Let us now consider something that resembles the mechanical
prospection problem introduced in the introduction to matrix methods
for mechanical systems. In the figure below we offer a
crude mechanical model of a planar tissue, say,
e.g., an excised sample of the wall of a
vein.
Elastic fibers, numbered 1 through 20, meet at nodes, numbered 1
through 9. We limit our observation to the motion of the nodes
by denoting the horizontal and vertical displacements of node
jj by
x
2j-1
x
2j-1
(horizontal) and
x
2j
x
2j
(vertical), respectively. Retaining the convention that down
and right are positive we note that the elongation of fiber 1 is
e
1
=
x
2
−
x
8
e
1
x
2
x
8
while that of fiber 3 is
e
3
=
x
3
−
x
1
.
e
3
x
3
x
1
.
As fibers 2 and 4 are neither vertical nor horizontal their
elongations, in terms of nodal displacements, are not so easy to
read off. This is more a nuisance than an obstacle however, for
noting our
discussion of elongation in the small planar truss
module, the elongation is approximately just the stretch along
its undeformed axis. With respect to fiber 2, as it makes the
angle
-π4
4
with respect to the positive horizontal axis, we find
e
2
=
x
9
−
x
1
cos-π4+
x
10
−
x
2
sin-π4=
x
9
−
x
1
+
x
2
−
x
10
2
.
e
2
x
9
x
1
4
x
10
x
2
4
x
9
x
1
x
2
x
10
2
2
.
Similarly, as fiber 4 makes the angle
-3π4
3
4
with respect to the positive horizontal axis, its elongation is
e
4
=
x
7
−
x
3
cos-3π4+
x
8
−
x
4
sin-3π4=
x
3
−
x
7
+
x
4
−
x
8
2
.
e
4
x
7
x
3
3
4
x
8
x
4
3
4
x
3
x
7
x
4
x
8
2
2
.
These are both direct applications of the general formula
e
j
=
x
2
n
-
1
−
x
2
m
-
1
cos
θ
j
+
x
2n
−
x
2m
sin
θ
j
e
j
x
2
n
-
1
x
2
m
-
1
θ
j
x
2n
x
2m
θ
j
(1)
for fiber
jj, as
depicted in
Figure 2 below,
connecting node
mm
to node
nn and
making the angle
θ
j
θ
j
with the positive horizontal axis when node
mm is assumed to lie at
the point (0,0). The reader should check that our expressions
for
e
1
e
1
and
e
3
e
3
indeed conform to this general formula and that
e
2
e
2
and
e
4
e
4
agree with ones intuition. For example, visual inspection of
the specimen suggests that fiber 2 can not be supposed to
stretch (
i.e., have positive
e
2
e
2
)
unless
x
9
>
x
1
x
9
x
1
and/or
x
2
>
x
10
.
x
2
x
10
.
Does this jive with
Equation 1?
Applying Equation 1 to each of the
remaining fibers we arrive at
e=Ax
e
A
x
where AA is 20-by-18, one row for each
fiber, and one column for each degree of freedom. For systems of
such size with such a well defined structure one naturally hopes
to automate the construction. We have done just that in the
accompanying M-file
and
diary. The M-file begins with a matrix of raw data that
anyone with a protractor could have keyed in directly from Figure 1:
data = [ % one row of data for each fiber, the
1 4 -pi/2 % first two columns are starting and ending
1 5 -pi/4 % node numbers, respectively, while the third is the
1 2 0 % angle the fiber makes with the positive horizontal axis
2 4 -3*pi/4
...and so on... ]
This data is precisely what Equation 1
requires in order to know which columns of AA receive
the proper cos or sin. The final AA matrix is displayed in the
diary.
The next two steps are now familiar. If KK denotes
the diagonal matrix of fiber stiffnesses and ff denotes
the vector of nodal forces then
y=Ke
and
ATy=f
y
K
e
and
A
y
f
and so one must solve
Sx=f
S
x
f
where
S=ATKA
S
A
K
A
.
In this case there is an entire three--dimensional class of
zz
for which
Az=0
A
z
0
and therefore
Sz=0
S
z
0
.
The three indicates that there are three independent unstable
modes of the specimen, e.g., two translations
and a rotation. As a result SS is singular and x =
S\f in MATLAB will get us nowhere. The way out is to
recognize that SS has
18−3=15
18
3
15
stable modes and that if we restrict SS to 'act'
only in these directions then it `should' be invertible. We will
begin to make these notions precise in discussions on the Fundamental Theorem of Linear
Algebra. For now let us note that every matrix possesses
such a pseudo-inverse and that it may be computed
in MATLAB via the pinv command. Supposing
the fiber stiffnesses to each be one and the edge traction to be
of the form
f=-110111-100010-1-10-11-1T
,
f
-110
111
-100
010
-1-10
-11-1
,
we arrive at xx via
x=pinv(S)*f and offer below its graphical
representation.