Total load = Area of the load diagram ABEC
= Rectangle ABED + Triangle DEC
= (AB X BE) + (1/2 X DE X DC) = (9 X 125) + [1/2 X 9 X (250-125)]
= 1125N + 562.5N
Centroid of the load of 1125N(rectangular load) is at a distance of 9/2 = 4.5m from AD and the centroid of the load of 562.5N (Triangular load) is at a distance of 1/3 X DE = 1/3 X 9 = 3m from point A.
Let support reaction at A and B be RA and RB. For finding the support reaction, Taking moment about point A
1125 X 4.5 + 562.5 X 3 - RB X 9 = 0
RB = 750N ...(2)
Now,
RV = 0 RA + RB = 1125 + 562.5 = 1687.5
RA = 937.5N ...(3)
Calculation for the Shear force Diagram
Draw the section line, here total 1-section line, which break the point A and B
Let
Distance of section 1-1 from point B is X
Consider right portion of the beam
Consider section 1-1
Forces on right of section 1-1 is RB and Load of PBEF and Load of EFH
SF1-1 = RB - load on the area PBEF - load on the area EFH
= RB - X.125 - 1/2.X.FH
In the equiangular triangles DEC and FEH
DC/DE = FH/FE or, 125/9 = FH/X
FH = 125X/9
*.F. between B and A = 750 - 125X - 125X2/18 (Equation of Parabola)
Parabola means a parabolic curve is formed, value of bending moment at both nearest point of the section is varies with X = 0 to X = 9
At X = 0 SFB = 750N ...(4)
At X = 9
SFA = –937.5N
Since the value of SF changes its sign, which is between the point A and B we get max. BM For the point of zero shear,
750 – 125X – 125X2/18 = 0
On solving we get, X = 4.75m
That is BM is max. at X = 4.75 from point B
Calculation for the Bending moment Diagram
Consider section 1-1
BM1–1 = 750X – PB.BE.X/2 – 1/2.FE.FH.1/3.FE
= 750X – X.125.(X/2) – 1/2.X.(125 X /9)(X/3)
= 750x – 125 x 2/2 – 125X2/54 (Equation of Parabola)
Parabola means a parabolic curve is formed, value of bending moment at both nearest point of the section is varies with X = 0 to X = 9
At X = 0
BMB = 0 ...(6)
At X = 4.75
BMmax = 1904N-m ...(7)
At X = 9
BMA = 0 ...(8)
