Benchmark solution ===================== Benchmark results ---------------------- .. _RefHeading__34027_2146554684: ZE200a ~~~~~ .. _RefHeading__26212_1839245846: Calculation of Sn ^^^^^^^^^^^^ The parameter :math:`\mathit{Sn}` represents the amplitude of variation of the linear stress (mean stress :math:`\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 :math:`{S}_{n}=120` and for situation 2 :math:`{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 |:math:`{\mathrm{\sigma }}^{\mathit{moyen}}`|:math:`{\mathrm{\sigma }}^{\mathit{flexion}}`|:math:`{\mathrm{\sigma }}_{0}^{\mathit{lin}}`|:math:`{\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 | +-------+-------------------+---+----+-------------------------------------------+---------------------------------------------+---------------------------------------------+---------------------------------------------+ .. csv-table:: "Instant 1", "Instant 2"," :math:`{\mathit{Sn}}_{0}` "," :math:`{\mathit{Sn}}_{L}`" "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, :math:`{\mathit{Sn}}_{0}=120+90=210` and :math:`{\mathit{Sn}}_{L}=120+235=355`. *Situation* 2 +-------+-------------------+---+----+-------------------------------------------+---------------------------------------------+---------------------------------------------+---------------------------------------------+ |Instant|Thermal constraints |:math:`{\mathrm{\sigma }}^{\mathit{moyen}}`|:math:`{\mathrm{\sigma }}^{\mathit{flexion}}`|:math:`{\mathrm{\sigma }}_{0}^{\mathit{lin}}`|:math:`{\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 | +-------+-------------------+---+----+-------------------------------------------+---------------------------------------------+---------------------------------------------+---------------------------------------------+ .. csv-table:: "Instant 1", "Instant 2"," :math:`{\mathit{Sn}}_{0}` "," :math:`{\mathit{Sn}}_{L}`" "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, :math:`{\mathit{Sn}}_{0}=60+95=155` and :math:`{\mathit{Sn}}_{L}=60+220=280`. 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 {}})\ Vert\ end {tran}}} ^ {\ text {tran}}} ^ {\ text {tran}} ^ {\ text {tran}}} ^ {\ text {}})\ Green\ end {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} =1\ ast\ frac {\ mathrm {0.5}} {1} |201-1|+2\ ast\ frac {\ mathrm {0.5}} {1} {1}\ sqrt {1}\ sqrt {({(21-1\ pm 2\ ast 21)}} ^ {2})} =100+62=162` at the origin and at the end. *For the situation* *2without constraints in the form of a transitory,* :math: `{\ mathit {Sn}} _ {S} =1\ ast\ frac {\ mathrm {0.5}} {1} |0-0|+2\ ast\ frac {\ mathrm {0.5}} {1} {1}\ sqrt {1}}\ sqrt {({(1-61\ pm 2\ ast 21)} {1} |0-0|+2\ ast\ frac {\ mathrm {0.5}}} {1}} {1}} {1}} {1}} {1}} {1}} {\ mathrm {0.5}}} {1}} {1}} {1}} {1}} {1}} {1}} {5}} {1}} {1}} {1}} { 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 :math:`{\mathit{Sn}}_{0}=210` (part :ref:`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 {}})\ Green\ end {array}` *Without thermal constraints*, for situation 1 :math:`{S}_{p}^{0}=120`. *With thermal constraints,* we calculate the maximum amplitudes of thermal stresses linearized at the origin: .. csv-table:: "Instant 1", "Instant 2"," :math:`{\mathit{Sp}}_{0}`" "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 :math:`{\mathit{Sp}}_{0}=120+100=220`. So for :math:`\mathit{Sm}=2000\mathit{MPa}`, we have :math:`\mathit{Ke}=1` and :math:`{\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 :math:`{\mathit{Nadm}}_{0}=\frac{500000}{{\mathit{Salt}}_{0}}=4545` or :math:`{\mathit{FU}}_{0}=\mathrm{2,2}{10}^{-4}`. *Situation* 2 Similarly for situation 2, we have :math:`{\mathit{Sn}}_{0}=155`, :math:`{\mathit{Sp}}_{0}=60+100=160`, or :math:`\mathit{Ke}=1` and :math:`{\mathit{Salt}}_{0}=80\mathit{MPa}` or :math:`{\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 :math:`{\mathit{Sn}}_{0}=95+\mathrm{140,5}=\mathrm{235,5}` for moments 4,5 and 1. So :math:`\mathit{Ke}=1`. Without thermal :math:`{\mathit{Sp}}_{0}^{1}=\mathrm{140,5}` then for example by combining the moments 2.5 and 3 :math:`{\mathit{Sp}}_{0}^{1}=\mathrm{240,5}` or :math:`{\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 :math:`\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 .. csv-table:: "", "Situation 1", "Situation 2" "Situation 1", ":math:`\mathrm{2,2}{10}^{-4}` "," :math:`{\mathrm{3,41.10}}^{-4}`" "Situation 2", "", ":math:`{\mathrm{1,6.10}}^{-4}`" As we have :math:`{\mathit{Nocc}}_{1}=1` and :math:`{\mathit{Nocc}}_{2}=1` we have :math:`{\mathit{FU}}_{\mathit{TOTAL}}^{\mathit{ORI}}={\mathrm{3,41.10}}^{-4}`. ZE200b ~~~~~ Calculation of Sn ^^^^^^^^^^^^ The parameter :math:`\mathit{Sn}` represents the amplitude of variation of the linear stress (mean stress :math:`\pm` flexural stress) between two moments of the transient in question. :math:`{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 :math:`{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 |:math:`{\mathrm{\sigma }}^{\mathit{moyen}}`|:math:`{\mathrm{\sigma }}^{\mathit{flexion}}`|:math:`{\mathrm{\sigma }}_{0}^{\mathit{lin}}`|:math:`{\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 | +-------+-----------------------+----+----+-------------------------------------------+---------------------------------------------+---------------------------------------------+---------------------------------------------+ .. csv-table:: "Instant 1", "Instant 2"," :math:`{\mathit{Sn}}_{0}` "," :math:`{\mathit{Sn}}_{L}`" "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, :math:`{\mathit{Sn}}_{0}=20+180=200` and :math:`{\mathit{Sn}}_{L}=20+470=490`. *Situation*2* *Without thermal constraints* *or pressure*, for situation 2 :math:`{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 :ref:`2.1.1.1 `). For situation 2 with thermal constraints, :math:`{\mathit{Sn}}_{0}=60+95=155` and :math:`{\mathit{Sn}}_{L}=60+220=280`. 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}_{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,* :math:`{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,* :math:`{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. Uncertainty about the solution --------------------------- Analytical solution.