Design Examples of Steel and Steel-Concrete Composite Bridges - Finite Element Analysis and Design of Steel and Steel-Concrete Composite Bridges - page 428

4 m

A

40 m

R A

g vk = 77.1 kN/m

450 kN 450 kN

q vk = 45.65 kN/m

1.2 m
1.0

0.97

+

Figure 4.189 Determination of the reaction R A using the influence line method.

4.6.3.11 Calculation of Force in the Vertical Members
The force in the vertical truss members V 2 and V 4 can be calculated, as
shown in Figure 4.190 , as follows:
A ve V ðÞ¼ 0 : 5 10 1 : 0 ¼ 5

F D : L : V ðÞ¼ 5 77 : 1 ¼ 385 : 5kN

F L : L : V ðÞ negative ð Þ ¼ 450 1 ð : 0+0 : 76 Þ +5 45 : 65 ¼ 1020 : 3kN

F Ed D ð maximum ¼F D : L : g g + F L : L : g q

F Ed D ð maximum ¼ 385 : 5 1 : 3 1020 : 3 1 : 35

¼ 1878 : 6 kN Compression force ð Þ:

It should be noted that from the truss equilibrium, the forces in members
V 3 and V 5 are zero under the applied dead and live loads, while the force in
member V 1 is equal to the reaction at supports.
Figure 4.191 summarizes the calculated forces in the truss members and
presents the commonly known distribution of forces in the Warren main
truss under the dead and live cases of loading.

4.6.3.12 Design of the Maximum Compression Upper Chord Members
U 4 and U 3
After the calculation of the design forces in the main truss members, we can
now design different members of the main truss. Let us start by designing
the maximum compression upper chord members U 4 and U 3 ,shown
in Figure 4.192 , carrying a compressive design force of 10,366.4 kN.