The degradation of elastic stiffness is investigated systematically within the framework of continuum damage mechanics. Consistent equations are obtained showing how the degradation of elastic stiffness can be decomposed into a part due to cracks and another part due to voids. For this purpose, the hypothesis of elastic energy equivalence of order n is utilized. In addition, it is shown that the hypothesis of elastic strain equivalence is obtained as a special case of the hypothesis of elastic energy equivalence of order n. In the first part of this work, the formulation is scalar and applies to the one-dimensional case. The tensorial formulation for the decomposition is also presented that is applicable to general states of deformation and damage. In this general case, one cannot obtain a single explicit tensorial decomposition equation for elastic stiffness degradation. Instead, one obtains an implicit system of three tensorial decomposition equations (called the tensorial decomposition system). Finally, solution of the tensorial decomposition system is illustrated in detail for the special case of plane stress.

# Decomposition of Elastic Stiffness Degradation in Continuum Damage Mechanics OPEN ACCESS

**George Z. Voyiadjis**

**Peter I. Kattan**

Contributed by the Materials Division of ASME for publication in the JOURNAL OF ENGINEERING MATERIALS AND TECHNOLOGY. Manuscript received May 29, 2016; final manuscript received September 10, 2016; published online February 1, 2017. Assoc. Editor: Taehyo Park.

*J. Eng. Mater. Technol*139(2), 021005 (Feb 01, 2017) (15 pages) Paper No: MATS-16-1152; doi: 10.1115/1.4035292 History: Received May 29, 2016; Revised September 10, 2016

In 1958, Kachanov [1] pioneered the concept of effective stress and introduced the topic of continuum damage mechanics. This development was followed by Rabotnov [12] and by others later [4–6,10,11,13,14]. In the framework of continuum damage mechanics, a scalar damage variable $\phi $ is introduced, which has values in the range $0\u2264\phi \u22641$. Thus, the value of the damage variable is zero when the virgin material is undamaged, while the value approaches one upon complete rupture. However, practically the damage value cannot exceed 0.3–0.4 without violating the concept of a continuum mechanics.

Research on damage mechanics accelerated rapidly in the past few years [2,3,7,9,15–25]. Another approach to damage characterization is to use an entropy generation rate as a damage metric rather than the damage potential surface [26,27]. Practical examples of elastic stiffness degradation in highly jointed rock masses are given in Ref. [28]. This work consists of three major sections that deal with full theoretical characterization of the degradation of elastic stiffness in materials within the framework of continuum damage mechanics.

Section 2 presents a simple scalar decomposition for the one-dimensional case. In this case, one is able to obtain one single scalar equation that describes the decomposition of the elastic stiffness degradation in materials in terms of the elastic stiffness degradation due to the presence of cracks and the elastic stiffness degradation due to the present of voids.

In Sec. 3, where the general case of three-dimensional deformation and damage is investigated, one resorts to the use of tensors and their matrix representations. In this case, the situation is not simple as in the scalar case. One cannot obtain one single tensorial decomposition equation. Instead, an implicit tensorial system of coupled simultaneous equations is obtained. Two such decomposition systems are obtained: one is of the first kind, while the other one is of the second kind. The difference between them is in the order of the multiplication of the various involved matrices (two different systems are obtained because matrix multiplication is not commutative).

An example is investigated in Sec. 4 using the tensorial decomposition systems of Sec. 3. The special case of plane stress is selected mainly because the tensors can be represented by 3 × 3 matrices. The solution of the tensorial decomposition system is illustrated for this case in detail. Section 5 concludes the paper.

In this section, two major scalar damage variables used by researchers today are discussed. The first scalar damage variable is defined in terms of cross-sectional area reduction, while the second scalar damage variable is defined in terms of the reduction in the elastic modulus or elastic stiffness [6].

Consider a body (in the form of a cylinder) in the initial undeformed and undamaged configuration. Consider also the configuration of the body that is both deformed and damaged after a set of external agencies act on it (see Fig. 1). Next, consider a fictitious configuration of the body obtained from the damaged configuration by removing all the damage that the body has undergone, i.e., this is the state of the body after it had only deformed without damage (see Fig. 1). Therefore, in defining a damage variable $\varphi $, its value must vanish in the fictitious configuration.

The first damage variable $\varphi $ is usually defined as follows:

where *A* is the cross-sectional area in the damaged configuration, while $A\xaf$ is the cross-sectional area in the fictitious configuration with $A>A\xaf$. It is clear that when a body is undamaged, i.e., when $A=A\xaf$, then $\varphi =0$.

The stress in the fictitious configuration is called the effective stress and is denoted by $\sigma \xaf$. The value of the effective stress $\sigma \xaf$ may be obtained using the relation $\sigma \xafA\xaf=\sigma A$, where *σ* is the stress in the damaged configuration. Therefore, using this relation along with the definition in Eq. (1), one obtains

It should be mentioned that the equilibrium condition in the paragraph above reflects a mean-field type of assumption on the stress redistribution (uniform over the resistive section) and therefore appears to be appropriate only in the dilute damage regime, away from the stress–strain peak where cooperate effects dominate and damage localization takes place.

The second scalar damage variable $\u2113$ may be defined in terms of the reduction in the elastic modulus as follows:

where *E* is the elastic modulus in the damaged state, while $E\xaf$ is the effective elastic modulus (in the fictitious state) with $E\xaf>E$ (see Fig. 2). This damage variable was used recently in Refs. [15,29,30]. It should also be mentioned that Voyiadjis [24] used a similar relation but in the context of elastoplastic deformation (see also Ref. [6]).

The definition of the alternative damage variable of Eq. (3) may be rewritten in the following more appropriate form:

It is clear from the definition in Eq. (3) that $\u2113=0$ when the body is undamaged, i.e., when $E\xaf=E$.

In order to establish the exact relationship between the two damage variables $\varphi $ and $\u2113$, one needs to utilize one of the equivalence hypotheses that are used in the literature of damage mechanics. For this purpose, one introduces the following generalized hypothesis of elastic strain equivalence of order *n*. The expression for the generalized hypothesis was introduced recently by the authors within the context of the theory of undamageable materials [31–35]. It will be shown that all the equivalence hypotheses of damage mechanics can be obtained as special cases of this generalized hypothesis

where $n=0,1,2,...$. The classical hypothesis of elastic energy equivalence is obtained as a special case of Eq. (5) when *n* = 1. Next, one investigates what happens to Eq. (5) when $n\u2192\u221e$. To investigate this extreme and hypothetical case, one raises both sides of Eq. (5) to the power $(1/n)$ to obtain

Substituting $n\u2192\u221e$ into Eq. (6) and noting that $\sigma \xaf1\u221e\u21921$ and $\sigma 1\u221e\u21921$, Eq. (6) reduces to

It is noted that the expression given in Eq. (7) represents exactly the hypothesis of elastic strain equivalence that is used frequently in the literature. The simple derivation in Eqs. (5)–(7) establishes the *missing link* in damage mechanics and shows that the hypothesis of elastic strain equivalence is a special case of the hypothesis of elastic energy equivalence of order $n$ when $n\u2192\u221e$. This extreme and special case is hypothetical, is not physically based, and should not be used in practical applications. Table 1 shows a summary of the results thus far.

Next, one proceeds with the derivation of the relationship between the two scalar damage variables $\varphi $ and $\u2113$ utilizing the generalized hypothesis of elastic energy equivalence of order $n$ as expressed in Eq. (5). Using the following constitutive relations (once for general $n$, and once for Hooke's law where $n=1$) in the damaged and fictitious/undamaged configurations, respectively (the nature of the unspecified function $f$ as well as its detailed derivation was presented in detail by the authors elsewhere, Voyiadjis and Kattan [31–35])

one then substitutes Eqs. (8) and (9) into Eq. (5), while noting that $n=1$ for the second case and simplifies the result to obtain

where in Eq. (10), two formulas appear (the first one is for a general value of $n$, while the second one is for $n=1$). Note that for values of $n>1$, one deals with higher-order materials for which there is no well-defined elastic stiffness (for more details, see the work of the authors elsewhere, Voyiadjis and Kattan [31–35]). Note that in Eqs. (8) and (9), the following special case holds: $f(\sigma ,1)=\sigma /E$ and $f(\sigma \xaf,1)=\sigma \xaf/E$.

Next, one substitutes Eqs. (2) and (4) into Eq. (10) to obtain

The expression given by Eq. (11) represents the exact relationship between the two scalar damage variables $\varphi $ and $\u2113$. Note that $\u2113$ is not defined for general values of $n>1$.

In the remaining part of the derivation, one limits the equations to the case of linear elastic materials with $n=1$. Solving Eq. (11) for $\u2113$, one obtains

On the other hand, solving Eq. (11) for $\varphi $, one obtains

It should be noted that when one substitutes the special cases of $n=1$ and $n\u2192\u221e$ for the hypothesis of elastic energy equivalence and the hypothesis of elastic strain equivalence, respectively, into Eq. (13), one obtains the following relations:

It should be noted that the two expressions of Eqs. (14) and (15) were derived previously and directly by the authors without resort to the generalized hypothesis [6] using the two separate hypothesis of equivalence. It is now noted that both expressions are special cases of the general expression of Eq. (13).

One next assumes that the damaged cross-sectional area of Fig. 1 consists of damage in the form of cracks and voids only (other forms of damage may be included but these two will suffice for simplicity). Kattan and Voyiadjis [10] showed that in this case, the damage variable $\varphi $ can be decomposed a follows:

where the two scalar damage variables $\phi c$ and $\phi v$ represent the damage due to cracks and the damage due to voids, respectively. Next, one substitutes the decomposition equation (16) into Eq. (12) to obtain

In an analogous way to the two scalar damage variables $\varphi c$ and $\varphi v$, one introduces two scalar damage variables $\u2113c$ and $\u2113v$ for cracks and voids, respectively, based on Eq. (4) as follows:

where $Ec$ is the elastic modulus of the damaged material due to cracks only, $Ev$ is the elastic modulus of the damaged material due to voids only, $E\xafc$ is the effective/undamaged elastic modulus when voids are removed (only cracks are present), and $E\xafv$ is the effective/undamaged elastic modulus when cracks are removed (only voids are present).

Equation (13) can be written individually for the individual scalar damage variables for cracks and voids as follows:

Next, one substitutes Eqs. (20) and (21) into Eq. (17) to obtain

Equation (22) is simplified using Wolfram Alpha^{1} and one obtains the following simplified result:

Rewriting Eq. (23) in the following form:

Expanding the expression under the square root in Eq. (24) and using the fist two terms of the Taylor series expansion of the square root function ($1+x\u22481+(1/2)x+\cdots $, applicable for small values of $x$), the decomposition formula of Eq. (24) is approximate as follows for small values of damage:

Second, for the hypothesis of elastic strain equivalence, one arrives at the following simple decomposition:

Comparing the two simplified decomposition of Eqs. (25) and (26), one notices that the elastic stiffness degradation based on decomposition using the hypothesis of elastic energy equivalence results in half of that using the hypothesis of elastic strain equivalence. One also should compare Eqs. (25) and (26) with Eq. (16).

Finally, substituting the appropriate expressions for $\u2113$, $\u2113c$, and $\u2113v$ from Eqs. (4), (18), and (19), respectively, into Eq. (23) and simplifying the result, one obtains the sought decomposition equation as follows:

Since $E\xaf>E$ (see Fig. 2), the elastic stiffness degradation ratio is actually $E/E\xaf<1$ but the formulation above is written in terms of the ration $E\xaf/E$ with the condition that $E/E\xaf=1/(E\xaf/E)$.

^{(1)}When there is no elastic stiffness degradation due to cracks $((E\xafc/Ec)=1)$ and no elastic stiffness degradation due to voids $((E\xafv/Ev)=1)$, then there is no degradation in total elastic stiffness, i.e., $(E\xaf/E)=1$.^{(2)}When there are no voids, i.e., the total degradation of elastic stiffness is then due to cracks only. In this case, $(E\xafv/Ev)=1$ and $(E\xaf/E)=(E\xafc/Ec)12$.^{(3)}When there are no cracks, i.e., the total degradation of elastic stiffness is then due to voids only. In this case, $(E\xafc/Ec)=1$ and $(E\xaf/E)=(E\xafv/Ev)12$.^{(4)}For the case of the hypothesis of elastic energy equivalence, i.e., when $n=1$, the decomposition formula of Eq. (27) becomes $(E\xaf/E)=(E\xafc/Ec)(E\xafv/Ev)$. In this case, the decomposition is the square root of a multiplicative expression.^{(5)}For the case of the hypothesis of elastic strain equivalence, i.e., when $n\u2192\u221e$, the decomposition formula of Eq. (27) becomes $(E\xaf/E)=(E\xafc/Ec)(E\xafv/Ev)$. In this case, the decomposition is clearly multiplicative.

Next, one investigates the decomposition formula for elastic stiffness degradation further. One assumes that the part of elastic stiffness degradation due to cracks is given by the following formula where $\alpha $ is called the *crack damage fraction* ($0\u2264\alpha \u22641$):

Similarly, one assumes that the part of elastic stiffness degradation due to voids is given by the following formula where $\beta $ is called the *void damage fraction* ($0\u2264\beta \u22641$):

Substituting Eqs. (28) and (29) into Eq. (27), one obtains

The expression given in Eq. (30) represents the degradation of the elastic stiffness written in terms of the crack damage ratio $\alpha $ and the void damage ratio $\beta $. For the special case of elastic energy equivalence, Eq. (30) reduces to the following expression:

while for the special case of elastic strain equivalence, one obtains the following expression:

In Eq. (30), one can deduce a special simplified case when the effect of crack damage and void damage is equal. In this case, one substitutes $\alpha =\beta $ into Eq. (30) to obtain the following simplified formula for the decomposition:

It is noted from Eq. (33) that the elastic stiffness degradation is inversely proportional to the crack/void damage fraction. It is noted that degradation of elastic stiffness increases as the value of the ratio $E\xaf/E$ increases beyond one. When $E\xaf/E=1$, there is no degradation in elastic stiffness and the material is undamaged—in this case, the value of the crack/void damage fraction is one.

It is true that in the mathematical formulation presented in this work, there does not seem to be a difference between the two types of damage, i.e., cracks and voids. Indeed, when one looks at Eqs. (16) and (26), one finds that the decomposition is symmetric with respect to cracks and voids. Several methods can be used to induce asymmetry in the decomposition, but this is beyond the scope of this work. The main result of the decomposition of elastic stiffness degradation of Eq. (27) is also symmetric. The authors believe that they have achieved their goals in writing this manuscript, mainly to provide a way to derive a mathematical decomposition of elastic stiffness degradation. The issue how the two parts of the decomposition (cracks and voids) differ does not come into play in this regard. It is left to each researcher to include specific models of crack damage and void damage into the decomposition equations (whether scalar or tensorial). Specifically, one can use the void damage model of Gurson [36] in this regard.

The coverage of the current paper is limited to the elastic range. It is made very clear in the paper that the elastic stiffness degradation due to damage is the one that is being investigated. Coverage of plasticity and degradation to plasticity are separate topics and need a separate paper for its investigation.

Finally, the whole decomposition procedure is generalized (using tensors) to the state of general deformation and damage. An attempt is made to obtain an explicit formula for the tensorial decomposition of elastic stiffness degradation. Instead of using indicial notation to represent tensors, use is made of the appropriate vector and matrix representation of tensors in order to simplify the resulting equations. Thus throughout this section, all the equations are matrix equations (i.e., matrix representations of tensors). For this purpose, a vector is denoted by braces, while a tensor is denoted by brackets.

where ${\sigma}$ and ${\sigma \xaf}$ are the vector representations of the second-rank Cauchy stress tensor and its effective counterpart, respectively, while $[M]$ is the matrix representation of the fourth-rank damage effect tensor. The damage effect tensor is a fourth-rank tensor that is extensively defined and used in the literature [2,3,7,9,18,24,25] keeping in mind that for the one-dimensional scalar case, $M1111=(1/1\u2212\phi )$.

The scalar damage variable $\u2113$ of Eq. (3) can be easily generalized to three dimensions applicable to general states of deformation and damage. This generalization can be achieved using two different methods as follows [6] using matrix notation (see Fig. 4)—it is assumed that $n=1$ throughout the following formulation which is applicable to linear elastic materials:

All the matrices appearing in Eqs. (36) and (37) are representations of fourth-rank tensors. The matrices $[E]$ and $[E\xaf]$ represent the fourth-rank elasticity tensor and its effective counterpart, respectively. The matrices $[L](1)$ and $[L](2)$ are two different possible representations of the fourth-rank generalized damage tensor corresponding to the scalar damage variable $\u2113$. It is seen from Eqs. (36) and (37) that both $[L](1)$ and $[L](2)$ are defined in terms of the reduction of the elastic stiffness as was done with the scalar damage variable $\u2113$. Furthermore, both Eqs. (36) and (37) reduce to the single scalar equation (3) once the appropriate substitutions are made for the one-dimensional case, i.e., $E1111=E$, $E\xaf1111=E\xaf$, $L1111(1)=\u2113$, and $L1111(2)=\u2113$. In this work, $[L](1)$ and $[L](2)$ are called fourth-rank damage tensors of the first kind and of the second kind, respectively. It should be noted that the two definitions of $[L](1)$ and $[L](2)$ of Eqs. (36) and (37) are identical except for the order of multiplication of the matrices $[E\xaf]\u2212[E]$ and $[E\xaf]\u22121$. This is so because matrix multiplication is generally not commutative.

One can also define a fourth-rank damage tensor associated with elastic stiffness reduction by taking the average value of $[L](1)$ and $[L](2)$ as follows:

where $[L]$ is the matrix representation of the new generalized fourth-rank damage tensor. Again for the one-dimensional case, Eq. (38) reduces to $L1111=\u2113$.

Substituting the expressions of Eqs. (36) and (37), into Eq. (38), and simplifying, one obtains the following general explicit expression for $[L]$:

where $[I]$ is the matrix representation of the fourth-rank identity tensor, noting that for the one-dimensional case $I1111=1$.

Next, one studies the relationship between the fourth-rank damage tensors $[L]$ and $[M]$. In particular, one needs to find the exact relationship between $[L](1)$ and $[M]$ on the one hand, and between $[L](2)$ and $[M]$ on the other hand. For this purpose, use is made of the generalization of the elastic constitutive equations (8) and (9) written in the matrix form as follows:

The definition of the two fourth-rank damage tensors associated with elastic stiffness degradation of Eqs. (36) and (37) may be rewritten in the following more appropriate form (actually, one can obtain the exact relationship between the two fourth-rank damage tensors $[L](1)$ and $[L](2)$ by solving Eqs. (42) and (43) simultaneously while eliminating either $[E]$ or $[E\xaf]$ as is done in Ref. [6]):

The formulation will be presented below utilizing the fourth-rank damage tensor $[L](1)$ of the first kind in detail where use is made of Eqs. (35), (36), and (42). A similar formulation can be followed to derive the corresponding equations for the fourth-rank damage tensor $[L](2)$ of the second kind using Eqs. (35), (37), and (43). The subsequent derivation is based on the hypothesis of elastic energy equivalence ($n=1$). However, the formulation starts with the general hypothesis of elastic energy equivalence of order $n$ of Eq. (5), then it is reduced to the case $n=1$. The scalar representation of the hypothesis can be generalized as follows in both indicial and matrix forms:

where the number of strain terms in Eq. (44) that are multiplied on both sides is $n$. In the remaining part of this work, use is made exclusively of matrix notation to represent tensors for the sake of simplicity. It should be noted from Eq. (45) that the operation of raising a vector to the power $n$ is not defined mathematically. What is implied by the operation ${\epsilon}n$ is raising the associated matrix representation of the strain tensor to the power $n$. The same holds also for ${\epsilon \xaf}n$. This peculiar operation is illustrated in detail in the Appendix.

The following part of the formulation is limited to linear elastic materials with $n=1$. Substituting Eqs. (40) and (41) into Eq. (45) and performing the relevant matrix operations and simplifications, one obtains

Substituting Eq. (35) into Eq. (46) and simplifying, one obtains the following transformation equation:

Next, one uses the following tensorial decomposition of the fourth-rank damage effect tensor $[M]$ (which is a generalization of the scalar decomposition equation (16) [10]):

where $[M]c$ and $[M]v$ are the fourth-rank damage effect tensors due to cracks and voids, respectively (based on cross-sectional area reduction). It should be noted that $M1111c=(1/1\u2212\varphi c)$ and $M1111v=(1/1\u2212\varphi v)$ (see Fig. 5).

Substituting Eq. (48) into Eq. (47), one obtains

Substituting Eq. (42) into Eq. (47) and solving the resulting expression for $[L](1)$, one obtains

It is clear from Eq. (50) that one cannot obtain an explicit expression for $[M]$ in terms of $[L](1)$.

Based on the matrix form of Eqs. (42) and (43), one introduces the fourth-rank damage tensors $[Lc](1)$ and $[Lv](1)$ of the first kind for cracks and voids, respectively, and writes the following two expressions for them (directly based on Eqs. (42) and (43)):

It should be noted that the fourth-rank damage tensor of the first kind $[Lc](1)$ represents the degradation of the fourth-rank elasticity tensor when cracks are present only (voids are removed), while the fourth-rank damage tensor of the first kind $[Lv](1)$ represents the degradation of the fourth-rank elasticity tensor when voids are present only (cracks are removed). Equation (50) can be rewritten for cracks and voids, separately as follows (see Fig. 6):

Before continuing with the derivation, the symbols used for the various fourth-rank elasticity tensors are summarized in Table 3.

Next, one can generalize the scalar decomposition of elastic stiffness degradation of Eq. (23) using tensors as follows:

The proof that the above generalization of the tensorial decomposition of elastic stiffness degradation works is not shown here but can be formulated by the reader by performing it backward, i.e., start from the final decomposition of Eq. (57) and go backward to deduce the original starting Eq. (42) or (43).

In order to solve a specific example of the decomposition of elastic stiffness degradation, one needs to use all the three Eqs. (55), (56), and (57) simultaneously. These three equations comprise an implicit system of simultaneous tensorial decomposition equations for the elastic stiffness degradation of the first kind. These three equations are called a *tensorial decomposition system* for elastic stiffness degradation. This system is summarized in the boxed equations below followed by the steps needed to solve a specific example using this system:
Tensorial decomposition system of the first kind

^{(1)}Substitute the appropriate matrix elements for $[Lc](1)$ into Eq. (A) and obtain the appropriate matrix elements for $[Mc]$.^{(2)}Substitute the appropriate matrix elements for $[Lv](1)$ into Eq. (B) and obtain the appropriate matrix elements for $[Mv]$.^{(3)}Substitute the appropriate matrix elements for $[Lc](1)$ and $[Lv](1)$ into Eq. (C) to obtain the appropriate matrix elements of $[L](1)$.

The above three steps in using the decomposition system ((A)–(C)) are now applied for the simple example of the scalar one-dimensional case. For this purpose, one substitutes $M1111c=(1/1\u2212\varphi c)$, $E1111c=Ec$, $L1111c(1)=\u2113c$, and $I=11111$ into Eq. (A) to obtain $1+\u2113c=(1/(1\u2212\varphi c)2)$. Similarly, one substitutes $M1111v=(1/1\u2212\varphi v)$, $E1111v=Ev$, $L1111v(1)=\u2113v$, and $I=11111$ into Eq. (B) to obtain $1+\u2113v=(1/(1\u2212\varphi v)2)$. The two scalar expressions thus obtained conform with the scalar expression of Eq. (12) for the one-dimensional case of Sec. 2. Finally, substituting $L1111(1)=\u2113$, $(L1111c)(1)=\u2113c$, and $(L1111v)(1)=\u2113v$ into Eq. (C) to recover the scalar decomposition of Eq. (23) of the one-dimensional case of Sec. 2.

The above tensorial derivation of Eqs. (A)–(C) may now be repeated for the fourth-rank damage tensor of the second kind $[L](2)$. In this case, the following tensorial decomposition system of the second kind is obtained and shown in the boxed equations below:
Tensorial decomposition system of the second kind

^{(1)}Substitute the appropriate matrix elements for $[Lc](2)$ into Eq. (D) and obtain the appropriate matrix elements for $[Mc]$.^{(2)}Substitute the appropriate matrix elements for $[Lv](2)$ into Eq. (E) and obtain the appropriate matrix elements for $[Mv]$.^{(3)}Substitute the appropriate matrix elements for $[Lc](2)$ and $[Lv](2)$ into Eq. (F) to obtain the appropriate matrix elements of $[L](2)$.

One can show that the tensorial decomposition system of the second kind of Eqs. (D)–(F) also reduces to the single scalar decomposition equation (23), once the appropriate substitutions are made. In particular, both Eqs. (D) and (E) reduce to the scalar equation (12) of the one-dimensional case of Sec. 2, while Eq. (F) reduces to the scalar decomposition equation (23).

As an illustrative example, the decomposition system equations are solved below for the special case of plane stress with $n=1$. The linear elasticity tensors $[E]$ and $[E\xaf]$ of Eqs. (40) and (41), respectively, can be written in the following matrix form for the case of plane stress [6,16]:

where $E$, $G$, and $\upsilon $ are the elastic modulus, shear modulus, and Poisson's ratio in the deformed/damaged configuration, and $G=(E/2\u2009(1+\upsilon ))$. One would want to solve the tensorial decomposition system of the first kind of Eqs. (A)–(C) (or alternatively, the tensorial decomposition system of the second kind of Eqs. (D)–(F)) in order to find an implicit relationship between the elastic stiffness degradation $E\xaf/E$, the shear modulus degradation $G\xaf/G$, and the Poisson's ratio's degradation $\upsilon \xaf/\upsilon $ on the one hand, and the respective degradations in the associated ratio for cracks and voids, i.e., $E\xafc/Ec$, $E\xafv/Ev$, $G\xafc/Gc$, $G\xafv/Gv$, $\upsilon \xafc/\upsilon c$, and $\upsilon \xafv/\upsilon v$, on the other hand. In the notation used here, a superscript *c* is used to denote a quantity that is valid when cracks are present only, while a superscript *v* is used to denote a quantity that is valid when voids are present only. When there is no superscript used, it is assumed that the quantity is valid for the total damage system when both cracks and voids are present.

The solution of the special example of plane stress is performed in this section using the fourth-rank damage tensor of the first kind $[L](1)$, along with the tensorial decomposition system of the first kind. A somewhat similar solution but a bit more complicated can be performed using the fourth-rank damage tensor of the second kind $[L](2)$, along with the tensorial decomposition system of the second kind, and is presented in Sec. 4.2.

The fourth-rank damage effect tensor of the first kind $[L](1)$ can be written in the following general matrix representation for the case of plane stress:

where the superscript (1) is dropped in the elements of the matrix above in the remaining part of this section.

Next, one considers the damage transformation equation for the Cauchy stress tensor of Eq. (35) where the matrix representation $[M]$ of the fourth-rank damage tensor takes the following form for the special case of plane stress [5,6]:

and $\phi 11$, $\phi 22$, and $\phi 12$ are the nonzero components of the damage tensor for the case of plane stress. As can be seen from Eq. (51), the case of plane stress can be represented by a simple 3 × 3 matrix of nonzero elements.

Finally, the use of the three steps to apply/solve the tensorial decomposition system of the first kind (Eqs. (A)–(C)) is illustrated below. In order to be able to obtain reasonable results, one needs to use a numerical value for the integer exponent $n$. In this case, one solves the plane stress example for the classical case of the hypothesis of elastic energy equivalence where $n=1$. Thus, this value of the exponent is used in all the subsequent substitutions. The three steps of the tensorial decomposition of the first kind are illustrated next for plane stress.

^{(1)}Substitute the following matrices for $[Mc]$, $[Ec]$, $[I]$, and $[Lc](1)$ into Eq. (A) for the plane stress case (based on the matrices of Eqs. (58)–(61) but with a superscript*c*to denote cracks):Display Formula (65)$[Ec]=Ec1\u2212\upsilon c2[1\upsilon c0\upsilon c10001\u2212\upsilon c2]$Display Formula (66)$[Ec]\u22121=[1Ec\u2212\upsilon cEc0\u2212\upsilon cEc1Ec0001Gc]$Display Formula (67)$[Mc]=1\Delta c[\psi 22c0\phi 12c0\psi 11c\phi 12c\phi 12c2\phi 12c2\psi 11c+\psi 22c2]$Display Formula (68)$\psi 11c=1\u2212\phi 11c$Display Formula (69)$\psi 22c=1\u2212\phi 22c$Display Formula (70)$\Delta c=\psi 11c\psi 22c\u2212(\phi 12c)2$Display Formula (71)$[Lc](1)=[\u211311c\u211312c\u211313c\u211321c\u211322c\u211323c\u211331c\u211332c\u211333c]$Display Formula (72)$[I]=[100010001]$while noting thatDisplay Formula $Gc=(Ec/2\u2009(1+\upsilon c))$ in Eq. (66).Carrying out the matrix operations of Eq. (A) using Wolfram Alpha (see footnote 1), one obtains the following unexpected diagonal matrix:(73)$[Lc](1)=1(\Delta c)2[(\psi 22c)21\u2212(\upsilon c)2\u22121000(\psi 11c)21\u2212(\upsilon c)2\u22121000(\psi 11c+\psi 22c2)2\u22121]$where it is clear from Eq. (73) that the following components are obtained for the damage tensor of the first kind $[Lc](1)$ with respect to elastic stiffness degradation under plane stress:Display Formula (74)$\u211311c=(\psi 22c)21\u2212(\upsilon c)2\u22121$Display Formula (75)$\u211322c=(\psi 11c)21\u2212(\upsilon c)2\u22121$Display Formula (76)$\u211333c=(\psi 11c+\psi 22c2)2\u22121$Display Formula (77)$\u2113ijc=0,\u2003i\u2260j$It is observed that the fact that the resulting matrix representation of the damage tensor of the first kind is diagonal means that the elastic stiffness degradation due to the presence of cracks (only) occurs along the principal directions. This simple result is indeed unexpected.^{(2)}Substitute the following matrices for $[Mv]$, $[Ev]$, $[I]$, and $[Lv](1)$ into Eq. (B) for the plane stress case (based on the matrices of Eqs. (58)–(61) but with a superscript*v*to denote voids):Display Formula (78)$[Ev]=Ev1\u2212\upsilon v2[1\upsilon v0\upsilon v10001\u2212\upsilon v2]$Display Formula (79)$[Ev]\u22121=[1Ev\u2212\upsilon vEv0\u2212\upsilon vEv1Ev0001Gv]$Display Formula (80)$[Mv]=1\Delta v[\psi 22v0\phi 12v0\psi 11v\phi 12v\phi 12v2\phi 12v2\psi 11v+\psi 22v2]$Display Formula (81)$\psi 11v=1\u2212\phi 11v$Display Formula (82)$\psi 22v=1\u2212\phi 22v$Display Formula (83)$\Delta v=\psi 11v\psi 22v\u2212(\phi 12v)2$Display Formula (84)$[Lv](1)=[\u211311v\u211312v\u211313v\u211321v\u211322v\u211323v\u211331v\u211332v\u211333v]$Display Formula (85)$[I]=[100010001]$while noting that $Gv=(Ev/2\u2009(1+\upsilon v))$ in Eq. (79).Carrying out the matrix operations of Eq. (B) using Wolfram Alpha (see footnote 1), one obtains the following unexpected diagonal matrix:Display Formula (86)$[LV](1)=1(\Delta v)2[(\psi 22v)21\u2212(\upsilon v)2\u22121000(\psi 11v)21\u2212(\upsilon v)2\u22121000(\psi 11v+\psi 22v2)2\u22121]$where it is clear from Eq. (86) that the following components are obtained for the damage tensor of the first kind with respect to elastic stiffness degradation under plane stress:Display Formula (87)$\u211311v=(\psi 22v)21\u2212\upsilon 2\u22121$Display Formula (88)$\u211322v=(\psi 11v)21\u2212(\upsilon v)2\u22121$Display Formula (89)$\u211333v=(\psi 11v+\psi 22v2)2\u22121$Display Formula (90)$\u2113ijv=0,\u2003i\u2260j$It is observed that the fact that the resulting matrix representation of the damage tensor of the first kind is diagonal means that the elastic stiffness degradation due to the presence of voids (only) occurs along the principal directions. This simple result is indeed unexpected.^{(3)}Substitute Eqs. (60), (71), (72), and (84) into Eq. (C) in order to obtain the following explicit decomposition form of the 3 × 3 matrix representation of $\u230aL\u230b(1)$ in terms of the components of $[Lc](1)$ and $[Lv](1)$ (derived using Wolfram Alpha (see footnote 1):Display Formula (91)$[\u211311\u211312\u211313\u211321\u211322\u211323\u211331\u211332\u211333]=[(\u211311c+1)\u2009(\u211311v+1)\u22121\u211312c\u211312v\u211313c\u211313v\u211321c\u211321v(\u211322c+1)\u2009(\u211322v+1)\u22121\u211323c\u211323v\u211331c\u211331v\u211332c\u211332v(\u211333c+1)\u2009(\u211333v+1)\u22121]$

It is noted from Eq. (91) that the superscript (1) is dropped from the components of the various damage tensors. Comparing the matrices of Eqs. (73), (86), and (91), one concludes that the matrix of Eq. (91) above must be a diagonal matrix. Thus, one concludes that the following off-diagonal components vanish, i.e., $\u211312c=\u211321c=\u211313c=\u211331c=\u211323c=\u211332c=0$ and $\u211312v=\u211321v=\u211313v=\u211331v=\u211323v=\u211332v=0$. Therefore, the matrix representation of Eq. (91) reduces to the following diagonal form: