Module 8 · Lesson 8.1
Newmark's method
One assumption about how acceleration varies inside a step, one insistence that equilibrium holds at its end — and everything else is algebra.
Why this matters
Every method so far has needed the loading to have a convenient shape. Harmonic gave a closed form. Standard pulses gave closed forms. Convolution handled anything at all, and then turned out to cost n² operations and to collapse the moment the structure yields.
What is actually used, in every analysis package you will ever open, is step-by-step integration — and overwhelmingly Newmark's method. It costs n operations, it handles any loading, and it extends to nonlinear response with only an iteration added.
It is worth knowing what it assumes, because everything that can go wrong with it follows from that assumption.
By the end of this lesson you should be able to
- State the two Newmark parameters and what each controls
- Derive the update equations from the assumed acceleration variation
- Explain why the method is implicit and what that costs
- Identify the two standard parameter choices and when each is used
What you should already know
- The equation of motion (Module 3)
- The idea of a numerical time step
- Integration of a constant and of a linear function
The idea in one sentence
Guess how acceleration varies across one small step, integrate that guess to get velocity and displacement at the end of the step, then force the equation of motion to be satisfied there.
That is the whole method. Everything else is bookkeeping.
The two parameters
Newmark wrote the assumption in a form general enough to cover several earlier methods at once, using two numbers:
ẋᵢ₊₁ = ẋᵢ + [(1 − γ)ẍᵢ + γẍᵢ₊₁]Δt
xᵢ₊₁ = xᵢ + ẋᵢΔt + [(½ − β)ẍᵢ + βẍᵢ₊₁]Δt²
γ decides how the acceleration is weighted between the start and end of the step when computing velocity. At γ = ½ the two ends are weighted equally, which is symmetric and introduces no numerical damping. Above ½ the end of the step is weighted more, which bleeds energy out of the solution.
β decides the same thing for displacement. Two values matter:
- β = ¼ corresponds to assuming the acceleration is CONSTANT across the step, at the average of its two end values. Hence constant average acceleration.
- β = 1/6 corresponds to assuming the acceleration varies LINEARLY across the step. Hence linear acceleration.
Why it is implicit
Notice that ẍᵢ₊₁ appears on the right-hand side of both update equations. The displacement at the end of the step depends on the acceleration at the end of the step, which depends on the displacement at the end of the step.
So the equations cannot simply be evaluated; they must be SOLVED. For a linear SDOF system that means one division. For a linear MDOF system it means one matrix factorisation, done once and reused. For a nonlinear system it means an iteration at every step.
That is the price of the method, and it buys unconditional stability, which is worth a great deal.
From first principles
The Newmark update equations
We want to show: Derive the recurrence that advances the solution one step, starting from the assumed acceleration variation and the requirement that equilibrium holds at the end of the step.
Over a short enough interval, the acceleration cannot do anything very complicated. Assume it holds at some weighted average of its values at the two ends. Integrating that assumption once gives the velocity at the end of the step, and integrating it twice gives the displacement — both expressed in terms of the unknown end-of-step acceleration. Then demand that the equation of motion is satisfied at the end of the step, and there are exactly enough equations to find it.
Try it
Newmark's method against an answer we already know
Free vibration has an exact solution, so the numerical error can be plotted rather than guessed at.
Scheme
Problem
The only parameter that decides whether the answer means anything.
- Newmark, average acceleration
- Exact solution
Show the numbers behind this plot
- Time step Δt
- 0.1 s
- Steps in the analysis
- 81
- Stability limit
- unconditionally stable
- Worst error
- 53.11%
- Period elongation
- 3.32%
- Amplitude decay per cycle
- 10.94%
Halving the step doubles the cost.
Should be zero for γ = 1/2.
53.1% out. The error is almost entirely PHASE — the numerical solution has the right amplitude at the wrong time. That is the signature of too coarse a step in an average-acceleration scheme.
Things worth doing here
- Choose central difference and raise Δt/Tn past 0.318. It does not become slightly wrong; it diverges to infinity within a few steps.
- Choose average acceleration and try Δt/Tn = 0.5. It stays bounded — unconditional stability means the answer will not explode — but a bounded answer is not a correct one, and the error is enormous.
- Compare linear acceleration and average acceleration at the same modest step. Linear acceleration is the more accurate, which is why it is worth having despite the stability limit.
- With γ = 0.7, watch the amplitude decay even though ζ is set to zero. That damping is entirely numerical and does not exist in the structure.
What this shows: Newmark makes one assumption — how acceleration varies within a step — and then enforces equilibrium at the end of the step. Everything about accuracy and stability follows from that assumption and from the size of the step relative to the period.
Two parameter choices, and why both exist
Constant average acceleration, γ = ½ and β = ¼. Unconditionally stable for a linear system — the answer stays bounded whatever the step. No numerical damping: the amplitude is preserved exactly. It does elongate the period, by about (π²/3)(Δt/T)², which is 3.3% at Δt/T = 0.1.
This is the default in essentially every commercial package, and it is the default in this course.
Linear acceleration, γ = ½ and β = 1/6. More accurate than average acceleration at the same step. Conditionally stable, requiring Δt/T ≤ 0.551.
It is worth having because it is more accurate, and it is worth teaching because the stability limit can be crossed in a lab and watched.
γ greater than ½. Deliberately introduces numerical damping, which bleeds amplitude out of the solution. Used in large models to suppress spurious high-frequency response that comes from the discretisation rather than from the structure.
The cost is that it damps the modes you wanted as well, and the effect grows with Δt/T — so it hits high modes hardest, which is usually the intention, but the fact that it is happening must be a decision rather than an accident.
Stability is not accuracy
These are different properties and confusing them is the commonest misunderstanding in this subject.
Stability asks: does the solution stay bounded? An unstable scheme does not degrade gracefully — it diverges to absurd numbers within a handful of steps, and there is no possibility of mistaking the output for an answer.
Accuracy asks: is the bounded answer close to the truth? An unconditionally stable scheme at a coarse step produces a smooth, plausible, entirely bounded response that is badly out of phase with reality.
Unconditional stability is a guarantee that the answer will not explode. It is not a guarantee that the answer is right — and it is the more dangerous of the two failure modes precisely because nothing announces it.
Worked example
Two steps of Newmark by hand
Given
- m = 1 000 kg, k = 100 kN/m, c = 0 (undamped, to keep the arithmetic visible)
- A constant force of 5 kN applied from t = 0, structure at rest
- Constant average acceleration: γ = ½, β = ¼
- Δt = 0.05 s
Find
The displacement after two steps, and comparison with the exact answer.
Assumptions
- Undamped, so the exact solution is available for comparison
Predict first
An analysis using constant average acceleration is run at Δt = T/2 — a very coarse step. What happens?
Practice
A structure has m = 2 000 kg, k = 500 kN/m and no damping. Using γ = ½, β = ¼ and Δt = 0.02 s, what is the effective stiffness k̂, in MN/m?
Practice
A structure of mass 4 000 kg and stiffness 900 kN/m is at rest when a force of 18 kN is suddenly applied. What is the initial acceleration, in m/s²?
Check yourself
What does the Newmark parameter β = ¼ correspond to physically?
Worked example
Choosing a time step for the right reason
Given
- A structure with a fundamental period of 0.45 s
- The highest mode of interest has a period of 0.08 s
- Newmark average acceleration is to be used
Find
A defensible time step
Summary
- Assume how acceleration varies in a step; integrate; enforce equilibrium at the END
- γ weights velocity; β weights displacement; γ = ½ gives no numerical damping
- β = ¼ is constant average acceleration; β = 1/6 is linear acceleration
- The method is implicit — k̂xᵢ₊₁ = F̂ᵢ₊₁, so each step is a solve
- k̂ = k + γc/(βΔt) + m/(βΔt²), computed once for a linear system
- Cost is O(n) against convolution's O(n²)
- ẍ₀ comes from equilibrium at t = 0 and is not zero when a force is applied
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