Establishment method of high-temperature alloy constitutive model considering hardness
By modifying the parameters of the Ludwik model to a hardness function, a constitutive model of high-temperature alloys considering hardness was established, which solved the problem of finite element simulation accuracy for gradient hardness materials and achieved accurate stress-strain prediction under gradient hardness conditions.
Patent Information
- Application Number
- CN202210836542.7
- Authority / Receiving Office
- CN · China
- Patent Type
- Patents(China)
- Current Assignee / Owner
- Filing Date
- 2022-07-15
- Publication Date
- 2025-12-12
- Estimated Expiration
- 2042-07-15
AI Technical Summary
The existing Ludwik model cannot effectively analyze materials with gradient hardness distribution, resulting in a decrease in the accuracy of finite element simulation.
The yield strength, strength coefficient, and strain hardening exponent in the Ludwik model were modified to be functions of hardness. True stress-true strain data were obtained through compression tests, and a constitutive model of high-temperature alloys considering hardness was established.
It enables accurate prediction of stress and strain data under gradient hardness conditions, improves the accuracy of finite element simulation, and is applicable to the machining and mechanical analysis of gradient hardness parts.
Smart Images
Figure CN115206467B_ABST
Abstract
Description
Technical Field
[0001] This invention relates to a method for establishing a constitutive model of high-temperature alloy materials, and more particularly to a method for establishing a constitutive model of high-temperature alloys that takes hardness into account. Background Technology
[0002] Constitutive models of materials are a fundamental theoretical basis for numerical simulation of plastic deformation and an important reference element for process specification formulation. The Ludwik model, as a conventional constitutive model for high-temperature alloys, only considers work hardening effects. Although its equations are simple and model parameters are easy to obtain, it does not consider the influence of material hardness on material stress. In finite element simulations, it can only provide stress-strain data for materials with a single hardness. In certain specific situations, such as when a part exhibits a gradient hardness distribution, the conventional Ludwik model based on the material cannot perform effective simulation analysis because it can only provide stress-strain data for a single hardness of the material, failing to provide corresponding stress-strain data for other hardnesses within the gradient. If the conventional Ludwik model is still used to provide stress-strain data for the entire gradient hardness region, the simulation accuracy will be significantly reduced. Summary of the Invention
[0003] To overcome the shortcomings of existing technologies, this invention provides a method for establishing a constitutive model of high-temperature alloys that takes hardness into account. The method modifies the yield strength σ0 (constant), strain hardening exponent n (constant), and strength coefficient K (constant) in the conventional Ludwik model into functions of hardness to obtain a new model. In the simulation of gradient hardness parts, this model can be used directly to provide the required stress-strain data for different hardnesses.
[0004] The technical solution adopted by this invention to solve its technical problem is: a method for establishing a constitutive model of a high-temperature alloy considering hardness, based on the conventional Ludwik constitutive model σ=σ0+K(ε). n A new model is established, where σ is the true stress (MPa); σ0 is the yield strength (MPa); K is the strength coefficient; ε is the true strain; and n is the strain hardening coefficient. The specific steps of the establishment method are as follows:
[0005] Step 1: Perform compression tests on m groups of high-temperature alloy samples with different hardness to obtain the corresponding engineering stress-strain data, and then use the formula σ = σ eng ×(1+ε eng ) and the formula ε=ln(1+ε eng The m sets of engineering stress-engineering strain data are converted into m sets of true stress-true strain data, and then the elastic modulus values of each hardness specimen are obtained through the compression test.
[0006] In the formula, σ is true stress, MPa; ε is true strain; σ eng is engineering stress, MPa; ε eng is engineering strain.
[0007] Step two: based on m groups of true stress-true strain data, the yield strength σ0 corresponding to each hardness is extracted, and a quadratic polynomial regression equation between the yield strength and the hardness is established by regression analysis:
[0008] σ 0m = a0 + a1 x h + a2 x h 2
[0009] In the formula, σ 0m is the yield stress considering hardness, MPa; h is the hardness value of the material, HV; a0, a1, a2 are coefficients in the established regression equation.
[0010] Step three: based on m groups of true stress-true strain data, the true stress in the plastic deformation stage is subtracted from the corresponding yield strength σ0, and then the logarithm Ln of both sides is taken, linear regression processing is performed, and the linear regression formula of Ln(σ-σ0) and Ln(ε) is obtained. The intercept and slope are taken as the LnK value and the n value respectively, then the corresponding K value is calculated according to LnK, and finally the regression analysis of the K value and the n value corresponding to different hardnesses is performed respectively, and a quadratic polynomial regression equation between K and the hardness h is established:
[0011] K m = b0 + b1 x h + b2 x h 2
[0012] And a quadratic polynomial regression equation between n and the hardness h:
[0013] n m = c0 + c1 x h + c2 x h 2
[0014] In the formula, K m is the strength coefficient considering hardness; n m is the strain hardening coefficient considering hardness; b0, b1, b2, c0, c1, c2 are coefficients in the corresponding regression equation.
[0015] Step four: the σ0, K and n in the conventional Ludwik constitutive model are replaced by the quadratic polynomial regression equations between the yield strength and the hardness, between K and the hardness, and between n and the hardness respectively, and the high-temperature alloy constitutive model considering hardness is obtained:
[0016]
[0017] In the formula, σ p is the predicted stress, MPa.
[0018] As a further preferred embodiment, the number of high-temperature alloy samples during the compression test is greater than or equal to 5, i.e. m≥5.
[0019] As a further preferred embodiment, in step two, based on the m sets of true stress-true strain data, the plastic stress corresponding to the occurrence of 0.2% plastic strain is taken as the yield strength to extract the yield strength σ0corresponding to each hardness,
[0020] As a further preferred embodiment, the high-temperature alloy is A286 alloy.
[0021] Advantages of the present application: compared with the conventional Ludwik model, the present application modifies the three key parameters in the conventional Ludwik model into functions of hardness in a corresponding manner respectively, so that a new model is established with an additional hardness variable. By using the new model, the corresponding stress can be calculated by inputting different hardness values, i.e. the stress-strain curves of different hardnesses are obtained, so that the required stress-strain data of different hardnesses can be obtained directly by using the model in the simulation of gradient hardness parts. Compared with the conventional single hardness model, the new model established by the model method can more accurately predict the stress of different hardnesses, which ensures the simulation accuracy of workpiece machining and construction mechanics analysis under the condition of gradient hardness from the root, and has good engineering application prospect. BRIEF DESCRIPTION OF DRAWINGS
[0022] Figure 1 is the true stress-true strain curve of A286 alloy with different hardnesses;
[0023] Figure 2 is a schematic diagram for determining the yield strength;
[0024] Figure 3a is a schematic diagram for regression processing of the strength coefficient and the strain hardening coefficient;
[0025] Figure 3b is a graph of the regression processing results of the strength coefficient and the strain hardening coefficient;
[0026] Figure 4a is a comparison of the stress prediction value and the test value under 370HV hardness;
[0027] Figure 4b is a comparison of the stress prediction value and the test value under 251HV hardness;
[0028] Figure 4c is a comparison of the stress prediction value and the test value under 235HV hardness;
[0029] Figure 4d is a comparison of the stress prediction value and the test value under 198HV hardness;
[0030] Figure 4eIt is a comparison between the predicted and experimental values of stress at a hardness of 175HV;
[0031] Figure 4f This is a comparison between the predicted stress value and the experimental value at a hardness of 297HV. Detailed Implementation
[0032] The preferred embodiments of the present invention will now be described in detail with reference to the accompanying drawings.
[0033] A preferred embodiment of the present invention provides a method for establishing a constitutive model of a high-temperature alloy that considers hardness, based on the conventional Ludwik constitutive model σ=σ0+K(ε). n A new model is established, where σ is the true stress (MPa); σ0 is the yield strength (MPa); K is the strength coefficient; ε is the true strain; and n is the strain hardening coefficient. The specific steps of the establishment method are as follows:
[0034] Step 1: Obtain true stress-true strain data for different hardness levels
[0035] Compression tests were conducted on five groups of A286 high-temperature alloy specimens (made into cylindrical specimens φ10×13mm) with hardnesses of 370HV, 251HV, 235HV, 198HV, and 175HV, respectively. The corresponding engineering stress-strain data were obtained and analyzed using the formula σ=σ eng ×(1+ε eng ) and the formula ε=ln(1+ε eng Convert 5 sets of engineering stress-strain data into m sets of true stress-strain data (e.g.) Figure 1 As shown in Table 1), the elastic modulus values of each hardness specimen are obtained through this compression test.
[0036] In the formula, σ is the true stress (MPa); ε is the true strain; eng For engineering stress, MPa; ε eng For engineering contingency;
[0037] Table 1 Elastic modulus corresponding to different hardness levels
[0038] Hardness value / HV 370 251 235 198 175 Elastic modulus / Mpa 110000 110000 82000 81000 45000
[0039] Step 2: Modeling the yield strength considering hardness
[0040] (1) Extraction of yield strength σ0
[0041] Based on 5 sets of true stress-true strain data, such as Figure 1 As shown, since no obvious yielding phenomenon was observed in the obtained true stress-true strain curves, the plastic stress corresponding to 0.2% plastic strain was set as the yield strength, i.e., the nominal yield limit, and σ was used as the yield strength. 0.2(MPa) to extract the yield strength corresponding to each hardness. The yield strength calculation method is:
[0042] σ 0.2 = (ε1-0.002) x E
[0043] wherein: E is the elastic modulus, MPa; ε1 is the yield strength σ 0.2 corresponding strain value. Its geometric form is as shown in Figure 2 When the yield strength slope in Figure 2 intersects the true stress-true strain curve, the intersection point ordinate value is the yield strength. In the actual calculation process, different strain values are substituted into the above yield strength calculation formula, and when the calculated stress value is equal to the stress value in the true stress-true strain data in Figure 1 , the stress value is taken as the yield strength corresponding to the hardness, and the finally obtained yield strength values corresponding to each hardness (corresponding to the yield strength test values in Table 2) are shown in Table 2.
[0044] Table 2 Yield strength test values and predicted values
[0045]
[0046] (2) Yield strength prediction mathematical model establishment
[0047] A quadratic polynomial regression equation between yield strength and hardness is established by regression analysis, that is, the hardness values and yield strength test values in Table 2 are subjected to regression processing, and a quadratic curve regression mathematical model between yield strength and hardness values is established as follows:
[0048] σ 0m = -506.527 + 4.94375689370756 x h - 0.0032904389770899 x h 2
[0049] wherein σ 0m is the yield stress considering hardness, MPa; h is the hardness value of the material, HV;
[0050] Using the formula to predict the yield strength of the above 5 groups of hardness, as shown in Table 2, the maximum prediction error is 2.06%, and the minimum prediction error is 0.02%, indicating that it has good prediction accuracy.
[0051] Step three: modeling considering hardness strength coefficient and strain hardening coefficient
[0052] Based on the true stress-true strain data of the 5 groups, the true stress in the plastic deformation stage is subtracted from the corresponding yield strength σ0, and the logarithm Ln is taken on both sides of the formula, that is, the formula:
[0053] σ = σ0+ K(ε)n
[0054] The formula obtained after deformation is:
[0055] σ-σ0=K(ε) n
[0056] Taking the logarithm of Ln on both sides, we obtain the following expression:
[0057] Ln(σ-σ0)=Ln[K(ε) n ]=LnK+nLn(ε)
[0058] This yields a linear function in one variable with Ln(ε) as the independent variable and Ln(σ-σ0) as the dependent variable, where Ln(σ-σ0) and Ln(ε) have a linear relationship (see...). Figure 3a ).
[0059] Specifically, will Figure 1 The logarithm Ln was taken after subtracting the yield strength from the true stress during the intermediate plastic deformation stage, and the logarithm Ln was taken directly from the true strain data. Linear regression was then performed (results are shown in...). Figure 3b The linear regression mathematical model of Ln(σ-σ0) and Ln(ε) was obtained. The intercept and slope were used as LnK and n values, respectively, and the corresponding K value was calculated based on LnK. Finally, the coefficient values of the regression model and the transformed strength coefficient values for each group of hardness are shown in Table 3.
[0060] Table 3 Values of Strength Coefficient and Strain Hardening Coefficient
[0061] Serial number Hardness h / HV Ln K n K 1 370 8.2 0.739 3640.95 2 251 8.083 0.813 3238.935 3 235 8.365 0.921 4294.112 4 198 8.525 1.021 5039.187 5 175 8.865 1.227 7079.793
[0062] Finally, regression analysis was performed on the K and n values corresponding to different hardnesses, that is, the hardness values in Table 3 were regressed with the strength coefficient and the strain hardening coefficient, respectively. The quadratic curve regression mathematical model between the strength coefficient and the hardness value is as follows:
[0063] K m =25519.4167005472-149.370245237023×h+0.243947401620425×h 2
[0064] In the formula, K m The strength coefficient taking hardness into account;
[0065] The established prediction model between the strain hardening coefficient and the hardness value is as follows:
[0066] n m= 3.04025730292675 - 0.0143116912632314 x h + 0.0000218762045448684 x h 2
[0067] wherein n m is the strain hardening coefficient considering hardness.
[0068] Step four: establishing the Ludwik constitutive model considering hardness
[0069] The σ0, K and n in the conventional Ludwik constitutive model are respectively replaced by the quadratic polynomial regression equations between the yield strength and hardness, K and hardness, and n and hardness obtained, to obtain the high-temperature alloy constitutive model considering hardness:
[0070]
[0071] wherein σ p is the predicted stress.
[0072] wherein:
[0073] σ 0m = -506.527 + 4.94375689370756 x h - 0.0032904389770899 x h 2
[0074] K m = 25519.4167005472 - 149.370245237023 x h + 0.243947401620425 x h 2
[0075] n m = 3.04025730292675 - 0.0143116912632314 x h + 0.0000218762045448684 x h 2
[0076] The strain value and hardness value are substituted into the obtained prediction model, and the corresponding stress prediction value can be directly calculated. The comparison between the processed plastic stage stress prediction value and the test value is shown in Figure 4a to Figure 4f It can be seen from the figure that the constitutive model can well predict the stress in the plastic deformation stage of a specific hardness, and from the root, ensure the simulation accuracy of workpiece machining and construction mechanics analysis under the condition of gradient hardness, and has better engineering application significance.
[0077] The above merely describes the preferred embodiments of the present application, and it should be understood that the above description of the embodiments is only used to help understand the method of the present application and its core idea, and is not used to limit the protection scope of the present application, and any modification, equivalent replacement, etc. within the idea and principle of the present application should be included in the protection scope of the present application.
Claims
1. A method for establishing a high-temperature alloy constitutive model considering hardness, based on a conventional Ludwik constitutive model σ = σ0 + K(ε) n a new model is established, wherein σ is true stress; σ0 is yield strength; K is strength coefficient; ε is true strain; and n is strain hardening coefficient; and The establishing method specifically comprises the following steps: Step one: compression test is carried out on m groups of high-temperature alloy samples with different hardnesses to obtain corresponding engineering stress-engineering strain data, and m groups of true stress-true strain data are converted from the engineering stress-engineering strain data through the formula σ = σ eng × (1 + ε eng ) and the formula ε = ln (1 + ε eng ), and then the elastic modulus values of the samples of each hardness are obtained through the compression test; In the formula, sigma is true stress, MPa; ε is true strain; σ eng is engineering stress, MPa; ε eng ε is the engineering strain; Step two: based on m groups of true stress-true strain data, the yield strength sigma0 corresponding to each hardness is extracted, and a quadratic polynomial regression equation between the yield strength and the hardness is established through regression analysis: σ 0m = a0+ a1x h + a2x h 2 where σ is the yield stress, MPa; h is the hardness value, HV; a0, a1, a2 are coefficients in the regression equation established; and 0m where σ is the yield stress, MPa; h is the hardness value, HV; a0, a1, a2 are coefficients in the regression equation established; and Step three: based on m groups of true stress-true strain data, the true stress in the plastic deformation stage is subtracted by the corresponding yield strength sigma0, and then the logarithm Ln of both sides of the formula is taken, linear regression processing is carried out, and a linear regression formula of Ln(sigma-sigma0) and Ln(epsilon) is obtained; the intercept and the slope are taken as the LnK value and the n value respectively, then the corresponding K value is calculated according to the LnK, finally, the K value and the n value corresponding to different hardnesses are respectively subjected to regression analysis, and a quadratic polynomial regression equation between K and the hardness h is established: K m = b0+ b1x h + b2x h 2 And a quadratic polynomial regression equation between n and the hardness HV is established: n m = c0+ c1x h + c2x h 2 wherein K m is a strength coefficient accounting for hardness; n m is a strain hardening coefficient accounting for hardness; b0, b1, b2, c0, c1, c2 are coefficients in the regression equation. Step four: the sigma0, K and n in the conventional Ludwik constitutive model are respectively replaced by the quadratic polynomial regression equations between the yield strength and the hardness, between K and the hardness and between n and the hardness, and a high-temperature alloy constitutive model considering the hardness is obtained: where σ p is the predicted stress.
2. The method of claim 1, wherein the method further comprises: determining a stress-strain relationship of the superalloy material based on the determined material properties. When the compression test is carried out, the number of high-temperature alloy samples is greater than or equal to 5, that is, m is greater than or equal to 5.
3. The method of claim 1, wherein the method further comprises: determining a stress-strain relationship of the superalloy material based on the determined material properties. In step two, based on m groups of true stress-true strain data, the plastic stress corresponding to the occurrence of 0.2% plastic strain is taken as the yield strength, and the yield strength sigma0 corresponding to each hardness is extracted.
4. The method of claim 1, wherein the method further comprises: The high-temperature alloy is A286 alloy.