**Compute the maximum bending moment:**

A simply supported beam of 7 m length carries point loads 2 kN, 4 kN and 6 kN at distances of 1 m, 2 m & 4 m from the fixed end respectively. Draw BMD & SFD and also compute the maximum bending moment that will take place.

**Solution**

By taking moments around A to find RB,

R_{B} × 7 - (6 × 4) - (4 × 2) - (2 × 1) = 0

Therefore, R_{B} = 4.857 kN, and

R_{A} = 12 - 4.857 = 7.143 kN

Shear Force (Starting from left end A)

SF at A, F_{A} = + 7.143 kN

SF just left of C = + 7.143 kN

SF just right of C = + 7.143 - 2 = + 5.143 kN

SF just left of D = + 5.143 kN

SF just right of D = + 5.143 - 4 = + 1.143 kN

SF just left of E = + 1.143 kN

SF just right of E = + 1.143 - 6 = - 4.857 kN

SF just left of E = - 4.857 kN = Reaction at B

**Bending Moment** (Starting from right end B) BM at A, MA = 0

BM at E, M_{E} = + 4.857 × 3 = + 14.571 kN m

BM at D, M_{D} = + (4.857 × 5) - (6 × 2) = + 12.285 kN m

BM at C, M_{C} = + (4.857 × 6) - (6 × 3) - (4 × 1) = + 7.142 kN m

Or M_{C} = 7.143 × 1 = + 7.143 kN m (considering left side)

BM at A, M_{A} = 0

**Figure **

Maximum bending moment shall occur at a point where the shear force changes sign. Here, SF alters from positive to negative at E. The Bending Moment at E shall be maximum bending moment.

Thus, M_{max} = + 14.571 kN m