**2. Numerical simulation analysis**

## **2.1 The elastic: plastic model**

The yield function adopted is given in Eq. 1 [6]:

$$F = \sqrt{J\_{2D}} - \left(\kappa + H\gamma\_{\rm act}^P\right) = \mathbf{0} \tag{1}$$

where *J*2*<sup>D</sup>* is the second invariant of deviatoric stress; *κ* is the size of the initial yield surface; *γ p oct* is the plastic octahedral shear strain; and *H* is the strain-hardening (or strain-softening) parameter. Differentiating Eq. 1 leads to:

$$dF = \left(\frac{\partial F}{\partial \sigma}\right)^T d\sigma + \left(\frac{\partial F}{\partial \gamma\_{\rm act}^p}\right)^T d\gamma\_{\rm act}^p = \left(\frac{\partial F}{\partial \sigma}\right)^T d\sigma - H d\gamma\_{\rm act}^p = 0 \tag{2}$$

where *dσ* is the incremental stress vector; the transpose of vector *<sup>∂</sup><sup>F</sup> <sup>∂</sup><sup>σ</sup>* is *<sup>∂</sup><sup>F</sup> ∂σ* � �*<sup>T</sup>* ; the transpose of vector *<sup>∂</sup><sup>F</sup> ∂γ p oct* is *<sup>∂</sup><sup>F</sup> ∂γ p oct* � �*<sup>T</sup>* ; and *dγ p oct* is the incremental plastic octahedral shear strain. The flow rule is:

$$d\mathbf{c}^p = \lambda \frac{\partial F}{\partial \sigma} \tag{3}$$

where *dε<sup>p</sup>* is the incremental plastic strain vector. Therefore, the elastic–plastic stress–strain matrix *Dep* can be expressed as follows:

$$\mathbf{D}^{\rm p} = \mathbf{D}^{\varepsilon} - \mathbf{D}^{p} = \mathbf{D}^{\varepsilon} - \mathbf{D}^{\varepsilon} \left(\frac{\partial \mathbf{F}}{\partial \sigma}\right)^{T} \frac{\partial \mathbf{F}}{\partial \sigma} \mathbf{D}^{\varepsilon} \left[\frac{1}{\sqrt{6}} H + \left(\frac{\partial \mathbf{F}}{\partial \sigma}\right)^{T} \mathbf{D}^{\varepsilon} \frac{\partial \mathbf{F}}{\partial \sigma}\right]^{-1} \tag{4}$$

where super indexes e and p mean elastic and plastic, respectively.

#### **2.2 Conditions of stability**

For strain-softening materials, under the action of external force, when the strain goes deep into the plastic range, to obtain the stable incremental finite-element solutions, Prevost [7] used the variational approach based on the internal strain energy due to the stresses of the existing state and the stress increments, thereby obtaining the failure mechanism; then, define "A failure mechanism develops when the incremental external work put into the system, plus the incremental work reduced by the plastic, strain-softening zone, equals or exceeds the work that may be absorbed by the surrounding unyielded and/or strain-hardening material." The total volume of the body, V, can be divided into *Vps* and V- *Vps*, where *Vps* is the total volume of the strain-softening regions. Under this consideration, the uniqueness of the incremental solutions is proved.

For physical materials, if the total external incremental applied energy is positive, the total induced internal incremental strain energy must be positive. On the other hand, if the total external incremental applied energy is negative, the total induced internal incremental strain energy must be negative. All stable numerical solutions must obey such a law; otherwise, the numerical solution will be unstable.

If *πp*, is the total potential energy in static analyses under a variational approach, then the total incremental potential energy, *dπ<sup>p</sup>* is:

$$d\pi\_p = \frac{1}{2} \int\_V d\sigma^T d\mathbf{e}dV - \sum d\mathbf{u}^T d\mathbf{F} \tag{5}$$

where *dσ<sup>T</sup>* is the transpose of the incremental stress vector, *dε* is the incremental strain vector, *du<sup>T</sup>* is the transpose of the incremental displacement vector, and *dF* is the incremental force vector.

In the finite-element approximation, Eq. 5 can be expressed as

$$d\pi\_p = \frac{1}{2}\int\_V (d\mathbf{u}^T \mathbf{B}^T \mathbf{D}^{ep} \mathbf{B} d\mathbf{u})dV - \sum (d\mathbf{u}^T d\mathbf{F})\tag{6}$$

where *B<sup>T</sup>* is the transpose of the strain–displacement matrix. The global stiffness matrix *K* is:

$$\mathbf{K} = \int\_{V} (\mathbf{B}^T \mathbf{D}^{ep} \mathbf{B})dV\tag{7}$$

Substitution of Eq. 7 into 6 leads to:

$$d\pi\_p = \frac{1}{2} d\mathfrak{u}^T \mathbf{K} d\mathfrak{u} - d\mathfrak{u}^T d\mathbf{F} \tag{8}$$

It can be seen from Eq. 8 that, for the condition of the prescribed displacements and since all of the forces are induced, both terms on the right-hand side of Eq. 8 will always have the same sign. This means that the condition of stability for the incremental solutions for the prescribed displacement case is guaranteed. When the forces are prescribed and if the total external incremental energy is positive, the stability of the incremental solutions is guaranteed only when the global stiffness matrix is positive definite. When the global stiffness matrix is negative definite, the solution will be unstable [6].

### **2.3 Conditions of uniqueness**

For a strain-hardening material or perfectly plastic material, since these types of materials meet the Drucker stability postulates, the solution is unique because the two potential energy increments on the right-hand side of Eq. 8 are both positive.

Based on the need to prove a unique solution, first let us assume two sets of solutions, *du*<sup>1</sup> and *du*2, satisfy the equilibrium condition at the same time. Under this assumption, both *du*<sup>1</sup> and *du*<sup>2</sup> should provide the same minimum total incremental potential energy. By substituting *du*<sup>1</sup> and *du*<sup>2</sup> into Eq. 8 and minimizing the total incremental potential energy with respect to *du*<sup>1</sup> and *du*2, one obtains:

$$\mathbf{K}d\mathfrak{u}\_1 = d\mathbf{F} \tag{9}$$

$$\mathbf{K}d\mathbf{u}\_2 = d\mathbf{F} \tag{10}$$

Eq. 9 minus Eq. 10 equals:

$$\mathbf{K}(d\mathfrak{u}\_1 - d\mathfrak{u}\_2) = \mathbf{0} \tag{11}$$

Therefore, one of the following two conditions must be true:

$$du\_1 = du\_2\tag{12}$$

(when the determinant of *K* is not equal to zero) or

$$\det(\mathbf{K}) = \mathbf{0} (\text{when } \mathbf{du}\_1 \neq \mathbf{du}\_2) \tag{13}$$

#### *Earthquakes - Recent Advances, New Perspectives and Applications*

**Figure 5.** *Finite element mesh, boundary conditions, and prescribed lateral displacements.*

#### **2.4 Finite element solutions**

For the 5.08 cm � 2.54 cm plate shown in **Figure 5** under plane strain conditions loaded at both ends and where the movement in the direction perpendicular to the loading is constrained, a uniform 50 � 25 mesh was used to analyze its behavior under uniformly prescribed loading conditions. The material properties used were: (1) the initial size of the yield surface, *κ*, was equal to 24 kPa, (2) Young's modulus, *E*, was equal to 1200 kPa, (3) Poisson's ratio, *ν*, was equal to 0.3, (4) the shear modulus, *G*, was equal to 462 kPa, and (5) the strain softening parameter, *H*/2*G*, was equal to 0.0 to model perfectly plastic behavior and equal to �0.05 to model strain softening behavior. To obtain shear bands in a numerical solution close to 45 degrees, it will be better to use square with the five-node elements. For the plate problem analyzed, the sequence for the application of the uniform prescribed displacements in the main load steps are 0.1143 cm, 0.0635 cm, 0.0635 cm, 0.0635 cm, 0.0635 cm, and 0.0635 cm, respectively. Each main load step is further subdivided into five sub-steps of load. Therefore, after applying all the load, the width of the plate remains unchanged at 2.54 cm, and the length is shortened to 4.6482 cm.

For elastic–perfectly plastic materials, when the determinant of the structural stiffness matrix *K* is not equal to zero, then *du*<sup>1</sup> ¼ *du*2. In this case, the results of finite element analysis including the deformed mesh, velocity vector distribution map, and contour lines of incremental strain energy density (see **Figure 6**) do not include shear failure bands and maintain the original symmetry because Drucker stability postulates continue to be satisfied.

#### **Figure 6.**

*Results of the finite-element analysis of a plate using an elastic–perfectly plastic model [6]: (a) deformed mesh; (b) velocity vector distribution map; (c) contours of incremental strain energy density. Note: Dimensions for the plates shown in Figure 6b and c are 5.08 cm* � *2.54 cm.*

*Plasticity Model Required to Prevent Geotechnical Failures in Tectonic Earthquakes DOI: http://dx.doi.org/10.5772/intechopen.107223*

#### **Figure 7.**

*Results of the finite-element analysis of a plate using an elastic–plastic strain softening model [6]: (a) deformed mesh; (b) velocity vector distribution map; (c) contours of incremental strain energy density. Note: Dimensions for the plates shown in Figure 7b and c are 5.08 cm* � *2.54 cm.*

For elastic–plastic strain-softening materials, the determinant of the structural stiffness matrix *K* is equal to zero. In this case, external agencies must be applied as specified displacements instead of specified loads to obtain the results of the finite element analysis, including the deformed mesh, velocity vector distribution map, and contour lines of incremental strain energy density (see **Figure 7**), such that they do include asymmetric shear failure bands.

#### **2.5 Loss of symmetry**

It can be seen from **Figure 7** that the condition for the loss of symmetry of the foundation soil or the soil specimen is the appearance of shear failure bands induced by plastic strain softening.

As far as the structural stiffness matrix *K* is concerned, its determinant will be equal to zero because of the plastic strain softening, which will induce shear failure bands under prescribed displacements. In the elastic range, the entries of the structural stiffness matrix *K* have a symmetric condition, that is *kij* ¼ *kji*, and in the plastic strain-softening range, the condition for the determinant of *K* to be equal to zero is that the entries in the two adjacent rows of the structural stiffness matrix *K* correspond, that is *kmi* ¼ *knj* for *n* = *m*+1. The structure matrix is therefore asymmetric [3].
