In the principle of virtual work a unit virtual force imparts virtual forces on the structure internally.

When real loading is then applied the additional work done by these forces is their magnitude x distance delta.

I don’t get why we assume the virtual forces are constant as the shape deforms. Would they not change with the changing structure?

This makes me think this method is approximate.


1 Answer 1


Virtual work is exact, the problem is everything else

Virtual work is what is known as an "energy method". It finds the structure's stable configuration by simply stating that the work done by internal forces must be equal to that done by external forces. And since that's basically the law of conservation of energy, it must be exact, since that's a rule actually followed by the real world.

The problem is that getting the exact result involves lots of integrals and I'm not even sure if they're analytically solvable.

I'm not sure because I'm a structural engineer, and that means I'm lazy. I don't need the exact result, I need something that's close enough to get the job done.

So I don't just use virtual work to get my solutions. I use virtual work and a bunch of simplifying assumptions. I assume that materials are linear elastic, meaning that if I double the stress, I double the strain. I assume that we're dealing with small deflections and rotations, which lets me simplify a bunch of trigonometric functions which are otherwise a headache. And since the deflections and rotations are small, they probably won't change the line of action of applied forces in any meaningful way, so I can assume geometric linearity.

The sum of all these assumptions (and there are many others I haven't gotten into) is what you're talking about: so long as these assumptions hold, I can state that if I apply a force $P$ and get a stress/strain/deflection $x$, a force $2P$ will give me $2x$.

Now, to be clear, all of these assumptions are false.

No material is perfectly linear elastic, but those we use are reasonably close to being so (prior to yielding, of course), so it gets the job done.

Our "small deflections and rotations" simplification leads to errors (compared to using the true trigonometric functions, but they're very small, so it gets the job done.

Lastly, the subsequent assumption of geometric linearity. Sure, if a column bends a bit to the side, the axial force will start creating a bending moment, but the deflection is so small... taking that into consideration would just be a headache and I'd rather get the job done.

That being said, geometric linearity is the first assumption to fail catastrophically. There are cases where a structure is so flexible that the deflections lead to non-trivial changes to the line of action of applied forces, which increases internal stresses and therefore strains, which means greater deflections, which further changes the line of action... you get a feedback loop which may or may not be stable (that is, the structure might deform and suffer greater stresses than anticipated but eventually reach a stable configuration... or not).

These are what we call "second-order effects" and are something worth keeping an eye out during design. Thankfully there are rules of thumb which indicate when those are relevant, in which case we pull out the big computational guns and ensure they are properly handled.

You'll learn these methods and rules of thumb eventually, but for now, rest easy knowing that these assumptions are your best friends. Because without them, everything's a headache (thankfully computers don't get headaches, though).

  • $\begingroup$ The real point here is that none of these assumptions matter, because Virtual Work assumes the displacements are infinitesimally small. Therefore in any change in load, stress, etc because of deformation of the structure, nonlinear material behaviour, etc, is guaranteed to be a second-order effect, and "infinitesimally small squared" is negligible compared with "infinitesimally small." To really understand this mathematically you need a rigorous course in calculus, which is normally only studied by mathematicians and called "real analysis" not "calculus 1 and 2". $\endgroup$
    – alephzero
    Commented Apr 26, 2020 at 18:47
  • $\begingroup$ @alephzero I get that the displacements can be as small as you would like and therefore do not effect the movement. What I do not get is how we know that what started at a virtual force of .2 won’t become a virtual force of .3 as the structure deforms. $\endgroup$
    – jrudd
    Commented Apr 26, 2020 at 20:52
  • $\begingroup$ Thanks for this in depth answer, I am confused because it seems like virtual work would even be a approximation for a given model: Let’s take a linear elastic model of a material. I agree that the work done by the virtual force is equal to the internal work done by the virtual force. I guess my issue is with the calculation of the internal work done by the virtual force. We calculate the internal work by multiplying the internal virtual forces at time zero by the change in lengths. But the virtual forces might change as the structure deforms.. $\endgroup$
    – jrudd
    Commented Apr 26, 2020 at 21:07
  • $\begingroup$ @alephzero I am actually a mathematician by training and have done real analysis via follands book. Do you know where I could find a rigorous treatment of structural analysis? $\endgroup$
    – jrudd
    Commented Apr 27, 2020 at 2:02
  • $\begingroup$ @user521247 The external work is calculated as an integral over the trajectory of the load from its original point of application in the original structure to its final stable configuration in the deformed structure. Therefore, if treated accurately (without all the simplifications), that trajectory will have to include the deformations. If this means more external work was done, then an equal amount of internal work (greater stresses/strains) must also ensue. $\endgroup$
    – Wasabi
    Commented Apr 27, 2020 at 2:15

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