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Some useful techniques for solving shear force and bending moment diagrams.bilal

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Determine reaction in supports Construct Q1 Mx diagrams Determine Solution First of all we have to convert the distributed load into point load. And please note that length l is missing in the data so I am taking it as variable only. The distributed load having intensity q and it is distributed over length of l therefore we can assume that total load \"ql\"will act at centre that is at a distance of l/2 from left end. similalrly the load which is distributed over length k1l is also represent by point load but in upward direction. Now on applying the equilibrium condition we will calculate the reaction. R1+R3= 20ql - 20qkl + F1 since thre is no horizontal force therefore the reaction R2 = 0 Now third equation of equilibrium that is moment about point O. hence we can calculate the value of remaining reaction. Since the value of l is missing so I am unable to calculate the actual value. Once we know the reaction it is very easy to draw shear force and bending moment diagram I am unable to post picture due to less space. For Shear force we consider all the point where point load is acting and analyse the shear force on left side of that point. For e.g. at point O left side of O = 0 (since no force is there) Right side of O = R1 similarly we consider other points also and draw the obtained value on the graph just above the marked point of beam. and finally joinn them./ Same steps have to perform to draw bending moment diagram only one difference is there that is instead of shear force we have to ccalculate the bending moment..

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- Determine reaction in supports Construct Q1 Mx diagrams Determine Solution First of all we have to convert the distributed load into point load. And please note that length l is missing in the data so I am taking it as variable only. The distributed load having intensity q and it is distributed over length of l therefore we can assume that total load "ql"will act at centre that is at a distance of l/2 from left end. similalrly the load which is distributed over length k1l is also represent by point load but in upward direction. Now on applying the equilibrium condition we will calculate the reaction. R1+R3= 20ql - 20qkl + F1 since thre is no horizontal force therefore the reaction R2 = 0 Now third equation of equilibrium that is moment about point O. hence we can calculate the value of remaining reaction. Since the value of l is missing so I am unable to calculate the actual value. Once we know the reaction it is very easy to draw shear force and bending moment diagram I am unable to post picture due to less space. For Shear force we consider all the point where point load is acting and analyse the shear force on left side of that point. For e.g. at point O left side of O = 0 (since no force is there) Right side of O = R1 similarly we consider other points also and draw the obtained value on the graph just above the marked point of beam. and finally joinn them./ Same steps have to perform to draw bending moment diagram only one difference is there that is instead of shear force we have to ccalculate the bending moment.

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