A PENNY-SHAPED CRACK IN AN INHOMOGENEOUS ELASTIC MATERIAL UNDER AXISYMMETRIC TORSION

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1 A PENNY-SHAPED CRACK IN AN INHOMOGENEOUS ELASTIC MATERIAL UNDER AXISYMMETRIC TORSION W. T. Ang Department of Applied Mathematics University of Adelaide, South Australia Abstract The problem of a penny-shaped crack in an inhomogeneous elastic material under axisymmetric torsion is considered here. The shear modulus of the material is assumed to exhibit a slight variation in the direction perpendicular to the crack. A solution to the problem in series form is proposed, and the first two terms of the series are obtained analytically by using a Hankel transform technique. Note. This has been a draft of the published paper (by WT Ang) in the journal Solid Mechanics Archives Vol. 12 (No. 2) (1987) pp (Solid Mechanics Archives was edited by Professor John Roorda of the Solid Mechanics Division at the University of Waterloo, Canada, before its last issue was published around 199.) The analysis in this draft is essentially the one given in the published paper, although there may be differences in the numbering of the equations as well as other descriptive details. The published paper also contains slightly more analytical details including an appendix containing used results taken from the table of Hankel transforms in Erdélyi et al [2]. 1 Introduction The present paper is concerned with an extension of the recent work in Ang and Clements [1] to the case of a penny-shaped crack in an inhomogeneous 1

2 elastic material under axisymmetric torsion. The shear modulus of the material is assumed to exhbit a slight variation in the direction perpendicular to the crack. A solution to the problem in series form is proposed, and the first two terms of the series are derived analytically by using a Hankel transform technique. The truncated series obtained by retaining only the first two terms of the series is then used to derive an approximate expression for the relevant stress intensity factor. Specific cases of the problem, such as the case in which the shear modulus varies linearly, are considered. 2 Basic equations for axisymmetric torsion With reference to a cylindrical coordinate system (r, θ, z), consider an elastic material under axisymmetric torsion. The only non-zero component of the displacement is u θ which is a function of r and z. The only non-zero stresses are σ θz = µ u θ z and σ θr = µ[ u θ r u θ ], (1) r where µ is the shear modulus of the material. In the absence of body forces, the equilibrium equation is given by r [σ θr]+ z [σ θz]+ 2 r σ θr =. (2) Following Ang and Clements [1], we take the shear modulus to be given by µ = µ + ²µ 1 (z), (3) where µ is a constant, ² is some positive real constant such that ² << 1 and µ 1 is a continuous and differentiable function of z in the solution domain of interest. Substutition of (1) and (3) into (2) yields [µ + ²µ 1 (z)][ 2 u θ 1 r u θ]+²µ 1(z) u θ =. (4) 2 z Note that the operator 2 is defined by 2 = 2 r r r + 2 z. (5) 2 2

3 Assume that the displacement u θ may be written in the form X u θ = ² n Φ n (r, z). (6) where where n= From (1), the stress σ θz may be written as σ () θz = µ Φ z Substitution of (6) into (4) gives σ θz = σ () θz + ²σ(1) θz + O(²2 ), (7) 2 Φ 1 r 2 Φ = and σ(1) θz = µ Φ 1 z + µ Φ 1 z. (8) 2 Φ n 1 r 2 Φ n = f n 1 (r, z) forn 1, (9) f n (r, z) = 1 [µ µ 1 (z)( 2 Φ n 1 r Φ n)+µ 2 1 (z) Φ n ]. (1) z 3 A penny-shaped crack problem 3.1 Statement of the problem Consider an infinite elastic material with a penny-shaped crack in its interior. The crack lies in the region r<a, θ < 2π, z=, where a is a positive constant. The shear modulus is given by (3) with µ 1 being an even function of z. The material is subject to a small axisymmetric torsion so that an internal stress σ θz = s (r) acts on the crack. The elastic field generated by the crack vanishes at infinity. The problem is to find the displacement and stress fields throughout the material. Specifically, we are interested in finding out how the inhomogeneity of the material affects the stress intensity factor K defined by K = lim r a +(r a)1/2 σ θz (r, ). (11) 3

4 where to and From (7), K may be rewritten as K = K () + ²K (1) + O(² 2 ), (12) K (i) = lim r a +(r a)1/2 σ (i) θz (r, ) for i =, 1. (13) Mathematically, the problem is to solve (4) in the half-space z>subject u θ (r, ) = for r>a, (14) σ θz (r, ) = s (r) for r<a. (15) Assume that the problem admits a series solution of the form (6) and that we are interested in finding only the first two terms of the series. The task of solving (4) in the half-space z> subject to (14) and (15) can then be reduced to following two problems. Problem 1. Solve (in the half-space z>) subject to 2 Φ 1 r 2 Φ =, (16) Φ (r, ) = for r>a, (17) and σ () θz (r, ) = s (r) for r<a. (18) Problem 2. Solve (in the half-space z>) subject to and 2 Φ 1 1 r 2 Φ 1 = µ 1 (z) µ Φ z, (19) Φ 1 (r, ) = for r>a, (2) σ (1) θz (r, ) = for r<a. (21) 4

5 3.2 Solution of problem 1 A solution of (16) is given by Φ (r, z) = ψ (ξ)exp( ξz)j 1 (ξr)dξ, (22) where J 1 is a Bessel function of order 1 and ψ is an arbitrary function yet to be determined. Note Φ as given above tends to as r 2 + z 2. Now, let ψ(ξ) =ξ m g(t)j ν (ξt)dt, (23) where J ν is a Bessel function of order ν and g is an arbitrary function to be determined. From (22) and (23), we find that Φ (r, z) = r 1/2 g(t) ξ m 1/2 J ν (ξt)j 1 (ξr)(ξr) 1/2 dξdt. (24) From the table of Hankel transforms in Erdélyi et al [2], we find that Φ (r, ) = forr>t,if we select m =1/2 andν =3/2. The inequality r>tis always true if r>a(since <t<a). This implies that the condition (17) is satisfied if we take ψ to be ψ(ξ) =ξ 1/2 g(t)j 3/2 (ξt)dt. (25) It follows that σ () θz (r, ) = µ τ r 2 Z d min(r,a) dr r r 3/2 g(t)dt 2 (τ = ), (26) (r 2 t 2 ) 1/2 π if we make use of some results from the table of Hankel transforms in Erdélyi et al [2]. Condition (18) now becomes d dr Z r which can be inverted to give r 3/2 g(t)dt r2 = τs (r 2 t 2 ) 1/2 (r) for r<a, (27) µ g(t) = 2 µ 2πt Z t u 2 s (u)du for <t<a. (28) (t 2 u 2 ) 1/2 5

6 From (13) and (26), the stress intensity factor K () may be given by (after integration by parts) K () = µ g(a) a π. (29) Note that g(a) can be evaluated either analytically or numerically from (28). 3.3 Solution of problem 2 Let Φ 1 (r, z) = From (22), we find that (3) is a solution of (19) if ω(ξ,z)j 1 (ξr)dξ. (3) (ξr) 2 J1 (ξr)+ξrj1(ξr)+[ r2 2 ω ω z r2 ξ µ 2 ωµ 1(z)ψ(ξ)exp( ξz) 1]J 1 (ξr) =. (31) Equation (31) holds if ω satisfies The general solution of (32) is 2 ω z 2 ξ2 ω = ξ µ µ 1(z)ψ(ξ)exp( ξz). (32) ω(ξ,z)=α(ξ)exp(ξz)+[β(ξ)+φ(ξ,z)] exp( ξz), (33) where α and β are arbitrary functions to be determined and φ(ξ,z)= ξ µ ψ(ξ)exp(2ξz) Z z µ 1 (t)exp( 2ξt)dt. (34) We set α(ξ) =sincewerequireφ 1 asz. Thus, from (3) and (33), we find that Φ 1 (r, z) = [β(ξ)+φ(ξ,z)] exp( ξz)j 1 (ξr)dξ. (35) 6

7 Condition (2) is satisfied if β is chosen to be β(ξ) =ξ 1/2 h(t)j 3/2 (ξt)dt φ(ξ, ), (36) where h is yet to be determined. From (8), (35) and (36), we find that σ (1) θz (r, ) = µ φ z (ξ, )J 1 (ξr)dξ τ Z min(r,a) u 3/2 [µ h(u)+µ 1 ()g(u)]du. r 2 (r 2 u 2 ) 1/2 (37) Thus, by using (27), condition (21) may be rewritten as Z d r dr t 3/2 Z h(t)dt (r 2 t 2 ) = 1/2 τr2 [ Equation (38) can be inverted to obtain h(t)+ µ 1() g(t) = 2 Z t µ 2πt φ z (ξ, )J 1 (ξr)dξ + µ 1() s µ 2 (r)] for r<a. (38) u 2 J 1 (ξu) (t 2 u 2 ) 1/2 φ z(ξ, )dξdu for t<a. (39) It follows that the stress intensity factor K (1) defined in (13) is given by K (1) = µ lim r a +(r a)1/2 + 2µ πa 2a 4 Uniform torsion φ z (ξ, )J 1 (ξr)dξ u 2 J 1 (ξu) (a 2 u 2 ) 1/2 φ z(ξ, )dξdu. (4) Take s (r) = σ (constant), that is, the crack is acted upon by a constant torsion. From (28), we obtain g(t) = t2 σ 2µ τ t. (41) 7

8 From (29), K () is then given by K () = aσ 2 2a. (42) Through (25) and (41) together with the result (in Abramowitz and Stegun [3]) we obtain J 3/2 (z) = τ z [ sin(z) z 2 cos(z) ], (43) z ψ(ξ) = σ 2 [2(1 cos(aξ)) aξ sin(aξ)]. (44) 2µ ξ Consider the following cases. Case 1. µ 1 = k z (k is a positive constant). From (34), we obtain φ(ξ,z)= kψ(ξ) [2ξz +1], (45) 4µ ξ and hence φ(ξ,z)= kψ(ξ). (46) 2µ Using (44) and (46) together with the fact that J ν (z) behaves as O(z 1/2 ) for large z, we find that the integrand of the first integral in (4) behaves as O(ξ 3/2 )forlargeξ. The integrand is also bounded everywhere within the interval ξ <. Thus, the integral is well defined for all r and does not contribute to the stress intensity factor K (1). It follows that K (1) = kσ 2µ π u 2 du 2a (a 2 u 2 ) 1/2 1 ξ [2(1 cos(aξ)) aξ sin(aξ)]j 1(ξu)dξ. (47) 2 With the help of the table of Hankel transforms, the integral in (47) can be evaluated to obtain K (1) = kσ a 2 (1 λ), (48) 4πµ 2a 8

9 where λ = 4(ln(2) + 1/2)/3 > 1. Thus, for this particular case in which µ = µ + ² z, the stress intensity factor K is lower than the corresponding stress intensity factor for a material with constant shear modulus µ. It is interesting to note that as µ decreases the difference between these stress intensity factors becomes more pronounced. This observation is consistent with that in Ang and Clements [1] for the case in which the variation of the elastic modulus is linear. Case 2. µ 1 = kz 2 (k is a positive constant). From (34), we obtain φ(ξ,z)= kψ(ξ) 4µ ξ 2 [2ξ2 z 2 +2ξz +1]. (49) As in Case 1, the first integral in (4) is finite for r and does not contribute to the stress intensity factor K (1). Hence K (1) = kσ 2µ π 2a u 2 du (a 2 u 2 ) 1/2 [ 2 ξ 3 (1 cos(aξ)) a ξ 2 sin(aξ)]j 1(ξu)dξ, (5) which eventually gives (after some analytical manipulations using results from the table of Hankel transforms) K (1) = kσ a 3. (51) 48µ 2a As in Case 1, the variation of the shear modulus considered here has the effect of reducing the stress intensity factor K. Also, decreasing µ has the effect of increasing the magnitude of K (1). Case 3. µ 1 = k(z δ) 2 if <z δ and µ 1 =ifz>δ (k and δ are positive constants). The results in Cases 1 and 2 may be used to derive K (1) = kσ a 2 ( a 2µ 2a λ δ ). (52) π From (52), it is obvious that K (1) > ifa<24 1 λ δ/π, that is, the crack is less stable in the particular inhomogeneous material considered here than in a homogeneous material with shear modulus µ if the crack length a is shorter than a critical value which depends on δ. Note that if δ is very large it is more likely for the crack to be less stable. 9

10 References [1]W.T.AngandD.L.Clements,Onsomecrackproblemsforinhomogeneous elastic materials, International Journal of Solids Structures 23 (1987) [2] A. Erdélyi et al, Tables of Integral Transforms, Vol. 2, 1954, McGraw-Hill. [3] M. Abramowitz and A. Stegun, Handbooks of Mathematical Functions, 197, Dover. 1