Direct stiffness method

The direct stiffness method is the displacement method written in matrix form. The unknowns are the displacements and rotations of the nodes; equilibrium at every node yields a system of linear equations. Statically determinate and indeterminate structures follow exactly the same path, which is why the method underlies practically every frame analysis program – including this calculator. This page derives it step by step for plane frames and works through a complete example.

1. Basic idea

A frame structure consists of straight members connected at nodes. Once the displacements and rotations of all nodes are known, so is the position of both ends of every member. For a single member it is possible to state exactly which forces are needed at its ends to move it into that position. That is precisely what its element stiffness matrix describes.

The method therefore proceeds in three steps:

1. For every member, set up the relation between the displacements and the forces at its ends.
2. At every node, require the forces of all connected member ends to be in equilibrium with the external load. This gives $$\mathbf K \, \mathbf u = \mathbf F$$ with the global stiffness matrix $\mathbf K$, the vector $\mathbf u$ of all nodal displacements and the load vector $\mathbf F$.
3. Introduce the supports, solve the system of equations, and recover the member end forces, the support reactions and the internal force diagrams from the nodal displacements.

Compatibility is satisfied automatically: all members rigidly connected at a node share its displacements and rotation. Unlike the force method, there is no need to know the degree of static indeterminacy or to choose a statically determinate primary structure.

2. Degrees of freedom and sign convention

Every node of a plane frame has three degrees of freedom: the displacements $u_x$ and $u_y$ and the rotation $\varphi$. With $n$ nodes this gives $3n$ unknowns, collected in the vector

Node $i$ occupies positions $3i-2$, $3i-1$ and $3i$. On this page and in the calculator, the following sign convention applies:

• Global system: $x$ to the right, $y$ upwards. A downward load therefore has $F_y < 0$.
• Rotations and moments are positive counter-clockwise.
• Every member has a direction from its start node $a$ to its end node $b$. Its local $x$ axis points in that direction; the local $z$ axis follows by rotating it 90° counter-clockwise. For a horizontal member running from left to right, $z$ therefore points upwards.
• The member inclination $\alpha$ is the angle from the global to the local $x$ axis, with $c = \cos\alpha = \Delta x / L$ and $s = \sin\alpha = \Delta y / L$.

Many textbooks let the local $z$ axis point downwards. That changes the signs of individual matrix entries, not the method. The internal forces output by the calculator follow the usual structural engineering convention in any case, see section 8.

3. Element stiffness matrix in local coordinates

In local coordinates, each member end has three displacement quantities: the axial displacement $u$, the transverse displacement $w$ and the rotation $\varphi$. The member end forces act in the same directions:

$\mathbf f$ are the forces exerted by the nodes on the member. For a straight, unloaded member with constant axial stiffness $EA$ and bending stiffness $EI$, Euler–Bernoulli beam theory (first-order theory, no shear deformation) gives exactly $\mathbf f = \mathbf k\,\mathbf d$ with

Each column is a unit displacement state: column $j$ contains the member end forces needed to impose $d_j = 1$ while all other member end quantities remain zero.

Where the entries come from

• Axial direction: a member lengthened by $\Delta u$ carries the axial force $N = EA\,\Delta u / L$. This gives the entries $\pm EA/L$. In local coordinates the axial and the bending parts are independent of each other.
• Rotation of one end: if the start of the member is rotated by $\varphi_a = 1$ while the other end stays fixed, the moment $4EI/L$ develops at the start and half of it, $2EI/L$, at the fixed end (carry-over factor ½). The shear forces $\pm 6EI/L^2$ keep these moments in equilibrium.
• Transverse displacement of one end: $w_a = 1$ with both end rotations held produces the end moments $6EI/L^2$ and the shear forces $\pm 12EI/L^3$.

Formally, the bending entries follow from the differential equation $EI\,w'''' = q_z$. For an unloaded member the deflection curve is a cubic polynomial, which can be written in terms of the four end values using Hermite polynomials:

The stiffness entries are then $k_{ij} = \int_0^L EI\,H_i''\,H_j''\,\mathrm dx$; for instance, $\int_0^L EI\,(H_1'')^2\,\mathrm dx$ gives exactly $12EI/L^3$.

4. Transformation into the global system

The nodal displacements are expressed in the global system, while the element stiffness matrix applies in local coordinates. The conversion is a rotation by $\alpha$, separately for both member ends; the rotation $\varphi$ is the same in both systems:

$\mathbf u_e$ contains the six global displacement quantities of the two member nodes. $\mathbf T$ is orthogonal, $\mathbf T^{-1} = \mathbf T^{\mathsf T}$, so forces are rotated back with $\mathbf f_{\text{global}} = \mathbf T^{\mathsf T}\,\mathbf f$. Substituting both into $\mathbf f = \mathbf k\,\mathbf d$ gives the element stiffness matrix in the global system:

For a horizontal member running from left to right, $\mathbf T = \mathbf I$. For a vertical member running upwards ($c = 0$, $s = 1$), the axial stiffness $EA/L$ becomes the stiffness in the global $y$ direction and the bending stiffness $12EI/L^3$ the one in the $x$ direction – exactly what one expects of a column.

5. Loads on the member

Loads acting on a member between the nodes enter the system of equations as work-equivalent nodal loads. They are chosen so that they do the same work on the member end displacements as the actual load:

This applies to the transverse load; in the axial direction the same is done with the linear shape functions $1-\xi$ and $\xi$. For a uniformly distributed load $q_z$ over the whole member and for a point load $F_z$ at a distance $a$ from the start, with $b = L - a$:

With the opposite sign, these are the end forces of the member fixed at both ends – the fixed-end forces known from tables, such as $qL^2/12$ or $F a b^2/L^2$ and $F a^2 b/L^2$. For a loaded member the element equation therefore reads

$\mathbf Q$ is also rotated into the global system with $\mathbf T^{\mathsf T}$. Trapezoidal, triangular and partial distributed loads are handled with the same integral.

6. Assembly of the global stiffness matrix

Equilibrium at a node requires the external nodal load $\mathbf P$ to equal the sum of the forces the node exerts on the connected member ends. For all nodes together:

In practice this means: every member connects two nodes, and its location vector lists the six global numbers of its degrees of freedom. For a member from node $i$ to node $j$:

Each entry of the global element matrix is added to $\mathbf K$ at its position, and each entry of $\mathbf Q_e$ to $\mathbf F$:

Nodal loads go straight to their position in $\mathbf F$. $\mathbf K$ is symmetric and sparse: an entry $K_{pq}$ is non-zero only if the degrees of freedom $p$ and $q$ belong to the same node or are connected by a member. Where several members meet at a node, their stiffnesses add up on the main diagonal.

7. Supports and boundary conditions

Without supports, $\mathbf K$ is singular – the whole structure could move as a rigid body. Supports fix individual degrees of freedom. The unknowns are split into free ($f$) and restrained ($s$) degrees of freedom, and the system of equations is partitioned accordingly:

The restrained displacements $\bar{\mathbf u}_s$ are known: usually zero, or the settlement value for a prescribed support settlement. What is unknown there instead are the support reactions $\mathbf R$. The first row contains only the free displacements as unknowns:

The second row then gives the support reactions:

A spring support does not restrain a degree of freedom. Its spring stiffness $k_F$ is added to the main diagonal entry of that degree of freedom, and the support reaction is then $-k_F\,u$. More on this, including support settlements, on the page Support types.

8. Solution and recovery

The reduced system of equations is solved by Gaussian elimination. This gives all nodal displacements, and for every member the following are obtained:

1. the local member end displacements $\mathbf d = \mathbf T\,\mathbf u_e$,
2. the member end forces $\mathbf f = \mathbf k\,\mathbf d - \mathbf Q$,
3. the internal forces at any position $x$ from equilibrium of a cut-off part of the member.

The internal forces follow the structural engineering convention: $N$ positive in tension, $M$ positive when the bottom side (the side facing away from the local $z$ axis) is in tension, and $Q$ such that $\mathrm dM/\mathrm dx = Q$. At the start of the member this gives

and along the member, with the axial load $p_x$ and the transverse load $q_z$,

Point loads on the member cause jumps in $N$ and $Q$, point moments jumps in $M$. At the end of the member, the result must again be $N(L) = f_{x,b}$, $Q(L) = -f_{z,b}$ and $M(L) = m_b$ – a useful check for any hand calculation.

9. Worked example: two-span beam

10. What the calculator does

The frame calculator follows exactly this path:

• the same order of degrees of freedom, the same element matrix and the same transformation as above,
• assembly via the location vectors, with nodal loads and work-equivalent member loads in the load vector,
• trapezoidal, triangular, parabolic and partial distributed loads integrated exactly,
• supports by removing the restrained degrees of freedom, springs on the main diagonal, support settlements on the right-hand side,
• solution by Gaussian elimination with partial pivoting,
• internal forces and deflection curve computed exactly from the member end forces and the load, with extreme values at the zeros of the shear force.

Releases at member ends are taken into account by static condensation of the element matrix before assembly. After every analysis the calculator checks global equilibrium of loads and support reactions; for this check the distributed loads are integrated independently of the solution.

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