EXAMPLE
The 20x2 mm centre-cracked strip of the previous examples is loaded by a stress which varies cyclically between 150 and 250 MPa. The material's Paris index is 4.4, and a stress intensity range of 5.7 MPa√m causes a crack growth rate of 1 mm/Mc.
Integrate the Paris equation ( 5) to determine the number of cycles required for a crack to grow from 0.2 to 2, and from 2 to 20 mm.
Repeat, with a more realistic provision for instability.
The integral of the basic Paris equation ( 5) is ( 5b) with the last term negligible ( ac --> ∞ ).
With normalised crack size limits of 0.01 and 0.1, the integral ( 5b) may be evaluated numerically, noting that w = 10 mm or 0.01 m and that each term in { } is dimensionless :-
{ 1 ∗ ΔN12 / 10 } x { ( 250 -150) √( 0.010 π ) / 5.7 }4.4 | = | ∫0.010.1 ( ( cos πα/2 )/α )2.2 dα | |
{ ( mm/Mc) Mc / mm } { MPa . √m) / MPa√m } | = | 195.7 --> ΔN12 = 13 Mc | |
In the second case, with limits of 0.1 and 1, the integral is | = | 10.25 --> ΔN12 = 0.70 Mc |
A prior R6 analysis of this model, with a peak stress of 250 MPa, has shown that the critical normalised crack size is 0.63, with corresponding elastic configuration factor of 1.35.
So, applying ( 5b) to the first case, the RHS becomes ; 195.7 - ( 1.35 √ 0.63 ) -4.4 x ( 0.1 - 0.01)
Clearly the integral and hence the period is unaffected by the modification - because the period lies wholly within stage II.
For the second case, the upper limit must be reduced to the critical 0.63. The RHS of ( 5b) becomes :
∫0.10.63( ( cos πα/2 )/α ) 2.2 dα - ( 1.35 √0.63 ) -4.4 x ( 0.63 - 0.1) = 10.2 - 0.39 --> ΔN12 = 0.67 Mc
The introduction of the more realistic upper bound has not made much difference to the period - simply because the advance is intrinsically so rapid in the last phase of failure.
In practice, closed form integration is usually impossible; graphical techniques may be used.