Crack closure modeling

The Constraint Factor You Inherited

Newman’s crack opening equation contains a single user input, α, that most analysts take from a material file default without ever deciding on it. At the negative stress ratios random vibration produces, it is worth a factor of two to four in life — and the sensitivity is largest precisely where vibration problems live.

A previous post worked through the choice between full-range and clipped stress intensity range at R = −1, and put the difference at a factor of $2^n$ — eight to eleven in life for common structural metals. The conclusion was that clipping is a relaxation requiring justification, not a conservatism.

The obvious next step is to model closure explicitly rather than clipping. NASGRO does that natively, through Newman’s crack opening equation. What is less obvious is that doing so does not eliminate the judgment call. It relocates it.

Inside the closure model sits a parameter called the constraint factor. It has a defensible physical meaning, it spans a factor of three in the values people actually use, and in most decks it arrives as whatever the material file happened to contain. This post is about what it does and how to pick it.

The equation

Newman’s crack opening function gives the ratio of opening stress to maximum stress:

The two stresses in that ratio come from different places, and keeping them straight is the whole point of the model.

$S_{max}$ is applied. It is the peak remote stress in the cycle — whatever the load spectrum delivers. You know it before any fracture mechanics happens.

$S_{op}$ is a response. It is the applied stress level at which the crack faces finally separate at the tip, and it is a property of the crack’s own history rather than of the loading. The plastic wake left behind by the advancing tip — material stretched at the tip, then unloaded and abandoned along the flanks — props the faces apart. The crack does not open when the load turns tensile. It opens when the applied stress becomes large enough to overcome that propping.

So $S_{op}$ is not a load you apply. It is the load level the wake demands before the tip can feel anything, and it depends on the stress ratio, the applied stress level and the through-thickness constraint. Computing it is what the equation below is for.

Everything between $S_{min}$ and $S_{op}$ is spent prying the faces apart rather than driving the tip. Only the portion above $S_{op}$ produces a crack tip singularity, which is why the effective range starts there instead of at $S_{min}$.

Expressing the result as the dimensionless ratio $f$ rather than as $S_{op}$ itself is what lets one function serve every stress level. Because $f$ is a fraction of the peak, the same equation applies whether the structure is cycling at 5 ksi or 50.

The function is evaluated piecewise in the stress ratio R:

with coefficients

Here $\alpha$ is the constraint factor, $\sigma_0$ the flow stress — taken as the average of yield and ultimate — and $S_{max}/\sigma_0$ the ratio of maximum applied stress to flow stress. That last ratio appears because the size of the plastic wake scales with how hard the material is being worked: a structure cycling near its flow stress lays down a larger wake, and a larger wake props the faces apart at a higher load.

The effective range follows as

and the growth law is written against $\Delta K_{eff}$ rather than the nominal range.

What α means physically

The constraint factor represents the through-thickness stress state at the crack tip, expressed as the ratio of the effective flow stress under constraint to the uniaxial flow stress.

• α ≈ 1 — plane stress. Thin section, plastic zone large relative to thickness, material free to contract through the thickness. Large plastic wake, more closure, higher $f$.
• α ≈ 3 — plane strain. Thick section, plastic zone small relative to thickness, through-thickness contraction suppressed. Smaller wake, less closure, lower $f$.

Real cracks are usually somewhere between, and the state can shift along the crack front — plane strain at the interior, plane stress at the free surfaces — and shift again as the crack grows and the plastic zone enlarges. Newman is explicit that the constraint factor in a two-dimensional model is averaging three-dimensional plastic stress states at the crack front and in the plastic wake. It is a lumped parameter, not a measured property.

Worth knowing that 1 and 3 are the mathematical bounds, not the working range. In his own analyses of thin-sheet 2024-T3, Newman used α = 2 at low growth rates and 1.15 at high rates, with 1.73 for rates below 1×10^-7 m/cycle. Across that realistic 1.15 to 2.0 band the life ratio is about 1.8 — smaller than the full bracket, but still large enough to matter.

Why negative R is the clean case

Look again at the piecewise form. For $R < 0$ the cubic branch never activates; only $A_0$ and $A_1$ are in play. The equation collapses to a straight line in R with two coefficients, both carrying α.

This matters for vibration work specifically. A zero-mean random response puts most of its damaging cycles at negative R, and a fully reversed response sits at R = −1. So the stress ratios that dominate a random vibration crack growth analysis are exactly the ones where the α dependence is simplest to trace — and, as it happens, strongest.

At R = −1 the expression reduces to

and if the applied stress is modest relative to flow stress, the cosine term sits near unity and $A_1$ is small. Then $f \approx A_0$, and the whole answer rides on the polynomial $\left(0.825 – 0.34\alpha + 0.05\alpha^2\right)$.

Working the numbers

Newman’s own Figure 3 in the 1999 technical memorandum plots opening-stress ratio against R for three constraint levels at $S_{max}/\sigma_0 = 0.2$. Evaluating the equation above at R = −1 reproduces it:

With the formulation confirmed, sweep α at that same stress level. The effective fraction at R = −1 is $(1-f)/2$, and relative life is referenced to plane strain:

Across the full plane-stress to plane-strain range the life ratio is 2.8 at this stress level. That is the same order as the full-range versus clipped decision from the previous post, which was worth 8 — and which that post argued deserves a written basis. α rarely gets one.

Sensitivity grows as applied stress falls

The α spread is not fixed. It depends on how hard the structure is working relative to its flow stress, and it moves in the direction that matters for vibration:

At low applied stress the cosine term approaches unity and $A_1$ vanishes, leaving the polynomial $\left(0.825 – 0.34\alpha + 0.05\alpha^2\right)$ carrying the α dependence undiluted. As applied stress rises, the $1/\alpha$ exponent on the cosine and the growing $A_1$ both compress the spread.

The limits of f, and what they mean

Before trusting an opening function it is worth knowing what values it can and cannot take. Three cases matter, and only one of them is a trap.

f cannot exceed 1

The coefficient constraint $A_2 = 1 – A_0 – A_1 – A_3$ is not arbitrary. Substituting it at R = 1 collapses the polynomial to exactly 1, so $f(1) = 1$ by construction. A numerical sweep across every α, every stress level and R from 0 to 1 returns a maximum of precisely 1.0 — the equation cannot overshoot.

That is a useful diagnostic. If your code reports $f > 1$, the problem is upstream of the closure model. Look for a flow stress entered in the wrong units, a negative or zero constraint factor, or an applied stress exceeding $\sigma_0$.

f = 1 is arrest, and only occurs at R = 1

Setting $f = 1$ makes $S_{op} = S_{max}$: the crack opens at the instant of peak load and at no other point in the cycle. The effective range

is indeterminate in form but zero in substance, since $\Delta K$ is itself zero at R = 1. There is no cycling and no growth. It is a degenerate corner of the formulation rather than a condition you encounter in a spectrum.

The floor at f = R is the closure-free condition

On the $R \ge 0$ branch the $\max(R, \ldots)$ operator enforces $S_{op} \ge S_{min}$. An opening stress below the minimum load would be meaningless — the crack would never close, so there would be nothing for an opening stress to describe.

When that floor binds, the arithmetic is tidy:

The model degrades gracefully to no closure at all rather than producing a nonsense value. At high stress ratios, where the crack stays open throughout, this is the branch that governs.

f can go negative, and this one is a trap

The $R < 0$ branch has no floor. It is a straight line, $f = A_0 + A_1 R$, and at high applied stress the $A_1 R$ term can outrun $A_0$:

A negative $f$ means $S_{op} < 0$: the crack remains open below zero load, so part of the compressive excursion is effective. This is the only place in the entire formulation where compression can drive crack growth.

For most vibration work this is moot. At $S_{max}/\sigma_0$ below about 0.6 the sign change never arrives and the question is academic. But high-stress, thin-section problems can reach it, and nothing in the output announces the crossing.

Where α lives in NASGRO

In NASGRO the constraint factor is an input to the crack growth equation, sitting alongside the Paris-region constants, the threshold parameters, and the fracture toughness. It is stored in the material file.

That last point is the one worth dwelling on. α arrives with the material, but it is a property of the geometry. A 7075-T6 material file carries a constraint factor that was appropriate for whatever configuration the file was assembled against. Your lug is not that configuration.

Choosing it

The physical basis is the size of the crack tip plastic zone relative to the section thickness. The plane strain plastic zone radius is approximately

Compare $2r_p$ against thickness $B$:

• $B \gg 2r_p$ — plastic zone small relative to thickness. Constraint is high, plane strain dominates, α toward 3.
• $B \lesssim 2r_p$ — plastic zone comparable to or larger than thickness. Constraint relaxes, plane stress dominates, α toward 1.

The ASTM E399 plane strain validity criterion, $B \ge 2.5(K/\sigma_{ys})^2$, is the same comparison written as a threshold. If your section is well below that, you are not in plane strain and an α of 3 is not defensible.

Two complications worth acknowledging rather than papering over. First, $K$ grows as the crack grows, so the plastic zone grows with it, and a crack can start under plane strain constraint and finish under plane stress. A single α is an average over that history.

Second, and more usefully, there is a physical marker for the transition. Constraint loss accompanies the change from flat to slant crack growth, and Newman models it as α decreasing with increasing growth rate through that transition — 2 down to 1.15 across a constraint-loss regime spanning roughly 1×10^-7 to 2×10^-6 m/cycle for thin-sheet 2024. If your material file supports variable constraint, that is a better representation than any fixed value, and the fracture surface itself tells you where the transition occurred.

The double-credit problem

This deserves its own section because it silently invalidates otherwise careful work, and it is the same trap flagged in the previous post in a different guise.

The $da/dN$ data in the material file was itself reduced using a closure model. Test data taken at several stress ratios is collapsed onto a single effective-range curve by applying an opening function — with some assumed constraint factor. That assumed value is part of the data reduction, not an independent choice.

If you then run the analysis with a different α than the file was reduced under, the closure correction is applied inconsistently: once during reduction with one value, once during analysis with another. The result is not wrong in an obvious way. It produces a number, the number looks reasonable, and nothing in the output indicates a problem.

Before changing α, establish what the material file was built against. In NASGRO’s supplied material database this is documented; in a locally assembled file it may not be. If nobody can say, that is itself the finding, and it bounds how much confidence the analysis can carry.

What to ask your deck

1. What is α set to right now? Not what it should be — what it is. Find the number in the material file before forming an opinion.
2. Was it chosen for this geometry, or inherited? If nobody on the program can say who set it and why, treat it as unjustified.
3. What is $S_{max}/\sigma_0$? This governs how much α sensitivity exists in your problem. Low applied stress means more spread, not less — a high-cycle vibration case is the worst place to leave α unexamined.
4. Does your implementation floor f at zero for negative R? Only matters above S_max/σ₀ ≈ 0.6, but there it is worth about 1.7× in growth rate and nothing in the output flags it.
5. What α was the material data reduced under? Changing the analysis value away from the reduction value applies the closure correction twice with different constants.
6. What does the answer look like at α = 1 and α = 3? Two runs bracket the physically plausible range and cost almost nothing. If the spread is large, the analysis is not converged on an input nobody decided.