Principle of virtual work

Why it matters

Some equilibrium problems are awkward with ΣF = 0 and ΣM = 0 because you would have to find many internal pin forces you do not care about: the torque on an engine crank from the gas force, the screw force in a car's scissor jack, the effort on a toggle clamp. The principle of virtual work relates the applied forces directly through the geometry of motion and skips the internal and reaction forces. The same idea, in its "virtual force" form, is the unit-load method used to find beam and truss deflections.

Key ideas

Virtual displacement. An imagined, infinitesimally small displacement δr (or rotation δθ) of the system that is consistent with its constraints, made at an instant with all forces held at their current values. It is a mathematical test, not a real motion, so time does not pass and nothing accelerates.

Virtual work. The work done by the actual forces during a virtual displacement: δW = F·δr for a force, M·δθ for a couple. Only the component of the displacement along the force counts.

Workless constraints. Ideal constraints do no virtual work: smooth surfaces (the reaction is perpendicular to any allowed motion), fixed frictionless pins (the point does not move), rigid links and inextensible cables (internal forces cancel in pairs) and rolling without slip (the contact point is instantaneously at rest). Their reactions therefore drop out of the equation, which is the whole advantage of the method. Friction is not workless: if present, include the friction force as an applied force.

The principle. A system with ideal constraints is in equilibrium if and only if the total virtual work of the applied (active) forces is zero for every virtual displacement consistent with the constraints: δW = 0. For a one-degree-of-freedom system, express every displacement in terms of one coordinate (an angle θ or a length x), write δW = (…)·δθ = 0, and since δθ ≠ 0 the bracket must vanish. A system with n degrees of freedom gives n independent equations.

Finding a reaction. To find a support reaction, release that support (replace it by its unknown force), which turns the structure into a mechanism with one degree of freedom, and apply the principle. This gives the reaction in one line.

Velocity view. Dividing by a small time step shows δW = 0 is equivalent to "power in = power out" for an ideal mechanism: F_in·v_in = F_out·v_out. Instantaneous centres of rotation give the velocity ratios quickly.

Potential energy and stability. If all active forces are conservative (weights, springs), δW = −δV, where V is the potential energy. Equilibrium then means dV/dθ = 0. The equilibrium is stable if d²V/dθ² > 0 (a minimum of V), unstable if d²V/dθ² < 0 and neutral if it is zero.

Virtual forces (unit-load method). The complementary form applies a virtual force system that is in equilibrium (a unit load at the point and in the direction of the deflection you want) to the real compatible displacements. It gives 1·Δ = Σ (n·N·L)/(A·E) for trusses and 1·Δ = ∫ m·M/(E·I) dx for beams, where n and m are the internal forces from the unit load. You will use this with strain energy and Castigliano's theorem.

Formulas

δW = Σ F_i·δr_i + Σ M_j·δθ_j = 0

• F_i: applied forces (N); δr_i: virtual displacement of each point of application, component along F_i (m); M_j: applied couples (N·m); δθ_j: virtual rotations (rad). Equilibrium of a system with ideal (workless) constraints.

δW = Q·δq = 0, so Q = 0

• q: the single generalised coordinate of a one-degree-of-freedom system; Q: generalised force.

dV/dq = 0 (equilibrium), d²V/dq² > 0 (stable)

• V: potential energy of gravity and springs (J), with V_gravity = W·h and V_spring = ½·k·x².

1·Δ = Σ n·N·L / (A·E) (trusses), 1·Δ = ∫ m·M / (E·I) dx (beams)

• N, M: real member forces and bending moments; n, m: those due to a unit load; L: member length (m); A: area (m²); E: modulus (Pa); I: second moment of area (m⁴). Linear elastic structures.

Worked examples

Example 1 (standard): beam reaction by virtual work. Given: simply supported beam AB, span 8 m, UDL 2 kN/m over the whole span and a 12 kN point load 2 m from A. Find R_B.

1. Release support B and replace it by the upward force R_B. Give the beam a small virtual rotation δθ about A.
2. Virtual displacements (upward positive): B rises 8·δθ; the UDL resultant (16 kN at 4 m) rises 4·δθ; the point load rises 2·δθ.
3. δW = R_B·(8δθ) − 16·(4δθ) − 12·(2δθ) = 0.
4. R_B = (64 + 24) / 8 = 11 kN. Answer: R_B = 11 kN (and R_A = 28 − 11 = 17 kN).

Example 2 (GATE level): crank torque in an engine mechanism. Given: slider-crank with crank OA = r = 50 mm, connecting rod AB = l = 200 mm, crank angle θ = 45° from inner dead centre, gas force on the piston F = 10 kN. Neglect friction and inertia. Find the crank torque T that holds the mechanism in equilibrium.

1. Piston position from O: x = r·cos θ + √(l² − r²·sin²θ).
2. Differentiate: δx = −[r·sin θ + r²·sin θ·cos θ / √(l² − r²·sin²θ)]·δθ.
3. Numbers: r·sin θ = 0.03536 m; r²·sin θ·cos θ = 0.00125 m²; √(0.04 − 0.00125) = 0.19685 m, so the second term is 0.00635 m. Hence |δx/δθ| = 0.04171 m/rad.
4. Virtual work: the gas force does F·|δx| as the piston moves outward while the resisting torque does −T·δθ, so T = F·|δx/δθ|.
5. T = 10 000 × 0.04171 = 417 N·m. Answer: T ≈ 417 N·m. No connecting-rod force or bearing reaction had to be found.

Example 3: scissor jack screw force. Given: a scissor jack is a rhombus of four equal links of length L, with the load W at the top joint, the base fixed and a horizontal screw joining the two side joints. Each link makes angle θ with the horizontal. W = 6 kN, θ = 30°.

1. Height of the top joint: y = 2L·sin θ, so δy = 2L·cos θ·δθ.
2. Side-joint separation: x = 2L·cos θ, so δx = −2L·sin θ·δθ.
3. The screw tension T pulls the side joints together: δW = −W·δy − T·δx = 0.
4. −W·2L·cos θ + T·2L·sin θ = 0, so T = W·cot θ = 6 / tan 30° = 10.39 kN. Answer: T ≈ 10.4 kN. The screw force rises steeply as the jack is lowered (small θ), which is why jacks are hardest to turn at the start.

Common mistakes

• Including reactions at fixed pins or smooth surfaces in δW. They do no virtual work, which is the point of the method.
• Leaving out friction. Friction does work and must be included as an applied force.
• Wrong signs: the virtual work of a force is positive only when the point moves in the direction of the force.
• Using a displacement the constraints do not allow, for example translating a simply supported beam bodily.
• Mixing degrees and radians in δθ, or forgetting that a couple does M·δθ.
• Finding δx by geometry at the wrong configuration; differentiate first, then substitute the angle.

For GATE ME

Expect one-degree-of-freedom mechanisms (levers, toggles, scissor and four-bar linkages, slider-crank torque from a piston force), reactions of beams and compound beams found by releasing one support, and stability of a rigid bar held by a spring. Practise writing the coordinates of loaded points in terms of one angle, differentiating cleanly and checking the answer with a quick statics equation.

Quick check

1. Which forces drop out of the virtual work equation for a frictionless mechanism?
2. A lever pivoted at O carries 5 kN at 0.2 m on one side. What effort at 0.8 m on the other side holds it?
3. For a scissor jack, what happens to the screw tension as θ approaches zero?
4. What condition on potential energy means a stable equilibrium?

Answers: 1. Reactions at fixed frictionless pins and smooth surfaces, and internal forces in rigid links. 2. P·0.8·δθ = 5·0.2·δθ, so P = 1.25 kN. 3. T = W·cot θ grows without limit. 4. d²V/dq² > 0 (V is a minimum).