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The corresponding files can be obtained from:

- Jupyter Notebook:
`thermoelasticity_transient.ipynb`

- Python script:
`thermoelasticity_transient.py`

# Thermo-elastic evolution problem (full coupling)¶

## Introduction¶

In this tour, we will solve a transient thermoelastic evolution problem in which both thermomechanical fields are fully coupled, we will however assume that the evolution is quasi-static and will hence neglect inertial effects. Note that a staggered approach could also have been adopted in which one field is calculated first (say the temperature for instance) using predicted values of the other field and similarly for the other field in a second step (see for instance FAR91).

Static elastic computation with thermal strains is treated in the Linear thermoelasticity (weak coupling) tour.

## Problem position¶

The problem consists of a quarter of a square plate perforated by a circular hole. Thermo-elastic properties are isotropic and correspond to those of aluminium (note that stress units are in \(\text{MPa}\), distances in \(\text{m}\) and mass in \(\text{kg}\)). Linearized thermoelasticity will be considered around a reference temperature of \(T_0 = 293 \text{ K}\). A temperature increase of \(\Delta T=+10^{\circ}\text{C}\) will be applied on the hole boundary. Symmetry conditions are applied on the corresponding symmetry planes and stress and flux-free boundary conditions are adopted on the plate outer boundary.

We first import the relevant modules and define the mesh and material parameters (see the next section for more details on the parameters).

```
In [1]:
```

```
from __future__ import print_function
from dolfin import *
from mshr import *
import numpy as np
import matplotlib.pyplot as plt
%matplotlib notebook
L = 1.
R = 0.1
N = 50 # mesh density
domain = Rectangle(Point(0.,0.), Point(L, L)) - Circle(Point(0., 0.), R, 100)
mesh = generate_mesh(domain, N)
T0 = Constant(293.)
DThole = Constant(10.)
E = 70e3
nu = 0.3
lmbda = Constant(E*nu/((1+nu)*(1-2*nu)))
mu = Constant(E/2/(1+nu))
rho = 2700. # density
alpha = 2.31e-5 # thermal expansion coefficient
kappa = Constant(alpha*(2*mu + 3*lmbda))
cV = Constant(910e-6)*rho # specific heat per unit volume at constant strain
k = Constant(237e-6) # thermal conductivity
```

We now define the relevant FunctionSpace for the considered problem.
Since we will adopt a monolithic approach i.e. in which both fields are
coupled and solved at the same time, we will need to resort to a Mixed
FunctionSpace for both the displacement \(\boldsymbol{u}\) and the
temperature \(T\). For an introduction on the use of Mixed
FunctionSpace, check out the FEniCS tutorials on the mixed Poisson
equation
or the Stokes
problem.
Let us just point out that the constructor using `MixedFunctionSpace`

has been deprecated since version 2016.2 (see this
post).
In the following code, `MixedElement([Vue, Vte])`

could also be
replaced by `Vue * Vte`

.

```
In [2]:
```

```
d = 1 # interpolation degree
Vue = VectorElement('CG', mesh.ufl_cell(), d) # displacement finite element
Vte = FiniteElement('CG', mesh.ufl_cell(), d) # temperature finite element
V = FunctionSpace(mesh, MixedElement([Vue, Vte]))
def inner_boundary(x, on_boundary):
return near(x[0]**2+x[1]**2, R**2, 1e-3) and on_boundary
def bottom(x, on_boundary):
return near(x[1], 0) and on_boundary
def left(x, on_boundary):
return near(x[0], 0) and on_boundary
bc1 = DirichletBC(V.sub(0).sub(1), Constant(0.), bottom)
bc2 = DirichletBC(V.sub(0).sub(0), Constant(0.), left)
bc3 = DirichletBC(V.sub(1), DThole, inner_boundary)
bcs = [bc1, bc2, bc3]
```

Dirichlet boundary conditions must be defined from the full
FunctionSpace `V`

using the appropriate subspaces that is `V.sub(0)`

for the displacement (and `.sub(0)`

or `.sub(1)`

for the
corresponding x/y component) and `V.sub(1)`

for the temperature. Note
also that in the following, we will in fact work with the temperature
variation \(\Delta T = T-T_0\) as a field unkwown instead of the
total temperature. Hence, the boundary condition on the hole boundary
reads indeed as \(\Delta T=+10^{\circ}\text{C}\) .

Note:The definition of the`inner_boundary`

region used the syntax`near(expr, value, tolerance)`

to define the region where`expr = value`

up to the specified`tolerance`

. The given tolerance is quite large given that`mshr`

generates a faceted approximation of a Circle (here using 100 segments). Mesh nodes may therefore not lie exactly on the true circle.

## Variational formulation and time discretization¶

The linearized thermoelastic constitutive equations are given by:

- \(\lambda,\mu\) the Lamé coefficients
- \(\rho\) material density
- \(\alpha\) thermal expansion coefficient
- \(\kappa = \alpha(3\lambda+2\mu)\)
- \(C_{\varepsilon}\) the specific heat at constant strain (per unit of mass).
- \(s\) (resp. \(s_0\)) the entropy per unit of mass in the current (resp. initial configuration)

These equations are completed by the equilibrium equation which will later be expressed in its weak form (virtual work principle) and the linearized heat equation (without source terms):

where the heat flux is related to the temperature gradient through the isotropic Fourier law: \(\boldsymbol{q} = - k\nabla T\) with \(k\) being the thermal conductivity. Using the entropy constitutive relation, the weak form of the heat equation reads as:

with \(V_T\) being the FunctionSpace for the temperature field.

The time derivatives are now replaced by an implicit Euler scheme, so that the previous weak form at the time increment \(n+1\) is now:

where \(T\) and \(\boldsymbol{\varepsilon}\) correspond to the
*unknown* fields at the time increment \(n+1\). For more details on
the time discretization of the heat equation, see also the Heat
equation FEniCS
tutorial.

In addition to the previous thermal weak form, the mechanical weak form reads as:

where \(V_U\) is the displacement FunctionSpace and \(W_{ext}\) the linear functional corresponding to the work of external forces.

The solution of the coupled problem at \(t=t_{n+1}\) is now \((\boldsymbol{u}_{n+1},T_{n+1})=(\boldsymbol{u},T)\in V_U\times V_T\) verifying \((1)\) and \((2)\). These two forms are implemented below with zero right-hand sides (zero Neumann BCs for both problems here). One slight modification is that the temperature unknown \(T\) is replaced by the temperature variation \(\Delta T=T-T_0\) which appears naturally in the stress constitutive relation.

Note: We will later make use of the`lhs`

and`rhs`

functions to extract the corresponding bilinear and linear forms.

```
In [3]:
```

```
U_ = TestFunction(V)
(u_, T_) = split(U_)
dU = TrialFunction(V)
(du, dT) = split(dU)
Uold = Function(V)
(uold, Told) = split(Uold)
def eps(v):
return sym(grad(v))
def sigma(v, dT):
return (lmbda*tr(eps(v)) - kappa*dT)*Identity(2) + 2*mu*eps(v)
dt = Constant(0.)
mech_form = inner(sigma(du, dT), eps(u_))*dx
therm_form = (cV*(dT-Told)/dt*T_ + kappa*T0*tr(eps(du-uold))/dt*T_ + dot(k*grad(dT), grad(T_)))*dx
form = mech_form + therm_form
```

## Resolution¶

The problem is now solved by looping over time increments. Because of the typical exponential time variation of temperature evolution of the heat equation, time steps are discretized on a non-uniform (logarithmic) scale. \(\Delta t\) is therefore updated at each time step. Note that since we work in terms of temperature variation and not absolute temperature all fields can be initialized to zero, otherwise \(T\) would have needed to be initialized to the reference temperature \(T_0\).

```
In [4]:
```

```
Nincr = 10
t = np.logspace(1, 4, Nincr+1)
Nx = 100
x = np.linspace(R, L, Nx)
T_res = np.zeros((Nx, Nincr+1))
U = Function(V)
(u, T) = split(U)
for (i, dti) in enumerate(np.diff(t)):
print("Increment " + str(i+1))
dt.assign(dti)
solve(lhs(form) == rhs(form), U, bcs)
Uold.assign(U)
T_res[:, i+1] = [U(xi, 0.)[2] for xi in x]
```

```
Increment 1
Increment 2
Increment 3
Increment 4
Increment 5
Increment 6
Increment 7
Increment 8
Increment 9
Increment 10
```

At each time increment, the variation of the temperature increase
\(\Delta T\) along a line \((x, y=0)\) is saved in the `T_res`

array. This evolution is plotted below. As expected, the temperature
gradually increases over time, reaching eventually a uniform value of
\(+10^{\circ}\text{C}\) over infinitely long waiting time.

```
In [5]:
```

```
plt.figure()
plt.plot(x, T_res[:, 1::Nincr/10])
plt.xlabel("$x$-coordinate along $y=0$")
plt.ylabel("Temperature variation $\Delta T$")
plt.legend(["$t={:.0f}$".format(ti) for ti in t[1::Nincr/10]], ncol=2)
plt.show()
```

The final stresses and temperature variation are plotted using
`matplotlib`

. Note that we can add colorbar and change the plot axis
limit.

```
In [8]:
```

```
plt.figure()
p = plot(sigma(u, T)[1,1], title="$\sigma_{yy}$ stress near the hole")
plt.xlim((0, 3*R))
plt.ylim((0, 3*R))
plt.colorbar(p)
plt.show()
plt.figure()
p = plot(T, title="Temperature variation")
plt.xlim((0, 3*R))
plt.ylim((0, 3*R))
plt.colorbar(p)
plt.show()
```

## References¶

[FAR91]: Farhat, C., Park, K. C., & Dubois-Pelerin, Y. (1991). An unconditionally stable staggered algorithm for transient finite element analysis of coupled thermoelastic problems. Computer Methods in Applied Mechanics and Engineering, 85(3), 349-365. https://doi.org/10.1016/0045-7825(91)90102-C