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.