#Part-13/#Overhangingbeam which is subject to point load and uniformly d...

Problem:

                  Draw the shear force and bending moment diagram for the beam shown in Fig. 1.1 Indicate on the diagram the values of shear force and bending moment at significant point. Also show the location and magnitude of the maximum bending moment.

Video solution: 

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Solution: Let the reaction at the left and right supports be Vb and Vd respectively.

Taking moment about the left support we have,

Vd × 12 = 2×12×3+8×9

        Vd = 12 kN

        Vb = 2×12+8–12

             = 20 kN

Shear force calculation:

                                 Shear force at A = 0

S.F. just on the left hand side of 

B = –2×3 = –6 kN

S.F. just on thr right hand side of

B = –6 + 20 = +14 kN

S.F. just on the right hand side of

C = –12 kN

S.F. just on the left hand side of 

C = –12+8 = –4 kN

Let the S.F. be zero at x metres from A (between B and C)

Equating the S.F. to zero, we have 

                            20 – 2x = 0

                                      x = 10 metres

Bending moment calculation:

                            B.M. at A = Ma = 0

         B.M. at B = Mb = – 2×3³/2 = –9 kNm

         B.M. at C = Mc = + 12 ×3 = +36 kNm

         B.M. at D = Md = 0

B.M. at 10 metres from 

A = Mc = 20×7– 2×10²/2 = 40 kNm

Point if contraflexure :

 Let the B.M. be zero at x metres from A ( between the two supports). 

Equating the beginning moment to zero, we get,

                        20(x–3) –2×(x²/2) = 0

                        x² – 20x +60 = 0

The partical value of x should be between 3 and 12.

Solving the equation we get x = 3.67 metres.

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