Example of Shear force and bending moment diagram:
Q: Draw the Shear Force and bending moment diagram for the simply supported beam loaded as shown in the figure given below.
![104_Example of Shear force and bending moment diagram1.png](https://www.expertsmind.com/CMSImages/104_Example of Shear force and bending moment diagram1.png)
![1080_Example of Shear force and bending moment diagram.png](https://www.expertsmind.com/CMSImages/1080_Example of Shear force and bending moment diagram.png)
Sol.: Let reaction at support A and B be, RA and RB First find support reaction
For that,
![1873_Example of Shear force and bending moment diagram2.png](https://www.expertsmind.com/CMSImages/1873_Example of Shear force and bending moment diagram2.png)
Taking moment about the point A,
ΣMA = 0
2 X 1 + 4 X 2 + 2 X 3 - RB X 4 = 0
RB = 4KN ...(2)
From equation (1), RA = 4KN ...(3)
Calculation for Shear force Diagram
Draw the section line, here total four section line, which break the load RA and 2KN(Between A and C),
2KN and 4KN(Between C and D),
4KN and 2KN (Between D and E) and
2KN and RB(Between E and B) Consider left portion of beam Take section 1-1
Force on left of section 1-1 is RA
SF1-1 = 4KN (constant value)
Constant value means value of shear force at both the nearest point of section is equal that is
SFA = SFC = 4KN ...(4)
Take section 2-2
Forces on left of section 2-2 is RA and 2KN
SF2-2 = 4 - 2 = 2KN (constant value)
Constant value means value of shear force at both the nearest point of section is equal that is
SFC = SFD = 2KN ...(5)
Take section 3-3
Forces on left of section 3-3 is RA, 2KN, 4KN
SF3-3 = 4 - 2 - 4 = -2KN (constant value)
The constant value means that value of shear force at both the nearest point of the section is equal that is
SFD = SFE = -2KN ...(6)
Take section 4-4
Forces on left of the section 4-4 is RA, 2KN, 4KN, 2KN
SF4-4 = 4 - 2 - 4 - 2 = - 4KN (constant value)
The constant value means that value of shear force at both the nearest point of section is equal that is
SFE = SFB = -4KN ...(7)
Plot the SFD with the help of above shear force values.
Calculation for bending moment Diagram
Let
The distance of section 1-1 from A is X1
The distance of section 2-2 from A is X2
The distance of section 3-3 from A is X3
The distance of section 4-4 from A is X4
Take left portion of the beam
Take section 1-1, taking moment about section 1-1
BM1-1 = 4.X1
It is the Equation of straight line (Y = mX + C), inclined linear.
Inclined linear means that the value of bending moment at both the nearest point of section is varies with
![1982_Example of Shear force and bending moment diagram3.png](https://www.expertsmind.com/CMSImages/1982_Example of Shear force and bending moment diagram3.png)
It is Equation of the straight line (Y = mX + C), inclined linear.
Inclined linear means that the value of bending moment at both the nearest point of section varies with
![2107_Example of Shear force and bending moment diagram4.png](https://www.expertsmind.com/CMSImages/2107_Example of Shear force and bending moment diagram4.png)
Plot the BMD with the help of obtained bending moment values.