 So, today I am going to discuss about a numerical example on analysis and analysis of multi span continuous RCC beam, learning outcomes. At the end of this session the viewers will be able to analyze multi span RCC continuous beam that is they will be able to determine the bending moment and shear force at critical sections of a multi span continuous beam. Example analyze a continuous rectangular beam having 5 spans of each 6 meter to carry a dead load of 10 kilo Newton per meter and a live load of 13 kilo Newton per meter as shown in figure below this is the figure. The beam is continuous over 6 columns use M20 concrete and Fe 4 and 5 steel. So, this is a 5 span 6 meter each a continuous beam over which we are having a dead load and a live load acting on it both are uniform distributed loads. Solution given data span is 6 meter each span is 6 meter WD is 10 kilo Newton per meter WL that is live load is 13 kilo Newton per meter FCK that is characteristic strength is 20 Newton per Newton square and the yield strength of steel FY is 4 and 5 Newton per Newton square. Step 1 cross sectional dimensions assume the effective depth D. So, first of all we are supposed to finalize the cross assume some cross sectional dimensions. So, assume effective depth D which is taken usually 115 to 120th of the span span is 6000 year. So, therefore, 6000 mm therefore, D was sort to be 430 mm the overall depth D capital D it is equal to 430 mm plus 50 mm effective cover effective cover means it is up to the center of steel clear cover plus half the diameter of the bar is effective cover. So, here we have taken 50 mm. So, therefore, the overall depth work out to be 480 mm and assume a breadth of 230 mm. Then second step bending moment at critical sections we are going to determine the bending moment at different critical sections. The self weight of the beam is to be calculated which is equal to breadth into overall depth of the beam into density of concrete. So, density of concrete is 25 kilo Newton per qubit meter. So, self weight of beam is equal to 0.23 this is the breadth in meter 0.48 is the overall depth in meter and 25 kilo Newton per cubic meter is density of concrete. So, therefore, the self weight of the beam was sort to be 2.76 kilo Newton per meter dead load 10 kilo Newton per meter assume a finish a finishing load of 0.6 kilo Newton per meter. So, total dead load WD is equal to 13.36 kilo Newton per meter whereas, WL is given as 13 kilo Newton per meter. So, referring table number 12 of IS 456 for bending moment coefficients. So, this is the table number 12 of the IS 456 2000. So, bending moment coefficients are given for both first column it deals with dead load and live load types of loads and we get span moments that is near middle of end span we have the coefficients and at middle of interior span we have the coefficients. So, that means, we have to find out what is the bending moment by using these coefficients. So, that is near middle of end span. So, by using these coefficients at the middle of interior span similarly, so for support moments we are having at the support next to the end support we are having coefficients for this. So, this negative sign indicated is hugging moment and here the positive sign indicates it is sagging moment. So, at the other interior supports we are having again negative moment. So, here is the bending moment diagram I have drawn. So, here you find this is a 5 span continuous beam. So, having each 6 meter span having WD and WL. So, first of all we have seen the span moments there are two columns in the span moment that is at the middle of the end span and at the interior middle of the interior spans. So, therefore, this is M1 and this is M2 both are sagging and next we have seen span moments span moments means that over the support. So, here this is span moment M3 this is maximum. So, it is at the support to the next to end support and these are M3 and M4 it is at the interior supports. So, that means we are supposed to calculate a totally four moments. So, two span moments that is M1 and M2 and two support moments that is M3 and M4 out of these three highest is M3. So, here I have calculated the bending moment at critical section that is span moment M1 near middle of the end span. So, it is the coefficient one-twelfth of WDL square and plus one-tenth of WLL square. So, it was sort to be 86.88 kilo Newton meter then M2 it is again at the middle span of the interior spans that is one-sixteenth of WDL square plus one-twelfth of WLL square that was sort to be 69.06 kilo Newton per meter where WDL and WL are dead load and live load. So, next bending moment hogging that is negative with negative coefficient we have seen at the support next to end support here one-tenth of WDL square one-tenth of WLL into L square wherein we get it as 100.096 kilo Newton meter then M4 that is at the interior support. So, we get the one-twelfth of WDL square plus one-tenth of WLL into L square that is 98.08 kilo Newton meter. So, that means, so support moments are higher in comparison with the span moments where we will be the maximum bending moment and what is its numerical value. So, overall we are supposed to find out the highest bending moment acting in the continuous beam. So, therefore, where will be the maximum bending moment and what is its numerical value. So, please answer it the moment is maximum at support next to end support as per table number 12 of IS 4562000 M3 that is the maximum moment at the support next to end support which is given by one-tenth of WDL square plus one-tenth of WLL into L square. So, that was sort to be 100.096 kilo Newton meter. The maximum moment is usually considered for assuming the sections. So, therefore, the maximum factored bending moment Mu it is to be multiplied 100 point M is to be multiplied by 1.5 that is 1.5.100 into 100.096 that also to be 150.144 kilo Newton meter. Next step shear force at critical sections total dead load WD is 13.36 kilo Newton per meter live load WL is 13 kilo Newton per meter. So, we are supposed to refer table 13 of IS 456 for shear force coefficients. So, this is table number 13 of IS 456. So, here first column it indicates types of loads that is dead load and imposed load then next second column it is the coefficients at end support. So, it is 0.4 for dead load 0.45 for live load then at support next to the end support. So, outer side it is 0.6.6 and inner side it is 0.55 and 0.6 and at all interior supports it is 0.5 and 0.6. So, this is to be multiplied by the WD into L that is for dead load and coefficient coefficient into WD into L and for imposed load it is coefficient into WD into L. So, therefore, as for the this particular coefficients we get we are supposed to determine the shear force at these four sections. So, this is the first section at the support outer support then next here that is at support next to end support this is highest which is having 0.6 WD into L plus 0.6 WL into L then next this is at interior support and again this is again at last support. So, here are the shear forces shear force at critical sections V 1 at the end support we get 0.4 WD into L plus 0.45 WL into L that was sort to be 67.164 kilo Newton at the support next to the end support this is highest V 2 that is outer side we get 0.6 WD into L plus 0.6 WL into L it is 94.896 kilo Newton then next to the V 3 that is at all interior support it is 86.88 kilo Newton where will be the maximum shear force and what is its numerical value the maximum shear force occurs outer at outer side of the support next to end support at the next at the support next to end support we get V 2 that is outer side it is 0.6 WD into L plus 0.6 WL into L which was sort to be 94.896 the maximum factored shear force it is V U it is equal to 1.5 times 94.896 that is 142.344 kilo Newton. So, these are the references used for the preparing this particular presentation. Thank you and end all.