# Study on Model of Penetration into Thick Metallic Targets with Finite Planar Sizes by Long Rods

Authors: Juan Wang; Junhai Zhao; Jianhua Zhang; Yuan Zhou
Journal: Shock and Vibration
 (2021)

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## Abstract

A finite cylindrical cavity expansion model for metallic thick targets with finite planar sizes, composed of ideal elastic-plastic materials, with penetration of high-speed long rod is presented by using the unified strength theory. Considering the lateral boundary and mass abrasion of the target, the penetration resistance and depth formulas are proposed, solutions of which are obtained by MATLAB program. Then, a series of different criteria-based analytical solutions are obtained and the ranges of penetration depth of targets with different ratios of target radius to projectile radius (rt/rd) are predicted. Meanwhile, the numerical simulation is performed using the ANSYS/LS-DYNA finite element code to investigate the variations in residual projectile velocity, length, and mass abrasion. It shows that various parameters have influences on the antipenetration performance of the target, such as the strength coefficient b, rt/rd, the shape of the projectile nose, and the impact velocity of the projectile, among which the penetration depth has increased by 18.95% as b = 1 decreases to b = 0 and has increased by 32.28% as rt/rd = 19.88 decreases to rt/rd = 4.9.


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## Body

## 1. Introduction

The cavity expansion theory (CET) is the main theoretical method to study the penetration problem [1–5]. Since a medium-low speed projectile penetrating into a target travels along an approximately straight trajectory, its mass consumption can be ignored, and the projectile can be considered a rigid projectile [6, 7]. However, when a high-speed long rod penetrates into a target, the projectile-target interface stress is greater than the yield strength of the material and is sufficient to deform and erode the projectile [8]. To properly reflect the main physical process of penetration, Tate’s eroding rod model is developed [9]. For years, investigators have successfully applied the combination of Tate’s model and the CET to penetration problems with high-speed projectiles and have achieved many results by adopting different constitutive models to study different targets [10]. To consider all stress components of a unit and make the results be applied to all kinds of materials, the unified strength theory (UST) has been used to analyze the dynamic response of long rods penetrating into different targets [10–12] and to establish the calculation model of penetration depth, whose results are the closest to test results. However, most of the existing theoretical studies on the problems of high-speed impact have assumed that the planar size of the target is infinite, not considering the influence of the lateral free boundary of the target. The few investigations of penetration into thick targets that are finite in radial extent by projectiles at high speed are limited to experiments [13, 14]. Given this, Jiang et al. [15] have proposed the finite cylindrical cavity expansion (FCCE) theory and the penetration model of thick metallic targets with finite planar sizes by long rod systematically. Nevertheless, this model is applied only to materials whose ultimate shear strength is 0.577 times the tension or compression yield strength because it has adopted the von Mises yield criterion, and it has not made comprehensive parametric analysis.To further study the mass erosion of a long rod at a high-speed impact and the effects of the lateral free boundaries, expand the scope of application for the solution, and ensure the material antipenetration properties to be played fully, using the UST to consider the effect of the second principal stress, a new FCCE model is built to deduce the radial pressure solution for the cavity wall of a thick metallic target composed of an ideal elastic-plastic material and with finite planar sizes when it is penetrated by a high-speed long rod. Furthermore, the resistance formula and penetration depth formula are obtained, which are also adapted to the semi-infinite metallic target. The penetration process is simulated with the explicit finite element software LS-DYNA to study the variation in the motion behavior of the projectiles. Comparisons of the presented model, the previous test, and the model in other documents show that the presented model in this paper has higher precision and incorporates the case in [15]. In the MATLAB program, the penetration depth ranges of metallic targets with different rt/rd are predicted. In addition, the factors influencing the terminal ballistic effects are studied. The conclusions of this study can ensure the material properties to be played fully and will be helpful for the design of protective structures.


## 2. UST

The UST was established in 1991 which considers the effects of all stress components and can be applied to all kinds of materials. Ifσ1, σ2, and σ3 are maximum principal stress, intermediate principal stress, and minimum principal stress of a unit, respectively, the mathematical expression of the UST is as [16](1a)σ1−α1+bbσ2+σ3=σt,σ2≤σ1+ασ31+α,(1b)11+bσ1+bσ2−ασ3=σt,σ2≥σ1+ασ31+α.Here,α is the tension-compression strength ratio; the influence coefficient b0≤b≤1 shows intermediate principal stress which has an influence on material yield, which is called yield criterion coefficient.


## 3. FCCE Model Based on UST

### 3.1. Computation Model

The FCCE model [15] is shown in Figure 1. The target has radius rt, its cavity radius is rc at time t (r˙c is constant), and its elastic-plastic boundary radius is rp. The whole expansion that rc increases from 0 to rcf (the end value) can be divided into two stages. The first stage is the elastic-plastic stage rp<rt, its cylinder consists of the cavity region 0<r≤rc, plastic region rc<r≤rp, elastic region rp<r≤re, and undisturbed region re<r≤rt. When re=rt, the undisturbed region disappears, and when rp=rt, this stage ends. The second stage is the plastic stage rp≡rt, and the cylinder consists only of the cavity region and plastic region. When there appear cracks on the cylinder, the plastic stage ends.Figure 1
FCCE model.The FCCE model is an axisymmetric plane strain problem;σzσz=m/2σr+σθ is the intermediate principal stress, where m is this stress parameter (2υ≤m≤1, υ is Poisson’s ratio of the material) and m=1 in the plastic region [17]. Because of σ1≥σ2≥σ3, if the pressure is positive, σ1=σr, σ2=σz, and σ3=σθ, then σ2≤σ1+ασ3/1+α. For metallic materials, we generally take α=1, B=21+bσt/2+b. With equation (1a), we obtain(2)σr−σθ=21+bσt2+b=B.For ideal elastic-plastic materials, the strain-stress relation can be written as(3a)σ=Eεelastic region,(3b)σ=σoyplastic region.Combining equations (2) and (3a)-(3b), we can get(4a)σr−σθ=E1+υεr−εθelastic region,(4b)σr−σθ=21+bσoy2+bplastic region,where σr and σθ are the radial and hoop stress, respectively; εr and εθ are the radial and hoop strain, respectively; and E is the elastic modulus of the material.If the displacement of the particle iss, whose spatial coordinates are r at time t, the geometric relations are as follows [15]:(5)εr=−∂s∂r,εθ=−srelastic region,(6)εr=−∂s/∂r1−∂s/∂r,εθ=−sr−splastic region.In the elastic zone, with equation (5) and the particle displacement field relation, one obtains [15](7)εr−εθ≈rc2r2.Combining equations (7) and (4a)-(4b), the stress-strain relationship is given as(8a)σr−σθ=Erc21+υr2elastic region,(8b)σr−σθ=21+bσoy2+bplastic region.The velocity field of particles for the FCCE model is as follows [15]:(9)v=rcrr˙c.Based on the momentum conservation, we can get(10)∂σr∂r+σr−σθr=−ρ∂v∂t+v∂v∂r,where ρ is the density of the material.Equations (5), (6), and (8a)∼(10) form the fundamental equations of the FCCE model for ideal elastic-plastic materials based on UST.


### 3.2. Cavity Wall Radial Stress Calculation

#### 3.2.1. Elastic-Plastic Stagerp<rt

Whenrc<r≤rp, substituting equations (2) and (9) into (10) yields(11)∂σr∂r=−Br−ρr˙c21r−rc2r3.Integrating equation (11) yields(12)σr=−Blnr−12ρr˙c22lnr+rc2r2+M1,where M1 is a constant.According to the boundary conditionσr|r=rc=σrc, M1 is formulated as follows:(13)M1=σrc+Blnrc+12ρr˙c22lnrc+1.Combining equations (13) and (12), the radial stress is as follows:(14)σr=σrc+B2lnrc2r2+12ρr˙c21+lnrc2r2−rc2r2.Substituting the plastic region side condition of the elastic-plastic boundaryσr|r=rp=σrp− into equation (14), the radial stress on the cavity wall σrc can be formulated as(15)σrc=σrp−B2lnrc2rp2−12ρr˙c21+lnrc2rp2−rc2rp2,where σrp and rc/rp are determined by the continuity conditions and yield condition of the elastic-plastic boundary.Substituting equations (9) and (8a) into (10), we can get(16)∂σr∂r=−E1+υ×rc2r3−ρr˙c21r−rc2r3.Whenrc<rtr˙c/c, re<rt. Integrating equation (16) and combining equation (9) and the momentum conservation condition [15] σr|r=re=crc/r˙c=ρcv, we have(17)σr=E21+υrc2r2−r˙c2c2+12ρr˙c2×2+r˙c2c2+2lncrcr˙cr−rc2r2.Whenrc≥rtr˙c/c, re=rt. Integrating equation (17) and considering σr|r=rt=0, we have(18)σr=E21+υrc2r2−rc2rt2+12ρr˙c2×lnrt2r2+rc2rt2−rc2r2.Because of the continuity of the stresses and particle velocity at the elastic-plastic boundary [18], according to equations (8a)-(8b), one obtains(19)rcrp=B1+υE.Whenrp=rt, the cavity radius rc1 at the end of the elastic-plastic stage can be expressed as(20)rc1=rtB1+υE.In the elastic-plastic stage,rc≤rc1.For the plane strain problems, the elastic wave velocityc can be expressed as [18,19].(21)c=E1−υρ1+υ1−2υ.Combining equations (15) and (17)∼(21), the radial stress of the cavity wall in this stage σrc1 is derived through the stress continuity conditions σr|r=rp=σrp:(22)σrc1=σrc0+Δσrc0+12ρr˙c2×n.Here, when rc<rtr˙c/c, σrco=B/2+B/2lnE/B1+υ, Δσrc0=0, and d=υ/1−υ+ρr˙c1+υ1−2υ/E1−υ+lnE1−υ/ρr˙c1+υ1−2υ; when rtr˙c/c≤rc≤rc1, σrco=B/2+B/2lnE/B1+υ, Δσrc0=−Erc2/21+υrt2, and n=lnrt2/rc2+rc2/rt2−1,where σrco is the strength term of the CET for a semi-infinite target, Δσrco is the term reflecting the effect of planar size on the strength, and n is the flow resistance coefficient.From equation (22), the elastic wave does not reach the lateral boundary of the target when rc<rtr˙c/c, so there is an undisturbed zone existing at the target edge, and Δσrco=0 so that the planar size has no effect on the target strength and flow resistance coefficient.


#### 3.2.2. Plastic Stagerp≡rt

Because ofσr|r=rt=0 and equation (14) (the stress field in the plastic region), we can get the radial stress on the cavity wall in this stage σrc2:(23)σrc2=Blnrtrc+12ρr˙c2lnrt2rc2+rc2rt2−1,rc1≤rc≤rc2,where rc2 is the cavity radius at the end of the plastic stage, which is gotten by the fracture criterion in [20], where εeq and εf are the equivalent strain and the uniaxial tension fracture strain, respectively.Due to the boundary conditionr=rt, that is, σr=0, we obtain σ2≤σ1+ασ3/1+α from equations (1a)-(1b), and we can get the equivalent stress based on UST σeq:(24)σeq=2+b21+bσr−2+b21+bσθ=−2+b21+bσθ.For the FCCE problem, on the basis of the plastic work hypothesis [21, 22] and the stress-strain relationship, we have(25)σeqεeq=σε=σθεθ,where εeq is the equivalent strain based on UST. According to equations (24) and (25), εeq can be given as(26)εeq=−21+bεθ2+b.Considering equation (26), the strain components εr=−εθ≈1/2rc/r2 [15], and εeq=εf when r=rt [20], we can get the cavity radius at the end of the plastic stage rc2:(27)rc2=rt2+bεf1+b.Combining equations (20) and (27) and considering rc2>rc1, the conclusion in this paper is applied to materials as(28)εf>B21+υ2E,which are generally satisfied by metal materials.The theory quoted by this paper assumes that the elastic-plastic boundary movement velocity is less than the elastic wave velocity,r˙p≤c. According to equations (19) and (21), when r˙p=c, the maximum cavity expansion velocity r˙cmax applied for materials in this paper meets the following conditions:(29)r˙cmax=B1−υρ1−2υ.Whenrc1≤rcf<rc2, the expansion process experiences the elastic-plastic stage and the plastic stage, but when rcf≤rc1, the elastic wave has not reached or has just reached the target lateral boundary, and the undisturbed region exists or has just disappeared, at which point the expansion process experiences only the elastic-plastic stage; when rcf<rtr˙c/c, the target size shows no effect on the expansion process; when rcf>rc2, the target has been fractured.


### 3.3. Energy Consuming by Expanding and Mean Stress on Cavity Wall

According to [15], the energy consumed by expanding W and the work done by the radial pressure on the cavity wall σrc are equal during the whole process of the cavity radius rc increasing from 0 to rcf. Taking a target of unit thickness, we can obtain(30)W=2π∫0rcfσrcrcdrc.Supposing the mean radial stress of the cavity wall isσ¯rc and considering that the energy consumed by expansion remains unchanged, we obtain [15](31)σ¯rc=Wπrcf2=2rcf2∫0rcfσrcrcdrc.Substituting equations (31) into (18) and (19), one obtains(32)σ¯rc=A1+B1×12ρr˙c2,where A1=B/21+lnE/B1+υ and B1=υ/1−υ+1+υ1−2υρr˙c2/E1−υ+lnE1−υ/1+υ1−2υρr˙c2, when rcf<rtr˙c/c; A1=B/21+lnE/B1+υ−Ercf2/41+υrt2 and B1=lnrt2/rcf2+rcf2/2rt2, when rtr˙c/c≤rcf<rc1; A1=B/2lnrt2/rcf2+1 and B1=rcf2/2rt2+lnrt/rcf−1/2, when rc1≤rcf<rc2.


## 3.1. Computation Model

The FCCE model [15] is shown in Figure 1. The target has radius rt, its cavity radius is rc at time t (r˙c is constant), and its elastic-plastic boundary radius is rp. The whole expansion that rc increases from 0 to rcf (the end value) can be divided into two stages. The first stage is the elastic-plastic stage rp<rt, its cylinder consists of the cavity region 0<r≤rc, plastic region rc<r≤rp, elastic region rp<r≤re, and undisturbed region re<r≤rt. When re=rt, the undisturbed region disappears, and when rp=rt, this stage ends. The second stage is the plastic stage rp≡rt, and the cylinder consists only of the cavity region and plastic region. When there appear cracks on the cylinder, the plastic stage ends.Figure 1
FCCE model.The FCCE model is an axisymmetric plane strain problem;σzσz=m/2σr+σθ is the intermediate principal stress, where m is this stress parameter (2υ≤m≤1, υ is Poisson’s ratio of the material) and m=1 in the plastic region [17]. Because of σ1≥σ2≥σ3, if the pressure is positive, σ1=σr, σ2=σz, and σ3=σθ, then σ2≤σ1+ασ3/1+α. For metallic materials, we generally take α=1, B=21+bσt/2+b. With equation (1a), we obtain(2)σr−σθ=21+bσt2+b=B.For ideal elastic-plastic materials, the strain-stress relation can be written as(3a)σ=Eεelastic region,(3b)σ=σoyplastic region.Combining equations (2) and (3a)-(3b), we can get(4a)σr−σθ=E1+υεr−εθelastic region,(4b)σr−σθ=21+bσoy2+bplastic region,where σr and σθ are the radial and hoop stress, respectively; εr and εθ are the radial and hoop strain, respectively; and E is the elastic modulus of the material.If the displacement of the particle iss, whose spatial coordinates are r at time t, the geometric relations are as follows [15]:(5)εr=−∂s∂r,εθ=−srelastic region,(6)εr=−∂s/∂r1−∂s/∂r,εθ=−sr−splastic region.In the elastic zone, with equation (5) and the particle displacement field relation, one obtains [15](7)εr−εθ≈rc2r2.Combining equations (7) and (4a)-(4b), the stress-strain relationship is given as(8a)σr−σθ=Erc21+υr2elastic region,(8b)σr−σθ=21+bσoy2+bplastic region.The velocity field of particles for the FCCE model is as follows [15]:(9)v=rcrr˙c.Based on the momentum conservation, we can get(10)∂σr∂r+σr−σθr=−ρ∂v∂t+v∂v∂r,where ρ is the density of the material.Equations (5), (6), and (8a)∼(10) form the fundamental equations of the FCCE model for ideal elastic-plastic materials based on UST.


## 3.2. Cavity Wall Radial Stress Calculation

### 3.2.1. Elastic-Plastic Stagerp<rt

Whenrc<r≤rp, substituting equations (2) and (9) into (10) yields(11)∂σr∂r=−Br−ρr˙c21r−rc2r3.Integrating equation (11) yields(12)σr=−Blnr−12ρr˙c22lnr+rc2r2+M1,where M1 is a constant.According to the boundary conditionσr|r=rc=σrc, M1 is formulated as follows:(13)M1=σrc+Blnrc+12ρr˙c22lnrc+1.Combining equations (13) and (12), the radial stress is as follows:(14)σr=σrc+B2lnrc2r2+12ρr˙c21+lnrc2r2−rc2r2.Substituting the plastic region side condition of the elastic-plastic boundaryσr|r=rp=σrp− into equation (14), the radial stress on the cavity wall σrc can be formulated as(15)σrc=σrp−B2lnrc2rp2−12ρr˙c21+lnrc2rp2−rc2rp2,where σrp and rc/rp are determined by the continuity conditions and yield condition of the elastic-plastic boundary.Substituting equations (9) and (8a) into (10), we can get(16)∂σr∂r=−E1+υ×rc2r3−ρr˙c21r−rc2r3.Whenrc<rtr˙c/c, re<rt. Integrating equation (16) and combining equation (9) and the momentum conservation condition [15] σr|r=re=crc/r˙c=ρcv, we have(17)σr=E21+υrc2r2−r˙c2c2+12ρr˙c2×2+r˙c2c2+2lncrcr˙cr−rc2r2.Whenrc≥rtr˙c/c, re=rt. Integrating equation (17) and considering σr|r=rt=0, we have(18)σr=E21+υrc2r2−rc2rt2+12ρr˙c2×lnrt2r2+rc2rt2−rc2r2.Because of the continuity of the stresses and particle velocity at the elastic-plastic boundary [18], according to equations (8a)-(8b), one obtains(19)rcrp=B1+υE.Whenrp=rt, the cavity radius rc1 at the end of the elastic-plastic stage can be expressed as(20)rc1=rtB1+υE.In the elastic-plastic stage,rc≤rc1.For the plane strain problems, the elastic wave velocityc can be expressed as [18,19].(21)c=E1−υρ1+υ1−2υ.Combining equations (15) and (17)∼(21), the radial stress of the cavity wall in this stage σrc1 is derived through the stress continuity conditions σr|r=rp=σrp:(22)σrc1=σrc0+Δσrc0+12ρr˙c2×n.Here, when rc<rtr˙c/c, σrco=B/2+B/2lnE/B1+υ, Δσrc0=0, and d=υ/1−υ+ρr˙c1+υ1−2υ/E1−υ+lnE1−υ/ρr˙c1+υ1−2υ; when rtr˙c/c≤rc≤rc1, σrco=B/2+B/2lnE/B1+υ, Δσrc0=−Erc2/21+υrt2, and n=lnrt2/rc2+rc2/rt2−1,where σrco is the strength term of the CET for a semi-infinite target, Δσrco is the term reflecting the effect of planar size on the strength, and n is the flow resistance coefficient.From equation (22), the elastic wave does not reach the lateral boundary of the target when rc<rtr˙c/c, so there is an undisturbed zone existing at the target edge, and Δσrco=0 so that the planar size has no effect on the target strength and flow resistance coefficient.


### 3.2.2. Plastic Stagerp≡rt

Because ofσr|r=rt=0 and equation (14) (the stress field in the plastic region), we can get the radial stress on the cavity wall in this stage σrc2:(23)σrc2=Blnrtrc+12ρr˙c2lnrt2rc2+rc2rt2−1,rc1≤rc≤rc2,where rc2 is the cavity radius at the end of the plastic stage, which is gotten by the fracture criterion in [20], where εeq and εf are the equivalent strain and the uniaxial tension fracture strain, respectively.Due to the boundary conditionr=rt, that is, σr=0, we obtain σ2≤σ1+ασ3/1+α from equations (1a)-(1b), and we can get the equivalent stress based on UST σeq:(24)σeq=2+b21+bσr−2+b21+bσθ=−2+b21+bσθ.For the FCCE problem, on the basis of the plastic work hypothesis [21, 22] and the stress-strain relationship, we have(25)σeqεeq=σε=σθεθ,where εeq is the equivalent strain based on UST. According to equations (24) and (25), εeq can be given as(26)εeq=−21+bεθ2+b.Considering equation (26), the strain components εr=−εθ≈1/2rc/r2 [15], and εeq=εf when r=rt [20], we can get the cavity radius at the end of the plastic stage rc2:(27)rc2=rt2+bεf1+b.Combining equations (20) and (27) and considering rc2>rc1, the conclusion in this paper is applied to materials as(28)εf>B21+υ2E,which are generally satisfied by metal materials.The theory quoted by this paper assumes that the elastic-plastic boundary movement velocity is less than the elastic wave velocity,r˙p≤c. According to equations (19) and (21), when r˙p=c, the maximum cavity expansion velocity r˙cmax applied for materials in this paper meets the following conditions:(29)r˙cmax=B1−υρ1−2υ.Whenrc1≤rcf<rc2, the expansion process experiences the elastic-plastic stage and the plastic stage, but when rcf≤rc1, the elastic wave has not reached or has just reached the target lateral boundary, and the undisturbed region exists or has just disappeared, at which point the expansion process experiences only the elastic-plastic stage; when rcf<rtr˙c/c, the target size shows no effect on the expansion process; when rcf>rc2, the target has been fractured.


## 3.2.1. Elastic-Plastic Stagerp<rt

Whenrc<r≤rp, substituting equations (2) and (9) into (10) yields(11)∂σr∂r=−Br−ρr˙c21r−rc2r3.Integrating equation (11) yields(12)σr=−Blnr−12ρr˙c22lnr+rc2r2+M1,where M1 is a constant.According to the boundary conditionσr|r=rc=σrc, M1 is formulated as follows:(13)M1=σrc+Blnrc+12ρr˙c22lnrc+1.Combining equations (13) and (12), the radial stress is as follows:(14)σr=σrc+B2lnrc2r2+12ρr˙c21+lnrc2r2−rc2r2.Substituting the plastic region side condition of the elastic-plastic boundaryσr|r=rp=σrp− into equation (14), the radial stress on the cavity wall σrc can be formulated as(15)σrc=σrp−B2lnrc2rp2−12ρr˙c21+lnrc2rp2−rc2rp2,where σrp and rc/rp are determined by the continuity conditions and yield condition of the elastic-plastic boundary.Substituting equations (9) and (8a) into (10), we can get(16)∂σr∂r=−E1+υ×rc2r3−ρr˙c21r−rc2r3.Whenrc<rtr˙c/c, re<rt. Integrating equation (16) and combining equation (9) and the momentum conservation condition [15] σr|r=re=crc/r˙c=ρcv, we have(17)σr=E21+υrc2r2−r˙c2c2+12ρr˙c2×2+r˙c2c2+2lncrcr˙cr−rc2r2.Whenrc≥rtr˙c/c, re=rt. Integrating equation (17) and considering σr|r=rt=0, we have(18)σr=E21+υrc2r2−rc2rt2+12ρr˙c2×lnrt2r2+rc2rt2−rc2r2.Because of the continuity of the stresses and particle velocity at the elastic-plastic boundary [18], according to equations (8a)-(8b), one obtains(19)rcrp=B1+υE.Whenrp=rt, the cavity radius rc1 at the end of the elastic-plastic stage can be expressed as(20)rc1=rtB1+υE.In the elastic-plastic stage,rc≤rc1.For the plane strain problems, the elastic wave velocityc can be expressed as [18,19].(21)c=E1−υρ1+υ1−2υ.Combining equations (15) and (17)∼(21), the radial stress of the cavity wall in this stage σrc1 is derived through the stress continuity conditions σr|r=rp=σrp:(22)σrc1=σrc0+Δσrc0+12ρr˙c2×n.Here, when rc<rtr˙c/c, σrco=B/2+B/2lnE/B1+υ, Δσrc0=0, and d=υ/1−υ+ρr˙c1+υ1−2υ/E1−υ+lnE1−υ/ρr˙c1+υ1−2υ; when rtr˙c/c≤rc≤rc1, σrco=B/2+B/2lnE/B1+υ, Δσrc0=−Erc2/21+υrt2, and n=lnrt2/rc2+rc2/rt2−1,where σrco is the strength term of the CET for a semi-infinite target, Δσrco is the term reflecting the effect of planar size on the strength, and n is the flow resistance coefficient.From equation (22), the elastic wave does not reach the lateral boundary of the target when rc<rtr˙c/c, so there is an undisturbed zone existing at the target edge, and Δσrco=0 so that the planar size has no effect on the target strength and flow resistance coefficient.


## 3.2.2. Plastic Stagerp≡rt

Because ofσr|r=rt=0 and equation (14) (the stress field in the plastic region), we can get the radial stress on the cavity wall in this stage σrc2:(23)σrc2=Blnrtrc+12ρr˙c2lnrt2rc2+rc2rt2−1,rc1≤rc≤rc2,where rc2 is the cavity radius at the end of the plastic stage, which is gotten by the fracture criterion in [20], where εeq and εf are the equivalent strain and the uniaxial tension fracture strain, respectively.Due to the boundary conditionr=rt, that is, σr=0, we obtain σ2≤σ1+ασ3/1+α from equations (1a)-(1b), and we can get the equivalent stress based on UST σeq:(24)σeq=2+b21+bσr−2+b21+bσθ=−2+b21+bσθ.For the FCCE problem, on the basis of the plastic work hypothesis [21, 22] and the stress-strain relationship, we have(25)σeqεeq=σε=σθεθ,where εeq is the equivalent strain based on UST. According to equations (24) and (25), εeq can be given as(26)εeq=−21+bεθ2+b.Considering equation (26), the strain components εr=−εθ≈1/2rc/r2 [15], and εeq=εf when r=rt [20], we can get the cavity radius at the end of the plastic stage rc2:(27)rc2=rt2+bεf1+b.Combining equations (20) and (27) and considering rc2>rc1, the conclusion in this paper is applied to materials as(28)εf>B21+υ2E,which are generally satisfied by metal materials.The theory quoted by this paper assumes that the elastic-plastic boundary movement velocity is less than the elastic wave velocity,r˙p≤c. According to equations (19) and (21), when r˙p=c, the maximum cavity expansion velocity r˙cmax applied for materials in this paper meets the following conditions:(29)r˙cmax=B1−υρ1−2υ.Whenrc1≤rcf<rc2, the expansion process experiences the elastic-plastic stage and the plastic stage, but when rcf≤rc1, the elastic wave has not reached or has just reached the target lateral boundary, and the undisturbed region exists or has just disappeared, at which point the expansion process experiences only the elastic-plastic stage; when rcf<rtr˙c/c, the target size shows no effect on the expansion process; when rcf>rc2, the target has been fractured.


## 3.3. Energy Consuming by Expanding and Mean Stress on Cavity Wall

According to [15], the energy consumed by expanding W and the work done by the radial pressure on the cavity wall σrc are equal during the whole process of the cavity radius rc increasing from 0 to rcf. Taking a target of unit thickness, we can obtain(30)W=2π∫0rcfσrcrcdrc.Supposing the mean radial stress of the cavity wall isσ¯rc and considering that the energy consumed by expansion remains unchanged, we obtain [15](31)σ¯rc=Wπrcf2=2rcf2∫0rcfσrcrcdrc.Substituting equations (31) into (18) and (19), one obtains(32)σ¯rc=A1+B1×12ρr˙c2,where A1=B/21+lnE/B1+υ and B1=υ/1−υ+1+υ1−2υρr˙c2/E1−υ+lnE1−υ/1+υ1−2υρr˙c2, when rcf<rtr˙c/c; A1=B/21+lnE/B1+υ−Ercf2/41+υrt2 and B1=lnrt2/rcf2+rcf2/2rt2, when rtr˙c/c≤rcf<rc1; A1=B/2lnrt2/rcf2+1 and B1=rcf2/2rt2+lnrt/rcf−1/2, when rc1≤rcf<rc2.


## 4. Penetration Effect on Metallic Thick Target with Finite Planar Sizes for Long Rod

### 4.1. Penetration Model Analysis for Long Rod

Assume that the long rod has a radius ofrd, a density of ρd, a yield stress of σd, a feature strength of Yd, a length of l (l0 is the initial length) at time t, and a speed of v (v0 is the impact speed) at time t. The target has a penetration velocity of u, a penetration depth of x, and a penetration resistance of R. According to Tate’s model [9],(33)dxdt=u,dvdt=−σdρl,(34)dldt=−v−u,(35)0.5ρdv−u2+Yd=0.5ρu2+R.This paper assumes thatR=A1 and rcf is approximately equal to the channel radius as [15](36)rcf=rd1+0.287v0+0.148v02,where v0 is in units of km/s.IfYd<R, the projectile velocity is vc1 when the penetration stops u=0; after this, the abrasion of the projectile continues until the projectile velocity v equals 0 or the projectile length l equals 0, and vc1 can be obtained by equation (35):(37)vc1=2R−Ydρd.However, the abrasion stops first whenYd>R, and the residual projectile can be considered to be rigid and will continue to penetrate with vc2=u until v=0; then, vc2 can also be obtained by equation (35):(38)vc2=2Yd−Rρ.


### 4.2. Penetration Depth Calculation

IfYd<R, the total penetration depth D can be calculated to u=0 by substituting it into the formula given by [15]:(39)D=−l0ρdσd∫u00expρdσdfu−fu0×f′uρ/ρdu2+vc12du,where fu=ρu2/2ρd+u/2ρ/ρdu2+vc12+ρd/ρ×vc12/2×uρ/ρd+ρ/ρdu2+vc12.Substitutingu=u0 and v=v0 into equation (35) and rearranging the equation yield(40)u0=v0ρd+ρd2v02−ρd−ρ2Yd−2A1+ρdv02ρd−ρ.IfYd>R, the total penetration depth D includes the penetration depth x1 of the abrasion penetration stage and the penetration depth x2 of the rigid penetration stage; that is,(41)D=x1+x2.Considering the ending conditionu=v=vc2 of x1 and energy conservation, x1, x2, and the residual length of the projectile l1 when x1 ends can be evaluated by the following equations proposed in [15]:(42)x1=−l0ρdσdρdρ∫u0vc2expgu−gu0×g′uuu2−vc22du,x2=ρdl1Advc222πRrcf2,l1=l0expρdσdgvc2−gu0,where gu=ρu2/2ρd+ρd/ρu/2u2−vc22−vc22/2×lnu+u2−vc22.Integral items of the penetration depth formula can be calculated in the MATLAB program.


## 4.1. Penetration Model Analysis for Long Rod

Assume that the long rod has a radius ofrd, a density of ρd, a yield stress of σd, a feature strength of Yd, a length of l (l0 is the initial length) at time t, and a speed of v (v0 is the impact speed) at time t. The target has a penetration velocity of u, a penetration depth of x, and a penetration resistance of R. According to Tate’s model [9],(33)dxdt=u,dvdt=−σdρl,(34)dldt=−v−u,(35)0.5ρdv−u2+Yd=0.5ρu2+R.This paper assumes thatR=A1 and rcf is approximately equal to the channel radius as [15](36)rcf=rd1+0.287v0+0.148v02,where v0 is in units of km/s.IfYd<R, the projectile velocity is vc1 when the penetration stops u=0; after this, the abrasion of the projectile continues until the projectile velocity v equals 0 or the projectile length l equals 0, and vc1 can be obtained by equation (35):(37)vc1=2R−Ydρd.However, the abrasion stops first whenYd>R, and the residual projectile can be considered to be rigid and will continue to penetrate with vc2=u until v=0; then, vc2 can also be obtained by equation (35):(38)vc2=2Yd−Rρ.


## 4.2. Penetration Depth Calculation

IfYd<R, the total penetration depth D can be calculated to u=0 by substituting it into the formula given by [15]:(39)D=−l0ρdσd∫u00expρdσdfu−fu0×f′uρ/ρdu2+vc12du,where fu=ρu2/2ρd+u/2ρ/ρdu2+vc12+ρd/ρ×vc12/2×uρ/ρd+ρ/ρdu2+vc12.Substitutingu=u0 and v=v0 into equation (35) and rearranging the equation yield(40)u0=v0ρd+ρd2v02−ρd−ρ2Yd−2A1+ρdv02ρd−ρ.IfYd>R, the total penetration depth D includes the penetration depth x1 of the abrasion penetration stage and the penetration depth x2 of the rigid penetration stage; that is,(41)D=x1+x2.Considering the ending conditionu=v=vc2 of x1 and energy conservation, x1, x2, and the residual length of the projectile l1 when x1 ends can be evaluated by the following equations proposed in [15]:(42)x1=−l0ρdσdρdρ∫u0vc2expgu−gu0×g′uuu2−vc22du,x2=ρdl1Advc222πRrcf2,l1=l0expρdσdgvc2−gu0,where gu=ρu2/2ρd+ρd/ρu/2u2−vc22−vc22/2×lnu+u2−vc22.Integral items of the penetration depth formula can be calculated in the MATLAB program.


## 5. Example and Discussion

### 5.1. Experimental Verification

For comparison, the ballistic test data in [13] were substituted into formulas in this paper to calculate and analyze the total penetration depth and penetration resistance. The targets were machined from 4340 steel with different radius, which had a density of ρ=7850kg/m3 and an elastic modulus of E=200.6GPa, and its Poisson’s ratio is υ=0.29 and its yield stress is σoy=1.365GPa. The long rod projectile made of tungsten had a rod length of L0=77.9mm, a striking velocity of v=1500m/s, a density of ρd=17730kg/m3, a feature strength of Yd=1.33GPa, a yield stress of σs=1.3GPa, and a limit stress σst=1.33GPa. In addition, other material parameters, the test results, and the theoretical value are shown in Table 1.Table 1
Penetration data for long rod penetration into thick metallic targets with finite planar sizes.
rt (mm)rd (mm)rt/rdRt (MPa)Dshi (mm)Dmax (mm)δ19.103.89804.902556.8587.2593.670.073619.103.89804.902556.8588.4993.670.058525.403.89576.523036.7286.0086.590.006825.453.89746.533039.3086.3186.550.002851.103.894813.124211.4974.8672.16−0.036051.053.896913.104208.9270.9772.190.017377.453.895919.884337.5066.7670.810.060777.103.895919.794334.7169.7270.840.016177.253.895619.834335.9669.2570.830.0227111.003.890628.534494.8870.7369.17−0.0221304.803.924877.664623.37–67.86–Notes:Rt is the theoretical penetration resistance calculation result in this paper; Dmax is the theoretical penetration depth calculation result in this paper; Dshi is the test result of penetration depth in [13]; and δ is the relative error of Dmax and Dshi.To model the penetration process and transient response of the projectile and target which cannot be got by other ways [23,24], we set up 3D solid models to simulate the penetrations by LS-DYNA. The elastic-plastic hydrodynamic model and Johnson-Cook (J-C) constitutive model are used to describe the penetrator and the target, respectively.J-C model has simple expression and easy access to data from tests, as follows [25]:(43)σy=A0+B0ε¯pn01+C0lnε˙∗1−T0∗m0,where A0,B0,C0,m0,n0 are the material parameters; ε¯p and ε˙∗ are effective plastic strain and its rate ε˙∗=ε¯⋅p/ε˙0; T0∗=T0−Tr/Tm−Tr is relative temperature; T0 is absolute temperature; Tr and Tm are indoor temperature and melt temperature, respectively.The damage model is expressed as(44)εf=D1+D2expD3σ∗1+D4lnε˙∗1+D5T0∗,where σ∗=p/σeff is the stress triaxiality. The fracture occurs when D0=∑Δε¯p/εf=1.The elastic-plastic hydrodynamic model is material type 10. If EPS (effective plastic strain) and ES (effective stress) are defined, the yield stress expression is [25](45)σy=σ0+EtE0E−Etε¯p+a1+pa2maxp,0,where E0 is Young’s modulus; Et is the tangent modulus; p is the pressure.ε¯p is calculated as(46)ε¯p=∫0t32DijpDijp1/2dt.If EPS and ES are undefined, the yield stress expression is(47)σ¯=32sijsij1/2,where sij and Dijp are the deviatoric stress tensor and the plastic part of the deviatoric strain tensor, respectively.If EH (parameter of plastic hardening modulus) is undefined, the yield stress expression isσy=gε¯p, which is gained by linear interpolation.Referring to the literature [13,26], model parameters for calculation are determined after several trial calculations according to the analysis results of parameters influence in the literature [26]; see Table 2 for related parameters.Table 2
Material model parameters.
MaterialEH (MPa)C1S1S2S3GAMAOaTungsten alloy87130.4041.23001.540MaterialA0 (MPa)B0(MPa)n0C0m0a—4340 steel7354730.260.0141.030.43—A Lagrange solver is applied in the simulation of the projectile and target, and surface-to-surface erosion contact is set between them. Then, a quarter model is established, as shown in Figure2, disassembled evenly by hexahedron elements in the direction of the cross section and thickness. The numbers of elements and nods for projectile are 7500 and 9191, respectively; and these numbers for target are 144000 and 159221, respectively. Symmetric boundary constraint is imposed on the symmetry surface. Because grid discretization will cause a mass error, four times the mass of the projectile in the model is 65.55 g, differing by 1.31% from the mass of the projectile in the test (64.7 g), which can be ignored.Figure 2
The calculation model of penetration.
(a)(b)(c)Figure3 shows the comparisons of the maximum penetration depth theoretical calculation in this paper when b=0.6, the simulation results in this paper, the test results in [13], and the formula results in [13–15]. During the calculation, when rt≤10rcf, rtr˙c/c≤rcf<rc1, and the penetration expansion process only experiences the elastic-plastic stage, while when rt>10rcf, rc1≤rcf<rc2, and the penetration expansion process experiences the elastic-plastic stage and the plastic stage. Figure 3 indicates that the simulation results are in good agreement with the test data, and the effectiveness and correctness of parameter selection and model establishment are proved. Moreover, the average relative error between the theoretical formula calculation in this paper, in [15], and in [13] and the test results are 2.00%, 5.47%, and 8.27%, respectively. The maximum error of calculation is 7.36% in this paper but is 10.26% for [15] and more than 20% for [13]. It is indicated that the theoretically calculated values based on UST are more accurate than others.Figure 3
Comparison of the penetration depth.


### 5.2. Projectile Erosion Change

The time course of the projectile tail velocity, penetration velocity, projectile length, and projectile mass based on the simulations can be seen from Figures4 and 5. From these figures, we can intuitively observe that the projectile tail velocity is much faster than the projectile nose velocity at the beginning of the penetration. The projectile begins to erode at t = 0.01 ms, and the erosion stops at t = 0.12 ms when there are more mass residues of the projectile. Thereafter, the curve of the projectile mass is steady, remaining at approximately 13.863 g, which is 85.71% of the projectile mass in the early stage of penetration (16.175 g), and the projectile length is 3.198 cm, which is 41.05% of the initial projectile length (7.79 cm). Because the length loss is much larger than the mass loss, the projectile becomes a short and thick pestle. The period t = 0.01∼0.12 ms is the erosion stage, in which the difference between the projectile tail velocity and penetration velocity is 32 m/s. At t = 0.15 ms, the projectile length can be considered unchanging because the projectile tail velocity and the penetration velocity are essentially equal; thus, 0.12∼0.15 ms is the deformation without erosion stage. The stage 0.15∼0.25 ms is the rigid penetration stage, in which the projectile tail velocity and the penetration velocity are essentially equal, and their curves basically coincide. The period after 0.25 ms is the elastic resilience stage.Figure 4
Time course changes of the projectile tail velocity, penetration velocity, and projectile length.Figure 5
Time course changes of the projectile tail velocity, penetration velocity, and projectile mass.


### 5.3. Effect of Strength Coefficient

Since different materials take the values of the strength coefficientb differently, we can gain a range of calculations that the UST is applicable to a variety of materials. On this basis, the ranges of penetration depth for metallic targets with different rt/rd are predicted, and the results are summarized in Table 3. Calculated by the theoretical formulas in this paper, the relationship between penetration depth Dmax and rt/rd values for different b is shown in Figure 6. From this, we can see that the coefficient b has a large effect on Dmax: the larger b is, the greater the effect of σ2 is and the smaller Dmax is. Therefore, appropriate values of b should be adopted for different materials during the penetration analysis. Meanwhile, b is also called yield criterion coefficient that when b takes a different value, the UST degenerates into different yield criterion. Consequently, UST includes infinity yield criteria that a range of different penetration results based on different strength criteria can work out. For instance, when rt/rd≥20, a comparison of b=1 (the Tresca yield criterion) and b=0 (the twin shear yield criterion) shows that Dmax is higher by up to 17.09%. When rt/rd<20, this effect becomes more striking that Dmax can be up to 18.95% higher. It indicates that, to make the members’ antipenetration performance be played fully, the effect of the intermediate principal stress σ2 cannot be ignored. The result in [15] (von Mises yield criterion result), which is only applicable for materials with τs=0.577σs, is only a special case of the results in this paper when b=0.366.Table 3
Penetration depth ranges.
rt/rdPenetration depth range (mm)rt/rdPenetration depth range (mm)77.6664.79∼75.7213.1268.48∼81.4528.5366.03∼77.2213.168.50∼81.4819.8867.58∼79.126.5283.17∼95.0919.7967.60∼79.156.5383.13∼95.0519.8367.59∼79.144.990.42∼101.68Figure 6
Comparison of penetration depth.


### 5.4. Effect of Target Radius

The relationship betweenDmax and rt/rd is also shown in Figure 6 (in the case of b = 0.6): as rt/rd increases, Dmax decreases. When rt/rd≥20, Dmax decreases so slowly with the increase of rt/rd that it has decreased by only 4.19% between rt/rd=19.83 and rt/rd=77.66. When rt/rd≥30, as rt⟶r∞, Dmax has decreased by just 2%. However, when rt/rd<20, Dmax decreases dramatically with the increase of rt/rd. When rt/rd=4.9 increases to rt/rd=19.88, Dmax decreases by 32.28%. It indicates that the impact of the target planar size cannot be ignored when rt/rd<20, according well with the result in [15].


### 5.5. Effect of Other Parameters

Models of long rod projectiles with flat-shaped, hemisphere-shaped, and cone-shaped noses are constructed as shown in Figure7. Then, penetration models of three nose shapes with v0 = 500 m/s, v0 = 700 m/s, v0 = 1500 m/s, and v0 = 2500 m/s are numerically calculated and the penetration depth curves are shown in Figure 8. Figure 8 clearly shows that the penetration depth of flat-shaped projectiles is 22.3% less than that of hemisphere-shaped projectiles and 25.2% less than that of cone-shaped projectiles when v0 = 500 m/s. Nevertheless, when v0 = 1500 m/s and v0 = 2500 m/s, the disparity of the penetration depth among three projectile shapes is less than 2%, which indicates that the projectile nose shape has little effect on the penetration process at high speeds because the stress produced by a high-speed impact leads to erosional failure of the long rod nose in the initial stages of contact with the target.Figure 7
Three nose shapes for the long rod.Figure 8
Penetration depth versus striking velocity of projectiles.It also can be seen from Figure8 that the striking velocity of projectile v0 has some effect on the penetration: the higher v0 is, the greater Dmax is, and the increase of Dmax for cone-shaped projectiles is much more obvious than others.


## 5.1. Experimental Verification

For comparison, the ballistic test data in [13] were substituted into formulas in this paper to calculate and analyze the total penetration depth and penetration resistance. The targets were machined from 4340 steel with different radius, which had a density of ρ=7850kg/m3 and an elastic modulus of E=200.6GPa, and its Poisson’s ratio is υ=0.29 and its yield stress is σoy=1.365GPa. The long rod projectile made of tungsten had a rod length of L0=77.9mm, a striking velocity of v=1500m/s, a density of ρd=17730kg/m3, a feature strength of Yd=1.33GPa, a yield stress of σs=1.3GPa, and a limit stress σst=1.33GPa. In addition, other material parameters, the test results, and the theoretical value are shown in Table 1.Table 1
Penetration data for long rod penetration into thick metallic targets with finite planar sizes.
rt (mm)rd (mm)rt/rdRt (MPa)Dshi (mm)Dmax (mm)δ19.103.89804.902556.8587.2593.670.073619.103.89804.902556.8588.4993.670.058525.403.89576.523036.7286.0086.590.006825.453.89746.533039.3086.3186.550.002851.103.894813.124211.4974.8672.16−0.036051.053.896913.104208.9270.9772.190.017377.453.895919.884337.5066.7670.810.060777.103.895919.794334.7169.7270.840.016177.253.895619.834335.9669.2570.830.0227111.003.890628.534494.8870.7369.17−0.0221304.803.924877.664623.37–67.86–Notes:Rt is the theoretical penetration resistance calculation result in this paper; Dmax is the theoretical penetration depth calculation result in this paper; Dshi is the test result of penetration depth in [13]; and δ is the relative error of Dmax and Dshi.To model the penetration process and transient response of the projectile and target which cannot be got by other ways [23,24], we set up 3D solid models to simulate the penetrations by LS-DYNA. The elastic-plastic hydrodynamic model and Johnson-Cook (J-C) constitutive model are used to describe the penetrator and the target, respectively.J-C model has simple expression and easy access to data from tests, as follows [25]:(43)σy=A0+B0ε¯pn01+C0lnε˙∗1−T0∗m0,where A0,B0,C0,m0,n0 are the material parameters; ε¯p and ε˙∗ are effective plastic strain and its rate ε˙∗=ε¯⋅p/ε˙0; T0∗=T0−Tr/Tm−Tr is relative temperature; T0 is absolute temperature; Tr and Tm are indoor temperature and melt temperature, respectively.The damage model is expressed as(44)εf=D1+D2expD3σ∗1+D4lnε˙∗1+D5T0∗,where σ∗=p/σeff is the stress triaxiality. The fracture occurs when D0=∑Δε¯p/εf=1.The elastic-plastic hydrodynamic model is material type 10. If EPS (effective plastic strain) and ES (effective stress) are defined, the yield stress expression is [25](45)σy=σ0+EtE0E−Etε¯p+a1+pa2maxp,0,where E0 is Young’s modulus; Et is the tangent modulus; p is the pressure.ε¯p is calculated as(46)ε¯p=∫0t32DijpDijp1/2dt.If EPS and ES are undefined, the yield stress expression is(47)σ¯=32sijsij1/2,where sij and Dijp are the deviatoric stress tensor and the plastic part of the deviatoric strain tensor, respectively.If EH (parameter of plastic hardening modulus) is undefined, the yield stress expression isσy=gε¯p, which is gained by linear interpolation.Referring to the literature [13,26], model parameters for calculation are determined after several trial calculations according to the analysis results of parameters influence in the literature [26]; see Table 2 for related parameters.Table 2
Material model parameters.
MaterialEH (MPa)C1S1S2S3GAMAOaTungsten alloy87130.4041.23001.540MaterialA0 (MPa)B0(MPa)n0C0m0a—4340 steel7354730.260.0141.030.43—A Lagrange solver is applied in the simulation of the projectile and target, and surface-to-surface erosion contact is set between them. Then, a quarter model is established, as shown in Figure2, disassembled evenly by hexahedron elements in the direction of the cross section and thickness. The numbers of elements and nods for projectile are 7500 and 9191, respectively; and these numbers for target are 144000 and 159221, respectively. Symmetric boundary constraint is imposed on the symmetry surface. Because grid discretization will cause a mass error, four times the mass of the projectile in the model is 65.55 g, differing by 1.31% from the mass of the projectile in the test (64.7 g), which can be ignored.Figure 2
The calculation model of penetration.
(a)(b)(c)Figure3 shows the comparisons of the maximum penetration depth theoretical calculation in this paper when b=0.6, the simulation results in this paper, the test results in [13], and the formula results in [13–15]. During the calculation, when rt≤10rcf, rtr˙c/c≤rcf<rc1, and the penetration expansion process only experiences the elastic-plastic stage, while when rt>10rcf, rc1≤rcf<rc2, and the penetration expansion process experiences the elastic-plastic stage and the plastic stage. Figure 3 indicates that the simulation results are in good agreement with the test data, and the effectiveness and correctness of parameter selection and model establishment are proved. Moreover, the average relative error between the theoretical formula calculation in this paper, in [15], and in [13] and the test results are 2.00%, 5.47%, and 8.27%, respectively. The maximum error of calculation is 7.36% in this paper but is 10.26% for [15] and more than 20% for [13]. It is indicated that the theoretically calculated values based on UST are more accurate than others.Figure 3
Comparison of the penetration depth.


## 5.2. Projectile Erosion Change

The time course of the projectile tail velocity, penetration velocity, projectile length, and projectile mass based on the simulations can be seen from Figures4 and 5. From these figures, we can intuitively observe that the projectile tail velocity is much faster than the projectile nose velocity at the beginning of the penetration. The projectile begins to erode at t = 0.01 ms, and the erosion stops at t = 0.12 ms when there are more mass residues of the projectile. Thereafter, the curve of the projectile mass is steady, remaining at approximately 13.863 g, which is 85.71% of the projectile mass in the early stage of penetration (16.175 g), and the projectile length is 3.198 cm, which is 41.05% of the initial projectile length (7.79 cm). Because the length loss is much larger than the mass loss, the projectile becomes a short and thick pestle. The period t = 0.01∼0.12 ms is the erosion stage, in which the difference between the projectile tail velocity and penetration velocity is 32 m/s. At t = 0.15 ms, the projectile length can be considered unchanging because the projectile tail velocity and the penetration velocity are essentially equal; thus, 0.12∼0.15 ms is the deformation without erosion stage. The stage 0.15∼0.25 ms is the rigid penetration stage, in which the projectile tail velocity and the penetration velocity are essentially equal, and their curves basically coincide. The period after 0.25 ms is the elastic resilience stage.Figure 4
Time course changes of the projectile tail velocity, penetration velocity, and projectile length.Figure 5
Time course changes of the projectile tail velocity, penetration velocity, and projectile mass.


## 5.3. Effect of Strength Coefficient

Since different materials take the values of the strength coefficientb differently, we can gain a range of calculations that the UST is applicable to a variety of materials. On this basis, the ranges of penetration depth for metallic targets with different rt/rd are predicted, and the results are summarized in Table 3. Calculated by the theoretical formulas in this paper, the relationship between penetration depth Dmax and rt/rd values for different b is shown in Figure 6. From this, we can see that the coefficient b has a large effect on Dmax: the larger b is, the greater the effect of σ2 is and the smaller Dmax is. Therefore, appropriate values of b should be adopted for different materials during the penetration analysis. Meanwhile, b is also called yield criterion coefficient that when b takes a different value, the UST degenerates into different yield criterion. Consequently, UST includes infinity yield criteria that a range of different penetration results based on different strength criteria can work out. For instance, when rt/rd≥20, a comparison of b=1 (the Tresca yield criterion) and b=0 (the twin shear yield criterion) shows that Dmax is higher by up to 17.09%. When rt/rd<20, this effect becomes more striking that Dmax can be up to 18.95% higher. It indicates that, to make the members’ antipenetration performance be played fully, the effect of the intermediate principal stress σ2 cannot be ignored. The result in [15] (von Mises yield criterion result), which is only applicable for materials with τs=0.577σs, is only a special case of the results in this paper when b=0.366.Table 3
Penetration depth ranges.
rt/rdPenetration depth range (mm)rt/rdPenetration depth range (mm)77.6664.79∼75.7213.1268.48∼81.4528.5366.03∼77.2213.168.50∼81.4819.8867.58∼79.126.5283.17∼95.0919.7967.60∼79.156.5383.13∼95.0519.8367.59∼79.144.990.42∼101.68Figure 6
Comparison of penetration depth.


## 5.4. Effect of Target Radius

The relationship betweenDmax and rt/rd is also shown in Figure 6 (in the case of b = 0.6): as rt/rd increases, Dmax decreases. When rt/rd≥20, Dmax decreases so slowly with the increase of rt/rd that it has decreased by only 4.19% between rt/rd=19.83 and rt/rd=77.66. When rt/rd≥30, as rt⟶r∞, Dmax has decreased by just 2%. However, when rt/rd<20, Dmax decreases dramatically with the increase of rt/rd. When rt/rd=4.9 increases to rt/rd=19.88, Dmax decreases by 32.28%. It indicates that the impact of the target planar size cannot be ignored when rt/rd<20, according well with the result in [15].


## 5.5. Effect of Other Parameters

Models of long rod projectiles with flat-shaped, hemisphere-shaped, and cone-shaped noses are constructed as shown in Figure7. Then, penetration models of three nose shapes with v0 = 500 m/s, v0 = 700 m/s, v0 = 1500 m/s, and v0 = 2500 m/s are numerically calculated and the penetration depth curves are shown in Figure 8. Figure 8 clearly shows that the penetration depth of flat-shaped projectiles is 22.3% less than that of hemisphere-shaped projectiles and 25.2% less than that of cone-shaped projectiles when v0 = 500 m/s. Nevertheless, when v0 = 1500 m/s and v0 = 2500 m/s, the disparity of the penetration depth among three projectile shapes is less than 2%, which indicates that the projectile nose shape has little effect on the penetration process at high speeds because the stress produced by a high-speed impact leads to erosional failure of the long rod nose in the initial stages of contact with the target.Figure 7
Three nose shapes for the long rod.Figure 8
Penetration depth versus striking velocity of projectiles.It also can be seen from Figure8 that the striking velocity of projectile v0 has some effect on the penetration: the higher v0 is, the greater Dmax is, and the increase of Dmax for cone-shaped projectiles is much more obvious than others.


## 6. Conclusions

(1)
Based on the UST to consider the effect ofσ2, combined with Tate’s eroding rod model, the FCCE model of high-speed penetration by long rods into metallic targets with finite planar sizes is obtained. Additionally, the penetration process under different conditions is simulated by LS-DYNA. These results are compared with test data and other calculations, showing that calculated results in this paper have higher precision than those in other studies. On this basis, the ranges of penetration depth for metallic targets with different rt/rd are predicted.(2)
The strength coefficientb has a large effect on the penetration: the larger b is, the greater the effect of σ2 is and the smaller Dmax is. For metallic materials, there are large differences among the calculated results because of different b. For instance, as b=1 decreases to b=0, Dmax increases by up to 18.95%. Therefore, appropriate values of b should be adopted for different materials during penetration analysis. The result in [15], which is only applicable for materials with τs=0.577σs, is only a special case of the results in this paper when b=0.366.(3)
Whenrt/rd≥20, Dmax decreases slowly with the increase of rt/rd: as rt/rd=19.83 increases to rt/rd=77.66, Dmax decreases by only 4.19%. When rt/rd<20, Dmax decreases dramatically with the increase of rt/rd. When rt/rd=4.9 increases to rt/rd=19.88, Dmax decreases by 32.28%. Therefore, the influence of the target planar size is obvious when rt/rd<20, and the target cannot be considered as a semi-infinite target anymore.(4)
The striking velocity of projectilev0 has some effect on the penetration: the higher v0 is, the greater Dmax is. The projectile nose shape has little effect on the penetration performance at high speeds because of the erosional failure of the long rod nose.



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*Source: 6666770-2021-03-13.xml*