Virtual Work

Why it matters

Virtual work converts equilibrium into a scalar work equation. Carefully chosen virtual motions can eliminate unknown ideal constraint reactions and relate forces in mechanisms.

Key ideas

A virtual displacement is an infinitesimal imagined change in configuration at a fixed instant, compatible with the constraints being retained. It is not necessarily the actual displacement over time.

For a rigid system in equilibrium, the total virtual work vanishes for every admissible virtual displacement:

δW = Σ(F_i · δr_i) + Σ(M_i · δθ_i) = 0.

For planar motion, a couple contributes M_i δθ_i with consistent signs. Forces are in newtons, displacements in metres, moments in newton metres, and rotations in radians. Each work term is in joules.

Ideal constraint reactions may be omitted only when they do no work in the chosen virtual motion: for example, a fixed pin has zero displacement, while a smooth surface reaction is perpendicular to allowed motion. Friction forces generally require explicit treatment.

For a deformable body, external virtual work equals internal virtual work. The rigid-body expression alone does not account for strain. Statically indeterminate reactions require deformation compatibility and constitutive information; equilibrium alone cannot determine them.

Worked example: beam reactions by releasing a constraint

A rigid beam AB of length 6 m has a pin at A, a vertical roller reaction at B, and a downward 10 kN point load at its midpoint. Find the vertical reactions.

To determine R_B, replace the vertical support at B with its unknown upward force R_B. Retain the pin at A. The released system permits an infinitesimal counterclockwise rotation δθ about A.

• A has zero displacement, so its reactions do no virtual work.
• B moves upward by 6δθ m to first order.
• The midpoint moves upward by 3δθ m.

Virtual work gives R_B(6δθ) − 10(3δθ) = 0, hence R_B = 5 kN. Vertical force equilibrium then gives R_A = 10 − 5 = 5 kN; the horizontal reaction is zero.

The release is essential: with both original rigid supports retained, an arbitrary nonzero midpoint displacement is incompatible with rigid-body motion. For an elastic beam, a bending virtual displacement instead requires internal virtual work.

Generalized-coordinate form

For a system described by one coordinate q, write each force application point as r_i(q). Its virtual displacement is (∂r_i/∂q)δq. Collecting terms gives δW = Qδq, where Q is the generalized force. Equilibrium requires Q = 0 when δq can have either sign.

For conservative forces and smooth ideal constraints, Q = −dΠ/dq, so equilibrium requires dΠ/dq = 0. A strict local minimum of total potential energy indicates stable equilibrium under small admissible disturbances; a stationary point alone does not establish stability.

Common mistakes

• Choosing displacements inconsistent with the retained constraints.
• Dropping a reaction that actually does work after a support is released.
• Omitting couple work or using inconsistent signs.
• Applying a rigid-body work equation to a deforming structure without internal work.
• Claiming equilibrium alone resolves static indeterminacy.

Quick check

1. Why does the pin reaction at A disappear from the example's work equation? Its application point has zero virtual displacement.
2. What is B's virtual displacement? 6δθ upward, to first order.
3. Does zero work for one arbitrarily selected motion prove equilibrium? No. The condition must hold for all admissible independent virtual motions.