The point the mechanism turns about

A plastic collapse calculation is a work equation: set the work done by the loads equal to the work absorbed at the hinges, and solve for the load factor. Writing it down needs two things — how much each hinge rotates, and how far each load moves — and for a rectangular portal both are obvious from the drawing.

For anything with a slope in it they are not. A pitched portal’s rafter is neither pinned to the ground nor moving parallel to itself, and working out how far its apex descends when the frame sways by a given amount is a piece of trigonometry with an easy sign error in it. The alternative is to notice that the rafter, like every other rigid piece of a mechanism, is rotating about some point, and to find where.

Two rules, and there are no others #

A rigid body moving in a plane is either translating or rotating about a point. If it is rotating about $O$, then every point $P$ of it moves by $\theta \cdot |P - O|$ in a direction perpendicular to $OP$ — so knowing $O$ and one rotation is knowing the whole motion of that body.

For a chain of bodies there are exactly two rules for finding those points.

A body pinned to the ground rotates about the pin. That covers every column with a hinge at its base, and it is where every one of these calculations starts.

Two bodies joined by a hinge have that hinge in common, so the hinge’s displacement computed from the first body’s centre must equal its displacement computed from the second’s. Two displacements are equal only if they are parallel, and both are perpendicular to a radius — so the second body’s centre lies on the line through the first body’s centre and the hinge, extended as far as it needs to go.

One hinge gives a line. Two hinges give two lines, and the centre is where they cross. That is the whole method.

Which free body produced the number #

Take the classic case: a pitched portal of span $L$, eaves height $h$ and apex rise $f$, with plastic hinges at both column bases, at one eaves and at the apex.

The left column and the rafter up to the apex form one rigid body, hinged to the ground at $A$, so it turns about $A$. The right column is a body hinged to the ground at $E$ and turns about $E$. The right-hand rafter is the third body, joined to the first at the apex $C$ and to the third at the eaves $D$.

Apply the second rule twice. Because body one turns about $A$ and shares $C$ with body two, body two’s centre is on the line $A$–$C$ extended. Because body three turns about $E$ and shares $D$ with body two, body two’s centre is also on the line $E$–$D$ extended — and $E$–$D$ is the right-hand column, so that line is vertical at $x = L$.

The line from $A(0,0)$ through $C(L/2,\, h+f)$ reaches $x = L$ at $y = 2(h+f)$. So

$$I = \big(L,\ 2(h+f)\big)$$

twice the ridge height, above the far column. For a twelve-metre frame with four-metre eaves and a metre and a half of rise, that is eleven metres up — the frame itself is 5.5 metres tall, and the centre is above the top of any sensible drawing of it.

And the answer, in two ratios #

With the centre found, everything else is division. The rotation of body two follows from matching the apex displacement:

$$\theta_2 = \theta_1 \frac{|A - C|}{|C - I|}$$

and body three’s from matching the eaves:

$$\theta_3 = \theta_2 \frac{|D - I|}{|D - E|} = \theta_2 \frac{h + 2f}{h}$$

The hinge rotations are the differences between the rotations of the bodies either side, the internal work is $M_p$ times their sum, and the external work is each load times the component of its own displacement in its own direction — where every displacement is $\theta$ times a radius, perpendicular to it.

Nothing in that has a sine or a cosine in it. For this frame the apex descends by $\theta_2 \cdot L/2$ and the eaves sways by $\theta_1 h$, both of them a rotation times a horizontal or vertical offset from the relevant centre, because a rotation moves a point vertically by $\theta \Delta x$ and horizontally by $-\theta \Delta y$. The trigonometry that seemed to be needed was an artefact of not having found the centre.

The check that the method is right is the flat case. Set $f = 0$ and the same construction gives $I = (L, 2h)$, the rotations come out equal, the hinge rotations sum to $6\theta$, and

$$\lambda = \frac{6 M_p}{WL/2 + Hh}$$

which is the closed form for a fixed-base rectangular portal’s combined mechanism, arrived at from somewhere entirely different. With pinned bases the two base hinges cost nothing and the same construction gives $4M_p$ over the same denominator.

The hinge is an unknown, not a location #

Everything above assumed the rafter hinge is at the apex, and for an unsymmetrical load it is not.

The upper-bound theorem says that every mechanism gives a load factor at or above the true collapse load, so the honest calculation is to sweep the hinge position and take the lowest. Doing that on the frame drawn here, with the vertical load replaced by a uniform load on the rafters and a substantial horizontal load at the eaves, puts the hinge at 63 to 77 per cent of the way from the eaves to the apex — and assuming the apex overstates the collapse load by up to about three per cent.

Three per cent is not much and the direction is the problem: it is on the unsafe side, because every mechanism is an upper bound. A calculation that puts the hinge where the drawing has a corner has not made a small approximation, it has computed the collapse load of a structure that will collapse a different way. The instantaneous centre makes the sweep cheap, because moving the hinge moves one line and everything else follows.

The geometry of the migration is worth reading off the centre. Push the frame harder sideways and the hinge slides towards the eaves, because a mechanism that sways more wants its hinge where the sway does most work. Steepen the roof and it slides further, because the rafter’s own geometry gives the apex less vertical travel per unit rotation.

Where the two bounds meet #

A mechanism is an upper bound; an equilibrium moment field that nowhere exceeds $M_p$ is a lower one. The collapse load is where they meet, and the instantaneous centre belongs entirely to the first half.

That asymmetry is worth stating plainly. The kinematic route always produces a number: pick any admissible mechanism, do the work equation, get a load factor. The static route often does not — constructing an admissible moment field is real work — which is why practical plastic design is nearly always a search over mechanisms and why the search has to be exhaustive rather than plausible. A mechanism nobody thought of is not caught by checking the arithmetic of the ones that were.

A worked reading of one number #

It is worth putting the two ratios to work once, because the arithmetic is short enough to follow and the answer is not obvious from looking at the frame.

Take the twelve-metre frame with four-metre columns, a metre and a half of rise, fixed feet and a plastic moment of 200 kNm throughout. The centre is at $(12,\,11)$ metres. Rotate the left body by one unit about $A$: the apex, six metres out and 5.5 metres up, moves perpendicular to $AC$, and its vertical component is one unit times the horizontal offset, six metres. The apex descends by six.

Body two turns about $I$. The apex is six metres to the left of $I$ and 5.5 metres below it, and $|C - I|$ is the same as $|C - A|$ because $C$ sits exactly midway along the line — so $\theta_2$ equals $\theta_1$. The eaves $D$ is at $(12,\,4)$, seven metres below $I$, so it moves horizontally by seven units; the right column, turning about $E$ four metres below $D$, must move it by $4\theta_3$, so $\theta_3 = 7/4$.

The hinge rotations are then $1$ at $A$, $2$ at the apex, $1 + 7/4$ at $D$ and $7/4$ at $E$: seven and a half units, absorbing $1{,}500$ kNm of work. The loads do $200 \times 6 + 60 \times 4 = 1{,}440$ kNm at unit load factor, so $\lambda = 1.042$ and the frame has four per cent in hand against this mechanism.

Every number in that is a length read off the drawing. What makes it worth doing by this route rather than any other is that a reader can check it by measuring the figure.

The centre at infinity, which is a translation #

The construction has a degenerate case and it is the most useful thing in it.

If the two lines are parallel, they do not cross, and the body has no centre. What it has instead is a translation: every point moves the same distance in the same direction. That is not a failure of the method, it is the method telling the truth — a rectangular portal’s beam mechanism has its beam segments rotating about points, but the sway mechanism has the whole beam translating sideways, and the construction gives parallel lines because both columns are vertical.

So the centre’s position is a continuous description of the mechanism’s character. Far away, and the body is nearly translating; close in, and it is nearly pivoting. Watching where it goes as a geometry is varied says more about how a frame collapses than the collapse load does.

The drawing that was the calculation #

The instantaneous centre is not a plastic-analysis idea. It is a kinematics idea, it predates plasticity by a century, and it arrived in structural engineering by the same route as the force polygon: as a piece of graphical mechanics.

Its structural home before plasticity was influence lines. The Müller-Breslau principle says that the influence line for a redundant force is the deflected shape of the structure released in that force’s direction — and for a structure released into a mechanism, that shape is a set of rigid-body rotations, drawn by finding the centres. Every influence line for a bridge girder in the nineteenth century was constructed that way, with a pair of dividers and no arithmetic at all.

The plastic version is the same construction with the hinges chosen rather than released, and the same two rules govern both. That is why the method reads as being older than the theory it is used for: it is.

What it says about a real portal shed #

The frame in these figures is not an abstraction — it is the commonest structure in the built world. A single-storey portal with a shallow pitch, pinned or nominally fixed feet, haunched at the eaves, at six-metre centres, is what most of the industrial floor area on the planet is made of, and it is designed plastically almost everywhere.

Two features of that design come straight out of the construction above.

The haunch exists to move the hinge. Its structural purpose is to make the eaves stronger than the rafter so that the plastic hinge forms at the end of the haunch rather than at the corner, where the connection is. The mechanism that governs is therefore not the one drawn in a textbook: the effective eaves hinge is a metre or more out along the rafter, the rafter body is shorter, and its centre moves accordingly. A frame analysed with hinges at its corners has a mechanism a little too generous, on the unsafe side again.

And the collapse load is barely the point. A portal shed of ordinary proportions is almost never governed by first-order plastic collapse. It is governed by sway stability — the frame is slender, it carries gravity load, and the load factor at which the second-order effects run away is lower than the plastic one. The plastic mechanism supplies the shape the frame is going to fail in and the number gets reduced. Which is a common relationship in this subject: a calculation that is exact about a mechanism, embedded in a check that is approximate about a load.

Where the model stops #

Every piece is rigid. The whole method assumes that all the deformation happens at the hinges and none along the members, which is exactly true only at the collapse limit and never before it. The load–deflection curve up to collapse is not a straight line and is not what this calculation is about.

The hinges are points. A plastic hinge is really a region a section deep or more, and for a haunched frame it is not even where the moment is largest — it forms at the end of the haunch, where the section changes. That moves the geometry the whole construction depends on, and a mechanism drawn with hinges at the corners of a haunched frame is the wrong mechanism.

And the rotations are small. The centre is instantaneous: it is where the body is turning about at this instant, and it moves as the mechanism develops. Every displacement in the work equation is a first-order one, and a frame that has swayed far enough for the geometry to change is in a second-order problem that this calculation does not describe. For a sway mechanism carrying substantial gravity load, that correction is not academic — it can be the thing that decides the frame.

The generalisation #

The habit here is a general one about mechanisms, and it applies well beyond frames.

Any assembly with one degree of freedom has, at each instant, one centre per rigid piece, and the centres are found by intersecting lines rather than by solving equations. That is the same statement for a truss that has lost a member, for a determinacy count that comes out at zero but describes a structure that folds anyway, for a linkage, and for a retaining wall’s failure wedge sliding on a plane.

In each of those the useful question is the same: what is turning about what? A mechanism nobody can draw is a mechanism nobody has checked, and the centres are how it gets drawn.

There is one more thing the construction is quietly good for, and it is not about collapse at all. Virtual work needs a compatible displacement field and does not care whether the structure is really moving; a mechanism is the cheapest compatible field there is, because it costs no strain energy anywhere except at the hinges. So the centres are a way of generating admissible displacement fields for any virtual-work argument — including the reciprocal theorem and the influence-line construction above — and the plastic application is a special case of a much older habit.