Computational Engineering / Nonlinear and explicit analysis
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.
Lesson review: expert verification remains open
This lesson has received a full-context editorial review and a check of its technical explanations. Selected arithmetic has been checked separately. This is not independent engineering sign-off.
Checks needed for project use
- A computational engineering reviewer must check algorithms, convergence claims, statistical assumptions and numerical examples beyond the cases explicitly recorded.
- Review the arch limit-point numerical example and all lab convergence controls independently.
- Contact, plasticity and post-buckling instruction requires specialist confirmation of loading history and stability.
These are educational models and exercises. Check their assumptions before using the results. The full technical content and every diagram have not yet been independently verified.
Concrete and steel calculations retain their stated first-generation Eurocode teaching basis. Use the standards and amendments specified for your project; do not mix generations without an appropriate design basis.
Read the lesson register and outstanding review notesWhy 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
Equilibrium paths
An equilibrium path plots a selected load parameter against a displacement measure. A load limit point occurs where the load has a stationary value along the path. A descending branch can contain valid equilibrium states that are unstable under dead-load control. A physical snap-through event is dynamic; a static solver jumping between branches does not resolve that motion.
A bifurcation is a point where another equilibrium branch meets the primary one. Stability, imperfections and the loading history determine which response is realised; a simple instruction to choose the lower-energy branch is insufficient. A symmetric model may continue along an unstable primary branch unless a stability check or branch-following procedure detects the bifurcation.
Control methods
Displacement control prescribes a selected displacement and solves for the load parameter. It can follow a descending load branch where that displacement remains a suitable path parameter. It does not guarantee a unique solution and can fail at snap-back.
Arc-length methods add a constraint involving increments of load and displacement. They can follow many limit points and snap-back paths, but still require appropriate stepping, scaling and branch selection. Their path parameter is numerical; it is not physical time. Check the reached load level, stability and configuration before reporting a design result.
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
Dynamic relaxation uses an artificial dynamic system to seek static equilibrium. Mass, damping and pseudo-time steps are chosen for numerical performance. For the explicit form used here, internal forces are updated without factorising a global tangent matrix. Only a converged, appropriately checked state represents the intended static equilibrium; the intermediate motion has no physical time interpretation.
Explicit time integration
With a diagonal mass matrix, an explicit method can update accelerations without a coupled global solve. Its stable step depends on the highest numerical frequencies. A useful one-dimensional estimate is h/c, with c = √(E/ρ) for an axial rod. Three-dimensional solids, shell formulations, distortion, damping and contact can change the governing limit; use the solver’s formulation-specific estimate.
For the rod estimate, steel with E = 210 GPa and density 7850 kg/m³ gives c ≈ 5172 m/s. A 50 mm length then gives h/c ≈ 9.67 μs. Small elements can greatly increase step count, so inspect the controlling elements and any mass scaling. Explicit methods are widely used for impact and contact, and can also be used for quasi-static work when inertial effects and energy balance are controlled.
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