FINITE ELEMENT METHOD 325ke = λλλλλTkλλλλλ (11.9j)withk and λλλλλ defined in Eqs. (11.9h) and (11.9i), respectively.
Example 11.3. Consider a slightly involved example of an over-bridge modeled using frame elements.
The horizontal and vertical lengths are of 10 m each except the vertical element at the center of length
20 m. The flexural rigidity EI is taken for all elements as 10^7 Nm^2 and cross section areas Aare
10 –2 m^2. The example is solved using a frame finite element implementation in MATLABTM for
displacements. The assembly procedure for the global stiffness matrix is similar to Example 11.2 with
element stiffness matrices computed using Eqs. (11.9h) and (11.9j). The displaced configuration
(dashed lines) is given in Figure 11.11(b) with the maximum downward displacement at the center
node on bottom edge as 0.0012 m.
100 kN 100 kN 100 kN20 m40 m
(a)(b)
Figure 11.11 (a) An over-bridge modeled with frame elements and (b) displacement profile (scaled)11.6 Continuum Triangular Elements
Truss, beam and frame elements are often considered discrete elements as they approximate a region
only partially. A more comprehensive discretization for a two-dimensional region is performed using
triangular or quadrilateral elements. Consider, for instance, a triangular element shown in Figure
11.12 with three nodes i,jandk, each having two degrees of freedom (ux,vx) along xandydirections
respectively, and the same number of external forces (fx,fy) as shown.
To interpolate the displacements (u,v) at point P in the element, we assume that u is dependent
only on the nodal displacements in the xdirection. Since there are three such unknowns ui,uj and uk,
displacementu at P can be interpolated as
u = a 0 + a 1 x + a 2 y (11.10a)