Compatibility Equations Expressed in terms of the Stress Components for an Isotropic, Homogeneous, Linear Elastic Solid

In static problems where the surface tractions are prescribed on the entire boundary, it is convenient to solve the equations of equilibrium

$$T_{ij,j} + \rho _0g_i = 0$$ (5.6.1)

in terms of the stress components. Having obtained $T_{ij}$ which satisfy (5.6.1) and the assigned boundary conditions, we solve for strains from eqn. (5.2.8). In order that $e_{ij}$ give a unique displacement field $u_i,\ e_{ij}$ must satisfy the compatibility conditions (3.14.32). Substitution for $e_{ij}$ from (5.2.8) into (3.14.32) yields

$$\frac{1}{2\mu} [T_{ij,kp} + T_{kj,ij} - T_{ip,jk} - T_{jk,ip}]$$

$$\hspace*{.3in} -\frac{\lambda}{2\mu (3\lambda + 2\mu )}[\delta_{i... ...kp} + T_{mm,ij} \delta_{kp} - T_{mm,ip}\delta_{jk} - T_{mm,jk}\delta_{ip}] = 0.$$ (5.6.2)

There are only 6 independent equations as there were only 6 independent equations expressed by (3.14.32). In (5.6.2) set $k = p$ and sum with respect to the common index to obtain

$$T_{ij,kk} + T_{kk,ij} - T_{ik,jk} - T_{jk,ik} = \frac{\lambda}{3\lambda + 2\mu} [T_{mm,kk} \delta_{ij} + T_{mm,ij}].$$ (5.6.3)

Out of these 9 equations only six are independent because $T_{ij} = T_{ji}$. Consequently, in combining linearly some of the equations (5.6.2), the number of independent equations has not been reduced, and hence the resulting set (5.6.3) of equations is equivalent to the original set (5.6.2).

In solid mechanics problems, it is usual to neglect the effect of gravity and write (5.6.1) as

$$T_{ij,j} = 0.$$ (5.6.4)

Whenever (5.6.4) holds, the third and fourth terms on the left hand side of (5.6.3) vanish and it simplifies to

$$T_{ij,kk} + \frac{2(\lambda + \mu )}{3\lambda + 2\mu} T_{kk,ij} - \frac{\lambda}{3\lambda + 2\mu} T_{mm,kk}\delta_{ij} = 0.$$ (5.6.5)

Setting $i=j$ and summing over the repeated index, we get $T_{mm,kk} = 0$ and hence (5.6.5) reduces to

$$T_{ij,kk} + \frac{2(\lambda + \mu )}{3\lambda + 2\mu} T_{kk,ij} = 0.$$ (5.6.6)

These are the compatibility equations in the absence of body forces. Thus for a stress field to be a solution of a static problem with zero body force for a homogeneous and isotropic linear elastic body, it must satisfy

(a)

equations of equilibrium (5.6.4),

(b)

compatibility conditions (5.6.6), and

(c)

the appropriate boundary conditions.

Note that if $T_{ij}$ satisfies (a) and (c) but not (b), then that $T_{ij}$ will not correspond to a stress field in a linear elastic body since $e_{ij}$ corresponding to such a $T_{ij}$ will not result in a unique $u_i$.

Exercise: Can the following stress field represent a possible solution of a static problem with zero body force for a homogeneous and isotropic linear elastic body?

$$T_{11} =\ c(X^2_2 + \nu (X^2_1 - X^2_2)),\ c = \ne 0,$$

$$T_{22} =\ c(X^2_1 + \nu (X^2_2 - X^2_1)),$$

$$T_{33} =\ c\nu (X^2_1 + X^2_2),$$

$$T_{12} =\ -2c\nu X_1X_2,\ T_{23} = T_{31} = 0.$$

Example: For a plane stress state, express the compatibility conditions in terms of a stress function.

Solution: Recall the second exercise given on page 4-9. For a plane stress state,

$$\begin{split}T_{11} =\ & \phi ,_{22}\ ,\ T_{22} = \phi ,_{11}... ...\ & \phi(X_1,X_2)\ ,\ T_{13} = T_{23} = T_{33} = 0. \end{split}$$ (5.6.7)

As was proved in that exercise, the stress field thus obtained from $\phi$ satisfies the equilibrium equations (5.6.4) for every choice of the function $\phi$. For this state of stress there is only one independent compatibility condition. This we obtain from (5.6.6) by setting $i=j$ and summing over the repeated index. The result is

$$T_{ii,kk} = 0.$$ (5.6.8)

Now substitution from (5.6.7) into (5.6.8) gives

$$(\phi_{,22} + \phi_{,11})_{,11} + (\phi_{,22} + \phi_{,11})_{,22} = 0$$

or

$$\nabla^2 (\nabla^2\phi) = \nabla^4\phi = 0.$$ (5.6.9)

Thus the problem of solving a static plane stress problem reduces to finding a solution of (5.6.9) under the appropriate boundary conditions.

Exercise: Show that for plane strain problems (i.e., those for which $u_1 = u_1 (X_1,X_2),\ u_2 = u_2 (X_1,X_2), u_3 = 0)$ with zero body force there is only one independent compatibility condition in terms of the components of the stress and this can be written as

$$\nabla^2 (T_{11} + T_{22}) = 0.$$ (5.6.10)

Exercise: (a) Simplify the equations of equilibrium with zero body force for plane strain problems. (b) Introduce a function

$$\phi (X_1,X_2)\ \ T_{11} = \phi_{,22};\ T_{22} = \phi_{,11};\ T_{12} = -\phi_{,12}.$$ (5.6.11)

Will this stress distribution satisfy the equilibrium equations obtained in part (a)?

Note that for plane strain case $T_{33}$ need not and, in general, will not vanish. If we substitute the stress field given by (5.6.11) into (5.6.10) we arrive at (5.6.9). Thus the task of finding a solution of a static plane strain problem also reduces to that of finding a solution of (5.6.9) which satisfies the pertinent boundary conditions.