Work done by External Forces

Recall that we have two types of forces acting on a continuous body: surface tractions acting on the boundary and body forces acting at every interior point of the body. Thus during an incremental displacement $\Delta\mathbf{u}$of the body, the incremental work, $\Delta U$, done by external forces is given by

$$\Delta U = \int_\Gamma f_i\Delta u_i d\Gamma + \int_\Omega \rho g_i\Delta u_id\Omega . \tag{8.1}$$ (8.1)

Here f is the surface traction (force/area) acting at a point on the boundary $\Gamma$ of the body $\Omega$ and g is the gravitational constant. If nonzero displacements are prescribed on a part of the boundary $\Gamma$, then f represents the reactions, i.e., forces exerted by the loading device on the body. Substitution for f from (7.2) into (8.1) gives

$$\Delta U = \int_\Gamma\sigma_{ji}n_j\Delta u_id\Gamma + \int_\Gamma\rho g_i\Delta u_id\Omega . \tag{8.2}$$ (8.2)

Using the Divergence Theorem, $$\left(\int_\Gamma A_in_id\Gamma = \int_\Omega\frac{\partial A_i}{\partial X_i}d\Omega\right)$$, we obtain
$$\begin{align}\Delta U =\ & \int_\Omega\frac{\partial (\sigma_{ji}\Delta u_i)}{\p... ...\sigma_{ji}\frac{\partial\Delta u_i}{\partial X_j} d\Omega,\tag{8.3} \end{align}$$
where we have also used the chain rule of differentiating the product of two functions, and equation (3.11). We now write
$$\begin{align}\frac{\partial\Delta u_i}{\partial X_j} =\ & \frac{1}{2}\left(\frac... ...ight),\nonumber\\ =\ & \Delta e_{ij} + \Delta \omega_{ij},\tag{8.4} \end{align}$$
where

$$\omega_{ij} = \frac{1}{2}\left(\frac{\partial u_i}{\partial X... ...c{\partial u_j}{\partial X_i}\right) = - \omega_{ji}.\tag{8.5}$$ (8.5)

Thus $\omega_{ij}$ is skew symmetric which implies that $\omega_{11} = \omega_{22} = \omega_{33} = 0$, $\omega_{12} = -\omega_{21}$, $\omega_{23} = -\omega_{32}$, and $\omega_{31} = -\omega_{13}$. Relation (8.4) implies that

$$\sigma_{ji}\frac{\partial\Delta u_i}{\partial x_j} = \sigma_{ji}\Delta e_{ji} + \sigma_{ij}\Delta \omega_{ij}.\tag{8.6}$$ (8.6)

Note that each term in (8.6) represents the sum of nine terms since indices i and j are repeated. The skew symmetry of $\omega_{ij}$gives
$$\begin{align}\sigma_{ij}\Delta\omega_{ij} =\ & \sigma_{12}\Delta\omega_{12} + \s... ...(\Delta\omega_{31} + \Delta\omega_{13}),\nonumber\\ =\ & 0\tag{8.7} \end{align}$$
where we have used the symmetry $(\sigma_{ij} = \sigma_{ji})$ of ${\mbox{\boldmath {$\sigma$ }}}$. Substitution from (8.6) and (8.7) into (8.3) gives
$$\begin{align}\Delta U =\ & \int_\Omega \sigma_{ij}\Delta e_{ij}d\Omega = \int_\O... ...ta \Delta e_\alpha d\Omega,\ \alpha,\beta = 1,2,\ldots , 6.\tag{8.8} \end{align}$$
Thus

$$U = \int_\Omega \frac{1}{2}C_{\alpha\beta}e_\alpha e_\beta d\... ...a Wd\Omega\hspace{.3in} \alpha,\beta = 1,2,\ldots , 6\tag{8.9}$$ (8.9)

where $W = \frac{1}{2}C_{\alpha\beta}e_\alpha e_\beta = \frac{1}{2}\sigma_\alpha e_\alpha$ is called the strain-energy density (strain energy/volume). U equals the total energy stored in the body. For an elastic body, the work done by external forces equals the strain energy of the body. Note that

$$\frac{\partial W}{\partial e_\alpha} = C_{\alpha\beta}e_\beta = \sigma_\alpha, \tag{8.10}$$ (8.10)

where we have used the symmetry of the $6\times 6$ matrix C. Thus W serves as the potential for stresses.