Principle of virtual work

Why it matters

For a mechanism with one degree of freedom, such as a toggle clamp, a scissor lift or a slider-crank, writing equilibrium for every link brings in many pin forces that you do not care about. Virtual work skips them: one equation relates the input actuator force or motor torque to the output load. The same idea gives the static relation between joint torques and tip force of a robot arm (τ = Jᵀ·F) and underlies energy methods for deflection later in this subject.

Key ideas

Virtual displacement. An imagined, infinitesimally small change δq in the configuration of a system, made at an instant, consistent with the constraints (pins stay pinned, rollers stay on their surfaces, links keep their length). It is not a real motion and takes no time, so forces are treated as constant during it.

Virtual work. The work a force does during a virtual displacement: δW = F·δr = F·δs·cos α, where δs is the virtual displacement of its point of application and α the angle between them. A couple M does M·δθ.

Principle of virtual work. A system of rigid bodies with ideal constraints is in equilibrium if and only if the total virtual work of the active (applied) forces is zero for every virtual displacement consistent with the constraints: δW = 0.

Why reactions drop out. Ideal constraints do no virtual work: a fixed pin does not move, a smooth surface's normal force is perpendicular to the allowed sliding, internal pin forces between connected links act equally and oppositely on points that move together, and the internal forces of a rigid body do no work. Friction is not workless; include it as an active force.

Degrees of freedom. A system with n degrees of freedom gives n independent equations, one for each independent virtual displacement. Most textbook mechanisms have one, so one equation answers the question.

Finding a reaction or an internal force. To find a support reaction or a member force with virtual work, release the constraint that carries it (replace the support by its force, or cut the member and show its force on both sides), which gives the system one degree of freedom. Then give that virtual displacement and solve δW = 0. This is the basis of influence lines.

Using coordinates. Express the position of each loaded point in terms of the generalised coordinate (often an angle θ), then differentiate: if y = l·sin θ, then δy = l·cos θ·δθ. Signs come out automatically if each force is written with its component along the positive coordinate direction.

Potential energy and stability. If all active forces are conservative (gravity, linear springs), δW = −δV, so equilibrium is where dV/dq = 0. The equilibrium is stable if d²V/dq² > 0 (a minimum of V), unstable if d²V/dq² < 0, and neutral if it is zero.

Mechatronics link. For a robot arm or any linkage driven by actuators, the input power equals the output power in an ideal mechanism: τ·δθ = F·δx. So the actuator torque needed is τ = F·(δx/δθ), the force times the velocity ratio. For a multi-joint arm this becomes τ = Jᵀ·F, where J is the arm's Jacobian.

Formulas

δW = Σ F_i·δs_i·cos α_i + Σ M_j·δθ_j = 0

• F_i: active force (N); δs_i: virtual displacement of its point of application (m); α_i: angle between force and displacement; M_j: applied couple (N·m); δθ_j: virtual rotation (rad). Condition for equilibrium with ideal constraints.

δy = (dy/dθ)·δθ

• y: coordinate of a loaded point as a function of the generalised coordinate θ. Used to relate virtual displacements.

δW = −δV, dV/dq = 0, d²V/dq² > 0 (stable)

• V: potential energy (J) of conservative forces, gravity V = W·y, spring V = ½·k·x²; q: generalised coordinate.

τ = F·(δx/δθ)

• τ: input torque (N·m); F: output force (N); δx/δθ: output displacement per unit input rotation (m/rad). Ideal (frictionless) mechanism.

Worked examples

Example 1 (standard): beam reaction by virtual work. Given: a simply supported beam AB of span 6 m carries 10 kN at 2 m from A and 4 kN at 5 m from A. Find R_B.

1. Release support B and replace it by an upward force R_B. The beam can now rotate about A.
2. Give a small virtual rotation δθ about A. Points at distance x rise by x·δθ.
3. δW = R_B·(6δθ) − 10·(2δθ) − 4·(5δθ) = 0.
4. R_B = (20 + 20)/6 = 6.667 kN.
5. Repeat with B as the pivot to find R_A: R_A·6 = 10 × 4 + 4 × 1 = 44, so R_A = 7.333 kN. Check: 6.667 + 7.333 = 14 kN, the total load. Answer: R_A ≈ 7.33 kN, R_B ≈ 6.67 kN.

Example 2 (GATE level): toggle (scissor) mechanism. Given: two equal light links AB and BC, each of length l, are pinned together at B. A is pinned to the base and C slides on a smooth horizontal guide, with both links at angle θ = 30° to the horizontal. A vertical load W = 2 kN acts downward at B. A horizontal actuator force P at C, pointing towards A, holds the system. Find P.

1. Coordinates with origin at A: y_B = l·sin θ, x_C = 2l·cos θ.
2. Virtual displacements: δy_B = l·cos θ·δθ, δx_C = −2l·sin θ·δθ.
3. Active forces: W in −y at B; P in −x at C. The pin at A and the normal force at C do no work.
4. δW = −W·δy_B + (−P)·δx_C = −W·l·cos θ·δθ + 2P·l·sin θ·δθ = 0.
5. P = W / (2·tan θ) = 2 / (2 × 0.5774) = 1.732 kN.
6. Check by statics: by symmetry the vertical reaction at C is W/2 = 1 kN. Moments about B for link BC: 1 × l·cos 30° = P·l·sin 30°, so P = 1/tan 30° = 1.732 kN. Answer: P ≈ 1.73 kN. As θ falls towards zero, P = W/(2·tan θ) grows without limit: the toggle gives a very large clamping force near the flat position, which is why toggle clamps and presses use it.

Common mistakes

• Using virtual displacements that break a constraint without adding the corresponding reaction to the equation.
• Getting the sign of a virtual displacement wrong; write coordinates and differentiate instead of guessing.
• Forgetting that a couple does work M·δθ, or measuring δθ in degrees.
• Treating friction as workless. Only ideal (smooth, frictionless) constraints do no virtual work.
• Using the finite height of a point instead of its small change; the method uses derivatives.
• Assuming every equilibrium found from dV/dq = 0 is stable; check d²V/dq².

For GATE ME

Expect one-degree-of-freedom linkages (toggle, scissor, slider-crank, ladder or rod with springs) where the force or torque to hold equilibrium is asked, beam reactions by released constraints, and stability of equilibrium using potential energy. Practise writing positions as functions of a single angle and differentiating cleanly.

Quick check

1. Why do the pin reactions at a fixed support not appear in the virtual-work equation?
2. A couple of 20 N·m turns through a virtual rotation of 0.01 rad. What virtual work does it do?
3. For the toggle in Example 2 at θ = 45° and W = 2 kN, what is P?
4. What condition on potential energy indicates stable equilibrium?
5. Does friction do virtual work?

Answers: 1. The support point does not move, so its force does no work. 2. 0.2 J. 3. P = 2/(2 × 1) = 1 kN. 4. d²V/dq² > 0 (V is a minimum). 5. Yes, it must be included as an active force.