E1. Column subjected to bending and axial action, Non-dissipative material

Transcription

1 E1. Column subjected to bending and axial action, Non-dissipative material Problem definition In this exercise, calculate roughly the maximum load, due to the material behavior, of a column structure made of a common railway steel, as shown in the Figure below. s the Figure, indicates, the column structure is loaded in both axial and transversal directions. The cross section of the column is singly symmetric around the z-axis. Figure 1: Column problem considered, use H 10 h, b u h, b l h 3 and h 5mm in the assignment. Based in the Euler-Bernoulli beam theory, the uniaxial strain of each "beamfiber" can be expressed as a linear function in terms of the strain 0 due to the uniaxial elongation of the beam and the beam curvature. We thus obtain the linear relationship z 0 z Buwith B 1, z and u 0 where B is the strain displacement transformation and u is the generalized strain measure (comprising both the uniaxial strain and the curvature) of the cross section. The stress resultants of the beam consists of the bending moment and the axial force which are defined in the usual as h z N z b z z z h z M z z zb z z z where is the uniaxial stress. The material behavior is described by the experimentally determined cyclic stress-strain curve shown in the Figure below. In the present example we are not interested in the actual cyclic material behavior. Instead, we extract data from the first monotonic loading branch up to the uniaxial strain e=0.9%. non-linear elastic law, named the modified Ramberg-Osgood law, is used to pick up the material behavior. This law defines the non-linear elastic stress-strain relationship E 1 bs 0 n 1 n with 0 0 E

2 2 CE1.nb where the material parameters 0 and n have been used to calibrate the material parameters as shown in the Figure below. The material parameters in the elastic law are obtained after manual curve fitting as: MPa, n 2.66, with E MPa It may be shown that the tangent stiffness E t can be written as E t d d E 1 bs 0 n 1 n n Figure 1: Cyclic behavior of a typical railway steel in strain-controlled uniaxial test. Results taken from Johan hlström, Birger Karlsson, Wear 258, , (2005). s MPa e Figure 2: Non-linear elastic response of modified Ramberg-Osgood law, calibrated from the first branch of the experimental curve above. Exercise problems s alluded to in the problem definition, let us consider the assignment problems è Calculate roughly the maximum load of the column with respect to the non-linear material law. è Discuss possible deficiencies of the proposed material modelling. What will happen if the loading is removed? How about the behavior at cyclic loading? How can the residual stress state be modeled, etc.? è Extra problem (for those of you who are specially interested) consider the same analysis using the standard Ramberg-Osgood law written in a flexibility format as

3 CE1.nb 3 E 0 bs 0 n Sign s MPa e Figure 2: Non-linear elastic response of true Ramberg-Osgood law, calibrated from the first branch of the experimental curve shown above. The following paramters where used: MPa, n 11.2, E , 1, Numerical procedures ü Equilibrium condition The equilibrium conditions for the considered problem may be formulated as the balance relation of internal forces b u and external forces f t defined as g u b u f t 0 with b 1 z B t N u M u, f N 0 M 0 and u 0 where N u and M u are the stress resultants of the internal stress state in the cross section, and N 0 10 P t and M 0 P t H are the prescribed stress resultants due to equilibrium of the column in the Figure above. The problem is to find the total "displacement" (or total generalized strain measure) u 0, so that g u 0. To this end, the nonlinear solution strategy is used as discussed below. ü Spatial discretization of beam cross section In order to integrate numerically the stress resultants involved in the vectort b, the beam cross is discretized into "nlay" number of finite slabs z at distances z i from the geometrical center so that the stress resultants can the integrated numerically as b N M 1 nlay z 1 z z i b z i z i i i 1 where it is assumed that the stress z i is constant within each lamella z i. The adopted spatial discretization of the beam cross section is shown in the Figure below.

4 4 CE1.nb ü Non-linear solution strategy - Newton's method In order to solve the non-linear problem, let us consider the balance of internal and extermal forces for each given time step as g u b u f t 0 with u n u v t and t n 1 t where n u is the displacement from the previous time step n t and v is the "displacement" velocity pertinent to the current time step t n 1 t. For each time t n 1 t, consider the algorithm defined as: for each iterate i, evaluate the improved displacement increment i 1 from the Newton step t i K i 1 g n u i v t, i 1 v i v i 1 where K is the Jacobian (or tangent stiffness) matrix of the beam cross section. It is defined from the linearization dg u db u tkdv with K b u E t B t B The spatial discretization of the iteration matrix becomes and E t 1 K E t z z z 2 nlay E t z i i 1 1 z i z i z i 2 b z i z i ü Prelimimary results Preliminary result from the analyses are given the figures below in terms "bending moment" versus curvature and stress distribution aling the z-axis of the beam cross section. M t Bending moment versus curvature k t

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6 6 CE1.nb Implementation: Pseudo code ü Pseudo code of main module h 10; H 10 h; bu h; bl h ; 3 gp bl, bu, h ; geometry parameters array ; E ; n 2.66; mp E, 0, n ; material parameters array nlay 100; number of layers for integration of cross section dz h nlay ; cg 2bl bu h ; geometrical center of gravity 3 bl bu z Range dz,h, h cg; define vector with z values based on dz 2 nlay mtim 150; number of timesteps T 1; t T mtim; define time step size defining "displacement" increment P xx; M0 0; dm0 PH mtim; define bending moment and axialforce N0 0; dn0 10 P mtim; sig = Constantrray[0, nlay];(*zero initial continuum stress state*) = Constantrray[0, mtim];(*curvature for output*) u = {0, 0};(*total displacement*) v = {0, 0};(*displacement velocity*) Do[(*for all time steps {itim=1, mtim}*) {M0 = itim dm0; N0 = itim dn0;(*increment the load*) pra[mp, gp, t, nlay, z, dz, N0, M0];(*Evaluate stress state in cross section for given load N0, M0*) u += tv;(*update total displacement*) [itim] = u[2]}(*save curvature for data representation*), {itim, 1, mtim} ]

7 CE1.nb 7 ü Pseudo code of non-linear solver module pra mp, gp, t, nlay, zval, dz, N0, M0 : Module mite 10; 10 4 ; bl, bu, h gp; Do for all Newton iterations iite 1,mite S Constantrray 0, 2, 2 ; N 0; M 0; Do for all layers ilay 1,nlay z zval ilay ; sig ilay, E t praa mp, u tv,z ; Stress and tangent stiffness in layer sig ilay ; cg z bl cg h z bu b ; h N bdz;m zbdz; S t E t Outer Times, 1, z, 1, z dz b, ilay, 1, nlay ; g N N0, M M0 ; out of balance forces or residual error Norm g ; If iite 1, ref error ; reps error ;If reps, Break ; ref Inverse S.g; Evaluate the iterative improvement v ; update improved solution, iite, 1, mite ; ü Pseudo code of stress evaluation module praa mp,u,z : Module E, 0, n mp; 1, z.u; calculate total fiber strain at the actual layer 0 0 ; E ; establish the stress E t ; establish the tangent stiffness