Structural Analysis Methods II: Structural Systems - Mechanics of Structures: Variational and Computational Methods - page 297

Element 3:
The distributed loading vector p 30 = 0 , as there is no distributed load on this element.

m 2 , and EA Use = 3 . 0m, EI = 10 490 kN · =∞ , giving

∞ 0 0 −∞ 0 0

0 4662 . 2 − 6993 . 3 0 − 4662 . 2 − 6993 . 3

0 − 6993 . 3 13987 0 6993 . 3 6993 . 3

k 3

= (4)

−∞ 0 0 ∞ 0 0

0 − 4662 . 2 6993 . 3 0 4662 . 2 6993 . 3

0 − 6993 . 3 6993 . 3 0 4662 . 2 6993 . 3

Use the same transformation employed in Example 5.5 to convert this to global coordinates,

4662 . 2 0 6993 . 3 − 4662 . 2 0 6993 . 3

0 ∞ 0 0 −∞ 0

6993 . 3 0 13987 − 6993 . 3 0 6993 . 3

k 3 = (5)

− 4662 . 20 − 6993 . 3 4662 . 20 − 6993 . 3

0 −∞ 0 0 ∞ 0

6993 . 3 0 6993 . 3 − 6993 . 3 0 13987

Formation of the Global Equations
The global stiffness m a trix K is as sembled as it was in Example 5.5. The global equilibrium
equations are KV P ∗ P 0 = + = P . F or the first l o ading case, with p 0 = 2 . 0 on bea m 1 only

and no concentrated applied forc es , P ∗ 0 and P 0 is taken simply from p 10 Thus, P 0

j

p 10
j

= . =

c or d , P 0
j

with j = a, b . For j = = 0 . For the second loading case, with no loading

distributed between nodes, P 0 0 , and P ∗ is formed from the applied forces concentrated

at the nodes. Since the beams are fixed at the bases, all displacements are equal to zero at
nodes a and d

=

Delete the columns corresponding to these displacements, and ignore the
rows corresponding to the reactions at a and d

.

. Then

47 210 +∞ − 79 412 2040 . 4 −∞ 0 0 U Xb

U Zb
b
U Xc
U Zc

− 79 412 140 994 . 2 − 1963 . 2 0 − 2094 . 2 − 3141 . 3

2040 . 4 − 1963 . 2 11 723 . 8 0 3141 . 3 3141 . 3

−∞ 0 0 4662 . 2 +∞ 0 − 6993 . 3

0 − 2094 . 2 3141 . 3 0 2094 . 2 +∞ 3141 . 3

0 − 3141 . 3 3141 . 3 6993 . 3 3141 . 3 20 269 . 7

c

K V

Load Load

Case 1 Case 2

.

3 . 0 0

√ 3 .

20

.

2 . 0 0

=

(6)

0 .

10

.

0 0

0 .

0

.

P 0 P ∗