Module 8 · Lesson 8.2
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.
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 highest period carrying meaningful mass, 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.
- Relative error
- 5% line
| 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/Tn = 0.25
- Within 1% at
- Δt/Tn = 0.02
- Cost of the 1% answer
- 601 steps
Available here only because the forcing is harmonic.
12.3× the cost of the 5% answer
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 highest period carrying meaningful mass, 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
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