Appendix ===== Uniform rotation loading around :math:`\mathrm{OZ}` ------------------------------------------------------------- 2D axisymmetric model ~~~~~~~~~~~~~~~~~~~~~~~~~ The centrifugal force density is: :math:`\rho {\Omega }^{2}r{e}_{r}`. The following boundary conditions are considered: :math:`{u}_{z}(r,z)=0` in :math:`z=0` and :math:`z=L` Displacement is postulated in the form: :math:`{u}_{r}=u(r)` :math:`{u}_{z}={u}_{\theta }=0` So: :math:`{\varepsilon }_{\mathrm{rr}}=u\text{'}`; :math:`{\varepsilon }_{\theta \theta }=\frac{u}{r}`; :math:`{\varepsilon }_{\mathrm{zz}}={\varepsilon }_{\mathrm{rz}}={\varepsilon }_{\theta z}={\varepsilon }_{r\theta }=0` .. image:: images/100010AE000010F300001ADFB4542473AA5068E3.svg :width: 218 :height: 346 .. _RefImage_100010AE000010F300001ADFB4542473AA5068E3.svg: Hollow cylinder geometry: :math:`R=20\mathrm{mm}` :math:`h=1\mathrm{mm}` Elastic stresses are expressed as: :math:`{\sigma }_{\mathrm{rr}}=\frac{E}{(1+\nu )(1-2\nu )}[(1-\nu )u\text{'}+\nu \frac{u}{r}]` :math:`{\sigma }_{\theta \theta }=\frac{E}{(1+\nu )(1-2\nu )}[(1-\nu )\frac{u}{r}+\nu u\text{'}]` :math:`{\sigma }_{\mathrm{zz}}=\frac{\nu E}{(1+\nu )(1-2\nu )}[\frac{u}{r}+u\text{'}]` The radial equilibrium equation is written as: :math:`{(r{\sigma }_{\mathrm{rr}})}_{,r}-{\sigma }_{\theta \theta }=-\rho {\Omega }^{2}{r}^{2}` So: :math:`(\frac{(\mathrm{ru})\text{'}}{r})\text{'}=\frac{-(1+\nu )(1-2\nu )}{(1-\nu )E}\rho {\Omega }^{2}r` :math:`\mathrm{éq}1.1-1` **Note:** :math:`\frac{u}{r}+u\text{'}=\frac{(\mathrm{ru})\text{'}}{r}` Hence the general solution: :math:`u(r)=\frac{-(1+\nu )(1-2\nu )}{(1-\nu )E}\rho {\Omega }^{2}\frac{{r}^{3}}{8}+\mathrm{Ar}+\frac{B}{r}` :math:`\mathrm{éq}1.1-2` The constraints are then: :math:`{\sigma }_{\mathrm{rr}}(r)=\frac{-3-2\nu }{1-\nu }\rho {\Omega }^{2}\frac{{r}^{2}}{8}+\frac{E}{(1+\nu )(1-2\nu )}(A-(1-2\nu )\frac{B}{{r}^{2}})` :math:`{\sigma }_{\theta \theta }(r)=\frac{-1+2\nu }{1-\nu }\rho {\Omega }^{2}\frac{{r}^{2}}{8}+\frac{E}{(1+\nu )(1-2\nu )}(A-(1-2\nu )\frac{B}{{r}^{2}})` :math:`\mathrm{éq}1.1-3` :math:`{\sigma }_{\mathrm{zz}}(r)=\frac{-\nu }{1-\nu }\rho {\Omega }^{2}\frac{{r}^{2}}{2}+\frac{2\nu E}{(1+\nu )(1-2\nu )}A` The stress boundary conditions are: :math:`{\sigma }_{\mathrm{rr}}=0` in :math:`r=R\pm \frac{h}{2}` We note: :math:`x=\frac{h}{\mathrm{2R}}` Thanks to :math:`[\mathrm{éq}1.1-3]`, we get: :math:`B=\frac{(3-2\nu )(1+\nu )}{8(1-\nu )E}\rho {\Omega }^{2}{R}^{4}{(1-{x}^{2})}^{2}` then: :math:`A=\frac{(3-2\nu )(1+\nu )(1-2\nu )}{4(1-\nu )E}\rho {\Omega }^{2}{R}^{2}(1-{x}^{2})` **Digital application:** :math:`\rho =8.10–\mathrm{6 }\mathrm{kg}/{\mathrm{mm}}^{3}` :math:`\Omega =\mathrm{1 }{s}^{-1}` :math:`E=2.105N/{\mathrm{mm}}^{2}` :math:`\nu =0.3` From where: :math:`A={\mathrm{7.13588.10}}^{-9}{\mathrm{mm}}^{2}` :math:`B={\mathrm{3.561258.10}}^{-\mathrm{6 }}{\mathrm{mm}}^{2}` **Note:** :math:`\frac{(1+\nu )(1-2\nu )}{(1-\nu )E}\rho \frac{{\Omega }^{2}}{8}=3.714286E-12{\mathrm{mm}}^{2}` :math:`\frac{\nu }{1-\nu }\rho \frac{{\Omega }^{2}}{2}=1.714286E-6{\mathrm{MPa.mm}}^{2}` So: * in inner skin: :math:`\{\begin{array}{}{u}_{r}=2.9424{E}^{-7}\mathrm{mm}\\ {\sigma }_{\mathrm{zz}}=0.99488{E}^{-3}\mathrm{Mpa}\end{array}` * in outer skin: :math:`\{\begin{array}{}{u}_{r}=2.8801{E}^{-7}\mathrm{mm}\\ {\sigma }_{\mathrm{zz}}=0.92631{E}^{-3}\mathrm{Mpa}\end{array}` Axisymmetric shell model ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Centrifugal force is equivalent to **distributed pressure**: :math:`p=\rho {\Omega }^{2}\mathrm{hR}(1+\frac{{h}^{2}}{{\mathrm{12R}}^{2}})` The solution is membrane-bound, the normal balance is written as: :math:`{N}_{\theta \theta }=\mathrm{pR}` Membrane deformation is: :math:`{E}_{\theta \theta }=\frac{w}{R}`, while :math:`{E}_{\theta \theta }=0={K}_{\theta \theta }={K}_{\mathrm{zz}}`. In elasticity: :math:`{N}_{\theta \theta }=\frac{\mathrm{Eh}}{1-{\nu }^{2}}{E}_{\theta \theta }`; :math:`{N}_{\mathrm{zz}}=\nu {N}_{\theta \theta }`; :math:`{M}_{\alpha \beta }=0` Hence the solution (deflection and normal circumferential force): :math:`p=\rho {\Omega }^{2}\mathrm{hR}(1+\frac{{h}^{2}}{{\mathrm{12R}}^{2}})`; :math:`{N}_{\theta \theta }=\rho {\Omega }^{2}{R}^{2}h(1+\frac{{h}^{2}}{{\mathrm{12R}}^{2}})` The axial stress is equal to: :math:`{\sigma }_{\mathrm{zz}}=\nu \rho {\Omega }^{2}{R}^{2}(1+\frac{{h}^{2}}{{\mathrm{12R}}^{2}})` (constant in thickness) If you do not take into account the metric correction, you must remove the term :math:`(1+\frac{{h}^{2}}{{\mathrm{12R}}^{2}})` in the previous expressions. **Digital application** (without metric correction) **:** :math:`p=\mathrm{1,600000}.10–4\mathrm{MPa}` :math:`w=\mathrm{2,912000}.10–7\mathrm{mm}` :math:`{N}_{\mathrm{zz}}=\mathrm{0,96000}.10–3N/\mathrm{mm}` :math:`{\sigma }_{\mathrm{zz}}=\mathrm{0,96000}.10–3\mathrm{MPa}` Gravity loading ----------------------- 2D axisymmetric model ~~~~~~~~~~~~~~~~~~~~~~~~~ The force density is: :math:`-\rho g{e}_{z}` (vertical gravity). The following boundary conditions are considered: :math:`{U}_{z}(r,z)=0` in :math:`r=R` and :math:`z=0` (circle of support) with uniform traction: :math:`{\sigma }_{\mathrm{zz}}(r,z)=\rho gL` in :math:`z=L`, balancing the weight. We postulate the elastic solution of the type: :math:`\sigma =(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& {\sigma }_{\mathrm{zz}}\end{array})` so that: :math:`{\sigma }_{\mathrm{zz}}(r,z)=\rho gL`; :math:`{\varepsilon }_{\mathrm{rz}}=0={\varepsilon }_{r\theta }={\varepsilon }_{\theta z}` We observe as follows: :math:`{u}_{r,r}=\frac{{u}_{r}}{r}\iff {u}_{r}(r,z)=-\nu A\text{'}(z)r` Then: :math:`\begin{array}{}-\nu A\text{'}(z)={\varepsilon }_{\mathrm{rr}}=-\nu {\varepsilon }_{\mathrm{zz}}\iff {u}_{z,z}(r,z)=A\text{'}(z)\\ \iff {u}_{r}(r,z)=A(z)+B\end{array}` From :math:`{\varepsilon }_{\mathrm{rz}}=0`, we get: :math:`B\text{'}(r)-\nu \mathrm{rA}\text{'}\text{'}(z)=0` either: :math:`A\text{'}\text{'}(z)=\mathrm{cste}=\alpha`; :math:`B\text{'}(r)=\alpha \nu r` From the stress limit conditions, we obtain: :math:`A(z)=\frac{\rho g{z}^{2}}{2E}+\beta`; :math:`B(r)=\nu \frac{\rho g{r}^{2}}{2E}` Finally, :math:`\beta` checks: :math:`\beta =-\nu \rho g\frac{{R}^{2}}{2E}` So: :math:`\begin{array}{}{u}_{r}(r,z)=\frac{-\nu \rho gzr}{E};{u}_{z}(r,z)=\frac{\rho g}{2E}({z}^{2}+\nu ({r}^{2}–{R}^{2}))\\ {\sigma }_{\mathrm{zz}}(r,z)=\rho gz\end{array}` :math:`\mathrm{éq}2.1-1` **Digital app** :math:`g=\mathrm{10 }N/\mathrm{kg}` :math:`\rho ={8.10}^{-\mathrm{6 }}\mathrm{kg}/{\mathrm{mm}}^{3}` :math:`R=\mathrm{20 }\mathrm{mm}` :math:`L=\mathrm{10 }\mathrm{mm}` :math:`E={2.10}^{\mathrm{5 }}N/{\mathrm{mm}}^{2}` :math:`\nu =0.3` :math:`h=\mathrm{1 }\mathrm{mm}` * in inner skin: :math:`\{\begin{array}{}{u}_{r}(L)=-2.34000{E}^{-8}\mathrm{mm}\\ {\sigma }_{\mathrm{zz}}(L)=8.0000{E}^{-4}\mathrm{Mpa}\\ {u}_{z}(O)=-1.185000{E}^{-9}\mathrm{mm}\end{array}` * in outer skin: :math:`\{\begin{array}{}{u}_{r}(L)=-2.46000{E}^{-8}\mathrm{mm}\\ {\sigma }_{\mathrm{zz}}(L)=8.0000{E}^{-4}\mathrm{Mpa}\\ {u}_{z}(O)=1.215000{E}^{-9}\mathrm{mm}\end{array}` Axisymmetric shell model ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ A vertical pull is exerted in :math:`z=L`: :math:`F=\rho ghL` Gravity leads to vertical force: :math:`f=-\rho gh{e}_{z}` The boundary condition on the support circle is: :math:`{u}_{z}(z)=0` in :math:`z=0` The solution is membrane-bound, the vertical balance is written as: :math:`{N}_{\mathrm{zz},z}=\rho gh` Plus: :math:`{N}_{\theta \theta }=0`. In elasticity, we then deduce: :math:`{E}_{\theta \theta }=\frac{w}{R}=\frac{-\nu {N}_{\mathrm{zz}}}{\mathrm{Eh}}=\frac{-\nu \rho gz}{E}` :math:`{E}_{\mathrm{zz}}={u}_{z,z}=\frac{{N}_{\mathrm{zz}}}{\mathrm{Eh}}\Rightarrow {u}_{z}(z)=\frac{\rho g}{2E}{z}^{2}` The axial stress is: :math:`{\sigma }_{\mathrm{zz}}=\rho gz` (constant in thickness) **Digital application:** :math:`F={8.10}^{-4}{\mathrm{N.mm}}^{-1}` :math:`w(L)=-{\mathrm{2.4000.10}}^{-8}\mathrm{mm}` :math:`{N}_{\mathrm{zz}}(L)={\mathrm{8.0000.10}}^{-4}{\mathrm{N.mm}}^{-1}` :math:`{\sigma }_{\mathrm{zz}}(L)={\mathrm{8.0000.10}}^{-4}{\mathrm{N.mm}}^{-1}` Thermomechanical loading -------------------------- 2D axisymmetric model ~~~~~~~~~~~~~~~~~~~~~~~~~ :math:`T(r)–{T}_{\mathrm{réf}}(r)=\frac{{T}_{s}+{T}_{i}}{2}+\frac{({T}_{s}–{T}_{i})}{h}(r-R)` :math:`\mathrm{éq}\mathrm{3.1.}-1` .. image:: images/10000A4600001F1C0000170C534B0B7595A77027.svg :width: 218 :height: 346 .. _RefImage_10000A4600001F1C0000170C534B0B7595A77027.svg: Displacement is postulated in the form: :math:`{u}_{r}=u(r)`; :math:`{u}_{z}={u}_{\theta }=0` with the appropriate boundary conditions. Thus, elastic stresses are expressed: :math:`{\sigma }_{\mathrm{rr}}=\frac{E}{(1+\nu )(1-2\nu )}[(1-\nu )u\text{'}+\nu \frac{u}{r}]-\frac{\alpha E}{1-2\nu }(T-{T}_{\mathrm{réf}})` :math:`{\sigma }_{\theta \theta }=\frac{E}{(1+\nu )(1-2\nu )}[(1-\nu )\frac{u}{r}+\nu u\text{'}]-\frac{\alpha E}{1-2\nu }(T-{T}_{\mathrm{réf}})` :math:`\mathrm{éq}1.1-3` :math:`{\sigma }_{\mathrm{zz}}=\frac{\nu E}{(1+\nu )(1-2\nu )}[\frac{u}{r}+u\text{'}]-\frac{\alpha E}{1-2\nu }(T-{T}_{\mathrm{réf}})` The radial equilibrium equation :math:`{(r,{\sigma }_{\mathrm{rr}})}_{,r}-{\sigma }_{\theta \theta }=0` gives: :math:`(\frac{(\mathrm{ru})\text{'}}{r})\text{'}=\frac{\alpha (1+\nu \text{'})}{(1-\nu )}(T-{T}_{\mathrm{réf}})\text{'}` :math:`\mathrm{éq}3.1-2` Hence the general solution: :math:`u(r)=\frac{\alpha (1+\nu )}{(1-\nu )}\frac{({T}_{s}-{T}_{i})}{h}\frac{{r}^{2}}{3}+\mathrm{Ar}+\frac{B}{r}` :math:`\mathrm{éq}3.1-3` The constraints are then: :math:`{\sigma }_{\mathrm{rr}}(r)=\frac{\alpha E({T}_{s}-{T}_{i})}{h}(\frac{R}{1-2\nu }-\frac{r}{3(1-\nu )})-\frac{\alpha E}{1-2\nu }({T}_{s}-{T}_{i})+\frac{E}{(1+\nu )(1-2\nu )}(A-(1-2\nu )\frac{B}{{r}^{2}})` :math:`{\sigma }_{\theta \theta }(r)=\frac{\alpha E({T}_{s}-{T}_{i})}{h}(\frac{R}{1-2\nu }-\frac{\mathrm{2r}}{3(1-\nu )})-\frac{\alpha E}{1-2\nu }({T}_{s}-{T}_{i})+\frac{E}{(1+\nu )(1-2\nu )}(A-(1-2\nu )\frac{B}{{r}^{2}})` :math:`{\sigma }_{\mathrm{zz}}(r)=\frac{\alpha E({T}_{s}-{T}_{i})}{h}(\frac{R}{1-2\nu }-\frac{r}{3(1-\nu )})-\frac{\alpha E}{1-2\nu }({T}_{s}-{T}_{i})+\frac{2\nu E}{(1+\nu )(1-2\nu )}A` :math:`\mathrm{éq}3.1-4` The limit conditions in effort are: in :math:`r=R\pm \frac{h}{2},{\sigma }_{\mathrm{rr}}=0`. Note: :math:`x=\frac{h}{\mathrm{2R}}`. Thanks to :math:`[\mathrm{éq}3.1-4]`, we get: :math:`B=\frac{\alpha ({T}_{s}-{T}_{i})}{\mathrm{6h}(1-\nu )}(1+\nu ){R}^{3}{(1-{x}^{2})}^{2}` then: :math:`A=\alpha (1+\nu )[\frac{-({T}_{s}-{T}_{i})R}{\mathrm{6h}(1-\nu )}(3-(1-2\nu ){x}^{2})+\frac{({T}_{s}+{T}_{i})}{2}]` **Digital application:** :math:`F={8.10}^{-4}{\mathrm{N.mm}}^{-1}` :math:`w(L)=-{\mathrm{2.4000.10}}^{-8}\mathrm{mm}` :math:`{N}_{\mathrm{zz}}(L)={\mathrm{8.0000.10}}^{-4}{\mathrm{N.mm}}^{-1}` :math:`{\sigma }_{\mathrm{zz}}(L)={\mathrm{8.0000.10}}^{-4}{\mathrm{N.mm}}^{-1}` From where: :math:`A=-{\mathrm{0.18569881.10}}^{-3}{\mathrm{mm}}^{2}` :math:`B=0.02473096{\mathrm{mm}}^{2}` **Note:** :math:`\frac{\alpha (1+\nu )}{1-\nu }\frac{{T}_{s}–{T}_{i}}{\mathrm{3h}}=0.61904762E-5` * in inner skin: :math:`\{\begin{array}{}{u}_{r}=1.056145{E}^{-6}\mathrm{mm}\\ {\sigma }_{\mathrm{zz}}=1.4321427\mathrm{Mpa}\end{array}` * in outer skin: :math:`\{\begin{array}{}{u}_{r}=1.110317{E}^{-6}\mathrm{mm}\\ {\sigma }_{\mathrm{zz}}=-1.4250001\mathrm{Mpa}\end{array}` :math:`\begin{array}{}{\sigma }_{\mathrm{zz}}=1.449319\mathrm{Mpa}\\ \mathrm{ou}\\ {\sigma }_{\mathrm{zz}}=1.428671\mathrm{Mpa}\text{( sans correction métrique )}\end{array}` In case we take :math:`{T}_{s}={T}_{i}=0.1°C`: :math:`A=\mathrm{0,00130000}{.10}^{-3}` :math:`B=\mathrm{0,0 }{\mathrm{mm}}^{2}` So: * in inner skin: :math:`\begin{array}{}{u}_{r}=25.350000{E}^{-6}\mathrm{mm}\\ {\sigma }_{\mathrm{zz}}=-0.200000\mathrm{Mpa}\end{array}` * in outer skin: :math:`\begin{array}{}{u}_{r}=26.650000{E}^{-6}\mathrm{mm}\\ {\sigma }_{\mathrm{zz}}=-0.200000\mathrm{Mpa}\end{array}` Axisymmetric shell model ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ For the temperature field in the thickness given by :math:`[\mathrm{éq}3.1-1]`, the following expression for the law of behavior is obtained: :math:`\begin{array}{}{N}_{\theta \theta }=\frac{\mathrm{Eh}}{1-{\nu }^{2}}({E}_{\theta \theta }+\nu {E}_{\mathrm{zz}})-\frac{\alpha Eh}{1-\nu }[\frac{{T}_{s}+{T}_{i}}{2}+\frac{{T}_{s}-{T}_{i}}{12}\frac{h}{R}]\\ {N}_{\mathrm{zz}}=\frac{\mathrm{Eh}}{1-{\nu }^{2}}(\nu {E}_{\theta \theta }+{E}_{\mathrm{zz}})-\frac{\alpha Eh}{1-\nu }[\frac{{T}_{s}+{T}_{i}}{2}+\frac{{T}_{s}-{T}_{i}}{12}\frac{h}{R}]\end{array}` :math:`\mathrm{éq}3.2-1` and: :math:`\begin{array}{}{N}_{\theta \theta }=\frac{\mathrm{Eh}}{1-{\nu }^{2}}({E}_{\theta \theta }+\nu {E}_{\mathrm{zz}})-\frac{\alpha Eh}{1-\nu }[\frac{{T}_{s}+{T}_{i}}{2}+\frac{{T}_{s}-{T}_{i}}{12}\frac{h}{R}]\\ {N}_{\mathrm{zz}}=\frac{\mathrm{Eh}}{1-{\nu }^{2}}(\nu {E}_{\theta \theta }+{E}_{\mathrm{zz}})-\frac{\alpha Eh}{1-\nu }[\frac{{T}_{s}+{T}_{i}}{2}+\frac{{T}_{s}-{T}_{i}}{12}\frac{h}{R}]\end{array}` :math:`\mathrm{éq}3.2-2` According to these expressions, the thermal terms in .. image:: images/Object_1196.svg :width: 218 :height: 346 .. _RefImage_Object_1196.svg: are to be overlooked if we do not consider the metric correction in thickness, that is to say in the case of the usual models. In our situation: :math:`{E}_{\theta \theta }=\frac{w}{R}` :math:`{E}_{\mathrm{zz}}=0` :math:`{K}_{\theta \theta }={K}_{\mathrm{zz}}=0` The normal balance in the shell is written as: :math:`{N}_{\theta \theta }=0` Hence the arrow: :math:`w=\alpha (1+\nu )[\frac{{T}_{s}+{T}_{i}}{2}+\frac{{T}_{s}+{T}_{i}}{12}\frac{h}{R}]R` and: :math:`{N}_{\mathrm{zz}}=\alpha \mathrm{Eh}[\frac{{T}_{s}+{T}_{i}}{2}+\frac{{T}_{s}+{T}_{i}}{12}\frac{h}{R}]` :math:`{M}_{\mathrm{zz}}=\frac{-\alpha {\mathrm{Eh}}^{2}}{12(1-\nu )}[({T}_{s}-{T}_{i})+\frac{{T}_{s}+{T}_{i}}{2}\frac{h}{R}]` Since the second member of dilation does not take into account the metric correction, the terms in .. image:: images/Object_1201.svg :width: 218 :height: 346 .. _RefImage_Object_1201.svg: above are overlooked. **Digital app** :math:`R=\mathrm{20 }\mathrm{mm}` :math:`h=\mathrm{1 }\mathrm{mm}` :math:`\alpha ={10}^{-5}°{C}^{-1}` :math:`{T}_{s}={T}_{i}=0.5°C` :math:`\nu =0.3` :math:`E={2.10}^{5}N/{\mathrm{mm}}^{2}` From where: :math:`{M}_{\mathrm{zz}}=-0.2380952N` * in inner skin: :math:`\begin{array}{}{\sigma }_{\mathrm{zz}}=1.449319\mathrm{Mpa}\\ \mathrm{ou}\\ {\sigma }_{\mathrm{zz}}=1.428671\mathrm{Mpa}\text{( sans correction métrique )}\end{array}` In case we take :math:`{T}_{s}={T}_{i}=\mathrm{0,1}°C`: :math:`w={\mathrm{26.00000.10}}^{-6}\mathrm{mm}` :math:`{N}_{\mathrm{zz}}=–0.2{\mathrm{N.mm}}^{-1}` :math:`{M}_{\mathrm{zz}}=–0.001190476N` * in inner skin: :math:`\begin{array}{}{\sigma }_{\mathrm{zz}}=-0.2122466\mathrm{Mpa}\\ \mathrm{ou}\\ {\sigma }_{\mathrm{zz}}=-0.200000\mathrm{Mpa}\text{( sans correction métrique )}\end{array}` **Thickness constraints with metric correction are given by:* :math:`{\sigma }_{\mathrm{zz}}({x}_{3})=\frac{{N}_{\mathrm{zz}}–\frac{{M}_{\mathrm{zz}}}{R}}{h(1-\frac{{h}^{2}}{{\mathrm{12R}}^{2}})}+({M}_{\mathrm{zz}}-{N}_{\mathrm{zz}}\frac{{h}^{2}}{\mathrm{12R}})\frac{{\mathrm{12x}}_{3}}{{h}^{3}(1-\frac{{h}^{2}}{{\mathrm{12R}}^{2}})}`