Mechanical Initiation and Propagation Mechanism of a Thrust Fault: A Case Study of the Yima Section of the Xiashi-Yima Thrust (North Side of the Eastern Qinling Orogen, China)

Abstract

Thrust faults exist extensively in nature, and their activities often cause earthquakes and disasters involving underground engineering, such as the May 12, 2008 Wenchuan Earthquake; the April 20, 2013 Ya’an Earthquake; and the Nov. 3, 2011 Yima Qianqiu Coal-Mining Accident in China. In this paper, the initiation and propagation of a thrust are discussed from a mechanical viewpoint using fault mechanics and fault-slip analysis, taking as an example the Yima section of the Xiashi-Yima thrust (north side of the eastern Qinling Orogen, China). The research primarily focuses on the stress field and the formation trajectory of the thrust and the genesis of the large-scale inversion thrust sheet. The results show that the thrust results from failures in the compressive deformation state and that its stress state is entirely compressive shear. The rupture trajectory of the thrust develops upward, and the fault fracture zone forms similarly to a listric fault, up-narrow and down-wide. The model results and the genesis of the large-scale inversion thrust sheet are consistent with in situ exploration observations. This investigation can be extended to other thrust faults with similar characteristics, particularly for the design of mining operations in tectonic-active areas. Moreover, this research can be used to further study the mechanism of thrust faults and provide support for the feasibility of using fault-slip analysis to assess fault stability.

Appendices

Appendix A: Model of Material Mechanics-Linear Elastic Fracture Mechanics (MM-LEFM) (Ma et al. 2003)

Assuming an overhanging beam with an asymmetrical crack and two opposite forces (P ₁ and P ₂) at the free end (see Fig. 21), the section (B–C) can be approximated as an overhanging beam with equivalent force (P = P ₂ − P ₁) acting on the end region (B-B’), and thereby, the stresses can be described by material mechanics theory:

$$\sigma_{x} = \frac{{3\left( {P_{2} - P_{1} } \right)}}{{2h_{{}}^{3} b}}\left( {x - a} \right)y,\;\sigma_{y} = 0,\;\tau_{xy} = \frac{{3\left( {P_{2} - P_{1} } \right)}}{4bh}\left( {1 - \frac{{y^{2} }}{{h_{{}}^{2} }}} \right)$$

(9)

where b is the width of the beam and h is the height of the beam.

Let 2h ₁ and 2h ₂ be the distances from the crack to the lower and upper free surfaces, respectively. Accordingly, the upper and lower section of AB can be, respectively, approximated as the overhanging beam with end region (B-B’). Thus, the stresses can be expressed as follows:

$$\sigma_{x1} = \frac{{3P_{1} }}{{2h_{1}^{3} b}}xy,\;\sigma_{y1} = 0,\;\tau_{xy1} = \frac{{3P_{1} }}{{4bh_{1} }}\left( {1 - \frac{{y^{2} }}{{h_{1}^{2} }}} \right)$$

(10)

$$\sigma_{x2} = \frac{{3P_{2} }}{{2h_{2}^{3} b}}xy,\;\sigma_{y2} = 0,\;\tau_{xy2} = \frac{{3P_{2} }}{{4bh_{2} }}\left( {1 - \frac{{y^{2} }}{{h_{2}^{2} }}} \right)$$

(11)

By substituting $y = H - 2h_{2}$ and $h = H$ into Eq. (9), the shear stress ($\tau_{xy}$) of the crack tip can be expressed as follows:

$$\tau_{xy} = \frac{{3\left( {P_{2} - P_{1} } \right)}}{4bH}\left( {1 - \frac{{\left( {H - 2h_{2} } \right)^{2} }}{{H_{{}}^{2} }}} \right)$$

(12)

The relative tension stress of the crack tip can be expressed as

$$\sigma_{x(a)} = \sigma_{x2(a)} - \sigma_{x1(a)} = \frac{{3a\left( {P_{2} h_{1}^{2} - P_{1} h_{2}^{2} } \right)}}{{2b\left( {h_{1} h_{2} } \right)^{2} }}$$

(13)

The force (P ₁) acting on the lower section is usually very small or approximately zero, and the thickness is usually very large. For this reason, the lower section can be equivalent to a large, thick basement that largely stretches downward. Therefore, $\sigma_{y} = 0$, $\tau_{xy} \approx 0$, and Eq. (13) is approximated as

$$\sigma_{x(a)} = \frac{{3aP_{2} }}{{2bh_{2}^{2} }}$$

(14)

According to the material mechanics theory, the maximum tension stress (or normal stress) exists around the outward surface on the fixed end of beam. If the stress ($\sigma_{x(a)}$) is considered only, a vertical crack developing upward will likely be generated in the crack end of the upper branch beam.

Additionally, the crack, under the stress state shown in Fig. 21, can be considered a mode I crack from the perspective of fracture mechanics. Therefore, the stress components are expressed as follows:

$$\left\{ \begin{gathered} \bar{\sigma }_{x} = \frac{{K_{\text{I}} }}{{\sqrt {2\pi r} }}\cos \frac{{\bar{\theta }}}{2}\left( {1 - \sin \frac{{\bar{\theta }}}{2}\sin \frac{{3\bar{\theta }}}{2}} \right) \hfill \\ \bar{\sigma }_{y} = \frac{{K_{\text{I}} }}{{\sqrt {2\pi r} }}\cos \frac{{\bar{\theta }}}{2}\left( {1 + \sin \frac{{\bar{\theta }}}{2}\sin \frac{{3\bar{\theta }}}{2}} \right) \hfill \\ \bar{\tau }_{xy} = \frac{{K_{\text{I}} }}{{\sqrt {2\pi r} }}\sin \frac{{\bar{\theta }}}{2}\cos \frac{{\bar{\theta }}}{2}\cos \frac{{3\bar{\theta }}}{2} \hfill \\ \end{gathered} \right.$$

(15)

where $K_{\text{I}}$ is the stress intensity factor of the mode I crack. $r$ is the distance parameter in polar coordinates, in which the crack end is defined as the coordinate origin. $\bar{\theta }$ is the cleavage angle (between the cleavage direction and the direction of the original crack).

Using the stress components as axes, the following coordinate system can be derived:

$$\left\{ \begin{gathered} \sigma_{r} = \frac{{\bar{\sigma }_{x} + \bar{\sigma }_{y} }}{2} + \frac{{\bar{\sigma }_{x} - \bar{\sigma }_{y} }}{2}\cos 2\bar{\theta } + \bar{\tau }_{xy} \sin 2\bar{\theta } \hfill \\ \sigma_{{\bar{\theta }}} = \frac{{\bar{\sigma }_{x} + \bar{\sigma }_{y} }}{2} - \frac{{\bar{\sigma }_{x} - \bar{\sigma }_{y} }}{2}\cos 2\bar{\theta } - \bar{\tau }_{xy} \sin 2\bar{\theta } \hfill \\ \tau_{{r\bar{\theta }}} = \bar{\tau }_{xy} \cos 2\bar{\theta } - \frac{{\bar{\sigma }_{x} - \bar{\sigma }_{y} }}{2}\sin 2\bar{\theta } \hfill \\ \end{gathered} \right.$$

(16)

According to fracture mechanics theory, the cleavage direction of a mode I crack is in the direction of the original crack ($\bar{\theta } = 0$). In this context, the crack propagation is dominated by $\sigma_{r}$ (see Eq. (17)) in the stress field of the crack end.

$$\sigma_{r} = \frac{{K_{I} }}{{\sqrt {2\pi r} }},\;K_{I} = k\left( {P_{1} + P_{2} } \right)a\left( {\frac{{h_{1} + h_{2} }}{{2h_{1} h_{2} }}} \right)^{3/2}$$

(17)

where $k$ is a correction factor.

It should be noted that the stress values ($\sigma_{r}$, $\sigma_{{\bar{\theta }}}$, and $\tau_{{r\bar{\theta }}}$) are all zero near the crack end when $\bar{\theta } = \pi$. However, the stress ($\sigma_{x(a)}$) is not zero (see Eq. 14). Therefore, crack propagation should be affected by the resultant stress ($\sigma$) of $\sigma_{r}$ and $\sigma_{x(a)}$, and the direction of crack propagation is the vertical direction of $\sigma$, which can be expressed as follows:

$$\sigma = \sqrt {\sigma_{r}^{2} + \sigma_{x(a)}^{2} } ,\;\bar{\theta } = \arctan \left( {\frac{{\sigma_{x(a)}^{{}} }}{{\sigma_{r} }}} \right)$$

(18)

By substituting Eqs. (14) and (17) into Eq. (18), and letting P ₁ = 0, the cleavage angle ($\bar{\theta }$) can be expressed as

$$\bar{\theta } = \arctan \left( {\frac{{6\sqrt {\pi r} }}{bk}\frac{1}{{\sqrt {h_{2} } }}} \right) \propto \frac{1}{{\sqrt {h_{2} } }}$$

(19)

The parameter (b) can be regarded as the unit length, and r is not equal to zero. Irwin’s theory states that the radius of the plastic zone at the crack end can be considered as the value of r, and as such, it is usually constant for a given material. As a result, with the propagation of the crack, h ₂ decreases, and in this context, the stratum on the crack face can be considered to be equivalent to a beam with the thickness gradually thinning. The foregoing discussion suggests that the cleavage angle ($\bar{\theta }$) becomes increasingly large, and ultimately, a listric trajectory with a steeply dipping shallow depth and a gently dipping deeper end can be formed.

It remains for these results to be applied to shear faults or inverse faults. Even if compressive stress is applied, shear fracture can still occur. In that case, the fracture and the propagation of the micro-crack occur simultaneously. The compressive stress in the end of the micro-crack is adjusted to the tensional stress, which makes the micro-crack develop similarly as a mode I crack. As shown in Fig. 22, a force (P ₂) parallel to the fault plane is applied to the end of a shear fault. Consequently, bending deformation near the crack end will occur based on the buckling of struts theory. Ultimately, the stress state of Fig. 22 can be adjusted to the state shown in Fig. 21.

Appendix B: Maximum Effective Moment (MEM) Criterion

As shown in Fig. 23, a unit square from a potential cleavage zone, the foliation of which has not yet rotated, is cut to analyze the stress state. The normal stress $\sigma_{\alpha }$ and the shear stress $\tau_{\alpha }$ acting on cleavage boundaries are given, respectively, by

$$\sigma_{\alpha } = \frac{1}{2}\left( {\sigma_{1} + \sigma_{3} } \right) + \frac{1}{2}\left( {\sigma_{1} - \sigma_{3} } \right)\cos 2\alpha$$

(20)

$$\tau_{\alpha } = \frac{1}{2}\left( {\sigma_{1} - \sigma_{3} } \right)\sin 2\eta = - \frac{1}{2}\left( {\sigma_{1} - \sigma_{3} } \right)\sin 2\alpha$$

(21)

where $\alpha$ is the angle between $\sigma_{1}$ and the potential deformation zone.

To create a deformation band, there must be an effective moment $M_{\text{eff}}$ that drives the foliation in the cleavage zone to rotate away from its initial orientation. The effective moment ($M_{\text{eff}}$) driving the foliation to rotate is given by the force (F) acting tangentially on the boundaries of the potential cleavage and its arm H: $M_{\text{eff}} = FH$. Because F = stress × area, the stress is equivalent to the force if we substitute the stress for the force acting on the side of a unit area. The force is an effective shear stress $\tau_{\text{eff}}$. Thus, (Zheng et al. 2004)

$$M_{\text{eff}} = \tau_{\text{eff}} H = \tau_{\alpha } L\sin \alpha$$

(22)

where L is the side of the unit square or H _max in the $\sigma_{1}$ direction (Fig. 23a). Substituting Eq. (21) into Eq. (22) thus yields

$$M_{\text{eff}} = - \frac{1}{2}\left( {\sigma_{1} - \sigma_{3} } \right)L\sin 2\alpha \sin \alpha$$

(23)

Recently, Tong (2012) developed the theoretical MEM criterion (Fig. 23b), and Eq. (23) was revised to

$$M_{{G - {\text{eff}}}} = \frac{1}{2}\left( {\sigma_{1} - \sigma_{3} } \right)L\sin 2\alpha \sin \left( {\alpha - \beta } \right)$$

(24)

Appendix C: Stress Field and Distribution Derived by the Mechanical Model of a Gravitational Gliding Layer and the Parabolic Mohr Failure Criterion

Based on the mechanical model of a gravitational gliding layer (see Fig. 24) (Hubbert and Rubey 1959; Chen 1986), the analytical expression can be expressed by the effective stress as follows:

$$\left\{ \begin{gathered} \bar{\tau }_{zx} = \gamma z\sin \theta \hfill \\ \bar{\sigma }_{z} = \gamma z\cos \theta (1 - \lambda ) \hfill \\ \end{gathered} \right.$$

(25)

where $\gamma$ is the stress gradient of fluid-filled rocks; $\lambda$ is the ratio of the pore fluid pressure to the geostatic stress.

Herein, the parabolic Mohr failure criterion (Cai et al. 2002) is employed (see Fig. 25), which can be expressed as follows:

$$\tau^{2} = n(\sigma + \sigma_{t} )$$

(26)

where $\sigma_{t}$ is the uniaxial tensile strength; $n = c^{2} /\sigma_{t}$ is the undetermined coefficient; and $c$ is the cohesion.

An expression for the parabolic Mohr envelope using the stress components as axes can be derived:

$$(\bar{\sigma }_{x} - \bar{\sigma }_{z} )^{2} = 2n(\bar{\sigma }_{x} + \bar{\sigma }_{z} ) + 4n\sigma_{t} - n^{2} - 4\bar{\tau }_{zx}^{2}$$

(27)

By substituting Eq. (25) into Eq. (27), the ultimate stresses ($\tilde{\sigma }_{z}$, $\tilde{\tau }_{zx}$, and $\tilde{\sigma }_{x}$) can be obtained as follows:

$$\left\{ \begin{gathered} \tilde{\sigma }_{z} = \gamma z\cos \theta (1 - \lambda ) \hfill \\ \tilde{\tau }_{zx} = \gamma z\sin \theta \hfill \\ \tilde{\sigma }_{x} = (\tilde{\sigma }_{z} + n) \pm 2\sqrt {n(\sigma_{t} + \tilde{\sigma }_{z} ) - \tilde{\tau }_{zx}^{2} } \hfill \\ \end{gathered} \right.$$

(28)

Equation (28) shows that the sliding system can reach the state of ultimate stress only if $n(\sigma_{t} + \tilde{\sigma }_{z} ) - \tilde{\tau }_{zx}^{2} \ge 0$. When $n(\sigma_{t} + \tilde{\sigma }_{z} ) - \tilde{\tau }_{zx}^{2} = 0$, the mass of the sliding system reaches limit equilibrium at $z = z_{h}$. Substituting $n(\sigma_{t} + \tilde{\sigma }_{z} ) - \tilde{\tau }_{zx}^{2} = 0$ and $z = z_{h}$ into Eq. (28) yields the following:

$$\left\{ \begin{gathered} \tilde{\tau }_{zx} = \gamma z_{h} \sin \theta \hfill \\ \tilde{\sigma }_{z} = \gamma z_{h} \cos \theta (1 - \lambda ) \hfill \\ z_{h} = \frac{{n\gamma \cos \theta (1 - \lambda ) + \sqrt {n^{2} \gamma^{2} \cos^{2} \theta (1 - \lambda )^{2} + 4\gamma^{2} \sin^{2} \theta \sigma_{t} n} }}{{2\gamma^{2} \sin^{2} \theta }} \hfill \\ \cos \theta = \frac{{ - n(1 - \lambda ) + \sqrt {n^{2} (1 - \lambda )^{2} + 4(\gamma^{2} z_{h}^{2} - \sigma_{t} n} )}}{{2\gamma z_{h} }} \hfill \\ \end{gathered} \right.$$

(29)

There are two solutions for $\tilde{\sigma }_{x}$ ($\tilde{\sigma }_{x}^{ + }$ and $\tilde{\sigma }_{x}^{ - }$) in Eq. (28); thus, there are two states of ultimate stress: $\tilde{\sigma }_{x}^{ + } = (\tilde{\sigma }_{z} + n) + 2\sqrt {n(\sigma_{t} + \tilde{\sigma }_{z} ) - \tilde{\tau }_{zx}^{2} }$, $\tilde{\sigma }_{x}^{ - } = (\tilde{\sigma }_{z} + n) - 2\sqrt {n(\sigma_{t} + \tilde{\sigma }_{z} ) - \tilde{\tau }_{zx}^{2} }$.