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 1 jgs 110 % $Id$ 2 \chapter{Bending Beam under Uniform Load} 3 \label{BEAM CHAP} 4 \sectionauthor{Jannine Eisenmann}{} 5 6 Given is a two-dimension bending beam which is fixed in all directions 7 at the left end and free at the other, see \fig{BEAM FIG 1}. Furthermore the beam is loaded 8 with a uniform load $p$. 9 10 \begin{figure} 11 % \centerline{\includegraphics[width=\figwidth]{DiffusionRes1}} 12 \caption{Loaded Beam.} 13 \label{BEAM FIG 1} 14 \end{figure} 15 16 17 18 For emphasizing the displacement this picture is drawn (uebertrieben), 19 cause since we use the linear theory otherwise no displacements would 20 be visible. 21 There are two ways of solving this problem: on the one hand one can 22 use the differential equation for the deformed neutral fibre of the 23 beam. This classical differential equation is based on several simplifying 24 assumptions concerning the statics and kinematics of the problem. 25 However the results are known to be highly accurate at least for slender 26 beams with length to hight ratios $> 10$. Alternatively, in connection 27 with finite element based differential equation toolkits one may simply 28 consider the beam as a special case of a 2D or 3D elastic continuum 29 problem and solve the stress equilibrium equations combined with Hooke's 30 law for the specific boundary conditions of a beam. Both cases assume 31 isotropic and linear elastic material properties. 32 33 The beam equation is easily solved analytically. The analytical solutions 34 can be used for benchmarkung finite element solutions. In section 35 1.2 we formluate a finite element code for general isotropic elasticity 36 problems including thin and deep beams under arbitrary loading conditiond 37 in 2D or 3D. 38 39 40 \section{2-dimensional} 41 As the stress equilibrium equation is a partial differential equation 42 (PDE), we choose to use Finley to solve it, since Finley is a finite 43 element kernel library, that has been incorporated as a differential 44 equation solver into the python based numerical toolkit called escript. 45 We divided the beam into ten square elements of the size 1x1. Each 46 element consists of 8 nodes, which leads to a quadratic interpolation 47 of the model point displacements \\ 48 49 The key ingredients is \textbf{Hooks Law}. We use Hooks Law because 50 we are dealing with \textbf{linear elastic material} \textbf{behaviour}. 51 We have \\ 52 53 54 $\sigma_{ik}=2G\left(\varepsilon_{ik}+\frac{\nu}{1-2\nu}\cdot e\cdot\delta_{ik}\right)$\hfill{}(1)\\ 55 where the engineering strain$\varepsilon_{ik}$is defined as: 56 57 $\varepsilon_{ik}=\frac{1}{2}\cdot\left(u_{k,i}+u_{i,k}\right)$\hfill{}(2)\\ 58 59 60 with 61 62 \begin{enumerate} 63 \item e= Volume strain = $\varepsilon_{kk}$ 64 \item $\delta_{ik}$= Kronecker symbol 65 \end{enumerate} 66 Inserting equation (2) in (1) (and with further mathematical conversions) 67 leads to the following partial differential equation :\\ 68 69 70 $\sigma_{ij}=\left[\mu\left(\delta_{ik}\delta_{jl}+\delta_{il}\delta_{jk}\right)+\lambda\left(\delta_{ij}\delta_{kl}\right)\right]u_{k,l}$\\ 71 72 73 For tension equilibrium we require:\\ 74 75 76 $\sigma_{ij,j}=0$\\ 77 78 79 which leads to the following expression:\\ 80 81 82 $83 \left(\left[\mu\left(\delta_{ik}\delta_{jl}+\delta_{il}\delta_{jk}\right)+\lambda\left(\delta_{ij}\delta_{kl}\right)\right]u_{k,l}\right)_{,j}=0$ 84 85 86 with the natural boundary condition: 87 88 $89 n_{j}\sigma_{ij}=-p_{i}$ 90 \\ 91 $p_{i}$ representing a uniform load on top of the beam. 92 93 A Dirichlet Boundary condition is assumed on the left end of the beam 94 for which no displacements can occure.\\ 95 \\ 96 \includegraphics[% 97 width=0.60\linewidth,bb = 0 0 200 100, draft, type=eps]{/home/jeannine/sandbox/report/draws/dir_cond_beam.eps}\\ 98 This is described in the code with the setting a mask for the left 99 end and setting values to that mask: 100 101 \begin{python} 102 q = xNodes{[}0{]}.whereZero(){*}{[}1.0,1.0{]} 103 104 r = Vector({[}0.0, 0.0{]}, where = nodes) 105 \end{python} 106 The Finley template PDE reads:\\ 107 108 109 $110 -(A_{ijkl}u_{k,l})_{,j}-(B_{ijk}u_{k})_{,j}+C_{ikl}u_{k,l}+D_{ik}u_{k}=-X_{ij,j}+Y_{i}$ 111 \\ 112 with the natural boundary condition: 113 114 $115 n_{j}(A_{ijkl}u_{k,l}+B_{ijkl}u_{k})+d_{ik}u_{k}=n_{j}X_{ij}+y_{i}on\Gamma_{i}^{D}$ 116 117 118 Yields by comparing the coefficients : 119 120 \begin{enumerate} 121 \item $A_{ijkl}$= $\left[\mu\left(\delta_{ik}\delta_{ij}+\delta_{jl}\delta_{il}\right)+\lambda\left(\delta_{ij}\delta_{kl}\right)\right]$ 122 \item $y_{i}$= $-p_{i}$ 123 \item $u_{k}$= displacement $u$ 124 \end{enumerate} 125 $B_{ijk,}=C_{ikl}=D_{ik}=X_{ij}=Y_{i}=d_{ik}=0$\\ 126 127 128 Where 0 in the last line is taken as a scalar, vector or tensor, depending 129 on what the belonging coefficient is. 130 131 These equations are the base for the \textbf{Finley Script}: 132 133 \begin{python} 134 from ESyS import {*} 135 import Finley 136 137 138 139 \#mu lamda 140 141 def mu (E, nu): \#= shear modul G 142 143 return E/(2{*}(1+nu)) 144 145 def lamda (E, nu): 146 147 return (nu{*}E)/((1-2{*}nu){*}(1+nu)) 148 149 150 151 def main() 152 153 \#material, beam PROPERTIES 154 155 L = 10.0 \#length of beam {[}m{]} 156 157 h = 1 \#height of beam {[}m{]} 158 159 p = 1 \#outer uniform load {[}kN/m{]} 160 161 E0 = 210000 \#Young's modulus {[}kN/m\textasciicircum{}2{]} 162 163 nu = 0.4 \#Poisson ratio {[}-{]} 164 165 166 167 print \char\"{}L=\char\"{}, L 168 169 print \char\"{}h=\char\"{}, h 170 171 print \char\"{}p=\char\"{}, p 172 173 print \char\"{}E=\char\"{}, E0 174 175 print \char\"{}nu=Poisson =\char\"{}, nu 176 177 print \char\"{}mu=\char\"{}, mu (E0,nu) 178 179 print \char\"{}lamda=\char\"{}, lamda (E0,nu) 180 181 182 183 \#SET MESH for FE 184 185 mesh= Finley.Rectangle(n0=10 ,n1=1 ,order=2, l0=L, l1=h) 186 187 nodes = mesh.Nodes() 188 189 xNodes = nodes.getX() 190 191 elements = mesh.Elements() 192 193 faceElements = mesh.FaceElements() 194 195 xFaceElements = faceElements.getX() 196 197 198 199 \#DISPLACEMENT boundary 200 201 q = xNodes{[}0{]}.whereZero(){*}{[}1.,1.0{]} \#setting the mask for r 202 203 r = Vector({[}0.0, 0.0{]}, where = nodes) \#fixed end 204 205 206 207 \#STRESS boundary 208 209 ymask = xFaceElements{[}1{]}.whereEqualTo(h) 210 211 y = Vector({[}0, -p{]}, where = faceElements) 212 213 y = y{*}ymask 214 215 216 217 \#Finley coeff. 218 219 A = Tensor4(0, where = elements) 220 221 for i in range (2) : 222 223 for j in range (2) : 224 225 A{[}i,i,j,j{]}+= lamda (E0,nu) 226 227 A{[}j,i,j,i{]}+= mu (E0,nu) 228 229 A{[}j,i,i,j{]}+= mu (E0,nu) 230 231 232 233 M,b = mesh.assemble(A=A, B=B, q=q, 234 235 y=y,r=r,type=CSR,num\_equations=2) 236 237 u= M.iterative(b, tolerance=1e-8,iter\_max=50000, 238 239 iterative\_method=PCG) 240 241 print \char\"{}u{[}0{]}=\char\"{},u{[}0{]} 242 243 print \char\"{}u{[}1{]}=\char\"{},u{[}1{]} 244 245 main() 246 \end{python} 247 The finer the mesh the more exact is the solution. E.g. a 20x2 mesh 248 is more exact than a 10x1 mesh. 249

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