Flaw size

Insurance in pressure boundary integrity of NPP unit is strongly influenced by technical capabilities and efficiency of metal examination system. Ordinary ultrasonic examination tools and procedures have limitations in flaw sizing and positioning. The problems arise for welds and repair zones of welds made by filler materials of austenitic type. 

AUGUR-family expert NDE tools, developed in Russia, showed good capabilities of flaw sizing and characterization during on-site inspections at numerous Russian NPPs. 

RDIPE, VNIIAES. Procedure for Allowable Flaw Sizes Evaluation in Metal of Equipment and Piping during NPP Operation, M-02-91, (in Russian). 

Hanke, R. B5bel, F. Determination of material flaw size by intensity evaluation of polychromatic X-ray transmission. NDT-I-E International, Vol. 25(2), 1992, pp. 87-93. 

Fracture markings can be used to locate the failure origin, which is the discontinuity or flaw that caused the appHed stress to be amplified locally. Once the failure origin has been located, the failure stress can be estimated using the flaw size and equation 6, or the distances to the boundaries of the mirror, mist, and hackle (whichever is most evident) and the foUowing relation (63)... 

Fig. 5c. Relation of T-curve to crack growth and strength (c) fracture strength as a function of crack size showing region of flaw size tolerance.

Figure 16.6 shows the general yield and fast fracture loci for a pressure-vessel steel and an aluminium alloy. The critical flaw size in the steel is =9 mm that in the aluminium alloy is =1 mm. It is easy to detect flaws of size 9 mm by ultrasonic testing, and pressure-vessel steels can thus be accurately tested non-destructively for safety -vessels with cracks larger than 9 mm would not be passed for service. Raws of 1 mm size cannot be measured so easily or accurately, and thus aluminium is less safe to use. 

If the critical flaw size for fast fracture is less than the wall thickness (t) of the vessel, then fast fracture can occur with no warning. But suppose the critical size (2a jt) is... 

Fig. 18.1. If small samples are cut from a large block of o brittle ceramic, they will show o dispersion of strengths because of the dispersion of flaw sizes. The average strength of the small samples is greater than that of the large sample.

Most ceramics (as we have seen) contain flaws holes and cracks left by processing, cracks caused by thermal stress, corrosion or abrasion. Even if there are no cracks to start with, differences in elastic moduli between phases will nucleate cracks on loading. And most of these flaws have a size which is roughly that of the powder particles from which the ceramic was made. If the flaw size can be reduced, or if samples containing abnormally large flaws can be detected and rejected, the mean strength of the ceramic component is increased. 

The decreasing value of K with decreasing flaw size for fracture of IG-11 graphite suggests that the near crack tip failure criterion would be better expressed by combining the near crack tip stress intensification estimated by K with the far field applied stress, Sfraa(Eq. 28). Using simple superposition of the stress perpendicular... 

The second approach to fracture is different in that it treats the material as a continuum rather than as an assembly of molecules. In this case it is recognised that failure initiates at microscopic defects and the strength predictions are then made on the basis of the stress system and the energy release processes around developing cracks. From the measured strength values it is possible to estimate the size of the inherent flaws which would have caused failure at this stress level. In some cases the flaw size prediction is unrealistically large but in many cases the predicted value agrees well with the size of the defects observed, or suspected to exist in the material. 

Fig. 2.69 Effect of varying stress field on flaw size for ductile/brittle transition (AT = constant)...

Example 2.22 A certain grade of acrylic has a Kc value of 1.6 MN and the fatigue crack growth data as shown in Fig. 2.77. If a moulding in this material is subjected to a stress cycle which varies from 0 to 15 MN/m, estimate the maximum internal flaw size which can be tolerated if the fatigue endurance is to be at least 1(P cycles. 

Solution The first step is to calculate the critical flaw size which will cause brittle failure to occur in one cycle. This may be obtained from equation (2.100) assuming Y =. ... 

In contrast, the burst pressure or the required fracture toughness of steel or weld can be estimated when the maximum flaw size or the desired MAOP are given, respectively. The pipeline parameters used for the calculation are shown in Table 10.2. The flaws for the calculation are assumed to be semiellipse with the depth-to-length ratio of 1 10. 

Pipeline with the higher fracture toughness fCIH can tolerate larger flaw size (depth). 

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