Civil Engineering Reference
In-Depth Information
The internal forces in the members may be found from the axial displacements of the
nodes. Thus, for a member
ij
, the internal force
F
ij
is given by
AE
L
(
F
ij
=
w
j
− ¯
¯
w
i
)
(v)
But
w
j
=
¯
λ
w
j
+
µv
j
w
i
=
λ
w
i
+
µv
i
¯
Hence
w
j
− ¯
¯
w
i
=
λ
(
w
j
−
w
i
)
+
µ
(
v
j
−
v
i
)
Substituting in Eq. (v) and rewriting in matrix form,
L
[
λµ
]
w
j
−
AE
w
i
F
ij
=
(vi)
v
j
−
v
i
Thus, for the members of the framework
.
/
!
−
WL
AE
−
0
AE
L
[1
F
12
=
0]
=−
W
(compression)
0
−
3
·
828
WL
AE
"
−
0
1]
0
AE
L
[0
−
0
F
13
=
=
0 (obvious from inspection)
0
−
0
.
/
0
!
"
WL
AE
0
+
AE
1
.
414
L
[
F
23
=
−
0
.
707
0
.
707]
=
1
.
414
W
(tension)
3
.
828
WL
AE
0
+
The matrix method of solution for the statically determinate truss of Ex. 17.1 is com-
pletely general and therefore applicable to any structural problem. We observe from
the solution that the question of statical determinacy of the truss did not arise. Statically
indeterminate trusses are therefore solved in an identical manner with the stiffness
matrix for each redundant member being included in the complete stiffness matrix as
described above. Clearly, the greater the number of members the greater the size of
the stiffness matrix, so that a computer-based approach is essential.
The procedure for the matrix analysis of space trusses is similar to that for plane
trusses. Themain difference lies in the transformation of themember stiffnessmatri
ce
s
from
l
ocal to global coordinates since, as we see from Fig. 17.5, axial nodal forces
F
x
,
i
and
F
x
,
j
have each, now, three global components
F
x
,
i
,
F
y
,
i
,
F
z
,
i
and
F
x
,
j
,
F
y
,
j
,
F
z
,
j
,
respectively. The member stiffness matrix referred to global coordinates is therefore
