CN113297767A - Deformation decomposition method for any triangular prism unit in space structure - Google Patents

Abstract

The invention belongs to the technical field of mechanical analysis, and relates to a deformation decomposition method of any triangular prism unit in a space structure, which comprises the following steps: constructing basic displacement and deformation basis vectors of any triangular prism unit aiming at any triangular prism unit under a space rectangular coordinate system, and further obtaining a complete orthogonal mechanics basis matrix; establishing a space structure model, and dividing the space structure model by adopting triangular prism units to obtain node coordinate displacement vectors of the triangular prism units; projecting the node coordinate displacement vector of the triangular prism unit onto a complete orthogonal mechanics basis matrix to obtain the basic displacement and deformation projection coefficient vector of the triangular prism unit; and judging the primary deformation and the secondary deformation of the triangular prism unit according to the size of the projection coefficient in the projection coefficient vector, so that the deformation decomposition and the deformation identification of any space structure can be realized. The method can decompose any triangular prism unit without the limitation of unit division size and direction, thereby greatly reducing the calculation workload.

Description

Deformation decomposition method for any triangular prism unit in space structure

Technical Field

The invention belongs to the technical field of mechanical analysis, and relates to a deformation decomposition method of any triangular prism unit in a space structure.

Background

The shape of the triangular prism unit in the rectangular spatial coordinate system can change along with the height of the prism and the change of the internal angle of the two triangular faces, so that the flexibility and variability of the triangular prism unit are often used for discrete spatial structures. Compared with a regular unit, the triangular prism unit can finely divide a spatial irregular area without being limited by boundary conditions. Therefore, finite element analysis of any spatial structure can be realized by adopting the triangular prism units.

At present, a common finite element strain analysis method can realize the analysis of the deformation performance of a space structure on a microscopic level, but the engineering structure design not only needs to consider the information of basic macroscopic deformation according to microscopic deformation information. According to the linear superposition principle, the macroscopic comprehensive deformation of the space structure can be obtained by combining and superposing the basic macroscopic deformations of the discrete triangular prism units, so that a mathematical mechanics theory for quantitatively identifying the basic macroscopic deformations is constructed, a space structure deformation decomposition method based on the triangular prism units is formed, the method has important theoretical significance and engineering value, and the engineering structure design can be effectively guided.

Disclosure of Invention

The invention aims to provide a deformation decomposition method of any triangular prism unit in a space structure, which can identify main basic deformation and secondary basic deformation of the triangular prism unit, thereby more accurately showing the area occupied by each basic deformation after the space structure is stressed and reflecting the correctness and superiority of the deformation decomposition method of any triangular prism unit in a rectangular space coordinate system.

In order to achieve the purpose, the invention adopts the following technical scheme:

the invention provides a deformation decomposition method of any triangular prism unit in a space structure, which comprises the following steps:

step 1: constructing basic displacement and deformation basis vectors of any triangular prism unit according to the geometric characteristics, stress balance and orthogonal theoretical conditions of any triangular prism unit under a space rectangular coordinate system, and further obtaining a complete orthogonal mechanics basis matrix P;

step 2: establishing a space structure model under a space rectangular coordinate system, dividing the space structure model by adopting triangular prism units to obtain node coordinate values of the triangular prism units in the space structure model and node coordinate values of the triangular prism units after being subjected to any load working condition, and further obtaining node coordinate displacement vectors s of the triangular prism units;

and step 3: projecting the node coordinate displacement vector s of the triangular prism unit onto a complete orthogonal mechanics basis matrix P to obtain a basic displacement and deformation projection coefficient vector r of the triangular prism unit;

and 4, step 4: and judging the primary deformation and the secondary deformation of the triangular prism unit according to the size of the projection coefficient in the projection coefficient vector r, so that the deformation decomposition and the deformation identification of any space structure can be realized.

Preferably, the vertexes of the triangular prism unit are marked as node 1, node 2, node 3, node 4, node 5 and node 6, and the spatial comprehensive deformation of the triangular prism unit is formed by stacking 18 basic displacements and deformations of X-axis rigid body linear displacement, Y-axis rigid body linear displacement, Z-axis rigid body linear displacement, X-axis tension-compression deformation, Y-axis tension-compression deformation, Z-axis tension-compression deformation, XOZ-plane X-direction bending deformation, YOZ-plane Y-direction bending deformation, XOZ-plane Z-direction bending deformation, XOY-plane shear deformation, XOZ-plane shear deformation, YOZ-plane shear deformation, XOY-plane reverse shear deformation, XOY-plane torsion deformation, XOY-plane rigid body rotational displacement, XOZ-plane rigid body rotational displacement and YOZ-plane rigid body rotational displacement.

Preferably, the step 1 specifically comprises the following steps:

step 1.1: aiming at any triangular prism unit in a space rectangular coordinate system, 18 basic displacement and deformation basis vectors p of the any triangular prism unit are constructed according to the geometric characteristics, the stress balance and the orthogonal theoretical conditions of the any triangular prism unit₁～p₁₈The method comprises the following steps:

p₁is XAxial rigid body linear displacement base vector:

p₁＝[0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0]^T；

p₂is a Y axial rigid body linear displacement base vector:

p₂＝[0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0]^T；

p₃is a Z-axis rigid body linear displacement base vector:

p₃＝[0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082]^T；

p₄is an X axial tension-compression deformation base vector:

p₄＝α[d-f,0,0,-b+f,0,0,b-d,0,0,d-f,0,0,-b+f,0,0,b-d,0,0]^T；

p₅the Y axial tension-compression deformation base vector is as follows:

p₅＝β[0,c-e,0,0,-a+e,0,0,a-c,0,0,c-e,0,0,-a+e,0,0,a-c,0]^T；

p₆the Z-axis tension-compression deformation base vector is as follows:

p₆＝χ[0,0,de-cf,0,0,af-be,0,0,bc-ad,0,0,cf-de,0,0,be-af,0,0,ad-bc]^T；

p₇is the XOZ plane X-direction bending deformation base vector:

p₇＝α[-d+f,0,0,b-f,0,0,d-b,0,0,d-f,0,0,-b+f,0,0,b-d,0,0]^T；

p₈the base vector of the Y-direction bending deformation of the YOZ plane is:

p₈＝β[0,-c+e,0,0,a-e,0,0,-a+c,0,0,c-e,0,0,-a+e,0,0,a-c,0]^T；

p₉is the XOZ plane Z-direction bending deformation base vector:

p₁₀is a Z-direction bending deformation base vector of a YOZ plane:

p₁₁is XOY plane shear deformation basis vector:

p₁₂is XOZ plane shear deformation basis vector:

p₁₃is a base vector of YOZ plane shear deformation:

p₁₄is XOY plane reverse shear deformation basis vector:

p₁₅is XOY plane torsional deformation base vector:

p₁₆as XOY plane rigid body rotation displacement basis vector:

p₁₇is XOZ plane rigid body rotation displacement base vector:

p₁₈is a rigid body rotation displacement base vector of a YOZ plane:

wherein:

u＝a²+c²+e²-ac-ae-ce；v＝-b²-d²-f²+bd+bf+df；w＝2(bc-ad+af-be-cf+de)；

a is the difference value of the node 1 or the node 4 in the triangular prism unit and the centroid O' of the triangular prism unit in the X axial direction;

b is the difference value of the Y-axis direction of the node 1 or the node 4 in the triangular prism unit and the centroid O' of the triangular prism unit;

c is the difference value of the X axial direction of the node 2 or the node 5 in the triangular prism unit and the centroid O' of the triangular prism unit;

d is the difference value of the Y axial direction of the node 2 or the node 5 in the triangular prism unit and the centroid O' of the triangular prism unit;

e is the difference value of the X axial direction of the node 3 or the node 6 in the triangular prism unit and the centroid O' of the triangular prism unit;

f is the difference value of the Y axial direction of the node 3 or the node 6 in the triangular prism unit and the centroid O' of the triangular prism unit;

l is the height of the triangular prism unit in the Z-axis direction;

step 1.2: 18 basic displacement and deformation basis vectors p₁～p₁₈Constructing a complete orthogonal mechanical basis matrix P:

P＝[p₁,p₂,p₃,p₄,p₅,p₆,p₇,p₈,p₉,p₁₀,p₁₁,p₁₂,p₁₃,p₁₄,p₁₅,p₁₆,p₁₇,p₁₈]；

preferably, step 2 specifically comprises the following steps:

step 2.1: establishing a space structure model under a space rectangular coordinate system, and dividing the space structure model by adopting triangular prism units, so that the node coordinate value d of the ith triangular prism unit in the dispersed space structure model_i：

d_i＝[x_i-1,y_i-1,z_i-1,x_i-2,y_i-2,z_i-2,x_i-3,y_i-3,z_i-3,x_i-4,y_i-4,z_i-4,x_i-5,y_i-5,z_i-5,x_i-6,y_i-6,z_i-6]；

Wherein:

d_ithe coordinates of the node of the ith triangular prism unit in the spatial structure model are (i ═ 1,2,3,4, … …, m);

x_i-jcoordinate values representing the node j in the ith triangular prism unit in the X axial direction;

y_i-jcoordinate values representing a node j in the ith triangular prism unit in the Y-axis direction;

z_i-jcoordinate values (j ═ 1,2,3,4,5,6) in the Z-axis direction indicating the node j in the ith triangular prism unit;

step 2.2: d 'is a node coordinate value of the ith triangular prism unit in the space structure model after being subjected to any displacement and deformation under any load working condition'_i：

d'_i＝[x'_i-1,y'_i-1,z'_i-1,x'_i-2,y'_i-2,z'_i-2,x'_i-3,y'_i-3,z'_i-3,x'_i-4,y'_i-4,z'_i-4,x'_i-5,y'_i-5,z'_i-5,x'_i-6,y'_i-6,z'_i-6]；

Wherein:

d'_ithe node coordinates of the ith triangular prism unit in the space structure model after being subjected to any displacement and deformation under any load working condition are shown, (i is 1,2,3,4, … …, m);

x'_i-jthe coordinate value of the ith triangular prism unit after the node j generates any displacement and deformation in the X axial direction under any load working condition is shown;

y'_i-jthe coordinate value of the ith triangular prism unit after the node j generates any displacement and deformation in the Y-axis direction under any load working condition is shown;

z'_i-jcoordinate values (j is 1,2,3,4,5 and 6) which indicate that the ith triangular prism unit generates any displacement and deformation in the Z-axis direction under any load working condition;

step 2.3: d'_iMinus d_iObtaining the node coordinate displacement vector s of the ith triangular prism unit_i：

Preferably, the step 3 specifically includes the following steps:

step 3.1: the node coordinate displacement vector s of the ith triangular prism unit_iProjected to a corresponding complete orthonormal mechanical basis matrix P_iTo obtain:

s_i＝r_i·P_i ^T；

step 3.2: the above formula is converted to obtain the basic displacement and deformation projection coefficient vector r of the ith triangular prism unit in the space structure model_i：

r_i＝s_i·(P_i ^T)^-1＝s_i·P_i，

Wherein s is_iThe node coordinate displacement vector of the ith triangular prism unit in the spatial structure model is obtained; p_iIs a complete orthonormal mechanical basis matrix, P, of the ith triangular prism unit in the spatial structure model_i ^TIs P_iThe transposed matrix of (P)_i ^T)^-1Is P_i ^TThe inverse matrix of (d); r is_iIs a complete orthonormal mechanical base matrix P of the ith triangular prism unit in the space structure model_iA corresponding projection coefficient vector;

r_i＝[r_i-1,r_i-2,r_i-3,r_i-4,r_i-5,r_i-6,r_i-7,r_i-8,r_i-9,r_i-10,r_i-11,r_i-12,r_i-13,r_i-14,r_i-15,r_i-16,r_i-17,r_i-18]；

wherein:

p_i-kthe k-th basic displacement and deformation basis vector of the ith triangular prism unit in the space structure model (k is 1,2,3, … …, 18);

r_i-kis a complete orthonormal mechanical base matrix P of the ith triangular prism unit in the space structure model_iCorresponding projection coefficient vector r_iThe kth basic displacement and deformation projection coefficients (k ═ 1,2,3, … …, 18).

Preferably, the step 4 specifically includes:

complete orthonormal mechanics basis matrix P for ith triangular prism unit in space structure model_iCorresponding projection coefficient vector r_iMiddle 12 basic deformation projection coefficients r_i-4～r_i-15The absolute value of the three-dimensional triangular prism unit is compared, and the basic deformation corresponding to the projection coefficient with the maximum absolute value is judged as the main deformation of the triangular prism unit; determining the basic deformation corresponding to the projection coefficient with the second largest absolute value as the secondary deformation of the triangular prism units, and obtaining the basic deformation component information of all triangular prism units in the space structure according to the secondary deformation, thereby finally realizing the deformation decomposition and the deformation identification of the space structure; wherein the projection coefficient r_i-4、r_i-5、r_i-6When the deformation is a positive value, the corresponding basic deformation is the tensile deformation of the ith triangular prism unit in the space structure model; projection coefficient r_i-4、r_i-5、r_i-6And when the deformation is a negative value, the corresponding basic deformation is the compression deformation of the ith triangular prism unit in the spatial structure model.

Compared with the prior art, the invention has the beneficial effects that:

according to the geometric characteristics, stress balance and orthogonal theoretical conditions of any triangular prism unit under a space rectangular coordinate system, basic displacement and deformation basis vectors of any triangular prism unit are constructed, and a complete orthogonal mechanics basis matrix is further constructed; after the spatial structure model is dispersed by the triangular prism units, compared with a finite element analysis method, the method can only obtain the analysis result of the comprehensive deformation of the triangular prism units, and can identify the primary basic deformation and the secondary basic deformation of the triangular prism units, so that the area of each basic deformation occupied by the spatial structure after stress is more accurately displayed, and the correctness and the superiority of any triangular prism unit deformation decomposition method under a spatial rectangular coordinate system can be reflected. In addition, the method can carry out deformation decomposition on any triangular prism unit without the limitation of unit division size and direction, can be also suitable for dividing large units in a space structure, and greatly reduces the calculation workload compared with the traditional finite element stress analysis method.

Drawings

Fig. 1 is a schematic flow chart of a method for decomposing deformation of any triangular prism unit in a spatial structure according to the present invention.

Fig. 2 is a schematic diagram of an arbitrary triangular prism unit in a rectangular spatial coordinate system.

Fig. 3 is a schematic diagram of deformation of any triangular prism unit after being stressed in a rectangular spatial coordinate system.

FIG. 4 is a schematic diagram of the X-axis rigid body linear displacement of any triangular prism unit in a rectangular spatial coordinate system.

FIG. 5 is a schematic diagram of the Y-axis rigid body linear displacement of any triangular prism unit in a rectangular spatial coordinate system.

FIG. 6 is a schematic diagram of the Z-axis rigid body linear displacement of any triangular prism unit in a rectangular spatial coordinate system.

Fig. 7 is a schematic diagram of X-axis tension-compression deformation of any triangular prism unit in a rectangular spatial coordinate system.

Fig. 8 is a schematic diagram of Y-axis tension-compression deformation of any triangular prism unit in a rectangular spatial coordinate system.

Fig. 9 is a schematic diagram of the Z-axis tension-compression deformation of any triangular prism unit in a rectangular spatial coordinate system.

Fig. 10 is a schematic diagram of the XOZ plane X-direction bending deformation of any triangular prism unit in a space rectangular coordinate system.

Fig. 11 is a schematic diagram of the YOZ plane Y-direction bending deformation of any triangular prism unit in a rectangular spatial coordinate system.

Fig. 12 is a schematic diagram of the XOZ plane Z-direction bending deformation of any triangular prism unit in a space rectangular coordinate system.

Fig. 13 is a schematic diagram of the YOZ plane Z-direction bending deformation of any triangular prism unit in a space rectangular coordinate system.

Fig. 14 is a schematic diagram of XOY plane shear deformation of an arbitrary triangular prism unit in a spatial rectangular coordinate system.

Fig. 15 is a schematic diagram of XOZ plane shear deformation of an arbitrary triangular prism unit in a spatial rectangular coordinate system.

Fig. 16 is a schematic diagram of the YOZ plane shear deformation of any triangular prism unit in a rectangular spatial coordinate system.

Fig. 17 is a schematic diagram of XOY plane reverse shear deformation of an arbitrary triangular prism unit in a spatial rectangular coordinate system.

Fig. 18 is a schematic diagram of XOY plane torsional deformation of an arbitrary triangular prism unit in a rectangular spatial coordinate system.

Fig. 19 is a schematic diagram of XOY plane rigid body rotational displacement of an arbitrary triangular prism unit in a rectangular spatial coordinate system.

Fig. 20 is a schematic diagram of XOZ plane rigid body rotational displacement of an arbitrary triangular prism unit in a rectangular spatial coordinate system.

Fig. 21 is a schematic diagram of the rotational displacement of the YOZ plane rigid body of an arbitrary triangular prism unit in a rectangular spatial coordinate system.

Fig. 22 is a schematic diagram of coordinate differences between each node of any triangular prism unit and the centroid O' in a rectangular spatial coordinate system.

Fig. 23 is a schematic diagram of distances from each node of any triangular prism unit to the centroid O' along the coordinate axis in the rectangular spatial coordinate system.

FIG. 24 is a schematic diagram of cell size and node coordinates of selected cells in rigid body rotational displacement error analysis according to the present invention.

Fig. 25 is a schematic diagram of a sloping plate structure according to an embodiment of the present invention.

Fig. 26 is a schematic diagram illustrating a triangular prism unit in a tilted plate structure according to an embodiment of the present invention.

FIG. 27 is a schematic diagram of finite element analysis of the upper and lower surfaces σ of the tilted plate structure according to the embodiment of the present invention_xStress cloud pictures.

FIG. 28 is a schematic diagram of finite element analysis of the upper and lower surfaces σ of the tilted plate structure according to the embodiment of the present invention_yStress cloud pictures.

FIG. 29 is a schematic diagram of finite element analysis of the upper and lower surfaces τ of a tilted plate structure according to an embodiment of the present invention_xyStress cloud pictures.

FIG. 30 is a schematic view of a second apertured plate in accordance with embodiments of the present invention.

FIG. 31 is a schematic view of a triangular prism unit in a second aperture plate according to an embodiment of the present invention.

Fig. 32 is a diagram illustrating a calculation result of deformation decomposition of a second aperture plate according to an embodiment of the present invention.

Detailed Description

The following examples are intended to illustrate the invention, but are not intended to limit the scope of the invention. Unless otherwise specified, the technical means used in the examples are conventional means well known to those skilled in the art. The test methods in the following examples are conventional methods unless otherwise specified.

Fig. 1 shows a flow chart of a method for decomposing the deformation of any triangular prism unit in the space structure according to the present invention. The schematic diagram of any triangular prism unit under a space rectangular coordinate system is shown in fig. 2, and the deformation schematic diagram of any triangular prism unit after being stressed is shown in fig. 3. Fig. 4 to 21 are schematic diagrams of basic displacement and deformation of any triangular prism unit, which are sequentially X-axis rigid body linear displacement, Y-axis rigid body linear displacement, Z-axis rigid body linear displacement, X-axis tension and compression deformation, Y-axis tension and compression deformation, Z-axis tension and compression deformation, XOZ plane X-direction bending deformation, YOZ plane Y-direction bending deformation, XOZ plane Z-direction bending deformation, YOZ plane Z-direction bending deformation, XOY plane shear deformation, XOZ plane shear deformation, YOZ plane shear deformation, XOY plane reverse shear deformation, XOY plane torsional deformation, XOY plane rigid body rotational displacement, XOZ plane rigid body rotational displacement, and YOZ plane rigid body rotational displacement.

A deformation decomposition method for any triangular prism unit in a space structure comprises the following steps:

step 1.1: aiming at any triangular prism unit in a space rectangular coordinate system, 15 mutually independent and orthogonal basic displacement and basic deformation basis vectors p of the any triangular prism unit are constructed according to the geometric characteristics, stress balance and orthogonal theoretical conditions of the any triangular prism unit₁～p₁₅Obtaining an XOY plane rigid body rotation displacement base vector, an XOZ plane rigid body rotation displacement base vector and a YOZ plane rigid body rotation displacement base vector of any triangular prism unit by a Schmidt orthogonalization method, which specifically comprises the following steps:

p₁is an X axial rigid bodyLinear displacement basis vector:

p₁＝[0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0]^T；

p₂is a Y axial rigid body linear displacement base vector:

p₂＝[0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0]^T；

p₃is a Z-axis rigid body linear displacement base vector:

p₃＝[0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082]^T；

p₄is an X axial tension-compression deformation base vector:

p₄＝α[d-f,0,0,-b+f,0,0,b-d,0,0,d-f,0,0,-b+f,0,0,b-d,0,0]^T；

p₅the Y axial tension-compression deformation base vector is as follows:

p₅＝β[0,c-e,0,0,-a+e,0,0,a-c,0,0,c-e,0,0,-a+e,0,0,a-c,0]^T；

p₆the Z-axis tension-compression deformation base vector is as follows:

p₆＝χ[0,0,de-cf,0,0,af-be,0,0,bc-ad,0,0,cf-de,0,0,be-af,0,0,ad-bc]^T；

p₇is the XOZ plane X-direction bending deformation base vector:

p₇＝α[-d+f,0,0,b-f,0,0,d-b,0,0,d-f,0,0,-b+f,0,0,b-d,0,0]^T；

p₈the base vector of the Y-direction bending deformation of the YOZ plane is:

p₈＝β[0,-c+e,0,0,a-e,0,0,-a+c,0,0,c-e,0,0,-a+e,0,0,a-c,0]^T；

p₉is the XOZ plane Z-direction bending deformation base vector:

p₁₀is a Z-direction bending deformation base vector of a YOZ plane:

p₁₁is XOY plane shear deformation basis vector:

p₁₂is XOZ plane shear deformation basis vector:

p₁₃is a base vector of YOZ plane shear deformation:

p₁₄is XOY plane reverse shear deformation basis vector:

p₁₅is XOY plane torsional deformation base vector:

the 15 groups of mutually independent and orthogonal base vectors are adopted to obtain an XOY plane rigid body rotation displacement base vector, an XOZ plane rigid body rotation displacement base vector and a YOZ plane rigid body rotation displacement base vector of any triangular prism unit by a Schmidt orthogonalization method.

p₁₆As XOY plane rigid body rotation displacement basis vector:

p₁₇is XOZ plane rigid body rotation displacement base vector:

p₁₈is a rigid body rotation displacement base vector of a YOZ plane:

wherein:

u＝a²+c²+e²-ac-ae-ce；v＝-b²-d²-f²+bd+bf+df；w＝2(bc-ad+af-be-cf+de)；

a is the difference value of the node 1 or the node 4 in the triangular prism unit and the centroid O' of the triangular prism unit in the X axial direction;

b is the difference value of the Y-axis direction of the node 1 or the node 4 in the triangular prism unit and the centroid O' of the triangular prism unit;

c is the difference value of the X axial direction of the node 2 or the node 5 in the triangular prism unit and the centroid O' of the triangular prism unit;

d is the difference value of the Y axial direction of the node 2 or the node 5 in the triangular prism unit and the centroid O' of the triangular prism unit;

e is the difference value of the X axial direction of the node 3 or the node 6 in the triangular prism unit and the centroid O' of the triangular prism unit;

f is the difference value of the Y axial direction of the node 3 or the node 6 in the triangular prism unit and the centroid O' of the triangular prism unit;

and l is the height of the triangular prism unit in the Z-axis direction.

As shown in FIG. 22, the coordinates of the centroid O' of the triangular prism unit in the X-axis direction, the Y-axis direction, and the Z-axis direction are X₀、y₀、z₀Then, the difference between each node and the centroid in the X-axis direction and the Y-axis direction is:

the Z-axis difference from each node to the centroid is respectively as follows: -l/2, l/2.

Step 2.1: establishing a space structure model under a space rectangular coordinate system, and dividing the space structure model by adopting triangular prism units, so that the node coordinate value d of the ith triangular prism unit in the dispersed space structure model_i：

d_i＝[x_i1,y_i1,z_i1,x_i2,y_i2,z_i2,x_i3,y_i3,z_i3,x_i4,y_i4,z_i4,x_i5,y_i5,z_i5,x_i6,y_i6,z_i6]；

Wherein:

d_ithe coordinates of the node of the ith triangular prism unit in the spatial structure model are (i ═ 1,2,3,4, … …, m);

x_ijcoordinate values representing the node j in the ith triangular prism unit in the X axial direction;

y_ijcoordinate values representing a node j in the ith triangular prism unit in the Y-axis direction;

z_ijcoordinate values in the Z-axis direction of a node j in the ith triangular prism unit, (j ═ 1,2,3,4,5, 6);

step 2.2: d 'is a node coordinate value of the ith triangular prism unit in the space structure model after being subjected to any displacement and deformation under any load working condition'_i：

d'_i＝[x'_i1,y'_i1,z'_i1,x'_i2,y'_i2,z'_i2,x'_i3,y'_i3,z'_i3,x'_i4,y'_i4,z'_i4,x'_i5,y'_i5,z'_i5,x'_i6,y'_i6,z'_i6]；

Wherein:

d'_ithe node coordinates of the ith triangular prism unit in the space structure model after being subjected to any displacement and deformation under any load working condition are shown, (i is 1,2,3,4, … …, m);

x'_ijthe coordinate value of the ith triangular prism unit after the node j generates any displacement and deformation in the X axial direction under any load working condition is shown;

y'_ijthe coordinate value of the ith triangular prism unit after the node j generates any displacement and deformation in the Y-axis direction under any load working condition is shown;

z'_ijcoordinate values which represent that the node j generates any displacement and deformation in the X axial direction under any load working condition when the ith triangular prism unit is subjected to any load working condition, (j is 1,2,3,4,5 and 6);

step 2.3: d'_iMinus d_iObtaining the node coordinate displacement vector s of the ith triangular prism unit_i：

Step 3.1: the node coordinate displacement vector s of the ith triangular prism unit_iProjected to a corresponding complete orthonormal mechanical basis matrix P_iTo obtain:

s_i＝r_i·P_i ^T；

step 3.2: the above formula is converted to obtain the basic displacement and deformation projection coefficient vector r of the ith triangular prism unit in the space structure model_i：

r_i＝s_i·(P_i ^T)^-1＝s_i·P_i；

Wherein s is_iThe node coordinate displacement vector of the ith triangular prism unit in the spatial structure model is obtained; p_iIs a complete orthonormal mechanical basis matrix, P, of the ith triangular prism unit in the spatial structure model_i ^TIs P_iThe transposed matrix of (P)_i ^T)^-1Is P_i ^TThe inverse matrix of (d); r is_iIs a complete orthonormal mechanical base matrix P of the ith triangular prism unit in the space structure model_iA corresponding projection coefficient vector;

r_i＝[r_i-1,r_i-2,r_i-3,r_i-4,r_i-5,r_i-6,r_i-7,r_i-8,r_i-9,r_i-10,r_i-11,r_i-12,r_i-13,r_i-14,r_i-15,r_i-16,r_i-17,r_i-18]；

wherein:

p_i-kthe k-th basic displacement and deformation basis vector of the ith triangular prism unit in the space structure model (k is 1,2,3, … …, 18);

r_i-kis a complete orthonormal mechanical base matrix P of the ith triangular prism unit in the space structure model_iCorresponding projection coefficient vector r_iThe k-th basic displacement and deformation projection coefficients ( k 1,2,3,… …,18), see in particular table 1.

And 4, step 4: complete orthonormal mechanics basis matrix P for ith triangular prism unit in space structure model_iCorresponding projection coefficient vector r_iMiddle 12 basic deformation projection coefficients r_i4～r_i15The absolute value of the three-dimensional triangular prism unit is compared, and the basic deformation corresponding to the projection coefficient with the maximum absolute value is judged as the main deformation of the triangular prism unit; determining the basic deformation corresponding to the projection coefficient with the second largest absolute value as the secondary deformation of the triangular prism units, and obtaining the basic deformation component information of all triangular prism units in the space structure according to the secondary deformation, thereby finally realizing the deformation decomposition and the deformation identification of the space structure; wherein the projection coefficient r_i4、r_i5、r_i6When the deformation is a positive value, the corresponding basic deformation is the tensile deformation of the ith triangular prism unit in the space structure model; projection coefficient r_i4、r_i5、r_i6And when the deformation is a negative value, the corresponding basic deformation is the compression deformation of the ith triangular prism unit in the spatial structure model.

TABLE 1 basic displacement and deformation of ith triangular prism unit and projection coefficient corresponding table

It should be noted that the X-axis rigid body linear displacement, the Y-axis rigid body linear displacement, the Z-axis rigid body linear displacement, the XOY plane rigid body rotational displacement, the XOZ plane rigid body rotational displacement, and the YOZ plane rigid body rotational displacement belong to rigid body displacements, and no stress strain is generated, so that the projection coefficients on the basis vectors thereof do not need to be considered, and only the projection coefficients on the other 12 basic deformation basis vectors in the complete orthogonal mechanics basis matrix need to be analyzed.

Rigid body rotational displacement error analysis

The rigid body rotational displacement is nonlinear displacement, and an error is generated when linear decomposition is carried out, namely, the base vector of the rigid body rotational displacement of the triangular prism unit has a projection coefficient on the base vector of the rigid body rotational displacement, and also has a projection coefficient on other displacement and deformation base vectors, and an error exists with a theoretical condition, so that error analysis is carried out on the numerical value of the projection coefficient of the rigid body rotational displacement of the triangular prism unit on other deformation and displacement base vectors, and whether the calculation precision is influenced by the rigid body rotational displacement is judged.

As shown in fig. 23, in any triangular prism unit in the rectangular spatial coordinate system, the distances from each node to the centroid O' in the X-axis direction, the Y-axis direction, and the Z-axis direction are calculated and are respectively expressed as: l_1x，l_1y，l_1z，l_2x，l_2y，l_2z，l_3x，l_3y，l_3z，l_4x，l_4y，l_4z，l_5x，l_5y，l_5z，l_6x，l_6y，l_6z. Obviously: l_1x＝l_4x＝|a|，l_1y＝l_4y＝|b|，l_2x＝l_5x＝|c|，l_2y＝l_5y＝|d|，l_3x＝l_6x＝|e|，l_3y＝l_6y＝|f|，l_1z＝l_2z＝l_3z＝l_4z＝l_5z＝l_6zL/2. When the triangular prism unit rotates clockwise around the centroid O' along three different planes (XOY plane, XOZ plane and YOZ plane) by an angle theta, error analysis is carried out on the rigid body rotation displacement of the triangular prism unit. Because the basic displacement and deformation base vector of any triangular prism unit can change along with the change of the node coordinates, error analysis can be carried out only by using parameter variables (such as a, b, c, d, e, f, l and theta) contained in the node coordinates, and then a general solution of the error analysis result of any triangular prism unit is obtained. However, the calculation process is very complicated due to more parameters, and the error analysis process cannot be clearly expressed due to the fact that the calculation process needs to be solved by a computer; to more clearly express the error analysis calculation process, a triangular prism unit is arbitrarily selected for error analysis, and the specific geometric parameters of the unit are shown in fig. 24. The node coordinates are respectively: node 1(5mm,5mm,5mm), node 2(7mm,5mm,5mm), node 3(6mm,6.732mm,5mm), node 4(5mm,5mm,7mm), node 5(7mm,5mm,7mm), node 6(6mm,6.732mm,7 mm); at this time: a-1 mm, b-0.5773 mm, c-1 mm, d-0.5773 mm, e-0, f-0.5773 mm1.1547mm，l＝2mm。

The unit obtains 18 basic displacement and deformation basis vectors p₁～p₁₈Comprises the following steps:

p₁＝(0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0)^T；

p₂＝(0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0)^T；

p₃＝(0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082)^T；

p₄＝(-0.5,0,0,0.5,0,0,0,0,0,-0.5,0,0,0.5,0,0,0,0,0)^T；

p₅＝(0,0.2887,0,0,0.2887,0,0,-0.5774,0,0,0.2887,0,0,0.2887,0,0,-0.5774,0)^T；

p₆＝(0,0,-0.4083,0,0,-0.4083,0,0,-0.4083,0,0,0.4083,0,0,0.4083,0,0,0.4083)^T；

p₇＝(-0.5,0,0,0.5,0,0,0,0,0,0.5,0,0,-0.5,0,0,0,0,0)^T；

p₈＝(0,0.2887,0,0,0.2887,0,0,-0.5774,0,0,-0.2887,0,0,-0.2887,0,0,0.5774,0)^T；

p₉＝(0,0,-0.5,0,0,0.5,0,0,0,0,0,0.5,0,0,-0.5,0,0,0)^T；

p₁₀＝(0,0,0.2887,0,0,0.2887,0,0,-0.5774,0,0,-0.2887,0,0,-0.2887,0,0,0.5774)^T；

the complete orthogonal mechanical base P is as follows:

P＝[p₁,p₂,p₃,p₄,p₅,p₆,p₇,p₈,p₉,p₁₀,p₁₁,p₁₂,p₁₃,p₁₄,p₁₅,p₁₆,p₁₇,p₁₈]。

(1) XOY plane rigid body rotation displacement error analysis

After the triangular prism unit rotates clockwise by an angle theta along the XOY plane, six node coordinate displacements s_XOYAs shown in table 2. Rewriting six-node coordinate displacement into vector form s_XOYThen the node coordinate displacement vector is:

s_XOY＝[s_1x,s_1y,s_1z,s_2x,s_2y,s_2z,s_3x,s_3y,s_3z,s_4x,s_4y,s_4z,s_5x,s_5y,s_5z,s_6x,s_6y,s_6z]；

wherein: s_jkRepresents the axial displacement of the jth node k; j is a number of bits of 1,2,3,4,5,6；k＝X、Y、Z。

TABLE 2 node coordinate Displacement generated by triangular prism Unit rotating clockwise theta along XOY plane

The node coordinate displacement vector s of the triangular prism unit_XOYProjecting the vector onto the complete orthonormal mechanical basis matrix P of the triangular prism unit to obtain a projection coefficient vector r_XOY：

r_XOY＝s_XOYP。

Projection coefficient vector r_XOYThe calculation results are shown in Table 3:

TABLE 3 calculation results of various displacement and deformation projection coefficients after the prism unit rotates clockwise theta along XOY plane

As can be seen from table 3, after the triangular prism unit is rotated clockwise by θ along the XOY plane, the projection coefficients exist only on the XOY plane rigid body rotational displacement basis vector and the X, Y axial tension-compression deformation basis vector, and the projection coefficients on the other basic displacement and deformation basis vectors are equal to 0.

Obviously, the projection coefficient r of the X-axis tension-compression deformation solved by the unit under the condition of small deformation (when theta is close to 0)₄And the projection coefficient r of Y-axis tension-compression deformation₅Projection coefficient r relative to XOY plane rigid body rotation displacement₁₆The order is infinitesimal, so the projection coefficient of the rotation displacement of the XOY plane rigid body on the X, Y axial tension-compression deformation basis vector is negligible, and the error is in the allowable range of the method.

(2) XOZ plane rigid body rotation displacement error analysis

After the triangular prism unit is rotated clockwise by an angle theta along the XOZ plane, six node coordinate displacements are shown in Table 4.

TABLE 4 nodal coordinate Displacement resulting from a clockwise rotation of the prismatic element by θ along the XOZ plane

Rewriting six-node coordinate displacement into vector form s_XOZThen the node coordinate displacement vector is:

s_XOZ＝[s_1x,s_1y,s_1z,s_2x,s_2y,s_2z,s_3x,s_3y,s_3z,s_4x,s_4y,s_4z,s_5x,s_5y,s_5z,s_6x,s_6y,s_6z]；

wherein: s_jkRepresents the axial displacement of the jth node k; j is 1,2,3,4,5, 6; k is X, Y, Z.

The node coordinate displacement vector s of the triangular prism unit_XOZProjecting the vector onto the complete orthonormal mechanical basis matrix P of the triangular prism unit to obtain a projection coefficient vector r_XOZ：

r_XOZ＝s_XOZP。

The results of the projection coefficient vector r calculation are shown in table 5:

as can be seen from table 5, after the triangular prism unit is rotated by θ clockwise along the XOZ plane, the projection coefficients exist only on the XOZ plane rigid body rotational displacement basis vector and the X, Z axial tension-compression deformation basis vector, and the projection coefficients on the other basic displacement and deformation basis vectors are equal to 0.

Obviously, the projection coefficient r of the X-axis tension-compression deformation solved by the unit under the condition of small deformation (when theta is close to 0)₄And projection coefficient r of Z-axis tension-compression deformation₆Projection coefficient r relative to XOZ plane rigid body rotation displacement₁₇The high order is infinitesimal, so the projection coefficient of the XOZ plane rigid body rotation displacement on the X, Z axial tension-compression deformation basis vector is negligible, and the error is within the allowable range of the method.

TABLE 5 calculation results of various displacement and deformation projection coefficients after column unit rotates clockwise theta along XOZ plane

(3) YOZ plane rigid body rotation displacement error analysis

After the triangular prism unit is rotated clockwise by an angle theta along the YOZ plane, six node coordinate displacements are shown in Table 6.

TABLE 6 node coordinate Displacement generated by the column Unit rotating clockwise theta along the YOZ plane

Rewriting six-node coordinate displacement into vector form s_YOZThen the node coordinate displacement vector is:

s_YOZ＝[s_1x,s_1y,s_1z,s_2x,s_2y,s_2z,s_3x,s_3y,s_3z,s_4x,s_4y,s_4z,s_5x,s_5y,s_5z,s_6x,s_6y,s_6z]，

wherein: s_jkRepresents the axial displacement of the jth node k; j is 1,2,3,4,5, 6; k is X, Y, Z.

The node coordinate displacement vector s of the triangular prism unit_YOZProjecting the vector onto the complete orthonormal mechanical basis matrix P of the triangular prism unit to obtain a projection coefficient vector r_YOZ：

r_YOZ＝s_YOZP。

Projection coefficient vector r_YOZThe calculation results are shown in Table 7.

TABLE 7 calculation results of various displacement and deformation projection coefficients after column unit rotates clockwise theta along YOZ plane

As can be seen from table 7, when the triangular prism unit is rotated clockwise by θ along the YOZ plane, the projection coefficients exist only on the YOZ plane rigid body rotational displacement basis vector and the Y, Z axial tension-compression deformation basis vector, and the projection coefficients on the other basic displacement and deformation basis vectors are equal to 0.

Obviously, the projection coefficient r of the Y-axis tension-compression deformation solved by the unit under the condition of small deformation (when theta is close to 0)₅And projection coefficient r of Z-axis tension-compression deformation₆Projection coefficient r of rigid body rotation displacement relative to YOZ plane₁₈The high order is infinitesimal, so the projection coefficient of the rotation displacement of the YOZ plane rigid body on the Y, Z axial tension-compression deformation basis vector is negligible, and the error is in the allowable range of the method.

The remaining triangular prism units at any position in the spatial rectangular coordinate system can perform rigid body rotation displacement error analysis in three directions according to the triangular prism units at the positions shown in fig. 24.

Example one

As shown in FIG. 25, four inclined plates each having a side length of 1200mm and a thickness of 60mm were provided, the oblique angle was 30 °, and the elastic modulus was 2.1X 10¹¹pa, poisson's ratio of 0.3. Constraint conditions are as follows: the two ends are simply supported and free. The inclined plate structure is divided by the triangular prism units, and then stress conditions of obtuse angles, acute angles and midspan positions of the inclined plate structure under self weight are analyzed. Selecting a triangular prism unit in the acute angle region, the obtuse angle region and the midspan region respectively, marking the unit number at the acute angle as the unit number 1, marking the unit number at the obtuse angle as the unit number 2, and marking the unit number at the midspan as the unit number 3, and performing deformation decomposition on the selected triangular prism unit as shown in fig. 26.

Obtaining the node coordinates d contained by No. 1-3 triangular prism units in the inclined plate structure by utilizing finite element analysis₁～d₃And node coordinate displacement vector s₁～s₃Finite element analysis result sigma of inclined plate structure_x、σ_y、τ_xyUpper and lower ofSurface stress clouds are shown in FIGS. 27-29.

Node coordinate d of No. 1-3 triangular prism unit in the embodiment₁～d₃: the calculation can be directly carried out according to the coordinate axes and the geometric relationship or the calculation and extraction can be carried out by utilizing finite element software.

Node coordinate d of No. 1-3 triangular prism unit in the embodiment₁～d₃And node coordinate displacement vector s₁～s₃Extracting by using finite element software;

acute angle 1 unit node coordinate d₁And node coordinate displacement vector s₁Respectively as follows:

d₁＝(50,86.6025,0,100,0,0,0,0,0,50,86.6025,60,100,0,60,0,0,60)；

s₁＝(0,0,0,-0.0565,0.0142,-0.0907,0,0,0,0,0,0,0.0565,-0.0142,-0.0907,0,0,0)。

no. 2 unit node coordinate d at obtuse angle₂And node coordinate displacement vector s₂Respectively as follows:

s₂＝(0.1386,-0.0460,-0.2784,0,0,0,0,0,0,-0.1386,0.0460,-0.27834,0,0,0,0,0,0)。

node coordinates d of unit No. 3 at midspan₃And node coordinate displacement vector s₃Respectively as follows:

according to node coordinates d₁～d₃Respectively obtaining complete orthogonal mechanical basis matrixes corresponding to 3 triangular prism units, and recording the complete orthogonal mechanical basis matrixes as P₁、P₂、P₃；

Unit number 1 gets18 basic displacement and deformation basis vectors p_1-1～p_1-18Comprises the following steps:

P₁＝[p_1-1,p_1-2,p_1-3,p_1-4,p_1-5,p_1-6,p_1-7,p_1-8,p_1-9,p_1-10,p_1-11,p_1-12,p_1-13,p_1-14,p_1-15,p_1-16,p_1-17,p_1-18]；

p_1-1＝[0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0]^T；

p_1-2＝[0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0]^T；

p_1-3＝[0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082]^T；

p_1-4＝[0,0,0,0.5000,0,0,-0.5000,0,0,0,0,0,0.5000,0,0,-0.5000,0,0]^T；

p_1-5＝[0,0.5774,0,0,-0.2887,0,0,-0.2887,0,0,0.5774,0,0,-0.2887,0,0,-0.2887,0]^T；

p_1-6＝[0,0,-0.4082,0,0,-0.4082,0,0,-0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082]^T；

p_1-7＝[0,0,0,0.5000,0,0,-0.5000,0,0,0,0,0,-0.5000,0,0,0.5000,0,0]^T；

p_1-8＝[0,-0.5774,0,0,0.2887,0,0,0.2887,0,0,0.5774,0,0,-0.2887,0,0,-0.2887,0]^T；

p_1-9＝[0,0,-0.5000,0,0,0,0,0,0.5000,0,0,0.5000,0,0,0,0,0,-0.5000]^T；

p_1-10＝[0,0,-0.1890,0,0,-0.3780,0,0,0.5670,0,0,0.1890,0,0,0.3780,0,0,-0.5670]^T；

p_1-12＝[0.3280,0,0,0.3280,0,-0.2961,0.3290,0,0.2961,-0.3280,0,0,-0.3280,0,-0.2961,-0.3280,0,0.2961]^T；

p_1-14＝[0.4082,0,0,-0.2041,0.3535,0,-0.2041,-0.3535,0,-0.4082,0,0,0.2041,-0.3535,0,0.2041,0.3535,0]^T；

p_1-15＝[-0.4082,0,0,0.2041,0.3535,0,0.2041,-0.3535,0,0.4082,0,0,-0.2041,-0.3535,0,-0.2041,0.3535,0]^T；

p_1-16＝[0.4082,0,0,-0.2041,-0.3535,0,-0.2041,0.3535,0,0.4082,0,0,-0.2041,-0.3535,0,-0.2041,0.3535,0]^T；

p_1-17＝[-0.2417,0,0,-0.2417,0,-0.4029,-0.2417,0,0.4029,0.2417,0,0,0.2417,0,-0.4029,0.2417,0,0.4029]^T；

18 basic displacement and deformation basis vectors p obtained by No. 2 unit_2-1～p_2-18Comprises the following steps:

p_2-1＝[0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0]^T；

p_2-2＝[0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0]^T；

p_2-3＝[0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082]^T；

p_2-4＝[0.5,0,0,-0.5,0,0,0,0,0,0.5,0,0,-0.5,0,0,0,0,0]^T；

p_2-5＝[0,-0.2887,0,0,-0.2887,0,0,0.5774,0,0,-0.2887,0,0,-0.2887,0,0,0.5774,0]^T；

p_2-6＝[0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,-0.4082,0,0,-0.4082,0,0,-0.4082]^T；

p_2-7＝[-0.5,0,0,0.5,0,0,0,0,0,0.5,0,0,-0.5,0,0,0,0,0]^T；

p_2-8＝[0,-0.2887,0,0,-0.2887,0,0,0.5774,0,0,0.2887,0,0,0.2887,0,0,-0.5774,0]^T；

p_2-9＝[0,0,-0.2887,0,0,-0.2887,0,0,0.5774,0,0,0.2887,0,0,0.2887,0,0,-0.5774]^T；

p_2-10＝[0,0,0.5,0,0,-0.5,0,0,0,0,0,-0.5,0,0,0.5,0,0,0]^T；

p_2-11＝[0.2041,-0.3535,0,0.2041,0.3535,0,-0.4082,0,0,0.2041,-0.3535,0,0.2041,0.3535,0,-0.4082,0,0]^T；

p_2-12＝[0.3290,0,0.2961,0.3290,0,-0.2961,0.3290,0,0,-0.3290,0,0.2961,-0.3290,0,-0.2961,-0.3290,0,0]^T；

p_2-14＝[0.2041,-0.3535,0,0.2041,0.3535,0,-0.4082,0,0,-0.2041,0.3535,0,-0.2041,-0.3535,0,0.4082,0,0]^T；

p_2-15＝[-0.2041,-0.3535,0,-0.2041,0.3535,0,0.4082,0,0,0.2041,0.3535,0,0.2041,-0.3535,0,-0.4082,0,0]^T；

p_2-16＝[0.2041,0.3535,0,0.2041,-0.3535,0,-0.4082,0,0,0.2041,0.3535,0,0.2041,-0.3535,0,-0.4082,0,0]^T；

p_2-17＝[-0.2417,0,0.4029,-0.2417,0,-0.4029,-0.2417,0,0,0.2417,0,0.4029,0.2417,0,-0.4029,0.2417,0,0]^T；

18 basic displacement and deformation basis vectors p obtained by No. 3 unit_3-1～p_3-18Comprises the following steps:

p_3-1＝[0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0]^T；

p_3-2＝[0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0]^T；

p_3-3＝[0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082]^T；

p_3-3＝[0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082]^T；

p_3-4＝[0,0,0,0.5,0,0,-0.5,0,0,0,0,0,0.5,0,0,-0.5,0,0]^T；

p_3-5＝[0,0.5774,0,0,-0.2887,0,0,-0.2887,0,0,0.5774,0,0,-0.2887,0,0,-0.2887,0]^T；

p_3-6＝[0,0,-0.4082,0,0,-0.4082,0,0,-0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082]^T；

p_3-7＝[0,0,0,0.5,0,0,-0.5,0,0,0,0,0,-0.5,0,0,0.5,0,0]^T；

p_3-8＝[0,-0.5774,0,0,0.2887,0,0,0.2887,0,0,0.5774,0,0,-0.2887,0,0,-0.2887,0]^T；

p_3-9＝[0,0,-0.5,0,0,0,0,0,0.5,0,0,0.5,0,0,0,0,0,-0.5]^T；

p_3-10＝[0,0,-0.1890,0,0,-0.3780,0,0,0.5670,0,0,0.1890,0,0,0.3780,0,0,-0.5670]^T；

p_3-12＝[0.3280,0,0,0.3280,0,-0.2961,0.3290,0,0.2961,-0.3290,0,0,-0.3290,0,-0.2961,-0.3290,0,0.2961]^T；

p_3-14＝[0.4082,0,0,-0.2041,0.3535,0,-0.2041,-0.3535,0,-0.4082,0,0,0.2041,-0.3535,0,0.2041,0.3535,0]^T；

p_3-15＝[-0.4082,0,0,0.2041,0.3535,00.2041,-0.3535,0,0.4082,0,0,-0.2041,-0.3535,0,-0.2041,0.3535,0]^T；

p_3-16＝[0.4082,0,0,-0.2041,-0.3535,0,-0.2041,0.3535,0,0.4082,0,0,-0.2041,-0.3535,0,-0.2041,0.3535,0]^T；

p_3-17＝[-0.2417,0,0,-0.2417,0,-0.4029,-0.2417,0,0.4029,0.2417,0,0,0.2417,0,-0.4029,0.2417,0,0.4029]^T；

displacement vector s of node coordinate₁、s₂、s₃Respectively projected to corresponding complete orthogonal mechanical basis matrixes P₁、P₂、P₃Then 3 sets of projection coefficient vectors r are obtained₁、r₂、r₃Thus, the proportion of each unit deformation is calculated, and the results are shown in tables 8-10.

From the finite element calculation result graphs 27-29, it can be seen that the acute angle, the obtuse angle, and the mid-span middle sigma_x、σ_y、τ_xyThe stress is obvious, and the finite element method cannot distinguish which deformation is the main; tables 8-10 show that the proportion of each basic deformation can be accurately obtained by the method of the present invention based on finite elements, and the acute angle, the obtuse angle and the mid-span are mainly bending deformation in the X direction of the XOZ plane, and the XOY is secondarily deformed by reverse shearing. Further verifying the reasonability and superiority of the method.

TABLE 8 proportion of the deformation projection coefficients of No. 1 unit at an acute angle

Projection coefficient r1	In proportion of	Projection coefficient r1	In proportion of
				X axial tension and compression r1-4	0.00％	Z-direction curvature r of YOZ plane1-10	0.00％
Y-axis tension and compression r1-5	0.00％	XOY plane shear r1-11	0.00％
				Z-axial tension and compression r1-6	0.00％	XOZ plane shear r1-12	11.02％
XOZ plane X-direction curvature r1-7	37.99％	YOZ plane shear r1-13	14.22％
				YOZ plane Y-direction curvature r1-8	5.77％	XOY plane reverse shear r1-14	22.57％
XOZ plane Z-direction bending r1-9	0.00％	XOY plane torsion r1-15	8.44％

TABLE 9 proportion of deformation projection coefficients of No. 2 unit at obtuse angle

Projection coefficient r2	In proportion of	Projection coefficient r2	In proportion of
				X axial tension and compression r2-4	0.00％	Z-direction curvature r of YOZ plane2-10	0.00％
Y-axis tension and compression r2-5	0.00％	XOY plane shear r2-11	0.00％
				Z-axial tension and compression r2-6	0.00％	XOZ plane shear r2-12	17.30％
XOZ plane X-direction curvature r2-7	33.52％	YOZ plane shear r2-13	15.35％
				YOZ plane Y-direction curvature r2-8	6.47％	XOY plane reverse shear r2-14	21.61％
XOZ plane Z-direction bending r2-9	0.00％	XOY plane torsion r2-15	5.76％

TABLE 10 proportion of deformation projection coefficients of No. 3 cells at first span

Projection coefficient r3	In proportion of	Projection coefficient r3	In proportion of
				X axial tension and compression r3-4	0.00％	Z-direction curvature r of YOZ plane3-10	0.00％
Y-axis tension and compression r3-5	0.00％	XOY plane shear r3-11	0.00％
				Z-axial tension and compression r3-6	0.00％	XOZ plane shear r3-12	13.73％
XOZ plane X-direction curvature r3-7	39.17％	YOZ plane shear r3-13	10.28％
				YOZ plane Y-direction curvature r3-8	5.79％	XOY plane reverse shear r3-14	30.78％
XOZ plane Z-direction bending r3-9	0.00％	XOY plane torsion r3-15	0.25％

Example two

As shown in fig. 30, a flat plate with four sides of 1200mm, a thickness of 60mm and a central opening is provided, and the radius of the hole is 200 mm; the modulus of elasticity of the apertured plate is 2.1X 10¹¹pa, poisson's ratio of 0.3; and analyzing the stress characteristics of the bearing under the self-weight load under the condition of simply supporting four sides.

Modeling by means of finite element software ANSYS, adopting triangular prism grids to freely divide, and respectively extracting node coding information of all triangular prism units and coordinate and displacement information of all nodes after loading and solving; and importing the two files into mathematic software MATLAB for further analysis, and performing reassembly on each unit information and node information to enter a deformation decomposition calculation process.

The perforated plate has good symmetry, only the area 1/4 at the right lower end needs to be selected for analysis, a triangular prism unit is randomly selected at a support and a corner for analyzing the stress form, the support is selected as the No. 1 unit, and the corner is selected as the No. 2 unit (as shown in FIG. 31); the results of the entire analysis of the apertured plate are shown in FIG. 32.

Node coordinates d of No. 1 and No. 2 triangular prism units in this embodiment₁、d₂And node coordinate displacement vector s₁、s₂Extracting by adopting finite element software, which respectively comprises the following steps:

s₁＝(0,0,0,0.0583,-0.0062,-0.0696,0,0,0,0,0,0,-0.0583,0.0062,-0.0696,0,0,0)；

according to node coordinates d₁Calculating a complete orthogonal mechanical basis matrix corresponding to the No. 1 triangular prism unit, and recording the complete orthogonal mechanical basis matrix as P₁(ii) a 18 basic displacement and deformation basis vectors p obtained by No. 1 unit_1-1～p_1-18Comprises the following steps:

P₁＝[p_1-1,p_1-2,p_1-3,p_1-4,p_1-5,p_1-6,p_1-7,p_1-8,p_1-9,p_1-10,p_1-11,p_1-12,p_1-13,p_1-14,p_1-15,p_1-16,p_1-17,p_1-18]；

p_1-1＝(0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0)^T；

p_1-2＝(0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0)^T；

p_1-3＝(0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082)^T；

p_1-4＝(-0.2739,0,0,0.5771,0,0,-0.3032,0,0,-0.2739,0,0,0.5771,0,0,-0.3032,0,0)^T；

p_1-5＝(0,0.5,0,0,0,0,0,-0.5,0,0,0.5,0,0,0,0,0,-0.5,0)^T；

p_1-6＝(0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,-0.4082,0,0,-0.4082,0,0,-0.4082)^T；

p_1-7＝(0.2739,0,0,-0.5771,0,0,0.3032,0,0,-0.2739,0,0,0.5771,0,0,-0.3032,0,0)^T；

p_1-8＝(0,0.5,0,0,0,0,0,-0.5,0,0,-0.5,0,0,0,0,0,0.5,0)^T；

p_1-9＝(0,0,0.3780,0,0,0.1890,0,0,-0.5670,0,0,-0.3780,0,0,-0.1890,0,0,0.5670)^T；

p_1-10＝(0,0,0.0163,0,0,0.4916,0,0,-0.5080,0,0,-0.0163,0,0,-0.4916,0,0,0.5080)^T；

displacement vector s of node coordinate₁Projected to the corresponding complete orthonormal mechanical basis matrix P₁Then the corresponding No. 1 triangular prism unit projection coefficient vector r can be obtained₁Thus, the ratio of each deformation was calculated, and the results are shown in Table 11.

According to node coordinates d₂Calculating a complete orthogonal mechanical basis matrix corresponding to No. 2 triangular prism unit, and recording the complete orthogonal mechanical basis matrix as P₂(ii) a 18 basic displacement and deformation basis vectors p obtained by No. 2 unit_2-1～p_2-18Comprises the following steps:

p_2-1＝(0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0)^T；

p_2-2＝(0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0)^T；

p_2-3＝(0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082)^T；

p_2-4＝(0.3332,0,0,0.2418,0,0,-0.5749,0,0,0.3332,0,0,0.2418,0,0,-0.5749,0,0)^T；

p_2-5＝(0,0.4884,0,0,-0.5109,0,0,0.02254,0,0,0.4884,0,0,-0.5109,0,0,0.02254,0)^T；

p_2-6＝(0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,-0.4082,0,0,-0.4082,0,0,-0.4082)^T；

p_2-7＝(-0.3332,0,0,-0.2418,0,0,0.5749,0,0,0.3332,0,0,0.2418,0,0,-0.5749,0,0)^T；

p_2-8＝(0,0.4884,0,0,-0.5109,0,0,0.02254,0,0,-0.4884,0,0,0.5109,0,0,-0.02254,0)^T；

p_2-9＝(0,0,0.4630,0,0,-0.5302,0,0,0.0672,0,0,-0.4630,0,0,0.5302,0,0,-0.0672)^T；

p_2-10＝(0,0,0.3038,0,0,0.2732,0,0,-0.5771,0,0,-0.3038,0,0,-0.2732,0,0,0.5771)^T；

displacement vector s of node coordinate₂Projected to the corresponding complete orthonormal mechanical basis matrix P₂Then the corresponding No. 2 triangular prism unit projection coefficient vector r can be obtained₂Thus, the ratio of each deformation was calculated, and the results are shown in Table 12.

TABLE 11 proportion of deformation projection coefficients of No. 1 unit at the second right end support

TABLE 12 example ratio of the deformation projection coefficients of No. 2 unit at two corners

Projection coefficient r2	In proportion of	Projection coefficient r2	In proportion of
				X axial tension and compression r2-4	0.00％	Z-direction curvature r of YOZ plane2-10	0.00％
Y-axis tension and compression r2-5	0.00％	XOY plane shear r2-11	0.00％
				Z-axial tension and compression r2-6	0.00％	XOZ plane shear r2-12	0.22％
XOZ plane X-direction curvature r2-7	8.91％	YOZ plane shear r2-13	22.80％
				YOZ plane Y-direction curvature r2-8	8.91％	XOY plane reverse shear r2-14	58.56％
XOZ plane Z-direction bending r2-9	0.00％	XOY plane torsion r2-15	0.60％

As can be seen from fig. 32, under the self-weight of the four short simply supported perforated plates, the XOZ plane X-direction bending deformation (dark gray area) is dominant at the edge support positions of X0 and X1200 mm, and the YOZ plane Y-direction bending deformation (white area) is dominant at the edge support positions of Y0 and Y1200 mm; near the opening, the stress is complex, the upper and lower parts of the round hole mainly bend and deform in the X direction of the XOZ plane, and the left and right parts of the round hole mainly bend and deform in the Y direction of the YOZ plane; four corner regions, mainly reverse shear deformation (light gray regions) in the XOY plane occurred. FIG. 32 more clearly shows the areas occupied by the various major fundamental deformations than the finite element method, providing a sufficient theoretical basis for how much structural failure occurs.

The above-mentioned embodiments are merely preferred embodiments of the present invention, which are merely illustrative and not restrictive, and it should be understood that other embodiments may be easily made by those skilled in the art by replacing or changing the technical contents disclosed in the specification, and therefore, all changes and modifications that are made on the principle of the present invention should be included in the scope of the claims of the present invention.

Claims (6)

1. A deformation decomposition method for any triangular prism unit in a space structure is characterized by comprising the following steps:

step 1: constructing basic displacement and deformation basis vectors of any triangular prism unit according to the geometric characteristics, stress balance and orthogonal theoretical conditions of any triangular prism unit under a space rectangular coordinate system, and further obtaining a complete orthogonal mechanics basis matrix P;

step 2: establishing a space structure model under a space rectangular coordinate system, dividing the space structure model by adopting triangular prism units to obtain node coordinate values of the triangular prism units in the space structure model and node coordinate values of the triangular prism units after being subjected to any load working condition, and further obtaining node coordinate displacement vectors s of the triangular prism units;

and step 3: projecting the node coordinate displacement vector s of the triangular prism unit onto a complete orthogonal mechanics basis matrix P to obtain a basic displacement and deformation projection coefficient vector r of the triangular prism unit;

and 4, step 4: and judging the primary deformation and the secondary deformation of the triangular prism unit according to the size of the projection coefficient in the projection coefficient vector r, so that the deformation decomposition and the deformation identification of any space structure can be realized.

2. The method of claim 1, wherein the method further comprises the step of decomposing the deformation of any triangular prism unit in the space structure, the space comprehensive deformation of the triangular prism unit is formed by superposing 18 basic displacements and deformations of X-axis rigid body linear displacement, Y-axis rigid body linear displacement, Z-axis rigid body linear displacement, X-axis tension-compression deformation, Y-axis tension-compression deformation, Z-axis tension-compression deformation, XOZ plane X-direction bending deformation, YOZ plane Y-direction bending deformation, XOZ plane Z-direction bending deformation, YOZ plane Z-direction bending deformation, XOY plane shear deformation, XOZ plane shear deformation, YOZ plane shear deformation, XOY plane reverse shear deformation, XOY plane torsion deformation, XOY plane rigid body rotation displacement, XOZ plane rigid body rotation displacement and YOZ plane rigid body rotation displacement.

3. The method for decomposing the deformation of any triangular prism unit in the spatial structure according to claim 1, wherein the step 1 specifically comprises the following steps:

step 1.1: aiming at any triangular prism unit under a space rectangular coordinate system, according to the geometrical characteristics, the stress balance and the orthogonal theory of the unitCondition, constructing 18 basic displacement and deformation basic vectors p of arbitrary triangular prism unit₁～p₁₈The method comprises the following steps:

p₁is X axial rigid body linear displacement base vector:

p₁＝[0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0]^T；

p₂is a Y axial rigid body linear displacement base vector:

p₂＝[0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0]^T；

p₃is a Z-axis rigid body linear displacement base vector:

p₃＝[0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082,0,0,0.4082]^T；

p₄is an X axial tension-compression deformation base vector:

p₄＝α[d-f,0,0,-b+f,0,0,b-d,0,0,d-f,0,0,-b+f,0,0,b-d,0,0]^T；

p₅the Y axial tension-compression deformation base vector is as follows:

p₅＝β[0,c-e,0,0,-a+e,0,0,a-c,0,0,c-e,0,0,-a+e,0,0,a-c,0]^T；

p₆the Z-axis tension-compression deformation base vector is as follows:

p₆＝χ[0,0,de-cf,0,0,af-be,0,0,bc-ad,0,0,cf-de,0,0,be-af,0,0,ad-bc]^T；

p₇is the XOZ plane X-direction bending deformation base vector:

p₇＝α[-d+f,0,0,b-f,0,0,d-b,0,0,d-f,0,0,-b+f,0,0,b-d,0,0]^T；

p₈the base vector of the Y-direction bending deformation of the YOZ plane is:

p₈＝β[0,-c+e,0,0,a-e,0,0,-a+c,0,0,c-e,0,0,-a+e,0,0,a-c,0]^T；

p₉is the XOZ plane Z-direction bending deformation base vector:

p₁₀is a Z-direction bending deformation base vector of a YOZ plane:

p₁₁is XOY plane shear deformation basis vector:

p₁₂is XOZ plane shear deformation basis vector:

p₁₃is a base vector of YOZ plane shear deformation:

p₁₄is XOY plane reverse shear deformation basis vector:

p₁₅is XOY plane torsional deformation base vector:

p₁₆as XOY plane rigid body rotation displacement basis vector:

p₁₇is XOZ plane rigid body rotation displacement base vector:

p₁₈is a rigid body rotation displacement base vector of a YOZ plane:

wherein:

u＝a²+c²+e²-ac-ae-ce；v＝-b²-d²-f²+bd+bf+df；w＝2(bc-ad+af-be-cf+de)；

a is the difference value of the node 1 or the node 4 in the triangular prism unit and the centroid O' of the triangular prism unit in the X axial direction;

b is the difference value of the Y-axis direction of the node 1 or the node 4 in the triangular prism unit and the centroid O' of the triangular prism unit;

c is the difference value of the X axial direction of the node 2 or the node 5 in the triangular prism unit and the centroid O' of the triangular prism unit;

d is the difference value of the Y axial direction of the node 2 or the node 5 in the triangular prism unit and the centroid O' of the triangular prism unit;

e is the difference value of the X axial direction of the node 3 or the node 6 in the triangular prism unit and the centroid O' of the triangular prism unit;

f is the difference value of the Y axial direction of the node 3 or the node 6 in the triangular prism unit and the centroid O' of the triangular prism unit;

l is the height of the triangular prism unit in the Z-axis direction;

step 1.2: 18 basic displacement and deformation basis vectors p₁～p₁₈Constructing a complete orthogonal mechanical basis matrix P:

P＝[p₁,p₂,p₃,p₄,p₅,p₆,p₇,p₈,p₉,p₁₀,p₁₁,p₁₂,p₁₃,p₁₄,p₁₅,p₁₆,p₁₇,p₁₈]。

4. the method for decomposing the deformation of any triangular prism unit in the spatial structure according to claim 1, wherein the step 2 specifically comprises the following steps:

step 2.1: establishing a space structure model under a space rectangular coordinate system, and dividing the space structure model by adopting triangular prism units, so that the node coordinate value d of the ith triangular prism unit in the dispersed space structure model_i：

d_i＝[x_i-1,y_i-1,z_i-1,x_i-2,y_i-2,z_i-2,x_i-3,y_i-3,z_i-3,x_i-4,y_i-4,z_i-4,x_i-5,y_i-5,z_i-5,x_i-6,y_i-6,z_i-6]；

Wherein:

d_ithe coordinates of the node of the ith triangular prism unit in the spatial structure model are (i ═ 1,2,3,4, … …, m);

x_i-jcoordinate values representing the node j in the ith triangular prism unit in the X axial direction;

y_i-jcoordinate values representing a node j in the ith triangular prism unit in the Y-axis direction;

z_i-jcoordinate values (j ═ 1,2,3,4,5,6) in the Z-axis direction indicating the node j in the ith triangular prism unit;

step 2.2: d 'is a node coordinate value of the ith triangular prism unit in the space structure model after being subjected to any displacement and deformation under any load working condition'_i：

d′_i＝[x′_i-1,y′_i-1,z′_i-1,x′_i-2,y′_i-2,z′_i-2,x′_i-3,y′_i-3,z′_i-3,x′_i-4,y′_i-4,z′_i-4,x′_i-5,y′_i-5,z′_i-5,x′_i-6,y′_i-6,z′_i-6]；

Wherein:

d′_ithe node coordinates of the ith triangular prism unit in the space structure model after being subjected to any displacement and deformation under any load working condition are shown, (i is 1,2,3,4, … …, m);

x′_i-jthe coordinate value of the ith triangular prism unit after the node j generates any displacement and deformation in the X axial direction under any load working condition is shown;

y′_i-jthe coordinate value of the ith triangular prism unit after the node j generates any displacement and deformation in the Y-axis direction under any load working condition is shown;

z′_i-jcoordinate values (j is 1,2,3,4,5 and 6) which indicate that the ith triangular prism unit generates any displacement and deformation in the X-axis direction under any load working condition;

step 2.3: d'_iMinus d_iObtaining the node coordinate displacement vector s of the ith triangular prism unit_i：

5. The method for decomposing the deformation of any triangular prism unit in the spatial structure according to claim 3 or 4, wherein the step 3 specifically comprises the following steps:

step 3.1: the node coordinate displacement vector s of the ith triangular prism unit_iProjected to a corresponding complete orthonormal mechanical basis matrix P_iTo obtain:

s_i＝r_i·P_i ^T；

step 3.2: the above formula is converted to obtain the ith triangular prism unit basic displacement and deformation projection coefficient vector r in the space structure model_i：

r_i＝s_i·(P_i ^T)^-1＝s_i·P_i；

Wherein s is_iThe node coordinate displacement vector of the ith triangular prism unit in the spatial structure model is obtained; p_iIs a complete orthonormal mechanical basis matrix, P, of the ith triangular prism unit in the spatial structure model_i ^TIs P_iThe transposed matrix of (P)_i ^T)^-1Is P_i ^TThe inverse matrix of (d); r is_iIs a complete orthonormal mechanical base matrix P of the ith triangular prism unit in the space structure model_iA corresponding projection coefficient vector;

wherein:

p_i-kthe k basic displacement and deformation basic vector of the ith triangular prism unit in the space structure model，(k＝1,2,3,……,18)；

r_i-kIs a complete orthonormal mechanical base matrix P of the ith triangular prism unit in the space structure model_iCorresponding projection system

Number vector r_iThe kth basic displacement and deformation projection coefficients, (k ═ 1,2,3, … …, 18).

6. The method according to claim 5, wherein the step 4 specifically comprises:

complete orthonormal mechanics basis matrix P for ith triangular prism unit in space structure model_iCorresponding projection coefficient vector r_iMiddle 12 basic deformation projection coefficients r_i-4～r_i-15The absolute value of the three-dimensional triangular prism unit is compared, and the basic deformation corresponding to the projection coefficient with the maximum absolute value is judged as the main deformation of the triangular prism unit; determining the basic deformation corresponding to the projection coefficient with the second largest absolute value as the secondary deformation of the triangular prism units, and obtaining the basic deformation component information of all triangular prism units in the space structure according to the secondary deformation, thereby finally realizing the deformation decomposition and the deformation identification of the space structure; wherein the projection coefficient r_i-4、r_i-5、r_i-6When the deformation is a positive value, the corresponding basic deformation is the tensile deformation of the ith triangular prism unit in the space structure model; projection coefficient r_i-4、r_i-5、r_i-6And when the deformation is a negative value, the corresponding basic deformation is the compression deformation of the ith triangular prism unit in the spatial structure model.

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