3D Lattice Model for Post-Yield and Fracture Behaviour of Timber.

3D Lattice Model for Post-Yield and Fracture Behaviour of Timber.
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  3D Lattice Model for Post-Yield and Fracture Behaviour of Timber Thomas REICHERTResearch StudentCentre for Timber Engineering, Napier UniversityEdinburgh, UKDr Daniel RIDLEY-ELLISPrincipal Research FellowCentre for Timber Engineering, Napier UniversityEdinburgh, UK   Summary The paper describes the development of a bespoke Finite Element program to model timber with athree dimensional lattice of single spring elements. These springs mimic meso-scale timberbehaviour, namely the crushing and separation of fibre bunches by following a tri-linear load-displacement curve. Strength and stiffness parameters for longitudinal, lateral and diagonalelements are randomised for heterogeneity. To save computing time, two specialised algorithmshave been implemented to perform a nonlinear analysis faster than an iterative Newton-Raphsonalgorithm. The algorithms have been adopted and extended to suit a 3D lattice model for timber.Furthermore lattice elements have only been used in areas where plasticity and fracture is expected,with transverse isotropic continuum elements elsewhere. The general calibration procedure of thishybrid model to tested timber specimens of Sitka spruce ( Picea sitchensis ) is described. 1.   Concept Lattice Model Lattice models have been used widely for concrete, but have only recently been applied to timber,e.g. [1][2][3]. A clear advantage of modelling timber with a lattice is the possibility to predict brittlefailure without prior knowledge of the failure location and therefore with no need for re-meshingthe Finite Element (FE) model.The basic unit cell in a lattice has to be constructed to be periodically repeating in space (Fig. 1). Inthis lattice, each cell consists of six different types of elements. Longitudinally orientated springstransfer load in the X direction (grain direction) and lateral springs in the Y and Z direction.Diagonal springs resist shear in the XY, XZ and YZ plane, as well as providing additional X, Y andZ components. This can be simplified by the assumption of transverse isotropy, to four independentelements by equating the Y and the Z direction. Thus, elements in the XY plane are the same as Fig. 1 Lattice structure with longitudinal (red), lateral (green/blue) and diagonal (grey)“half” springs in one unit cell (left), load-displacement curve for spring i (top,right) and a definition of c v with mean and standard deviation (bottom, right) dydzdzdx x z y tension δ compression +S T,j,i -S C,j,i K   j,i 1 γ T,j ·K   j,i 1 γ C,j ·K   j,i 1 with  j = x,z,y,xz,xy,zy (direction)and spring iS T,j , S C,j , K   j φ µ + σ  - σ   µ σ   = v c S  their respective elements in the XZ plane. A complete nomenclature can be found at the end of thispaper.In order to create a lattice with as few nodes as possible, they have been arranged in a diagonalchecked pattern. Thus, instead of constructing nodes at each potential junction of springs only everysecond one is used. Each element follows a tri-linear load-displacement curve with limit strengthand yield strength values respectively under tension and compression, followed by a softening orfracture line.Material heterogeneity can be implemented by assigning randomised strength ( S C,j and S T,j ) andstiffness ( K   j ) properties to springs based on a mean value for each spring type  j (x, y/z, xy/xz andyz) and a coefficient of variation ( c v ). This coefficient is assumed to be 0.2 since it has only minorinfluence on the bulk model behaviour [2]. The growth rings can be taken into account as structuredvariation of properties in the lattice. This is implemented by mapping generated growth rings on thelattice and changing the mean strength and stiffness properties of lattice members according to theirassumed position within these rings (section 1.2). 1.1   Nonlinear Solution Former lattice models for timber, e.g. [1][2][3], adopted a simple technique to solve for thenonlinear solution: After assembling the global stiffness matrix, this system of equations is solvedfor a fixed displacement step. The resulting stress for each element is computed and checked if itexceeds its predefined maximum strength. Elements are removed accordingly and the process isrepeated until no element fails. Then the next displacement is assigned. This algorithm is repeateduntil the final displacement step assigned or the system becomes singular. However, with thistechnique any accumulated elastic work stored in the lattice before breaking occurs is neglected.Since the model described in this paper will be used to perform contact and geometric nonlinearanalysis in the future, the solution algorithm required a more general approach, as for example theNewton-Raphson algorithm.To further save computation time, a specialised technique [4] has been adopted. Jirásek and Bažantcall it the “Method of Inelastic Forces” (MIF) and the “Step Size Control” (SSC) algorithm. Thelatter allows for faster computation by following the solution path through single linear steps fromone element changing its stiffness to the next element changing. Thus, no additional iteration isnecessary. Further, the MIF treats any change that would occur in the matrix due to a change in theelement’s stiffness, as an added inelastic force that represents the difference between the systemwith changed stiffness and the elastic one. Thus, only the force vector has to be modified and it isnot necessary to solve the global stiffness matrixagain. The interested reader is referred to a moredetailed description of this algorithm in thesrcinal paper [4]. For this research, the SSCmethod has been modified to allow for a tri-linearload-displacement definition of the springelements as depicted in Fig. 1. 1.2   Structured Heterogeneity Heterogeneity, on the level of the growth ringstructure, is mapped on a lattice of the cell sizeof 2 x 1 x 1 mm ( dx x dy x dz ). This size resultsfrom a balance between acceptable computationaleffort for larger lattices and represented detail of the growth ring structure. The mean ring widthmeasured from test specimens has been 5.47 mm,with a c v of 24.8%. Specimens with ring width lessthan 2 mm were discarded. This ensures that onegrowth ring encompasses at least two lattice cells.Several measurements were taken from the testedspecimens. The cross section of each one (frontand back) was scanned with an ordinary flatbed Fig. 2 Measured parameters for recreatinga growth ring structure in the model r  var   ∆ r  i+1  ∆ r  i-1  y zr  i+1 r  i r  i-1 α  r  shift   ∆ r  i =  ∆ r  centre 1...2 1 −=∆−∆=∆ − nir r r  iidiff  nir r r  iii ...2 1 =−=∆ − centreshift shift  r r r  ∆= r   pith rings, frontrings, back   scanner. A programme was written that enables the user to draw 3-point circles onto the latewoodof each growth ring in these images. By averaging the centre points of each circle, the assumedposition of the pith can be determined. With this information the following parameters can beobtained from one specimen (Fig. 2): α , r   pith , shift  r  , mean    ∆ r  diff  and mean r  var  .Mean values are calculated along with their coefficients of variation from the specimens of one testseries. This serves then as input parameters to create a random ring structure for the lattice model,based on the characteristics of tested specimens. 1.3   Mapping the Density Profile In order to map the ring structure on the lattice, thesimplest assumption would be to correlate stiffnessand strength variation directly with density variationwithin a growth ring. Therefore, densitymeasurements from Sitka spruce samples were takenand have been normalised. The experimental work was done by the chemistry department in theUniversity of Glasgow, which used an Itrax densityscanner [7]. Fig. 3 shows a density profile for oneradial strip, plotted from pith to bark (blue line). Eachpeak represents the end of one growth ring. A goodapproximation of these lines is a fitted power functioncurve (red line) that encompasses one growth ringfrom one peak to the next one. The equation for theseapproximated curves is given in the left box of Fig. 4.The right box depicts the resulting curve for thisequation for three rings. From several of these radialspecimens mean values of  min,i  ρ  , idiff  ,  ρ  ,   i exp,  ρ    and their c v can be calculated, serving as further inputparameters to generate a density profile for the model.Each individual  full spring encompasses an area of the cross section of  dy · dz for longitudinal anddiagonal springs and 2 · dy · dz for lateral springs, as shown in Fig. 4. The average normalised densityof this area from a randomly generated profile is calculated. Finally, the mean strength and stiffnessparameters for this particular spring are simply adjusted by multiplying this value with the originalmean parameters. Fig. 4 Density profile mapped on lattice dy dz  y zr  1 r  2 r  3 r  4 r  α  P c,i Fig. 3 Density profile with fitted curve of power functions in each growth ring ( ) imin,)1(1idiff, exp,  ρ  ρ  ρ   ρ  +      −−⋅= ++ i iii r r r r r  r r  1 r  2 r  3 r  4  ρ  1.00.0 min,1  ρ  diff,1min,1 ρ  ρ  + 2min,  ρ  2diff,2min, ρ  ρ  + 3min,  ρ   3diff,3min, ρ  ρ  + Distance r  from Pith [mm]    D  e  n  s   i   t  y   V  a  r   i  a   t   i  o  n   (  n  o  r  m  a   l   i  s  e   d  p  e  r  s  a  m  p   l  e   )  2.   Methodology As shown in the load-displacement curve definition of one spring i in Fig. 1, three types of meanparameters have to be adjusted: firstly, the mean stiffness values K  that can be directly (withlimitations) derived from E-moduli, secondly, the mean strength parameters S C  and S T  for eachspring type that will be determined by trial and error, and thirdly, the parameters γ T  and γ C  thatdefine the softening curve.All tested specimens came from one timber species, Sitka spruce ( Picea sitchensis ).Given that the timber behaves transverse isotropic on the small scale that it is modelled, it can beassumed that parameters in the Y and Z direction are the same. Thus, in summary, there are fourindependent mean elastic parameters ( K   x  , K   y = K   z  , K   yz  , K   xz = K   xy  , K   yz ), six independent meanstrength parameters ( S C,x  , S T,x  , S C,y = S C,z  , S T,y = S T,z  , S CT,xy = S CT,xz  , S CT,yz ) and two softening stiffnessparameters in compression ( γ C,x and   γ C,y =   γ C,z ). The remaining represent a very steep softeningcurve, thus ( γ T,x =   γ T,y = γ T,z =   γ CT,xy =   γ CT,xz =   γ CT,yz ≈ − ∞ ) .  While the stiffness parameters can be obtained from given  E  -moduli, the strength parameters areadjusted by means of comparisons between tested small clear specimens under various loadingconditions and their respective FE models. Fig. 5 demonstrates the methodology of the calibration. 2.1   Elastic Parameters The K  values for a lattice can not be adjusted entirely freely to represent full anisotropic or simplytransverse isotropic behaviour. The geometry imposes certain limitations. These could be overcomeby introducing angular springs that act in between the existing springs in one unit cell. With this, itwould be possible to adjust, for example, the elastic stiffness K   x and K   z independently from theshear modulus G  xz .However, this has not been done for this research as it adds considerably to the computationalproblem. In this model (without angular springs), calculating the possible elastic parameters fromthe independent spring stiffness of a lattice cell can be performed according to [5] and shall bepresented here briefly. This can be achieved by equating strain energy stored in a unit lattice celland energy stored in the respective continuum of the same volume. Fig. 5 Flowchart for the calibration process mean of:and c v mean of: α , r   pith , ,  ∆ r  centre mean  ∆ r  diff   , mean r  var  and c v Growth RingStructuresInput Parametersfor LinksDensityProfilesElasticParametersStrength Parameters S C,x  , S T,x  , S C,y =S C,z  , S T,y =S T,z  , S CT,xy =S CT,xz S CT,yz  , γ C,x  , γ C,y and c v =0.2 Five Experimental Test Series inTension (x), Shear (xy), Cleavage (y/z)Compression (x), Compression (y/z) LatticeModels Load–DisplacementCurvesmeanof: K   x  ,K   y =K   z K   xy =K   xz  ,K   yz and c v =0.2 shift  r  diff min , ρ  ρ  K   Exp S max,Exp K   Model S max,Model Load–DisplacementCurves Trial &Error   Trial &Error Geometries exp  ρ   ( ) ( ) ∫∑ ∑ ⋅=⋅== V continuumb N bbbcell dV U uF  E U  b ε σ   212121 ( ) ( ) ε ε  ⋅⋅=⋅= ∑ C U uKuU  continuum N bbcell b 2121 ∑ =⋅⋅= b  N bbmbk b jbibbijkm mk  jinnnnK l V C  3..1,,, 1 )()()()( )()( 2 (1)The strain energy can further be written as:(2)The former can be arranged as:(3)A subsequent step involves equating both strain energies and connecting displacement u with strain ε  , thus deriving the stiffness tensor C  . At the final stage C  can be written as:(4) V  represents the volume of the unit cell repeating in space ( V=2·dx·dy·dz ). The resulting stiffnesstensor C  ijkm of the size 3x3x3x3 can be transferred due to symmetry to the more widely used VoigtNotation with C  ij of the size 6x6. From this, the  E  -moduli and Poisson coefficients can be directlyobtained by calculating the inverse C  -1 . Thus, it is possible to calculate the elastic constants fromassumed spring stiffness. However, as mentioned before, due to the geometry of the lattice onlylimited elastic moduli can be obtained with certain K  s. Therefore, a program was written thatoptimises the K  values to find relatively close  E  -moduli and Poisson coefficient predictions.As input values, the  E  -modulus in the longitudinal direction was measured from tension test data(  E   x = 9792 N/mm²). The remaining  E  -moduli and Poisson ratios were then determined with ratiostaken from the Wood Handbook [6]. Since it is assumed that the material behaves transverseisotropic on the small scale of the lattice cells, several elastic parameters are the same. For theseinstances the mean value is taken as shown in Table 1.The best fit was achieved by optimising a target function, which is the sum of squared, normaliseddifferences between the calculated elastic parameter and the target parameter (E-modulus, shear-modulus and Poisson ratio). The optimisation routine resulted in the following parameters. Table 1 Determination of elastic parameters ElasticContinuumParametersTarget[N/mm²], [-]Result[N/mm²], [-]LatticeStiffnessParametersResult[N/mm]    E   x   9792 9608 K   x   1423.5  E   y = E   z   592 681 K   y = K   z 357.3 G  xy = G  xz   612 557 K   xy = K   xz 1392.2 G  yz not fitted to 325 K   yz 1297.9 ν  xy = ν  xz   0.43 0.4862 ν  yz = ν  zy   0.34 0.3719 ν  yx = ν  zx not fitted to 0.0345fit   Φ K    0.0635 Using these resulting K  values for the lattice and resulting  E, G, ν for solid elements, which adjointhe latter, they will both behave in the same way as far as bulk elastic properties are concerned 2.2   Strength Parameters Five different calibration tests have been undertaken to obtain load-displacement data for simplestress states: a tensile, shear and cleavage test along with compression tests in the longitudinal andlateral direction. These results serve as an input to calibrate the lattice’s strength parameters. continuumcell U U  =
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