1.16.2 Contour integral evaluation: three-dimensional case

Product: Abaqus/Standard  

This example is an illustration of contour integral evaluation in a fully three-dimensional crack configuration. The example provides validation of the method (for linear elastic response), because comparative results are available for this geometry.

Abaqus provides values of the J-integral; stress intensity factor, ; and T-stress as a function of position along a crack front in three-dimensional geometries. Several contours can be used; and, since the integral should be path independent, the scatter in the values obtained with different contours can be used as an indicator of the quality of the results. The domain integral method used to calculate the contour integral in Abaqus generally gives accurate results even with rather coarse models, as is shown in this case. Abaqus offers the evaluation of these parameters for fracture mechanics studies based on either the conventional finite element method or the extended finite element method (XFEM).

Problem description

Results and discussion

Python scripts

Input files

References

Tables

Table 1.16.2–1 J-integral estimates for semi-elliptic crack (× 10–3 N/mm (top); × 10–3 lb/in (bottom)).

Crack Front Location, (deg)ContourAverage Value
123
0.000.80810.82320.82220.8178
4.60994.69644.69074.6656
11.250.78170.78180.78400.7825
4.45974.45994.47274.4641
22.500.87030.88140.88340.8783
4.96475.0285.03975.0108
33.751.03001.04581.04851.0415
5.87615.96625.98175.9413
45.001.22361.22291.22611.2242
6.98016.97626.99476.9836
56.251.38081.38001.38361.3815
7.87717.87257.89337.8809
67.501.44881.47231.47621.4658
8.26498.39918.42138.3617
78.751.57461.57451.57861.5759
8.98278.98189.00538.8989
90.001.55721.57831.58251.5727
8.88329.9.02758.9715

Table 1.16.2–2 Comparison of computed Mode I stress intensity factors (N/mm2–mm1/2 (top); lb/in2–in1/2 (bottom)).

Crack Front Location, (deg)Newman and RajuAverage Value with Conventional Method (calculated from J-integral) Average Value with Conventional Method (calculated directly by Abaqus)
0.0012.9913.6413.23
373.60392.19380.47
11.2513.2313.3413.48
380.50383.62387.66
22.5014.2614.1314.06
410.20406.43404.52
33.7515.6315.3915.31
449.60442.56440.37
45.0016.9016.6816.91
486.20479.82486.35
56.2517.9217.7217.96
515.40509.71516.65
67.5018.6818.2618.16
537.30525.03522.23
78.7519.1218.9318.79
554.40544.40540.48
90.0019.2718.9318.79
554.40543.84546.17


Figures

Figure 1.16.2–1 Semi-elliptic surface crack in a half-space.

Figure 1.16.2–2 Mesh for semi-elliptic surface crack problem with conventional finite element method.

Figure 1.16.2–3 Normal to the crack front is used to define the crack extension direction.

Figure 1.16.2–4 Mesh for semi-elliptic surface crack problem with extended finite element method.

Figure 1.16.2–5 Crack front for the semi-elliptic surface crack problem with extended finite element method.

Figure 1.16.2–6 Semi-elliptic surface crack in a rectangular plate.

Figure 1.16.2–7 Mesh for semi-elliptic surface crack in a rectangular plate with conventional finite element method.

Figure 1.16.2–8 Mesh profile on the crack surface with conventional finite element method.

Figure 1.16.2–9 Stress intensity factors computed for a semi-elliptic crack.

Figure 1.16.2–10 Stress intensity factors computed for a semi-elliptic crack in a rectangular plate.

Figure 1.16.2–11 T-stresses computed for a semi-elliptic crack in a rectangular plate.