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vector p i . To derive a stiffness matrix with a loading vector appended, begin by writing
a transfer matrix in the notation of Sign Convention 2, giving
v b
U vv
v a
F 0
v
U v p
=
+
(4.15)
p b
U p v
U pp
p a
F p
where
w
V b
0
b
F 0
F p =
v =
b
M b
θ
Expressions for the entries of the vectors F 0
v
and F p will be given later. Follow the procedure
utilized for forming the stiffness matrix of Eq. (4.11) to rearrange Eq. (4.15) into stiffness
matrix form. This gives
=
+
.
U 1
v
p F 0
p a
U 1
v
U 1
v
v a
···
p U vv
v
p
··················
······
(4.16)
.
U pp U 1
v
U pp U 1
v
F p
U pp U 1
v
p F 0
U p v
p U vv
p b
v b
p
v
The contents of this matrix are easily identified with the nomenclature of the matrix of
Eq. (4.14).
4.3.3
Flexibility Matrix
A flexibility matrix relates the forces p at a and b of an element to the displacements v at a
and b . From this definition, it would appear that the flexibility matrix is simply the inverse
of the stiffness matrix, which relates the displacement v at a and b to the forces p at a and
b . Certainly, this would be the case if the stiffness matrix were nonsingular. However, it
will be shown in this section that the stiffness matrix for an unconstrained beam element
is singular, and, hence, the corresponding flexibility matrix does not exist.
The stiffness matrices defined in Section 4.3.2 relate all of the degrees of freedom (dis-
placements) to all of the end forces, without regard as to how the element is supported or
constrained. Thus, the beams are treated as being free or unconstrained bodies, and rigid
body displacements can occur. Although rigid body displacements may not affect the de-
formation of the beam, they cause the rows or columns of the stiffness matrix to be linearly
dependent.
We wish to observe the problems arising from the unconstrained beam element. From
Eqs. (4.7a), (4.7b), and (4.7c), written in the nomenclature of Sign Convention 2 (replace s
by p ), the fundamental relations for a beam element can be written as
=
v
U v p p a
(4.17a)
v b
U vv v a
=
v
(4.17b)
p b
=
U pp p a
(4.17c)
From Eq. (4.17a) for given forces p a ,
v can be determined uniquely, as can p b from
Eq. (4.17c). It follows from Eq. (4.17b) that a particular combination of displacements v b
and v a can be determined uniquely, but not v b and v a themselves. Thus, v b and v a are
not single-valued, as one depends on the other. In other terms, the deformation of the
beam as characterized by
v is not influenced by certain combinations of displacements.
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