Problems for Chapter 5

Analytical Techniques and Solutions for Linear Elastic Solids

5.8.  Energy Methods

5.8.1.      A planet that deforms under its own gravitational force can be idealized as a linear elastic sphere with radius a, Young’s modulus E and Poisson’s ratio $\nu$ that is subjected to a radial gravitational force $b=-\left(g\text{\hspace{0.17em}}R/a\right){e}_{R}$, where g is the acceleration due to gravity at the surface of the sphere, and R is the radial coordinate. Use the reciprocal theorem, together with a hydrostatic stress distribution ${\sigma }_{ij}=p{\delta }_{ij}$ as the reference solution, to calculate the change in volume of the sphere, and hence deduce the radial displacement of its surface.

5.8.2.      Consider an isotropic, linear elastic solid with Young’s modulus E, mass density $\rho$, and Poisson’s ratio $\nu$, which is subjected to a body force distribution ${b}_{i}$ per unit mass, and tractions ${t}_{i}$ on its exterior surface.   By using the reciprocal theorem, together with a state of uniform stress ${\sigma }_{ij}^{*}$ as the reference solution, show that the average strains in the solid can be calculated from

$\frac{1}{V}\underset{V}{\int }{\epsilon }_{ij}dV=\frac{1+\nu }{EV}\underset{A}{\int }{x}_{j}{t}_{i}dA-\frac{\nu }{EV}\underset{A}{\int }{\delta }_{ij}{x}_{k}{t}_{k}dA+\frac{1+\nu }{EV}\underset{V}{\int }{x}_{j}\rho {b}_{i}dV-\frac{\nu }{EV}\underset{V}{\int }{\delta }_{ij}{x}_{k}\rho {b}_{k}dV$

5.8.3.      A cylinder with arbitrary cross-section rests on a flat surface, and is subjected to a vertical gravitational body force $b=-\rho g{e}_{3}$, where ${e}_{3}$ is a unit vector normal to the surface. The cylinder is a linear elastic solid with Young’s modulus E, mass density $\rho$, and Poisson’s ratio $\nu$. Define the change in length of the cylinder as

$\delta L=\frac{1}{A}\underset{A}{\int }{u}_{3}\left({x}_{1},{x}_{2},L\right)dA$

where ${u}_{3}\left({x}_{1},{x}_{2},L\right)$ denotes the displacement of the end ${x}_{3}=L$ of the cylinder.  Show that $\delta L=W/2EA$, where W is the weight of the cylinder, and A its cross-sectional area.

5.8.4.      In this problem, you will calculate an expression for the change in potential energy that occurs when an inelastic strain ${\epsilon }_{ij}^{T}$ is introduced into some part B of an elastic solid.  The inelastic strain can be visualized as a generalized version of the Eshelby inclusion problem $–$ it could occur as a result of thermal expansion, a phase transformation in the solid, or plastic flow.   Note that B need not be ellipsoidal.

The figure illustrates the solid of interest.  Assume that:

The solid has elastic constants ${C}_{ijkl}$

No body forces act on the solid (for simplicity)

Part of the surface of the solid ${S}_{1}$ is subjected to a prescribed displacement ${u}_{i}^{*}$

The remainder of the surface of the solid  ${S}_{2}$ is subjected to a prescribed traction ${t}_{i}^{*}$

Let ${u}_{i}^{0},{\epsilon }_{ij}^{0},{\sigma }_{ij}^{0}$ denote the displacement, strain, and stress in the solid before the inelastic strain is introduced.  Let ${V}_{0}$ denote the potential energy of the solid in this state.

Next, suppose that some external process introduces an inelastic strain ${\epsilon }_{ij}^{T}$ into part of the solid. Let $\Delta {u}_{i},\Delta {\epsilon }_{ij},\Delta {\sigma }_{ij}$ denote the change in stress in the solid resulting from the inelastic strain.  Note that these fields satisfy

The strain-displacement relation $\Delta {\epsilon }_{ij}=\left(\partial \Delta {u}_{i}/\partial {x}_{j}+\partial \Delta {u}_{i}/\partial {x}_{j}\right)/2$

The stress-strain law  $\Delta {\sigma }_{ij}={C}_{ijkl}\left(\Delta {\epsilon }_{kl}-{\epsilon }_{kl}^{T}\right)$ in B, and $\Delta {\sigma }_{ij}={C}_{ijkl}\Delta {\epsilon }_{kl}$ outside B

Boundary conditions $\Delta {u}_{i}=0$ on ${S}_{1}$, and $\Delta {\sigma }_{ij}{n}_{j}=0$ on ${S}_{2}$

5.8.4.1.            Write down an expressions for ${V}_{0}$ in terms of ${u}_{i}^{0},{\epsilon }_{ij}^{0},{\sigma }_{ij}^{0}$

5.8.4.2.            Suppose that ${u}_{i}^{0},{\epsilon }_{ij}^{0},{\sigma }_{ij}^{0}$ are all zero (i.e. the solid is initially stress free).  Write down the potential energy ${V}_{S}$ due to $\Delta {u}_{i},\Delta {\epsilon }_{ij},\Delta {\sigma }_{ij}$.  This is called the “self energy” of the eigenstrain $–$ the energy cost of  introducing the eigenstrain ${\epsilon }_{ij}^{T}$ into a stress-free solid.

5.8.4.3.            Show that the expression for the self-energy can be simplified to

${V}_{S}=-\frac{1}{2}\underset{B}{\int }\Delta {\sigma }_{ij}{\epsilon }_{ij}^{T}dV$

5.8.4.4.            Now suppose that ${u}_{i}^{0},{\epsilon }_{ij}^{0},{\sigma }_{ij}^{0}$ are all nonzero.  Write down the total potential energy of the system ${V}_{TOT}$, in terms of ${u}_{i}^{0},{\epsilon }_{ij}^{0},{\sigma }_{ij}^{0}$ and $\Delta {u}_{i},\Delta {\epsilon }_{ij},\Delta {\sigma }_{ij}$.

5.8.4.5.            Finally, show that the total potential energy of the system can be expressed as

${V}_{TOT}={V}_{0}+{V}_{S}-\underset{B}{\int }{\sigma }_{ij}^{0}{\epsilon }_{ij}^{T}dV$

Here, the last term is called the “interaction energy” of the eigenstrain with the applied load. The steps in this derivation are very similar to the derivation of the reciprocal theorem.

5.8.5.      An infinite, isotropic, linear elastic solid with Young’s modulus E and Poisson’s ratio $\nu$ is subjected to a uniaxial tensile stress ${\sigma }_{0}$.  As a result of a phase transformation, a uniform dilatational strain ${\epsilon }_{ij}^{T}=\beta {\delta }_{ij}$ is then induced in a spherical region of the solid with radius a

5.8.5.1.            Using the solution to problem 1, and the Eshelby solution, find an expression for the change in potential energy of the solid, in terms of $\beta ,{\sigma }_{0}$ and relevant geometric and material parameters.

5.8.5.2.            Assume that the interface between the transformed material an the matrix has an energy per unit area $\gamma$.  Find an expression for the critical stress at which the total energy of the system (elastic potential energy + interface energy) is decreased as a result of the transformation