Degree of freedom and static determinacy

Degree of freedom

The degree of freedom (DOF) of a system is the number of independent displacement components (translations and rotations) by which the position of the system can change. For a plane structure a free body has 3 DOF: two translations ($x$ and $y$) and one rotation about the axis perpendicular to the plane. A support removes the DOF that it restrains; the number of reactions of a support equals the number of DOF it removes.

Can a statically determinate or indeterminate structure be geometrically unstable?

Yes. Static determinacy only compares the number of reactions $r$ with the equations of equilibrium (3 for a plane structure). Geometric stability depends on how the supports are arranged. Even if the count is right, the structure is unstable if the reactions cannot resist every possible load.

Example 1: determinate but unstable. A beam on three rollers (all vertical reactions), $r = 3 =$ equations, so it looks determinate. But there is no horizontal restraint; $\Sigma F_x = 0$ cannot be satisfied for a horizontal load, so the beam can slide. Another case: a beam with three reactions that are parallel or meet at one point (concurrent) cannot resist a moment about that point.

 ○──────────────○──────────────○   all reactions vertical:
 ▲              ▲              ▲   horizontal load → beam moves

Example 2: indeterminate but unstable. A beam supported on four vertical rollers has $r = 4 > 3$, so it is indeterminate by one degree (vertical reactions only), yet it is horizontally unstable.

Hence $r \ge 3$ is a necessary condition for stability, not a sufficient one. The supports must also be arranged so that the reactions are neither all parallel nor all concurrent, and no part of the structure forms a mechanism.