2. Benchmark solution#
2.1. Benchmark results#
2.1.1. ZE200a#
2.1.1.1. Calculation of 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.
- math:
`{S} _ {n} =begin {array} {c} {c} {C} {C} _ {1}frac {R} {e} | {A} - {P} - {B} |+ {C}} {B} |+ {C} _ {B} |+ {C} _ {B} |+ {C} _ {B} |+ {C} _ {2}}frac {R} {I} _ {I}}sqrt {({M}} _ {mathit {xB}})} ^ {2} + {({M}} + {({M} _ {mathit {yA}}} - {M} _ {mathit {yB}})} ^ {2} + {({M}} _ {mathit {zA}} _ {mathit {zB}}})} ^ {2})} ^ {2}) + {({M} _ {M} _ {mathit {zB}})} ^ {2}) + {(M} _ {M} _ {mathit {zB}})} ^ {2}) + {(M} _ {M} _ {mathit {zB}})} ^ {2})} + {{M} _ {{M} _ {m {sigma}} _ {text {tran}}} ^ {text {tran}}} ^ {text {tran}}} ^ {text {}}) - {mathrm {sigma}} _ {text {tran}}}} ^ {text {tran}}} ^ {text {tran}}} ^ {text {tran}}} ^ {text {tran}}} ^ {text {tran}}} ^ {text {tran}}} ^ {text {tran}}} ^ {text {tran}}} ^ {text {tran}}} ^ {text {tran}}} ^ {text {tran}}} ^ {text {tran}}} ^ {text {tran}}}
Without thermal constraints, for situation 1 \({S}_{n}=120\) and for situation 2 \({S}_{n}=60\) (at the origin and at the end).
With thermal constraints, we calculate the maximum amplitudes of thermal stresses linearized at the origin and then at the end.
Situation 1
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, 5 |
90 |
100 |
110 |
100 |
10 |
90 |
110 |
2, 5 |
0 |
100 |
-90 |
27.5 |
-45 |
72.5 |
-17.5 |
3, 5 |
100 |
-50 |
-100 |
-25 |
-100 |
75 |
-125 |
4, 5 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
Instant 1 |
Instant 2 |
|
|
1, 5 |
2, 5 |
17.5 |
127.5 |
1, 5 |
3, 5 |
15 |
235 |
1, 5 |
4, 5 |
90 |
110 |
2, 5 |
3, 5 |
2.5 |
107.5 |
2, 5 |
4, 5 |
72.5 |
17.5 |
3, 5 |
4, 5 |
75 |
125 |
For situation 1 with thermal constraints, \({\mathit{Sn}}_{0}=120+90=210\) and \({\mathit{Sn}}_{L}=120+235=355\).
Situation 2
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 |
90 |
100 |
90 |
95 |
0 |
95 |
95 |
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 |
22.5 |
12.5 |
1 |
3 |
20 |
2 20 |
1 |
4 |
9 5 |
95 |
2 |
3 |
2.5 |
107.5 |
2 |
4 |
72.5 |
17.5 |
3 |
4 |
75 |
125 |
For situation 2 with thermal constraints, \({\mathit{Sn}}_{0}=60+95=155\) and \({\mathit{Sn}}_{L}=60+220=280\).
2.1.1.2. Calculation of Sn with earthquake#
We have just added the contribution of the earthquake to the magnitude Sn such that
- math:
{S} _ {mathit {nS}} =begin {array} {c} {C} _ {1}frac {R} {e} | {P} _ {A} - {A} - {P} _ {B} | {B} |\ + {B}} |\ + {B}} |\ + {C} _ {B} |\ + {C} _ {B} |\ + {C} _ {2}frac {R} {I}}sqrt {({M}} _ {mathit {xA} _ {mathit {xA}}} - {M} _ {mathit {xB}}}pm 2 {M}}pm 2 {M} _ {mathit {xS}})} ^ {2} + {mathit {yA}} - {mathit {yA}}} - {M}} - {mathit {xA}}}} - {mathit {yA}}}} - {mathit {yA}}}} - {M}}} - {M}}} - {M}}} - {M}}} - {M}}} - {M}}} - {M}}} - {M}}} - {M}}} - {M}}} - {M}}} - {M}}} - {M}}} - {M}}} mathit {zA}} - {M} _ {mathit {zB}}}pm 2 {M}} _ {mathit {zS}})} ^ {2})}\ +Vert {mathrm {sigma}}} _ {mathrm {sigma}}} _ {text {sigma}}} _ {text {sigma}}} _ {text {sigma}}} _ {text {sigma}}} _ {text {sigma}}} _ {mathrm {}}) - {mathrm {sigma}}} _ {mathrm {}}} m {sigma}} _ {text {tran}}} ^ {text {lin}} ^ {text {tran}}} ^ {text {}})Vertend {tran}}} ^ {text {tran}}} ^ {text {tran}} ^ {text {tran}}} ^ {text {}})Greenend {array}
For the situation 1 without constraints in the form of a transitory,
- math:
{S} _ {mathit {nS}} =begin {array} {c} {C} _ {1}frac {R} {e} | {P} _ {A} - {A} - {P} _ {B} | {B} |+ {B}} |+ {C}} |+ {C}} | + {C} _ {B} |+ {C}} | + {C} _ {B} |+ {C} _ {B} |+ {C} _ {B} |+ {C} _ {B} |+ {C} _ {B} |+ {C} _ {B} |+ {C} _ {B} |+ {C} _ {B} |+ {C} _ {B} |+ {C} _ {B} |+ {M} _ {mathit {xB}}}pm 2 {M} _ {mathit {xS}})} ^ {2})}}end {array}
- math:
{mathit {Sn}} _ {S} =1astfrac {mathrm {0.5}} {1} |201-1|+2astfrac {mathrm {0.5}} {1} {1}sqrt {1}sqrt {({(21-1pm 2ast 21)}} ^ {2})} =100+62=162 at the origin and at the end.
For the situation 2without constraints in the form of a transitory,
2.1.1.3. 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*originally*.
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 with thermal constraints \({\mathit{Sn}}_{0}=210\) (part 2.1.1). The quantity Sp is calculated at the origin.
- math:
{S} _ {p} =begin {array} {c} {c} {K} _ {1} {C} _ {1}frac {R} {e} | {P} _ {A} - {P} - {P} - {P} - {P} _ {P} - {P} - {P} - {P} - {P} - {P} - {P} - {P} - {P} - {P} - {P} - {P} - {P} - {P} - {P} - {P} - {P} - {P} _ {P} - {P} - {P} - {P} - {P} - {P} - {P} - {P} - {P} - {P} - {P} - {P} - {mathit {xA}} - {M} _ {mathit {xB}})})} ^ {2}} + {({M} _ {mathit {yA}} - {M} _ {mathit {yB}}})})})} ^ {2}})} ^ {2} + {({M}} _ {mathit {zB}}})}) ^ {2})}\ +Vert {mathrm {sigma}}} _ {text {tran}}} ^ {text {}} ({t} _ {1} ^ {text {}}) - {mathrm {sigma}}) - {mathrm {sigma}}}} _ {text {sigma}}} _ {text {}})Greenend {array}
Without thermal constraints, for situation 1 \({S}_{p}^{0}=120\).
With thermal constraints, we calculate the maximum amplitudes of thermal stresses linearized at the origin:
Instant 1 |
Instant 2 |
|
1, 5 |
2, 5 |
90 |
1, 5 |
3, 5 |
10 |
1, 5 |
4, 5 |
90 |
2, 5 |
3, 5 |
100 |
2, 5 |
4, 5 |
0 |
3, 5 |
4, 5 |
100 |
So for situation 1, we have \({\mathit{Sp}}_{0}=120+100=220\).
So for \(\mathit{Sm}=2000\mathit{MPa}\), we have \(\mathit{Ke}=1\) and \({\mathit{Salt}}_{0}=\frac{1}{2}\frac{{E}_{c}}{E}\mathit{Ke}{\mathit{Sp}}_{0}=110\mathit{MPa}\). According to the Wöhler curve we therefore have \({\mathit{Nadm}}_{0}=\frac{500000}{{\mathit{Salt}}_{0}}=4545\) or \({\mathit{FU}}_{0}=\mathrm{2,2}{10}^{-4}\).
Situation 2
Similarly for situation 2, we have \({\mathit{Sn}}_{0}=155\), \({\mathit{Sp}}_{0}=60+100=160\), or \(\mathit{Ke}=1\) and \({\mathit{Salt}}_{0}=80\mathit{MPa}\) or \({\mathit{FU}}_{0}={\mathrm{1,6.10}}^{-4}\).
Combination of situations 1 and 2
For the combination of situations 1 and 2 we have without the thermal
- math:
{S} _ {n} ^ {0} =begin {array} {0} =begin {array} {0} =frac {R} {e} |201-0|+ {C} _ {2}frac {R} {0}}frac {R} {0}} =frac {R} {0} {0}} =frac {R} {0} {0} =frac {R} {0} {0} =frac {R} {0} {0} =frac {R} {0} {0} =frac {R} {0} {0} =frac {R} {0} {0} =frac {R} {0} {0} =frac {R} {0} {mathrm {140.5}
With thermal we have \({\mathit{Sn}}_{0}=95+\mathrm{140,5}=\mathrm{235,5}\) for moments 4,5 and 1. So \(\mathit{Ke}=1\).
Without thermal \({\mathit{Sp}}_{0}^{1}=\mathrm{140,5}\) then for example by combining the moments 2.5 and 3 \({\mathit{Sp}}_{0}^{1}=\mathrm{240,5}\) or \({\mathit{FU}}_{0}^{1}=\mathrm{2,405.}{10}^{-4}\).
We calculate the second fictional transient, without the thermal on the moments and the pressure we take the complementary states either:math: {mathit {Sp}} _ {0} ^ {2} =begin {array} {c} {C} {C} {C} {C} {C} {C} {C} {2} =begin {array} {c} {2} =begin {array} {c} {C} {2} =begin {array} {c} {C} {2} =begin {array} {c} {C} {C} {2} =begin {array} {c} {C} {C} {2} =begin {array} {c} {C} {C} {2} =begin {array} {c} {C} {C} {{(1-1)} ^ {2})}end {array} =mathrm {0.5}. Then by combining the moments 3.5 and 2:math: {mathit {Sp}}} _ {0} ^ {2} =100+mathrm {0.5} =mathrm {100.5} either:math: {mathit {FU}}} _ {0} ^ {2} =mathrm {1,005.} {10} ^ {-4}.
So the use factor of combining situations 1 and 2 is \(\mathit{FU}={\mathit{FU}}_{0}^{1}+{\mathit{FU}}_{0}^{2}=\mathrm{2,405.}{10}^{-4}+\mathrm{1,005.}{10}^{-4}=\mathrm{3,41.}{10}^{-4}\)
So the table of elementary factors of use at the origin is
Situation 1 |
Situation 2 |
|
Situation 1 |
\(\mathrm{2,2}{10}^{-4}\) |
|
Situation 2 |
\({\mathrm{1,6.10}}^{-4}\) |
As we have \({\mathit{Nocc}}_{1}=1\) and \({\mathit{Nocc}}_{2}=1\) we have \({\mathit{FU}}_{\mathit{TOTAL}}^{\mathit{ORI}}={\mathrm{3,41.10}}^{-4}\).
2.1.2. ZE200b#
2.1.2.1. Calculation of 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.
\({S}_{n}=\begin{array}{c}{C}_{2}\frac{R}{I}\sqrt{({({M}_{\mathit{xA}}-{M}_{\mathit{xB}})}^{2}+{({M}_{\mathit{yA}}-{M}_{\mathit{yB}})}^{2}+{({M}_{\mathit{zA}}-{M}_{\mathit{zB}})}^{2})}+\Vert {\mathrm{\sigma }}_{\text{tran}}^{\text{lin}}({t}_{1}^{\text{}})-{\mathrm{\sigma }}_{\text{tran}}^{\text{lin}}({t}_{2}^{\text{}})\Vert \end{array}\)
Without thermal constraints or pressure, for situation 1 \({S}_{n}=20\) (at the origin and at the end).
With stresses in the form of a transient, we calculate the maximum amplitudes of thermal stress+pressure linearized at the origin and then at the end.
Situation 1
Instant |
Thermal stress+pressure |
\({\mathrm{\sigma }}^{\mathit{moyen}}\) |
\({\mathrm{\sigma }}^{\mathit{flexion}}\) |
\({\mathrm{\sigma }}_{0}^{\mathit{lin}}\) |
\({\mathrm{\sigma }}_{L}^{\mathit{lin}}\) |
||
Abscissa |
|||||||
0 |
1 |
2 |
|||||
1, 5 |
180 |
200 |
220 |
200 |
20 |
180 |
220 |
2, 5 |
0 |
200 |
-180 |
55 |
-90 |
145 |
-35 |
3.5 |
200 |
-100 |
-200 |
-50 |
-200 |
150 |
-250 |
4, 5 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
Instant 1 |
Instant 2 |
|
|
1, 5 |
2, 5 |
35 |
255 |
1, 5 |
3, 5 |
30 |
470 |
1, 5 |
4, 5 |
18 0 |
220 |
2, 5 |
3, 5 |
5 |
215 |
2, 5 |
4, 5 |
145 |
35 |
3, 5 |
4, 5 |
150 |
250 |
For situation 1 with constraints in the form of a transitory, \({\mathit{Sn}}_{0}=20+180=200\) and \({\mathit{Sn}}_{L}=20+470=490\).
Situation*2
Without thermal constraints or pressure, for situation 2 \({S}_{n}=60\) (at the origin and at the end).
Situation 2 has no pressure constraints so the calculation of the part in the form of a transient is the same as in ZE200a (part 2.1.1.1).
For situation 2 with thermal constraints, \({\mathit{Sn}}_{0}=60+95=155\) and \({\mathit{Sn}}_{L}=60+220=280\).
2.1.2.2. Calculation of Sn with earthquake#
We have just added the contribution of the earthquake to the magnitude Sn such that
\({S}_{\mathit{nS}}=\begin{array}{c}{C}_{2}\frac{R}{I}\sqrt{({({M}_{\mathit{xA}}-{M}_{\mathit{xB}}\pm 2{M}_{\mathit{xS}})}^{2}+{({M}_{\mathit{yA}}-{M}_{\mathit{yB}}\pm 2{M}_{\mathit{yS}})}^{2}+{({M}_{\mathit{zA}}-{M}_{\mathit{zB}}\pm 2{M}_{\mathit{zS}})}^{2})}\\ +\Vert {\mathrm{\sigma }}_{\text{tran}}^{\text{lin}}({t}_{1}^{\text{}})-{\mathrm{\sigma }}_{\text{tran}}^{\text{lin}}({t}_{2}^{\text{}})\Vert \end{array}\)
For the situation 1 without constraints in the form of a transitory,
\({S}_{\mathit{nS}}=\begin{array}{c}{C}_{2}\frac{R}{I}\sqrt{({({M}_{\mathit{xA}}-{M}_{\mathit{xB}}\pm 2{M}_{\mathit{xS}})}^{2})}\end{array}=2\ast \frac{\mathrm{0,5}}{1}\sqrt{({(21-1\pm 2\ast 21)}^{2})}=62\) at the origin and at the end.
For the situation 2without constraints in the form of a transitory,
\({S}_{\mathit{nS}}=\begin{array}{c}{C}_{2}\frac{R}{I}\sqrt{({({M}_{\mathit{xA}}-{M}_{\mathit{xB}}\pm 2{M}_{\mathit{xS}})}^{2})}\end{array}=2\ast \frac{\mathrm{0,5}}{1}\sqrt{({(1-61\pm 2\ast 21)}^{2})}=102\) at the origin and at the end.
2.2. Uncertainty about the solution#
Analytical solution.