Design Examples of Steel and Steel-Concrete Composite Bridges - Finite Element Analysis and Design of Steel and Steel-Concrete Composite Bridges - page 280

F D : L : L ðÞ¼ 0 : 5 60 1 : 68 78 : 1 ¼ 3936 : 2kN

F L : L : L ðÞ¼ 375 1 ð : 68 + 1 : 632 Þ +0 : 5 60 1 : 68 43 : 8 ¼ 3449 : 5kN

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

F Ed L ð maximum ¼ 3936 : 2 1 : 3 + 3449 : 5 1 : 35

¼ 9773 : 9 kN Tension force ð Þ

F L : L : L ðÞ negative ð Þ ¼ 0 : 5 60 1 : 68 0 : 83 ¼ 41 : 8kN

F Ed L ð minimum ¼ 3936 : 2 1 : 3 41 : 8 1 : 35

¼ 5060 : 6 kN Tension force ð Þ

It should be noted that, from the equilibrium of the truss (see
Figure 4.57 ) , the force in upper chord truss member U 3 is equal to that
of the calculated lower chord member L 4 but with a negative sign (a com-
pression force of 9773.9 kN).

4.3.3.5 Calculation of Force in the Lower Chord Member L 3
The force in member L 3 due to the dead and live loads can be calculated, as
shown in Figure 4.58 , as follows:
F D : L : L ðÞ¼ 0 : 5 60 1 : 28 78 : 1 ¼ 2999 kN

F L : L : L ðÞ¼ 375 1 ð : 28 + 1 : 248 Þ +0 : 5 60 1 : 28 43 : 8 ¼ 2629 : 9kN

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

s

U 2

a

B

L 3

A s

12 m 48 m

g vk = 78.1 kN/m

375 kN

375 kN

q vk = 43.8 kN/m

1.2 m

+

1.248

12 × 48/(60 × 7.5) = 1.28

Figure 4.58 Determination of the tensile force in lower chord member L 3 using the
influence line method.