Civil Engineering Reference
In-Depth Information
and (17.19) we have
[
K
ij
][
T
]
[
T
]
{
F
}=
{
δ
}
Hence
[
T
−
1
][
K
ij
][
T
]
{
F
}=
{
δ
}
(17.20)
It may be shown that the inverse of the transformation matrix is its transpose, i.e.
[
T
−
1
]
[
T
]
T
=
Thus we rewrite Eq. (17.20) as
[
T
]
T
[
K
ij
][
T
]
{
F
}=
{
δ
}
(17.21)
The nodal force system referred to the global axes,
{
F
}
, is related to the corresponding
nodal displacements by
{
F
}=
[
K
ij
]
{
δ
}
(17.22)
in which [
K
ij
] is the member stiffness matrix referred to global coordinates. Compari-
son of Eqs (17.21) and (17.22) shows that
[
T
]
T
[
K
ij
][
T
]
[
K
ij
]
=
Substituting for [
T
] from Eq. (17.17) and [
K
ij
] from Eq. (17.16) we obtain
λ
2
λ
2
−
−
λµ
λµ
µ
2
µ
2
AE
L
−
−
λµ
λµ
[
K
ij
]
=
(17.23)
λ
2
λ
2
−
−
λµ
λµ
−
µ
2
µ
2
−
λµ
λµ
Evaluating
λ
(
sin
θ
) for each member and substituting in Eq. (17.23)
we obtain the stiffness matrix, referred to global axes, for each member of the
framework.
=
cos
θ
) and
µ
(
=
E
XAMPLE
17.1
Determine the horizontal and vertical components of the deflection
of node 2 and the forces in the members of the truss shown in Fig. 17.4. The product
AE
is constant for all members.
We see from Fig. 17.4 that the nodes 1 and 3 are pinned to the foundation and are
therefore not displaced. Hence, referring to the global coordinate system shown,
w
1
=
v
1
=
w
3
=
v
3
=
0
