Q. A simply supported beam of span 4 m carries a udl of 2 kN/m over its left half of the span. Draw SFD and BMD.

SOLUTION:

Total load= 2×2=4 kN
To find reactions
Taking moments about A
RB×4=(2×2)2/2=4
RB=4/4 = 1kN
RA = Total load – RB = 4 – 1 = 3kN

SFD ( SHEAR FORCE DIAGRAM)
FA= 3kN
FC = -1 kN
FB = -1kN
Max S.F occurs at support A = F{max} = 3kN

BMD (BENDING MOMENT DIAGRAM)
M{A}=M{B}=0
M{C}= 1×2 = 2kNm
max BM occurs at D, where SF changes its sign
x/3 = (2 – x)/1
x = 6 – 3 x ;4 x = 6;
x = 6/4 = 1.5m
MD = Mmax = 3 × 1.5 – ( 2 × 1.5) 1.5/2
= 4.5 – 2.25 = 2.25 kNm