## 4 Numerical Simulations of Frost-Heave Failure of the Slope Rock Mass
### Numerical Model of a Slope with an Ice Crack
Figure 11b shows the distribution of the model's minimum principal stress when a \(-5\)\({}^{\circ}\)C temperature load is applied to the slope surface. Because each element is heterogeneous, the stress around f3 vibrates under the drive of frost-leave force, and the tensile stress of elements 30-40 at the bottom of f3 is concentrated. The model's stress contour also shows that there is a tensile stress zone (shown in white) at the bottom of the ice-containing crack f3, and the element stress response of line C is earlier than those of line A and line B. When the temperature load on the slope surface is maintained at \(-10\)\({}^{\circ}\)C, the frost-leave force begins to damage the rock mass around the crack f3, as illustrated in Figure 11c. At the bottom of the crack, some failure elements develop, and the crack progressively expands to the slope's surface. The element tensile stress increases near the slope surface at the bottom of crack f3. Figure 11d shows that when the temperature drops, the oscillation amplitude of the minimum principal stress on both sides of f3 increases, and tensile stress zones develop in many locations. The frost-leave crack at the bottom continues to develop and gradually develops downward. It has a connecting trend with the weak interlayer f2, and the failure surface is initially formed. When the temperature drops further, the bottom of crack f3 extends downward, joins with f2 to form a linked crack, and then gradually expands along the weak interlayer f2, as shown in Figure 11e. Figure 11: _Cont._  
To summarize, the water stored in the cracks or weak interlayers of the rock slope undergoes a water-ice phase change under negative temperature conditions. The frost-heave force is generated at this point when the volume of the ice expands but is confined by the surrounding rock mass. The rock mass is microcracked when the frost-heave force exceeds the crack propagation threshold value. The nearby weak interlayer and fracture  
Figure 11: Evolution diagrams of the minimum principal stress \(\sigma_{3}\) for the slope model: (**a**) 1 h; (**b**) 5 h; (**c**) 10 h; (**d**) 15 h; (**e**) 20 h.  
zone connect to each other when frost-heavy damage accumulates, severely reducing the slope's stability and resulting in geological disasters such as landslides and collapses. Influence of Fissure Water Freezing Rate on Damage and Deterioration of Slope Rock Mass under Negative Temperature  
The freezing rate of fissure water is a main factor affecting the frost-heave failure of fractured rock mass. Four slope models with various fissure water freezing rates have been constructed, which are 25%, 50%, 75%, and 100%, respectively; the equivalent thermal expansion coefficients are, respectively, \(-37.5\), \(-75.0\), \(-112.5\), and \(-150.0\). Figure 12 shows the distribution of the model's minimum principal stress after 20 h of frost-heavy expansion. The tensile stress zone is initiated at the bottom of the crack, accompanied by a small number of failure elements when the freezing rate of the fissure water is 25%. There is no further crack growth due to the small frost-heave force. The final crack propagation mode of the rock mass with 50%, 75%, and 100% fissure water freezing rates is essentially the same. However, as the freezing rate increases, the crack propagation length increases, and the model's maximum tensile stress decreases. In order to study the evolution process of the tensile stress at the bottom of the fracture, the minimum principal stress \(\sigma_{3}\) of the calculation element at the bottom of the crack f3 under different water freezing rates is extracted, as shown in Figure 13. The time it takes to  
Figure 12: Minimum principal stress distributions under different freezing rates: (**a**) freezing rate = 25%, \(\sigma_{3}=-6.18\)–\(4.51\) MPa; (**b**) freezing rate = 50%, \(\sigma_{3}=-8.27\)–\(5.98\) MPa; (**c**) freezing rate = 75%, \(\sigma_{3}=-10.9\)–\(9.06\) MPa; (**d**) freezing rate = 100%, \(\sigma_{3}=-13.6\)–\(10.08\) MPa.