Design Examples of Steel and Steel-Concrete Composite Bridges - Finite Element Analysis and Design of Steel and Steel-Concrete Composite Bridges - page 281

F Ed L ð maximum ¼ 2999 1 : 3 + 2629 : 9 1 : 35

¼ 7449 : 1 kN Tension force ð Þ

F L : L : L ðÞ negative ð Þ ¼ 0 : 5 60 1 : 28 0 : 83 ¼ 31 : 9kN

F Ed L ð minimum ¼ 2999 1

:

3 31

:

9 1

:

35

¼ 3855 : 6 kN Tension force ð Þ

It should be noted that, from the equilibrium of the truss (see
Figure 4.58 ), the force in upper chord truss member U 2 is equal to that
of the calculated lower chord member L 3 but with a negative sign (a com-
pression force of 7449.1 kN).

4.3.3.6 Calculation of Force in the Lower Chord Member L 2
The force in member L 2 due to the dead and live loads can be calculated,
as shown in Figure 4.59 , as follows:
F D : L : L ðÞ¼ 0 : 5 60 0 : 72 78 : 1 ¼ 1687 kN

F L : L : L ðÞ¼ 375 0 ð : 72 + 0 : 704 Þ +0 : 5 60 0 : 72 43 : 8 ¼ 1480 : 1kN

F Ed L ðÞ¼F D : L : g g + F L : L : g q

F Ed L ð maximum ¼ 1687 1 : 3 + 1480 : 1 1 : 35

¼ 4191 : 2 kN Tension force ð Þ

F L : L : L ðÞ negative ð Þ ¼ 0 : 5 60 0 : 72 0 : 83 ¼ 17 : 9kN

F Ed L ð minimum ¼ 1687 1 : 3 17 : 9 1 : 35

¼ 2168 : 9 kN Tension force ð Þ

s

U 2

a

B

L 1 L 2

A s

54 m

6 m

g vk = 78.1 kN/m

375 kN 375 kN

q vk = 43.8 kN/m

1.2 m

+

0.704

6 × 54/(60 × 7.5) = 0.72

Figure 4.59 Determination of the tensile force in lower chord member L 2 using the
influence line method.