### Some updates of Reissner DG

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 ... ... @@ -10,16 +10,18 @@ Introduction ------------- This program solves the Reissner-Mindlin plate equations on the unit square with uniform transverse loading and simply supported boundary conditions. square with uniform transverse loading and clamped boundary conditions. The corresponding file can be obtained from :download:reissner_mindlin_dg.py. It uses a Discontinuous Galerkin interpolation for the rotation field to remove shear-locking issues in the thin plate limit. Details of the formulation can be found in *P. Hansbo et al., Comput. Methods Appl. Mech. Engrg. 200 (2011) 638–648*, https://doi.org/10.1016/j.cma.2010.09.009 can be found in [HAN2011]_. The solution for :math:w in this demo will look as follows: The solution for :math:\theta_x on the middle line of equation :math:y=0.5 will look as follows for 10 elements and a stabilization parameter :math:s=1: .. image:: dg_rotation_N10_s1.png :scale: 15% ... ... @@ -30,6 +32,7 @@ Implementation Material properties and loading are the same as in :ref:ReissnerMindlinQuads:: from __future__ import print_function from fenics import * E = Constant(1e3) ... ... @@ -45,37 +48,37 @@ The unit square mesh is here divided in triangles and we get the facet MeshFunct mesh = UnitSquareMesh(N, N) facets = MeshFunction("size_t", mesh, 1) facets.set_all(0) ds = Measure("ds")[facets] ds = Measure("ds", subdomain_data=facets) Continuous interpolation using of degree 2 is chosen for the deflection :math:w whereas the rotation field :math:\underline{\theta} is discretized using discontinuous linear polynomials:: We = FiniteElement("Lagrange", mesh.ufl_cell(), 2) Te = VectorElement("DG", mesh.ufl_cell(), 1) V = FunctionSpace(mesh,MixedElement([We,Te])) V = FunctionSpace(mesh, MixedElement([We, Te])) Clamped boundary conditions on the lateral boundary are defined as:: def border(x, on_boundary): return on_boundary bc = [DirichletBC(V.sub(0),Constant(0.), border)] bc = [DirichletBC(V.sub(0), Constant(0.), border)] Standard part of the variational form is the same (without full integration):: def strain2voigt(eps): return as_vector([eps[0,0],eps[1,1],2*eps[0,1]]) return as_vector([eps[0, 0], eps[1, 1], 2*eps[0, 1]]) def voigt2stress(S): return as_tensor([[S,S],[S,S]]) return as_tensor([[S, S], [S, S]]) def curv(u): (w,theta) = split(u) (w, theta) = split(u) return sym(grad(theta)) def shear_strain(u): (w,theta) = split(u) (w, theta) = split(u) return theta-grad(w) def bending_moment(u): DD = as_tensor([[D,nu*D,0],[nu*D,D,0],[0,0,D*(1-nu)/2.]]) DD = as_tensor([[D, nu*D, 0], [nu*D, D, 0],[0, 0, D*(1-nu)/2.]]) return voigt2stress(dot(DD,strain2voigt(curv(u)))) def shear_force(u): return F*shear_strain(u) ... ... @@ -84,10 +87,10 @@ Standard part of the variational form is the same (without full integration):: u_ = TestFunction(V) du = TrialFunction(V) L = f*u_*dx a = inner(bending_moment(u_),curv(du))*dx + dot(shear_force(u_),shear_strain(du))*dx a = inner(bending_moment(u_), curv(du))*dx + dot(shear_force(u_), shear_strain(du))*dx We then add the contribution of jumps in rotation across all internal facets plus a stabilization term involing a user-defined parameter :math:s:: ... ... @@ -96,40 +99,48 @@ a stabilization term involing a user-defined parameter :math:s:: h = CellVolume(mesh) h_avg = (h('+')+h('-'))/2 stabilization = Constant(10.) (dw,dtheta) = split(du) (w_,theta_) = split(u_) a -= dot(avg(dot(bending_moment(u_),n)),jump(dtheta))*dS + dot(avg(dot(bending_moment(du),n)),jump(theta_))*dS \ - stabilization*D/h_avg*dot(jump(theta_),jump(dtheta))*dS (dw, dtheta) = split(du) (w_, theta_) = split(u_) a -= dot(avg(dot(bending_moment(u_), n)), jump(dtheta))*dS + dot(avg(dot(bending_moment(du), n)), jump(theta_))*dS \ - stabilization*D/h_avg*dot(jump(theta_), jump(dtheta))*dS Because of the clamped boundary conditions, we also need to add the corresponding contributions of the external facets (the imposed rotation is zero on the boundary contributions of the external facets (the imposed rotation is zero on the boundary so that no term arise in the linear functional):: a -= dot(dot(bending_moment(u_),n),dtheta)*ds + dot(dot(bending_moment(du),n),theta_)*ds \ - 2*stabilization*D/h*dot(theta_,dtheta)*ds a -= dot(dot(bending_moment(u_), n), dtheta)*ds + dot(dot(bending_moment(du), n), theta_)*ds \ - 2*stabilization*D/h*dot(theta_, dtheta)*ds We then solve for the solution and export the relevant fields to XDMF files :: solve(a == L, u, bc) (w,theta) = split(u) Vw = FunctionSpace(mesh,We) Vt = FunctionSpace(mesh,Te) (w, theta) = split(u) Vw = FunctionSpace(mesh, We) Vt = FunctionSpace(mesh, Te) ww = Function(Vw, name="Deflection") tt = Function(Vt, name="Rotation") ww.assign(project(w, Vw)) tt.assign(project(theta, Vt)) file_results = XDMFFile("RM_DG_results.xdmf") file_results.parameters["flush_output"] = True file_results.parameters["functions_share_mesh"] = True file_results.write(ww, 0.) file_results.write(tt, 0.) The solution is compared to the Kirchhoff analytical solution:: print "Kirchhoff deflection:", -1.265319087e-3*float(f/D) print "Reissner-Mindlin FE deflection:", ww(0.5,0.5) \ No newline at end of file print("Kirchhoff deflection:", -1.265319087e-3*float(f/D)) print("Reissner-Mindlin FE deflection:", -ww(0.5, 0.5)) For :math:h=0.001 and 50 elements per side, one finds :math:w_{FE} = 1.38322\text{e-5} against :math:w_{\text{Kirchhoff}} = 1.38173\text{e-5} for the thin plate solution. ----------- References ----------- .. [HAN2011] Peter Hansbo, David Heintz, Mats G. Larson, A finite element method with discontinuous rotations for the Mindlin-Reissner plate model, *Computer Methods in Applied Mechanics and Engineering*, 200, 5-8, 2011, pp. 638-648, https://doi.org/10.1016/j.cma.2010.09.009. \ No newline at end of file
 ... ... @@ -10,13 +10,13 @@ Introduction ------------- This program solves the Reissner-Mindlin plate equations on the unit square with uniform transverse loading and fully clamped boundary conditions. square with uniform transverse loading and fully clamped boundary conditions. The corresponding file can be obtained from :download:reissner_mindlin_quads.py. It uses quadrilateral cells and selective reduced integration (SRI) to remove shear-locking issues in the thin plate limit. Both linear and quadratic interpolation are considered for the transverse deflection :math:w and rotation :math:\underline{\theta}. remove shear-locking issues in the thin plate limit. Both linear and quadratic interpolation are considered for the transverse deflection :math:w and rotation :math:\underline{\theta}. The solution for :math:w in this demo will look as follows: ... ... @@ -32,11 +32,12 @@ Implementation Material parameters for isotropic linear elastic behavior are first defined:: from __future__ import print_function from fenics import * E = Constant(1e3) nu = Constant(0.3) Plate bending stiffness :math:D=\dfrac{Eh^3}{12(1-\nu^2)} and shear stiffness :math:F = \kappa Gh with a shear correction factor :math:\kappa = 5/6 for a homogeneous plate of thickness :math:h:: ... ... @@ -60,38 +61,38 @@ Continuous interpolation using of degree :math:d=\texttt{deg} is chosen for bo deg = 1 We = FiniteElement("Lagrange", mesh.ufl_cell(), deg) Te = VectorElement("Lagrange", mesh.ufl_cell(), deg) V = FunctionSpace(mesh,MixedElement([We,Te])) V = FunctionSpace(mesh, MixedElement([We, Te])) Clamped boundary conditions on the lateral boundary are defined as:: def border(x, on_boundary): return on_boundary bc = [DirichletBC(V,Constant((0.,0.,0.)), border)] bc = [DirichletBC(V, Constant((0., 0., 0.)), border)] Some useful functions for implementing generalized constitutive relations are now defined:: def strain2voigt(eps): return as_vector([eps[0,0],eps[1,1],2*eps[0,1]]) return as_vector([eps[0, 0], eps[1, 1], 2*eps[0, 1]]) def voigt2stress(S): return as_tensor([[S,S],[S,S]]) return as_tensor([[S, S], [S, S]]) def curv(u): (w,theta) = split(u) (w, theta) = split(u) return sym(grad(theta)) def shear_strain(u): (w,theta) = split(u) (w, theta) = split(u) return theta-grad(w) def bending_moment(u): DD = as_tensor([[D,nu*D,0],[nu*D,D,0],[0,0,D*(1-nu)/2.]]) DD = as_tensor([[D, nu*D, 0], [nu*D, D, 0],[0, 0, D*(1-nu)/2.]]) return voigt2stress(dot(DD,strain2voigt(curv(u)))) def shear_force(u): return F*shear_strain(u) The contribution of shear forces to the total energy is under-integrated using a custom quadrature rule of degree :math:2d-2 i.e. for linear (:math:d=1) a custom quadrature rule of degree :math:2d-2 i.e. for linear (:math:d=1) quadrilaterals, the shear energy is integrated as if it were constant (1 Gauss point instead of 2x2) and for quadratic (:math:d=2) quadrilaterals, as if it were quadratic (2x2 Gauss points instead of 3x3):: ... ... @@ -99,37 +100,37 @@ and for quadratic (:math:d=2) quadrilaterals, as if it were quadratic (2x2 Gau u_ = TestFunction(V) du = TrialFunction(V) dx_shear = dx(metadata={"quadrature_degree":2*deg-2}) dx_shear = dx(metadata={"quadrature_degree": 2*deg-2}) L = f*u_*dx a = inner(bending_moment(u_),curv(du))*dx + dot(shear_force(u_),shear_strain(du))*dx_shear a = inner(bending_moment(u_), curv(du))*dx + dot(shear_force(u_), shear_strain(du))*dx_shear We then solve for the solution and export the relevant fields to XDMF files :: solve(a == L, u, bc) (w,theta) = split(u) Vw = FunctionSpace(mesh,We) Vt = FunctionSpace(mesh,Te) (w, theta) = split(u) Vw = FunctionSpace(mesh, We) Vt = FunctionSpace(mesh, Te) ww = Function(Vw, name="Deflection") tt = Function(Vt, name="Rotation") ww.assign(project(w, Vw)) tt.assign(project(theta, Vt)) file_results = XDMFFile("RM_results.xdmf") file_results.parameters["flush_output"] = True file_results.parameters["functions_share_mesh"] = True file_results.write(ww, 0.) file_results.write(tt, 0.) The solution is compared to the Kirchhoff analytical solution:: print "Kirchhoff deflection:", -1.265319087e-3*float(f/D) print "Reissner-Mindlin FE deflection:", -min(ww.vector().get_local()) # point evaluation for quads # is not implemented in fenics 2017.2 print("Kirchhoff deflection:", -1.265319087e-3*float(f/D)) print("Reissner-Mindlin FE deflection:", -min(ww.vector().get_local())) # point evaluation for quads # is not implemented in fenics 2017.2 For N=50 quads per side, one finds :math:w_{FE} = 1.38182\text{e-5} for linear quads For :math:h=0.001 and 50 quads per side, one finds :math:w_{FE} = 1.38182\text{e-5} for linear quads and :math:w_{FE} = 1.38176\text{e-5} for quadratic quads against :math:w_{\text{Kirchhoff}} = 1.38173\text{e-5} for the thin plate solution. \ No newline at end of file the thin plate solution. \ No newline at end of file
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