Q:

Strong force and metric expansion

If the expansion of the universe (the metric) continues (and perhaps is accelerating), in a very large but finite time the expansion of the metric will clash with the effects of the strong nuclear force. What happens then? Would the energy required halt the expansion and if so what is the slowing effect of the "stretch" on the strong force today - too small to be measured?
Update: Thanks for the answers so far - I hadn't even considered the big rip scenario. I was merely thinking that as the metric expanded the strong force inside baryons would require unlimited resources of energy to be overcome - but I now understand that is wrong however it still looks to me like a potential perpetual motion machine - continually creating matter/anti-matter pairs. Perhaps that should be a separate question though...

A:

Definitely too small to measure. The expansion isn't even enough to measurably impact galaxy superclusters.
What you are otherwise thinking of is known as the Big Rip. This is not guaranteed to happen, though. It depends on a constant describing the density of dark energy over time, usually denoted $w$. 
When $w<-1$ the Big Rip occurs in finite time, separating all parts of a structure in the sense that lightspeed signals can no longer be transmitted between the parts. This includes hadrons like protons, so in the very last moments the quarks will no longer be able to communicate and will thus cease to be bound together.
Disclaimer: That ignores quantum gravity and QCD in highly curved spacetime. We can barely manage QCD in the flat case, and quantum gravity has been a holy grail for decades. Since a definitive answer to your question requires both of these things, the only definitive answer is "nobody knows for sure".
When $w=-1$ we have constant density and structures smaller than galaxies survive. Our solar system, assuming it survives a very long time, will remain bound together. Star clusters will remain together.
Note, however, that we would still, technically, be able to see everything we could see before. Telescopes can already see galaxies which are currently receding at superluminal speeds. We just see those last moments before such speeds were hit get smeared over ever longer time frames, and heavily redshifted. But we will never receive new signals from the moment it hit superluminal speeds. Only ever the faint remnants of those last moments. The redshifts will eventually be so extreme that everything at galactic distances and beyond becomes effectively impossible to detect these remnants from. At which point it's just you and your closest neighbors.
Current observations put $w$ close to -1, but we cannot currently distinguish between $w<-1$, $w=-1$, and $w>-1$

A:

Excepting a Big Rip scenario, there is no eventual 'clash'. 
Consider a Friedmann–Lemaître–Robertson–Walker universe:
$$\mathrm{d}s^2 = -\mathrm{d}t^2 + a^2(t)\left[\frac{\mathrm{d}r^2}{1-kr^2} + r^2\left(\mathrm{d}\theta^2 + \sin^2\theta\,\mathrm{d}\phi^2\right)\right]\text{,}$$
where $a(t)$ is the scale factor and $k\in\{-1,0,+1\}$ corresponds to a spatially open, flat, or closed cases, respectively. In a local orthonormal frame, the nonzero Riemann curvature components are, up to symmetries:
$$\begin{eqnarray*}
R_{\hat{t}\hat{r}\hat{t}\hat{r}} = R_{\hat{t}\hat{\theta}\hat{t}\hat{\theta}} = R_{\hat{t}\hat{\phi}\hat{t}\hat{\phi}} =& -\frac{\ddot{a}}{a} &= \frac{4\pi G}{3}\left(\rho + 3p\right)\text{,} \\
R_{\hat{r}\hat{\theta}\hat{r}\hat{\theta}} = R_{\hat{r}\hat{\phi}\hat{r}\hat{\phi}} = R_{\hat{\theta}\hat{\phi}\hat{\theta}\hat{\phi}} =& \frac{k+\dot{a}^2}{a^2} &= \frac{8\pi G}{3}\rho\text{,}
\end{eqnarray*}$$
where overdot denotes differentiation with respect to coordinate time $t$ and the Friedmann equations were used to rewrite them in terms of density $\rho$ and pressure $p$. From the link, note in particular that $\dot{\rho} = -3\frac{\dot{a}}{a}\left(\rho+p\right)$.
If dark energy is described by a cosmological constant $\Lambda$, as it is in the standard ΛCDM model, then it contributes a constant density and pressure $\rho = -p = \Lambda/8\pi G$, and so no amount of cosmic expansion would change things. Locally, things look just the same as they ever did: for a universe dominated by dark energy, the curvature stays the same, so the gravitational tidal forces do too. Through those tiny tidal forces, the dark energy provides some immeasurably tiny perturbation on the behavior of objects, including atomic nuclei, forcing a slightly different equilibrium size than would be otherwise. But it is so small that that it has no relevance to anything on that scale, nor do those equilibrium sizes change in time. The cosmological constant holds true to its name.
On the other hand, if dark energy has an equation of state $p = w\rho$, then a flat expanding universe dominated by dark energy has
$$\dot{\rho} = -3\frac{\dot{a}}{a}\left(\rho+p\right) = -3\sqrt{\frac{8\pi G}{3}}\left(1+w\right)\rho^{3/2}\text{,}$$
and immediately one can see that there is something special about $w<-1$, leading to an accumulation of dark energy, while the cosmological constant's $w = -1$ leads to no change. This leads to a Big Rip more generally, as zibadawa timmy's answer explains.

? If the metric expands surely objects get further away from one another and that would include the stars inside galaxaies as well as the galaxies themselves?

Not at all. It wouldn't even make any sense: if you have an object like an atom or a star in gravitational freefall, by the equivalence principle only tidal forces across it are relevant. The tidal forces stretch the object until the internal forces balance them. But for a Λ-driven accelerated expansion, dark energy contribution to tidal forces is constant. Hence, an object already in equilibrium has no reason to further change its size, no matter how long the cosmic acceleration occurs. This also applies to galaxies, only that the internal forces are also gravitational and balance the dark energy contribution.
Looking at this in more detail, imagine a test particle with four-velocity $u$, and a nearby one with the same four-velocity, separated by some vector $n$ connecting their worldlines. If they're both in gravitational freefall, then their relative acceleration is given by the geodesic deviation equation, $\frac{D^2n^a}{d\tau^2} = -R^\alpha{}_{\mu\beta\nu}u^\mu u^\nu n^\beta$. The gravitoelectric part of the Riemann tensor, $\mathbb{E}^\alpha{}_\beta = R^\alpha{}_{\mu\beta\nu}u^\mu u^\nu$, represents the static tidal forces in a local inertial frame comoving with $u$, which will drive those particles apart (or together, depending). Hence, keeping those particles at the same distance would require a force between them, but it's not necessary for this force to change unless the tidal forces also change.
Galaxies don't change size through cosmological expansion. Stars don't either, nor do atoms. Not for Λ-driven expansion, at least. It would take an increase in tidal forces, such as those provided by a Big Rip, for them to do so.
A related way of looking at the issue is this: according to the Einstein field equation, the initial acceleration of a small ball of initially comoving test particles with some four-velocity $u$ (in the comoving inertial frame), which is given by the Ricci tensor, turns out to be:
$$\lim_{V\to 0}\left.\frac{\ddot{V}}{V}\right|_{t=0}\!\!\!\!= -\underbrace{R_{\mu\nu}u^\mu u^\nu}_{\mathbb{E}^\alpha{}_\alpha} = -4\pi G(\rho + 3p)\text{,}$$
where $\rho$ is density and $p$ is the average of the principal stresses in the local inertial frame. For a positive cosmological constant, $\rho = -p>0$, and correspondingly a ball of test particles will expand. That's cosmic expansion on a local scale; because of uniformity of the FLRW universe, the same proportionality works on the large scale as well, as we can think of distant galaxies as themselves test particles if their interactions are too minute to make a difference.
Again, we are led to the same conclusion: if the ball has internal forces that prevent expansion, then those forces don't need to change in time unless the dark energy density also changes, which doesn't happen for cosmological constant. If they're not 'clashing' now, then they won't need to in the future either.

