Structural Dynamics / Numerical time integration

Stability, accuracy and demonstrating convergence

Two different failure modes, two different limits — and the one procedure that establishes a time step is small enough rather than asserting it.

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Why this matters

A time step has to be chosen. Rules of thumb exist — T/10, T/20, T/50 — and they get you into the right region and no further.

The difficulty is that the two ways a step can be wrong look completely different. One announces itself unmistakably; the other does not announce itself at all. Knowing which is which, and having a procedure that catches the silent one, is the practical content of this lesson.

By the end of this lesson you should be able to

State the stability limits and where they come from
Distinguish period elongation from amplitude decay
Say why the step must resolve the highest mode that matters, not the first
Carry out and report a convergence study

The stability limits

Scheme	γ, β	Stability
Constant average acceleration	½, ¼	Unconditional
Linear acceleration	½, 1/6	Δt/T ≤ 0.551
Central difference	explicit	Δt/T < 1/π ≈ 0.318
Newmark, general	γ ≥ ½	Unconditional if 2β ≥ γ

The general Newmark condition, when it is not unconditional, is

Δt/T ≤ 1/(π√2 · √(γ − 2β))

and substituting γ = ½, β = 1/6 gives 0.551, which is where that number comes from.

What crossing a stability limit looks like

Not a gradual degradation. The solution grows without bound, doubling every step or two, and reaches absurd numbers — 10²⁰ metres of displacement — within a few dozen steps.

This is the SAFE failure mode. Nobody has ever mistaken 10²⁰ metres for an answer.

What an inadequate but stable step looks like

A smooth, bounded, plausible response with the wrong phase.

For constant average acceleration the error is almost entirely period elongation: the numerical period is longer than the true one by approximately

ΔT/T ≈ (π²/3)(Δt/T)²

which is 0.13% at Δt/T = 0.02, 3.3% at Δt/T = 0.1, and 13% at Δt/T = 0.2.

The amplitude is preserved exactly. So the response has the right size at the wrong time, and it drifts further out of phase as the analysis proceeds. Early in the record the two solutions agree; late in the record they can be in antiphase.

For γ > ½ there is also amplitude decay — a deliberate bleeding of energy which is the point of choosing γ > ½, and which affects the modes you wanted along with the ones you did not.

The step must resolve the highest mode that matters

This is the trap that catches most people, and it is worth stating baldly.

A ten-storey building might have T₁ = 1.4 s and T₈ = 0.07 s. A step of 0.02 s is T₁/70 — which sounds generous — and T₈/3.5, which is hopeless.

Higher modes contribute little DISPLACEMENT and a great deal of ACCELERATION. So a displacement result can look perfectly converged while the floor accelerations are 40% out. If the analysis exists to check equipment or occupant comfort, that is the quantity that matters and it is the one that has not converged.

Choose Δt against the shortest period relevant to the response quantity, and check convergence on the quantity you are actually going to use.

The convergence study

The only thing that establishes a step is adequate:

1.Run the analysis at Δt.
2.Run it again at Δt/2.
3.Compare the quantity you will actually use.
4.If it has not changed materially, the coarser step was adequate. If it has, halve again.

It doubles the run time and it is the difference between an answer that is converged and an answer that is merely plausible. Report the result — 'peak drift changed by 0.4% between Δt = 0.005 s and Δt = 0.0025 s′ is a statement a reviewer can check.

Try it

How small does the time step need to be?

Run the same problem at fourteen time steps and watch the answer stop changing.

Scheme

Quantity checked

Acceleration converges more slowly than displacement — always check the one you need.

Natural period T_ns

Damping ratio ζ

Relative error
5% line

Error in the peak against Time step Δt / T_n. Relative error reaches a peak magnitude of 95.1 %. 5% line reaches a peak magnitude of 5 %.

Δt / T_n	Steps	Peak	Error	Change from previous
0.500	25	4.43	95.13%	—
0.350	35	28.9	68.24%	84.67%
0.250	49	89.4	1.67%	67.70%
0.180	68	136	49.71%	34.32%
0.125	97	115	26.43%	18.41%
0.090	134	102	11.96%	12.92%
0.063	193	95.6	5.16%	6.47%
0.045	268	93.3	2.60%	2.49%
0.030	401	91.9	1.10%	1.49%
0.020	601	91.4	0.49%	0.60%
0.014	858	91.1	0.20%	0.30%
0.010	1201	91	0.11%	0.08%
0.007	1715	91	0.06%	0.05%
0.005	2401	91	0.02%	0.04%

Exact steady-state amplitude: 75.3 mm
Within 5% at: Δt/T_n = 0.25
Within 1% at: Δt/T_n = 0.02
Cost of the 1% answer: 601 steps

The commonly quoted rule Δt ≤ T/10 gets you into the right region and no further. Notice how much the answer is still moving at that step, and how little it moves once you are at T/50.

What this shows: A time step is adequate when halving it stops changing the answer — not when a rule of thumb says so. The rule of thumb is where you START the convergence study.

Worked example

Choosing and justifying a time step

Given

A twelve-storey frame; modal analysis gives T₁ = 1.52 s
The eighth mode, at T₈ = 0.068 s, is the highest with meaningful effective mass
Ground-motion record sampled at 0.02 s
The analysis is required to produce floor accelerations for equipment qualification

Find

An adequate time step, and how to demonstrate that it is.

Assumptions

Constant average acceleration, as used by the package
Linear response

Predict first

A time-history analysis is run twice. Peak displacement changes by 3% between the two steps; peak floor acceleration changes by 35%. What does this tell you?

Practice

A structure has a natural period of 0.8 s. What is the largest time step, in seconds, for which the central-difference method is stable?

Practice

Using constant average acceleration at Δt/T = 0.15, what is the approximate period elongation, as a percentage?

Practice

A building has T₁ = 1.2 s and its highest significant mode has T₆ = 0.09 s. Using a T/20 rule on the mode that governs, what time step should be used, in seconds?

Practice

Using the general Newmark stability condition Δt/T ≤ 1/(π√2·√(γ − 2β)), what is the limit for γ = 0.5 and β = 1/6?

Check yourself

Which failure mode is more dangerous in practice?

Check yourself

An explicit central-difference run of a finely meshed model needs an extremely small time step. Which period governs it?

Worked example

What an unstable run looks like

Given

Central difference integration, Δt = 0.03 s
A model whose highest mode has a period of 0.08 s

Find

Whether the run will be stable, and how a failure would appear

Summary

Constant average acceleration is unconditionally stable; linear acceleration needs Δt/T ≤ 0.551
Central difference is explicit and needs Δt/T < 1/π ≈ 0.318
Instability diverges obviously; an inadequate stable step fails silently
Average acceleration elongates the period by about (π²/3)(Δt/T)² and keeps the amplitude
Choose Δt against the shortest period relevant to the response quantity, not the first
Acceleration converges last, because it is dominated by the higher modes
A step is adequate when halving it stops changing the quantity you will use