2. Benchmark solution#
2.1. Benchmark results#
2.1.1. Calculating \(\mathrm{Pm}\) and \(\mathrm{Pb}\)#
The parameters \(\mathrm{Pm}\) and \(\mathrm{Pb}\) represent the primary membrane stress and the flexural stress respectively. Criteria must also be verified on quantity (\(\mathrm{Pm}\pm \mathrm{Pb}\)), at the origin and at the end of the analysis segment.
Each of these parameters can be calculated analytically from the data of the stress tensor on the segment. Only primary constraints need to be taken into account. The user can either directly provide the mechanical stresses alone, or provide the total thermo-mechanical stresses and the constraints related to thermal loading alone, in which case these are automatically subtracted with “EVOLUTION”.
The signed value of the parameters \(\mathrm{Pm}\) and \(\mathrm{Pb}\) is indicated in the tables below, even if it is the Tresca norm for these quantities that should be finally retained.
Situation 1
Instant |
Total mechanical stresses (pressure+efforts/moments) |
Pm |
Pb |
PmpB (0) |
PmpB (L) |
||
Abscissa |
|||||||
0 |
1 |
2 |
|||||
1 |
40 |
0 |
-40 |
0 |
│ -40│ |
40 |
│ -40│ |
2 |
0 |
50 |
10 |
27.5 |
5 |
22.5 |
32.5 |
3 |
100 |
-50 |
-150 |
609 | -37, 5, | |
│ -125│ |
87.5 |
609, 5, 5 |
4 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
For situation 1, \(\mathit{Pm}=\mathrm{37,5}\) \(\mathit{Pb}=125\) \({\mathit{PmPb}}_{0}=\mathrm{87,5}\) and \({\mathit{PmPb}}_{L}=\mathrm{162,5}\).
Situation 2
Instant |
Total mechanical stresses (pressure+efforts/moments) |
Pm |
Pb |
PmpB (0) |
PmpB (L) |
||
Abscissa |
|||||||
0 |
1 |
2 |
|||||
1, 5 |
40 |
0 |
-60 |
609 - 5 |
│ -50│ |
45 |
│ -55│ |
2, 5 |
0 |
50 |
10 |
27.5 |
5 |
22.5 |
32.5 |
3, 5 |
100 |
-50 |
-150 |
609 | -37, 5, | |
│ -125│ |
87.5 |
609, 5, 5 |
For situation 2, \(\mathit{Pm}=\mathrm{37,5}\) \(\mathit{Pb}=125\) \({\mathit{PmPb}}_{0}=\mathrm{87,5}\) and \({\mathit{PmPb}}_{L}=\mathrm{162,5}\).
Situations 3 and 4
For situation 3, \(\mathit{Pm}=\mathrm{37,5}\) \(\mathit{Pb}=125\) \({\mathit{PmPb}}_{0}=\mathrm{87,5}\) and \({\mathit{PmPb}}_{L}=\mathrm{162,5}\).
For situation 4, \(\mathit{Pm}=\mathrm{37,5}\) \(\mathit{Pb}=125\) \({\mathit{PmPb}}_{0}=\mathrm{87,5}\) and \({\mathit{PmPb}}_{L}=\mathrm{162,5}\).
2.1.2. Calculation of \(\mathit{Sn}\)#
The parameter \(\mathit{Sn}\) represents the amplitude of variation of the linear stress (mean stress \(\pm\) flexural stress) between two moments of the transient in question.
Situation 1
Instant |
Total constraints |
\({\mathrm{\sigma }}^{\mathit{moyen}}\) |
\({\mathrm{\sigma }}^{\mathit{flexion}}\) |
\({\mathrm{\sigma }}_{0}^{\mathit{lin}}\) |
\({\mathrm{\sigma }}_{L}^{\mathit{lin}}\) |
||
Abscissa |
|||||||
0 |
1 |
2 |
|||||
1 |
90 |
100 |
110 |
100 |
10 |
90 |
110 |
2 |
0 |
100 |
-90 |
27.5 |
-45 |
72.5 |
-17.5 |
3 |
100 |
-50 |
-100 |
-25 |
-100 |
75 |
-125 |
4 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
Instant 1 |
Instant 2 |
|
|
1 |
2 |
17.5 |
127.5 |
1 |
3 |
15 |
235 |
1 |
4 |
90 |
110 |
2 |
3 |
2.5 |
107.5 |
2 |
4 |
72.5 |
17.5 |
3 |
4 |
75 |
125 |
For situation 1, \({\mathit{Sn}}_{0}=90\) and \({\mathit{Sn}}_{L}=235\).
Situation 2
For situation 2, \({\mathit{Sn}}_{0}=\mathrm{22,5}\) and \({\mathit{Sn}}_{L}=220\).
Situation 3
State |
Total mechanical stresses |
\({\mathrm{\sigma }}^{\mathit{moyen}}\) |
\({\mathrm{\sigma }}^{\mathit{flexion}}\) |
\({\mathrm{\sigma }}_{0}^{\mathit{meca},\mathit{lin}}\) |
\({\mathrm{\sigma }}_{L}^{\mathit{meca},\mathit{lin}}\) |
||
Abscissa |
|||||||
0 |
1 |
2 |
|||||
A |
100 |
-50 |
-150 |
-37.5 |
-125 |
87.5 |
-162.5 |
B |
0 |
50 |
10 |
27.5 |
5 |
22.5 |
32.5 |
State 1 |
State 2 |
\({\mathit{Sn}}_{0}\) |
|
A |
B |
65 |
195 |
Instant |
Thermal constraints |
\({\mathrm{\sigma }}^{\mathit{moyen}}\) |
\({\mathrm{\sigma }}^{\mathit{flexion}}\) |
\({\mathrm{\sigma }}_{0}^{\mathit{lin}}\) |
\({\mathrm{\sigma }}_{L}^{\mathit{lin}}\) |
||
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 |
Instant 1 |
Instant 2 |
|
|
1 |
2 |
0 |
200 |
1 |
3 |
62.5 |
112.5 |
1 |
4 |
50 |
150 |
2 |
3 |
62.5 |
87.5 |
2 |
4 |
50 |
50 |
3 |
4 |
12.5 |
37.5 |
For situation 3, \({\mathit{Sn}}_{0}=65+\mathrm{62,5}=\mathrm{127,5}\) and \({\mathit{Sn}}_{L}=195+200=395\).
Situation 4
For situation 4, \({\mathit{Sn}}_{0}=\mathrm{42,5}+\mathrm{62,5}=105\) and \({\mathit{Sn}}_{L}=\mathrm{107,5}+200=\mathrm{307,5}\).
2.1.3. Calculation of \({\mathrm{Sn}}^{\text{*}}\)#
The parameter \({\mathit{Sn}}^{\text{*}}\) represents the amplitude \(\mathrm{Sn}\) calculated without taking into account thermal bending stresses. Only the calculation of the quantity \({\mathit{Sn}}^{\text{*}}\) for situation 1 is detailed.
Situation 1
Instant |
\({\mathrm{\sigma }}^{\mathit{moyen}}\) |
|
|
|
|
|
|
|||
1 |
100 |
10 |
90 |
90 |
90 |
110 |
50 |
140 |
60 |
|
2 |
27.5 |
-45 |
72.5 |
72.5 |
-17.5 |
-50 |
22.5 |
32.5 |
||
3 |
-25 |
-100 |
-100 |
75 |
75 |
-125 |
+25 |
100 |
-150 |
|
4 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
Instant 1 |
Instant 2 |
|
|
1 |
2 |
117.5 |
27.5 |
1 |
3 |
40 |
2 10 |
1 |
4 |
140 |
60 |
2 |
3 |
77.5 |
182.5 |
2 |
4 |
22.5 |
32.5 |
3 |
4 |
100 |
150 |
For situation 1, \(\mathit{Sn}{\text{*}}_{0}=140\) and \(\mathit{Sn}{\text{*}}_{L}=210\).
The calculation is not detailed for the other three situations
For situation 2, \(\mathit{Sn}{\text{*}}_{0}=\mathrm{122,5}\) and \(\mathit{Sn}{\text{*}}_{L}=195\).
For situation 3, \(\mathit{Sn}{\text{*}}_{0}=165\) and \(\mathit{Sn}{\text{*}}_{L}=295\).
For situation 4, \(\mathit{Sn}{\text{*}}_{0}=\mathrm{142,5}\) and \(\mathit{Sn}{\text{*}}_{L}=\mathrm{207,5}\) with the “TOUT_INST” method and \(\mathit{Sn}{\text{*}}_{0}=130\) and \(\mathit{Sn}{\text{*}}_{L}=\mathrm{207,5}\) with the “TRESCA” method. In fact, when the method for selecting the moments” TRESCA “is selected, we do not recalculate the moments that maximize \(\mathit{Sn}\text{*}\) but we take those that have maximized \(\mathit{Sn}\), which may be non-conservative.
2.1.4. Thermal ratchet calculation#
The calculation is detailed for situation 1 only.
First, the membrane stress due to pressure \({\mathrm{\sigma }}_{m}\) is calculated.
Instant |
Stress due to pressure |
\({\mathrm{\sigma }}_{m}\) |
||
Abscissor |
||||
0 |
1 |
2 |
||
1 |
40 |
0 |
40 |
20 |
2 |
0 |
50 |
0 |
25 |
3 |
10 |
-10 |
-200 |
-52.5 |
4 |
0 |
0 |
0 |
0 |
\({\mathrm{\sigma }}_{m}\) is worth 52.5 MPa.
Based on the equations of RCC -M, \(x=\frac{{\mathrm{\sigma }}_{m}}{\mathit{Sy}}=\frac{\mathrm{52,5}}{200}=\mathrm{0,2625}\) and \(y{\text{'}}_{\mathit{LINE}}=\frac{1}{x}=\frac{1}{\mathrm{0,2625}}\).
Then. \({\mathrm{\sigma }}_{\mathrm{\theta },\mathit{LINE}}=y{\text{'}}_{\mathit{LINE}}\ast \mathit{Sy}=\frac{1}{\mathrm{0,2625}}\ast 200=\mathrm{761,905}\mathit{MPa}\)
According to the previous calculations \({\mathit{Sn}}_{\mathit{ther},\mathit{ORI}}^{\mathit{max}}=\mathrm{62,5}\) and \({\mathit{Sn}}_{\mathit{ther},\mathit{EXT}}^{\mathit{max}}=200\), we therefore respect the criterion for situation 1.
2.1.5. Fatigue calculation for situations 1 and 2 in the same group#
The calculation is detailed for the combination of situations 1 and 2 only and original.
We are trying to fill in the table of elementary use factors.
We first calculate the quantities by situations and then the combination.
Situation 1
For situation 1, remember that \({\mathit{Sn}}_{0}=90\) (part 2.1.2). The quantity Sp is calculated.
Instant 1 |
Instant 2 |
|
1 |
2 |
90 |
1 |
3 |
10 |
1 |
4 |
90 |
2 |
3 |
100 |
2 |
4 |
0 |
3 |
4 |
100 |
So for situation 1, we have \({\mathit{Sp}}_{0}=100\).
So for \(\mathit{Sm}=200\mathit{MPa}\), we have \(\mathit{Ke}=1\) and \({\mathit{Salt}}_{0}=\frac{1}{2}\frac{{E}_{c}}{E}\mathit{Ke}{\mathit{Sp}}_{0}=50\mathit{MPa}\). According to the Wöhler curve we therefore have \({\mathit{Nadm}}_{0}=\frac{500000}{{\mathit{Salt}}_{0}}=10000\) or \({\mathit{FU}}_{0}=\frac{1}{10000}={10}^{-4}\).
Situation 2
Similarly for situation 2, we have \({\mathit{Sn}}_{0}=\mathrm{22,5}\), \({\mathit{Sp}}_{0}=100\), or \(\mathit{Ke}=1\) and \({\mathit{Salt}}_{0}=50\mathit{MPa}\) or \({\mathit{FU}}_{0}={10}^{-4}\).
Combination of situations 1 and 2
Instant |
Total constraints |
\({\mathrm{\sigma }}^{\mathit{moyen}}\) |
\({\mathrm{\sigma }}^{\mathit{flexion}}\) |
\({\mathrm{\sigma }}_{0}^{\mathit{lin}}\) |
\({\mathrm{\sigma }}_{L}^{\mathit{lin}}\) |
||
Abscissa |
|||||||
0 |
1 |
2 |
|||||
1 |
90 |
100 |
110 |
100 |
10 |
90 |
110 |
2 |
0 |
100 |
-90 |
27.5 |
-45 |
72.5 |
-17.5 |
3 |
100 |
-50 |
-100 |
-25 |
-100 |
75 |
-125 |
4 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
1.5 |
90 |
100 |
90 |
95 |
0 |
95 |
95 |
2.5 |
0 |
100 |
-90 |
27.5 |
-45 |
72.5 |
-17.5 |
3.5 |
100 |
-50 |
-100 |
-25 |
-100 |
75 |
-125 |
For the combination of situations 1 and 2 we therefore have \({\mathit{Sn}}_{0}=95-0=95\) for the moments 4 and 1.5. So \(\mathit{Ke}=1\). \({\mathit{Sp}}_{0}=100\) (for example by combining moments 2 and 3) or \({\mathit{FU}}_{0}={10}^{-4}\).
So the table of elementary use factors with “B3200” is
Situation 1 |
Situation 2 |
|
Situation 1 |
\({10}^{-4}\) |
|
Situation 2 |
\({10}^{-4}\) |
In B3200, if \({\mathit{Nocc}}_{1}=1\) and \({\mathit{Nocc}}_{2}=1\) we have \({\mathit{FU}}_{\mathit{TOTAL}}^{\mathit{ORI}}={2.10}^{-4}\).
With EVOLUTION, the operator combines all the moments together, such as loading states
So the table of elementary use factors with “EVOLUTION” is
Instants |
1 |
2 |
3 |
3 |
3 |
4 |
1.5 |
2.5 |
3.5 |
1 |
\({9.10}^{-5}\) |
|
|
|
|
|
|||
2 |
\({1.10}^{-4}\) |
|
|
|
|
||||
3 |
\({1.10}^{-4}\) |
|
|
|
|||||
4 |
\({9.10}^{-5}\) |
|
|
||||||
1.5 |
\({9.10}^{-5}\) |
|
|||||||
2.5 |
\({1.10}^{-4}\) |
||||||||
3,5 |
And with EVOLUTION, if \({\mathit{Nocc}}_{1}=1\) and \({\mathit{Nocc}}_{2}=1\) we have \({\mathit{FU}}_{\mathit{TOTAL}}^{\mathit{ORI}}={\mathrm{2,9.10}}^{-4}\).
2.1.6. Fatigue calculation for situations 3 and 4 in the same group#
The calculation is detailed for the combination of situations 3 and 4 only and at the extreme.
We are trying to fill in the table of elementary use factors.
We first calculate the quantities by situations and then the combination.
Situation 3
For situation 3, we recall that \({\mathit{Sn}}_{L}=395\) .We calculate the quantity Sp.
Instant 1 |
Instant 2 |
|
1 |
2 |
250 |
1 |
3 |
100 |
1 |
4 |
150 |
2 |
3 |
150 |
2 |
4 |
100 |
3 |
4 |
50 |
So for situation 3, we have \({\mathit{Sp}}_{L}={\mathit{Sp}}_{L}^{\mathit{meca}}+{\mathit{Sp}}_{L}^{\mathit{ther}}=160+250=410\).
For \(\mathit{Sm}=200\mathit{MPa}\), we have \(\mathit{Ke}=1\) and \({\mathit{Salt}}_{L}=\frac{1}{2}\frac{{E}_{c}}{E}\mathit{Ke}{\mathit{Sp}}_{L}=205\mathit{MPa}\). According to the Wöhler curve we therefore have \({\mathit{Nadm}}_{L}=\frac{500000}{{\mathit{Salt}}_{L}}=2439\) or \({\mathit{FU}}_{L}=\frac{1}{2439}={\mathrm{4,1.10}}^{-4}\).
Situation 4
Similarly for situation 4, we have \({\mathit{Sn}}_{L}=\mathrm{307,5}\) and \({\mathit{Sp}}_{L}={\mathit{Sp}}_{L}^{\mathit{meca}}+{\mathit{Sp}}_{L}^{\mathit{ther}}=90+250=340\).
So \(\mathit{Ke}=1\) and \({\mathit{Salt}}_{L}=170\mathit{MPa}\) is \({\mathit{FU}}_{L}={\mathrm{3,4.10}}^{-4}\).
Combination of situations 3 and 4
For the combination of situations 3 and 4 we therefore have \({\mathit{Sn}}_{L}=440\). So \(\mathit{Ke}=1\), we have \({\mathit{Sp}}_{L}^{1}=410\) and \({\mathit{Sp}}_{L}^{2}=340\) is \({\mathit{FU}}_{L}={\mathrm{4,1.10}}^{-4}+{\mathrm{3,4.10}}^{-4}={\mathrm{7,5.10}}^{-4}\).
So the table of elementary use factors with “B3200” is
Situation 3 |
Situation 4 |
|
Situation 3 |
|
|
Situation 4 |
\({\mathrm{3,4.10}}^{-4}\) |
In B3200, if \({\mathit{Nocc}}_{3}=1\) and \({\mathit{Nocc}}_{4}=1\) we have \({\mathit{FU}}_{\mathit{TOTAL}}^{\mathit{EXT}}={\mathrm{7,5.10}}^{-4}\).
In B3200, if \({\mathit{Nocc}}_{3}=10\) and \({\mathit{Nocc}}_{4}=7\) we have \({\mathit{FU}}_{\mathit{TOTAL}}^{\mathit{EXT}}=7\ast {\mathrm{7,5.10}}^{-4}+3\ast {\mathrm{4,1.10}}^{-4}={\mathrm{6,48.10}}^{-3}\).
2.1.7. Calculation of environmental fatigue for situations 3 and 4 in the same group#
The calculation is detailed for the combination of situations 3 and 4 only and at the end with \({\mathit{Nocc}}_{3}=1\) and \({\mathit{Nocc}}_{4}=1\).
Environmental fatigue is only applied after classical fatigue. The combinations that occurred in classical fatigue are therefore used again to calculate the total use factor.
In classical fatigue, \({\mathit{FU}}_{\mathit{TOTAL}}^{\mathit{EXT}}={\mathrm{7,5.10}}^{-4}\) involves the combination of situations 3 and 4. So you have to calculate the FEN of this combination.
Stress increment \(\mathrm{\Delta }\mathrm{\sigma }\) is easy to calculate because the tensor is uniaxial, so there is no need to diagonalize. Only thermal factors are involved in this increment because the other quantities are not a function of time.
When the stress increment is negative, it is considered that the environment has no effect, so the other quantities are not calculated.
Otherwise \(\mathrm{\Delta }\mathrm{ϵ}=\frac{\mathit{Ke}\ast \mathrm{\Delta }\mathrm{\sigma }}{E(T)}\). In this example, we entered a constant Young’s modulus as a function of temperature and of the order of 200,000 MPa. And the \(\mathit{Ke}\) is the one that was used for the combination of situations 3 and 4, that is \(\mathit{Ke}=1\).
Then we calculate \(\dot{\mathrm{ϵ}}=\frac{\mathrm{\Delta }\mathrm{ϵ}}{{t}_{i}-{t}_{i-1}}\). In this case, it is below threshold \({\mathrm{ϵ}}_{\mathit{seul},\mathit{inf}}\) so
\(\dot{\mathrm{ϵ}}\text{*}=\mathrm{ln}(\frac{{\mathrm{ϵ}}_{\mathit{seuilinf}}}{{\mathrm{ϵ}}_{\mathit{seuilsup}}})=\mathrm{ln}(\frac{1}{2})\)
\(F=\mathrm{exp}[(A+B\dot{\mathrm{ϵ}}\text{*})S\text{*}O\text{*}T\text{*}+C]=\mathrm{exp}[\dot{\mathrm{ϵ}}\text{*}T\text{*}]\) knowing that T* is a function of T, with
\(T=\frac{T({t}_{i})+T({t}_{i-1})}{2}\). The user has obviously entered the temperature during each situation under the keyword TABL_TEMP.
Time increment processed |
\(\mathrm{\Delta }\mathrm{\sigma }\) |
|
|
\(\dot{\mathrm{ϵ}}\text{*}\) |
T |
T* |
F |
||
1-2 |
-250 |
||||||||
2-3 |
+150 |
|
|
|
|
2.2 |
|
||
3-4 |
-50 |
||||||||
1.5-2.5 |
-250 |
||||||||
2.5-3.5 |
+150 |
|
|
|
|
2.2 |
|
Finally, \({\mathit{FEN}}_{\mathit{EXTREMITE}}=\frac{{F}_{2-3}\mathrm{\Delta }{\mathrm{ϵ}}_{2-3}+{F}_{\mathrm{2,5}-\mathrm{3,5}}\ast \mathrm{\Delta }{\mathrm{ϵ}}_{\mathrm{2,5}-\mathrm{3,5}}}{\mathrm{\Delta }{\mathrm{ϵ}}_{2-3}+\mathrm{\Delta }{\mathrm{ϵ}}_{\mathrm{2,5}-\mathrm{3,5}}}=\mathrm{0,218}\).
We can calculate the partial use factor as well as the total use factor with environmental effect \({\mathit{FU}}_{\mathit{partiel},\mathit{env}}^{\mathit{EXT}}={\mathit{FEN}}_{\mathit{EXTREMITE}}\ast {\mathit{FU}}_{\mathit{partiel}}=\mathrm{0,218}\ast \mathrm{7,5}\ast {10}^{-4}={\mathrm{1,6323.10}}^{-4}\)
We then compare \({\mathit{FEN}}_{\mathit{EXTREMITE}}\) to the parameter FEN_INTEGRE.
If \({\mathit{FEN}}_{\mathit{EXTREMITE}}\le {\mathit{FEN}}_{\mathit{INTEGRE}}\) (case where FEN_INTEGRE = 50) then:
\({\mathit{FU}}_{\mathit{TOTAL},\mathit{env}}^{\mathit{EXT}}={\mathit{FU}}_{\mathit{TOTAL}}^{\mathit{EXT}}={\mathrm{7,5.10}}^{-4}\).
If \({\mathit{FEN}}_{\mathit{EXTREMITE}}>{\mathit{FEN}}_{\mathit{INTEGRE}}\) (case where FEN_INTEGRE = 0.1) then:
\({\mathit{FU}}_{\mathit{TOTAL},\mathit{env}}^{\mathit{EXT}}={N}_{\mathit{occ}}\ast {\mathit{FU}}_{\mathit{partiel},\mathit{env}}^{\mathit{EXT}}/{\mathit{FEN}}_{\mathit{INTEGRE}}=1\ast {\mathrm{1,6323.10}}^{-4}/\mathrm{0,1}\).
2.2. Uncertainty about the solution#
Analytical solution.