Member Stiffness in Local Axes

Everything in the method rests on one small matrix: what a single straight member does when its ends are moved. It is written in the member's own axes — along it and across it — because in those axes the terms are the standard beam results, and inclination is somebody else's problem (the next guide's).

A column of the stiffness matrix is a physical experiment: displace one end degree of freedom by one unit, hold the other five at zero, and write down the six forces needed to hold that shape.

The local axes and the six freedoms

Call the member's ends i and j. The local x̄ axis runs from i to j, along the member; the local ȳ axis is perpendicular to it. Each end has three degrees of freedom, numbered in this order:

ū₁, ū₄	Displacement along the member at end i and end j — stretching or shortening it.
ū₂, ū₅	Displacement across the member at end i and end j — bending it.
ū₃, ū₆	Rotation of end i and end j about the z axis.

The forces at the ends use the same order and the same directions: axial force, shear force, and moment, at i then at j.

Axial action: the member as a bar

Stretch a prismatic bar by δ and the force needed is EA δ/L — Hooke's law with the bar's own constants. So a unit displacement of end j along the member needs EA/L pushed at j, and −EA/L held at i for equilibrium.

k_axial = EA / L

axial stiffness

Axial action does not interact with bending in a linear analysis, so these terms sit alone in their rows and columns.

Bending action: the four beam results

The bending terms come from the same slope-deflection results used by hand methods — a prismatic member with both ends held against rotation, one end given a unit movement. Four cases give every bending term in the matrix.

4EI/L	Moment at an end rotated by one radian, with everything else held.
2EI/L	The moment that appears at the far end while it is held — the carry-over.
6EI/L²	The coupling term: a moment produces end shear, and a sideways movement produces end moments.
12EI/L³	The shear needed to displace one end sideways by one unit with both ends held against rotation.

Where 12EI/L³ comes from

Take a member fixed at both ends and push end j across by Δ. Both ends stay horizontal, so the member takes an S-shape with a point of contraflexure at mid-span. Each half is then a cantilever of length L/2 carrying the end shear V, deflecting Δ/2:

Δ/2 = V(L/2)³ / 3EI ⟹ V = 12EIΔ / L³

The end moments that go with it are V·L/2 = 6EIΔ/L², which is the coupling term in the table above. Every bending entry in the matrix is one of these four numbers with a sign.

The member stiffness matrix

Put the six columns together and this is the whole of a plane frame member:

k̄ =

	ū₁	ū₂	ū₃	ū₄	ū₅	ū₆
f̄₁	EA/L	0	0	−EA/L	0	0
f̄₂	0	12EI/L³	6EI/L²	0	−12EI/L³	6EI/L²
f̄₃	0	6EI/L²	4EI/L	0	−6EI/L²	2EI/L
f̄₄	−EA/L	0	0	EA/L	0	0
f̄₅	0	−12EI/L³	−6EI/L²	0	12EI/L³	−6EI/L²
f̄₆	0	6EI/L²	2EI/L	0	−6EI/L²	4EI/L

Read a column, not the matrix. Column 2, for instance, is the member with end i pushed one unit across it: shears of 12EI/L³ up at i and down at j, and a moment of 6EI/L² at each end. Those four numbers are one loading case in equilibrium, which is a useful check on any matrix you write out: each column sums to zero in force, and to zero in moment about either end.

Three properties worth knowing

It is symmetric. k̄_ab = k̄_ba, which is Maxwell–Betti reciprocity: the force at a from a unit displacement at b equals the force at b from a unit displacement at a. Symmetry survives assembly, which is why the structure matrix is symmetric too, and why programs store half of it.
It is singular. Move both ends the same way and nothing resists — the member simply travels. Those three rigid-body motions (two translations and a rotation) make the 6×6 matrix rank 3. A member alone is not a structure; the supports are what remove them.
Its diagonal is positive. Displacing a freedom always takes force in the direction you displaced it. A negative or zero diagonal term in an assembled matrix means an error somewhere.

Two variants that come up constantly

The truss member

Delete every row and column that involves bending and what is left is a bar with only axial stiffness:

k̄_truss =(EA/L)

+1	−1
−1	+1

In practice a program does not usually keep a separate truss element: it releases the moments at both ends of a frame member instead, which comes to the same thing and lets one member type serve everywhere.

A member with a released end

A pinned connection means the member can rotate at that end while the joint does something else, so that end carries no moment. The freedom is condensed out of the 6×6 matrix — the row and column for that rotation are eliminated, and their effect redistributed into the rest. Releasing end j, the flexural terms become:

3EI/L	Rotational stiffness of the remaining fixed end i — down from 4EI/L.
3EI/L²	The coupling between sway and end moment — down from 6EI/L².
3EI/L³	The sway stiffness — a quarter of the fully fixed 12EI/L³.
0	Every term involving the released rotation. There is no carry-over to a hinge.

This is the same reduction hand methods make when they use a modified stiffness for a member pinned at the far end, and it is why a portal frame on pinned bases sways so much more than the same frame on fixed ones.

A worked member

A steel beam, E = 200 GPa, I = 100×10⁶ mm⁴ = 100×10⁻⁶ m⁴, A = 5000 mm² = 5×10⁻³ m², spanning L = 6 m:

EA/L	200×10⁶ kN/m² × 5×10⁻³ m² / 6 m = 166 667 kN/m
12EI/L³	12 × 200×10⁶ × 100×10⁻⁶ / 6³ = 1 111 kN/m
6EI/L²	6 × 200×10⁶ × 100×10⁻⁶ / 6² = 3 333 kN/rad
4EI/L	4 × 200×10⁶ × 100×10⁻⁶ / 6 = 13 333 kN·m/rad

Note the ratio: this member is 150 times stiffer along itself than across itself. That is why the axial shortening of columns is usually negligible next to frame sway, and why a matrix full of numbers that differ by orders of magnitude has to be solved carefully.

Next: this matrix describes a member lying along its own axis. A real frame has members at every angle, and they cannot be added together until they are all described in the same axes.