30. Appendix

30.1. Presentation

The problem studied is such that the solution is uniform in space, without any external effort given, so as to test only the treatment of the behavioral relationship.

The following solid is thus considered:

• height \(H\),

• axisymmetric (with radii \(a\) and \(b\)),

• or parallelepipedic (thickness \(b-a\)).

It is placed between two rigid lubricated plates.

The material is homogeneous thermoelastoplastic (see below) with isotropic work hardening and Von Mises criterion.

The temperature is assumed to be uniform in space, and increasing.

30.2. Kinematics, balance

30.2.1. Axisymmetric case (2D)

Move fields: \(u={u}_{r}(r){e}_{r}\) (blocked in \(z\))

deformation fields: \(\varepsilon (u)=(\begin{array}{ccc}{u}_{r}\text{'}& 0& 0\\ 0& 0& 0\\ 0& 0& \frac{{u}_{r}}{r}\end{array})\) (according to \((\begin{array}{c}r\\ z\\ \theta \end{array})\))

deformation fields: \(\sigma (u)={\sigma }_{L}(\begin{array}{ccc}0& 0& 0\\ 0& 1& 0\\ 0& 0& 0\end{array})\) (cf. boundary conditions) (according to \((\begin{array}{c}r\\ z\\ \theta \end{array})\))

30.2.2. Parallelepipedic case

Move fields: \(u={u}_{x}(x){e}_{x}+{u}_{y}(y){e}_{y}\) (blocked in \(z\))

deformation fields: \(\varepsilon (u)=(\begin{array}{ccc}{u}_{x}\text{'}& 0& 0\\ 0& 0& 0\\ 0& 0& {u}_{y}\text{'}\end{array})\) (according to \((\begin{array}{c}x\\ z\\ y\end{array})\))

deformation fields: \(\sigma (u)={\sigma }_{L}(\begin{array}{ccc}0& 0& 0\\ 0& 1& 0\\ 0& 0& 0\end{array})\) (cf. boundary conditions) (according to \((\begin{array}{c}x\\ z\\ y\end{array})\))

The case can be studied in D_ PLAN and in 3D.

30.3. Behavioral relationship

Isotropic, linear work hardening (constant tangent module \({E}_{T}\)).

VonMises criterion.

The elastic coefficients, \(E\) and \(\nu\), as well as the tangent module \({E}_{T}\), are invariant depending on the temperature.

The elastic limit \({\sigma }_{y}\) varies according to the temperature \(T\):

\({\sigma }_{y}(T)={\sigma }_{y}^{o}(1-s(T-{T}^{o}))\)

(for the temperature range studied, \({\sigma }_{y}\) is positive!).

The thermal expansion coefficient \(\alpha\) is constant.

\(2\mu =\frac{E}{1+\nu }\) \(\mathrm{3K}=\frac{E}{1-2\nu }\)

The law of behavior can be written (scalar internal variable \(P\)):

\(\varepsilon =\frac{1}{\mathrm{9K}}\mathrm{tr}\sigma \mathrm{Id}+\frac{1}{2\mu }{\sigma }^{D}+{\varepsilon }^{p}=\alpha (T-{T}^{o})\mathrm{Id}\)

with: \({\sigma }^{D}=\sigma –\frac{1}{3}\mathrm{tr}\sigma \mathrm{Id}\) (constraint deviator)

\(\dot{{\sigma }^{p}}=\frac{3}{2}\dot{p}\frac{{\sigma }^{D}}{∥{\sigma }_{\mathrm{éq}}∥}\), with \(\sqrt{\frac{3}{2}}\sqrt{{\sigma }^{D}{\sigma }^{D}}\)

\(\dot{p}=0\) if \(f(\sigma ,p)=∥{\sigma }_{\mathrm{éq}}∥–R(p)<0\)

\(\dot{p}\ge 0\) if \(f(\sigma ,p)=0\)

\(R(p)\) refers to the work hardening function:

\(R(p)={\sigma }_{y}+\frac{{\mathrm{EE}}_{T}}{E–{E}_{T}}p\)

The \(\dot{p}\) rate can be expressed, when \(f(\sigma ,p)=0\). Indeed, from \(\dot{p}f\) that is identical to zero, we draw: \(\dot{p}\dot{f}+\ddot{p}f=0\). So, when we are on criterion \(f=0\), necessarily \(\dot{f}=0\). That is to say:

\(\frac{3}{2}\frac{{\sigma }^{D}\dot{{\sigma }^{D}}}{∥{\sigma }_{\mathrm{éq}}∥}-{R}_{,T}\dot{T}–{R}_{,p}\dot{p}=0\)

\(\iff \frac{3}{2}\frac{{\sigma }^{D}\dot{{\sigma }^{D}}}{∥{\sigma }_{\mathrm{éq}}∥}-{\sigma }_{y}^{o}s\dot{T}-\frac{{\mathrm{EE}}_{T}}{E-{E}_{T}}\dot{p}=0\)

From where:

\(\dot{p}=\frac{E-{E}_{T}}{{\mathrm{EE}}_{T}}(\frac{3}{2}\frac{{\sigma }^{D}\dot{{\sigma }^{D}}}{∥{\sigma }_{\mathrm{éq}}∥}+{\sigma }_{y}^{o}s\dot{T})\) if \(\dot{p}\ge 0\), for \(∥{\sigma }_{\mathrm{éq}}∥=R(p)\)

(criterion met, in « load »)

30.4. Thermal charging

Uniform temperature in space

\(T(t)=\theta t+{T}_{o},\theta >0\)

\(t\in [\mathrm{0,}{t}_{\mathrm{fin}}]\) with \({t}_{\mathrm{fin}}<\frac{1}{s\theta }\)

Initial blank status: \(\begin{array}{}{\sigma }_{L}=0\\ p=0\end{array}\)

30.5. Solution

Since the stress field is uniaxial, we have:

\({\sigma }^{D}=\frac{{\sigma }_{L}}{3}(\begin{array}{ccc}-1& 0& 0\\ 0& 2& 0\\ 0& 0& -1\end{array})\)

So:

\(∥{\sigma }_{\mathrm{éq}}∥=∣{\sigma }_{L}∣\)

and:

\(\dot{{\varepsilon }^{p}}=\frac{\dot{p}}{2}\mathrm{sgn}({\sigma }_{L})(\begin{array}{ccc}-1& 0& 0\\ 0& 2& 0\\ 0& 0& -1\end{array})\)

The behavioral relationship leads to

\(\dot{{\varepsilon }_{\mathrm{rr}}}=\dot{{\varepsilon }_{\theta \theta }}=\frac{-\nu }{E}\dot{{\sigma }_{L}}-\frac{\dot{p}}{2}\mathrm{sgn}({\sigma }_{L})+\alpha \dot{T}\) (\(\dot{{\varepsilon }_{\mathrm{xx}}}=\dot{{\varepsilon }_{\mathrm{yy}}}\) for the case of the parallelepiped)

\(\dot{{\varepsilon }_{\mathrm{zz}}}=0=\frac{1}{E}\dot{{\sigma }_{L}}+\dot{p}\mathrm{sgn}({\sigma }_{L})+\alpha \dot{T}\)

From where:

\(\dot{{\varepsilon }_{\mathrm{rr}}}=\dot{{\varepsilon }_{\theta \theta }}=\frac{3}{2}\alpha \dot{T}+\frac{1-2\nu }{2E}\dot{{\sigma }_{L}}\)

\(\dot{p}=\mathrm{sgn}({\sigma }_{L})(-\alpha \dot{T}–\frac{\dot{{\sigma }_{L}}}{E})=0\) if \(∣{\sigma }_{L}∣<R(p)\)

\(\dot{p}=\mathrm{Max}[0;\frac{E-{E}_{T}}{{\mathrm{EE}}_{T}}(\frac{3}{2}\frac{{\sigma }^{D}{\sigma }^{D}}{∥{\sigma }_{\mathrm{éq}}∥}+{\sigma }_{y}^{o}s\dot{T})]\) otherwise

That is to say, in case \(∣{\sigma }_{L}∣=R(p)\) (criterion met)

\(\dot{p}=\mathrm{Max}[0;\frac{E-{E}_{T}}{{\mathrm{EE}}_{T}}(\mathrm{sgn}({\sigma }_{L})\dot{{\sigma }_{L}}+{\sigma }_{y}^{o}s\dot{T})]\)

30.5.1. Elastic phase

At the start of thermal loading, \(∣{\sigma }_{L}∣\) being less than \({\sigma }_{y}\), \(\dot{p}\) is zero.

From where:

\(\dot{{\sigma }_{L}}=-E\alpha \dot{T}\)

\(\dot{{\varepsilon }_{\mathrm{rr}}}=\dot{{\varepsilon }_{\theta \theta }}=\alpha \dot{T}(1+\nu )\)

So:

\({\sigma }_{L}=-E\alpha \theta t\) (\({\sigma }_{L}<0\) compression)

\({\varepsilon }_{\mathrm{rr}}={\varepsilon }_{\theta \theta }=\alpha \theta (1+\nu )t\)

Validity of the elastic solution

The criterion is:

\(∣{\sigma }_{L}(t)∣-{\sigma }_{y}(t)=E=\theta t-{\sigma }_{y}^{o}(1-s\theta t)\le 0\)

The criterion is not met for \(t=[\mathrm{0,}{t}_{y}]\), with:

\({t}_{y}=\frac{{\sigma }_{y}^{o}}{\theta (E\alpha +{\sigma }_{y}^{o}s)}\)

At moment \({t}_{y}\):

\({\sigma }_{L}({t}_{y})=\frac{-E\alpha {\sigma }_{y}^{o}}{E\alpha +{\sigma }_{L}^{o}s}\)

30.5.2. Elastoplastic phase

\(t\ge {t}_{y}\). We are on the criteria. So:

\(\dot{p}=\mathrm{Max}[0;\frac{E-{E}_{T}}{{\mathrm{EE}}_{T}}(\dot{{\sigma }_{L}}\mathrm{sgn}({\sigma }_{L})+{\sigma }_{y}^{o}s\dot{T})]\)

Assuming that you are « in charge » \((\dot{p}>0)\), then you eliminate \(\dot{p}\) to have

\(\dot{{\sigma }_{L}}=-{E}_{T}\dot{T}(\alpha +\mathrm{sgn}({\sigma }_{L})\frac{E-{E}_{T}}{{\mathrm{EE}}_{T}}s{\sigma }_{y}^{o})\)

then:

\(\dot{p}=\frac{E-{E}_{T}}{E}\dot{T}(-\alpha \mathrm{sgn}({\sigma }_{L})+\frac{s{\sigma }_{y}^{o}}{E})\)

to \(t={t}_{y}\), \({\sigma }_{L}=-E\alpha \theta {t}_{y}<0\); we then integrate these expressions for \(t\ge {t}_{y}(\dot{T}=\theta )\):

\({\sigma }_{L}(t)=-{E}_{T}\theta (t-{t}_{y})[\alpha –\frac{E-{E}_{T}}{{\mathrm{EE}}_{T}}s{\sigma }_{y}^{o}]-{\sigma }_{L}({t}_{y})\)

\(p(t)=\frac{E-{E}_{T}}{{E}^{2}}\theta [\alpha E+s{\sigma }_{y}^{o}](t-{t}_{y})\)

Or, after rearrangement, \((t>{\mathrm{0t}}_{y})\):

\({\sigma }_{L}(t)={\sigma }_{y}^{o}(s\theta t-1+\frac{{E}_{T}}{E}(1-\frac{t}{{t}_{y}}))\)

\(p(t)=\frac{{\sigma }_{y}^{o}(E-{E}_{T})}{{E}^{2}}(\frac{t}{{t}_{y}}–1)\)

Validity of this elastoplastic solution

You have to make sure that \({\sigma }_{L}(t)\) stays negative. Knowing that \(s\theta t<1\), and \(t>{t}_{y}\), the previous result confirms that \({\sigma }_{L}(t)<0\).

Finally, we note that:

\(\mathrm{sgn}({\sigma }_{L})\frac{1-2\nu }{2}\dot{p}+\dot{{\varepsilon }_{\mathrm{rr}}}=\alpha (1+\nu )\dot{T}\)

from where:

\({\varepsilon }_{\mathrm{rr}}(t)={\varepsilon }_{\theta \theta }(t)=\alpha \theta (1+\nu )t+\frac{1-2\nu }{2}p(t),\forall t\in [{t}_{y},{t}_{\mathrm{fin}}]\)

(since \({\sigma }_{L}(t)<0\)).

30.6. Digital application

\(E=200000\mathrm{MPa}\); \(\nu =0.3\); \(\alpha =10.0E-5°{C}^{-1}\); \(\theta =1.0{s}^{-1}\)

\({\sigma }_{y}^{o}=400\mathrm{MPa}\); \({T}^{o}=0°C\); \(s=10.0E-2°{C}^{-1}\); \({t}_{\mathrm{fin}}<100s\)

\({E}_{T}=50000\mathrm{MPa}\);

From which we obtain in the elastic phase:

\({t}_{y}=66.6666s\)

\({\sigma }_{L}({t}_{y})=-133.333\mathrm{MPa}\)

\({\varepsilon }_{\mathrm{rr}}({t}_{y})={\varepsilon }_{\theta \theta }({t}_{y})=0.866666E-3\)

Then, elastoplastic phase:

to \(t=\mathrm{80s}\): \({\sigma }_{L}(80)=-100\mathrm{MPa}\)

\(p(80)=0.30E-3\)

\({\varepsilon }_{\mathrm{rr}}(80)={\varepsilon }_{\theta \theta }(80)=1.100E-3\)

to \(t=\mathrm{90s}\): \({\sigma }_{L}(90)=-75\mathrm{MPa}\)

\(p(90)=0.525E-3\)

\({\varepsilon }_{\mathrm{rr}}(90)={\varepsilon }_{\theta \theta }(90)=1.275E-3\)