How to Use Virtual Work Due to Axial Strains?

Set-Up Virtual Structures

The first step is to represent each deflection component with a unit load and apply it to a corresponding virtual structure. We would also need to assume the direction of each element:

• Horizontal translation at \(E\) is to the right; hence, the applied unit load also acts to the right.
• Vertical translation at \(E\) is downward; hence, the applied unit vertical load also acts downward.

To find the rotation of member \(CE\), we represent it with two equal parallel forces applied perpendicularly at both ends of member \(CE\); so that the effect resembles a unit couple. We'll assume this couple acts clockwise to represent a clockwise rotation.

Formulate S, u, A, and E Equations.

After creating the virtual structures, it's time to find \(S\), \(u\), \(L\), \(A\), and \(E\) of each member for both the virtual and superimposed structures. As seen below, creating a matrix that summarizes all these variables is best for this approach.

We first identify \(L\), \(A\), and \(E\) for each truss member. If these variables vary along the member's length, we must express it in terms of \(x\) (member's local axis). Since our example deals with constant axial rigidity \(AE\), we do not have to worry about those instances.

Next, we solve for the bar forces \(S\) of the superimposed structure, as well as the bar forces of the virtual equivalent \(u_h\), \(u_v\), and \(u_{\alpha_{CE}}\). We can do this exactly using the method of joints, the method of sections, or a combination of both, depending on the truss type. 

We summarise our results for all bar forces \(S\), \(u_h\), \(u_v\), and \(u_{\alpha_{CE}}\) in the table. Positive bar forces are in tension, while negative-sensed bar forces are in compression.

Apply Virtual Work Equations

The final step of the process is to apply the virtual work equations:

\(1 \times \Delta_{E_h}=\sum_{i=0}^n \frac{S_i u_{E_{h_i}} L_i}{A_i E_i}\)

\(1 \times \Delta_{E_v}=\sum_{i=0}^n \frac{S_i u_{E_{v_i}} L_i}{A_i E_i}\)

\(1 \times \theta_{C E}=\sum_{i=0}^n \frac{S_i u_{C E_{\alpha_i}} L_i}{A_i E_i}\)

We can use the table above to add the corresponding values for each column to determine the deflection. For horizontal and vertical translation:

\(1 \times \Delta_{E_h}=0+0-\frac{45}{E}-\frac{22.5}{E}+0+0\)

\(\Delta_{E_h}=-\frac{67.5}{E}\)

\(1 \times \Delta_{E_v}=\frac{16.875}{E}+\frac{62.5}{E}+\frac{67.5}{E}+\frac{16.875}{E}+\frac{156.25}{E}+\frac{80}{E}\)

\(\Delta_{E_v}=\frac{400}{E}\)

The negative sign of the horizontal translation means that the movement of joint \(E\) is to the left instead of the right. On the other hand, the positive sign of vertical translation means that the movement is indeed downward. With these two translation components, we can finally solve for the resultant deflection:

\(\Delta_E=\sqrt{\left(-\frac{67.5}{E}\right)^2+\left(\frac{400}{E}\right)^2}=\frac{405.66}{E}\)

The only remaining component we need to answer is the rotation of member \(CE\). By adding all the values in the table, we have:

\(1 \times \theta_{C E}=\frac{0.5625}{E}+\frac{0.75}{E}+\frac{1.125}{E}+\frac{0.5625}{E}+0+0\)

\(\theta_{C E}=\frac{3}{E}\)

The positive direction confirms our assumption that member \(CE\) rotated clockwise from its original position.