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Deflections - Method of Virtual Work
Deflection of a Truss
The virtual work method can be used to determine the deflection of trusses. We
know from the principle of virtual work for trusses that the deflection can be calculated
by the equation 
with n equal to the virtual force in the member and 
equal to the
change in length of the member. Therefore, the deflection of a truss due to any
condition that causes a change in length of the members can be calculated. This
change in length can be caused by the applied loads acting on each member, temperature
changes, and by fabrication errors.
Axial Deformation:
From statics we know how to determine member forces in a truss by using either the method of joints or the method of sections. Once these forces are known we can determine the axial deformation of each member by using the equation:
The equation for the deflection can be modified with this value for .
where m is equal to the number of members, n is the force in the member due to the virtual load, N is the force in the member due to the applied load, L is the length, A is the area, and E represents Young's Modulus of Elasticity.
Temperature Changes:
The axial deformation of a truss member of length L due to a change in temperature of 
is given by:
where 
is
the coefficient of thermal expansion.
The equation for the deflection is then modified with this value for .
where j is the number of members experiencing temperature change and n is the force in the member due to the virtual load.
Fabrication Errors:
In the case of fabrication errors, the deformation of each member is known. Therefore, the original equation for deflection of a truss can be modified.
where k is the number of members undergoing fabrication errors and n is the force in
the member due to the virtual load and 
is the change in length of the member due to
fabrication errors.
The total deflection of a truss is made up of the sum of all of these cases.
This equation is now used to find the deflection of a truss. Please refer to an introductory text book on structural analysis for a complete description of this approach.
problem statement
Using the method of virtual work, determine the vertical deflection at joint G in the truss below, under the loading conditions show in figures i), ii), and iii).
The member properties are A=2 in2 and E=29x103 ksi.
The truss is subjected to the following applied loads:
i) 
Figure 1 - Truss structure to analyze
And the following fabrication errors are present:
ii) 
Figure 2 - Fabrication errors present
NOTE: TO PREVENT ERRORS, CALCULATE THE INFLUENCE OF EACH CASE INDEPENDENTLY AND THEN ADD THE RESULTS AT THE END
Figure 3 - Frame structure with applied loads
Calculate the support reactions (caused by the applied loads) by summing the moments about A and E: (answers in Kips)
Check these reactions by summing vertical and horizontal forces:
The resulting system,
Figure 4 - Support reactions due to applied loads
For equilibrium at joint A,
Figure 5 - Joint equilibrium at A
Sum vertical and horizontal forces to determine the force in each member, (Kips)
Remember that in the method of joints, a joint reaction is in the opposite direction to how the force acts on the member. Therefore, member AB is in compression.
Continue this method for each joint in the structure.
Truss diagram with internal forces due to applied loads,
Figure 6 - Truss member reactions
Apply the virtual load at the point of interest in the desired direction. In this case, we want to know the deflection at point G. Therefore, apply a unit load at point G.
Figure 7 - Truss with virtual unit load applied
Following the same procedure used previously, calculate the support reactions (caused by the virtual load).
The resulting system,
Figure 8 - Support reactions due to virtual unit load
Use the method of joints as illustrated in Step 2 to determine the member results due to the unit virtual load. Add the results to your existing table:
Truss diagram with internal forces due to virtual load,
Figure 9 - Internal forces due to virtual unit load
The deflection of the truss can now be determined by completing the equation:
For the case of Axial Deformation
| Member | n(k) | N(k) | L(in) | AE (in2-ksi) | nNL/AE (in-k) |
|---|---|---|---|---|---|
| AB | -0.67 | -33.33 | 48 | 58000 | 0.0184 |
| BC | -0.67 | -33.33 | 48 | 58000 | 0.0184 |
| CD | -0.67 | -46.66 | 48 | 58000 | 0.0257 |
| DE | -0.67 | -46.66 | 48 | 58000 | 0.0257 |
| AF | 0.83 | 41.67 | 60 | 58000 | 0.0359 |
| BF | 0 | -10 | 36 | 58000 | 0 |
| CF | -0.83 | -25 | 60 | 58000 | 0.0216 |
| FG | 1.33 | 53.33 | 48 | 58000 | 0.0589 |
| CG | 1 | 0 | 36 | 58000 | 0 |
| CH | -0.83 | -8.33 | 60 | 58000 | 0.0072 |
| GH | 1.33 | 53.33 | 48 | 58000 | 0.0589 |
| DH | 0 | -30 | 36 | 58000 | 0 |
| HE | 0.83 | 58.33 | 60 | 58000 | 0.0503 |
| Total | 0.3209 | ||||
For the case of Fabrication Error
| Member | n(k) | Change in Length ()(in) |
n()(k-in) |
|---|---|---|---|
| AB | -0.67 | + 0.4 | -0.268 |
| FG | 1.33 | + 0.6 | 0.798 |
| HE | 0.83 | - 0.3 | -0.249 |
| Sum | 0.281 |
Since there were no temperature effects included in this example, the total deflection at point G is the sum of these two results.
(1 k)(
) = 0.281 in-k + 0.321 in-k = 0.602 in-k
= 0.602 in-k / 1 k = 0.602 in 
The positive answer of 0.602 in indicates that the structure will deflect down in the direction of the virtual load.
Contact Dr. Fouad Fanous for more information.