thermoelastic stresses in an anisotropic slab

THERMOELASTIC STRESSES 114 AN ANISOPROPICSLAB

BY HARINDER SINGH* AND A. SINGH **

Received June 10, 1969

(Communicated by Dr. B. R. Stth, F.A.sc.)

ABSTRACT

This paper deals with the thermoelastic stresses in an infinite slabof an anisotropic material of constant finite thickness, which arise dueto temperature field varying along the thickness and length cf the slab.One face of the slab is rigidly fixed and held at a constant temperaturewhile on the other face, which is held stress-free, a continuous temperaturefield is applied. The differential equations involved are solved by thegeneral operator method and the stress, displacement and temperaturedistributions in the slab are numerically evaluated for zinc. Resultsare represented in graphs.

1. INTRODUCTION

THIS paper deals with the problem of steady state thermoelastic stresses inan anisotropic infinite flat slab of finite constant thickness due to a conti-nuous surface temperature, varying only in one direction, applied to oneface of the slab. The face of the slab which is heated is taken stress-freeand the other face is assumed to be rigidly fixed. The material of the slabis assumed to obey generalized Hooke's law and Fourier heat conductionequation. While solving the problem it has been assumed that the thermaand mechanical problems are uncoupled.

Many papers have appeared concerning problems of similar naturesuch as Charles J. Marton and Robert J. Payton' who have solved the prob-lem of thermal stresses in an isotropic slab using integral transforms andtaking surface temperature in terms of 'Heaviside step function. In thepresent paper the method employed to solve the problem is the generaloperator method, which is analogous to the method of S. G. Lekhnitskii 2

and A. I. Lur'e 3 employing which elegant solutions of simple form havebeen obtained. To be more realistic the surface temperature has been taken

• U.G.C. Research Scholar, Mathematics Department, Punjabi University, Patiala, India•* Reader, Mathematics Department, Punjabi University, Patiala, India.

167

168 HARINDER SINGH AND A. SINGH

to be continuous. The general case considered is that of the material withorthorhombic symmetry. A special case of hexagonal symmetry has alsobeen considered. The final results for displacements and stress compo-nents are obtained in series in powers of an operator, operating on the sur-face temperature. In the numerical example the material considered iszinc (hexagonal).

From the numerical examples considered in [1] and the present case, itbecomes clear that the general operator method is shorter, easy to handleand applicable in large number of cases when surface temperature is conti-nuous and satisfies certain other conditions. In the Fourier transformmethod, as in 113 , finding inverse transforms is often extremely involved andthe numerical work is often more laborious and needs the help of a com-puter for summing the truncated series consisting of complex terms whichappear in the inverse transforms. It has been found that the series involvedin the present case are very rapidly converging and give quite accurateresults on taking only fifteen terms whereas in [1] the convergence has beenattained after hundred terms near the points of discontinuity of surfacetemperature.

2. STATEMENT OF THE PROBLEM

We consider an infinite elastic anisotropic fiat slab of finite constant-

thickness h which is held rigidly on one face and is stress-free on the other.On the stress-free face we impose a temperature distribution which variesonly in one direction and is independent of time. We use the rectangula rcartesian co-ordinates x, y, z with y-axis along the axis of geometric sym-metry and x-, z-axes in the rigid face.

A. Thermal Problem

The thermal boundary-value problem is formulated in terms of tempe-rature T = T (x, y). The heat conduction equation is

(K1+K22)T=o, (1)

and the boundary conditions are

T(x,0)=0; T(x,h)=f(x) (2)

where Kl and K2 are coefficients of thermal conductivity along x and ydirections respectively and f (x) is a known even function of x, remainsfinite as x —'P oo and is of class- O°°.

Thermoelastic Stresses in an Anisotropic Slab 169

B. Mechanical Problem

The displacement components in x, y, z directions are u = u (x, y) ;v = v (x, y) and w = 0 respectively. The strain displacement relations4 are

^uex = x̂ — aT = Ex — aT

ey = ^y—T= Ev —/3T

ez = — yT

2e _ 4 + ^x = 2Exv

eyx = ezx = 0. (3)

For orthorhombic material and materials of higher order of symmetry,the elastic moduli satisfy

C14 = C15 = C16 = C24 = C25 = C26 — C34 = C35 = C36 = ("45 = (' 46

= C56 = 0.

Thus the generalized Hooke's law becomes

Qx = Cll Ex + C12 Ey — [C11a + C12$ + C13Y] T

= C12 Ex + C22 €y — [c12a + C229 + e23y] TUIZ = C13 Ex + C23Ey — [C13a + C299 + c33y] T

TXU = 2c86Ex1; Tyz — Tzx = 0 (4)

where a, P and y are coefficients of thermal expansion in x, y and z directionrespectively.

The equilibrium equations are

+ ^Txv — 0^x ^y

+ Y==Øby (5)

These are to be solved under the boundary conditions:

u (x, 0) = v (x, 0) = 0 (rigid constraint)

Qy (x, h) = TXy (x, h) = 0 (stress-free) (6)

17O HARINDER SINGH AND A. SINGH

It is convenient to use non-dimensional quantities

_yx U Vy-_ h X= 7 U 71u ^ 71 v = h .

It is clear that the introduction of the dimensionless co-ordinates anddisplacements in equations (1) to (6) does not change their form except thatnow the co-ordinates and displacements are primed. For convenience, wedrop the primes in the following discussion.

3. SOLUTION BY OPERATOR METHOD

A. Thermal Problem

Rewriting the heat conduction equation (1) in the operator form wehive

jy -I- K 8 PB T = 0 (7)

where

8 K1Kz P ^x

Solving (7) as an ordinary differential equation of order two under boundaryconditions (2) we get

T (x, y) — sin KYi' f(x) (8)sin K,6

This gives the temperature distribution throughout the slab,

B. Mechanical Problem

Equations (5) in terms of displacements are

btu bs u a$ v ^Tal x + as -T + a. bxby a° az (9)

a$u b2v a'v bT (10

)a4^x^y+a8 x -{-as b? =aeTy

where

a1= c11; as — cas; as = EH; ad = c13 + c51

as = ca + cis + c a9 and a9 — C19a + cysB + csa).


In the simultaneous partial differential equations (9) and (10)_ in u and vwe substitute p for b/ x and solving the resulting equations as ordinarydifferential equations we get

4

u (x, Y) _ e°4VP A, (x)+ A sin Kyp .f (x) (11)p—sin p4=i

4

v (x, y) _ CteaEtm At (x) + B coKyp f(x) (12)p sin Kp-1

where a{'s are the roots of the auxiliary equation

"14 + a^a2 + ass — a4' m2 + a^ 0, (13)a$a8 ae

and

Cs— (alas — a4 2) ai + asasais

asaa

A = alas + (aaae — a2as) K2a2asK4 — (aias + as ' — as K2 + alas

B = K (alas — a4a6 — aa8K2)

a2a3K4 — (a1a2 + as — a42 K 8

and A; (x) are arbitrary functions of x or constants to be determined fromboundary conditions (4). Out of a{'s at least two are complex or imaginaryand the other two may be real, imaginary or complex. In case of hexa-gonal and isotropic materials ai are all imaginary and are + i, E i.

From stress-strain relations we have

4 i Kyp {ay — P 4 (a 4 — as + a2Ctai) ee'tVP At (x) — M

snn p ./ (x)

{=1

4

C66 rxv — P Z (Ct a4) ^°°'iID Ai (x) — N cs p .f (x)

where

M= — clA+ca:3K+ (Ca +c22 9 + c,Y)

N= — AK — B.


Using the boundary condition (6) we get

4

Ai(x)=0

4

I, CiAi (x) = _ p siB f(x)

4T (a4 — as + a2Ciai) eap Ai (x) = p .f (x)

4

r (Ci + ai) eoip Ai (x) = N o t P. f (x)• (14)

Solving these simultaneous equations in four unknowns Ai (x)'s we get

A3 (x) 11(P) .f (x)

where I (p) denotes the determinant with coefficients of Ai (x)'s in the setof equations (14) as its elements and Ii (p) is the determinant obtained fromI (p) by replacing the coefficients of Ai (x) by corresponding terms on theright-hand side of each equation. All the determinants are functions of theoperator

P'^fix '

Taking

G(x)=1– f(x)

We get

4

u (x, Y) = Ii (P) a°4vp+ Apls nl K I (p) G (x) (15)

ax (x, Y) _ (cii + e12 Ciai) Ii (p) a°{vP


+ I (p) [c11 (A — a) — c12 (BK — ^) — c13Y]

sin KypX sin Kp G(x) (16)

etc.

4. CASE OF HEXAGONAL MATERIAL

If the material of the slab is hexagonal, such as zinc, magnesium, etc.,there is one axis of elastic symmetry which we take to be the z-axis. Thenon-zero elastic moduli in this case are:

C11 = C22 ; C33 ; C44 = C55 ; C66 = (C11 — c12)/2; C13 = C23 ; C12•

By Neumann principle the material exhibits the same type of symmetry inthermal properties as it does in elastic properties, i.e., a = P and K l andK2 are nearly equal; therefore in further analysis we take

K==1.

The results obtained in the previous section cannot be applied to hexa-gonal case because on making the necessary alterations for A and B theybecome infinite. Solving the differential equations (9) and (10), after makingthe necessary alterations, under the boundary conditions (6) we get

u (x, y) = y cos yp A2 + sin yp (A3 + yA 4)

v(x,y)_ —(y sin yp+P cos yp)A2 + cos ypA3

+ Cy cos yp — P sin yp) A4 + dP onYp .f (x) (17)

where

a

c12 ; a = — 2 (cll + c12) a + clsr

c11 + cl2 Cu + C12

and A2 , A3 and A 4 are arbitrary functions of x to be determined from theboundary conditions.

From stress-strain relations we get

2= 1 a) sin 2 cos A2 2p sin A3

y [( + ) yp — 2yp yp] 2 — P ypC11 — C12

Q

—[(1 +a)cos YP+2YP sin yp]A4 —dssin p.f(x)'


2 — 1— a) cos 2 sin A 2p cos AC11—Cis xu — [( ) yp 2yp YP^ E+ P yp s

+ [(1 — a) sin yp + 2yp cos yp] Aa + d co p f (x).

(18)

Applying the boundary conditions on v, au and r and solving the result-sing linear equations in three unknowns A{ (i = 2, 3, 4) we get

At (x) = 1 + a Ii (P) G (x)

where

G(x)"[1+a) — a (1 pa) sinp+a ms (l1())$f(x)

00

12 (P) (1 +a) + ' (— 1)" n pan

-i

w

Is (P) = — 1 pa + 2(2 + a) p — a (— 1)n 2nn j p sn-i

0-3

28ri+1Is (P) "^ r (— 1)"

(2n + po+t+i . (19)

Thus we get

u + a2 {Yls (P) cos yp + [Is(p) + YI4 (p)] sin yp} G (x)

11 _ cis ax — 1 -}- a)2 {[2ypcosyp + (1 + a) sin yp] 13 (p)

+ 2p sin YP Is (P) + [2yp sin yp — (1 + a) cos yp)

X I4(P) + si P I (P)} G (x) (20)

and similar expressions for other displacements and stress components.

The final form of the result depends upon the form of f (x). If f (x)is of the exponential form or if it can be expressed in terms of trigono-

Thermbeiastic Stresses' in aw Anisotropic Slab 15

metric functions, then we can determine the displacements and stress dimbution from the relations (20) by using the formulae:

sin ap x eax = sin aa.eax ; cos ap x eax = cos aa.eax

By a and a is always meant a non-zero arbitrary constant or a function of y.

If the function f (x) is not expressible in exponential form, we takeG (x) in the form of series involving f (x) and its integrals and derivatives, i.e.,

G (x) = f f (x) dx + E D 1 (x)(x) (21)

where

x' + "(x)

and

Drn+i—(— 1)n+2 (1 +a) 2 +(32n+s _.3)a .+ 8 (n+ 1)(2x+3)n+ 3) !

V(^ l)' (1_+ a)2 +_ (32+1 — 3) a + 8i(2i + 1 )i+1)!

x Dc+,._ ^+i•

Substituting (21) and (14) in (8) and (20) we get infinite series for tempe-rature displacements and stresses involving the function f (x) and its deri-vatives, i.e.,

(x, Y) - S Ten (Y) f *" (x)

00

d (x, Y) (1 u (x, Y) U2n+1 (.v).f'n+' (x)

n.0

e0

S11 (x, Y) = 2 i +. )2 ex (x, Y) = L2n (Y). n (x) (22)w-0

etc.,

176 HARINDER SINGH - AND A. SINGH •

where

To (Y) —y

T2 (y) = — [integral of TSn_$ (y) twice w.r. to y such that

Tan (0) =T2 (1)==O]

U1 (y) = 2 (1 +a)y -2 ( 1 +a),3r

u2n+,(Y)= 2 ( 1 +a)Dan-i+(a — 1)

222+ 2X E( 1)t (2i + 2) ! D2 cn-i>-i Y

2at+i

1)i+i(2i --1)! Dz (n-i)-i Y QIT"

} (— 1)2n+s (2n (2 2) (1 i a) yzn+s

1.0 (Y) = 4 ( 1 a) Y

22n+1Lan (Y) _ (3 — a)

[,,tj (— 1)i+l (2i {- 1) D2 (n-i)-i

+ 4(1 +a)D2n-i -2(3—a)Dan-s +(5—a)

N 22%x(— 1)i (2i)1 D2(n-i)-1 Y

I-(— 1)2n+1 72n

+a) zn+1(2n + 1) ! y

n =1,2,3......

Thermoelastic Stresses in an Anisotropic Stab 177

5. NUMERICAL EXAMPLE

We take f (x) = lxi e' where 0 <b < 1. This function is equi-valent to the function x.e bx for 0 <x < oo. In the numerical calculationwe take b=05.

Although the function is of the form eax fl(x) where f1(x) is a polyno-mial in x and we can get accurate values by applying the standard resultsfor such functions, we have made the calculations using the relations (22)to illustrate the method. However, for temperature, calculations havealso been made by using the result

e-bxT (x, y) = sin b [(x -f- cot b) sin yb — y cos yb] (23)

for different values of y corresponding to x = 0 and x = 1. It has beenfound that in both cases values agree upto 3 places of decimal at y = 0.2and upto 4 places of decimal after y = 0.5 as is clear from the tablebelow:

From relations T (0, • 2) T (0, • 8) T (1, • 2) T(1, • 8)

(22) —0•033908 —0. 050106 0.105729 0.462270

(23) —0034015 —0.050156 0.105629 0.462233

The results of numerical calculations have been represented in graphsfor zinc.5 The calculations have been made with the help of a desk cal-culator. While making the calculations only six places after decimal wereconsidered. The procedure adopted was: first the values of a and Den+1'swere calculated. Using these T2n's, u2n+,'s and L2n 's, etc., were calculatedfor different values of y. The final results were obtained from relations (22)for different values of x and y. The graphs have been given only for tem-perature, displacement u and normal stress S11 for want of space.

6. DISCUSSION OF RESULTS

The data have been calculated throughout the thickness of the slaband for 0 < x < 5. The displacements are dimensionless and so are thestresses. The graphs have been drawn for only one half of the slab asy-axis is the axis of symmetry-so that a reflection about y-axis is the imageof what is shown.

In Fig. 1 temperature curves have been drawn versus x for differentvalues of y. It shows that temperature increases upto about x = 2 andafter that it decreases and tends to zero as x —> oo.

HARIN ER SnNOB Am A. SINOH•s

4

•a

q

^•e

V='6

^t='4

0

2 1 _J

o I 2 3 4 5 6x ----s

FiG. 1. Plot of temperature T (x, y) versus x for different values of y.

12

10

6

4 Y6

2

Y:

o %

-20

.ti4

x ---s

1. 2. Plot of displacement A (x, y) versus x for different values of y.


Figure 2 shows that u (0, y), y 0 0 is maximum for y = 1. It decreasesrapidly becoming zero between x = 3.15 and 4.0 after which it is negativeand attains maximum magnitude for y = 1 near x = 5 which is less than1/12th of previous maximum. After that it decreases in magnitude andwithout changing sign tends to zero as x —*00.

6

4

O

-4

' e

-^6

-zo

-2 40

•=J

7m'4Y=• a7;0

2 3 4 5 6x ---+

Fio. 3. Plot of normal stress S u (x, y) versus x for different values of y.

Figure 3 shows that the normal stress S11 (x, y) is negative for x 3 0and for all y, and has its maximum magnitude there. For 0.2 < y < 1magnitude of S11 (x, y) decreases becoming zero between x = 1.3 andx = 1.95. After this it increases again and after attaining maximum bet-ween x = 3•3 and x = 4 it starts decreasing and vanishes for x — ^ oo.S11 (x, 0) decreases numerically and vanishes as x.


ACKNOWLEDGEMENT

The first author wishes to express his thanks to UGC for the awardof research grant.

REFERENCES

1. Marton, C. J. and Payton, "Thermoelastic stresses in a slab," Jour. Math. andR. G. Mech., 1964, 13, 1.

2. Lekhnitskii, S. G. .. "On the problem of elastic equilibrium of an anisotropicstrip," PMM, 1963, 27, 1.

3. Lur'e, A. I. .. Three-Dimensional Problems of the Theory of Elasticity,Interscience Publishers.

4. Sokolnikof, I. S. .. The Mathematical Theory of Elasticity, 2nd Ed., McGraw-Hill Book Co., Inc., 1956.

5. American Institute of Physics Hand—Book, McGraw-Hill Book Co., Inc.

thermoelastic stresses in an anisotropic slab

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