Since hinged at point A and D, suppose reaction at support A and D be, RAH, RAV and RDH, RDV first find the support reaction. For finding the support reaction, convert UDL and UVL in to
point load and, Point load of UDL equal to 10 X 2 = 20KN, acting at mid point of UDL i.e. 1m from point A.
Point load of UVL equal to 1/2 X 20 X 2 = 20KN, acting at a distance 1/3 of total distance i.e. 1/3m from point D.
For that,
∑V = 0
RAV + RDV – 20 – 20 = 0, RA + RB = 40 ...(1)
Taking moment about point A,
∑MA = 0
20 X 1 + 20 X 5.33 – RDV X 6 = 0
RDV = 21.1 KN
From equation (1), RAV = 18.9KN ...(3)
Calculation for the Shear force Diagram
Draw the section line, here total 3-section line, which break the
load RAV and UDL (Between Point A and B),
No load (Between Point B and C) and UVL (Between Point C and D).
Let
Distance of section 1-1 from point A is X1
Distance of section 2-2 from point A is X2
Distance of section 3-3 from point A is X3
Consider left portion of the beam Consider section 1-1
Force on left of section 1-1 is RAV and UDL (from point A to the section line i.e. UDL on total distance of X1
SF1-1 = 18.9 -10X1 KN (Equation of straight line)
It is Equation of straight line (Y = mX + C), inclined linear
Inclined linear means value of shear force at both nearest point of the section is varies with X1 = 0 to X1 = 2
At X = 0
SFA = 18.9 ...(4)
At X1 = 2
SFB = –1.1 ...(5)
i.e. inclined line 18.9 to - 1.1
Since here shear force changes the sign so at any point shear force will be zero and at that point bending moment is maximum.
For finding the position of zero shear force equate the shear force equation to zero, i.e. 18.9 –10X1 = 0; X1 = 1.89m, i.e. at 1.89m from point A bending moment is maximum. Consider section 2-2.
Forces on left of section 2-2 is RAV & 20KN
SF2-2 = 18.9 – 20 = – 1.1KN (constant value)
Constant value means value of shear force at both nearest point of the section is equal i.e.
SFB = SFC = – 1.1KN ...(6)
Consider section 3-3 Forces on left of section 3-3 is RAV & 20KN and UVL of 20KN/m over (X3 – 4) m length,
First calculate the total load of UVL over length of (X3 – 4)
Consider triangle CDE and CGF
DE/GF = CD/CG
Since DE = 20
20/GF = 2/(X3 – 4) GF = 10(X3 – 4)
Now load of triangle CGF = 1/2 X CG X GF = 1/2 X (X3 – 4) X 10(X3 – 4)
