Design Examples of Steel and Steel-Concrete Composite Bridges - Finite Element Analysis and Design of Steel and Steel-Concrete Composite Bridges - page 288

s

a

D 1

V 1

Fsin a

s

b

A

6 m 48 m

g vk = 78.1 kN/m

-4195.1

375 kN 375 kN

q vk = 43.8 kN/m

1.2 m

+

1.281

1.153 1.127

Figure 4.64 Determination of the force in diagonal member D 1 using the influence line
method.

It should be noted that, from the equilibrium of joint J 1 (see Figure 4.64 ),
the force in the vertical truss member V 1 is equal to that of D 1 multiplied
by sin a but with a negative sign (a compression force of 6711.5 sin
51.34 ¼ 5240.8 kN).

4.3.3.12 Calculation of the Reactions at Supports
The reactions at supports can be also calculated using the influence line
method, as shown in Figure 4.65 , as follows:
A +ve ðÞ¼A net D ðÞ¼ 0 : 5 60 1 : 0 ¼ 30 : 0

F D : L : ðÞ¼ 30 : 0 78 : 1 ¼ 2343 kN

F L : L : ðÞ positive ð Þ 375 1 ð : 0+0 : 98 Þ +30 : 0 43 : 8 ¼ 2056 : 5kN

F Ed ðÞ¼F D : L : g g + F L : L : g q
F Ed ð maximum ¼ 2343 1 2kN

Figure 4.66 summarizes the calculated forces in the truss members and
presents the commonly known distribution of forces in the N-shaped main
truss under the dead and live cases of loading.

: 3 + 2056 : 5 1 : 35 ¼ 5822 :

4.3.3.13 Design of the Maximum Compression Upper Chord Member U 5
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