CN113297767B - Deformation decomposition method of any triangular prism unit in space structure - Google Patents
Deformation decomposition method of any triangular prism unit in space structure Download PDFInfo
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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 base vectors of any triangular prism unit aiming at any triangular prism unit under a space rectangular coordinate system, and further obtaining a complete orthogonal mechanical base 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 orthomechanical base matrix to obtain a basic displacement and deformation projection coefficient vector of the triangular prism unit; and judging the main deformation and the secondary deformation of the triangular prism unit according to the magnitude of the projection coefficient in the projection coefficient vector, so that deformation decomposition and deformation identification of any spatial structure can be realized. The method can deform and decompose any triangular prism unit without being limited by the size and the direction of unit division, thereby greatly reducing the calculation workload.
Description
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 space rectangular coordinate system varies with the height of the column and the internal angles of the two triangular faces, so that the flexible variability of the triangular prism unit is often used for the discrete space structure. Compared with regular units, the triangular prism units can finely divide the space irregular area without being limited by boundary conditions. Therefore, finite element analysis of any spatial structure can be realized by adopting triangular prism units.
The conventional finite element strain analysis method can realize the deformation performance analysis of the spatial structure at the microscopic level, but the engineering structure design is not only based on microscopic deformation information, but also needs to consider basic macroscopic basic deformation information. According to the linear superposition principle, the macroscopic comprehensive deformation of the space structure can be obtained by superposition of basic macroscopic deformation combinations of discrete triangular prism units, so that a mathematical mechanics theory for quantitatively identifying the basic macroscopic deformation is constructed, a space structure deformation decomposition method based on the triangular prism units is formed, and the method has important theoretical significance and engineering value and can effectively guide engineering structure design.
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, so that the area occupied by each basic deformation of the space structure after being stressed is more accurately displayed, and the correctness and superiority of the deformation decomposition method of any triangular prism unit under a space rectangular coordinate system can be embodied.
In order to achieve the above purpose, the present 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: aiming at any triangular prism unit under a space rectangular coordinate system, constructing basic displacement and deformation basis vectors of the any triangular prism unit according to geometric characteristics, stress balance and orthogonal theoretical conditions of the triangular prism unit, and further obtaining a complete orthogonal mechanical 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 receiving any load working conditions, and further obtaining node coordinate displacement vectors s of the triangular prism units;
step 3: projecting the node coordinate displacement vector s of the triangular prism unit onto a complete orthomechanical base matrix P to obtain a basic displacement and deformation projection coefficient vector r of the triangular prism unit;
step 4: and judging the main deformation and the secondary deformation of the triangular prism unit according to the magnitude of the projection coefficient in the projection coefficient vector r, so that deformation decomposition and deformation identification of any spatial structure can be realized.
Preferably, the vertexes of the triangular prism units are denoted as node 1, node 2, node 3, node 4, node 5 and node 6, and the spatial integrated deformation of the triangular prism units is formed by superposition of 18 basic displacements and deformations of X-axial rigid body linear displacement, Y-axial rigid body linear displacement, Z-axial rigid body linear displacement, X-axial tension-compression deformation, Y-axial tension-compression deformation, Z-axial tension-compression deformation, XOZ-plane X-direction bending deformation, YOZ-plane Y-direction bending deformation, XOZ-plane Z-direction bending deformation, YOZ-plane shearing deformation, XOZ-plane shearing deformation, YOZ-plane shearing deformation, XOY-plane reverse shearing deformation, XOY-plane torsional 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 includes the following steps:
step 1.1: for any triangular prism unit under 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 triangular prism unit 1 ～p 18 The method is characterized by comprising the following steps:
p 1 the linear displacement base vector is an X-axis rigid body:
p 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 the linear displacement base vector is the rigid body in the Y axial direction:
p 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 the base vector is the linear displacement of the Z-axis rigid body:
p 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 4 the deformation base vector is drawn and pressed in the X-axis direction:
p 4 ＝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 ；
p 5 the deformation base vector is pulled and pressed in the Y axial direction:
p 5 ＝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 ；
p 6 the deformation base vector is drawn and pressed in the Z axial direction:
p 6 ＝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 ；
p 7 the deformation base vector is the X-direction bending deformation of the XOZ plane:
p 7 ＝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 ；
p 8 the Y-direction bending deformation base vector of the YOZ plane is as follows:
p 8 ＝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 ；
p 9 the deformation base vector is the Z-direction bending deformation of the XOZ plane:
p 10 the Z-direction bending deformation base vector of the YOZ plane is as follows:
p 11 for the XOY plane shear deformation basis vector:
p 12 is an XOZ plane shearDeformation basis vector:
p 13 the shear deformation basis vector is YOZ plane:
p 14 for the XOY plane reverse shear deformation basis vector:
p 15 the basis vector is the XOY plane torsional deformation:
p 16 the base vector is the XOY plane rigid body rotational displacement:
p 17 the rotational displacement basis vector is an XOZ plane rigid body:
p 18 the rotational displacement basis vector is YOZ plane rigid body:
wherein:
u＝a 2 +c 2 +e 2 -ac-ae-ce；v＝-b 2 -d 2 -f 2 +bd+bf+df；w＝2(bc-ad+af-be-cf+de)；
a is the difference value between the node 1 or the node 4 in the triangular prism unit and the X axis of the centroid O' of the triangular prism unit;
b is the difference value between the node 1 or the node 4 in the triangular prism unit and the Y-axis of the centroid O' of the triangular prism unit;
c is the difference value between the node 2 or the node 5 in the triangular prism unit and the X axis of the centroid O' of the triangular prism unit;
d is the difference value between the node 2 or the node 5 in the triangular prism unit and the Y-axis of the centroid O' of the triangular prism unit;
e is the difference value between the node 3 or the node 6 in the triangular prism unit and the X axis of the centroid O' of the triangular prism unit;
f is the difference value between the node 3 or the node 6 in the triangular prism unit and the Y-axis of the centroid O' of the triangular prism unit;
l is the height of the triangular prism unit Z in the axial direction;
step 1.2: the 18 basic displacements and the deformation base vector p 1 ～p 18 Constructed as a perfect orthomechanical basis matrix P:
P＝[p 1 ,p 2 ,p 3 ,p 4 ,p 5 ,p 6 ,p 7 ,p 8 ,p 9 ,p 10 ,p 11 ,p 12 ,p 13 ,p 14 ,p 15 ,p 16 ,p 17 ,p 18 ]；
preferably, step 2 specifically comprises the following steps:
step 2.1: establishing a space structure model under a space rectangular coordinate system, dividing the space structure model by adopting triangular prism units, and obtaining a node coordinate value d of an ith triangular prism unit in the discrete 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 i the node coordinates of the ith triangular prism unit in the spatial structure model, (i=1, 2,3,4, … …, m);
x i-j representing coordinate values of the node j in the ith triangular prism unit in the X-axis direction;
y i-j representing the coordinate value of the node j in the ith triangular prism unit in the Y-axis direction;
z i-j coordinate values (j=1, 2,3,4,5, 6) indicating the Z-axis direction of the node j in the ith triangular prism unit;
step 2.2: the coordinate value of the node of the ith triangular prism unit in the space structure model after any displacement and deformation under any load working condition is d' 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' i the node coordinates of the ith triangular prism unit in the space structure model after any displacement and deformation under any load working condition are generated, (i=1, 2,3,4, … …, m);
x' i-j the coordinate value of the ith triangular prism unit after the node j generates any displacement and deformation in the X-axis direction under any load working condition is shown;
y' i-j the 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-j coordinate values (j=1, 2,3,4,5, 6) indicating that the ith triangular prism unit generates any displacement and deformation in the Z axis direction under any load working condition;
step 2.3: will d' i Subtracting d i Obtaining 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 i Projection onto corresponding complete orthomechanical basis matrix P i And (3) obtaining:
s i ＝r i ·P i T ；
step 3.2: converting the above 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 i The node coordinate displacement vector of the ith triangular prism unit in the space structure model; p (P) i Is the complete orthodynamic matrix, P, of the ith triangular prism unit in the space structure model i T Is P i Transposed matrix of (P) i T ) -1 Is P i T An inverse matrix of (a); r is (r) i Complete orthodynamic basis matrix P for ith triangular prism unit in space structure model i A 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-k the kth basic displacement and deformation basis vector (k=1, 2,3, … …, 18) for the ith triangular prism unit in the spatial structure model;
r i-k complete orthodynamic basis matrix P for ith triangular prism unit in space structure model i Corresponding projection coefficient vector r i The k-th basic displacement and deformation projection coefficients (k=1, 2,3, … …, 18).
Preferably, the step 4 specifically includes:
complete orthodynamic basis matrix P for ith triangular prism unit in space structure model i Corresponding projection coefficient vector r i Projection coefficient r of 12 basic deformations i-4 ～r i-15 Comparing the absolute values of the three prism units, and judging the basic deformation corresponding to the projection coefficient with the maximum absolute value as the main deformation of the three prism units; the projection coefficient with the next largest absolute value corresponds toThe basic deformation is judged to be secondary deformation of the triangular prism units, so that basic deformation component information of all the triangular prism units in the space structure is obtained, and deformation decomposition and deformation identification of the space structure are finally realized; wherein the projection coefficient r i-4 、r i-5 、r i-6 When the deformation is positive, 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-6 And when the deformation is negative, the corresponding basic deformation is the compressive deformation of the ith triangular prism unit in the space 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 base vectors of the any triangular prism unit are constructed, and then a complete orthogonal mechanical base matrix is constructed; after the triangular prism units are adopted to disperse the space structure model, compared with the finite element analysis method, the analysis result of the comprehensive deformation of the triangular prism units can be obtained only, and the main basic deformation and the secondary basic deformation of the triangular prism units can be identified, so that the area occupied by each basic deformation of the space structure after being stressed is more accurately displayed, and the correctness and superiority of the deformation decomposition method of any triangular prism unit under a space rectangular coordinate system can be embodied. In addition, the method of the invention can deform and decompose any triangular prism unit without being limited by the dividing size and direction of the unit, and is applicable to dividing the triangular prism unit into large units in a space structure.
Drawings
Fig. 1 is a flow chart of a deformation decomposition method of any triangular prism unit in the space structure of the present invention.
Fig. 2 is a schematic view of an arbitrary triangular prism unit in a space rectangular coordinate system.
Fig. 3 is a schematic diagram showing deformation of any triangular prism unit after being stressed under a space rectangular coordinate system.
Fig. 4 is a schematic diagram of the linear displacement of the rigid body in the X-axis direction of any triangular prism unit under a space rectangular coordinate system.
Fig. 5 is a schematic diagram of the linear displacement of the Y-axis rigid body of any triangular prism unit under a space rectangular coordinate system.
Fig. 6 is a schematic view of the Z-axis rigid body linear displacement of any triangular prism unit under a space rectangular coordinate system.
Fig. 7 is a schematic drawing of the X-axis tensile compression deformation of any triangular prism unit in a space rectangular coordinate system.
Fig. 8 is a schematic diagram of Y-axis tensile compression deformation of an arbitrary triangular prism unit in a space rectangular coordinate system.
Fig. 9 is a schematic drawing of Z-axial tension-compression deformation of an arbitrary triangular prism unit in a space rectangular coordinate system.
Fig. 10 is a schematic view of X-direction bending deformation of the XOZ plane of any triangular prism unit in a space rectangular coordinate system.
Fig. 11 is a schematic view showing Y-direction bending deformation of the YOZ plane of any triangular prism unit in a space rectangular coordinate system.
Fig. 12 is a schematic view showing the bending deformation of the XOZ plane of any triangular prism unit in the space rectangular coordinate system.
Fig. 13 is a view showing the Z-bending deformation of the YOZ plane of any triangular prism unit in a space rectangular coordinate system.
Fig. 14 is a schematic view of XOY plane shear deformation of an arbitrary triangular prism unit under a space rectangular coordinate system.
Fig. 15 is a schematic view of XOZ plane shear deformation of an arbitrary triangular prism unit in a space rectangular coordinate system.
Fig. 16 is a view showing the YOZ-plane shear deformation of an arbitrary triangular prism unit in a space rectangular coordinate system.
Fig. 17 is a schematic view of the XOY plane reverse shear deformation of any triangular prism unit under a space rectangular coordinate system.
Fig. 18 is a schematic view of the XOY plane torsional deformation of any triangular prism unit in a space rectangular coordinate system.
Fig. 19 is a schematic view of XOY plane rigid body rotational displacement of any triangular prism unit under a space rectangular coordinate system.
Fig. 20 is a schematic diagram of XOZ plane rigid body rotational displacement of any triangular prism unit under a space rectangular coordinate system.
Fig. 21 is a diagram showing the rotational displacement of the YOZ plane rigid body of any triangular prism unit in a space rectangular coordinate system.
Fig. 22 is a schematic diagram showing the coordinate differences between each node of any triangular prism unit and the centroid O' in a space rectangular coordinate system.
Fig. 23 is a schematic diagram showing distances between each node of any triangular prism unit along a coordinate axis and a centroid O' in a space rectangular coordinate system.
FIG. 24 is a schematic diagram of cell sizes and node coordinates of selected cells in the rigid body rotational displacement error analysis of the present invention.
Fig. 25 is a schematic view of a swash plate according to an embodiment of the invention.
Fig. 26 is a schematic diagram illustrating selection of triangular prism units in a swash plate structure according to an embodiment of the invention.
FIG. 27 shows a finite element analysis of the upper and lower surfaces sigma of a swash plate structure according to an embodiment of the present invention x Stress cloud.
FIG. 28 shows a finite element analysis of the upper and lower surfaces sigma of a swash plate structure according to an embodiment of the present invention y Stress cloud.
FIG. 29 shows a finite element analysis of the upper and lower surfaces τ of a swash plate structure according to an embodiment of the present invention xy Stress cloud.
FIG. 30 is a schematic view of a second embodiment of an apertured plate.
Fig. 31 is a schematic view illustrating selection of triangular prism units in a two-aperture plate according to an embodiment of the invention.
FIG. 32 is a graph showing the deformation decomposition calculation results of a second perforated plate according to an embodiment of the present invention.
Detailed Description
The following examples are illustrative of the present invention and are not intended to limit the scope of the invention. The technical means used in the examples are conventional means well known to those skilled in the art unless otherwise indicated. The test methods in the following examples are conventional methods unless otherwise specified.
Fig. 1 shows a flow chart of a deformation decomposition method of any triangular prism unit in the spatial structure of the present invention. The triangular prism unit comprises six nodes, the vertexes of the nodes are marked as node 1, node 2, node 3, node 4, node 5 and node 6, the schematic diagram of the triangular prism unit under the space rectangular coordinate system is shown in figure 2, and the deformation schematic diagram of the triangular prism unit after being stressed is shown in figure 3. Fig. 4 to 21 are schematic views of basic displacement and deformation of any triangular prism unit, which are in turn 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, 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, YOZ-plane rigid body rotational displacement.
A deformation decomposition method of any triangular prism unit in a space structure comprises the following steps:
step 1.1: for any triangular prism unit under a space rectangular coordinate system, 15 mutually independent and orthogonal basic displacement and basic deformation base 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 unit 1 ～p 15 Obtaining 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, wherein the method comprises the following specific steps of:
p 1 the linear displacement base vector is an X-axis rigid body:
p 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 the linear displacement base vector is the rigid body in the Y axial direction:
p 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 the base vector is the linear displacement of the Z-axis rigid body:
p 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 4 the deformation base vector is drawn and pressed in the X-axis direction:
p 4 ＝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 ；
p 5 the deformation base vector is pulled and pressed in the Y axial direction:
p 5 ＝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 ；
p 6 the deformation base vector is drawn and pressed in the Z axial direction:
p 6 ＝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 ；
p 7 the deformation base vector is the X-direction bending deformation of the XOZ plane:
p 7 ＝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 ；
p 8 the Y-direction bending deformation base vector of the YOZ plane is as follows:
p 8 ＝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 ；
p 9 the deformation base vector is the Z-direction bending deformation of the XOZ plane:
p 10 the Z-direction bending deformation base vector of the YOZ plane is as follows:
p 11 for the XOY plane shear deformation basis vector:
p 12 is the shear deformation basis vector of the XOZ plane:
p 13 the shear deformation basis vector is YOZ plane:
p 14 for the XOY plane reverse shear deformation basis vector:
p 15 the basis vector is the XOY plane torsional deformation:
the 15 groups of mutually independent and orthogonal basis vectors are adopted to obtain an XOY plane rigid body rotational displacement basis vector, an XOZ plane rigid body rotational displacement basis vector and a YOZ plane rigid body rotational displacement basis vector of any triangular prism unit through a Schmidt orthogonalization method.
p 16 The base vector is the XOY plane rigid body rotational displacement:
p 17 the rotational displacement basis vector is an XOZ plane rigid body:
p 18 the rotational displacement basis vector is YOZ plane rigid body:
wherein:
u＝a 2 +c 2 +e 2 -ac-ae-ce；v＝-b 2 -d 2 -f 2 +bd+bf+df；w＝2(bc-ad+af-be-cf+de)；
a is the difference value between the node 1 or the node 4 in the triangular prism unit and the X axis of the centroid O' of the triangular prism unit;
b is the difference value between the node 1 or the node 4 in the triangular prism unit and the Y-axis of the centroid O' of the triangular prism unit;
c is the difference value between the node 2 or the node 5 in the triangular prism unit and the X axis of the centroid O' of the triangular prism unit;
d is the difference value between the node 2 or the node 5 in the triangular prism unit and the Y-axis of the centroid O' of the triangular prism unit;
e is the difference value between the node 3 or the node 6 in the triangular prism unit and the X axis of the centroid O' of the triangular prism unit;
f is the difference value between the node 3 or the node 6 in the triangular prism unit and the Y-axis of the centroid O' of the triangular prism unit;
l is the height of the triangular prism unit Z in the axial direction.
As shown in FIG. 22, the coordinates of the centroid O' of the triangular prism unit in the X-axis, Y-axis and Z-axis are X 0 、y 0 、z 0 The differences between the X axis and the Y axis from each node to the centroid are respectively as follows:
the differences between each node and the Z axis of the centroid are respectively as follows: -l/2, l/2.
Step 2.1: establishing a space structure model under a space rectangular coordinate system, dividing the space structure model by adopting triangular prism units, and obtaining a node coordinate value d of an ith triangular prism unit in the discrete 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 i the node coordinates of the ith triangular prism unit in the spatial structure model, (i=1, 2,3,4, … …, m);
x ij representing coordinate values of the node j in the ith triangular prism unit in the X-axis direction;
y ij representing the coordinate value of the node j in the ith triangular prism unit in the Y-axis direction;
z ij representing the coordinate value of the node j in the Z-axis direction in the ith triangular prism unit, (j=1, 2,3,4,5, 6);
step 2.2: the coordinate value of the node of the ith triangular prism unit in the space structure model after any displacement and deformation under any load working condition is d' 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' i the node coordinates of the ith triangular prism unit in the space structure model after any displacement and deformation under any load working condition are generated, (i=1, 2,3,4, … …, m);
x' ij the coordinate value of the ith triangular prism unit after the node j generates any displacement and deformation in the X-axis direction under any load working condition is shown;
y' ij the 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' ij coordinate values (j=1, 2,3,4,5, 6) after the node j generates any displacement and deformation in the X axis direction under any load working condition of the ith triangular prism unit;
step 2.3: will d' i Subtracting d i Obtaining 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 i Projection onto corresponding complete orthomechanical basis matrix P i And (3) obtaining:
s i ＝r i ·P i T ；
step 3.2: converting the above 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 i The node coordinate displacement vector of the ith triangular prism unit in the space structure model; p (P) i Is the complete orthodynamic matrix, P, of the ith triangular prism unit in the space structure model i T Is P i Transposed matrix of (P) i T ) -1 Is P i T An inverse matrix of (a); r is (r) i Complete orthodynamic basis matrix P for ith triangular prism unit in space structure model i A 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-k the kth basic displacement and deformation basis vector (k=1, 2,3, … …, 18) for the ith triangular prism unit in the spatial structure model;
r i-k complete orthodynamic basis matrix P for ith triangular prism unit in space structure model i Corresponding projection coefficient vector r i The k-th basic displacement and deformation projection coefficients (k=1, 2,3, … …, 18) are specifically shown in table 1.
Step 4: complete orthodynamic basis matrix P for ith triangular prism unit in space structure model i Corresponding projection coefficient vector r i Projection coefficient r of 12 basic deformations i4 ～r i15 Comparing the absolute values of the three prism units, and judging the basic deformation corresponding to the projection coefficient with the maximum absolute value as the main deformation of the three prism units; the basic deformation corresponding to the projection coefficient with the next largest absolute value is judged to be the secondary deformation of the triangular prism unit, so that the cyclic damage is achieved, and all the triangular prism units in the space structure are obtainedThe basic deformation component information of the spatial structure is finally realized; wherein the projection coefficient r i4 、r i5 、r i6 When the deformation is positive, 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 i6 And when the deformation is negative, the corresponding basic deformation is the compressive deformation of the ith triangular prism unit in the space structure model.
Table 1 table of the i th triangular prism unit basic displacement and deformation versus projection coefficient
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 are not required to be considered, and only the projection coefficients on the other 12 basic deformation basis vectors in the complete orthogonal mechanical basis matrix are required to be analyzed.
Rigid body rotational displacement error analysis
The rigid body rotation displacement is nonlinear displacement, and errors can be generated when linear decomposition is carried out, namely, the triangular prism unit rigid body rotation displacement base vector not only has projection coefficients on the rigid body rotation displacement base vector, but also has projection coefficients on other displacement and deformation base vectors, and errors exist in the theoretical situation, so that error analysis is carried out on the projection coefficient values of the triangular prism unit rigid body rotation displacement on other deformation and displacement base vectors, and whether the rigid body rotation displacement affects the calculation precision is judged.
As shown in fig. 23, in any triangular prism unit in the space rectangular 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 as: l (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 (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 6z =l/2. When the triangular prism units rotate clockwise along three different planes (XOY plane, XOZ plane and YOZ plane) around the centroid O', the error analysis is carried out on the rotation displacement of the rigid body. Since the basic displacement and deformation base vector of any triangular prism unit can change along with the change of the node coordinates, the error analysis can be performed only by using the parameter variables (such as a, b, c, d, e, f, l and theta) contained in the node coordinates, and then the general solution of the error analysis result of any triangular prism unit can be obtained. However, the calculation process is extremely complex due to more parameters, and the error analysis process cannot be clearly expressed due to the fact that the calculation process is solved by a computer; in order to more clearly express the error analysis and calculation process, any triangular prism unit is arbitrarily selected for error analysis, and specific geometric parameters of the unit are shown in fig. 24. The coordinates of the nodes are respectively as follows: node 1 (5 mm ), node 2 (7 mm,5 mm), node 3 (6 mm,6.732mm,5 mm), node 4 (5 mm,7 mm), node 5 (7 mm,5mm,7 mm), node 6 (6 mm,6.732mm,7 mm); at this time: a= -1mm, b= -0.5773mm, c= 1mm, d= -0.5773mm, e= 0, f= 1.1547mm, l= 2mm.
The unit obtains 18 basic displacements and deformation basis vectors p 1 ～p 18 The method comprises the following steps:
p 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 ＝(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 ＝(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 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 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 6 ＝(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 ；
p 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 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 9 ＝(0,0,-0.5,0,0,0.5,0,0,0,0,0,0.5,0,0,-0.5,0,0,0) T ；
p 10 ＝(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:
P＝[p 1 ,p 2 ,p 3 ,p 4 ,p 5 ,p 6 ,p 7 ,p 8 ,p 9 ,p 10 ,p 11 ,p 12 ,p 13 ,p 14 ,p 15 ,p 16 ,p 17 ,p 18 ]。
(1) XOY plane rigid body rotational displacement error analysis
After the triangular prism unit rotates clockwise along the XOY plane by an angle theta, six nodes coordinate displacement s XOY As shown in table 2. Rewriting six node coordinate displacements into vector form s XOY 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 is(s) jk Representing the displacement of the j-th node k in the axial direction; j=1, 2,3,4,5,6; k= X, Y, Z.
Table 2 node coordinate Displacement produced after the triangular prism units are rotated clockwise by θ along the XOY plane
The node coordinate displacement vector s of the triangular prism unit XOY Projecting onto the complete orthomechanical base matrix P of the triangular prism unit to obtain projection coefficient vector r XOY ：
r XOY ＝s XOY P。
Projection coefficient vector r XOY The calculation results are shown in Table 3:
TABLE 3 calculation of the projection coefficients for each displacement and deformation after clockwise rotation of the prism units along the XOY plane by θ
As shown in table 3, after the triangular prism unit rotates 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 deformation basis vector, and the projection coefficients on the other basic displacement and deformation basis vectors are equal to 0.
Obviously, under the condition of small deformation (when theta approaches 0), the projection coefficient r of the X-axis tensile compression deformation is solved 4 And projection coefficient r of Y-axis tension-compression deformation 5 Projection coefficient r relative to rigid body rotational displacement of XOY plane 16 Is of a high order infinitely small, so that the projection coefficient of the XOY plane rigid body rotation displacement on the X, Y axial tension-compression deformation basis vector is negligible, and the error is within the allowable range of the method.
(2) XOZ plane rigid body rotational displacement error analysis
Six node coordinate displacements of the triangular prism units after the triangular prism units are rotated clockwise by an angle theta along the XOZ plane are shown in table 4.
TABLE 4 node coordinate Displacement produced after clockwise rotation of prism units along the XOZ plane by θ
Rewriting six node coordinate displacements into vector form s XOZ 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 is(s) jk Representing the displacement of the j-th node k in the axial direction; j=1, 2,3,4,5,6; k= X, Y, Z.
The node coordinate displacement vector s of the triangular prism unit XOZ Projecting onto the complete orthomechanical base matrix P of the triangular prism unit to obtain projection coefficient vector r XOZ ：
r XOZ ＝s XOZ P。
The calculation result of the projection coefficient vector r is shown in table 5:
as shown in table 5, after the triangular prism unit rotates clockwise by θ 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 deformation basis vector, and the projection coefficients on the other basic displacement and deformation basis vectors are equal to 0.
Obviously, under the condition of small deformation (when theta approaches 0), the projection coefficient r of the X-axis tensile compression deformation is solved 4 And projection coefficient r of Z-axis tensile compression deformation 6 Projection coefficient r relative to XOZ plane rigid body rotational displacement 17 Is of a high order infinitely small, so that 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 projection coefficients for each displacement and deformation after clockwise rotation of the column unit by θ along the XOZ plane
(3) YOZ plane rigid body rotation displacement error analysis
Six node coordinate displacements of the triangular prism unit after clockwise rotation by an angle θ along the YOZ plane are shown in table 6.
Table 6 node coordinate Displacement produced after clockwise rotation of column unit along YOZ plane by θ
Rewriting six node coordinate displacements into vector form s YOZ 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 is(s) jk Representing the displacement of the j-th node k in the axial direction; j=1, 2,3,4,5,6; k= X, Y, Z.
The node coordinate displacement vector s of the triangular prism unit YOZ Projecting onto the complete orthomechanical base matrix P of the triangular prism unit to obtain projection coefficient vector r YOZ ：
r YOZ ＝s YOZ P。
Projection coefficient vector r YOZ The calculation results are shown in Table 7.
TABLE 7 calculation results of projection coefficients of each displacement and deformation after clockwise rotation of the column unit along YOZ plane by θ
As shown in table 7, after the triangular prism unit rotates clockwise by θ along the YOZ plane, the projection coefficients are only present on the YOZ plane rigid body rotational displacement basis vector and the Y, Z axial tension deformation basis vector, and the projection coefficients on the other basic displacement and deformation basis vectors are equal to 0.
Obviously, the unit is in a small deformation conditionDown (when θ approaches 0), the projection coefficient r of the solved Y-axis tension-compression deformation 5 And projection coefficient r of Z-axis tensile compression deformation 6 Projection coefficient r relative to YOZ plane rigid body rotation displacement 18 Is of a high order infinitely small, so the projection coefficient of the YOZ plane rigid body rotation displacement on the Y, Z axial tension-compression deformation basis vector is negligible, and the error is within the allowable range of the method.
All other triangular prism units at any positions in the space 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 1
As shown in FIG. 25, inclined plates each having a length of 1200mm on four sides and a thickness of 60mm were provided, the skew angle was 30℃and the elastic modulus was 2.1X10 11 pa, poisson's ratio is 0.3. Constraint conditions: the two ends are simply supported and the two ends are free. And after the triangular prism units are adopted for dividing the inclined plate structure, the stress condition of the inclined plate structure at the obtuse angle, the acute angle and the midspan position under the dead weight is analyzed. And selecting a triangular prism unit in an acute angle, an obtuse angle and a midspan region respectively, wherein the unit number at the acute angle is marked as a No. 1 unit, the unit number at the obtuse angle is marked as a No. 2 unit, the midspan unit number is marked as a No. 3 unit, and performing deformation decomposition on the selected triangular prism unit as shown in fig. 26.
Obtaining node coordinates d contained in the No. 1-3 triangular prism units in the inclined plate structure by utilizing finite element analysis 1 ～d 3 Node coordinate displacement vector s 1 ～s 3 Finite element analysis result sigma of sloping plate structure x 、σ y 、τ xy The upper and lower surface stress cloud diagrams of (a) are shown in fig. 27 to 29.
Node coordinates d of the triangular prism units 1 to 3 in the present embodiment 1 ～d 3 : the extraction can be directly calculated according to the coordinate axis and geometric relation or calculated by finite element software.
Node coordinates d of the triangular prism units 1 to 3 in the present embodiment 1 ～d 3 And node coordinate displacement vector s 1 ～s 3 Extracting by adopting finite element software;
acute angle position No. 1 unit node coordinate d 1 And nodeCoordinate displacement vector s 1 The method comprises the following steps of:
d 1 ＝(50,86.6025,0,100,0,0,0,0,0,50,86.6025,60,100,0,60,0,0,60)；
s 1 ＝(0,0,0,-0.0565,0.0142,-0.0907,0,0,0,0,0,0,0.0565,-0.0142,-0.0907,0,0,0)。
obtuse angle position No. 2 unit node coordinate d 2 And node coordinate displacement vector s 2 The method comprises the following steps of:
s 2 ＝(0.1386,-0.0460,-0.2784,0,0,0,0,0,0,-0.1386,0.0460,-0.27834,0,0,0,0,0,0)。
according to the node coordinates d 1 ～d 3 Respectively obtaining complete orthogonal mechanical base matrixes corresponding to the 3 triangular prism units, and marking the complete orthogonal mechanical base matrixes as P 1 、P 2 、P 3 ；
18 basic displacements and deformation basis vectors p obtained by unit No. 1 1-1 ～p 1-18 The method comprises the following steps:
P 1 ＝[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 displacements and deformation basis vectors p obtained by No. 2 unit 2-1 ～p 2-18 The method comprises 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 displacements and deformation basis vectors p obtained by No. 3 unit 3-1 ～p 3-18 The method comprises 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 coordinates 1 、s 2 、s 3 Respectively projecting to corresponding complete orthogonal mechanical base matrix P 1 、P 2 、P 3 Obtaining 3 corresponding projection coefficient vectors r 1 、r 2 、r 3 The proportion of each unit deformation was calculated, and the results are shown in tables 8 to 10.
From the finite element calculation results shown in FIGS. 27 to 29, it is apparent that σ is present in the acute angle, the obtuse angle, and the midspan x 、σ y 、τ xy The stress is obvious, and the finite element method cannot distinguish which deformation is the main; tables 8-10 can accurately obtain the proportion of each basic deformation by using the method on the basis of finite elements, and the acute angle position, the obtuse angle position and the midspan position are mainly formed by X-direction bending deformation of an XOZ plane and are subjected to reverse shear deformation of XOY. And further verifies the rationality and superiority of the method.
Table 8 example ratio of projection coefficients of deformation of No. 1 element at acute angle
| Projection coefficient r 1 | Proportion of occupied | Projection coefficient r 1 | Proportion of occupied |
| X-axis pulling and pressing r 1-4 | 0.00％ | YOZ plane Z-direction bend r 1-10 | 0.00％ |
| Y-axis pulling and pressing r 1-5 | 0.00％ | XOY plane shear r 1-11 | 0.00％ |
| Z-axis pulling and pressing r 1-6 | 0.00％ | XOZ plane shear r 1-12 | 11.02％ |
| xOZ plane X-direction bend r 1-7 | 37.99％ | YOZ plane shear r 1-13 | 14.22％ |
| Y-direction bending r of YOZ plane 1-8 | 5.77％ | xOY plane reverse shear r 1-14 | 22.57％ |
| xOZ plane Z-direction bend r 1-9 | 0.00％ | XOY plane twist r 1-15 | 8.44％ |
Table 9 example ratio of deformation projection coefficients of No. 2 element at obtuse angle
| Projection coefficient r 2 | Proportion of occupied | Projection coefficient r 2 | Proportion of occupied |
| X-axis pulling and pressing r 2-4 | 0.00％ | YOZ plane Z-direction bend r 2-10 | 0.00％ |
| Y-axis pulling and pressing r 2-5 | 0.00％ | XOY plane shear r 2-11 | 0.00％ |
| Z-axis pulling and pressing r 2-6 | 0.00％ | XOZ plane shear r 2-12 | 17.30％ |
| xOZ plane X-direction bend r 2-7 | 33.52％ | YOZ plane shear r 2-13 | 15.35％ |
| Y-direction bending r of YOZ plane 2-8 | 6.47％ | xOY plane reverse shear r 2-14 | 21.61％ |
| xOZ plane Z-direction bend r 2-9 | 0.00％ | XOY plane twist r 2-15 | 5.76％ |
Table 10 example ratio of deformation projection coefficients of unit No. 3 at midspan
| Projection coefficient r 3 | Proportion of occupied | Projection coefficient r 3 | Proportion of occupied |
| X-axis pulling and pressing r 3-4 | 0.00％ | YOZ plane Z-direction bend r 3-10 | 0.00％ |
| Y-axis pulling and pressing r 3-5 | 0.00％ | XOY plane shear r 3-11 | 0.00％ |
| Z-axis pulling and pressing r 3-6 | 0.00％ | XOZ plane shear r 3-12 | 13.73％ |
| xOZ plane X-direction bend r 3-7 | 39.17％ | YOZ plane shear r 3-13 | 10.28％ |
| Y-direction bending r of YOZ plane 3-8 | 5.79％ | xOY plane reverse shear r 3-14 | 30.78％ |
| xOZ plane Z-direction bend r 3-9 | 0.00％ | XOY plane twist r 3-15 | 0.25％ |
Example two
As shown in FIG. 30, a flat plate with 1200mm four sides, 60mm thick and a central opening is arranged, and the radius of the hole is 200mm; the elastic modulus of the perforated plate is 2.1 multiplied by 10 11 pa, poisson's ratio of 0.3; analysis of the dead weight load of the steel under the condition of simple four sidesIs a stress feature of (a).
With the help of finite element software ANSYS modeling, triangular prism grids are adopted for free division, and node coding information of all triangular prism units and coordinate and displacement information of all nodes are respectively extracted after loading and solving; and importing the two files into MATLAB (math software) for further analysis, and reassembling the unit information and the node information to enter a deformation decomposition calculation process.
The opening plate has good symmetry, only the 1/4 area analysis at the right lower end is needed, one triangular prism unit is selected at the supporting position and the corner position at will to analyze the stress form of the triangular prism unit, the No. 1 unit is selected at the supporting position, and the No. 2 unit is selected at the corner position (as shown in fig. 31); the results of the overall analysis of the aperture plate are shown in fig. 32.
Node coordinates d of triangular prism units No. 1 and No. 2 in the present embodiment 1 、d 2 And node coordinate displacement vector s 1 、s 2 Finite element software is adopted for extraction, and the extraction methods are respectively as follows:
s 1 ＝(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 the node coordinates d 1 Obtaining a complete orthodynamic base matrix corresponding to the No. 1 triangular prism unit, and marking the complete orthodynamic base matrix as P 1 The method comprises the steps of carrying out a first treatment on the surface of the 18 basic displacements and deformation basis vectors p obtained by unit No. 1 1-1 ～p 1-18 The method comprises the following steps:
P 1 ＝[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 coordinates 1 Projected to corresponding complete orthomechanical basis matrix P 1 The corresponding projection coefficient vector r of the No. 1 triangular prism unit can be obtained 1 The ratio of each deformation was calculated, and the results are shown in Table 11.
According to the node coordinates d 2 Obtaining a complete orthodynamic base matrix corresponding to the No. 2 triangular prism unit, and marking the complete orthodynamic base matrix as P 2 The method comprises the steps of carrying out a first treatment on the surface of the 18 basic displacements and deformation basis vectors p obtained by No. 2 unit 2-1 ～p 2-18 The method comprises 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 coordinates 2 Projected to corresponding complete orthomechanical basis matrix P 2 The projection coefficient vector r of the corresponding No. 2 triangular prism unit can be obtained 2 The ratio of each deformation was calculated, and the results are shown in Table 12.
Table 11 example ratio of deformation projection coefficients of No. 1 cell at two right end supports
Table 12 example ratio of deformation projection coefficients of No. 2 cell at two corners
| Projection coefficient r 2 | Proportion of occupied | Projection coefficient r 2 | Proportion of occupied |
| X-axis pulling and pressing r 2-4 | 0.00％ | YOZ plane Z-direction bend r 2-10 | 0.00％ |
| Y-axis pulling and pressing r 2-5 | 0.00％ | XOY plane shear r 2-11 | 0.00％ |
| Z-axis pulling and pressing r 2-6 | 0.00％ | XOZ plane shear r 2-12 | 0.22％ |
| xOZ plane X-direction bend r 2-7 | 8.91％ | YOZ plane shear r 2-13 | 22.80％ |
| Y-direction bending r of YOZ plane 2-8 | 8.91％ | xOY plane reverse shear r 2-14 | 58.56％ |
| xOZ plane Z-direction bend r 2-9 | 0.00％ | XOY plane twist r 2-15 | 0.60％ |
As can be seen from fig. 32, under the dead weight, the four simple aperture plates are mainly bent and deformed in the X-direction of the XOZ plane (dark gray area) at the edge support positions of x=0 and x=1200 mm, and are mainly bent and deformed in the Y-direction of the YOZ plane (white area) at the edge support positions of y=0 and y=1200 mm; near the open hole, the stress is more complex, the round hole is mainly bent and deformed in the X direction of the XOZ plane up and down, and the round hole is mainly bent and deformed in the Y direction of the YOZ plane left and right; four corner regions, mainly those with an XOY plane reverse shear deformation (light gray regions), occur. Fig. 32 more clearly shows the area occupied by various principal basic deformations than the finite element method, providing sufficient theoretical basis for what kind of destruction of the structure occurs.
The above-mentioned embodiments are merely preferred embodiments of the present invention, which are not intended to limit the scope of the present invention, and other embodiments can be easily made by those skilled in the art through substitution or modification according to the technical disclosure in the present specification, so that all changes and modifications made in the principle of the present invention shall be included in the scope of the present invention.
Claims (5)
1. The deformation decomposition method of any triangular prism unit in the space structure is characterized by comprising the following steps of:
step 1: aiming at any triangular prism unit under a space rectangular coordinate system, constructing basic displacement and deformation basis vectors of the any triangular prism unit according to geometric characteristics, stress balance and orthogonal theoretical conditions of the triangular prism unit, and further obtaining a complete orthogonal mechanical basis matrix P; 18 basic displacement and deformation basis vectors p of any triangular prism unit 1 ～p 18 The method comprises the following steps:
p 1 the linear displacement base vector is an X-axis rigid body:
p 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 the linear displacement base vector is the rigid body in the Y axial direction:
p 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 the base vector is the linear displacement of the Z-axis rigid body:
p 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 4 the deformation base vector is drawn and pressed in the X-axis direction:
p 4 ＝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 ；
p 5 the deformation base vector is pulled and pressed in the Y axial direction:
p 5 ＝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 ；
p 6 the deformation base vector is drawn and pressed in the Z axial direction:
p 6 ＝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 ；
p 7 the deformation base vector is the X-direction bending deformation of the XOZ plane:
p 7 ＝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 ；
p 8 the Y-direction bending deformation base vector of the YOZ plane is as follows:
p 8 ＝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 ；
p 9 the deformation base vector is the Z-direction bending deformation of the XOZ plane:
p 10 the Z-direction bending deformation base vector of the YOZ plane is as follows:
p 11 for the XOY plane shear deformation basis vector:
p 12 is the shear deformation basis vector of the XOZ plane:
p 13 the shear deformation basis vector is YOZ plane:
p 14 for the XOY plane reverse shear deformation basis vector:
p 15 the basis vector is the XOY plane torsional deformation:
p 16 the base vector is the XOY plane rigid body rotational displacement:
p 17 the rotational displacement basis vector is an XOZ plane rigid body:
p 18 the rotational displacement basis vector is YOZ plane rigid body:
wherein:
u＝a 2 +c 2 +e 2 -ac-ae-ce；v＝-b 2 -d 2 -f 2 +bd+bf+df；w＝2(bc-ad+af-be-cf+de)；
a is the difference value between the node 1 or the node 4 in the triangular prism unit and the X axis of the centroid O' of the triangular prism unit;
b is the difference value between the node 1 or the node 4 in the triangular prism unit and the Y-axis of the centroid O' of the triangular prism unit;
c is the difference value between the node 2 or the node 5 in the triangular prism unit and the X axis of the centroid O' of the triangular prism unit;
d is the difference value between the node 2 or the node 5 in the triangular prism unit and the Y-axis of the centroid O' of the triangular prism unit;
e is the difference value between the node 3 or the node 6 in the triangular prism unit and the X axis of the centroid O' of the triangular prism unit;
f is the difference value between the node 3 or the node 6 in the triangular prism unit and the Y-axis of the centroid O' of the triangular prism unit;
l is the height of the triangular prism unit Z in the axial direction;
step 1.2: the 18 basic displacements and the deformation base vector p 1 ～p 18 Constructed as a perfect orthomechanical basis matrix P:
P＝[p 1 ,p 2 ,p 3 ,p 4 ,p 5 ,p 6 ,p 7 ,p 8 ,p 9 ,p 10 ,p 11 ,p 12 ,p 13 ,p 14 ,p 15 ,p 16 ,p 17 ,p 18 ]；
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 receiving any load working conditions, and further obtaining node coordinate displacement vectors s of the triangular prism units;
step 3: projecting the node coordinate displacement vector s of the triangular prism unit onto a complete orthomechanical base matrix P to obtain a basic displacement and deformation projection coefficient vector r of the triangular prism unit;
step 4: and judging the main deformation and the secondary deformation of the triangular prism unit according to the magnitude of the projection coefficient in the projection coefficient vector r, so that deformation decomposition and deformation identification of any spatial structure can be realized.
2. The deformation decomposition method of any triangular prism unit in a spatial structure according to claim 1, wherein the vertices of the triangular prism unit are designated as node 1, node 2, node 3, node 4, node 5, and node 6, and the spatial integrated deformation of the triangular prism unit is formed by superposition of 18 basic displacements and deformations of X-axial rigid body linear displacement, Y-axial rigid body linear displacement, Z-axial rigid body linear displacement, X-axial tension-compression deformation, Y-axial tension-compression deformation, Z-axial tension-compression deformation, XOZ-plane X-direction bending deformation, YOZ-plane Y-direction bending deformation, YOZ-plane Z-direction bending deformation, XOY-plane shear deformation, XOZ-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.
3. The method for decomposing deformation of any triangular prism unit in a spatial structure according to claim 1, wherein step 2 specifically comprises the steps of:
step 2.1: establishing a space structure model under a space rectangular coordinate system, dividing the space structure model by adopting triangular prism units, and obtaining a node coordinate value d of an ith triangular prism unit in the discrete 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 i i=1, 2,3,4, … …, m, which is the node coordinates of the ith triangular prism unit in the spatial structure model;
x i-j representing coordinate values of the node j in the ith triangular prism unit in the X-axis direction;
y i-j representing the coordinate value of the node j in the ith triangular prism unit in the Y-axis direction;
z i-j representing the coordinate value of the node j in the Z-axis direction in the ith triangular prism unit, j=1, 2,3,4,5,6;
step 2.2: the coordinate value of the node of the ith triangular prism unit in the space structure model after any displacement and deformation under any load working condition is d 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' i the method is characterized in that the node coordinates of the ith triangular prism unit in the space structure model after any displacement and deformation are generated under any load working condition, i=1, 2,3,4, … … and m;
x' i-j the coordinate value of the ith triangular prism unit after the node j generates any displacement and deformation in the X-axis direction under any load working condition is shown;
y' i-j the 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-j the coordinate value of the ith triangular prism unit after the node j generates any displacement and deformation in the X axis direction under any load working condition is represented, j=1, 2,3,4,5 and 6;
step 2.3: will d' i Subtracting d i Obtaining the node coordinate displacement vector s of the ith triangular prism unit i ：
4. A method for deformation decomposition of any triangular prism unit in a spatial structure according to claim 3, wherein said step 3 specifically comprises the steps of:
step 3.1: the node coordinate displacement vector s of the ith triangular prism unit i Projection onto corresponding complete orthomechanical basis matrix P i And (3) obtaining:
s i ＝r i ·P i T ；
step 3.2: converting the above 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 i The node coordinate displacement vector of the ith triangular prism unit in the space structure model; p (P) i Is the complete orthodynamic matrix, P, of the ith triangular prism unit in the space structure model i T Is P i Transposed matrix of (P) i T ) -1 Is P i T An inverse matrix of (a); r is (r) i Complete orthodynamic basis matrix P for ith triangular prism unit in space structure model i A 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-k k=1, 2,3, … …,18, which is the kth basic displacement and deformation basis vector of the ith triangular prism unit in the spatial structure model; r is (r) i-k Complete orthodynamic basis matrix P for ith triangular prism unit in space structure model i Corresponding projection coefficient vector r i K=1, 2,3, … …,18.
5. The method for decomposing deformation of any triangular prism unit in a spatial structure according to claim 4, wherein the step 4 specifically comprises:
complete orthodynamic basis matrix P for ith triangular prism unit in space structure model i Corresponding projection coefficient vector r i Projection coefficient r of 12 basic deformations i-4 ～r i-15 Comparing the absolute values of the three prism units, and judging the basic deformation corresponding to the projection coefficient with the maximum absolute value as the main deformation of the three prism units; the basic deformation corresponding to the projection coefficient with the next largest absolute value is judged to be the secondary deformation of the triangular prism units, so that the basic deformation component information of all the triangular prism units in the space structure is obtained, and finally, the deformation decomposition and the deformation identification of the space structure are realized; wherein the projection coefficient r i-4 、r i-5 、r i-6 When the deformation is positive, 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-6 And when the deformation is negative, the corresponding basic deformation is the compressive deformation of the ith triangular prism unit in the space structure model.
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