A method for testing the fracture toughness of an anisotropic eccentrically loaded single-edge notched tensile specimen

Abstract

This invention relates to a method for testing the fracture toughness of anisotropic eccentrically loaded single-sided notched tensile specimens, comprising: S1, establishing a two-dimensional plane stress model of anisotropic material ESET specimens; S2, mesh generation; S3, calculating displacement, strain, and stress field to obtain the stress intensity factor K and compliance C under different initial crack lengths and anisotropic material parameters; S4, establishing crack length calculation formulas and stress intensity factor calculation formulas based on compliance under different eccentric loading rates, initial crack lengths, and anisotropic material parameters. This invention allows for rapid and accurate measurement of crack length simply by measuring the deformation of the material under external force; it reliably extends the application range of standardized eccentrically loaded single-sided notched tensile specimens from traditional isotropic metals to modern advanced materials such as composite materials, additive manufacturing alloys, and rolled plates, while ensuring the calculation accuracy under anisotropic conditions.

Description

Technical Field

[0001] This invention belongs to the field of fracture toughness testing technology, specifically relating to a method for testing the fracture toughness of an anisotropic eccentrically loaded single-sided notched tensile specimen. Background Technology

[0002] Fracture toughness testing is a core means of assessing the ability of defective materials or components to resist crack instability and propagation. It is the cornerstone of modern structural damage tolerance design and integrity evaluation. It goes beyond traditional strength indicators and provides an indispensable quantitative basis for predicting critical crack size, determining safe maintenance cycles, and preventing sudden brittle fracture. It is a key technology for ensuring the structural safety and reliability of critical fields such as aviation, energy, and transportation.

[0003] Based on standard recommendations, the specimens currently used for fracture toughness testing include compact tensile specimens, single-sided notched tensile (SENT) specimens, three-point bending specimens, eccentrically loaded single-sided notched tensile (ESET) specimens, and centrally cracked plate specimens. Among them, the eccentrically loaded single-sided notched tensile (ESET) specimen has a unique design concept and mechanical advantages compared to the traditional SENT specimen. The difference lies in the position of the loading line relative to the crack surface, which directly changes the stress state at the crack tip, i.e., the bending moment component. In addition, the eccentric loading of ESET effectively suppresses crack surface rotation, stabilizes the crack propagation path, and has better geometric flexibility. It can generate sufficiently high crack tip stress intensity under a relatively small total load, thus making it suitable for the testing needs of thin plates and brittle non-metallic materials.

[0004] To better utilize the aforementioned specimens, fracture testing standards have developed corresponding calculation formulas for crack length, stress intensity factor, compliance, etc., for different specimens. However, it should be noted that the calculation methods established in the current standards are all based on the isotropic assumption and cannot be used to describe the fracture mechanical behavior of anisotropic materials, such as alloys, composites, and biomaterials.

[0005] The applicant team filed a patent application in 2023 entitled "Test Method for Crack Length and Fracture Performance of Anisotropic Materials Based on Pin-Type Single-Sided Notch Tensile Specimens" and was granted authorization, with authorization announcement number: CN117371271B. This patent is based on single-sided notch tensile (SENT) specimens to test the crack length and fracture performance of anisotropic materials. However, for anisotropic materials, due to the inconsistency between the elastic principal axis direction and the loading / geometric axis direction, the effect of eccentric loading becomes more complex: the anisotropy of the material and the geometric / loading eccentricity will produce a coupling effect, making it impossible to predict the stress intensity factor at the crack tip through simple linear superposition. At the same time, the geometry and loading method of the ESET specimen are very different from those of SENT, so it cannot be directly used for ESET fracture toughness testing. Summary of the Invention

[0006] The purpose of this invention is to address the shortcomings and deficiencies of existing technologies by providing a method for testing the fracture toughness of anisotropic eccentrically loaded single-sided notched tensile specimens. Based on numerical analysis and data fitting, this method develops calculations for SENT crack length, compliance, and stress intensity factors that consider anisotropic characteristics, establishes corresponding fracture parameter calculation formulas, develops corresponding fracture toughness tests, and achieves accurate testing of fracture toughness based on ESET specimens.

[0007] The technical problem solved by this invention is achieved through the following technical solution: A method for testing the fracture toughness of anisotropic eccentrically loaded single-sided notched tensile specimens is provided. The method is designed for eccentrically loaded single-sided notched tensile (ESET) specimens and establishes formulas for calculating crack length and stress intensity factor based on compliance under different eccentric loading rates, initial crack lengths, and anisotropic material parameters. This enables accurate assessment of the fracture toughness of ESET specimens considering the anisotropic characteristics of the material. The steps of the method are as follows: S1. Using the finite element analysis software Abaqus, a two-dimensional plane stress model of the anisotropic material ESET specimen was established, and the elastic modulus of the anisotropic material was input. E 1 、E 2. Poisson's ratio n and shear modulus G 12 Reference points were established at the centers of the two pin holes of the ESET specimen. A coupling constraint was established between the reference points and the inner walls of the pin holes using motion coupling constraints. External loads were applied. P Apply them to two reference points respectively; S2. The two-dimensional plane stress model is meshed using 8-node quadrilateral elements with reduced integral (CPS8R). The mesh at the crack tip is refined and singular elements are introduced. A focused ring mesh is used. S3. Using the implicit solver built into the finite element analysis software Abaqus, calculate the displacement, strain, and stress field to obtain the stress intensity factor K and compliance C under different initial crack lengths and anisotropic material parameters. S4. Based on the finite element simulation results, establish the normalized stress intensity factor. With normalized initial crack length α、 Anisotropic material parameters ( l , r The relationship between the two, and the normalized initial crack length. α With normalized flexibility Relationship: Stress intensity factor applicable to anisotropic materials K The calculation formula is: ; (1) in: α=a / W , a The initial crack length is... W The width of the sample. B The sample thickness is set to 1 mm. e To load the eccentricity; Different eccentricities e Normalized stress intensity factor (a, l, p) The calculation formula is: ; (2) in: f(a, λ, p) The fitting coefficients are derived from the crack length ratio. α and material parameters l、r The decision is made using the following formula: ; (3) ; (4) ; (5) ; (6) in: P ij (X) ( i =1,2,3,4; j =0,1,2,3; X =A, B, C, D) are coefficients; q i for r The function; Q ij ( i =1,2,3,4; j =0,1,2,3) are coefficients; R i For coefficients; a 0 It is a constant; e =0.2, 0.3, 0.4 and 0.5 correspond to a 0 The values ​​are 1.289404, 1.011369, 0.639576, and 0.536189, respectively. The crack length prediction formula based on compliance is: ; (7) ; (8) ; (9) ; (10) in: ; d 0 Let be the fitting constant. e =0.2, 0.3, 0.4 and 0.5 correspond to d 0 The values ​​are 0.973427, 0.973194, 0.963342, and 0.967718, respectively. K ij (X) , M ij and N i For parameters.

[0008] Furthermore, the structural dimensions of the ESET specimen conform to the ASTM E647 standard, specifically as follows: Specimen width W It is 50 mm; The overall height of the sample is 185 mm; The height of the sample between the two pins is 150 mm; The crack was located in the middle of the sample, and the analysis included six different normalized crack lengths. α ,Right now α=a / W = 0.2, 0.3, 0.4, 0.5, 0.6, 0.7; The pin holes are symmetrically distributed along the crack, with the center of each pin hole being 75 mm from the crack and 10 mm, 15 mm, 20 mm, and 25 mm from the sample opening boundary, respectively. These correspond to eccentricities of [missing information]. e= 0.2, 0.3, 0.4 and 0.5, with a pin hole diameter of 10 mm.

[0009] Moreover, the elastic modulus of anisotropic materials E 1 、E 2. Poisson's ratio n 12 and shear modulus G 12 Using parameters to characterize in-plane anisotropic materials l、r Definition: ; (11) ;(12) Finite element model analysis studies a wide range of orthogonal anisotropic parameters, including 16. l Value, that is l = 0.02, 0.04, 0.06, 0.08, 0.1, 0.2, 0.4, 0.6, 0.8, 1, 2, 4, 6, 10, 20, 40 and 5 r Value, that is r= 0.1, 1, 4, 8, 10, a total of 80 different combinations of anisotropic material parameters.

[0010] Furthermore, the specific calculation steps for the compliance and stress intensity factors in S3 are as follows: The stress intensity factor along the crack tip is calculated using the contour integral method and output directly from the Abaqus .dat file. To determine the displacement field, the output of the crack apex node is considered. Compliance is achieved through the opening displacement at the two points above and below the crack apex. V With the applied force P It is obtained by the ratio of, i.e. C=V / P ; The normalized compliance in S4 The specific expression is as follows: ; (13) in Let be the equivalent elastic modulus of the material. For anisotropic materials, . ; B Where B is the sample thickness, B = 1 mm. C This refers to the compliance output by the finite element analysis software.

[0011] The advantages and beneficial effects of this invention are as follows: 1. This invention systematically solves the problem of stress intensity factor and compliance of anisotropic materials under eccentrically loaded single-sided notched tensile specimens for the first time, and develops a simple and universal calculation formula, filling the technical gap in this field.

[0012] 2. This invention develops a crack length prediction formula based on compliance for eccentrically loaded single-sided notched tensile specimens. The crack length can be quickly and accurately measured simply by measuring the deformation of the material under external force.

[0013] 3. This invention reliably expands the application scope of standardized eccentrically loaded single-sided notched tensile specimens from traditional isotropic metals to modern advanced materials such as composite materials, additive manufacturing alloys, and rolled plates, while ensuring the calculation accuracy under anisotropic conditions and reducing safety and economic risks caused by unreasonable evaluation. Attached Figure Description

[0014] Figure 1 This is a geometric schematic diagram of the eccentrically loaded single-sided notched tensile specimen used in the experiment of this invention; Figure 2 This is a schematic diagram of the fracture testing device of the present invention; Figure 3 This is a test load-displacement curve diagram for the present invention; Figure 4 This is the finite element mesh diagram of the eccentrically loaded single-sided notched tensile specimen of the present invention. Detailed Implementation

[0015] The present invention will be further described in detail below through specific embodiments. The following embodiments are merely descriptive and not limiting, and should not be used to limit the scope of protection of the present invention.

[0016] An innovative method for testing the fracture toughness of anisotropic eccentrically loaded single-sided notched tensile specimens is as follows: the method establishes crack length calculation formulas and stress intensity factor calculation formulas based on compliance under different eccentric loading rates, initial crack lengths, and anisotropic material parameters for eccentrically loaded single-sided notched tensile (ESET) specimens, thereby achieving accurate evaluation of the fracture toughness of ESET specimens considering the anisotropic characteristics of the material. The steps of the method are as follows: S1. Using the finite element analysis software Abaqus, a two-dimensional plane stress model of the anisotropic material ESET specimen was established, and the elastic modulus of the anisotropic material was input. E 1 、E 2. Poisson's ratio n and shear modulus G 12 Reference points were established at the centers of the two pin holes of the ESET specimen. A coupling constraint was established between the reference points and the inner walls of the pin holes using motion coupling constraints. External loads were applied. P Apply them to two reference points respectively; S2. The two-dimensional plane stress model is meshed using 8-node quadrilateral elements with reduced integral (CPS8R). The mesh at the crack tip is refined and singular elements are introduced. A focused ring mesh is used. S3. Using the implicit solver built into the finite element analysis software Abaqus, calculate the displacement, strain, and stress field to obtain the stress intensity factor K and compliance C under different initial crack lengths and anisotropic material parameters. S4. Based on the finite element simulation results, establish the normalized stress intensity factor. With normalized initial crack length α、 Anisotropic material parameters ( l , r The relationship between the two, and the normalized initial crack length. α With normalized flexibility Relationship: Stress intensity factor applicable to anisotropic materials K The calculation formula is: ; (1) in: α=a / W , a The initial crack length is... W The width of the sample.B The sample thickness is set to 1 mm. e To load the eccentricity; Different eccentricities e Normalized stress intensity factor (a, l, p) The calculation formula is: ; (2) in: f(a, λ, p) The fitting coefficients are derived from the crack length ratio. α and material parameters l、r The decision is made using the following formula: ; (3) ; (4) ; (5) ; (6) in: P ij (X) ( i =1,2,3,4; j =0,1,2,3; X =A, B, C, D) are coefficients; q i for r The function; Q ij ( i =1,2,3,4; j =0,1,2,3) are coefficients; R i For coefficients; a 0 It is a constant; e =0.2, 0.3, 0.4 and 0.5 correspond to a 0 The values ​​were 1.289404, 1.011369, 0.639576, and 0.536189, respectively. In Table 1, Formula (4) e Fitting parameters corresponding to 0.2 and 0.3 P ij (X)

[0017] In Table 2, Formula (4) e Fitting parameters corresponding to 0.4 and 0.5 P ij (X)

[0018] In Table 3, Formula (5) e Fitting parameters corresponding to 0.2 and 0.3 Q ij

[0019] In Table 4, Formula (5) e Fitting parameters corresponding to 0.4 and 0.5 Q ij

[0020] In Table 5, Formula (5) e Fitting parameters corresponding to 0.2-0.5 R i

[0021] The crack length prediction formula based on compliance is: ; (7) ; (8) ; (9) ; (10) in: ; d 0 Let be the fitting constant. e =0.2, 0.3, 0.4 and 0.5 correspond to d 0 The values ​​are 0.973427, 0.973194, 0.963342, and 0.967718, respectively. K ij (X) , M ij and N i For parameters.

[0022] In Table 6, Formula (8) e Fitting parameters corresponding to 0.2 and 0.3 K ij (X)

[0023] In Table 7, Formula (8) e Fitting parameters corresponding to 0.4 and 0.5 K ij(X)

[0024] In Table 8, Formula (9) e Fitting parameters corresponding to 0.2 and 0.3 M ij

[0025] Table 9 Formula (9) e Fitting parameters corresponding to 0.4 and 0.5 M ij

[0026] Table 10 Formula (10) e Fitting parameters corresponding to 0.2-0.5 N i

[0027] Furthermore, the structural dimensions of the ESET specimen conform to the ASTM E647 standard, specifically as follows: Specimen width W It is 50 mm; The overall height of the sample is 185 mm; The height of the sample between the two pins is 150 mm; The crack was located in the middle of the sample, and the analysis included six different normalized crack lengths. α ,Right now α=a / W = 0.2, 0.3, 0.4, 0.5, 0.6, 0.7; The pin holes are symmetrically distributed along the crack, with the center of each pin hole being 75 mm from the crack and 10 mm, 15 mm, 20 mm, and 25 mm from the sample opening boundary, respectively. These correspond to eccentricities of [missing information]. e= 0.2, 0.3, 0.4 and 0.5, with a pin hole diameter of 10 mm.

[0028] Moreover, the elastic modulus of anisotropic materials E 1 、E 2. Poisson's ratio n 12 and shear modulus G 12 Using parameters to characterize in-plane anisotropic materials l、r Definition: ; (11) ;(12) Finite element model analysis studies a wide range of orthogonal anisotropic parameters, including 16. l Value, that is l = 0.02, 0.04, 0.06, 0.08, 0.1, 0.2, 0.4, 0.6, 0.8, 1, 2, 4, 6, 10, 20, 40 and 5 r Value, that is r = 0.1, 1, 4, 8, 10, a total of 80 different combinations of anisotropic material parameters.

[0029] Furthermore, the specific calculation steps for the compliance and stress intensity factors in S3 are as follows: The stress intensity factor along the crack tip is calculated using the contour integral method and output directly from the Abaqus .dat file. To determine the displacement field, the output of the crack apex node is considered. Compliance is achieved through the opening displacement at the two points above and below the crack apex. V With the applied force P It is obtained by the ratio of, i.e. C=V / P ; The normalized compliance in S4 The specific expression is as follows: ; (13) in Let be the equivalent elastic modulus of the material. For anisotropic materials, . ; B Where B is the sample thickness, B = 1 mm. C This refers to the compliance output by the finite element analysis software.

[0030] This invention takes unidirectional T300 12K carbon fiber reinforced epoxy resin material produced by Guangwei Composites Co., Ltd. as an example, and uses the method proposed in this invention to obtain its accurate crack length and fracture properties. Through uniaxial tensile and shear tests, the basic mechanical properties of the composite material are measured: the elastic modulus in the fiber direction (X direction). E 1 is the elastic modulus of 86300 MPa, perpendicular to the fiber direction (Y direction). E 2 is 7370 MPa, Poisson's ratio n 12 The shear modulus is 0.266. G 12 The in-plane orthotropic parameter is 6517 MPa. l and r It can be calculated from equations (9) and (10) to obtain l =0.0854, r =1.8571.

[0031] The following section describes the accurate prediction of crack length in composite materials using the method of this invention.

[0032] (1) Sample design and processing An eccentrically loaded single-sided notched tensile specimen was used, and the normalized initial crack length was taken. α =0.5, its geometric diagram is as follows Figure 1 As shown. Specimen width W The initial crack length is 25 mm. a It is 12.5 mm thick. B The diameter is 3 mm. The crack direction is at 0°, 15°, and 30° to the fiber orientation direction, respectively. The pin hole diameter is 5 mm, and the distance from the center of the pin hole to the sample opening boundary is 5 mm, corresponding to the eccentricity. e =0.2.

[0033] (2) Fracture test process Fracture tests were conducted on the MTS 370.02 testing machine. Figure 2 Quasi-static tensile loading was applied at room temperature using a displacement-controlled method (0.1 mm / min). Crack opening displacement was measured using DIC technology, and the applied load and crack opening displacement of the specimen were recorded during the test. Crack length was monitored in real time using a high-magnification camera, which took one photograph of the specimen every 1 second to monitor the crack length. The test was stopped after the specimen completely fractured.

[0034] (3) Experimental data processing Two types of data will be collected during the experiment: photographs of the sample surface taken by the camera and text data collected by the MTS testing machine, including time, load, and opening displacement. The load-displacement curves under different orientation angles are shown in the figure. Figure 3 As shown. The compliance ratio is obtained by the ratio of displacement to load. Finally, the compliance ratio and crack length at each time point are obtained.

[0035] (4) Establish the relationship between stress intensity factor, compliance and crack length using finite element method. A two-dimensional plane stress model of a pin-type single-sided notched tensile specimen was established using the finite element software Abaqus, where the specimen width... W The crack length was 25 mm, and the analysis included six different normalized crack lengths. α (Right now a / W = 0.2, 0.3, 0.4, 0.5, 0.6, 0.7), pin hole diameter R It is 5 mm.

[0036] Input the parameters of the carbon fiber composite material, including the elastic modulus. E 1 = 86300 MPa E2 = 7370 MPa, Poisson's ratio n 12 =0.266, shear modulus G 12 =6517 MPa. The model was meshed using 8-node quadrilateral elements with reduced integrals. The mesh at the crack tip was refined and singular elements were introduced. A focused ring mesh was used. The finite element mesh model is as follows: Figure 4 As shown. The stress intensity factor and compliance are calculated for different anisotropic material parameters, eccentricity, and crack length. The empirical formulas for fitting are shown in Equation (1) and Equation (7).

[0037] The parameters of the anisotropic material can be obtained through equation (7). l =0.0854, r =1.8571, the crack lengths corresponding to various compliance levels. Through fracture tests, the normalized compliance levels corresponding to the crack initiation time of the samples with orientation angles of 0°, 15° and 30° were obtained as 20.65871, 20.81868 and 21.98939, respectively. Equation (7) yields the predicted crack length, which is compared with the initial crack length captured by the camera. α= As shown in Table 11, the current crack length prediction method based on flexibility is found to be effective when compared with 0.50.

[0038] Table 11 Comparison of normalized crack lengths from experiments and predictions

[0039] Finally, by substituting the measured normalized initial crack length into equation (1), the crack initiation fracture toughness under different orientations can be successfully calculated. K IC As shown in Table 12, Table 12 Fracture toughness values ​​calculated by formula (1)

[0040] Although embodiments and drawings of the present invention have been disclosed for illustrative purposes, those skilled in the art will understand that various substitutions, variations and modifications are possible without departing from the spirit and scope of the present invention and the appended claims. Therefore, the scope of the present invention is not limited to the contents disclosed in the embodiments and drawings.

Claims

1. A method for testing the fracture toughness of an anisotropically eccentrically loaded single-sided notched tensile specimen, characterized in that: The method is designed for eccentrically loaded single-sided notched tensile (ESET) specimens. It establishes formulas for calculating crack length and stress intensity factor based on compliance under different eccentric loading rates, initial crack lengths, and anisotropic material parameters, thereby enabling accurate assessment of the fracture toughness of ESET specimens considering the anisotropic characteristics of the material. The steps of the method are as follows: S1. Using the finite element analysis software Abaqus, a two-dimensional plane stress model of the anisotropic material ESET specimen was established, and the elastic modulus of the anisotropic material was input. E 1 、E 2. Poisson's ratio ν and shear modulus G 12 Reference points were established at the centers of the two pin holes of the ESET specimen. A coupling constraint was established between the reference points and the inner walls of the pin holes using motion coupling constraints. External loads were applied. P Apply them to two reference points respectively; S2. The two-dimensional plane stress model is meshed using 8-node quadrilateral elements with reduced integral (CPS8R). The mesh at the crack tip is refined and singular elements are introduced. A focused ring mesh is used. S3. Using the implicit solver built into the finite element analysis software Abaqus, calculate the displacement, strain, and stress field to obtain the stress intensity factor K and compliance C under different initial crack lengths and anisotropic material parameters. S4. Based on the finite element simulation results, establish the normalized stress intensity factor. With normalized initial crack length α、 Anisotropic material parameters ( λ , ρ The relationship between the two, and the normalized initial crack length. α With normalized flexibility Relationship: Stress intensity factor applicable to anisotropic materials K The calculation formula is: ； （1） in: α=a / W , a The initial crack length is... W The width of the sample. B The sample thickness is set to 1 mm. e To load the eccentricity; Different eccentricities e Normalized stress intensity factor (α, λ, ρ) The calculation formula is: ； （2） in: f(α, λ, ρ) The fitting coefficients are derived from the crack length ratio. α and material parameters λ, ρ The decision is made using the following formula: ； （3） ； （4） ； （5） ； （6） in: P ij (X) ( i =1,2,3,4; j =0,1,2,3; X =A, B, C, D) are coefficients; q i for ρ The function; Q ij ( i =1,2,3,4; j =0,1,2,3) are coefficients; R i For coefficients; a 0 It is a constant; e =0.2, 0.3, 0.4 and 0.5 correspond to a 0 The values ​​are 1.289404, 1.011369, 0.639576, and 0.536189, respectively. Based on normalized flexibility The formula for predicting crack length is: ； （7） ； （8） ； （9） ； （10） in: ; d 0 Let be the fitting constant. e =0.2, 0.3, 0.4 and 0.5 correspond to d 0 The values ​​are 0.973427, 0.973194, 0.963342, and 0.967718, respectively. K ij (X) , M ij and N i For parameters.

2. The method for testing the fracture toughness of an anisotropic eccentrically loaded single-sided notched tensile specimen according to claim 1, characterized in that: The structural dimensions of the ESET specimen are based on ASTM E647 standard, specifically: Specimen width W It is 50 mm; The overall height of the sample is 185 mm; The height of the sample between the two pins is 150 mm. The crack was located in the middle of the sample, and the analysis included six different normalized crack lengths. α ,Right now α=a / W = 0.2, 0.3, 0.4, 0.5, 0.6, 0.7; The pin holes are symmetrically distributed along the crack, with the center of each pin hole being 75 mm from the crack and 10 mm, 15 mm, 20 mm, and 25 mm from the sample opening boundary, respectively. These correspond to eccentricities of [missing information]. e= 0.2, 0.3, 0.4 and 0.5, with a pin hole diameter of 10 mm.

3. The method for testing the fracture toughness of an anisotropic eccentrically loaded single-sided notched tensile specimen according to claim 1, characterized in that: Elastic modulus of anisotropic materials E 1 、E 2. Poisson's ratio ν 12 and shear modulus G 12 Using parameters to characterize in-plane anisotropic materials λ, ρ Definition: ； （11） ； （12） Finite element model analysis studies a wide range of orthogonal anisotropic parameters, including 16. λ Value, that is λ = 0.02, 0.04, 0.06, 0.08, 0.1, 0.2, 0.4, 0.6, 0.8, 1, 2, 4, 6, 10, 20, 40 and 5 ρ Value, that is ρ = 0.1, 1, 4, 8, 10, a total of 80 different combinations of anisotropic material parameters.

4. The method for testing the fracture toughness of an anisotropic eccentrically loaded single-sided notched tensile specimen according to claim 1, characterized in that: The specific calculation steps for the compliance and stress intensity factors in S3 are as follows: The stress intensity factor along the crack tip is calculated using the contour integral method and output directly from the Abaqus .dat file. To determine the displacement field, the output of the crack apex node is considered. Compliance is achieved through the opening displacement at the two points above and below the crack apex. V With the applied force P It is obtained by the ratio of, i.e. C=V / P ; The normalized compliance in S4 The specific expression is as follows: ； （13） in Let be the equivalent elastic modulus of the material. For anisotropic materials, . ; B Where B is the sample thickness, B = 1 mm. C This refers to the compliance output by the finite element analysis software.

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