Design Examples of Steel and Steel-Concrete Composite Bridges - Finite Element Analysis and Design of Steel and Steel-Concrete Composite Bridges - page 422

F Ed U ðÞ¼ 3614 : 1 1 : 3 4198 : 6 1 : 35

¼ 10,366 : 4 kN Compression force ð Þ

It should be noted that, from the equilibriumof the truss (see Figure 4.183 ) ,
the force in upper chord truss member U 3 is equal to that of the calculated
upper chord member U 4 .

4.6.3.5 Calculation of Force in the Lower Chord Member U 2
The force in member U 2 due to the dead and live loads can be calculated, as
shown in Figure 4.184 , as follows:

F D : L : U ðÞ¼ 0 : 5 40 1 : 09375 77 : 1 ¼ 1686 : 6kN

F L : L : U ðÞ¼ 450 1 ð : : Þ 0 :

5 40 1

:

09375 45

:

09375 + 1 05625 65

¼ 1966 : 1kN

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

F Ed U ðÞ¼ 1686 : 6 1 : 3 1966 : 1 1 : 35

¼ 4846 : 8 kN Compression force ð Þ

It should be noted that, from the equilibriumof the truss (see Figure 4.184 ) ,
the force in upper chord truss member U 2 is equal to that of the upper chord
member U 1 .

s

U 1 U 2

J 2

4 m

a

A

s

5 m 35 m

g vk = 77.1 kN/m

450 kN 450 kN

q vk = 45.65 kN/m

1.2 m

-

1.05625

5 × 35/(40 × 4) = 1.09375

Figure 4.184 Determination of the tensile force in lower chord member L 3 using the
influence line method.