Module 15 · Lesson 15.2
Virtual work and the unit-load method
One idea that finds the deflection of any pin-jointed structure.
Why this matters
The energy balance in the last lesson had two serious limitations: only one load, and only the deflection at that load in its own direction. Virtual work removes both. It lets you ask for the deflection anywhere, in any direction, under any combination of loads — and the answer comes out of a sum you can do in a table.
By the end of this lesson you should be able to
- State the principle of virtual work
- Explain what makes a virtual force system valid
- Derive the unit-load expression for truss deflection
- Apply δ = Σ NnL/AE to a pin-jointed frame
What you should already know
- Method of joints for trusses (Module 4)
- Axial extension δ = PL/AE (Module 7)
- Strain energy (this module, lesson 1)
The principle of virtual work says something that sounds almost too simple to be useful:
If a structure in equilibrium is given any small compatible deformation, the work done by the external forces equals the work done by the internal forces.
The word doing all the heavy lifting is virtual. The forces and the displacements do not have to belong to the same problem. They can come from two entirely different situations, provided each satisfies its own requirement:
- The force system must be in equilibrium. It need not be the real loading, and it need not cause the displacements.
- The displacement system must be compatible — a geometrically possible deformation of the structure. It need not be caused by the forces.
That separation is the whole trick. It means you may invent a convenient force system, apply it to the real deformation, and get a true equation relating them.
Here is how that becomes a method. Suppose you want the vertical deflection of one joint of a truss.
- 1.Solve the truss under the real loads. Call the member forces N.
- 2.Remove the real loads. Apply a single unit load at the joint you care about, pointing in the direction of the deflection you want. Solve again. Call these member forces n.
- 3.Now apply virtual work, using the unit-load system as the equilibrium set and the real deformation as the compatible set.
The external virtual work is the unit load moving through the real deflection: 1 × δ.
The internal virtual work is each member's virtual force n moving through that member's real extension NL/AE, summed over the truss. Setting them equal:
What it calculates: the deflection of one joint, in one chosen direction
- N
- member force under the real loads (N)
- n
- member force under a unit load at the point and direction wanted (dimensionless)
- L
- member length (mm)
- A
- member area (mm²)
- E
- Young's modulus (N/mm²)
This assumes
- Pin-jointed members carrying axial force only
- Linearly elastic material
- Small deflections, so the geometry used to find N and n stays valid
In plain terms: The answer comes out with the sign of the assumed unit load: positive means the joint moves the way the unit load points, negative means the other way. Because both N and n carry signs, a member in compression under both systems contributes positively — which is correct, since it stores energy either way.
Predict first
You have found the vertical deflection of a joint using a downward unit load. You now want the horizontal deflection of the same joint under the same real loading. What has to be redone?
Worked example
Vertical deflection of a bracket by the unit-load method
Given
- Two-member pin-jointed bracket, pinned to a wall at A (0, 0) and B (0, 3 m)
- Both members meet at C (4 m, 0), where a 30 kN vertical load hangs
- Member AC is horizontal, 4.00 m; member BC is the diagonal, 5.00 m
- Both members: A = 500 mm², E = 205 000 N/mm²
Find
The vertical deflection of joint C.
Assumptions
- Pin joints throughout, so members carry axial force only
- Deflections small enough that the geometry is unchanged
Practice
For the bracket in the worked example, what is the force in the diagonal member BC, in kN?
Practice
For the same bracket, what is the magnitude of the force in the horizontal member AC, in kN?
Practice
The bracket is rebuilt with both members enlarged to A = 1000 mm², everything else unchanged. What is the vertical deflection of joint C now, in mm?
Summary
- Virtual work pairs an equilibrium force system with an unrelated compatible displacement system
- The result is exact, not approximate
- For trusses: solve for N under the real loads, n under a unit load, then δ = Σ NnL/AE
- The unit load's position and direction decide which deflection you get
- The N column is reused when you change the direction you are asking about
This is educational material. It uses simplified examples to teach principles, and must not be relied on for real design or safety-critical decisions. Module overview and checkpoint