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Intuitive description: In the 2D plane, there are $m$ bars connected by $n$ joints. The length of each bar is fixed. These joints and bars can be viewed as a graph (see the figures below). Denote $s_i$ as the static stress of bar $i$.

  1. For some graphs (see figure 1), it is clear all $s_i$ must be zero. Otherwise, the stress applied on each joint is non-zero and the graph cannot be balanced in the plane.
  2. For other graphs (see figure 2), some $s_i$ can be non-zero and the graph can be well balanced in the plane.

My question: does the later kind of graphs have a name? Any theoretical discussions on them in the literature?

Web link to the figures

Mathematical description: I am studying graph rigidity. Denote $$R=\mathrm{blkdiag}(e_1^T,\cdots,e_m^T)(H\otimes I_2)$$ as the rigidity matrix of a graph, where $e_i\in\mathbb{R}^2$ denotes the edge of the graph, $H\in\mathbb{R}^{m\times n}$ is the incidence matrix, and $I_2$ is the 2x2 identity matrix. The left null space of $R$ actually is the space of all non-zero stresses. So mathematically my question can be rephrased as: when is the rigidity matrix of full row rank?

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You could check the book J. Graver, B. Servatius, H. Servatius "Combinatorial rigidity"; see if they have a name for it. I think, your class of linkages consists of those whose configuration space is (locally) smooth and has dimension equal the number of edges. I'd call this class "fully flexible". – Misha Apr 25 '13 at 18:42
@Misha: Thank you so much for the suggestions. In fact, I just borrowed the book you mentioned. There is one subsections talking about this. But not much useful information. Regarding the "fully flexible", I'm not sure if you are meaning "non-rigid". In fact, for rigid graph (or framework), it is still possible to have no nonzero stresses. For example, in figure 2, if I remove one diagonal linkage, the graph is still rigid, but it cannot have any nonzero stresses. – Shiyu Apr 26 '13 at 2:47
@Misha: I'm not sure if this kind problem has been discussed in the field of mechanics. – Shiyu Apr 26 '13 at 2:50
up vote 3 down vote accepted

The term used by Connelly and Whiteley is self-stressed, and they discuss this notion extensively in their paper "Second-Order Rigidity and Prestress Stability for Tensegrity Frameworks."

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Not sure how helpful this will be, but: when the graph in question is infinitesimally rigid, then the existence of an equilibrium stress corresponds to the graph being "not minimally rigid". Also, graphs that are generically minimally rigid are called isostatic.

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Answer my own question: After searching a lot of papers, I find this problem is related to the Maxwell-Cremona theorem. Related papers include: 1993 - Crapo - Plane stress and projected polyhedra i the basic pattern 1983 - The algebraic geometry of stresses in frameworks 1982 - Whiteley - Motions and stresses of projected polyhedra

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The Maxwell-Cremona theorem is primarily concerned with planar graphs. – Steven Gortler May 8 '13 at 19:50

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