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Engineering

Paris' Law Crack Growth: Integrated Cycles and Critical Size

Build a constant-amplitude Paris-law spreadsheet with separate maximum stress and stress range, a logarithmic m = 2 limit, and explicit model boundaries.

Separate the two stress inputs

In this educational constant-geometry model, Delta K = Y * Delta sigma * SQRT(PI() * a). Critical size instead uses maximum stress: ac = (KIc / (Y * sigma_max))^2 / PI(). They coincide only when minimum stress is zero. This example accepts tension-tension cycles with 0 < Delta sigma <= sigma_max.

B2 is initial crack size in metres; B3 is dimensionless Y; B4 and B8 are stress range and maximum stress in MPa; B5 is toughness in MPa sqrt(m); B6 and B7 are paired Paris coefficients C and m. C has units m/cycle divided by (MPa sqrt(m))^m. The illustrative Y = 1.12 is a constant approximation, not a universal surface-crack factor. Use geometry and coefficients appropriate to the specimen and load ratio.

Integrate without a singularity at m = 2

For p = 1 - m/2, integration gives (ac^p - a0^p)/(C*(Y*Delta sigma*sqrt(pi))^m*p). At m = 2 the numerator quotient becomes LN(ac/a0). The formula below uses z = p*LN(ac/a0) and a fourth-order series for (EXP(z)-1)/z near zero, avoiding cancellation. B10 stores unrounded model cycles; B11 stores critical size in metres.

With the starting inputs, critical size is about 35.68 mm and integrated cycles about 252501. Changing only a0 to 8 mm gives about 87091 cycles. Increasing maximum stress reduces critical size even when stress range is unchanged.

=IF(OR(COUNT(B2:B8)<>7,MIN(B2:B8)<=0,B4>B8),"Use positive numeric inputs; stress range must not exceed maximum stress",LET(ac,(B5/(B3*B8))^2/PI(),IF(B2>=ac,"At or beyond model critical size",LET(powr,1-B7/2,logratio,LN(ac/B2),z,powr*logratio,ratio,IF(ABS(z)<0.0001,1+z/2+z^2/6+z^3/24+z^4/120,(EXP(z)-1)/z),B2^powr*logratio*ratio/(B6*(B3*B4*SQRT(PI()))^B7)))))

Critical size output

Display B11 * 1000 for millimetres. If a0 reaches this model boundary, the cycle output reports a status instead of a negative remaining life.

=IF(OR(COUNT(B2:B8)<>7,MIN(B2:B8)<=0,B4>B8),"Use positive numeric inputs; stress range must not exceed maximum stress",(B5/(B3*B8))^2/PI())

Where this calculation stops being a life prediction

Paris law describes the intermediate crack-growth region. It does not capture the near-threshold regime or rapid growth near fracture. Integrating it all the way to ac is a formal extrapolation, not a validated remaining service life or inspection interval. Constant amplitude, constant Y and linear-elastic conditions are assumptions. Real assessments require applicable fracture-control procedures, geometry corrections, load history and qualified material data.

Try the example

Use the original values or change them to test your understanding.

Further reading