# Why does the displacement have to be small to use the infintesimal strain elasticity equations?

In all the sources I have that discuss the elasticity equations, they start by saying that the strain tensor, $\epsilon$, is related to the displacement gradient, $\nabla u$, where $u$ is the (vector valued) displacement as

$$\epsilon_{ij} = \frac{1}{2}\left(\frac{\partial u_j}{\partial x_i} + \frac{\partial u_i}{\partial x_j} + \frac{\partial u_k}{\partial x_i} \frac{\partial u_k}{\partial x_j}\right).$$

Then they say something like "if $\left\|u\right\|$ is small and $\left\|\nabla u\right\|$ is small we ignore the third term in the parentheses and we get the infinitesimal strain elasticity equations..."

\begin{align*} \sigma_{kl} &= C_{ijkl}\frac{1}{2}\left(\frac{\partial u_j}{\partial x_i} + \frac{\partial u_i}{\partial x_j}\right)\\ -g_i &= \frac{\partial}{\partial x_j}\sigma_{ij} \end{align*}

Question: I understand why $\left\|\nabla u\right\|$ has to be small, but where in the theory does it require that $u$ is small? I can't find where that is required.

Reason for the question: If we consider a cantilevered shaft and apply a torque at the unsupported end, the resulting displacement isn't necessarily small compared to shaft's cross section. Does this mean that this very simple case has to be analyzed with the finite strain model?

• When you linearize the Green strain you don't assume anything about the magnitude of the displacement. The reference and current positions of a material point may be quite different, but the difference between the reference and current material directions is negligible for small strains (i.e., material rotations are small). For the torque example, anything beyond small rotations beyond a few degrees has to be analyzed with finite strains. – Biswajit Banerjee Jan 30 '17 at 4:29
• I am not sure I know what you mean by reference and current material directions. Do you know where the theory requires that $u$ has to be small? Thanks! – fred Jan 30 '17 at 15:22
• $u$ doesn't have to be small, but hard to find situations where $u$ is large, $\nabla u$ is small, but the orientation of a reference set of axes after deformation is not significantly different from the original direction. You can find detailed discussions of the issue in any good mathematical book on elasticity. – Biswajit Banerjee Jan 30 '17 at 21:31
• If $u$ is large, the material response will be nonlinear. This question is probably more appropriate for physics.SE. – David Ketcheson Jan 31 '17 at 0:04
• Can you name a book for example, that assumes displacements are small, in addition to displacement gradient ? – me10240 Feb 6 '17 at 12:06

In general, in the linearized theory of elasticity, it is only the displacement gradients that need to be small compared to unity. The displacements don't need to be small. (Note that when we say displacements are small, we usually mean displacements nondimensionalized with some quantity such as the characteristic length of the structure. The length of a beam in a beam problem would be a candidate for nondimensionalizing, for example.)

However, in the context of bars subjected to bending moments and torsion (Saint Venant's Problem), the assumption of small displacements (nondimensionalized appropriately) is made in addition to small displacement gradients. Without this assumption, the boundary conditions would involve terms of the form $\int_\mathcal{B}\mathbf{u} \times \mathbf{t} dS$ where $\mathbf{u}$ is the displacement vector $\mathbf{t}$ is the traction vector on one face of the bar. This is not an easy term to deal with as a boundary condition and therefore, we require $\mathbf{u}$ to be small so that the integral is of the same order as the order of the terms neglected in the linearized theory of elasticity.

Let us consider the uncoupled mechanical problem. Hence, we have 10 unknowns $$\{\rho, \mathbf{v}, \underline{\underline{\sigma}}\}$$, assuming that the stress and strains are symmetric (this might not be true in Cosserat solids, for example). And we have 10 equations, namely:

• Continuity;
• Conservation of density of momentum; and
• Constitutive equations.

That are given by

\begin{align} &\frac{\partial \rho}{\partial t} + \nabla \cdot \mathbf{\rho v} = 0 \tag1\\ &\nabla \cdot \underline{\underline{\sigma}} + \rho \mathbf{b} = \rho \frac{\partial \mathbf{v}}{\partial t} \tag2\\ &f_i(\underline{\underline{\sigma}}, \underline{\underline{\epsilon}}) = 0 \quad i=1, \cdots 6 \tag3 \end{align}

When you enforce that displacements are small you are saying that the material and spatial references of your problem are the same, i.e.,

$$\mathbf{x} = \mathbf{X} + \underbrace{\mathbf{u}}_{\approx 0} \Rightarrow \mathbf{x} \approx \mathbf{X}\, , \tag4$$

thus

$$\underline{\underline{F}} = \frac{\partial \mathbf{x}}{\partial \mathbf{X}} \approx \underline{\underline{I}} \Rightarrow |\underline{\underline{F}}| \approx 1\, . \tag5$$

As a consequence of (5) and (1), you obtain that the density in the initial configuration and the final one is the same, i.e.

$$\rho_0 = \rho_t |\underline{\underline{F}}| \approx \rho_t\, ,$$

and the density is not an unknown anymore.

Now, regarding a Finite Element formulation, when you have an integral of the form

$$\int_{\Omega_t} \underline{\underline{\sigma}}: \underline{\underline{\epsilon}}\, \mathrm{d}\Omega_t\, ,$$

you face the problem of not-knowing $$\Omega_t$$. When you assume that the displacements are small you are also saying that your initial configuration/domain is the same through your analysis.

This discussion should be present in several continuum mechanics books. That I remember it is presented in

• Olivella, Xavier Oliver, and Carlos Agelet de Saracíbar Bosch. Mecánica de medios continuos para ingenieros. Vol. 92. Univ. Politèc. de Catalunya, 2002.