Resistor surge rating
Heat flow takes time. In cases where nearly all of the energy is devoted to raising the temperature and where little useful portion of the heat is allowed to have time needed to significantly flow into its surroundings, then you can use the "action integral of the pulse" to estimate failures. If you can find a specification in "Joules per Ohm" or "\$I^2\cdot s\$ for the resistor, then you could apply it. If not, you'll have to use those curves to make estimates.

The above kinds of specifications are more commonly found for fuses, because that's the job they do and are therefore specified to do. Resistors, on the other hand, are actually designed to dissipate. So this adds another factor to consider.

Instead, let's look at your 2512 curve. It's flat until about \$t=100\:\mu\textrm{s}\$. At the corner, I'm guessing it can handle a pulse of about \$4000\:\textrm{W}\cdot 100\:\mu\textrm{s}=400\:\textrm{mJ}\$. This increases linearly (on a log scale) to about \$18\:\textrm{W}\cdot 1\:\textrm{s}=18\:\textrm{J}\$ for a pulse of \$1\:\textrm{s}\$. Given the log scales here, I get the following equation for the resistor's ability to absorb one pulse of energy over time:

$$\begin{split}E_{limit}&amp;=4000\:\textrm{W}\cdot t\\E_{limit}&amp;=1.91089572\:\textrm{J}\cdot \ln \left(t\right)+18\:\textrm{J}\end{split}\quad\begin{split}&amp;\textrm{ where}\quad t \le 100\:\mu\textrm{s}\\&amp;\textrm{ where}\quad 100\:\mu\textrm{s}\le t \le 10\:\textrm{s}\end{split}$$

This is a hot-spot calculation and it's probably only good to a few times the chart duration, where other factors allow the dissipation to stabilize at the rated power. They only show the curve going out to a second. But the above equation might work for a bit past the end of that curve. Regardless, it gives you an idea.



If I did the integral right, the energy delivered into your R, by your RC circuit, is the following function of time:

$$E_{decay}=\frac{V_0^2\cdot C}{2}\cdot \left(1-e^{-\cfrac{2\cdot t}{R\cdot C}}\right)$$

If this value exceeds \$E_{limit}\$ at any time, you might have a problem. Given that you are measuring up to \$40\:\textrm{A}\$, I'm going to say that your \$V_0=132\:\textrm{V}\$ for the above purposes. So if you look at the case for \$t=100\:\mu\textrm{s}\$, you get \$\approx 462\:\textrm{mJ}\$ which exceeds the rating curve you have. To be safe, you'd probably want to be substantially under it, I think. Not over.

The curve does indicate that, given a little more time, there should be enough time and therefore no remaining problems. But this does seem to suggest a corner case problem when using a single device.

I gather you are using two of them and still having problems. (I'm not sure how all this is mounted and that could also be important.) In any case, if you plug in the \$1.65\:\Omega\$ paired-resistor equivalent, you get \$812\:\textrm{mJ}\$ for both. Which, divided between the two still exceeds the spec (by only a little.)



Just an added note because I had to make a correction to the first equation above, for \$t\lt 100\:\mu\textrm{s}\$. I had just made it a constant before, but it really is a function of time. Less time? Less delivered energy. The curve's flat line there makes that apparent. I'd just failed to account for it in the equation.

So with the correction, you can more easily see that for an even smaller period of time, say \$t=10\:\mu\textrm{s}\$, that the \$E_{decay}\$ equation supplies (using my \$V_0=132\:\textrm{V}\$ figure based on the \$40\:\textrm{A}\$ you had written) for about \$100\:\textrm{mJ}\$ of energy into \$1.65\:\Omega\$. But \$4000\:\textrm{W}\cdot 10\:\mu\textrm{s}=40\:\textrm{mJ}\$ as a limit by the curve. So the curve is far exceeded when considering shorter times like this. Even using \$V_0=100\:\textrm{V}\$, I get \$60\:\textrm{mJ}\$ of energy in that short time. So, still exceeds the specification.

I can see why you are having troubles.