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): $$\begin{equation*} \delta U = - \delta W_i \end{equation*}$$

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• 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: $$\begin{equation*} \delta U + \delta V=0 \end{equation*}$$ $$\begin{equation*} \delta U = \delta W_e \end{equation*}$$

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11.5.1 Definitions

• $\delta W ( = \Delta W)$ “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.

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• $\delta F ( = \Delta F)$ “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.
  • $\delta W=0$

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11.5.2 Rigid body example

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.

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Rigid body virtual work

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

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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.) $$\begin{equation*} \begin{split} \delta y_P &= \, +\frac{l\,\cos \theta\,\delta \theta}{2} \\ \delta y_k &= \, -\frac{l\,\cos \theta\,\delta \theta}{2} \\ \delta y_Q &= \, +\frac{l\,\cos \theta\,\delta \theta}{2} \\ \end{split} \end{equation*}$$

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Rigid body virtual work

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

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Rigid body virtual work

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

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

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

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• 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 $\theta$: $$\begin{equation*} \theta=\frac{2\,\left(Q-P\right)}{l\,k} \end{equation*}$$

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Sanity checks?