2. Benchmark solutions#

2.1. Calculation method#

In simple traction and with the hypothesis of small deformations, the tensile stress

as well as the plastic multiplier

Just now

are given in the case under consideration by:

: \(\mathrm{\sigma }(u)=E\frac{l(u)-{L}_{0}}{{L}_{0}}\), \(\dot{p}(u)=0\), \(l({t}_{1}^{p})={L}_{0}(1+\frac{{\mathrm{\sigma }}_{Y}(-50°C)}{E})\)

: \(\mathrm{\sigma }(u)={E}_{t}\frac{l(u)-{L}_{0}}{{L}_{0}}+\frac{E-{E}_{t}}{E}{\mathrm{\sigma }}_{Y}(-50°C)\), \(\dot{p}(u)=(1-\frac{{E}_{t}}{E})\frac{\dot{l}(u)}{{L}_{0}}\)

: \(\mathrm{\sigma }(u)=\mathrm{\sigma }(u=10)-E\frac{l(u=10)-l(u)}{{L}_{0}}\), \(\dot{p}(u)=0\)

: \(\mathrm{\sigma }(u)=\mathrm{\sigma }(u=20)\), \(\dot{p}(u)=0\)

: \(\mathrm{\sigma }(u)=\mathrm{\sigma }(u=20)+{E}_{t}\frac{l(u)-l(u=20)}{{L}_{0}}\), \(\dot{p}(u)=(1-\frac{{E}_{t}}{E})\frac{\dot{l}(u)}{{L}_{0}}\)

2.2. Weibull#

The cumulative probability of breakage

Just now

is given by (cf. POST_ELEM and POST_BEREMIN):

\({P}_{f}(t)=1-\mathrm{exp}(-\sum _{\mathit{dV}}{[\begin{array}{c}\mathit{max}\\ {t}^{p}\le u\le t\end{array}(\frac{{\mathrm{\sigma }}_{I}(u)}{{\mathrm{\sigma }}_{u}(\mathrm{\theta }(u))})]}^{m}\frac{\mathit{dV}}{{V}_{0}})\).

The summation relates to the volumes of matter.

laminated (from now on

and

designating the maximum principal stress and the temperature in each of these volumes at the various times

. Here, the volume

The reference value is equal to \(50\mu {m}^{3}\). The Weibull module

is equal to 24 while the cleavage stress

depends on the temperature according to:

Temperature \(\mathrm{[}°C\mathrm{]}\)	—50	—100	—150	—150
\({\sigma }_{u}\mathrm{[}\mathit{MPa}\mathrm{]}\)	2800	2700	2600

The cumulative probability of breakage varies according to (

) according to:

\({P}_{f}(t)=1-\mathrm{exp}(-{[\begin{array}{c}\mathit{max}\\ {t}^{p}\le u\le t\end{array}(\frac{\mathrm{\sigma }(u)}{{\mathrm{\sigma }}_{u}(\mathrm{\theta }(u))})]}^{m}\frac{V}{{V}_{0}})\).

2.3. Rice and Tracey#

In simple traction, the natural logarithm of the cavity growth rate at the moment

is given by (cf. POST_ELEM):

\(\mathrm{log}(\frac{R(t)}{{R}_{0}})=0.283\times \mathrm{exp}(0.5)\times \underset{0}{\overset{t}{\int }}\dot{p}(u)\mathit{du}\)

2.4. Reference quantities and results#

and

for the following couples (temperature, travel = \((l\mathrm{-}{l}_{0})\)): \((–\mathrm{50,0}°C,\mathrm{20,35}\mathit{mm})\);

\((–\mathrm{50,0}°C,\mathrm{20,30}\mathit{mm})\); \((–\mathrm{150,0}°C,\mathrm{20,30}\mathit{mm})\) and \((–\mathrm{150,0}°C,\mathrm{32,53}\mathit{mm})\).

2.5. Uncertainties about the solution#

Analytical solution.