Deflections - Method of Virtual Work | Index of Examples | CCE Homepage |

*Deflections - Method of Virtual Work*

Rotation of a Beam - Superposition

problem statement

Using the same structure as used in the Beam
Deflection examples, determine the rotation at *A* of the beam shown in the
figure below using the method of Superposition. The modulus of elasticity (E) and the
moment of inertia (I) are constant for the entire beam.

**Note: The colors of the loads and moments are used to help indicate the
contribution of each force to the deflection or rotation being calculated. The moment
diagrams show the moments induced by a load using the same color as the load. **

Figure 1 - Beam structure to analyze

- calculate the support reactions

Calculate the support reactions (caused by the applied loads) using the following relationships:

Check these reactions by summing the vertical forces.

The resulting system,

Figure 2 - Beam structure with support reactions

- draw moment diagrams
*(M)*for the structure under applied "real" loads

Using the method of superposition, draw a moment diagram for each separate load applied to the beam.

The resulting moment diagram can then be calculated by solving for each applied load separately and adding the results.

Note: The centroid of each area is indicated by the numbered arrow and dot.

i) Moment diagram due to the 56 ft-k concentrated moment at *A*,

Figure 3 - Moment diagram due to 56 ft-k moment

ii) Moment diagram due to the 2 k/ft applied load,

Figure 4 - Moment diagram due to 2 k/ft applied load

iii) Moment diagram due to the 6k applied load at end *C,*

Figure 5 - Moment diagram due to 6 k applied load

Notice that the resultant moment diagram is equal to the sum of these three diagrams.

Figure 6 - Resultant moment diagram

- apply virtual load

Apply the virtual load at the point of interest in the desired direction. In this case,
we want to know the rotation at point *A*. Therefore, apply a unit moment at point *A*
in the positive (clockwise) direction.

Figure 7 - Beam structure with virtual unit load applied

- solve the support reactions due to the virtual load

Following the same procedure used previously, calculate the support reactions (caused by the virtual load) using the following relationships:

Check these reactions by summing the vertical forces.

The resulting system,

Figure 8 - Reactions due to virtual unit load

- draw virtual moment diagram
*(m)*

Determine the moment diagram due to the virtual load using the same method as used to find the moment diagrams for the applied loads.

Moment diagram due to the virtual load using superposition.

Figure 9 - Moment diagram due to virtual unit load

- calculate areas and centroids

Once the "real" moment diagrams are determined, calculate the area enclosed by each moment diagram and determine the location of the centroid of each of these areas.

Area No. | Area/EI (k-ft^{2}/EI) |
Location of centroid from support (ft) |

1. | 1/2x-56x20/EI=-560/EI | X_{1} = 1/3x20 = 6.67 |

2. | 2/3x20x100/EI=1333.33/EI | X_{2} = 1/2x20 = 10 |

3. | 1/2x20x-36/EI=-360/EI | X_{3} = 1/3x20 = 13.33 |

4. | 1/2x6x-36/EI=-108/EI | X_{4} = 1/3x6 = 2 |

- determine heights of virtual moment diagram at centroids

Determine the values - heights (h_{i}) - on the virtual moment diagram (*m*)
at the centroids of the moments due to the real loads. This is needed to carry out the
integration by using the equation given in the introduction,

Heights and locations by superposition.

Figure 10 - Heights of virtual moment diagram

- integrate

Integrate the equation by using the visual integration approach.

Multiply the areas of the "real" moment diagram by the heights found in the virtual moment diagram and add them together.

Area No. | Area (a) from M diagram (k-ft ^{2}/EI) |
Height (h) from m diagram (ft-k) |
A |

1. | -560/EI | 2/3 | -373.33/EI |

2. | 1333.33/EI | 1/2 | 666.67/EI |

3. | -360/EI | 1/3 | -120/EI |

4. | -108/EI | 0 | 0/EI |

Total | 173.33/EI |

Since EI is constant throughout the structure, the total rotation at *A* equals **+173.33 k ^{2}-ft^{3}/EI**.

The positive sign indicates that the rotation is in the same direction as the unit
moment applied at *A* - therefore the rotation is in the clockwise direction.

If values of E and I are specified, the vertical deflection at C in inches can be
determined. For example, let E = 29,000 ksi, I = 144 in^{4}, and Q = 1 ft-k, then

Contact Dr. Fouad Fanous for more information.