Q.Draw S.FD and B.M.D for the following beam and also calculate the position of the point of contra flexure.

SOLUTION:
Total load = 30×11+80= 330 + 80 = 410
To find reactions:
Taking  moments about A:
RB * 9 = (30 * 11) * 11/2 + 80 * 11 = 1815 + 880 = 2695
R{B} = 2695/9 = 299.44kN
R{A} = 410 – 299.44 = 110.56kN
To find x:
x/110.56 = (9 – x)/159.44
159.44x= (110.56 * 9) -110.56* x
x = 3.69m
BMD: ( BENDING MOMENT DIAGRAM)
M{A} = M{C} = 0
M max (-ve)=M B
= – 80 * 2 – (30 * 2) * 2/2 = – 220kNm
M 4.5 =M D =110.56 * 4.5-30 * 4.5 * 4.5 /2 = 193.77 kNm
M max (+ve)=M max =M 3.69 m
= (110.56 * 3.69) – (30 * 3.69) * 3.69/2
M max =203.73 kNm
M{E} = 0 = 110.56a – (30*a) * a/2
= 110.56a – 15a²
15a² = 110.56a
a = 110.56/15 = 7.37m
… Point of contraflexure occurs where B.M changes its sign= at E= 7.37 m from left support