Deformation energy decomposition method of planar structures based on isotropic square elements

Abstract

The present invention belongs to the field of mechanical analysis and material technology, and discloses a method for decomposing the deformation energy of a planar structure based on anisotropic square units. The method comprises the following steps: constructing a planar deformation of the anisotropic square units in a plane rectangular coordinate system to obtain basic deformation energy and basic displacement basis vectors of the anisotropic square units; establishing a planar structure model, dividing the structure using the anisotropic square units and performing finite element solution to obtain node displacement vectors, basic deformation energy and basic displacement projection coefficients of the anisotropic square units after arbitrary displacement and deformation under arbitrary load conditions; obtaining basic deformation information of the anisotropic square units under arbitrary load conditions, distinguishing the main deformation and secondary deformation of the anisotropic square units under arbitrary load conditions, and thereby realizing deformation energy decomposition and quantitative analysis of deformation performance of the anisotropic planar structure.

Description

Technical Field

[0001] The invention belongs to the field of mechanical analysis and materials, and relates to a method for decomposing deformation energy of a plane structure based on isotropic square units. Background Art

[0002] Classical elasticity theory assumes that materials have identical elastic properties under tension and compression. However, concrete, graphite, ceramics, new polymers, and composites commonly used in engineering often exhibit different elastic properties under identical tensile and compressive stresses. These materials are known as isotropic materials, exhibiting different tensile and compressive elastic moduli. Using classical elasticity theory to analyze and calculate isotropic structures makes it difficult to describe their true mechanical behavior. Therefore, it is important to fully consider the influence of the tensile and compressive elastic moduli of materials in isotropic structures.

[0003] With the development of computer technology, finite element simulation technology has become an important means to study the anisotropic and orthotropic problems of structures. However, the normal strain and shear strain of the structure calculated by the finite element method belong to the microscopic deformation level, while macroscopic deformations such as bending deformation are also the focus of attention in engineering structure design. The deformation decomposition method is a method that can decompose the comprehensive deformation of the structure into basic deformations such as tension, compression, bending and shear deformation, and is applied to the mechanical performance analysis of the structure. However, the existing deformation decomposition method is mainly for isotropic elements and cannot be applied to elements with anisotropic material properties. Therefore, it is necessary to propose a new deformation decomposition method for isotropic square elements. By introducing the physical parameters of the material, an in-depth mechanical analysis of the anisotropic structure is carried out from the perspective of basic deformation energy, and quantitative information of the macroscopic deformation of the anisotropic planar structure is identified. At present, there is no report on the quantitative analysis method of planar structure performance based on the deformation energy decomposition of isotropic square elements. Summary of the Invention

[0004] The purpose of the present invention is to provide a method for decomposing the deformation energy of a planar structure based on anisotropic square units, which can identify the main basic deformation and secondary basic deformation of the anisotropic square units, and thus realize the quantitative analysis of the deformation performance of the anisotropic planar structure.

[0005] To achieve the above object, the present invention adopts the following technical solutions:

[0006] A method for decomposing deformation energy of a planar structure based on isotropic square elements comprises the following steps:

[0007] Step 1: Construct the plane deformation of the isotropic square element in a rectangular coordinate system. Based on mathematical orthogonality and mechanical equilibrium and considering the material properties of the element, the basic deformation energy and basic displacement basis vectors of the isotropic square element are obtained.

[0008] Step 2: Establish a plane structure model, divide the plane structure using isotropic square elements and perform finite element solution to obtain the node coordinate values of the isotropic square elements in the plane rectangular coordinate system and the node coordinate values of the isotropic square elements after arbitrary displacement and deformation under any load condition. Then, obtain the node displacement vectors of the isotropic square elements after arbitrary displacement and deformation under any load condition.

[0009] Step 3: Express the node displacement vector of the isotropic square element after any displacement and deformation under any load condition as a linear combination of the basic deformation energy and basic displacement basis vector of the isotropic square element in step 1, and obtain the basic deformation energy and basic displacement projection coefficient vector of the isotropic square element after any displacement and deformation under any load condition;

[0010] Step 4: Based on the basic deformation energy projection coefficient of the isotropic square element after any displacement or deformation under any load condition, the basic deformation information of the isotropic square element under any load condition is obtained, and the main deformation and secondary deformation of the isotropic square element under any load condition are identified, thereby realizing the deformation energy decomposition and performance quantitative analysis of the isotropic planar structure.

[0011] Furthermore, the isotropic square unit has four nodes, and the coordinates of the four nodes in the X and Y directions are x1, y1, x2, y2, x3, y3, x4, and y4. The side length of the isotropic square unit is 2l, and the thickness is b. The plane deformation of the isotropic square unit is composed of the superposition of X-axial tensile and compressive deformation, Y-axial tensile and compressive deformation, X-axial bending deformation, Y-axial bending deformation, shear deformation in the XOY plane, X-axial rigid body translation, Y-axial rigid body translation, and XOY plane rigid body rotation.

[0012] For the tensile deformation of the isotropic square element, let its node loads be F X1 , F X2 , F X3 , F X4 , according to the force balance condition:

[0013]

[0014] The solution is:

[0015]

[0016] Let F X4 If it is unity, the nodal displacement basis vector d1 of the isotropic square element under X-axis tensile deformation can be obtained as:

[0017] d1=(10-10-1010) T ;

[0018] Then the unit energy q1 corresponding to the X-axial tensile deformation of the isotropic square element can be calculated as follows:

[0019]

[0020] Where: B is the strain matrix of the element; D t is the elastic matrix of the element:

[0021]

[0022]

[0023] Where: E t and μ t are the elastic modulus and Poisson's ratio of the isotropic square element under tension, respectively;

[0024] From this, the unit energy q1 corresponding to the X-axial tensile deformation of the isotropic square unit can be obtained as follows:

[0025]

[0026] Then the X-axis tensile deformation energy basis vector u1 of the anisotropic square element is obtained as:

[0027]

[0028] Elastic matrix D of anisotropic square element under compression c for:

[0029]

[0030] Where: E c and μ c are the elastic modulus and Poisson’s ratio of the isotropic square element under compression; let F X4 = -1, from which the x-axis compression deformation energy basis vector u2 of the anisotropic square element can be obtained as:

[0031]

[0032] According to the force balance, the displacement basis vector d3 of the node of the isotropic square element in the Y-axis tensile deformation can be obtained:

[0033] d3=(01010-10-1) T ;

[0034] Then the unit energy q3 corresponding to the Y-axis tensile deformation of the isotropic square element is:

[0035]

[0036] Then the Y-axis tensile deformation energy basis vector u3 of the anisotropic square element is obtained as:

[0037]

[0038] Similarly, the Y-axis compression deformation energy basis vector u4 of the anisotropic square element can be obtained as follows:

[0039]

[0040] Similarly, the remaining deformation energy basis vectors and displacement basis vectors u5-u of the isotropic square element can be obtained 10 As follows: u5 is the x-axis bending deformation energy basis vector of the anisotropic square element:

[0041]

[0042] u6 is the basis vector of the Y-axis bending deformation energy of the isotropic square element:

[0043]

[0044] u7 is the basis vector of the shear deformation energy of the isotropic square element in the XOY plane:

[0045]

[0046] u8 is the rigid body translation basis vector of the isotropic square unit along the X axis:

[0047] u8=(10101010) T ;

[0048] u9 is the Y-axis rigid body translation basis vector of the isotropic square unit:

[0049] u9=(01010101) T ;

[0050] u 10 The XOY plane rigid body rotation basis vectors of the isotropic square unit are:

[0051] u 10 =(-11-1-11-111) T ;

[0052] Furthermore, the node coordinate vector of the isotropic square unit in the plane rectangular coordinate system is u p ,

[0053] u p =(x1y1x2y2x3y3x4y4),

[0054] The node coordinate vector of the anisotropic square element after any displacement and deformation under any load condition is u q ,

[0055] u q =(x1'y1'x'2y'2x'3y'3x'4y'4),

[0056] by u q -u p The node displacement vector u of the anisotropic square element after any displacement and deformation under any load condition can be obtained: r ,

[0057] u r =(x1'-x1y1'-y1x'2-x2y'2-y2x'3-x3y'3-y3x'4-x4y'4-y4).

[0058] Furthermore, step 3 is specifically as follows:

[0059] The node coordinate displacement vector u after any displacement and deformation of any isotropic square element under any load condition is obtained r It is expressed as a linear combination of the basic deformation energy of the isotropic square element and the basic displacement basis vector, that is:

[0060]

[0061] Where: u k is the kth basic deformation energy or basic displacement basis vector, α k , are the projection coefficients of the basic deformation energy and basic displacement of the isotropic square unit, that is, α1 is the projection coefficient of the tensile deformation energy of the isotropic square unit in the X-axis direction, α2 is the projection coefficient of the compressive deformation energy of the X-axis direction, α3 is the projection coefficient of the tensile deformation energy of the Y-axis direction, α4 is the projection coefficient of the compressive deformation energy of the Y-axis direction, α5 is the projection coefficient of the bending deformation energy of the X-axis direction, α6 is the projection coefficient of the bending deformation energy of the Y-axis direction, α7 is the projection coefficient of the shear deformation energy, α8 is the projection coefficient of the rigid body translation in the X-axis direction, α9 is the projection coefficient of the rigid body translation in the Y-axis direction, α 10 is the rigid body rotation projection coefficient.

[0062] Furthermore, the step 4 specifically includes:

[0063] The decomposition results are separated into rigid and flexible parts, i.e. the basic displacement projection coefficient α8-α is ignored. 10In order to avoid the influence of the above-mentioned influence, only the absolute values of the basic deformation energy projection coefficients α1-α7 in the above-mentioned component information are compared. The deformation corresponding to the maximum absolute value of deformation energy is determined as the main deformation of the isotropic square element. Similarly, the second largest one is determined as the secondary deformation of the isotropic square element. Then, the proportion of arbitrary deformation under arbitrary load conditions of the isotropic plane structure p can be calculated. k , to achieve quantitative analysis of structural deformation performance:

[0064]

[0065] Where: n k is the number of orthotropic rectangular elements that undergo primary or secondary deformation, n t is the total number of orthotropic rectangular elements.

[0066] Compared with the prior art, the present invention has the following beneficial effects:

[0067] Based on mathematically complete orthogonality, mechanical force equilibrium conditions, and consideration of the unit's material properties, this paper proposes a method for decomposing the deformation energy of anisotropic square units. This method simultaneously considers the effects of both tensile and compressive moduli on the structural deformation performance. By decomposing the comprehensive deformation energy of anisotropic square units into basic deformation energies, quantitative deformation information for the units is obtained. This enables detailed analysis of the deformation performance of anisotropic structures, guiding targeted reinforcement and optimization designs. BRIEF DESCRIPTION OF THE DRAWINGS

[0068] Figure 1 Schematic diagram of the process of the deformation energy decomposition method of the anisotropic square unit of the present invention.

[0069] Figure 2 Schematic diagram of a four-node isotropic square unit in a plane rectangular coordinate system in the present invention.

[0070] Figure 3 Schematic diagram of the X-axial tensile deformation and stress conditions of the isotropic square unit in the plane rectangular coordinate system in the present invention.

[0071] Figure 4 Schematic diagram of the X-axial compression deformation and stress conditions of the isotropic square unit in the plane rectangular coordinate system in the present invention.

[0072] Figure 5 Schematic diagram of the Y-axis tensile deformation and stress conditions of the isotropic square unit in the rectangular coordinate system of the present invention.

[0073] Figure 6 Schematic diagram of the Y-axis compression deformation and stress conditions of the isotropic square unit in the rectangular coordinate system of the present invention.

[0074] Figure 7 Schematic diagram of the X-axis bending deformation and stress conditions of the isotropic square unit in the plane rectangular coordinate system in the present invention.

[0075] Figure 8 Schematic diagram of the Y-axis bending deformation and stress conditions of the isotropic square unit in the plane rectangular coordinate system in the present invention.

[0076] Figure 9 Schematic diagram of the shear deformation and stress conditions of the isotropic square unit in the XOY plane in the rectangular coordinate system in the present invention.

[0077] Figure 10 Schematic diagram of the X-axis rigid body translation and force conditions of the isotropic square unit in the plane rectangular coordinate system in the present invention.

[0078] Figure 11 Schematic diagram of the Y-axis rigid body translation and force conditions of the isotropic square unit in the plane rectangular coordinate system in the present invention.

[0079] Figure 12 Schematic diagram of the XOY plane rigid body rotation and force working conditions of the isotropic square unit in the plane rectangular coordinate system in the present invention.

[0080] Figure 13 Schematic diagram of the counterclockwise rotation of the isotropic square unit in the plane rectangular coordinate system in the present invention.

[0081] Figure 14 It is a schematic diagram of an isotropic simply supported beam in a plane rectangular coordinate system in the present invention. DETAILED DESCRIPTION

[0082] The following examples are used to illustrate the present invention, but are not intended to limit the scope of protection of the present invention. Unless otherwise specified, the technical means used in the examples are conventional means well known to those skilled in the art.

[0083] Figure 1 The flow chart of the deformation energy decomposition method of an isotropic square unit of the present invention is shown. Assume that any four-node isotropic square unit has a schematic diagram in a plane rectangular coordinate system as shown below: Figure 2 As shown, its basic deformation and displacement in the plane rectangular coordinate system are as follows Figures 3 to 12The coordinates of the four nodes in the X and Y directions are x1, y1, x2, y2, x3, y3, x4, and y4. The side length of the isotropic square unit is 2l, and the thickness is b. The plane deformation of the isotropic square unit is a linear superposition of the X-axis tensile and compressive deformation, the Y-axis tensile and compressive deformation, the X-axis bending deformation, the Y-axis bending deformation, the shear deformation in the XOY plane, the X-axis rigid body translation, the Y-axis rigid body translation, and the XOY plane rigid body rotation.

[0084] For the tensile deformation of the isotropic square element, let its node loads be F X1 , F X2 , F X3 , F X4 , according to the force balance condition:

[0085]

[0086] The solution is:

[0087]

[0088] Let F X4 If it is unity, the nodal displacement basis vector d1 of the isotropic square element under X-axis tensile deformation can be obtained as:

[0089] d1=(10-10-1010) T ;

[0090] Then the unit energy q1 corresponding to the X-axial tensile deformation of the isotropic square element can be calculated as follows:

[0091]

[0092] Where: B is the strain matrix of the element; D t is the elastic matrix of the element:

[0093]

[0094]

[0095] Where: E t and μ t are the elastic modulus and Poisson's ratio of the isotropic square element in tension, respectively. From this, the unit energy q1 corresponding to the X-axis tensile deformation of the isotropic square element can be obtained as follows:

[0096]

[0097] Then the X-axis tensile deformation energy basis vector u1 of the anisotropic square element is obtained as:

[0098]

[0099] Elastic matrix D of anisotropic square element under compression c for:

[0100]

[0101] Where: E c and μ c are the elastic modulus and Poisson’s ratio of the isotropic square element under compression; let F X4 = -1, from which the x-axis compression deformation energy basis vector u2 of the anisotropic square element can be obtained as:

[0102]

[0103] According to the force balance, the displacement basis vector d3 of the node of the isotropic square element in the Y-axis tensile deformation can be obtained:

[0104] d3=(01010-10-1) T ;

[0105] Then the unit energy q3 corresponding to the Y-axis tensile deformation of the isotropic square element is:

[0106]

[0107] Then the Y-axis tensile deformation energy basis vector u3 of the anisotropic square element is obtained as:

[0108]

[0109] Similarly, the Y-axis compression deformation energy basis vector u4 of the anisotropic square element can be obtained as follows:

[0110]

[0111] Similarly, the remaining deformation energy basis vectors and displacement basis vectors u5-u of the isotropic square element can be obtained 10 As follows: u5 is the x-axis bending deformation energy basis vector of the anisotropic square element:

[0112]

[0113] u6 is the basis vector of the Y-axis bending deformation energy of the isotropic square element:

[0114]

[0115] u7 is the basis vector of the shear deformation energy of the isotropic square element in the XOY plane:

[0116]

[0117] u8 is the rigid body translation basis vector of the isotropic square unit along the X axis:

[0118] u8=(10101010) T ;

[0119] u9 is the Y-axis rigid body translation basis vector of the isotropic square unit:

[0120] u9=(01010101) T ;

[0121] u 10 The XOY plane rigid body rotation basis vectors of the isotropic square unit are:

[0122] u 10 =(-11-1-11-111) T ;

[0123] The node coordinate vector of the isotropic square element in the above plane rectangular coordinate system is u p ,

[0124] u p =(x1y1x2y2x3y3x4y4),

[0125] The node coordinate vector of the isotropic square element after any displacement and deformation under any load condition is u q ,

[0126] u q =(x1'y1'x'2y'2x'3y'3x'4y'4),

[0127] by u q -u p The node displacement vector u of the anisotropic square element after any displacement and deformation under any load condition can be obtained: r ,

[0128] u r =(x1'-x1y1'-y1x'2-x2y'2-y2x'3-x3y'3-y3x'4-x4y'4-y4).

[0129] The node coordinate displacement vector u after any displacement and deformation of any isotropic square element under any load condition is obtained r It is expressed as a linear combination of the basic deformation energy of the isotropic square element and the basic displacement basis vector, that is:

[0130]

[0131] Where: u kis the kth basic deformation energy or basic displacement basis vector, α k , are the projection coefficients of the basic deformation energy and basic displacement of the isotropic square unit, that is, α1 is the projection coefficient of the tensile deformation energy of the isotropic square unit in the X-axis direction, α2 is the projection coefficient of the compressive deformation energy of the X-axis direction, α3 is the projection coefficient of the tensile deformation energy of the Y-axis direction, α4 is the projection coefficient of the compressive deformation energy of the Y-axis direction, α5 is the projection coefficient of the bending deformation energy of the X-axis direction, α6 is the projection coefficient of the bending deformation energy of the Y-axis direction, α7 is the projection coefficient of the shear deformation energy, α8 is the projection coefficient of the rigid body translation in the X-axis direction, α9 is the projection coefficient of the rigid body translation in the Y-axis direction, α 10 is the rigid body rotation projection coefficient.

[0132] The decomposition results are separated into rigid and flexible parts, i.e. the basic displacement projection coefficient α8-α is ignored. 10 In order to avoid the influence of the above-mentioned influence, only the absolute values of the basic deformation energy projection coefficients α1-α7 in the above-mentioned component information are compared. The deformation corresponding to the maximum absolute value of deformation energy is determined as the main deformation of the isotropic square element. Similarly, the second largest one is determined as the secondary deformation of the isotropic square element. Then, the proportion of arbitrary deformation under arbitrary load conditions of the isotropic plane structure p can be calculated. k , to achieve quantitative analysis of structural deformation performance:

[0133]

[0134] Where: n k is the number of orthotropic rectangular elements that undergo primary or secondary deformation, n t is the total number of orthotropic rectangular elements.

[0135] Analysis of rigid body rotation displacement error

[0136] Since rotational displacement is a nonlinear displacement, errors will occur during linear decomposition. That is, the unit rotational displacement vector not only has projection coefficients on the rigid body rotation basis vectors, but may also have projection coefficients on other basic deformation energies and rigid body displacement basis vectors. Therefore, it is necessary to analyze and calculate the errors caused by the rigid body rotational displacement to determine whether they affect the calculation accuracy.

[0137] like Figure 13 As shown in the figure, let the side length of the four-node isotropic square element m be 2l, and rotate the element counterclockwise around the centroid by an angle θ. The coordinate displacement vectors of the four nodes of the element are:

[0138]

[0139] The nodal displacement vectors when the unit undergoes rigid body rotation displacement are linearly expressed using basic deformation energy and basic displacement basis vectors. The unit's rotation displacement vector is projected only on the rigid body rotation basis vector, the X-axis tensile deformation energy basis vector, and the Y-axis tensile deformation energy basis vector. The projection coefficients on other basic deformation energies and basic displacement basis vectors are 0. Therefore, the solutions to the 10 constraint equations obtained by projecting the unit's rigid body rotation coordinate displacement vector are as follows:

[0140]

[0141] For α1, α3, α 10 Perform Taylor expansion at θ = 0:

[0142]

[0143] It can be seen that when the unit undergoes rigid body rotation, there are still projection coefficients on the tensile deformation energy basis vectors in the X-axis and Y-axis directions, resulting in a certain error. However, when θ approaches 0, α1 and α3 are α 10 Therefore, when small deformation occurs, the error caused by the rotation displacement of the rigid body can be ignored, that is, the constructed deformation energy decomposition basis matrix has sufficient accuracy.

[0144] Implementation Cases

[0145] like Figure 14 As shown, taking the isotropic simply supported beam as an example, the beam length is 2.25m, the beam height is 0.45m, the beam thickness is 0.05m, and E t =15GPa, E c =30GPa,μ t =0.1, μ c =0.2, density is 2700kg / m 3 .

[0146] Apply three-point loading to the beam. Figure 14 The nodal displacement vectors of the isotropic square elements 1-3 are:

[0147] u r1 =(2.35E-01 -8.20E-01 2.26E-01 -8.09E-01 2.15E-01 -8.09E-01 2.26E-01 -8.20E-01)×mm;

[0148] u r2 =(2.60E-01 -8.20E-01 2.62E-01 -8.10E-01 2.50E-01 -8.10E-01 2.51E-01 -8.20E-01)×mm;

[0149] u r3 =(2.79E-01 -3.27E-01 2.80E-01 -2.76E-01 2.34E-01 -2.76E-01 2.34E-01 -3.27E-01)×mm.

[0150] The deformation energy decomposition results of the anisotropic square elements No. 1-3 are shown in Table 1-3.

[0151] Table 1 Decomposition results of deformation energy of isotropic square element No. 1 (unit: J)

[0152]

[0153]

[0154] Table 2 Decomposition results of deformation energy of isotropic square element No. 2 (unit: J)

[0155] X-axial tensile deformation energy X-axial compression deformation energy Y-axis tensile deformation energy Y-axis compression deformation energy 0 1.54E-02 0 1.07E-02 X-axial bending deformation energy Y-axis bending deformation energy Shear deformation energy -1.05E+00 2.69E-03 2.10E-03

[0156] Table 3 Decomposition results of deformation energy of isotropic square element No. 3 (unit: J)

[0157] X-axial tensile deformation energy X-axial compression deformation energy Y-axis tensile deformation energy Y-axis compression deformation energy 0 4.50E-01 1.91E-02 0 X-axial bending deformation energy Y-axis bending deformation energy Shear deformation energy -5.08E-02 7.15E-03 -9.07E+00

[0158] We can further obtain the basic deformation ratios p of the simply supported beam under three-point loading. k As shown in Table 4.

[0159] Table 4 Basic deformation ratios of isotropic simply supported beams

[0160] X-axis tensile deformation / % X axial compression deformation / % Y-axis tensile deformation / % Y-axis compression deformation / % 33.58 32.10 0 3.46 X-axis bending deformation / % Y-axis bending deformation / % Shear deformation / % 5.19 0.49 25.19

[0161] It can be seen from Table 1 that the X-axial tensile deformation energy of unit 1 is the largest, which is 7.41E+01J, that is, the X-axial tensile deformation is the main deformation in the area where unit 1 is located; the X-axial bending deformation energy is the second largest, that is, the Y-axial tensile deformation is the secondary deformation in the area where unit 1 is located.

[0162] Similarly, it can be seen that the area where unit 2 is located has X-axial bending as the main deformation, and X-axial compression as the secondary deformation; the area where unit 3 is located has shear as the main deformation, and X-axial compression and as the secondary deformation.

[0163] As can be seen from Table 4, under the action of three-point loading, the main deformation of the simply supported beam is the X-axial tension and X-axial compression deformation, which are close in proportion, 33.58% and 32.10% respectively; while shear is the secondary deformation, accounting for 25.19%.

[0164] The embodiments described above are only preferred embodiments of the present invention and are only used to explain the present invention, not to limit the scope of implementation of the present invention. For those skilled in the art, it is of course possible to easily make other implementation methods by replacing or changing the technical content disclosed in this specification. Therefore, all changes and improvements made on the principles of the present invention should be included in the scope of the patent application of the present invention.

Claims

1. A method for decomposing deformation energy of a planar structure based on isotropic square units, characterized in that: The method comprises the following steps: step 1: constructing a plane deformation of an isotropic square unit in a plane rectangular coordinate system, and obtaining a basic deformation energy and a basic displacement basis vector of the isotropic square unit based on mathematical orthogonality and mechanical equilibrium and considering the material properties of the unit; the isotropic square unit has four nodes, and the coordinates of the four nodes in the X and Y directions are x1, y1, x2, y2, x3, y3, x4, and y4, the side length of the isotropic square unit is 21, and the thickness is b; the plane deformation of the isotropic square unit is a superposition combination of X-axial tensile and compressive deformation, Y-axial tensile and compressive deformation, X-axial bending deformation, Y-axial bending deformation, shear deformation in the XOY plane, X-axial rigid body translation, Y-axial rigid body translation, and rigid body rotation in the XOY plane; The basis vectors of the basic deformation energy and basic displacement of the isotropic square unit are as follows: u1 is the basis vector of the X-axis tensile deformation energy of the isotropic square element: u2 is the basis vector of the X-axis compression deformation energy of the isotropic square element: u3 is the basis vector of the Y-axis tensile deformation energy of the isotropic square element: u4 is the basis vector of the Y-axis compression deformation energy of the isotropic square element: u5 is the basis vector of the bending deformation energy of the isotropic square element in the X-axis direction: u6 is the basis vector of the Y-axis bending deformation energy of the isotropic square element: u7 is the basis vector of the shear deformation energy of the isotropic square element in the XOY plane: u8 is the rigid body translation basis vector of the isotropic square unit along the X axis: u8＝(1 0 1 0 1 0 1 0) T ； u9 is the Y-axis rigid body translation basis vector of the isotropic square unit: u9＝(0 1 0 1 0 1 0 1) T ； u 10 The XOY plane rigid body rotation basis vectors of the isotropic square unit are: u 10 ＝(-1 1-1-1 1-1 1 1) T ； Where: E t and E c are the elastic moduli of the isotropic square element in tension and compression, μ t and μ c are the Poisson's ratios of the isotropic square element in tension and compression, respectively; Step 2: Establish a plane structure model, divide the plane structure using isotropic square elements and perform finite element solution to obtain the node coordinate values of the isotropic square elements in the plane rectangular coordinate system and the node coordinate values of the isotropic square elements after arbitrary displacement and deformation under any load condition. Then, obtain the node displacement vectors of the isotropic square elements after arbitrary displacement and deformation under any load condition. Step 3: Express the node displacement vector of the isotropic square element after any displacement and deformation under any load condition as a linear combination of the basic deformation energy and basic displacement basis vector of the isotropic square element in step 1, and obtain the basic deformation energy and basic displacement projection coefficient of the isotropic square element after any displacement and deformation under any load condition; Step 4: Based on the basic deformation energy projection coefficient of the isotropic square element after any displacement or deformation under any load condition, the basic deformation information of the isotropic square element under any load condition is obtained, and the main deformation and secondary deformation of the isotropic square element under any load condition are identified, thereby realizing the deformation energy decomposition and performance quantitative analysis of the isotropic planar structure.

2. The method for decomposing deformation energy of a planar structure based on isotropic square units according to claim 1, characterized in that: The node coordinate vector of the isotropic square element in the plane rectangular coordinate system is u p , u p =(x1 y1 x2 y2 x3 y3 x4 y4), The node coordinate vector of the anisotropic square element after any displacement and deformation under any load condition is u q , u q =(x1'y1'x'2y'2x'3y'3x'4y'4), by u q -u p The node displacement vector u of the anisotropic square element after any displacement and deformation under any load condition can be obtained: r , u r =(x1'-x1 y1'-y1 x'2-x2 y'2-y2 x'3-x3 y'3-y3 x'4-x4 y'4-y4)。 3. The method for decomposing deformation energy of a planar structure based on isotropic square units according to claim 2, characterized in that: Step 3 is as follows: The node coordinate displacement vector u after any displacement and deformation of any isotropic square element under any load condition is obtained r It is expressed as a linear combination of the basic deformation energy of the isotropic square element and the basic displacement basis vector, that is: Where: u k is the kth basic deformation energy or basic displacement basis vector, α k , are the projection coefficients of the basic deformation energy and basic displacement of the isotropic square unit, that is, α1 is the projection coefficient of the tensile deformation energy of the isotropic square unit in the X-axis direction, α2 is the projection coefficient of the compressive deformation energy of the X-axis direction, α3 is the projection coefficient of the tensile deformation energy of the Y-axis direction, α4 is the projection coefficient of the compressive deformation energy of the Y-axis direction, α5 is the projection coefficient of the bending deformation energy of the X-axis direction, α6 is the projection coefficient of the bending deformation energy of the Y-axis direction, α7 is the projection coefficient of the shear deformation energy, α8 is the projection coefficient of the rigid body translation in the X-axis direction, α9 is the projection coefficient of the rigid body translation in the Y-axis direction, α 10 is the rigid body rotation projection coefficient.

4. The method for decomposing deformation energy of a planar structure based on isotropic square units according to claim 1 or 3, characterized in that: The step 4 specifically includes: The decomposition results are separated into rigid and flexible parts, i.e. the basic displacement projection coefficient α8-α is ignored. 10 In order to avoid the influence of the above factors, only the absolute values of the basic deformation energy projection coefficients α1-α7 in the component information are compared. The deformation corresponding to the maximum absolute value of deformation energy is determined as the main deformation of the isotropic square element. Similarly, the second largest one is determined as the secondary deformation of the isotropic square element. Then, the proportion of arbitrary deformation under any load condition of the isotropic plane structure p can be calculated. k , to achieve quantitative analysis of structural deformation performance: Where: n k is the number of orthotropic rectangular elements that undergo primary or secondary deformation, n t is the total number of orthotropic rectangular elements.

Images