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Static and kinematic determinacy

Before analysing a structure one asks: does it stand at all, and is equilibrium alone enough to find the forces? The answer is given by the degree of static indeterminacy. This page derives the counting criterion, shows how to count releases correctly and why the criterion is not always enough, and contrasts it with the kinematic indeterminacy that governs the stiffness method – with a fully worked T-frame.

1. Three cases

Statically determinate: all reactions and internal forces follow from equilibrium alone. The stiffnesses play no part in the forces, and support settlements or temperature produce no internal forces.
Statically indeterminate: there are more unknown forces than equilibrium conditions. The redundant quantities only follow from compatibility of the deformations; the forces then depend on the stiffness ratios, and imposed deformations produce internal forces.
Kinematically unstable (mechanism): the structure can move entirely or in part without any member deforming. For an arbitrary load there is then no state of equilibrium.

2. The counting criterion

One counts unknowns and equations. A straight member has three independent unknown internal forces: once $N$ , $Q$ and $M$ are known at one end, equilibrium of the member gives them at the other. Every support restraint brings one unknown reaction. Against these stand three equilibrium conditions at every node, and every release provides an additional condition “internal force equals zero”. With

$m$ : number of members,
$r$ : number of support restraints, i.e. of restrained degrees of freedom over all supports,
$k$ : number of nodes,
$c$ : number of conditions from releases,

the degree of static indeterminacy is

n = \underbrace{3m + r}_{\text{unknowns}} - \underbrace{(3k + c)}_{\text{equations}}

$n < 0$ : restraints are missing, the structure is a mechanism. $n = 0$ : as many equations as unknowns, the structure is statically determinate unless it is a mechanism. $n > 0$ : statically indeterminate to degree $n$ , with the same proviso, see section 6.

A spring support counts in $r$ like a rigid support: it brings one unknown force. The spring law $R = -k_F\,u$ is, like the condition $u = 0$ at a rigid support, a statement about deformations and not an equilibrium condition; it therefore does not enter the count.

3. Counting releases correctly

A release at a member end sets an internal force to zero there (see Releases and degrees of freedom). For axial and shear releases each release counts as one condition. For moment releases there is one exception to observe:

If $j$ member ends meet at a node, all with a moment release, and the rotation there is not held by a support, only $j - 1$ conditions are independent. The last one is already satisfied by moment equilibrium of the node: if $j - 1$ member end moments are zero and no moment acts on the node, the last one must be zero as well.
If the rotation at the node is held, every release counts. A moment release directly at a fixed support cancels its effect for that member.

It also follows that a moment release at the free end of a cantilever does not count ( $j = 1$ , so $j - 1 = 0$ ): the moment there is zero anyway.

4. Other forms of the criterion

Textbooks give further forms, all of which follow from the same balance.

Trusses

In a truss all members are hinged at both ends, and no node is held against rotation. At a node with $j_i$ members, $j_i - 1$ conditions therefore count. Since every member has two ends, $\sum j_i = 2m$ and $c = \sum (j_i - 1) = 2m - k$ . Substituted:

n = 3m + r - 3k - (2m - k) = m + r - 2k

This is the familiar truss criterion: member forces and reactions against two equilibrium conditions per node.

Rigid-body form

If rigidly connected members are combined into rigid bodies, one counts bodies instead of members. Each body must then be statically determinate in itself, i.e. contain no closed frame. With the number of bodies $s$ , of support reactions $a$ and of connecting forces between the bodies $z$ (a moment hinge between two bodies transmits two forces):

n = a + z - 3s

For the Gerber beam from section 5, two bodies joined by one hinge: $a = 4$ , $z = 2$ , $s = 2$ , so $n = 4 + 2 - 6 = 0$ . The rigid-body form is often shorter by hand; the member form suits a program, because it only counts members, nodes, supports and releases.

5. Examples

Structure	$m$	$r$	$k$	$c$	$n$
Simply supported beam (pinned and roller)	1	3	2	0	0
Cantilever	1	3	2	0	0
Two-span beam	2	4	3	0	1
Gerber beam with one hinge	3	4	4	1	0
Triangular truss	3	3	3	3	0
Frame, both columns fixed	3	6	4	0	3
Cantilever with a spring at its end	1	4	2	0	1
T-frame with cantilever arm	3	5	4	0	2

In the triangular truss each of the three nodes has two hinged member ends and so counts one condition: $c = 3$ . All structures in the table are in the calculator's example list, which shows the count at the top of the properties panel.

6. Necessary, but not sufficient

The counting criterion only tells whether there are enough restraints, not whether they are properly arranged. A structure with $n \ge 0$ can still be a mechanism; this is called a critical form. Three typical cases, all with $n = 0$ :

Two-span beam on three rollers: $m = 2$ , $r = 3$ , $k = 3$ . All reactions are parallel; nothing holds the beam in the $x$ direction.
Three hinges on a straight line: a three-hinged structure whose middle hinge lies on the line joining the two pinned supports. Both halves rotate about their support; the middle hinge moves perpendicular to that line for both, so the motion is compatible. By first-order theory the structure is a mechanism; in reality it only carries load after large deformations.
Pinned support and a support along the axis: a horizontal beam with a pinned support at $A$ and a support at $B$ that restrains only $u_x$ . The reaction at $B$ passes through $A$ and cannot prevent rotation about $A$ .

By hand, instability is checked with a pole diagram (instantaneous centres): if a consistent pole diagram can be found for the bodies of the structure, it is a mechanism. In the stiffness method the same shows up at no extra cost: the matrix $\mathbf K_{ff}$ is singular (see Direct stiffness method, section 7).

7. Kinematic indeterminacy

The degree of kinematic indeterminacy is the number of unknown, mutually independent node displacements and rotations. It is the number of unknowns in the displacement method, just as $n$ is the number of unknowns in the force method. The two quantities are unrelated:

Beam fixed at both ends: statically indeterminate to degree three, but kinematically determinate – both nodes are fully held, there is no unknown displacement. The internal forces are directly the fixed-end forces, such as $qL^2/12$ at both ends.
Cantilever: statically determinate, but kinematically indeterminate to degree three: $u_x$ , $u_y$ and $\varphi$ at the free end.
Two-span beam: statically indeterminate to degree one. The three rotations and the axial displacements of $B$ and $C$ are free, five unknowns in total.

How many unknowns there are also depends on the model. The calculator's stiffness method includes axial deformation and carries all free node degrees of freedom. Working by hand with axially rigid members, as usual in the slope-deflection method, the displacements fixed by inextensible members drop out: for the two-span beam the three rotations remain, for the frame with two fixed columns the two corner rotations and one common horizontal sway of the beam.

For the stiffness method the degree of static indeterminacy does not matter: determinate and indeterminate structures follow the same path. The counting criterion is still useful, as a quick check while building a model and as a hint whether imposed deformations can produce internal forces.

8. Worked example: T-frame with cantilever arm

Structure

Column F–K, $H = 4$ m, fixed at $F$ . Rigidly connected at its top $K$ : to the right the beam K–B, $\ell_2 = 2$ m, to the pinned support $B$ ; to the left a cantilever arm K–L, $\ell_1 = 1$ m, with a point load $P = 10$ kN downwards at its free end $L$ . All members with $EI = 20\,250$ kNm².

Static indeterminacy

n = 3m + r - 3k - c = 3 \cdot 3 + (3 + 2) - 3 \cdot 4 - 0 = 2

With the force method two redundants would have to be found, for instance the two forces at the pinned support $B$ .

Kinematic indeterminacy

In the calculator the degrees of freedom of $L$ (three), $K$ (three) and the rotation at the pinned support $B$ are free: seven unknowns. With axially rigid members $K$ cannot move – the column holds it vertically, the beam to the pinned support horizontally. The cantilever arm is a statically determinate appendage; it passes its load to $K$ as a force and a moment. A single unknown remains, the rotation $\varphi_K$ .

Solution with one unknown

The cantilever arm loads the node with the moment $P\,\ell_1 = 10$ kNm counter-clockwise. The rotational stiffnesses of the two other members at the node:

\begin{aligned} \text{beam, free to rotate at } B \text{:} &\quad \frac{3EI}{\ell_2} = 1.5\,EI \\ \text{column, fixed at } F \text{:} &\quad \frac{4EI}{H} = 1.0\,EI \end{aligned}

\varphi_K = \frac{10}{2.5\,EI} = \frac{4}{EI} = 1.975 \cdot 10^{-4}\ \text{rad}

The node moment is distributed in proportion to the stiffnesses: $6$ kNm into the beam and $4$ kNm into the column; half of it is carried over to the fixed support, $M_F = 2$ kNm. From this the reactions:

$B$ : vertically $6/2 = 3$ kN downwards, horizontally $(4 + 2)/4 = 1.5$ kN,
$F$ : vertically $10 + 3 = 13$ kN upwards, horizontally $1.5$ kN in the opposite direction, fixed-end moment $2$ kNm.

Two static redundants, but only one kinematic unknown: here the displacement method is clearly shorter by hand.

Comparison with the calculator

The calculator includes axial deformation with $EA = 2.7 \cdot 10^6$ kN. Under the column force node $K$ drops by about $0.02$ mm, and the results differ by 2 to 3 %: $B_y = -2.944$ kN, $F_y = 12.944$ kN, $H = 1.541$ kN and $M_F = 2.051$ kNm. With a very large $EA$ it gives exactly the values above.

9. What the calculator does

The top of the properties panel always shows the count $n = 3m + r - 3k - c$ with the numbers substituted and the result: kinematically unstable, statically determinate or statically indeterminate to degree $n$ .
Every restrained degree of freedom and every spring counts in $r$ . The rules of section 3 apply to $c$ . Axial and shear releases always count individually; in special cases with members in line the count therefore reports $n$ one too low.
The reliable check is the analysis itself. If $\mathbf K_{ff}$ is singular, the calculator reports the structure as a mechanism and names the node and degree of freedom at which this first shows up – even when the count gives $n \ge 0$ .
If every degree of freedom is restrained, as for the beam fixed at both ends, the calculator notes that there is nothing to solve and outputs the fixed-end forces.

Try it in the calculator

The T-frame is listed as “T-frame with cantilever arm” in the example list. The top of the properties panel shows “n = 3·3 + 5 − 3·4 − 0 = 2”, and the “Reactions” tab lists −2.944 kN at $B$ and 12.944 kN and 2.051 kNm at $F$ . Swapping a roller for a pinned support or adding a hinge updates the count immediately.

Open the T-frame in the calculator

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