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Structural analysis: the stiffness method

When a structure has more supports or members than strictly necessary (a continuous beam, a fixed portal frame), equilibrium is not enough to find its forces: you have to add how it deforms. The stiffness method does this systematically for any frame, beam or grid, in 2D or 3D, and it is what every structural analysis program runs inside. Here you will see its idea, its steps and how to read what the Structural analysis tool gives.

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Degree of indeterminacy

In a plane frame each member has three unknown internal forces and each joint gives three equilibrium equations. The degree of indeterminacy (GI) compares unknowns and equations:

G I = 3 m + r - 3 j - c (en 3D: 6 m + r - 6 j - c)

$m$ is the number of members, $r$ the reactions, $j$ the joints and $c$ the rotations released by internal hinges. With $G I = 0$ the structure is statically determinate; with $G I > 0$ , indeterminate to that degree; with $G I < 0$ , a mechanism. For example, a two-span continuous beam has $G I = 1$ and a portal fixed at both bases, $G I = 3$ .

The degree of indeterminacy tells how many compatibility equations you would have to write by hand with the force method. The stiffness method does not need them: it solves a determinate structure the same way as one of degree 20.

The idea: unknowns in the displacements

Instead of looking for forces, the stiffness method looks for the displacements of the joints: in 2D, two translations and one rotation per joint. If the end displacements of a member are known, its forces follow directly from its stiffness, and the equilibrium of each joint gives as many equations as unknown displacements.

The six degrees of freedom of a plane frame member, in its local axes.

The relation between the end forces and those six displacements is the member's stiffness matrix. For bending (degrees $v$ and $θ$ ) and axial force ( $u$ ):

k = \frac{E A}{L} 00 - \frac{E A}{L} 00 0 \frac{12 E I}{L ^{3}} \frac{6 E I}{L ^{2}} 0 - \frac{12 E I}{L ^{3}} \frac{6 E I}{L ^{2}} 0 \frac{6 E I}{L ^{2}} \frac{4 E I}{L} 0 - \frac{6 E I}{L ^{2}} \frac{2 E I}{L} - \frac{E A}{L} 00 \frac{E A}{L} 00 0 - \frac{12 E I}{L ^{3}} - \frac{6 E I}{L ^{2}} 0 \frac{12 E I}{L ^{3}} - \frac{6 E I}{L ^{2}} 0 \frac{6 E I}{L ^{2}} \frac{2 E I}{L} 0 - \frac{6 E I}{L ^{2}} \frac{4 E I}{L}

Each column has a physical meaning: it is the forces you must apply at the ends to produce a unit displacement in that degree of freedom with all the others held. Rotating one end by one unit, with the other one fixed, takes a moment $4 E I / L$ at that end and produces $2 E I / L$ at the other (the carry-over factor is 1/2).

Two columns of the matrix: the deformed member and the forces that hold it that way.

The steps of the method

Matrix of each member in its local axes, like the one above.
Transformation to global axes: with the direction cosines of the member, $k_{g} = T^{T} k T$ . A vertical column and a horizontal beam have the same local $k$ but different $k_{g}$ .
Loads on the members: they are replaced by the fixed-end forces (the reactions the member would have with both ends fixed), applied at the joints with the opposite sign.
Assembly: the $k_{g}$ are added into the global matrix $K$ according to the joints each member connects, and the loads into the vector $P$ .
Support conditions: the restrained degrees of freedom are removed and $K_{ff} d_{f} = P_{f}$ is solved.
Member forces: $f = k d + f_{emp}$ with the displacements of its ends; the reactions come from the restrained rows. Between the ends, the diagrams are completed with the equilibrium of a segment.

Load on a member fixed at both ends	Fixed-end moments
uniform $w$ over the whole span	$\frac{w L ^{2}}{12}$ at A and at B
point $P$ at midspan	$\frac{P L}{8}$ at A and at B
point $P$ at distances $a$ and $b$ from the ends	$\frac{P a b ^{2}}{L ^{2}}$ at A, $\frac{P a ^{2} b}{L ^{2}}$ at B
triangular from 0 (at A) to $w$ (at B)	$\frac{w L ^{2}}{30}$ at A, $\frac{w L ^{2}}{20}$ at B

The simulator computes the fixed-end forces by integrating the load with the shape functions, so it accepts any load (partial, trapezoidal, parabolic) and not just those in the table.

Details that change the result

Axial deformation. Hand methods (slope-deflection, moment distribution) usually assume members that do not shorten. The stiffness method includes $E A / L$ , so its results differ slightly (example 2). With real sections the difference is a few percent or less.
Internal hinges. A hinged end transmits no moment: the rotation of that end is removed from the member (static condensation) and the member stays connected to the joint in translation.
In 3D each joint has six degrees of freedom, and torsion ( $G J / L$ ) and bending in two planes appear. An open section (I, C) has a very small $J$ : it barely resists torsion (example 3).

Simulator convention in 2D:

N

positive in tension;

M

positive when it puts the bottom fiber (local

- y

side) in tension, drawn on the tension side;

V = d M / d x

. The end-force tables are in local axes, acting on the member.

Common mistakes

Forgetting the fixed-end forces when going from $k d$ to the actual member forces.
Applying the fixed-end forces to the joints with the same sign instead of the opposite one.
Mixing units in $E I$ (kN and m versus N and mm): the displacements come out a thousand or a million times wrong, even though the moments of a structure with a single section may come out right.
Reading local-axis forces as global ones in the columns: in a vertical member, the local $V$ is horizontal.
Expecting realistic displacements from an open section under torsion.

Worked examples

Example 1Two-span continuous beam

A beam with two equal 6 m spans (IPE 300,

E I = 16710

kN·m²), pinned at A and on rollers at B and C, carries 10 kN/m along its whole length. Find the rotations, the moment over the middle support and the reactions.

Solution

$G I = 3 \cdot 2 + 4 - 3 \cdot 3 = 1$ : indeterminate. Unknowns of the stiffness method: the rotations $θ_{A}$ , $θ_{B}$ and $θ_{C}$ (the vertical translations are restrained). By symmetry, $θ_{B} = 0$ .
Fixed-end moments of each span: $\frac{w L ^{2}}{12} = \frac{10 \cdot 36}{12} = 30$ kN·m.
Equilibrium of joint A (zero moment at the pin): $\frac{4 E I}{L} θ_{A} + 30 = 0 \Rightarrow θ_{A} = - \frac{30 \cdot 6}{4 \cdot 16 710} = - 0.002693$ rad, and $θ_{C} = + 0.002693$ rad.
Moment at B: $M_{B A} = \frac{2 E I}{L} θ_{A} - 30 = - 15 - 30 = - 45$ kN·m $= - \frac{w L ^{2}}{8}$ (top fiber in tension).
Reactions of span AB: $A_{y} = \frac{60}{2} - \frac{45}{6} = 22.5$ kN and B's share, $37.5$ kN; by symmetry $B_{y} = 2 \cdot 37.5 = 75$ kN and $C_{y} = 22.5$ kN. Sum: $120$ kN ✓.

θ_{A} = - 0.002693

rad,

M_{B} = - 45

kN·m,

A_{y} = C_{y} = 22.5

kN and

B_{y} = 75

kN; the maximum positive moment is 25.31 kN·m 2.25 m from each end.

Solve it in Simuveliq

The button opens the finished model; to build it: in Engineering → Structural analysis, joints at x = 0, 6 and 12, two members, a pin at A, rollers (vertical reaction) at B and C and a 10 kN/m distributed load on each member.
The Determinacy card says «Indeterminate, degree 1». The «Step by step · stiffness method» shows each member's matrix, the system $K_{ff} d_{f} = P_{f}$ with $P_{f} = (- 30, 0, 0, 0, 30)$ and the rotations ±0.002693.
With the M diagram you will see −45 over B; the reactions table gives 22.5, 75 and 22.5 kN.

Open in the simulator →

Bending moment of example 1 (positive downward, on the tension side).

Example 2Fixed portal frame with a symmetric load

A single-section portal frame (IPE 300) has 4 m columns fixed at the base and a 6 m beam with 20 kN/m. Find the moments at the corners, at the bases and at midspan, and the horizontal reaction.

Solution

By hand, without axial deformation (slope-deflection): by symmetry there is no sway and $θ_{C} = - θ_{B} = - θ$ . Fixed-end moment of the beam: $\frac{20 \cdot 6 ^{2}}{12} = 60$ kN·m.
At B: column $M_{B A} = \frac{4 E I}{4} θ = E I θ$ ; beam $M_{BC} = \frac{2 E I}{6} (2 θ - θ) - 60 = \frac{E I θ}{3} - 60$ .
Equilibrium of joint B: $E I θ + \frac{E I θ}{3} - 60 = 0 \Rightarrow E I θ = 45$ . Corner: 45 kN·m; base: $M_{A B} = \frac{2 E I}{4} θ = 22.5$ kN·m; midspan: $\frac{w L ^{2}}{8} - 45 = 45$ kN·m.
Horizontal reaction (equilibrium of the column): $H = \frac{45 + 22.5}{4} = 16.875$ kN; vertical, $60$ kN at each base.
The simulator includes axial deformation: the beam, compressed by $H$ , shortens $H L / E A \approx 0.094$ mm and the columns lean slightly inward. That is why it gives 44.93 at the corners, 22.32 at the bases, 45.07 at midspan and $H = 16.81$ kN: less than 1 % difference.

By hand: 45, 22.5 and 45 kN·m,

H = 16.875

kN. With axial deformation (simulator): 44.93, 22.32 and 45.07 kN·m,

H = 16.81

kN.

Solve it in Simuveliq

The button opens the frame with its three IPE 300 members, the bases fixed and 20 kN/m on the beam.
The Determinacy card says «degree 3» and M max 45.07 kN·m at the middle of BC. Step «6. End forces of AB» gives 22.32 and −44.93 kN·m, and the reactions table 16.81 and 60 kN.
The displacements of B and C (±0.0469 mm in x) are the shortening of the beam. Try multiplying the section area A by 1000 in the Sections and material panel: the results approach the hand ones.

Open in the simulator →

Bending moment of example 2 from the simulator, drawn on the tension side: on top at the corners and on the outside, at the bottom at midspan and on the inside at the bases.

Example 3L-shaped bracket in space: torsion

A horizontal L-shaped bracket (IPE 300) is fixed at A. Segment AB is 3 m long in the

x

direction and BC, 2 m in

y

. BC carries 2 kN/m and 10 kN hang at C. Find the reactions and the internal forces at the fixed end.

Solution

It is statically determinate ( $G I = 6 \cdot 2 + 6 - 6 \cdot 3 = 0$ ): equilibrium of the whole bracket is enough.
Vertical force: $R_{z} = 10 + 2 \cdot 2 = 14$ kN.
Moments about A with $M = r \times F$ : the load at C, at $(3, 2)$ , gives $(- 20, 30, 0)$ and the 4 kN resultant at $(3, 1)$ gives $(- 4, 12, 0)$ kN·m. The reaction balances the sum: $M_{x} = 24$ and $M_{y} = - 42$ kN·m.
For member AB, which runs along $x$ , $M_{x}$ is a torque of 24 kN·m and $M_{y}$ a bending moment of 42 kN·m in the vertical plane. Segment BC has no torsion: its load acts in its own plane.

R_{z} = 14

kN, torque

T = 24

kN·m and bending moment

M = 42

kN·m at the fixed end.

Solve it in Simuveliq

3D structure; in Examples, «Load structure» → Horizontal L-bracket (torsion).
The cards give T max −24 kN·m and Mz max −42 kN·m in member AB, x = 0; the reactions table, $R_{z} = 14$ , $M_{x} = 24$ and $M_{y} = - 42$ .
Look at the displacement of C: several meters, impossible. The I-section has $G J = 15.5$ kN·m², almost nothing, and AB twists freely. Increase J in Sections and material (as if it were a tube) and the displacement drops to millimeters: the forces do not change, because the bracket is statically determinate.

Open in the simulator →

Structural analysis: the stiffness method

Degree of indeterminacy

The idea: unknowns in the displacements

The steps of the method

Details that change the result

Common mistakes

Worked examples

Example 1Two-span continuous beam

Solution

Solve it in Simuveliq

Example 2Fixed portal frame with a symmetric load

Solution

Solve it in Simuveliq

Example 3L-shaped bracket in space: torsion

Solution

Solve it in Simuveliq

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