Numerical Limit Analysis of Rigid Plastic Structures

Transcription

1 Numerical Limit Analysis of Rigid Plastic Structures Hector Andres Tinoco Navarro Institute of Technology and Innovation University of Southern Denmark Supervisors: Prof. Ph.D. Linh Cao Hoang University of Southern Denmark Prof. Ph.D. Kristian Krabbenhoft University of Newcastle July 7, 2012

2 1

3 Contents 1 Introduction Introduction Rigid Plastic Bar Rigid Plastic Beam Plastic Moment Linear Programming Introduction to Optimization Problems Linear Programming Problems Simplex Method Interior Point Methods Analysis of Rigid Truss and Beam Structures Introduction Trusses Statically Determinate Force Method for the Analysis of Statically Determinate Structures Generalization of the Force Method for the Analysis of Statically Determinate Structures Trusses Statically Indeterminate Force Method for the Analysis of Statically Indeterminate Structures Rigid Beam Structures Equilibrium Equations for a Elemental Beam Structures with Elemental Beams Lower Bound Method Applied to Structures Lower Bound Theorem Application of the Lower Bound Theorem to Rigid Truss Structures Application of the Lower Bound Theorem to Rigid Beam Structures Estimation of the Collapse Mode with Sensitivity Analysis Numerical Examples for Truss Structures Example: Statically Determinate Truss Structures Examples: Statically Indeterminate Truss Structures

4 4.6 Numerical Examples for Beam Structures Analysis of Truss Structures with Material and Energy Minimization Energy Minimization for Structural Analysis in Trusses Elastic Structural Analysis as an Optimization Problem Applied to Trusses Example: Example [21] Material Minimization with the Collapse Load λα Example 1: Example of Figure Example 2: Example of Figure 25 Modied Conclusions 59 3

5 Chapter 1 Introduction 1.1 Introduction In the analysis of structures elastic calculations based on the nite element method (FEM), are often used to determine the displacements, strains and stresses. However, FEM depends on the elastic properties or constitutive relations in its formulation for determining the general state of stresses in structural elements. With this methodology, it is possible to analyze dierent combinations of external loading due to the rules of superposition. But, those rules are not applicable to the plastic region. To carry out the plastic analysis with FEM, it is necessary to perform a nonlinear analysis (Newton-Raphson method, arc length method, Piccard method, etc.) that resolves the dependency of the stiness matrix with the stress level. When FEM is applied, it is possible to obtain the load-displacement curve searching the ultimate load on the plastic region. The external load that plasties a structure is called the ultimate load or collapse load and this is considered an essential design criterion for the structural analysis. In structural design the process of plastic analysis by means of FEM can be very slow, and computationally complex to implement. The plastic analysis of structures can be carried out with limit state analysis [2], which is based on the idealization of the stress-strain curve as rigid, perfectly plastic (see Figure 1b)). For rigid perfectly plastic behavior, the linear equations of elasticity are not applicable. Essentially, limit state analysis of structures rests on three extremal principles of plastic theory. These are: lower bound theorem, upper bound theorem and the uniqueness theorem. The lower bound and upper bound theorems are based on static and kinematic principles and can be used to obtain the load capacity of structures as is shown by [3] and [4]. Those theorems were formulated initially by [5] and 4

6 independently by [6]. Mathematically, the applicationn of these theorems will require solution of constrained optimization problems. Recently, many applications have been developed with these theorems, which are combined with FEM to carry out plastic analysis in structures (see [7, 8]). In this report the lower bound theorem is applied to truss and beam structures using a methodology of nite elements as is detailed in Chapter 3. Some methods for resolving linear programming problems applying optimization techniques are described in Chapter Rigid Plastic Bar Figure 1a) shows a bar axially loaded by a force β = σa, where σ is the stress caused by the force β on the cross sectional area A, which is considered uniformly distributed. An axially loaded bar can have two states of load, tension and compression as shown in Figure 1a). Figure 1: a) Bar subject to an axial force β. b) Uni-axial stressstrain relation for a bar of rigid-plastic material. Figure 1b) shows the behavior of a rigid perfectly plastic material, which is assumed for the bar of Figure 1a). This behavior means that for a yield stress state σ y, it is possible to have innite strains in tension or compression, respectively. In this gure the stress σ applied to the bar cannot exceed the limits given by the yield stress in tension σ y +, and the yield stress in compression σy. All the values of σ that satisfy this 5

7 condition are statically admissible, i.e. Then, it is possible to establish a condition of plastic admissibility such that σ y σ σ + y. (1.1) 1.3 Rigid Plastic Beam For understanding the plastic behavior in a beam, the concept of plastic moment is introduced in this section. Figure 2: a) Rectangular cross section of a beam, b) strain distribution, b) stress by bending in a limit state, c) stress at an elastoplastic state, d) stress at the plastic limit state. Figure 2 shows a plastied beam cross section, where the applied moment m plasties (see Figure2d)) the section completely, when it reaches the limit state of stress Plastic Moment The stress and strain state in a rectangular beam cross section subjected to a bending moment m as shown in Figure 2. This loading conguration is called pure bending. In a beam, the stress state is linear along z axis, if the beam is in the elastic region (see Figure 2b) and 2c)). The strain distribution, on the other hand, is assumed to vary linearly regardless of the state of stress. The strain in a beam can be expressed 6

8 as follows, ε = zk, (1.2) where k is the curvature of the cross section. The bending moment when the cross section reaches the plastic limit state is determined by of M p. M p = 1 2 h 1 2 h σ y bzdz. (1.3) For the rectangular cross section of Figure 2 the plastic moment is obtained by M p = σ y bh 4. (1.4) Figure 2e) shows that the cross section is totally plastied when m reaches the value 7

9 Chapter 2 Linear Programming 2.1 Introduction to Optimization Problems An optimization problem is basically, a problem where a function can be minimized or maximized while at the same time satises a set of constrains. In mathematical programming this problem is expressed in a standard form as minimize f(x) (or maximize) (LP ) subject to : g(x) 0, (g(x) 0 or g(x) = 0) (2.1) x R n. Where f : R n R is called objective function. The constraints g : R n R p limit the space of the solutions, and these are called feasible solutions. x is the vector of decision of the feasible solution. Depending on the type of optimization problem (see Equation (2.1)), an objective function and its constraints can assume dierent forms, such as dierent solutions. These are described by the following statements: If f(x) and/or g(x) are nonlinear, where x = (x 1, x 2,..., x n ), x i R, i = 1, 2,..., n. This is a problem of nonlinear programming (NL). If f(x) and g(x) are linear, where x = (x 1, x 2,..., x n ), x i R, i = 1, 2,..., n. This is a problem of linear programming (LP). If f(x) and g(x) are linear, where x = (x 1, x 2,..., x n ), x i is integer i = 1, 2,..., n. This is a problem of integer linear programming (IPL). 8

10 If f(x) and g(x) are linear, where x = (x 1, x 2,..., x n ), and any x i is integer i = 1, 2,..., n. This is a problem of mixed integer linear programming (MIPL). 2.2 Linear Programming Problems A linear programming problem is the one in the form of Equation (2.1), where the objective function and the constraints are formed by linear equations. For solving a linear programming problem a standard form is established as maximize z = a T x (LP ) subject to Ax = b, (2.2) x 0. where A R m n and m n, x R n 1, a R n 1, b R n 1 and z is a scalar value. The solution of a linear programming problem consists in determining the vector x that satises the constraints, in which x maximizes or minimizes the objective function. The techniques used to determine an optimal solution numerically in a linear programming problem are based on two methods; simplex method and interior point methods. 2.3 Simplex Method The simplex method was presented by George Dantzig in 1947 (see [9]) as a systematic form of solution of linear programming problems. The optimal solution is found in one vertex dened by the intersection of the constraint functions. Figure 3 shows a basic description of the simplex method using two variables, i.e. n = 2. The method uses basically solutions that follow the path of the constrain functions until nding the vertice that represents the optimal solution. Actually, there are many modications of this algorithm. However, the principal algorithm that was initially proposed is easy to implement in programming languages. In this section a brief description of the method is shown. 9

11 Figure 3: Representation of the simplex method. To explain how the simplex method works, a linear programming problem can be represented in the format shown in Figure 4. The sets of variables related to the constraints should be dened, as follows: i : set of indexes of m basic variables. Example: i = {3, 4, 5}, then x 3, x 4, x 5 are basic variables. j : set of indexes of n m non basic variables. Example: j = {1, 2}, then x 1, x 2 are non basic variables. Note: It is veried that dim(i) + dim(j) = m + (n m) = n. Figure 4: Representation of a linear programming problem. 10

12 The problem of linear programming can be divided with relation to basic and non basic variables as is shown in Figure 5. Hence, it is considered a base i as a feasible base, then it is possible to establish that c T i x i + c T j x j = z A i x i + A j x j = b x i 0, x j 0., (2.3) Figure 5: Separation of LP in function of basic and non basic variables. The basic variables i can be described in function of non basic variables j, this means that x i = A 1 i b A 1 i A j x j. (2.4) Substituting x i in the objective function, it is obtained that Reorganizing Equation (2.5), it is solved that c T i (A 1 i b A 1 i A j x j ) + c T j x j = z. (2.5) c T i (A 1 i b) (c T i A 1 i A j + c T j )x j = z. (2.6) Dening κ T = c T i A 1 i i as and z 0 = κ T b, Equation (2.6) is rewritten in relation to base From Equation (2.4) is determined that 0x i + (κa j c T j )x j = z z 0. (2.7) 11

13 Ix i + A 1 i A j x j = A 1 i b. (2.8) Equations (2.7) and (2.8) can be organized in relation to the base i as is shown in Figure 6. Figure 6: Prepared form in relation to the base i. A set of new variables are dened for the problem such that  j = A 1 i A j, (2.9) ˆb = A 1 i b, (2.10) and ĉ = c j κ T A j, (2.11) where ĉ represents the relative cost to the base i. With the new variables, the principal problem (see Equation (2.3)) can be reorganized as ĉ T i x j = z z 0 Ix j + Âj x j = b x i 0, x j 0., (2.12) For a basic solution, x j = 0. Therefore, x i = ˆb. This represents a normal basic solution. Hence, it is noted that x i = ˆb = A 1 i b 0, This means that there is a basic feasible solution, such that z z 0 is the obtained value of the objective function. 12

14 2.4 Interior Point Methods From an introduction of an algorithm proposed by Karmarkar (see [10]), the interior point methods are the most notable for resolving problems of linear programming. Interior point methods are more ecient than the simplex method, since these are a modication of the simplex method. Figure 7 shows how a feasible solution is searched from an interior point of the feasible region in a linear programming problem dened. Figure 7: Representation of the interior point method. as: The interior point methods can be classied in four categories, respectively. Such 1. Ane Scaling Methods: These methods show a good performance in problems of large dimensions, despite that they are the easiest of implementing of the interior point methods. Ane scaling method was initially suggested by [11]. However, the method was presented with a modication such that a simplication of it was done using Karmarkar (see [10]) projective transformations. Those modications were proposed by the researchers [12, 13]. 2. Methods Based on Projective Transformations : These methods use projective transformations of the type x = X 1 x e T X 1 x, (2.13) where x R n at the current point, x represents the image of x in the new space of variables, X is a diagonal matrix that has as components the terms of 13

15 x (X = diag(x 1, x 2, x 3,..., x n )) and e T is a n dimensional vector of ones (e T = (1 1, 1 2, 1 3,..., 1 n ). The algorithm of Karmarkan (see [10]) belongs to this category. 3. Path-following methods: These methods use a technique called logarithmic barrier. The problems are solved by means of a sequence of sub-problems of such a way that the original problem is rewritten as n minimize z = a T x µ ln(x j ) j=1 (LP ) subject to Ax = b, x 0. (2.14) Hence, it can be proved that if µ 0, the sequence of solutions for LP µ is more close to the solution of LP. The set of solutions of LP µ for dierent µ provide the central path to search the optimal solution. The path-following method follows a path close to the central path to search the optimal value. The primal dual algorithm belongs to this category. 4. Potential Reduction Method: This method makes use of a potential function. The potential function is approached to + when x i s i 0, i = 1, 2, 3,..., n. On the other hand if µ 0, then µ = xt s. The solutions are close to the boundary n of the feasible region, however they do not approach to optimal solutions. In contrast, if the potential function is approached to, the solutions found for (x, y, s) are approximated to optimal solutions. The potential reduction method is based on the dual problem (see Equation (2.15)) of Equation (2.2). maximize b T y (LP ) subject to A T y + s = a. s 0. (2.15) This chapter showed a short description about linear programming. The details of the cited methods are not shown because the methods are not implemented in this study. Nevertheless, in this study are used two types of solvers for resolving the linear and nonlinear problems applied to structural problems shown. The solver used are: 14

16 MOSEK solver (see [18]) for MATLAB and MATAB solvers (functions linprog.m and fmincon.m). 15

17 Chapter 3 Analysis of Rigid Truss and Beam Structures 3.1 Introduction Rigid trusses are structures composed of straight rigid bars connected by means of pinned joints, i.e. the joints can be modelled as hinges, that cannot resist bending moments. Truss structures are subjected to loads at the joints. With this assumption the truss member works in tension or compression only (see Figure 1a)). In the case of beams the connections can be hinges as well as rigid connections. The rigid connections are used to transmit moments. If there are hinges between beam elements, the moments are not transmitted. 3.2 Trusses Statically Determinate A structure is statically determinate if all independent parameters (internal and reaction forces) can be solved by means of the equilibrium equations. In other words, there is the same number of equations as unknown parameters. Truss structures can be veried if they are statically determinate with the following expression N d = e + R 2J, (3.1) where e is the number of members in the structure, R is the number of reactions in 16

18 horizontal and vertical direction, J is the number of joints. Then, if N d = 0, the structure is statically determinate. But if N d > 0, the structure is statically indeterminate (see section 2.3). A particular case to Equation (3.1) is the following; if N d < 0, the structure is statically overdeterminate. It is due to the fact that there are more equations than unknowns parameters. A statically overdetermin ate structure could be a mechanism. Figure 8: a) Statically determinate truss. b) Internal and external forces applied to the joints. Consider the structure of Figure 8a) as a statically determin ate structure. Since, it has 5 joints which are connecting 6 members. Constraints in the nodes 1 and 5 are considered as pinned supports in both directions. Equation (3.1) is applied to verify the statement of determinacy, such that e = 6, R = 4, J = 5. It is veried that N d = 0, then the structure shown in the Figure 8 is statically determin ate Force Method for the Analysis of Statically Determinate Structures The force method is basically based on equilibrium equations. From equilibrium a number of equations are established for each node of the structure. The method aims to determine the independent parameters of a statically determin ate structure. For the load state of Figure 8b), the external forces f x3 and f y3 are applied to the node 3, the reactions R ix and R iy correspond to the constraints applied to the node i: i = 1, 5 and the magnitude of the internal forces N j associated to the element (truss member) j: j = 1, 2, 3, 4, 5, 6. The nodes k: k = 1, 2, 3, 4, 5, those have coordinates (x k, y k ), which are associated to the coordinate system x and y as is shown in Figure 8. Then, 17

19 the length of each element can be determined taking the nodes that correspond to each element of the following form l j = (x m x n ) 2 + (y m y n ) 2, (3.2) where m and n are the numerations of the nodes associated to the element j; m, n k. The direction for the structural elements is dened by a unitary vector such that nj = n jx î + n jy ĵ, (3.3) where n jx and n jy are the magnitudes of the unitary vector n j in x (unitary vector î) and y (unitary vector ĵ) directions. The components of n j represent the direction cosine and sine such that; n jx = cos(θ j ) and n jy = sin(θ j ). θ j is the angle with respect to the x axis of each element. Equation (3.3) can be rewritten in function of nodal coordinates as nj = (x m x n ) l j î + (y m y n ) l j ĵ. (3.4) Then, the components x and y of the internal forces N j in truss members can be obtained by N j = N j nj. (3.5) It is important to point out that a truss member generates two reactions in the nodes associated to the member, which are of equal magnitude but opposite direction as is shown in Figure 9. Applying the equilibrium equations on each node in well organized form (from node 1 until node 5) we can establish the following relations in matricial form H e β p + C e η p + αe p = 0, (3.6) T n 1x n 1y n 1x n 1y n 2x n 2y n 2x n 2y n 3x n 3y n 3x n 3y 0 0 where H e =, 0 0 n 4x n 4y 0 0 n 4x n 4y 0 0 n 5x n 5y n 5x n 5y n 6x n 6y n 6x n 6y [ ] T [ ] T β p = N 1 N 2 N 3 N 4 N 5 N 6, η p = R 1x R 1y R 5x R 5y, 18

20 Figure 9: Directions of internal forces in truss members, see Figure 8. [ αe p = f x3 f y ] T and Ce = T. The matrix H e can be called extended exibility matrix, it contains all direction cosines and sines associated to each element. α p e is a vector that contains all applied forces. η p is a vector that contains all reaction forces. The matrix H e can be simplied in such a way that the rank in its rows and columns are equal to the dimensional space of internal forces β p. The rows that correspond to the nodes with constraints (two rows for each node) are extracted of the matrix H e and of the vector α p e. The rows extracted are selected from C e, where the rows of C e that contains a one are eliminated of H e and α p e. For this case, 4 rows are extracted of H e and α p e ( in the node 1, rows 1 and 2 are removed; in the node 5, rows 9 and 10 are removed). The superscript p indicates particular example in the variables. Hence, we can eliminate and reorganize Equation (3.6) in two system of equations as H p β p + α p = 0, (3.7) and h p β p + η p = 0. (3.8) From Equation (3.7) the matrix H p is called exibility matrix obtained of H e (after removing rows) R 6 6 (dimensional space of β p in rows and columns, square matrix). 19

21 α p is the external force vector obtained of αe p (dimensional space of β p ), h p contains the rows and columns extracted of the matrix H e. First, Equation (3.7) can be resolved like a system of linear equations to obtain the values of β p. Then, Equation (3.8) is solved in direct form to obtain the values of η p. Another type of restrictions can be imposed in statically determinate problems, for instance roller support. The results of the force method should be the same Generalization of the Force Method for the Analysis of Statically Determinate Structures Equations (3.7) and (3.8) are obtained for a statically determinate structure (N d = 0). The structure presents p nodes, n elements, m reactions (reaction forces by pinned and roller supports) and s external forces as is shown in Figure 10. The assembly process for the matrix H t (equivalent to H p in section 3.2.1) should be the following; rst one, the matrix H e should be assembled based on the connectivity shown in Table 1. Figure 10: Generalized truss structure statically determinate. The process to assembly H e is dened in such a way that the contribution of each element (unitary vectors in x and y) is located in the matrix H e. Considering that in the matrix H e, the rows 1 and 2 represent the node 1, the rows 3 and 4 represent the node 2 until the rows 2p 1 and 2p that represent the node p, respectively. 20

22 Element Node 1 Node l r 1 r l + 1 r r + 1 l + 2 r + 1 r n 1 p 2 p 1 n p 1 p Table 1: Connectivity table for the generalized structure of Figure 10. We can obtain the matrix H e in general form as H e = n 1x n 1y n 1x n 2x 0 0 n 1y n 2y n 2x n 2y n (n 1)x n (n 1)y n (n 1)x n nx 0 0 n (n 1)y n ny n nx n ny. (3.9) Each column of the matrix H e represents each element, for example, the column 1 is the element 1, the column 2 is the element 2 until the column n is the element n. This means that the dimensional space of H e R 2p n. The unitary vectors are located in 21

23 H e depending on the numbering of the nodes that connect the elements (see Table 1). Therefore, each column (element) of H e can have only four components n ix, n iy, n ix and n iy for i-th element. The vector α e is assembled depending on the number of the node in which the external forces are applied. For instance, in Figure 10 the subscripts of the external forces indicate direction (x or y) and number of the node. The position of external forces in the vector α e is calculated as; force applied at the node r, it has position 2r 1 in x direction and position 2r in y direction. For the force applied at the node p 2, it has position 2(p 2) 1 in x direction and position 2(p 2) in y direction, etc. Then in general form we have that [ α e = ] 0 0 f x2r f y2r 0 0 f x(p 2) f y(p 2) 0 0. The size of the matrix C e is dened by m columns (number of reactions) and 2p rows. This matrix only contains ones and they are placed depending on the number of the node where the contraints are located. For example node 1 (one restriction): C e(1,1) = 1, node 2 (two restrictions): C e(3,3) = 1 and C e(4,4) = 1, etc. In the matrix C e the nodes are represented in its rows, as in the matrix H e. And so, it is obtained as C e = (3.10) Usually, the matrix C e is not used for more calculations, only it is used to express in general form the equilibrium equations (see Equation (3.6)). The matrix H e and the vector α e are reduced to H t and α t extracting rows from the matrix C e, where they contain the value of one, such that H t R n n and α t R n 1. The matrix h t contains 22

24 the coecients removed from the m rows of H e, such that h R m n. The procedures used are the same of the section to obtain the expressions (3.7) and (3.8). So we can obtain a set of general equations such as H t β t + α t = 0, (3.11) and h t β t + η t = 0. (3.12) where, the vectors η t and β t contain the variables of the reactions and internal forces, such that η t R m 1 and β t R n 1. It is important to point out that for statically determinate structures, the matrix H t is square. Equation (3.11) is referred to equilibrium equations that relate external forces with internal forces and Equation (3.12) is referred to equilibrium equations that relate reaction forces with internal forces. Those equations present unique solution and they can be solved like a system of linear equations. 3.3 Trusses Statically Indeterminate A structure is statically indeterminate if all independent parameters (internal forces and reaction supports) cannot be solved only by means of the equilibrium equations Here compatibility conditions are necessary for resolving all independent and redundant parameters. From Equation (3.1) we know that, if N d > 0, the structure is statically indetermin ate. The structure of Figure 11 is considered a statically indetermin ate structure. Since, it has 5 joints that are connected by 7 elements. It does present constraints in the nodes 1 and 5 of pinned supports. Equation (3.1) is applied to verify the statement of determinacy, such that e = 7, R = 4, J = 5. It is veried that N d = 1, then the structure shown in Figure 11 is statically indetermin ate. 23

25 Figure 11: a) Statically indeterminate truss. joints. b) Internal and external forces in the Force Method for the Analysis of Statically Indeterminate Structures The force method for statically indeterminate structures is applied basically with the same procedures shown in section In Figure 11 one bar or element is added between the nodes 5 and 2, as is shown in Figure 3.1. We obtain Equation (3.6) in the same form that is expressed for a statically determinate structure. Nevertheless, the matrix H e will have one column added correspond to element added. As well as the vector β t will have one additional row. The procedures to obtain H t from H e are the same applied to statically determinate structures, those are shown in section Consequently, with this idea in general form, we will add x columns to H e and rows to β t depending on the number of determinacy, such that N d = x. Therefore, for a statically indeterminate general structure (see section 3.2.2), H t R (2p m) (n+x) and β t R (n+x) 1, where (2p m) < (n+x). This means that there are more variables than equations and the system of equations shown in the equations (3.11) and (3.12) cannot be solved. To resolve this problem, it is necessary to use another type of conditions such as compatibility conditions or to use numerical techniques. 24

26 3.4 Rigid Beam Structures Equilibrium Equations for a Elemental Beam The beam of Figure 12 is subjected to external forces and moments in its nodes 1 (x = 0) and 2 (x = L). Such that q 1, q 2, q 4 and q 5 are external loads in x and y directions; q 3 and q 6 are external moments applied to the nodes 1 and 2; and p is a constant distributed load applied to the beam. β 1, β 2 and β 3 are internal moments and β 4 is a constant axial force produced by applying the external loads. Figure 12: Beam with a distributed load p. For the beam shown in Figure 12, the internal moment along x can be represented by the following approximation M(x) = ax 2 + bx + c, (3.13) where x (0, L). Let M(0) = β 1, M(L/2) = β 2 and M(L) = β 3. It is assumed that the moments β 1, β 2 and β 3 are determined from equilibrium equations. Replacing β 1, β 2 and β 3 in Equation (3.13), it is obtained that β 1 β 2 β a = L 2 L b. (3.14) L 2 L 1 c From Equation (3.14) the unknown coecients are determined by a b c = L 2 L 2 L L L L β 1 β 2 β 3. (3.15) The moment along beam can be determined as function of β 1, β 2 and β 3 as follows: 25

27 ( 2 M(x) = L β L β ) ( L β 2 3 x L β L β 2 1 ) L β 3 x + β 1. (3.16) Reorganizing Equation (3.19), the moment along the beam is solved as M(x) = ( 2 L 2 x2 3 ) L x + 1 β 1 + ( 4L 2 x2 + 4L ) ( 2 x β 2 + L 2 x2 1 ) L x β 3. (3.17) For a beam the shear forces are obtained from equilibrium equations, such that these are related with the moment as dm(x) dx = V (x). (3.18) Applying dierentiation to the moment, it is determined that dm(x) dx = V (x) = ( 4 L x 3 ) ( β L L x + 4 ) ( 4 β L L x 1 ) β 2 3. (3.19) L Shear forces in the nodes can be estimated by values known, V (0) = v 1 and V (L) = v 2, such that and v 1 = v 2 = ( 3 ) β 1 + L ( ) ( 1 β ) β 2 + L L ( ) ( 4 β ) β 3, (3.20) L L ( ) 3 β 3. (3.21) L Figure 13: Free body diagrams for a beam. 26

28 For the free body diagrams shown in Figure 13, equilibrium equations are established such that they can be expressed in matricial form as [ where β p = hβ p = q, (3.22) ] T [ ] T β 1 β 2 β 3 β 4, q = q 1 q 2 q 3 q 4 q 5 q 6 p and h = L 4 L 1 0 L L 4 L 3 0 L L 2 8 L 2 4 L 2 0. (3.23) The matrix h is considered as the elemental exibility matrix for a non oriented beam similar to the matrix obtained for bars. Figure 14: Oriented beam. If the external forces q are known the solution for β p exists as unique solution. Equation (3.22) satises and represents the equilibrium equations for an elemental beam. For an oriented beam an θ angle (see Figure 14) Equation (3.22) is modied in such a way that 27

29 T hβ p = q, (3.24) where cosθ sinθ sinθ cosθ T = cosθ sinθ 0 0, (3.25) sinθ cosθ and T is a transformation matrix of coordinates. This transformation is necessary due to the fact that the external forces are located at the x ȳ coordinate system. But they should be transformed to the x y coordinate system to apply the equilibrium equations. For beams with hinges the moments are zero in its nodes, this means that there is no transference of moments. Then, Equation (3.24) can be rewritten as H p β p + α p = 0, (3.26) where H p = T h and α p = q. Where H p is called elemental exibility matrix for a elemental beam. Equation (3.26) represents the equilibrium equations for an elemental beam similar to that of bars Structures with Elemental Beams For a structure composed by elemental beams, it is determined that H p i is the elemental exibility matrix, β p i is the vector of internal moments and axial forces, α p i is the vector of external forces and moments for the i-th element. Considering that the structure has p nodes, n elements and m reactions (rigid connections, hinge connections and roller supports). The elemental beams are connected by means of rigid or hinge connections in the structure. The placing in a global matrix depends on the connectivity of each elemental beam. For an oriented elemental beam, the matrix H p i is determined by 28

30 H p i = T i h i = 3 4 L i sinθ i L i sinθ i 1 L i sinθ i cosθ i 3 L i cosθ i 4 1 L i cosθ i L i cosθ i sinθ i L i sinθ i L i sinθ i 3 L i sinθ i cosθ i 1 L i sinθ i 4 3 L i sinθ i L i cosθ i sinθ i L 2 i 8 L 2 i 4 L 2 i 0. (3.27) The local matrix H p i should be placed depending on connectivity of elemental beams. Figure 15 shows a procedure to assembly H e with H p i matrices, those obtained for elemental beams. In the matrix H e, three rows represent one node and four columns represent one elemental beam. Therefore, the global matrix H e R (3p + n) 4n including reaction equations. Figure 15: Assembly of the matrix H e. 29

31 The matrices H p i are placed in H e of the following way; the three rst rows of H p i (local equilibrium equations of the local node 1) are located in three rows of H e that correspond to the global node of structure (see Figure 15) taking into account its corresponding column associated to the global element. The local equilibrium equations of the node 2 are located with the same procedure. The seventh row of the local matrix H p i (element i-th ) is placed in the row 3p + i and column i (element i) of the global matrix H e. This procedure is similar to that used in the force method for truss structures. Global equilibrium equations are established in general form such that H e β b + α e = 0, (3.28) where the global vector of external forces and moments is assembled as [ α e = q1 1 q2 1 q3 1 q4 2 q5 2 q6 2 q7 3 q8 3 q3p 2q p 3p 1q p p 3p p1 3p+1 p 2 3p+2 ] T p n 3p+n. (3.29) The superscript of the elements of α e represents the number of the global node and the subscript of the elements of α e represents the position in α e. The vector of internal moments and axial forces is assembled such that [ β b = β 1 1 β 1 2 β 1 3 β 1 4 β 2 5 β 2 6 β 2 7 β 2 8 β n 4n 3β n 4n 2β n 4n 1β n 4n ] T. (3.30) The superscript of the elements of β b represents the local position and the subscript represents the global position in β b. The procedure to remove restriction equations on the matrix H e and to obtain the matrix H b is the same used for truss elements shown in section α e is reduced to α b with the same procedure. In general form we can reorganize Equation (3.28) to establish a set of general equations for beam structures as H b β b + α b = 0, (3.31) and h b β b + η b = 0. (3.32) where H b R (3p+n m) 4n, and this is the exibility matrix for structures with elemental beams. α b R (3p+n m) 1 and β b R 4n 1. h b R m 4n, and this is the matrix obtained by reduction of H e. η b R m 1, and this is a vector that contains all variables 30

32 of the reactions. Equations (3.29) and (3.32) are similar to those obtained for truss structures. 31

33 Chapter 4 Lower Bound Method Applied to Structures 4.1 Lower Bound Theorem The lower bound theorem states that if a state of generalized stresses caused by an applied loads satises the equilibrium and those loads are delimited inside a yield criterion, a structure will not collapse by the application of such loads. More details of this theorem can be seen in [2, 4]. 4.2 Application of the Lower Bound Theorem to Rigid Truss Structures Consider a statically indeterminate structure of n elements, p nodes and m reactions (similar to Figure 10) in which its equilibrium equations are H t β t + λα t = 0, (4.1) where λ is a proportional multiplier to the external load α t. Then, we suppose that λ and internal loads β t are unknown. Consequently, we can apply the following reasoning, if the structure has rigid-perfectly plastic elements (see section 1.2) the internal loads β t are subjected to βy < β t < β y +. (4.2) 32

34 So, we can apply the lower bound theorem to nd a solution to λ and β t in such a way that it satises the equilibrium equations shown in Equation (4.1) and the yield criterion of Equation (4.2). For this case, it is assumed that βy = β y + = β y. The solution of this problem is projected as an optimization problem with linear constraints which is described as a linear programming problem (LP ), such that maximize λ (LP ) subject to H t β t + λα t = 0, (4.3) β y β t β y. Equation (4.3) means that a maximum value should be found for λ which is restricted by the equilibrium equations and the yield criterion. For resolving Equation (4.3), this should be organized in the standard form of a linear programming problem of the following way maximize a T t x t (LP ) subject to A t x t = 0, (4.4) C t x t c ty. [ ] [ ] T [ ] H where x t = β t λ, a T t α [ t t = , A t =, c ty = β y 0 β y I 0 C t = 0 0 I 0. I is a identity matrix, such that I Rn n, c ty R 2(n+1) and 0 0 A t R (n+1) (n+1). ] T, It is important to mention that β y can have dierent values for each element, since these values depend on a yield criterion for each element. The maximization problem shown in Equation (4.4) can be solved with dierent numerical techniques used for linear programming, such as simplex method, interior-point method, etc. The method shown in this section is the same applied by [19] 33

35 4.3 Application of the Lower Bound Theorem to Rigid Beam Structures Consider a structure of elemental beams with n elements, p nodes and m reactions, for which the equilibrium equations are described by H b β b + λα b = 0. (4.5) Here λ is a proportional multiplier to the external moments and loads α b. We assume that external loads λ and the internal moments and axial loads β b are unknown. We can apply the following reasoning, if the structure is composed by rigid-perfectly plastic beams (see section 1.2); for the element k, k = 1, 2, 3,..., n; the internal axial loads β k 4k are subject to β k y < β k 4k < β +k y. (4.6) The internal moments β4k+(i 4) k, i = 1, 2, 3; then, each elemental beam is subjected to a yield criterion such that [ where M y ±k = m ±k y m ±k y m ±k y M k y < β k 4k+(i 4) < M +k y, (4.7) ] T. So, we can apply the lower bound theorem to nd a solution to λ and β b (internal moments and forces) in such a way that it satises the equilibrium equations shown in Equation (4.5) and the yield criteria of the equations (4.6) and (4.7). The solution for this problem is projected as an optimization problem with linear constraints which is formulated by maximize λ subject to H b β b + λα b = 0, (LP ) My k β4k+(i 4) k M y +k, β k β4k k β+k y. (4.8) Equation (3.31) means that a maximum value should be found for λ which is restricted by equilibrium equations and yield criteria. To resolve Equation (4.8), this 34

36 should be organized in the standard form of a problem of linear programming as maximize a T b x b (LP ) subject to A b x b = 0, C b x b c by. (4.9) [ where x b = β b [ ] T [ ] T λ, ab = and Ab = [ ] T [ c by = D 1y 0 D 2y 0, where D1y = H b α b 0 0 M y +1 β y +1 M y +2 β y +2 M y +n β y +n and [ ] T D 2y = My 1 βy 1 My 2 βy 2 My n βy n. Ab R (4n+1) (4n+1). I 0 C = 0 0 I 0. I is a identity matrix, I R4n 4n and c y R 2(4n+1) The yield criteria My k, M y +k, β k and β +k can be dened with dierent values for each element k. The hinges (the moments are zero in it) are added to the matrix A b adding one equation of moment, such that β4k+(i 4) k = 0, i = 1, 3. This equation is placed in the column k related to the row 4k + (i 4) of the hinge node i, which corresponds to the element k in the matrix A b. Equation (4.9) is solved using MOSEK solver for MATLAB. The maximization problem shown ]. ] T 4.4 Estimation of the Collapse Mode with Sensitivity Analysis In linear programming problem it is possible to carry out sensitivity analysis of the objective function by changing the values of the constraints. When the constraints are increased, the values obtained for the objective function are dierent, and these values show which constraint has a higher signicance in the objective function. The change in objective function from the initial problem until the modied problem is called shadow price. The shadow price is dened as the change in the objective function by increasing a constraint by one. [19] showed that the shadow prices obtained for λ determine the 35

37 collapse mode of a structure, when λ is maximized for each modied constraint. The shadow prices can be obtained for truss structures considering the following problem, given Equation (4.3) is possible to determine that maximize λ (LP ) subject to H i tβt i + λαt i = 0, (4.10) β y β t β y. where i denotes the degree of freedom in the directions x and y for each established equilibrium equation. Being i = 1, 2, 3, 4,, 2p m and λ the maximum value obtained for Equation (4.10), we can determine each shadow price as Sp i = λ i λ, (4.11) where maximize λ i [ (LP ) subject to H i tβ t + λαt i = β y β t β y i 0 0 ] T, (4.12) Each value of Sp i represents the nodal displacement associated to the direction of the equilibrium equation. Such that, the displacement of the node j in x is determined by u x = Sp 2j 1 ; and the displacement in y, u y = Sp 2j. These values should be added to the nodal coordinates of the initial structure. For beams the strategy of obtaining the collapse mode is dierent, since the equilibrium equations have contributions of moment. Then for beams is given that maximize λ i [ ] T subjectto H i b (LP ) β b + λαb i = i 0 0, My k β4k+(i 4) k M y +k, β k β4k k β+k y. (4.13) Equation (4.11) is applied to beams in the same form that in truss structures, such that i = 1, 2, 3, 4,, (3p + n m). The displacement of the node j in direction x is determined by u x = Sp 3j 2 ; and the displacement in y, u y = Sp 3j 1. 36

38 It is important to note that the displacements obtained for the nodes are quantities that indicate the direction of the displacement of the node and they do not indicate the absolute value of this displacement. These values should be scaled with respect to the structural dimensions for showing the displacement in graphic form. 4.5 Numerical Examples for Truss Structures In this section, there are shown numerical examples of truss structures. The examples present dierent structural congurations. The results obtained are compared using dierent techniques of solution for optimization problems. Moreover, there are used two dierent types of solver, optimization solver of MATLAB version R2011b and MOSEK solver [18]. The characteristics of numerical computation used for the calculation of the examples are described in Table 2. It is important to mention that for the examples unit systems are not considered. Therefore, there are shown only numerical values. Computer reference Processor Memory (RAM) System type Lenovo T420 Intel(M) Core(TM) i5-2540m CPU 2.60 GHz 4 GB (3.89 GB usable) 64 bits Table 2: Computer characteristics. An algorithm was carried out in MATLAB for resolving the examples of this section, it is called trusslb.m. The solutions obtained by trusslb.m are: collapse load λ, internal forces β and reaction forces η. The following sub-functions are included, assemb.m (it is used to assembly the matrix H e, see section 3.2.2), applyc.m (used to obtain H t from H e and h t ) and graphtruss.m (subfunction created to illustrate structures, this function contains more sub-functions). The algorithms carried out are described in Appendix A. trusslb.m uses dierent types of algorithms for resolving the optimization problem shown in Equation (4.4). 37

39 4.5.1 Example: Statically Determinate Truss Structures Example 1 Figure 16 shows examples of statically determinate structures. Two cases of restriction are proposed, with pinned support and roller support. The origin of the coordinate system is located at the node 5 for both cases. To resolve the cases a) and b) shown in Figure 16 with the force method (see section 3.2.2), it is necessary to construct the connectivity table as preliminar information (see Table 3). Figure 16: Statically determinate structures: a) Pinned supports. b) Roller support and pinned support. The yield criterion in this example is β y = 200 and it is the same for all elements. The information of the example a) and b) is collected in two les carried out in MAT- LAB; Example1a.m and Example1b.m (see Appendix A). The les contain detailed information for the preprocessing of the force method implemented in trusslb.m. Structure a) Structure b) Element Node 1 Node 2 Element Node 1 Node Table 3: Connectivity for structures of Figure

40 Table 4 shows the results obtained for the collapse load λ and time of processing spent in trusslb.m using dierent optimization solvers. Results shows that MOSEK [18] optimization solver is faster than solver provided by MATLAB. Figure 17 shows the internal forces (optimization results) in a graphic format obtained by the algorithm graphtruss.m realized. This Figure shows a load state for tension, compression, zero loads (smaller than 0.05β y ) and plastied bars (bigger than 0.95β y ). Internal forces for Figure 16a) are β 1 = 100, β 2 =125, β 3 = 100, β 4 = 75, β 5 =125, β 6 = 200 and Figure 16b) are β 1 = , β 2 = , β 3 = 76.92, β 4 = 92.30, β 5 = , β 6 = 200, β 7 = The reaction forces for Figure 16a) are calculated from Equation (3.12), such that R 1x = 200, R 1y = 75, R 5x = 200 and R 5y = 0. From Figure 16b), R 1x = , R 1y = 92.30, R 5x = 200. In Figure 17b), we can observe that the added bar presents a zero load for the statically indeterminate structures. The collapse mode for the examples a) and b) is shown in Figure 18. Method used in trusslb.m Example a) Example b) λ Time (s) λ Time (s) Active Set Method (linprog MATLAB, see [14, 15]) Simplex Method (linprog MATLAB, see [16]) Linear Interior Point (linprog MATLAB, see [17]) Active Set Method (linprog provided by MOSEK [18]) Simplex Method (linprog provided by MOSEK [18]) Linear Interior Point (linprog provided by MOSEK [18]) Table 4: Results of Figure 16 using dierent optimization solvers. Figure 17: Graphic results obtained by graphtruss.m in MATLAB: a) Example of Figure 16a). b) Example of Figure 16b). 39

41 Figure 18: Collapse mode: a) Example of Figure 16a). b) Example of Figure 16b) Examples: Statically Indeterminate Truss Structures Example 1 Figure 19 is an example of a statically indeterminate structure proprosed by [19]. The information of the connectivity and restrictions are written in example2.m which is called from trusslb.m algorithm done for MATLAB. The yield criterion in this example is β y = 200 and it is the same for all elements. Table 5 shows the results obtained for the collapse load λ. The time of processing is faster with MOSEK [18] optimization solver than the solver provided by MATLAB. The results are shown graphically in Figure 20. In the results, it is seen that the criterion of plastication and zero load are the same described in Example 1 of section Figure 19: Truss example proposed by [19]. 40

42 Method used in trusslb.m Example 1 λ Time (s) Active Set Method (linprog MATLAB, see [14, 15]) Simplex Method (linprog MATLAB, see [16]) Linear Interior Point (linprog MATLAB, see [17]) Active Set Method (linprog provided by MOSEK [18]) Simplex Method (linprog provided by MOSEK [18]) Linear Interior Point (linprog provided by MOSEK [18]) Table 5: Results of Figure 19 using dierent optimization solvers. The reaction forces for r1 and r2 are calculated from Equation (3.12), such that R 1x = 240, R 1y = 40, R 2x = 240 and R 2y = 40. In Figure 20, we can observe that there are two bars that have zero load. This means that they are not necessary in the initial structural conguration. Therefore, the structure can be optimized. Figure 21 shows the collapse mode of the structure. Figure 20: Graphic results obtained by graphtruss.m in MATLAB for Example 1. Figure 21: Collapse mode for Example 1. Example 2 This example shows a structure in arc form with n elements connected as is shown in Figure 22. A methodology is used to create the structure in a general form, such 41

43 that a semi-circle of radios r is divided in p sections to obtain the coordinates and the connectivity. Each section is composed by three elements and two elements are shared to the previous and subsequent section, excluding the 1-st and p-th section. The structure is described in the algorithm example4.m which is called from trusslb.m. For this example, 36 sections were created and r = 10 was chosen. The elements 2, 4, 6, 8,..., n 3, n 1 are considered straight (bar elements) in the algorithm. This example was taken from [20]. Figure 22: Example 2 proposed by [20]. The collapse load obtained is λ = The elements that are on the path of the circle were plastied. The internal elements (from 3 until n 2) are in tension. The elements 1 and n are not loaded (zero load). Using the values of β, the reaction forces for r1 and r2 were calculated with Equation (3.12), such that R 1x = 8.72, R 1y = , R 2x = 8.72 and R 2y = The collapse mode of the structure is shown in Figure 24. Figure 23: Graphic results obtained by graphtruss.m in MATLAB for Example 2. 42

44 Figure 24: Collapse mode of Example 2. Example 3 Figure 25 shows a mesh generated with an algorithm carried out for this example (see example5.m in Appendix A). The generated mesh presents 20 divisions in each side a and b, such that a = 10 and b = 10. However, the algorithm can generate a mesh of n divisions. The structure of Figure 25 has 1640 elements and 441 nodes. The collapse load obtained by trusslb.m is λ = The results are shown graphically in Figure 26. The notation for the color of the bars shown in Figure 26 is the same used in Figure 23. Figure 25: Mesh generated by Example5.m for 20 divisions in a and b. Using the values of β obtained by trusslb.m, the reaction forces are determined for R ix and R iy, i = 1, 2,..., 8 and these are calculated from Equation (3.12), such that R 1y = , R 2y = , R 3y = , R 4y = , R 5y = , R 6y = , R 7y = , R 8x = 0 and R 8y = The values of β are shown in Figure 26, the 43

45 bars in tension are blue and the bars in compression are black. The bars that presents a 5% of β y are considered zero bars, which are shown in gray color. The collapse mode of the structure is shown in Figure 27. Figure 26: Graphic results obtained by graphtruss.m in MATLAB for Example 3. Figure 27: Collapse mode for Example Numerical Examples for Beam Structures In this section, numerical examples are shown. They are proposed to beams and the examples are solved using the lower bound theorem. The computation characteristics used for the calculations of the numerical examples are the same to those used in Section

46 To resolve the numerical examples, the lower bound method applied to beam structures was implemented in MATLAB. The principal program is called beamlb.m, it applies lower bound theorem on each example. Numerical examples are preprocessed in other subprograms. The solutions obtained are: collapse load λ, β b and η b. The collapse mode obtained for the beams was determined with sensitivity analysis, the procedure is described in section 4.4. The following sub-functions are include, assembeam.m (it is used to assembly the matrix H e, see section 3.4.2), applycbeam.m (it is used to obtain H b from H e and h b ), Blocal.m (it is used to obtain H p i ), BCon.m (it is used to add hinges), grapmo.m (subfunction created to illustrate the internal moments) and grapmo2.m (subfunction created to illustrate the internal moments in horizontal beams located in series). The algorithms carried out are detailed in Appendix A. Example 1 The example shown in Figure 28 is composed of three elemental beams connected in series, each elemental beams have dierent orientation. The joints are considered rigid (black point in Figure 28), this means that there is transference of moments between elements (continuity). The yield criteria for each elemental beam are βy = β y + = 200 N, My = 50 N m and M y + = 20 N m. The information of the example is found in a le called Example1ReBeam.m written for MATLAB. Figure 28: Example 1. 45

47 Figure 29: Internal moments of each elemental beam. Figure 30: Collapse mode. In Figure 29, there are shown internal moments obtained by optimization for each element of structure. The collapse load is λ = 2.8 N/m. The reactions calculated with Equation (3.30) for the restrictions imposed to R 1 are: R 1x = 21 N, R 1y = 7 N and M 1 = 50 N m; and imposed to R 2 are: R 2x = 7 N, R 2y = 21 N and M 2 = 20 N m. The black points in Figure 29 indicate plastied zone. For this example was obtained the collapse mode using the shadow prices as is shown in Figure 30. Example 2 The example shown in Figure 28 is composed of three elemental beams connected in series by four nodes, the elemental beams are in horizontal orientation. In the example 46

48 there are two proposed problems, a continuous beam (see Figure 28a)) and a beam with a hinge placed at the node 3 (see Figure 28b)). The yield criteria for this example are βy = β y + = 200 N, My = 50 N m, M y + = 20 N m and L = 5 m. The information of the example is found in a le called Example2ReBeam.m written for MATLAB. Figure 31: Example 2, a) beam without hinges, 2b) beam with hinges. The collapse load for the case a) is λ = 7.2 N/m and for the case b) is λ = 4 N/m. Figure 32 shows the internal moments along the beam for the cases a) and b). For the case a), the node 3 is plastied and in the case b) the hinge avoid that it become plastied. The axial forces through the beams are zero. Table 6 shows the results of the reactions for the cases a) and b). The collapse mode of the examples a) and b) is shown in Figure 33. Figure 32: Results for Example 2, a) and b). 47

49 Reactions N or N m a) b) R 1x 0 0 R 1y 7 4 M R 2y R 4y 4 0 Table 6: Results of reactions. Figure 33: Collapse mode for Example 2, a) and b). Example 3 The example shown in Figure 28 is composed of four elemental beams which represent a continuous beam. For this example the yield criteria are dierent for each elemental beam. In Table 7 is shown the yield criterion for each beam. The information of the example is detailed in a le called Example3ReBeam.m written for MATLAB. For this example L = 5 m. element β y + = βy M y + My Table 7: Yield criteria. 48

50 Figure 34: Example 3: Continuous beam. The collapse load is λ = 9.6 N/m and the reactions are R 1y = N, R 2y = N, R 3y = N, R 4y = N and R 5y = 20 N. Figure 35 shows the internal moment along the beam and the limit of the yield criterion for each elemental beam. In this gure the yield criteria for the moments are denoted by dashed lines. Figure 35: Results of Example 3 obtained by graphmo2.m. Example 4: Collapse Mode for a Continuous Beam with n Elements The beam shown in Figure 38 is divided by n elements, each element is dened as an elemental beam. The length of the beam is L = 10. For this example are considered two cases, beam with n = 10 and beam with n = 70 elements. The information of the example is detailed in a le called Example6ReBeam.m written for MATLAB. 49

51 Figure 36: Example 4: Continuous beam. Figure 37: Moment along the beam. The collapse load determined for this example is λ = and the reactions are R 1x = 0, R 1y = 21.48, M 1 = 50 and R (n+1)y = Figure 35 shows the internal moment along the beam and the yield criterion that delimits the moment. In this gure the yield criterion for the moments are denoted by dashed lines. Figure 38 shows the mode collapse of the beam for the case n = 10 and n = 70 elements. Figure 38: Collapse mode of the beam for n = 10 and n = 70 elements. In this chapter, there was shown a methodology which permits to obtain the collapse 50

52 load for rigid-perfectly plastic truss and beam structures using linear programming. The examples were solved and compared in some cases with two types of solvers commonly used in linear programming, ie., MATLAB solver and MOSEK [18] solver. By means of an algorithm carried out in MATLAB (see graphtruss.m in Appendix A), the results obtained by optimization can be shown in a graphic format. Those results help to identify which elements were not loaded and which were plastied for the case of truss structures. For the case of beams the graphic function (grapmo.m and grapmo2.m in Appendix A) identify the zones that are plastied, which are dierentiated from other zones. 51

53 Chapter 5 Analysis of Truss Structures with Material and Energy Minimization 5.1 Energy Minimization for Structural Analysis in Trusses Consider the strain energy of the elastic bar of Figure 1a) : e = 1 2 V σɛdv, (5.1) where ɛ is the strain. Using Hooke's law σ = Eɛ and the denition of stress σ = β/a, Equation (5.1) can be rewritten in function of β as e = 1 2 l 0 β 2 dx, (5.2) AE where β is the force applied to constant cross section of area A. For a bar of length l, the elastic energy is obtained by e = 1 l. (5.3) 2 AE In order to obtain the total energy stored for n elements of the structure, we can establish in general form that e T (β) = 1 2 β 2 n βi 2 l i. (5.4) A i E i For resolving Equation (5.4), we should know the values of β. Those values are determined from the equilibrium, if a load state is established for the structure to be studied. i=1 52

54 5.1.1 Elastic Structural Analysis as an Optimization Problem Applied to Trusses It is well known that the stress distribution in an elastic structure can be determined by minimizing the elastic strain energy. That is, among all possible stress distributions that satisfy the equilibrium equations, the distribution that minimize the eleastic energy will be the true solution. So, we establish a problem of energy minimization as an optimization problem of the following way minimize e T (β) (NOP ) subject to Hβ + λα = 0. (5.5) The function e T (β) is modied such that if all the elements are of the same material, Young's modulus is excluded of the objective function as fe T (β) = Then, Equation (5.5) is redened as, n βi 2 l i. (5.6) A i i=1 minimize fe T (β) (NOP ) subject to Hβ + λα = 0. (5.7) Equation (5.7) is a nonlinear optimization problem (NOP), since the objective function is nonlinear. The β values can be solved by dierent methods, but in this study Sequential Quadratic Programming (SQP) is used for resolving the problem. Equation (5.5) is another form of obtaining the equilibrium of a structure, the results should be the same to those obtained by Finite Element Method (FEM). The examples shown in this section are solved with the algorithm called TrussEquiLB.m. It presents some sub-functions as assemb.m, applyc2.m, applycf.m and fmincon.m (optimization solver provided by MATLAB). The code of the algorithms is presented in Appendix A Example: Example [21] Figure 39 shows an example extracted from [21]. The equilibrium equations are established and solved by means of Finite Element Method (FEM) applied to truss elements. In [21] an algorithm is proposed for resolving this example in MATLAB by 53

55 FEM. The algorithm ex742.m was implemented in MATLAB and adequated for obtaining the internal forces of the structure shown in Figure 39. This algorithm contains the following sub-functions feapplyc2.m, feassembl1.m, feeldof.m and fetruss2.m. Example is dened in example3.m and is called from ex742.m. The results obtained by FEM are compared with those obtained by optimization (see Equation (5.7)) and these are shown in Table 8. International system of units is used in this example. Figure 39: Example 1 proposed by [21]. The time of processing using fmincon.m optimization solver with SQP algorithm provided by MATLAB called from TrussEquiLB.m is s. Using FEM (see [21]) the time of processing is s. Force FEM [N] SQP [N] β β β β β β β β β Table 8: Results for internal forces obtained by FEM [21] and optimization. This example was solved using the computational characteristics shown in section

56 5.2 Material Minimization with the Collapse Load λα A problem of minimization of material is dened if given a structure, the volume of the structure can be reduced. First we consider that the collapse load λα is assumed to be known and this is obtained by means of Equation (4.4). This load is applied to a structure to obtain an internal load state such that we assume for a bar the following σ + y = β+ A, σ y = β A, (5.8) where σ y and σ + y are the absolute values of the yield stress in tension and compression, respectively. Then, we can obtain the volume for a bar of length l from Equation (5.8) as V e = β+ σ + y l, V = β l. (5.9) σy Obtaining the total volume for all structure we can determine that β + i σ y + β i σy, if β 0, (5.10), if β<0 where n and m are the bars subject to tension and compression, respectively. We can establish a Nonlinear Optimization Problem with the aim to minimize the volume of the structure. The problem is restricted to obey the equilibrium equations, so we can establish that minimize V (β) (NOP ) subject to Hβ + λα = 0. (5.11) The function V (β) can be minimized by SQP method, which is used in this study for resolving nonlinear pptimization problems. The examples shown in this section are calculated with trussminmat.m. It presents some sub-functions as assemb.m, applyc2.m, applycf.m, fmincon.m and graphtrussm.m. The examples are solved using the same computational characteristics shown in section

57 5.2.1 Example 1: Example of Figure 19 In this example, we consider the strutucture shown in Figure 19. The external load is λ = 80 (collapse load). The yield stress is σ y = σ + y = σ y = β y /A, where β y = 200 (see example 1 section 4.5.2) and A = 0.1. The system of units for this example is not considered, since it can be anyone (I.S or U.S.). The objective function obtained for the volume is V (β ) = Table 9 shows the values of β which correspond to the new volume determined. OV and NOV means optimized volume and non optimized volume. The initial total volume is V NO (β) = , with material optimization the volume was reduced to 65.6% of the initial volume. The results are shown graphically in Figure 40. The colors blue and black represent tension and compression in the bars. The width of the bars is associated with the new determined volume. element β OV NOV Table 9: Results for internal forces, optimized volume and non optimized volume. 56

58 Figure 40: Graphic results for Example of Figure 19 with material minimization Example 2: Example of Figure 25 Modied This example considers the structure shown in Figure 25, however the restriction r1 is a pinned support for this case. The external load applied is λ = (collapse load). The yield stress is σ y = σ y + = σy = β y /A, where β y = 200 (see example 1 section 4.5.2) and A = 0.1. The system of units for this example is not considered, since it can be any one (I.S or U.S.). The initial volume is determined by V NO (β) = and the new calculated volume after applying material minimization (see Equation (5.11)) is V (β ) = This result shows that the volume was reduced in 79.71% of the initial volume. The results are shown graphically in Figure 41. The volume of the bars is shown by the size of the lines in the gure. The colors blue and black represent tension and compression in the bars, respectively. Figure 41 shows how the structure can be optimized in terms of material. Figure 41: Graphic results for Example 2 with material minimization. In this chapter, two methods that permit to carry out structural analysis and mate- 57