Determinate and Indeterminate Spatial Structures and Trusses::Fundamentals::Knowledgebase::SAFAS

Using the “Joint Method” we can write three equilibrium equations for each joint of a pin-connected spatial structure, therefore, we have ‘3j’ total number of equations (j = number of joints). As the members are pin-connected and the spatial structure is loaded at its joints only, there are a total of ‘3m’ unknown internal axial forces (m = number of members). If the number of unknown support reactions is ‘r’, the total number of unknown forces to compute for the entire structure is ‘m+r’. Therefore, for a pin-connected spatial structure with loads acting at its joints:

If m+r = 3j or m = 3j-r, we can compute all the internal forces using the equilibrium equations. In this case, the structure is stable and is called a “determinate system”.
If m+r > 3j or m > 3j-r, we have more unknowns than the number of equations and, therefore, we cannot find the internal forces by satisfying the internal equilibrium only. However, the structure is stable and is called an “indeterminate system”.
If m+r < 3j or m < 3j-r, we have more equations than the number of unknowns. This means that the internal forces cannot be computed due to the instability of the structure and the structure is called an “unstable system”.

It is important to note that the stability versus instability condition mentioned here is only a “necessary” and not “sufficient” condition. This means that if m < 3j-r, the structure is unstable. However, we may have structures with m ≥ 3j-r that are still unstable. To understand this better, consider the case of a two-dimensional (plane) truss. For this structure, the above conditions are changed as the total number of equations is ‘2j’ (two equations for each joint):

Joint and Member Numbering of A Truss

Joint and Member Numbering of a Truss

m=33, j=18, r=3 ==> 2j-r = 2(18)-3 = 33 and m = 33
Therefore, this truss is a determinate system.
Now, we remove member 24 and place it at a different location as shown below:

In this case j, m, and r have not changed, therefore, the structure seems to be stable and determinate. However, since the truss is not triangulated it is unstable. From this example it is clear that the stability requirements given above are only ‘necessary’ but not ‘sufficient’ conditions.