Q. A simply supported beam of span 7 m, carries a concentrated load of 70 kN at a distance of 3 m from right support. Draw S.F and BM diagrams.

SOLUTION:

Total load= 70kN

Taking moments about A
RB × 7 = 70 × 4
RB = (70 × 4)/7 = 40kN

RA = Total load – RB
= 70 – 40
= 30kN

SFD (SHEAR FORCE DIAGRAM)
FA = 30kN (Same in AC)
FC (just left of C)= 30kN
FC (just right of C) = – 40kN
FB (just left of B) = – 40
FB (just right of B) = 0 kN
Max + ve SF occurs at A = 30kN (in AC)
Max – ve SF occurs at B = – 40kN (in CB)

BMD (BENDING MOMENT DIAGRAM)

MA = MB = 0
M C = M max =30 × 4= 120 kNm
Max + ve BM occurs at C = 120kNm