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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.)

A diagram showing a beam loaded with a concentrated force and applied moment
Figure 1 - Beam structure to analyze


Solution:

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

Equations

Check these reactions by summing the shear forces.

shear equation

The resulting system,

Loaded beam with reactions
Figure 2 - Beam structure with support reactions


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,

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

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

For 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,

For the 6 k applied load
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.

resultant moment
Figure 6 - Resultant moment diagram


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)

Beam with an applied unit force at point D
Figure 7 - Beam structure with virtual unit load applied


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

Sum moments about A and B.

Equations

Check these reactions by summing the vertical forces.

shear equations

The resulting system,

Beam diagram with reactions
Figure 8 - Support reactions due to virtual unit load


The moment diagram due to the virtual load.

 
Figure 9 - Moment diagram due to virtual unit load


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 X1, X2, X3 & X4 were determined previously)

Area No. Area/EI (k-ft2) Location of centroid
from support (ft)
1. 1/2x-56x20/EI=-560/EI X1 = 1/3x20 = 6.67
2. 2/3x20x100/EI=1333.33/EI X2 = 1/2x20 = 10
3. 1/2x20x-36/EI=-360/EI X3 = 1/3x20 = 13.33
4. 1/2x6x-36/EI=-108/EI X4 = 1/3x6   = 2

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 by using the equation given in the introduction,

 virtual work equation.


Figure 10 - Heights of virtual moment diagram

The heights (hi) are shown in the figure above at the locations of the centroids of the corresponding areas from the moment diagrams (M).


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 (ft2-k)
Height (h)
from m diagram (ft-k)

Ai*hi (k2-ft3)

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 k2-ft3/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 in4, and Q = 1 k, then

equation


Contact Dr. Fouad Fanous for more information.