Q. A simply supported beam of span 6 m subjected to a udl of 2 kN/m over the entire span. Find the support reactions.

SOLUTION:
Total load= 6 × 2 = 12kN
As the given loading is symmetrical
hence RA =RB = Total load / 2 = 12/2 = 6kN .

Q. A simply supported beam of length 6 m is carrying a udl 2 kN/m on complete length. Draw SFD and BMD.

SOLUTION:
Total load = 2 × 6 = 12kN Total load
As the load on the beam symmetrical
hence RA =RB = Total Load / 2 = 12/2 = 6kN

SFD (SHEAR FORCE DIAGRAM)
FA = 6kN
FB = – 6kN
FC = 0
max SF occurs at support A = 6 kN

BMD (BENDING MOMENT DIAGRAM)
M{A} = M{B} = 0
M{C} = 6 × 3 – (3 × 2) 3/2 = 18 – 9 = 9kN
M max = wl² / 8 = 2 × 6²/ 8 =9kNm
Max B.M occurs at C = Mmax = 9 kNm

Note: While drawing S.F.D. where udl is there draw slope line and in BMD where udl is there draw parabolic curve