# Potential energy in continuum physics: Why is this $0$ in the surface?

At the start of the continuum mechanics, we were calculating the potential energy. Here is the part that's not clear to me:
Let $V$ be the potential energy and $U$ be its density function. Then: $$V=\int U \mathrm{d}V$$ We have that $$\nabla V=-\underline{F}^e$$ $$\nabla U = -\underline{f}^e$$ $$\delta U=-\underline{f}^e \delta\underline u$$ Where $\underline F^e$ is the elastic force, $\underline f^e$ it's density and $\underline u$ is the displacement vector. $$\delta V = -\int \underline f ^e \delta \underline u \mathrm{d}V$$ $$= -\int \partial_l\,\sigma_{kl}\,\delta u_k\, \mathrm{d}V$$ $$= -\int \left[\partial_l \, (\sigma_{kl} \,\delta u_k)-\sigma_{kl} \, \partial_l \, \delta u_k\right] \mathrm{d}V$$ $$= -\oint \sigma_{kl} \,\delta u_k \mathrm{d}A_l+\int \sigma_{kl} \, \partial_l \, \delta u_k \mathrm{d}V$$ Because $\sigma_{kl}=0$ on the surface: $$=\int \sigma_{kl} \, \partial_l \, \delta u_k \mathrm{d}V$$

Q1: What does exactly the $\delta U$ and $\delta \underline u$ mean?
Q2: Why is the $\sigma_{kl}=0$ on the surface?

## 1 Answer

The quantities $\delta U$ and $\delta V$ are the infinitesimal variations of the potential density and potential respectively. There is an alternative derivation (from Landau & Lifshitz) that one can use that may help your understanding with the second question.

Consider the work required to deform a volume $\Omega$ (enclosed by surface S) by some infinitesimal displacement $\delta \textbf{u}$,

$$W = \iiint_\Omega \delta W d\Omega = \iiint_\Omega \textbf{b} \cdot \delta \textbf{u} d\Omega + \iint_S \left(\sigma : \hat{n} \right) \cdot \delta \textbf{u} dS.$$

Note we are just simply writing this as a sum of body and surface work. Here $\textbf{b}$ is the body force density and $\sigma$ is the stress tensor. The notation $\sigma : \hat{n}$ means to contract the tensor onto the surface normal ($\sigma_{ij} \hat{n}_j$). Using the divergence theorem, one finds

\begin{align} W &= \iiint_\Omega \textbf{b} \cdot \delta \textbf{u} d\Omega + \iiint_\Omega \nabla \cdot \left[\sigma : \delta \textbf{u} \right] \cdot \delta \textbf{u} d\Omega \\ \nonumber &= \iiint_\Omega \textbf{b} \cdot \delta \textbf{u} d\Omega + \iiint_\Omega \left(\nabla : \sigma \right) \cdot \delta \textbf{u} d\Omega +\iiint_\Omega \sigma : \delta \left(\nabla \cdot \textbf{u} \right) d\Omega\\\ \nonumber &= \iiint_\Omega \left(\textbf{b} + \nabla : \sigma \right) \cdot \delta \textbf{u} +\iiint_\Omega \sigma : \delta \left(\nabla \cdot \textbf{u} \right) d\Omega.\\ \nonumber \end{align}

Note immediately that the first term $\textbf{b} + \nabla : \sigma$ is zero since that is identically the condition of static mechanical equilibrium. This is also called Cauchy's first law of momentum or the stress divergence equation. So, therefore

$$W = \iiint_\Omega \sigma : \delta \left(\nabla \cdot \textbf{u} \right)d\Omega.$$

which is exactly what you have above. Relating this back to $W = \int_\Omega \delta W d\Omega$ shows that $\delta W = \sigma_{ij} \delta \varepsilon_{ij}$

The constraint $\sigma_{kl}|_S = 0$ is sometimes called a stress-free boundary condition. In continuum mechanics, this is a natural surface condition which indicates you do not have any shearing forces because they would violate mechanical equilibrium. Note that

$$\sigma_{kl} = C_{klij} \varepsilon_{ij} = \frac{1}{2} C_{klij} \left(\frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i} \right) = 0$$

where $C_{ijkl}$ is the elastic stiffness tensor.

• What is the body force density? And how do you/they introduce the $C$ tensor? One of my teachers (who did this derivation) did that $\sigma_{kl}(\epsilon)=\sigma_{kl}(0)+\frac{\partial \sigma_{kl}(\epsilon=0)}{\partial \epsilon_{pq}}\epsilon_{pq}+\frac{1}{2}\frac{\partial^2 \sigma_{kl}(\epsilon=0)}{\partial \epsilon_{pq}\epsilon_{ts}}\epsilon_{pq}\epsilon_{ts}+\dots$, and named the first derivate as $C_{klpq}$, and neglected the higher order terms (he did this after this derivation). Another tacher just assumed linear connection between them, and set $\sigma{ij}=C_{ijkl}\epsilon_{kl}$. – Botond Jun 4 '18 at 9:46
• The body force density is just a general force that exists within the volume. As for stiffness tensor, the first is correct. The first term is zero because in the absence of strain, you have zero stress. The second term is the definition of $C_{klpq}$. Note that in non-linear elastic (plastic) theories, the expansion to higher order becomes important. – John M Jun 4 '18 at 10:28
• Is that the same as the elastic force density? We were using $2$ forces: external and elastic. With $\underline F^{\text{elastic}}=\oint \underline P \mathrm{d}A$ – Botond Jun 4 '18 at 10:32
• Generally, it is thought of as an external force. For example, $\textbf{b}$ could be the gravitational force vector $\textbf{g}$. – John M Jun 4 '18 at 12:42
• Thank you! Do you mean $\varrho g$? – Botond Jun 4 '18 at 13:29