Oblique Penetration of Spherical Projectile into Low-Carbon Steel Target: Experiment, Theory, 3D Penetration Model

Abstract

To solve the ballistic limit and trajectory deflection angle of a spherical projectile after oblique penetration of a finite-thickness mild steel plate, a ballistic penetration model and its construction method are proposed, which combine the theory of cavity expansion penetration, the practice of projectile and target separation modeling, the technique of shooting line projectile-target intersection. The critical algorithms and calculation steps required to construct the penetration model in three-dimensional space are systematically described. Subsequently, the experiments of 93W spherical projectiles with 6mm diameter penetrating 4, 6, and 8 mm Q345 steel targets at angles of 0°, 20°, and 40° are carried, and obtain the penetration ballistic limit and trajectory. Finally, according to the experiment results, the calculation accuracy of the model is checked, and the results show that the ballistic limit calculation maximum errors for 4, 6, and 8 mm Q345 steel targets are 6.69, 18.24, and 16%. The main work of this paper shows that the penetration model established in this paper can accurately calculate the problem of spherical projectile penetrating finite thickness targets and can provide a new solution method for the problem of projectile penetration.

APPENDIX

The resistance decay model in this paper is derived based on the Forrestal’s spherical dynamic cavity expansion model. The derivation process [44] of Forrestal cavity expansion model is introduced before the derivation of the resistance decay model.

If r is the Euler spherical coordinates, u is the outward positive radial displacement of the particle, the radius of the cavity is $r - u$. Then, the equilibrium equation of axisymmetric sphere during expansion is:

$$\frac{{\partial {{\sigma }_{{rr}}}}}{{\partial r}} + \frac{2}{r}({{\sigma }_{{rr}}} - {{\sigma }_{{\theta \theta }}}) = {{\rho }_{t}}\frac{{d{v}}}{{dt}} = - {{\rho }_{t}}\left( {\frac{{\partial {v}}}{{\partial t}} + {v}\frac{{\partial {v}}}{{\partial r}}} \right).$$

(A.1)

Stresses are assumed to be positive in compression, the continuity equation is:

$$\frac{{\partial {v}}}{{\partial r}} + 2\frac{{v}}{r} = 0.$$

(A.2)

When the cavity expands, the target plate material satisfies the mass conservation equation so:

$${{(r - u)}^{3}} = {{r}^{3}} - {{a}^{3}}.$$

(A.3)

The radial displacement is:

$$u = r\left[ {1 - {{{\left( {1 - \frac{{{{a}^{3}}}}{{{{r}^{3}}}}} \right)}}^{{1/3}}}} \right].$$

(A.4)

In the instantaneous state, the strain of the cavity wall is ignored, as follows:

$$\frac{{\partial u}}{{\partial t}} = \frac{{du}}{{dt}} = \frac{r}{3}{{\left( {1 - \frac{{{{a}^{3}}}}{{{{r}^{3}}}}} \right)}^{{ - \frac{2}{3}}}}\left( { - 3\frac{{{{a}^{2}}\dot {a}}}{{{{r}^{3}}}}} \right) = {{\left( {1 - \frac{{{{a}^{3}}}}{{{{r}^{3}}}}} \right)}^{{ - \frac{2}{3}}}}\frac{{{{a}^{2}}\dot {a}}}{{{{r}^{2}}}}$$

(A.5)

The equation of strain is:

$$\frac{{\partial u}}{{\partial r}} = - 2{{\left( {1 - \frac{{{{a}^{3}}}}{{{{r}^{3}}}}} \right)}^{{ - \frac{2}{3}}}}\frac{{{{a}^{2}}\dot {a}}}{{{{r}^{3}}}} + \frac{{{{a}^{2}}\dot {a}}}{{{{r}^{2}}}}\left( {2{{{\left( {1 - \frac{{{{a}^{3}}}}{{{{r}^{3}}}}} \right)}}^{{ - \frac{5}{3}}}}\frac{{{{a}^{3}}}}{{{{r}^{4}}}}} \right).$$

(A.6)

Ignoring small deformation, radial strain rate can be written as:

$${{\dot {\varepsilon }}_{{rr}}} = - \frac{{\partial {v}}}{{\partial r}} = \frac{{2{{a}^{2}}\dot {a}}}{{{{r}^{3}}}}.$$

(A.7)

And the circumferential and tangential strain rates are:

$${{\dot {\varepsilon }}_{{\theta \theta }}} = {{\dot {\varepsilon }}_{{\varphi \varphi }}} = - \frac{{v}}{r}.$$

(A.8)

The radial engineering strain is:

$${{\varepsilon }_{{rr}}}{\text{ = }}\ln \left( {1 - \frac{{du}}{{dr}}} \right)$$

(A.9)

Tangential engineering strain is:

$${{\varepsilon }_{{\theta \theta }}} = \ln \left( {1 - \frac{u}{r}} \right).$$

(A.10)

Considering that the constitutive model of target plate material is power hardening model, we can get:

$$\sigma = \left\{ \begin{gathered} E\varepsilon ,\quad \sigma < Y \hfill \\ Y{{\left( {\frac{{E\varepsilon }}{Y}} \right)}^{n}},\quad \sigma \geqslant Y \hfill \\ \end{gathered} \right..$$

(A.11)

By substituting Eqs. (A.5)–(A.11) into equilibrium Eq. (A.1), we can get:

$$\frac{{\partial {{\sigma }_{{rr}}}}}{{\partial r}} = - \frac{{2Y}}{r}{{\left( {\frac{{2E}}{{3Y}}} \right)}^{n}}{{\left[ { - \ln \left( {1 - {{{\left( {\frac{a}{r}} \right)}}^{3}}} \right)} \right]}^{n}} - {{\rho }_{t}}\left( {\frac{{\ddot {a}{{a}^{2}} + 2a{{{\dot {a}}}^{2}}}}{{{{r}^{2}}}} - \frac{{2{{a}^{4}}{{{\dot {a}}}^{2}}}}{{{{r}^{5}}}}} \right).$$

(A.12)

By integrating the above Eq. (A.12), we can get:

$${{\sigma }_{{rr}}}(a) = \int\limits_a^c {\frac{{2Y}}{r}{{{\left( {\frac{{2E}}{{3Y}}} \right)}}^{n}}{{{\left[ { - \ln \left( {1 - {{{\left( {\frac{a}{r}} \right)}}^{3}}} \right)} \right]}}^{n}} - {{\rho }_{t}}\left( {\frac{{\ddot {a}{{a}^{2}} + 2a{{{\dot {a}}}^{2}}}}{{{{r}^{2}}}} - \frac{{2{{a}^{4}}{{{\dot {a}}}^{2}}}}{{{{r}^{5}}}}} \right)dr} .$$

(A.13)

Complete integration of the above equation can be obtained:

$${{\sigma }_{{rr}}}(a) = {{\sigma }_{{rr}}}(c) + 2Y{{\left( {\frac{{2E}}{{3Y}}} \right)}^{n}}\int\limits_a^c {{{{\left[ { - \ln \left( {1 - {{{\left( {\frac{a}{r}} \right)}}^{3}}} \right)} \right]}}^{n}}\frac{{dr}}{r} + {{\rho }_{r}}\left[ { - (\ddot {a}{{a}^{2}} + 3a{{{\dot {a}}}^{2}})\left( {\frac{1}{c} - \frac{1}{a}} \right) + \frac{{{{a}^{4}}{{{\dot {a}}}^{2}}}}{2}\left( {\frac{1}{{{{c}^{4}}}} - \frac{1}{{{{a}^{4}}}}} \right)} \right]} ,$$

(A.14)

where ${{\sigma }_{{rr}}}(c)$ is the elastic-plastic boundary pressure.

At first, we solve for the stress in the elastic region. According to the elastic equilibrium equation. Eq. (A.3) can be obtained as follows:

$${{r}^{3}} - 3{{r}^{2}}u + 3r{{u}^{2}} - {{u}^{3}} = {{r}^{3}} - {{a}^{3}}.$$

(A.15)

Since the elastic region displacement u is very small, the quadratic term u in Eq. (A.15) is removed, and the deformation can be obtained as follows:

$$u = \frac{{{{a}^{3}}}}{{3{{r}^{2}}}}.$$

(A.16)

Then the radial strain at the elastic region is:

$${{\varepsilon }_{{rr}}} = \frac{{du}}{{dr}} = \frac{{2{{a}^{3}}}}{{3{{r}^{3}}}}.$$

(A.17)

The circumferential strain in the elastic region is:

$${{\varepsilon }_{{\theta \theta }}} = {{\varepsilon }_{{\varphi \varphi }}} = \frac{{du}}{{dr}} = \frac{{2{{a}^{3}}}}{{3{{r}^{3}}}}.$$

(A.18)

According to generalized Hooke’s law of elastic stage:

$${{\sigma }_{{rr}}} - {{\sigma }_{{\theta \theta }}} = 2G({{\varepsilon }_{{rr}}} - {{\varepsilon }_{{\theta \theta }}}).$$

(A.19)

By substituting Eq. (A.18) into Eq. (A.19), we can get:

$${{\sigma }_{{rr}}} - {{\sigma }_{{\theta \theta }}} = 2G\left( {\frac{{2{{a}^{3}}}}{{3{{r}^{3}}}} - \left( { - \frac{{{{a}^{3}}}}{{3{{r}^{3}}}}} \right)} \right) = 2G\left( {\frac{{{{a}^{3}}}}{{{{r}^{3}}}}} \right).$$

(A.20)

The relationship between shear modulus G and elastic modulus E is as follows:

$$G = \frac{E}{{2(1 + \nu )}}.$$

(A.21)

For metal materials, Poisson’s ratio is 0.5. Then, Eqs. (A.16)–(A.21) are substituted into the stress balance Eq. (A.1) to obtain:

$$\frac{{d{{\sigma }_{{rr}}}}}{{dr}} = - \frac{2}{r}({{\sigma }_{{rr}}} - {{\sigma }_{{\theta \theta }}}) - {{\rho }_{t}}\left( {\frac{{\partial u}}{{\partial t}} + {v}\frac{{\partial u}}{{\partial r}}} \right) = - \frac{{4E{{a}^{3}}}}{{3{{r}^{2}}}} - {{\rho }_{t}}\left( {\frac{{\ddot {a}{{a}^{2}} + 2a{{{\dot {a}}}^{2}}}}{{{{r}^{2}}}} - \frac{{2{{a}^{4}}{{{\dot {a}}}^{2}}}}{{{{r}^{5}}}}} \right).$$

(A.22)

Integral:

$${{\sigma }_{{rr}}}(c) = \frac{{4E{{a}^{3}}}}{{9{{c}^{3}}}} + {{\rho }_{t}}(\ddot {a}{{a}^{2}} + 2a{{\dot {a}}^{2}})\left( {\frac{1}{k} - \frac{1}{c}} \right) - \frac{{{{a}^{4}}{{{\dot {a}}}^{2}}}}{2}\left( {\frac{1}{{{{k}^{4}}}} - \frac{1}{{{{c}^{4}}}}} \right).$$

(A.23)

According to the position relation of the elastic-plastic region shown in Fig. 4, for the case of penetration of the semi-infinite target, there is: $k \to \infty$. Therefore, the dynamic resistance term of the elastic region is:

$${{\rho }_{t}}(\ddot {a}{{a}^{2}} + 2a{{\dot {a}}^{2}})\left( { - \frac{1}{c}} \right) + \frac{{{{a}^{4}}{{{\dot {a}}}^{2}}}}{2}\left( {\frac{1}{{{{c}^{4}}}}} \right).$$

(A.24)

Meanwhile, according to Eq. (A.14), the dynamic resistance term of the plastic region is:

$$ - {{\rho }_{t}}(\ddot {a}{{a}^{2}} + 2a{{\dot {a}}^{2}})\left( {\frac{1}{c} - \frac{1}{a}} \right) - \frac{{{{a}^{4}}{{{\dot {a}}}^{2}}}}{2}\left( {\frac{1}{{{{c}^{4}}}} - \frac{1}{{{{a}^{4}}}}} \right).$$

(A.25)

The sum of the above two formulas is the dynamic resistance term of the dynamic cavity expansion:

$${{\rho }_{t}}(\ddot {a}{{a}^{2}} + 2a{{\dot {a}}^{2}})\left( {\frac{1}{k} - \frac{1}{a}} \right) - \frac{{{{a}^{4}}{{{\dot {a}}}^{2}}}}{2}\left( {\frac{1}{{{{k}^{4}}}} - \frac{1}{{{{a}^{4}}}}} \right).$$

(A.26)

According to Eq. (A.14), the static resistance term of the plastic region is:

$$2Y{{\left( {\frac{{2E}}{{3Y}}} \right)}^{n}}\int\limits_a^c {{{{\left[ { - \ln \left( {1 - {{{\left( {\frac{a}{r}} \right)}}^{3}}} \right)} \right]}}^{n}}\frac{{dr}}{r}} .$$

(A.27)

According to Eqs. (A.23) and (A.27), the sum of static resistance terms of the elastic-plastic interface and the plastic region is:

$${{\sigma }_{{rr}}}(a) = \frac{{4E{{a}^{3}}}}{{9{{c}^{3}}}} + 2Y{{\left( {\frac{{2E}}{{3Y}}} \right)}^{n}}\int\limits_a^c {\frac{{{{{\left[ { - \ln {{{\left( {1 - \frac{a}{r}} \right)}}^{3}}} \right]}}^{n}}}}{r}dr} .$$

(A.28)

Let $x = 1 - {{\left( {\frac{a}{r}} \right)}^{3}}$, substitute into Eq. (A.28) to get:

$${{\sigma }_{{rr}}}(a) = \frac{{4E{{a}^{3}}}}{{9{{c}^{3}}}} + \frac{{2Y}}{3}{{\left( {\frac{{2E}}{{3Y}}} \right)}^{n}}\int\limits_0^{1 - {{{\left( {\frac{a}{c}} \right)}}^{3}}} {\frac{{{{{[ - \ln x]}}^{n}}}}{{1 - x}}} dx.$$

(A.29)

If the moving velocity of the cavity wall is ${{{v}}_{c}}$, and the displacement of the cavity wall is: $a = {{{v}}_{c}}t$, then the displacement of the elastic-plastic interface is:

$$c = {{\left( {\frac{{2E}}{{3Y}}} \right)}^{{1/3}}}{{{v}}_{c}}t.$$

(A.30)

We can get:

$$\frac{a}{c} = {{\left( {\frac{{3Y}}{{2E}}} \right)}^{{1/3}}}.$$

(A.31)

Then the static resistance of the projectile and target contact surface is:

$${{\sigma }_{{rr}}}(a) = \frac{{2Y}}{3} + \frac{{2Y}}{3}{{\left( {\frac{{2E}}{{3Y}}} \right)}^{n}}\int\limits_0^{1 - {{{\left( {\frac{a}{c}} \right)}}^{3}}} {\frac{{{{{[ - \ln x]}}^{n}}}}{{1 - x}}} dx.$$

(A.32)

Equation (A.32) is hill’s static cavity expansion resistance model, corresponding to Eq. (2.6) in paper. Eq. (A.32) and Eq. (A.26) are superimposed and summed to obtain the dynamic cavity expansion resistance model:

$${{\sigma }_{{rr}}}(a) = \frac{{2Y}}{3} + \frac{{2Y}}{3}{{\left( {\frac{{2E}}{{3Y}}} \right)}^{n}}\int\limits_0^{1 - {{{\left( {\frac{a}{c}} \right)}}^{3}}} {\frac{{{{{[ - \ln x]}}^{n}}}}{{1 - x}}} dx + {{\rho }_{t}}\left[ {(\ddot {a}{{a}^{2}} + 2a{{{\dot {a}}}^{2}})\left( {\frac{1}{k} - \frac{1}{a}} \right) - \frac{{{{a}^{4}}{{{\dot {a}}}^{2}}}}{2}\left( {\frac{1}{{{{k}^{4}}}} - \frac{1}{{{{a}^{4}}}}} \right)} \right].$$

(A.33)

When the semi-infinite target is penetrated, $k \to \infty$, $\frac{1}{k}$ and $\frac{1}{{{{k}^{4}}}}$ is zero, then the above equation is simplified as:

$${{\sigma }_{{rr}}}(a) = \frac{{2Y}}{3} + \frac{{2Y}}{3}{{\left( {\frac{{2E}}{{3Y}}} \right)}^{n}}\int\limits_0^{1 - {{{\left( {\frac{a}{c}} \right)}}^{3}}} {\frac{{{{{[ - \ln x]}}^{n}}}}{{1 - x}}} dx + {{\rho }_{t}}\left( {\ddot {a}a + \frac{3}{2}{{{\dot {a}}}^{2}}} \right).$$

(A.34)

Acceleration term $\ddot {a}$ and displacement term a are included in the above equation. Since we cannot measure the acceleration term $\ddot {a}$, the acceleration term $\ddot {a}$ is removed in practical work, and then the model becomes:

$${{\sigma }_{{rr}}}(a) = \frac{{2Y}}{3}\left[ {1 + {{{\left( {\frac{{2E}}{{3Y}}} \right)}}^{n}}\int\limits_0^{1 - \frac{{3Y}}{{2E}}} {\frac{{{{{[ - \ln x]}}^{n}}}}{{1 - x}}} dx} \right] + \beta {{\rho }_{t}}{{\dot {a}}^{2}}.$$

(A.35)

When $\beta$ takes as 1.5, Eq. (A.35) is the Forrestal resistance model, see Eq. (2.7).

The Warren resistance decay model [33] is based on the position relation between the d distance from any point on the surface of the projectile to the expansion elastic-plastic boundary of the target cavity and b the distance to the free boundary of the target plate. We adopt the same judgment method in this paper, but the difference is that in Warren’s paper [33], the target plate material constitutive model is an ideal plastic material model, which satisfies ${{\sigma }_{{rr}}} - {{\sigma }_{{\theta \theta }}} = Y$. In order to ensure the consistency of the resistance model and the resistance decay model, the power hardening model is adopted in this paper, as shown in Eq. (A.11).

When the distance between the projectile surface and the elastic-plastic interface of the target plate is greater than the distance between the projectile surface and the free boundary of the target plate, i.e. $d > b$, according to Eq. (A.33), the acceleration term $\ddot {a}$ is ignored, thus:

$$\sigma (d) = \frac{{2Y}}{3}\left( {1 - {{{\left( {\frac{c}{d}} \right)}}^{3}}} \right) + \frac{{2Y}}{3}{{\left( {\frac{{2E}}{{3Y}}} \right)}^{n}}\int\limits_0^{1 - \frac{{3Y}}{{2E}}} {\frac{{{{{[ - \ln x]}}^{n}}}}{{1 - x}}dx} ] + {{\rho }_{t}}{{\dot {a}}^{2}}\left[ {\frac{3}{2} - \frac{{2a}}{d} + \frac{1}{2}\frac{{{{a}^{4}}}}{{{{d}^{4}}}}} \right].$$

(A.36)

In this case, the elastoplastic interface has not reached the free boundary of the target plate, and the stress in the plastic deformation region has not been affected. When the distance between the projectile surface and the elastic-plastic interface of the target plate is less than the distance between the projectile surface and the free boundary of the target plate, i.e. $d < b$. According to Eq. (A.33), ignoring the acceleration term $\ddot {a}$, we can get:

$$\sigma (d) = \frac{{2Y}}{3}{{\left( {\frac{{2E}}{{3Y}}} \right)}^{n}}\int\limits_0^{1 - {{{\left( {\frac{a}{d}} \right)}}^{3}}} {\frac{{{{{[ - \ln x]}}^{n}}}}{{1 - x}}dx} ] + {{\rho }_{t}}{{\dot {a}}^{2}}\left[ {\frac{3}{2} - \frac{{2a}}{d} + \frac{1}{2}\frac{{{{a}^{4}}}}{{{{d}^{4}}}}} \right].$$

(A.37)

In this case, the stress in the elastic deformation region is 0, and the stress in the plastic deformation region is affected.

For the semi-infinite target penetration problem, the Forrestal model is applied to the resistance model, as follows:

$$\sigma = \frac{{2Y}}{3}\left[ {1 + {{{\left( {\frac{{2E}}{{3Y}}} \right)}}^{n}}\int\limits_0^{1 - \frac{{3Y}}{{2E}}} {\frac{{{{{[ - \ln x]}}^{n}}}}{{1 - x}}dx} } \right] + \frac{3}{2}{{\rho }_{t}}{{\dot {a}}^{2}}.$$

(A.38)

Therefore, the resistance decay coefficient can be written as: $\sigma (d){\text{/}}\sigma$. Written as a piecewise function:

$$\left\{ \begin{gathered} f(d,a,\ddot {a}) = \left\{ \begin{gathered} A{\text{/}}C\begin{array}{*{20}{c}} {}&{}&{d > b} \end{array} \hfill \\ B{\text{/}}C\begin{array}{*{20}{c}} {}&{}&{d < b} \end{array} \hfill \\ \end{gathered} \right. \hfill \\ A = \frac{{2Y}}{3}\left( {1 - {{{\left( {\frac{c}{d}} \right)}}^{3}}} \right) + \frac{{2Y}}{3}{{\left( {\frac{{2E}}{{3Y}}} \right)}^{n}}\int\limits_0^{1 - \frac{{3Y}}{{2E}}} {\frac{{{{{[ - \ln x]}}^{n}}}}{{1 - x}}dx} ] + {{\rho }_{t}}{{{\dot {a}}}^{2}}\left[ {\frac{3}{2} - \frac{{2a}}{d} + \frac{1}{2}\frac{{{{a}^{4}}}}{{{{d}^{4}}}}} \right] \hfill \\ B = \frac{{2Y}}{3}{{\left( {\frac{{2E}}{{3Y}}} \right)}^{n}}\int\limits_0^{1 - {{{\left( {\frac{a}{d}} \right)}}^{3}}} {\frac{{{{{[ - \ln x]}}^{n}}}}{{1 - x}}dx} ] + {{\rho }_{t}}{{{\dot {a}}}^{2}}\left[ {\frac{3}{2} - \frac{{2a}}{d} + \frac{1}{2}\frac{{{{a}^{4}}}}{{{{d}^{4}}}}} \right] \hfill \\ C = \frac{{2Y}}{3}\left[ {1 + {{{\left( {\frac{{2E}}{{3Y}}} \right)}}^{n}}\int\limits_0^{1 - \frac{{3Y}}{{2E}}} {\frac{{{{{[ - \ln x]}}^{n}}}}{{1 - x}}dx} } \right] + \frac{3}{2}{{\rho }_{t}}{{{\dot {a}}}^{2}} \hfill \\ \end{gathered} \right..$$

(A.39)

The above equation is Eqs. (2.9)–(2.12) in Section 2.4. Derivation is completed