On the proof of the Principle of Virtual Work

Question

I'm having some trouble understanding the proof of the Principle of Virtual Work for deformable bodies. I'll give below the proof that I've read, and, next, I'll remark what I'm not understanding.

The first thing to remember before going through the proof is that the virtual work done by a system of virtual forces in equilibrium as a rigid body undergoes a small, compatible displacement is zero.

PROOF:

Suppose that a deformable body is in static equilibrium under the external loads of a virtual $Q$-force system.

Since the body as a whole is in equilibrium, any particular particle can be isolated and will be in equilibrium under the internal virtual $Q$ stresses developed by the external virtual $Q$ forces.

Now suppose that the body is subjected to a small change in shape caused by some other source than the virtual $Q$-force system. Owing to this change in shape, any particle might be deformed as well as translated and rotated as a rigid particle. Hence, the boundaries of such a particle would move and hence do virtual work. Let the virtual work done by the $Q$ stresses on the boundaries of the differential particle be designated by $\text{d}W_s$. Part of this virtual work will be done because of the movements of the boundaries of the particle caused by the deformation of the particle itself; this part will be called $\text{d}W_d$. The remaining part of $\text{d}W_s$ will be the virtual work done by the $Q$ stresses during the remaining part of the displacement of the boundaries and will be equal to $\text{d}W_s-\text{d}W_d$. However, this remaining is caused by the translation and rotation of the particle as a rigid body, and, as reminded above, the virtual work done in such a case is equal to zero. Hence

$$\text{d}W_s=\text{d}W_d$$

If the virtual work done by the $Q$ stresses on all particles of the body is now added, this equation becomes

$$W_s=W_d$$

To evaluate first $W_s$, we recognize that this term represents the total virtual work done by the virtual $Q$ stresses on all the boundaries of all the particles. However, for every internal boundary of a particle there is an adjoining particle whose adjacent boundary is actual the same line on the body as whole, and therefore these adjacent boundaries are displaced exactly the same amount. Since the forces acting on the two adjacent internal boundaries are numerically equal but opposite in direction, the total virtual work done on the pair of adjoining internal boundaries is zero. Hence, since all internal boundaries occur in pairs, there is no net virtual work done by the forces on all the internal boundaries. $W_s$ therefore consists only of the work done by the external $Q$ forces on the external boundaries.

$W_d$, on the other hand, was obtained by integrating the virtual work associated with deformation of the element. This work includes the effects of all forces on the element, both stress resultants and external forces. However, when an element deforms, only the stress resultants perform any work. Thus, Wd represents the virtual work done by the stress resultants alone.

END OF PROOF

I understood the part concerning $W_s$.

However, I don't understand why "when an element deforms, only the stress resultants perform any work".

Next, it is said that "$W_d$ represents the virtual work done by the stress resultants alone". But, according to the paragraph pertaining to $W_s$, this work should be zero, since all particle's internal boundaries occur in pairs and that one side of an internal boundary of a particle undergoes the same displacement as the common side of a neighboring particle and such two sides are subjected to equal and opposite stresses, respectively.

If we were to apply this principle according to my (flawed) conclusion, all points of the deformable body would undergo no displacement at all: a complete nonsense.

Where's the flaw in my reasoning?

Answer 1

Imagine a one dimensional beam with its left edge fixed at $x=0$ and its right edge at $x=L$. Now imagine you push on the right edge. What happens? The beam will contract. Let's concentrate on a infitissemal region at some point in the middle of the bar.

What happens to this region? The proof you quote breaks the movement of this region into two pieces: a rigid motion, and a strain. The rigid motion cannot lead to any work over the little region, basically because the stress is uniform throughout the beam, so you have opposite forces on the left and right sides of the region, but the same displacement so total work is zero (this is the reminder before the proof).

But the strain can cause work. The strain component of the motion corresponds to the center of the region remaining stationary and the ends moving in. In this case, the force is opposite at both ends, but the displacement is oppositie as well, so you get a net work. When we look at the work this way, the left boundary of our region moves to the right, while the right boundary of hte region to the left moves to the left, so there is no obvious cancellation.

Hopefully this explains why $W_d$ is non-zero, but to make things even more obvious, lets consider a beam of length $8L$ that gets compressed to a length of $4L$ upon application of a force $F$. Let's break this beam into four peices originally of length $2L$. In each piece's center of mass frame, both the ends move in by $L/2$, so the work done is $2 F L/2=FL$. and so the total work done is $4FL$. This is how you calculate things the $W_d$ way.

Instead of looking at things in the frame of each piece, you can look at things in the lab frame. For the leftmost piece, the left end doens't move, and the right end moves by $L$, so the work done on it is $FL$. For the next piece, the left end moves to the left by $L$, while the right end moves to the left by $2L$, so thet net work is $-FL+2FL$, which equals $FL$. Proceding in a similar way, the work ion the third and fourth pieces are $-2L+3L$ and $-3L + 4L$ respectively. So again we have that the work done on all pieces is $FL$.But we notice that the work at the interior interfaces cancel when we do the sum for the total work, $0 + FL - FL + 2FL-2FL+3FL-3FL+4FL$, thus the full work can be written just in terms of forces on the surface, and the interior contributions cancel.

The above answer was copied from the physics stack exchange site.

Answer 2

The proof is, as should be expected, correct. But it can be made clearer if broken down in additional steps.

Consider, again, a small particle of the deformable body. The total work done by the virtual forces, $\text{d}W_t,$ system is given by the equality $$\text{d}W_t=\text{d}W_r+\text{d}W_d,$$where $\text{d}W_r$ is the work done due to the displacement of the particle as a rigid body, and $\text{d}W_d$ is the work done due to the displacement of the points on the boundaries of the particle as they deform. As said before, $\text{d}W_r=0$, and so $$\text{d}W_t=\text{d}W_d.$$ Integration of the last equation over the entire volume gives $$W_t=W_d.$$

To evaluate $W_t$ let's consider one of the internal boundaries of the particle in question (one should imagine the particle as having a cubic or square shape). Call it boundary $B$. The total work done at such boundary by the internal virtual stresses, $\text{d}W_t^B,$ is given by the equality $$\text{d}W_t^B=\text{d}W_r^B+\text{d}W_d^B,$$ where $\text{d}W_r^B$ is the work done by the internal virtual stresses due to the displacement of the points of boundary $B$ as the particle undergoes a displacement as a rigid body, and $\text{d}W_d^B$ is the work done by the internal virtual stresses as boundary $B$ deforms. It should be noted that $\text{d}W_r^B$, $\text{d}W_d^B$ and $\text{d}W_r^B+\text{d}W_d^B$ are generally not equal to zero.

Now, for every internal boundary of a particle there is an adjoining particle that has a common boundary to the particle initially considered, and therefore these adjacent boundaries are displaced exactly the same amount. Since the forces acting on the two adjacent internal boundaries are numerically equal but opposite in direction, the total virtual work done on the pair of adjoining internal boundaries is zero. Hence, since all internal boundaries occur in pairs, there is no net virtual work done by the forces on all the internal boundaries. Hence, since all internal boundaries occur in pairs, there is no net virtual work done by the forces on all the internal boundaries. $W_t$ therefore consists only of the work done by the external forces on the external boundaries.

So, what should be noted in the proof is that it is the total work done at boundary $B$ at the original particle plus the total work done at boundary $B$ of the adjoining particle that adds up to zero. If we were to consider the sum $\text{d}W_d^B + \text{{d}W_d^B}'$ (where $\text{{d}W_d^B}'$ is the work done by the virtual stresses on the adjoining particle as its boundary $B$ deforms), then, for the reasons pointed by Brian Moths in his answer, such sum would not be zero.