I have been working on a program to solve frame problems, and I have used the following formula for determining whether a frame is statically determinate:

$$RD=3m + r-3j-s,$$


  • $m$ is the number of members,
  • $r$ is the number of support reactions,
  • $j$ the number of joints, and
  • $s$ the number of equations of condition.

The structure is:

  • unstable if $0<RD$,
  • determinate if $RD=0$, and
  • statically indeterminate if $RD<0$.

It seems to work, for the most part, but I came across a problem that is deemed to be statically indeterminate by this method, which nevertheless can be solved using only the sum of forces and moments. The frame is as follows:


For the frame above, $RD=3$, so the frame is supposedly statically indeterminate. However, it can be solved by first solving te support reactions, then solving for the normal and shear forces of each member (8 variables using 8 equations from nodes A, B, C, and D) and then using these to solve for the internal moments. Is the method I used to determine determinacy invalid?

  • $\begingroup$ Your equation seems to be for a 3d Truss. Could you tell us what you used for m,r,j and s and why? What I would use is m=4,j=4,r=4, and s=0. In which case RD=0. $\endgroup$
    – NMech
    Aug 27, 2020 at 12:21

1 Answer 1


Your equation is correct but your calculation is wrong .

We have $m = 4$

$ r = 4 $

$ j = 4 $

And $ S = 4 $

There are 4 rods , 4 reactions from the supports 4 joints and 4 rotations .

S can be understood as the number Of Relative rotations at each joint .

For example at joint B there are 2 relative rotations

Take an attention to the word relative . If two rotate the third doesn’t matter .

Joint A C D each have 1 relative rotation


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