Experimental and Theoretical Study on Crack Growth Using Rock-Like Resin Samples Containing Inherent Fissures and its Numerical Assessment

Abstract

The initiation and propagation of microcracks along preexisting fissures is one of the typical failure modes for inherently fractured rocks in geotechnical engineering. To better capture the growth of such microcracks, an improved transparent rock-like resin sample with preexisting fissures is first proposed and applied in three-dimensional uniaxial compression test to vividly reproduce the dynamic growth of microcracks due to fissures reactivation. An analytical mathematical solution for calculating the growth length of wing-shaped cracks is derived with the help of the composite fracture strength criterion. The theoretical values agree well with the experimental results. In addition, further investigations on the microcracking mechanism as well as its influence factors induced by the combined interaction of double parallel fissures are carried out under the framework of particle flow simulations. Results suggest that the microcracks induced by fissure reactivation, they begin to appear when the loading stress reaches 50% of the peak stress (σ_p), develop rapidly after surpassing 70% σ_p and expand more slowly beyond 80% σ_p. Especially in the case of double-fissures, some counter-wing and petal-shaped cracks will finally form in the later stages of microcracks propagation, with a length of approximately 1/3–1/2 of the minor axis of the preexisting fissures.

Highlights

• The newly improved rock-like resin specimen has the ability to reproduce the mechanical behaviors of red sandstone sample in terms of deformation and rupture;

• The dynamic growth of wing-shaped cracks induced by fissure reactivation is experimentally studied using rock-like resin sample containing initial preset fissures;

• A mathematical solution for calculating the growth length of wing-shaped cracks under the combined interaction of double parallel fissures is derived theoretically;

• The growth mechanism of wing-shaped cracks as well as their effect factors is further explored numerically by capturing the initiation and propagation of induced microcracks.

Appendix 1

Derivation of the total displacement.

When the initial microcracks begin to slide, the total energy induced by the driven force on the fissure surface can be obtained from Eq. (11) and that is:

$$W = 4\alpha_1 T_s aB\delta \cdot \left\{ {1 - \frac{1}{{6\left( {1 + L} \right)}}} \right\} + 2\alpha_2 \sigma_3 aLB\delta$$

(A-1)

where the first component denotes the work by the shear sliding force on the initial crack plane, and the second component represents the work done as the wing crack overcomes σ₃.

When L < 1, we can further get the following simplified formula as:

$$W = 4\alpha_1 T_s aB\delta + 2\alpha_2 \sigma_3 aLB\delta$$

(A-2)

The displacement δ in Eq. (A-2) can be decomposed into two components in tangential and normal directions (δ_s, δ_n), and that is:

$$\left\{ {\begin{array}{*{20}c} {\delta_n = \delta \sin \theta = \frac{\delta L\sin \psi }{{\left( {1 + L^2 + 2L\cos \psi } \right)^\frac{1}{2} }}} \\ {\delta_s = \delta \cos \theta = \frac{\delta (1 + L\cos \psi )}{{\left( {1 + L^2 + 2L\cos \psi } \right)^\frac{1}{2} }}} \\ \end{array} } \right.$$

(A-3)

According to the reference of (Ashby et al. 1986), the opening displacement δn at the center of a straight crack is assumed to only affect a stress field in a zone of (l + a) and thus results in elastic energy accumulation. The corresponding elastic energies variation can be expressed by:

$$\left\{ {\begin{array}{*{20}l} {U_n = 2\alpha_3 E_0 B\delta_n^2 } \hfill \\ {U_s = 2\alpha_3 E_0 B\delta_s^2 } \hfill \\ \end{array} } \right.$$

(A-4)

where α₃ is a constant related to experimental material. The total strain energy can be obtained by:

$$U = U_n + U_s = 2\alpha_3 E_0 B\delta^2$$

(A-5)

According to the assumption above and keeping the minimum energy of the sliding displacement, we can get the following formula as:

$$\frac{d}{d\delta }\left\{ {U_0 + U - W} \right\} + 0$$

(A-6)

Then, by combining Eqs. (A-1), (A-5) and (A-6) and neglecting the initial elastic stain energy U₀, the total displacement can be calculated and expressed by:

$$\delta = \frac{a}{E_0 }\left\{ {\frac{\alpha_1 }{{\alpha_3 }}T_s \left( {1 - \frac{1}{{6\left( {1 + L} \right)^2 }}} \right) + \frac{\alpha_2 }{{2\alpha_3 }}\sigma_3 L} \right\}$$

(A-7)