60 3 Multispan Beams and Trusses
Influence lines for reactions are constructed in the same manner as in the case of
a simple truss. These influence lines are not shown.
The construction of influence lines for internal forces is presented in tabulated
form below.
ForceV 2 - 3
Vertical member 2-3 belongs to the main truss only; therefore, its influence line is
obtained by considering the main truss only. Consider the equilibrium of joint 3 of
the main truss:
PD 1 is applied at joint 3 PD 1 is applied at any joint except 3
V 2 - 3!P
YD 0 !V 2 - 3 PD0; V 2 - 3!P
YD 0 !V 2 - 3 D0;
V 2 - 3 DP!IL.V 2 - 3 /D 1 IL.V 2 - 3 /D 0ForcesD 40 - 7 ,U 5 sec (^0) - 7
To calculate these forces we need to consider secondary truss 5- 40 -7- 50 (Fig. 3.19).
Fig. 3.19 Secondary truss
loading 0.5
7
Usec
5 ′− 7
D 4 ′–7
4 ′
5 ′
(^57)
P=1
a
Assume that loadPD 1 is located at point 50. Then the reactions at points 5 and
7 will equal 0.5.
The equilibrium condition of joint 7 (Fig.3.19) leads to the following results:
D 40 - 7 D
1
2 sin ̨
;U 5 sec (^0) - 7 D
1
2 tan ̨
:
These influence lines are shown in Fig.3.18.
It is obvious that IL.D 40 - 7 /DIL
Dsec 40 - 5
and IL
U 5 sec (^0) - 7
DIL
U 5 sec (^0) - 5
.
ForceD 40 - 5
Diagonal member 40 -5 belongs simultaneously to the main and to a secondary truss.
Therefore, an influence line should be constructed considering the main truss and
secondary truss 5- 40 -7- 50 together:
D 40 - 5 DD 5 main- 6 CDsec 40 - 5 !IL.D 40 - 5 /DIL
Dmain 5 - 6
CIL
D 4 sec (^0) - 5
: