Q. A horizontal beam of 12 m long simply supported at its end, it is subjected to vertical loads of 10 kN, 20 kN and 25 kN at 3 m, 7 m and 10 m from left support respectively. Draw shear force and bending moment diagrams indicating values at salient points.

SOLUTION:
Total load = 10 + 20 + 25 = 55kN
To find reactions
Taking moments about A
RB × 12 = 10 × 3 + 20 × 7 + 25 × 10
= 30 + 140 + 250 = 420
RB = 420/12 = 35kN
RA =Total load – RB = 55 – 35 = 20kN

SFD (SHEAR FORCE DIAGRAM)
FA = +20 kN(same inAC )
FC = +20 kN (just left of C) (same in CD)
FC = +10 kN (just right of C)
FD = + 10 kN (Just left of D
F D = – 10kN (just right of D) (Same in DE)
F E = – 10kN (just left of E)
FE = – 35 kN (just right of E) (same in EB)
FB = – 35kN
Max + ve SFD occurs at A = 20kN and
Max – ve SFD occurs at B = – 35kN

BMD (BENDING MOMENT DIAGRAM)
M{A} = M{B} = 0
M{C} = 20 × 3 = 60kNm
M{D} = 20 × 7 – 10 × 4 = 100kNm
M{E} = 35 × 2 = 70kNm
Max BM occurs at D=M max =100 kNm