336 COMPUTER AIDED ENGINEERING DESIGN
ke =^16
15
10(1) (2) (3) (4) (5) (6) (7) (8)
0.9333 0.3500 –0.0933 –0.0700 –0.4667 –0.3500 –0.3733 0.0700 (1)
0.3500 1.6333 0.0700 0.6067 –0.3500 –0.8167 –0.0700 –1.4233 (2)
–0.0933 0.0700 0.9333 –0.3500 –0.3733 –0.0700 –0.4667 0.3500 (3)
–0.0700 0.6067 –0.3500 1.6333 0.0700 –1.4233 0.3500 –0.8167 (4)
–0.4667 –0.3500 –0.3733 0.0700 0.9333 0.3500 –0.0933 –0.0700 (5)
–0.3500 –0.8167 –0.0700 –1.4233 0.3500 1.6333 0.0700 0.6067 (6)
–0.3733 –0.0700 –0.4667 0.3500 –0.0933 0.0700 0.9333 –0.3500 (7)
0.0700 –1.4233 0.3500 –0.8167 –0.0700 0.6067 –0.3500 1.6333 (8)×^7⎡⎣⎢
⎢⎢
⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎤⎦⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥That the degrees of freedom (1), (2), (7) and (8) are fixed, removing the corresponding rows and
columns gives a 4 × 4 system as
16
15
100.9333 –0.3500 –0.3733 –0.0700
–0.3500 1.6333 0.0700 –1.4233
–0.3733 0.0700 0.9333 0.3500
–0.0700 –1.4233 0.3500 1.6333(3)
(4)
(5)
(6)=2000- 1000
0
0
×^7⎡⎣⎢
⎢
⎢
⎢⎤⎦⎥
⎥
⎥
⎥⎡⎣⎢
⎢
⎢
⎢⎤⎦⎥
⎥
⎥
⎥⎡⎣⎢
⎢
⎢
⎢⎤⎦⎥
⎥
⎥
⎥U
U
U
UThe non-zero displacements are
U
U
U
U(3)
(4)
(5)
(6)=^15
16
100.1786- 0.2381
0.1786 - 0.2381
–3⎡⎣⎢
⎢
⎢
⎢⎤⎦⎥
⎥
⎥
⎥×⎡⎣⎢
⎢
⎢
⎢⎤⎦⎥
⎥
⎥
⎥which compare well with those in Example 11.4.
The discussion in this chapter is restricted to elementary finite elements to demonstrate the principles
in the finite element displacement and stress analysis. The reader may note that numerous advanced
books are available on this subject which extend the method to heat transfer and fluid dynamics, and
stress and displacement analysis for geometrically and materially nonlinear problems, both in two-
and three-dimensions.
Exercises
- Compute the displacements at nodes 2 (treat node 4 the same as node 2) and 5 treating springs as finite
elements as in section 11.2. Verify the result using Newtonian equilibrium.
Figure P11.1132
k 1k 14k (^25) F