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If
some of the displacements are dependent on other displacements, the partial derivatives of
(8) cannot be taken because this implies that
V
j
other than
V
k
and
V
h
are to be held constant.
Then the stiffness matrix
K
does not exist. Dependent displacements are treated in some
detail in Chapter 4.
Equation (8) applies only if
U
i
is expressed in terms of independent displacements
V
k
.
EXAMPLE 3.4 Properties of Stiffness Matrices
The definitions of Examples 3.2 and 3.3 permit some important properties of stiffness ma-
trices to be identified. In summation form, the strain energy of Eq. (4) of Example 3.3 would
appear as
n
n
1
2
U
i
=
K
kj
V
k
V
j
(1)
k
=
1
j
=
1
This double summation is said to be a
quadratic form
in variables
V
k
,
provided that coeffi-
cients
K
kj
are symmetric—which they are for a stiffness matrix.
A quadratic form such as (1) is defined as being
positive definite
if
U
i
is positive for
arbitrary non-zero values of
V
k
and
V
j
and is zero if all
V
k
and
V
j
are zero. The strain
energy satisfies this definition, and, hence, it is positive definite, and the associated stiffness
matrix
K
is referred to as a
positive definite matrix
. In Example 3.3, the positive definite
K
considered is defined only if the displacements
V
k
are independent variables. It will be
shown in Chapter 4 that this corresponds to a system for which rigid body motion has been
eliminated. A stiffness matrix from which the rigid body motion has not been removed is
positive semi-definite.
Another property of
K
is that the elements on the principal diagonal,
K
ii
,
are always
positive because, by definition,
K
ii
is the force at
i
corresponding to a displacement
i
. The
force is in the same direction as the displacement and hence
K
ii
is positive. This property
of
K
ii
is not valid if the structure becomes unstable.
EXAMPLE 3.5 Stiffness Matrix for a Truss
Find a relationship between the displacements and the applied loads of the truss of Fig. 3.2a.
All of the bars are elastic and have the same cross-sectional areas.
The desired load-displacement relation can be obtained by calculating the stiffness coef-
ficients using Eq. (8) of Example 3.3. The elongation
v
of the
j
th bar is (Fig. 3.2b)
v
=
(
u
xb
−
u
xa
)
cos
θ
+
(
u
zb
−
u
xa
)
sin
θ
(1)
The strain energy of the
j
th bar with extension
v
is
V
2
L
EA
2
U
i
=
(2)
Invoke the displacement boundary conditions and the conditions of compatibility as is
required by the principle of the virtual work method.
=
=
=
=
=
=
=
=
u
xd
|
5
u
zd
|
5
u
xd
|
3
u
zd
|
3
0
,
u
xa
|
1
u
za
|
1
u
xa
|
4
u
za
|
4
0
u
xb
|
5
=
u
xb
|
1
=
u
xb
|
2
=
V
xb
=
V
1
u
zb
|
5
=
u
zb
|
1
=
u
zb
|
2
=
V
zb
=
V
2
(3)
u
xc
|
4
=
u
xc
|
3
=
u
xc
|
2
=
V
xc
=
V
3
u
zc
|
4
=
u
zc
|
3
=
u
zc
|
2
=
V
zc
=
V
4
where
u
xd
|
5
=
u
xd
of bar 5, etc.
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