15. Appendix#

15.1. Uniform rotation loading around \(\mathrm{OZ}\)#

15.1.1. 2D axisymmetric model#

The centrifugal force density is: \(\rho {\Omega }^{2}r{e}_{r}\).

The following boundary conditions are considered:

\({u}_{z}(r,z)=0\) in \(z=0\) and \(z=L\)

Displacement is postulated in the form:

\({u}_{r}=u(r)\)

\({u}_{z}={u}_{\theta }=0\)

So:

\({\varepsilon }_{\mathrm{rr}}=u\text{'}\); \({\varepsilon }_{\theta \theta }=\frac{u}{r}\); \({\varepsilon }_{\mathrm{zz}}={\varepsilon }_{\mathrm{rz}}={\varepsilon }_{\theta z}={\varepsilon }_{r\theta }=0\)

_images/100010AE000010F300001ADFB4542473AA5068E3.svg

Hollow cylinder geometry:

\(R=20\mathrm{mm}\)

\(h=1\mathrm{mm}\)

Elastic stresses are expressed as:

\({\sigma }_{\mathrm{rr}}=\frac{E}{(1+\nu )(1-2\nu )}[(1-\nu )u\text{'}+\nu \frac{u}{r}]\)

\({\sigma }_{\theta \theta }=\frac{E}{(1+\nu )(1-2\nu )}[(1-\nu )\frac{u}{r}+\nu u\text{'}]\)

\({\sigma }_{\mathrm{zz}}=\frac{\nu E}{(1+\nu )(1-2\nu )}[\frac{u}{r}+u\text{'}]\)

The radial equilibrium equation is written as:

\({(r{\sigma }_{\mathrm{rr}})}_{,r}-{\sigma }_{\theta \theta }=-\rho {\Omega }^{2}{r}^{2}\)

So:

\((\frac{(\mathrm{ru})\text{'}}{r})\text{'}=\frac{-(1+\nu )(1-2\nu )}{(1-\nu )E}\rho {\Omega }^{2}r\) \(\mathrm{éq}1.1-1\)

Note:

\(\frac{u}{r}+u\text{'}=\frac{(\mathrm{ru})\text{'}}{r}\)

Hence the general solution:

\(u(r)=\frac{-(1+\nu )(1-2\nu )}{(1-\nu )E}\rho {\Omega }^{2}\frac{{r}^{3}}{8}+\mathrm{Ar}+\frac{B}{r}\) \(\mathrm{éq}1.1-2\)

The constraints are then:

\({\sigma }_{\mathrm{rr}}(r)=\frac{-3-2\nu }{1-\nu }\rho {\Omega }^{2}\frac{{r}^{2}}{8}+\frac{E}{(1+\nu )(1-2\nu )}(A-(1-2\nu )\frac{B}{{r}^{2}})\)

\({\sigma }_{\theta \theta }(r)=\frac{-1+2\nu }{1-\nu }\rho {\Omega }^{2}\frac{{r}^{2}}{8}+\frac{E}{(1+\nu )(1-2\nu )}(A-(1-2\nu )\frac{B}{{r}^{2}})\) \(\mathrm{éq}1.1-3\)

\({\sigma }_{\mathrm{zz}}(r)=\frac{-\nu }{1-\nu }\rho {\Omega }^{2}\frac{{r}^{2}}{2}+\frac{2\nu E}{(1+\nu )(1-2\nu )}A\)

The stress boundary conditions are:

\({\sigma }_{\mathrm{rr}}=0\) in \(r=R\pm \frac{h}{2}\)

We note:

\(x=\frac{h}{\mathrm{2R}}\)

Thanks to \([\mathrm{éq}1.1-3]\), we get:

\(B=\frac{(3-2\nu )(1+\nu )}{8(1-\nu )E}\rho {\Omega }^{2}{R}^{4}{(1-{x}^{2})}^{2}\)

then:

\(A=\frac{(3-2\nu )(1+\nu )(1-2\nu )}{4(1-\nu )E}\rho {\Omega }^{2}{R}^{2}(1-{x}^{2})\)

Digital application:

\(\rho =8.10–\mathrm{6 }\mathrm{kg}/{\mathrm{mm}}^{3}\)

\(\Omega =\mathrm{1 }{s}^{-1}\)

\(E=2.105N/{\mathrm{mm}}^{2}\)

\(\nu =0.3\)

From where: \(A={\mathrm{7.13588.10}}^{-9}{\mathrm{mm}}^{2}\)

\(B={\mathrm{3.561258.10}}^{-\mathrm{6 }}{\mathrm{mm}}^{2}\)

Note:

\(\frac{(1+\nu )(1-2\nu )}{(1-\nu )E}\rho \frac{{\Omega }^{2}}{8}=3.714286E-12{\mathrm{mm}}^{2}\)

\(\frac{\nu }{1-\nu }\rho \frac{{\Omega }^{2}}{2}=1.714286E-6{\mathrm{MPa.mm}}^{2}\)

So:

  • in inner skin: \(\{\begin{array}{}{u}_{r}=2.9424{E}^{-7}\mathrm{mm}\\ {\sigma }_{\mathrm{zz}}=0.99488{E}^{-3}\mathrm{Mpa}\end{array}\)

  • in outer skin: \(\{\begin{array}{}{u}_{r}=2.8801{E}^{-7}\mathrm{mm}\\ {\sigma }_{\mathrm{zz}}=0.92631{E}^{-3}\mathrm{Mpa}\end{array}\)

15.1.2. Axisymmetric shell model#

Centrifugal force is equivalent to distributed pressure:

\(p=\rho {\Omega }^{2}\mathrm{hR}(1+\frac{{h}^{2}}{{\mathrm{12R}}^{2}})\)

The solution is membrane-bound, the normal balance is written as:

\({N}_{\theta \theta }=\mathrm{pR}\)

Membrane deformation is: \({E}_{\theta \theta }=\frac{w}{R}\), while \({E}_{\theta \theta }=0={K}_{\theta \theta }={K}_{\mathrm{zz}}\).

In elasticity:

\({N}_{\theta \theta }=\frac{\mathrm{Eh}}{1-{\nu }^{2}}{E}_{\theta \theta }\); \({N}_{\mathrm{zz}}=\nu {N}_{\theta \theta }\); \({M}_{\alpha \beta }=0\)

Hence the solution (deflection and normal circumferential force):

\(p=\rho {\Omega }^{2}\mathrm{hR}(1+\frac{{h}^{2}}{{\mathrm{12R}}^{2}})\); \({N}_{\theta \theta }=\rho {\Omega }^{2}{R}^{2}h(1+\frac{{h}^{2}}{{\mathrm{12R}}^{2}})\)

The axial stress is equal to:

\({\sigma }_{\mathrm{zz}}=\nu \rho {\Omega }^{2}{R}^{2}(1+\frac{{h}^{2}}{{\mathrm{12R}}^{2}})\) (constant in thickness)

If you do not take into account the metric correction, you must remove the term \((1+\frac{{h}^{2}}{{\mathrm{12R}}^{2}})\) in the previous expressions.

Digital application (without metric correction) :

\(p=\mathrm{1,600000}.10–4\mathrm{MPa}\)

\(w=\mathrm{2,912000}.10–7\mathrm{mm}\)

\({N}_{\mathrm{zz}}=\mathrm{0,96000}.10–3N/\mathrm{mm}\)

\({\sigma }_{\mathrm{zz}}=\mathrm{0,96000}.10–3\mathrm{MPa}\)

15.2. Gravity loading#

15.2.1. 2D axisymmetric model#

The force density is: \(-\rho g{e}_{z}\) (vertical gravity).

The following boundary conditions are considered:

\({U}_{z}(r,z)=0\) in \(r=R\) and \(z=0\) (circle of support)

with uniform traction: \({\sigma }_{\mathrm{zz}}(r,z)=\rho gL\) in \(z=L\), balancing the weight.

We postulate the elastic solution of the type:

\(\sigma =(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& {\sigma }_{\mathrm{zz}}\end{array})\)

so that:

\({\sigma }_{\mathrm{zz}}(r,z)=\rho gL\); \({\varepsilon }_{\mathrm{rz}}=0={\varepsilon }_{r\theta }={\varepsilon }_{\theta z}\)

We observe as follows:

\({u}_{r,r}=\frac{{u}_{r}}{r}\iff {u}_{r}(r,z)=-\nu A\text{'}(z)r\)

Then:

\(\begin{array}{}-\nu A\text{'}(z)={\varepsilon }_{\mathrm{rr}}=-\nu {\varepsilon }_{\mathrm{zz}}\iff {u}_{z,z}(r,z)=A\text{'}(z)\\ \iff {u}_{r}(r,z)=A(z)+B\end{array}\)

From \({\varepsilon }_{\mathrm{rz}}=0\), we get:

\(B\text{'}(r)-\nu \mathrm{rA}\text{'}\text{'}(z)=0\)

either:

\(A\text{'}\text{'}(z)=\mathrm{cste}=\alpha\); \(B\text{'}(r)=\alpha \nu r\)

From the stress limit conditions, we obtain:

\(A(z)=\frac{\rho g{z}^{2}}{2E}+\beta\); \(B(r)=\nu \frac{\rho g{r}^{2}}{2E}\)

Finally, \(\beta\) checks: \(\beta =-\nu \rho g\frac{{R}^{2}}{2E}\)

So:

\(\begin{array}{}{u}_{r}(r,z)=\frac{-\nu \rho gzr}{E};{u}_{z}(r,z)=\frac{\rho g}{2E}({z}^{2}+\nu ({r}^{2}–{R}^{2}))\\ {\sigma }_{\mathrm{zz}}(r,z)=\rho gz\end{array}\) \(\mathrm{éq}2.1-1\)

Digital app

\(g=\mathrm{10 }N/\mathrm{kg}\)

\(\rho ={8.10}^{-\mathrm{6 }}\mathrm{kg}/{\mathrm{mm}}^{3}\)

\(R=\mathrm{20 }\mathrm{mm}\)

\(L=\mathrm{10 }\mathrm{mm}\)

\(E={2.10}^{\mathrm{5 }}N/{\mathrm{mm}}^{2}\)

\(\nu =0.3\)

\(h=\mathrm{1 }\mathrm{mm}\)

  • in inner skin: \(\{\begin{array}{}{u}_{r}(L)=-2.34000{E}^{-8}\mathrm{mm}\\ {\sigma }_{\mathrm{zz}}(L)=8.0000{E}^{-4}\mathrm{Mpa}\\ {u}_{z}(O)=-1.185000{E}^{-9}\mathrm{mm}\end{array}\)

  • in outer skin: \(\{\begin{array}{}{u}_{r}(L)=-2.46000{E}^{-8}\mathrm{mm}\\ {\sigma }_{\mathrm{zz}}(L)=8.0000{E}^{-4}\mathrm{Mpa}\\ {u}_{z}(O)=1.215000{E}^{-9}\mathrm{mm}\end{array}\)

15.2.2. Axisymmetric shell model#

A vertical pull is exerted in \(z=L\):

\(F=\rho ghL\)

Gravity leads to vertical force:

\(f=-\rho gh{e}_{z}\)

The boundary condition on the support circle is: \({u}_{z}(z)=0\) in \(z=0\)

The solution is membrane-bound, the vertical balance is written as:

\({N}_{\mathrm{zz},z}=\rho gh\)

Plus: \({N}_{\theta \theta }=0\). In elasticity, we then deduce:

\({E}_{\theta \theta }=\frac{w}{R}=\frac{-\nu {N}_{\mathrm{zz}}}{\mathrm{Eh}}=\frac{-\nu \rho gz}{E}\)

\({E}_{\mathrm{zz}}={u}_{z,z}=\frac{{N}_{\mathrm{zz}}}{\mathrm{Eh}}\Rightarrow {u}_{z}(z)=\frac{\rho g}{2E}{z}^{2}\)

The axial stress is:

\({\sigma }_{\mathrm{zz}}=\rho gz\) (constant in thickness)

Digital application:

\(F={8.10}^{-4}{\mathrm{N.mm}}^{-1}\)

\(w(L)=-{\mathrm{2.4000.10}}^{-8}\mathrm{mm}\)

\({N}_{\mathrm{zz}}(L)={\mathrm{8.0000.10}}^{-4}{\mathrm{N.mm}}^{-1}\)

\({\sigma }_{\mathrm{zz}}(L)={\mathrm{8.0000.10}}^{-4}{\mathrm{N.mm}}^{-1}\)

15.3. Thermomechanical loading#

15.3.1. 2D axisymmetric model#

\(T(r)–{T}_{\mathrm{réf}}(r)=\frac{{T}_{s}+{T}_{i}}{2}+\frac{({T}_{s}–{T}_{i})}{h}(r-R)\) \(\mathrm{éq}\mathrm{3.1.}-1\)

_images/10000A4600001F1C0000170C534B0B7595A77027.svg

Displacement is postulated in the form:

\({u}_{r}=u(r)\); \({u}_{z}={u}_{\theta }=0\)

with the appropriate boundary conditions. Thus, elastic stresses are expressed:

\({\sigma }_{\mathrm{rr}}=\frac{E}{(1+\nu )(1-2\nu )}[(1-\nu )u\text{'}+\nu \frac{u}{r}]-\frac{\alpha E}{1-2\nu }(T-{T}_{\mathrm{réf}})\)

\({\sigma }_{\theta \theta }=\frac{E}{(1+\nu )(1-2\nu )}[(1-\nu )\frac{u}{r}+\nu u\text{'}]-\frac{\alpha E}{1-2\nu }(T-{T}_{\mathrm{réf}})\) \(\mathrm{éq}1.1-3\)

\({\sigma }_{\mathrm{zz}}=\frac{\nu E}{(1+\nu )(1-2\nu )}[\frac{u}{r}+u\text{'}]-\frac{\alpha E}{1-2\nu }(T-{T}_{\mathrm{réf}})\)

The radial equilibrium equation \({(r,{\sigma }_{\mathrm{rr}})}_{,r}-{\sigma }_{\theta \theta }=0\) gives:

\((\frac{(\mathrm{ru})\text{'}}{r})\text{'}=\frac{\alpha (1+\nu \text{'})}{(1-\nu )}(T-{T}_{\mathrm{réf}})\text{'}\) \(\mathrm{éq}3.1-2\)

Hence the general solution:

\(u(r)=\frac{\alpha (1+\nu )}{(1-\nu )}\frac{({T}_{s}-{T}_{i})}{h}\frac{{r}^{2}}{3}+\mathrm{Ar}+\frac{B}{r}\) \(\mathrm{éq}3.1-3\)

The constraints are then:

\({\sigma }_{\mathrm{rr}}(r)=\frac{\alpha E({T}_{s}-{T}_{i})}{h}(\frac{R}{1-2\nu }-\frac{r}{3(1-\nu )})-\frac{\alpha E}{1-2\nu }({T}_{s}-{T}_{i})+\frac{E}{(1+\nu )(1-2\nu )}(A-(1-2\nu )\frac{B}{{r}^{2}})\)

\({\sigma }_{\theta \theta }(r)=\frac{\alpha E({T}_{s}-{T}_{i})}{h}(\frac{R}{1-2\nu }-\frac{\mathrm{2r}}{3(1-\nu )})-\frac{\alpha E}{1-2\nu }({T}_{s}-{T}_{i})+\frac{E}{(1+\nu )(1-2\nu )}(A-(1-2\nu )\frac{B}{{r}^{2}})\)

\({\sigma }_{\mathrm{zz}}(r)=\frac{\alpha E({T}_{s}-{T}_{i})}{h}(\frac{R}{1-2\nu }-\frac{r}{3(1-\nu )})-\frac{\alpha E}{1-2\nu }({T}_{s}-{T}_{i})+\frac{2\nu E}{(1+\nu )(1-2\nu )}A\) \(\mathrm{éq}3.1-4\)

The limit conditions in effort are: in \(r=R\pm \frac{h}{2},{\sigma }_{\mathrm{rr}}=0\). Note: \(x=\frac{h}{\mathrm{2R}}\). Thanks to \([\mathrm{éq}3.1-4]\), we get:

\(B=\frac{\alpha ({T}_{s}-{T}_{i})}{\mathrm{6h}(1-\nu )}(1+\nu ){R}^{3}{(1-{x}^{2})}^{2}\)

then:

\(A=\alpha (1+\nu )[\frac{-({T}_{s}-{T}_{i})R}{\mathrm{6h}(1-\nu )}(3-(1-2\nu ){x}^{2})+\frac{({T}_{s}+{T}_{i})}{2}]\)

Digital application:

\(F={8.10}^{-4}{\mathrm{N.mm}}^{-1}\)

\(w(L)=-{\mathrm{2.4000.10}}^{-8}\mathrm{mm}\)

\({N}_{\mathrm{zz}}(L)={\mathrm{8.0000.10}}^{-4}{\mathrm{N.mm}}^{-1}\)

\({\sigma }_{\mathrm{zz}}(L)={\mathrm{8.0000.10}}^{-4}{\mathrm{N.mm}}^{-1}\)

From where: \(A=-{\mathrm{0.18569881.10}}^{-3}{\mathrm{mm}}^{2}\)

\(B=0.02473096{\mathrm{mm}}^{2}\)

Note:

\(\frac{\alpha (1+\nu )}{1-\nu }\frac{{T}_{s}–{T}_{i}}{\mathrm{3h}}=0.61904762E-5\)

  • in inner skin: \(\{\begin{array}{}{u}_{r}=1.056145{E}^{-6}\mathrm{mm}\\ {\sigma }_{\mathrm{zz}}=1.4321427\mathrm{Mpa}\end{array}\)

  • in outer skin: \(\{\begin{array}{}{u}_{r}=1.110317{E}^{-6}\mathrm{mm}\\ {\sigma }_{\mathrm{zz}}=-1.4250001\mathrm{Mpa}\end{array}\)

\(\begin{array}{}{\sigma }_{\mathrm{zz}}=1.449319\mathrm{Mpa}\\ \mathrm{ou}\\ {\sigma }_{\mathrm{zz}}=1.428671\mathrm{Mpa}\text{( sans correction métrique )}\end{array}\)

In case we take \({T}_{s}={T}_{i}=0.1°C\):

\(A=\mathrm{0,00130000}{.10}^{-3}\)

\(B=\mathrm{0,0 }{\mathrm{mm}}^{2}\)

So:

  • in inner skin: \(\begin{array}{}{u}_{r}=25.350000{E}^{-6}\mathrm{mm}\\ {\sigma }_{\mathrm{zz}}=-0.200000\mathrm{Mpa}\end{array}\)

  • in outer skin: \(\begin{array}{}{u}_{r}=26.650000{E}^{-6}\mathrm{mm}\\ {\sigma }_{\mathrm{zz}}=-0.200000\mathrm{Mpa}\end{array}\)

15.3.2. Axisymmetric shell model#

For the temperature field in the thickness given by \([\mathrm{éq}3.1-1]\), the following expression for the law of behavior is obtained:

\(\begin{array}{}{N}_{\theta \theta }=\frac{\mathrm{Eh}}{1-{\nu }^{2}}({E}_{\theta \theta }+\nu {E}_{\mathrm{zz}})-\frac{\alpha Eh}{1-\nu }[\frac{{T}_{s}+{T}_{i}}{2}+\frac{{T}_{s}-{T}_{i}}{12}\frac{h}{R}]\\ {N}_{\mathrm{zz}}=\frac{\mathrm{Eh}}{1-{\nu }^{2}}(\nu {E}_{\theta \theta }+{E}_{\mathrm{zz}})-\frac{\alpha Eh}{1-\nu }[\frac{{T}_{s}+{T}_{i}}{2}+\frac{{T}_{s}-{T}_{i}}{12}\frac{h}{R}]\end{array}\) \(\mathrm{éq}3.2-1\)

and:

\(\begin{array}{}{N}_{\theta \theta }=\frac{\mathrm{Eh}}{1-{\nu }^{2}}({E}_{\theta \theta }+\nu {E}_{\mathrm{zz}})-\frac{\alpha Eh}{1-\nu }[\frac{{T}_{s}+{T}_{i}}{2}+\frac{{T}_{s}-{T}_{i}}{12}\frac{h}{R}]\\ {N}_{\mathrm{zz}}=\frac{\mathrm{Eh}}{1-{\nu }^{2}}(\nu {E}_{\theta \theta }+{E}_{\mathrm{zz}})-\frac{\alpha Eh}{1-\nu }[\frac{{T}_{s}+{T}_{i}}{2}+\frac{{T}_{s}-{T}_{i}}{12}\frac{h}{R}]\end{array}\) \(\mathrm{éq}3.2-2\)

According to these expressions, the thermal terms in

_images/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:

\({E}_{\theta \theta }=\frac{w}{R}\)

\({E}_{\mathrm{zz}}=0\)

\({K}_{\theta \theta }={K}_{\mathrm{zz}}=0\)

The normal balance in the shell is written as:

\({N}_{\theta \theta }=0\)

Hence the arrow:

\(w=\alpha (1+\nu )[\frac{{T}_{s}+{T}_{i}}{2}+\frac{{T}_{s}+{T}_{i}}{12}\frac{h}{R}]R\)

and:

\({N}_{\mathrm{zz}}=\alpha \mathrm{Eh}[\frac{{T}_{s}+{T}_{i}}{2}+\frac{{T}_{s}+{T}_{i}}{12}\frac{h}{R}]\)

\({M}_{\mathrm{zz}}=\frac{-\alpha {\mathrm{Eh}}^{2}}{12(1-\nu )}[({T}_{s}-{T}_{i})+\frac{{T}_{s}+{T}_{i}}{2}\frac{h}{R}]\)

Since the second member of dilation does not take into account the metric correction, the terms in

_images/Object_1201.svg

above are overlooked.

Digital app

\(R=\mathrm{20 }\mathrm{mm}\)

\(h=\mathrm{1 }\mathrm{mm}\)

\(\alpha ={10}^{-5}°{C}^{-1}\)

\({T}_{s}={T}_{i}=0.5°C\)

\(\nu =0.3\)

\(E={2.10}^{5}N/{\mathrm{mm}}^{2}\)

From where:

\({M}_{\mathrm{zz}}=-0.2380952N\)

  • in inner skin: \(\begin{array}{}{\sigma }_{\mathrm{zz}}=1.449319\mathrm{Mpa}\\ \mathrm{ou}\\ {\sigma }_{\mathrm{zz}}=1.428671\mathrm{Mpa}\text{( sans correction métrique )}\end{array}\)

In case we take \({T}_{s}={T}_{i}=\mathrm{0,1}°C\):

\(w={\mathrm{26.00000.10}}^{-6}\mathrm{mm}\)

\({N}_{\mathrm{zz}}=–0.2{\mathrm{N.mm}}^{-1}\)

\({M}_{\mathrm{zz}}=–0.001190476N\)

  • in inner skin: \(\begin{array}{}{\sigma }_{\mathrm{zz}}=-0.2122466\mathrm{Mpa}\\ \mathrm{ou}\\ {\sigma }_{\mathrm{zz}}=-0.200000\mathrm{Mpa}\text{( sans correction métrique )}\end{array}\)

**Thickness constraints with metric correction are given by:*

\({\sigma }_{\mathrm{zz}}({x}_{3})=\frac{{N}_{\mathrm{zz}}–\frac{{M}_{\mathrm{zz}}}{R}}{h(1-\frac{{h}^{2}}{{\mathrm{12R}}^{2}})}+({M}_{\mathrm{zz}}-{N}_{\mathrm{zz}}\frac{{h}^{2}}{\mathrm{12R}})\frac{{\mathrm{12x}}_{3}}{{h}^{3}(1-\frac{{h}^{2}}{{\mathrm{12R}}^{2}})}\)