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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 XD in Figure 1).


Figure 1 - Beam structure to analyze


Solution:

Calculate the support reactions using statics.
Positive moment is in the clockwise direction

Sum Moment and Shear Equations

Sum Moment and Shear Equations

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

Sum Moment and Shear Equations

The resulting system,

Frame with reactions
Figure 2 - Support reactions due to applied loads


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

The final moment diagram under the applied loads is;

a diagram showing the final moment diagram
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,

20 k load
Figure 4 - Frame structure with 20 k load applied

The resulting moment diagram,

moment diagram due to 20 k load
Figure 5 - Moment diagram due to 20 k load

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

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

The resulting moment diagram,

moment diagram due to 1 k/ft load
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.;

parts moment diagram due to 1 k/ft load
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 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)

applied unit load
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

Sum Moment and Shear Equations

shear equations

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

Sum Moment and Shear Equations

The resulting system,


Figure 11 - Support reactions due to virtual unit load


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 the values - heights (hi) - 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 (hi) on the virtual moment diagram (m) at these locations.

Area No. Location from B Height (hi) (ft-k)
1. 7.5 ft -7.5
2. 5 ft -10
3. 15 ft -7.5
4. 10 ft -10

Locations of heights (hi);

location of heights
Figure 13 - Heights on virtual moment diagram


Integrate the equation , by using the visual integration approach.

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

Area No. Area (A) from M diagram (k-ft2) Height (hi) from m diagram (ft-k) Ai*hi (k2-ft3)/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 k2-ft3/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 in4, and Q = 1 k then;

equation


Contact Dr. Fouad Fanous for more information.