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Equilibrium of system of piecewise linear springs

This problem describes a mechanical system to find a hanging chain consisting of $ N$ links. The first and last nodes are fixed. Each node has a weight $ m_j$ hanging from it. The problem is to minimize the energy of the system (see [5,15]):

\begin{displaymath}\begin{array}{rcl} \mbox{min}& \sum_j m_j g y_j+\frac{k}{2}\V...
...,a_2),& \\  & (x_N,y_N)=(b_1,b_2),& \\  & t\geq 0,& \end{array}\end{displaymath} (3-10)

where $ g$ is the acceleration due to gravity, $ k$ is the stiffness constant of the springs, $ l_0$ is the rest length of each spring, and $ t_j$ is an upper bound on the spring energy of the $ j$-th spring. Consider $ \Vert t\Vert^2\leq 2uv$, $ u=1$, $ v\geq 0$, $ s_j=t_j+l_0,$, $ e_j=x_j-x_{j-1}$, $ f_j=y_j-y_{j-1}$, $ j=1:N$. Then $ \sum_j f_j=b_2-a_2$ and $ \sum_j e_j=b_1-a_1$. The objective function becomes $ \sum_j m_j g y_j+kv.$ Furthermore,

\sum_j m_j g y_j+kv&=&g\sum_jm_j(\sum_{i=...

The linear and cone constraints are

\sum f_j=b_2-a_2, &\\
\sum e_j=b_1-a_1,...
\Vert t\Vert^2\leq 2uv,& u,v\geq 0.

Example 6. The data is the same as in springs.mod in [16]. We have $ k=100,$ $ l_o=2\sqrt{(b_1-a_1)^2+(b_2-a_2)^2}/N,$ $ g=9.8,$ $ m=1,$ $ (a_1,a_2)=(0,0),$ and $ (b_1,b_2)=(2,-1)$. We use $ N=10,20,40,60$ as test cases.

Hans D. Mittelmann 2003-09-10