this will give us that is support moment at support next to the end support will give us the maximum winning moment. So therefore we have used the maximum support moment at the support next to end support it is hogging winning moment it is over the support EMU is 1.5 times AWD into L square upon 10 plus 1.5 times WL into L square upon 9. So this 1.5 is to convert EMU into MU. So the maximum span moment at the near of the middle span that is the highest of span moment we have taken. So because this is sagging in nature this is sagging and above one is hogging. So therefore EMU for sagging it will be WD 1.5 to WD into L square upon 12 1.5 to WL into L square upon 10. So next the shear coefficients as per IS 456 2000 if you see the table the first column it gives dead load and live load and if you see the coefficients so this is the highest coefficient 0.6 for WD and 0.6 for WL. So to find out the highest shear force so the highest shear force is at the outer side at the support next to the end support that will give us the highest shear force and for design we require highest shear force. So therefore we can calculate what is the shear force here. So the maximum shear force by using table number 13 of IS 456 2000 it is at the support next to end support it is 1.5 times this is 1.5 it is for conversion of shear force into VU that is the ultimate value. So WD into 0.6 plus 1.5 into WL into 0.6 into L. So this will give us the maximum shear force. So design of continuous beam for bending moment at critical section now first we have determined bending moment now we have to design for bending moment at critical section. So depending upon the grade of steel used the ME limit is given by following equations. So it is 0.148 FCK BD square if it is FE 250 that is my steel it is ME limit is equal to 0.138 FCK BD square if FE used the steel used is FE 4 and 5 if ME limit is equal to 0.133 FCK BD square if the steel used is FE 500 where FCK is characteristic strength and ME limit is the limiting moment of resistance of the section without compression reinforcement that is for a singly reinforcement. The maximum moment ME is usually hogging and it is compared with ME limit if ME is less than ME limit it is under reinforcement it can be designed as singly reinforcement. So find area of steel by equation G 0.1 0.1 B of IS 456 2000 that is ME limit is equal to 0.87 FY ASTD into 1 minus AST FY upon BD FCK. So in this equation ME is already known so therefore AST you have to determine remaining all values are given in the example. If ME is greater than ME limit it is over reinforcement IS does not allow the design of over reinforcement because of brittle failures certain failures. So hence it is to be designed as a doubly reinforcement. So therefore using equilibrium equation of a singly reinforcement find the tensile steel required for singly reinforcement that is EST 1 is the steel required for singly reinforcement. So 0.87 FY AST 1 into D minus 0.4 to EXU limit so that is equated with EXU limit of a singly reinforcement which is already determined. So therefore so we have to take EXU limit this is depth of neutral axis limiting value of depth of neutral axis it is 0.53 D for FE 250 and 0.48 D for FE 4 and 5 and 0.46 D for FE 500. Equating moment of resistance offered by the forces in the compression steel with additional tensile steel AST 2 we get the following equation 0.87 FY AST into D minus D dash. So 0.87 FY into AST 2 this is the force taken by compression steel D minus D dash this is lever arm is equal to MU minus MU limit this is MU limit of a singly reinforcement section. Find AST 2 the total tensile steel required it is AST 1 plus AST 2. The strain in the compression steel is given by epsilon SC is equal to EXU limit minus D dash divided by EXU limit into 0.0035 this is from the strain diagram. So find force in steel in compression zone by referring figure 23 of IS 456 2000 and equating compressive stress in steel to the tensile force in additional steel which we have provided we get FSC into ASC is equal to 0.87 FY into AST 2. Find ASC ASC is area of compression steel steel in compression required. Determine the steel determination of steel is to be carried out for maximum hugging moment which is over the support as well as for maximum sagging moment which is at mid span separately and sectional details near support and at mid span are shown in the figure below. So this is the sectional detail near the support wherein we get hugging moment therefore we get more steel at top and for sagging moment minimum 50 percent of the steel should go up to end therefore we are having less number of bars. So next sectional details at mid span wherein we have sagging moment that is at bottom so bending tension at bottom therefore more steel at bottom and top we are having only hanger bars. Then design of shear reinforcement. So we have to find F shear stress nominal shear stress tau V is equal to V upon BD. So V is already determined B is known D is known therefore find percentage steel we have already provided steel for bending therefore we have to find out what is percentage steel provided using table number 19 of IS 456 2000 find tau C tau C is the stress that can be taken by concrete alone for percentage steel provided over there. So V us is equal to V u minus tau C BD where V us is the shear to be taken by stirrups and tau V is the nominal shear stress and tau C is the design shear strength of concrete. Find the spacing of the stirrups SV is equal to 0.87 FY ASV D divided by V us where ASV is the area of stirrups. So the last one is check for deflection so L by D max is equal to L by D max given in the table of IS 456 23.2.1 as of IS 456 2000 and into F1 into F2 into F3. So F1 from figure number 4 of IS 456 2000 F2 from figure number 5 of IS 456 2000 and F3 from figure number 6 of IS 456 2000. Here L by D provided should be less than L by D max. Now here you will find this is the figure number 4 wherein from this figure number 4 we are supposed to find out F1 and from figure number 5 we have to find out here percentage compression steel we know it already therefore we can find out modification factor for compression steel if it is provided then this is if it is a t-section if you are designing the beam as a t-section in that particular case here you will get this particular constant also. So therefore we are supposed to use the constant F1 F2 F3 from these 3 graphs given in IS 456 2000. So by using the constants F1 F2 F3 we can calculate L by D max so L by D max provided should be less than L by D max as we have calculated from this particular equation. So these are the references used for this particular preparation of OER video. Thank you Vanindal.