# 0.4 A look at a paper by strang and kohn

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This module attempts to present the paper "Hencky-Prandtl Nets and Constrained Michell Trusses" by Gilbert Strang and Robert Kohn in a way that is accessible to undergraduate mathematics students.

Hencky-Prandtl Nets and Constrained Michell Trusses

## Introduction

The paper “Hencky-Prandtl Nets and Constrained Michell Trusses” by Gilbert Strang and Robert Kohn [link] provides many interesting insights into the problem posed by Michell in his 1904 paper “The limits of economy of material in framed-structures” [link] . This module attempts to present the former paper in a manner that can be understood by undergraduate students in mathematics.

Michell's problem assumes a material with given tensile and compressive yield limits, say $±{\sigma }_{0}$ , and allows concentrated forces. A force $F$ can only be carried by beams with cross-sectional area $F/{\sigma }_{0}$ , so if the points of application of the forces are too close together, there will be difficulty in finding enough space. Because there is no restriction on the magnitude or the spacing of the external loads, which are surface tractions $f$ distributed along the boundary $\Gamma$ of the given simply connected domain $\Omega$ , one needs to be careful to ensure Michell's problem is self-consistent. To do this, we remove the limitation imposed by ${\sigma }_{0}$ and allow increasingly strong beams whose cost is proportional to their strength. This way, the cross-sectional area doesn't get too large, so there isn't a problem with finding enough space.

## Variational forms of the stress problems

The following is a brief derivation of the Michell problem, where the stress constrains come from the equilibrium equations:

$\text{div}\phantom{\rule{4.pt}{0ex}}\sigma =0\phantom{\rule{4.pt}{0ex}}\text{in}\phantom{\rule{4.pt}{0ex}}\Omega ,\sigma :n=f\phantom{\rule{4.pt}{0ex}}\text{on}\phantom{\rule{4.pt}{0ex}}\Gamma$

where

$\sigma =\left[\begin{array}{cc}{\sigma }_{x}& {\tau }_{xy}\\ {\tau }_{xy}& {\sigma }_{y}\end{array}\right]$
$\text{div}\sigma =\left[\begin{array}{cc}\frac{d}{dx}{\sigma }_{x}& \frac{d}{dy}{\tau }_{xy}\\ \frac{d}{dx}{\tau }_{xy}& \frac{d}{dy}{\sigma }_{y}\end{array}\right]$

and its eigenvalues, or principal stresses, are

${\lambda }_{1,2}=\frac{1}{2}\left({\sigma }_{x}+{\sigma }_{y}±{\left({\left({\sigma }_{x}-{\sigma }_{y}\right)}^{2}+4{{\tau }^{2}}_{xy}\right)}^{1/2}\right)$

The formula for [link] comes from the quadratic formula for the 2-D Lagrangian strain tensor. Bars are placed in the directions of principle stress, which, by definition, come from the orthogonal eigenvectors of $\sigma$ . The required cross-sectional areas at any point are proportional (scaled by ${\sigma }_{0}$ ) to $|{\lambda }_{1}|$ and $|{\lambda }_{2}|$ and the total volume of the truss is proportional to

$\Phi \left(\sigma \right)=\underset{\Omega }{\phantom{\rule{0.277778em}{0ex}}\int \int \phantom{\rule{0.277778em}{0ex}}}\left(|{\lambda }_{1}|+|{\lambda }_{2}|\right)\phantom{\rule{4pt}{0ex}}dx\phantom{\rule{4pt}{0ex}}dy$

Michell's optimal design can be determined from the solution to

$\begin{array}{cccc}& \text{Minimize}\hfill & & \Phi \left(\sigma \right)\hfill \\ & \text{subject}\phantom{\rule{4.pt}{0ex}}\text{to}\hfill & & \text{div}\phantom{\rule{4.pt}{0ex}}\sigma =0,\phantom{\rule{0.277778em}{0ex}}\sigma :n=f\hfill \end{array}$

To express the problem in a way that suggests a computational algorithm, we introduce a stress function $\psi \left(x,y\right)$ . For any statically admissible stress tensor $\sigma$ there is a function $\psi$ such that

$\sigma =\left(\begin{array}{cc}{\psi }_{yy}& -{\psi }_{xy}\\ -{\psi }_{xy}& {\psi }_{xx}\end{array}\right)$

where the subscripts indicate partial derivatives. It is easy to see that $\text{div}\phantom{\rule{4.pt}{0ex}}\sigma =0$ . In the first column, for example,

$\frac{\partial }{\partial x}{\sigma }_{x}+\frac{\partial }{\partial y}{\tau }_{xy}=\frac{\partial }{\partial x}{\psi }_{yy}+\frac{\partial }{\partial y}\left(-{\psi }_{xy}\right)={\psi }_{yyx}-{\psi }_{xyy}=0.$

$\psi$ is determined up to a linear function, because all second derivatives of a linear function are zero. Given the unit normal $n=\left({n}_{x},{n}_{y}\right)$ , the unit tangent $t=\left(-{n}_{y},{n}_{x}\right),$ and the condition

$\left(\begin{array}{cc}{\sigma }_{x}& {\tau }_{xy}\\ {\tau }_{xy}& {\sigma }_{y}\end{array}\right)\left(\begin{array}{c}{n}_{x}\\ {n}_{y}\end{array}\right)=\left(\begin{array}{c}{f}_{1}\\ {f}_{2}\end{array}\right),$

${\psi }_{yy}{n}_{x}-{\psi }_{xy}{n}_{y}={f}_{1}$ or $\left(\text{grad}\phantom{\rule{4.pt}{0ex}}{\psi }_{y}\right)·t={f}_{1}$ and similarly for ${f}_{2}$ . Therefore the tangential derivative of ${\psi }_{y}$ is, by definition, $\frac{\partial {\psi }_{y}}{\partial t}={f}_{1}$ or ${\psi }_{y}\left(P\right)={\int }_{{P}_{0}}^{P}{f}_{1}\phantom{\rule{4pt}{0ex}}ds$ by the fundamental theorem of calculus, and similarly for ${f}_{2}$ . In order for the constraints to be compatible, the external loads ${f}_{1}$ and ${f}_{2}$ must be self-equilibrating, and therefore the boundary values ${\psi }_{x}$ and ${\psi }_{y}$ are well defined: the integrals around the closed curve $\Gamma$ are zero. Provided that there is no net torque from the external forces, $\psi$ itself can be determined. Write $\psi =g$ , ${\psi }_{n}=h$ , where ${\psi }_{n}$ is the normal derivative of $\psi$ .

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