Civil Engineering Reference
In-Depth Information
are computed via Equations 3.29 and 3.30, respectively. Specification of the
cross-sectional area
A
and modulus of elasticity
E
of each element allows com-
putation of the element stiffness matrix in the global frame using Equation 3.28.
Finally, the element stiffness matrix terms are added to the global stiffness matrix
using the element displacement location vector.
In the context of the current example, the reader is to imagine a 6
×
6 array
of mailboxes representing the global stiffness matrix, each of which is originally
empty (i.e., the stiffness coefficient is zero). We then consider the stiffness ma-
trix of an individual element in the (2-D) global reference frame. Per the location
vector (addresses) for the element, the individual values of the element stiffness
matrix are placed in the appropriate mailbox. In this fashion, each element is
processed in sequence and its stiffness characteristics added to the global matrix.
After all elements are processed, the array of mailboxes contains the global stiff-
ness matrix.
3.5 BOUNDARY CONDITIONS,
CONSTRAINT FORCES
Having obtained the global stiffness matrix via either the equilibrium equations
or direct assembly, the system displacement equations for the example truss of
Figure 3.2 are of the form
U
1
U
2
U
3
U
4
U
5
U
6
F
1
F
2
F
3
F
4
F
5
F
6
[
K
]
=
(3.42)
As noted, the global stiffness matrix is a singular matrix; therefore, a unique so-
lution to Equation 3.42 cannot be obtained directly. However, in developing
these equations, we have not yet taken into account the constraints imposed on
system displacements by the support conditions that must exist to preclude rigid
body motion. In this example, we observe the displacement boundary conditions
U
1
=
0
(3.43)
leaving only
U
5
and
U
6
to be determined. Substituting the boundary condition
values and expanding Equation 3.42 we have, formally,
K
15
U
5
+
U
2
=
U
3
=
U
4
=
K
16
U
6
=
F
1
K
25
U
5
+
K
26
U
6
=
F
2
K
35
U
5
+
K
36
U
6
=
F
3
(3.44)
K
45
U
5
+
K
46
U
6
=
F
4
K
55
U
5
+
K
56
U
6
=
F
5
K
56
U
5
+
K
66
U
6
=
F
6
