10.7 Solutions to simple problems involving membranes, plates and shells

 

In this section, we derive solutions to several initial and boundary value problems for plates and shells to illustrate applications of the general theories derived in the preceding sections.

 

 

 

10.7.1 Thin circular plate bent by pressure applied to one face

 

A thin circular plate, with radius R and thickness h is made from a linear elastic solid with Young’s modulus $E$ and Poisson’s ratio $\nu$, as shown in the figure. It is subjected to a pressure $p=p_3{e}_3$ acting perpendicular to the plate, and is simply supported at its edge.  The solution can be derived using the simplified version of plate theory described in Section 10.6.1.  Although the plate is circular, the problem can be solved by expressing all vector and tensor quantities as components in a Cartesian basis $\{e_1,e_2,e_3\}$ shown in the figure.

 

The deflection of the plate is given by

$u_3={\displaystyle \frac{3(1-{\nu }^2)p}{16E{h}^3}\left(R^2-r^2\right)\left({\displaystyle \frac{5+\nu }{1+\nu }{R}^2-r^2}\right)}$

where $r=\sqrt{x_{\alpha }{x}_{\alpha }}$.The internal force and moment in the plate are

$\begin{array}{l}{M}_{\alpha \beta }=-{\displaystyle \frac{1}{8}(1-\nu )p{x}_{\alpha }{x}_{\beta }-{\displaystyle \frac{1}{16}p\left(r^2(1+3\nu )-R^2(3+\nu )\right){\delta }_{\alpha \beta }}}\\ V_{\beta }={\displaystyle \frac{\partial M_{\alpha \beta }}{\partial x_{\alpha }}=-p{x}_{\beta }/2}\end{array}$

 

 

Derivation:

 

1. The transverse deflection must satisfy the static equilibrium equation

${\displaystyle \frac{E{h}^3}{12(1-{\nu }^2)}{\displaystyle \frac{{\partial }^4{u}_3}{\partial x_{\alpha }\partial x_{\alpha }\partial x_{\beta }\partial x_{\beta }}=p}}$

 

2. The solution must be axially symmetric, so that $u_3=w\left(r\right)$ where $r=\sqrt{x_{\alpha }{x}_{\alpha }}$.  Substituting this expression into the equilibrium equation, and using $\partial r/\partial x_{\alpha }=x_{\alpha }/r$, reduces the governing equation to

${\displaystyle \frac{E{h}^3}{12(1-{\nu }^2)}\left({\displaystyle \frac{d^{4}w}{d{r}^4}+{\displaystyle \frac{2}{r}{\displaystyle \frac{d^{3}w}{d{r}^3}-{\displaystyle \frac{1}{r^2}{\displaystyle \frac{d^{2}w}{d{r}^2}+{\displaystyle \frac{1}{r^3}{\displaystyle \frac{dw}{dr}}}}}}}}\right)={\displaystyle \frac{E{h}^3}{12(1-{\nu }^2)}{\displaystyle \frac{1}{r}{\displaystyle \frac{d}{dr}\left(r{\displaystyle \frac{d}{dr}\left({\displaystyle \frac{1}{r}{\displaystyle \frac{d}{dr}\left(r{\displaystyle \frac{dw}{dr}}\right)}}\right)}\right)=p}}}}$

 

3. This equation can be integrated repeatedly to give

$w={\displaystyle \frac{1-{\nu }^2}{E{h}^3}\left({\displaystyle \frac{3}{16}p{r}^4+A{r}^2\log r+B\log r+C{r}^2+D}\right)}$

where A,B,C,D are constants of integration.

 

4. The curvature is related to w by

${\kappa }_{\alpha \beta }=-{\displaystyle \frac{{\partial }^2{u}_3}{\partial x_{\alpha }\partial x_{\beta }}=-\left({\displaystyle \frac{d^{2}w}{d{r}^2}-{\displaystyle \frac{1}{r}{\displaystyle \frac{dw}{dr}}}}\right){\displaystyle \frac{x_{\alpha }{x}_{\beta }}{r^2}-{\displaystyle \frac{dw}{dr}{\displaystyle \frac{{\delta }_{\alpha \beta }}{r}}}}}$

 

5. Substituting this result into the curvature-moment equations gives the following equation for the internal moment distribution

$M_{\alpha \beta }={\displaystyle \frac{-E{h}^3}{12\left(1-{\nu }^2\right)}\left((1-\nu )\left({\displaystyle \frac{d^{2}w}{d{r}^2}-{\displaystyle \frac{1}{r}{\displaystyle \frac{dw}{dr}}}}\right){\displaystyle \frac{x_{\alpha }{x}_{\beta }}{r^2}+\nu \left({\displaystyle \frac{d^{2}w}{d{r}^2}+{\displaystyle \frac{1}{r}{\displaystyle \frac{dw}{dr}}}}\right){\delta }_{\alpha \beta }}\right)}$

 

6. The displacement and curvature of the plate must be finite at r=0, which is only possible if A=B=0 To see this, note that the Blog(r) term in the formula for w is inifite at r=0; similarly, substituting the expression for w into the curvature formula produces a term involving Alog(r) which is also infinite at r=0.    The remaining constants must be determined from the boundary conditions at the edge of the plate.  For a simply supported plate, the boundary conditions are $w=0$ and $M_{\alpha \beta }{n}_{\alpha }{n}_{\beta }=M_{\alpha \beta }{x}_{\alpha }{x}_{\beta }/r^2=0$ on r=R, where $n_{\alpha }=x_{\alpha }/r$ is a unit vector normal to the edge of the plate.  This yields two equations that can be solved for C and D.  Substituting the results back into the formulas in (3) and (5) then gives the solution.

 

 

 

10.7.2 Vibration modes and natural frequencies for a circular membrane

 

A thin circular membrane with thickness h, radius R and mass density $\rho$ is subjected to a uniform radial force per unit length $T_0$  acting on its edge, as shown in the figure   Our goal is to calculate the mode shapes and natural frequencies of vibration of the membrane (the physical significance of natural frequencies and mode shapes for a continuous system are described in more detail in Section 10.4.1)

 

The natural frequencies of vibration and mode shapes are identified by two integers (m,n) that characterize the mode shape.  The index m=1,2,3… corresponds to the number of circumferential lines (with r=constant) on the membrane that have zero displacement, while n=0,1,2… corresponds to the number of diametral lines (with $\theta =$ constant) that have zero displacement.

 

The natural frequencies of vibration are given by the solutions to the equation $J_n({\omega }_{(m,n)}R\sqrt{\rho h/T_0})=0$,  where $J_n(x)$ is a Bessel function of the first kind of order n  (This sounds scary, but symbolic manipulation programs have predefined functions that compute roots of Bessel functions).  A few zeros for Bessel functions of order n=0,1,2 are listed in Table 10.3.

 

The first few natural frequencies are ${\omega }_{(1,0)}=2.4048\sqrt{T_0/(\rho h{R}^2)}$, ${\omega }_{(1,1)}=3.8317\sqrt{T_0/(\rho h{R}^2)}$ ${\omega }_{(1,2)}=5.1356\sqrt{T_0/(\rho h{R}^2)}$, and so on.

 

 

The mode shapes are $U_{(m,n)}(r,\theta )=A{J}_n(r{\omega }_{(m,n)}\sqrt{\rho h/T_0})\sin (n\theta +{\theta }_0)$

 

 

Contour plots of the displacements for the first few vibration modes are shown in the figure.

 

 

Derivation:

 

1. The equation that governs transverse motion of a membrane under equibiaxial tension was derived in Section 10.6.5 as

$T_0\left({\displaystyle \frac{{\partial }^2{u}_3}{\partial r^2}+{\displaystyle \frac{1}{r}{\displaystyle \frac{\partial u_3}{\partial r}+{\displaystyle \frac{1}{r^2}{\displaystyle \frac{{\partial }^2{u}_3}{\partial {\theta }^2}}}}}}\right)=\rho h{\displaystyle \frac{{\partial }^2{u}_3}{\partial t^2}}$

 

2. The general solution to this equation (which can be found by separation of variables, or if you are lazy, using a symbolic manipulation program) is

$u_3(r,\theta )=\left(A{J}_n(k_{(m,n)}r)\sin (n\theta +{\theta }_0)+B{Y}_n(k_{(m,n)}r)\sin (n\theta +{\theta }_1)\right)\cos ({\omega }_{(m,n)}t+\varphi )$

where $k_{(m,n)}={\omega }_{(m,n)}\sqrt{\rho h/T_0}$, A, B, n, ${\theta }_0,{\theta }_1,\varphi$ are arbitrary constants, and $J_n,Y_n$ are Bessel functions of the first and second kinds, with order n, respectively. 

 

3. The solution must satisfy $u_3(\theta )=u_3(2\pi +\theta )$, which is only possible if n is an integer.

 

4. The Bessel function of the second kind is infinite at r=0, so B=0.

 

5. The transverse displacement must satisfy the boundary condition $u_3=0$ on the edge of the membrane, which leads to the condition $J_n({\omega }_{(m,n)}R\sqrt{\rho h/T_0})=0$.

 

 

 

10.7.3 Estimate for the fundamental frequency of vibration of a simply supported rectangular flat plate

 

The figure shows an initially flat plate, which lies in the $e_1,e_2$ plane and is free of external force.  The plate has Young’s modulus $E$, mass density $\rho$, and thickness h.  Its edges are simply supported.  We wish to calculate the lowest natural frequency of vibration for the plate.

 

The exact natural frequencies and modes of vibration for a rectangular plate are best calculated using a numerical method (e.g. finite elements).   However, it is very straightforward to estimate the lowest natural frequency of vibration using the Rayleigh-Ritz method described in Section 5.9.

 

The Rayleigh-Ritz method proceeds as follows:

 

1. Select a suitable estimate for the lowest frequency mode of vibration. $–$ which must satisfy all displacement boundary conditions.  For present purposes, the following mode shape is reasonable

${\widehat{U}}_3(x_1,x_2)=(x_1-a)(x_1+a)(x_2-b)(x_2+b)+C{(x_1-a)}^2{(x_1+a)}^2{(x_2-b)}^2{(x_2+b)}^2$

where C is a parameter that can be adjusted to obtain the best estimate for the natural frequency.  More terms could be added to obtain a more accurate solution.

 

2. Calculate the kinetic energy measures

$\widehat{V}={\displaystyle \frac{E{h}^3}{24(1-{\nu }^2)}{\displaystyle \underset{A}{\int }\left((1-\nu ){\displaystyle \frac{{\partial }^2{U}_3}{\partial x_{\alpha }\partial x_{\beta }}{\displaystyle \frac{{\partial }^2{U}_3}{\partial x_{\alpha }\partial x_{\beta }}+\nu {\left({\displaystyle \frac{{\partial }^2{U}_3}{\partial x_{\alpha }\partial x_{\alpha }}}\right)}^2}}\right)}dA\widehat{T}={\displaystyle \underset{A}{\int }\left({\displaystyle \frac{h}{2}\rho U_3{U}_3}\right)}dA}$

 

3. The frequency is estimated as ${\omega }^2\le \widehat{V}/\widehat{T}$ - we therefore need to choose C to minimize $\widehat{V}/\widehat{T}$.  Although an exact formula can be calculated for the resulting upper bound to the natural frequency, the expression is very long, and is best displayed graphically.  The figure shows the variation of normalized natural frequency as a function of the aspect ratio of the plate $b/a$.  As a guide to the accuracy of the solution, an exact solution can be calculated for the natural frequency in the limit $b/a\to \infty$ (in this limit the plate is a beam), following the procedure described in Section 10.4.1.  The result is $12(1-{\nu }^2){\omega }^2\rho a^4/E{h}^2={(\pi /2)}^4$.  It is clear that the Rayleigh-Ritz method gives an excellent estimate of the natural frequency in this limit.

 

 

 

10.7.4 Bending induced by inelastic strain in a thin film on a substrate  

 

The figure illustrates the problem to be solved.   A thin film, with Young’s modulus $E_f$, Poisson’s ratio ${\nu }_f$ and thickness $h_f$ is deposited onto the surface of an initially flat, circular wafer, which has Young’s modulus $E_s$, Poisson’s ratio ${\nu }_s$, radius $R$, and thickness $h_s$.   An inelastic strain ${\epsilon }_{11}^p={\epsilon }_{22}^p={\epsilon }_0$ is introduced into the film by some external process, which generates stresses in the film, and also causes the substrate to bend.  Provided the inelastic strain is not too large, the plate adopts a state of uniform curvature $\kappa$ (its deformed shape can be visualized as a spherical cap, with large radius of curvature $1/\kappa$ ).  Our goal is to relate the curvature $\kappa$ to the inelastic strain ${\epsilon }_0$, and to calculate the stress in the film.  The results are important because stresses in thin films are often determined by measuring the curvature of the substrate. 

 

The inelastic strain may be caused by a number of different processes, including

 

1. A mismatch in thermal expansion between the film and the substrate.  In this case the inelastic strain is related to the thermal expansion coefficients ${\alpha }_f$, ${\alpha }_s$ of the film and substrate and the temperature T by ${\epsilon }_0=({\alpha }_f-{\alpha }_s)(T-T_0)$, where $T_0$ is the temperature at which the system is stress free (many films are approximately free of stress at deposition temperature)

 

2. The film may grow epitaxially on the substrate, so that the spacing between atoms in the film is forced to match that of the substrate.  In this case the inelastic strain can be calculated as follows.  Suppose that, in their stress free states, the film and substrate have lattice spacing  $a_f$ and $a_s$, respectively.  Then ${\epsilon }_0=(a_f-a_s)/a_s$.

 

3. Mismatch strain may develop in the film as a result of the deposition process. 

 

4. Mismatch strain may be developed as a result of interdiffusion and possibly chemical reactions between the film and substrate.

 

 

Solution: The inelastic strain in the film is related to the curvature of the substrate by

${\epsilon }_0={\displaystyle \frac{\kappa h_s(1+{\rho }^4{\eta }^2+4\rho \eta +4\rho {\eta }^3+6{\rho }^2\eta )}{6\rho \eta (1+\rho )}}$

where $\rho =h_f/h_s$, and $\eta =E_f(1-{\nu }_s)/(E_s(1-{\nu }_f))$.  Note that with the sign convention adopted here, the substrate has a positive curvature if the film is on the convex side of the bent plate.

 

The stress in the film is related to the curvature by

${\sigma }_{11}={\sigma }_{22}={\displaystyle \frac{-\kappa E_s}{(1-{\nu }_s)}{\displaystyle \frac{[(1+{\rho }^3\eta )h_s-6\eta \rho z(1+\rho )]}{6\rho (1+\rho )}}}$

where $z$ is the distance above the mid-plane of the film.

 

In most practical situations the thickness of the substrate greatly exceeds the thickness of the film, in which case these results can be approximated by

${\epsilon }_0\approx {\displaystyle \frac{\kappa h_s^2{E}_s(1-{\nu }_f)}{6h_f{E}_f(1-{\nu }_s)}{\sigma }_{11}={\sigma }_{22}\approx {\displaystyle \frac{-\kappa h_s^2{E}_s}{6h_f(1-{\nu }_s)}}}$

These are known as the Stoney equations.

 

Derivation:  It is simplest to derive these results by using the general equations of shell theory to write down the potential energy of the bent plate, and then calculating the values of mid-plane strain and curvature that minimize the potential energy. To this end:

 

1. We consider the plate to consist of the film and substrate together, with combined thickness $h=h_f+h_s$.  The mid-plane of the plate is at height $h/2$ above the base of the substrate.

 

2. We assume that the deformed plate has a small, uniform curvature ${\kappa }_{11}={\kappa }_{22}=\kappa {\kappa }_{12}=0$, and mid-plane strain ${\gamma }_{11}={\gamma }_{22}=\gamma {\gamma }_{12}=0$.  As long as the curvature of the plate is small, the in-plane strain is a function only of the in-plane displacement components of the plate, while the curvature is a function only of the out-of-plane displacement (see Sect 10.6.2).  This means that $\gamma$ and $\kappa$ can be taken as independent variables that describe the deformed shape of the plate.

 

3. The total strain in the substrate follows as ${\epsilon }_{11}={\epsilon }_{22}=\gamma +x_3\kappa$, where $x_3$ is the distance from the mid-plane of the plate. 

 

4. The stress in the substrate is proportional to the total strain.  We assume that the plate is in a state of plane stress, so that the stress components in the substrate are

${\sigma }_{11}={\sigma }_{22}={\displaystyle \frac{E_s}{\left(1-{\nu }_s\right)}\left(\gamma +x_3\kappa \right)}$

The strain energy density in the substrate can then be calculated as

$U_s={\displaystyle \frac{1}{2}{\sigma }_{ij}{\epsilon }_{ij}={\displaystyle \frac{E_s}{\left(1-{\nu }_s\right)}{\left(\gamma +x_3\kappa \right)}^2}}$

 

5. In the film, the total strain includes contributions from an elastic distorsion of the lattice, together with the inelastic strain, so ${\epsilon }_{ij}={\epsilon }_{ij}^e+{\epsilon }_{ij}^p$.  The stress in the film is proportional to the elastic strain ${\epsilon }_{ij}^e={\epsilon }_{ij}-{\epsilon }_{ij}^p$. The nonzero components of elastic strain follow as ${\epsilon }_{11}^e={\epsilon }_{22}^e=\gamma +x_3\kappa -{\epsilon }_0$ and the stress in the film is

${\sigma }_{11}={\sigma }_{22}={\displaystyle \frac{E_f}{\left(1-{\nu }_f\right)}\left(\gamma +x_3\kappa -{\epsilon }_0\right)}$.

The strain energy density in the film is

$U_f={\displaystyle \frac{1}{2}{\sigma }_{ij}{\epsilon }_{ij}^e={\displaystyle \frac{E_f}{\left(1-{\nu }_f\right)}{\left(\gamma +x_3\kappa -{\epsilon }_0\right)}^2}}$

 

6. The total potential energy of the system is the integral of the strain energy density

$V={\displaystyle \underset{0}{\overset{R}{\int }}{\displaystyle \underset{-h/2}{\overset{h/2-h_f}{\int }}2\pi r{U}_{s}d{x}_{3}dr}+}{\displaystyle \underset{0}{\overset{R}{\int }}{\displaystyle \underset{h/2-h_f}{\overset{h/2}{\int }}2\pi r{U}_{f}d{x}_{3}dr}}$

The resulting expression is lengthy and will not be written out here.

 

7. Finally, the equilibrium values of $\gamma$ and $\kappa$ can be calculated from the condition that the potential energy must be a minimum, which requires that

${\displaystyle \frac{\partial V}{\partial \kappa }=0{\displaystyle \frac{\partial V}{\partial \gamma }=0}}$

Solving the resulting linear equations for $\gamma$ and ${\epsilon }_0$ in terms of $\kappa$ gives the formula relating ${\epsilon }_0$ and $\kappa$; substituting the results into the formula for stress in (5) and setting $x_3=(h-h_f)/2+z$ gives the formula for stress. 

 

 

 

10.7.5 Bending of a circular plate caused by a through-thickness temperature gradient

 

The figure illustrates the problem to be solved.  An initially flat, circular plate, which lies in the $e_1,e_2$ plane is free of external force.  The plate has Young’s modulus $E$, thermal expansion coefficient $\alpha$, radius R and thickness h.  Its edges are free.  The plate is heated on one face, and cooled on the other, so as to establish a temperature distribution $T=T_0+\Delta T{x}_3/h$ through the thickness of the plate.  Here, $T_0$ is the temperature of the mid-plane of the plate, while $\Delta T$ is the drop in temperature across the plate.  The thermal expansion of the plate causes it to bend: our objective is to estimate the curvature of the plate as a function of the temperature gradient $\Delta T/h$.  The solution will account for large out-of-plane deflections, and will predict that the plate buckles when the temperature gradient reaches a critical value.

 

We will derive an approximate solution, by assuming that the curvature of the plate is uniform.  Since we are interested in calculating the plate’s shape after buckling, the solution is obtained by means of the Von-Karman theory described in Section 10.6.3. We denote the two principal curvatures of the deformed plate by ${\kappa }_1,{\kappa }_2$.   There are three possible equilibrium configurations, as follows

 

1. For temperature gradients satisfying

${\displaystyle \frac{\alpha \Delta T{R}^2{(1+\nu )}^{3/2}}{4h^2}<2}$

the plate bends into a spherical cap shape, with two equal principal curvatures.  The curvatures are related to the temperature gradient by

${\displaystyle \frac{\alpha \Delta T{R}^2{(1+\nu )}^{3/2}}{4h^2}=\overline{\kappa }\left({\overline{\kappa }}^2(1-\nu )+(1+\nu )\right){\kappa }_1={\kappa }_2={\displaystyle \frac{4h\overline{\kappa }}{R^2\sqrt{1+\nu }}}}$

 

2. For temperature gradients

${\displaystyle \frac{\alpha \Delta T{R}^2{(1+\nu )}^{3/2}}{4h^2}>2}$

the solution (1) is still a possible equilibrium configuration, but is unstable.  There are infinitely many additional stable configurations, which have two unequal principal curvatures.  One of these solutions can be related to the temperature gradient by

${\displaystyle \frac{\alpha \Delta T{R}^2{(1+\nu )}^{3/2}}{4h^2}={\displaystyle \frac{{\widehat{\kappa }}^2+1}{\widehat{\kappa }}{\kappa }_1={\displaystyle \frac{4h}{\widehat{\kappa }{R}^2\sqrt{1+\nu }}{\kappa }_2={\displaystyle \frac{4h\widehat{\kappa }}{R^2\sqrt{1+\nu }}}}}}$

The other solutions have the same principal curvatures, but the principal directions are different.

 

 

These results are displayed by plotting the normalized curvature ${\kappa }_1{R}^2\sqrt{1+\nu }/(4h)$ as a function of the dimensionless temperature gradient $\alpha \Delta T{R}^2{(1+\nu )}^{3/2}/(4h^2)$ in the figure, for a Poisson’s ratio $\nu =0.3$ (the graph is virtually identical for other values of $\nu$ ). To visualize the significance of the graph, suppose that the temperature drop across the plate is gradually increased from zero.  The plate will first deform with two equal principal curvatures, which are related to the temperature by the formula given in (1).  At the critical temperature, the plate will buckle, and assume one of the two possible equilibrium configurations, with two unequal principal curvatures.

 

 

Derivation: The solution is derived by approximating the shape of the plate, and selecting the deformed shape that minimizes the potential energy. 

 

1. The displacement of the mid-plane of the plate will be approximated as

$u_3={\kappa }_1{x}_1^2/2+{\kappa }_2{x}_2^2/2u_1=A_1{x}_1+A_2{x}_1^3+A_3{x}_1{x}_2^2{u}_2=B_1{x}_2+B_2{x}_2^3+B_3{x}_2{x}_1^2$

where ${\kappa }_1$ and ${\kappa }_2$ are the two principal curvatures of the plate, and $A_i,B_i$ are six adjustable parameters that must be selected to minimize the potential energy of the plate.

 

2. The total strain in the plate must be calculated using the Von-Karman formulas in Section 10.6.3, which yield

${\epsilon }_{\alpha \beta }={\displaystyle \frac{1}{2}\left({\displaystyle \frac{\partial u_{\alpha }}{\partial x_{\beta }}+{\displaystyle \frac{\partial u_{\beta }}{\partial x_{\alpha }}+{\displaystyle \frac{\partial u_3}{\partial x_{\alpha }}{\displaystyle \frac{\partial u_3}{\partial x_{\beta }}}}}}\right)+x_3{\kappa }_{\alpha \beta }\Rightarrow \left\{\begin{array}{l}{\epsilon }_{11}=A_1+3A_2{x}_1^2+A_3{x}_2^2+{\kappa }_1^2{x}_1^2/2+x_3{\kappa }_1\\ {\epsilon }_{22}=B_1+3B_2{x}_2^2+B_3{x}_1^2+{\kappa }_2^2{x}_2^2/2+x_3{\kappa }_2\\ {\epsilon }_{12}=(A_3+B_3)x_1{x}_2+{\kappa }_1{\kappa }_2{x}_1{x}_2/2\end{array}\right.}$

 

3. The plate is assumed to be in a state of plane stress: the stress components can be calculated using the plane stress version of the linear elastic constitutive equations

${\sigma }_{\alpha \beta }={\displaystyle \frac{E}{1+\nu }\left\{{\epsilon }_{\alpha \beta }+{\displaystyle \frac{\nu }{1-\nu }{\epsilon }_{\gamma \gamma }{\delta }_{\alpha \beta }}\right\}-{\displaystyle \frac{E\alpha (T_0+\Delta T{x}_3/h)}{1-\nu }{\delta }_{\alpha \beta }}}$

 

4. The strain energy density in the plate can be calculated using the formulas given in Section 3.1.7 as $U=((1+\nu ){\sigma }_{\alpha \beta }{\sigma }_{\alpha \beta }-\nu {\sigma }_{\gamma \gamma }{\sigma }_{\alpha \alpha })/2E$.  The result is lengthy and is best calculated using a symbolic manipulation program.

 

5. The total strain energy of the plate follows by integrating the strain energy density over the volume of the plate as

$\Phi ={\displaystyle \underset{0}{\overset{R}{\int }}{\displaystyle \underset{0}{\overset{2\pi }{\int }}{\displaystyle \underset{-h/2}{\overset{h/2}{\int }}U(r,\theta ,x_3)d{x}_3}d\theta }r}dr$.

 To evaluate the integral, the strain energy density can be expressed in polar coordinates by substituting $x_1=r\cos \theta$, $x_2=r\sin \theta$ into the results of (4).  Again, a symbolic manipulation program makes the algebra painless.

 

6. The coefficients $A_i,B_i$ and the curvatures must now be determined by minimizing the potential energy $\Phi$.  To proceed, we first calculate the coefficients $A_i,B_i$ in terms of the temperature gradient and curvature by solving the six simultaneous equations $\partial \Phi /\partial A_i=0\partial \Phi /\partial B_i=0$.  Substituting the resulting formulas for $A_i,B_i$ back into the results of (5), and using the two remaining conditions $\partial \Phi /\partial {\kappa }_1=0\partial \Phi /\partial {\kappa }_2=0$ yields two equations for ${\kappa }_1$ and ${\kappa }_2$

$\begin{array}{l}{\kappa }_1{\kappa }_2^2(1-{\nu }^2)R^4+16h^2{\kappa }_1+16h^2\nu {\kappa }_2-16(1+\nu )\alpha \Delta Th=0\\ {\kappa }_2{\kappa }_1^2(1-{\nu }^2)R^4+16h^2{\kappa }_2+16h^2\nu {\kappa }_1-16(1+\nu )\alpha \Delta Th=0\end{array}$

 

7. Eliminating the temperature from these equations and simplifying the result gives the expression

$({\kappa }_1-{\kappa }_2)({\kappa }_1{\kappa }_2{R}^4(1+\nu )-16h^2)=0$

This shows that there are two possible equilibrium configurations: in the first, ${\kappa }_1={\kappa }_2$; in the second, the two curvatures are related by ${\kappa }_1{R}^4(1+\nu )=16h^2/{\kappa }_2$.  Finally, these two possible relationships can be substituted back into either of the two equations in (6) to relate the temperature gradient to the curvatures.

 

 

 

10.7.6 Buckling of a cylindrical shell subjected to axial loading

 

The figure illustrates a thin-walled cylinder, with radius a, height L and wall thickness h.  The shell is made from a linear elastic solid with Young’s modulus $E$ and Poisson’s ratio $\nu$.  It is loaded in compression by subjecting its ends to a prescribed axial displacement $\Delta$.   We wish to estimate the critical axial strain $\Delta /L$ or axial force $P$ that will cause the cylinder to buckle.

 

We will derive an approximate solution, by assuming a shape for the buckled cylinder, and calculating the shape that minimizes the potential energy of the system.   Specifically, we assume that the radial displacement of the surface of the cylinder at the instant of buckling has the form $u_r=C+A\sin (\lambda \pi z/L)$.  The solution shows that:

 

1. Buckling occurs at a critical axial strain $\Delta /L=h/\left(\sqrt{3}a(1-{\nu }^2)\right)$

 

2. The corresponding axial load is $P=2\pi E{h}^2/\left((1-{\nu }^2)\sqrt{3}\right)$

 

3. The wavelength of the buckling mode is $L/\lambda =\pi \sqrt{ah}/{(12)}^{1/4}$.  Note that the buckling mode describes the shape of the cylinder at the instant when buckling begins; it does not correspond to the shape of the cylinder after buckling.

 

The exact buckling strain and load are a factor $\sqrt{1-{\nu }^2}$ smaller than this approximate result.

 

 

Derivation: We must calculate, and then minimize, the potential energy of the cylinder.  It is convenient to work through this problem using curvilinear coordinates: we shall use cylindrical-polar coordinates  ${\xi }_1\equiv z,{\xi }_2\equiv \theta$ to identify a point on the mid-plane of the shell.   

 

1. The position vector of a point in the undeformed shell is $\overline{r}=a\cos \theta e_1+a\sin \theta e_2+z{e}_3$

 

2. The natural basis vectors for the undeformed shell are therefore

${\overline{m}}_1={\displaystyle \frac{\partial \overline{r}}{\partial z}=e_3{\overline{m}}_2={\displaystyle \frac{\partial \overline{r}}{\partial \theta }=-a\sin \theta e_1+a\cos \theta e_2{\overline{m}}_3=-(\cos \theta e_1+\sin \theta e_2)}}$

and the reciprocal basis vectors are ${\overline{m}}^1=e_3{\overline{m}}^2=(-\sin \theta e_1+\cos \theta e_2)/a{\overline{m}}^3={\overline{m}}_3$

 

3. The components of the metric tensor are

${\overline{g}}^{\alpha \beta }={\overline{m}}^{\alpha }\cdot {\overline{m}}^{\beta }\Rightarrow {\overline{g}}^{11}=1,{\overline{g}}^{12}=0,{\overline{g}}^{22}=1/a^2$

${\overline{g}}_{\alpha \beta }={\overline{m}}_{\alpha }\cdot {\overline{m}}_{\beta }\Rightarrow {\overline{g}}_{11}=1{\overline{g}}_{12}=0{\overline{g}}_{22}=a^2$

 

4. The covariant components of the curvature tensor for the undeformed shell are

${\overline{\kappa }}_{\alpha \beta }=-{\overline{m}}_3\cdot {\displaystyle \frac{\partial {\overline{m}}_{\alpha }}{\partial {\xi }_{\beta }}\Rightarrow {\overline{\kappa }}_{11}={\overline{\kappa }}_{12}=0{\overline{\kappa }}_{22}=-a}$

 

5. The position vector of the mid-plane of the deformed shell is approximated as

$r=\left(a+C+A\sin \lambda \pi z/L\right)\left(\cos \theta e_1+\sin \theta e_2\right)+(1-\Delta /L)z{e}_3$

It will greatly simplify subsequent calculations to assume a priori that $\lambda >>1$

 

6. The natural basis vectors for the deformed shell are therefore

$\begin{array}{l}{m}_1={\displaystyle \frac{\partial r}{\partial z}=\left(A\lambda \pi /L\right)\sin \lambda \pi z/L\left(\cos \theta e_1+\sin \theta e_2\right)+(1-\Delta /L)e_3}\\ m_2={\displaystyle \frac{\partial r}{\partial \theta }=\left(a+C+A\sin \lambda \pi z/L\right)\left(-\sin \theta e_1+\cos \theta e_2\right)}\\ m_3\approx \left(A\lambda \pi /L\right)\cos (\lambda \pi z/L)e_3-\left(\cos \theta e_1+\sin \theta e_2\right)\end{array}$

 

7. The covariant components of the metric tensor and curvature for the deformed shell follow as

$\begin{array}{l}{g}_{\alpha \beta }=m_{\alpha }\cdot m_{\beta }\Rightarrow g_{11}=\left(\lambda \pi A/L\right){}^2{\cos }^2(\lambda \pi z/L)+{\left(1-\Delta /L\right)}^2\\ g_{12}=0g_{22}={\left(a+C+A\sin (\lambda \pi z/L)\right)}^2\end{array}$

$\begin{array}{l}{\kappa }_{\alpha \beta }=-m_3\cdot {\displaystyle \frac{\partial m_{\alpha }}{\partial {\xi }_{\beta }}\Rightarrow {\kappa }_{11}\approx {\left(\pi \lambda /L\right)}^{2}A\sin (\lambda \pi z/L)}\\ {\kappa }_{12}=0{\kappa }_{22}\approx -(a+C+A\sin (\lambda \pi z/L))\end{array}$

 

8. The in-plane strain tensor and curvature change tensor may be approximated as

$\begin{array}{l}{\gamma }_{\alpha \beta }=(g_{\alpha \beta }-{\overline{g}}_{\alpha \beta })/2\Rightarrow {\gamma }_{11}=-\Delta /L+{\left(\lambda \pi A/L\right)}^2{\cos }^2(\pi \lambda z/L)/2\\ {\gamma }_{12}=0{\gamma }_{22}\approx aC+Aa\cos (\pi \lambda z/L)\end{array}$

$\Delta {\kappa }_{\alpha \beta }={\kappa }_{\alpha \beta }-{\overline{\kappa }}_{\alpha \lambda }{\overline{g}}^{\lambda \mu }{g}_{\mu \beta }\Rightarrow \Delta {\kappa }_{11}\approx {(\lambda \pi /L)}^{2}A\sin (\lambda \pi z/L)\Delta {\kappa }_{12}=0\Delta {\kappa }_{22}\approx 0$

 

9. The strain energy of the deformed shell can be calculated from

$\Phi =\left({\displaystyle \underset{A}{\int }{\displaystyle \frac{h}{2}{D}^{\alpha \beta \rho \mu }{\gamma }_{\alpha \beta }{\gamma }_{\rho \mu }+}}{\displaystyle \frac{h^3}{24}{D}^{\alpha \beta \rho \mu }\Delta {\kappa }_{\alpha \beta }\Delta {\kappa }_{\rho \mu }}\right)dA$ 

with

 $D^{\alpha \beta \rho \mu }={\displaystyle \frac{E}{2(1-{\nu }^2)}\left(({\overline{g}}^{\alpha \rho }{\overline{g}}^{\beta \mu }+{\overline{g}}^{\alpha \mu }{\overline{g}}^{\beta \rho })(1-\nu )+2\nu {\overline{g}}^{\alpha \beta }{\overline{g}}^{\rho \mu }\right)}$

Substituting for ${\overline{g}}^{\alpha \beta }$ and noting that ${\gamma }_{12}=\Delta {\kappa }_{12}=0$ reduces this result to

$\Phi ={\displaystyle \frac{\pi ahE}{(1-{\nu }^2)}{\displaystyle \underset{0}{\overset{L}{\int }}\left({\gamma }_{11}^2+{\gamma }_{22}^2/a^4+2\nu {\gamma }_{11}{\gamma }_{22}/a^2\right)+{\displaystyle \frac{h^2}{12}\left(\Delta {\kappa }_{11}^2+\Delta {\kappa }_{22}^2/a^4+2\nu \Delta {\kappa }_{11}\Delta {\kappa }_{22}/a^2\right)}}}$

 

10. The potential energy can be evaluated exactly, but the resulting expression is too long to write out in full.  To proceed, we eliminate C by finding the value of C that minimizes the potential energy (set $\partial \Phi /\partial C=0$ and solve for C, then substitute the result back into $\Phi$ ).  For $\lambda >>1$ the resulting expression for  $\Phi$ may be simplified to

$\begin{array}{l}\Phi \approx E\pi ahL{\left({\displaystyle \frac{\Delta }{L}}\right)}^2+{\displaystyle \frac{{\pi }^{2}Eha\lambda A^2}{4(1-{\nu }^2)L}\left({\varphi }_1-{\varphi }_2\left({\displaystyle \frac{\Delta }{L}}\right)\right)+O(A^4)}\\ {\varphi }_1={\displaystyle \frac{{\pi }^3{\lambda }^3{h}^2}{6L^2}+{\displaystyle \frac{2L^2}{\pi \lambda a^2}{\varphi }_2=(1-{\nu }^2)2\pi \lambda }}\end{array}$

 

11. The buckling load can now be deduced from this result.  Note that both ${\varphi }_1>0,{\varphi }_2>0$, so the potential energy is minimized with $A=0$ (no buckling) if $\Delta /L<{\varphi }_1/{\varphi }_2$, and with $A>0$ if $\Delta /L>{\varphi }_1/{\varphi }_2$.  The critical axial strain for which buckling is first possible corresponds to the minimum value of ${\varphi }_1/{\varphi }_2$ with respect to $\lambda$.  The minimum occurs for $\lambda ={(12)}^{1/4}L/(\pi \sqrt{ah})$, which gives a critical axial strain of $\Delta /L=h/\left(a(1-{\nu }^2)\sqrt{3}\right)$.

 

12. The axial force at buckling can be computed trivially by noting that the cylinder is in a state of uniaxial axial stress.  The axial force is therefore $2\pi ahE\Delta /L=2\pi E{h}^2/\left((1-{\nu }^2)\sqrt{3}\right)$

 

 

 

10.7.7 Torsion of an open-walled circular cylinder

 

The figure shows a thin-walled tube, with radius $a$ and wall-thickness h, which has been slit along a line parallel to its axis.  The tube is made from a linear elastic solid with Young’s modulus $E$ and Poisson’s ratio $\nu$, and is subjected to a twisting moment $\Lambda =\Lambda e_z$ parallel to its axis.  The moment causes the end of the tube at $z=L$ to twist through an angle $\varphi$ relative to the end at $z=0$.

 

The displacement field and the internal forces in the shell can be expressed as components in a cylindrical-polar basis $\{e_r,e_{\theta },e_z\}$ as follows

$\begin{array}{l}u={\displaystyle \frac{3\Lambda (1+\nu )}{\pi h^{3}aE}\left(a\theta e_z+z{e}_{\theta }\right)}\\ T={\displaystyle \frac{\Lambda }{4\pi a^2}{e}_z\otimes e_{\theta }M={\displaystyle \frac{\Lambda }{4\pi a}\left(e_z\otimes e_{\theta }+e_{\theta }\otimes e_z\right)}}\end{array}$

Note that this is one of the rare shell geometries for which the internal force tensor T is not symmetric.

 

 

Derivation: We choose the cylindrical-polar coordinates  ${\xi }_1\equiv z,{\xi }_2\equiv \theta$ as our coordinate system.  

 

1. The position vector of a point in the undeformed shell is $\overline{r}=a\cos \theta e_1+a\sin \theta e_2+z{e}_3$

 

2. The natural basis vectors for the undeformed shell are therefore

${\overline{m}}_1={\displaystyle \frac{\partial \overline{r}}{\partial z}=e_3{\overline{m}}_2={\displaystyle \frac{\partial \overline{r}}{\partial \theta }=-a\sin \theta e_1+a\cos \theta e_2{\overline{m}}_3=-(\cos \theta e_1+\sin \theta e_2)}}$

and the reciprocal basis vectors are ${\overline{m}}^1=e_3{\overline{m}}^2=(-\sin \theta e_1+\cos \theta e_2)/a{\overline{m}}^3={\overline{m}}_3$

 

3. The components of the metric tensor are

${\overline{g}}^{\alpha \beta }={\overline{m}}^{\alpha }\cdot {\overline{m}}^{\beta }\Rightarrow {\overline{g}}^{11}=1,{\overline{g}}^{12}=0,{\overline{g}}^{22}=1/a^2$

${\overline{g}}_{\alpha \beta }={\overline{m}}_{\alpha }\cdot {\overline{m}}_{\beta }\Rightarrow {\overline{g}}_{11}=1{\overline{g}}_{12}=0{\overline{g}}_{22}=a^2$

 

4. The covariant components of the curvature tensor for the undeformed shell are

$\begin{array}{l}{\overline{\kappa }}_{\alpha \beta }=-{\overline{m}}_3\cdot {\displaystyle \frac{\partial {\overline{m}}_{\alpha }}{\partial {\xi }_{\beta }}\Rightarrow {\overline{\kappa }}_{11}={\overline{\kappa }}_{12}=0{\overline{\kappa }}_{22}=-a}\\ {\overline{\kappa }}_{\beta }^{\alpha }={\overline{\kappa }}_{\beta \lambda }{\overline{g}}^{\alpha \lambda }\Rightarrow {\overline{\kappa }}_1^1={\overline{\kappa }}_2^1={\overline{\kappa }}_1^2=0{\overline{\kappa }}_2^2=-1/a\end{array}$

 

5. The Christoffel symbols for the undeformed shell are ${\overline{\Gamma }}_{\beta \alpha }^{\lambda }=0$

 

6. To proceed, we assume that the internal stresses and moments in the cylinder are uniform; in addition, we assume small strains, so that the geometric terms in the equilibrium equations can be approximated using the geometry of the undeformed shell.  The equilibrium equations therefore reduce to

$V^2=0T^{22}a=0$         $-V^{\beta }=0T^{12}-T^{21}+M^{21}/a=0$

 

7. The boundary conditions on $\theta =0,\theta =2\pi$ are  $T^{21}=0T^{22}-M^{22}/a=0M^{22}=0$

 

8. The boundary conditions on $z=0,z=L$ are $T^{11}=0T^{12}-M^{12}/a=P^2{M}^{11}=0$

 

9. The only nonzero components of internal force are the in-plane shear forces $T^{12},T^{21}$ and the twisting moments $M^{12},M^{21}$.  We therefore assume that the shell deforms in shear, so that the position vector of a point in the deformed shell is

$\overline{r}=a\cos \theta e_1+a\sin \theta e_2+z{e}_3+Ca\theta e_3+(az\varphi /L)(-\sin \theta e_1+\cos \theta e_2)$

 

10. The natural basis vectors for the deformed shell are

$\begin{array}{l}{m}_1=e_3+(a\varphi /L)(-\sin \theta e_1+\cos \theta e_2)\\ m_2=-a\sin \theta e_1+a\cos \theta e_2+Ca{e}_3-(z\varphi a/L)(\cos \theta e_1+\sin \theta e_2)\\ m_3\approx -\left(\cos \theta e_1+\sin \theta e_2\right)+(z\varphi /L)\left(\sin \theta e_1-\cos \theta e_2\right)\end{array}$

11. The metric tensor for the deformed shell can be approximated by

$g_{\alpha \beta }=m_{\alpha }\cdot m_{\beta }\Rightarrow g_{11}\approx 1g_{12}=Ca-(a^2\varphi /L)g_{22}\approx a^2$

 

12. The strain and curvature components follow as

${\gamma }_{11}={\gamma }_{22}=0,{\gamma }_{12}=\left(Ca-(a^2\varphi /L)\right)/2,\Delta {\kappa }_{11}=0,\Delta {\kappa }_{12}=(a\varphi /L),\Delta {\kappa }_{22}\approx 0$

 

13. The constitutive equations can be reduced to

$\begin{array}{l}{T}^{12}={\displaystyle \frac{Eh}{(1+\nu )a^2}{\gamma }_{12}-{\displaystyle \frac{E{h}^3}{12(1+\nu )a^4}{\gamma }_{12}{T}^{21}={\displaystyle \frac{Eh}{(1+\nu )a^2}\left({\gamma }_{12}-{\displaystyle \frac{h^2}{12a}\Delta {\kappa }_{12}}\right)}}}\\ M^{12}={\displaystyle \frac{E{h}^3}{12(1+\nu )a^2}\Delta {\kappa }_{12}{M}^{21}={\displaystyle \frac{E{h}^3}{12(1+\nu )a^2}\left(\Delta {\kappa }_{12}-{\gamma }_{12}/a\right)}}\end{array}$

Note that this is one of the rare shell geometries for which the full coupled constitutive equations must be used.

 

14. The equations listed in 6-8 and 12,13 can be solved to show that $C=(a\varphi /L)(1+h^2/12a^2)$,

$\begin{array}{l}{T}^{12}={\displaystyle \frac{E{h}^3}{12(1+\nu )a^3}{\displaystyle \frac{a\varphi }{L}\left(1-{\displaystyle \frac{h^2}{12a^2}}\right)}}\\ M^{12}={\displaystyle \frac{E{h}^3}{12(1+\nu )a^2}{\displaystyle \frac{a\varphi }{L}{M}^{21}={\displaystyle \frac{E{h}^3}{12(1+\nu )a^2}{\displaystyle \frac{a\varphi }{L}\left(1-{\displaystyle \frac{h^2}{12a^2}}\right)}}}}\end{array}$

 

15. The components of internal force and moment may be expressed in terms of cylindrical-polar coordinates by noting that $m_1=e_z{m}_2=a{e}_{\theta }$, whence

$\begin{array}{l}T=T^{\alpha \beta }{m}_{\alpha }\otimes m_{\beta }=a{T}^{12}{e}_z\otimes e_{\theta }\\ M=M^{\alpha \beta }{m}_{\alpha }\otimes m_{\beta }=a{M}^{12}{e}_z\otimes e_{\theta }+a{M}^{21}{e}_{\theta }\otimes e_z\end{array}$

 

16. Finally, external force and couple per unit length acting on the end of the cylinder at $z=L$ are $P=e_z\cdot TQ=e_z\cdot M$.  The resultant moment about the axis of the cylinder due to these tractions is

$\Lambda =\Lambda e_z={\displaystyle \underset{0}{\overset{2\pi }{\int }}\left(a{e}_r\times P+Q\right)}ad\theta ={\displaystyle \frac{2\pi E{h}^3}{12(1+\nu )}{\displaystyle \frac{a\varphi }{L}\left(2-{\displaystyle \frac{h^2}{12a^2}}\right)\approx {\displaystyle \frac{\pi E{h}^3}{3(1+\nu )}{\displaystyle \frac{a\varphi }{L}}}}}$

It follows that the twist per unit length is related to the twisting moment by $\varphi /L=3\Lambda (1+\nu )/\left(\pi a{h}^{3}E\right)$.  Substituting this result back into the formulas for T and M and neglecting the terms of order $h^2/a^2$ gives the result stated.

 

 

 

10.7.8 Membrane shell theory analysis of a spherical dome under gravitational loading

 

The figure shows a thin-walled, spherical dome with radius R, thickness h and mass density $\rho$.  We wish to calculate the internal forces induced by gravitational loading of the structure.   Shells that are used for structural applications are usually modeled using a simplified version of general shell theory, known as `Membrane Shell Theory.’   The theory simplifies the governing equations by neglecting internal moments, so that the structure is supported entirely by in-plane forces $T^{\alpha \beta }$.   The theory is intended to be applied to masonry or concrete structures, which can generally support substantial (compressive) in-plane forces but are weak in bending.  Of course, some bending resistance is critical to ensure stability against buckling; in addition, significant bending moments may develop near the edges of the structure if the boundary conditions constrain the rotation or transverse motion of the shell; so membrane theory must be used with caution.

 

The internal forces are best expressed as components in a spherical-polar basis of unit vectors $\{e_R,e_{\varphi },e_{\theta }\}$ shown in Figure 10.40.  The solution shows that the in-plane membrane forces are

$T=\rho ghR\left({\displaystyle \frac{1}{(1+\cos \theta )}-\cos \theta }\right)e_{\varphi }\otimes e_{\varphi }-{\displaystyle \frac{\rho ghR}{(1+\cos \theta )}{e}_{\theta }\otimes e_{\theta }}$

where g is the gravitational acceleration.

 

A concrete dome should be designed so that the membrane forces are compressive everywhere.  The solution shows that the hoop forces $T_{\theta \theta }$ are always compressive, but the circumferential forces $T_{\varphi \varphi }$ are compressive only if $\cos \theta >1/(1+\cos \theta )$.   The dome should therefore be designed with ${\theta }_0<{51.8}^0$

 

Derivation: We adopt as curvilinear coordinates the spherical-polar coordinates $(\varphi ,\theta )$ shown in Figure 10.40.  Let $\{e_1,e_2,e_3\}$ be the Cartesian basis that is used to provide reference directions for $(\varphi ,\theta )$ as indicated in the figure.  Then

 

1. The position vector of a point in the undeformed shell can be expressed as

$\overline{r}=R\cos \varphi \sin \theta e_1+R\sin \varphi \sin \theta e_2+R\cos \varphi e_3$

 

2. The natural basis vectors follow as

$\begin{array}{l}{\overline{m}}_1={\displaystyle \frac{\partial \overline{r}}{\partial \varphi }=-R\sin \varphi \sin \theta e_1+R\cos \varphi \sin \theta e_2}\\ {\overline{m}}_2={\displaystyle \frac{\partial \overline{r}}{\partial \theta }=R\cos \varphi \cos \theta e_1+R\sin \varphi \cos \theta e_2-R\sin \theta e_3}\\ {\overline{m}}_3=-\left(\cos \varphi \sin \theta e_1+\sin \varphi \sin \theta e_2+\cos \theta e_3\right)\end{array}$

 

3. The reciprocal base vectors are

$\begin{array}{l}{\overline{m}}^1=\left(-\sin \varphi e_1+\cos \varphi e_2\right)/R\sin \theta \\ {\overline{m}}^2=\left(\cos \varphi \cos \theta e_1+\sin \varphi \cos \theta e_2-\sin \theta e_3\right)/R\\ {\overline{m}}^3=-\left(\cos \varphi \sin \theta e_1+\sin \varphi \sin \theta e_2+\cos \theta e_3\right)\end{array}$

 

4. The Christoffel symbols for the undeformed shell and its curvature components can be calculated as

$\begin{array}{l}{\Gamma }_{\beta \gamma }^{\alpha }={\overline{m}}^{\alpha }\cdot {\displaystyle \frac{{\partial }^2\overline{r}}{\partial {\xi }_{\beta }\partial {\xi }_{\gamma }}\Rightarrow {\overline{\Gamma }}_{11}^1={\overline{\Gamma }}_{22}^1=0{\overline{\Gamma }}_{12}^1={\overline{\Gamma }}_{21}^1={\displaystyle \frac{\cos \theta }{\sin \theta }}}\\ {\overline{\Gamma }}_{11}^2=-\sin \theta \cos \theta {\overline{\Gamma }}_{12}^2={\overline{\Gamma }}_{21}^2={\overline{\Gamma }}_{22}^2=0\end{array}$

${\overline{\kappa }}_{\alpha \beta }=-{\overline{m}}^3\cdot {\displaystyle \frac{\partial {\overline{m}}_{\alpha }}{\partial {\xi }_{\beta }}\Rightarrow {\overline{\kappa }}_{11}=-R{\sin }^2\theta {\overline{\kappa }}_{12}=0{\overline{\kappa }}_{22}=-R}$

 

5. We now introduce two assumptions (i) the bending resistance of the shell is zero, so that the internal moment components $M^{\alpha \beta }=0$; (ii) the deformations are small enough so that the Christoffel symbols and curvature terms in the equilibrium equations can be approximated using the values for the undeformed shell (this amounts to enforcing equilibrium in the undeformed configuration of the shell). In addition, the shell is in static equilibrium, and the external couples are zero. The equations of motion of Section 10.5.8 can therefore be reduced to

$\begin{array}{l}{\displaystyle \frac{\partial T^{\alpha \beta }}{\partial {\xi }_{\alpha }}+T^{\alpha \beta }{\Gamma }_{\alpha \gamma }^{\gamma }+T^{\alpha \gamma }{\Gamma }_{\gamma \alpha }^{\beta }+V^{\alpha }{\kappa }_{\alpha }^{\beta }+p^{\beta }=0V^{\beta }=0}\\ {\displaystyle \frac{\partial V^{\alpha }}{\partial {\xi }_{\alpha }}+V^{\alpha }{\Gamma }_{\alpha \beta }^{\beta }-T^{\alpha \beta }{\kappa }_{\alpha \beta }+p^3=0T^{12}-T^{21}=0}\end{array}$ 

 

 

6. We note that $T^{12}=0$ by symmetry, and the external force acting on unit area of the shell is $p=-\rho gh{e}_3=p^i{m}_i$.  The contravariant components of p can therefore be computed as $p^i=-\rho gh{m}^i\cdot e_3$.   Substituting these, as well as the Christoffel symbols and curvature components reduces the equilibrium equations to

$\begin{array}{l}{\displaystyle \frac{\partial T^{11}}{\partial \varphi }=0}\\ {\displaystyle \frac{\partial T^{22}}{\partial \theta }+T^{22}{\displaystyle \frac{\cos \theta }{\sin \theta }-T^{11}\sin \theta \cos \theta +{\displaystyle \frac{\rho gh}{R}\sin \theta =0}}}\\ T^{11}R{\sin }^2\theta +T^{22}R+\rho gh\cos \theta =0\end{array}$

 

7. Eliminating $T^{11}$ from the second and third equations gives

${\displaystyle \frac{\partial T^{22}}{\partial \theta }+2T^{22}{\displaystyle \frac{\cos \theta }{\sin \theta }={\displaystyle \frac{\rho gh}{R\sin \theta }}}}$

 

8. This equation can be integrated directly by substituting $T^{22}=\Psi /{\sin }^2\theta$, with the solution

$T^{22}={\displaystyle \frac{\rho gh}{R{\sin }^2\theta }(\cos \theta -1)+C={\displaystyle \frac{-\rho gh}{R(1+\cos \theta )}+C}}$

where C is a constant of integration. 

 

9. The constant of integration can be found using the boundary condition at the edge of the shell at $\theta ={\theta }_0$.  The reaction force must act in the plane of the shell, and the vertical component of the force must balance the shell’s weight, so that the force per unit length is $P=\rho ghR(1-\cos {\theta }_0)e_{\theta }/{\sin }^2{\theta }_0$.  The boundary condition requires that $e_{\theta }\cdot T=P$ at $\theta ={\theta }_0$, which shows that C=0.

 

10. $T^{11}$ can be calculated from the third equation in (6), giving

$T^{11}={\displaystyle \frac{\rho gh\left({(1+\cos \theta )}^{-1}-\cos \theta \right)}{R{\sin }^2\theta }}$

 

11. Finally, the components of T in the cylindrical-polar basis can be calculated by noting that $m_1=R\sin \theta e_{\varphi }{m}_2=R{e}_{\theta }$, whence

$T=T^{11}{m}_1\otimes m_1+T^{22}{m}_2\otimes m_2=T^{11}{R}^2{\sin }^2\theta e_{\varphi }\otimes e_{\varphi }+T^{22}{R}^2{e}_{\theta }\otimes e_{\theta }$