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Releases and degrees of freedom

A release sets one internal force to zero at a point of the structure and in return allows a jump in the corresponding deformation. In the direct stiffness method this raises a question: every node has only one rotation, yet the members meeting at a moment hinge rotate by different amounts. This page shows how hinges are described as releases at member ends, how static condensation builds them into the element matrix and what this means for the degrees of freedom – with a fully worked Gerber beam.

1. Three kinds of release

A moment-resisting connection transmits all three internal forces $N$ , $Q$ and $M$ , and the deformations $u$ , $w$ and $\varphi$ are continuous across it. A release frees exactly one of these connections. The internal force is zero there, and the corresponding deformation may jump:

Moment release: $M = 0$ ; the deflected shape has a kink, the rotation jumps by $\Delta\varphi$ . Drawn in the calculator as an open ring. This is the hinge in the usual sense, as in a Gerber beam, a three-hinged frame or every truss node.
Shear release: $Q = 0$ ; the two parts may slide against each other transversely to the axis but stay parallel, the transverse displacement jumps by $\Delta w$ . In practice this is a parallel guide. Drawn in the calculator as two strokes along the member axis.
Axial release: $N = 0$ ; the parts may slide against each other along the axis, the axial displacement jumps by $\Delta u$ . In practice a sliding sleeve. Drawn in the calculator as two strokes across the member axis.

Releases can be combined. A member end that releases $M$ and $Q$ transmits only an axial force; a member with $M$ released at both ends is a truss member, which carries only axial force as long as there is no load on the member itself.

Every release removes one connection and adds the condition “internal force equals zero”. The degree of static indeterminacy drops by one as a result, with one exception explained in section 2. If it becomes negative, the structure is a mechanism.

2. Releases belong to member ends

In the stiffness method every node has three degrees of freedom $u_x$ , $u_y$ and $\varphi$ , shared by all members connected to it rigidly (see Direct stiffness method, section 2). A moment hinge between two members, on the other hand, means that their ends rotate differently at the same point. The two ideas fit together once the release is assigned not to the node but to a member end:

A rigidly connected member end has the rotation of the node.
A member end with a moment release has a rotation of its own that is not coupled to the node. The moment at that member end is zero.

Where $n$ members meet at a hinged node, $n - 1$ releases are therefore enough. In a Gerber beam a moment release at the end of one member suffices; the other remains rigidly connected to the node, and the node rotation is its end rotation.

If all $n$ member ends are released, no stiffness is left on the node rotation: the node could spin freely without any member noticing. This does not change the result, but it makes the system of equations singular. Every truss node is in this situation. Such a node cannot take a moment load.

The same reasoning applies at supports. A moment release at the member end directly at a fixed support cancels the fixity for that member; it is then only pinned (see Support types, section 3).

3. Two ways in the stiffness method

There are two equivalent ways to capture the separate rotation of a released member end:

An additional degree of freedom. Two nodes are placed on top of each other at the hinge, their displacements $u_x$ and $u_y$ are coupled and their rotations kept separate. Every moment hinge then adds one unknown to the system of equations.
Static condensation. The released degree of freedom is eliminated already in the element equation. The member enters the assembly with a modified stiffness matrix and a modified load vector; the number of global unknowns stays the same. The rotation at the released end is recovered from the element equation after solving.

Both ways give exactly the same results. Condensation has the advantage of working the same way for all three kinds of release and of leaving the node untouched. The calculator uses it.

4. Static condensation

The starting point is the element equation in local coordinates, $\mathbf f = \mathbf k\,\mathbf d - \mathbf Q$ . The six end quantities are split into the released ones ( $c$ ) and the retained ones ( $r$ ), and the equation is reordered accordingly:

\begin{bmatrix} \mathbf f_r \\ \mathbf f_c \end{bmatrix} = \begin{bmatrix} \mathbf k_{rr} & \mathbf k_{rc} \\ \mathbf k_{cr} & \mathbf k_{cc} \end{bmatrix} \begin{bmatrix} \mathbf d_r \\ \mathbf d_c \end{bmatrix} - \begin{bmatrix} \mathbf Q_r \\ \mathbf Q_c \end{bmatrix}

The release requires the corresponding end forces to vanish, $\mathbf f_c = \mathbf 0$ . The second row then gives the released displacements:

\mathbf d_c = \mathbf k_{cc}^{-1} \left( \mathbf Q_c - \mathbf k_{cr}\,\mathbf d_r \right)

Substituted into the first row, this leaves

\mathbf f_r = \mathbf k^{*}\,\mathbf d_r - \mathbf Q^{*} \qquad\text{with}\qquad \mathbf k^{*} = \mathbf k_{rr} - \mathbf k_{rc}\,\mathbf k_{cc}^{-1}\,\mathbf k_{cr}, \qquad \mathbf Q^{*} = \mathbf Q_r - \mathbf k_{rc}\,\mathbf k_{cc}^{-1}\,\mathbf Q_c

$\mathbf k^{*}$ and $\mathbf Q^{*}$ are written back as a $6 \times 6$ matrix and a vector with six entries, with zeros in the rows and columns of the releases. They then pass through transformation and assembly exactly like an ordinary element matrix. At the released position the member passes no force to the node, and the node displacement does not act back on the member.

The load vector must be condensed as well. Taking the fixed-end forces of the member fixed at both ends for a released member under a uniform load misses the redistribution of moment to the fixed end: there would be $qL^2/12$ instead of $qL^2/8$ .

5. Moment release at the member end

The most common case is a moment release at member end $b$ . Only $\varphi_b$ is released, so $\mathbf k_{cc} = 4EI/L$ is a single number and the condensation can be carried out by hand. For the first bending entry, for instance:

k^{*}_{22} = \frac{12EI}{L^3} - \frac{6EI}{L^2} \cdot \frac{L}{4EI} \cdot \frac{6EI}{L^2} = \frac{12EI}{L^3} - \frac{9EI}{L^3} = \frac{3EI}{L^3}

Altogether:

\def\arraystretch{2.1} \mathbf k^{*} = \begin{bmatrix} \dfrac{EA}{L} & 0 & 0 & -\dfrac{EA}{L} & 0 & 0 \\ 0 & \dfrac{3EI}{L^3} & \dfrac{3EI}{L^2} & 0 & -\dfrac{3EI}{L^3} & 0 \\ 0 & \dfrac{3EI}{L^2} & \dfrac{3EI}{L} & 0 & -\dfrac{3EI}{L^2} & 0 \\ -\dfrac{EA}{L} & 0 & 0 & \dfrac{EA}{L} & 0 & 0 \\ 0 & -\dfrac{3EI}{L^3} & -\dfrac{3EI}{L^2} & 0 & \dfrac{3EI}{L^3} & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}

This is the stiffness of the propped cantilever, fixed at one end and pinned at the other: $3EI/L$ instead of $4EI/L$ for the rotation of the start, with no carry-over to the released end. The axial part is unaffected by the release. For a uniform load $q_z$ over the whole member the condensation gives

\def\arraystretch{2.1} \mathbf Q^{*} = \begin{bmatrix} 0 & \dfrac{5\,q_z L}{8} & \dfrac{q_z L^2}{8} & 0 & \dfrac{3\,q_z L}{8} & 0 \end{bmatrix}^{\mathsf T}

which again are the familiar table values of this member: fixed-end moment $qL^2/8$ , end reactions $5qL/8$ and $3qL/8$ . After solving, recovery gives the rotation of the released end:

\varphi_b = -\frac{q_z L^3}{48\,EI} - \frac{3\,(w_a - w_b)}{2L} - \frac{\varphi_a}{2}

The kink at the hinge is the difference between $\varphi_b$ and the rotation of the node the member end is attached to.

6. Inadmissible combinations

Condensation requires $\mathbf k_{cc}$ to be invertible. That fails exactly when the releases allow the member a rigid-body motion in which nothing happens at any of the retained end quantities:

$N$ at both ends: $\mathbf k_{cc} = \tfrac{EA}{L}\left[\begin{smallmatrix} 1 & -1 \\ -1 & 1 \end{smallmatrix}\right]$ is singular. The member could slide any distance along its axis.
$Q$ at both ends: likewise, the member could move transversely as a whole.
$M$ at both ends plus $Q$ : the member could rotate about the end at which the shear force is still held.

Such members have no defined position. The calculator checks for this before the analysis and reports the member together with the releases involved. Admissible, by contrast, is for instance $M$ and $Q$ at the same end: that member end then transmits only the axial force.

7. Worked example: Gerber beam

Structure

A beam over $12$ m with a pinned support $A$ at $x = 0$ , roller supports $B$ at $x = 6$ m and $C$ at $x = 12$ m and a moment hinge $G$ at $x = 8$ m. Uniform load $q = 10$ kN/m downwards over the whole length, $EI = 20\,250$ kNm². Three members from left to right: A–B ( $L = 6$ m), B–G ( $L = 2$ m) and G–C ( $L = 4$ m). The hinge is modelled as a release of $M$ at the end of member B–G; G–C is rigidly connected at $G$ .

The structure is statically determinate: the suspended span G–C is a simply supported beam and passes $20$ kN to $G$ . The stiffness method does not need this insight; the calculation runs exactly as it would for an indeterminate structure.

Degrees of freedom

As in the two-span beam, the axial displacements are uncoupled from bending and remain zero. The unknowns are the rotations $\varphi_A$ , $\varphi_B$ , $\varphi_G$ , $\varphi_C$ and the deflection $u_{y,G}$ at the hinge, which has no support. The rotation of member end B–G at $G$ is not an unknown: it has been condensed out.

Member contributions

A–B, $L = 6$ m, no release: $4EI/L = \tfrac{2}{3}EI$ , $2EI/L = \tfrac{1}{3}EI$ ; load terms $q_z L^2/12 = -30$ kNm at $A$ and $+30$ kNm at $B$ .
B–G, $L = 2$ m, condensed as in section 5: $3EI/L = 1.5\,EI$ , $3EI/L^2 = 0.75\,EI$ , $3EI/L^3 = 0.375\,EI$ ; load terms $q_z L^2/8 = -5$ kNm at $B$ and $3 q_z L/8 = -7.5$ kN at $G$ .
G–C, $L = 4$ m, no release: $12EI/L^3 = 0.1875\,EI$ , $6EI/L^2 = 0.375\,EI$ , $4EI/L = EI$ , $2EI/L = 0.5\,EI$ ; load terms $q_z L/2 = -20$ kN and $q_z L^2/12 = -13.33$ kNm at $G$ , $+13.33$ kNm at $C$ .

System of equations

With the factor $EI/48$ in front of the matrix all entries become integers:

\frac{EI}{48} \begin{bmatrix} 32 & 16 & 0 & 0 & 0 \\ 16 & 104 & -36 & 0 & 0 \\ 0 & -36 & 27 & 18 & 18 \\ 0 & 0 & 18 & 48 & 24 \\ 0 & 0 & 18 & 24 & 48 \end{bmatrix} \begin{bmatrix} \varphi_A \\ \varphi_B \\ u_{y,G} \\ \varphi_G \\ \varphi_C \end{bmatrix} = \begin{bmatrix} -30 \\ 30 - 5 \\ -7.5 - 20 \\ -13.33 \\ 13.33 \end{bmatrix}

The entry $104 = 32 + 72$ at $\varphi_B$ is made up of $4EI/6$ from A–B and $3EI/2$ from B–G, the entry $27 = 18 + 9$ at $u_{y,G}$ of $3EI/8$ from B–G and $3EI/16$ from G–C. There is no coupling between $\varphi_G$ and B–G: no moment is transmitted across the hinge.

Solution

EI \begin{bmatrix} \varphi_A \\ \varphi_B \\ u_{y,G} \\ \varphi_G \\ \varphi_C \end{bmatrix} = \begin{bmatrix} -30 \\ -30 \\ -133.33 \\ 6.667 \\ 60 \end{bmatrix} \quad\Longrightarrow\quad \begin{aligned} \varphi_A = \varphi_B &= -1.481 \cdot 10^{-3}\ \text{rad} \\ u_{y,G} &= -6.584 \cdot 10^{-3}\ \text{m} \\ \varphi_G &= 3.292 \cdot 10^{-4}\ \text{rad} \\ \varphi_C &= 2.963 \cdot 10^{-3}\ \text{rad} \end{aligned}

Check by substitution, first row: $\bigl(32 \cdot (-30) + 16 \cdot (-30)\bigr)/48 = -30$ . The other rows work out the same way.

Kink at the hinge

The recovery formula from section 5 gives for the end of member B–G, with $w_a = 0$ , $w_b = u_{y,G}$ and $\varphi_a = \varphi_B$ :

EI\,\varphi_b = \frac{10 \cdot 2^3}{48} - \frac{3 \cdot 133.33}{2 \cdot 2} + \frac{30}{2} = 1.667 - 100 + 15 = -83.33

Member B–G therefore arrives at the hinge with $\varphi_b = -4.115 \cdot 10^{-3}$ rad, while member G–C starts there with $\varphi_G = +3.292 \cdot 10^{-4}$ rad. The kink is $\Delta\varphi = 90/EI = 4.444 \cdot 10^{-3}$ rad.

Internal forces

From $\mathbf f = \mathbf k\,\mathbf d - \mathbf Q$ for A–B:

\begin{aligned} m_b &= \tfrac{2EI}{L}\,\varphi_A + \tfrac{4EI}{L}\,\varphi_B - 30 = -10 - 20 - 30 = -60\ \text{kNm} \\ f_{z,a} &= \tfrac{6EI}{L^2}\,(\varphi_A + \varphi_B) + 30 = -10 + 30 = 20\ \text{kN} \\ f_{z,b} &= -\tfrac{6EI}{L^2}\,(\varphi_A + \varphi_B) + 30 = 10 + 30 = 40\ \text{kN} \end{aligned}

and with the condensed matrix for B–G:

\begin{aligned} m_a &= \tfrac{3EI}{L}\,\varphi_B - \tfrac{3EI}{L^2}\,u_{y,G} - (-5) = -45 + 100 + 5 = 60\ \text{kNm} \\ f_{z,a} &= \tfrac{3EI}{L^2}\,\varphi_B - \tfrac{3EI}{L^3}\,u_{y,G} - (-12.5) = -22.5 + 50 + 12.5 = 40\ \text{kN} \end{aligned}

Support moment $M_B = -60$ kNm, the same from both sides: $m_b = -60$ at the end of A–B, $M(0) = -m_a = -60$ at the start of B–G.
Moment at the hinge $M_G = 0$ , as required.
Reactions $A = 20$ kN, $B = 40 + 40 = 80$ kN and, from the suspended span, $C = 20$ kN; together $120$ kN $= q \cdot 12$ m.
Maximum span moments $+20$ kNm at $x = 2$ m in A–B and at mid-span of G–C.

Check with table formulas

The displacements can be checked independently. Span A–B as a simply supported beam with uniform load and the support moment $-60$ kNm at $B$ :

EI\,\varphi_B = \frac{qL^3}{24} - \frac{|M_B|\,L}{3} = 90 - 120 = -30

Cantilever B–G, $\ell = 2$ m, fixed with the rotation $\varphi_B$ , carrying the uniform load and the force $F = 20$ kN from the suspended span:

\begin{aligned} EI\,u_{y,G} &= EI\,\varphi_B\,\ell - \frac{q\ell^4}{8} - \frac{F\ell^3}{3} = -60 - 20 - 53.33 = -133.33 \\ EI\,\varphi_b &= EI\,\varphi_B - \frac{q\ell^3}{6} - \frac{F\ell^2}{2} = -30 - 13.33 - 40 = -83.33 \end{aligned}

8. Shear and axial releases

For the other two kinds of release only the condensed degree of freedom changes. Two short examples, both in the calculator's example list:

Shear release

Beam fixed at both ends, $L = 6$ m, uniform load $q = 10$ kN/m, shear force released at the start. Because $Q(0) = 0$ the whole load goes to the right support, $B = qL = 60$ kN, and $M(x) = M_0 - q x^2/2$ . The rotation is held at both ends, so the curvature must integrate to zero:

\int_0^L \frac{M(x)}{EI}\,\mathrm dx = 0 \quad\Longrightarrow\quad M_0\,L - \frac{q L^3}{6} = 0 \quad\Longrightarrow\quad M_0 = \frac{qL^2}{6} = 60\ \text{kNm}

and at the right end $M(L) = qL^2/6 - qL^2/2 = -qL^2/3 = -120$ kNm. The deflection jumps at the released end: the beam drops there without rotating, by $540/EI = 26.7$ mm; this follows from integrating the curvature $M/EI$ twice starting from $B$ .

Axial release

A horizontal beam made of two members A–C and C–B of $3$ m each, $A$ fixed, $B$ pinned, and a horizontal load of $12$ kN to the right at $C$ . Without the release the load is shared equally between both supports. If member A–C releases the axial force at $A$ , it carries no axial force any more: the whole load goes through C–B, $N_{CB} = -12$ kN (compression), $N_{AC} = 0$ .

9. What the calculator does

Releases are set per member end, either on the member under “Releases” or on the node under “Connections at this node”. $N$ , $Q$ and $M$ can be combined freely.
Before assembly, the element matrix and load vector of every member with releases are condensed as described in section 4, including trapezoidal, triangular and point loads on the member.
After solving, the released displacements are recovered. The end forces follow from the full, uncondensed element equation; the internal force at the released end comes out as zero, which is also a check. The deflected shape shows the kink or offset at the hinge.
Inadmissible combinations as in section 6 are reported before the analysis.
If a node rotation has no stiffness because every member end attached to it has a moment release, the calculator restrains it for the analysis and shows a note. If a moment load acts there, it stops with a message.

Try it in the calculator

The worked example is available as the Gerber beam in the example list. The bending moment diagram shows −60 kNm over $B$ and zero at the hinge, the deflected shape the kink at $G$ . The “Displacements” tab lists the deflection of $G$ as −6.584 mm. The truss, the axial release and the shear release are in the same list.

Open the Gerber beam in the calculator

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