Supports, Reactions and Determinacy

The rule underneath the table: prevent a translation and you get a force; prevent a rotation and you get a moment. Every support in the subject follows from it, so there is nothing to memorise.

A rigid body in a plane can slide horizontally, slide vertically, and rotate. Preventing each gives one equilibrium equation: ΣH = 0, ΣV = 0, ΣM = 0. That is the entire budget statics has.

Taking moments about a second point does not produce a fourth equation — it is a linear combination of the three you already had. This is worth internalising early, because trying to squeeze a fourth equation out of an indeterminate structure is a very common dead end.

1. 1Two rollers give r = 2. Nothing resists horizontal load, so ΣH = 0 cannot be satisfied and the beam simply moves.
2. 2Pin and roller give r = 3. Exactly solvable — the simply supported beam every textbook opens with.
3. 3Two pins give r = 4. One equation short: indeterminate to the first degree.
4. 4Fixed and pin give r = 5. Indeterminate to the second degree.

The roller in the determinate case is not a compromise. It deliberately allows the beam to expand with temperature; a beam pinned at both ends develops large thermal stresses instead, which is exactly why bridges have expansion joints.

Three reactions can still be a mechanism if they are arranged badly, and the arithmetic cannot see it.

• All parallel — the body slides perpendicular to them. Three vertical rollers pass the count and resist no horizontal load whatsoever.
• All concurrent — if the three lines of action meet at a point, the body can rotate about that point, since none of them produces a moment about it.
• Geometric instability within the structure — a truss can have the right number of members overall and still contain a floppy panel.

So the check is two-part: count the reactions, then confirm they are neither all parallel nor all concurrent. A structure that fails the second test is called geometrically unstable, and no amount of analysis will rescue it.

An indeterminate structure has infinitely many force distributions satisfying equilibrium. The real one is picked out by compatibility — a statement about how the structure deforms, such as "the deflection at this support is zero".

This has a sharp practical consequence. A determinate analysis needs no material properties at all — a steel truss and a timber truss of the same geometry carry identical member forces. An indeterminate one needs E and I, because the load distributes itself according to relative stiffness.

• Redundancy. Lose one member and the load finds another path. A determinate structure has no alternative path — every member is critical by definition.
• Stiffness. For the same material, an indeterminate structure deflects less.
• Smaller moments. Continuity over supports reduces the peak bending moment, so members can be lighter.

The price is that the structure now cares about things a determinate one ignores: support settlement, temperature change and fabrication error all induce stress, because there is nowhere for the structure to move to. This is why a continuous bridge needs its bearings designed carefully and a simply supported one does not.

Static indeterminacy counts unknown *forces* beyond equilibrium. Kinematic indeterminacy counts unknown *displacements* — the degrees of freedom of the joints.

The two are complementary, and they decide which method is less work. The force method solves for redundant forces, so it suits a structure with low static indeterminacy; the stiffness method solves for joint displacements, so it suits low kinematic indeterminacy. Since a computer does not mind how many unknowns there are, every structural analysis program uses the stiffness method.