#Part-12/#Overhangingbeam which is subject to point load at uniformly distributed

 Q. A beam ABCD, 4 m long is overhanging by 1 m and carries load as shown in below.  

   Draw the shear force and bending moment diagrams for the beam and locate the point of contraflexure.

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Solution: Given : Span (l) = 4m; Uniformly distributed load over AB(w) = 2 kN/m and point load at C (W) = 4kN.                                                                                                                                                       First of all, let us find out the reactions RB  and RD. Taking moments about B and equating the same,.                                                                                                                                                                                                                                                                                                                 RD × 3 = (4 × 1) – (2 ×1)× 1/2 = 3.                                                                                                   ∆.                         RD = 3/3 = 1kN.                                                                                             and.              RB = (2×1) + 4 – 1 = 5 kN.  

 

Shear force diagram:

                     The shear force diagram is shown in Fig. (b) and the values are tabulated here: 

               FA = 0

               FB = 0 – (2×1) + 5 = + 3 kN

               Fc = + 3 – 4 = – kN

               FD = 1 kN

Bending moment diagram

      The bending moment diagram is shown in Fig. (c) and the values are tabulated here:

               MA = 0

               MB = – (2 ×1) 0.5 = – 1 kN-m

               MC = 1 × 2 = + 2 kN

               MD = 0

    

    We know that maximum bending moment occurs either at B or C, where the shear force changes sign. From the geometry of the bending moment digram, we find that maximum nagetive bending moment occurs at B and maximum positive bending moment occurs at C.

Point of contraflexure:

       Let p be the point of contraflexure at a distance y from the support B. From the geometry of the figure between B and C, we find that

                 y/1.0 = 1–y/ 2.0

                 2y = 1–y.            Or.  3y = 1

                 y = 1/3 = 0.33 m.   Ans. 

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