Total force acting are 20KN, 40KN and 60KN (UDL),
Hence resultant of the system = √(∑H)2 + ∑H)2
∑H = 0 and ∑V = 20 + 40 + 60 = 120KN
R = 120KN
Here total two-section line, which cut AB, AC
Distance of section 1-1 from point A is X1
Distance of section 2-2 from point A is X2
Consider left portion of the beam
*.F. Calculations:
*.F.1-1 = – 20 – 20.X1 (Equation of inclined line)
At X1 = 0
SFA = –20KN
At X1 = 1
SFB = –40KN
*.F.2-2 = –20 – 40 – 20X2
At X2 = 1
SFB = –80KN
At X2 = 3
SFC = –120KN
Plot the SFD with the help of above value
B.M. Calculations:
B.M.1–1 = –20X1 – 20.X1(X1/2) (Equation of Parabola)
At X1 = 0
BMA = 0
At X1 = 1
BMB = –30KN-m
BM2-2 = –20X2 – 40(X2-1) – 20X2(X2/2)
At X2 = 1
BMB = –30KN-m
At X2 = 3
SFC = –230KN-m
Plot the BMD with the help of above value.
