Алгоритм метода сил в расчетах статически неопределимых ферм

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1. Introduction The duality of the displacement method and the force method in structural mechanics is well known, and the application of the methods for “manual” analysis structures is approximately equally labor-intensive. There are certain classes of problems where one or another method may be convenient, for example, in terms of the number of unknowns, but the methods can be considered to be on the same footing. However, the equality of the methods is lost when CAE packages are used to analyze structures. The absolute majority of commercial software packages are based on the displacement method. The advantage of the displacement method is the relative simplicity of its algorithmization, the matrix of governing equations is unambiguously determined by the numbering of the structure nodes. At the same time, the stiffness matrix has a band structure, is sparsely populated and, generally, is well-conditioned. In contrast, the matrix of the governing equations of the force method can be formed in a non-unique way. From the point of view of classical structural mechanics, this is explained by the non-uniqueness of the choice of the primary system. In turn, the force method is important in optimization problems and exhibits efficiency in comparison with the displacement method [1-3], nonlinear analysis of structures [4-6], analysis of prestressed structures [7], and design of adaptive structures [8]. The known ways of algorithmization of the force method can be divided into three main groups: algebraic, topological and mixed. Algebraic methods are generally reduced to particular operations on the matrix of nodal equilibrium equations. One of the first attempts to construct an algorithm for selecting redundant unknowns was the use of the Gauss-Jordan method [9; 10]. Subsequently, LU decomposition methods [11-13], singular value decomposition [14; 15] were proposed, as well as one of the methods for constructing the general solution presented in [16]. The mixed algebraic-topological methods are discussed in papers [17-20]. The main disadvantage of the algebraic methods is the necessity of preliminary application of complex operations on the matrix of equilibrium equations of the structure to form the matrix of governing equations of the force method. This disadvantage prevents from constructing an efficient algorithm in terms of the number of computational operations. Topological methods are based on the use of the geometric properties of the structure, such as periodicity and cyclicity. The methods proposed in [21-23] can be referred to topological methods. One of the varieties of such methods is discussed in studies [24-26] devoted to the use of the fundamental basis of cycles of a graph, which is equivalent for the considered structure. Algorithms that exploit the cyclic nature of the structure have been proposed [27; 28]. The issue with the methods based on periodicity or cyclicity of the structure is that they cannot be applied to problems with arbitrary geometry. The use of graph operations has the same disadvantage as the algebraic methods. The widely used integrated force method, first proposed in [29], can be highlighted. Currently, the integrated force method has been generalized to plane and spatial problems of elasticity theory and nonlinear problems [30-32]. The key idea of the method is to solve the joint system of equilibrium equations of the structure and strain compatibility equations. However, the structure of the obtained matrix and the number of unknowns do not indicate efficiency of the method in comparison with the displacement method. Thus, no algorithm for the force method comparable in complexity to the displacement method has been constructed so far. This paper presents a method for the analysis of statically indeterminate trusses. The key idea is to use the transposed strain compatibility matrix as the matrix of general solution of the homogeneous equilibrium equations. 2. Method 2.1. Problem Statement of Force Method Algorithmization The equations of structural mechanics of trusses can be written in the form of the following system of equations: ATN P= , (1а) AU=ε=ε +ε0 e, (1b) ε =e ΛN, (1c) where AT is the specified nodal equilibrium matrix; [ ]... T is the matrix transpose operation; N is the column of axial forces in the truss members; P is the column of specified nodal loads; U is the column of nodal displacements; ε is the column of axial strains of the members; ε0 is the column of specified initial strains of the members; εe is the column of elastic strains of the members; Λ=diag l( i / EAi ) is the flexibility coefficient matrix of the members of the system; li is the length of the i-th member; EAi is the axial stiffness of the i-th member. Equation (1a) represents the equilibrium equations of the system nodes, (1b) represents the geometric equations relating displacements and deformations, and (1c) represents the constitutive equations relating forces and deformations. It is known that the general solution of a non-homogeneous system of equations is the sum of some particular solution of this system and the general solution of the corresponding homogeneous system of equations. In statically indeterminate systems, the rank of matrix AT is equal to the number of its rows and is obviously less than the number of unknowns, and therefore, the system of the homogeneous equilibrium equations has a nontrivial solution. The construction of the general solution is the main difficulty in the algorithmization of the force method. Suppose that the fundamental system of solutions of the homogeneous system is constructed. The columns of the fundamental system are taken as the rows of some matrix B . By definition of the fundamental system: ABT T =0. (2) Therefore, for an arbitrary column F the following is valid: AB FT T =0. (3) Thus, BT F is the general solution of the system of the homogenous equilibrium equations. Considering an arbitrary particular solution and (1а): N = B FT + NP. (4) By transposing (2), one obtains: BA=0. (5) Multiplying (1b) by B yields: BAU = ε=B B(ε +ε0 e), (6) and taking into account (5), the following is valid for any column U : B(ε +ε =0 e ) 0. (7) By substituting (3) into (1c), one obtains: ε =e ΛBT F + ΛNp. (8) Substituting (8) into (7) yields the governing system of equations of the force method: B B FΛ T + Bε +0 B NΛ p = 0. (9) Similar to the method of displacements, BΛBT is the flexibility matrix of the structure. The solution of the problem is now reduced to the solution of system (9), the forces in the structural elements are recalculated according to (4). Expression (7) has the physical meaning of the strain compatibility equations. From the algebraic point of view, the transpose of compatibility matrix B produces the desired general solution of the homogeneous equilibrium equations. This is the essence of the proposed method, as it will be shown later, the strain compatibility equations can be constructed without using the nodal equilibrium matrix. The physical meaning of the unknown column F in system (9) remains unknown. There is no need to choose the primary system and “extra” unknowns, the vector of unknowns is formed “automatically”. 2.2. Construction of Strain Compatibility Equations The idea behind the proposed method of constructing the strain compatibility equations is the relationship between the strain of particular members constituting a loop and the change in area of this loop. For illustration, a truss cell, which is statically indeterminate to the first degree, is shown in Figure 1, a. Here, the members are numbered, arrows indicate their orientation, and letters i k m s, , , denote the nodes of the structure. It should be noted that the numbering and orientation of the members do not affect the final result. The cell under consideration consists of three independent loops 1, 2, 3 denoted in Figure 1, b. These three loops together constitute the fourth one, shown in Figure 1, c. а b c Figure. 1. Truss diagram: а - numbering of members and nodes; b - loops No. 1, 2, 3; c - loop No. 4 S o u r c e : made by V.V. Lalin, T.R. Ibragimov The following relationship is valid for the areas of the considered loops: S4 = S1 + S2 + S3, (10) where is the area of the j-th loop. After deformation of the structure due to external loads, the areas of the loops will change, but for the new values of the areas the same identity will be true due to the continuity of the structure: S'4 = +S1' S2' + S3' . (11) By denoting the change in area of the j-th loop as ΔS j = S 'j - S j ,the following relationship is obtained: ΔS4 =ΔS1 +ΔS2 +ΔS3. (12) Expression (12) has the meaning of the continuity condition of the loop area. If expressed through the member strains, equation (12) will be the desired equation of strain compatibility of the considered truss cell. 3. Results and Discussion 3.1. Strain Compatibility Equation of 6-member and 4-node Cell Before obtaining the strain compatibility equations of the truss cell in Figure 1, an arbitrary member in x, y plane and nodes i and k is considered. The member is oriented by unit vector tT = tx ,ty . Unit vector nT = n nx, y , which is normal to vector t , is introduced such that vectors t n z, , constitute a righthand vector system, similar to coordinate system x, , .y z The nodal displacements of the member are written as UiT = Uix ,Uiy , UkT = Ukx,Uky . Axial strains of the member can now be expressed as: ε= tT (Uk -Ui ). (13) The following notation is introduced: ω= nT (Uk -Ui ). (14) Thus, ω represents the relative displacement along the normal to the axis of the member, that is, the relative displacement of the nodes corresponding to the rotation of the member as a rigid body. The following expression follows from equations (13), (14), which relates the member strains and the displacements of its nodes: U k -Ui =ε +ωt n. (15) A convenient tool for evaluating the change in area is the outer product operation [33]. The outer product of two vectors a = a ax, y T , b = b bx, y T lying in the x, y plane can be written as: ax bx a ∧ =b det[a b, ]=, (16) ay by where det … is the matrix determinant. The main properties of the outer product [33]: a ∧ =- ∧b b a, a ∧ λ =λ ∧ = λ ∧( b) (a b) ( a) b, λ∈ , a ∧ + = ∧ + ∧(b c) a b a c, a ∧ = ↔b 0 a bP , a b, ≠ 0. (17) The outer product is the oriented area of the parallelogram constructed with the multiplied vectors, that is, it is equal to the area of the parallelogram with a positive or negative sign depending on whether the axis triples x, ,y z and a b z, , coincide in orientation or not. Now loop 4 is examined to determine its change in area. Unit vectors for each member (Figure 2) are introduced. Let r r r, , be the position vectors of the nodes of loop i k m Figure. 2. Loop No. 4 S o u r c e: made by V.V. Lalin, T.R. Ibragimov i k m, , having an arbitrary origin. By using outer product, the change in area may be written as: . (18) By expanding the brackets and grouping the terms, the following is obtained: 2ΔS4 =Ui ∧Uk +Uk ∧Um +Um ∧Ui +(Um -Uk )∧ +ri (Uk -Ui )∧ +rm (Ui -Um )∧rk . (19) It should be noted that the second group of terms consists of the products of the difference between the displacements of the nodes and the position vector of the opposite node. The strains of each of the three members of the loops are expressed according to (15): Uk -Ui =ε +ω11t 1 1n ; Um -Uk =ε +ω2 2t 2 2n ; (20) Ui -Um =ε +ω3 3t 3 3n . Thus, expression (19) will include the products of the member strains (and their rotation as a rigid body) and the position vectors of the opposite nodes. According to the properties of the outer product, the rotation component can be eliminated by requiring the position vector and vector normal to the member to be parallel. For any triangular loop this is possible if the orthocenter of the triangle (point in Figure 2) is taken as the origin of the position vector. By rewriting (19), leaving the non-zero terms, one obtains: 2ΔS4 =Ui ∧Uk +Uk ∧Um +Um ∧Ui +ε ∧1 1t rmikm +ε2 2t ∧riikm +ε ∧3 3t rkikm, (21) where the upper index ikm denotes that the origin of the position vector is the orthocenter of triangle ikm. Similarly, the changes in area of loops 1, 2, 3 are determined: 2ΔS1 =Us ∧Uk +Uk ∧Ui +Ui ∧Us +ε ∧1 1t rsski -ε4 4t ∧ rkski +ε ∧5 5t riski ; (22a) 2ΔS2 =Us ∧Um +Um ∧Uk +Uk ∧Us +ε2 2t ∧ rssmk -ε ∧5 5t rmsmk -ε ∧6 6t rksmk ; (22b) 2ΔS3 =Us ∧Um +Um ∧Ui +Ui ∧Us +ε ∧3 3t rssmi +ε4 4t ∧ rmsmi +ε ∧6 6t rismi . (22c) Substituting now (21)-(22) into (12), it can be seen that the quadratic displacement terms are identically eliminated. The obtained expression will be the strain compatibility equation for a statically indeterminate to the first degree truss cell: ε ∧11t rmikm +ε2 2t ∧ riikm +ε ∧3 3t rkikm = =ε ∧11t rsski -ε4 4t ∧ rkski +ε ∧5 5t riski + +ε2 2t ∧ rssmk -ε ∧5 5t rmsmk -ε6 6t ∧ rksmk + (23) +ε ∧3 3t rssmi +ε4 4t ∧ rmsmi +ε ∧6 6t rismi. Expression (23) is valid for any nondegenerate cell consisting of 6 members connected in a similar way to the considered case. Thus, for the cell shown in Figure 3, the outer product for the terms of the second loop will be obtained with a negative sign and expression (12) will be reduced to the following form: ΔSikm +ΔSksm =ΔSiks +ΔSism. (24) Figure 3. Topologically similar statically indeterminate cell Figure 4. Equilateral triangle truss panel S o u r c e: made by V.V. Lalin, T.R. Ibragimov S o u r c e: made by V.V. Lalin, T.R. Ibragimov Strain compatibility matrix B can now be constructed from the rows of the strain compatibility equations for each statically indeterminate elementary cell. Thus, structure flexibility matrix L = B BΛ T is uniquely determined by the numbering of statically indeterminate loops of the system. At the same time, it is not necessary to use matrix AT of nodal equilibrium equations to construct the structure flexibility matrix. The algorithm of analysis using the force method comes down to the construction of strain compatibility equations for independent statically indeterminate cells in order to form the strain compatibility matrix of the system, construct of the structure flexibility matrix and solve the governing system. Figure 4 demonstrates a structure in the form of an equilateral triangle with base a and node s in the center of mass of triangleikm. Construction of the strain compatibility equations for this system is presented below. Unit vectors ti for the members are expressed as: t1 = 1 , t2 = -1 , t =- 1 1 , 02 3 1 3 0 t4 = , t5 = , t6 = . 2 1 2 1 1 The position vectors originating from the orthocenter and pointing to the nodes of loop 4: riikm =-a 1 1 , rkikm = a 1 -1 , rmikm = a 0 . 2 3 / 3 2 3 / 3 3 / 3 As a result of calculating the outer products, the following is obtained: ε ∧ =ε11t rm 1 a , ε ∧ =ε2 2t ri 2 a , ε ∧ =ε3 3t rk 3 a . Thus, the left-hand side of expression (12): a (ε +ε +ε1 2 3). Similar procedure is applied to loops 1, 2, 3 to obtain: 1 loop iks: a ε +ε -1 5ε1 , 1 loop kms: a ε +ε -5 6ε2 , loop ism : a ε +ε -4 6ε3 . By substituting the obtained expressions into (23), expanding the brackets and grouping the terms, the strain compatibility equation is derived: a ε + ε +ε -1 2 3 3(ε + ε +ε4 5 6) = 0. (25) 3.2. Cross Brace Truss As mentioned earlier, expression (23) is suitable for any cell with 6 members, however, there is an important degenerate case for which equation (23) is not acceptable. Consider the design shown in Figure 5. The requirement of taking the orthocenter of the triangle as the origin of the position vector makes the terms accounting for the strains of members 5 and 6 equal to zero, since the orthocenter of a right triangle is at the apex of a right angle. Figure 5. Rectangular truss panel S o u r c e: made by V.V. Lalin, The strain compatibility equation can be obtained from the T.R. Ibragimov following equality for the areas: ΔSiks +ΔSism =ΔSiko +ΔSkso +ΔSmos +ΔSiom. (26) In this case, the following equalities must be used: Um - = - =Uo Uo Uk 1 (Um -Uk ); 2 1 Ui - = - =Uo Uo Us (Ui -Us ). (27) 2 For the orientation of the members shown in Figure 5, the following unit vectors are used: 1 0 -1 t1 = , t2 = , t3 = ; 0 1 0 t4 = -01 , t5 = 1c -ba , t6 = 1c ba , where c = a2 +b2. For the loop constructed with vertices i k s, , the orthocenter is point k , therefore a 0 ri =- ,rs = , 0 b and to the nearest quadratic terms: 2ΔSiks =ε2 2t ∧ +ε ∧ =ε +εri 1 1t rs 2a 1b. Similarly, for loop ism : 2ΔSism =ε +ε4a 3b. For the loop with vertices i, k, o: ro = b2 2-ba2 10 , ri = - a2 a b1/ , rk = a2 -a b1/ ; 2ΔSiko = ε1 1t ∧ ro + 1 ε5 5t ∧ ri + 1 ε6 6t ∧ rk = a c⋅ (ε + ε5 6 )-ε1 b2 - a2 . 2 2 2b 2b Similarly, the change in area of the loop with vertices m o s, , , 2ΔSmos = a c⋅ (ε + ε5 6 ) -ε3 b2 - a2 . 2b 2b For the loop with vertices i o m, , : ro = b2 - a2 1 , r = -b b a/ , rm = b -b a/ ; i 2a 0 2 1 2 1 2ΔSiom = ε4 4t ∧ ro - 1 ε t ∧ r + 1 ε t ∧ r = b c⋅ (ε + ε5 6 ) -ε4 a2 -b2 . 6 6 m 5 5 i 2 2 2a 2a Similarly, for loop kos : 2ΔSkos = b c⋅ (ε + ε5 6 ) -ε2 a2 -b2 . 2a 2a After substitution into condition (26), the following strain compatibility equation is obtained after simplifications: a(ε +ε + ε +ε -1 3) b( 2 4) a2 + b2 (ε +ε =5 6) 0. In the particular case of a square cell (a = b), the equation becomes: (28) a ε +ε +ε +ε -1 2 3 4 2(ε +ε5 6) = 0. (29) A similar expression is given in [34], where it was obtained by analyzing the matrix of nodal equilibrium equations of the structure. Using the obtained expression (29), the formation of the structure flexibility matrix of the example truss presented in Figure 6 is discussed below. The truss consists of n square cells, the axial stiffness of each member is EA. The members are numbered according to the scheme shown in Figure 6. The total number of members in the truss is 1+ 5n, the total number of nodes is 2(n +1). The diagonal matrix of the member flexibility coefficients: Λ= a diag 1,1,1,1, 2, 2,1,…,1, 2, 2 . EA Figure 6. Rectangular truss S o u r c e: made by V.V. Lalin, T.R. Ibragimov The strain compatibility matrix of the system, according to (29), will have the following form (only the first three rows are shown): 1 1 1 1 - 2 - 2 0 L B = a 0 0 0 1 0 0 1 1 1 - 2 - 2 0 L . L 0 1 0 0 1 1 1 - 2 - 2 0 L By multiplying out Λ , the following tridiagonal flexibility matrix of the system is obtained: 4 1( + 2) 1 L = a3 1 4 1( + 2) 1 . (30) EA O 1 4 1( + 2) Thus, the obtained flexibility matrix has the size of n n× . The displacement method stiffness matrix, in turn, will have the dimension of 4(n+1 .) 3.3. Externally Statically Indeterminate Trusses In this section, the problem of composing the strain compatibility equations for externally statically indeterminate trusses is considered. These are trusses, the support reactions of which cannot be determined from the equilibrium equations. Figure 7, a shows a statically indeterminate to the first degree cell with two independent loops 1 and 2, and the third loop denoted in Figure 7, b. An obvious equality is true for the areas of the loops: ΔS3 =ΔS1 +ΔS2. (31) In the case of no additional support, the cell would be statically determinate and the quadratic displacement terms in equation (31) would not reduce. The presence of supports leads to some constraints on the displacements of the nodes, as a result of which the quadratic terms are reduced and the equality can be expressed through the member strains. a b Figure 7. Externally statically indeterminate truss: a - loops No. 1, 2; b - loop No. 3 S o u r c e: made by V.V. Lalin, T.R. Ibragimov The changes in area of the loops taking into account that Ui -Uk =Ui -U s +U s -Uk : 2ΔS3 =Ui ∧Um +Um ∧Uk +Uk ∧ +Ui +(Uk -Um )∧ rimik +(Um -Ui )∧ rkmik +(Us -Uk )∧ rmmik ; 2ΔS1 =Ui ∧Um +Um ∧Us +Us ∧ +Ui (32a) +(Us -Um )∧ riism +(Um -Ui )∧ rsism +(Ui -Us )∧ rmism; 2ΔS2 =Um ∧Uk +Uk ∧Us +Us ∧Um + (32b) . (32c) For the given cell, the displacements of nodes , are parallel, hence by the properties of the outer product: U s ∧U k = 0 . (33) In turn, the displacement of node is zero and the remaining non-zero terms are: 2ΔS3 =Ui ∧Um +(Uk -Um )∧ rimik +(Um -Ui )∧ rkmik +(Us -Uk )∧ rmmik ; (34a) 2ΔS1 =Um ∧Us +(Us -Um )∧ riism +(Um -Ui )∧ rsism +(Ui -Us )∧ rmism ; (34b) 2ΔS2 =Um ∧Uk +Us ∧Um +(Us -Uk )∧ rmskm +(Um -Us )∧ rkskm +(Um -Uk )∧ rsskm . (34c) As seen from expressions (34), the quadratic terms are identically eliminated when substituted into expression (31). The remaining ones, written in terms of member strains in accordance with (15), represent the strain compatibility equation for the considered externally statically indeterminate truss. By taking, for example, the lengths of members 1, 2, 3 equal to , and correspondingly the lengths of members 4, 5 equal to 2a , it is possible to obtain the following strain compatibility equation using (31): a ε +ε + ε -1 2 2 3 2(ε +ε4 5) = 0. (35) 4. Conclusion 1. The main problem in the algorithmization of the force method is finding the general solution to the homogeneous equilibrium equations of the structure A NT = 0 . The method of obtaining the strain compatibility equations completes the construction of the algorithm for solving the problems of statically indeterminate trusses using the force method. 2. The proposed formulation of the force method allows to not have to select the “primary system” and the unknowns of the force method. The proposed method automatically “selects” the vector of unknowns . The numbering of statically indeterminate loops unambiguously determines the structure of the flexibility matrix of the system. 3. The advantage of the proposed method is that the equilibrium equations of the structure are not required. There is no need to store in the computer memory and use the matrix of nodal equilibrium equations of the structure AT .

Об авторах

Владимир Владимирович Лалин

Санкт-Петербургский государственный архитектурно-строительный университет; Российский университет дружбы народов

Email: vllalin@yandex.ru
ORCID iD: 0000-0003-3850-424X
SPIN-код: 8220-6921

доктор технических наук, профессор Высшей школы промышленно-гражданского и дорожного строительства Инженерно-строительного института, Санкт-Петербургский политехнический университет Петра Великого; профессор кафедры технологий строительства и конструкционных материалов инженерной академии, Российский университет дружбы народов

Санкт-Петербург, Россия; Москва, Россия

Тимур Равилевич Ибрагимов

Санкт-Петербургский государственный архитектурно-строительный университет

Автор, ответственный за переписку.
Email: timuribragimov.ra@gmail.com
ORCID iD: 0000-0002-2742-1345
SPIN-код: 5342-2799

аспирант Высшей школы промышленно-гражданского и дорожного строительства Инженерно-строительного института

Санкт-Петербург, Россия

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