Stability Analysis - Mechanics of Structures: Variational and Computational Methods - page 687

Formation of the System Matrix
To form the global stiffness matrix, assemble the stiffness matrices of the elements with the
usual process of the displacement method as described in Chapter 5. The geometric matrices
are handled in the same manner as the linear matrices. The superposition is shown below,
starting with the topological description of the frame in the incidence table, followed by an
expression of the assembly as equilibrium at the nodes. For details, the reader is referred
to Chapter 5, Example 5.5, where the same procedure is described.

Incidence Table

Element ( u Xa ) (w Za ) (θ Ya ) ( u Xb ) (w Zb ) (θ Yb )

1 0 0 0 1 0 2

(8)

2 0 0 2 0 0 0

3 1 0 0 0 0 0

From the incidence table we obtain information as to which degrees of freedom of a partic-
ular element contribute to the formation of equilibrium for the unknown nodal displace-
ments. In this example, the unknown horizontal displacement U 2 and the unknown rota-
tion

2 at node 2 are determined by the equilibrium conditions for the conjugate horizontal
forces F X 2 and moments M Y 2 at this node.

1 2

U 2 2

1941 . 92 8325 . 86 Element 1

1 F X 2 Element 2

(9)

− 523 . 41 Element 3

. .

8325 86 43 809 20 Element 1

2 M Y 2 52 200 . 00 Element 2

Element 3

In this 2 2 matrix the contributions of the three elements can be clearly distinguished when

compared with the element matrices of (3), (5), and (7). The first row gives the equilibrium
of the forces in the horizontal direction at node 2, while the second describes the equilibrium
of the moments at the same node. Summation of the element stiffnesses renders the stiffness
matrix of the system:

×

1418

. 51 8325 . 86

K total = (10)

8325 . 86 96 009 . 20

From Eq. (5.100), the addition of the loading vectors takes the form

P ∗ P 0 P = + (11)

P ∗ P 0

− −

82 . 5 − 36 . 00 + 0 . 0 118 . 50

P total = = (12)

0 . 0 − 48 . 00 − 75 . 0 − 123 . 00

where P ∗ contains the loads at the nodes (

5, 0) applied directly at node 2 and the
vector P 0 contains the horizontal force and the sum of the two end moments (-36

− 82 .

0)
and (-48.00 - 75.0) arising from the distributed loads on the two bars adjacent to node 2.

. 00 + 0 .