Maximum Bending Moment:

The SF changes sign at A, among A and B and at B.

Let a section XX at a distance x from the end C as illustrated in Figure

SF at section XX, F _x  =+ 30 - 15.2 - ((½) ( x - 3) × 0.6 × ( x - 3) )

Equating this equation to zero, 14.8 - 0.3 ( x - 3)²  = 0

- 0.3x²  + 1.8x + 12.1 = 0

or         0.3x²  - 1.8x - 12.1 = 0

              x²  - 6 x - 40.333 = 0

On solving out by trial & error, we obtain x = 10 m.

M max =+ (30 × 7) + 24 - (15.2 × 10) - (( ½) × 7 × 7 × (1/3) × 7 × 0.6 ?

= + 47.7 kN-m

Maximum positive bending moment = + 47.7 kN-m

Maximum negative bending moment = - 21.6 kN-m