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Deflections - Method of Virtual Work
Horizontal Deflection of a Frame - Cantilever


The following example utilizes the cantilever 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 cantilever 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 the cantilever method can be used.

In the cantilever method, a point on the structure is selected where a single fixed support can be used to replace all the degrees of freedom in the structure. This method is only useful in statically determinate structures.

In this example, point B is selected and a fixed support is inserted at this location.

The loaded system with a fixed support at point B.

a diagram showing the fixed structure
Figure 4 - Frame structure fixed at B

The resulting moment diagram can be calculated by solving for each applied load separately and adding the results. This can be done by using superposition of the loads on the members acting as cantilevers off of support B.

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

i) Moment diagram due to the 1 k/ft distributed load,

 
Figure 5 - Moment diagram due to 1 k/ft load
Area: 1/3x(Base)x(Height) = 1/3 x 15 ft x -112.5 ft-k = -562.5 ft2-k
Location of centroid: 1/4x(Base) = 1/4 x 15 ft = 3.75 ft from support B

ii) Moment diagram due to the 15 k support reaction at A,


Figure 6 - Moment diagram due to 15 k support reaction

Area: 1/2x(Base)x(Height) = 1/2 x 15 ft x 225 ft-k = 1687.5 ft2-k
Location of centroid: 1/3x(Base) = 1/3 x 15 ft = 5 ft from support B

iii) Moment diagram due to the 20 k concentrated load at C,


Figure 7 - Moment diagram due to 20 k applied load
Area: 1/2x(Base)x(Height) = 1/2 x 15 ft x -300 ft-k = -2250 ft2-k
Location of centroid: 1/3x(Base) = 1/3 x 15 ft = 5 ft from support B

iv) Moment diagram due to the 13.75 k support reaction at D,


Figure 8 - Moment diagram due to 13.75 k support reaction
Area: 1/2x(Base)x(Height) = 1/2 x 30 ft x 412.5 ft-k = 6187.5 ft2-k
Location of centroid: 1/3x(Base) = 1/3 x 30 ft = 10 ft from support B

The sum of these four diagrams equals the total resultant for the structure.


Figure 9 - Resultant moment diagram

Notice that the answer at both sides of joint B is equal, i.e.;

225-112.5 = 112.5 ft-k
412.5-300 = 112.5 ft-k


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 structure 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

note: the value of XA is negative because the resulting force acts in the opposite direction to how it is drawn in Figure 10, above.

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 unit load


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

The cantilever structure;

 
Figure 12 - Cantilever structure with loads and reactions applied

i) Moment diagram due to the reaction load at A,

 
Figure 13 - Moment diagram due to reaction loads at A

ii) Moment diagram due to the reaction load at D,

 
Figure 14 - Moment diagram due to reaction loads at D

Again, 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 the 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. 3.75 ft -11.25
2. 5 ft -10
3. 5 ft -12.5
4. 10 ft -10

Locations of heights (hi);

location of heights
Figure 15 - 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 of the virtual moment diagram and add them together.

Area No. Area (A) from M diagram (k-ft2)/EI Height (hi) from m diagram (k-ft) Ai*hi (k2-ft3)/EI
1. -562.5 -11.25 6328.125/EI
2. 1687.5 -10 -16875/EI
3. -2250 -12.5 28125/EI
4. 6187.5 -10 -61875/EI
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.