Plane Crack Problem Airy Stress Function For the plane problem, the equations of equilibrium are satisfied when the stress components are expressed by the Airy stress function through x , , 11 22 xxx rr 21 , , 222 r r rr rr . 1 x rr r Using these definitions for the stresses and Hooke’s law, the strains can be expressed in terms of . x It can be shown that the compatibility equation, when expressed in terms of the Airy stress function, satisfies the biharmonic equation: . 11 ,0 2 2 22 2 222 rrrr The boundary conditions for this plane crack problem 22 The boundary conditions for this plane crack problem are: for 0 r . These conditions express the fact that the crack is traction-free (no loads applied to crack face). Note: there is no condition on . rr A choice of the Airy stress function for the present crack problem should be such that x has a singularity at the crack tip, and is single-valued. We try a solution of the form: ,,, 2 rqrpr 23 Where p and p are harmonic functions of r and θ (i.e. and 0 2 p ).0 2 q Now consider a separate solution, of the following form (Williams, 1957): , rR ,sincos 21 rArAp .2sin2cos 2 2 2 1 rBrBq This form leads to the following expression for x: 2coscos 11 2 BArx . 2 sin sin 2 B A r 24 . 2 sin sin 22 2 B A r Note that we have expressed x as a symmetric part and an anti-symmetric part. The symmetric part provides the Mode I solution while the anti-symmetric part provides the Mode II solution. We will derive the Mode I solution here. 2 2 r 2coscos12 11 BAr rx rr r 1 25 2sin2sin1 11 BAr Apply the boundary conditions: ,0cos 11 BA .0sin2 11 BA The admissible cases are: (i) cos λ π = 0, hence where Z is an integer including zero, and thus B 1 = A 1 λ /(λ + 2) or (ii) sin λπ = 0 and hence λ = Z and B 1 = A 1 . Since the governing equations are linear, any linear combination of the admissible solutions provides a solution, hence λ can have any satisfying: 2 12 Z 26 a solution, hence λ can have any satisfying: , 2 Z Where Z is a positive or negative integer, including zero. Out of all the possible values of λ, how do we decide the appropriate value of λ? The value of λ cannot be found from any mathematical argument. We need to use a physical, based on the total strain energy around the crack tip. From the expressions for the stresses, and Therefore the r ~ . ~ r 27 for the stresses, and Therefore the strain energy density is given by r ij ~ . ~ r ij .~ 2 1 2 r ijij The total strain energy within an annular region,with inner and outer radii r 0 and R, respectively, centered at the crack tip, with unit thickness is .~ 2 1 00 12 2 0 2 0 drdrrdrd R r ijij R r 28 We assert that the strain energy should be bounded (Φ < ∞) as r 0 0. Using this physical argument, we see that λ > -1. (λ = 1 gives ). If λ < 1, the strain energy will not be bounded. 0 ij 2 Thus the physically admissible values of λ are , 2 )( ,,2, 2 3 ,1, 2 1 ,0, 2 1 Z or where Z is –1, 0, or a positive integer. Taking the most dominant singular term (λ = 1/2 and thus B 1 =A 1 /3) we find that: 29 we find that: 2 5 2 1 2 3 2 3 cos 3 1 2 cos rrAr ~ 2/10 2 1 1 rrrA ijij I ijij The higher order terms, with exponents greater than zero, vanish as r 0. We write where K I is the stress intensity factor. Thus we have that: 2/ 1 I KA . ~ 2 jxix I ij I ij r K 30 The first term is the leading singular term for linear elastic mode I crack problems. is a function of θ alone (no r dependence). The second term, generally referred to as the ‘ T term’, is a non-singular term which can be important in some situations involving fatigue. is the Kronecker delta function. I ij ~ ij . B 1 =A 1 /3) we find that: 29 we find that: 2 5 2 1 2 3 2 3 cos 3 1 2 cos rrAr ~ 2/10 2 1 1 rrrA ijij I ijij The higher order. values of λ are , 2 )( ,,2, 2 3 ,1, 2 1 ,0, 2 1 Z or where Z is –1, 0, or a positive integer. Taking the most dominant singular term (λ = 1/2 and thus B 1 =A 1 /3) we find that: 29 we find. of λ? The value of λ cannot be found from any mathematical argument. We need to use a physical, based on the total strain energy around the crack tip. From the expressions for the stresses,