Chapter 11

Virtual Work

Chapter Objectives

Introduce the principle of virtual work and apply it to finding equilibrium configurations of pin-connected systems.
Establish the potential-energy function and use the potential-energy method to investigate equilibrium type and stability.

11.1

Definition of Work

The principle of virtual work was introduced by the Swiss mathematician Jean Bernoulli in the eighteenth century. It provides an alternative method for solving equilibrium problems involving particles, rigid bodies, or systems of connected rigid bodies. Before applying this principle, we must carefully define what it means for a force or couple moment to do work.

Work of a Force

A force does work only when it undergoes a displacement in the direction of its line of action. If a force \(\mathbf{F}\) acts at angle \(\theta\) to a differential displacement \(d\mathbf{r}\), the work done is the product of the force's component along the displacement and the displacement magnitude. Work is a scalar quantity — it can be positive or negative depending on the relative sense of the force and its displacement.

Work of a Force

\[ dU = F \, dr \cos\theta \]

Equivalently using the dot product: \( dU = \mathbf{F} \cdot d\mathbf{r} \)

In the SI system, the unit of work is the joule (J) — the work done by a 1-N force displacing 1 m in the direction of the force (\(1\text{ J} = 1\text{ N}\!\cdot\!\text{m}\)). In the FPS system, the unit is the foot-pound (ft·lb).

Work of a Couple Moment

When a rigid body rotates through a differential angle \(d\theta\), a couple moment \(M\) of magnitude \(Fr\) does work. The couple forces produce no net work during translation (they cancel), but during rotation through \(d\theta\), the work contributed is positive when \(M\) and \(d\theta\) share the same sense, and negative otherwise.

Work of a Couple Moment

\[ dU = M \, d\theta \]

Virtual Work

Virtual work is the work done by actual forces and moments through imaginary (virtual) displacements or rotations. These virtual displacements are first-order differential quantities, denoted \(\delta r\) and \(\delta\theta\), that do not actually occur — they are simply assumed for the purpose of analysis. For a force with virtual displacement \(\delta r\):

Virtual Work — Force and Couple

\[ \delta U = F \cos\theta \; \delta r \qquad \text{(force)} \] \[ \delta U = M \; \delta\theta \qquad \text{(couple moment)} \]

A force F at angle θ to a displacement dr — only the component along the displacement direction does work.

Fig. 11–1 (original)

11.2

Principle of Virtual Work

If a body is in equilibrium, the algebraic sum of all virtual work done by every force and couple moment acting on it is zero for any virtual displacement of the body.

\[ \delta U = 0 \]

This principle is equivalent to the equations of static equilibrium. For a rigid body in a coplanar force system, virtual translations in the \(x\) and \(y\) directions and a virtual rotation about any point O correspond to \(\Sigma F_x = 0\), \(\Sigma F_y = 0\), and \(\Sigma M_O = 0\) respectively.

Internal Forces Do No Net Work

When applying the virtual-work equation to a rigid body, internal forces need not be considered. Because the body does not deform, internal forces appear as equal-and-opposite collinear pairs, so their virtual work contributions cancel perfectly.

For simple single-body problems, the principle of virtual work offers no computational advantage over direct equilibrium equations. Its real power emerges when analyzing systems of connected rigid bodies, as explored in Section 11.3.

11.3

Principle of Virtual Work for a System of Connected Rigid Bodies

Virtual work is especially powerful for systems of multiple connected rigid bodies — linkages, mechanisms, and frames. When such a system has a single degree of freedom, its entire configuration is determined by one coordinate (typically an angle \(\theta\)). A single virtual-work equation can directly yield an unknown force or the equilibrium position without dismembering the mechanism.

One Degree of Freedom

A system has one degree of freedom if its geometry can be completely described by a single independent coordinate \(q\). Knowing \(q\) and the member lengths allows all force positions to be located. Only systems with one degree of freedom are treated here.

Important Points

A force does work when it displaces along its line of action. A couple moment does work when it rotates. Positive work results when the force/couple and its displacement share the same sense.
Virtual work is most useful for finding the equilibrium configuration of a mechanism or computing a reactive force in a connected system.
A virtual displacement is imaginary — it is a differential displacement given in the positive direction of a position coordinate; it does not actually occur.
Forces or couple moments that experience no virtual displacement contribute zero virtual work.

Procedure for Analysis — Virtual Work Method

Step 1: Free-Body Diagram

Draw the FBD of the entire connected system and define the single coordinate \(q\).
Sketch the "deflected position" of the system when it undergoes a positive virtual displacement \(\delta q\).

Step 2: Virtual Displacements

Define position coordinates \(s_i\), each measured from a fixed point to each active force, and oriented parallel to that force's line of action.
Express each \(s_i\) as a function of \(q\), then differentiate to obtain \(\delta s_i\) in terms of \(\delta q\).

Step 3: Virtual-Work Equation

Write \(\delta U = 0\), treating each \(\delta s_i\) as positive. Assign positive work when force and displacement share the same sense; negative otherwise.
Express all terms in \(\delta q\) and factor it out. Solve for the unknown.

📝 Course Note

The virtual work method really shines for mechanisms — you bypass writing three equations per link and skip free-body diagrams for every internal pin. Once you commit the position coordinates and take derivatives, \(\delta U = 0\) gives you everything in one shot. Practice finding position coordinates geometrically; this is the step where most mistakes happen. Always check sign convention: if an increase in \(q\) causes a coordinate to decrease, its virtual displacement is negative.

11.4

Conservative Forces

Conservative Force

A force is conservative if the work it does depends only on the initial and final positions of the body — not on the path taken. The two most important examples in statics are the weight of a body and the force of a spring.

Weight

For a body of weight \(W\) that moves from point A to point B through a vertical displacement \(h\), the work done by gravity depends only on the vertical distance, not on the curved or inclined path traveled:

Work Done by Weight

\[ U = -Wh \]

Negative because weight acts downward while the body moves upward through \(h\).

Spring Force

A linearly elastic spring with stiffness \(k\) stores and releases energy conservatively. The work done by the spring force as the block moves from position \(s_1\) to \(s_2\) (measured from the unstretched length) is:

Work Done by Spring Force

\[ U = -\left(\tfrac{1}{2}ks_2^2 - \tfrac{1}{2}ks_1^2\right) \]

Friction — A Non-Conservative Force

Non-Conservative (Path-Dependent) Forces

Unlike weight and spring forces, friction is non-conservative. The work done by kinetic friction depends on the length of the path traversed — a longer path dissipates more energy as heat. Friction cannot be incorporated into a potential-energy function, so the potential-energy methods of Sections 11.5–11.7 apply only to systems without friction.

11.5

Potential Energy

A conservative force gives a body the capacity to do work by virtue of its position. This capacity is called potential energy.

Gravitational Potential Energy \(V_g\)

If a body of weight \(W\) is at height \(y\) above a chosen datum, it has gravitational potential energy \(V_g = Wy\). Above the datum, \(V_g > 0\); below the datum, \(V_g < 0\); at the datum, \(V_g = 0\). The datum location is arbitrary — only differences in \(V_g\) matter.

Gravitational Potential Energy

\[ V_g = Wy \quad \text{(11-4)} \]

\(y\) is measured positive upward from the datum.

Elastic Potential Energy \(V_e\)

When a spring is stretched or compressed by amount \(s\) from its unstretched (datum) position, it stores elastic potential energy. This quantity is always positive, regardless of whether the spring is elongated or compressed.

Elastic Potential Energy

\[ V_e = \tfrac{1}{2}ks^2 \quad \text{(11-5)} \]

Potential Function \(V\)

When a system is subjected to both gravitational and elastic forces, the total potential energy is their sum. For a single-degree-of-freedom system described by coordinate \(q\), the potential function \(V(q)\) completely characterizes the energy state. The work done by all conservative forces moving the system from \(q_1\) to \(q_2\) equals the drop in potential energy.

Potential Function and Work–Energy Relation

\[ V = V_g + V_e \quad \text{(11-6)} \] \[ U_{1\to 2} = V(q_1) - V(q_2) \quad \text{(11-7)} \]

A block supported by a spring. Datum at the spring's unstretched length. Potential function: \( V = -Wy + \tfrac{1}{2}ky^2 \)

Fig. 11–14 (original)

11.6

Potential-Energy Criterion for Equilibrium

For a frictionless single-degree-of-freedom system with potential function \(V(q)\), the principle of virtual work (\(\delta U = 0\)) translates directly into a condition on the derivative of \(V\):

Potential-Energy Equilibrium Criterion

A frictionless connected system of rigid bodies is in equilibrium when the first derivative of its potential function equals zero: \[ \frac{dV}{dq} = 0 \quad \text{(11-9)} \]

This is a powerful result: rather than writing virtual-work expressions with position coordinates and differentials, we only need to express \(V\) in terms of \(q\) and differentiate once. Each root \(q_{eq}\) of \(dV/dq = 0\) is an equilibrium position.

Example — Block on Spring

For a block of weight \(W\) on a spring of stiffness \(k\), with \(y\) measured downward from the unstretched length as datum: \(V = -Wy + \tfrac{1}{2}ky^2\). Setting \(dV/dy = -W + ky = 0\) gives the equilibrium position \(y_{eq} = W/k\), which is the same result as \(\Sigma F_y = 0\).

11.7

Stability of Equilibrium Configuration

Once an equilibrium position is found, the second (and higher) derivative of \(V\) reveals whether that equilibrium is stable, neutral, or unstable.

Stable Equilibrium

A system is in stable equilibrium when a small perturbation causes it to return to its original position. Potential energy is at a local minimum at this configuration. A ball resting at the bottom of a bowl is a familiar example.

Neutral Equilibrium

A system is in neutral equilibrium when small perturbations leave it in the new position — it neither returns nor moves further away. Potential energy is constant in the neighborhood of the equilibrium. A ball on a flat surface illustrates this.

Unstable Equilibrium

A system is in unstable equilibrium when a small perturbation causes it to move farther from the original position. Potential energy is at a local maximum. A ball balanced at the top of a hill represents this case.

Equilibrium Type	Condition	Potential Energy
Stable	\(\dfrac{dV}{dq}=0, \quad \dfrac{d^2V}{dq^2} > 0\)	Minimum
Unstable	\(\dfrac{dV}{dq}=0, \quad \dfrac{d^2V}{dq^2} < 0\)	Maximum
Neutral	\(\dfrac{dV}{dq}=\dfrac{d^2V}{dq^2}=\dfrac{d^3V}{dq^3}=\cdots=0\)	Constant

If \(d^2V/dq^2 = 0\), examine higher-order derivatives. The equilibrium is stable if the first non-zero derivative is even-order and positive; unstable if that derivative is odd-order, or even-order and negative.

Three equilibrium types illustrated by the shape of the potential-energy function \(V(q)\). The filled circle marks the equilibrium position.

Fig. 11–15/16 (original)

Procedure for Analysis — Potential-Energy Method

Step 1: Potential Function

Sketch the system in the arbitrary position defined by coordinate \(q\). Set a horizontal datum through a convenient fixed point.
Express \(V_g = Wy\) for each member's weight and \(V_e = \tfrac{1}{2}ks^2\) for each spring.
Form \(V = V_g + V_e\) and write all position variables in terms of \(q\).

Step 2: Equilibrium Positions

Differentiate \(V\) with respect to \(q\) and set \(dV/dq = 0\). Solve for all \(q_{eq}\).

Step 3: Stability

Evaluate \(d^2V/dq^2\) at each \(q_{eq}\). Positive → stable; negative → unstable; zero → check higher derivatives.

📝 Course Note

The stability criterion is intuitive once you connect it to the shape of \(V(q)\): a bowl-shape (minimum) means stable; a hill-shape (maximum) means unstable. In practice, always check stability after finding equilibrium — a structure can be in equilibrium yet tip over the moment you load it. Engineers use this in machine design, vehicle dynamics, and tall structure analysis. Example 11.7 is a beautiful result: it shows analytically why a tall block on a curved surface is risky.

Physical meaning

Virtual work measures only the force component along an allowed infinitesimal motion. Ideal pins and smooth constraints often disappear because their reactions have no displacement along their lines of action.

Problem-solving strategy

Isolate the whole one-degree-of-freedom mechanism and choose a positive coordinate.
Express every active-force position as a function of that coordinate.
Differentiate positions, assign work signs, and set total virtual work to zero.
Check a reaction does no work before omitting it and test equilibrium stability if needed.

EXAMPLE 1Holding force on a rotating arm

A 0.60 m weightless arm is pinned at one end, supports a 300 N downward load at its tip, and makes \(30^\circ\) above horizontal. Find the horizontal leftward tip force that holds equilibrium.

Recognition strategy

One angle describes the arm, and both applied forces act at the moving tip, making virtual work a direct alternative to a moment equation.

Given information

\(L=0.60\text{ m}\), \(W=300\text{ N}\), \(\theta=30^\circ\); positive virtual rotation increases \(\theta\).

Model and interpretation

The pin is ideal and stationary, so its reactions do zero work; \(x=L\cos\theta\) and \(y=L\sin\theta\).

Step-by-step solution

1. Differentiate geometry. \(\delta x=-L\sin\theta\,\delta\theta\) and \(\delta y=L\cos\theta\,\delta\theta\).

2. Assign work. A leftward \(P\) contributes \(-P\delta x\), while the downward weight contributes \(-W\delta y\).

3. Enforce equilibrium. \(\delta U=PL\sin\theta\,\delta\theta-WL\cos\theta\,\delta\theta=0\).

4. Solve. \(P=W\cot30^\circ=300(1.732)=519.6\text{ N}\).

Verification

Units: Each work coefficient is force times length, N·m, and the common dimensionless \(\delta\theta\) cancels.

Signs and direction: Increasing angle moves the tip left, so a leftward force supplies positive work and the weight supplies negative work.

Plausibility: Near horizontal, a large horizontal holding force is expected because its moment arm approaches zero; 520 N exceeds the 300 N load.

Final result: \(\boxed{P=520\text{ N leftward}}\)

Engineering interpretation

The required actuator force rises sharply near the horizontal dead position; linkage travel limits and pin capacities require separate design checks.

EXAMPLE 2Spring-loaded slider mechanism

A 0.50 m massless link drives a slider horizontally while its tip compresses a vertical spring. At \(\theta=40^\circ\), the spring has extension \(0.50\sin\theta-0.15\text{ m}\) and \(k=1200\text{ N/m}\). Find the leftward slider force for equilibrium.

Recognition strategy

The link has one coordinate and the spring force is configuration-dependent, so derive both displacement ratios before writing work.

Given information

\(L=0.50\text{ m}\), \(k=1200\text{ N/m}\), \(\theta=40^\circ\), with spring force acting downward.

Model and interpretation

Use \(x=L\cos\theta\) for the slider and \(y=L\sin\theta\) for spring extension; ideal guide reactions do no virtual work.

Step-by-step solution

1. Evaluate spring force. \(F_s=1200[0.50\sin40^\circ-0.15]=205.1\text{ N}\).

2. Differentiate motion. \(\delta x=-0.50\sin\theta\delta\theta\) and \(\delta y=0.50\cos\theta\delta\theta\).

3. Write work. \(-P\delta x-F_s\delta y=0\), because leftward \(P\) opposes the negative \(\delta x\).

4. Solve force. \(P=F_s\cot40^\circ=205.1(1.192)=244.5\text{ N}\).

Verification

Units: \(k\) times extension gives N, and each virtual-work term has N·m multiplied by \(\delta\theta\).

Signs and direction: Increasing angle moves the slider left and spring point upward, so leftward force and downward spring correctly oppose in work.

Plausibility: The force exceeds the 205 N spring load by \(\cot40^\circ\), consistent with the less favorable horizontal motion ratio.

Final result: \(\boxed{P=245\text{ N leftward}}\)

Engineering interpretation

Motion ratio converts spring load into actuator demand; real mechanisms additionally need friction, spring travel, and force-rating allowances.

EXAMPLE 3Torsional support equilibrium and stability

A 0.50 m arm is acted on by an upward applied actuator/test force \(P=180\text{ N}\) at its tip and a torsional spring with \(k_\theta=240\text{ N}\!\cdot\!\text{m/rad}\). With \(V=\tfrac12k_\theta\theta^2-PL\sin\theta\), find the positive equilibrium angle and classify it.

Recognition strategy

The prescribed upward force has a position-dependent potential and the spring is conservative, so potential energy gives equilibrium and a local stability test in one coordinate.

Given information

Upward applied \(P=180\text{ N}\), \(L=0.50\text{ m}\), \(k_\theta=240\text{ N}\!\cdot\!\text{m/rad}\); angle is measured in radians.

Model and interpretation

The actuator force remains vertically upward as the arm moves, the spring is linear, and pins are ideal; damping and stops are excluded from this local static prediction.

Step-by-step solution

1. Differentiate potential. The upward applied force gives \(-PL\sin\theta\), so \(dV/d\theta=240\theta-90\cos\theta\text{ N}\!\cdot\!\text{m}\).

2. Set equilibrium. Set the spring torque against the applied-force effect and solve \(240\theta=90\cos\theta\) numerically for the positive root.

3. Evaluate root. At \(\theta=0.352\text{ rad}\), \(240\theta-90\cos\theta\approx0\text{ N}\!\cdot\!\text{m}\).

4. Test curvature. The applied-force potential gives \(d^2V/d\theta^2=240+90\sin0.352=271\text{ N}\!\cdot\!\text{m/rad}>0\).

Verification

Units: Both potential terms are N·m, and the second derivative correctly has N·m/rad for the radian coordinate.

Signs and direction: The upward applied force lowers potential as positive angle increases its vertical tip position, while the spring restoring term is positive.

Plausibility: \(0.352\text{ rad}=20.2^\circ\) is below the small-angle applied-force estimate \(PL/k_\theta= 90/240=0.375\text{ rad}\).

Final result: \(\boxed{\theta=0.352\text{ rad }(20.2^\circ),\ \text{stable}}\)

Engineering interpretation

Positive curvature indicates a restoring tendency against small disturbances under this upward actuator/test force, not guaranteed system safety; stops, fatigue, and nonlinear spring data need review.

Chapter recap

Use compatible virtual displacement rather than arbitrary body motion.
Differentiate geometry to obtain force motion ratios.
Before moving on, verify work signs, inactive reactions, and potential-energy curvature.