Q. A simply supported beam of span 5 m carries a udl of 3 kN/m over its left half of the span and also a point load of 10 kN, 1 m from right end. Draw SFD.

SOLUTION:

Total load = 3 * 2.5 + 10 = 17.5kN
Taking moments about A
R{B} * 5 = 10 * 4 + (3 * 2.5) 2.5/2 = 49.375
R{B} = 49.375/5 = 9.875kN
R{A} = 17.5 – 9.875 = 7.625kN
SFD ( SHEAR FORCE DIAGRAM)
F{A} = 7.625kN
F{B} = – 9.875kN
F{C} = 0.125kN
F{D} = 0.125kN
Max +veSF occurs at A = 7.625kN
Max -veSF occurs at B = – 9.875kN (Same in DB)