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# A Uniaxial Truss

Module by: Doug Daniels. E-mail the author

Summary: In this module we expand the discussion of matrix methods for modeling and solving problems by invesigating the mechanical prospection of tissue. Modeling the tissue with a uniaxial truss, we use its mechanical properties to set up a linear system. We then introduce and use Gaussian Elimination to solve this system. We also show an alternate path to a solution, using the Gauss-Jordan inverse method to generate an inverse. Finally, we briefly explore which matrices are invertible and which are not.

## Introduction

We now investigate the mechanical prospection of tissue, an application extending techniques developed in the electrical analysis of a nerve cell. In this application, one applies traction to the edges of a square sample of planar tissue and seeks to identify, from measurement of the resulting deformation, regions of increased hardness' or stiffness.' For a sketch of the associated apparatus, visit the Biaxial Test site .

## A Uniaxial Truss

As a precursor to the biaxial problem let us first consider the uniaxial case. We connect 3 masses with four springs between two immobile walls, apply forces at the masses, and measure the associated displacement. More precisely, we suppose that a horizontal force, f j f j , is applied to each m j m j , and produces a displacement x j x j , with the sign convention that rightward means positive. The bars at the ends of the figure indicate rigid supports incapable of movement. The k j k j denote the respective spring stiffnesses. The analog of potential difference (see the electrical model) is here elongation. If e j e j denotes the elongation of the jjth spring then naturally, e 1 = x 1 e 1 x 1 e 2 = x 2 x 1 e 2 x 2 x 1 e 3 = x 3 x 2 e 3 x 3 x 2 e 4 = x 3 e 4 x 3 or, in matrix terms, e=Ax e A x , where A=( 100 -110 0-11 00-1 ) A 100 -110 0-11 00-1 We note that e j e j is positive when the spring is stretched and negative when compressed. This observation, Hooke's Law, is the analog of Ohm's Law in the electrical model.

Definition 1: Hooke's Law
1. The restoring force in a spring is proportional to its elongation. We call the constant of proportionality the stiffness, k j k j , of the spring, and denote the restoring force by y j y j .
2. The mathematical expression of this statement is: y j = k j e j y j k j e j , or,
3. in matrix terms: y=Ke y K e where K=( k 1 000 0 k 2 00 00 k 3 0 000 k 4 ) K k 1 000 0 k 2 00 00 k 3 0 000 k 4
The analog of Kirchhoff's Current Law is here typically called force balance.'
Definition 2: force balance
1. Equilibrium is synonymous with the fact that the net force acting on each mass must vanish.
2. In symbols, y 1 y 2 f 1 =0 y 1 y 2 f 1 0 y 2 y 3 f 2 =0 y 2 y 3 f 2 0 y 3 y 4 f 3 =0 y 3 y 4 f 3 0
3. or, in matrix terms, By=f B y f where f= f 1 f 2 f 3 and B=( 1-100 01-10 001-1 ) f f 1 f 2 f 3 and B 1-100 01-10 001-1

As in the electrical example we recognize in BB the transpose of AA. Gathering our three important steps:

e=Ax e A x
(1)
y=Ke y K e
(2)
ATy=f A y f
(3)
we arrive, via direct substitution, at an equation for xx. Namely, (ATy=f)(ATKe=f)(ATKAx=f) A y f A K e f A K A x f Assembling ATKAx A K A x we arrive at the final system:
( k 1 + k 2 k 2 0 k 2 k 2 + k 3 k 3 0 k 3 k 3 + k 4 ) x 1 x 2 x 3 = f 1 f 2 f 3 k 1 k 2 k 2 0 k 2 k 2 k 3 k 3 0 k 3 k 3 k 4 x 1 x 2 x 3 f 1 f 2 f 3
(4)

## Gaussian Elimination and the Uniaxial Truss

Although Matlab solves systems like the one above with ease our aim here is to develop a deeper understanding of Gaussian Elimination and so we proceed by hand. This aim is motivated by a number of important considerations. First, not all linear systems have solutions and even those that do do not necessarily possess unique solutions. A careful look at Gaussian Elimination will provide the general framework for not only classifying those systems that possess unique solutions but also for providing detailed diagnoses of those defective systems that lack solutions or possess too many.

In Gaussian Elimination one first uses linear combinations of preceding rows to eliminate nonzeros below the main diagonal and then solves the resulting triangular system via back-substitution. To firm up our understanding let us take up the case where each k j =1 k j 1 and so Equation 4 takes the form

( 2-10 -12-1 0-12 ) x 1 x 2 x 3 = f 1 f 2 f 3 2-10 -12-1 0-12 x 1 x 2 x 3 f 1 f 2 f 3
(5)
We eliminate the (2,1) (row 2, column 1) element by implementing new row 2=old row 2+12row 1 new row 2 old row 2 1 2 row 1 bringing ( 2-10 032-1 0-12 ) x 1 x 2 x 3 = f 1 f 2 + f 1 2 f 3 2-10 0 3 2 -1 0-12 x 1 x 2 x 3 f 1 f 2 f 1 2 f 3 We eliminate the current (3,2) element by implementing new row 3=old row 3+23row 2 new row 3 old row 3 2 3 row 2 bringing the upper-triangular system ( 2-10 032-1 0043 ) x 1 x 2 x 3 = f 1 f 2 + f 1 2 f 3 +2 f 2 3+ f 1 3 2-10 0 3 2 -1 00 4 3 x 1 x 2 x 3 f 1 f 2 f 1 2 f 3 2 f 2 3 f 1 3 One now simply reads off x 3 = f 1 +2 f 2 +3 f 3 4 x 3 f 1 2 f 2 3 f 3 4 This in turn permits the solution of the second equation x 2 =2( x 3 + f 2 + f 1 2)3= f 1 +2 f 2 + f 3 2 x 2 2 x 3 f 2 f 1 2 3 f 1 2 f 2 f 3 2 and, in turn, x 1 = x 2 + f 1 2=3 f 1 +2 f 2 + f 3 4 x 1 x 2 f 1 2 3 f 1 2 f 2 f 3 4 One must say that Gaussian Elimination has succeeded here. For, regardless of the actual elements of ff, we have produced an xx for which ATKAx=f A K A x f .

## Alternate Paths to a Solution

Although Gaussian Elimination remains the most efficient means for solving systems of the form Sx=f S x f it pays, at times, to consider alternate means. At the algebraic level, suppose that there exists a matrix that undoes' multiplication by SS in the sense that multiplication by 2-1 2 undoes multiplication by 2. The matrix analog of 2-12=1 2 2 1 is S-1S=I S S I where II denotes the identity matrix (all zeros except the ones on the diagonal). We refer to S-1 S as:

Definition 3: Inverse of S
Also dubbed "S inverse" for short, the value of this matrix stems from watching what happens when it is applied to each side of Sx=f S x f . Namely, (Sx=f)(S-1Sx=S-1f)(Ix=S-1f)(x=S-1f) S x f S S x S f I x S f x S f
Hence, to solve Sx=f S x f for xx it suffices to multiply ff by the inverse of SS.

## Gauss-Jordan Method: Computing the Inverse of a Matrix

Let us now consider how one goes about computing S-1 S . In general this takes a little more than twice the work of Gaussian Elimination, for we interpret S-1S=I S S I as nn (the size of SS) applications of Gaussian elimination, with ff running through nn columns of the identity matrix. The bundling of these nn applications into one is known as the Gauss-Jordan method. Let us demonstrate it on the SS appearing in Equation 5. We first augment SS with II. ( 2-10100 -12-1010 0-12001 ) 2-10 100 -12-1 010 0-12 001 We then eliminate down, being careful to address each of the three ff vectors. This produces ( 2-10100 032-11210 004313231 ) 2-10 100 0 3 2 -1 1 2 1 0 00 4 3 1 3 2 3 1 Now, rather than simple back--substitution we instead eliminate up. Eliminating first the (2,3) element we find ( 2-10100 0320343234 004313231 ) 2-10 100 0 3 2 0 3 4 3 2 3 4 00 4 3 1 3 2 3 1 Now, eliminating the (1,2) element we achieve ( 20032112 0320343234 004313231 ) 200 3 2 1 1 2 0 3 2 0 3 4 3 2 3 4 00 4 3 1 3 2 3 1 In the final step we scale each row in order that the matrix on the left takes on the form of the identity. This requires that we multiply row 1 by 12 1 2 , row 2 by 32 3 2 , and row 3 by 34 3 4 , with the result ( 100341214 01012112 001141234 ) 100 3 4 1 2 1 4 0 1 0 1 2 1 1 2 00 1 1 4 1 2 3 4

Now in this transformation of SS into II we have, ipso facto, transformed II to S-1 S ; i.e., the matrix that appears on the right after applying the method of Gauss-Jordan is the inverse of the matrix that began on the left. In this case, S-1=( 341214 12112 141234 ) S 3 4 1 2 1 4 1 2 1 1 2 1 4 1 2 3 4 One should check that S-1f S f indeed coincides with the xx computed above.

## Invertibility

Not all matrices possess inverses:

Definition 4: singular matrix
A matrix that does not have an inverse.

### Example

A simple example is: ( 11 11 ) 11 11

Alternately, there are
Definition 5: Invertible, or Nonsingular Matrices
Matrices that do have an inverse.

### Example

The matrix SS that we just studied is invertible. Another simple example is ( 01 11 ) 01 11

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