Example of shear force and bending moment diagram, Mechanical Engineering

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

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

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

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

Plot the BMD with the help of obtained bending moment values.

Posted Date: 10/20/2012 1:38:18 AM | Location : United States







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