Total load = 20 * 8 = 160 kN
To find the reactions
Taking moments about A
R{B} * 6 = (20 * 8) * 8/2 = 640
R{B} = 640/6 = 106.67kN
R{A} = 160 – 106.67 = 53.33kN
SFD (SHEAR FORCE DIAGRAM)
F{A} = 53.33kN
FB =just left ofB = – 66.67kN
FB =(just right ofB )=40 kN
F{c} = 0
Max +veSF occurs at A = 53.33kN
Max – ve SF occurs at B = 66.67kN and
BMD ( BENDING MOMENT DIAGRAM)
M{A} = M{C} = 0
M{B} = – (20 * 2) * 2/2 = – 40kNm
To find x
x/53.33 = (6 – x)/66.67
66.67x =53.33 * 6 – 53.33 * x
120x = 53.33 * 6
x = (53.33 * 6)/120 =2.67 m from A
Mmax =M 2.67 = 53.33 * 2.67 – (20 * 2.67)* 2.67/ 2 = 71.1021 kNm To find point of contraflexure
Find BM at xx and equate it to zero
53.33 a – (20×a) * a/2 = 0
a (53.33 – 10a) = 0
10a = 53.33
a = 53.33/10 = 5.3333m
point of contra flexure occurs at 5.3333 m from support A.
