Recognitions:

## What is the Theory of Elasticity?

1. INTRODUCTION.

This thread is an experiment. I think the public discussion between myself, pervect, and Greg Egan in the rotating hoop thread has given students and the interested public and glimpse into how people actually think about a challenging problem, but I guess the discussion went over the heads of all but a handful of trained physicists and mathematicians who happen to have prior familiarity with the theory of elasticity.

I feel that prevect, Greg, and I have by no means exhausted the possibilities in the other thread, but my intention is that this is the thread where anyone can ask questions about all this elasticity stuff.

I'll start by offering a synopsis of the theory of nonrelativistic elasticity, which is already highly nontrivial even for a nonrelativistic hoop. My synopsis covers standard stuff but is tailored toward relativistic applications in various ways.

2. REFERENCES:

I only list some of the books I consulted while preparing my synopsis; this list is weighted toward books I judge should be easily available. In particular, large university bookstores may have the classic textbook by Landau and Lifschitz in stock, and any good library will have multiple copies.

Some excellent references for the nonrelativistic theory of elasticity:

L. D. Landau and E. M. Lifschitz,
Theory of Elasticity, 3rd Edition.
Pergamon, 1986.

Chi-Teh Wang,
Applied Elasticity.
McGraw-Hill, 1953.

Karl F. Graff,
Wave Motion in Elastic Solids.
Dover reprint, 1991 (org. 1975).

S. P. Timoshenko and J. N. Goodier,
Theory of Elasticity, 3rd Edition.
McGraw Hill reprint 1970 (org. 1934).

An excellent survey of continuum mechanics generally:

Alexander L. Fetter and John Dirk Walecka,
Theoretical Mechanics of Particles and Continua.
McGraw-Hill, 1980.

An engineering-oriented book which offers an introduction to engineering applications of the theory of elasticity:

Donald F. Young, William F. Riley, Kenneth G. McConnell, Thomas R. Rogge,
Essentials of Mechanics, Iowa State University Press, 1974.

One of the best overall surveys of PDEs is:

Ronald B. Guenther and John W. Lee.
Partial Differential Equations of Mathematical Physics and Integral Equations.
Dover reprint, 1996 (org. 1988)

Two of the best introductions to symmetry methods for solving DEs:

Peter J. Olver,
Applications of Lie Groups to Differential Equations, 2nd Edition.
Springer, 1993.

Brian J. Cantwell,
Introduction to Symmetry Analysis.
Cambridge University Press, 2002.

An excellent introduction to the classical theory of curves and surfaces, needed here only for the result that the highest derivatives appearing in the curvature and torsion of a curve are two, three respectively:

Dirk J. Struik,
Lectures on Classical Differential Geometry, 2nd Edition.
Dover reprint 1988 (org. 1950).

3. ELASTIC PROPERTIES OF MATERIALS

Imagine we have a cylindrical steel shaft firmly clamped in a vise at either end. If we adjust the vise so that the shaft is under tension, we find that the shaft is elongated and also shrinks in diameter. If we plot how this axial elongation and orthogonal compression depends on the applied tension, we find that for sufficiently small tension, the relation is roughly linear, but with different slopes. If we repeat the experiment with a steel shaft of different size, we find the same two slopes. If we repeat with shafts of different materials, we obtain similar results but with new pairs of slopes. This suggests that these slopes characterize the response of the material to tension. They are called Young's modulus E (for extension), and the product of Young's modulus with Poisson's ratio $\nu$ (many authors write this $\sigma$; this topic has only partially standardized notation). It turns out that E, nu suffice to characterize the elastic properties (valid for small deformations) of any homogeneous isotropic material, but as we will see it is often useful to define alternative "elastic moduli"; these will be rational combinations of each other.

4. THE STRAIN TENSOR

We must begin by describing the deformation of a material. To do this we imagine that the deformation is obtained by applying a diffeomorphism,
i.e. a smooth map E^3 -> E^3 whose inverse is also smooth. Indeed, let us assume this has the form
$$x^{i'} = x^i + u^i (x^1, x^2, x^3)$$
where the vector field $\vec{u}[/tex] is the displacement vector, so that $$du^i = \frac{\partial u^i}{\partial x^k} \, dx^k = {u^i}_{,k} \, dx^k$$ We wish to compute how the distances between initially nearby material points change. But the new distances are given by $$(ds')^2 = \delta_{i' j'} \, dx^{i'} \, dx^{j'} = \delta_{ij} \, \left( dx^i + du^i \right) \, \left( dx^j + du^j \right)$$ $$\approx \delta_{ij} \, dx^i \, dx^j + 2 \, u_{i,j} \, dx^i \, dx^j$$ where we neglected terms which are second order in du^j. But the LHS is symmetric in i,j so we can and should replace u_{i,j} by its symmetrization, giving $$(ds')^2 = ds^2 + 2 \, \varepsilon_{ij} \, dx^i \, dx^j$$ where $$\varepsilon_{ij} = u_{(i,j)}$$ is the strain tensor. We can see from this that the components of the strain tensor are dimensionless. Notice that by definition, rigid motions of the material, which do not change any distances between material points, are precisely those diffeomorphisms which have vanishing strain. The antisymmetric part of the gradient of the displacement vector field, which automatically dropped out above (this is why we can replace [itex]u_{i,j}$ by $u_{(i,j)}$, is
$$\omega_{ij} = u_{[i,j]} [/itex] This represents rigid rotations of the body. Similarly, rigid translations of the body correspond to constant $u^i$, so they also dropped out. Since it is often convenient to adopt a coordinate chart adapted to any symmetries, it will be useful to work out some general formulas, so let us see how this works for a cylindrical chart [tex] ds^2 = dz^2 + dr^2 + r^2 \, d\phi^2, \; -\infty < z < \infty, 0 < r < \infty, -\pi < \phi < \pi$$
with the coframe field
$$\sigma^1 = dz, \; \sigma^2 = dr, \, \sigma^3 = r \, d\phi$$
and its dual frame field
$$\vec{e}_1 = \partial_z, \; \vec{e}_2 = \partial_r, \; \vec{e}_3 = \frac{1}{r} \, \partial_\phi$$
Let us take
$$\vec{u} = A \, \vec{e}_1 + B \, \vec{e}_2 + C \, \vec{e}_3$$
where A, B, C are functions of $z,r,\phi$. Then we have
$$\varepsilon_{\hat{m}\hat{n}} = \left[ \begin{array}{ccc} A z & \frac{1}{2} \left( A_r+B_z \right) & \frac{1}{2} \left( \frac{A_\phi}{r} + C_z \right) \\ \cdot & B_r & \frac{1}{2} \left( \frac{B_\phi - C}{r} + C_r \right) \\ \cdot & \cdot & \frac{ B + C_\phi}{r} \end{array} \right]$$
where the dots invite the reader to fill in redundant entries using the symmetry of the tensor, where subscripts on functions denote partial derivatives, and where the hats on the indices emphasize that these are the components in the frame. For comparison, the coordinate basis components are
$$\varepsilon_{mn} = \left[ \begin{array}{ccc} A z & \frac{1}{2} \left( A_r+B_z \right) & \frac{1}{2} \left( A_\phi + r\, C_z \right) \\ \cdot & B_r & \frac{1}{2} \left( B_\phi - C + r\, C_r \right) \\ \cdot & \cdot & r \, \left( B + C_\phi \right) \end{array} \right]$$
In the sequel I will drop the hats and only use specified frame fields.

Note: as with their other books, if one can get past the first few pages of the classic textbook by Landau and Lifschitz, one can see how clear and beautifully organized it is. Unfortunately, judging from my experience at PF and elsewhere, many readers are likely to get stuck on p. 3, where in (1.8), both the components of the vector and the strain tensor are written in terms of the frame, not the coordinate basis! This procedure is so sensible that Landau and Lifschitz did not consider it worthy of mention, but of course if the student is expecting to see results reported in terms of a coordinate basis, he will be unable to reconcile his own computation with the result stated by Landau and Lifschitz!

Many books dealing with elasticity note that some vector analysis formulae are only valid in a Cartesian chart, which seems sure to confuse readers in our context. The neccessity of such remarks are avoided by simply using frame fields! Also, strain components (and in the next section, stress components) computed with respect to the appropriate frame field, rather than the coordinate basis components, are the ones which would be measured in tension, compression, or torsion tests.

Alert readers will have already noticed that our computation above didn't depend upon using the metric of Euclidean space! In fact we have
$$(ds')^2 = g_{i'j'} \, dx^{i'} \, dx^{j'} \approx \left( g_{ij} + 2 \, \varepsilon_{ij} \right) \, dx^i \, dx^j$$
or
$$(ds')^2 = ds^2 + 2 \, \varepsilon_{ij} \, dx^i \, dx^j$$
where the strain tensor
$$\varepsilon = u_{(i;j)}$$
is the symmetrized covariant gradient of the displacement vector field. Just as before, rigid rotations correspond to displacements vector fields whose antisymmetrized gradient
$$\omega = u_{[i,j]}$$
vanishes. (What about translations?) All such rigid motions drop out of computations of material deformations.

For example, let us take the hyperbolic plane in the upper half plane chart familiar from elementary complex analysis:
$$ds^2 = \frac{dx^2 + dy^2}{y^2}, \; 0 < y < \infty, -\infty < x < \infty$$
Let us adopt the obvious coframe field
$$\sigma^1 = \frac{dx}{y}, \; \sigma^2 = \frac{dy}{y}$$
with dual frame field
$$\vec{e}_1 = y \, \partial_x, \; \vec{e}_2 = y \, \partial_y$$
Writing the displacement vector field $\vec{u} = A \, \vec{e}_1 + B \, \vec{e}_2[/tex] where A, B are functions of x,y, we have $$\varepsilon_{ij} = \left[ \begin{array}{cc} y \, A_x - B \; & \; \frac{ y \; \left( A_y + B_x \right) }{2} + \frac{A}{2} \\ \cdot & y B_y \end{array} \right]$$ Which displacement vector fields correspond to rigid motions? The ones whose strain tensor vanishes. Solving, we find a three-dimensional Lie algebra of solutions, $$\vec{X}_1 = \partial_x, \; \vec{X}_2 = x \, \partial_x + y \, \partial_y, \; \vec{X}_3 = \frac{1 + x ^2 - y^2}{2} \, \partial_x + x \, y \, \partial_y$$ This Lie algebra of vector fields is of course precisely the Killing vector fields of H^2, i.e. the rigid motions, the diffeomorphisms which preserve hyperbolic distances. Students of differential geometry know that the distance preserving condition can be formulated in a different way, in terms of Lie dragging of the metric tensor itself, which is very important since otherwise one might worry that the condition is somehow circular, since the metric enters into the definition of the covariant derivative we used, more precisely the notion of covariant differentiation which arises from the Levi-Civita connection. Lie differentiation on the other hand is defined at a lower level of structure, independently of any metric tensor we might place on our manifold. Note that [itex]\vec{X}_2$ represents a dilation with respect to the obvious euclidean metric on the upper half plane.

Exercise: compute the antisymmetrized covariant gradient of the three (rigid motion) vector fields listed above. Find this is nonzero constant for one (which therefore represents a rotation about a certain point in H^2). The other two represent translations in H^2.

(To be continued...)
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 Recognitions: Science Advisor 5. THE STRESS TENSOR Physically speaking, a body will not deform without reason, or more precisely, unless bits of material experience some force. There are two ways in which this might happen: we might apply a force to the surface of the body, which is then transmitted to the interior because bits of the body press on their neighbors, or we might have a force, such as the force of gravity (aha!) which applies directly to bits of material. The distinction here is between a force resulting from the immediate presence of an actor as opposed to a long range interaction which penetrates the body, such as the gravitational interaction. (Textbooks on elasticity also consider piezoelectric forces and other phenomena which do not fit into this simple dichotomy.) Consider a bit of material. The net force due to "internal stresses" (other bits of material pressing on it) is given by an integral over the region $\Omega$, which can be rewritten as a surface integral exactly in case this force is the gradient of a second rank tensor: $$\int_\Omega \vec{F} \, dV = -\int_\Omega {S^{jk}}_{,k} \, dV = -\int_{\partial \Omega} S^{jk} n_k dA$$ Here $S^{jk} \, n_k \, dA$ is the j-th component of force on an infinitesimal surface element with normal $\vec{n}$, and $S^{jk}$ is the stress tensor. Forces applied at the surface, and body forces such as gravity will be treated separately. Thus, the theory of elasticity incorporates at the outset a kind of "holographic principle", reflecting the fact that we are considering forces which are transmitted internally entirely by bits of matter pressing on each other's surface. Consider the moment of such an internal force. These should also be surface integrals $$M^{jk} = -\int_\Omega \left( {S^{jm}}_{,m} \, x^k - {S^{kn}}_{,n} \, x^j \right) dV$$ which reduces to $$-\int_{\partial \Omega} \left( S^{j \ell} \, x^k - S^{k \ell} \, x^j \right) n_\ell \, dA + \int_\Omega \left( S^{jk} - S^{kj} \right)$$ and the volume integral vanishes exactly if the stress tensor is symmetric. For future reference, we note that strictly speaking, in the theory of elasticity, the stress tensor is defined only up to a divergence term ${\Psi^{jk\ell}}_{,\ell}$ where $\Psi_{jk\ell} = -\Psi_{kj\ell}$. Adding such a term won't change the final answer in our computations. This is useful to know in the context of the theory of symmetry of PDEs (or systems of PDEs), where apply Noether's theorem typically leads to nonsymmetric stress tensors. Then it is good to know a procedure for obtaining the equivalent symmetric stress tensor: put $$S_{[jk]} = 2 \, {\phi_{jk}}^\ell}_,\ell$$ Then put $$\chi_{jk\ell} = \phi_{k \ell j} + \phi_{j \ell k} - \phi_{j k \ell}$$ Then $$\tilde{S}^{jk} = S^{jk} + {\chi^{jk\ell}}_{,\ell}$$ is the desired symmetrization. We defined the stress tensor by saying that its gradient is the force density on a bit of matter, which has units of mass density times acceleration. The relativistic units of mass density are $L^{-2}$ while acceleration is a path curvature so has units $L^{-1}$. Thus, the components of the gradient of the stress tensor have units $L^{-3}$, so the stress tensor components have units $L^{-2}$. Thus, stresses are sectional curvatures, just like mass density and energy density. 6. STRAIN ENERGY Let us consider the work done by a very small displacement $$F^j \, u_j = -{S^{jk}}_{,k} \, {\delta u}_j$$ Integrating this over the volume of the body, we obtain $$W = -\int_{\partial \Omega} {S^{jk}} \, {\delta u}_j \, n_k dA - \int_\Omega S^{jk} {\delta u}_{j,k} \, dV$$ By taking the surface far away we can make the surface term vanish, leaving $$W = -\frac{1}{2} \, \int_\Omega S^{jk} \varepsilon_{jk} \, dV$$ The (density of) Helmholtz free energy is $F = E -TS$, where T is the temperature, S is the entropy, and E the internal energy of our body. The density of free energy which is due to strain, or strain energy density, is given by $$F = -\frac{1}{2} \, S^{jk} \, \varepsilon_{jk}$$ The expression $dE = T \, dS -\frac{1}{2} \, S^{jk} \, d\varepsilon_{jk}$, where dS is the change in entropy and T is the temperature, generalizes $dE = T \, dS - p \, dV$ from elementary thermodynamics. In other words, $dF = -\frac{1}{2} \, S^{jk} \, d\varepsilon_{jk}$, so with temperature held fixed we have $$S^{jk} = -\frac{\partial F}{\partial \varepsilon_{jk}}$$ The (density of) Gibbs free energy is $$\Phi = E - T \, S + S^{jk} \, \varepsilon_{jk}$$ With temperature held fixed we have $$\varepsilon_{jk} = -\frac{\partial \Phi}{S^{jk}}$$ In these expressions, note that I have chosen the sign for the stress tensor which is appropriate for considering elastic deformation as subject to restoring forces. In particular, isotropic compression, as in a pressure vessel, will yield positive stresses called pressures. Many authors, including Landau and Lifschitz, choose the opposite sign. This choice makes pressure terms appear with a negative sign in the stress tensor! This convention disagrees with the sign convention used in the study of perfect fluids in gtr, so I have adopted the sign as given in the expressions above. Many textbooks consider heating of bodies which results from their deformation due to internal stresses and/or applied forces. In this synopsis I will ignore these thermal effects. 7. HOOKE'S LAW Recall that at the outset we noted that for sufficiently forces applied to their surfaces, bodies exhibit a linear stress-strain relationship. Since we are representing both the strain and stress as second rank symmetric tensors, it follows that the natural expression of this law, valid for small strains, should be $$S^{jk} = -C^{jkmn} \, \varepsilon_{mn}$$ where $$C_{jkmn} = C_{kjmn} = C_{jknm} = C_{mnjk}$$ so that the fourth rank elastic tensor $C_{jkmn}$, which depends only on the properties of the material of which our body is made, has only 21 algebraically independent components (one more than the Riemann curvature tensor, with which it should not be confused; note that the Riemann tensor has similar but distinct symmetries, with some antisymmetries instead of symmetries which ensures that it has only 20 algebraically independent components). Note that the components of the elastic tensor must have relavistic units of curvature, the same as the stree tensor components. From the preceding section we immediately obtain the fact that the strain energy is quadratic in the strain: $$F = \frac{1}{2} \, C_{jkmn} \, \varepsilon^{jk} \, \varepsilon^{mn}$$ 8. HOMOGENEOUS ISOTROPIC MATERIALS Typically we are interested in computing the response to applied forces and/or body forces of a body made of a material which is homogeneous, meaning that the material properties do not change throughout the body, and isotropic, meaning that these properties do not depend upon orientation inside the body. Now the low order invariants of the strain tensor, which are of course independent of orientation, are its trace $$\varepsilon = {\varepsilon^n}_n$$ and the trace of the square of the operator ${\varepsilon_j}^k$, namely $${\varepsilon_j}^k \, {\varepsilon_k}^j = \varepsilon_{mn} \, \varepsilon^{mn}$$ So the free energy due to strain should be expressible as a linear combination of the square of the trace and the trace of the square: $$F = \frac{\lambda}{2} \, \varepsilon^2 + \mu \, \varepsilon_{jk} \, \varepsilon^{jk}$$ Then, since F is a homogeneous function of the strain components (having degree two), one can apply Euler's theorem on homogeneous functions: $$2 \, F = \varepsilon_{jk} \frac{\partial F}{\partial \varepsilon_{jk}} = -\varepsilon_{jk} \, S^{jk}$$ or $$F = -\frac{1}{2} \, \varepsilon_{jk} \, S^{jk}$$ in agreement with the result obtained in the previous section. This argument shows that, in the case of a homogeneous isotropic material, the 21 algebraically components of the elastic tensor must reduce to just 2. Indeed, algebraically speaking the vector space of fourth rank isotropic constant tensors is just three dimensional. A basis for this space is $$a_{jkmn} = \delta_{jk} \, \delta_{mn},$$ $$b_{jkmn} = \delta_{jm} \, \delta_{kn} + \delta_{jn} \, \delta_{km} - \frac{2}{3} \, \delta_{jk} \, \delta_{mn},$$ $$c_{jkmn} = \delta_{jm} \, \delta_{kn} - \delta_{jn} \, \delta_{km}$$ Plugging $$C_{jkmn} = \lambda \, a_{jkmn} + \mu \, b_{jkmn} + \gamma \, c_{jkmn}$$ into the definition, we find that the last term drops out by the symmetry of the strain tensor, and we are left with $$S_{jk} = -\lambda \, \varepsilon \, \delta_{jk} - 2 \mu \, \varepsilon_{jk}$$ Taking the trace, we find $S = -( 3 \lambda + 2 \mu) \, \varepsilon$, so that the inverse relationship is $$\varepsilon_{jk} = \frac{\lambda}{3 \lambda + 2 \mu} \, S \, \delta_{jk} - \frac{1}{2 \mu} \, S_{jk}$$ 9. ELASTIC MODULI As we'll see, in various contexts it can be convenient to introduce some additional parameters describing the physical properties of a material, which are defined in terms of $\lambda, \, \mu$. For ease of reference I'll collect them here. Modulus of incompressibility (first Lame coefficient): $$\lambda = \frac{\nu \, E}{(1 - 2 \nu) \; (1+\nu)} = \mu \, \frac{E-2\mu}{3 \mu-E}$$ Shear modulus (second Lame coefficient; written G in many books) $$\mu = \frac{1}{2} \, \frac{E}{1+\nu}$$ Bulk modulus $$K = \frac{1}{3} \, \frac{E}{1-2 \nu} = \frac{ E}{3} \, \frac{\mu}{3 \mu - E}$$ Poisson's ratio (written $\sigma$ in many books) $$\nu = \frac{1}{2} \, \frac{\lambda}{\lambda + \mu} = \frac{E}{2 \mu}-1$$ Young's modulus (modulus of extension) $$E = \mu \, \frac{3 \lambda + 2 \mu}{\lambda + \mu} = \frac{9 K \mu}{3 K + \mu}$$ Any two of the quantities $\lambda, \, \mu, \, \nu, \, E, \, K$ can be used to express any of the others. For example, we can rewrite Hooke's law for a homogeneous isotropic material in the form $$S_{jk} = - \frac{E}{1+\nu} \; \left( \varepsilon_{jk} + \frac{\nu}{1-2\nu} \, \varepsilon \, \delta_{jk} \right)$$ $$\varepsilon_{jk} = \frac{\nu}{E} \, S \, \delta_{jk} - \frac{1+\nu}{E} \, S_{jk}$$ These moduli are all defined to be positive for real materials. In relativistic units $\lambda, \, \mu, \, K, \, E$ have the dimensions of sectional curvature, while $\nu$ is dimensionless. To convert an energy density, pressure, or force expressed in ordinary units into relativistic units, multiply by the constant $G/c^4$ expressed in the appropriate units. Here are some examples in nonrelativistic units: For stainless steel, we have $\rho = 7.91 \; {\rm g}/{\rm cm}^3}$, $\nu = 0.30$, and $$E=19.6, \, K=16.3, \, \mu = 7.57 \; {\rm dyn}/{\rm cm}^2$$. For tungsten carbide, we have $\rho = 13.8 \; {\rm g}/{\rm cm}^3}$, $\nu = 0.22$, and $$E=53.4, \, K=31.7, \, \mu = 21.95 \; {\rm dyn}/{\rm cm}^2$$. For pyrex, we have $\rho = 2.32 \; {\rm g}/{\rm cm}^3}$, $\nu = 0.24$, and $$E=6.2, \, K=4.0, \, \mu = 2.5 \; {\rm dyn}/{\rm cm}^2$$. For lucite, we have $\rho = 1.18 \; {\rm g}/{\rm cm}^3}$, $\nu = 0.4$, and $$E=0.40, \, K=0.66, \, \mu = 0.143 \; {\rm dyn}/{\rm cm}^2$$. (To be continued...)
 Recognitions: Science Advisor 10. SOME BASIC EXAMPLES Let's see how this machinery works for some particularly simple examples. EXAMPLE ONE: Consider a displacement vector of form $$\vec{u} = \frac{p}{E} \, \left( - \nu \, \left( x \, \partial_x + y \, \partial_y \right) + \, z \, \partial_z \right)$$ where of course we are using a Cartesian chart (the only example where the coordinate basis is also a frame field). Then the strain tensor is (think: components wrt the frame) $$\varepsilon_{ab} = - \frac{p}{E} \, \left[ \begin{array}{ccc} \nu & 0 & 0 \\ 0 & \nu & 0 \\ 0 & 0 & -1 \end{array} \right]$$ and the stress tensor is $$S^{ab} = p \; \left[ \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{array} \right]$$ where -p is the tension along $\partial_z$, this is negative since this is a tension. Exponentiating $\vec{u}$, that is, computing the integral curves of this congruence, gives $$(x,y,z) \mapsto \left( x \, \exp \left(-p \, \frac{\nu}{E} \, \tau \right), y \, \exp \left(-p \, \frac{\nu}{E} \, \tau \right), z \, \exp \left( p \, \frac{1}{E} \, \tau \right) \right)$$ To first order in the parameter $\tau$, this is $$(x,y,z) \mapsto \left( x \, \left( 1-\frac{p \, \nu \, \tau}{E} \right), y \, \left( 1-\frac{p \, \nu \, \tau}{E} \right), z \, \left( 1+\frac{p \tau}{E} \right) \right)$$ This is a diffeomorphism which shrinks orthogonally and expands axially, so we have shown that a constant purely axial tension produces the pattern noted experimentally in section 1. Rearranging terms gives our diffeomorphism in the form $$(x,y,z) \mapsto (x,y,z) + \tau \, \vec{u}$$ so our assumption of small displacements amounts to assuming a small parameter when we exponentiate the displacement vector field to obtain our diffeomorphism. The strain energy is $F = \frac{p^2}{2 E}$. Here, Young's modulus specifies the strain energy, for the case of purely axial tension, in a particularly simple fashion. Notice that Poisson's ratio does not appear in this expression! For those who haven't previously encountered the phrase "exponentiate a vector field", let me pause to briefly explain that. To find the integral curves of a vector field such as $$-y \, \partial_x + x \, \partial_y$$ we write the corresponding system of ODEs $$\dot{x} = -y, \; \dot{y} =x, \; x(0) = x_0, \; y(0) = y_0$$ where dot refers to differentiation with respect to a parameter (the parameter of the "unidimensional subgroup") and solve, giving (in this example) $$x(s) = x_0 \, \cos(s) - y_0 \, \sin(s), \; \; y(s) = -x_0 \, \sin(s) + y_0 \, \cos(s)$$ This describes a one dimensional subgroup of diffeomorphisms of the euclidean plane to itself, indeed these are isometries (distance preserving diffeomorphisms), in fact rotations. Here, the original vector field is the generator of the (one-dimensional) Lie subalgebra corresponding to this one-dimensional Lie subgroup. The term "exponentiate" comes from the fact that in intermediate courses on systems of ODEs one learns how to solve such initial value problems by exponentiating a matrix. Here, we define the exponential of a matrix times a scalar parameter as $$\exp( \lambda \, M) = I + \lambda \, M + \frac{\lambda^2}{2} \, M^2 + \frac{\lambda^3}{6} M^3 + \dots$$ just as for real numbers. Often this infinite series terminates after a finite number of steps or is periodic, or otherwise easy to evaluate. See section 1.4 of Frankel, Geometry of Physics, Cambridge University Press, 1997, for more about integral curves and one-parameter subgroups. EXAMPLE TWO: Consider an isotropic compression $$\vec{u} = -\frac{p}{3 \, K} \left( x \, \partial_x + y \, \partial_y + z \, \partial_z \right)$$ Then the strain tensor is $$\varepsilon_{ab} = -\frac{p}{3 K} \, \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right]$$ and the stress tensor is $$S^{ab} = p \; \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right]$$ Note that the diagonal components are positive, in agreement with the convention used in studying perfect fluids in gtr. Exponentiating our displacement vector field gives $$(x,y,z) \mapsto \exp(- \frac{p}{K} \, \tau) \; \left( x, y, z \right) \approx (x,y,z) - \frac{p}{K} \; (x,y,z)$$ The strain energy is $F = \frac{p^2}{2 K}$. Here, K is called the bulk modulus because it specifies the strain energy, for the case of isotropic compression, in a particularly simple fashion. EXAMPLE THREE: Consider a displacement vector of form $$\vec{u} = \frac{q}{2 \mu} \, x \, \partial_y$$ Then the strain tensor is $$\varepsilon_{ab} = - \frac{q}{2 \mu} \; \left[ \begin{array}{ccc} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right]$$ and the stress tensor is $$S^{ab} = q \; \left[ \begin{array}{ccc} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right]$$ Exponentiating our displacement vector field gives the pure shear $$(x,y,z) \mapsto (x, y-\frac{q}{2 \mu} \, \tau, z)$$ The strain energy is $F = \frac{q^2}{2 \mu}$. Here, $\mu$ is called the shear modulus because it specifies the strain energy, for the case of a pure shear, in a particularly simple fashion. EXAMPLE FOUR: Consider a nonconstant displacement vector of form $$\vec{u} = \frac{q z}{\mu} \, \left( -y \, \partial_x + x \, \partial_y \right)$$ Then the strain tensor is $$\varepsilon_{ab} = \frac{q}{2\mu} \; \left[ \begin{array}{ccc} 0 & 0 & -y \\ 0 & 0 & x \\ -y & x & 0 \end{array} \right]$$ and the stress tensor is $$S^{ab} = q \; \left[ \begin{array}{ccc} 0 & 0 & -y \\ 0 & 0 & x \\ -y & x & 0 \end{array} \right]$$ Exponentiating the displacement vector gives $$(x,y,z) \mapsto (x_0 \, \cos ((q \, z/\mu) \, \tau) + y_0 \, \sin((q \, z/\mu) \, \tau), -x_0 \, \sin ((q \, z/\mu) \, \tau) + y_0 \, \cos ((q \, z/\mu) \, \tau), z )$$ This is an axially symmetric deformation, a pure torsion about the z-axis. It has the properties that there is no axial elongation, cross sections remain parallel but are rotated by an angle depending linearly on z, and diameters in cross sections are mapped to rotated diameters. The strain energy is $F = \frac{q \, (x^2+y^2)}{8 \mu}$. Thus, the energy of a shaft experiencing simple torsion is concentrated near the surface of the shaft. Exercise: rewrite Examples One, Three, and Four in terms of a cylindrical chart on E^3, using the frame used in section 3. Rewrite Example Two in terms of a polar spherical chart on E^3, using a suitable frame field. 11. STATIC EQUILIBRIUM From the definition of the stress tensor, we see that if no body forces act, the internal stresses on each bit of material must balance, i.e. the net force must be zero, which gives the condition ${S^{jk}}_{,k} = 0$, i.e. the divergence of the stress tensor must vanish. If a body force $\rho \vec{a}$ acts, where $\rho$ is the density of matter (which will in general be a function of position), this becomes $${S^{jk}}_{,k} = \rho \, a^j$$ Alternatively, recall that static equilibrium occurs at a local minimum of the free energy. Taking the variation confirms the result we just obtained. The boundary conditions are these: on any surface of our body with an applied force, we must have $$S^{jk} \, n_k = F^j$$ where $\vec{n}$ is a unit normal to the surface. On any surface of the body which is regarded as immovably fixed (e.g. in the case of a beam with clamped endpoints), the displacement must vanish. For a homogenous isotropic body, plugging in our expression for the stress tensor gives $$\frac{2 \, (1+\nu)}{E} \, \rho \, \vec{a} = \Delta \vec{u} + \frac{1}{1-2 \nu} \, \nabla \left( \nabla \cdot \vec{u} \right)$$ Using the vector calculus identity $$\Delta \vec{u} = \nabla \left( \nabla \cdot \vec{u} \right) - \nabla \times \left( \nabla \times \vec{u} \right)$$ this can be rewritten $$\frac{(1+\nu)}{1-\nu} \; \frac{1-2 \nu}{E} \; \rho \vec{a} = \nabla \left( \nabla \cdot \vec{u} \right) - \frac{1-2 \nu}{2 (1-\nu)} \; \nabla \times \left( \nabla \times \vec{u} \right)$$ When no body forces act, $\vec{a}=0$ and this reduces to $$(1-2 \nu) \, \Delta \vec{u} + \nabla \left( \nabla \cdot \vec{u} \right) = 0$$ Taking the divergence of both sides and using our vector calculus identity, this becomes $\Delta \left( \nabla \cdot \vec{u} \right) = 0$ or $$\Delta \, \Delta \, \vec{u} = 0$$ In other words, the components of the displacement vector (with respect to a frame field) must be (three-dimensional) biharmonic functions. An important special case arises when we assume that the displacement vector field is planar, i.e., in a suitable Cartesian chart, $$\vec{u}= a(x,y) \, \partial_x + b(x,y) \, \partial_y$$ Then it can be shown (see Landau and Lifschitz) that the stress tensor can be written in the form $$S^{ab} = \left[ \begin{array}{ccc} f_{yy} & -f_{xy} & 0 \\ -f_{xy} & f_{xx} & 0 \\ 0 & 0 & -\nu \, \Delta f \end{array} \right]$$ where $f(x,y)$ is a two-dimensional biharmonic function. Then the strain tensor is also planar: $$\varepsilon_{ab} = \frac{1}{2 \, \mu} \; \left[ \begin{array}{ccc} \nu \, \Delta f - f_{yy} & f_{xy} & 0 \\ f_{xy} & \nu \, \Delta f - f_{xx} & 0 \\ 0 & 0 & 0 \end{array} \right]$$ For those who have never seen a biharmonic function of any dimension, a rational two-dimensional example is $f(x,y) = x^3 \, y + x^2 \, y^2 - x^4/3$, for which $\Delta f = 6 \, x \, y + 2 \, y^2 - 2 \, x^2$, from which it is easy to see that $\Delta \, \Delta \, f = 0$. Further examples of two-dimensional biharmonic functions can be readily generated using the fact that $f(x,y) = (x^2+y^2) \, u(x,y) + v(x,y)$ is biharmonic whenever $u,v$ are harmonic. The close relationship between biharmonic and harmonic functions is partially explained by considering the "triangularized" system of linear PDEs: $$\Delta u = v, \; \Delta v = 0$$ Here, if u,v is any solution, u must be biharmonic. Those of you who already know how to compute the point symmetry group of a system of PDEs may wish to compare the symmetry of this (second order) system with the point symmetry group of the (fourth order) biharmonic equation itself. In making models of static equilibria, we typically chose a coordinate chart and frame adapted to any symmetries of the problem, then write down an appropriate displacement vector with some undetermined functions. Then, plugging into the equilibrium equation, we immediately obtain equations which can often be solved. Plugging the result into the stress tensor and comparing with the appropriate boundary conditions (most of which will say that any normal stresses at the surface of our body must vanish, unless balanced by a force applied to part of the surface of the body) typically then determines the displacement in terms of the physical parameters of the problem. From this we can compute the strain and stress tensor. From the strain tensor we can compute the approximate total change in the surface shape of the body in its strained condition. From the stress tensor we can sometimes also read off conditions under which the equilibrium becomes unstable, in which case we typically conclude that, regarding some externally applied force involved in the statement of the problem, if some critical value of this force is exceeed, then the stressed body will "jump" to a new equilibrium. An important example of this phenomenon is the Euler buckling of a loaded beam. 12. EXAMPLES OF STATIC EQUILIBRIA EXAMPLE ONE: Consider a spherical pressure vessel with radii $0 < R_1 < R_2 < \infty$, with internal pressures $p_1$ pressing on the internal surface, and external pressure $p_2$ pressing on the outer surface, but with no body forces acting. Adopt a polar spherical chart with coframe $$\sigma^1 = dr, \; \sigma^2 = r \, d\theta, \; \sigma^3 = r \, \sin(\theta) \, d\phi$$ Assume a purely radial displacement vector field $$f(r) \, \vec{e}_1$$ Plugging into the equilibrium conditions gives $f(r) = a \, r + b/r^2$. The boundary conditions state $S^{11} = p_1$ at $r=R_1$ and $S^{11} = p_2$ at $r=R_2$. (This illustrates what I said about externally applied surface forces entering into a problem only via the boundary conditions.) From these we obtain the constants of integration a,b, in terms of the pressures and dimensions of the vessel: $$a = \frac{1-2\nu}{E} \; \frac{p_1 \, R_1^3 - p_2 \, R_2^3}{R_2^3-R_1^3}$$ $$b = \frac{1+\nu}{2 \, E} \; \frac{R_1^3 \, R_2^3 \, (p_1 - p_2)}{R_2^3-R_1^3}$$ If $p_2=0$ and $R_1 = R, \; R_2 = R + h, \; h \ll R$, then to first order in h, $$\vec{u} \approx \frac{p \, R^2}{3 h} \, \frac{1-\nu}{E} \; \vec{e}_1$$ The radial stress is dominated by the tangential stresses, which are of the order $p \, R/(2 \, h)$. EXAMPLE TWO: Consider a rigidly rotating shaft (solid cylinder) with radius R and constant angular velocity $\Omega$. Adopt a cylindrical chart with coframe $$\sigma^1 = dz, \; \sigma^2 = dr, \; \sigma^3 = r \, d\phi$$ The appropriate body acceleration is $\vec{a}=\Omega^2 \, r$. Assume a purely radial displacement $\vec{u}=f(r) \, \vec{e}_2$. Plugging into the equation of equilibrium we find $$f(r) = c_1 r + c_2/r^2 + \Omega^2 \, \rho \, \frac{(1+\nu) \, (1-2\nu)}{8 \, E \, (1-\nu)} \, r^3$$ To avoid a singularity in f on the axis, we must take $c_2 = 0$. Computing the stress tensor, the normal stress $S^{22}=0$ at r=R. This gives the displacement vector $$\vec{u} = \frac{1+\nu}{1-\nu} \; \frac{1-2\nu}{8 \, E} \; \rho \, \Omega^2 \, r \left( (3-2\nu) R^2-r^2 \right) \; \partial_r$$ The strain tensor (components wrt the coframe!) is $$\varepsilon_{ab} = \frac{1+\nu}{1-\nu} \; \frac{1-2\nu}{8 \, E} \; \rho \, \Omega^3 \left[ \begin{array}{ccc} 0 & 0 & 0 \\ 0 & (3-2\nu) \, R^2 - 3 \, r^2 & 0 \\ 0 & 0 & (3-2\nu) \, R^2 - r^2 \end{array} \right]$$ Notice that this is a purely planar strain (there is no axial strain). The stress tensor (components wrt the dual frame!) is $$S^{ab} = \frac{\rho \, \Omega^2}{8 \, (1-\nu)} \; \left[ \begin{array}{ccc} 2 \nu \, (3-2\nu) R^2 -2 \, r^2 & 0 & 0 \\ 0 & (3-2\nu) \, (R^2-r^2) & 0 \\ 0 & 0 & (3-2\nu) \, R^2 -(1+2\nu) \, r^2 \end{array} \right]$$ The internal stress in the shaft is maximal on the axis of rotation. This is important to know in the design of turbines, for example! Exercise: generalize this to a cylindrical shell with inner and outer radii $) < R_1< R_2 < \infty$. What do you notice happens in the limit $R_1 \rightarrow 0$? (Hint: very small holes can have large effects!) EXAMPLE THREE: Consider a shaft (solid cylindrical body) with radius R which is subjected to a torque externally applied at one end. Adopt a cylindrical chart with the usual frame field. Assume $$\vec{u} = \tau \; \left( \psi \, \vec{e}_2 + z \, r \vec{e}_3 \right)$$ where $\psi$ is a function of $r, \phi$ only, with no body forces acting. The equilibrium condition gives $$\Delta \psi = \psi_{rr} + \frac{\psi_r}{r} + \frac{\psi_{\phi \phi}}{r^2} = 0$$ This gives the strain tensor $$\varepsilon_{ab} = \frac{\tau}{2} \left[ \begin{array}{ccc} 0 & \psi_r & r + \frac{\psi_\phi}{r} \\ \cdot & 0 & 0 \\ \cdot & 0 & 0 \end{array} \right]$$ and stress tensor $$S^{ab} = -\frac{\tau}{4 \, \mu} \, \left[ \begin{array}{ccc} 0 & \psi_r & r + \frac{\psi_\phi}{r} \\ \cdot & 0 & 0 \\ \cdot & 0 & 0 \end{array} \right]$$ The strain energy is $$F = \frac{\tau^2}{2\, \mu} \; \left( \psi_r^2 + \left( \frac{\psi_\phi}{r} + r \right)^2 \right)$$ We require $\psi$ to be regular on the axis. The normal stresses on the cylindrical surface and at the far end of the shaft vanish automatically. This time the internal stresses are (most likely) maximal near the surface of the shaft, with $$\operatorname{max} S^{23} \approx - \frac{\tau \,R}{4 \mu}$$ Thus, the distribution of stresses during spinup of the shaft may be very different from the distribution during steady-state rotation. In the last example, it is useful to know a bit more about biharmonic functions. In particular, the boundary problem $$\Delta \, \Delta \, w = 0 \; {\rm inside} \; r=R, \; w(R,\phi)=f(\phi), \, w_r(R,\phi)= g(\phi)$$ can be solved by evaluating certain integral transform of the boundary values, as follows: $$w(r,\phi) = \frac {(R^2-r^2)^2}{2\pi R} \; \int_0^{2\pi} \frac{(R-r \cos(\eta-\phi) ) \, f(\eta) \, d\eta} {(r^2+R^2-2 R r \cos(\eta-\phi))^2}$$ $$\hspace{1in} - \frac{(R^2-r^2)^2}{4\pi R} \; \int_0^{2\pi} \frac {g(\eta) \, d\eta} {r^2+R^2-2 R r \cos(\eta-\phi)}$$ Note that this expression is convenient for numerical solutions obtained using numerical integration. (To be continued...)

Recognitions:

## What is the Theory of Elasticity?

 Quote by cristo Hi Chris, This is a very interesting thread indeed.
Thanks!

 Quote by cristo I have taken a course in classical elasticity a few years back, however it was taught somewhat differently to the way in which this thread is progressing-- I imagine since this is clearly going to lead into a relativistic discussion! The most obvious difference is that we did not study frame fields (nor have I encountered them in any GR study thus far). From what I have read above, I gather that the frame field is locally chosen so that the basis vectors are unit vectors.
Exactly.

 Quote by cristo However, I imagine there is more to it than this! Could you suggest a good reference from which I can learn more? (I've looked through the few GR texts I've got to hand, and they do not appear to be covered in those).
Not much to say, if you compute tensors in the coordinate basis and then report results using the frame. (But you can also compute directly in the frame using appropriate formulas for covariant differentiation.) "To be safe" I'll point you and anyone else with a similar concern to the last version I edited of a Wikipedia article I wrote http://en.wikipedia.org/w/index.php?...oldid=42117350
(If you spot any typos, don't edit that version or you'll revert any subsequent work. Naturally I think that would be an improvement but it would be sure to annoy active Wikipedians.)

 Quote by cristo As an aside, you have a typo in the third/fourth paragraph under section 4. There is an r^2 missing in front of the dphi^2 term in the line element for the cylindrical coordinates.
Yes, indeed! Thanks much--- I've corrected that. If you spot any more please let me know.

 Quote by Stingray Chris, Thank you very much for such a useful post. While I was aware of much of this material, it's nice to have it all in one short summary.
Thanks! It's a lovely subject, and great fun in that it connects physics with stuff we observe in daily life.

 Quote by Stingray I've always been interested in the theory of elasticity, but have never given it enough time to learn properly. Continuum mechanics of all kinds seems to be almost entirely ignored in physics courses, unfortunately. It's kind of strange considering how most everyday phenomena depend on it.
Yes, exactly! In math too unfortunately there are wonderful and important subjects which are terribly undertaught. Yet when I propose to start students on math earlier and keep them learning longer, I hear howls of dismay. ("Eight years for an undergraduate degree?!!!!") I think this is perfectly sensible on many grounds. After all, it is widely appreciated that earning a college degree has nothing to do with job-related skills, as all college students are quick to remind us whenever they feel unhappy with the syllabus! Rather, requiring all ambitious young adults to spend time in college is a socially acceptable way of delaying the entry of the scions of the middle class into the job market as long as possible. From this point of view, the universities are simply a kind of glorified daycare, providing busywork to prevent young people from turning to crime out of sheer boredom. This social function of (nearly) universal college education outrages some students who see it as a kind of universal scam which simply wastes their time, and indeed, this scheme does exhibit some characteristics of a scam. But I hasten to add that as technology advances, we are quickly approaching a world in which citizens are capable of working only (except in low quality service jobs) only for a dozen years at most, so that life is framed by increasingly long college training on one end and increasingly long retirement at the other. Given this, I think it makes perfect sense to train students much more carefully for much longer periods. The added cost of all that school would be offset by added productivity.

 Quote by Stingray I tried learning from Marsden and Hughes' book a few years ago, but it's not exactly bedtime reading. I'll check out the other references you've given.
Marsden is responsible for a very interesting theorem in gtr, to the effect that the vacuum solutions which possess Killing vectors are exceptional, so that if you want to understand gtr, pursuing exact solutions is terribly wrongheaded. (Naturally this interpretation is deliberately misleading--- like everything else in this subject, it's all more subtle than that!)

Recognitions:
 Quote by Chris Hillman Marsden is responsible for a very interesting theorem in gtr, to the effect that the vacuum solutions which possess Killing vectors are exceptional, so that if you want to understand gtr, pursuing exact solutions is terribly wrongheaded. (Naturally this interpretation is deliberately misleading--- like everything else in this subject, it's all more subtle than that!)
I didn't know that. Do you have a reference? I'd like to know more precisely what that means.

I do realize that Marsden has contributed many things in several different fields. He seems like a very smart guy. I regret not taking any of his courses when I had the chance.
 Recognitions: Science Advisor Marsden is indeed a very smart guy. (For those who don't know, he is rather legendary for proving many theorems, mostly related to Hamiltonian dynamical systems, which are of great interest both in pure mathematics and for their applications to a vast array of subjects in physics and other disciplines. He's also legendary for--- oh, mathematicians are such gossips!--- never mind.) I am thinking of a body of work due to Marsden, Arthur Fischer, and Marsden's former student Judith Arms, on linearization stability of vacuum solutions and related to the Hamiltonian formulation of gtr and "superspace"--- hopefully this brief description is not too misleading! (In Wikipedia I'd be required to disclose that I am slightly acquainted with Marsden and Arms, but in my experience WP:COI is more honored in the breach than the observance.) From http://hcr3.isiknowledge.com/ these are among the most relevant papers: A. E. Fischer and J. E. Marsden [1976] A new Hamiltonian structure for the dynamics of general relativity. Gen. Rel. and Grav., 7, 915-920.. J. Arms, A. Fischer and J. E. Marsden [1975] Une approche symplectique pour des theoremes de decomposition en geometrie et relativite general. C. R. Acad. Sci., Paris, 281, 517-520.. A. E. Fischer and J. E. Marsden [1975] Linearization stability of nonlinear partial equations. Proc. Symposia in Pure Math., 27, 219-263.. A. E. Fischer and J. E. Marsden [1974] Global analysis and general relativity. Gen. Relativity and Grav.. 5, 71-76.. A. E. Fischer and J. E. Marsden [1974] General relativity as a Hamiltonian system. Symposia Math. XIV, 193-205.. A. E. Fischer and J. E. Marsden [1973] New theoretical techniques in the study of gravity. Gen. Relativity and Grav., 4, 309-317.. IIRC, there is a review of this stuff in the Einstein cententary volume. The techniques they used have led to further developments: http://www.arxiv.org/abs/gr-qc/9304004 Robert Bartnik and Gyula Fodor, "On the restricted validity of the thin sandwich conjecture" Phys. Rev. D 48, 3596 - 3599 (1993) C. Torre, "Is general relativity an "already parametrized" theory?" Phys. Rev. D 46, R3231 - R3234 (1992) In happier days, Charlie Torre often posted at sci.physics.research and until the recent reorganization of the Google search tool, one could find a discussion between he and I of this stuff which took place there c. 1999.
 Recognitions: Science Advisor 13. STATIC EQUILIBRIA WITH GRTENSORII GRTensorII is an extremely useful tool for students of gtr; see Eric Poisson, A Relativist's Toolkit, Cambridge University Press, 2004 for some other applications. GRTensorII runs under Maple and is available at no cost at http://grtensor.phy.queensu.ca/ Assuming you have installed GRTensorII, have begun a Maple session and have loaded GRTensorII (see the above cited website for instructions), here are some definitions which you can paste (or load from a file) into your Maple session. Code: # NOTE WELL: We use sign convention of Petter and Walecka in which stress # has opposite sign to Landau and Lifschitz, Graff, etc., but agrees # with MTW sign conventions for gtr context. # # Input stress tensor grdef(stress{((a) (b))}): # Compute traceless stress tensor grdef(TLS{((a) (b))} := stress{((a) (b))} - eta{(a) (b)}*stress{(m) ^(m)}): # Compute strain tensor (symmetric): # nu = Poisson's ration, E = Young's modulus grdef(epsilon{((j) (k))} := -(1+nu)/E*stress{(j)(k)}+nu/E*eta{(j)(k)}*stress{(m)^(m)}): grdef(t1epsilon := epsilon{(m) ^(m)}): # Compute elastic stored energy grdef(ele := -1/2*epsilon{(m) (n)}*stress{^(m) ^(n)}): # Compute divergence of stress tensor grdef(divstress{^(a)} := stress{^(a) ^(n) ;(n)}): # # COMPUTING STRAIN, STRESS, STRAIN ENERGY AND EQUILIBRIUM CONDITIONS # FROM GIVEN DISPLACEMENT AND BODY ACCELERATION # Enter a displacement vector U (usually coeffs have undetermined functions): grdef(U{^(a)}): # Enter a body specific force vector (i.e. acceleration vector): grdef(F{^(a)}): # Compute strain tensor for this U grdef(strainU{((a) (b))} := (U{(a) ;(b)}+U{(b) ;(a)})/2): # Compute stress tensor for this U (homogeneous isotropic ONLY!) grdef(stressU{((a) (b))} := -E/(1+nu)*( strainU{(a) (b)} + nu/(1-2*nu)*strainU{^(n) (n)}*eta{(a) (b)})): # Compute equation of equilibrium for given U and F: grdef(equilU{^(a)} := -2*(1+nu)/E*rho*F{^(a)} + strainU{^(a) ^(n) ;(n)} + 1/(1-2*nu)*strainU{^(n) ^(a) ;(n)}): # Compute strain energy for given U grdef(eleU{} := -1/2*strainU{(m) (n)}*stressU{^(m) ^(n)}): We can define a coframe field written in a cylindrical chart on E^3 by: Code: Ndim_ := 3: x1_ := z: x2_ := r: x3_ := phi: eta11_ := 1: eta22_ := 1: eta33_ := 1: bd11_ := 1: bd22_ := 1: bd33_ := r: Info_ := Cylindrical coordinates for E^3: # Chart covers -infty < z < infty, 0 < r < infty, -Pi < phi < Pi I usually put this in a text file called something like E2_cyl.mpl, and use the qload command of GRTensorII to load this coframe field. Suppose you wish to compute the strain and strain energy from a given stress tensor. These commands will accomplish that: Code: grcalcalter(stress(bdn,bdn),1):grdisplay(_); grcalcalter(epsilon(bdn,bdn),ele,stress(bup,bup),divstress(bup),1):grdisplay(_); The first causes GRTensorII to prompt you to enter the six algebraically independent components of the desired stress tensor and the second line then computes the corresponding strain and strain energy density. In solving static equilibrium problems, it is often more convenient to work in the opposite direction, by assuming a displacement vector field of a certain form. Typically that entails writing down an Ansatz for $\vec{u}$ in terms of a suitable frame field written in a chart adapted to the symmetries of the problem. Usually this involves some undetermined functions for which we will solve. The following sequence of commands will cause GRTensorII to prompt you for your displacement vector Ansatz, then for any body force per unit mass (i.e. acceleration), and it will then compute some quantities in terms of the undetermined functions. The quantity equilU(bup) must vanish, so we write these equations and use the Maple casesplit command to "triangularize" them using Groebner basis type methods of computational algebra. (These methods are analogous to Gaussian reduction, but the computations occur in a differential ring rather than a vector space.) Code: grcalcalter(U(bup),equilU(bup),stressU(bup,bup),eleU,1):grdisplay(_); [seq(grcomponent(equilU(bup), [grcomponent(x(up),[j])]),j=1..3)]: casesplit(%); The output of casesplit is one or more sets of equations and conditions under which the equations are valid. If there is only one set, we can strip off the equations using Code: op(%)[1]; If there is more than one, we can excise the second set of equations (say) using Code: op(%[2])[1]; Now we can use dsolve or pdsolve to try to solve these equations. With some combination of luck, experience, and good judgement this often results in a solution with some undetermined constants of integration. We substitute this into the results we previously computed: Code: grmap(U(bup),F(bup),strainU(bdn,bdn),stressU(bup,bup),eleU,subs,%,'x'): grmap(_,simplify,'x'):grdisplay(_); Now, typically we will need to satisfy some boundary conditions stating that the normal stresses vanish at the surface of the body, except where they are balanced by some externally applied force (such as the pressure inside a pressure vessel). So to make the radial component of the stress vanish at the surface of a solid cylinder, we could write Code: subs(r=R,grcomponent(stressU(bdn,bdn),[2,2])); and then solve for one of the integration constants. If the right hand side is not a constant expression in terms of the physical parameters such as dimensions of the body and density and elastic moduli characterizing the material of which it is made, something went wrong! In a well-formulated problem, all the integration constants should be determined in this way. As a GRTensorII exercise, assiduous readers can try to duplicate the results reported in the previous section. 14. STATIC EQUILIBRIUM OF BEAMS AND COLUMNS In this section, we study in more detail the static equilibrium of beams and columns under various externally applied surface forces, and possibly also body forces such as gravity. Examples include the loading of a cantilever beam or a vertical column or cable, or the sagging of a horizontal cable under its own weight. (TO APPEAR--- I'M SKIPPING AHEAD FOR NOW; should include torsion viz. path curvature of a curve, symmetry Euler beam equation, buckling, application of Krasnoselskii-Rabinowitz bifurcation to buckling of a loaded column, and beams in hyperbolic geometry) 15. STATIC EQUILIBRIUM OF PLATES In this section, we study in more detail static equilibrium of plates under various externally applied surface forces, in addition to (possibly) gravitational or centrifugal body forces. The equation of equilibrium for a plate can be written $$D \, \Delta \, \Delta \zeta = P$$ where $$D = \frac{E \, h^3}{12 \, (1-\nu^2)}$$ Here, h is the thickness of the plate, $\zeta$ is the "vertical" component of the displacement vector field $\vec{u}$, and P is the "vertical" component of any body force. Note that P vanishes if only surface forces act on the plate. The appearance of the biharmonic equation should be no suprise by now, but the coefficient is new. It is called the flexural rigidity modulus, and depends on the dimensions of the plate as well as on the material of which it is made. (TO APPEAR--- I'M SKIPPING AHEAD FOR NOW; should include symmetry analysis of plate equation, plates in hyperbolic three-space) 16. ELASTIC WAVES The general (nonrelativistic!) equation governing the displacement of bits of material in an elastic body is $$\rho \ddot{u^j} + {S^{jk}}_{,k} = \rho \, a^j$$ For a homgeneous isotropic material, this can be rewritten $$\rho \, \ddot{\vec{u}} = \rho \, \vec{a} + \mu \, \Delta \vec{u} + \left( \lambda + \mu \right) \; \nabla \left( \nabla \cdot \vec{u} \right)$$ Once again, our boundary conditions are that normal components of stress at the surface of the body must vanish or match any applied surface forces, and in addition the displacement vector field (and possibly some of its derivatives) must vanish at any parts of the body which are clamped in place. The Helmholtz decomposition of an arbitary vector field in R^3 can be expressed as saying that any vector field can be written uniquely as the sum of an irrotational vector field plus an incompressible vector field, plus a harmonic vector field (which I am disposed to blithely ignore without offering any justification): $$\vec{u} = \vec{v} + \vec{w}$$ where $$\nabla \times \vec{v} = 0, \; \; \nabla \cdot \vec{w} = 0$$ Similarly, $$\vec{a} = \vec{b} + \vec{c}$$ where $$\nabla \times \vec{b} = 0, \; \; \nabla \cdot \vec{c} = 0$$ Plugging this decomposition into our equation of motion, we obtain an expression which looks like the sum of two separate equations, one involving only the incompressible part, and the other involving only the irrotational part. Appealing to the Fourier analysis of wave solutions, one can argue that our wave equation really split into two separate equations, one describing longitudinal waves (p-waves or "push waves" in seismology), and the other transverse waves (s-waves or "shear waves" in seismology). After applying the vector calculus identity $$\Delta \vec{q} = \nabla \left( \nabla \cdot \vec{q} \right) - \nabla \times \left( \nabla \times \vec{q} \right)$$ the irrotational component of our equation of motion becomes: $$\rho \, \ddot{\vec{v}} = \left( \lambda + 2 \, \mu \right) \; \nabla \left( \nabla \cdot \vec{v} \right) + \rho \, \vec{b}$$ In the absence of irrotational body forces, this becomes the wave equation $$\frac{1}{c_p^2} \; \ddot{\vec{v}} = \Delta \vec{v}$$ where $$c_p^2 = \frac{\lambda + 2 \, \mu}{\rho} = \frac{3 \, K + 4 \, \mu}{3 \, \rho} = \frac{1-\nu}{1+\nu} \; \frac{E}{1-2 \nu} \; \frac{1}{\rho}$$ This describes longitudinal waves of displacement propagating at speed $c_p$. These waves change the trace of the strain tensor as they propagate, i.e. they change the volume enclosed by a small sphere of material particles. The incompressible component of our equation of motion becomes, after applying the noted identity twice: $$\rho \, \ddot{\vec{w}} = \mu \, \Delta \vec{w} \right) + \rho \, \vec{c}$$ In the absence of incompressible body forces, this becomes the wave equation $$\frac{1}{c_s^2} \; \ddot{\vec{w}} = \Delta \vec{w}$$ where $$c_s^2 = \frac{\mu}{\rho}$$ This describes shear waves propagating at speed $c_s$. Note that the speed of s-waves depends only on the shear modulus (and the density) of the material, and is always less than the speed of p-waves. These s-waves do not change the trace of the strain tensor, so they do not affect the volume enclosed by a small sphere of material particles. These results have been derived under the tacit assumption that the wavelengths of the waves are small compared to the size of the object. If this condition is violated, they may not hold true. Some examples in nonrelativistic units: For stainless steel, we have $c_p = 5.79, \, c_s = 3.10 \, 10^5 \, {\rm cm/s}$ For tungsten carbide, we have $c_p = 6.66, \, c_s = 3.98 \, 10^5 \, {\rm cm/s}$ For pyrex, we have $c_p = 5.64, \, c_s = 3.28 \, 10^5 \, {\rm cm/s}$ For lucite, we have $c_p = 2.68, \, c_s = 1.10 \, 10^5 \, {\rm cm/s}$ These speeds are on the order of $10^{-5}$ times the speed of light, so sound waves in steel or other materials belong firmly to the nonrelativistic domain. The analysis of elastic waves propagating inside objects can become quite complicated. One reason for this is that when an p-wave propagating inside a body encounters the surface of the body, it is not only reflected from the surface but partially converted into an s-wave; this is called mode conversion. Another reason is that in addition to p-waves and s-waves propagating in the interior (and reflecting off the surface, scattering off internal defects, and so on), there is another type of wave, called Rayleigh waves, which propagate largely on the surface of the body. All these topics are treated in Graff and in Landau and Lifschitz. (To be continued...)

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 Quote by Voltage Thanks for all this Chris. I have some familiarity with Elastodynamics. Can you say how this Theory of Elasticity relates to it? Is the former a subset of the latter?
Gee, you're asking me? OK: elastodynamics is simply that part of the theory of elasticity which deals with nonstatic situations, such as p-waves, s-waves, Rayleigh waves, and (aha!) modeling the spinup of an elastic hoop and studying questions such as: is the spinup stable wrt small vibrations? (These vibrations would be governed by the dynamical equation in Section 16). In contrast, Sections 14,15 on static equilibria belong to elastostatics. So yes, both elastostatics and elastodynamics are subsumed within the theory of elasticity.

Most textbooks on elasticity only study linear elasticity, but the literature does contain stuff on nonlinear stress-strain relations (e.g. to model materials beyond their elastic limit).

Besides the theory of elasticity, another major branch of continuum mechanics is the theory of plasticity, in which one studies the flow of solids under pressure, e.g. in manufacturing wire one needs to model the process of extrusion. And I haven't even mentioned subjects concerning the properties of materials in which quantum mechanics plays an essential role!

I probably could have mentioned some on-line resources, but I am concerned that the intent at these schools might not have been to provide course notes permanently available to all on-line. (Although, personally, I think it would only make ambitious students more willing to pay an Ivy League tuition to see how demanding the courses at our best universities are...) However, FWIW:

Course notes at EN222: Mechanics of Solids - Simple Elastodynamics at Brown University:

http://www.engin.brown.edu/courses/E...todynamics.htm

And just look who's teaching elasticity at Cal Tech! Whee!

Chapter 10: Elastostatics
http://www.pma.caltech.edu/Courses/p...p10/0210.1.pdf

Chapter 11: Elastodynamics
http://www.pma.caltech.edu/Courses/p...p11/0211.1.pdf

 In the previous chapter we considered elastostatic equilibria in which those forces acting on elements of an elastic solid were balanced so that the solid remained at rest. When this equilibrium is disturbed, the solid will undergo acceleration. This is the subject of elastodynamics.
Just in case you doubted my word
 Recognitions: Science Advisor I hope no-one went out and designed a bridge based on the figures I gave, because in the course of editing I inadvertently deleted a factor of $10^{11}$. Here is a corrected table, with some more materials added for good measure: $${\rm stainless} \; {\rm steel} \! : \rho = 7.91 \; {\rm g}/{\rm cm}^3}, \; E=19.6, \, K=16.3, \, \mu = 7.57 \; 10^{11} \, {\rm dyn}/{\rm cm}^2, \; \nu = 0.30$$ $${\rm structural} \; {\rm steel} \! : \rho = 7.91 \; {\rm g}/{\rm cm}^3}, \; E=20.0, \, K=16.7, \, \mu = 7.69 \; 10^{11} \, {\rm dyn}/{\rm cm}^2, \; \nu = 0.30$$ $${\rm titanium} \! : \rho = 4.76 \; {\rm g}/{\rm cm}^3}, \; E=10.3, \, K=10.7, \, \mu = 3.84 \; 10^{11} \, {\rm dyn}/{\rm cm}^2, \; \nu = 0.34$$ $${\rm silver} \! : \rho = 10.4 \; {\rm g}/{\rm cm}^3}, \; E=7.5, \, K=10.3, \, \mu = 2.7 \; 10^{11} \, {\rm dyn}/{\rm cm}^2, \; \nu = 0.38$$ $${\rm copper} \! : \rho = 8.96 \; {\rm g}/{\rm cm}^3}, \; E=11.0, \, K=14.0, \, \mu = 4.8 \; 10^{11} \, {\rm dyn}/{\rm cm}^2, \; \nu = 0.34$$ $${\rm aluminum} \! : \rho = 2.83 \; {\rm g}/{\rm cm}^3}, \; E=7.3, \, K=7.2, \, \mu = 2.74 \; 10^{11} \, {\rm dyn}/{\rm cm}^2, \; \nu = 0.33$$ $${\rm tungsten} \; {\rm carbide} \! : \rho = 13.8 \; {\rm g}/{\rm cm}^3}, \; E=53.4, \, K=31.7, \, \mu = 21.95 \; 10^{11} \, {\rm dyn}/{\rm cm}^2, \; \nu = 0.22$$ $${\rm pyrex} \; {\rm glass} \! : \rho = 2.32 \; {\rm g}/{\rm cm}^3}, \; E=6.2, \, K=4.0, \, \mu = 2.5 \; 10^{11} \, {\rm dyn}/{\rm cm}^2, \; \nu = 0.24$$ $${\rm window} \; {\rm glass} \! : \rho = 2.53 \; {\rm g}/{\rm cm}^3}, \; E=7.0, \, K=4.7, \, \mu = 2.8 \; 10^{11} \, {\rm dyn}/{\rm cm}^2, \; \nu = 0.25$$ $${\rm lucite} \! : \rho = 1.18 \; {\rm g}/{\rm cm}^3}, \; E=0.40, \, K=0.66, \, \mu = 0.143 \; 10^{11} \, {\rm dyn}/{\rm cm}^2, \; \nu = 0.4$$ To convert from ${\rm psi}$ to ${\rm dyn}/{\rm cm}^2$, multiply by $6.895 \times 10^4$. To convert from ${\rm dyn}/{\rm cm}^2$ to ${\rm Pa} = {\rm N}/{\rm m}^2$, divide by ten. To convert from ${\rm N}/{\rm m}^2$ to relativistic units of ${\rm m}^{-2}$, multiply by $G/c^4 = 8.26 \times 10^{-45} 1/{\rm N}$. I should stress that the elastic moduli (again, any two form an complete set of elastic moduli from which any other can be determined) do not give a complete description of the properties of material in continuum mechanics. For example, adding carbon to steel increases the range over which it can be treated as an elastic material--- and increases the forces required to break a steel blade, hence the importance of Damascus steel in the history of warfare--- but doesn't affect the moduli themselves. Sorry for the awkward interruption, but PF doesn't allow me to edit my own posts after a day or so!
 Recognitions: Science Advisor 16. ELASTIC WAVES: Addendum Note that these are nondispersive wave equations; that is, the speed of propagation is independent of the frequency of the wave. This pleasant feature breaks down in a finite object, but it is approximately true for wavelengths much smaller than the size of our object. The energy transported by a p-wave is the sum of the kinetic and strain energies. These two terms are equal, so we have equipartition of energy, just as for the theory of springs. For those comparing with the treatment in Blandford and Thorne, note that I avoided appealing to the "potential form" of the Helmholtz decomposition.
 Recognitions: Science Advisor 4. THE STRAIN TENSOR: Addendum Given any second rank tensor, we can decompose it uniquely into its irreducible parts. That is, we can write any such tensor as the sum of its trace part (a scalar multiple of the identity), its traceless symmetric part, and its antisymmetric part. In the case of the divergence of the displacement vector field, we obtain: $$u_{j,k} = {\varepsilon^n}_n \, \delta_{jk} + \tilde{\varepsilon}_{jk} + \omega_{jk}$$ where the scalar part of the strain tensor, ${\varepsilon^n}_n \, \delta_{jk}$, is called the dilitation, and the traceless part of the strain tensor, $$\tilde{\varepsilon}_{jk} = \varepsilon_{jk} - \frac{1}{3} \, {\varepsilon^n}_n \, \delta_{jk}$$ is called the shear tensor. The trace of the strain tensor is simply the divergence of the displacement vector field, ${u^j}_{,j} = {\varepsilon^j}_j$. Let us integrate the radial component of the displacement vector over a very small sphere: $$\int_{\partial \Omega} \vec{u} \cdot \vec{n} \, dA = \int_\Omega \nabla \cdot \vec{u} \, dV = \int_\Omega {\varepsilon^n}_n \, dV$$ This is the change in volume enclosed by a small sphere of material points which have been subjected to a displacement. Taking the limit in which the volume tends to zero, we obtain $${\varepsilon^n}_n = \frac{dV}{V} = -\frac{d\rho}{\rho}$$ which gives a notable interpretation of the dilitation. The trace of any linear operator ${H^j}_k$ is the first coefficient of its characteristic polynomial. To compute this polynomial, we first compute the traces of its powers: $$t_1 = {H^\ell}_\ell, \; t_2 = {H^\ell}_m \, {H^m}_\ell, \; t_3 = {H^n}_m \, {H^m}_n \, {H^n}_\ell, \dots$$ Newton's recurrence states $$t_1 = a_1, \; t_2 = a_1 \, t_1 -2 \, a_2, \; t_3 = a_1 \, t_2 - a_2 \, t_1 + 3 \, a_3, \dots$$ Inverting this gives $$a_1 = t_1, \; a_2 = \frac{t_1^2 - t_2}{2}, \; a_3 = \frac{t_1^3 - 3 \, t_1 \, t_2 + 2 \, t_3}{6}, \dots$$ This is valid for any dimension, but we only need the first three formulae to obtain the characteristic polynomial of a three-dimensional linear operator: $$\chi_H (\lambda) = \lambda^3 - a_1 \, \lambda^2 + a_2 \, \lambda - a_3$$ As the reader no doubt is well aware, the roots of the characteristic polynomial are the eigenvalues of the operator. In general, these eigenvalues need not be real. But recall that the principle axis theorem in the theory of inner product spaces, which is valid for a positive definite inner product. states that every symmetric real linear operator possesses an orthonormal basis of eigenvectors. But ${\varepsilon^j}_k$ is just such a symmetric linear operator, so by rotating our frame at each point, we can make all the off-diagonal terms of the strain vanish. (Does this mean that the strain tensor has only three algebraically independent components at each point? No, because it takes three parameters to specify the rotation of our frame neccessary to kill the off-diagonal components.) The real eigenvalues, the roots of the characteristic equation of the strain tensor, namely $\lambda^3 - a_1 \, \lambda^2 + a_2 \, \lambda - a_3$ where $$a_1 = {\varepsilon^\ell}_\ell$$ $$a_2 = \frac{ {\varepsilon^\ell}_\ell \, {\varepsilon^m}_m - \varepsilon_{\ell m} \, \varepsilon^{\ell m} }{2}$$ $$a_3 = \frac{ {\varepsilon^\ell}_\ell \, {\varepsilon^m}_m \, {\varepsilon^n}_n - 3 \, \varepsilon_{\ell m} \, \varepsilon^{\ell m} \, {\varepsilon^n}_n + 2 \, {\varepsilon^\ell}_m \, {\varepsilon^m}_n \, {\varepsilon^n}_\ell }{6}$$ (the last is the determinant of the strain tensor), are called the principle stresses. Let us consider very small displacements along the principle axes. Assume for convenience that we have written the strain tensor in a Cartesian chart such that it is already diagonalized by the coordinate basis vectors at the point of interest. Then from our distance formula we find that the quantities $dL/L$ are given by the principle stresses.
 Blog Entries: 47 Recognitions: Gold Member Homework Help Science Advisor Chris, Do you have any comments on the use and interpretation of "principal invariants" in elasticity [i.e., coefficients appearing in the Cayley-Hamilton theorem, where the trace and the determinant are the first and last examples of such invariants]?

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Hi, Rob,

 Quote by robphy Chris, Do you have any comments on the use and interpretation of "principal invariants" in elasticity [i.e., coefficients appearing in the Cayley-Hamilton theorem, where the trace and the determinant are the first and last examples of such invariants]?
I am sure that I do, in fact at this very moment I am struggling to continue the above Addendum to address this very point!

(EDIT: done, more or less. It was kinda cool that you asked me that right when I was composing an addendum addressing that very point!)
 Blog Entries: 47 Recognitions: Gold Member Homework Help Science Advisor It might be useful to say that these principal invariants can be written as elementary symmetric functions of the eigenvalues. Tensorially, they can be written neatly as $$\varepsilon^a{}_{[a}\cdots \varepsilon^n{}_{n]}$$. FYI, I'm most interested in $$a_2$$ and its geometrical and physical interpretation in elasticity.

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 Quote by robphy It might be useful to say that these principal invariants can be written as elementary symmetric functions of the eigenvalues.
That's pretty much what I said (Newton's recursion and all that), but yeah, I'll try to rewrite it tommorrow. About $a_2$, have you read Fekete, Real Linear Algebra. This is an idiosyncratic but perfectly sane math book, but Walt Pohl pointed out to me that Fekete dropped out of math and has become known in Economics as something of a... maverick. Be this as it may, this book contains some cool stuff about $a_2$. Another thing I should mention is the interleaving eigenvalue theorem.

Blog Entries: 47
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Gold Member
Homework Help
 Quote by Chris Hillman That's pretty much what I said (Newton's recursion and all that), but yeah, I'll try to rewrite it tommorrow. About $a_2$, have you read Fekete, Real Linear Algebra. This is an idiosyncratic but perfectly sane math book, but Walt Pohl pointed out to me that Fekete dropped out of math and has become known in Economics as something of a... maverick. Be this as it may, this book contains some cool stuff about $a_2$. Another thing I should mention is the interleaving theorem.
If I recall correctly, he refers to $a_2$ as an "invection". However, his treatment of the principal invariants seems tied to three dimensions. I am looking for something more dimensionally-independent.
 Recognitions: Science Advisor 11. STATIC EQUILIBRIUM: Correction The sign on the left hand side of both versions of the equation of static equilibrium should be reversed: $$-\frac{2 \, (1+\nu)}{E} \, \rho \, \vec{a} = \Delta \vec{u} + \frac{1}{1-2\nu} \, \nabla \left( \nabla \cdot \vec{u} \right)$$ 12. EXAMPLES OF STATIC EQUILIBRIUM: Corrections and Addendum Correction: in EXAMPLE TWO, the sign of the stress tensor components should be reversed. Ade: The stresses are tensile stresses, as is shown by their negative sign. (Warning!: in books which use the opposite sign convention, tensile stresses have positive signs!) Correction: the remark after EXAMPLE THREE was intended for section 15 and should be moved there to complement text not yet written. This remark (and the "most likely") should be replaced with: Notice that on $z=0, \, 0 \leq r < R$ the displacement vector is purely axial. This is the end of the shaft which is held fixed while the other end $z=z_0, \, 0 \leq r < R$ is twisted. External forces must be applied to the surface of the shaft on both of these disks in order to effect the twist, but the normal stress on the cylindrical part of the surface of the shaft vanishes automatically. Here, it helps to know a bit about harmonic functions. In particular, the solution of the boundary value problem $$\Delta w = 0 \; {\em on} \; 0 \leq r < R, \; w(R,\phi) = f(\phi)$$ may be written as Poisson's integral: $$w(r,\phi) = \frac{1}{2\pi} \int_0^{2\pi} \frac{R^2-r^2}{R^2 -2 \, R \, r \, \cos(\phi-\eta) + r^2} \, d\eta$$ This is guaranteed by the mean value theorem on harmonic functions to be regular at $r=0$; the same theorem guarantees that the stresses resulting from this torsion will be maximal at $r=R$. This contrasts with what we found for the distribution of stresses in a rigidly rotating shaft. In particular, we can't choose R to large, or the resulting strains will exceed the rough limit $10^{-3}$ (recall that strains are dimensionless ratios) at which we can expect a material to break. A simple nontrivial example is $\psi(r,\phi) = a \, r^m \, \cos (2 \pi \, m \phi)$, where $m$ is some integer. This corresponds to applied forces at each cap which exhibit cosine variation. Note that the trivial case, $\psi=0$, is precisely our fourth "Basic example" from section 10. Addendum: EXAMPLE FOUR: Consider a spinning shaft (solid cylinder) which is subjected in the comoving frame to surface forces applied at either end which introduce a torsion. Following the now familiar procedure, we postulate a reasonable displacement vector, of form $$\vec{u} = a \, \vec{e}_1 + b \, \vec{e}_2 + \tau \, z \, r \vec{e}_3$$ where a,b are undetermined functions. Requiring the radial component of the stress to vanish on $r=R$ and requiring the displacement vector to be purely radial on the "untwisted cap" $z=0, \, 0 \leq < R$, we find that our solution is the sum of the trivial case of the last example plus the rigidly rotating shaft solution: $$\vec{u} = \frac{1+\nu}{1-\nu} \, \frac{1-2\nu}{8 \, E} \, \rho \, \Omega^2 \, r \, \vec{e}_2 + \tau \, z \, r \, \vec{e}_3$$ Note that this gives a nonzero rotation tensor because we are progressively twisting cross sections of the shaft as we increase the z coordinate. The stress tensor is (components wrt the frame!) $$S^{13} = \frac{\tau \, r}{\mu}$$ $$S^{11} = -\nu \, \frac{\rho \, \Omega^2}{4 \, (1-\nu)} \, \left( (3-2\nu) \, R^2 - 2 \, r^2 \right)$$ $$S^{22} = -\frac{\rho \, \Omega^2}{4 \, (1-\nu)} \, \left( (3-2\nu) \, R^2 - r^2 \right)$$ $$S^{33} = -\frac{\rho \, \Omega^2}{8 \, (1-\nu)} \, \left( (3-2\nu) \, R^2 - (1+2\nu) \, r^2 \right)$$ Here, the diagonal components are maximal on the axis, while the torsion component is maximal at $r=R$ (it need not vanish on the cylindrical "free surface" of the shaft because it is not a normal stress). EXAMPLE FIVE: Consider an isolated solid ball made of a homogeneous isotropic material, such as steel, which is compressed by its own weight. Naturally we use the polar spherical chart with the frame $$\vec{e}_1 = \partial_r, \; \vec{e}_2 = \frac{1}{r} \, \partial_\theta, \; \vec{e}_3 = \frac{1}{r \, \sin(\theta)} \, \partial_\phi$$ The displacement vector will clearly be purely radial, so we need only compute the stress and then demand that the radial component vanish at the surface $r=R$, together with imposing the equilibrium condition with body acceleration $$\vec{a} = -\frac{g \,r}{R} \, \vec{e}_1$$ where g is the surface gravity of our solid ball, plus the condition that the displacement be regular at the center. The solution we obtain is $$\vec{u} = -\frac{g \rho}{10 E} \, \frac{1-2\nu}{1-\nu} \, \frac{r}{R} \, \left( (3-\nu) \, R^2 - (1+\nu) \, r^2) \right) \; \vec{e}_2$$ The nonvanishing components of the strain tensor are: $$\varepsilon_{11} = -\frac{g \rho}{10 E} \, \frac{1-2\nu}{1-\nu} \, \left( (3-\nu) \, R^2 - 3 \, (1+\nu) \, r^2) \right)$$ $$\varepsilon_{22} = \varepsilon_{33} = -\frac{g \rho}{10 E} \, \frac{1-2\nu}{1-\nu} \, \left( (3-\nu) \, R^2 - (1+\nu) \, r^2) \right)$$ The nonvanishing components of the stress tensor are: $$S^{11} = \frac{g \rho}{10 R} \, \frac{3-\nu}{1-\nu} \, (R^2-r^2)$$ $$S^{22} = S^{33} = \frac{g \rho}{10 R} \, \frac{1}{1-\nu} \left( (3-\nu) \, R^2 - (1+3\nu) \, r^2) \right)$$ As we would expect, the stresses are compressive (as shown by the positive sign of the components). They are maximal at the center, where they all agree (as demanded by the spherical symmetry!), so that our ball looks like a perfect fluid at the center. For $\nu=1/3$, the maximal stress is of the order $\frac{2}{5}\, g \, \rho \, R$. The inward radial displacement of the surface of the ball is, for $\nu=1/3$, of magnitude $\frac{g \rho R^2}{15 E}$. EXAMPLE SIX: Consider a block (solid cuboid) hanging under its own weight from a fixed ceiling. Adopting a Cartesian chart with the coordinate basis as our frame, we can solve for the displacement in two steps. First, assume the stress is purely axial, $S^{33} = f(z)$. Compute the corresponding strain tensor. Demand that these vanish at $z=z_0$, the free end of the hanging block. Together with the equilibrium equation with body acceleration $\vec{a} = -g \, \vec{e}_3$, this determines the stress and strain tensors. Second, compute the strain for an arbitrary displacement and solve the equations which result when we equate these strain components to the ones obtained in the previous step. This gives components with six integration constants. Demanding that the displacement vector vanish at $z=z_1$, where $z_1 > z_0[/tex], determines three of these, and demanding that the rotation tensor vanish determines the other three. The resulting solution is $$\vec{u} = g \, \rho \frac{1+\nu}{E} \, \frac{1-2\nu}{1-\nu} \, \left( z-z_1 \right) \, \left( z-z_1 - 2 \, z_0 \right) \; \vec{e}_3$$ The nonvanishing components of the strain tensor are $$\varepsilon_{11} = \varepsilon_{22} = -\nu \, g \, \rho \, \frac{1+\nu}{E} \, \frac{1-2\nu}{1-\nu} \, \left( z-z_0 \right)$$ $$\varepsilon_{33} = g \, \rho \, \frac{1+\nu}{E} \, \frac{1-2\nu}{1-\nu} \, \left( z-z_0 \right)$$ The nonvanishing component of the stress tensor is $$S^{33} = -g \, \rho \, \frac{1+\nu}{E} \, \frac{1-2\nu}{1-\nu} \, \left( z-z_0 \right) = -\frac{g \, \rho}{2 \mu} \, \frac{1-2\nu}{1-\nu} \, \left( z-z_0 \right)$$ Note that as promised, both the stress and strain tensors vanish at [itex]z=z_0$, while the displacement vector vanishes at $z=z_1$. The stress and strain components are maximal at the upper end of the block, as we would expect. It is well worth connecting this computation with a simpler order of magnitude computation. The mass of hanging block of length L, area A, and mass density $\rho$ is $\rho \, L \, A$, and a typical stress at the upper end would be $g \, \rho \, L$ (in relativistic units this has the dimensions of sectional curvature). The typical strain would then be this stress divided by twice Young's modulus: $\frac{g \, \rho \, L}{2 \, E}$. This is a dimensionless ratio. Finally, multiplying by L gives the rough extension $\frac{g \, \rho \, L^2}{2 \, E}$. This agrees with the result of integrating the strain over z. Exercise: similarly, for a vertically hanging rod (solid cylinder). Do you actually need to compute anything? Exercise: using the figures given for structural steel in (the addendum to Section 9), show that by our order of magnitude estimate, a 2000 ft length of steel cable should extend about 2.8 inches under its own weight. Next, estimate the length of the longest steel cable which can hang under its own weight without breaking. It is estimated that a nanocarbon fiber cable might have $E \approx 1.2 \times 10^{14} \; {\rm dyn}/{\rm cm}^3$. Assuming (for lack of available estimates) that $\nu = 0$, estimate the length of the longest nanocarbon fiber cable which can hang under its own in a uniform gravitational field without breaking. Is the result sufficiently large to justify an analysis using the obvious nonconstant spherically symmetric gravitational gravitational field? Saint Venant's principle is often useful when solving static equilibrium problems. It says that if we alter the boundary conditions to new conditions which are however "statically equivalent" to the original conditions (meaning that the resulant forces and moments are the same), the stresses sufficiently far from where we made the alterations will be little affected. For example, Landau and Lifschitz give a not-quite-solution for a solid block standing on a fixed floor (dual to Example Six above), which doesn't satisfy the vanishing displacement condition over the entire floor. But this doesn't affect the main conclusions. SECTION 13: Correction: the same correction noted above for the equation of equilibrium; the GRTensorII definition should read: Code: # Compute equation of equilibrium for given U and F: grdef(equilU{^(a)} := 2*(1+nu)/E*rho*F{^(a)} + strainU{^(a) ^(n) ;(n)} + 1/(1-2*nu)*strainU{^(n) ^(a) ;(n)}):