0.4 A look at a paper by strang and kohn  (Page 2/3)

If the signs of ${\lambda }_1$ and ${\lambda }_2$ are the same, then

because the trace of the matrix $\sigma$ is equal to the sum of the eigenvalues.

If the ${\lambda }_1$ and ${\lambda }_2$ are of opposite signs, then

Therefore, in order to minimize the largest principle stress, the Michell problem becomes

In real life, unlike in Michell's problem, beams in a truss cannot be arbitrarily strong. Thus, if $\Lambda$ denotes the upper bound on the stresses, the design problem for a constrained Michell truss is

The stress tensor is made up of two tensors: the mean hydrostatic stress tensor and the stress deviator tensor. The hydrostatic stress is the average of the normal stress components, ${\sigma }_x$ and ${\sigma }_y$ . The deviatoric stress is what's left over, namely ${\sigma }^D=\sigma -\frac{1}{2}({\sigma }_x+{\sigma }_y)I$ . The constraint on the strength of the beams can act only on ${\sigma }^D$ instead of $\sigma$ , therefore ignoring the hydrostatic component of the force. This is useful for working with a composite material that can only withstand pure pressure. The constraint then applies to the eigenvalues of ${\sigma }^D$ . Using the same quadratic formula as in [link] , we get that

The matrix ${\sigma }^D$ has trace zero, so the eigenvalues are of equal magnitude and opposite sign. Therefore the plain strain redesign problem is

Rewriting this in terms of the stress function $\psi$ , and denoting it by $\parallel \psi \parallel$ , we get

Therefore the constrained problem above takes the form

If the $k$ is too small, or $f_1$ and $f_2$ are too large, then the material may be too weak to withstand the load, and no truss can exist without collapsing.

The dual problems and optimality conditions

Now we give an informal derivation of the dual problem, which is a maximization over displacements $u=(u_1(x,y),u_2(x,y))$ . No boundary conditions are imposed, because $\sigma :n=f$ was prescribed on all of $\Gamma$ . To solve for the maximum over the constraint $\text{div}\sigma =0$ , we introduce a Lagrange multiplier. This is a system of two equations, $\frac{\partial {\sigma }_x}{\partial x}+\frac{\partial {\tau }_{xy}}{\partial y}=0$ and $\frac{\partial {\tau }_{xy}}{\partial x}+\frac{\partial {\sigma }_y}{\partial y}=0$ , and we denote the multipliers by $u_1$ and $u_2$ . Then

where $\epsilon$ is the deformation tensor with components ${\epsilon }_{ij}=\frac{1}{2}(\frac{\partial u_i}{\partial j}+\frac{\partial u_j}{\partial i})$ and the inner product $⟨\epsilon ,\sigma ⟩$ is $\sum \sum {\epsilon }_{ij}{\sigma }_{ij}=\text{trace}\left(\epsilon \sigma \right)$ .

The next step is minimization over $\sigma$ , so $\epsilon$ remains fixed.

First, we must diagonalize $\epsilon$ by a principle axis transformation, $\epsilon =R^{-1}{\epsilon }^{\text{'}}R$ where $R^{-1}=R^t$ , so that the eigenvalues ${\epsilon }_1$ and ${\epsilon }_2$ (the principal strains) appear in the diagonal matrix ${\epsilon }^{\text{'}}$ . The inner product $⟨ϵ,\sigma ⟩$ and the eigenvalues $\lambda \left(\sigma \right)$ are invariant if this transformation is applied at the same time to $\sigma$ :

Thus the minimization simplifies to