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Real work and virtual work, limitations

Short answer

Real work method equates external work ½PΔ to strain energy U = ∫ M²/(2EI)dx for a single load, giving deflection only at its point. Virtual work method applies a unit load, using 1·Δ = ∫ (M m/EI)dx for any loading. Real work limitations: only one load, only deflection at load point, cannot handle temperature/settlement, requires linear elasticity.

Define real work method and virtual work method for deformable structures with neat sketches. What are the limitations of the real work method?

Real work method

The real work method equates the external work done by the actual loads (applied gradually) to the strain energy stored in the structure:

We=U  ⇒  12P Δ=∫M22EI dxW_e = U \;\Rightarrow\; \tfrac12 P\,\Delta = \int \frac{M^2}{2EI}\,dx

The work is "real" because the loads move through their own actual displacements. Example: a cantilever with end load PP: 12PΔ=P2L36EI\tfrac12P\Delta = \dfrac{P^2L^3}{6EI}, so Δ=PL33EI\Delta = \dfrac{PL^3}{3EI}.

 P |    /|
   |  /  |  Real work = area = 1/2 P*D
   |/____|
   +-----+--- Displacement
        D

Virtual work method

In the virtual work method, an imaginary (virtual) force system, usually a unit load at the point where the displacement is wanted, is applied. The external virtual work of the virtual forces moving through the real displacements equals the internal virtual work of the virtual internal forces on the real deformations:

1⋅Δ=∫M mEI dx(beams, frames)1\cdot\Delta = \int \frac{M\,m}{EI}\,dx \qquad\text{(beams, frames)} 1⋅Δ=∑F k LAE(trusses)1\cdot\Delta = \sum \frac{F\,k\,L}{AE} \qquad\text{(trusses)}

Here the virtual load is constant while the real displacement occurs, so the virtual work is 1×Δ1\times\Delta (no factor of 12\tfrac12).

 Real loads          Virtual load
 P1  P2              1 (at the point)
 v   v               v
 ______________      ______________
 -> M, F             -> m, k

Limitations of the real work method

  1. Only one load can act on the structure; with several loads the energy equation has several unknown displacements.
  2. It gives the displacement only at the point of application of the load, and only in the direction of the load.
  3. It cannot give the displacement of a point where no load acts, or a rotation, unless a fictitious load is added.
  4. It cannot account for temperature change, lack of fit or support settlement.
  5. It is applicable only to linearly elastic structures.

✓ Answer checked by Claude Opus 5.5 (high effort)

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