Draw the shear force diagram of beam:

Draw the shear force diagram (SFD) & bending moment diagram (BMD) for the beam illustrated in Figure .

Solution

Reaction at the support A, R_A = + 4 + (1.5 × 6) = + 13 kN.

Shear Force

SF at A, F_A = + 13 kN (considering right side)

SF at C, F_C = + 13 - 1.5 × 6

                      = + 4 kN

SF just left of D = + 13 - 1.5 × 6

                          = + 4 kN

SF just right of D = + 4 - 4 = 0

Figure

Bending Moment

BM at B, M_B = 0  (As, the moment at the free end is equal to zero)

BM at D, M_D = 0 (letting right side of D, there is no loading on BD)

BM at C, M_C = - 4 × 2 = - 8 kN m (- ve sign because of right anticlockwise)

BM at A, MA = - 4 × 8 - 1.5 × 6 ×(6/2) = - 59 kN m.