I am trying to solve Linear thermoelasticity but i recieve those 2 errors avascript Error: IPython is not defined / Object cannot be plotted directly, projecting to piecewise linears.

I already installed Ipython

#!/usr/bin/env python

## coding: utf-8

## # Linear thermoelasticity (weak coupling)

## ## Introduction

## In this tour, we will solve a linear thermoelastic problem. In permanent regime, the temperature field is uncoupled from the mechanical fields whereas the latter depend on the temperature due to presence of thermal strains in the thermoelastic constitutive relation. This situation can be described as

weakthermomechanical coupling.

Fullthermomechanical coupling for a transient evolution problem is treated in this tour.## In the present setting, the temperature field can be either given as a

`Constant`

or an`Expression`

throughout the domain or obtained as the solution of a steady-state heat (Poisson) equation. This last case will be implemented in this tour.## ## Problem position

## We consider the case of a rectangular 2D domain of dimensions L\times H fully clamped on both lateral sides and subjected to a self-weight loading. The bottom side is subjected to a uniform temperature increase of \Delta T = +50^{\circ}C while the top and lateral boundaries remain at the initial temperature T_0. The geometry and boundary regions are first defined.

## In[1]:

from dolfin import *

from mshr import *

import matplotlib.pyplot as plt

get_ipython().run_line_magic(‘matplotlib’, ‘notebook’)L, H = 5, 0.3

mesh = RectangleMesh(Point(0., 0.), Point(L, H), 100, 10, “crossed”)def lateral_sides(x, on_boundary):

return (near(x[0], 0) or near(x[0], L)) and on_boundary

def bottom(x, on_boundary):

return near(x[1], 0) and on_boundary

def top(x, on_boundary):

return near(x[1], H) and on_boundary## Because of the weak coupling discussed before, the thermal and mechanical problem can be solved separately. As a result, we don’t need to resort to Mixed FunctionSpaces but can just define separately both problems.

## ## Resolution of the thermal problem

## The temperature is solution to the following equation \text{div}(k\nabla T) = 0 where k is the thermal conductivity (here we have no heat source). Since k is assumed to be homogeneous, it will not influence the solution. We therefore obtain a standard Poisson equation without forcing terms. Its formulation and resolution in FEniCS is quite standard with the temperature variation \Delta T as the main unknown.

## >

Note: We needed to define the linear form`LT`

in order to use the standard`solve(aT == LT, T, bcT)`

syntax for linear problems. We could also have equivalently defined the residual`res = dot(grad(T), grad(T_))*dx`

and used the nonlinear-like syntax`solve(res == 0, T, bcT)`

.## In[2]:

VT = FunctionSpace(mesh, “CG”, 1)

T_, dT = TestFunction(VT), TrialFunction(VT)

Delta_T = Function(VT, name=“Temperature increase”)

aT = dot(grad(dT), grad(T_))*dx

LT = Constant(0)*T_*dxbcT = [DirichletBC(VT, Constant(50.), bottom),

DirichletBC(VT, Constant(0.), top),

DirichletBC(VT, Constant(0.), lateral_sides)]

solve(aT == LT, Delta_T, bcT)

plt.figure()

p = plot(Delta_T, mode=“contour”)

plt.colorbar(p)

plt.show()## ## Mecanical problem

## The linearized thermoelastic constitutive equation is given by:

## $$\begin{equation}

## \boldsymbol{\sigma} = \mathbb{C}:(\boldsymbol{\varepsilon}-\alpha(T-T_0)\boldsymbol{1}) = \lambda\text{tr}(\boldsymbol{\varepsilon})\boldsymbol{1}+2\mu\boldsymbol{\varepsilon} -\alpha(3\lambda+2\mu)(T-T_0)\boldsymbol{1}

## \end{equation}$$

## where \lambda,\mu are the Lamé parameters and \alpha is the thermal expansion coefficient. As regards the current problem, the last term corresponding to the thermal strains is completely known. The following formulation can thus be generalized to any kind of known initial stress or eigenstrain state such as pre-stress or phase changes.

## In[3]:

E = Constant(50e3)

nu = Constant(0.2)

mu = E/2/(1+nu)

lmbda = Enu/(1+nu)/(1-2nu)

alpha = Constant(1e-5)

f = Constant((0, 0))

def eps(v):

return sym(grad(v))

def sigma(v, dT):

return (lmbdatr(eps(v))- alpha(3lmbda+2mu)dT)eps(v)Identity(2) + 2.0mu

Vu = VectorFunctionSpace(mesh, ‘CG’, 2)

du = TrialFunction(Vu)

u_ = TestFunction(Vu)

Wint = inner(sigma(du, Delta_T), eps(u_))*dx

aM = lhs(Wint)

LM = rhs(Wint) + inner(f, u_)*dx

bcu = DirichletBC(Vu, Constant((0., 0.)), lateral_sides)

u = Function(Vu, name=“Displacement”)## First, the self-weight loading is deactivated, only thermal stresses are computed.

## In[4]:

solve(aM == LM, u, bcu)

plt.figure()

p = plot(1e3*u[1],title=“Vertical displacement [mm]”)

plt.colorbar(p)

plt.show()

plt.figure()

p = plot(sigma(u, Delta_T)[0, 0],title=“Horizontal stress [MPa]”)

plt.colorbar(p)

plt.show()## We now take into account the self-weight.

## In[5]:

rho_g = 2400*9.81e-6

f.assign(Constant((0., -rho_g)))

solve(aM == LM, u, bcu)plt.figure()

p = plot(1e3*u[1],title=“Vertical displacement [mm]”)

plt.colorbar(p)

plt.show()

plt.figure()

p = plot(sigma(u, Delta_T)[0, 0],title=“Horizontal stress [MPa]”)

plt.colorbar(p)

plt.show()## In: