Let reaction at support A be RAV, RAH and M(anti clock wise), First find the support reaction For that,
∑V = 0
RAV –2 –3 –3 = 0, RAV = 8 ...(1)
Taking moment about point A,
∑MA = 0
–M + 2 X 1 + 3 X 3 + 3 X 5 = 0
M = 26KNm ...(2)
∑H = 0
RAH = 0 ...(3)
Calculation for the Shear force Diagram
Draw the section line, here total 3 section line, which break the load RAV and 2KN(Between Point A and B), 2KN and 3KN(Between Point B and C),
3KN and 3KN (Between Point C and D).
Consider left portion of the beam
Consider section 1-1
Force on left of section 1-1 is RAV
SF1-1 = 8KN (constant value)
Constant value means value of shear force at both nearest point of the section is equal i.e.
SFA = SFB = 8KN ...(4)
Consider section 2-2
Forces on left of section 2-2 is RAV & 2KN
SF2-2 = 8 – 2 = 6KN (constant value)
Constant value means value of shear force at both nearest point of the section is equal i.e.
SFB = SFC = 6KN ...(5)
Consider section 3-3
Forces on left of section 3-3 is RA, 2KN, 3KN
SF3-3 = 8 – 2 – 3 = 3KN (constant value)
Constant value means value of shear force at both nearest point of the section is equal i.e.
SFC = SFD = 3KN ...(6)
Calculation for the Bending moment
Diagram 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 section 1-1, taking moment about section 1-1
BM1-1 = 8.X1
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 X1 = 0
to X1 = 1
At X1 = 0
BMA = 0 ...(8)
At X1 = 1
BMB = 8 ...(9)
i.e. inclined line 0 to 8 Consider section 2-2,taking moment about section 2-2
BM2-2 = 8.X2 – 2.(X2 – 1)
= 6.X2 + 2
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 to X2 = 3
At X2 = 1
BMB = 8 ...(10)
At X2 = 3
BMC = 20 ...(11)
i.e. inclined line 8 to 20
Consider section 3-3, taking moment about section 3-3
BM3-3 = 8.X3 – 2.(X3 – 1) – 3.(X3 – 3)
= 3.X3 + 11
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 = 3 to X3 = 5
At X3 = 3 BMC = 20 ...(12)
At X3 = 5 BMD = 26 ...(13)
i.e. inclined line 20 to 26 Plot the BMD with the help of above bending moment values.
The SFD and BMD is shown in fig,
