Computer Aided Engineering Design

322 COMPUTER AIDED ENGINEERING DESIGN

Consistent with the definition of U for this example, the external force vector is given as

F = 10

0

–5

0

–10

0

1 2 3 4 5 6 7 8

4

1

7

8

F

F

F

⎡

⎣

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥

withF 1 ,F 7 and F 8 as the unknown reaction forces and moments at the supports for which U(1),U(7)
andU(8) are zero, respectively. To solve only for the unknown displacements, eliminating the 1st, 7th
and 8th rows and columns, we have the linear system of size 5 × 5 as

12 34 5 678

10

4–62 00

–6 18 –3 –6 3

2–36–31

0–6–3180

03103

1 2 3 4 5 6 7 8

0

(2)

(3)

(4)

(5)

(6)

0

0

6

⎡

⎣

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥

⎡

⎣

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥

⎡

⎣

⎢ ⎢ ⎢ ⎢ ⎢ ⎢

U

U

U

U

U

⎢⎢

⎢

⎢

⎢

⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥

⎡

⎣

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥

= 10

0

–5

0

–10

0

4

1

7

8

F

F

F

solving which we get

U = 10–2× [0 –3.58 –3.00 –1.84 –1.86 3.62 0 0]T

Post-multiplying the above displacement vector with the original 8 × 8 stiffness matrix, we get the
force vector as

F = 10^3 × [34.87 0 –50 0 –100 –0 115.13 –37.83]T

and thus the reaction forces are F 1 = 34.87 kN, F 7 = 115.13 kN and F 8 = –37.83 kNm.
It is left as an exercise for the reader to solve this example analytically. Note that the beam is
statically indeterminate (with 3 reactions and two equations) and that the equilibrium and deflection
equations should be solved together.

11.5 Frame elements

Frame elements are extended beam elements wherein axial displacements are incorporated as well.
A frame element has three degrees of freedom per node, namely, the horizontal and vertical displacements