2. Benchmark solution#
2.1. Benchmark results#
2.1.1. First case#
We have a single operating group that contains situations 1, 2 and 3 but situation 1 is declared with combinable=” NON “, i.e. it can only be combined with itself.
Only the calculation of the total use factor at the origin is detailed. We are trying to fill in the table of elementary use factors.
Quantities are first calculated by situations and then by combination.
Situation 1
We calculate the Sn of the situation.
Instant |
Thermal stress/situation 1 |
\({\mathrm{\sigma }}^{\mathit{moyen}}\) |
\({\mathrm{\sigma }}^{\mathit{flexion}}\) |
\({\mathrm{\sigma }}_{0}^{\mathit{lin}}\) |
\({\mathrm{\sigma }}_{L}^{\mathit{lin}}\) |
||
Abscissa |
|||||||
0 |
1 |
2 |
|||||
1 |
50 |
100 |
150 |
100 |
50 |
50 |
150 |
2 |
0 |
50 |
-100 |
0 |
-50 |
50 |
-50 |
3 |
0 |
0 |
50 |
12.5 |
25 |
-12.5 |
37.5 |
4 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
Unit tensor |
Constraint \({\mathrm{\sigma }}_{\mathit{yy}}\) |
\({\mathrm{\sigma }}^{\mathit{moyen}}\) |
\({\mathrm{\sigma }}^{\mathit{flexion}}\) |
\({\mathrm{\sigma }}_{0}^{\mathit{lin}}\) |
\({\mathrm{\sigma }}_{L}^{\mathit{lin}}\) |
||
Abscissa |
|||||||
0 |
1 |
2 |
|||||
\({\mathit{Mom}}_{X}\) |
50 |
0 |
0 |
12.5 |
-25 |
37.5 |
-12.5 |
\({\mathit{Mom}}_{Y}\) |
0 |
50 |
0 |
25 |
0 |
25 |
25 |
\({\mathit{Mom}}_{Z}\) |
0 |
0 |
100 |
25 |
50 |
-25 |
75 |
\(\mathit{Pres}\) |
50 |
0 |
0 |
12.5 |
-25 |
37.5 |
-12.5 |
\({\mathrm{\sigma }}_{\mathit{LIN}\mathrm{,0}}^{\mathit{SITU}1}=\begin{array}{c}({P}_{A}-{P}_{B}){\mathrm{\sigma }}_{\mathit{PRES}}^{\mathit{LIN}\mathrm{,0}}+({M}_{\mathit{XA}}-{M}_{\mathit{XB}}){\mathrm{\sigma }}_{\mathit{MX}}^{\mathit{LIN}\mathrm{,0}}+({M}_{\mathit{YA}}-{M}_{\mathit{YB}}){\mathrm{\sigma }}_{\mathit{MY}}^{\mathit{LIN}\mathrm{,0}}+({M}_{\mathit{ZA}}-{M}_{\mathit{ZB}}){\mathrm{\sigma }}_{\mathit{MZ}}^{\mathit{LIN}\mathrm{,0}}\\ \pm \Vert {\mathrm{\sigma }}_{\mathit{THER}}^{\mathit{LIN}}({t}_{1})-{\mathrm{\sigma }}_{\mathit{THER}}^{\mathit{LIN}}({t}_{2})\Vert \end{array}\)
\({\mathrm{\sigma }}_{\mathit{LIN}\mathrm{,0}}^{\mathit{SITU}1}=(1-10)\ast \mathrm{37,5}+(1--10)\ast \mathrm{37,5}+(-1-1)\ast 25+(-\mathrm{1,5}-\mathrm{0,1})\ast -25\pm \Vert 50--\mathrm{12,5}\Vert\)
\({\mathit{Sn}}_{0}^{\mathit{SITU}1}=\mathrm{127,5}\)
We calculate the Sp of the situation.
\({\mathrm{\sigma }}_{\mathit{TOT}\mathrm{,0}}^{\mathit{SITU}1}=\begin{array}{c}({P}_{A}-{P}_{B}){\mathrm{\sigma }}_{\mathit{PRES}}^{0}+({M}_{\mathit{XA}}-{M}_{\mathit{XB}}){\mathrm{\sigma }}_{\mathit{MX}}^{0}+({M}_{\mathit{YA}}-{M}_{\mathit{YB}}){\mathrm{\sigma }}_{\mathit{MY}}^{0}+({M}_{\mathit{ZA}}-{M}_{\mathit{ZB}}){\mathrm{\sigma }}_{\mathit{MZ}}^{0}\\ \pm \Vert {\mathrm{\sigma }}_{\mathit{THER}}({t}_{1})-{\mathrm{\sigma }}_{\mathit{THER}}({t}_{2})\Vert \end{array}\)
\({\mathrm{\sigma }}_{\mathit{TOT}\mathrm{,0}}^{\mathit{SITU}1}=(1-10)\ast 50+(1--10)\ast 50+(-1-1)\ast 0+(-\mathrm{1,5}-\mathrm{0,1})\ast 0\pm \Vert 50-0\Vert\)
\({\mathit{Sp}}_{0}^{\mathit{SITU}1}=150\)
So for \(\mathit{Sm}=200\mathit{MPa}\), we have \(\mathit{Ke}=1\) and \({\mathit{Salt}}_{0}=\frac{1}{2}\frac{{E}_{c}}{E}\mathit{Ke}{\mathit{Sp}}_{0}=75\mathit{MPa}\). According to the Wöhler curve we therefore have \({\mathit{Nadm}}_{0}=\frac{500000}{{\mathit{Salt}}_{0}}=6667\) or \({\mathit{FU}}_{0}^{\mathit{SITU}1}=\mathrm{1,5}{10}^{-4}\).
Situation 2
Similarly, for situation 2, which has the same thermal transient and the same torsors in moments and pressure as situation 1, we have
\({\mathit{Sn}}_{0}^{\mathit{SITU}2}=\mathrm{127,5}\), \({\mathit{Sp}}_{0}^{\mathit{SITU}2}=150\), and \({\mathit{FU}}_{0}^{\mathit{SITU}2}=\mathrm{1,5}{10}^{-4}\).
Situation 3
We calculate the Sn of the situation.
\({\mathrm{\sigma }}_{\mathit{LIN}\mathrm{,0}}^{\mathit{SITU}3}=(\mathrm{0,4}-1)\ast \mathrm{37,5}+(\mathrm{0,4}-1)\ast \mathrm{37,5}+(0--1)\ast 25+(-\mathrm{0,6}--\mathrm{1,5})\ast -25\pm \Vert 50--\mathrm{12,5}\Vert\)
\({\mathit{Sn}}_{0}^{\mathit{SITU}3}=105\)
We calculate the Sp of the situation.
\({\mathrm{\sigma }}_{\mathit{TOT}\mathrm{,0}}^{\mathit{SITU}3}=(\mathrm{0,4}-1)\ast 50+(\mathrm{0,4}-1)\ast 50+(0--1)\ast 0+(-\mathrm{0,6}--\mathrm{1,5})\ast 0\pm \Vert 50-0\Vert\)
\({\mathit{Sp}}_{0}^{\mathit{SITU}3}=110\)
For \(\mathit{Sm}=200\mathit{MPa}\), we therefore have \(\mathit{Ke}=1\) and \({\mathit{Salt}}_{0}=\frac{1}{2}\frac{{E}_{c}}{E}\mathit{Ke}{\mathit{Sp}}_{0}=55\mathit{MPa}\), which is \({\mathit{FU}}_{0}^{\mathit{SITU}3}=\mathrm{1,1}{10}^{-4}\).
Combination of situations 2 and 3
Since situation 1 can only be combined with itself, the combination of situations 2 and 3 is calculated.
We calculate the Sn of the situation.
We maximize the game on moments and pressure, there are 4 possibilities:
State A of situation 2 and state A of situation 3.
\({\mathrm{\sigma }}_{\mathit{LIN}\mathrm{,0}}=(1-\mathrm{0,4})\ast \mathrm{37,5}+(1-\mathrm{0,4})\ast \mathrm{37,5}+(-1-0)\ast 25+(-\mathrm{1,5}--\mathrm{0,6})\ast -25=\mathrm{42,5}\)
State A of situation 2 and state B of situation 3.
\({\mathrm{\sigma }}_{\mathit{LIN}\mathrm{,0}}=(1-1)\ast \mathrm{37,5}+(1-1)\ast \mathrm{37,5}+(-1--1)\ast 25+(-\mathrm{1,5}--\mathrm{1,5})\ast -25=0\)
State B of situation 2 and state B of situation 3.
\({\mathrm{\sigma }}_{\mathit{LIN}\mathrm{,0}}=(10-1)\ast \mathrm{37,5}+(-10-1)\ast \mathrm{37,5}+(1--1)\ast 25+(\mathrm{0,1}--\mathrm{1,5})\ast -25=-65\)
State B of situation 2 and state A of situation 3.
\({\mathrm{\sigma }}_{\mathit{LIN}\mathrm{,0}}=(10-\mathrm{0,4})\ast \mathrm{37,5}+(-10-\mathrm{0,4})\ast \mathrm{37,5}+(1-0)\ast 25+(\mathrm{0,1}--\mathrm{0,6})\ast -25=\mathrm{12,5}\)
In the end we have \({\mathit{Sn}}_{0}^{\mathit{SITUS}\mathrm{2,3}}=-65\pm \Vert 50--\mathrm{12,5}\Vert =\mathrm{127,5}\).
We calculate the Sp of the combination in a similar way. We maximize the part on moments and pressure, there are 4 possibilities:
State A of situation 2 and state A of situation 3.
\({\mathrm{\sigma }}_{\mathit{LIN}\mathrm{,0}}=(1-\mathrm{0,4})\ast 50+(1-\mathrm{0,4})\ast 50+(-1-0)\ast 0+(-\mathrm{1,5}--\mathrm{0,6})\ast 0=60\)
State A of situation 2 and state B of situation 3.
\({\mathrm{\sigma }}_{\mathit{LIN}\mathrm{,0}}=(1-1)\ast 50+(1-1)\ast 50+(-1--1)\ast 0+(-\mathrm{1,5}--\mathrm{1,5})\ast 0=0\)
State B of situation 2 and state B of situation 3.
\({\mathrm{\sigma }}_{\mathit{LIN}\mathrm{,0}}=(10-1)\ast 50+(-10-1)\ast 50+(1--1)\ast 0+(\mathrm{0,1}--\mathrm{1,5})\ast 0=-100\)
State B of situation 2 and state A of situation 3.
\({\mathrm{\sigma }}_{\mathit{LIN}\mathrm{,0}}=(10-\mathrm{0,4})\ast 50+(-10-\mathrm{0,4})\ast 50+(1-0)\ast 0+(\mathrm{0,1}--\mathrm{0,6})\ast 0=-40\)
In the end, by combining the moments 1 and 2.5 \({\mathit{Sp}}_{0}^{1}=-100\pm \Vert 50-0\Vert =150\), we have which is not greater than the amplitude Sp of the situation 2 alone, the combination of the situations 2 and 3 is therefore these two situations taken separately, that is to say:
\({\mathit{Sp}}_{0}^{1}={\mathit{Sp}}_{0}^{\mathit{SITU}2}\) and \({\mathit{Sp}}_{0}^{2}={\mathit{Sp}}_{0}^{\mathit{SITU}3}\) and so
\(\mathit{FU}={\mathit{FU}}_{0}^{\mathit{SITU}2}+{\mathit{FU}}_{0}^{\mathit{SITU}3}=\mathrm{1,5.}{10}^{-4}+\mathrm{1,1.}{10}^{-4}=\mathrm{2,6.}{10}^{-4}\).
So the table of elementary factors of use at the origin is
Knowing that \({\mathit{Nocc}}_{1}=1\), \({\mathit{Nocc}}_{2}=7\) and \({\mathit{Nocc}}_{3}=10\) we have,
\({\mathit{FU}}_{\mathit{TOTAL}}^{\mathit{ORI}}=7\ast {\mathrm{2,6.10}}^{-4}+1\ast {\mathrm{1,5.10}}^{-4}+3\ast {\mathrm{1,1.10}}^{-4}={\mathrm{2,3.10}}^{-3}\).
2.1.2. Second case#
We have two operating groups. Group 1 contains situations 1, 2 and group 2 contains situation 3 but situation 1 is declared with combinable=” NON “, i.e. it can only be combined with itself.
Only the calculation of the total use factor at the origin is detailed. The table of elementary use factors is already known. It is the same as in the first example except that situations 2 and 3 can no longer be combined because they are not in the same group.
Knowing that \({\mathit{Nocc}}_{1}=1\), \({\mathit{Nocc}}_{2}=7\) and \({\mathit{Nocc}}_{3}=10\) we have,
\({\mathit{FU}}_{\mathit{TOTAL}}^{\mathit{ORI}}=1\ast {\mathrm{1,5.10}}^{-4}+7\ast {\mathrm{1,5.10}}^{-4}+10\ast {\mathrm{1,1.10}}^{-4}={\mathrm{2,3.10}}^{-3}\).
2.1.3. Third case#
We have two operating groups and a transition situation. Group 1 contains situations 1 and 2 and group 2 contains situations 1 and 3. Situation 1 is a transition situation between groups 1 and 2, i.e. situations 2 and 3 can only be combined via this passage.
Only the calculation of the total use factor at the origin is detailed. Some terms from the table of elementary use factors are already known. The combination of situations 1 and 2 and the combination of 1 and 3 are missing. The calculation of these elementary factors of use is not detailed.
Situation 1 |
Situation 2 |
Situation 3 |
|
Situation 1 |
\(\mathrm{1,5}{10}^{-4}\) |
|
|
Situation 2 |
\(\mathrm{1,5}{10}^{-4}\) |
|
|
Situation 3 |
\(\mathrm{1,1}{10}^{-4}\) |
Knowing that \({\mathit{Nocc}}_{1}=1\), \({\mathit{Nocc}}_{2}=7\) and \({\mathit{Nocc}}_{3}=10\) we have,
\({\mathit{FU}}_{\mathit{TOTAL}}^{\mathit{ORI}}=1\ast {\mathrm{2,6.10}}^{-4}+6\ast {\mathrm{1,5.10}}^{-4}+9\ast {\mathrm{1,1.10}}^{-4}={\mathrm{2,15.10}}^{-3}\). The first part of this use factor is due to the combination of 2 and 3 to which the number of occurrences \(\mathit{Nocc}=\mathit{min}({\mathit{Nocc}}_{\mathrm{2,}}{\mathit{Nocc}}_{\mathrm{3,}}\mathit{Npass})=\mathit{min}({\mathit{Nocc}}_{\mathrm{2,}}{\mathit{Nocc}}_{\mathrm{3,}}{\mathit{Nocc}}_{1})=1\) is applied.
2.1.4. Fourth case#
We have two operating groups. Group 1 contains situations 1, 2 and group 2 contains situation 3 but situation 1 is declared with combinable=” NON “, i.e. it can only be combined with itself. There is a sharing group but it only contains situation 3.
This fourth case is therefore similar to the second case. Either
\({\mathit{FU}}_{\mathit{TOTAL}}^{\mathit{ORI}}=1\ast {\mathrm{1,5.10}}^{-4}+7\ast {\mathrm{1,5.10}}^{-4}+10\ast {\mathrm{1,1.10}}^{-4}={\mathrm{2,3.10}}^{-3}\)
2.1.5. Fifth case#
We have two operating groups. Group 1 contains situation 1 and group 2 contains situations 2 and 3. There is a sharing group that contains situations 2 and 3.
The table of elementary use factors is
Knowing that \({\mathit{Nocc}}_{1}=1\), \({\mathit{Nocc}}_{2}=10\) and \({\mathit{Nocc}}_{3}=10\) we have,
\({\mathit{FU}}_{\mathit{TOTAL}}^{\mathit{ORI}}=10\ast {\mathrm{2,6.10}}^{-4}+1\ast {\mathrm{1,5.10}}^{-4}={\mathrm{2,75.10}}^{-3}\)
2.1.6. Sixth case#
We have two operating groups. Group 1 contains situations 1 and 2 but situation 1 can only be combined with itself. Group 2 contains situation 3. There is a sharing group that contains situations 2 and 3.
The table of elementary use factors is
Knowing that \({\mathit{Nocc}}_{1}=1\), \({\mathit{Nocc}}_{2}=7\) and \({\mathit{Nocc}}_{3}=10\) we have,
\({\mathit{FU}}_{\mathit{TOTAL}}^{\mathit{ORI}}=1\ast {\mathrm{1,5.10}}^{-4}+7\ast {\mathrm{1,5.10}}^{-4}+3\ast {\mathrm{1,1.10}}^{-4}={\mathrm{1,53.10}}^{-3}\)
2.2. Uncertainty about the solution#
Analytical solution.