Module 11 · Lesson 11.3
Equilibrium paths and limit points
Where load control runs out, what displacement and arc-length control can see that it cannot, and what snap-through actually is.
Why this matters
A load-controlled analysis answers 'what happens at this load?'. For most structures that is the right question. For a shallow arch, a shallow dome, a snap-fit, or a member well into its post-buckling range, it has no answer — because at that load there is no nearby state of equilibrium at all.
Understanding why is what separates 'the analysis failed' from 'the structure collapses here'.
By the end of this lesson you should be able to
- Read an equilibrium path and identify limit points and negative-tangent branches
- Explain why load control cannot follow a descending branch
- Distinguish a limit point from a bifurcation
- Say what displacement control and arc-length control each buy
- Recognise snap-through in an analysis output
The path
Plot load against displacement for a nonlinear structure and you get its equilibrium path — every state at which it is in equilibrium. For a hardening structure the path rises forever. For a yielding one it flattens. For a shallow arch it rises, turns over, descends, and rises again.
That turning over is the whole subject of this lesson.
Limit points
A limit point is where the tangent to the path is horizontal: the structure has reached the most load it can carry in that configuration.
Past it, the path descends: the structure continues to deflect while carrying less load. It is still in equilibrium at every point along that branch — it is a perfectly real set of states — but it is unstable, in the sense that a load-controlled system cannot sit there.
Why not? Because load control asks 'at load P, where is the structure?' and on the descending branch there are three answers for some values of P, or none for others. The solver is looking for a root that either does not exist nearby or is not the one it is walking towards.
Snap-through
Past the limit point, a real structure under a real load does not follow the descending branch. It jumps — dynamically, suddenly — to the far branch where the path has risen again to meet the applied load. That is snap-through, and it is what a shallow arch or a dome does when it inverts.
In an analysis, this shows up in one of two ways, and both are worth recognising:
- the solver fails to converge, which at least announces itself;
- the solver converges — onto the far branch, at a displacement far beyond where it was, with no indication that it jumped.
The second is more dangerous. A converged result on the far branch is a legitimate equilibrium state and the output will not say that the structure got there by inverting.
Bifurcation is different
A limit point is where the path turns over. A bifurcation is where two paths cross: the structure has a choice, and it takes the one with lower energy.
A perfectly straight strut under axial load is the classic case. Up to the critical load it shortens; at the critical load a second path — bending — becomes available, and the structure takes it. A model built perfectly straight has no reason to prefer either branch, which is why real analyses introduce an imperfection to break the symmetry and tell the structure which way to go.
A limit point needs no imperfection to find. A bifurcation usually does, and a perfectly symmetric model can sail past one entirely.
Displacement control
Instead of prescribing the load and finding the displacement, prescribe the displacement and find the load.
This works on the descending branch, because the question 'at this displacement, what load holds the structure?' has exactly one answer everywhere on the path. Nothing about the structure has changed; the question has been rephrased into one that always has a unique answer.
Arc-length control
Displacement control fails in its turn where the path turns back on itself in displacement — snap-back, seen in some composite and delamination problems.
Arc-length control prescribes neither. It advances a fixed distance along the path, letting both the load and the displacement be found. Geometrically it is walking along the curve at constant speed rather than stepping in x or in y. It follows anything the other two can and some things they cannot.
The cost is that the load at the end of an increment is not something you chose, so 'analyse at the factored design load' becomes 'analyse until you reach it' — which needs a stopping criterion and some care about what to report.
Try it
Equilibrium path explorer
The full path, traced by displacement control. Then try to reach a load under load control and see where it stops.
System
- Limit points
- 2
- Highest load on the rising branch
- 24.12 kN
- Snap-through?
- yes
- Load control at this load
- converged at u = 0.506 m
load control cannot pass this
Reading the plot
- Red dots mark the negative-tangent branch: real equilibrium states that load control cannot sit on.
- Orange markers are limit points, where the tangent is horizontal and the structure carries the most it can in that configuration.
- Move the attempted load above the first limit point and watch what load control does. Then note that displacement control traced the whole curve without difficulty.
What this shows: A limit point is a property of the structure. Load control cannot follow a descending branch; displacement control can.
Worked example
Finding the limit point of a shallow arch
Given
- A shallow-arch spring characterised by F = 60u − 45u² + 9u³
- So the tangent stiffness is kT = 60 − 90u + 27u²
Find
The limit points, the maximum load carried on the rising branch, and what happens above it
Dynamic relaxation
A different approach entirely: pretend the static problem is a dynamic one, give the structure a fictitious mass, apply the load, and damp the resulting motion until it stops. Where it stops is the static equilibrium.
There is no stiffness matrix and no factorisation, which makes it attractive for very large problems and for form-finding, where the target is an equilibrium shape and the route there is irrelevant.
Two things must be said about it every time:
The mass and damping are fictitious. They are chosen for convergence speed, not from the structure. Nothing about the path is physical.
Only the converged state means anything. The intermediate states overshoot, oscillate and settle. Reading a displacement off an intermediate step is reading a number the method invented.
It is also conditionally stable: too large a pseudo-time step and it diverges rather than converging, in exactly the way an explicit dynamic analysis does.
Explicit solvers
An explicit solver marches forward in time, computing each step directly from the last with no equations to solve. That makes each step extremely cheap and extremely numerous.
The stable time step is the time a stress wave takes to cross the smallest element:
Δt < h / c, where c = √(E/ρ) is the material wave speed
For steel, c is about 5 170 m/s. A 50 mm element gives a critical step of about 10 microseconds; a practical run uses perhaps 90 % of that.
The consequence is the one that decides whether an explicit analysis is feasible: the step is set by the smallest element in the model, not the average. One sliver element among a million multiplies the run time by however many times smaller it is. Explicit analysis rewards mesh discipline more than any other analysis type, and a mesh that would merely be inelegant elsewhere is here a schedule risk.
Explicit solvers earn their place on problems implicit methods handle badly: impact, blast, fragmentation, contact between many bodies, progressive collapse. They are not a general alternative to a static solver, and the course does not pretend that watching an animation teaches you to run one.
Try it
Dynamic relaxation explorer
Solve a static problem by pretending it is dynamic, then damping the motion until it stops.
Above the stability limit it diverges rather than converging.
- Stable?
- yes
- Converged?
- yes
- Steps taken
- 640
- Settled at
- 1.7725 m
- Newton–Raphson gives
- 1.7725 m
- Peak overshoot
- 1.9620 m
the same static problem, solved a completely different way
an intermediate state, and physically meaningless
Settled on the same answer as Newton–Raphson, without ever assembling or factorising a stiffness matrix — which is why the method is used for very large problems and for form-finding.
What the picture is not
- The mass is fictitious. It is chosen for convergence speed, not from the structure, so nothing about the transient is physical.
- Reduce the damping and the path oscillates for far longer before settling — on exactly the same answer.
- The peak overshoot is larger than the final displacement. Reading a value off an intermediate step would be reading a number the method invented.
What this shows: The mass and damping are fictitious and the intermediate states mean nothing. Only where it settles is real.
Try it
Explicit time-step explorer
The stable step is the time a stress wave takes to cross the smallest element. Everything else follows from it.
Material
The SMALLEST, not the average. That is the whole point.
- Wave speed √(E/ρ)
- 5172 m/s
- Critical step h/c
- 7.734 μs
- Used step (90 %)
- 6.960 μs
- Steps required
- 8,621
- Element updates
- 1.72e+9
- If the smallest were 30 mm
- 11,494 steps
steps × elements — the real cost driver
0.8× the current run
The wave speed is a property of the material alone: 5172 m/s. Steel is a little over 5 km/s, concrete about 3.5, timber about 4.7 along the grain — worth knowing as a sanity check on any reported step.
The three things this controls
- The step is set by the smallest element in the model, not the average. Four sliver elements out of a million set it for all of them.
- A softer material has a slower wave and therefore a longer step. Timber tolerates a coarser time discretisation than steel at the same element size.
- Halving the element size doubles the step count AND doubles the element count, so the cost grows faster than element count alone would suggest.
What this shows: One small element sets the step for the entire model, and therefore the run time — which is why explicit analysis rewards mesh discipline more than any other.
Practice
Steel has E = 210 × 10⁶ kN/m² and ρ = 7.85 t/m³. Compute the elastic wave speed in m/s. (In these units, kN/m² ÷ t/m³ gives m²/s² directly.)
Practice
Using that wave speed, compute the critical explicit time step for a 40 mm steel element, in microseconds.
Check yourself
An explicit model of 800 000 elements contains four elements 2 mm across; the rest average 30 mm. What is the effect on run time?
Check yourself
Why can load control not pass a limit point?
Check yourself
What does arc-length control add over displacement control?
Summary
- A limit point is where the tangent vanishes and the structure can carry no more in that configuration
- Past it the path descends: real equilibrium states that load control cannot reach
- Snap-through is the dynamic jump to the far branch, and a converged far-branch result will not say it jumped
- A bifurcation is two paths crossing and usually needs an imperfection to find
- Displacement control follows descending branches; arc-length follows snap-back too
- Dynamic relaxation's intermediate states are fictitious; only the converged state is real
- The explicit time step is set by the smallest element in the model, not the average
This is educational material. It uses simplified examples to teach principles, and must not be relied on for real design or safety-critical decisions. Module overview and checkpoint