# Converting impact speed to pressure magnitude

In explosive safety and stability testing, a drop test is commonly used to determine the sensitivity to impact. In the test, an impactor of known mass is dropped. The initial height varies throughout the experiment. The detonation status is recorded in the material and various techniques are used to convert the heights into metrics used to characterize the stability; for instance, a commonly used metric is $h_{50}$ or the height of the impactor that generates a detonation 50% of the time.

The explosive material being tested is typically in a rod or puck shape whose radius is considerably smaller than the impactor and anvil. The impact generates a shock wave through the material sample.

Since I am using an Eulerian code to simulate this, I need to know the pressure of the resulting shock wave in the material. Is there a way to determine this pressure?

Let's assume that I know all of the material properties (density, speed of sound, Youngs Modulus, etc), that the impactor is perfectly rigid, and that the impact is within the elastic limits of the material sample.

The only approach I can think of would be to treat the problem as 1D rod and assign an initial velocity (say in the $-x$ direction) to the rod such that the momentum is the same as that of the impactor used in the experiment. Then impose that the velocity is zero on the left edge of the rod. This would require some numerical work and is less than ideal. Perhaps there is an analytical or empirical relation that exists between impactor momentum and the resulting pressure wave magnitude?

Background
There is a good reference1 on the physics of sound/shock waves in solids (look at Chapter XI). I found the following (on page 688) very interesting and relevant to your question:

In a solid or liquid, a shock wave with a strength of even a hundred thousand atmospheres is regarded as weak. Such a wave differs little from an acoustic wave: it travels with a speed close to the speed of sound, compresses the material by only a few percent or perhaps of the order of ten percent, and imparts a velocity to the material behind the front which is of the order of a tenth the velocity of the wave itself... then a strong shock wave for condensed media is one whose pressure is not less than tens or hundreds of millions of atmospheres.

Let us define $P$ as the pressure and $\varepsilon$ as the internal energy of a solid material. These can be divided into two parts: an elastic (subscript $c$) and thermal part. $P_{c}$ and $\varepsilon_{c}$ depend only upon the density of the material, $\rho$, or the specific volume, $V$ = $1/\rho$. These are equal to the total pressure and specific internal energy at absolute zero or $T = 0 \ K$. Let us assume that the specific volume at $T = 0$ and $P = 0$ is given by $V_{oc}$, which is only ~1-2% smaller than the specific volume at STP, $V_{o}$, for most metals.

The potential energy curve, or curve defining $\varepsilon_{c}$, is qualitatively similar to the potential energy curve describing the interaction between two atoms as a function of the intranuclear distance, $\Delta x_{n}$. When $V > V_{oc}$, the attractive forces dominate but fall off rapidly as the intranuclear distances increase (e.g., as $T$ increases). In other words, when the atoms move further apart $\varepsilon_{c}$ will asymptotically increase to some value $U$, which is roughly the binding energy of the atoms in the body. Thus, $U$ represents the energy required to remove all atoms from the object to infinity, which is roughly equal to the heat of vaporization for the material (I wrote some more details on the heat of vaporization and provided several useful links in this answer). For instance, the heat of atomization (similar to vaporization) for iron is roughly $415 kJ mol^{-1}$ or ~4.3 eV/atom. Thus, $\varepsilon_{c} (V) \rightarrow U$ as $\Delta x_{n} \rightarrow \sim 2$.

Conversely, the repulsive forces dominate if $V < V_{oc}$. We can define this quantitatively by considering that the work done by compressing the material will be equal to the increase in internal energy. In other words: $$P_{c} = - \left( \frac{ d \varepsilon_{c} }{ d V } \right)_{T = 0}$$ which is equivalent to saying it is the isothermic/isentropic equation for cold compression. The negative sign shows that if a tensile force were applied to the body, the binding forces between atoms would act as a restoring force. The slope of the $P_{c}$ curve at $P = 0$ (or 1 atm) defines the compressibility of the material under normal conditions (i.e., $T = T_{o} \sim 300 \ K$). This is given by: $$\kappa_{o} = - \frac{ 1 }{ V_{o} } \left( \frac{ \partial V }{ \partial P } \right)_{T_{o}}$$ Note that the slope of $\kappa_{o}$ defines the speed of elastic waves within the object. Thus, let us define the speed of sound in the solid as this speed, given by: $$C_{o} = \sqrt{ - V^{2} \left( \frac{ \partial P }{ \partial V } \right)_{S} }$$ where the subscript $S$ indicates an isentropic derivative and the partial derivative will be negative to avoid imaginary speeds of sound.

Simple Zeroth Order Approximation
My knee-jerk assumption is that the simplest approach, given that you assume elastic collision relations, is to just approximate the $\Delta \varepsilon_{c}$ by the final kinetic energy of your impacting object, assuming that the impactee(?) does not move after impact.

First Order Approximation
[The following comes from Chapter XI, Sections 3.14-3.16 in Reference 1]

Below we will consider the effects on a cylindrical rod (used for symmetry and simplicity).

For small deformations, the relative change in volume, $\Delta V/V$, is given by: $$\frac{ \Delta V }{ V } = - \kappa \ P = - \frac{ P }{ K }$$ where $K = 1/\kappa$ is the bulk modulus.

Let us define $C_{1}$ as the speed of a compression wave in the material due to the application of a constant pressure, $P$, applied to one end of the rod at some initial time. The material between the wave front and the end of the rod contracts at a constant speed, $u$. Under these conditions, we can use Hooke's law and show that for small loads and deformations we have: $$\frac{ u }{ C_{1} } = \frac{ P }{ E }$$ where $E$ is Young's modulus. After some time, $t$, the mass of material encompassed by the wave will acquire a momentum $\rho \ C_{1} \ t \ u$, which must be equal to $P \ t$ from Newton's law, which gives us: $$P = \rho \ u \ C_{1}$$