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Support types

Supports connect the structure to its surroundings. In the stiffness method every support fixes individual degrees of freedom of a node – rigidly, with a prescribed value as in a support settlement, or elastically through a spring. This page relates the usual support types to the degrees of freedom, shows how the reactions arise and works through a support settlement and a spring support in full.

1. Supports as restraints

A node of a plane frame has three degrees of freedom: the displacements $u_x$ and $u_y$ and the rotation $\varphi$ (see Direct stiffness method, section 2). A support restrains one or more of them. Every restrained degree of freedom comes with exactly one support reaction, so the number of restrained degrees of freedom is also the number of reactions the support provides.

Support	restrained	free	restraints	reactions
Roller	$u_y$	$u_x$ , $\varphi$	1	$A_y$
Pinned	$u_x$ , $u_y$	$\varphi$	2	$A_x$ , $A_y$
Fixed	$u_x$ , $u_y$ , $\varphi$	–	3	$A_x$ , $A_y$ , $M_A$
Guided (sliding fixed)	$u_y$ , $\varphi$	$u_x$	2	$A_y$ , $M_A$

The table shows the usual orientation with a horizontal sliding direction. There is equally a roller that restrains only $u_x$ , or a guided support that slides in the $y$ direction. A spring support restrains no degree of freedom but holds it elastically; it is covered in section 5.

A support determines which movements are prevented, not how the member is connected. Whether a member end can rotate freely at a pinned support depends on the pinned support not restraining $\varphi$ – no hinge in the member is needed for that, see section 3.

2. Support reactions

The degrees of freedom are split into free ( $f$ ) and restrained ( $s$ ) ones and the system $\mathbf K\,\mathbf u = \mathbf F$ is reordered accordingly. At the restrained degrees of freedom the unknown reaction $\mathbf R$ is added to the load $\mathbf F_s$ :

\begin{bmatrix} \mathbf K_{ff} & \mathbf K_{fs} \\ \mathbf K_{sf} & \mathbf K_{ss} \end{bmatrix} \begin{bmatrix} \mathbf u_f \\ \bar{\mathbf u}_s \end{bmatrix} = \begin{bmatrix} \mathbf F_f \\ \mathbf F_s + \mathbf R \end{bmatrix}

The upper row gives the free displacements, the lower row then the reactions:

\mathbf K_{ff}\,\mathbf u_f = \mathbf F_f - \mathbf K_{fs}\,\bar{\mathbf u}_s, \qquad \mathbf R = \mathbf K_{sf}\,\mathbf u_f + \mathbf K_{ss}\,\bar{\mathbf u}_s - \mathbf F_s

$\mathbf R$ are the forces the support exerts on the structure, positive in the direction of the global axes, with the moment positive counter-clockwise. A support pushing upwards therefore has $A_y > 0$ . The term $\mathbf F_s$ contains loads acting directly on the restrained degree of freedom and the work-equivalent shares of member loads at that node: a point load directly above a roller goes straight into its reaction.

As a check, loads and reactions together must be in equilibrium: $\sum F_x = 0$ , $\sum F_y = 0$ and $\sum M = 0$ about any point.

3. Supports and member connections

At a pinned support the rotation $\varphi$ is a free degree of freedom and therefore an unknown of the system of equations. If only one member is attached there, it rotates with the node, and the moment at the member end is zero because nothing else at the node can take a moment. A moment release at the member end changes nothing about this: it only leaves the node rotation without stiffness (see Releases and degrees of freedom, section 2).

It is different in two cases:

Several members at a pinned support. Rigidly connected, they share the node rotation and can exchange moments with each other. Moment releases at the member ends are needed only if the members really are hinged – with $n$ members, $n - 1$ are enough.
Fixed support. Here $\varphi$ is restrained, and a rigidly connected member receives the fixed-end moment. A moment release at the member end cancels the fixity for that member; it is then only pinned.

4. Support settlement

A support settlement is a prescribed displacement $\bar u$ at a restrained degree of freedom. In the equation above it sits in $\bar{\mathbf u}_s$ and acts through $-\mathbf K_{fs}\,\bar{\mathbf u}_s$ like a load on the free degrees of freedom. A prescribed support rotation is treated in exactly the same way. A statically determinate structure follows a settlement without restraint, so no internal forces arise. Only redundant restraints resist the movement; the resulting restraint forces grow with the stiffness $EI$ .

Structure

Two-span beam as in the stiffness method example: two spans of $L = 6$ m, $EI = 20\,250$ kNm², pinned support $A$ , rollers $B$ and $C$ . No load; the middle support $B$ settles by $s = 10$ mm, so $\bar u_{y,B} = -0.01$ m.

System of equations

The unknowns are again the rotations $\varphi_A$ , $\varphi_B$ and $\varphi_C$ ; the matrix $\mathbf K_{ff}$ is the same as there. What is new is the right-hand side. The column of $\mathbf K_{fs}$ belonging to $u_{y,B}$ contains the entries $\mp 6EI/L^2$ from the element matrix: $-6EI/L^2$ at $\varphi_A$ (member A–B, end $b$ ), $-6EI/L^2 + 6EI/L^2 = 0$ at $\varphi_B$ and $+6EI/L^2$ at $\varphi_C$ (member B–C, start $a$ ). With $6EI/L^2 = 3\,375$ kN:

\frac{EI}{L} \begin{bmatrix} 4 & 2 & 0 \\ 2 & 8 & 2 \\ 0 & 2 & 4 \end{bmatrix} \begin{bmatrix} \varphi_A \\ \varphi_B \\ \varphi_C \end{bmatrix} = -\mathbf K_{fs}\,\bar u_{y,B} = \begin{bmatrix} 3\,375 \cdot (-0.01) \\ 0 \\ -3\,375 \cdot (-0.01) \end{bmatrix} = \begin{bmatrix} -33.75 \\ 0 \\ 33.75 \end{bmatrix} \text{kNm}

Solution

Symmetry gives $\varphi_B = 0$ and $\varphi_C = -\varphi_A$ . The first row with $4EI/L = 13\,500$ kNm gives

\varphi_A = -\frac{33.75}{13\,500} = -2.5 \cdot 10^{-3}\ \text{rad} = \frac{3\,\bar u_{y,B}}{2L}

Reactions and support moment

From $\mathbf R = \mathbf K_{sf}\,\mathbf u_f + \mathbf K_{ss}\,\bar{\mathbf u}_s$ , with $12EI/L^3 = 1\,125$ kN/m and, since both members add up at $B$ , $24EI/L^3 = 2\,250$ kN/m:

\begin{aligned} A_y &= \tfrac{6EI}{L^2}\,\varphi_A - \tfrac{12EI}{L^3}\,\bar u_{y,B} = -8.4375 + 11.25 = 2.8125\ \text{kN} \\ B_y &= -\tfrac{6EI}{L^2}\,\varphi_A + \tfrac{6EI}{L^2}\,\varphi_C + \tfrac{24EI}{L^3}\,\bar u_{y,B} = 8.4375 + 8.4375 - 22.5 = -5.625\ \text{kN} \end{aligned}

The middle support therefore pulls the beam down, the end supports push it up; by symmetry $C_y = A_y$ , and the sum is zero, as it must be without load. The support moment is the end moment of member A–B:

M_B = m_b = \tfrac{2EI}{L}\,\varphi_A - \tfrac{6EI}{L^2}\,\bar u_{y,B} = -16.875 + 33.75 = 16.875\ \text{kNm} = \frac{3EI\,s}{L^2}

Positive, meaning tension at the bottom: the beam sags over $B$ . The bending moment is linear in both spans, from zero at the end supports to $16.875$ kNm over $B$ . As a check: $A_y \cdot L = 2.8125 \cdot 6 = 16.875$ kNm.

5. Spring supports

A spring support holds a degree of freedom elastically: the reaction is proportional to the displacement, $R = -k_F\,u$ , with the spring stiffness $k_F$ in kN/m for displacements and in kNm/rad for the rotation (rotational spring). This models a flexible subsoil, a supporting neighbouring component or partial fixity.

In the stiffness method the degree of freedom stays free. The spring is an additional element attached to this single degree of freedom; its stiffness is added to the diagonal entry:

K_{ii} \mathrel{+}= k_F

The limiting cases are the free end ( $k_F \to 0$ ) and the rigid support ( $k_F \to \infty$ ). When counting supports, a spring support counts like a rigid one: it brings one unknown force and, with the spring law, one more equation.

Example: cantilever with a spring support

Cantilever $L = 6$ m, fixed at $A$ , resting at its free end $B$ on a vertical spring $k_F = 1\,000$ kN/m. Uniform load $q = 10$ kN/m downwards, $EI = 20\,250$ kNm².

The unknowns are the deflection $u_{y,B}$ and the rotation $\varphi_B$ . With $12EI/L^3 = 1\,125$ kN/m, $6EI/L^2 = 3\,375$ kN and $4EI/L = 13\,500$ kNm and the work-equivalent loads at member end $b$ , $q_z L/2 = -30$ kN and $-q_z L^2/12 = +30$ kNm:

\begin{bmatrix} 1\,125 + 1\,000 & -3\,375 \\ -3\,375 & 13\,500 \end{bmatrix} \begin{bmatrix} u_{y,B} \\ \varphi_B \end{bmatrix} = \begin{bmatrix} -30 \\ 30 \end{bmatrix}

The spring appears only on the diagonal. With the determinant $2\,125 \cdot 13\,500 - 3\,375^2 = 17\,296\,875$ :

u_{y,B} = \frac{-30 \cdot 13\,500 + 3\,375 \cdot 30}{17\,296\,875} = -0.017\,561\ \text{m}, \qquad \varphi_B = \frac{2\,125 \cdot 30 - 3\,375 \cdot 30}{17\,296\,875} = -2.168 \cdot 10^{-3}\ \text{rad}

The spring pushes upwards with $B_y = -k_F\,u_{y,B} = 17.561$ kN. Equilibrium of the whole beam gives $A_y = 60 - 17.561 = 42.439$ kN and the fixed-end moment $M_A = 10 \cdot 6 \cdot 3 - 17.561 \cdot 6 = 74.634$ kNm, counter-clockwise.

Check with the flexibility

The same spring force follows from compatibility at the free end: the deflection of the cantilever under $q$ , reduced by that due to $R$ , must equal the spring compression $R/k_F$ .

R = \frac{q L^4 / (8EI)}{1/k_F + L^3/(3EI)} = \frac{0.08}{0.001 + 0.003\,556} = 17.561\ \text{kN}

For $k_F \to \infty$ the formula gives $R = 3qL/8 = 22.5$ kN, the propped cantilever; for $k_F \to 0$ it gives $R = 0$ , and the fixed-end moment becomes $qL^2/2 = 180$ kNm.

Rotational spring and degree of fixity

A rotational spring $c$ (in kNm/rad) at one end of a simply supported beam under uniform load lies between a pinned support and full fixity. Compatibility at the spring end, end rotation of the simply supported beam equal to spring rotation, gives the end moment

|M_A| = \frac{qL^2}{8} \cdot \frac{1}{1 + \dfrac{3EI}{c\,L}}

For $c \to \infty$ this is $qL^2/8$ , the fixed-end moment of the propped cantilever, for $c \to 0$ it is zero. The fraction is often called the degree of fixity.

6. Inclined supports

A roller on an inclined surface restrains the displacement only perpendicular to that surface. With the unit vector $\mathbf n$ in the restrained direction, the condition reads

\mathbf n^{\mathsf T} \begin{bmatrix} u_x \\ u_y \end{bmatrix} = n_x\,u_x + n_y\,u_y = 0

It involves two global degrees of freedom at once. In the stiffness method the degrees of freedom of this node are rotated into a nodal coordinate system with one axis along $\mathbf n$ , and one degree of freedom is restrained there. The calculator does not offer inclined supports directly. They can be modelled with a short, very axially stiff pendulum member in the direction $\mathbf n$ , with moment releases at both ends and a pinned support at its far end: it holds the node exactly along its axis.

7. Enough supports, properly arranged

A plane structure can move as a rigid body in three ways: in $x$ , in $y$ and by rotation. The supports must prevent all three. At least three restraints are needed for that, but not every arrangement of three restraints is enough:

All reactions parallel. A beam on three rollers cannot take anything in the $x$ direction and can slide horizontally.
All lines of action through one point. A horizontal beam with a pinned support at $A$ and a support at $B$ that restrains only $u_x$ can rotate about $A$ : the reaction at $B$ acts along the beam axis through $A$ and has no lever arm.

In both cases $\mathbf K_{ff}$ is singular. The calculator detects this during the solution from a vanishing pivot and reports the node and the degree of freedom at which the motion first shows up. More on such critical forms on the page Static and kinematic determinacy.

8. What the calculator does

Every node has the quick choices free, roller, pinned, fixed and spring. Below them each of the three degrees of freedom can be restrained individually, which also gives a roller that restrains only $u_x$ , or a guided support.
Restrained degrees of freedom are removed from the system of equations, and the reactions are recovered as in section 2.
Prescribed displacements and rotations, in m and rad, act only in restrained degrees of freedom and enter the right-hand side through $-\mathbf K_{fs}\,\bar{\mathbf u}_s$ .
Springs are entered per degree of freedom and added to the diagonal; their force $-k_F\,u$ appears in the table of reactions. In a rigidly restrained degree of freedom a spring has no effect.
When a node becomes a support, the calculator offers once to connect the members arriving there with hinges. According to section 3 this only matters at a fixed support or when several members meet.
After every analysis the calculator checks equilibrium of loads, reactions and spring forces.

Try it in the calculator

Both worked examples are in the example list. For the support settlement the “Reactions” tab shows 2.813, −5.625 and 2.813 kN, and the bending moment diagram 16.875 kNm over $B$ . For the cantilever with a spring support the spring force of 17.561 kN is listed in the row of $B$ and the fixed-end moment of 74.634 kNm at $A$ .

Open the support settlement in the calculator Open the spring support in the calculator

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