Structural form

The frame that leans, and what stops it

A rectangle of pinned bars folds flat. Make the corners rigid instead of adding a diagonal and it does not — which buys an unobstructed opening and costs bending in every member of it.

Four bars pinned into a rectangle fold flat under the lightest sideways push. There are two ways to stop them. Put in a diagonal, which makes the rectangle two triangles and cannot fold. Or make the corners rigid, so that the bars cannot rotate relative to each other.

The first is a truss and is cheap. The second is a portal frame, and it costs a great deal more — bending in members that would otherwise carry only axial force, connections that must transmit moment, and an analysis statics cannot finish. It is built anyway, everywhere, because the diagonal would be in the doorway.

A portal frame swaying under 20A portal frame pushed sideways, solved by the stiffness method because statics cannot divide the load between two columns. The base shears come out at 10.0 and 10.0 and add to the applied 20; the peak moment is 22.2. The sway is drawn hugely exaggerated, and the moment diagram is plotted on each member's tension face.20H 10.0 M 22.2H 10.0 M 22.2the two base shears add to the applied 20 — the split came from stiffness, not staticsthe sway is exaggerated; a real frame at this load moves a fraction of a millimetre
Fig. 1 A portal frame pushed sideways. Nothing triangulates it; what resists the push is the refusal of the corners to change angle, and the price is a bending moment running continuously round both of them.

Rigidity is a restraint, and restraints are equations

A pinned joint transmits two forces and no moment. A rigid joint transmits a moment as well, and that third quantity is what stops the rectangle folding.

The consequence for the counting is immediate. Each member of a rigid-jointed plane frame carries three unknowns rather than one — an axial force, a shear and a moment — so the arithmetic that settles whether statics can finish the job has to be rewritten. A portal with two fixed bases has six reaction components against three equations: three times redundant. With two pinned bases, four against three: once redundant.

Either way it is redundant, and that is not incidental. The joint rigidity that provides the stability is itself the redundancy — a structure that resists by joint action rather than by triangulation cannot be determinate, because the moment at a rigid joint is a force that equilibrium alone does not determine.

So a portal frame is a structure whose method of standing up puts it beyond statics by construction. Every number in the figures on this page came from a stiffness calculation, and there is no way to get them without one.

Where the load actually goes

Push a portal sideways with a force HH and ask how much each column carries. Equilibrium says only that the two base shears add to HH. Any split satisfies it.

What decides is stiffness. The two columns are forced to sway by the same amount, because the beam connecting them is far too stiff axially to let them move differently — so each column takes load in proportion to how hard it resists that shared displacement. Two identical columns take half each. A column twice as stiff takes twice the share.

A portal frame swaying under 20A portal frame pushed sideways, solved by the stiffness method because statics cannot divide the load between two columns. The base shears come out at 10.0 and 10.0 and add to the applied 20; the peak moment is 40.0. The sway is drawn hugely exaggerated, and the moment diagram is plotted on each member's tension face.20H 10.0H 10.0the two base shears add to the applied 20 — the split came from stiffness, not staticsthe sway is exaggerated; a real frame at this load moves a fraction of a millimetre
Fig. 2 The same frame on pinned bases. The split between the columns is unchanged because the two are identical, but every moment in the structure is different — pinning the feet removes the base moments and pushes the whole demand up into the corners.

Comparing the two figures gives the practical rule that follows. Fixed bases halve the peak moment and take a substantial share of it into the foundations; pinned bases leave the corners carrying everything and need larger members. Which is chosen is nearly always a foundation question rather than a frame question, because a fixed base needs a foundation capable of resisting a moment, and that is a considerably larger and more expensive object than one resisting force alone.

The general principle is the one that runs through every redundant structure: load goes where the stiffness is. A frame with one stiff bay and several flexible ones sends nearly all the lateral load to the stiff bay, whether or not anyone intended it — which is why an infill wall that nobody counted as structural can attract enough load to fail, and why removing one during refurbishment is a structural act.

The moment runs round the corner

The moment diagram of a frame is the moment diagram of a beam with the corners turned, and drawing it correctly requires giving up the sagging-positive convention.

At a rigid corner, the moment in the beam and the moment in the column are the same moment, transmitted through the joint. It has to be: take the joint itself as a free body, and the moments on its two faces must balance. So the diagram is continuous round the corner, and any drawing in which it jumps at a joint has an equilibrium error in it.

The convention that survives the rotation is to plot the ordinate on the tension face, with no sign written. Under a sideways push the frame’s windward corner is in tension on the outside and the leeward corner is in tension on the inside — so the diagram runs up the outside of one column, round the top, and down the inside of the other. That is what the figures on this page draw, and the continuity at the corners is the check.

A portal frame swaying under 30A portal frame pushed sideways, solved by the stiffness method because statics cannot divide the load between two columns. The base shears come out at 15.0 and 15.0 and add to the applied 30; the peak moment is 33.0. The sway is drawn hugely exaggerated, and the moment diagram is plotted on each member's tension face.30H 15.0 M 33.0H 15.0 M 33.0the two base shears add to the applied 30 — the split came from stiffness, not staticsthe sway is exaggerated; a real frame at this load moves a fraction of a millimetre
Fig. 3 A wider frame with a stiffer beam. The base shears still add to the applied load — that much is statics — while every moment has changed, because a stiffer beam holds the column tops against rotation more firmly and drives more of the demand into the bases.

There is a point of contraflexure somewhere in each column, where the moment passes through zero and the tension changes face. For a fixed-base portal under sway it sits near mid-height, which is the observation the classical portal method turns into a hand calculation: assume hinges at the column mid-heights and at the beam mid-span, and the frame becomes determinate and solvable by a section cut through the storey. It is an assumption rather than a result, and for a regular frame it is a good one.

What symmetry gives back

The frame is redundant, so stiffness is needed. Symmetry can supply part of what stiffness would have, and the pinned-base portal is the cleanest instance.

A horizontal load on a symmetric frame is an antisymmetric load case: reflect the frame about its centreline and the load reverses. The response must have the same property, which forces the two horizontal reactions to be equal — and that single statement is exactly the extra equation the once-redundant frame was missing.

With it, the analysis is arithmetic. Each column carries H/2H/2 horizontally. Its base is a pin and carries no moment, so the moment at the top of the column is that shear times the height:

Mcorner=Hh2.M_{\text{corner}} = \frac{H h}{2}.

For the frame in the second figure — HH of twenty over a height of four — that is forty, which is the peak moment the solver returns, to the digit. No stiffness entered the calculation at all.

The trick evaporates as soon as the symmetry does. Give the two columns different sections, or different heights, or put the load anywhere other than at the eaves, and the split is a stiffness question again. It also fails for the fixed-base frame, where symmetry still gives equal shears but the base moments remain undetermined by it — which is why the fixed-base figure’s numbers could not have been obtained this way and the pinned one’s could.

Two lessons are worth taking. Symmetry arguments are equations, and they should be counted alongside the equilibrium ones when deciding whether a problem is answerable. And they are fragile in a way equilibrium is not: an equilibrium equation holds whatever the structure turns out to be, while a symmetry argument holds only while the structure remains what it was assumed to be, which on a real project it very often does not.

When the diagonal cannot be there

The whole case for the portal frame rests on the diagonal being unacceptable, so it is worth being specific about what is being bought.

A braced bay is dramatically cheaper. Its members carry axial force only, so they are sized by strength and slenderness rather than by bending; its connections carry force rather than moment, so they are simple plates and a few bolts; and it is determinate, so it can be checked by hand. Against a portal of the same size it is typically a third of the steel and rather less than a third of the fabrication.

A Warren truss of 6 panelsA Warren truss under equal panel-point loads. The joint equilibrium equations were assembled and solved; 10 members came out in tension, 11 in compression and 2 carrying nothing.tensioncompression2 carrying nothing
Fig. 4 A braced bay, which is a truss stood on end. Every member carries axial force alone and every connection is a simple one — and a doorway through the middle of it is not available at any price.

What the portal buys is the hole. An unobstructed opening the full width and height of the bay is worth an enormous amount in a factory, a showroom, a warehouse or a car park, and there is no version of a triangulated frame that provides it. So the portal is not competing with the braced bay on efficiency; it is the answer to a question the braced bay cannot answer at all.

The same reasoning explains the Vierendeel girder, which is a row of portals laid on its side. It has no diagonals, so its panels are open and can be walked through or glazed; it resists shear entirely by bending in its chords and posts; and it is heavier than an equivalent truss by a factor of two or more. It is used where the opening is worth that, and nowhere else.

And it explains the transfer structure, which is the same trade at a different scale: a building whose upper columns do not line up with its lower ones needs a deep member to carry the difference, and that member is expensive for exactly the reason a portal is — load being carried by bending where it could have been carried axially. Bending is the inefficient way to carry anything, and every one of these forms is a decision to pay for it in exchange for a space.

Sway is the design case

For a low-rise portal the vertical load usually sizes the beam and the horizontal load usually sizes everything else, and the reason is that sway is expensive in three separate ways at once.

Bending. The moments from sway add to those from gravity at two of the four corners and subtract at the other two, so the governing corner is a combination rather than either case alone.

Second-order amplification. A frame that has swayed carries its vertical load off the line it was acting on, which produces more sway. The amplification is one over one minus the load ratio, and a sway frame’s critical load is low precisely because nothing triangulates it — so the correction arrives at loads far below anything that looks like instability.

Effective length. A column whose top can move sideways buckles by leaning rather than by bowing, and its effective length exceeds its height — often substantially. Since capacity falls with the square of that length, a sway frame’s columns are worth a fraction of what the same columns are worth in a braced bay.

The ends decide the length that mattersFour columns of identical height and section, buckling under four sets of end conditions. The effective length factor is the fraction of the column that behaves like a pin-ended one, and the buckling load goes as its inverse square.K = 0.5both ends fixedK = 0.7one fixed, one pinnedK = 1both ends pinnedK = 2fixed at the base, free at the topsame column, same section, four ways of holding the endsthe load at which each buckles goes as 1 ÷ K² — a factor of sixteen across this row
Fig. 5 Four end conditions and the buckling length each implies. The rightmost case is a sway column, and the factor of four in capacity between it and the pinned reference is the price of having no bracing.

All three grow together, which is why the profession draws a hard line between sway and non-sway frames rather than treating sway as a matter of degree. A frame stiff enough that its elastic critical load factor exceeds about ten is treated as braced and the whole apparatus above is skipped; below it, all three effects have to be carried explicitly.

What the connection has to do

A portal frame’s difficulty is concentrated at four points, and it is a fabrication difficulty rather than an analytical one.

The moment at a corner is transmitted as a couple: tension in the outer flange, compression in the inner one, separated by the member’s depth. So the connection has to develop the full flange force in each — which for an ordinary frame is several hundred kilonewtons — and deliver it round a ninety-degree turn.

The standard solution is a bolted end plate, extended above the beam’s top flange to get more bolts into the tension zone, with a haunch beneath. The haunch is a wedge of extra depth at the corner, and it does two things at once: it increases the lever arm exactly where the moment is largest, and it lengthens the connection so that more bolts can share the tension. A haunched portal frame is a moment diagram made physical — deep where the diagram is deep, shallow where it is not.

Load, shear and moment — a cantileverThe applied load, the shear force it produces and the bending moment that follows, drawn one above another to the same horizontal scale. Shear is the integral of the load and moment is the integral of shear.4 per unit lengthshear32.0moment-128.0 at x = 0.00the moment peaks exactly where the shear passes through zero
Fig. 6 A cantilever’s moment diagram, growing toward its support. A portal’s column carries the same shape, and the haunch at the corner is that diagram converted into steel.

The stiffness of that connection matters as much as its strength, and this is where portal frames are most often mis-modelled. The analysis assumes the corner is rigid — that the angle between beam and column does not change. A real bolted connection has some flexibility, and if it has enough, the frame sways further than the calculation said and the moments redistribute toward mid-span. Connections are therefore classified as rigid, semi-rigid or nominally pinned, and using a connection of one class in a frame analysed as another is an error that no member check will find, because every member is correct for the frame that was described.

Where the model stops

Rigid joints, exactly. As above. Semi-rigid behaviour is real, and modelling it requires a rotational spring at each corner whose stiffness is a property of the bolts and plates rather than of the members.

Elastic behaviour. Every figure here is an elastic analysis. A steel portal loaded to collapse forms plastic hinges — typically at the two corners and in the beam — and the collapse load depends on those hinges rather than on the elastic distribution, which is why plastic design is the normal method for portal frames and elastic analysis is the exception.

Small deflections. The whole analysis is first-order, and for a sway-sensitive frame that is the assumption most likely to be inadequate.

In-plane only. A portal frame is a plane structure that must be prevented from falling over out of its plane by purlins, side rails and bracing in the roof. None of that appears in any figure here and none of it is optional.

Rigid foundations. A “fixed” base is fixed by a foundation with finite stiffness, and a base that rotates a little is somewhere between the two cases drawn. Since the two differ by a factor of two in peak moment, the interpolation matters.

The figures share a limitation this site keeps running into. The deflected shape is drawn at an exaggeration of a few hundred times — a real portal under working wind moves a few millimetres over a storey, which at the scale of these drawings is less than a line width. The exaggeration makes the sway mode legible and makes the frame look alarmingly flexible, and both the moment diagram and the base shears are for the undeformed geometry regardless of how far the drawn shape has moved.

The ladder from here

Later rungs on this anchor: the portal method and the cantilever method. Moment distribution applied to frames. Sway and non-sway classification. Plastic collapse mechanisms in portals — beam, sway and combined. Haunch design. Semi-rigid connections and their classification. Multi-bay and multi-storey frames. Frames with pitched roofs, where the sway mechanism changes shape. And the general displacement method, which is what every analysis program is doing and which portals are the smallest interesting example of.

The rigid-jointed portal became the standard industrial building frame in the 1950s, once welding made a full-strength corner cheap and moment distribution made the analysis routine. Before both, the same buildings were built as trusses on columns — with a diagonal somewhere, and a door somewhere else.