Let reaction at support A and B be, RA and RB First find the support reaction. For finding the support reaction, convert UDL in to point load and equal to 2 X 5 = 10KN, acting at mid point of UDL i.e. 2.5m from point A.
For that,
∑V = 0
RA + RB - 5.5 - 10 -2 = 0, RA + RB = 17.5 ...(1)
Taking moment about point A,
∑MA = 0
10 X 2.5 + 5.5 X 2 - RB X 5 + 2 X 7 = 0
RB = 10 KN ...(2)
From equation (1), RA = 7.5 KN ...(3)
Calculation for the Shear force Diagram
Draw the section line, here total 3-section line, which break the load RA, 5.5KN (Between Point A and E), 5.5KN and UDL (Between Point E and B),
Point B and 2KN (Between Point B and C).
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 RA and UDL (from point A to the section line i.e. UDL on total distance of X1
SF1-1 = 7.5 -2X1 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 X1 = 0 SFA = 7.5 ...(4)
At X1 = 2 SFE = 3.5 ...(5)
i.e. inclined line 7.5 to 3.5 Consider section 2-2
Forces on left of section 2-2 is RA, 5.5KN and UDL on X2 length
SF2-2 = 7.5 - 5.5 - 2X2 = 2 - 2X2 (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 X2 = 2 to X2 = 5
At X2 = 2 SFE = –2 ...(4)
At X2 = 5 SFB = –8 ...(5)
i.e. inclined line -2 to -8
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. 2 - 2X2; X2 = 1m, i.e. at 1m from point A bending moment is maximum.
Consider section 3-3
Forces on left of section 3-3 is RA, 5.5KN and 10KN and RB
SF3-3 = 7.5 - 5.5 -10 +10 = 2KN (constant value)
Constant value means value of shear force at both nearest point of the section is equal
i.e. SFB = SFC = 2KN ...(7)
Plot the SFD with the help of above shear force values.
