Q. A simply supported beam 6m span is carrying a udl of 4kN / m over its right half of its span and a point load of 10 kN at mid span. Draw SF and BM diagrams and find value of maximum hending moment.
Total load= 10 + 4 * 3 = 22kN
To find the reactions
Taking moments about A
R{B} * 6 =10 * 3 + (4 * 3) (3/2+3) = 84
R_{B} = 84/6 = 14kN
R{A} = 22 – 14 = 8kN
SFD SHEAR FORCE DIAGRAM
F{A} = 8kN (same in AC)
Fc (just left of c )=8 kN
F{C} = (just right of c)= – 2kN
F{B} = – 14kN
Max +veSF occurs at A = 8kN (Same in AC) and Max -veSF occurs at B = – 14 kN
BMD ( BENDING MOMENT DIAGRAM)
M_{A} = M_{B} = 0
M_{c} = 8 * 3 = 24kNm
Max B.M occurs at C (where SF changes its sign)= 24kNm
