 So, the topic of today's discussion is numerical example on design of beam subjected to torsion. Learning outcomes. At the end of this session, the viewers will be able to design a beam subjected to torsion. This is an example. The design reinforcement detail required for the rectangular beam section with following data size is given 300 by 600 mm, grade of concrete is M 20 and grade of steel is Fe 415. The factored shear force acting is 95 kilo Newton and factored torsional moment is 45 kilo Newton meter and factored bending moment Mu it is 1.15 kilo Newton meter. So, sketch the reinforcement details solution. First step find equivalent bending moment M EL and equivalent shear force with respect to the IS clauses. Assume we have to assume some cover we are assuming 50 mm effective cover. So, effective depth D is equal to overall depth minus 50 mm that is 550 mm as per IS 456 2000 clause number 41.4.2. So, MT is equal to Tu into 1 plus D by B whole divided by 1.7. So, we have substituted the values. So, we get MT as the torsional moment it is 79.41 kilo Newton meter. So, MT is less than Mu hence no need of providing steel on compression phase that means no need of we need not provide compression steel. So, M EL that is equivalent bending moment is Mu plus MT. So, Mu plus MT is 194.41 kilo Newton meter as per IS 456 2000 clause number 41.3.1 we have to find out equivalent shear force V EL. So, V EL is equivalent shear force which is equal to V U plus 1.6 into this that is M that is torsional moment divided by breadth. So, this 45 is Tu so, 45 divided by 0.3 that works out to be 335 kilo Newton meter. So, second step is to design the longitudinal reinforcement. So, axial limit for Fe 4 and 5 is 0.48 D so, that works out to be 264 mm. So, Mu limit it is 0.3 is F6 AB into axial limit D minus 0.4 to axial limit this is as per annexure clause number G 1.1 C of IS 456 we get 250.4 kilo Newton meter. So, M EL is less than M mu limit so, equivalent bending moment is less than M mu limit hence a singly reinforced section is can be designed for this. Now, as per IS 456 2000 clause number G 1.1 B the tensile reinforcement AST is given by M EL is equal to 0.87 fy into AST into 1 minus AST fy upon BDFC K. So, substituting the values here so, we get a square equation so, by solving that particular square equation we get the value of AST that is 1144 mm square provide 4 bars of 20 mm diameter. So, AST provided is 1256 which is greater than 1144 mm square. First step design of shear reinforcement so, we are supposed to find out equivalent nominal shear stress tau EL so, it is V EL upon BD so, it works out to be 2.03 Newton per mm square. So, for referring table number 19 of IS 456 we have to find out percentage steel provide we have already provided 4 bars of 12 mm torque so, that we have calculated percentage steel provide it works out to be 0.76. So, referring this table number 19 of IS 456 we can find out shear strength of concrete or shear stress in that can be taken by concrete that is tau C. So, it works out to be 0.56 Newton per mm square for from table number 20 of IS 456 tau C max is 2.8 Newton per mm square since, our value of tau V is between tau C and tau C max so, hence shear reinforcement is to be provided as per IS 456 clause number 40.4. Now, for that we have to determine V us, V us is the shear that it should be taken by stirrups that is V EL minus tau C V D. So, we are getting 24000242600 kilo Newton selecting 2 legged 8 mm H Y S D bars for shear reinforcement. So, V us this is as per clause number 40.4 A. So, V us is equal to 0.87 F Y S V into D divided by SV. So, from this equation we can find out SV so, this has worked out to be 82.29. So, this 82.29 mm it is very small so, therefore so, we have to increase the reinforcement sorry spacing of the stirrups spacing of the stirrups is too close so, therefore which may create problem during concreting so, hence provide 4 legged instead of 2 legged we will provide 4 legged. So, 4 legged means spacing can be doubled that is 160 mm centre to centre so, check for minimum shear reinforcement assuming 50 mm side cover as per IS 456 2000 clause number 41.4.3 B 1 is 300 minus 100 that is 200 mm D 1 is 600 minus 100 that means, we have to subtract cover that is from the reinforcement what is the spacing remaining it is 500 mm. So, ASV minimum as per IS is given by T U into SV divided by B 1 D 1 into 0.87 F Y plus V U into SV divided by 2.5 D 1 into 0.87 F Y. So, by substituting all these values in this we are getting ASV that is the minimum shear reinforcement required as 233 mm square. So, can you guess what is the exact or correct expression for minimum shear reinforcement as per IS 456 2000. So, we have given 4 choices out of these 4 choices you are supposed to select the correct one. Now, the 4th one is the right choice that is T U SV upon B 1 D 1 0.87 F Y plus V U SV upon 2.5 D 1 0.87 F Y. So, that is the correct option for expression for ASV as per IS 456 2000. Now, ASV provided is 4 into pi it is 4 legged therefore, it is 4 4 into area of 1 bar that is pi by 4 D square it is 201 mm square. So, which is less than ASV. So, ASV we have calculated just minimum reinforcement that is 233. So, therefore, this is less than ASV. So, reduce spacing that means, which we have taken 160 instead of 160 I have reduced the spacing to 150 mm. Then the ASV minimum I have calculated how much it was ought to be again by substituting the values by we get ASV minimum as 194 which is less than ASV provided that is 201 mm square. So, the maximum spacing provided as per 26.5 point 1.7 A X 1 is equal to 200 plus 8 that is 208. Then Y 1 required for finding maximum spacing it is 500 plus 8 that is 500 is your D 1 plus 8 mm that is your center to center of the leg that is shear strength that is 8 mm diameter we have added 5 0 8. So, therefore, X 1 plus Y 1 divided by 4 that works out to be 179 mm or 300 mm whichever is less this is what IS says. So, by selecting ASV is equal to 150 this requirement is also satisfied hence a 2 legged 8 mm stirrups at 150 mm center to center is satisfactory for this particular condition for the taking the beam torsion. Now, this is the side phase reinforcement. So, as per clause 26.5 point 1.7 A of IS 456. So, D 1 is greater than 450 mm hence side phase reinforcement required as per clause number 26.5 point 1.3 must be provided. So, therefore, as per this particular clause 26.5 point 1.3. So, 0.1 percent of web area should be provided as a side phase reinforcement. So, therefore, we have taken 0.1 divided by 100 into 300 into 600 that is your web area. So, that is equal to 180 mm square reinforcement required on each phase. So, 180 I have divided by 2. So, on both phase we have provided it is 90 mm square. Hence, provide 1 bar of 12 mm on each phase as shown in figure. So, area provided will be pi by 4 into 12 square it is 1 1 3 mm square. So, these are the reinforcement details. Now, this we have we are having the reinforcement this is the bottom reinforcement that is your tension reinforcement right. So, and this is a side phase reinforcement that is to take the torsion that is side phase reinforcement is 2 12 mm torque side phase reinforcement is 2 12 mm torque and the we have provided 4 12 mm bars here at top that is to anchor these particular stirrups because it is a 4 legged stirrups. So, since it is a 4 legged stirrups we have provided 4 bars at the top. So, that is 4 12 mm torque these are called as hanger bars. So, there is no design required for this this is just to hang these particular shear stirrups. Now, that your main steel at bottom that is tensile steel it is 4 20 mm torque which is provided bottom. So, this is total depth 600 and this is 300 and this is 200 mm that is the distance between these 2 bars and this way. So, that means rather you are supposed to you are supposed to find out to be by subtracting 50 50 mm cover we have taken d 1 x 1. So, how we have done it that I have tried to show. So, 50 50 mm cover. So, this is your b 1 200. So, 300 minus 100 that is 200 similarly here d 1 will be total 600 minus 50 mm cover 50 mm cover at top and bottom. So, that was sort to be 500 mm that we have used for the calculation. So, these are the references used for the this particular presentation. Thank you one and all.