2. Benchmark solution#
2.1. Calculation method used for the reference solution#
In plane \((\sigma ,\tau \sqrt{3})\), the von Mises norm is expressed by the classical distance in the octahedral plane between the projection of the stress state and the hydrostatic line, so that we can immediately predict the charging and discharging phases during the loading path, since these are respectively the phases in which the norm increases or decreases:
\({A}^{0}\) |
(123.8; |
76.23) |
\(A\) |
(151.2.; |
93.10) |
\({B}^{0}\) |
(158.23; |
89.12) |
\(B\) |
(257.2; |
33.10) |
\(O-{A}^{0}\) |
Elastic load |
\({A}^{0}-A\) |
Plastic load |
\(A-{B}^{0}\) |
Landfill |
\({B}^{0}-B\) |
Plastic load |
\(O-{A}^{0}\) |
Elastic load |
\({A}^{0}-A\) |
Plastic load |
\(A-{B}^{0}\) |
Landfill |
\({B}^{0}-B\) |
Plastic load |
The loading is done according to a curve parameterized by the moment:
Phase 1: plasticization from point \(O\) to point \(A\) (moments 0.0 to 1.0).
Phase 2: discharge from point \(A\) to point \(B\) (moments 1.0 to 2.0).
Phase 3: total discharge from point \(B\) to point \(C\) (moments 2.0 to 3.0).
Instant |
\(\sigma \text{[MPa]}\) |
|
Number of steps |
0,0 — Point \(O\) |
0 |
0 |
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0.1 |
1 |
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0.9 |
10 |
||
1.0 — Point \(A\) |
151.2 |
93.1 |
1 |
2.0 — Point \(B\) |
257.2 |
33.1 |
40 |
3.0 — Point \(C\) |
0 |
0 |
1 |
2.1.1. Resolution process#
Mechanically, this is a \(\mathrm{0D}\) stress-controlled test, the material being elastoplastic with the von Mises criterion and linear isotropic work hardening. For stress-controlled loading, the cumulative plastic deformation is easily determined:
\(F(\sigma ,p)={\sigma }_{\mathrm{éq}}-{\sigma }_{y}-R\text{'}p\le 0\Rightarrow p=\frac{{\sigma }_{\mathrm{éq}}-{\sigma }_{y}}{R\text{'}}\text{en charge}\) eq 2.1.1-1
The integration of plastic deformation is, of course, more difficult. The flow equation is written as:
\(\dot{{\varepsilon }^{p}}=\frac{3}{2}\dot{p}\frac{\tilde{\sigma }}{{\sigma }_{\mathrm{éq}}}\Rightarrow \dot{{\varepsilon }^{p}}=\frac{3}{2R\text{'}}\mathrm{.}\frac{\dot{{\sigma }_{\mathrm{éq}}}}{{\sigma }_{\mathrm{éq}}}\mathrm{.}\tilde{\sigma }\text{en charge}\) eq 2.1.1-2
Finally, we will deduce the deformation via the state relationship:
\(\varepsilon ={\varepsilon }^{p}+{E}^{-1}\text{:}\sigma \Rightarrow {\varepsilon }_{\mathrm{xx}}={\varepsilon }_{\mathrm{xx}}^{p}+\frac{\sigma }{E}\text{et}{\varepsilon }_{\mathrm{xy}}={\varepsilon }_{\mathrm{xy}}^{p}+\frac{\tau }{2\mu }\) eq 2.1.1-3
2.1.2. Radial loading phase treatment#
Note that during the radial loading phase, the flow law [éq 2.1.2-1] integrates directly:
\({\varepsilon }^{p}=\frac{3}{2}p\frac{\tilde{\sigma }}{{\sigma }_{\mathrm{éq}}}\) eq 2.1.2-1
The cumulative plastic deformation is then given by [éq 2.1.1-1], the plastic deformation by [éq2.1.2-1] and the total deformation by [éq 2.1.1-3]. With:
\(E\) |
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We get: |
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\(p(A)\) |
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\({\varepsilon }_{\mathrm{xy}}^{p}(A)\) |
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2.1.3. Treatment of the non-radial loading phase#
In the non-radial loading phase \({B}^{0}-B\), the stress path can be parameterized by:
\(\sigma (q)={\sigma }^{{B}^{0}}+q\underset{\text{direction fixe}}{\underset{\underbrace{}}{({\sigma }^{B}-{\sigma }^{{B}^{0}})}}\text{avec}0\le q\le 1\) eq 2.1.3-1
As the loading path remains confined in the traction-shear plane \((\sigma ,\tau )\), it will be useful to represent the stress state by a complex number:
\(\Sigma =\sigma +i\sqrt{(3)}\tau \Rightarrow {\sigma }_{\text{éq}}=\mid \Sigma \mid \text{et}\Sigma (q)={\Sigma }^{{B}^{0}}+q\underset{\text{direction fixe}}{\underset{\underbrace{}}{({\Sigma }^{{B}^{0}}-{\Sigma }^{0})}}\) eq 2.1.3-2
The integration of the flow law [éq 2.1.1-2], followed by an integration by part, makes it possible to express the plastic deformation:
\(\frac{\mathrm{2R}\text{'}}{3}{[{\varepsilon }^{p}]}_{0}^{1}={\int }_{0}^{1}\frac{{\dot{\sigma }}_{\text{éq}}}{{\sigma }_{\text{éq}}}\tilde{\sigma }\mathrm{dq}={\left[\mathrm{ln}({\sigma }_{\text{éq}})\tilde{\sigma }\right]}_{0}^{1}-\frac{1}{2}\underset{{\tilde{\sigma }}^{B}-{\tilde{\sigma }}^{{B}^{0}}}{\underset{\underbrace{}}{\dot{\tilde{\sigma }}}}{\int }_{0}^{1}\mathrm{ln}({\sigma }_{\text{éq}}^{2})\mathrm{dq}\)
The adoption of the complex plan allows an easy calculation of the last integral:
\({\int }_{0}^{1}\mathrm{ln}({\sigma }_{\text{éq}}^{2})\mathrm{dq}={\int }_{0}^{1}\mathrm{ln}(\Sigma \stackrel{ˉ}{\Sigma })\mathrm{dq}={\int }_{0}^{1}\mathrm{ln}(\Sigma )\mathrm{dq}+{\int }_{0}^{1}\mathrm{ln}(\stackrel{ˉ}{\Sigma })\mathrm{dq}=2\text{Re}\left[{\int }_{0}^{1}\mathrm{ln}(\Sigma )\mathrm{dq}\right]=2\text{Re}{\left[\frac{\Sigma \mathrm{ln}(\Sigma )-\Sigma }{{\Sigma }^{B}-{\Sigma }^{{B}^{0}}}\right]}_{0}^{1}\)
Finally, the plastic deformation increment on path \({B}^{0}-B\) is equal to:
\({\left[{\varepsilon }^{p}\right]}_{{B}^{0}}^{B}=\frac{3}{2R\text{'}}{\left[\mathrm{ln}({\sigma }_{\text{éq}})\tilde{\sigma }\right]}_{{B}^{0}}^{B}-\frac{3}{2R\text{'}}\text{Re}{\left[\frac{\Sigma \mathrm{ln}(\Sigma )-\Sigma }{{\Sigma }^{B}-{\Sigma }^{{B}^{0}}}\right]}_{{B}^{0}}^{B}({\tilde{\sigma }}^{B}-{\tilde{\sigma }}^{{B}^{0}})\) eq 2.1.3-4
2.2. Benchmark results#
By calculating the cumulative plastic deformation by [éq 2.1.1-1], the plastic deformation by [éq 2.1.3-4], and the total deformation by [éq 2.1.1-3], we get:
\(p(B)\) |
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We get: |
\({\varepsilon }_{\mathrm{xy}}^{p}(B)\) |
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We will focus on the values of the stresses, deformations and the cumulative plastic deformation at points \(A\) and \(B\) of the loading path.
2.3. Bibliographical references#
French Society of Mechanics. Guide to the validation of structural calculation software packages (VPCS). AFNOR Technical, 1990.