Benchmark solution
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Calculation method used for the reference solution
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Given the nature of the stresses, the solution (constraints :math:`\sigma`, deformations :math:`\varepsilon`) is homogeneous.

Let's say the load: :math:`(\begin{array}{ccc}0& 0& 0\\ 0& {\sigma }_{d}(t)& 0\\ 0& 0& 0\end{array})`

Granger's own creep model is such that the viscoelastic deformation corresponding to the case of a constant load :math:`{\sigma }_{0}` applied at time :math:`{t}_{0}` is equal to: (cf. [:external:ref:`R7.01.01 <R7.01.01>`])

:math:`{\varepsilon }^{\mathrm{fl}}(t)={\sigma }_{0}\sum _{k=1}^{8}{J}_{k}\text{.}(1-\text{exp}\left[-\frac{t-{t}_{0}}{{\tau }_{s}}\right])`

The model also depends on temperature and humidity as follows:

:math:`{\varepsilon }^{\mathrm{fl}}(t)={\sigma }_{0}h\cdot \frac{T-\text{248}}{\text{45}}\cdot \sum _{k=1}^{8}{J}_{k}\text{.}(1-\text{exp}\left[-\frac{t-{t}_{0}}{{\tau }_{s}}\right])`

but in this test the temperature and humidity fields are chosen to be constant and such that :math:`h` and :math:`\frac{T-248}{45}` equal 1 respectively.

When the constraint evolves over time then:

:math:`{\varepsilon }^{\text{fl}}(t)=\sum _{k=1}^{8}\underset{\tau =0}{\overset{t}{\int }}{J}_{k}\text{.}(1-\text{exp}\left[-\frac{t-\tau }{{\tau }_{s}}\right])\text{.}\frac{\partial \sigma }{\partial \tau }\text{.}d\tau`

So for the present case we have:

:math:`{\varepsilon }_{\mathrm{yy}}^{\text{fl}}=\sum _{k=1}^{2}\underset{\tau =0}{\overset{t={t}_{1}}{\int }}{J}_{k}\text{.}(1-\text{exp}\left[-\frac{t-\tau }{{\tau }_{s}}\right])\text{.}\frac{{\sigma }_{1}}{{t}_{1}}\text{.}d\tau`

the load remaining constant beyond t 1. That is:

:math:`{\varepsilon }_{\mathrm{yy}}^{\text{fl}}={\sigma }_{1}\sum _{k=1}^{k=2}{J}_{k}\text{.}(1-\text{exp}(-\frac{t-{t}_{1}}{{\tau }_{s}}))\underset{\text{}\approx 1}{\underset{\underbrace{}}{\frac{{\tau }_{k}}{{t}_{1}}(1-\text{exp}(\frac{-{t}_{1}}{{\tau }_{s}}))}}`

A longitudinal creep deformation is accompanied by a transverse deformation such as:

:math:`{\varepsilon }_{\mathrm{xx}}^{\mathrm{fl}}=-\nu {\varepsilon }_{\mathrm{yy}}^{\mathrm{fl}}`

The total uniaxial deformation is equal to: :math:`{\varepsilon }_{\mathrm{yy}}={\varepsilon }_{\mathrm{yy}}^{\mathrm{fl}}+\frac{{\sigma }_{\mathrm{yy}}}{E}`


Benchmark results
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We will focus on the values of creep deformations at 45 days, 245 days and 365 days.