axisymmetric_elasticity.py 6.62 KB
 Jeremy BLEYER committed May 28, 2018 1 2 3 4 5 6 7 8 9 10 11  # coding: utf-8 # # Axisymmetric formulation for elastic structures of revolution # # In this numerical tour, we will deal with axisymmetric problems of elastic solids. We will consider a solid of revolution around a fixed axis $(Oz)$, the loading, boundary conditions and material properties being also invariant with respect to a rotation along the symmetry axis. The solid cross-section in a plane $\theta=\text{cst}$ will be represented by a two-dimensional domain $\omega$ for which the first spatial variable (x[0] in FEniCS) will represent the radial coordinate $r$ whereas the second spatial variable will denote the axial variable $z$. # # ## Problem position # # We will investigate here the case of a hollow hemisphere of inner (resp. outer) radius $R_i$ (resp. $R_e$). Due to the revolution symmetry, the 2D cross-section corresponds to a quarter of a hollow cylinder.  Jeremy BLEYER committed Jun 12, 2018 12 # In[51]:  Jeremy BLEYER committed May 28, 2018 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67  from __future__ import print_function from dolfin import * from mshr import * import matplotlib.pyplot as plt get_ipython().magic(u'matplotlib notebook') Re = 11. Ri = 9. rect = Rectangle(Point(0., 0.), Point(Re, Re)) domain = Circle(Point(0., 0.), Re, 100) - Circle(Point(0., 0.), Ri, 100) domain = domain - Rectangle(Point(0., -Re), Point(-Re, Re)) - Rectangle(Point(0., 0.), Point(Re, -Re)) mesh = generate_mesh(domain, 40) plot(mesh) class Bottom(SubDomain): def inside(self, x, on_boundary): return near(x[1], 0) and on_boundary class Left(SubDomain): def inside(self, x, on_boundary): return near(x[0], 0) and on_boundary class Outer(SubDomain): def inside(self, x, on_boundary): return near(sqrt(x[0]**2+x[1]**2), Re, 1e-1) and on_boundary facets = MeshFunction("size_t", mesh, 1) facets.set_all(0) Bottom().mark(facets, 1) Left().mark(facets, 2) Outer().mark(facets, 3) ds = Measure("ds", subdomain_data=facets) # ## Definition of axisymmetric strains # # For axisymmetric conditions, the unkown displacement field is of the form: # # # \boldsymbol{u} = u_r(r,z)\boldsymbol{e}_r + u_z(r,z)\boldsymbol{e}_z # # # As a result, we will work with a standard VectorFunctionSpace of dimension 2. The associated strain components are however given by: # # # \boldsymbol{\varepsilon} = \begin{bmatrix} \partial_r u_r & 0 & (\partial_z u_r + \partial_r u_z)/2 \\ # 0 & u_r/r & 0 \\ # (\partial_z u_r + \partial_r u_z)/2 & 0 & \partial_z u_z\end{bmatrix}_{(\boldsymbol{e}_r,\boldsymbol{e}_\theta,\boldsymbol{e}_z)} # # # The previous relation involves explicitly the radial variable $r$, which can be obtained from the SpatialCoordinate x[0], the strain-displacement relation is then defined explicitly in the eps function. # # > **Note**: we could also express the strain components in the form of a vector of size 4 in alternative of the 3D tensor representation implemented below.  Jeremy BLEYER committed Jun 12, 2018 68 # In[52]:  Jeremy BLEYER committed May 28, 2018 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106  x = SpatialCoordinate(mesh) def eps(v): return sym(as_tensor([[v[0].dx(0), 0, v[0].dx(1)], [0, v[0]/x[0], 0], [v[1].dx(0), 0, v[1].dx(1)]])) E = Constant(1e5) nu = Constant(0.3) mu = E/2/(1+nu) lmbda = E*nu/(1+nu)/(1-2*nu) def sigma(v): return lmbda*tr(eps(v))*Identity(3) + 2.0*mu*eps(v) # ## Resolution # # The rest of the formulation is similar to the 2D elastic case with a small difference in the integration measure. Indeed, the virtual work principle reads as: # # \text{Find } \boldsymbol{u}\in V \text{ s.t. } \int_{\Omega} # \boldsymbol{\sigma}(\boldsymbol{u}):\boldsymbol{\varepsilon}(\boldsymbol{v}) d\Omega # = \int_{\partial \Omega_T} \boldsymbol{T}\cdot\boldsymbol{v} dS \quad \forall\boldsymbol{v} \in V # # where $\boldsymbol{T}$ is the imposed traction on some part $\partial \Omega_T$ of the domain boundary. # # In axisymmetric conditions, the full 3D domain $\Omega$ can be decomposed as $\Omega = \omega \times [0;2\pi]$ where the interval represents the $\theta$ variable. The integration measures therefore reduce to $d\Omega = d\omega\cdot(rd\theta)$ and $dS = ds\cdot(rd\theta)$ where $dS$ is the surface integration measure on the 3D domain $\Omega$ and $ds$ its counterpart on the cross-section boundary $\partial \omega$. Exploiting the invariance of all fields with respect to $\theta$, the previous virtual work principle is reformulated on the cross-section only as follows: # # # \text{Find } \boldsymbol{u}\in V \text{ s.t. } \int_{\omega} # \boldsymbol{\sigma}(\boldsymbol{u}):\boldsymbol{\varepsilon}(\boldsymbol{v}) rd\omega # = \int_{\partial \omega_T} \boldsymbol{T}\cdot\boldsymbol{v} rds \quad \forall\boldsymbol{v} \in V # # where the $2\pi$ constants arising from the integration on $\theta$ have been cancelled on both sides. As a result, the bilinear and linear form are similar to the plane 2D case with the exception of the additional $r$ term in the integration measures. # # The final formulation is therefore pretty straightforward. Since a uniform pressure loading is applied on the outer boundary, we will also need the exterior normal vector to define the work of external forces form.  Jeremy BLEYER committed Jun 12, 2018 107 # In[53]:  Jeremy BLEYER committed May 28, 2018 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127  n = FacetNormal(mesh) p = Constant(10.) V = VectorFunctionSpace(mesh, 'CG', degree=2) du = TrialFunction(V) u_ = TestFunction(V) a = inner(sigma(du), eps(u_))*x[0]*dx l = inner(-p*n, u_)*x[0]*ds(3) u = Function(V, name="Displacement") # First, smooth contact conditions are assumed on both $r=0$ (Left) and $z=0$ (Bottom) boundaries. For this specific case, the solution corresponds to the classical hollow sphere under external pressure with a purely radial displacement: # # u_r(r) = -\dfrac{R_e^3}{R_e^3-R_i^3}\left((1 − 2\nu)r + (1 + \nu)\dfrac{R_i^3}{2r^2}\right)\dfrac{p}{E}, # \quad u_z=0 #  Jeremy BLEYER committed Jun 12, 2018 128 # In[54]:  Jeremy BLEYER committed May 28, 2018 129 130 131 132 133 134 135 136 137  bcs = [DirichletBC(V.sub(1), Constant(0), facets, 1), DirichletBC(V.sub(0), Constant(0), facets, 2)] solve(a == l, u, bcs) print("Inwards radial displacement at (r=Re, theta=0): {:1.7f} (FE) {:1.7f} (Exact)".format(-u(Re, 0.)[0], float(Re**3/(Re**3-Ri**3)*((1-2*nu)*Re+(1+nu)*Ri**3/2/Re**2)*p/E))) print("Inwards radial displacement at (r=Ri, theta=0): {:1.7f} (FE) {:1.7f} (Exact)".format(-u(Ri, 0.)[0], float(Re**3/(Re**3-Ri**3)*((1-2*nu)*Ri+(1+nu)*Ri/2)*p/E)))  Jeremy BLEYER committed Jun 12, 2018 138 # In[55]:  Jeremy BLEYER committed May 28, 2018 139 140 141 142 143 144 145 146 147  plt.figure() plot(100*u, mode="displacement") plt.show() # The second loading case corresponds to a fully clamped condition on $z=0$, the vertical boundary remaining in smooth contact.  Jeremy BLEYER committed Jun 12, 2018 148 # In[56]:  Jeremy BLEYER committed May 28, 2018 149 150 151 152 153 154 155 156 157 158  bcs = [DirichletBC(V, Constant((0., 0.)), facets, 1), DirichletBC(V.sub(0), Constant(0), facets, 2)] solve(a == l, u, bcs) plt.figure() plot(mesh, linewidth=0.2) plot(200*u, mode="displacement") plt.show()