Computer Aided Engineering Design

FINITE ELEMENT METHOD 317

Example 11.1. An assemblage of truss members is shown in Figure 11.5. It is required to determine
the horizontal and vertical displacements at point A which is subjected to the loads of 200 kN and
100 kN as shown. Points B and C are fixed. Length of members AB and BC is 2 m each, and area of
cross section and elastic modulus are given in Table 11.1.
First, node numbers are assigned to each node as shown in the figure. The x axis is taken along
BA and y along BC. Coordinates of A, B and C (nodes 1, 2 and 3, respectively) are (2, 0), (0, 0) and
(0, 2). Table 11.1 provides the connectivity information for each element.

Figure 11.5 Truss assemblage for Example 11.1

2 m Element 1

C

3

Element 3

200 kN

2

B

1

A

2 m

Element 2

100 kN

Table 11.1 Element data for Example 11.1

Truss Element Node Cross-sectional Young’s

member number connectivity area (cm^2 ) modulus (GPa)

BC 1 2, 3 20 70

AB 2 2, 1 20 200

AC 3 1, 3 50 200

Note that the structure has 6 degrees of freedom, 3 along x and 3 along yaxes respectively. Let U
be the global displacement vector such that its odd entries are assigned x displacements while even
are assigned y displacements, that is, for node k,U(2k–1) = ukx while U(2k) = uky. Then, U will have
the form

U = [U(1) U(2) U(3) U(4) U(5) U(6)]T = [u 1 x u 1 y u 2 x u 2 y u 3 x u 3 y]T (11.8a)

and the global stiffness matrix K will be of size 6 × 6. To get K, we would assemble the local stiffness
matrices as follows:
For element 1, θ = 90°,l= 2 m, A = 20 × 10 –4 m^2 and E = 70 × 109 Nm–2. Using Eq. (11.7o), we have
3456

ke^17 = 7 10

0000

010–1

0000

0–10 1

3

4

5

6

×

⎡

⎣

⎢

⎢

⎢

⎢

⎤

⎦

⎥

⎥

⎥

⎥

(11.8b)