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

*Deflections - Method of Virtual Work*

Vertical Deflection of a Beam - Superposition

The following example illustrates the steps to be followed to calculate deflections of statically determinate structures using superposition and the method of virtual work.

**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. **

problem statement

Determine the vertical displacement at end *C* of the beam shown in the figure
below. The modulus of elasticity (E) and the moment of inertia (I) are constant for the
entire beam. *(This problem is identical to the Vertical
Deflection of a Beam - Cantilever example, except that the moment diagrams are
developed using the method of superposition.)*

Figure 1 - Beam structure to analyze

Solution:

- calculate the support reactions

Calculate the support reactions (caused by the applied "real" loads) using the following relationships.

Check these reactions by summing the shear forces.

The resulting system,

Figure 2 - Beam structure with support reactions

- draw moment diagram
*(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.

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, Q

Apply the virtual load at the point of interest in the desired direction. In this case,
apply a unit load at point *C* in the vertical direction. (see figure below)

Figure 7 - Beam structure with virtual unit load applied

- solve the support reactions due to the virtual load, Q

Following the same procedure used previously, calculate the support reactions (caused by the virtual load).

Sum moments about A and B.

Check these reactions by summing the vertical forces.

The resulting system,

Figure 8 - Support reactions due to virtual unit load

- draw virtual moment diagram
*(m)*

The moment diagram due to the virtual load.

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. *(the
locations of X _{1}, X_{2}, X_{3} & X_{4} were
determined previously)*

Area No. | Area/EI (k-ft^{2}) |
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,

.

Figure 10 - Heights of virtual moment diagram

The heights (h_{i}) are shown in the figure above at the locations of the
centroids of the corresponding areas from the moment diagrams (*M).*

- integrate

Integrate the equation , by using the visual integration approach.

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

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

1. | -560/EI | -2 | 1120/EI |

2. | 1333.33/EI | -3 | -4000/EI |

3. | -360/EI | -4 | 1440/EI |

4. | -108/EI | -4 | 432/EI |

Total | -1008/EI |

Since EI is constant throughout the structure, the total deflection at *C* equals **-1008 k ^{2}-ft^{3}/EI**.

The negative sign indicates that the displacement is opposite to the direction of the
unit load that was applied at *C* - therefore the deflection is upward.

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 k, then

Contact Dr. Fouad Fanous for more information.