Introduction to Fracture Mechanics phần 3 docx

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,