Solution
This question is designed to demonstrate why one uses the semi-minor axis to assess the criticality of surface defects and to indicate that cracks tend towards a stable aspect ratio (2c/a). It requires some thought in interpreting the table of geometry correction factors but is straightforward and should
take about 10 minutes to complete.
A thick leaf spring experiences unidirectional bending and develops a small semi-elliptic crack on the tensile surface. The crack has an initial aspect ratio of 0.2 when found. The plane of the crack is perpendicular to the direction of applied bending stress. As the spring undergoes repeated deflection, the crack grows by fatigue.
Using the K solution and geometry correction factors for tensile loading, given in the table below, determine whether the ellipticity ratio of the crack will increase or decrease. The figure below shows the crack geometry.
|
|
a/c |
Phi |
Y |
|
|
|
a/B |
|
|
|
0.2 |
0.4 |
0.6 |
0.8 |
1.051 |
0.2 |
0o
45o
90o |
0.617
0.990
1.173 |
0.724
1.122
1.359 |
0.899
1.384
1.642 |
1.190
1.657
1.851 |
1.151 |
0.4 |
0o
45o
90o |
0.767
0.998
1.138 |
0.896
1.075
1.225 |
1.080
1.247
1.370 |
1.318
1.374
1.447 |
1.277 |
0.6 |
0o
45o
90o |
0.916
1.024
1.110 |
1.015
1.062
1.145 |
1.172
1.182
1.230 |
1.353
1.243
1.264 |
1.571 |
1.0 |
0o
45o
90o |
1.174
1.067
1.049 |
1.229
1.104
1.062 |
1.355
1.181
1.107 |
1.464
1.193
1.112 |
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Theory
Answer: The crack
shape tends towards a stable value of a/c of around 0.8 for a/B ratios < 0.8.
This problem can be solved by inspection of the Y values in the table above. Essentially, a crack will grow fastest in fatigue where the value of the stress intensity factor is highest (this is easily visualised using a fatigue crack growth rate plot of log da/dN versus log (stress intensity factor range)). Although there are effects on K of both crack shape (through ) and of finite geometry (through Y), for all a/c ratios < 0.6 and for 0.2 < a/B < 0.6, the Y value is highest for phi = 90o. As a/B becomes closer to 1.0, this changes and the highest Y value corresponds to phi = 0o - this can be seen in the column where a/c = 0.6 and a/b = 0.8, and in all columns where a/c = 1.0. This occurs because the crack front is approaching a free surface (the back face of the specimen) with a corresponding loss of constraint. Thus the
interpretation of the Y value data is that the crack will tend towards a stable aspect ratio, equivalent to a/c of around 0.8 (by interpolating between 0.6 and 1.0).
That this occurs in practise has been shown by a number of empirical studies - see for example:
PM Scott and TW Thorpe (1981), A critical review of crack tip stress intensity factors for semi-elliptic cracks, Fatigue of Engineering Materials and Structures (now Fatigue and Fracture of Engineering Materials and Structures), Vol. 4 No. 4 pp. 291-309.
Note that the situation for cracks in bending is slightly different to that in tension - we have assumed the tension solution here because the crack is small (initially) compared with the spring thickness, and is therefore growing in a reasonably uniform stress field. As a/B tends towards 1.0, this is clearly untrue. However, the question illustrates why the critical crack length for
calculating K values is taken as the semi-minor axis.
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