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 20 X 1.5 = 30KN, acting at mid point of UDL i.e. 0.75m from point A. For that,
∑V = 0
RA + RB - 30 - 20 = 0, RA + RB = 50 ...(1)
Taking moment about point A,
∑MA = 0
30 X 0.75 + 30 + 20 X 3 – RB X 4 = 0
RB = 28.125 KN ...(2)
From equation (1), RA = 21.875KN ...(3)
Calculation for the Shear force Diagram
Draw the section line, here total 4-section line, which break the
load RA and UDL (Between Point A and E),
30KN/m and 20KN (Between Point E and D),
30KN/M and 20KN (Between Point D and C) and
20KN and RB (Between Point C and B)
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
Distance of section 4-4 from point A is X4
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 = 21.875 -20X1 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 = 1.5
At X1 = 0
SFA = 21.875 ...(4)
At X1 = 1.5 SFE = –8.125 ...(5)
i.e. inclined line 21.875 to – 8.125
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. 21.875 -20X1 = 0; X1 = 1.09375m, i.e. at 1.09375m from point A bending moment is maximum. Consider section 2-2.
Forces on left of section 2-2 is RA & 30KN
SF2-2 = 21.875 – 30 = – 8.125KN (constant value)
Constant value means value of shear force at both nearest point of the section is equal i.e.
SFE = SFD = – 8.125KN ...(6)
Consider section 3-3
Forces on left of section 3-3 is RA & 30KN, since forces are equal that of section 2-2,
so the value of shear force at section 3-3 will be equal that of section 2-2
SF3-3 = 21.875 – 30 = – 8.125KN (constant value)
Constant value means value of shear force at both nearest point of the section is equal i.e. SFD = SFC = – 8.125KN ...(7)
Consider section 4-4
Forces on left of section 4-4 is RA, 30KN, 20KN
SF4-4 = 21.875 – 30 -20 = -28.125KN (constant value)
Constant value means value of shear force at both nearest point of the section is equal
i.e. SFC = SFB = –28.125KN ...(8)
Plot the SFD with the help of above shear force values.
Calculation for the Bending moment Diagram
Consider left portion of the beam Consider section 1-1,
taking moment about section 1-1
BM1-1 = 21.875X1 -20X1(X1/2)
It is Equation of Parabola (Y = mX2 + C), 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 = 1.5
At X1 = 0 BMA = 0 ...(9) At X1 = 1.5 BMC = 10.3125
But B.M. is maximum at X1 = 1.09,
which **** between X1 = 0 to X1 = 1.5 So we also find the value of BM at X1 = 1.09
At X1 = 1.09
BMmax = 11.8 ...(11)
i.e. curve makes with in 0 to 11.8 to 10.3125 region.
Consider section 2-2,taking moment about section 2-2
BM2-2 = 21.875X2 – 30(X2 – 0.75)
= –8.125.X2 + 22.5 It is
Equation of straight line (Y = mX + C), inclined linear.
Inclined linear means value of bending moment at both nearest point of the section is varies with
X2 = 1.5 to X2 = 2 At
X2 = 1.5 BME = 10.3125 ...(12)
At X2 = 2 BMD = 6.25 ...(13)
i.e. inclined line 10.3125 to 6.25
Consider section 3-3, taking moment about section 3-3
BM3-3 = 21.875X3 – 30(X3 – 0.75) + 30
= –8.125.X2 + 52.5
It is Equation of straight line (Y = mX + C), inclined linear. Inclined linear means value of bending moment at both nearest point of the section is varies with
X3 = 2 to X3 = 3 At X3 = 2
BMD = 36.25 ...(14)
At X3 = 3
BMC = 28.125 ...(15)
Consider section 4-4, taking moment about section 4-4
BM4-4 = 21.875X4 – 30(X4 – 0.75) + 30 – 20 (X4 – 3)
= –28.125.X4 + 112.5
It is Equation of straight line (Y = mX + C), inclined linear. Inclined linear means value of bending moment at both nearest point of the section is varies with
X4 = 3 to X4 = 4
At X4 = 3
BMC = 28.125
At X4 = 4
BMB = 0
i.e. inclined line 28.125 to 0 Plot the
BMD with the help of above bending moment values.
