on temperature and stresses in a thermoelastic half space with temperature dependent properties

Meccanica DOI 10.1007/s11012-016-0610-0 On temperature and stresses in a thermoelastic half-space with temperature dependent properties Stanisław J Matysiak Dariusz M Perkowski Roman Kulchytsky-Zhyhailo Received: 25 May 2016 / Accepted: 29 December 2016 Ó The Author(s) 2017 This article is published with open access at Springerlink.com Abstract The paper deals with the axisymmetric problem of the thermoelastic half-space with temperature dependent properties The thermal coefficients: heat conductivity and coefficient of linear expansion are assumed to be functions of temperature The mechanical properties: Young modulus and Poisson ratio are taken into account as constants Two cases of boundary conditions are considered: a normal heat flux acting on a circle with given radius and two variants of the boundary conditions on the outside of the heated region: (1) a thermal insulation, or (2) a constant temperature, taken as reference The boundary is assumed to be free of mechanical loadings The linear dependences of thermal properties on temperature is considered as a special case The obtained exact results are presented in the forms of multiple integrals and the detailed analysis are derived for linear dependences of the thermal properties on temperature D M Perkowski Á R Kulchytsky-Zhyhailo Faculty of Mechanical Engineering, Białystok University of Technology, Wiejska Str 45C, 15-351 Białystok, Poland S J Matysiak (&) Institute of Hydrogeology and Engineering Geology, _ Faculty of Geology, University of Warsaw, Al Zwirki i Wigury 93, 02-089 Warsaw, Poland e-mail: s.j.matysiak@uw.edu.pl Keywords Temperature Á Heat flux Á Displacements Á Stresses Á Thermoelasticity Á Temperature dependent properties Introduction Nonhomogeneous materials, whose material properties vary continuously, have received considerable technical interest in the engineering applications The design of elements of structures, machines subjected to extremely high thermal loadings should consider changes of material properties under temperatures The solids, which in the isothermal state are characterized by constant thermal and mechanical parameters, can be treated as homogeneous bodies, but if they are subjected to high thermal loadings then their properties are dependent on temperature and indirectly vary continuously with respect to spatial variables and time The thermoelasticity of bodies with temperature dependent properties was developed by Nowin´ski [1–4] The monograph [4] includes some wide scientific descriptions of the author’s results as well as other investigators The papers [5, 6] deal with the problems of stress distributions in the thermoelastic plate with temperature dependent properties weakened by a Griffith crack The problem of stress distributions in an elastic layer with temperature dependent properties caused by concentrated loads is considered in [7] The review on thermal stresses in materials with temperature dependent properties for papers published after 123 Meccanica 1980 is presented in [8] The problems of an annular cylinder based on the finite element method is solved in [9] The paper [10] deals with the problem of SH harmonic wave propagation in an elastic layer whose shear modulus and mass density are linearly dependent on temperature In the paper [11] the wave fronts propagated in thermoelastic bodies with temperature dependent properties are analysed Some problems of thermoelasticity for thermosensitive bodies are investigated in papers [12–15] The authors assumed that the considered problems are axisymmetric or pointsymmetrical, so it is useful to introduce the cylindrical or spherical coordinates and to reduce the dimensions of the boundary value problems Boundary value problems of thermoelasticity with both thermal and mechanical properties dependent on temperature are rather too complicated for analytical approaches in the two-dimensional or three-dimensional cases So, in the paper [12] the stresses caused by thermal loadings in a layer with only mechanical properties dependent on temperature are investigated In this paper the axisymmetrical problem of thermal loadings of an elastic half-space with temperature dependent thermal properties is considered The mechanical properties are assumed to be independent of temperature (Young modulus and Poisson ratio are taken into account as constants) The elastic half-space is heated by a given normal heat flux on a circle and two cases of boundary conditions on the outside of the heated region: (1°) a thermal insulation, or (2°) a zero temperature, are investigated The boundary is assumed to be free of mechanical loadings The considered problem is stationary and axisymmetric The problem is solved for arbitrary given a priori functions dependent on temperature being the thermal conductivity and coefficient of linear expansion The linear dependences of thermal properties on temperature is analysed as a special case The obtained numerical results are presented in the form of figures for both boundary cases The influence of parameters that determine the thermal properties of the half-space on the stress distributions on the boundary is investigated coefficients being constants Let ðr; u; zÞ denote the cylindrical coordinate system, such that the plane z ¼ is the boundary surface of the half-space z [ Let T denote the temperature and q ẳ qr ; qu ; qz ị denote the heat flux vector Let K and a be the thermal conductivity and the linear expansion coefficients, respectively The mechanical properties will be denoted as follows: E be Young modulus, m be Poisson ratio In the paper the thermal and mechanical properties will be taken into account in the form: K T ị ẳ K0 f T ị; m ¼ const:; aðT Þ ¼ a0 gðT Þ; E ¼ const:; ð2:1Þ where K0 ; a0 are constants being the thermal properties of the body in the reference temperature The functions f ðT Þ; gðT Þ are a priori given functions describing changes of thermal properties under influence of temperature The functions are determined experimentally and are dependent on the kind of materials [16, 17] The half-space is heated by a normal heat flux on the circle with given radius a dependent only on variable r and two cases of the boundary conditions on the outside of heated region are considered: (1°) (2°) a thermal insulation, or zero temperature Moreover, the half-space is assumed to be free of mechanical loadings The considered problems are stationary and axisymmetric, independent on u and from the boundary conditions and symmetry of equation it follows that qu = The two following cases of the thermal boundary conditions will be taken into account: Problem qz r; 0ị ẳ q0 q r ị; for r\a and qz r; 0ị ẳ for r ! a; ð2:2Þ where qà ðÁÞ is a given function, q0 a given constant Moreover, the condition qr r; 0ị ẳ 0, qu r; 0ị ẳ that correspond to normal flux vector are considered Problem 2 Formulations of the problems qz r; 0ị ẳ q0 q r ị; for r \ a; and T r; 0ị ẳ 0; for r ! a: Consider a thermoelastic half-space with temperature dependent thermal coefficients and mechanical The solutions of both problems should satisfy the condition at infinity 123 ð2:3Þ Meccanica T ðr; zị ! for r ỵ z2 ! 1: 2:4ị Denote by ur; zị ẳ ur ; 0; uz Þ the displacement vector and by rðr; zÞ the stress tensor with nonzero components rrr ; ruu ; rzz ; rrz The boundary plane is assumed to be free of loadings, so the mechanical boundary conditions can be written: rrz r; 0ị ẳ 0; rzz r; 0ị ẳ 0; r ! 0: ð2:5Þ Solutions and analysis of results First, the temperature T satisfying Eq (2.7) with the boundary conditions (2.2) and (2.4) (for Problem 1) or (2.3) with (2.4) (for Problem 2) should be determined For this aim to a linearization of the considered problems the integral Kirchhoff’s transform will be applied (see [22]) W¼ The regularity conditions at infinity take the form: rðr; zÞ ! for r ỵ z2 ! 1: 2:6ị The temperature T and displacements ur ; uz besides the thermal and mechanical boundary conditions and the conditions at infinity should satisfy the following equations of thermoelasticity [4]: (a) the stationary equation of heat conduction     1o oT o oT K T ịr K T ị ỵ ẳ 0; r or or oz oz r ! 0; ð2:7Þ z [ 0; and (b) o ẳ 21 ỵ mị or ZT o2 ur o2 uz ỵ oz2 oroz a#ịd#; r ! 0; K ð#Þ d#: K0 ð3:1Þ Substituting (3.1) into (2.7) the thermal potential W should satisfy the linear partial differential equation   1o oW o2 W r ð3:2Þ þ ¼ 0: r or or oz Because the components of heat flux qr, qz are expressed by the potential W as follows oT oW ¼ ÀK0 ; or or oT oW ¼ ÀK0 ; qz ¼ ÀK oz oz qr ẳ K 3:3ị the boundary conditions (2.2)(2.4) can be rewritten in the form: the equilibrium equations 2ð1 À mịD21 ur ỵ 2mị ZT Problem K0 z [ 0; oWr; 0ị ẳ q0 q r ịH ða À r Þ; oz ð3:4Þ and ð1 À 2mịD20 uz ỵ 21 mị ẳ 21 ỵ mị o oz ZT Problem o2 uz o ỵ Dur oz2 oz að#Þd#; r ! 0; z [ 0; ð2:8Þ o2 o 2; ỵ r or r or o Dẳ ỵ : or r ð3:5Þ with the condition at infinity where m is Poisson’s ratio and D21 ẳ oWr; 0ị ẳ q0 q r Þ; oz for r\a; Wðr; 0Þ ¼ 0; for r [ a; K0 D20 ẳ Wr; zị ! 0; o2 o ; ỵ or r or 2:9ị for r ỵ z2 ! 1: 3:6ị The boundary value problems for potential W take the same form as for the well-known problem of temperature in the case of linear theory of heat 123 Meccanica conduction [19] The solution of Problem takes the form q0 Wr; zị ẳ K0 Z1 qà ðsÞeÀsz J0 ðsr Þds; ð3:7Þ 2uer r;zịẳ q sị ẳ f2ỵd1 d1 szịa1 sịỵ2a2 sịsgJ1 sr Þ Za rqà ðr ÞJ0 ðsr Þdr: ð3:8Þ ÂexpðÀszÞds; Z1 e 2uz r;zịẳ fd1 za1 sị2a2 sịgsJ0 sr ịexpszịds; 0 Problem is the well-known mixed boundary value problem which can be reduced to dual integral equations and next, to the Abel integral equation [20] The final solution for potential W is given by Wr; zị ẳ Z1 rerr r;zị ẳ l Z1 f2d1 ỵ1d1 szịa1 sịỵ2a2 sịsgsJ0 sr ị expszịdsỵ r Z1 f2ỵd1 d1 szịa1 sị AðsÞe Àsz J0 ðsr Þds; ð3:9Þ reuu ðr;zÞ l where A s ị ẳ Z1 where The general solution of the homogeneous equations [Eq (2.8) with the right hand side equals zero] takes the form [18, p 40]: ỵ2a2 sịsgJ1 sr ịexpszịds; Z1 ẳ f1d1 ịa1 sịgsJ0 sr ịexpszịdsỵ Za gtị sinstịdt; 3:10ị J1 sr ịexpszịds; Z1 rezz r;zị ẳ f1d1 szịa1 sịỵ2a2 sịsgsJ0 sr Þ l and q0 gð t Þ ¼ p K0 Zt xqà ð xÞdx pffiffiffiffiffiffiffiffiffiffiffiffiffiffi : t À x2 ð3:11Þ The displacements ur, uz should satisfy Eqs (2.8) together with conditions (2.5) and (2.6) The problem for displacements is linear, so the solution can be written in the form uz r; zị ẳ uer r; zị uez r; zị ỵ ỵ uth r r; zị; uth z ðr; zÞ: ð3:12Þ where uer , uez are the components of displacement vector for the problem of elasticity (under assumption that the temperature is zero—general solution) and uth r , uth are the displacements being a special solution of z Eq (2.8) 123 Z1 f2ỵd1 d1 szịa1 sịỵ2a2 sịsg 0 ur r; zị ẳ r expszịds; Z1 rerz r;zị ẳ f1ỵd1 d1 szịa1 sịỵ2a2 sịsgsJ1 sr Þ l ÂexpðÀszÞds; ð3:13Þ , l—shear modulus, and J0(Á), J1(Á) are where d1 ¼ 1À2m the Bessel functions of first kind, a1(s), a2(s) are unknowns which will be determined from mechanical boundary conditions (2.5) To obtain a special solution of Eqs (2.6) the following thermoelastic potential U is introduced [19]: uth r ¼ oU oU ; uth : z ẳ or oz 3:14ị Meccanica The following relations for the stress tensor components and potential U can be written   1o oUðr; zÞ r rth ð r; z Þ ¼ À2l ; zz r or or   o Ur; zị rth r; z ị ẳ 2l : rz oroz Zz ỵ m< expszị T s; nị sinhsnịdn ỵ sinhszị ur s; zị ẳ m: DUr; zị ẳ 1ỵm 1m Z1 z T s; nị expsnịdn ; ; Zz ỵ m< expðÀszÞ Tà ðs; nÞ sinhðsnÞdn À coshðszÞ uz ðs; zị ẳ m: a#ịd#: 3:16ị Z1 o2 oz2 where D ẳ ỵ ỵ Special solution of Eq (3.16) takes the form Ur; zị ẳ 1ỵm 1m = à o r or Z1 Tà s; nị expsnịdnỵỵ21 mị sz expszịị Z1 o2 or2 ð3:15Þ Substituting (3.14) into Eqs (2.8) we obtain ZT e th summing uer and uth r as well as uz and uz ) take the following form J0 ðsr Þds Z1 Tà ðs; nÞ expðÀsnÞdnÀð1 À 2m ỵ szị expszị z Z1 T s; nị sinhẵsn zފdn; = à T ðs; nÞ expðÀsnÞdn : ; z ð3:20Þ ð3:17Þ where Tà ðs; nÞ is the Hankel transform of the zero RT order of function a#ịd#, so  T s; nị ẳ Z1 xJ0 ðsxÞdx TZðx;nÞ að#Þd#: ð3:18Þ 0 th Knowing potential U displacements uth r , uz being the special solution of Eq (2.6) can be determined by using (3.14) and (3.15) Substituting obtained radial th and normal displacements uth r , uz into (3.12) and using (3.13)–(3.15) and (3.17) from the boundary conditions (2.5) we obtain the unknown functions a1 ðsÞ, a2 ðsÞ which are given in the general solution (3.13): a1 sị ẳ 21 2mịs 1ỵm 1m Z1 1ỵm 1m Z1 ur r; 0ị ẳ 21 ỵ mị Z1 sJ1 ðsr Þds Z1 xJ0 ðsxÞdx TZðx;nÞ Z1 Tà ðs; nÞfð1 À 2mÞ expðÀsnÞ À sinhðsnÞgdn: expðÀsnÞdn að#Þd#; Z1 sJ0 ðsr Þds Tà ðs; nÞ expðÀsnÞdn; Z1 0 uz r; 0ị ẳ 21 ỵ mị a sị ẳ The displacements ur , uz can be obtained from (3.20) by using inverse Hankel transforms of first and zero order, respectively In the future analysis we focus considerations on the stresses and displacements on the boundary plane z ¼ For this reason from Eq (3.20) and inverse transforms it follows that xJ0 ðsxÞdx TZðx;nÞ Z1 expðÀsnÞdn að#Þd#: ð3:21Þ ð3:19Þ The Hankel transforms of the first order in the case of radial displacement ur and zero order for normal displacement uz representing the final solution (after Because rzz r; 0ị ẳ 0, rrz r; 0ị ẳ we confine on the calculation of ruu ðr; 0Þ and rrr ðr; 0Þ Assuming that rzz r; 0ị ẳ from the constitutive relations [4] for z ¼ we have 123 Meccanica Fig The dimensionless stress tensor component rÃrr on boundary surface z ¼ as a function of parameter b (a) 0.5 ρ 1.5 -0.5 (b) 0.5 1.5 -1 -1.5 1: β = −0.001K −1 ; : β = −0.0005 K −1 ; -2.5 : β = K −1 ; -3 -3.5 σ rr* : β = 0.0005 K −1 ; −1 : β = 0.001K ; γ = 0.0005K −1 1 our m ur ỵ m rrr r; 0ị ẳ ỵ 2l À m or À m r À m ZT Z1 að#Þd#; m our ur ỵ m ruu r; 0ị ẳ ỵ 2l À m or À m r À m ZT að#Þd#: 1: β = −0.001K −1 ; -2 -2.5 -3 -3.5 -4 2 : β = −0.0005 K −1 ; : β = K −1 ; : β = 0.0005 K −1 ; : β = 0.001K −1 ; σ rr* γ = −0.0005K −1 J0 ðsr ÞJ0 sxị expsnịds ! 4x2 r ẳ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi F ; ; 1; À Á2 ; 4 n2 ỵ r ỵ x n2 ỵ x ỵ r 3:26ị 3:22ị From Eq (3.22) it follows that the stress components rrr ðr; 0Þ and ruu ðr; 0Þ are based on the displacement ur Taking into account Eq (3.21) and introducing the following notation K r; x; nị ẳ sJ1 sr ịJ0 sxị expsnịds; ð3:23Þ where F(Á,Á;Á;Á) is the hypergeometric function Substituting (3.26) into (3.25) we obtain ! r 4r x2 K r; x; nị ẳ q F ; ; 1; n ỵ r ỵ x2 n ỵ r ỵ x2 ! rx2 n2 ỵ x2 r 4r x2 À qÀffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Á2 : Á7ffi F ; ; 2; À 2 n ỵ r ỵ x2 n2 ỵ x2 ỵ r ð3:27Þ the radial displacement ur ðr; 0Þ can be written in the form ur r; 0ị ẳ 21 ỵ mị Z1 Z1 dn 0 B x@ TZðx;nÞ C að#Þd#AK ðr; x; nÞdx: r The derivative ou or will be calculated numerically The above presented solutions are derived for arbitrary forms of aðT Þ and K ðT Þ ð3:24Þ The integral in Eq (3.23) is calculated from the relation Z1 o K ðr; x; nị ẳ J0 sr ịJ0 sxị expsnịds: or ð3:25Þ The integral in (3.25) has the form [21] 123 -1.5 -2 Z1 ρ -0.5 -1 -4 Special case In the further analysis and numerical calculations the following coefficients of heat conduction and linear expansion are taken into account: a ẳ a0 ỵ cTị; K ẳ K0 ỵ bTị; where a0 , c, K0 , b are given constants From Eq (4.1) and (3.1) it follows that 4:1ị Meccanica Wẳ ZT Remark K #ị bT : d# ẳ T ỵ K0 4:2ị c ẳ b; 4:3ị ZT and Tẳ ỵ p ỵ 2bW : b 4:6ị then Knowing potential W from Eq (4.2) we obtain bT ỵ 2T 2W ¼ 0; It can be observed that in the case when a#ịd# ẳ a0 W; 4:7ị 4:4ị Having temperature and using (4.1) the following integral can be determined   ZT T2 a#ịd# ẳ a0 T ỵ c : ð4:5Þ what it means that the considered case presents the analogical problem to the temperature and stresses distributions for a homogeneous half-space investigated within the framework of the linear theory of thermal stresses with boundary conditions given in (2.2)–(2.6) Fig The dimensionless stress tensor component rÃuu on boundary surface z ¼ as a function of parameter b 0 0.5 -1 : β = −0.0005 K ; -2.5 -3 * σ φφ -0.2 -0.8 -1 : β = 0.0005 K −1 ; γ = 0.0005 K 0.5 1: β = −0.001K −1 ; -0.2 : β = K −1 ; : β = 0.0005 K −1 ; : β = 0.001K −1 ; σ -4 ρ 1.5 : β = −0.0005 K −1 ; -3.5 −1 (a) * φφ γ = −0.0005 K −1 (b) 0.5 1.5 ρ -0.4 σ rr* -0.6 -0.8 1: β = −0.001K −1 ; −1 : β = −0.0005 K ; −1 : β = 0K ; -1.4 -1.6 -2 ρ -3 : β = 0.001K −1 ; -1.2 -1.8 1.5 -2.5 : β = K −1 ; -0.4 -0.6 -2 −1 -4 0.5 -1.5 1: β = −0.001K −1 ; -3.5 -0.5 -1 -2 0 ρ -0.5 -1.5 Fig The dimensionless stress tensor component rÃrr on boundary surface z ¼ as a function of parameter b 1.5 1: β = −0.001K −1 ; -1 : β = −0.0005 K −1 ; -1.2 -1.4 −1 : β = 0.0005 K ; −1 : β = 0.001K ; γ = 0.0005K −1 -1.6 -1.8 -2 σ rr* : β = K −1 ; : β = 0.0005 K −1 ; : β = 0.001K −1 ; γ = −0.0005 K −1 123 Meccanica Fig The dimensionless stress tensor component rÃuu on boundary surface z ¼ as a function of parameter b (a) (b) 1 * σ φφ * σ φφ 0.5 0.5 0 0.5 ρ 1.5 1: β = −0.001K −1 ; -0.5 -1 : β = −0.0005 K −1 ; : β = K −1 ; 0.5 -0.5 34 -2 −1 : β = 0.001K ; γ = 0.0005 K −1 For further calculations the following heat flux qà ðr Þ is taken for both problems [boundary conditions (3.4)—Problem 1, and (3.5)—Problem 2]: rffiffiffiffiffiffiffiffiffiffiffiffiffi r2 à q ðr Þ ¼ À : ð4:8Þ a From Eqs (4.8) and (3.7) it follows that [21]:   p r2 > > 1À ; r\a q0 a <  2a  Wr; 0ị ẳ a 1 a2 K0 > > : F ; ; ; ; r [ a; 3r 2 r ð4:9Þ Problem From Eqs (3.9)–(3.11) and (4.8), using the following integral [21]: Zt pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi   x a2 À x2 a2 À t a þ t pffiffiffiffiffiffiffiffiffiffiffiffiffiffi dx ¼ ln at þ ; 2 aÀt t À x2 we obtain q0 Wr; 0ị ẳ K0 p  Za  a2 t a ỵ t dt p : ln tỵ aÀt 2a t À r2 r ð4:11Þ 123 : β = 0.0005 K −1 ; -1.5 -2 q0 a Wq; 0ị ẳ K0 p : β = 0.001K −1 ; γ = −0.0005 K  Z1  t2 ỵ t dt p ; ln tỵ 1t t q2 q ð4:12Þ and the following algorithm is applied to separate a singular (logarithmic) part of integral (4.12) Z1 q f tịdt p ẳ f qị 2 p t Àq p Z1 q dt pffiffiffiffiffiffiffiffiffiffiffiffiffiffi t À q2 Z1 dt ðf ðtÞ À f ðqÞÞ pffiffiffiffiffiffiffiffiffiffiffiffiffiffi t À q2 q h  i pffiffiffiffiffiffiffiffiffiffiffiffiffi ẳ f qị ln ỵ q2 ln q p Z1 dt ỵ f tị f qịị p : p t q2 ỵ ð4:10Þ The integral in (4.11) will be calculated numerically by using dimensionless variable q ¼ ar , so potential given in (4.11) can rewritten in the form p and ρ : β = −0.0005 K −1 ; r q¼ ; a Problem 1.5 : β = K −1 ; -1 : β = 0.0005 K −1 ; -1.5 1: β = −0.001K −1 ; q ð4:13Þ Meccanica Fig The maximal dimensionless tensile stress rmax ẳ 100ruu 1; 0ị=Tmax as a function of maximal temperature Tmax ẳ T 0; 0ị 0.34 0.34 * σ max 0.32 0.32 0.3 0.3 0.28 0.28 0.26 * σ max 0.26 −1 0.24 −1 1: β = −0.001K ; : β = −0.0005 K −1 ; 0.24 1: β = −0.001K ; : β = −0.0005 K −1 ; 0.22 : β = K −1 ; : β = 0.0005 K −1 ; 0.22 : β = K −1 ; : β = 0.0005 K −1 ; 0.2 : β = 0.001K −1 ; γ = −0.0005 K −1 100 200 0.34 Tmax 0.2 300 * σ max 0.32 100 200 Tmax 300 0.3 : β = 0.001K −1 ; γ = K −1 0.28 0.26 0.24 1: β = −0.001K −1 ; : β = −0.0005 K −1 ; 0.22 : β = K −1 ; : β = 0.0005 K −1 ; : β = 0.001K −1 ; γ = 0.0005 K −1 0.2 Knowing potential W from Eq (4.4) we have temperature T, what leads to determination of RT að#Þd# from (4.5) Next, using (3.24), (3.27) and (3.12) after numerical calculations the results obtained for dimensionless stress components rÃrr ðq; 0Þ, rÃuu ðq; 0Þ, where  rÃrr ; rÃuu  rrr ; ruu ; ẳ 2l1 ỵ mịa0 100K À ð4:14Þ are presented in the form of figures Further analysis of stresses will be derived numerically For this aim it can be concluded that the dimensionless stress components are dependent on four parameters qÃ0 ¼ q0 a=K0 ; b; c and m for calculations it will be taken m ¼ 0:3 and qÃ0 ¼ 500 (for Figs 1, 2, 3, 4) 100 200 Tmax 300 Problem Figure 1a presents the dimensionless stress component rÃrr on the boundary plane z ¼ for b ¼ À0:001; À0:0005; 0; 0:0005; 0:001 KÀ1 and c ¼ 0:0005 KÀ1 It can be observed that the values of rÃrr decrease together with decrease of parameter b The biggest differences between the values of rÃrr are in the centre of heating, for q ! the values of rÃrr tend to zero Figure 1b shows rrr q; 0ị for b ẳ 0:001; À0:0005; 0; 0:0005; 0:001 KÀ1 and c ¼ À0:0005 KÀ1 It is seen that for b ¼ À0:001 KÀ1 we have the smallest values of rÃrr Comparing Fig 1a with Fig 1b we observe some increase of rÃrr for the same b and small values of c The dimensionless stress component rÃuu is shown in Fig Figure 2a presents rÃuu for b¼ À1 and c ¼ À0:001; À0:0005; 0; 0:0005; 0:001 K 0:0005 KÀ1 , Fig 2b for c ¼ À0:0005 KÀ1 We 123 Meccanica observe analogical behaviour of rÃuu as rÃrr in the heating centre For q [ the differences between the curves for different values of b are very small and rÃuu ! for q ! Problem The results for the mixed boundary value problem are presented in Figs and Figures 3a, b presents dimensionless stress component rÃrr for b ¼ À0:001; À0:0005; 0; 0:0005; 0:001 KÀ1 and c ¼ 0:0005 KÀ1 as well as c ¼ 0:0005 KÀ1 , respectively The greater differences between the curves for adequate different values of b are observed in the heating region and rÃrr ! for q ! Figures 4a, b shows the dimensionless stress component rÃuu on the boundary plane for b ¼ À0:001; À0:0005; 0; 0:0005; 0:001 KÀ1 and c ¼ 0:0005 KÀ1 (Fig 4a) or c ¼ 0:0005 KÀ1 (Fig 4b) In these cases rÃuu changes sign for q % 0:9 and achieves maximal value for q ¼ (on the boundary of heated region) Moreover rÃuu tends to zero for q ! The dependences of rmax ẳ 100ruu 1; 0ị=Tmax with respect of Tmax ¼ T ð0; 0Þ are shown in Fig 5a, b, c Figure 5a presents the dimensionless stresses rÃmax for a c ¼ À0:0005 KÀ1 ; b ¼ À0:001; À0:0005; 0; 0:0005; 0:001 KÀ1 as a function of Tmax The dependences are almost linear and the highest values are obtained for b ¼ 0:001 KÀ1 Figure 5b shows the dimensionless stresses rÃmax for the same values of parameter b as Fig 5a, but different value of parameter c, namely c ¼ 0K À1 , as well as Fig 5c where it assumes that c ¼ 0:0005 KÀ1 From these figures it can be observed small differences of values rÃmax for the same values of b Young modulus and Poisson ratio The problems are solved for arbitrary forms of dependency of heat conductivity on temperature and arbitrary form of the boundary heat flux The obtained stress components in the half-space are presented in the exact forms by multiple integrals The detailed analysis of stresses on the boundary is presented for linear forms of dependencies of a and K on temperature and the boundary heat flux given by (4.8) For this case the multiple integrals are calculated partially analytically and by using numerical methods and the results are presented in the form of graphics It can be underlined that in the case of thermal conductivity K proportional to the coefficient of linear expansion the temperature and stresses distributions are analogous to the corresponding problems of homogenous half-space within the framework of the linear theory of thermal stresses Acknowledgements This work was carried out within the project ‘‘Selected problems of thermomechanics for materials with temperature dependent properties’’ The project was financed by the National Science Centre awarded based on the Decision Number DEC-2013/11/D/ST8/03428 Compliance with ethical standards Conflict of interest The authors declare that they have no conflict of interest Open Access This article is distributed under the terms of the Creative Commons 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investigated In this paper the axisymmetrical problem of thermal loadings of an elastic half- space. .. the thermal and mechanical boundary conditions and the conditions at infinity should satisfy the following equations of thermoelasticity [4]: (a) the stationary equation of heat conduction  ... the half- space is assumed to be free of mechanical loadings The considered problems are stationary and axisymmetric, independent on u and from the boundary conditions and symmetry of equation