11.5 Virtual Work

  • Recall our definition of a potential function:
  • Potential (V)–a function for relating a force-like quantity to a gradient. Fi=Vi
  • Therefore: define V as a potential function of the external forces (acting through external displacements). δV=δWe(Think of δ as an arbitrary increment.)
  • Call U a potential function of the internal forces (acting through internal displacements): δU=δWi

  • We can now examine a “work” equation such that: δW=+δWe+δWi=0
  • We can refer to this equation as “Virtual Work”.
  • Note that this means: δU+δV=0 δU=δWe

11.5.1 Definitions

  • δW(=ΔW) “Virtual work” - The “work” done by a force (moment) due to a “virtual displacement”.
  • δu(=Δu) “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.

  • δF(=ΔF) “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.
    • δW=0

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.

Rigid body virtual work

  • Step 1: Describe the current position of the forces: yP(θ)=+lsinθ2yk(θ)=lsinθ2yQ(θ)=+lsinθ2

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.) δyP=+lcosθδθ2δyk=lcosθδθ2δyQ=+lcosθδθ2

Rigid body virtual work

  • Step 3: Write the virtual work, ie the forces times the virtual displacements. δWe=δyQFQ+δyPFP+δykFk δWe=lFQcosθδθ2+lFPcosθδθ2lFkcosθδθ2

Rigid body virtual work

  • Step 4: Identify the forces: FP=PFQ=+Q Fk=kyk(θ)Fk=lksinθ2

  • Plug these forces into the virtual work equation: lFQcosθδθ2+lFPcosθδθ2lFkcosθδθ2=0

    l(FQ+FPFk)cosθδθ2=0lcosθ(QPlksinθ2)δθ2=0


  • Since we know that δθ is arbitrary: lcosθ(QPlksinθ2)2=0

  • Now linearize the problem with the small angle assumptions: sinθ=θcosθ=1

    l(QPlkθ2)2=0

  • We can now solve the problem for θ: θ=2(QP)lk


Sanity checks?