Equilibrium of Structures

We don’t usually think about how a roof works, but when we do it’s intriguing that it’s possible for a heavy weight to be suspended over a region of empty space.  It’s possible because we can use beams and cables to exert forces over substantial distances, and humanity’s ability to build larger and more sophisticated structures has advanced on the back of advances in beam and cable technology.   But beams and cables have an enormous limitation, which is that they can only exert their forces in a single direction, the direction along their length.  In order to make structures that resist combinations of external forces in different directions, people have devised some highly imaginative assemblies of these basic structural elements.

Inventing a support system is a creative, artistic process.  But calculating the tensile and compressive forces in the cables and beams when the structure resists a given distribution of external forces is a pretty simple problem in linear algebra.  It totally fits with the ideas and topics Strang presents in Chapter 2, and I can’t believe he left it out.  I would expect him to know all about it himself, and to agree with my feeling that it’s interesting and instructive.  I suspect a rival author sneaked into the editorial offices and tried to steal the manuscript.  Fortunately they didn’t get away with the whole thing, but unfortunately the absence of the part they did get away with was not noticed until it was too late.  Strang himself probably never needed to refer to the actual printed book, so never noticed the gap, and everyone else who has ever used the book never knew any different.  Except me, and now you also will benefit from what I am sure Prof. Strang wanted you to learn.

ANYWAY, a beam or cable is basically a long straight piece of material that transmits force along its length.  Since a beam can both push and pull, while a cable can only pull, we shall speak of beams only.  We usually idealize it as a massless line segment between two points  and .  Its length is , the usual formula for distance between two points.  If the element is under tension T , this means that the end at experiences force , while the end at experiences the equal and opposite force .  (see figure below, left.)

The division by L reflects the physical fact that the force has magnitude T regardless of the actual distance between  and .  For algebraic simplicity, I feel it’s easier to use as a working variable in the calculations, and to find the tension by multiplying by L at the end.

A node of the structure is a point where one or more beams have a common end, and where an external force may also be applied (see figure above, right).  Here, the total force on node 1 is just the sum of the external force and the forces along the beams: .  In the diagram and formula, I am using lower-case letters to label the beams, and numbers for the nodes.  Now, if the structure is in equilibrium, the total force on any node is in fact zero, so we must have In general, if there is a whole network of beams and nodes, then there will be an equation of this form at each node.  Typically we are given the positions of the nodes and the external forces (usually the weights to be supported at the nodes), and we want to find the tensions in the elements.  This is clearly a linear system, but whether or not any solution exists (and if so, how many) depends on the issues of null spaces and free variables that we have been discussing.

The overall linear system has a structure similar to an electrical circuit, except that the elements in the matrix and right-hand side are themselves vectors with two or three components, depending on whether we’re building our structures in 2 or 3 dimensions.  The “elements” of the overall right-hand side vector are the external forces acting on the nodes.  The overall matrix has columns corresponding to the edges (beams or cables) and rows corresponding to the nodes.  As in circuits, if a node is not one of the ends of an edge, the matrix entry is zero.  In contrast, if a node is an end of an edge, then the “matrix element” is the vector displacement from the other end of the edge to the node in question.  The elements of the vector of variables are not vectors, but the scalars representing the tensions in the beams.

For example, consider the simple 2d triangular roof on the top of the next page.  The linear system for the “tensions” is  , which looks as if it represents 3 equations in 3 unknowns.  But it’s really 6 equations, because each position vector  and each external force vector is itself a column vector with 2 elements.
 a

The ’s are still scalars, however, so we have 3 more equations than unknowns.  Our experience with such systems suggests that a solution will exist only under special circumstances.  Our knowledge of linear systems theory indicates that conditions under which solutions exist will come from finding null vectors of the matrix transpose.

It turns out that we can generally guess three null vectors: , , and .  (Check that these satisfy .)  In fact, any vector that corresponds to a motion of the node points that does not change the lengths of the beams is a null vector.  The vector corresponds to an upward shift of the whole structure,  to an upward shift of the whole structure, and   to an anticlockwise rotation of the whole structure.  If there are any other null vectors, they will correspond to deformations of the structure that don’t change the beam lengths, i.e. what is called “mechanical indeterminacy”.  Ideally a structure won’t have any such deformations – we don’t want the roof or whatever structure it is folding up on itself!

The solvability conditions associated with these null vectors are (letting represent the right-hand-side vector of external forces) , which means that the total external horizontal force must be zero, , which means that the total external vertical force must be zero, and , which means that the total external torque must be zero.  This is exactly what we learned many years ago in Physics 100.  If there is another null vector, this would imply that the forces that would deform the structure also have to cancel – but ideally this situation will not happen, as mentioned above.

Last but not least, it is possible that if the solvability conditions are all satisfied, there may be a free variable in the general solution, and therefore an infinite set of possible tensions in the beams.  This may happen if the structure is “overdetermined”, i.e. extra beams or cables added even after the structure is mechanically determinate.  The extra beams may be subjected to arbitrary (as far as this theory is concerned) tensions or compressions, which then are compensated by additional forces in the other beams.  For example, consider adding three more beams to the roof above, to get the structure below:

The tension in beam “e” can be anything (positive or negative), and the forces in the other beams adjust to balance whatever is in “e”, plus the external force on the new node.  (Note that the solvability conditions involve the force vector on the new node.)