58 3 Multispan Beams and Trusses
The internal forces in the elements of the secondary truss should be calculated
considering only load, which is applied at the joints of the secondary truss. In other
words, in order to find the internal forces in elements 2-4 and 3-4 (due to the load
applied at joint 2), we need to consider the secondary truss, so the following notation
is used:V 2 - 4 DV 2 sec- 4 andD 3 - 4 DD 3 sec- 4. Since joints 1 and 3 represent the supports
of the secondary truss, if the load is applied at these joints it has no affect on the
elements of the secondary truss. Therefore, this load impacts the elements belonging
only to the main truss.
Internal forces in the members of the first group should be calculated consid-
ering only the main truss. In this case, the load that acts at joint 2 should be
resolved into two forces applied at joints 1 and 3 of the main truss and then
V 3 - 5 DV 3 main- 5 ID 4 - 5 DDmain 1 - 5.
Internal forces in the members of the third group should be calculated by consid-
ering both the main and the secondary trusses and by summing the corresponding
internal forces:
D 1 - 4 DDmain 1 - 5 CD 1 sec- 4 IO 1 - 2 DOmain 1 - 3 CO 1 sec- 2 :The same principle will be used for the construction of influence lines.
Another type of this class of trusses is the subdivided Warren truss. This truss is
generated from a simple Warren truss using the same generation principle used for
a Baltimore truss.
Let us consider a detailed analysis of a subdivided Warren truss presented in
Fig.3.17. The moving load is applied to the lower chord. As usual, we start from
a kinematical analysis. This truss is geometrically unchangeable because the rigid
discs are connected by means of hinges and bars.
1
5013
94126 8 10Fig. 3.17 Kinematical analysis of the Warren truss
In order for the structure to be statically determinate the number of barsSand
jointsJmust satisfy the formulaSD2J 3. In our caseSD 49 ,J D 26 ,so
SD2J 3 D 52 3 D 49. Therefore, this structure is geometrically unchangeable
and statically determinate.
Once again the necessity of kinematical analysis must be emphasized, espe-
cially for trusses that are not simple and contain many elements. For example, what
happens if the entire structure in Fig.3.18is modified by removing the following
symmetrical set of members: case (1) 40 - 50 and 60 - 70 ; case (2) 40 - 50 , 40 -7, 60 - 70 ,and
60 - 70 ; or case (3) 40 -7 and 60 -7? Detailed kinematical analysis, as carried out above,
shows that this structure remains geometrically unchangeable for modifications (1)