R2/ R1 1.6 2.0 2.4 3.2 6.5 6 5.5 5 4.5 4 3.5 3 2.5 2 1.5 1 0 0 .1 0 .2 0 .3 0.4 0 .5 0.6 0.7 0 .8 J41 L a/ c * * * * * * * * * 1800 1600 1400 1200 1000 800 600 1 2 3 4 5 6 7 8 9 10 Crack Length a/m m Stress intensity factor K/Mpa· * Case 1 Case 2 Case 3 Figure 8. The configuration factor J41L curves Figure 9. Comparison of the stress intensity factors 3. Calculation of the SIF equations equations The geometric parameters of the lug are wrote to the APDL program text file and executed by ANSYS to solve the SIFs automatically and efficiently. The case of the straight attachment lug subjected to axial pin-load is taken as an example for illustrating the approach. Figure 10 shows the parameterized geometry of the lug. a 2R1 A A t P A-A 2R2 β Figure 10. Geometry of the straight attachment lug The influence law of the dimensionless crack length ( a/R1), the ratio of outer radius to inside radius ( R2 /R1) and the inside radius ( R1) on the SIF are studied, and then, the equation (4) is modified to fit the SIF equation for the straight attachment lug with a single through-the-thickness crack. Table 1 shows the solving idea. K, K1, K2, K3 stand for the SIFs calculated by equation (4) and represent the values that need to be modified by the influence law of a/R1, R2 /R1 and R1 respectively. K′ stands for the SIF calculated by ANSYS as the reference exact value for modifying K1, K2, K3. Table 1. Solving the SIF equation of the straight lug subjected to axial pin-load Variate Invariants K K′ ( K′)i /Ki Calculating process ( a/R1)1 R2 /R1, R1 ( K1)1 ( K′)1 ( K′)1/( K1)1 ( a/ R1)i is set as the variate, while ( K′)i/( K1)i as the dependent variable, function g( a/ R1) can be fitted as below. K2=K1· g( a/ R1) (10) ( a/ R1)n R2/ R1, R1 ( K1)n ( K′)n ( K′)n/( K1)n ( R2/ R1)1 a/ R1, R1 ( K2)1 ( K′)1 ( K′)1/( K2)1 ( R2/ R1)i is set as the variate, while ( K′)i/( K2)i as the dependent variable, function f( R2/ R1) can be fitted as below. K3=K2· f( R2/ R1)= K1· g( a/ R1) f( R2/ R1) (11) ( R2/ R1)n a/ R1, R1 ( K2)n ( K′)n ( K′)n/( K2)n ( R1)1 a/ R1, R2/ R1 ( K3)1 ( K′)1 ( K′)1/( K3)1 ( R1)i is set as the variate, while ( K′)i/( K3)i as the dependent variable, function h( R1) can be fitted. Finally the straight attachment lug’s SIF equation can be obtained as below. K=K3· h( R1)=K1 g( a/ R1)· f( R2/ R1)· h( R1) =K1F( a/ R1, R2/ R1, R1) (12) ( R1)n a/ R1, R2/ R1 ( K3)n ( K′)n ( K′)n/( K3)n � � � � � � � � � � � � � � �
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