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
- 0.3x^{2 } + 1.8x + 12.1 = 0
or 0.3x^{2} - 1.8x - 12.1 = 0
x^{2} - 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