Uncoupled thermoelastic analysis for a thick cylinder with radiation

Uncoupled thermoelastic analysis for a thick cylinder with radiation

THEORETICAL & APPLIED MECHANICS LETTERS 2, 021005 (2012) Uncoupled thermoelastic analysis for a thick cylinder with radiation Navneet Kumar Lamba,a) ...

424KB Sizes 1 Downloads 149 Views

THEORETICAL & APPLIED MECHANICS LETTERS 2, 021005 (2012)

Uncoupled thermoelastic analysis for a thick cylinder with radiation Navneet Kumar Lamba,a) and N. W. Khobragade P. G Department of Maths, RTM Nagpur University, Nagpur, India

(Received 15 November 2011; accepted 16 January 2012; published online 10 March 2012) Abstract An attempt has been made to study the uncoupled thermoelastic response of thick cylinder of length 2h in which heat sources are generated according to the linear function of the temperature, with boundary conditions of the radiation type. This approach is based upon integral transform techniques, to find out the thermoelastic solution. The results are obtained in terms of Bessel c 2012 The Chinese Society of Theoretical and Applied functions in the form of infinite series. Mechanics. [doi:10.1063/2.1202105] Keywords integral transform, Marchi-Fasulo transform, Marchi-Zgrablich transform, cylinder The uncoupled thermoelasticity of structural problems is frequently referred to in literature, where the assumption is used in the advanced engineering design problems for the structures under thermal shock loads. Numerical methods, such as the finite element and boundary element techniques, have also been applied to solve this class of problems. A number of analytical solutions of one dimensional uncoupled thermoelasticity problems in rectangular and cylindrical geometries have been published. The uncoupled thermomechanical phenomenon is of great importance in situations where the structures are subjected to highly various thermal conditions such as a sudden change in temperature. A number of practical examples can be seen in nuclear reactors and in structures of spacecraft and propulsion systems. However, there are only a few studies concerned with the two dimensional steady-state thermal stress. Study of the one dimensional thermoelastic waves produced by an instantaneous plane source of heat in homogeneous isotropic infinite and semi-infinite bodies of the Green–Lindsay type is presented by Hetnarski and Ignaczak.1 Also in generalized thermoelasticity analysis of the laser-induced waves propagating in an absorbing thermoelastic semi-space of the Green–Lindsay type is expressed by Hetnarski and Ignaczak.2 Bagri and Eslami3 presented a solution for the generalized thermoelasticity of a disk. They employed the Laplace transform and the Galerkin finite element method to solve the governing equations. Eraslan and Orean4 obtained the transient solution of the thermostatic-plastic deformation of internal heatgenerating tubes by using the partial differential solver PDECOL for this purpose. The PDECOL is based on the method of lines and uses a finite element collocation procedure for the discretization of the spatial variable. Ting and Chen,5 Li et al.,6 and Eslami and Salehzadeh7 applied the finite element method to solve thermoelastic problems by using governing equations formulated by Nowacki,8 and Boley and Weiner.9 Sharma and Sharma,10 developed a mathematical model for predicting the response of a thick thermoelastic axisymmetric

solid plate subjected to sudden lateral loading and thermal shock. Gosn and Sabbaghian,11 obtained the one dimensional quasi-static coupled problems of thermoelasticity region. This paper concerns with the uncoupled thermoelastic response of a thick cylinder with radiation type boundary conditions discussed by employing heat sources generated according to the linear function of temperature, further temperature distribution and stresses for thermoelastic problem presented by using integral transform technique defined in Refs. 13 and 14. We consider a thick cylinder of length 2h and the inside and outside radius a and b, respectively. The material of the cylinder is isotropic and all the material properties are assumed to be uniform. For a radially symmetric loading condition, the thermoelasticity equations reduce to the following dimensionless equations (Gosn and Sabbaghian11 )   ∂ 1 ∂ ∂T 0 0 0 (r , u ) = , (1) ∂r0 r0 ∂r0 ∂r0     1 ∂ ∂ ∂ C ∂ (r0 u˙ 0 ) , (2) r0 0 − T0 = r0 ∂r0 ∂r ∂t r ∂r0 where u0 = u0 (r0 , t0 ),

T 0 = T 0 (r0 , t0 ),

2

β 2 T0 ηβk (3λ + 2µ) α2 T0 = 2 2 = , ρ (λ + 2µ) c ρ c1 c ρc21 βT0 k β = (3λ + 2µ) α, η = , k= . k ρc C=

(3) (4)

Modifying system of Eqs. (1), (2) in two dimension form and employing heat sources Q which is generated according to the linear function of temperature, we get     ∂ 1 ∂θ0 ∂ 2 θ0 ∂θ0 0 0 0 0 k + + Q (r , z , t , θ ) = (5) ∂r0 r0 ∂r0 ∂z 02 ∂t0     ∂ ∂ 1 ∂ θ0 + Q (r0 , z 0 , t0 , θ0 ) = k 0 0 r0 0 − r ∂r ∂r ∂t C ∂ (r0 , θ0 , z 0 ) , (6) r ∂r0 where

a) Corresponding

author. Email: [email protected].

Q (r0 , z 0 , t0 , θ0 ) = Φ (r0 , z 0 , t0 ) + ψ(t0 )θ0 (r0 , z 0 , t0 ), (7)

021005-2

Theor. Appl. Mech. Lett. 2, 021005 (2012)

N. K. Lamba, and N. W. Khobragade



and θ0 (r0 , z 0 , t0 ) = T 0 (r0 , z 0 , t0 )e− 0

0

0

0

0

0

χ(r , z , t ) = Φ(r , z , t )e



Rt 0

Rt 0

ψ(ς)dς

ψ(ς)dς

,

T

0

  −(kt¯ λn,m )t0 − ℘ · e 0 n,m

Pm (z 0 ) S0 (k1 , k2 , µn r),

,

(8) where ℘n,m =

or for the sake of brevity, we consider

F (n,m) k(¯ λn,m −w)

(16)

and



χ (r0 , z 0 , t0 ) =

0

0

δ(r − r0 )δ(z − z0 ) −ω t e , 2πr0 0

(9)

where a ≤ r0 ≤ b, −h ≤ z0 ≤ h, ω > 0. Substituting Eqs. (7), (8) and (9) into system of Eqs. (5) and (6), we obtain     1 ∂T 0 ∂2T 0 ∂T 0 ∂ + + χ (r0 , z 0 , t0 ) = ,(10) k 0 0 0 02 ∂r r ∂r ∂z ∂t0     1 ∂ ∂ ∂ k 0 0 r0 0 − T 0 + χ (r0 , z 0 , t0 ) = r ∂r ∂r ∂t C ∂ (r0 , z 0 , T 0 ) . (11) r ∂r0 0 Dimensionless quantities r0 , u0 , t0 , T 0 , z 0 and σij are related to the dimensional quantities r , u , t, T, z and σij as

c1 c2 ρc3 r , r0 = t , u0 = u, k k βkT0 σij T − T0 0 c4 0 z , σij = T0 = ,z = . T0 k βT0

r0 =

(12)

Pm (h)k Pm (−h)k Pm (z0 ) − + k3 k4 2πr0 r0 S0 (k1 , k2 , µn r0 )

F (n, m) =

0

Lr (T , 1, k1 , a) = 0 Lr (T 0 , 1, k2 , b) = 0

Pm (z 0 ) S0 (k1 , k2 , µn r)e−

for all a ≤ r0 ≤ b, t0 > 0,

S0 (k1 , k2 , µn r)e

R t0 0

ψ(ς)dς

,

(

0

ψ(ς)dς

.

(17)

u0r =

∂2M ∂φ − , ∂r0 ∂r0 ∂z 0

(18)

u0z =

∂φ ∂2M + 2(1 + ν)∇2 M − 2 0 . 0 ∂r ∂ z

(19)

(14)

In which Goodier’s thermoelastic potential must satisfy the equation   1+ν 2 ∇ φ= αt T 0 , (20) 1−ν

(15)

and the Michell’s function M must satisfy the equation

where δ(r0 − r0 ) is the Dirac Delta function, a ≤ r0 ≤ b, ω > 0 is a constant; exp(−ωt0 )δ(r0 − r0 ) is the additional sectional heat available on its surface at z 0 = h , T0 is the reference temperature. In order to solve the system of uncoupled Eqs. (10) and (11), applying the integral transform defined in Appendix referred with Eqs. (A) and (B) and taking their inversion, we obtain X  ∞ ∞ X 1 1 0 0 0 0 ℘n,m e−wt + T (r , z , t ) = C λ m=1 m n=1 n

∞ (1 + ν) 0 X 1 φ (r0 , z 0 , t0 ) = αt (1 − ν) n=1 Cn

Rt

The function given in Eq. (17) represents the temperature at every instant and at all points of cylinder when there are conditions of radiation type. The displacement functions are represented by Goodier’s thermoelastic displacement potential φ(r0 , z 0 , t0 ) and Michell’s function M 12 given by the following relations

(13)



for all − h ≤ z0 ≤ h,  Lz (T 0 , 1, k3 , h) = exp(−ωt0 )δ(r0 − r0 ) Lz (T 0 , 1, k4 , −h) = exp(−ωt0 )δ(r0 − r0 )

·

Taking into account the first equation of (8), we finally represent the temperature distribution by  ∞ ∞ X 1 X 1 θ0 (r0 , z 0 , t0 ) = ℘n,m e−wt + C λ n m m=1 n=1   0  −(kt¯ λn,m )t0 T 0 − ℘n,m e ·

Subject to the boundary conditions Lt (T 0 , 1, 0, 0) = T0 for a ≤ r0 ≤ b, −h ≤ z0 ≤ h,



∇2 (∇2 M ) = 0,

(21)

where 1 ∂ ∇ = 0 0 r ∂r 2



∂ r 0 ∂r 0

 +

∂2 . ∂z 02

Substituting the value of T 0 (r, z, t) from Eq. (16) into Eq. (20), one obtains the thermoelastic displacement function φ(r, z, t) as

∞ X

) h  0  i 0 −1 ℘n,m e−wt + T 0 − ℘n,m e−(kt¯λn,m )t Pm (z 0 ) · λ (¯ λ ) m n,m m=1 (22)

021005-3

Uncoupled thermoelastic analysis for a thick cylinder with radiation

Theor. Appl. Mech. Lett. 2, 021005 (2012)

Similarly, the solution for Michell’s function M is assumed so as to satisfy the governing condition of Eq. (21) as ( ∞ ) ∞ h  0  i X X 0 −1 1 (1 + ν) ℘n,m e−wt + T 0 − ℘n,m e−(kt¯λn,m )t α0 M (r0 , z 0 , t0 ) = · (1 − ν) t n=1 Cn m=1 λm (¯ λn,m ) [Bnm sinh(µn z) + Cnm z cosh(µn z)] S0 (k1 , k2 , µn r)e

R t0 0

ψ(ς)dς

,

(23)

Using Eqs. (22) and (23) in Eqs. (18) and (19), one obtains the radial and axial displacement as ( ∞ ) ∞ h   0 i −1 (1 + ν) 0 X µn X −wt −(kt¯ λn,m )t0 0 ℘n,m e α + T 0 − ℘n,m e · ur = (1 − ν) t n=1 Cn m=1 λm (¯ λn,m ) [Pm (z) − (Bnm µn + Cnm ) cosh(µn z) + Cnm z sinh(µn z)] S00 (k1 , k2 , µn r)e u0z

R t0 0

ψ(ς)dς

,

(24)

( ∞ ) ∞ h  0  i X −1 (1 + ν) 0 X 1 −wt −(kt¯ λn,m )t0 ℘n,m e + T 0 − ℘n,m e α = · (1 − ν) t n=1 Cn m=1 λm (¯ λn,m )   −am [Qm sin(am z) + Wm cosh(an z)] + C z cosh(µ z) − 2 (−1 + 2v) C sinh(µ z) S0 (k1 , k2 , µn r) + nm n nm n −µ2n (−1 + 2v) [Bnm z sinh(µn z)]  R 0 t S 0 (k1 , k2 , µn r) e 0 ψ(ς)dς . (25) µn (2(1 − ν)) [Bnm sinh(µn z) + Cnm z cosh(µn z)] [µn S000 (k1 , k2 , µn r) 0 r

The stress functions on the traction free function are given by σz0 z0 = σr0 z0 = 0, at z 0 = ±h. The components of the stresses are represented by the use of the potential φ and Michell’s function M as     2 ∂ ∂2M ∂ φ 2 2 −∇ φ + 0 ν∇ M − , σr0 r0 = 2G ∂r02 ∂z ∂r02     ∂ 1 ∂2φ ∂M − ∇ 2 φ + 0 ν ∇2 M − σθ0 θ0 = 2G , 0 0 r ∂r ∂z ∂r0  2    ∂ φ ∂ ∂2M 2 2 σz0 z0 = 2G − ∇ φ + 0 (2 − ν ) ∇ M − , ∂z 02 ∂z ∂r02

(26)

(27) (28) (29)

and  σ

r0 z 0

= 2G

  ∂2φ ∂ ∂2M 2 + 0 (1 − ν ) ∇ M − . ∂z 0 ∂r0 ∂z ∂z 02

(30)

Using Eqs. (22) and (23) in Eqs. (27) to (30), the stress functions are obtained as ( ∞ ) ∞ h  0  i X (1 + ν) 0 X 1 −1 −wt0 −(kt¯ λn,m )t0 α ℘n,m e + T 0 − ℘n,m e σr0 r0 = 2G · (1 − ν) t n=1 Cn m=1 λm (¯ λn,m )   0  n  S (k1 , k2 , µn r0 ) 2 0 2 −Pm (z 0 ) 0 − a S (k , k , µ r ) + µ (ν − 1) µn Bnm cosh(µn z 0 ) + Cnm [z sinh(µn z 0 ) + n m 0 1 2 n r0 n o cosh(µn z 0 )] S000 (k1 , k2 , µn r0 ) + µn ν Bnm cosh(µn z 0 )µn +  o  S 0 (k , k , µ r0 ) n 0 1 2 0 Cnm [z cosh(µn z) + cosh(µn z)] + µn S0 (k1 , k2 , µn r ) · r0  R 0 t 2vCnm µ2n z cosh(µn z)S0 (k1 , k2 , µn r) e 0 ψ(ς)dς , (31) ( ∞ ) ∞ h  0  i X 0 0 (1 + ν) 0 X 1 −1 σθ0 θ0 = 2G α ℘n,m e−wt + T 0 − ℘n,m e−(kt¯λn,m )t · (1 − ν) t n=1 Cn m=1 λm (¯ λn,m )    −Pm (z 0 ) µ2n S 00 0 (k1 , k2 , µn r0 ) − a2m S0 (k1 , k2 , µn r0 ) +

021005-4

Theor. Appl. Mech. Lett. 2, 021005 (2012)

N. K. Lamba, and N. W. Khobragade

µn (ν − 1) [( µn Bnm + Cnm ) cosh(µn z 0 ) + µn Cnm z 0 sinh(µn z 0 ) ] S00 (k1 , k2 , µn r0 ) + r µ2n ν [(µn Bnm + Cnm ) cosh(µn z 0 ) + µn Cnm z 0 sinh(µn z 0 )] + [S 00 0 (k1 , k2 , µn r0 )+ µn S0 (k1 , k2 , µn r0 )] ·  R 0 t 2vCnm µ2n cosh(µn z)S0 (k1 , k2 , µn r0 ) e 0 ψ(ς)dς ,

(32)

( ∞ ) ∞ h  0  i X −1 (1 + ν) 0 X 1 −wt0 −(kt¯ λn,m )t0 ℘n,m e + T 0 − ℘n,m e α · σz0 z0 = 2G (1 − ν) t n=1 Cn m=1 λm (¯ λn,m )    0 S0 (k1 , k2 , µn r0 ) 00 0 + µ S (k , k , µ r ) + µ2n [Bnm cosh(µn z 0 ) + Cnm z sinh(µn z 0 )] (2 − ν) + −µn Pm (z 0 ) n 1 2 n 0 r0   µn S000 (k1 , k2 , µn r0 ) + r−1 S00 (k1 , k2 , µn r0 ) +  0   R 0 t S0 (k1 , k2 , µn r0 ) 00 0 2 0 (2 − ν) + µ S (k , k , µ r ) (1 − ν) µ S (k , k , µ r ) e 0 ψ(ς)dς , (33) n 1 2 n 0 1 2 n 0 n 0 r ( ∞ ) ∞ h  0  i X −1 (1 + ν) 0 X 1 −wt0 −(kt¯ λn,m )t0 α ℘n,m e + T 0 − ℘n,m e · σr0 z0 = 2G (1 − ν) t n=1 Cn m=1 λm (¯ λn,m ) n o −µn am [Qm sin (am z 0 ) + Wm sin (am z 0 )] S00 (k1 , k2 , µn r0 ) + [Bnm sinh(µn z 0 ) + Cnm z cosh(µn z 0 ) µ2n ·    (1 − ν) 2 vµn + − 2vµn Cnm sinh(µn z) S00 (k1 , k2 , µn r0 ) + r0  R 0  t S00 (k1 , k2 , µn r0 ) 0 0 3 000 0 e 0 ψ(ς)dς . (34) (1 − ν) [Bnm sinh(µn z ) + Cnm z cosh(µn z )] µn S0 (k1 , k2 , µn r ) − r0 To find Bnm and Cnm , applying boundary condition (26) to Eqs. (33) and (34), we obtain     Pm (h) A h cosh (µn h) B − C − am lµn S 00 0 (k1 , k2 , µn r0 )m (2 − ν) A + (1 − ν) µn S0 (k1 , k2 , µn r0 ) Bnm = · (2 − ν) A + (1 − ν) µn S0 (k1 , k2 , µn r0 )    (35) (hµn ) cosh (µn h) h cosh (µn h) B − C − m sinh (µn h) B, and     Pm (h) A sinh (µn h) B − am lµn S 00 0 (k1 , k2 , µn r0 ) (2 − ν) A + (1 − ν) µn S0 (k1 , k2 , µn r0 )

Cnm = (2 − ν) A + (1 − ν) µn S0 (k1 , k2 , µn r0 )    (hµn ) cosh (µn h) h cosh (µn h) B − C − m sinh (µn h) B,

· (36)

where S00 (k1 , k2 , µn r0 ) + S000 (k1 , k2 , µn r0 ), r0   S 0 (k1 , k2 , µn r0 ) S00 (k1 , k2 , µn r0 ) 3 000 0 B = µ2n [vµn + (1 − ν)] 0 , (1 − ν) µ S (k , k , µ r ) − 1 2 n n 0 r0 r02 A=

C = 2vµ2n sinh(µn h)S00 (k1 , k2 , µn r0 ), m = cosh(µn h) + (µn h) sinh(µn h).

l = Wm cos(am h) + Qm sin(am h),

To interpret the numerical computations, we consider material properties of aluminum metal, which can be commonly used in both wrought and cast forms. The low density of aluminum results in its extensive use in aerospace industry and other transportation fields. Its resistance to corrosion leads to its use in food and chemical handling (cookware, pressure vessels, etc.) and ar-

chitectural uses. Figures 1–3 give the plot of temperature distribution along the radial direction of uncoupled thermoelasticity of thick cylinder. It is observed that due to the thickness of the cylinder, a steep decrease in temperature can be found at the beginning of the transient period. As expected, temperature increases more and

021005-5

Uncoupled thermoelastic analysis for a thick cylinder with radiation

Theor. Appl. Mech. Lett. 2, 021005 (2012)

Table 1. Material properties and parameters used in this study. Property values are nominal. Modulus of elasticity, E/(N · cm−2 )

6.9 × 106

Shear modulus, G/(N · cm−2 )

2.7 × 106

Poisson ratio, υ

0.281

Thermal expansion coefficient, αt /(1 · C−1 ) 2.5 × 10−6 Thermal diffusivity, κ/(cm2 · s−1 )

0.86

Thermal conductivity, λ/[W · (mC)−1 ]

200.96

Outer radius, b/cm

0.5

Thickness, h/cm

1

Fig. 3. U versus r for different values of t.

form and Marchi-Zgrablich transform have been used to obtain the numerical results. The obtained results can be applied to the design of useful structures or machines in engineering applications. Any particular case of special interest can be derived by assigning suitable values to the parameters and functions in the expressions. The authors are thankful to anonymous referee for their valuable suggestions. They are also thankful to University Grant Commission, New Delhi for providing the partial financial assistance under major research project scheme Fig. 1. T (r, z, t) versus r for different values of t.

more gradually along the central portion of thickness direction. Finally temperature distribution further increases and attains zero value at the extreme end. The intensity of the waves of T (r, z, t)/β shows a decrease from the inner to the outer side along the radius, and the waves decrease with time t. The intensity of the wave pattern of thermoelastic displacement function increases along the radius from the centre towards the outer radius of the cylinder; but decreases with time. The radial displacement increases along the radius, but with the increase of time the radial displacement decreases and comes to zero. In this paper, the temperature distribution, displacement and thermal stresses of uncoupled thermoelastic response of cylinder due to axisymmetrical heating have been determined. The Marchi-Fasulo trans-

1. R. B. Hetnarski, and J. Ignaczak, J. Therm. Stresses 16, 473 (1993). 2. R. B. Hetnarski, and J. Ignaczak, J. Therm. Stresses 17, 377 (1994). 3. A. Bagri, and M. R. Eslami, J. Therm. Stresses 27, 691 (2004). 4. A. N. Eraslan, and Y. Orean, J. Therm. Stresses 25, 559 (2002). 5. E. C. Ting, and H. C. Chen, J. Compos. Struct. 15, 279 (1982). 6. Y. Y. Li, H. Ghoneim, and Y. Chen, J. Thermal Stresses 6, 353 (1982). 7. M. R. Eslami, and A. Salehzadeh, Application of Galerkin method to coupled thermoelasticity problem. In: Proc 5th Int Modal Anal Conf, April 6–9, 1987. 8. W. Nowacki, Thermoelasticity (Pregman Press, Oxford, 1962). 9. B. A. Boley, and J. H. Weiner, Theory of Thermal Stresses (Wiley, New York, 1960). 10. R. L. Sharma, and J. N. Sharma. Response of thermoelastic solid thick plate subjected to lateral loading. In: Proc 3rd Int Cong Thermal Stresses, June 13–17, 1999. 11. A. H. Gosn, and M. Sabbaghian, J. Therm. Stresses 5, 299 (1982). 12. N. Noda, and R. B. Hetnarski, and Y. Tanigawa, Thermal Stresses, 2nd Ed., (Taylor and Francis, New York, 2003) 13. E. Marchi, and G. Zgrablich. Bull. Calcutta Maths. 22, 73 (1946). 14. E. Marchi, and A. Fasulo. Sci. Torino CI. Sci. Fis. Mat. Natur. 101, 373 (1966/1967).

APPENDIX

Fig. 2. φ(r, z, t) versus r for different values of t.

We first introduce the integral transform13 of order n over the variable r0 . Let n be the parameter of the transform. Then the integral transform and its inver-

021005-6

Theor. Appl. Mech. Lett. 2, 021005 (2012)

N. K. Lamba, and N. W. Khobragade

sponding to the boundary condition of type (15)

sion are given by

Z Z f (n) = f (r0 ) =

b

r0 f (r0 )Sp (k1 ,k2 , µn r0 )dr0 ,

a ∞ X

f (z 0 , t0 ) =  f p (n)/Cn Sp (k1 ,k2 , µn r0 ).

(A)

n=1

The operational property is

Z

b



2

2



1 ∂f ∂ f p + − f Sp (k1 , k2 , µn x)dx = ∂x2 x ∂x x2   b ∂f Sp (k1 , k2 , µn b) f + k2 − k2 ∂x x=b   a ∂f Sp (k1 , k2 , µn a) f + k1 − µ2n f (n). k1 ∂x x=a

x a

h

f (m, t) =

We introduce another integral transform14 corre-

f (z 0 , t0 )Pm (z 0 )dz 0 ,

−h ∞ X

f (m, t0 ) Pm (z 0 ), λ m m=1

Pm (z 0 ) = Qm cos(am z 0 ) − wm sin(am z 0 ),

(B)

where Qm = am (k3 + k4 ) cos(am h), Wm = 2 cos(am h) + (k3 − k4 ) am sin(am h), Z h   2 2 λm = Pm (z)dz = h Q2m + Wm + −h

 sin(2am h)  2 2 Qm − Wm . 2am The eigen values am are the positive roots of the characteristic equation [k3 a cos(ah) + sin(ah)] [cos(ah) + k4 a sin(ah)] = [k4 a cos(ah) − sin(ah)] [cos(ah) − k3 a sin(ah)] .