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

*Deflections - Method of Virtual Work*

Horizontal Deflection of a Frame - Superposition

The following example utilizes the superposition method to determine the deflection of a frame.

**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 horizontal displacement at *D* of the frame shown (indicated by X_{D}
in Figure 1).

Figure 1 - Beam structure to analyze

Solution:

- calculate the support reactions

Calculate the support reactions using statics.

*Positive moment is in the clockwise direction*

Check these reactions by summing the moments at any point on the structure.

The resulting system,

Figure 2 - Support reactions due to applied loads

- determine moment diagrams
*(M)*for the structure under applied loads

Construct the moment diagrams caused by the applied loads utilizing the superposition method.

The final moment diagram under the applied loads is;

Figure 3 - Moment diagram due to applied loads

This diagram can be determined by using statics, or a simplified method using superposition can be used.

To get the final moment diagram by superposition, start by selecting one of the applied loads, determine the support reactions due to the load and drawing the resulting moment diagram.

i) for the 20 k applied load at *C*,

Figure 4 - Frame structure with 20 k load applied

The resulting moment diagram,

Figure 5 - Moment diagram due to 20 k load

ii) for the 1 k/ft applied load between *AB*,

Figure 6 - Frame structure with 1 k/ft load applied

The resulting moment diagram,

Figure 7 - Moment diagram due to 1 k/ft load

This diagram can be simplified by drawing the effects of each load separately, i.e.;

Figure 8 - Moment diagram due to 1 k/ft load (by parts)

The sum of these diagrams are equal the total resultant for the structure.

Figure 9 - Resultant moment diagram

- apply virtual load and solve support reactions

Apply the virtual load at the point of interest in the desired direction. In this case,
we want to know the horizontal deflection at point *D*. Therefore, apply a unit load
at point *D* in the horizontal direction. (see figure 3 below)

Figure 10 - Frame stucture with unit load applied

Following the same procedure used for the loaded structure, calculate the support
reactions (caused by the virutal load).

*Positive moment is in the clockwise direction*

Check these reactions by summing the moments at any point on the structure.

The resulting system,

Figure 11 - Support reactions due to virtual unit load

- draw virtual moment diagram
*(m)*

Determine the moment diagram due to the virtual load using the same procedure used to draw the "real" moment diagram.

i) Moment diagram due to the unit load at *D*,

Figure 12 - Moment diagram due to virtual unit load

Notice that the moments on both sides of joint *B* are equal.

- 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 to determine the deflection.

Using this locations of the centroids, determined previously, determine the heights (h_{i})
on the virtual moment diagram (*m*) at these locations.

Area No. | Location from B |
Height (h_{i}) (ft-k) |

1. | 7.5 ft | -7.5 |

2. | 5 ft | -10 |

3. | 15 ft | -7.5 |

4. | 10 ft | -10 |

Locations of heights (h_{i});

Figure 13 - Heights on virtual moment diagram

- integrate

Integrate the equation , by using the visual integration approach.

Multiply the areas of the "real" moment diagram by the heights (h_{i})
of the virtual moment diagram and add them together.

Area No. | Area (A) from M diagram (k-ft^{2}) |
Height (h_{i}) from m diagram (ft-k) |
A_{i}*h_{i} (k^{2}-ft^{3})/EI |

1. | 281.25 | -7.5 | -2109.375 |

2. | 843.75 | -10 | -8437.5 |

3. | 2250 | -7.5 | -16875 |

4. | 1687.5 | -10 | -16875 |

Total |
-44296.875/EI |

Since EI is constant throughout the structure, the total horizontal deflection at *D*
equals **-44296.875 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 *D* - therefore the deflection is to the right.

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.