Chapter 4

Solutions to simple boundary and initial value problems

4.3 Spherically symmetric solution to quasi-static large strain elasticity problems

4.3.1 Summary of governing equations of finite elasticity in Cartesian components

This section is intended to illustrate the nature of solutions to elasticity problems with large shape changes.

We are given the following information

1.      The geometry of the solid

2.      A constitutive law for the material (i.e. the hyperelastic strain energy potential)

3.      The body force density ${b}_{i}$ (per unit mass) (if any)

4.      Prescribed boundary tractions ${t}_{i}$ and/or boundary displacements ${u}_{i}$

To simplify the problem we will assume

The solid is stress free in its undeformed configuration;

Temperature changes during deformation are neglected;

The solid is incompressible

With these assumptions, we wish to calculate the displacement field ${u}_{i}$, the left Cauchy-Green deformation tensor ${B}_{ij}$ and the stress field ${\sigma }_{ij}$ satisfying the following equations:

Displacement$—$strain relation ${B}_{ij}={F}_{ik}{F}_{jk}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{F}_{ij}={\delta }_{ij}+\frac{\partial {u}_{i}}{\partial {x}_{j}}$

Incompressibility condition $J=\mathrm{det}\left(F\right)=1$

Stress$—$strain relation ${\sigma }_{ij}=2\left[\left(\frac{\partial U}{\partial {I}_{1}}+{I}_{1}\frac{\partial U}{\partial {I}_{2}}\right){B}_{ij}-\left({I}_{1}\frac{\partial U}{\partial {\overline{I}}_{1}}+2{I}_{2}\frac{\partial U}{\partial {\overline{I}}_{2}}\right)\frac{{\delta }_{ij}}{3}-\frac{\partial U}{\partial {I}_{2}}{B}_{ik}{B}_{kj}\right]+p\frac{{\delta }_{ij}}{3}$

where ${\sigma }_{ij}$ is the Cauchy stress tensor,  $U\left({I}_{1},{I}_{2}\right)$ is the strain energy potential for the elastic solid, p is the hydrostatic part of the stress (which must be determined as part of the solution) and ${I}_{1}={B}_{kk}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{I}_{2}=\left({I}_{1}^{2}-{B}_{ik}{B}_{ki}\right)/2$.

Equilibrium Equation $\frac{\partial {\sigma }_{ij}}{\partial {y}_{i}}+\rho {b}_{j}=0$

Traction boundary conditions ${\sigma }_{ij}{n}_{i}={t}_{j}$ on parts of the boundary where tractions are known.

Displacement boundary conditions ${u}_{i}={d}_{i}$ on parts of the boundary where displacements are known.

4.3.2 Simplified equations for incompressible spherically symmetric solids

A representative spherically symmetric problem is illustrated in the picture.  We consider a hollow, spherical solid, which is subjected to spherically symmetric loading (i.e. internal body forces, as well as tractions or displacements applied to the surface, are independent of $\theta$ and $\varphi$, and act in the radial direction only).

The solution is most conveniently expressed using a spherical-polar coordinate system, illustrated in the figure.  For a finite deformation problem, we need a way to characterize the position of material particles in both the undeformed and deformed solid.  To do this, we let $\left(R,\Theta ,\Phi \right)$ identify a material particle in the undeformed solid. The coordinates of the same point in the deformed solid is identified by a new set of spherical-polar co-ordinates $\left(r,\theta ,\varphi \right)$.  One way to describe the deformation would be to specify each of the deformed coordinates $\left(r,\theta ,\varphi \right)$ in terms of the reference coordinates $\left(R,\Theta ,\Phi \right)$. For a spherically symmetric deformation, points only move radially, so that

$r=f\left(R\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\theta =\Theta \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\varphi =\text{\hspace{0.17em}}\Phi$

In finite deformation problems vectors and tensors can be expressed as components in a basis $\left\{{e}_{R},{e}_{\Theta },{e}_{\Phi }\right\}$ associated with the position of material points in the undeformed solid, or, if more convenient, in a basis $\left\{{e}_{r},{e}_{\theta },{e}_{\varphi }\right\}$ associated with material points in the deformed solid.  For spherically symmetric deformations the two bases are identical $–$ consequently, we can write

Position vector in the undeformed solid       $x=R{e}_{r}$

Position vector in the deformed solid       $y=r{e}_{r}=f\left(R\right){e}_{r}$

Displacement vector $u=y-x=r{e}_{r}-R{e}_{r}=\left(f\left(R\right)-R\right){e}_{r}$

The stress, deformation gradient and deformation tensors tensors (written as components in $\left\{{e}_{r},{e}_{\theta },{e}_{\varphi }\right\}$ ) have the form

$\sigma \equiv \left[\begin{array}{ccc}{\sigma }_{rr}& 0& 0\\ 0& {\sigma }_{\theta \theta }& 0\\ 0& 0& {\sigma }_{\varphi \varphi }\end{array}\right]\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}F\equiv \left[\begin{array}{ccc}{F}_{rr}& 0& 0\\ 0& {F}_{\theta \theta }& 0\\ 0& 0& {F}_{\varphi \varphi }\end{array}\right]\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}B\equiv \left[\begin{array}{ccc}{B}_{rr}& 0& 0\\ 0& {B}_{\theta \theta }& 0\\ 0& 0& {B}_{\varphi \varphi }\end{array}\right]$

and furthermore must satisfy ${\sigma }_{\theta \theta }={\sigma }_{\varphi \varphi }$ ${F}_{rr}={F}_{\theta \theta }\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{B}_{\theta \theta }={B}_{\varphi \varphi }$.

For spherical symmetry, the governing equations reduce to

Strain Displacement Relations ${F}_{rr}=\frac{df}{dR}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{F}_{\varphi \varphi }={F}_{\theta \theta }=\frac{f\left(R\right)}{R}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{B}_{rr}={\left(\frac{df}{dR}\right)}^{2}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{B}_{\varphi \varphi }={B}_{\theta \theta }={\left(\frac{f\left(R\right)}{R}\right)}^{2}$

Incompressibility condition $\left(\frac{df}{dR}\right){\left(\frac{f\left(R\right)}{R}\right)}^{2}=1$

Stress$—$Strain relations

$\begin{array}{l}{\sigma }_{rr}=2\left[\left(\frac{\partial U}{\partial {I}_{1}}+{I}_{1}\frac{\partial U}{\partial {I}_{2}}\right){B}_{rr}-\frac{{I}_{1}}{3}\frac{\partial U}{\partial {I}_{1}}-\frac{2{I}_{2}}{3}\frac{\partial U}{\partial {\overline{I}}_{2}}-\frac{\partial U}{\partial {I}_{2}}{B}_{rr}^{2}\right]+p\\ {\sigma }_{\theta \theta }={\sigma }_{\varphi \varphi }=2\left[\left(\frac{\partial U}{\partial {I}_{1}}+{I}_{1}\frac{\partial U}{\partial {I}_{2}}\right){B}_{\theta \theta }-\frac{{I}_{1}}{3}\frac{\partial U}{\partial {I}_{1}}-\frac{2{I}_{2}}{3}\frac{\partial U}{\partial {\overline{I}}_{2}}-\frac{\partial U}{\partial {I}_{2}}{B}_{\theta \theta }^{2}\right]+p\end{array}$

Equilibrium Equations

$\frac{d{\sigma }_{rr}}{dr}+\frac{1}{r}\left(2{\sigma }_{rr}-{\sigma }_{\theta \theta }-{\sigma }_{\varphi \varphi }\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}+{\rho }_{0}{b}_{r}=0$

Boundary Conditions

Prescribed Displacements ${u}_{r}\left(a\right)={g}_{a}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{u}_{r}\left(b\right)={g}_{b}$

Prescribed Tractions ${\sigma }_{rr}\left(a\right)={t}_{a}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\sigma }_{rr}\left(b\right)={t}_{b}$

4.3.3 Pressurized hollow sphere made from an incompressible rubber

As an example, consider a pressurized hollow rubber shell, as shown in the picture. Assume that

Before deformation, the sphere has inner radius A and outer radius B

After deformation, the sphere has inner radius a and outer radius b

The solid is made from an incompressible Mooney-Rivlin solid, with strain energy potential

$U=\frac{{\mu }_{1}}{2}\left({I}_{1}-3\right)+\frac{{\mu }_{2}}{2}\left({I}_{2}-3\right)$

No body forces act on the sphere

The inner surface r=a is subjected to pressure ${p}_{a}$

The outer surface r=b is subjected to pressure ${p}_{b}$

The deformed radii a,b of the inner and outer surfaces of the spherical shell are related to the pressure by

$\frac{{p}_{a}-{p}_{b}}{{\mu }_{1}}=2\left(\frac{1}{\beta }-\frac{1}{\alpha }\right)+\frac{1}{2}\left(\frac{1}{{\beta }^{4}}-\frac{1}{{\alpha }^{4}}\right)-\frac{2{\mu }_{2}}{{\mu }_{1}}\left(\beta -\alpha \right)+\frac{{\mu }_{2}}{{\mu }_{1}}\left(\frac{1}{{\beta }^{2}}-\frac{1}{{\alpha }^{2}}\right)$

where $\alpha =a/A$, $\beta =b/B$, and $\alpha ,\beta$ are related by

$\frac{{B}^{3}}{{A}^{3}}=\frac{1-{\alpha }^{3}}{1-{\beta }^{3}}$

Provided the pressure is not too large (see below), the preceding two equations can be solved for $\alpha$ and $\beta$ given the pressure and properties of the shell (for graphing purposes, it is better to assume a value for $\alpha$, calculate the corresponding $\beta$, and then determine the pressure).

The position  r of a material particle after deformation is related to its position R before deformation by

$\frac{r}{A}={\left(\frac{{R}^{3}}{{A}^{3}}+{\alpha }^{3}-1\right)}^{1/3}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{R}{A}={\left(\frac{{r}^{3}}{{A}^{3}}-{\alpha }^{3}+1\right)}^{1/3}$

The deformation tensor distribution in the sphere is

${B}_{rr}={\left(R/r\right)}^{4}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{B}_{\theta \theta }={B}_{\varphi \varphi }={\left(r/R\right)}^{2}$

The Cauchy stress in the sphere is

${\sigma }_{rr}={\mu }_{1}\left(\frac{2R}{r}+\frac{{R}^{4}}{2{r}^{4}}\right)-{\mu }_{2}\left(\frac{2r}{R}-\frac{{R}^{2}}{{r}^{2}}\right)+C$

${\sigma }_{\theta \theta }={\mu }_{1}\left(\frac{2R}{r}-\frac{{R}^{4}}{2{r}^{4}}+\frac{{r}^{2}}{{R}^{2}}\right)-{\mu }_{2}\left(\frac{2r}{R}-\frac{{r}^{4}}{{R}^{4}}\right)+C$

$C=-\frac{{\mu }_{1}}{2}\left(\frac{2}{\alpha }+\frac{1}{2{\alpha }^{4}}+\frac{2}{\beta }+\frac{1}{2{\beta }^{4}}\right)+\frac{{\mu }_{2}}{2}\left(2\alpha -\frac{1}{{\alpha }^{2}}+2\beta -\frac{1}{{\beta }^{2}}\right)-\frac{{p}_{a}+{p}_{b}}{2}$

The variation of the internal radius of the spherical shell with applied pressure is plotted in the figure, for ${\mu }_{2}/{\mu }_{1}=0.04$ (a representative value for a typical rubber).  For comparison, the linear elastic solution (obtained by setting $E=3\left({\mu }_{1}+{\mu }_{2}\right)\text{\hspace{0.17em}}$ and $\nu =1/2$ in the formulas given in section 4.1.4) is also shown.  Note that:

1.      The small strain solution is accurate for $u/A<0.05$

2.      The relationship between pressure and displacement is nonlinear in the large deformation regime.

3.      As the internal radius of the sphere increases, the pressure reaches a maximum, and thereafter decreases (this will be familiar behavior to anyone who has inflated a balloon).  This is  because the wall thickness of the shell decreases as the sphere expands.

The stress distribution for various displacements in the shell is plotted in the figures below, for ${p}_{b}=0$, ${\mu }_{2}/{\mu }_{1}=0.04$ and B/A=3.  The radial stress remains close to the linear elastic solution even in the large deformation regime.  The hoop stress distribution is significantly altered as the deformation increases, however.

Derivation

1.      Integrate the incompressibility condition from the inner radius of the sphere to some arbitrary point R

$\underset{f\left(A\right)}{\overset{f\left(R\right)}{\int }}{\left[f\left(R\right)\right]}^{2}df=\underset{A}{\overset{R}{\int }}{R}^{2}dR\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}f{\left(R\right)}^{3}-f{\left(A\right)}^{3}={R}^{3}-{A}^{3}$

2.      Note that $f\left(R\right)=r$ by definition, and $f\left(A\right)=a$ since the point at R=A moves to r=a after deformation.  This gives the relationship between the position r of a point in the deformed solid and its position R before deformation

$r=f\left(R\right)=\text{\hspace{0.17em}}\sqrt[3]{{R}^{3}+{a}^{3}-{A}^{3}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}R={f}^{-1}\left(r\right)=\text{\hspace{0.17em}}\sqrt[3]{{r}^{3}+{A}^{3}-{a}^{3}}$

3.      The components of the Cauchy-Green tensor follow as ${B}_{rr}={\left(R/r\right)}^{4}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{B}_{\theta \theta }={B}_{\varphi \varphi }={\left(r/R\right)}^{2}$

4.      The stresses follow from the stress-strain equation as

${\sigma }_{rr}=\frac{2}{3}\left({\mu }_{1}+{\mu }_{2}{B}_{\theta \theta }\right)\left({B}_{rr}-{B}_{\theta \theta }\right)+p\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\sigma }_{\theta \theta }={\sigma }_{\varphi \varphi }=\frac{1}{3}\left({\mu }_{1}+{\mu }_{2}{B}_{\theta \theta }\right)\left({B}_{\theta \theta }-{B}_{rr}\right)+p$

5.      Substituting these stresses into the equilibrium equation leads to the following differential equation for ${\sigma }_{rr}$

$\frac{d{\sigma }_{rr}}{dr}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+\frac{2}{r}\text{\hspace{0.17em}}\left({\mu }_{1}+{\mu }_{2}{B}_{\theta \theta }\right)\left({B}_{rr}-{B}_{\theta \theta }\right)=0$

6.      After substituting for ${B}_{rr}$ and ${B}_{\theta \theta }$, and expressing R in terms of r, this equation can be integrated and simplified to see that

${\sigma }_{rr}={\mu }_{1}\frac{R\left(4{r}^{3}+{R}^{3}\right)}{2{r}^{4}}-{\mu }_{2}\frac{\left(2{r}^{3}-{R}^{3}\right)}{{r}^{2}R}+C$

7.      The boundary conditions require that ${\sigma }_{rr}=-{p}_{a}$ on (r=a,R=A), while ${\sigma }_{rr}=-{p}_{b}$ on (r=b,R=B), which requires

$\begin{array}{l}-{p}_{a}={\mu }_{1}\left(\frac{2}{\alpha }+\frac{1}{2{\alpha }^{4}}\right)-{\mu }_{2}\left(2\alpha -\frac{1}{{\alpha }^{2}}\right)+C\\ -{p}_{b}={\mu }_{1}\left(\frac{2}{\beta }+\frac{1}{2{\beta }^{4}}\right)-{\mu }_{2}\left(2\beta -\frac{1}{{\beta }^{2}}\right)+C\end{array}$

where $\alpha =a/A$ and $\beta =b/B$.  The expression that relates $\alpha$ and $\beta$ to the pressure follows by subtracting the first equation from the second.   Adding the two equations gives the expression for C.

8.      Finally, the hoop stress follows by noting that, from (4) ${\sigma }_{\theta \theta }-{\sigma }_{rr}=\left({\mu }_{1}+{\mu }_{2}{B}_{\theta \theta }\right)\left({B}_{\theta \theta }-{B}_{rr}\right)$