Design Examples of Steel and Steel-Concrete Composite Bridges - Finite Element Analysis and Design of Steel and Steel-Concrete Composite Bridges - page 279

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

F Ed L ð maximum ¼ 4498 : 6 1 : 3 + 3938 : 9 1 : 35

¼ 11,165 : 7 kN Tension force ð Þ

Since this member is a tensile member, we should check the minimum
force due to the negative distributed loads since they may change the force in
the member to compression:

F L : L : L ðÞ negative ð Þ ¼ 0 : 5 60 1 : 92 0 : 83 ¼ 47 : 8kN

F Ed L ð minimum ¼ 4498 : 6 1 : 3 47 : 8 1 : 35
¼ 5783

ð Þ

: 7 kN Tension force

It should be noted that, from the equilibrium of the truss (see
Figure 4.56 ), the force in upper chord truss member U 4 is equal to that
of the calculated lower chord member L 5 but with a negative sign (a com-
pression force of 11,165.7 kN).

4.3.3.4 Calculation of Force in the Lower Chord Member L 4
We can repeat the earlier procedures now and change the pole where the
moment is calculated to determine the force in the member, as shown in
Figure 4.57 . Hence, the forces due to the dead and live loads can be calcu-
lated as follows:

s

U 3

a

B

L 4

A

s

18 m 42 m

g vk = 78.1 kN/m

375 kN

375 kN

q vk = 43.8 kN/m

1.2 m

+

1.632

18 × 42/(60 × 7.5) = 1.68

Figure 4.57 Determination of the tensile force in lower chord member L 4 using the
influence line method.