Jtam.dvi JOURNAL OF THEORETICAL AND APPLIED MECHANICS 41, 2, pp. 241-269, Warsaw 2003 FUNDAMENTAL SOLUTIONS RELATED TO THERMAL STRESS INTENSITY FACTORS OF MODES I AND II – THE AXIALLY SYMMETRIC PROBLEM Bogdan Rogowski Mechanics of Materials Division, Technical University of Łódź, Poland e-mail: brogowsk@ck-sg.p.lodz.pl This elaboration considers the crack problems for infinite thermoelastic solids subjected to steady temperature or heat flux. The crack faces are assumed to be insulated. Green’s functions are obtained for the thermal stress intensity factors ofmodes I and II. TheGreen’s functions are defi- ned as a solution to the problemof a thermoelastic transversely isotropic solidwith a penny-shaped or an external crack under general axisymme- tric thermal loadings acting along a circumference on the plane parallel to the crack plane. Key words: thermoelasticity, anisotropy, crack problems, Green’s func- tions, stress intensity factors of mode I and II 1. Introduction The penny-shaped crack in a temperature field was treated by Olesiak and Sneddon (1960); the problem was symmetrical with respect to the crack plane. The features of antisymmetry were presented by Florence andGoodier (1963) in the linear thermoelastic problem of uniform heat flow disturbed by a penny-shaped insulated crack. In this paper, we consider the steady thermal stress in a cracked solid. The problems of the crack treated here are solved by using two types of axisymme- tric ring thermal loadings as fundamental solutions: a uniform heat flux and temperature. The research is aimed at the assessing of the effect of dissimilar thermal conditions on the stress intensity factors. The stress intensity factors of modes I and II are derived in this study in terms of elementary functions. The results presented for general cases are new, but some of those related 242 B.Rogowski to special cases of isotropic or transversely isotropic solids with crack surface thermal loadings have been already known (cf. Olesiak and Sneddon, 1960; Florence andGoodier, 1963; Rogowski, 1984). 2. Basic equations The basic equations of axisymmetric thermal stress problems for homo- geneous transversely isotropic bodies are the equilibrium equations (in the absence of body forces) σrr,r+σrz,z+ 1 r (σrr−σθθ)= 0 σrz,r+σzz.z+ 1 r σrz =0 (2.1) the strain-displacement relations err =ur,r eθθ = ur r ezz =uz,z 2erz =ur,z +uz,r (2.2) the constitutive equations σrr = c11err+ c12eθθ + c13ezz −β1T σθθ = c12err+ c11eθθ + c13ezz −β1T σzz = c13err+ c13eθθ+ c33ezz −β3T σrz =2c44erz (2.3) and the heat conduction equation (steady state without heat generation) T,rr+r −1T,r+s −2 0 T,zz =0 (2.4) where partial differentiation is indicated by the comma followed by the va- riables, cij are the elastic constants of a transversely isotropic material, β1 = (c11 + c12)αr + c13αz, β3 = 2c13αr + c33αz are the thermal stress coef- ficients, αr and αz are the coefficients of the linear thermal expansion in the radial and axial direction, s20 =λr/λz, λr and λz are the thermal conductivi- ties in the radial and axial direction.By substitutingEq. (2.3) into equilibrium equations (2.1) and using relations (2.2), we obtain c11 ( ur,rr+ 1 r ur,r− 1 r2 ur ) + c44ur,zz+(c13+ c44)uz,rz−β1T,r =0 (2.5) c44 ( uz,rr+ 1 r uz,r ) + c33uz,zz+(c13+ c44) ( ur,rz+ 1 r ur,z ) −β3T,z =0 Fundamental solutions related to thermal stress... 243 Tosolve partial differential equations (2.4) and (2.5)we introducepotential functions which relate to the displacements as follows (Rogowski, 1978) ur =(kϕ1+ϕ2+ϕ0),r uz =(ϕ1+kϕ2+ lϕ0),z (2.6) and the Hankel transforms defined as follows u∗r = ∞∫ 0 urrJ1(ξr) dr u ∗ z = ∞∫ 0 uzrJ0(ξr) dr (2.7) where ξ is the Hankel parameter and Jν(ξr) denotes the Bessel function of the first kind of order ν. The Hankel transform is its own inverse. The potential functions must satisfy the following equations ϕi,rr+ 1 r ϕi,r+ 1 s2i ϕi,zz =0 i=0,1,2 (2.8) ϕ0,zz =Ms 2 0T where s20 =λr/λz, s 2 i (i=1,2) are the roots of the equation c33c44s 4− [c11c33− c13(c13+2c44)]s2+ c11c44 =0 (2.9) and k, l,M are the material parameters defined as follows k= c33s 2 1− c44 c13+ c44 l= β1(c13+ c44)−β3(c11−s20c44) β1(c33s 2 0− c44)−β3s20(c13+ c44) (2.10) M = β1(c33s 2 0− c44)−β3s20(c13+ c44) s20(c13+ c44) 2− (c11−s20c44)(c33s20− c44) The thermal stresses components σzz and σrz are represented as follows σzz =Gz(k+1)(s −2 1 ϕ1+s −2 2 ϕ2),zz+GzM(1+ l)T (2.11) σrz =Gz(k+1)(ϕ1+ϕ2),rz +Gz(1+ l)ϕ0,rz where Gz = c44 is the shear modulus along the z-axis. The stress components σrr and σθθ may be similarly expressed. Consider an infinite transversely isotropic elastic solid containing a penny- shaped crack with its diameter 2a or an external crack covering the outside of a circle of the radius a, as shown in Fig.1. Denote by (r,θ,z) the cylindrical co-ordinate system with its origin at the middle point of the penny-shaped crack face or of the bonding region, respectively. The thermal loading con- ditions (Fig.2 and Fig.3) may be decomposed into symmetrical (Fig.4) and antisymmetrical (Fig.5) with respect to the crack plane. 244 B.Rogowski Fig. 1. Thermoelastic solid with a penny-shaped or external crack under thermal loadings Fig. 2. Temperature loading acting along a circle Fig. 3. Axial heat flux acting along a circle Fundamental solutions related to thermal stress... 245 Fig. 4. Symmetric thermal loadings Fig. 5. Antisymmetric thermal loadings 3. Temperature field For a uniform temperature and heat flux applied along the circumference r= b on the plane z=h, the thermal loading conditions are T(r,h+0)−T(r,h−0)= T0 4πr δ(r− b) (3.1) T,z(r,h+0)−T,z(r,h−0)= Q0 4πλzr δ(r− b) where δ(r− b) is the Dirac delta function and T0, Q0 are the constant tem- perature and heat flux, respectively. 246 B.Rogowski Applying the Hankel transforms to Eqs (2.4) and (3.1 a,b), we find the temperature as follows T(r,z) = ∞∫ 0 Aij(ξ)e −ξs0zJ0(ξr) dξ+ (3.2) + 1 2 ∞∫ 0 [ξν0H0(ξs0z)−ν1H1(ξs0z)]J0(ξb)J0(ξr) dξ z­ 0 where H0(ξs0z)= sgn(z−h)e−ξs0|z−h|− (−1)i+je−ξs0(z+h) H1(ξs0z)= e −ξs0|z−h|+(−1)i+je−ξs0(z+h) sgn(z−h)= { 1 for z >h −1 for z a (3.7) (iii) For the external crack and symmetric temperature field A12(ξ)=−(ν0ξ+ν1)e−ξs0hJ0(ξb) (3.8) (iv) For the external crack and antisymmetric temperature field ∞∫ 0 A22(ξ)J0(ξr) dξ= ∞∫ 0 (ν0ξ+ν1)e −ξs0hJ0(ξb)J0(ξr) dξ 0¬ ra (3.9) Both solutions (3.6) and (3.8) give the temperature field T(r,z) = 1 2 ∞∫ 0 J0(ξb)J0(ξr) · (3.10) · [ (ν0ξ sgn(z−h)−ν1)e−ξs0|z−h|− (ν0ξ+ν1)e−ξs0(z+h) ] dξ related to the symmetric thermal loading conditions of the solid with the penny-shaped or an external crack. Dual integral equations (3.7) are converted to the Abel integral equation bymeans of the following integral representation for A21(ξ) (Noble, 1963) A21 = √ 2 π a∫ 0 g0(x)sin(ξx) dx (3.11) on the assumption that g0(x)→ 0 as x→ 0+. This representation of A21(ξ) identically satisfiesEq. (3.7)2 (seeAppendix, Eqs (A.1) and (A.9)). 248 B.Rogowski Substituting A21(ξ) into Eq. (3.7)1 leads to the following Abel integral equation in an auxiliary function g0(x) √ 2 π r∫ 0 (dg0(x) dx 1√ r2−x2 ) dx=− ∞∫ 0 ξ(ν0ξ+ν1)J0(ξr)J0(ξb)e −ξs0h dξ (3.12) Applying the Abel solution method to invert the left hand side of Eq. (3.12) gives the formula for g0(x) g0(x)=− √ 2 π ∞∫ 0 (ν0ξ+ν1)sin(ξx)J0(ξb)e −ξs0h dξ (3.13) The improper integrals appearing in Eq. (3.13) are calculated analytically (see Appendix, Eqs (A.1) and (A.2)). Consequently, the auxiliary function g0(x) is obtained in terms of the oblate spheroidal co-ordinates ζ0 and η0, defined in the Appendix, as g0(x)= √ 2 π [ ν0 d dx ( ζ0 D0 ) −ν1 ( η0 D0 )] = (3.14) =− √ 2 π { ν0 ζ0 D20(ζ 2 0 +η 2 0) [(1−η20)(η20 −ζ20)+2η20(1+ ζ20)]+ν1 η0 D0 } where D0 =x(ζ 2 0 +η 2 0) Finally, the temperature field is obtained as T(r,z) = 2 π a∫ 0 [ ν0 d dx ( ζ0 D0 ) −ν1 ( η0 D0 )] η D dx+ (3.15) + 1 2 ∞∫ 0 { [ν0ξ sgn(z−h)−ν1]e−ξs0|z−h|+(ν0ξ+ν1)e−ξs0(z+h) } J0(ξb)J0(ξr)dξ where D=x(ζ2+η2) and where the oblate spheroidal co-ordinates ζ, η are associated with r, s0z and x, while ζ0, η0 are associated with b, s0h and x (see Appendix). Fundamental solutions related to thermal stress... 249 Dual integral equations (3.9) are converted to the Abel integral equations bymeans of the following integral representation of A22(ξ) A22(ξ)= √ 2 π a∫ 0 f0(x)cos(ξx) dx (3.16) In this representation the auxiliary function f0(x) is assumed to be conti- nuous over the interval [0,a]. This representation of A22(ξ) identically satisfies Eq. (3.9)2. Substituting A22(ξ) into Eq. (3.9)1 leads to the following Abel integral equation in an auxiliary function f0(x) √ 2 π r∫ 0 f0(x)√ r2−x2 dx= ∞∫ 0 (ν0ξ+ν1)e −ξs0hJ0(ξr)J0(ξb) dξ (3.17) Applying the Abel solution method, give the formula for f0(x) f0(x)= √ 2 π ∞∫ 0 (ν0ξ+ν1)e −ξs0hcos(ξx)J0(ξb) dξ (3.18) Substituting the integrals (A.1) and (A.2) (see Appendix), gives the final solution for f0(x) f0(x)= √ 2 π [ ν0 d dx ( η0 D0 ) +ν1 ζ0 D0 ] = (3.19) = √ 2 π { ν0 η0 D20(ζ 2 0 +η 2 0) [(1+ ζ20)(ζ 2 0 −η20)+2ζ20(1−η20)]+ν1 ζ0 D0 } For the external crack in the antisymmetric temperature field the tempe- rature is obtained as T(r,z) = 2 π a∫ 0 [ ν0 d dx ( η0 D0 ) +ν1 ζ0 D0 ] ζ D dx+ (3.20) + 1 2 ∞∫ 0 { [ν0ξ sgn(z−h)−ν1]e−ξs0|z−h|− (ν0ξ+ν1)e−ξs0(z+h) } J0(ξr)J0(ξb)dξ 250 B.Rogowski Byusing the superposition of two thermal fields (3.10) and (3.15) or (3.20), we obtain T(r,z) = 2 π a∫ 0 [ ν0 d dx ( ζ0 D0 ) −ν1 η0 D0 ] η D dx+ (3.21) + ∞∫ 0 [ν0ξ sgn(z−h)−ν1]e−ξs0|z−h|J0(ξb)J0(ξr) dξ z­ 0 for the penny-shaped crack and T(r,z) = 2 π a∫ 0 [ ν0 d dx ( η0 D0 ) +ν1 ζ0 D0 ] ζ D dx+ (3.22) + ∞∫ 0 { [ν0ξ sgn(z−h)−ν1]e−ξs0|z−h|− (ν0ξ+ν1)e−ξs0(z+h) } J0(ξb)J0(ξr) dξ where z­ 0, for the external crack. 4. Thermal stresses Considering Eqs (2.8) and (3.2), we find the potential functions (z­ 0) ϕ0(r,z) =M ∞∫ 0 ξ−2 { Aij(ξ)e −ξs0z + + 1 2 [ν0ξH0(ξs0z)−ν1H1(ξs0z)]J0(ξb) } J0(ξr) dξ ϕ1(r,z) = s2 Gz(k+1)(s1−s2) ∞∫ 0 ξ−1B1j(ξ)e −ξs1zJ0(ξr) dξ (4.1) ϕ2(r,z) =− s1 Gz(k+1)(s1−s2) ∞∫ 0 ξ−1B2j(ξ)e −ξs2zJ0(ξr) dξ Substituting Eqs (4.1) into Eqs (2.6) and (2.11), we obtain (z­ 0) Fundamental solutions related to thermal stress... 251 ur(r,z) =−M ∞∫ 0 ξ−1 { Aij(ξ)e −ξs0z + + 1 2 [ν0ξH0(ξs0z)−ν1H1(ξs0z)]J0(ξb) } J1(ξr) dξ− − 1 Gz(k+1)(s1−s2) ∞∫ 0 [ ks2B1j(ξ)e −ξs1z −s1B2j(ξ)e−ξs2z ] J1(ξr) dξ (4.2) uz(r,z) =−Ms0l ∞∫ 0 ξ−1 { Aij(ξ)e −ξs0z + + 1 2 [ν0ξH ′ 0(ξs0z)−ν1H′1(ξs0z)]J0(ξb) } J0(ξr) dξ− − s1s2 Gz(k+1)(s1−s2) ∞∫ 0 [ B1j(ξ)e −ξs1z −kB2j(ξ)e−ξs2z ] J0(ξr) dξ σzz(r,z) =GzM(1+ l) ∞∫ 0 { Aij(ξ)e −ξs0z + + 1 2 [ν0ξH0(ξs0z)−ν1H1(ξs0z)]J0(ξb) } J0(ξr) dξ+ + 1 s1−s2 ∞∫ 0 ξ [ s2B1j(ξ)e −ξs1z −s1B2j(ξ)e−ξs2z ] J0(ξr) dξ (4.3) σrz(r,z) =GzM(1+ l)s0 ∞∫ 0 { Aij(ξ)e −ξs0z + + 1 2 [ν0ξH ′ 0(ξs0z)−ν1H′0(ξs0z)]J0(ξb) } J1(ξr) dξ+ + s1s2 s1−s2 ∞∫ 0 ξ [ B1j(ξ)e −ξs1z −B2j(ξ)e−ξs2z ] J1(ξr) dξ where H′0(ξs0z)= e −ξs0|z−h|− (−1)i+je−ξs0(z+h) (4.4) H′1(ξs0z)= sgn(z−h)e−ξs0|z−h|+(−1)i+je−ξs0(z+h) 252 B.Rogowski The crack problemmust be solved under the following conditions σzr(r,0)= 0 r­ 0 σzz(r,0)= 0 r∈Ac uz(r,0)= 0 r∈ Ãc (4.5) for the symmetric thermal condition and σzz(r,0)= 0 r­ 0 σzr(r,0)= 0 r∈Ac ur(r,0)= 0 r∈ Ãc (4.6) for the antisymmetric thermal condition. Conditions (4.5)1 and (4.6)1 yield, respectively B2j(ξ)=B1j(ξ)+GzM(1+ l) (s0 s2 − s0 s1 ) ξ−1 · (4.7) · { A1j(ξ)+ 1 2 (ν0ξ+ν1)[1+(−1)j]J0(ξb)e−ξs0h } or B2j(ξ)= s2 s1 B1j(ξ)+GzM(1+ l) ( 1− s2 s1 ) ξ−1 · (4.8) · { A2j(ξ)− 1 2 (ν0ξ+ν1)[1+(−1)j]J0(ξb)e−ξs0h } Thedisplacements andstressesmeetingmixedboundaryconditions (4.5)2,3 and (4.6)2,3 on the plane where the crack appears are uz(r,0)= 1 GzC ∞∫ 0 B1j(ξ)J0(ξr) dξ+ Ms0(k− l) k+1 · · ∞∫ 0 ξ−1 { A1j(ξ)+ 1 2 (ν0ξ+ν1)[1+(−1)j]e−ξs0hJ0(ξb) } J0(ξr) dξ σzz(r,0)=− ∞∫ 0 ξB1j(ξ)J0(ξr) dξ+GzM(1+ l) ∞∫ 0 {( 1− s0 s2 ) A1j(ξ)− − 1 2 (ν0ξ+ν1) [ 1+ s0 s2 − ( 1− s0 s2 ) (−1)j ] e−ξs0hJ0(ξb) } J0(ξr) dξ (4.9) Fundamental solutions related to thermal stress... 253 ur(r,z) =− 1 GzCs1 ∞∫ 0 B1j(ξ)J1(ξr) dξ− − M(k− l) k+1 ∞∫ 0 ξ−1 { A2j(ξ)− 1 2 (ν0ξ+ν1)[1+(−1)j]e−ξs0hJ0(ξb) } J1(ξr) dξ σrz(r,z) = s2 ∞∫ 0 ξB1j(ξ)J1(ξr) dξ+GzM(1+ l) ∞∫ 0 { (s0−s2)A2j + + 1 2 (ν0ξ+ν1)[s0+s2− (s0−s2)(−1)j]e−ξs0hJ0(ξb) } J1(ξr) dξ where C = (k+1)(s1−s2) (k−1)s1s2 (4.10) 5. Mode I loading TheMode I crack problem corresponds to the symmetric thermal loading. The penny-shaped crack problem is obtained for j=1 and the external crack problem is obtained for j=2. 5.1. The penny-shaped crack problem Substituting Eqs (4.9)1,2 into boundary conditions (4.5)2,3 and using that A11(ξ)= 0, the following dual integral equations are obtained ∞∫ 0 ξB11(ξ)J0(ξr) dξ= =−GzM(1+ l) ∞∫ 0 (ν0ξ+ν1)e −ξs0hJ0(ξb)J0(ξr) dξ 0¬ r a (5.2) Dual integral equations (5.1), (5.2) are converted to theAbel integral equ- ation bymeans of the following integral representation of B11(ξ) 254 B.Rogowski B11(ξ)= √ 2 π a∫ 0 g(x)sin(ξx) dx (5.3) on the assumption that g(x)→ 0 as x→ 0+. This representation of B11(ξ) identically satisfies Eq. (5.2). Substituting B11(ξ) into Eq. (5.1) leads to the following Abel integral equation in an auxi- liary function g(x) √ 2 π r∫ 0 (dg(x) dx 1√ r2−x2 ) dx= (5.4) =−GzM(1+ l) ∞∫ 0 (ν0ξ+ν1)e −ξs0hJ0(ξb)J0(ξr) dξ Applying theAbel solutionmethod to invert the left hand side of Eq. (5.4) gives the formula for g(x) g(x)=− √ 2 π GzM(1+ l) ∞∫ 0 (ν0+ν1ξ −1)e−ξs0hJ0(ξb)sin(ξx) dξ (5.5) The improper integrals appearing in Eq. (5.5) are calculated analytically (see Appendix, Eqs (A.1) and (A.3)). Consequently, the auxiliary function g(x) is obtained explicitly in terms of the oblate spheroidal co-ordinates ζ0 and η0 (see Appendix) as g(x) =− √ 2 π GzM(1+ l) [ ν0 η0 x(ζ20 +η 2 0) +ν1 (π 2 − tan−1ζ0 )] (5.6) The singular part of the axial stress is given by the formula σzz(r,0)= √ 2 π g(a)√ r2−a2 as r→ a+ (5.7) Defining the stress intensity factor of Mode I as KI = lim r→a+ √ 2(r−a)σzz(r,0) (5.8) one obtains KI =− 2 π √ a GzM(1+ l) [ ν0 η0 a(ζ 2 0+η 2 0) +ν1 (π 2 − tan−1ζ0 )] (5.9) Fundamental solutions related to thermal stress... 255 where ζ0, η0 are obtained from ζ0, η0 for x= a (see Appendix). Solution (5.9) contains three other problems as special cases, namely: (i) h = 0 and b < a, (ii) h = 0 and b > a, (iii) b = 0. We can dedu- ce the results for these three cases from equations (5.9), (A.8) and (A.9) for x= a. The results are given in Table 1. 5.2. The external crack The dual integral equations of the external crack problem are ∞∫ 0 B12(ξ)J0(ξr) dξ=0 0¬ r a (5.11) For the temperature loading we use the integral representation of B12(ξ) B12(ξ)= √ 2 π a∫ 0 f(x)cos(ξx) dx−GzM(1+ l)ν0e−ξs0hJ0(ξb) (5.12) and find the Abel integral equation in an auxiliary function f(x) √ 2 π r∫ 0 ( f(x)√ r2−x2 ) dx=GzM(1+ l)ν0 ∞∫ 0 e−ξs0hJ0(ξb)J0(ξr) dξ (5.13) The solution for this equation is f(x)= √ 2 π GzM(1+ l)ν0 ∞∫ 0 e−ξs0hJ0(ξb)cos(ξx) dξ (5.14) Substituting the analytical expression for the improper integral (Eq. (A.2) in the Appendix), we get f(x)= √ 2 π GzM(1+ l)ν0 ζ0 x(ζ20 +η 2 0) (5.15) 256 B.Rogowski The stress transmitted through the neck is found to be σzz(r,0)=− √ 2 π f(a)√ a2−r2 + a∫ r df(x) dx dx√ x2−r2 (5.16) Defining the stress intensity factor of Mode I as KI = lim r→a− √ 2(a−r)σzz(r,0) (5.17) one obtains KI =− 2 π √ a GzM(1+ l)ν0 ζ0 a(ζ 2 0+η 2 0) (5.18) where ζ0, η0 are the values of ζ0, η0 for x= a (see Appendix). For the heat flux problemwe use the integral representation of B12(ξ) B12(ξ)= √ 2 π a∫ 0 f1(x) [sin(ξx) ξx − cos(ξx) ] dx−GzM(1+ l)ν1ξ−1e−ξs0hJ0(ξb) (5.19) This representation identically satisfies Eq. (5.11) associatedwith the heat flux and converts Eq. (5.10) to the Abel integral equation − √ 2 π r∫ 0 f1(x)√ r2−x2 dx+ √ 2 π a∫ 0 f1(u) u [∞∫ 0 sin(ξu) ξ J0(ξr) dξ ] du= (5.20) =GzM(1+ l)ν1 ∞∫ 0 ξ−1e−ξs0hJ0(ξb)J0(ξr) dξ Applying the Abel solution method we obtain −f1(x)+ 2 π a∫ 0 f1(u) u [∞∫ 0 sin(ξu)cos(ξx) ξ dξ ] du= (5.21) = √ 2 π GzM(1+ l)ν1 ∞∫ 0 ξ−1e−ξs0hJ0(ξb)cos(ξx) dξ We use the integral ∞∫ 0 sin(ξu)cos(ξx) ξ dξ= π 2 H(u−x) (5.22) Fundamental solutions related to thermal stress... 257 where H(·) is the Heaviside unit function. Then we have −f1(x)+ a∫ x f1(u) u du= √ 2 π GzM(1+l)ν1 ∞∫ 0 ξ−1e−ξs0hJ0(ξb)cos(ξx) dξ (5.23) It is seen that the integrand in the improper integral is unbounded as ξ→ 0. This improper behaviour at ξ→ 0 can be removed by adding to both sides of Eq. (5.23) the value of f1(0), obtained formally from this equation. After adjusting the improper behaviour at ξ→ 0, Eq (5.23) becomes x∫ 0 1 x d dx [xf1(x)] dx= √ 2 π GzM(1+ l)ν1 ∞∫ 0 1−cos(ξx) ξ e−ξs0hJ0(ξb) dξ (5.24) The improper integral in Eq. (5.24) has an analytic expression given by Eqs (A.5) and (A.6) in the Appendix. We use the following relationships 1− cos(ξx) ξ = x∫ 0 sin(ξx) dx= x∫ 0 1 x d dx [x ξ (sin(ξx) ξx − cos(ξx) )] dx (5.25) and integral (A.4) from the Appendix. Then, the solution to Eq (5.24) is obtained in the form f1(x)= √ 2 π GzM(1+ l)ν1η0 [ 1− ζ0 (π 2 − tan−1ζ0 )] (5.26) It is noted that f1(x) tends to zero as x→ 0+. The stress transmitted through the neck is found to be σzz(r,0)= √ 2 π [ r2f1(a) a2 √ a2−r2 −r2 a∫ r d dx (f1(x) x2 ) dx√ x2−r2 − (5.27) −2 a∫ r f1(x) x dx√ x2−r2 ] The stress intensity factor of Mode I is given by KI = 2 π √ a GzM(1+ l)ν1η0 [ 1− ζ0 (π 2 − tan−1ζ0 )] (5.28) where ζ0, η0 are the values of ζ0, η0 for x= a (see Appendix). In special cases, KI takes the values which are shown in Table 1. 258 B.Rogowski 6. Mode II loading 6.1. The penny-shaped crack The dual integral equations are: — for 0¬ r a ∞∫ 0 B11(ξ)J1(ξr) dξ=− GzM(k− l) k−1 (s1 s2 −1 ) ∞∫ 0 ξ−1A21(ξ)J1(ξr) dξ (6.2) The integral representation of B11(ξ) B11(ξ)= √ ξ a∫ 0 √ xh(x)J3/2(ξx) dx− GzM(k− l) k−1 (s1 s2 −1 ) ξ−1A21(ξ) (6.3) on the assumption that √ xh(x)→ 0 as x→ 0+, satisfies identically Eq. (6.2), while Eq. (6.1) is converted to the Abel integral equation √ 2 π r∫ 0 (d[xh(x)] dx 1√ r2−x2 ) dx= =GzMr{ [k− l k−1 (s1 s2 −1 ) − (1+ l) (s0 s2 −1 )] ∞∫ 0 A21(ξ)J1(ξr) dξ− (6.4) −(1+ l) s0 s2 ∞∫ 0 (ν0ξ+ν1)e −ξs0hJ0(ξb)J1(ξr) dξ } Fundamental solutions related to thermal stress... 259 The solution to this equation is h(x) = √ 2 π GzM {[k− l k−1 (s1 s2 −1 ) − (1+ l) (s0 s2 −1 )] · · ∞∫ 0 A21(ξ) d dξ (−sin(ξx) ξx ) dξ− (6.5) −(1+ l) s0 s2 ∞∫ 0 (ν0+ν1ξ −1)e−ξs0hJ0(ξb) [sin(ξx) ξx − cos(ξx) ] dξ } Integrating the first integral in Eq. (6.5) by parts, substituting A21(ξ) and g0(x) from Eqs (3.11) and (3.13) and substituting for the second integral the analytical formula (see Appendix, Eqs (A.2), (A.3) and (A.4)), lead to the following exact formula for h(x) h(x)= √ 2 π GzM(1+ l) κ s2 · (6.6) · {ν0 x (π 2 − tan−1ζ0− ζ0 ζ20 +η 2 0 ) +ν1η0 [ ζ0 (π 2 − tan−1ζ0 ) −1 ]} where κ= s2+ k− l k−1 s1−s2 1+ l (6.7) The singular part of the shear stress is given by σrz(r,0)=− √ 2 π s2ah(a) r √ r2−a2 as r→ a+ (6.8) The stress intensity factor of Mode II is obtained as follows KII =− 2 π √ a GzM(1+ l)κ · (6.9) · {ν0 a (π 2 − tan−1ζ0− ζ0 ζ 2 0+η 2 0 ) +ν1η0 [ ζ0 (π 2 − tan−1ζ0 ) −1 ]} In special cases KII, takes the values which are shown in Table 1. 260 B.Rogowski 6.2. The external crack The dual integral equations are:— for 0¬ r a ∞∫ 0 ξB12(ξ)J1(ξr) dξ= (6.11) =−GzM(1+ l) ∞∫ 0 [(s0 s2 −1 ) A22(ξ)+(ν0ξ+ν1)e −ξs0hJ0(ξb) ] J1(ξr) dξ The integral representation of B12(ξ) B12(ξ)= √ 2 π a∫ 0 t(x)sin(ξx) dx− (6.12) −GzM(1+ l)ξ−1 [(s0 s2 −1 ) A22(ξ)+(ν0ξ+ν1)e −ξs0hJ0(ξb) ] gives the Abel integral equation √ 2 π r∫ 0 xt(x)√ r2−x2 dx=−GzMr · · {[k− l k−1 (s1 s2 −1 ) − (1+ l) (s0 s2 −1 )] ∞∫ 0 ξ−1A22(ξ)J1(ξr) dξ− (6.13) − [k− l k−1 (s1 s2 −1 ) +1+ l ] ∞∫ 0 (ν0+ν1ξ −1)e−ξs0hJ0(ξb)J1(ξr) dξ } Fundamental solutions related to thermal stress... 261 The solution to this equation is t(x)=− √ 2 π GzM · · {[k− l k−1 (s1 s2 −1 ) − (1+ l) (s0 s2 −1 )] ∞∫ 0 ξ−1A22(ξ)sin(ξx) dξ− (6.14) − [k− l k−1 (s1 s2 −1 ) +1+ l ] ∞∫ 0 (ν0+ν1ξ −1)e−ξs0hJ0(ξb)sin(ξx) dξ } Substituting A22(ξ) fromEq. (3.16) and f0(x) fromEq. (3.19), integrating and using Eqs (A.1) and (A.3) from the Appendix, we obtain t(x)= √ 2 π GzM(1+ l) s0 s2 [ ν0 η0 x(ζ20 +η 2 0) +ν1 (π 2 − tan−1ζ0 )] (6.15) The singular part of the shear stress is σzr(r,0)= √ 2 π rt(a)s2 a √ a2−r2 as r→ a− (6.16) The stress intensity factor of Mode II is obtained in the form KII = 2 π √ a GzM(1+ l)s0 [ ν0 η0 a(ζ 2 0+η 2 0) +ν1 (π 2 − tan−1ζ0 )] (6.17) where the oblate spheroidal co-ordinates ζ0, η0 are calculated for x= a. In special cases KII, takes the values presented in Table 1. 7. Applications The exact solutions have been presented for the stress intensity factors of Mode I and II at the tips of the penny-shaped crack and external crack under thermal loadings. These solutions are obtained explicitly in terms of elementary functions. For any axisymmetrical distribution of thermal loadings of the medium with internal or external cracks the integration and/or simple superposition of the obtained results can yield the stress intensity factors. 262 B.Rogowski When the cracked solid is subjected to temperature T(r,z) = T0t(r,z) and/or heat flux Q(r,z) =Q0q(r,z), then the components Ki (i= I,II) of the stress intensity factor may be calculated as follows Ki = ∫ V [t(r,z)Ki0(r,z)+q(r,z)Ki1(r,z)] dV (7.1) where V denotes the domain volume of the thermally loaded region and Ki0(r,z), Ki1(r,z) denote the stress intensity factors when the temperatu- re and heat flux ring loading (index 0 or 1, respectively) act along a circle (r,z) of the radius r on the plane z (the co-ordinates b, h should be replaced by r, z in the obtained results). We now proceed to consider some specific cases of thermal loadings, when the temperature T0/2 and the heat flux Q0/2 are applied on the planes z= ±h in an annular region b ¬ r ¬ c symmetrically or asymmetrically with respect to z=0 plane. Then, equation (7.1) yields Ki =2π c∫ b [Ki0(r,h)+Ki1(r,h)]r dr (7.2) where r dr= a2(ζ2+η2) dζ ζ dζ ζ =−dη η in the oblate spheroidal co-ordinates r2 = a2(1+ ζ2)(1−η2), s0h= aζη and Ki0(r,h), Ki1(r,h) are presented in those co-ordinates. Example 1: Consider the case of the temperature loading T0/2 on the planes z=±h in the annular region b¬ r¬ c. From equation (5.8) and (5.18) we obtain: — for the penny-shaped crack (0¬ r¬ a) KI =− T0 √ a π GzM(1+ l) η(c)∫ η(b) η ζ2+η2 ( −ζ 2+η2 η ) dη= (7.3) =− T0 √ a π GzM(1+ l)[η(b)−η(c)] Fundamental solutions related to thermal stress... 263 where η(r)= 1 a √ 2 √√ (r2+s20h 2−a2)2+4a2s20h2− (r2+s20h2−a2) (7.4) — for the external crack (r­ a) KI =− T0 √ a π GzM(1+ l) ξ(c)∫ ξ(b) ζ ζ2+η2 ζ2+η2 ζ dζ = (7.5) =−T0 √ a π GzM(1+ l)[ζ(c)− ζ(b)] where ζ(r)= 1 a √ 2 √√ (r2+s20h 2−a2)2+4a2s20h2+r2+s20h2−a2 (7.6) Since for real materials GzM(1+ l)< 0, the cracks open if T0 > 0. In special cases, KI assumes the values: — for the penny-shaped crack KI =A    1 for b=0 c→∞ 1−η(c) for b=0 c finite 1 for h=0 b=0 c= a √ a2− b2− √ a2− c2 a for h=0 b< c¬ a 0 for h=0 b­ a c>a (7.7) — for the external crack KI =A    ζ(c)− s0h a for b=0 c finite √ c2−a2− √ b2−a2 a for h=0 a¬ b< c 0 for h=0 bx s0h x for b=0 (A.8) 268 B.Rogowski η0 =    √ 1− b 2 x2 for h=0 bx 1 for b=0 (A.9) The co-ordinates ζ0, η0 for x= a are denoted by ζ0, η0. The co-ordinates for b= r, h= z are denoted by ζ, η and those for x= a by ζ, η. References 1. Florence A.L., Goodier J.N., 1963, The linear thermoelastic problem of uniform heat flow disturbed by a penny-shaped crack, Int. J. Engng Sci., 1, 533-540 2. Noble B., 1963, Dual Bessel function integral equations, Proc. Camb. Phil. Soc., Math. Phys., 59, p.351 3. Olesiak Z., Sneddon I.N., 1960, The distribution of thermal stress in an infinite elastic solids containing a penny-shaped crack,Arch. Rat. Mech. Anal., 4, 238-254 4. Rogowski B., 1978, The generalized equations of thermoelastic problems of thick orthotropic plates (in Polish), Scient. Bull. of Lodz Tech. Univ. Build., 21, 209-222 5. Rogowski B., 1984, Thermal stresses in a transversely isotropic layer conta- ining an annular crack. Tensile- and shear-type crack, J. Theor. Appl. Mech., 22, 3-4, 473-492 Rozwiązania podstawowe dla termicznych współczynników intensywności naprężenia typów I i II. Zagadnienie osiowo symetryczne Streszczenie Wpracy rozpatrzono zagadnienia szczeliny dla nieograniczonego termosprężyste- go ciała stałego poddanego działaniu ustalonej temperatury lub strumienia ciepła. Założono, że powierzchnie szczeliny są termicznie izolowane.Otrzymano funkcje Gre- ena dla współczynników intensywności naprężenia typów I i II. Funkcje Greena zde- Fundamental solutions related to thermal stress... 269 finiowano jako rozwiązanie zagadnienia termosprężystego, poprzecznie izotropowego ciała z kołową lub zewnętrzną szczeliną, gdy na płaszczyźnie równoległej do płasz- czyzny szczeliny działają dowolne osiowo symetryczne termiczne obciążeniawpostaci ustalonej temperatury lub strumienia ciepła, rozłożonych na okręgu. Manuscript received October 7, 2002; accepted for print January 14, 2003