2.1 Stress cycle due to drop impact
The main building blocks of the existing model that is described in
detail in14,15 are the water –hammer pressure, which
originates from compression of the liquid on impact of the surface, see
Figure 1 and the resulting stress cycle from the Rayleigh wave. The
water-hammer pressure \(p_{\text{wh}}\) is given
by14,22:
\begin{equation}
p_{\text{wh}}=v_{d}\frac{\rho_{w}c_{w}}{\left(\frac{\rho_{w}c_{w}}{\rho_{m}c_{l,m}}+1\right)}\nonumber \\
\end{equation}\begin{equation}
c_{w}=c_{w0}+kv_{d}\nonumber \\
\end{equation}\(v_{d}\) = drop impact velocity,
\(\rho_{w}\) = density of water at 1 bar, 1000 kg/m3,
\(c_{w}\) = speed of sound in water at the pressure \(p_{\text{wh}}\),
\(c_{w0}\) = speed of sound in water at a pressure of 1 bar, 1647 m/s,
\(k\) = constant for the pressure influence on the speed of
sound22, 1.921,
\(\rho_{m}\) = density of metal,
\(c_{l,m}\) = longitudinal wave velocity of metal.
Due to the impact and the assumed round shape of the droplet, the
initial velocity of the boundary of the contact area with the solid
surface is infinitely high. This velocity \(v_{a}\) decreases rapidly to
respectively the longitudinal wave velocity (\(c_{l,m}\)), the
transverse wave velocity (\(c_{t}\)), the Rayleigh wave velocity
(\(c_{R}\)), and finally to the speed of sound in water for this
compressed state (\(c_{w}\), see Eq. 1). When \(v_{a}<c_{w}\) the
compressed water volume losses its high pressure (\(p_{\text{wh}}\)).
The radius of the contact area (\(a\)) of the droplet with the rigid
surface as a function of time (\(t\))23 is described
by:
\begin{equation}
a=\sqrt{\left({2R}_{d}v_{d}t-\left(v_{d}t\right)^{2}\right)}\nonumber \\
\end{equation}The radial velocity of this contact area boundary (\(v_{a}\)) is defined
by:
\begin{equation}
v_{a}=\frac{\text{da}}{\text{dt}}=\frac{\left(R_{d}v_{d}-{v_{d}}^{2}t\right)}{\sqrt{\left(2R_{d}v_{d}t-\left(v_{d}t\right)^{2}\right)}}\nonumber \\
\end{equation}With
\begin{equation}
R_{d}=\frac{d_{d}}{2}\nonumber \\
\end{equation}At the time when the surface Rayleigh wave starts \(v_{a}\) equals
cR and the radius of the contact area (\(r_{0}\)) reads:
\begin{equation}
r_{0}=\frac{d_{d}c_{R}}{2v_{d}}\left\{\sqrt{1+2\left(\frac{v_{d}}{c_{R}}\right)^{2}}-1\right\}\nonumber \\
\end{equation}The compressed water volume loses its high pressure when the velocity of
its contact area boundary decreases to the speed of sound in water for
this compressed state, \(c_{w}\). The maximum contact area for which the
water-hammer pressure acts follows from the condition:\(v_{a}=c_{w}\). Replacing \(c_{R}\) by \(c_{w}\) in Eq. 4 and using
Eq. 1b gives the related contact radius (\(r_{\text{wh}}\)):
\begin{equation}
r_{\text{wh}}=\frac{d_{d}\left(c_{w0}+kv_{d}\right)}{2v_{d}}\left\{\sqrt{1+2\left(\frac{v_{d}}{c_{w0}+kv_{d}}\right)^{2}}-1\right\}\nonumber \\
\end{equation}In this work, it is chosen to select one droplet diameter for all model
predictions, while the impact velocity is varied. A droplet diameter\(d_{d}\) = 1.8 mm serves as representative value, within the presented
range by Heymann10 of 1.2 to 2.0 mm. The drop impact
velocities are taken equal to 140, 210 and 400 m/s, similar to the
experimental work that is used by Heymann10.
For this droplet diameter, the impact velocities result for AISI 316 in
contact radii (\(r_{0}\)) of respectively 0.04, 0.07 and 0.13 mm, and
radii (\(r_{\text{wh}}\)) of respectively 0.07, 0.09 and 0.15 mm.
The water-hammer pressure (\(p_{\text{wh}}\)) gives rise to cyclic
stresses (\(S\)). Cyclic stresses are well known to cause fatigue, in
this case surface fatigue. The maximum stress of this cycle due to the
Rayleigh surface wave can be given by14:
\begin{equation}
S_{\max}=\frac{A}{r^{n}}\nonumber \\
\end{equation}In which \(A\) depends on the water-hammer pressure (\(p_{\text{wh}}\)),\(n\) is a constant and \(r\) is the radial coordinate. By assuming that
the stress cycle starts at \(r=r_{0}\) (see Eq. 4) and is attenuated
at \(r=r_{1}\) at a maximum stress level \(S_{\max}=S_{max,1},\)which is equal to the fatigue limit after which no further fatigue
damage occurs, the stress cycle is known. The value \(A\) in Eq. 6 is
determined by Slot et al.14,15 for \(d_{d}\) = 1.8 mm
and \(n\) = 0.5, and reads:
\begin{equation}
A=0.60p_{\text{wh}}\text{\ \ \ \ \ \ \ \ \ \ \ }\left(d_{d}=1.8\ mm\right)\nonumber \\
\end{equation}