Computational Engineering / Nonlinear and explicit analysis

Newton–Raphson and the residual

Guess, measure the out-of-balance, divide by the tangent, correct, repeat — and what each of those four words is doing.

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Why this matters

Nonlinear solvers are usually presented as a formula and an animation. Both miss the thing worth understanding, which is that the method is a measurement followed by a correction, and that the quantity being measured — the residual — is a physical force with a physical meaning.

Once the residual is clear, everything else follows: why convergence is a choice, why the tolerance matters, why the method sometimes diverges, and why the modified version is worth using despite taking longer.

By the end of this lesson you should be able to

Define the residual and say what it physically is
Follow one Newton–Raphson iteration through its four steps
Explain quadratic convergence and when it does not apply
Say when the modified method is the better choice
Explain why load increments help even though the final answer is the same

Residual and correction

For a static problem, R = F_applied − F_internal(u) measures the unbalanced generalised forces at the current iterate. Translational equations have force units and rotational equations have moment units. Newton’s method solves K_T Δu = R, updates u and repeats. For the scalar springs here, Δu = R/k_T.

Use stated, appropriately scaled convergence criteria. A small residual establishes approximate equilibrium of the model, not the accuracy of its assumptions or the stability of that equilibrium.

Convergence

Full Newton has local quadratic convergence near a regular root when the residual is sufficiently smooth and the consistent tangent is nonsingular. A poor starting point, a near-singular tangent, nonsmooth contact or an inconsistent tangent can slow or prevent convergence. Negative tangent stiffness alone does not prevent Newton from finding a root, although the equilibrium may be unstable under the specified loading control.

Modified Newton holds a tangent fixed over several iterations, commonly within a load step. It can reduce factorisation cost but often needs more iterations. Compare performance and robustness for the actual model.

Increments

Smaller increments can improve the starting estimate. For path-dependent material or contact models, step size can also affect integration accuracy and the recorded path, so demonstrate step-size convergence. Standard monotonic load control is poorly suited to passing a load limit point; displacement or arc-length control may be needed.

Newton–Raphson iteration

R_i = F_{\text{applied}} - F_{\text{internal}}(u_i), \qquad \Delta u_i = \frac{R_i}{k_T(u_i)}, \qquad u_{i+1} = u_i + \Delta u_i

What it calculates: The next estimate of the displacement that balances the applied load

R: Residual — the out-of-balance force (kN)
F_internal: Force the structure develops at the current displacement (kN)
k_T: Tangent stiffness at the current displacement (kN/m)
Δu: Correction applied this iteration (m)

This assumes

The tangent is a usable local predictor — false near a limit point
The tangent is non-zero; a zero tangent makes the correction infinite
Convergence is declared when |R| falls below a tolerance you chose

In plain terms: Every term is a physical quantity. The residual is a force you could measure; the tangent is a stiffness you could measure; the correction is the displacement that force would cause at that stiffness. Nothing in the method is abstract, which is why watching one iteration is worth more than reading three derivations of it.

Try it

Newton–Raphson explorer

Three nonlinear springs. Watch the residual, the tangent and the correction at every iteration.

System

Applied loadkN

Method

Modified holds the first tangent throughout: more iterations, each much cheaper.

Tension stiffening. The tangent only ever increases, so Newton–Raphson converges from any starting point and load control never fails.

Iteration trace. R is the residual; k_T the tangent used; Δu the correction.
#	u (m)	R (kN)	k_T (kN/m)	Δu (m)
1	0.0000	500.00	100.0	5.0000
2	5.0000	-5000.00	3100.0	-1.6129
3	3.3871	-1393.04	1476.7	-0.9434
4	2.4437	-328.13	816.6	-0.4018
5	2.0419	-44.75	600.3	-0.0745
6	1.9674	-1.34	564.5	-0.0024
7	1.9650	-0.00	563.4	-0.0000

Result: converged
Iterations: 7
Displacement: 1.9650 m
Check: internal force: 500.00 kN

Things worth trying

Hardening at 500 kN: the first correction is the LINEAR answer, and it overshoots by nearly three times. That is what nonlinearity means in practice.
Switch to modified Newton and every row shows the same tangent. It takes many more iterations and each is far cheaper.
Yielding at 200 kN: above the plateau of 150 kN there is no displacement that balances the load, and no number of iterations finds one.
Snap-through at 25 kN: above the limit point of 24.1 kN, load control either fails or lands on the far branch — a completely different state of the structure.

What this shows: The residual is a force in kilonewtons — the load the structure is not yet carrying — and convergence is a tolerance you chose.

Worked example

Three iterations on a hardening spring

Given

A cable-like spring: F = 100u + 40u³, so k_T = 100 + 120u²
Applied load 500 kN
Starting from u = 0
Tolerance on the residual: 10⁻⁶ kN

Find

The first three iterations, and how the residual falls

Predict first

A nonlinear analysis now needs 40 iterations per step. Which structural possibility should be investigated alongside the solver settings?

Practice

A nonlinear spring carries F = 150(1 − e^(−u/1.2)). At u = 0 the applied load is 100 kN. What is the residual at the first iteration, in kN?

Practice

The same spring has tangent stiffness k_T = 125e^(−u/1.2). At u = 0, what correction does the first Newton–Raphson iteration apply under a 100 kN load, in metres?

Check yourself

What is the residual in a Newton–Raphson iteration?

Check yourself

A nonlinear analysis converges in every increment and reports a final answer. What has that established?

Summary

The residual is the load the structure is not yet carrying, in kilonewtons
Convergence is a tolerance you chose, not an event that happens
The correction is R divided by the current tangent — the method's only approximation
Near the answer the error roughly squares each pass; far from it, nothing is guaranteed
The modified method trades more iterations for cheaper ones
Check increment sensitivity where path integration matters; use suitable control near limit points