Hi All
the stress intensity factor SIF of crack in the fIG. A attached has two components: for tension and shear
the SIF formula of this specimens derived in the literature and tabulated for different angles and crack length.
Is there a way to define the SIF of this specimen in term that one of crack perpendicular on the load (straight horizontal crack) in Fig. B
your comments will be helpful
| Attachment | Size |
|---|---|
| find SIF of crack in A in term that of B | 18.36 KB |
| find SIF of crack in A in term that of B | 18.36 KB |
Forums
solution
I checked the main idea I had and its work
similar to the formula of the line crack, the 2 components of SIF are computed
For K-1=F1*σ√(a*∏)cos^(β)
the hole effect is known for mode-I (Bowie, Norman, Hsu, etc..)
F1 for each angle is normalised by the ratio of the concentration factor at each angle (see Paterson-Book), for example, F-1 for(β-60)= Kt-(β-60)/Kt-(β-90)
For shear component, the SIF for each angle is the always (2*K11) where KII is the stress intensity factor for β-45
Thanks