Let Suppose reaction at support A and B be, RA and RB first find the support reaction. Due to symmetry, RA = RB = 6000/2 = 3000N ...(1)
Calculation for the Shear force Diagram
Draw the section line, here total 2-section line, which break the point A,D and Point D,B
Let 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
Consider section 1-1
Forces on left of section 1-1 is RA and UVL of 6000N/m over X1 m length,
Since Total load = 6000 = 1/2 X AB X CD
1/2 X 6 X CD = 6000, CD = 2000N ...(2)
First calculate the total load of UVL over length of X1 Consider triangle ADC and AFE
DC/EF = AD/AF
Since DC = 2000
2000/EF = 3/X1
EF = (2000X1)/3
Now load of triangle AEF
= 1/2 X EF × AF = (1/2 X 2000X1)/3 × (X1)
= 1000 -X12/3 a distance of X1/3 from F ...(3)
Parabola means a parabolic curve is formed, value of bending moment at both nearest point of the section is varies with X1 = 0 to X1 = 3
At X1 = 0 SFA = 3000N ...(4)
At X1 = 3
SFD = 0 ...(5)
Consider section 2-2
Forces on left of section 2-2 is RA and UVL of 2000N/m(At CD) and UVL over (X2 – 3) m length, First calculate the total load of UVL over length of (X2 – 3) Consider triangle CDB and BGH.
DC/GH = DB/BG
Since DC = 2000
2000/GH = 3/(6 - X2)
GH = 2000(6-X2)/3
Now load of triangle BGH = 1/2 X GH X BG
= [1/2 X 2000(6-X2)/3] X (6-X2)
= 1000(6 – X2)2/3, at a distance of X1/3 from F ...(6)
Load of CDB = 1/2 X 3 X 2000 = 3000
Now load of CDGH = load of CDB - load of BGH
= 3000 – 1000(6 – X2)2/3 ...(7)
SF2-2 = 3000 – 3000 – [3000 – 1000(6 – X2)2/3] (Parabola)
Parabola means a parabolic curve is formed, value of bending moment at both nearest point of the section is varies with X2 = 3 to X2 = 6
At X2 = 3 SFA = 0 ...(8)
At X2 = 6
SFD = –3000N ...(9)
Plot the SFD with the help of above value as shown in fig
Since SF change its sign at X2 = 3, that means at a distance of 3m from point A bending moment is maximum.
Calculation for the Bending moment Diagram
Consider section 1-1
BM1-1 = 3000X1 – [(1000X1 2)/3]X1/3 (Cubic)
Cubic means a parabolic curve is formed, value of bending moment at both nearest point of the section is varies with X1 = 0 to X1 = 3
At X1 = 0
BMA = 0 ...(10)
At X1 = 3 BMD = 6000 ...(11)
Consider section 2-2
Point of CG of any trapezium is = h/3[(b + 2a)/(a + b)]
i.e. Distance of C.G of the trapezium CDGH is given by
= 1/3 X DG X [(GH + 2CD)/(GH + CD)]
= 1/3.(X2-3).{[2000(6-X2)/3] + 2 X 2000)}/{[ 2000(6-X2)/3]+[ 2000]}
= {(X2 – 3)(12 – X2)}/{3(9 – X2)} ...(12)
BM2-2 = 3000X2-3000(X2-2)-[3000-1000(6 – X2) 2/3]{+ (X2 – 3)(12 – X2)}/{3(9 – X2)}
(Equation of Parabola) Parabola means a parabolic curve is formed, value of bending moment at both nearest point of the section is varies with X2 = 3 to X2 = 6
At X2 = 3 BMA = 6000 ...(13)
At X2 = 6
BMD = oN ...(14)
Plot the BMD with the help of above value.
