We return once again to the biaxial testing problem, introduced
in the uniaxial truss module. It turns out that
singular matrices are typical in the biaxial testing problem. As
our initial step into the world of such planar structures let us
consider the simple truss in the figure of a simple swing.
We denote by
x
1
x
1
and
x
2
x
2
the respective horizontal and vertical displacements of
m
1
m
1
(positive is right and down). Similarly,
f
1
f
1
and
f
2
f
2
will denote the associated components of force. The corresponding
displacements and forces at
m
2
m
2
will be denoted by
x
3
x
3
,
x
4
x
4
and
f
3
f
3
,
f
4
f
4
.
In computing the elongations of the three springs we shall make
reference to their unstretched lengths,
L
1
L
1
,
L
2
L
2
,
and
L
3
L
3
.
Now, if spring 1 connects
0−
L
1
0
L
1
to
01
0
1
when at rest and
0−
L
1
0
L
1
to
x
1
x
2
x
1
x
2
when stretched then its elongation is simply
e
1
=2
x
1
2+
x
2
+
L
1
2−
L
1
e
1
2
x
1
2
x
2
L
1
2
L
1
(1)
The price one pays for moving to higher dimensions is that lengths
are now expressed in terms of square roots. The upshot is that the
elongations are not linear combinations of the end displacements
as they were in the uniaxial case. If we presume, however,
that the loads and stiffnesses are matched in the sense that the
displacements are small compared with the original lengths, then
we may effectively ignore the nonlinear contribution in Equation 1. In order to make this precise we
need only recall the
The Taylor development of
21+t
2
1
t
about
t=0
t
0
is
21+t=1+t2+Ot2
2
1
t
1
t
2
O
t
2
where the latter term signifies the remainder.
With regard to
e
1
e
1
this allows
e
1
=2
x
1
2+
x
2
2+2
x
2
L
1
+
L
1
2−
L
1
=
L
1
21+
x
1
2+
x
2
2
L
1
2+2
x
2
L
1
−
L
1
e
1
2
x
1
2
x
2
2
2
x
2
L
1
L
1
2
L
1
L
1
2
1
x
1
2
x
2
2
L
1
2
2
x
2
L
1
L
1
(2)
e
1
=
L
1
+
x
1
2+
x
2
22
L
1
+
x
2
+
L
1
O
x
1
2+
x
2
2
L
1
2+2
x
2
L
1
2−
L
1
=
x
2
+
x
1
2+
x
2
22
L
1
+
L
1
O
x
1
2+
x
2
2
L
1
2+2
x
2
L
1
2
e
1
L
1
x
1
2
x
2
2
2
L
1
x
2
L
1
O
x
1
2
x
2
2
L
1
2
2
x
2
L
1
2
L
1
x
2
x
1
2
x
2
2
2
L
1
L
1
O
x
1
2
x
2
2
L
1
2
2
x
2
L
1
2
(3)
If we now assume that
x
1
2+
x
2
22
L
1
is small compared to
x
2
x
1
2
x
2
2
2
L
1
is small compared to
x
2
(4)
then, as the
OO term is even
smaller, we may neglect all but the first terms in the above and
so arrive at
e
1
=
x
2
e
1
x
2
To take a concrete example, if
L
1
L
1
is one meter and
x
1
x
1
and
x
2
x
2
are each one centimeter, then
x
2
x
2
is one hundred times
x
1
2+
x
2
22
L
1
x
1
2
x
2
2
2
L
1
.
With regard to the second spring, arguing as above, its elongation is
(approximately) its stretch along its initial direction. As its initial
direction is horizontal, its elongation is just the difference of the
respective horizontal end displacements, namely,
e
2
=
x
3
−
x
1
e
2
x
3
x
1
Finally, the elongation of the third spring is (approximately)
the difference of its respective vertical end displacements,
i.e.,
e
3
=
x
4
e
3
x
4
We encode these three elongations in
e=Ax
where
A=(
0100
-1010
0001
)
e
A
x
where
A
0100
-1010
0001
Hooke's
law is an elemental piece of physics and is not
perturbed by our leap from uniaxial to biaxial structures. The
upshot is that the restoring force in each spring is still
proportional to its elongation, i.e.,
y
j
=
k
j
e
j
y
j
k
j
e
j
where
k
j
k
j
is the stiffness of the jjth spring. In matrix
terms,
y=Ke
where
K=(
k
1
00
0
k
2
0
00
k
3
)
y
K
e
where
K
k
1
00
0
k
2
0
00
k
3
Balancing horizontal and vertical forces at
m
1
m
1
brings
(−
y
2
)−
f
1
=0
y
2
f
1
0
and
y
1
−
f
2
=0
y
1
f
2
0
while balancing horizontal and vertical forces at
m
2
m
2
brings
y
2
−
f
3
=0
y
2
f
3
0
and
y
3
−
f
4
=0
y
3
f
4
0
We assemble these into
By=f
where
B=(
0-10
100
010
001
)
,
B
y
f
where
B
0-10
100
010
001
,
and recognize, as expected, that BB is nothing more than
AT
A
. Putting the pieces together, we find that
xx must satisfy
Sx=f
S
x
f
where
S=ATKA=(
k
2
0−
k
2
0
0
k
1
00
−
k
2
0
k
2
0
000
k
3
)
S
A
K
A
k
2
0
k
2
0
0
k
1
00
k
2
0
k
2
0
000
k
3
Applying one step of Gaussian Elimination
brings
(
k
2
0−
k
2
0
0
k
1
00
0000
000
k
3
)
x
1
x
2
x
3
x
4
=
f
1
f
2
f
1
+
f
3
f
4
k
2
0
k
2
0
0
k
1
00
0000
000
k
3
x
1
x
2
x
3
x
4
f
1
f
2
f
1
f
3
f
4
and back substitution delivers
x
4
=
f
4
k
3
x
4
f
4
k
3
0=
f
1
+
f
3
0
f
1
f
3
x
2
=
f
2
k
1
x
2
f
2
k
1
x
1
−
x
3
=
f
1
k
2
x
1
x
3
f
1
k
2
The second of these is remarkable in that it contains no
components of xx.
Instead, it provides a condition on ff. In mechanical terms, it states
that there can be no equilibrium unless the horizontal forces on
the two masses are equal and opposite. Of course one could have
observed this directly from the layout of the truss. In modern,
three--dimensional structures with thousands of members meant to
shelter or convey humans one should not however be satisfied
with the `visual' integrity of the structure. In particular,
one desires a detailed description of all loads that can, and,
especially, all loads that can not, be equilibrated by the
proposed truss. In algebraic terms, given a matrix SS, one
desires a characterization of
-
all those ff for
which
Sx=f
S
x
f
possesses a solution
-
all those ff for
which
Sx=f
S
x
f
does not possess a solution
We will eventually provide such a characterization in our later
discussion of the
column
space of a matrix.
Supposing now that
f
1
+
f
3
=0
f
1
f
3
0
we note that although the system above is consistent it still
fails to uniquely determine the four components of xx. In
particular, it specifies only the difference between
x
1
x
1
and
x
3
x
3
.
As a result both
x=
f
1
k
2
f
2
k
1
0
f
4
k
3
and
x=0
f
2
k
1
−
f
1
k
2
f
4
k
3
x
f
1
k
2
f
2
k
1
0
f
4
k
3
and
x
0
f
2
k
1
f
1
k
2
f
4
k
3
satisfy
Sx=f
S
x
f
.
In fact, one may add to either an arbitrary multiple of
and still have a solution of
Sx=f
S
x
f
.
Searching for the source of this lack of uniqueness we observe
some redundancies in the columns of
SS. In particular, the third is
simply the opposite of the first. As
SS is simply
ATKA
A
K
A
we recognize that the original fault lies with
AA, where again, the first and
third columns are opposites. These redundancies are encoded in
zz in the sense that
Az=0
A
z
0
Interpreting this in mechanical terms, we view
zz as a displacement and
Az
A
z
as the resulting elongation. In
Az=0
A
z
0
we see a nonzero displacement producing zero elongation. One
says in this case that the truss deforms without doing any work
and speaks of
zz as
an
unstable mode. Again, this mode could have been
observed by a simple glance at
Figure 1. Such is not the case for more complex
structures and so the engineer seeks a systematic means by which
all unstable modes may be identified. We
shall see later that all these modes are captured by the
null space of
AA.
From
Sz=0
S
z
0
one easily deduces that SS is singular. More precisely,
if
S-1
S
were to exist then
S-1Sz
S
S
z
would equal
S-10
S
0
, i.e.,
z=0
z
0
, contrary to Equation 5. As
a result, Matlab will fail to solve
Sx=f
S
x
f
even when ff is a force that the truss can
equilibrate. One way out is to use the
pseudo-inverse, as we shall see in the General Planar Truss module.
