# Micromechanical Behavior of Single-Crystal Superalloy with Different Crystal Orientations by Microindentation

Authors: Jinghui Li; Fuguo Li; Junzhe Dong; Zhanwei Yuan; Shuo Zhang
Journal: Journal of Nanomaterials
 (2015)

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## Abstract

In order to investigate the anisotropic micromechanical properties of single-crystal nickel-based superalloy DD99 of four crystallographic orientations, (001), (215), (405), and (605), microindentation test (MIT) was conducted with different loads and loading velocities by a sharp Berkovich indenter. Some material parameters reflecting the micromechanical behavior of DD99, such as microhardnessH, Young’s modulus E, yield stress σy, strain hardening component n, and tensile strength σb, can be obtained from load-displacement relations. H and E of four different crystal planes evidently decrease with the increase of h. The reduction of H is due to dislocation hardening while E is related to interplanar spacing and crystal variable. σy of (215) is the largest among four crystal planes, followed by (605), and (001) has the lowest value. n of (215) is the lowest, followed by (605), and that of (001) is the largest. Subsequently, a simplified elastic-plastic material model was employed for 3D microindentation simulation of DD99 with various crystal orientations. The simulation results agreed well with experimental, which confirmed the accuracy of the simplified material model.


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## Body

## 1. Introduction

In recent years, single-crystal nickel-based superalloys are widely used as blade of modern gas turbine aeroengines, as they significantly raise the operation temperature and efficiency due to excellent mechanical properties in service [1–6]. Their excellent high-temperature properties are superior to conventional cast alloys, such as high-temperature creep and oxidation-resistant performance, which results from the elimination of grain boundaries in single-crystal alloys. However, the absence of grain boundaries also leads to orientation-dependent material response [7–9]. To date, a lot of researches concerned about different orientation of single-crystal nickel-based superalloys have been done. Caron et al. [10] investigated the anisotropic creep behavior of some advanced superalloys (CMSX-2, Alloy 454, MXON, and CMSX-4) in the temperature ranging from 1033 to 1323 K. He et al. [11] discussed the creep/fatigue damage characteristics of DD6 and the results show the capability of DD6 to avoid fatigue damage in [011] direction is better than that in [001] direction. Yi et al. [12] gave modeling process of tertiary creep of single-crystal superalloy along different orientations. Wu et al. [13] made extensile and compression tests at different temperatures on DD8 with various strain rates and the results indicate that the tendency to the plastic deformation inhomogeneity decreases in the order of [011], [001], and [111]. The previous investigations mainly concentrated on the fatigue life and creep properties with different orientations, while other fundamental performances of materials, such as hardness, Young’s modulus, and yield stress are rarely focused on.Conventional tensile tests are difficult to conduct at nano-and microscales to determine orientation dependent behavior. Microindentation can be an alternative approach to tension or compression to probe the micromechanical properties, such as elastic modulus and hardness. Due to its precise measurement and advantages of celerity, accuracy, and nondestructiveness [15], it will become increasingly popular in the future. However, thorough understanding of materials behavior under indenter cannot be achieved just through MIT; for example, the equivalent stress and strain cannot be obtained directly by indentation.FEM may be used as a supplement to solve complex 3D problems and more information can be extracted from microindentation simulation [16–18]. Lim and Munawar Chaudhri [19] investigated microindentation hardness of individual grains [(110) and (111) surfaces] of a polycrystalline copper using a spherical indenter and reported the indentation hardness in copper with different orientations to be very similar. Liu et al. [20] performed nanoindentation simulation using 3D elastic-plastic crystal plasticity FEM on single-crystal copper specimens on three orientations [(011), (100), and (111)] using a conical indenter and reported twofold, fourfold, and sixfold symmetries on (011), (100), and (111) faces, respectively. Fivel et al. [21] developed a 3D model to combine discrete dislocations with FEM for nanoindentation simulation on single-crystal copper. However, most of these researches were based on material properties of single-crystal data obtained from tensile testing at macroscale. In fact, it is essential to extract these parameters at micrometer scale.In this paper, MIT were conducted on crystal planes of nickel-based single-crystal DD99 to investigate its micromechanical properties. Subsequently, a FEM model based on the results of MIT was implemented in 3D ABAQUS/Explicit to prove the accuracy of this FEM model.


## 2. Calculation Methods

### 2.1. Elastic-Plastic Properties

The load-displacement response obtained by microindentation contains information about the elastic and plastic deformation of the indented materials. Therefore, it is often regarded as “fingerprint” of materials’ properties under identification [22]. Mechanical properties, such as the hardness and Young’s modulus, can be readily extracted from the load-displacement curves. In general, plastic behavior of metals can be characterized by a power-law description, as shown in Figure 1. A simplified elastic-plastic, true stress-strain behavior can be expressed as
(1)σ=Eε,forσ≤σyRεn,forσ≥σy.Figure 1
Power-law elastic-plastic, true stress-strain behavior [14].Whenσ is equal to σy, R can be deduced as follows:
(2)R=σyEσyn.Thus,1 can be rewritten as follows:
(3)σ=Eε,forσ≤σyσy1+Eσyεrn,forσ≥σy.In order to describe the mechanical properties of a power-law material, Young’s modulusE, yield stress σy, and strain hardening exponent n are needed.The typicalP-h curve of microindentation test is shown in Figure 2. Two important parameters (maximum indentation depth hm and maximum load Pm) can be obtained from the curve. According to Kick’s law, the loading curve of indentation can be expressed as
(4)P=Ch2.Figure 2
The typicalP-h curve in instrumented indentation test.According to the model proposed by Oliver and Pharr [23, 24], unloading curve can be interpreted as
(5)P=B×h-hft.The relationship between the apparent modulusE* and Young’s modulus E is as follows:
(6)1E*=1-ν2E+1-νi2Ei.Ei and νi are Young’s modulus and Poisson’s ratio of the indenter whose values are 1141 GPa and 0.07 for diamond indenter, respectively [25].According to King [26], E* has the form
(7)E*=π2βSQ.S and Q can be obtained by the following equations, respectively:
(8)S=dPdhh=hm=m×Bhm-hfm-1,(9)Q=πhm-γPmS2tan2θ=Khm-γPmS2,
where K is a constant related with indenter, whose value is 24.56 for Berkovich indenter, andγ is a coefficient with value of 0.75 for Berkovich indenter [27, 28].The next task is to find the yield stressσy and strain hardening exponent n. The method used to obtain σy and n is dimensional analysis proposed by Y. T. Cheng and C. M. Cheng and Tunvisut et al. [29–33]. For a sharp indenter (Berkovich indenter in this paper), the load P can be related with the following parameters [34]:
(10)P=Ph,E*,σc,n.According toΠ theorem in dimensional analysis, the equation above can be rewritten as
(11)P=σch2Π1E*σc,n.Π1 function is independent of n when strain is equal to 0.033 [34]. And σ0.033 can be obtained as
(12)Π1E*σ0.033=Cσ0.033=-1.131ln⁡E*σ0.0333+13.635ln⁡E*σ0.0332-30.594ln⁡E*σ0.033+29.267.Similarly, the unloading slope can be described as follows ath=hm:
(13)dPudhh=hm=S=E*hmΠ2E*σc,n,(14)Π2E*σc,n=SE*hm=-1.40557n3+0.77526n2+0.1583n-0.06831·ln⁡E*σ0.0333+17.93006n3-9.22091n2-2.37733n+0.86295·ln⁡E*σ0.0332+-79.99715n3+40.5562n2+9.00157n-2.54543·ln⁡E*σ0.033+122.65069n3-63.88418n2-9.58936n+6.20045.Finally, after plugging value ofσ0.033 and n into 3, σy can be calculated. Thus, one can obtain the elastic-plastic model characterized by a power-law function based on 1 and 2.


### 2.2. Tensile Strength

The ultimate tensile strength characterizes the resistance of largest uniform plastic deformation. The tensile strength is usually determined by uniaxial tensile test: the highest point of the stress-strain curve is the tensile strength. However, it can also be calculated by MIT.In the stage of uniform plastic deformation stage, loadF is as follows:
(15)F=σA.The result of differential calculation of15 is
(16)dFF=dσ-σ-+dAA.When nonuniform deformation such as necking occurs on certain part of materials,F reaches its maximum value and dF=0. According to 16, it can be induced that
(17)dσ-σ-=-dAA=dε,(18)dσ-dε=σ-.It has the following relationship, according to18:
(19)dσdεε=εb=σ-ε=εb.According to19, εb can be calculated and it is equal to n, which has been confirmed by [35]. Consider
(20)F=σbA=σ-bAb=RεbnAb=RnnAb,(21)σb=RnnAbA.Its formula is(22)ln⁡AAb=εb.According to21 and 22, σb can be calculated:
(23)σb=Rnne-εb=Rnne-n=Rnen.


### 2.3. Microhardness

Hardness is an ability of resistance to permanent (plastic) deformation. It represents the overall mechanical properties of materials. For example, hardness is related to other mechanics parameters, such as Young’s modulus, yield stress, and strain hardening component [36]. Besides, many references have shown that the intrinsic material length scale characterizing size scale can be identified from microhardness [37–41]. It has been found that the microhardness of materials is significantly higher that the microhardness and, furthermore, the correlation of microhardness and the indentation depth implies that the materials strength depends on both the absolute specimen size and strain gradients. Therefore, the results of the microhardness provide a new implication for the strain gradient-dependent constitutive equations in continuum plasticity theory. This has led to the development of phenomenological or mechanism-based strain gradient plasticity (SGP) theories, which have been used to interpret the size dependence of hardness from the micro- to nanoscales.Shim et al. [42] guessed the yield strength very roughly based on the hardness values from Berkovich indentations. Nix and Gao [43] showed that equivalent stress (σeq) has the following relationship with hardness: σeq=H/3. As a testing method under micro- and nanoscale, MIT plays an important role in the evaluation of microhardness and the microhardness can be obtained as follows:
(24)H=PmQ=Pmπhm-γPm/S2tan2θ.


## 2.1. Elastic-Plastic Properties

The load-displacement response obtained by microindentation contains information about the elastic and plastic deformation of the indented materials. Therefore, it is often regarded as “fingerprint” of materials’ properties under identification [22]. Mechanical properties, such as the hardness and Young’s modulus, can be readily extracted from the load-displacement curves. In general, plastic behavior of metals can be characterized by a power-law description, as shown in Figure 1. A simplified elastic-plastic, true stress-strain behavior can be expressed as
(1)σ=Eε,forσ≤σyRεn,forσ≥σy.Figure 1
Power-law elastic-plastic, true stress-strain behavior [14].Whenσ is equal to σy, R can be deduced as follows:
(2)R=σyEσyn.Thus,1 can be rewritten as follows:
(3)σ=Eε,forσ≤σyσy1+Eσyεrn,forσ≥σy.In order to describe the mechanical properties of a power-law material, Young’s modulusE, yield stress σy, and strain hardening exponent n are needed.The typicalP-h curve of microindentation test is shown in Figure 2. Two important parameters (maximum indentation depth hm and maximum load Pm) can be obtained from the curve. According to Kick’s law, the loading curve of indentation can be expressed as
(4)P=Ch2.Figure 2
The typicalP-h curve in instrumented indentation test.According to the model proposed by Oliver and Pharr [23, 24], unloading curve can be interpreted as
(5)P=B×h-hft.The relationship between the apparent modulusE* and Young’s modulus E is as follows:
(6)1E*=1-ν2E+1-νi2Ei.Ei and νi are Young’s modulus and Poisson’s ratio of the indenter whose values are 1141 GPa and 0.07 for diamond indenter, respectively [25].According to King [26], E* has the form
(7)E*=π2βSQ.S and Q can be obtained by the following equations, respectively:
(8)S=dPdhh=hm=m×Bhm-hfm-1,(9)Q=πhm-γPmS2tan2θ=Khm-γPmS2,
where K is a constant related with indenter, whose value is 24.56 for Berkovich indenter, andγ is a coefficient with value of 0.75 for Berkovich indenter [27, 28].The next task is to find the yield stressσy and strain hardening exponent n. The method used to obtain σy and n is dimensional analysis proposed by Y. T. Cheng and C. M. Cheng and Tunvisut et al. [29–33]. For a sharp indenter (Berkovich indenter in this paper), the load P can be related with the following parameters [34]:
(10)P=Ph,E*,σc,n.According toΠ theorem in dimensional analysis, the equation above can be rewritten as
(11)P=σch2Π1E*σc,n.Π1 function is independent of n when strain is equal to 0.033 [34]. And σ0.033 can be obtained as
(12)Π1E*σ0.033=Cσ0.033=-1.131ln⁡E*σ0.0333+13.635ln⁡E*σ0.0332-30.594ln⁡E*σ0.033+29.267.Similarly, the unloading slope can be described as follows ath=hm:
(13)dPudhh=hm=S=E*hmΠ2E*σc,n,(14)Π2E*σc,n=SE*hm=-1.40557n3+0.77526n2+0.1583n-0.06831·ln⁡E*σ0.0333+17.93006n3-9.22091n2-2.37733n+0.86295·ln⁡E*σ0.0332+-79.99715n3+40.5562n2+9.00157n-2.54543·ln⁡E*σ0.033+122.65069n3-63.88418n2-9.58936n+6.20045.Finally, after plugging value ofσ0.033 and n into 3, σy can be calculated. Thus, one can obtain the elastic-plastic model characterized by a power-law function based on 1 and 2.


## 2.2. Tensile Strength

The ultimate tensile strength characterizes the resistance of largest uniform plastic deformation. The tensile strength is usually determined by uniaxial tensile test: the highest point of the stress-strain curve is the tensile strength. However, it can also be calculated by MIT.In the stage of uniform plastic deformation stage, loadF is as follows:
(15)F=σA.The result of differential calculation of15 is
(16)dFF=dσ-σ-+dAA.When nonuniform deformation such as necking occurs on certain part of materials,F reaches its maximum value and dF=0. According to 16, it can be induced that
(17)dσ-σ-=-dAA=dε,(18)dσ-dε=σ-.It has the following relationship, according to18:
(19)dσdεε=εb=σ-ε=εb.According to19, εb can be calculated and it is equal to n, which has been confirmed by [35]. Consider
(20)F=σbA=σ-bAb=RεbnAb=RnnAb,(21)σb=RnnAbA.Its formula is(22)ln⁡AAb=εb.According to21 and 22, σb can be calculated:
(23)σb=Rnne-εb=Rnne-n=Rnen.


## 2.3. Microhardness

Hardness is an ability of resistance to permanent (plastic) deformation. It represents the overall mechanical properties of materials. For example, hardness is related to other mechanics parameters, such as Young’s modulus, yield stress, and strain hardening component [36]. Besides, many references have shown that the intrinsic material length scale characterizing size scale can be identified from microhardness [37–41]. It has been found that the microhardness of materials is significantly higher that the microhardness and, furthermore, the correlation of microhardness and the indentation depth implies that the materials strength depends on both the absolute specimen size and strain gradients. Therefore, the results of the microhardness provide a new implication for the strain gradient-dependent constitutive equations in continuum plasticity theory. This has led to the development of phenomenological or mechanism-based strain gradient plasticity (SGP) theories, which have been used to interpret the size dependence of hardness from the micro- to nanoscales.Shim et al. [42] guessed the yield strength very roughly based on the hardness values from Berkovich indentations. Nix and Gao [43] showed that equivalent stress (σeq) has the following relationship with hardness: σeq=H/3. As a testing method under micro- and nanoscale, MIT plays an important role in the evaluation of microhardness and the microhardness can be obtained as follows:
(24)H=PmQ=Pmπhm-γPm/S2tan2θ.


## 3. Experimental and Simulation Procedures

### 3.1. Experimental Procedure

The nominal chemical compositions of as-received single-crystal nickel-based superalloy DD99 (5 mm × 5 mm × 3 mm rectangle block) which has its crystal plane (001) marked in advance are shown in Table1.Table 1
Chemical composition of DD99 superalloy (mass fraction, %).
C
Cr
Co
W
Al
Ti
Ta
Ni
0.016
8.5
5.0
9.5
5.5
2.2
2.8
Bal.Although nickel-based single crystal has the optimum performance along crystal orientation [001] [14], there inevitably exist misorientations for actual engineering application. Moreover, as the structures of production, such as blade parts of aeroengine, are complicated, it is difficult to guarantee each plane of the product is parallel to [001]. In order to investigate the orientation dependent properties and provide performance reference of different planes for engineering applications in the future, the specimens of DD99 were subjected to wire-electrode cutting with angle of 30°, 45°, and 60° to crystal plane (001), as shown in Figures 3(a), 3(b), and 3(c). The red lines in Figure 3 represent molybdenum wire used in wire-electrode cutting and the cutting curve is parallel to [010].Figure 3
Schematic diagram of specimens with different cutting angles: (a) 30°; (b) 45°; (c) 60°.
(a)
(b)
(c)The cutting specimens along with the original one with crystal plane (001) were carefully ground with sand paper. Then, they were polished with 1.5μm diamond to mirror finish. Subsequently, the specimens were etched with corrosives (5 g CuSO4 + 20 mL HCl + 100 mL H2O) for 15 s to reduce the influence of surface hardening. Finally, MIT was performed using the commercial MCT W501 equipped with a Berkovich diamond indenter at room temperature. The test condition was shown in Table 2. There are four different loading velocities and under each of them nine different maximum loads were conducted. Notably, each test was conducted five times and the average values were calculated in order to eliminate the errors.Table 2
MIT condition.
Max load (mN)
500
1000
1500
2000
2500
3000
3500
4000
4500
Velocity (mN/s)
17.2842
34.5684
51.8527
103.7053


### 3.2. Simulation Procedure

In the researches of Li et al. [44] and Yuan et al. [45], the axisymmetric 2D was created by using the quadrilateral elements to simulate the indentation process and the P-h line of FEM is slightly higher than that of experiments. Walter and Mitterer [46] found that 2D model predicts higher scatter and relatively higher mean Young’s modulus compared to 3D model. According to researches from FEM simulation of indentation, it can be concluded that the main differences between 2D and 3D model are in the following two aspects.In terms of the shape of indenter, the Berkovich indenter of 2D model is represented by a straight with an angle of 70.3° to axis of symmetry, indicating the indenter itself as a whole is conical. In contrast, the real shape of Berkovich indenter is triangular pyramid. Considering the boundary conditions, circumferential displacement in 2D model is constrained and only the displacement in radial direction is permitted. By comparison, constrained circumferential displacement just appears on the symmetry plane and deformation can be expanded in both circumferential and radial direction within matrix in 3D model. Considering above variations and the fact that 2D model cannot be utilized to discuss the orientation dependent properties due to its rotational symmetry, a 3D simulation seems preferable, even though computational time is considerably higher.The Berkovich indenter is a triangular-based pyramid having a threefold symmetry. The load is applied along the axis of the indenter; thus the load symmetry is the same as the geometric one. For these considerations, a three-dimensional model is defined only by one-third of the entire system. The 3D model setup is shown in Figure4. In the simulations, the indenter is modeled as a rigid body. This is justified as the diamond indenter has a modulus of 1141 GPa. In this paper, the dimension of material in FEM model is 2 mm × 2 mm × 1 mm, which is much larger than maximum indentation displacement. As for the mesh section, size of mesh is generally a compromise between the computational cost and the solution accuracy. Based on various deformation levels, a finer mesh is used near the indenter tip and a coarser mesh for farther regions. And the element of material is C3D8R, while that of indenter is R3D3. The effect of friction coefficient on the nanoindentation behavior has been investigated by Liu et al. [47], illustrating that the friction does not change the load-displacement relationship. A lower friction contact pair is defined by two contact surfaces with associated nodes between the indenter and the material. In addition, boundary conditions are defined as an element on two symmetry planes parallel to indenter movement direction constrained, only able to expand along radical and axial direction with the help of cylindrical coordinate. But the elements on the bottom of material were defined as having no displacement in any direction.Figure 4
Microindentation model setup.


## 3.1. Experimental Procedure

The nominal chemical compositions of as-received single-crystal nickel-based superalloy DD99 (5 mm × 5 mm × 3 mm rectangle block) which has its crystal plane (001) marked in advance are shown in Table1.Table 1
Chemical composition of DD99 superalloy (mass fraction, %).
C
Cr
Co
W
Al
Ti
Ta
Ni
0.016
8.5
5.0
9.5
5.5
2.2
2.8
Bal.Although nickel-based single crystal has the optimum performance along crystal orientation [001] [14], there inevitably exist misorientations for actual engineering application. Moreover, as the structures of production, such as blade parts of aeroengine, are complicated, it is difficult to guarantee each plane of the product is parallel to [001]. In order to investigate the orientation dependent properties and provide performance reference of different planes for engineering applications in the future, the specimens of DD99 were subjected to wire-electrode cutting with angle of 30°, 45°, and 60° to crystal plane (001), as shown in Figures 3(a), 3(b), and 3(c). The red lines in Figure 3 represent molybdenum wire used in wire-electrode cutting and the cutting curve is parallel to [010].Figure 3
Schematic diagram of specimens with different cutting angles: (a) 30°; (b) 45°; (c) 60°.
(a)
(b)
(c)The cutting specimens along with the original one with crystal plane (001) were carefully ground with sand paper. Then, they were polished with 1.5μm diamond to mirror finish. Subsequently, the specimens were etched with corrosives (5 g CuSO4 + 20 mL HCl + 100 mL H2O) for 15 s to reduce the influence of surface hardening. Finally, MIT was performed using the commercial MCT W501 equipped with a Berkovich diamond indenter at room temperature. The test condition was shown in Table 2. There are four different loading velocities and under each of them nine different maximum loads were conducted. Notably, each test was conducted five times and the average values were calculated in order to eliminate the errors.Table 2
MIT condition.
Max load (mN)
500
1000
1500
2000
2500
3000
3500
4000
4500
Velocity (mN/s)
17.2842
34.5684
51.8527
103.7053


## 3.2. Simulation Procedure

In the researches of Li et al. [44] and Yuan et al. [45], the axisymmetric 2D was created by using the quadrilateral elements to simulate the indentation process and the P-h line of FEM is slightly higher than that of experiments. Walter and Mitterer [46] found that 2D model predicts higher scatter and relatively higher mean Young’s modulus compared to 3D model. According to researches from FEM simulation of indentation, it can be concluded that the main differences between 2D and 3D model are in the following two aspects.In terms of the shape of indenter, the Berkovich indenter of 2D model is represented by a straight with an angle of 70.3° to axis of symmetry, indicating the indenter itself as a whole is conical. In contrast, the real shape of Berkovich indenter is triangular pyramid. Considering the boundary conditions, circumferential displacement in 2D model is constrained and only the displacement in radial direction is permitted. By comparison, constrained circumferential displacement just appears on the symmetry plane and deformation can be expanded in both circumferential and radial direction within matrix in 3D model. Considering above variations and the fact that 2D model cannot be utilized to discuss the orientation dependent properties due to its rotational symmetry, a 3D simulation seems preferable, even though computational time is considerably higher.The Berkovich indenter is a triangular-based pyramid having a threefold symmetry. The load is applied along the axis of the indenter; thus the load symmetry is the same as the geometric one. For these considerations, a three-dimensional model is defined only by one-third of the entire system. The 3D model setup is shown in Figure4. In the simulations, the indenter is modeled as a rigid body. This is justified as the diamond indenter has a modulus of 1141 GPa. In this paper, the dimension of material in FEM model is 2 mm × 2 mm × 1 mm, which is much larger than maximum indentation displacement. As for the mesh section, size of mesh is generally a compromise between the computational cost and the solution accuracy. Based on various deformation levels, a finer mesh is used near the indenter tip and a coarser mesh for farther regions. And the element of material is C3D8R, while that of indenter is R3D3. The effect of friction coefficient on the nanoindentation behavior has been investigated by Liu et al. [47], illustrating that the friction does not change the load-displacement relationship. A lower friction contact pair is defined by two contact surfaces with associated nodes between the indenter and the material. In addition, boundary conditions are defined as an element on two symmetry planes parallel to indenter movement direction constrained, only able to expand along radical and axial direction with the help of cylindrical coordinate. But the elements on the bottom of material were defined as having no displacement in any direction.Figure 4
Microindentation model setup.


## 4. Experimental Results and Analysis

### 4.1. Microhardness and Tensile Strength

The load-displacement curves (P-h curves) of crystal plane (405) under loading velocity of 17.284 mN/s are shown in Figure 5. It can be found that curves of various loads show similar shape, which indicates a better repeatability of indentation tests. Different from the nanoindentation test, no “pop-in” size by the abrupt plastic flow generated by the high density of dislocation nucleation and propagation is observed in microindentation [48]. Based on these curves, micromechanical parameters (H, E, σy, σb, and n) can be obtained according to equations mentioned in Section 2.Figure 5
P-h curves of (405) under velocity of 17.284 mN/s.Based on24, microhardness of four different crystal planes ((001), (215), (405), and (605)) has been calculated. The results are shown in Figure 6.Figure 6
MicrohardnessH of different crystal planes under different velocities: (a) (001); (b) (215); (c) (405); (d) (605).
(a)
(b)
(c)
(d)It can be seen from Figure6 that H of each crystal plane under different velocities is almost the same although data of (001) with 51.8527 mN/s derives from others slightly, which indicates that microindentation loading velocities had little influence on the results of H. And the similar phenomenon was founded in Ti-6Al-4V alloy [49]. Therefore, loading velocity of 17.2842 mN/s was used in the following investigation. H of different crystal planes corresponding to 17.2842 mN/s was shown in Figure 7.Figure 7
H of different crystal planes under 17.2842 mN/s.According to Figure7, H of four crystal planes evidently decreases with the increase of maximum load. The phenomenon that indentation depth increases with increase of maximum load confirms the reduction of H. In addition, the apparent drop of H occurs when the loads are less than 2000 mN. When loads are above 2000 mN, their values reduce up to 8%. This phenomenon is attributed to indentation size effects (ISE) caused by geometrically necessary dislocations (GNDs). At the micro/nanoscales, GNDs are large enough and arranged periodically and regularly to cause strong obstacles to slip. GNDs have a strengthening effect on hardness and enhance indurations of material [50]. But values of GNDs have a decreasing tendency with the increase of h, which can be demonstrated by the equation of density of GNDs (ρG) for a Berkovich indenter:
(25)ρG=3tan2θ2bh,
where θ is the angle between surface of the indenter and plane of the surface with the value of 19.7° for Berkovich indenter [27].Moreover, Nix and Gao proposed a new model ofH and h, based on Taylor dislocation and geometrically necessary dislocation model [43, 51]:
(26)HH0=1+h*h,
where h* is a length that characterizes the depth dependence of hardness:
(27)H0=33αμbρS,(28)h*=812bα2tan2θμH02,
where α is a constant, whose value is 0.3–0.5.According to26 and 27, one can find that when P is low, h is of small value correspondingly. Thus ρG of region under indenter is considerable and the measured values of H are large. When load is relatively higher, ρG is in a lower amount and its hardening effect is relatively small on H, when H remains almost stable.As for each crystal plane,H of (001) is close to that of (405), while it is higher than those of (215) and (605). (405) has the largest values of H, followed by (001) and (215), which can be explained by dislocation hardening mechanism. It is well known that there are four mechanisms existing in metallic: solid solution hardening, dislocation hardening, boundary hardening, and precipitation hardening. In this paper, single crystal that has no crystal boundary was investigated and the solid solution and precipitation hardening can be ascribed to dislocation movement. Therefore, dislocation hardening is the dominant factor on H. In the literature, the density equation of dislocation (ρ) has been proposed:
(29)ρ=3tan2θ2b1h+1h*.The statistically stored dislocations density on four crystal planes can be calculated through29, as shown in Figure 8. It can be seen that (001) has the largest dislocation density among these crystal planes, and second for (405), the lowest for (215). Compared with H of four crystal planes in Figure 7, it is reasonable that H of (605) and (215) is lower than those of other two planes. Because dislocation density is lower, hardening effect caused by dislocation is weaker. However, with regard to H of (001) and (405), H of (405) is larger than that of (001), although ρ of (001) is higher. This phenomenon can be explained by the following facts. For metallic materials with FCC crystal structure, slip easily takes place on the close-packed {111} octahedral planes and in the 110 close-packed directions. Generally, for FCC lattice there are four slip planes and three corresponding slip directions, namely, posing twelve slip systems. But there are eight equivalent {111}  110 slip systems on (001) planes, while there is only one equivalent slip system on (405), (605), and (215). Compared with (405), dislocation may extend through different slip systems and hardly pile up due to more slip systems on (001). Therefore, H of (001) is a little lower than that of (405).Figure 8
Density of dislocation densityρ of four crystal planes.


### 4.2. Young’s ModulusE

Of the various unique mechanical properties of materials, Young’s modulus, which is a measure of elasticity, has attracted particular attention. Figure9 shows Young’s modulus of four different crystal planes ((001), (215), (405), and (605)) calculated according to 6.Figure 9
Young’s modulusE of different crystal planes under different speeds: (a) (001); (b) (215); (c) (405); (d) (605).
(a)
(b)
(c)
(d)It can be seen from Figure9 that the calculated E of each crystal plane under different velocities has similar values, illustrating that microindentation loading speeds had little influence on the results of E. And the similar phenomenon was also found in Ti-6Al-4V alloy [49]. Therefore, loading speed with value of 17.2842 mN/s was used for the following investigation.Figure10 gives the values of E from different crystal planes under the loading speed of 17.2842 mN/s. As shown in Figure 10, it can be concluded that E decreases with the increase of h on the whole.Figure 10
E of different crystal planes under 17.2842 mN/s.Compared withH of the same planes, E also decreases rapidly with loads below 2000 mN. And its values remain almost the same when subjected to the higher loads. The induced damage is responsible for this phenomenon. The reduction of Young’s modulus is typically regarded as a characterization of damage evolution and accumulation. When dislocations accumulate or pile up in a certain region, it is easy for stress to concentrate and eventually surpass its threshold. Thus, damage can be easily generated and extended under indentation load, resulting in elastic properties weakening and E decreasing. In addition, when loads are below 2000 mN, each of E curves falls quickly due to the high values of ρ (Figure 8) and the corresponding rapid damage accumulation. However, when the loads are around 4000 mN, fluctuation of ρ is steady and has no obvious effect on variation of E.For each crystal plane,E of (001) is larger than others and (215) has the lowest value. Crystalline structure and theory of metallic plasticity contribute qualitatively to the explanation of diverse E from different crystal plane. For example, for cubic system, the expression for the interplanar spacing (d) is as follows:
(30)1d2=j2+k2+l2a2,
where a is 0.358 nm for DD99.According to30, the interplanar spacing of (001), (215), (405), and (605) is 0.358 nm, 0.0703 nm, 0.06 nm, and 0.0493 nm, respectively. The greater the interplanar spacing is, the larger the density of atoms on this crystal plane is. The greater interplanar spacing is significantly efficient in driving the movement of atoms on the crystal plane, which poses a larger E. However, this explanation does not work for (405) in comparison with that of (215); this phenomenon needs further research.According to theory of metallic plasticity, the crystalE can be also obtained through stiffness coefficient and crystal indices [52]. For cubic system, E can be calculated as the following equation:
(31)1E=S11-2SAjk2+jl2+kl2j2+k2+l2,SA=S11-S12-S442.For the same material,E with different crystal orientation and planes mainly depends on
(32)jk2+jl2+kl2j2+k2+l22.For crystal nickel-based DD99, value ofSA is negative. According to 32, the value of (001) is 0, which is the smallest one among four crystal planes, while (605) is having the largest data of 0.24187. So (001) has the largest E, but E of (605) is relatively lower.


### 4.3. Elastic-Plastic Constitutive Model

According to equations in Section2.1, the values of σy and n of (405) under different loads are shown in Figure 11. It is evident that they remain constant regardless of the different loads.Figure 11
Values ofσy and n at different indentation depths h of (405).As shown in Figure12, values of σy and n are shown in bar graph. It can be found that σy of (215) is the largest, followed successively by (605) and (001). However, the values of n show the opposite tendency. That is, n of (215) is the lowest, followed by (605), and that of (001) is the largest.Figure 12
σ
y and n of four different crystal planes.It is well known thatσy is the strength assessing the ability to resist the plastic deformation. And for materials with power hardening law, n is used as a parameter evaluating a kind of ability that maintains homogeneous deformation. The higher n of certain material, the better compatibility of deformation of material. The differences of σy on four crystal planes can be explained by Schmid’s law.In Figure13, a typical uniaxial tensile stress σ exerted on a metal cylinder is shown. A0 is the area normal to the axial force F and A1 is the area of the slip plane on which the resolved shear force Fr is acting. ϕ is the angle between F and the normal to the slip plane area A1, and λ is the angle between F and the slip direction. For activating dislocations to move in the slip system, a sufficient resolved shear stress (τr) in the slip direction must be produced, and τr is computed as
(33)τr=FrA1=Fcos⁡λA0/cos⁡ϕ=FA0cos⁡λcos⁡ϕ=σcos⁡λcos⁡ϕ.Figure 13
Schematic diagram of uniaxial tension model.This is called Schmid’s law.m is called Schmid factor and defined as follows:
(34)m=cos⁡λcos⁡ϕ.m, σy, and n are shown in Table 3.Table 3
m, σy, and n of different crystal planes.
Crystal indices
n
σ
y (GPa)
m
(001)
0.3449
0.8248
0.408
(215)
0.2529
1.6701
0.435
(405)
0.3338
0.9667
0.448
(605)
0.3242
1.0585
0.442τ
r is a constant for a previous known lattice, and independent of the orientation of F, it can be inferred that σ is larger, if m is smaller according to 33 and 34. Therefore, the fact that σ of (215) is larger than that of (605), which is followed by (405), can be understandable, because m of (215), (605), and (405) increases sequentially. With particular emphasis on m of (001) which is the lowest, it has eight equivalent {111}  110 slip systems, while there is only one equivalent slip system on (405), (605), and (215). For FCC single-crystal structure, Schmid’s law is no longer valid, regardless of potential slip systems. When value and orientation of F are appropriate, τr of two or more slip systems can be achieved, which makes the situation complicated, and Schmid law does not work anymore.For (001) crystal having eight equivalent{111}  110 slip systems and a relatively lower σy, the dislocations can extend easily and the deformation resistance is low, resulting in better compatibility of uniform deformation. So, it has the largest n. With regard to (215), (605) and (405) who yield only one equivalent slip system, the yield strength decrease in order, namely, compatibility of deformation of (405) is the best, followed by (605). And that of (215) is the worst of all. Therefore, n of (405) is larger than that of (605).Based on2 and 23, R and σb along with other parameters representing elastic-plastic properties can be obtained in Table 4.Table 4
Elastic-plastic properties parameters.
Crystal indices
R
n
σ
y (GPa)
σ
b (GPa)
(001)
3.8941
0.3449
0.8248
1.9110
(215)
3.3090
0.2529
1.6701
1.8150
(405)
3.9661
0.3338
0.9667
1.9694
(605)
3.8123
0.3242
1.0585
1.9133Mechanical property parameters,E, R, n, and σy, were obtained from microindentation experimental data; thus the elastic-plastic equations of DD99 on different crystal planes can be obtained as
(35)Crystalplane(001)σ=137.9ε,σ≤824.83.8941ε0.3449,σ≥824.8,Crystalplane(215)σ=70.332ε,σ≤1670.13.309ε0.2529,σ≥1670.1,Crystalplane(405)σ=114.3465ε,σ≤966.73.9661ε0.3338,σ≥966.7,Crystalplane(605)σ=112.414ε,σ≤1058.53.8123ε0.3242,σ≥1058.5.


## 4.1. Microhardness and Tensile Strength

The load-displacement curves (P-h curves) of crystal plane (405) under loading velocity of 17.284 mN/s are shown in Figure 5. It can be found that curves of various loads show similar shape, which indicates a better repeatability of indentation tests. Different from the nanoindentation test, no “pop-in” size by the abrupt plastic flow generated by the high density of dislocation nucleation and propagation is observed in microindentation [48]. Based on these curves, micromechanical parameters (H, E, σy, σb, and n) can be obtained according to equations mentioned in Section 2.Figure 5
P-h curves of (405) under velocity of 17.284 mN/s.Based on24, microhardness of four different crystal planes ((001), (215), (405), and (605)) has been calculated. The results are shown in Figure 6.Figure 6
MicrohardnessH of different crystal planes under different velocities: (a) (001); (b) (215); (c) (405); (d) (605).
(a)
(b)
(c)
(d)It can be seen from Figure6 that H of each crystal plane under different velocities is almost the same although data of (001) with 51.8527 mN/s derives from others slightly, which indicates that microindentation loading velocities had little influence on the results of H. And the similar phenomenon was founded in Ti-6Al-4V alloy [49]. Therefore, loading velocity of 17.2842 mN/s was used in the following investigation. H of different crystal planes corresponding to 17.2842 mN/s was shown in Figure 7.Figure 7
H of different crystal planes under 17.2842 mN/s.According to Figure7, H of four crystal planes evidently decreases with the increase of maximum load. The phenomenon that indentation depth increases with increase of maximum load confirms the reduction of H. In addition, the apparent drop of H occurs when the loads are less than 2000 mN. When loads are above 2000 mN, their values reduce up to 8%. This phenomenon is attributed to indentation size effects (ISE) caused by geometrically necessary dislocations (GNDs). At the micro/nanoscales, GNDs are large enough and arranged periodically and regularly to cause strong obstacles to slip. GNDs have a strengthening effect on hardness and enhance indurations of material [50]. But values of GNDs have a decreasing tendency with the increase of h, which can be demonstrated by the equation of density of GNDs (ρG) for a Berkovich indenter:
(25)ρG=3tan2θ2bh,
where θ is the angle between surface of the indenter and plane of the surface with the value of 19.7° for Berkovich indenter [27].Moreover, Nix and Gao proposed a new model ofH and h, based on Taylor dislocation and geometrically necessary dislocation model [43, 51]:
(26)HH0=1+h*h,
where h* is a length that characterizes the depth dependence of hardness:
(27)H0=33αμbρS,(28)h*=812bα2tan2θμH02,
where α is a constant, whose value is 0.3–0.5.According to26 and 27, one can find that when P is low, h is of small value correspondingly. Thus ρG of region under indenter is considerable and the measured values of H are large. When load is relatively higher, ρG is in a lower amount and its hardening effect is relatively small on H, when H remains almost stable.As for each crystal plane,H of (001) is close to that of (405), while it is higher than those of (215) and (605). (405) has the largest values of H, followed by (001) and (215), which can be explained by dislocation hardening mechanism. It is well known that there are four mechanisms existing in metallic: solid solution hardening, dislocation hardening, boundary hardening, and precipitation hardening. In this paper, single crystal that has no crystal boundary was investigated and the solid solution and precipitation hardening can be ascribed to dislocation movement. Therefore, dislocation hardening is the dominant factor on H. In the literature, the density equation of dislocation (ρ) has been proposed:
(29)ρ=3tan2θ2b1h+1h*.The statistically stored dislocations density on four crystal planes can be calculated through29, as shown in Figure 8. It can be seen that (001) has the largest dislocation density among these crystal planes, and second for (405), the lowest for (215). Compared with H of four crystal planes in Figure 7, it is reasonable that H of (605) and (215) is lower than those of other two planes. Because dislocation density is lower, hardening effect caused by dislocation is weaker. However, with regard to H of (001) and (405), H of (405) is larger than that of (001), although ρ of (001) is higher. This phenomenon can be explained by the following facts. For metallic materials with FCC crystal structure, slip easily takes place on the close-packed {111} octahedral planes and in the 110 close-packed directions. Generally, for FCC lattice there are four slip planes and three corresponding slip directions, namely, posing twelve slip systems. But there are eight equivalent {111}  110 slip systems on (001) planes, while there is only one equivalent slip system on (405), (605), and (215). Compared with (405), dislocation may extend through different slip systems and hardly pile up due to more slip systems on (001). Therefore, H of (001) is a little lower than that of (405).Figure 8
Density of dislocation densityρ of four crystal planes.


## 4.2. Young’s ModulusE

Of the various unique mechanical properties of materials, Young’s modulus, which is a measure of elasticity, has attracted particular attention. Figure9 shows Young’s modulus of four different crystal planes ((001), (215), (405), and (605)) calculated according to 6.Figure 9
Young’s modulusE of different crystal planes under different speeds: (a) (001); (b) (215); (c) (405); (d) (605).
(a)
(b)
(c)
(d)It can be seen from Figure9 that the calculated E of each crystal plane under different velocities has similar values, illustrating that microindentation loading speeds had little influence on the results of E. And the similar phenomenon was also found in Ti-6Al-4V alloy [49]. Therefore, loading speed with value of 17.2842 mN/s was used for the following investigation.Figure10 gives the values of E from different crystal planes under the loading speed of 17.2842 mN/s. As shown in Figure 10, it can be concluded that E decreases with the increase of h on the whole.Figure 10
E of different crystal planes under 17.2842 mN/s.Compared withH of the same planes, E also decreases rapidly with loads below 2000 mN. And its values remain almost the same when subjected to the higher loads. The induced damage is responsible for this phenomenon. The reduction of Young’s modulus is typically regarded as a characterization of damage evolution and accumulation. When dislocations accumulate or pile up in a certain region, it is easy for stress to concentrate and eventually surpass its threshold. Thus, damage can be easily generated and extended under indentation load, resulting in elastic properties weakening and E decreasing. In addition, when loads are below 2000 mN, each of E curves falls quickly due to the high values of ρ (Figure 8) and the corresponding rapid damage accumulation. However, when the loads are around 4000 mN, fluctuation of ρ is steady and has no obvious effect on variation of E.For each crystal plane,E of (001) is larger than others and (215) has the lowest value. Crystalline structure and theory of metallic plasticity contribute qualitatively to the explanation of diverse E from different crystal plane. For example, for cubic system, the expression for the interplanar spacing (d) is as follows:
(30)1d2=j2+k2+l2a2,
where a is 0.358 nm for DD99.According to30, the interplanar spacing of (001), (215), (405), and (605) is 0.358 nm, 0.0703 nm, 0.06 nm, and 0.0493 nm, respectively. The greater the interplanar spacing is, the larger the density of atoms on this crystal plane is. The greater interplanar spacing is significantly efficient in driving the movement of atoms on the crystal plane, which poses a larger E. However, this explanation does not work for (405) in comparison with that of (215); this phenomenon needs further research.According to theory of metallic plasticity, the crystalE can be also obtained through stiffness coefficient and crystal indices [52]. For cubic system, E can be calculated as the following equation:
(31)1E=S11-2SAjk2+jl2+kl2j2+k2+l2,SA=S11-S12-S442.For the same material,E with different crystal orientation and planes mainly depends on
(32)jk2+jl2+kl2j2+k2+l22.For crystal nickel-based DD99, value ofSA is negative. According to 32, the value of (001) is 0, which is the smallest one among four crystal planes, while (605) is having the largest data of 0.24187. So (001) has the largest E, but E of (605) is relatively lower.


## 4.3. Elastic-Plastic Constitutive Model

According to equations in Section2.1, the values of σy and n of (405) under different loads are shown in Figure 11. It is evident that they remain constant regardless of the different loads.Figure 11
Values ofσy and n at different indentation depths h of (405).As shown in Figure12, values of σy and n are shown in bar graph. It can be found that σy of (215) is the largest, followed successively by (605) and (001). However, the values of n show the opposite tendency. That is, n of (215) is the lowest, followed by (605), and that of (001) is the largest.Figure 12
σ
y and n of four different crystal planes.It is well known thatσy is the strength assessing the ability to resist the plastic deformation. And for materials with power hardening law, n is used as a parameter evaluating a kind of ability that maintains homogeneous deformation. The higher n of certain material, the better compatibility of deformation of material. The differences of σy on four crystal planes can be explained by Schmid’s law.In Figure13, a typical uniaxial tensile stress σ exerted on a metal cylinder is shown. A0 is the area normal to the axial force F and A1 is the area of the slip plane on which the resolved shear force Fr is acting. ϕ is the angle between F and the normal to the slip plane area A1, and λ is the angle between F and the slip direction. For activating dislocations to move in the slip system, a sufficient resolved shear stress (τr) in the slip direction must be produced, and τr is computed as
(33)τr=FrA1=Fcos⁡λA0/cos⁡ϕ=FA0cos⁡λcos⁡ϕ=σcos⁡λcos⁡ϕ.Figure 13
Schematic diagram of uniaxial tension model.This is called Schmid’s law.m is called Schmid factor and defined as follows:
(34)m=cos⁡λcos⁡ϕ.m, σy, and n are shown in Table 3.Table 3
m, σy, and n of different crystal planes.
Crystal indices
n
σ
y (GPa)
m
(001)
0.3449
0.8248
0.408
(215)
0.2529
1.6701
0.435
(405)
0.3338
0.9667
0.448
(605)
0.3242
1.0585
0.442τ
r is a constant for a previous known lattice, and independent of the orientation of F, it can be inferred that σ is larger, if m is smaller according to 33 and 34. Therefore, the fact that σ of (215) is larger than that of (605), which is followed by (405), can be understandable, because m of (215), (605), and (405) increases sequentially. With particular emphasis on m of (001) which is the lowest, it has eight equivalent {111}  110 slip systems, while there is only one equivalent slip system on (405), (605), and (215). For FCC single-crystal structure, Schmid’s law is no longer valid, regardless of potential slip systems. When value and orientation of F are appropriate, τr of two or more slip systems can be achieved, which makes the situation complicated, and Schmid law does not work anymore.For (001) crystal having eight equivalent{111}  110 slip systems and a relatively lower σy, the dislocations can extend easily and the deformation resistance is low, resulting in better compatibility of uniform deformation. So, it has the largest n. With regard to (215), (605) and (405) who yield only one equivalent slip system, the yield strength decrease in order, namely, compatibility of deformation of (405) is the best, followed by (605). And that of (215) is the worst of all. Therefore, n of (405) is larger than that of (605).Based on2 and 23, R and σb along with other parameters representing elastic-plastic properties can be obtained in Table 4.Table 4
Elastic-plastic properties parameters.
Crystal indices
R
n
σ
y (GPa)
σ
b (GPa)
(001)
3.8941
0.3449
0.8248
1.9110
(215)
3.3090
0.2529
1.6701
1.8150
(405)
3.9661
0.3338
0.9667
1.9694
(605)
3.8123
0.3242
1.0585
1.9133Mechanical property parameters,E, R, n, and σy, were obtained from microindentation experimental data; thus the elastic-plastic equations of DD99 on different crystal planes can be obtained as
(35)Crystalplane(001)σ=137.9ε,σ≤824.83.8941ε0.3449,σ≥824.8,Crystalplane(215)σ=70.332ε,σ≤1670.13.309ε0.2529,σ≥1670.1,Crystalplane(405)σ=114.3465ε,σ≤966.73.9661ε0.3338,σ≥966.7,Crystalplane(605)σ=112.414ε,σ≤1058.53.8123ε0.3242,σ≥1058.5.


## 5. Verification of Elastic-Plastic Model

Simulation calculations have been performed by using the commercial finite element software ABAQUS. The experimental and FEM results of 500 mN, 2500 mN, and 4500 mN on crystal plane (405) were shown in Figure14. The experimental curves agreed well with computed P-h curves, which indicates that above elastic-plastic equations of DD99 obtained by MIT are valid and 3D FEM model of indentation can fully describe and simulate microindentation process.Figure 14
Experimental versus computedP-h curves of 500 mN, 2500 mN, and 4500 mN on (405).Although FEM results deviate from experimental results slightly, it is of great importance to focus on their difference. In order to illustrate cause of deviation to achieve better simulations in the future, experimental and FEM results of 2500 mN on different crystal planes were shown in Figure15.Figure 15
Experimental and FEM results of 2500 mN on different crystal planes: (a) (001); (b) (215); (c) (405); (d) (605).
(a)
(b)
(c)
(d)As shown in Figure15, it can be concluded that FEM results seem like experimental data moving right. To be more specific, FEM results have various degrees of deviation from experimental results during the loading process, especially in the middle process of the loading. This phenomenon can be explained by Kick’s law (Equation 4) and decrease of E along with increase of h.There is a relationship betweenH and σy originally suggested by Tabor [53]:
(36)H=Cθσy.According to9 and 24, Pm can be expressed as follows:
(37)Pm=ωσyhm2,
where ω is a constant.Therefore, Kick’s law seems to be reasonable. However, microindentation test is not an ideal plastic deformation process. ISE exists in the process of microindentation as discussed above, resulting in decrease ofH. As H is the function of h shown in 26, σy is also a function of h according to 37. And the relationship between Pm and h is no longer quadratic function. When the load is low, ISE is obvious [54] and error of results from Kick’s law is larger compared with those under high load. In order to prove it, the experimental and FEM results of (405) under 500 mN and 2500 mN were utilized, as shown in Figure 16. Besides, the lower load yields a poor accuracy between simulated and experimental results in comparison with the higher load, which may be due to the transition of contact modes from purely elastic under lower loads to elastic/plastic under higher loads. Jian et al. [22, 55, 56] have maintained that the behaviors during indentation can be roughly divided into two stages by the variation of microhardness. Namely, the hardness initially increases with the penetration depth due to the transition between purely elastic and elastic/plastic contact.Figure 16
Experimental and FEM results of (405) with different loads: (a) 500 mN; (b) 2500 mN.
(a)
(b)Also the decrease ofE along with increase of h may also be blamed for the difference between experimental and FEM results. In FEM model, the selection of E is a complicated question. As shown in Figure 10, E of four crystal planes decreases with the increase of h due to damage, E actually changes a little with high load, and the value of E with high load is used in FEM model. In the preliminary stage of indentation, E of materials is higher than that used in FEM model. According to stress-strain relationship in the elastic stage, the higher ε is corresponding to the smaller E. So h of FEM is larger than that of experimental data.With regard to unloading process, the slope of unloading curve is related withE according to 13. And it can be found that all the unloading curves of FEM results are parallel to that of experimental data (Figure 15), indicating that the difference of unloading process is attributed to the error of h in loading process and the selection of E in FEM model is reasonable.In order to interpret the matching degree ofP-h curves in FEM results and experimental data quantitatively, the predictability of the applied load is further quantified employing standard statistical parameters correlation coefficient (U). U was a commonly used statistic and provides information on the dispersion between the experimental and the computed values. It was expressed as [57, 58]
(38)U=∑i=1NPexp⁡i-P-exp⁡Ppi-P-p∑i=1NPexp⁡i-P-exp⁡2∑i=1NPpi-P-p2,
where Pexp⁡ is the experimental indenter load, Pp is the predicted indenter load, and P-exp⁡ and P-p are the mean values of Pexp⁡ and Pp, respectively.According to literature [59], the predictability of the matching degree of P-h curves can also be quantified by the average absolute relative error (AARE) and the results of U and AARE are as shown in Table 5:
(39)AARE%=100×1N∑i=1NPexp⁡i-PpiPexp⁡i.Table 5
Error analysis of four different crystal planes.
Crystal indices
Correlation coefficientU
Average absolute relative error (AARE %)
(001)
0.9884
15%
(215)
0.9914
14%
(405)
0.99024
14.3%
(605)
0.992515
12.8%


## 6. Conclusions

Microindentation measurements using a sharp Berkovich indenter on single-crystal nickel-based superalloy DD99 of four crystallographic orientations, that is, (001), (215), (405), and (605), were made to determine the load-displacement relations. Some material parameters reflecting the micromechanical behavior of DD99, such as microhardnessH, Young’s modulus E, yield stress σy, strain hardening component n, and tensile strength σb, can be obtained from load-displacement relations. Subsequently, the process of MIT is simulated using 3D FEM based on the above parameters. Eventually, the influence of crystal orientations on micromechanical properties can be concluded as follows.(1)
H of four different crystal planes evidently decreases with the increase of h. The crystal plane (405) has the largest micro-hardness values, followed by (001). H of (215) is the lowest. This phenomenon is related to dislocation hardening.(2)
E of these planes decreases with the increase of h. (001) has the largest E, followed by (405). And E of (215) is the lowest, which is attributed to the fact that (001) has the largest interplanar spacing and smallest crystal variable.(3)
n is inversely correlated with σy on all these planes. σy of (215) is the largest among four crystal planes, followed by (605), and (001) has the lowest value. However, n of (215) is the lowest, followed by (605), and that of (001) is the largest. It can be explained by Schmid’s factor (m) and the larger m crystal plane has and the lower σy it possesses. In addition, σb on four planes was calculated and it is similar except for (215).Although FEM results deviate slightly from experimental results, they can be used as sufficient evidences indicating the accuracy of 3D FEM model and material elastic-plastic model founded from MIT.



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