11.5 Virtual Work

Recall our definition of a potential function:
Potential (\(V\))–a function for relating a force-like quantity to a gradient. \[\begin{equation*} F_i = -{\frac{\partial V}{\partial i}} \end{equation*}\]
Therefore: define \(V\) as a potential function of the external forces (acting through external displacements). \[\begin{equation*} \delta V = - \delta W_e \end{equation*}\](Think of \(\delta\) as an arbitrary increment.)
Call \(U\) a potential function of the internal forces (acting through internal displacements): \[\begin{equation*} \delta U = - \delta W_i \end{equation*}\]

We can now examine a “work” equation such that: \[\begin{equation*} \delta W = + \delta W_e +\delta W_i =0 \end{equation*}\]
We can refer to this equation as “Virtual Work”.
Note that this means: \[\begin{equation*} \delta U + \delta V=0 \end{equation*}\] \[\begin{equation*} \delta U = \delta W_e \end{equation*}\]

\(\delta W ( = \Delta W)\) “Virtual work” - The “work” done by a force (moment) due to a “virtual displacement”.
\(\delta u ( = \Delta u)\) “Virtual displacement” - A kinematically admissible displacement
- A displacement that does not violate the kinematic boundary conditions
- The displacement is arbitrary
- It need not be related to the actual displacement that results from the applied loads.

\(\delta F ( = \Delta F)\) “Virtual force”
- Any force field that is zero on the boundary
  - It must not do work on the boundary
- The force is arbitrary
- It need not be related to the actual force that results from the displacements.
- Could also be a virtual stress
“The principle of virtual work”
- A system is in static equilibrium if (and only if) the virtual work is zero for all independent virtual displacements.
- \(\delta W=0\)

Rigid body virtual work

Note: since we are considering this as a rigid body and don’t want to track strain energy (spring energy), we can assume that the spring applies an external force based on the spring force formula.

Rigid body virtual work

Step 1: Describe the current position of the forces: \[\begin{equation*} \begin{split} y_P\left(\theta\right) &= \, +\frac{l\,\sin \theta}{2} \\ y_k\left(\theta\right) &= \, -\frac{l\,\sin \theta}{2} \\ y_Q\left(\theta\right) &= \, +\frac{l\,\sin \theta}{2} \\ \end{split} \end{equation*}\]

Rigid body virtual work

Step 2: Take the variation. (You are assuming that a small virtual displacement can be added to the equilibrium displacement. How does the position change assuming a small change in theta? This is very similar to taking a derivative.) \[\begin{equation*} \begin{split} \delta y_P &= \, +\frac{l\,\cos \theta\,\delta \theta}{2} \\ \delta y_k &= \, -\frac{l\,\cos \theta\,\delta \theta}{2} \\ \delta y_Q &= \, +\frac{l\,\cos \theta\,\delta \theta}{2} \\ \end{split} \end{equation*}\]

Rigid body virtual work

Step 3: Write the virtual work, ie the forces times the virtual displacements. \[\begin{equation*} \delta W_e=\delta y_Q\,F_Q+\delta y_P\,F_P+\delta y_k\,F_k \end{equation*}\] \[\begin{equation*} \delta W_e=\frac{l\,F_Q\,\cos \theta\,\delta \theta}{2}+\frac{l\,F_P\,\cos \theta\,\delta \theta}{2}-\frac{l\,F_k\,\cos \theta\,\delta \theta}{2} \end{equation*}\]

Rigid body virtual work

Step 4: Identify the forces: \[\begin{equation*} \begin{split} F_P &=\, -P \\ F_Q &=\, +Q \\ \end{split} \end{equation*}\] \[\begin{equation*} \begin{split} F_k &=\, - k\,y_k\left(\theta\right) \\ F_k &=\, \frac{l\,k\,\sin \theta}{2} \\ \end{split} \end{equation*}\]

Plug these forces into the virtual work equation: \[\begin{equation*} \frac{l\,F_Q\,\cos \theta\,\delta \theta}{2}+\frac{l\,F_P\,\cos \theta\,\delta \theta}{2}-\frac{l\,F_k\,\cos \theta\,\delta \theta}{2}=0 \end{equation*}\]

\[\begin{equation*} \begin{split} \frac{l\,\left(F_Q+F_P-F_k\right)\,\cos \theta\,\delta \theta}{2} &=\, 0 \\ \frac{l\,\cos \theta\,\left(Q-P-\frac{l\,k\,\sin \theta}{2}\right)\,\delta \theta}{2} &=\, 0 \\ \end{split} \end{equation*}\]

Since we know that \(\delta \theta\) is arbitrary: \[\begin{equation*} \frac{l\,\cos \theta\,\left(Q-P-\frac{l\,k\,\sin \theta}{2}\right)}{2}=0 \end{equation*}\]
Now linearize the problem with the small angle assumptions: \[\begin{align} \sin \theta =& \; \theta \\ \cos \theta =& \; 1\\ \end{align}\]

\[\begin{equation*} \frac{l\,\left(Q-P-\frac{l\,k\,\theta}{2}\right)}{2}=0 \end{equation*}\]
We can now solve the problem for \(\theta\): \[\begin{equation*} \theta=\frac{2\,\left(Q-P\right)}{l\,k} \end{equation*}\]

Sanity checks?