exact solution of the plane problem of elasticity in a

Journal of Computational and Applied Mechanics, Vol. 3., No. 2., (2002), pp. 193—206

EXACT SOLUTION OF THE PLANE PROBLEM OF ELASTICITY IN A RECTANGULAR REGION Vasyl Vihak, Yuriy Tokovyi, and Andriy Rychahivskyy Pidstryhach Institute for Applied Problems of Mechanics and Mathematics 79053 Lviv, Ukraine [email protected] [Received: January 24, 2002] Abstract. The exact solution of the quasi-static boundary value problem of plane elasticity in a rectangular region is constructed. This solution consisting of two parts satisfies both the conditions imposed on the boundary of the rectangle and the original relationships. The first part depends on the resultant vector and moment of the tractions and the second one depends on the self-equilibrated part of the tractions. Necessary existence conditions are deduced for the solution. Mathematical Subject Classification: 74B05 Keywords: plane problem of elasticity, rectangle, equilibrium and compatibility equations, eigenfunctions

1. Introduction In our opinion, the problem of constructing solutions to the quasi-static boundary value problems of elasticity and thermoelasticity in bounded regions with corner points is well elucidated by Grinchenko [1] and Grinchenko and Ulitko [2]. Also a method for solving such problems in a rectangular region has been proposed in [1, 2]. The method consists in superposing some solutions each of which is an exact one for the corresponding infinite regions without corner points. It is obvious that seeking solutions in such a form is complicated with regard to the requirement that the boundary conditions should be satisfied at the corner points of the region. In general, while constructing exact solutions to elasticity or thermoelasticity problems in regions having corner points, e.g. for a rectangle, a parallelepiped etc., one encounters great difficulties without a method for separating variables in the governing equations of elasticity — the Láme equations in terms of displacements or the Beltrami equations in terms of stresses. However, Saint-Venant’s principle makes it possible to represent solutions to the equations of elasticity in terms of stresses in the form of a superposition of two parts: the first one depending on the resultant vector and moment of the tractions, and the second one being the self-equilibrated part. This gives the idea that eigenfunctions, which should be found for the stress tensor components, could be helpful here. Unfortunately, separation of variables, e.g. in the

194

V. Vihak, Yu. Tokovyi, and A. Rychahivskyy

biharmonic equation for the plane elasticity problem in a rectangle, is complicated since there are two boundary conditions to be satisfied on each side of the rectangle. That is the reason why the two boundary conditions are equivalently replaced by an integral one in this paper when we integrate the equilibrium equations. This makes it possible to separate variables in the governing fourth-order integro-differential equation and to construct a complete set of functions for the appropriate component of the stress tensor, which consists of the eigenfunctions and the associated ones thus corresponding to Saint-Venant’s principle, and, finally, to construct the exact solution to the problem raised. 2. Formulation of the problem Let us consider the quasi-static boundary value problem of plane elasticity in terms of stresses for the rectangular region D = {(x, y) ∈ [−a, a] × [−b, b]} provided that the material is isotropic and homogeneous and there are no body forces. This problem is governed — see [3, 4] for details — by the equilibrium equations: ∂σx ∂σxy + = 0, ∂x ∂y

∂σxy ∂σy + = 0, ∂x ∂y

(x, y) ∈ D

(2.1)

and the compatibility equation: ∆(σx + σy ) = 0,

∆ = ∂ 2 /∂x2 + ∂ 2 /∂y 2 .

(2.2)

We shall assume that the normal stresses σx |x=a = −p1 (y),

σx |x=−a = −p2 (y),

σy |y=b = −p3 (x),

σy |y=−b = −p4 (x) (2.3)

and the shear stresses σxy |x=a = q1 (y),

σxy |x=−a = q2 (y),

σxy |y=b = q3 (x),

σxy |y=−b = q4 (x) (2.4)

are imposed on the boundary of the rectangle; the prescribed values are denoted by p1 , . . . , p4 and q1 , . . . , q4 . To determine the stress tensor components from the set of equations (2.1) and (2.2) under the boundary conditions (2.3) and (2.4), it is useful to replace boundary conditions (2.4) for shear stress by the corresponding conditions for derivatives of normal stresses ¯ ¯ ¯ ¯ dq1 dq2 dq3 dq4 ∂σx ¯¯ ∂σx ¯¯ ∂σy ¯¯ ∂σy ¯¯ = − = − = − =− , , , . ∂x ¯x=a dy ∂x ¯x=−a dy ∂y ¯y=b dx ∂y ¯y=−b dy (2.5) Such a replacement can be carried out with the help of equilibrium equations (2.1), which should be fulfilled on the boundary. Further on, both the boundary conditions for normal stresses (2.3) and (2.5) and those for the shear stress (2.4) will be used to construct the solution of equations (2.1)—(2.2).

Exact solution of the plane problem of elasticity in a rectangular region

195

By means of direct integration of the equilibrium equations (2.1) and separating the variables in the governing integro-differential equations [5, 6, 7, 8] formed separately for the stress components σy and σx on the basis of equation (2.2), we can construct complete sets of the orthogonal eigenfunctions and associated functions {1, x, cos(γn x/a), sin(λn x/a)} and {1, y, cos(γn y/b), sin(λn y/b)} (n = 1, 2, ...) in order to decompose the normal stresses ∞ ¡ ¢ P σx = X01 (x) + yX02 (x) + Xn1 (x) cos γn yb + Xn2 (x) sin λn yb , n=1 (2.6) ∞ ¡ ¢ P 1 2 Yn1 (y) cos γn xa + Yn2 (y) sin λn xa σy = Y0 (y) + xY0 (y) + n=1

and the tractions

pi = ai0 + ybi0 +

∞ ¡ ¢ P ain cos γn yb + bin sin λn yb ,

n=1

∞ ¡ ¢ P dqi cin cos γn yb + din sin λn yb , i = 1, 2, = ci0 + ydi0 + dy n=1 ∞ ¡ ¢ P j ajn cos γn xa + bjn sin λn xa , pj = a0 + xbj0 +

(2.7)

n=1

∞ ¡ ¢ P dqj cjn cos γn xa + djn sin λn xa , = cj0 + xdj0 + dx n=1

j = 3, 4 .

Here γn = nπ, λn > 0 are the roots of equation tan λ = λ (n = 1, 2, ...). The associated functions 1, y and 1, x define the so-called ”elementary” parts in decompositions (2.6). They depend on the parts of the tractions in (2.7) which are not self-equilibrated. The eigenfunctions cos(γn y/b), sin(λn y/b) and cos(γn x/a), sin(λn x/a) (n = 1, 2, ...) satisfying the homogeneous integral equilibrium conditions Zb

−b

σxs dy =

Zb

yσxs dy = 0,

−b

Za

σys dx =

−a

Za

xσys dx = 0

(2.8)

−a

respectively, determine in (2.7) the parts under the summation signs which correspond to self-equilibrated parts of tractions (2.7). Therefore we can present the normal stresses in the form σx = σx0 + σxs , σy = σy0 + σys . (2.9) Here and in the sequel the superscripts "0", "s" denote the elementary (not selfequilibrated) and self-equilibrated parts of solutions (2.6) or tractions, respectively. The latter should also be presented in the form pi = p0i + psi ,

qi = qi0 + qis ,

i = 1, 2, 3, 4.

(2.10)

In what follows, when no confusion can arise, the indices introduced will be omitted for the sake of brevity in writing. Now we shall consider how to derive the elementary and self-equilibrated parts of the normal stresses. Also the same parts of the shear stress will be constructed by using the method of direct integration of the equilibrium equations [7].

196

V. Vihak, Yu. Tokovyi, and A. Rychahivskyy 3. Elementary solutions

The elementary parts of the solution of the problem (2.1)—(2.3), (2.5) for the normal stresses σx0 = X01 (x) + yX02 (x),

σy0 = Y01 (y) + xY02 (y)

(3.1)

should be solutions of equations (2.1), (2.2), and should satisfy (as non-self-equilibrated tractions (2.7)) both the boundary conditions (2.3), (2.5), i.e., equations

µ

µ

¡ 1 ¢¯ 1 X0 + yX02 ¯x=(−1)i+1 a = − 2b

Zb

Zb

3y pi dy − 3 2b

pi ydy ,

−b

−b

−b Za

−b Za

¶¯ Zb Zb 1 dqi dX01 dX02 ¯¯ 3y dqi =− +y dy − 3 ydy , dx dx ¯x=(−1)i+1 a 2b dy 2b dy ¡

¢¯ 1 Y01 + xY02 ¯y=(−1)j+1 b = − 2a

pj dx −

−a Za

¶¯ dY01 1 dY02 ¯¯ =− +x dy dy ¯y=(−1)j+1 b 2a

−a

3x 2a3

i = 1, 2, (3.2)

pj xdx ,

−a

3x dqj dx − 3 dx 2a

Za

dqj xdx, dx

j = 3, 4 ,

−a

and the non-homogeneous integral equilibrium conditions [7] Za

2

σy dx = −

Za

xσy dx = −

Za

−a

Za

2

−a

−a

Zb

−a

σx dy = −

−b

2

Zb

Za

(p3 + p4 )xdx +

yσx dy = −

Zb Zb

−b

+

Zb

−b

Zb

(p1 − p2 )|y − ξ|dξ+

−b

Zb

((y − b)q3 + (y + b)q4 )dx − a

−b

−b

(q2 − q1 ) sign(y − ξ)dξ,

−b

−a

+

2

(p3 + p4 )dx +

Zb

(p1 + p2 )dy +

Za

−a

(p1 + p2 )ydy +

(q1 + q2 )sign(y − ξ)dξ,

−b

(3.3)

(q4 − q3 )sign(x − η)dη,

Za

−a

(p3 − p4 )|x − η|dη+

((x − a)q1 + (x + a)q2 )dy − b

Za

−a

(q3 + q4 )sign(x − η)dη .


197

The latter have been obtained by integrating the equilibrium equations (2.1) and taking the boundary conditions (2.3) and (2.4) into account. Also the constituents X0i , Y0i (i = 1, 2) of the solution (3.1) should satisfy the coordination conditions X01 =

1 2b

Zb

σx dy,

X02 =

−b

3 2b3

Zb

Y01 =

yσx dy,

1 2a

Za

Y02 =

σy dx,

−a

−b

3 2a3

Za

xσy dx

−a

(3.4) which follow directly from equations (3.1). It is obvious that the static equilibrium conditions Rb

−b Rb

−b

(p2 − p1 )dy +

Ra

−a

(p2 − p1 )ydy + b

(q3 − q4 )dx = Ra

Ra

−a

(q3 + q4 )dx =

−a

(p4 − p3 )dx + Ra

−a

Rb

−b

(q1 − q2 )dy = 0,

(p4 − p3 )xdx + a

Rb

(3.5)

(q1 + q2 )dy

−b

should also be satisfied by the tractions imposed on the boundary of the rectangle. Since the governing equations (3.2)—(3.5) are overdetermined, one should expect that for the existence of solutions (3.1) some necessary conditions should be laid upon the tractions. On the basis of equations (2.1), (2.2) the relations ∂ 2 σy0 ∂ 2 σx0 − =0 2 ∂x ∂y 2

∂ 2 σy0 ∂ 2 σx0 + =0 2 ∂x ∂y 2

follow for the solutions (3.1). Consequently ∂ 2 σy0 ∂ 2 σx0 = = 0. ∂x2 ∂y 2 According to these equations, by means of (3.1), (3.3), (3.4) and (2.10), we arrive at the following equations for the non-self-equilibrated part of the tractions: ¢ d 0 3y ¡ d (q30 + q40 ) = 0, (q4 − q30 ) + 2 p03 − p04 − b dx dx b ´ 3x ³ d 0 d (q10 + q20 ) = 0 . (q2 − q10 ) + 2 p01 − p02 − a dy dy a They obviously lead to the evident conditions d 0 d d 0 (q − q30 ) = (q − q10 ) = p03 − p04 − b (q30 + q40 ) = dx 4 dy 2 dx = p01 − p02 − a

d 0 (q + q20 ) = 0. (3.6) dy 1

198


In accordance with decompositions (2.7), the necessary conditions − aq43 = + =b abq34

Ra

−a Ra

(q4 − q3 )dx =

(q3 + q4 )dx +

−a

Rb

−b Ra

−a

(p2 − p1 )dy,

− bq21 =

Rb

−b

(q2 − q1 )dy =

+ (p3 − p4 )xdx, abq12 =a

q3 (a) − q3 (−a) = q4 (a) − q4 (−a),

Rb

Ra

−a

(q1 + q2 )dy +

−b

(p4 − p3 )dx, Rb

−b

(p1 − p2 )ydy,

q1 (b) − q1 (−b) = q2 (b) − q2 (−b),

(3.7)

to be satisfied by the tractions follow from equations (3.6). Here − q21 = (q2 − q1 )|y=b + (q2 − q1 )|y=−b = 2(q3 (−a) − q3 (a)) ,

− q43 = (q4 − q3 )|x=a + (q4 − q3 )|x=−a = 2(q1 (−b) − q1 (b)) ,

+ = (q1 + q2 )|y=b + (q1 + q2 )|y=−b , q12

+ q34 = (q3 + q4 )|x=a + (q3 + q4 )|x=−a .

+ + The equality q12 = q34 , which reflects the duality of the shear stress at the corner points of the rectangle, should also be fulfilled.

Consequently the elementary solutions (3.1), which satisfy equations (2.1), (2.2) and conditions (3.2)—(3.5), take the form σx0 =

1 4b

Zb ³ Zb ³ ´ ´ x x 3y (p2 − p1 ) − p1 − p2 dy + 3 (p2 − p1 ) − p1 − p2 ydy , a 4b a

−b Za

1 σy0 = 4a

−a

³

−b

´ y 3x (p4 − p3 ) − p3 − p4 dx + 3 b 4a

Za ³ ´ y (p4 − p3 ) − p3 − p4 xdx b

(3.8)

−a

if the necessary conditions (3.7) are fulfilled. The last formulae show that in the presence of conditions (3.7) the elementary solutions to the normal stresses are linear in their coordinates and depend only on pi (i = 1, 2) and pj (j = 3, 4), respectively. Now we can apply the principle of superposition. By direct integration of the equilibrium equations (2.1) and taking the boundary conditions (2.3) and (2.4) into account, one can determine the elementary part of shear stresses — for the normal stresses see (3.8) — as the sum of two constituents   Za h i 2 2 1 + 3(x − a )  + − x − y 0 σxy = aq34 − (q3 + q4 )dx + q − q21 − q43 + 4 12 a b 8a3 −a   Zb 3(y 2 − b2 )  + bq12 − (q1 + q2 )dy  . (3.9) + 8b3 −b

These constituents depend on the tractions qi (i = 1, 2) and qj (j = 3, 4), respectively and are polynomial functions of the coordinates with a degree not higher than two.

It should be noted that {1, x, x2 , sin(γn x/a), cos λn − cos(λn x/a)} (n = 1, 2, ...) is the complete set of functions for separating variables in the problem (2.1)—(2.4) for


199

σxy in respect of the variable x. Similarly, {1, y, y 2 , sin(γn y/b), cos λn − cos(λn y/b)} (n = 1, 2, ...) is the complete set in the variable y. This means that the solution to σxy can be presented in the form 1 2 3 (y) + Yxo (y)x + Yxo (y)x2 + σxy = Yxo

+

∞ X

1 [Yxn (y) sin γn

n=1

x x 2 (y)(cos λn − cos λn )] (3.10) + Yxn a a

or 1 2 3 σxy = Xyo (x) + Xyo (x)y + Xyo (x)y 2 +

+

∞ X

1 [Xyn (x) sin γn

n=1

y y 2 (x)(cos λn − cos λn )] , + Xyn b b

where the functions sin(γn x/a), cos λn −cos(λn x/a) and sin(γn y/b), cos λn −cos(λn y/b) (n = 1, 2, ...) are the eigenfunctions satisfying the homogeneous conditions s (a) σxy

=

s σxy (−a)

=

Za

−a

s σxy dx

= 0,

s σxy (b)

=

s σxy (−b)

=

Zb

s σxy dy = 0,

−b

respectively. It can be checked with ease that the necessary conditions (3.7) are also sufficient for the boundary conditions (3.2) to be satisfied by the stresses (3.8). The integral equilibrium conditions (3.7) set up for the tractions are special if they are compared with the general equilibrium conditions (3.5) since the latter are satisfied identically if conditions (3.7) are fulfilled. However, conditions (3.7) ensure the fulfillment of the principle of superposition for problem (2.1)—(2.5) in the sense that the solution can be considered as a sum of two other solutions — one for pi , qi 6= 0 (i = 1, 2) the other for pj = qj = 0 (j = 3, 4), and vice versa: pi = qi = 0 and pj , qj 6= 0.

The necessary equilibrium conditions (3.7) mean that in the absence of shear − − + + = q43 = q12 = q34 = 0, stresses at the corner points of the rectangle, i.e., when q21 for the existence of the solution of problem (2.1)—(2.5) in the form (2.6) satisfying the superposition and Saint-Venant’s principles, the tractions imposed on the boundary of the region should be equilibrated separately for pi , qi (i = 1, 2) and for pj , qj (j = 3, 4). This is so because within the general equilibrium conditions (3.5) only, superposition of the solution would be impossible. It should be finally emphasized that within the framework of the theory of elasticity, i.e., when σxy = Gexy (G is the shear modulus) due to the duality of the shear stresses at the corner points of the region D, fulfillment of conditions (3.7)5,6 is necessary for the total shear strains to be equal to zero at the corner points of the rectangle.

200

V. Vihak, Yu. Tokovyi, and A. Rychahivskyy 4. Solutions to self-equilibrated tractions

The self-equilibrated normal stresses (2.6) ∞ ¡ ¢ P Xn1 (x) cos γn yb + Xn2 (x) sin λn yb , n=1 ∞ ¡ ¢ P s Yn1 (y) cos γn xa + Yn2 (y) sin λn xa σy =

σxs =

(4.1)

n=1

should satisfy equations (2.1), (2.5) and the homogeneous integral equilibrium conditions (2.8). We shall construct both the normal stresses and the self-equilibrated shear stresses (3.10), the latter being presented in the form of the sum of two solutions s = σxy

∞ h X

1 (y) sin γn Yxn

n=1

x x i 2 (y)(cos λn − cos λn ) + + Yxn a a

∞ h X y y i 1 2 (x) sin γn + Xyn (x)(cos λn − cos λn ) , (4.2) Xyn + b b n=1

if we make use of the method proposed in [8] for constructing similar functions for the plane boundary value problem of thermoelasticity in a rectangle. Therefore σxs =

∞ X

σx(i) ,

σys =

i=1

∞ X

σy(i) ,

i=1

s σxy =

∞ X

(i) σxy ,

(4.3)

i=1

where Z a 1 ∂2 = σ (2i−1) |x − η|dη, 2 ∂y 2 −a y Z b 1 ∂2 (2i) σy = σ (2i) |y − ξ|dξ, 2 ∂x2 −b x

(2i−1) σx

Z a 1 ∂ =− σ (2i−1) sign(x − η)dη, 2 ∂y −a y Z b 1 ∂ =− σ (2i) sign(y − ξ)dξ. 2 ∂x −b x (4.4)

(2i−1) σxy (2i) σxy

Here and further on i = 1, 2, .... The constituents of normal stresses ´ ∞ ³ P (2i) (2i) (2i) σx = X1m (x) cos γm yb + X2m (x) sin λm yb , m=1 ³ ´ ∞ P (2i−1) (2i−1) (2i−1) = (y) cos γn xa + Y2n (y) sin λn xa Y1n σy

(4.5)

n=1

will be sought as the solutions of the integro-differential equations Z a (2i−1) (2i−1) ∂ 2 σy 1 ∂4 ∂ 2 l(2i−1) ∂ 2 l(2i−2) ∂ 2 σy (2i−1) + 2 + σ |x − η|dη = − ∂x2 ∂y 2 2 ∂y 4 −a y ∂y 2 ∂x2 Z b 2 (2i) 2 (2i) 2 (2i) 2 (2i−1) ∂ σx ∂ l ∂ l ∂ σx ∂4 +2 + 12 ∂x σx(2i) |y − ξ|dξ = − , 4 2 2 2 ∂y ∂x ∂x ∂y 2 −b (4.6)

Exact solution of the plane problem of elasticity in a rectangular region when the boundary conditions ½ ¯ −ps1 δi,1 , (2i) ¯ σx ¯ = −ps2 δi,1 , x=±a ¯ ½ (2i) ∂σx ¯¯ −dq1s /dyδi,1 , = ¯ −dq2s /dyδi,1 , ∂x ¯ x=±a

¯

(2i−1) ¯ σy ¯

¯

(2i−1) ¯ ∂σy ¯

¯ ¯

∂y

y=±b

=

½

½

=

y=±b

−ps3 δi,1 , −ps4 δi,1 ,

−dq3s /dxδi,1 , −dq4s /dxδi,1 ,

201

(4.7)

which follow from (2.3) and (2.5) due to equations (2.7), are satisfied. Here δi,j is the Kronecker delta; l(0) ≡ 0; " # ∞ 2 (2i−1) 2 (2i−1) X d d Y a cos γ Y cos λ n n 1n 2n +x = l(2i−1) = a dy 2 γn2 dy 2 λn n=1 ∞ n X y yo (2i−1) (2i−1) (2i−1) (2i−1) (˜ α1m + xβ˜1m ) cos γm + (˜ = α2m + xβ˜2m ) sin λm ; b b m=1 " # (4.8) ∞ (2i) (2i) X d2 X2m cos λm d2 X1m b cos γm + +y = l(2i) = b 2 dx2 γm dx2 λm m=1 ∞ n X x xo (2i) (2i) (2i) (2i) = α1n + yβ1n cos γn + (α2n + yβ2n ) sin λn . a a n=1

Thus, the possibility of separating variables in equations (4.6) has been established for representations (4.5) taking expressions (4.8) into account. For the functions in (2i−1) (2i) , Xjm (j = 1, 2), we arrive at the problem of solving the ordinary question, Yjn fourth-order differential equations · 4 ¸ ³ z ´2 2 ³ z ´4 ³ ´ ¡ zjn ¢4 d jn jn (2i−1) (2i−2) (2i−2) d − 2 + =− + yβjn Yjn αjn , 2 dy a 4 dy a a (4.9) · 4 ³ z ´2 2 ³ z ´4 ¸ ³ z ´4 ³ ´ d jm jm jm (2i) (2i−1) (2i−1) d −2 + xβ˜jm Xjm = − α ˜ jm dx2 + dx4 b b b with boundary conditions ¯

(2i) ¯ Xjm ¯

x=±a

¯

(2i) dXjm ¯¯

¯ dx ¯

=

=

x=±a

where j(i) P1m

j(i) S1n

δi,1 =− b δi,1 =− a

Zb

−b Za

−a

(

(

¯

1(i)

Pjm , 2(i) Pjm , 3(i)

Pjm , 4(i) Pjm ,

y pj cos γm dy, b x pj+2 cos γn dx, a

(2i−1) ¯ Yjn ¯ y=±b

¯ ¯ ¯ ¯

(2i−1) ¯

dYjn dy

j(i) P2m

=

1(i)

Sjn , 2(i) Sjn ,

=

(

Zb

y pj sin λm dy, b

y=±b

δi,1 =− b sin2 λm

j(i) S2n

(

3(i)

Sjn , 4(i) Sjn ,

−b

δi,1 =− a sin2 λn

Za

−a

x pj+2 sin λn dx, a

(4.10)

202

V. Vihak, Yu. Tokovyi, and A. Rychahivskyy Z b Z δi,1 b dqj δi,1 y dqj y j+2(i) =− =− cos γm dy , P2m sin λm dy , 2 b −b dy b dy b b sin λm −b Z a Z a δi,1 δi,1 dqj+2 x dqj+2 x j+2(i) = =− cos γn dx , S2n sin λn dx, 2 a −a dx a dx a a sin λn −a zjk = (2 − j)γk + (j − 1)λk , j = 1, 2; k = {n, m}.

j+2(i) P1m j+2(i)

S1n

The solution of problem (4.9), (4.10) can be given in the following form (2i)

Xjm =

4 X

∆k (

k(i) zjm b , a, x)Pjm

(2i−1)

+α ˜ jm

f1 (

zjm b , a, x)

(2i−1)

+ β˜jm

f2 (

zjm b , a, x),

k=1 (2i−1) Yjn

=

4 X

(4.11)

k(i) z ∆k ( jn a , b, y)Sjn k=1

+

(2i−2) z αjn f1 ( jn a , b, y)

+

(2i−2) z βjn f2 ( jn a , b, y),

where (2i−1)

α ˜ 1m

=

∞ h (−1)m X γn γm γn γm 3(i) 4(i) , a) K1 ( , , b)(S1n − S1n )+ K( , b a b a b n=1 i γn γm 1(i) 2(i) (2i−2) +K2 ( , , b)(S1n + S1n + 2α1n ) , a b

· ∞ (−1)m X λn γm λn γm (2i−1) 3(i) 4(i) , a) K1 ( , , b)(S2n − S2n )+ = λn K( , β˜1m ab n=1 a b a b +K2 ( (2i−1)

α ˜ 2m

=

· ∞ X 1 γn λm γn λm 1(i) 2(i) , , a) K3 ( , , b)(S1n − S1n )+ K( b2 sin λm n=1 a b a b +K4 (

(2i−1) = β˜2m

(2i)

(2i)

β1n =

¸ γn λm γn λm 3(i) 4(i) (2i−2) , , b)(S1n + S1n ) + K5 ( , , b)β1n ) , a b a b

· ∞ X 1 λn λm λn λm 1(i) 2(i) , , a) K3 ( , , b)(S2n − S2n )+ λ K( n b2 a sin λm n=1 a b a b +K4 (

α1n =

¸ λn γm 1(i) 2(i) (2i−2) , , b)(S2n + S2n + 2α2n ) , a b

¸ λn λm λn λm 3(i) 4(i) (2i−2) , , b)(S2n + S2n ) + K5 ( , , b)β2n ) , a b a b

∞ h (−1)n X γm γn γm γn 3(i) 4(i) , , b) K1 ( , , a)(P1m − P1m )+ K( a m=1 b a b a i γm γn 1(i) 2(i) (2i−1) , , a)(P1m + P1m + 2˜ +K2 ( α1m ) , b a

· ∞ (−1)n X λm γn λm γn 3(i) 4(i) , , b) K1 ( , , a)(P2m − P2m )+ λm K( ab m=1 b a b a +K2 (

¸ λm γn 1(i) 2(i) (2i−1) α2m ) , , , a)(P2m + P2m + 2˜ b a


(2i) α2n

(2i)

β2n

203

· ∞ X 1 γm λn γm λn 1(i) 2(i) , , b) K3 ( , , a)(P1m − P1m )+ = 2 K( a sin λn m=1 b a b a

¸ γm λn γm λn 3(i) 4(i) (2i−1) ˜ +K4 ( , , a)(P1m + P1m ) + K5 ( , , a)β1m ) , b a b a · ∞ X 1 λm λn λm λn 1(i) 2(i) , , b) K3 ( , , a)(P2m − P2m )+ = 2 λm K( a b sin λn m=1 b a b a ¸ λm λn λm λn 3(i) 4(i) (2i−1) ˜ +K4 ( , , a)(P2m + P2m ) + K5 ( , , a)β2m ) ; b a b a (sinh(ph) + ph cosh(ph)) cosh(pt) − pt sinh(ph) sinh(pt) −1, f1 (p, h, t) = 2 sinh(2ph) + 2ph f2 (p, h, t) = 2

h2 p sinh(ph) sinh(pt) + (sinh(ph) − ph cosh(ph))t cosh(pt) − t; sinh(2ph) − 2ph

∆j (p, h, t) = [(sinh(ph) + ph cosh(ph)) cosh(pt) − pt sinh(ph) sinh(pt)] / [sinh(2ph) + 2ph] − − (−1)j [(cosh(ph) + ph sinh(ph)) sinh(pt) − pt cosh(ph) cosh(pt)] / [sinh(2ph) − 2ph] ,

∆j+2 (p, h, t) = [t sinh(ph) cosh(pt) − h cosh(ph) sinh(pt)] / [sinh(2ph) − 2ph] − − (−1)j [t cosh(ph) sinh(pt) − h sinh(ph) cosh(pt)] / [sinh(2ph) + 2ph] ,

K(p, q, h) =

cos(ph) , (p2 + q 2 )2

K1 (p, q, h) =

j = 1, 2;

(p2 + 3q 2 ) sinh(2ph) + 2ph(p2 + q 2 ) , sinh(2ph) + 2ph

4pq2 sinh2 (ph) 8q 2 (sinh(ph) − ph cosh(ph))2 , K5 (p, q, h) = − , sinh(2ph) + 2ph p sinh(2ph) − 2ph 2 2 2 2 2 2 (q − p ) sinh(2ph) + 2(p + q )ph − 4pq h cosh (ph) , K3 (p, q, h) = sinh(2ph) − 2ph

K2 (p, q, h) = −

K4 (p, q, h) =

h(p2 + 3q 2 ) sinh(2ph) − 2(p2 + q 2 )ph2 − 4q 2 sinh2 (ph)/p . sinh(2ph) − 2ph

Having determined the constituents (4.11) of the decompositions (4.5), in accordance with formulae (4.3), (4.4) we get the self-equilibrated parts of the solution (4.1), (4.2): ∞ ¡ ¢ P X1m (x) cos γm yb + X2m (x) sin λm yb + σxs = m=1 i ∞ h 2 P d Y1n cos γn −cos(γn x/a) d2 Y2n λn (x/a) cos λn −sin(λn x/a) 2 + , +a 2 2 2 2 dy γ dy λ n=1

+b2

n

n

∞ ¡ ¢ P Y1n (y) cos γn xa + Y2n (y) sin λn xa + σys =

n=1 ∞ h 2 P d X1m cos γm −cos(γm y/b) dx2

m=1

s = −a σxy

−b Here Yjn =

∞ P

i=1

∞ P

m=1

2 γm

∞ h P dY1n sin(γn x/a)

dy γn n=1 h dX1m sin(γm y/b) dx γm

(2i−1)

Yjn

+

, Xjm =

∞ P

i=1

d2 X2m λm (y/b) cos λm −sin(λm y/b) dx2 λ2m

+

dY2n cos λn −cos(λn x/a) dy λn

i

i dX m y/b) + dx2m cos λm −cos(λ . λm

(2i)

Xjm (j = 1, 2).

−

i ,

(4.12)

204


Equivalence of representations (4.1), (4.2) to representations (4.12) is rigorously ensured by the completeness of the corresponding sets of functions. Finally solution to the problem (2.1)—(2.5) is written for normal stresses in the form (2.9), where the elementary and self-equilibrated parts are given by formulae (3.8), (4.12)1,2 . The solution to the shear stress is given by formulae (3.9), (3.10), and (4.12)3 . The solutions constructed correspond to Saint-Venant’s principle. Moreover, they correspond to it strictly, as it was Saint-Venant who pointed out the existence of the associated functions 1, y and 1, x for normal stresses σx , σy , which correspond to the tension and bending of a body with a rectangular cross-section. The eigenfunctions, which determine the parts of the solutions which depend on self-equilibrated tractions imposed on the boundary, have essential influence only not far from the ends x = ±a and y = ±b of the rectangular region. 5. Numerical results Let us consider some computational results providing a plane stress state if the rectangle is subjected to the loads p1 = p2 = p(b2 − y 2 ),

pj = qk = 0,

j = 3, 4;

k = 1, 2, 3, 4;

p = const.

Figure 1 shows how the normal stress σx divided by the traction p depends on the x-coordinate for the regions having constant width 2b = 2 while the lengths are

Figure 1. x—distribution of the stress σx /p for a = 10; 5; 3 and b = 1 2a = 6, 10, 20. It is clear from the graph that the stresses satisfy the boundary conditions even at the approximate distance (2 − 2, 5)b from the boundary, and tend to


205

Figure 2. Influence of the self-equilibrated tractions on distribution of the stress σx /p under x—shortening of the rectangle

Figure 3. Behavior of the stresses σy /p and σxy /p for a = 5 and b = 1 a constant compressive stress along the region’s width, which is equal to σx0 /p = −2/3 according to the elementary solution. Figure 2 presents the behavior of σx /p both in a square (a = b = 1) and in a rectangle with sides a = 0.5, b = 1 and for four different cross-sections y = 0; 0.3; 0.9; 1. So, if the regions have a short x-dimension compared to the width, the area of the disturbed stress caused by the boundary conditions captures the whole region. If

206


a = 0.5 and y tends to zero, the normal stress has a constant distribution in accordance with the boundary conditions, i.e., it is independent of x. The x-distribution of the stresses depicted in Figure 3 shows that the stresses are self-equilibrated. Self-equilibration of shear stress is caused by the fact that σxy is an odd function of x and y. 6. Conclusions By making use of the proposed method of direct integration, the exact solution of the boundary value problem of plane elasticity in a rectangle is constructed. This solution is the sum of the self-equilibrated and elementary parts. The last ones (3.8) determined by the associated functions 1, y and 1, x in the equations for the normal stresses σx , σy , respectively, correspond to the tension and bending. Naturally, they depend on non self-equilibrated tractions imposed on the boundary. The self-equilibrated parts (4.12) determined by the eigenfunctions, expressing the stresses caused by the self-equilibrated tractions, have an essential influence only not far from the boundary, tending to zero when moving away from it. Therefore, the decompositions (2.6) correspond to Saint-Venant’s principle. The elementary solutions (3.8), (3.9) of the problem (2.1)—(2.4) are either linear functions of the coordinates for normal stresses or parabolic ones for shear stresses. For the solutions to exist, the non self-equilibrated parts of the tractions should satisfy the equilibrium conditions (3.7). REFERENCES 1. Grinchenko, V. T.: Equilibrium and Settled Vibrations of Elastic Bounded Solids, Naukova Dumka, Kiev, 1978. (in Russian) 2. Grinchenko, V. T. and Ulitko, A. F.: Equilibrium of Elastic Solids of Canonical Form, Naukova Dumka, Kiev, 1985. (in Russian) 3. Papkovich, P. F.: Theory of Elasticity, Oborongiz, Moscow, 1939. (in Russian) 4. Timoshenko, S. P. and Goodier, J. N.: Theory of Elasticity, McGraw-Hill Book Company, New York, 1970. 5. Vihak, V. M.: Solution to the plane thermoelasticity problem in a rectangular region, Dopovidi NANU, 12, (1994), 58-61. (in Ukrainian) 6. Vihak, V. M.: Construction of the solution to the plane elasticity problem in a rectangular region, Dopovidi NANU, 10, (1995), 41-45. (in Ukrainian) 7. Vihak, V. M.: Solution of two-dimensional problems of elasticity and thermoelasticity for a rectangular region, Journal of Mathematical Sciences, 2(86), (1997), 2537-2542. 8. Vihak, V. M., Yuzvyak, M. Y., and Yasinsky, A. V.: The solution of the plane thermoelasticity problem for a rectangular domain, Journal of Thermal Stresses, 5(21), (1998), 545—561.