Paris Law Cycles to Failure

Also known as integrated Paris law · remaining life crack growth · cycles from initial to final crack · crack propagation life · damage tolerance life · inspection interval crack growth · residual life fatigue crack · Paris integration

N=af1m/2ai1m/2(1m2)C(YΔσπ)mN = \dfrac{a_f^{\,1 - m/2} - a_i^{\,1 - m/2}}{\left( 1 - \tfrac{m}{2} \right) C \left( Y \, \Delta\sigma \sqrt{\pi} \right)^{m}}

Enter your known values, leave one input blank, and solves for the missing one. Try different units for next level excitement!

Learning zone

Paris law gives the growth in one cycle. Separate the variables and integrate it from the flaw you found to the flaw that breaks the part, and you have the number damage-tolerant engineering actually turns on: how many more cycles there are.

The integration is straightforward if two things are held constant — the geometry factor YY and the stress range Δσ\Delta\sigma. Substituting ΔK=YΔσπa\Delta K = Y\Delta\sigma\sqrt{\pi a} into da/dN=C(ΔK)mda/dN = C(\Delta K)^m gives da/dN=C(YΔσπ)mam/2da/dN = C(Y\Delta\sigma\sqrt{\pi})^m a^{m/2}. Move the aa terms to one side, integrate both sides, and the result is a power law in the crack sizes.

Except at m=2m = 2, where it is not. Write q=1m/2q = 1 - m/2 and the closed form is (afqaiq)/(qCSm)(a_f^{\,q} - a_i^{\,q})/(qC S^m) with S=YΔσπS = Y\Delta\sigma\sqrt{\pi}. At m=2m = 2 exactly, qq is zero: the power rule for the integral divides by zero, and the antiderivative of 1/a1/a is a logarithm instead. The correct form there is N=ln(af/ai)/(CS2)N = \ln(a_f/a_i)/(C S^2), and physically it says the crack grows exponentially with cycles — each fixed number of cycles multiplies the crack length by the same factor. This is a genuinely different formula, not a limit that can be taken numerically, and a spreadsheet that plugs m=2m = 2 into the power form returns a division by zero or a silent nonsense. Both branches are implemented on this page and both are inverted in closed form, so nothing about m=2m = 2 is guarded away. It is not an exotic case: 2 sits in the middle of the range real steels occupy.

Almost all of the life is spent while the crack is small. Because the rate goes as ΔKm\Delta K^m and ΔK\Delta K goes as a\sqrt{a}, growth accelerates continuously and the last millimetres pass in a blink. A crack that takes 200 000 cycles to go from 1 mm to 20 mm may spend 150 000 of them getting to 3 mm. Two consequences follow, and they are the practical content of the whole page. First, the answer is far more sensitive to aia_i than to afa_f — halving the initial flaw size can double the life, while doubling the final crack size barely changes it. Second, better inspection buys more life than better material. The detection limit of your NDT method is therefore one of the most important numbers in the calculation, and it is the one most often assumed rather than measured. Probability-of-detection curves exist precisely because "we would have found a 2 mm crack" is a claim that needs evidence.

Where afa_f comes from is worth stating. Usually it is the critical crack size from the stress-intensity page with KICK_{IC} entered — the crack that fractures at the peak stress. Sometimes it is the wall thickness, when the design intent is leak-before-break and a through-wall crack that weeps is an acceptable, detectable end state rather than a rupture. Occasionally it is a serviceability limit that has nothing to do with fracture. Because the life is so insensitive to afa_f, being approximate about it costs little; being approximate about aia_i costs a great deal.

What the integral assumes, and how each assumption bites. YY is held constant, and real YY climbs as the crack approaches a boundary, so a real crack accelerates faster at the end than this predicts — unconservative, though only over the part of the life that is brief anyway. The loading is constant amplitude; a real spectrum needs cycle-by-cycle integration, and overload retardation can extend the true life well beyond a naive sum. Region II is assumed throughout, so if ΔK\Delta K at aia_i is below the threshold the crack is dormant and the calculated life is fiction in the safe direction, while if KmaxK_{max} at afa_f is near KICK_{IC} the real life is SHORTER than the number. And the environment is assumed benign; corrosion fatigue in seawater or sour gas can multiply growth rates several-fold and introduce a time dependence the cycle count cannot see.

This is why an inspection interval is set at a fraction of the calculated life — typically a half or a third — rather than at the whole of it. The interval is chosen so that a crack just below the detection limit at one inspection is still comfortably below critical at the next, giving two independent chances to find it before it matters. That structure, two opportunities and a margin, is what "damage tolerant" means in practice.

Paris Law Cycles to Failure
N=af1m/2ai1m/2(1m2)C(YΔσπ)mN = \dfrac{a_f^{\,1 - m/2} - a_i^{\,1 - m/2}}{\left( 1 - \tfrac{m}{2} \right) C \left( Y \, \Delta\sigma \sqrt{\pi} \right)^{m}}
ΔσNaiaf
Where
  • NN= Cycles to failure
  • aia_i= Initial crack size (mm)
  • afa_f= Final crack size (mm)
  • Δσ\Delta\sigma= Stress range (MPa)
  • YY= Geometry factor
  • CC= Paris coefficient C (m/cycle, ΔK in MPa·√m) ((m/cycle)/(MPa·√m)^m)
  • mm= Paris exponent m
Missing one of these? Work it out first, then come back