# Point Load Test of Half-Cylinder Core Using the Numerical Model and Laboratory Tests: Size Suggestion and Correlation with Cylinder Core

Authors: Fengyu Ren; Huan Liu; Rongxing He; Guanghui Li; Yang Liu
Journal: Advances in Civil Engineering
 (2018)

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## Abstract

The point load test (PLT) is intended as an index test for rock strength classification or estimations of other strength parameters because it is economical and simple to conduct in the laboratory and in field tests. In the literature, calculation procedures for cylinder cores, blocks, or irregular lumps can be found, but no study has researched such procedures for half-cylinder cores. This paper presents the numerical model and laboratory tests for half-cylinder and cylinder specimens. The results for half-cylinder and cylinder specimens are then presented, analysed, and discussed. A correlation of failure load between half-cylinder and cylinder specimens is established with a suitable size suggestion and correction factor. It is found that the failure load becomes stable when half-cylinder specimens have a length/diameter ratio higher than 0.9. In addition, the results show that the point load strength index (PLSI) of half-cylinder cores can be calculated using the calculation procedures for diametral testing on cylinder cores, and it is necessary to satisfy the conditions that the length/diameter ratio be higher than 0.9 and the failure load be multiplied by 0.8.


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## Body

## 1. Introduction

The point load test (PLT) is one of the primary tests to be applied during earthwork engineering practices, such as slope stabilization works, tunnelling construction, and the design of mining support and foundation [1]. The PLT can be performed either in the field or in the laboratory since the test equipment is portable and specimens can easily be prepared for testing. For these reasons, the PLT is a more practical, time-saving, and economical method compared to the uniaxial compressive strength (UCS) test.The PLT is based on breaking a rock specimen, which is subjected to an increasingly concentrated point load applied through a pair of conical platens [2]. Failure load, rock size, rock shape, and loading mode are used to calculate the point load strength index (PLSI or Is50) [3, 4]. The PLSI is an important strength parameter index for the classification of rock strength [4–7], determination of UCS [1–4, 8–14], and estimation of strength anisotropy [4, 11, 15–17].In general, rock specimens of PLT can be applied to cores (the diametral and axial tests), spheres, blocks, and irregular lumps. The rock specimens come in different shapes. Chau and Wei [18] and Russell and Wood [19] reported on theoretical analyses of spheres subjected to PLT. Several researchers reported the PLT of irregular lumps, but the results were very different for different rock types, and the data of tests were scattered. Wong et al. [1] studied the PLT of volcanic irregular lumps. Yin et al. [14] studied the PLT of granitic irregular lumps. Hobbs [20] studied the PLT of coal measure rocks and Portland limestone irregular lumps. Hiramatsu and Oka [21] investigated the stress state in an irregular test piece subjected to point load. Panek and Fannon [22] studied the size and shape effects in PLT of irregular lumps. Many researchers have studied the PLT of cylinder cores. Khanlari et al. [17] studied the PLT of metamorphic rocks from the cylindrical punch. Chau and Wong [23] studied the PLT (axial tests) of specimens from borehole rock cores. Palchik and Hatzor [24] studied the point load strength of cylindrical porous chalks. Tsiambaos and Sabatakakis [25] presented laboratory testing of cylindrical sedimentary rocks under point loading. Fener et al. [26] carried out the PLT on eleven Nigde rocks. Fener and Ince [27] studied the point load strength of cylindrical specimens on granitic rocks. Consequently, diametral and axial tests of cores are generally preferred for results compared to the PLT of irregular lumps. Thus, the shapes of rock specimens can be different, but calculation procedures of PLSI are different for different shapes.The testing and calculation procedures for PLSI have been standardized by the International Society for Rock Mechanics (ISRM) [3] and have been widely adopted. Failure load, rock size, rock shape, and loading mode affect the point load strength (Is) value; therefore, Is has been proposed to correct for a standard size of 50 mm (Is50). ISRM (1985) [3] suggests Is50 below:(1)Is50=F×Is=PDe2×De500.45,where F is the size correction factor (F=De/500.45), Is is the point load strength (Is=P/De2) in MPa, P is the failure load in kN, and De is the equivalent core diameter in mm. Depending on the specimen shape and loading direction, De varies.When the specimen of the core is diametrally loaded,De is defined as(2)De=D,where D is the distance between two contact points in mm. Under this condition, D is the diameter of the cylinder cores.When the specimen of the core is axially loaded, or the specimen consists of blocks or irregular lumps,De is defined as(3)De2=4WDπ,where W is the smallest specimen width of a plane through the contact points in mm.Although many researchers have studied the relationship between failure load andIs50 for different rock types, rock sizes, rock shapes, and loading modes, no study exists on half-cylinder cores. In this study, the numerical model and laboratory tests were carried out on the cylinder and half-cylinder specimens, and the correlations between the cylinder and the half-cylinder specimens were investigated. The main objective of the research reported in this paper is to determine Is50 of the half-cylinder cores.


## 2. Half-Cylinder Cores

In the design phase of the mining method, it was necessary to take several factors into consideration. The mechanical parameters of the rock were the basis index for evaluating the stability and caveability and determining the management method of stope ground pressure and the structural parameters of stope. Therefore, we obtained the detailed rock mechanics parameters in the mining design phase, which were significant in making the choice in the design of mining methods. However, there was no mining excavation engineering in the mining design phase, and a large number of rock specimens could not be obtained. The Luoboling Copper-Molybdenum mine was used in the design phase, and block caving was determined initially. One major problem for the mine was determining how to get more detailed rock mechanics parameters and their distribution to guide the mining engineering design.A total of 176 boreholes were completed at the exploration stage of the Luoboling Copper-Molybdenum mine. The cylinder core was cut into two pieces (the half-cylinder cores); one piece was analysed to determine the composition and content of the copper-molybdenum, and another piece was kept aside. Therefore, half-cylinder cores were kept (Figure1). The purpose of this study was to determine the PLSI of rock using these half-cylinder cores.Figure 1
Half-cylinder cores.
(a)
(b)First, 18 granodiorite specimens of the half-cylinder core were taken in the same borehole. According to the size requirements of ISRM, the loading direction of PLT was along the radial direction of the half-cylinder core specimens. The specimens after the break are shown in Figure2. Second, according to the calculation procedures of ISRM, we removed two test values of the fracture surfaces which passed through only one loading point, and we deleted the two highest and lowest values from the 16 valid values. Then, the mean value of Is50 was calculated using the 12 remaining values.Figure 2
The half-cylinder cores after breaking under PLT.
(a)
(b)The loading direction of half-cylinder specimens was along the radial direction, and the shape was not a cylinder. Therefore, we adopted two calculation procedures for the half-cylinder cores. Method 1 (M1) was based on the calculation procedure of diametral PLT on cylinder cores, formulae (1) and (2). Method 2 (M2) was based on the calculation procedure of PLT on irregular lumps, formulae (1) and (3). Figure 3 shows results by M1 and M2. According to M1, Is1(50) = 4.526 MPa, and the coefficient of variation (CV) (the standard deviation to the mean) was calculated as CV1 = 0.242. According to M2, Is2(50) = 2.444 MPa, CV2 = 0.442. There was a big difference in Is50 and CV for M1 and M2. However, CV1 was less than CV2, and the measured value of W was not accurate to subjective experience and external factors. Therefore, if the relationship between the cylinder and half-cylinder cores was determined in PLT, the test process could be simplified and test efficiency could be improved.Figure 3
The results ofIs50 using various calculation procedures.


## 3. Numerical Model

Point load tests in the field or laboratory are time-consuming, labourious, and lack reproducibility. In addition, because the rock specimens had multiple fractures and porosities, the variation coefficients of the test results were large. However, the numerical model ensured the consistency of mechanical properties.The numerical simulation was carried out using the distinct element method (DEM) [28]. The dynamical equation of the DEM was established using Newton’s laws of motion. The spherical particle was the basic element in the DEM. Particles interacted at pairwise contacts by means of an internal force and moment. Contact mechanics were embodied in particle-interaction laws that updated the internal forces and moments. The DEM model provided a synthetic material consisting of an assembly of rigid grains that interacted at contacts, and it included both granular and bonded materials. Moreover, it was possible to create bonded materials with half-cylinder and cylinder cores.


### 3.1. Contact Models

In this study, the half-cylinder and cylinder specimens of the numerical model were built through the flat-joint model (FJM) in DEM [29]. The flat-joint model could be applied to hard rock by introducing a polygonal grain structure to provide rotational restraint arising from intergranular interlock [29–31].Each spherical particle element carries a contact force (Fe). The element contact forces produce a statically equivalent force (Fc) at the centre of the interface given by [29](4)Fc=∑∀eFe.The element contact force (Fe) is decomposed into a normal and a shear force (Fne and Fse′) [29]:(5)Fne=∫eσdA=0,unbondedandge≥0,kngeAe,otherwise,Fse′=Fse−ksAeΔδ⌢se′,where σ is the interface normal stress, kn is the normal stiffness, ge is the gap at the element centroid, Ae is the area of each element, ks is the shear stiffness, and Δδ⌢se′ is the effective portion of the relative shear-displacement increment at the element centroid.The element normal and shear stresses (σe and τe′) are given by [29](6)σe=FneAe,τe′=Fse′Ae.The shear strengthτce follows the Coulomb criterion with a tension cutoff [29]:(7)τce=c−σetanϕ,where c is the bond cohesion and ϕ is the friction angle.If the element was bonded and the tensile-strength limit (σc) was exceeded (σe>σc), then the bond broke owing to tension, and a tension fracture was generated. If the shear-strength limit (τce) was exceeded (τe′>τce), then the bond broke in shear, and a shear fracture was generated.


### 3.2. Parameters of the Numerical Model

There is no mining excavation engineering in the Luoboling Copper-Molybdenum mine now, but mine carried out the UCS laboratory test for some rock types in the engineering geological investigation phase. The mechanical properties of different rock types are listed in Table1. As seen from Table 1, the mechanical properties of the same rock types also have a big difference. In this paper, the main purpose was to determine the relationship between cylinder and half-cylinder cores in PLT, and calculate Is50 of the half-cylinder cores. And Is50 was constantly being used as a determination of UCS; therefore, the UCS of rock was a mainly considered factor in PLT. At the same time, in order to reduce the complexity of the DEM micromechanical parameter calibration, we selected three kinds of rock as shown in Table 1 (number 2 porphyritic granodiorite, number 3 granodiorite, and number 10 granodiorite porphyry) as the calibration objects.Table 1
The mechanical properties of different rock types.
Number
Rock type
Sample depth (m)
UCS (MPa)
Young’s elastic modulus (GPa)
Poisson’s ratio
1
Granodiorite
696.6–701.7
67.0
33.7
0.24
2
Porphyritic granodiorite
873.2–885.4
89.6
43.7
0.22
3
Granodiorite
423.7–430.5
55.7
42.9
0.21
4
Granodiorite
729.4–739.4
103.8
41.5
0.21
5
Porphyritic granodiorite
855.5–861.3
93.5
38.8
0.23
6
Granodiorite porphyry
851.0–863.0
99.6
40.7
0.21
7
Granodiorite
471.0–478.0
24.2
41.0
0.20
8
Granodiorite
477.8–483.0
43.4
61.0
0.25
9
Granodiorite porphyry
503.8–507.9
90.8
36.0
0.21
10
Granodiorite porphyry
531.5–534.8
28.4
44.8
0.21
11
Granodiorite porphyry
463.1–474.6
71.6
33.5
0.24The DEM [28] of structures in rock or soil required calibration of the FJM parameters of materials using UCS tests [2, 32–38]. To simulate rocks with different UCS and PLSI, the tensile strength and cohesion of the FJM parameters were varied, while other parameters were kept constant.The properties of the rock samples, such as UCS, Young’s elastic modulus, and Poisson’s ratio, were obtained through UCS tests. Figure4 shows the stress-strain curves with three parameters (p1, p2, and p3) during the UCS tests. Figure 5 shows the load curves with three parameters (p1, p2, and p3) during the PLT, and the diameter of the cylinder specimen is 50 mm. The parameters of FJM with the properties of the rock are calibrated using UCS tests and PLT and are listed in Table 2. The ratios between UCS and Is50 were 21.76 (p1), 23.80 (p2), and 22.47 (p3), which confirmed the suggested values (the ratio between UCS and Is50 varies between 20 and 25) given by ISRM (1985). Comparing Tables 1 and 2, the micromechanical parameters (p1, p2, and p3) in Table 2 correspond to the mechanical properties (number 2 porphyritic granodiorite, number 3 granodiorite, and number 10 granodiorite porphyry) in Table 1.Figure 4
The stress-strain curves.Figure 5
The point load test curves.Table 2
Parameters applied to the numerical model.
Description
p1
p2
p3
Particle-based properties
Particle radius (mm)
2∼3
2∼3
2∼3
Particle density (g/cm3)
2.70
2.70
2.70
FJM parameters
Effective modulus (GPa)
15.00
15.00
15.00
Normal-to-shear stiffness ratio
0.36
0.36
0.36
Friction angle (°)
52.00
52.00
52.00
Friction coefficient
0.50
0.50
0.50
Tensile strength (MPa)
1.00
0.70
0.30
Cohesion (MPa)
20.00
12.00
5.80
Properties
UCS (MPa)
90.54
55.70
28.31
Young’s elastic modulus (GPa)
43.87
43.28
44.30
Poisson’s ratio
0.21
0.21
0.20
Failure load of PLT (kN)
10.40
6.59
3.14
I
s
50 (MPa)
4.16
2.64
1.26


### 3.3. Microfractures under the PLT

The specimen was composed of bonded particles and failed under the point load. A combination of tensile and shear failure occurred in the selected bonded contacts and resulted in the failure of the specimen, and tension fractures or shear fractures were generated in the specimen. For example, the parameter of the numerical model wasp1, and the distance D between two contact points was 30 mm. Figure 6 shows the fracture models of the half-cylinder (the length/diameter ratio is 2) and cylinder specimens (the length/diameter ratio is 1.2) under the point load.Figure 6
The fracture models of (a) half-cylinder specimens and (b) cylinder specimens.
(a)
(b)As seen from Figure6, the shear fractures generate around the two contact points at the beginning of loading. As loading increases, the tension fractures generate beyond a certain distance from the two contact points, and the number of tension fractures increases dramatically. The results are shown in Figure 6; the numbers of tension fractures and failure load of the half-cylinder specimens are higher than those of the cylinder specimen, but the number of shear fractures of the half-cylinder specimen is very close to that of the cylinder specimen. This was because the cross-sectional area of the half-cylinder specimen was larger than that of the cylinder specimen when they had the same D. To investigate the correlations of the failure load between cylinder and half-cylinder specimens, we carried out the PLT of the cylinder and half-cylinder specimens using the numerical model.


### 3.4. PLT of Cylinder Specimen

Diameters of boreholes were mainly 75 mm at the exploration stage of the Luoboling Copper-Molybdenum mine. Therefore, the diameters of the cylinder cores were from 50 mm to 70 mm. The cylinder cores were cut into two pieces (the half-cylinder cores) and so the radii of the half-cylinder core were 25–35 mm. And a large amount of the half-cylinder cores was 25–28 mm in radius that based on field measurement. According to the above analysis, the diameters of the cylinder specimens were determined. The diameters of the cylinder specimens were 18 mm, 22 mm, 26 mm, 30 mm, 34 mm, 38 mm, and 50 mm, and the length/diameter ratios were 1.2, according to the suggestion of ISRM (1985). Figure7 shows the failure load against different diameters on the cylinder specimens with three parameters. According to Figure 7, the failure load increases with the increase in diameters of cylinder specimens. Then, we used formulae (1) and (2) to calculate Is50, and the results are listed in Table 3. As seen from Table 3, the results of Is50 corresponding to different diameters were very close under the same parameters, and the CV of the results were 0.02 (p1), 0.04 (p2), and 0.07 (p3). These results also show that the numerical model of PLT was feasible.Figure 7
Results of the failure load with various diameters on the cylinder specimens.Table 3
I
s
50 of the cylinder specimens.
Diameter (mm)
p1Is50 (MPa)
p2Is50 (MPa)
p3Is50 (MPa)
18
4.16
2.78
1.32
22
4.06
2.64
1.18
26
4.10
2.80
1.43
30
4.32
2.57
1.16
34
4.14
2.49
1.20
38
4.11
2.76
1.34
50
4.16
2.63
1.26
Mean
4.15
2.67
1.27
Standard deviation
0.08
0.11
0.09
CV
0.02
0.04
0.07


### 3.5. PLT of Half-Cylinder Specimens

I
s
50 was affected by different failure loads, rock sizes, rock shapes, and loading modes. For the PLT of the half-cylinder specimens, the loading direction of PLT was only along the radial direction, and the length/diameter ratio and the distance D between two contact points were calculated. D was the radius of the half-cylinder specimens. Figure 8 shows the sizes and numerical model of the half-cylinder.Figure 8
The numerical model of half-cylinder specimens.


#### 3.5.1. Influence of the Length/Diameter Ratio for Failure Load

To obtain the stable value of the failure load for the half-cylinder specimens, the numerical PLT of the half-cylinder specimen was conducted. Figure9 shows the results of the failure load with the length/diameter ratio under the three parameters of the numerical model, when the radii of the half-cylinder specimens were 22 mm, 26 mm, and 30 mm. According to Figure 9, the failure load increases first and then tends to stabilize with the increase in the length/diameter ratio. The values of the failure load become stable when the length/diameter ratio was higher than 0.9.Figure 9
The results of the failure load with the length/diameter ratio.


#### 3.5.2. Influence of the DistanceD between Two Contact Points for Failure Load

To determine the influence of the distanceD between two contact points (the radius of the half-cylinder specimen) for the failure load, the numerical PLT of the half-cylinder specimen was conducted. Figure 10 shows the curves of the distance D with the failure load under the three parameters, when the radii of the half-cylinder specimens are 18 mm, 22 mm, 26 mm, 30 mm, 34 mm, 38 mm, and 50 mm and the length/diameter ratios are 1. According to Figure 10, the failure load increases with the increase of the distance D.Figure 10
Results of the failure load with the distanceD.


## 3.1. Contact Models

In this study, the half-cylinder and cylinder specimens of the numerical model were built through the flat-joint model (FJM) in DEM [29]. The flat-joint model could be applied to hard rock by introducing a polygonal grain structure to provide rotational restraint arising from intergranular interlock [29–31].Each spherical particle element carries a contact force (Fe). The element contact forces produce a statically equivalent force (Fc) at the centre of the interface given by [29](4)Fc=∑∀eFe.The element contact force (Fe) is decomposed into a normal and a shear force (Fne and Fse′) [29]:(5)Fne=∫eσdA=0,unbondedandge≥0,kngeAe,otherwise,Fse′=Fse−ksAeΔδ⌢se′,where σ is the interface normal stress, kn is the normal stiffness, ge is the gap at the element centroid, Ae is the area of each element, ks is the shear stiffness, and Δδ⌢se′ is the effective portion of the relative shear-displacement increment at the element centroid.The element normal and shear stresses (σe and τe′) are given by [29](6)σe=FneAe,τe′=Fse′Ae.The shear strengthτce follows the Coulomb criterion with a tension cutoff [29]:(7)τce=c−σetanϕ,where c is the bond cohesion and ϕ is the friction angle.If the element was bonded and the tensile-strength limit (σc) was exceeded (σe>σc), then the bond broke owing to tension, and a tension fracture was generated. If the shear-strength limit (τce) was exceeded (τe′>τce), then the bond broke in shear, and a shear fracture was generated.


## 3.2. Parameters of the Numerical Model

There is no mining excavation engineering in the Luoboling Copper-Molybdenum mine now, but mine carried out the UCS laboratory test for some rock types in the engineering geological investigation phase. The mechanical properties of different rock types are listed in Table1. As seen from Table 1, the mechanical properties of the same rock types also have a big difference. In this paper, the main purpose was to determine the relationship between cylinder and half-cylinder cores in PLT, and calculate Is50 of the half-cylinder cores. And Is50 was constantly being used as a determination of UCS; therefore, the UCS of rock was a mainly considered factor in PLT. At the same time, in order to reduce the complexity of the DEM micromechanical parameter calibration, we selected three kinds of rock as shown in Table 1 (number 2 porphyritic granodiorite, number 3 granodiorite, and number 10 granodiorite porphyry) as the calibration objects.Table 1
The mechanical properties of different rock types.
Number
Rock type
Sample depth (m)
UCS (MPa)
Young’s elastic modulus (GPa)
Poisson’s ratio
1
Granodiorite
696.6–701.7
67.0
33.7
0.24
2
Porphyritic granodiorite
873.2–885.4
89.6
43.7
0.22
3
Granodiorite
423.7–430.5
55.7
42.9
0.21
4
Granodiorite
729.4–739.4
103.8
41.5
0.21
5
Porphyritic granodiorite
855.5–861.3
93.5
38.8
0.23
6
Granodiorite porphyry
851.0–863.0
99.6
40.7
0.21
7
Granodiorite
471.0–478.0
24.2
41.0
0.20
8
Granodiorite
477.8–483.0
43.4
61.0
0.25
9
Granodiorite porphyry
503.8–507.9
90.8
36.0
0.21
10
Granodiorite porphyry
531.5–534.8
28.4
44.8
0.21
11
Granodiorite porphyry
463.1–474.6
71.6
33.5
0.24The DEM [28] of structures in rock or soil required calibration of the FJM parameters of materials using UCS tests [2, 32–38]. To simulate rocks with different UCS and PLSI, the tensile strength and cohesion of the FJM parameters were varied, while other parameters were kept constant.The properties of the rock samples, such as UCS, Young’s elastic modulus, and Poisson’s ratio, were obtained through UCS tests. Figure4 shows the stress-strain curves with three parameters (p1, p2, and p3) during the UCS tests. Figure 5 shows the load curves with three parameters (p1, p2, and p3) during the PLT, and the diameter of the cylinder specimen is 50 mm. The parameters of FJM with the properties of the rock are calibrated using UCS tests and PLT and are listed in Table 2. The ratios between UCS and Is50 were 21.76 (p1), 23.80 (p2), and 22.47 (p3), which confirmed the suggested values (the ratio between UCS and Is50 varies between 20 and 25) given by ISRM (1985). Comparing Tables 1 and 2, the micromechanical parameters (p1, p2, and p3) in Table 2 correspond to the mechanical properties (number 2 porphyritic granodiorite, number 3 granodiorite, and number 10 granodiorite porphyry) in Table 1.Figure 4
The stress-strain curves.Figure 5
The point load test curves.Table 2
Parameters applied to the numerical model.
Description
p1
p2
p3
Particle-based properties
Particle radius (mm)
2∼3
2∼3
2∼3
Particle density (g/cm3)
2.70
2.70
2.70
FJM parameters
Effective modulus (GPa)
15.00
15.00
15.00
Normal-to-shear stiffness ratio
0.36
0.36
0.36
Friction angle (°)
52.00
52.00
52.00
Friction coefficient
0.50
0.50
0.50
Tensile strength (MPa)
1.00
0.70
0.30
Cohesion (MPa)
20.00
12.00
5.80
Properties
UCS (MPa)
90.54
55.70
28.31
Young’s elastic modulus (GPa)
43.87
43.28
44.30
Poisson’s ratio
0.21
0.21
0.20
Failure load of PLT (kN)
10.40
6.59
3.14
I
s
50 (MPa)
4.16
2.64
1.26


## 3.3. Microfractures under the PLT

The specimen was composed of bonded particles and failed under the point load. A combination of tensile and shear failure occurred in the selected bonded contacts and resulted in the failure of the specimen, and tension fractures or shear fractures were generated in the specimen. For example, the parameter of the numerical model wasp1, and the distance D between two contact points was 30 mm. Figure 6 shows the fracture models of the half-cylinder (the length/diameter ratio is 2) and cylinder specimens (the length/diameter ratio is 1.2) under the point load.Figure 6
The fracture models of (a) half-cylinder specimens and (b) cylinder specimens.
(a)
(b)As seen from Figure6, the shear fractures generate around the two contact points at the beginning of loading. As loading increases, the tension fractures generate beyond a certain distance from the two contact points, and the number of tension fractures increases dramatically. The results are shown in Figure 6; the numbers of tension fractures and failure load of the half-cylinder specimens are higher than those of the cylinder specimen, but the number of shear fractures of the half-cylinder specimen is very close to that of the cylinder specimen. This was because the cross-sectional area of the half-cylinder specimen was larger than that of the cylinder specimen when they had the same D. To investigate the correlations of the failure load between cylinder and half-cylinder specimens, we carried out the PLT of the cylinder and half-cylinder specimens using the numerical model.


## 3.4. PLT of Cylinder Specimen

Diameters of boreholes were mainly 75 mm at the exploration stage of the Luoboling Copper-Molybdenum mine. Therefore, the diameters of the cylinder cores were from 50 mm to 70 mm. The cylinder cores were cut into two pieces (the half-cylinder cores) and so the radii of the half-cylinder core were 25–35 mm. And a large amount of the half-cylinder cores was 25–28 mm in radius that based on field measurement. According to the above analysis, the diameters of the cylinder specimens were determined. The diameters of the cylinder specimens were 18 mm, 22 mm, 26 mm, 30 mm, 34 mm, 38 mm, and 50 mm, and the length/diameter ratios were 1.2, according to the suggestion of ISRM (1985). Figure7 shows the failure load against different diameters on the cylinder specimens with three parameters. According to Figure 7, the failure load increases with the increase in diameters of cylinder specimens. Then, we used formulae (1) and (2) to calculate Is50, and the results are listed in Table 3. As seen from Table 3, the results of Is50 corresponding to different diameters were very close under the same parameters, and the CV of the results were 0.02 (p1), 0.04 (p2), and 0.07 (p3). These results also show that the numerical model of PLT was feasible.Figure 7
Results of the failure load with various diameters on the cylinder specimens.Table 3
I
s
50 of the cylinder specimens.
Diameter (mm)
p1Is50 (MPa)
p2Is50 (MPa)
p3Is50 (MPa)
18
4.16
2.78
1.32
22
4.06
2.64
1.18
26
4.10
2.80
1.43
30
4.32
2.57
1.16
34
4.14
2.49
1.20
38
4.11
2.76
1.34
50
4.16
2.63
1.26
Mean
4.15
2.67
1.27
Standard deviation
0.08
0.11
0.09
CV
0.02
0.04
0.07


## 3.5. PLT of Half-Cylinder Specimens

I
s
50 was affected by different failure loads, rock sizes, rock shapes, and loading modes. For the PLT of the half-cylinder specimens, the loading direction of PLT was only along the radial direction, and the length/diameter ratio and the distance D between two contact points were calculated. D was the radius of the half-cylinder specimens. Figure 8 shows the sizes and numerical model of the half-cylinder.Figure 8
The numerical model of half-cylinder specimens.


### 3.5.1. Influence of the Length/Diameter Ratio for Failure Load

To obtain the stable value of the failure load for the half-cylinder specimens, the numerical PLT of the half-cylinder specimen was conducted. Figure9 shows the results of the failure load with the length/diameter ratio under the three parameters of the numerical model, when the radii of the half-cylinder specimens were 22 mm, 26 mm, and 30 mm. According to Figure 9, the failure load increases first and then tends to stabilize with the increase in the length/diameter ratio. The values of the failure load become stable when the length/diameter ratio was higher than 0.9.Figure 9
The results of the failure load with the length/diameter ratio.


### 3.5.2. Influence of the DistanceD between Two Contact Points for Failure Load

To determine the influence of the distanceD between two contact points (the radius of the half-cylinder specimen) for the failure load, the numerical PLT of the half-cylinder specimen was conducted. Figure 10 shows the curves of the distance D with the failure load under the three parameters, when the radii of the half-cylinder specimens are 18 mm, 22 mm, 26 mm, 30 mm, 34 mm, 38 mm, and 50 mm and the length/diameter ratios are 1. According to Figure 10, the failure load increases with the increase of the distance D.Figure 10
Results of the failure load with the distanceD.


## 3.5.1. Influence of the Length/Diameter Ratio for Failure Load

To obtain the stable value of the failure load for the half-cylinder specimens, the numerical PLT of the half-cylinder specimen was conducted. Figure9 shows the results of the failure load with the length/diameter ratio under the three parameters of the numerical model, when the radii of the half-cylinder specimens were 22 mm, 26 mm, and 30 mm. According to Figure 9, the failure load increases first and then tends to stabilize with the increase in the length/diameter ratio. The values of the failure load become stable when the length/diameter ratio was higher than 0.9.Figure 9
The results of the failure load with the length/diameter ratio.


## 3.5.2. Influence of the DistanceD between Two Contact Points for Failure Load

To determine the influence of the distanceD between two contact points (the radius of the half-cylinder specimen) for the failure load, the numerical PLT of the half-cylinder specimen was conducted. Figure 10 shows the curves of the distance D with the failure load under the three parameters, when the radii of the half-cylinder specimens are 18 mm, 22 mm, 26 mm, 30 mm, 34 mm, 38 mm, and 50 mm and the length/diameter ratios are 1. According to Figure 10, the failure load increases with the increase of the distance D.Figure 10
Results of the failure load with the distanceD.


## 4. Laboratory Tests

To verify the accuracy and reliability of the numerical model, laboratory tests were carried out on granitic rock and sandstone. The selected granitic rock and sandstone had the same lithology and no obvious discontinuities. According to the results of the numerical model, the size parameters of the specimens were determined and are listed in Table4. According to Table 4, the specimens of the cylinder and half-cylinder were taken, and the number of tests per specimen was fixed at 10. Figure 11 shows the specimens of granitic rock and sandstone after failure and the PLT of the laboratory tests.Table 4
Size parameters of the laboratory tests.
Specimen
D (mm)
Length (mm)
Length/diameter ratio
Cylinder
50
60
1.200
40
50
1.250
25
30
1.200
Half-cylinder
50
110
1.100
40
90
1.125
25
70
1.400
25
55
1.100
25
45
0.900
25
35
0.700
25
25
0.500Figure 11
Laboratory point load tests: (a) the half-cylinder specimens, (b) PLT of the half-cylinder specimens, (c) the cylinder specimens, and (d) PLT of the cylinder specimens.The directions of failure surface of the half-cylinder specimens were along the axial direction (as shown in Figure12, where the length/diameter ratio is 0.5), when the length/diameter ratio was relatively small. With the increase of the length/diameter ratio, the directions of failure surface were along different directions (as shown in Figure 12 where the length/diameter ratio is 1.1). Thus, it can be seen that the length played a major role in the failure load when the length/diameter ratio was relatively small.Figure 12
The failure surface of the half-cylinder specimens.We processed the test according to the calculation procedures of ISRM. Figure13 shows the test results of the length/diameter ratio with the failure load. Figure 14 shows the test results of the distance D with the failure load.Figure 13
The results of the failure load with the length/diameter ratio from the laboratory tests.Figure 14
The results of the failure load with the distanceD from the laboratory tests: (a) granitic and (b) sandstone.
(a)
(b)As seen from Figure13, the calculation values of the failure load become stable when the length/diameter ratio is higher than 0.9. According to the results of Figure 14, the failure load increases with the increase of the distance D, and the failure loads of the half-cylinder specimens are larger than the cylinder specimens under the same D and lithology conditions.


## 5. Conclusions and Discussion

In this study, the PLT of half-cylinder specimens was investigated. For this purpose, the numerical model and laboratory tests were conducted on the cylinder and half-cylinder specimens. We determined the size (the length/diameter ratio) of the half-cylinder specimens and the correlation between cylinder and half-cylinder specimens. According to Figures9 and 13, the value of the length/diameter ratio needs to be higher than 0.9 in the PLT of the half-cylinder specimens. The failure load ratios between cylinder and half-cylinder specimens are presented in Figure 15, with the same distance D and lithology conditions. The average of ratios is 0.793 (approximately 0.8), and the CV is 0.048. The results show that Is50 of the half-cylinder cores could be calculated by formulae (1) and (2), and it was necessary to satisfy the conditions that the length/diameter ratio be higher than 0.9 and the failure load value be multiplied by 0.8. According to the above conclusions, Is50 of half-cylinder cores in Figure 2 is recalculated, and we remove two test values that the length/diameter ratio of specimens is less than 0.9. The calculation result of Is50 is 3.698 MPa, and the CV is 0.187. In conclusion, we determined Is50 by PLT of the half-cylinder cores.Figure 15
Ratio of the failure load against distanceD with various forms of numerical models and laboratory tests.The advantage of PLT for half-cylinder cores is that we combinedIs50 with ore body information. The ore body was achieved by the chemical analysis of drilling cores, and we selected some drilling cores that intersected with the ore body. We carried out the PLT for these drilling cores. According to the ISRM suggestions and the findings of the half-cylinder cores, the values of Is50 were calculated. Based on Is50 and the position coordinates of the cores, we built the spatial distribution maps (Figure 16) using the method of inverse distance weight. Figure 16 can be used as an index for the classification of the rock mass strength and a reference for mining engineering design.Figure 16
The spatial distribution map ofIs50.Data from the numerical model or laboratory tests exhibit some scatter. The scatter of data is due to the algorithm of DEM in the numerical model. In the dynamic simulation of DEM, the selected timestep is very small, and the velocity and acceleration of the particles are constant in a timestep. The movement of particles only affects the particles that are directly contacted. The contact forces acting on the particle or wall are determined by the directly contacted particles. In the numerical model of PLT, the load of the specimen is imposed by giving the two load points a constant velocity. The value of the load is calculated from the mean of the two load points in a timestep. However, the particles that are in contact with the two load points are different in a certain timestep during the point loads (the concentrated loads). In other words, the contact states between particles and load points are different. Accordingly, the contact forces at the two points are different (as shown in Figure17). However, the difference between the two contact points is relatively small at the same timestep. The results of the numerical model are still credible. The results of laboratory tests also exhibit some scatter, which is believed to have occurred because of the cracks or anisotropic nature of specimens. The cracks may be naturally present in the rock and may also occur during sample preparation. To decrease the scatter, the number of specimens was fixed at 10 in this test. However, the scatter of test results is still lower than that of other test results, especially during the field testing.Figure 17
The contact forces at the two points of the numerical model.In this study, cracks and joints in specimens are not considered either in the numerical models or in the laboratory tests. The cracks and joints in specimens will affect the failure load and the failure surface under the point load, which should be examined in future studies.



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*Source: 3870583-2018-11-04.xml*