marian ivan - theoretical geomechanics

8/3/2019 Marian Ivan - Theoretical Geomechanics

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THEORETICALGEOMECHANICS

Marian IVANUniversity of Bucharest

Department of Geophysics

6 Traian VUIA str.70138 BUCHAREST o.p.37

ROMANIAe-mail : [email protected]@math.math.unibuc.ro

Samizdat Press


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In memoriam

ProfessorMarin DOROBANU

INTRODUCTION

These course notes present some theoretical problems related to the Mechanics of a Continuum Solid Body, of particularimportance to Applied Geomechanics, Geological Engineering and Structural Geology. In most cases, only static aspects are

discussed, but some dynamic cases are also presented.

As a rule, the modern tensorial approach is used. Some basic elements are reviewed at the beginning of the notes. More detailscan be found into the excellent books of George Backus and Brian Kennett, also available at Samizdat Press. The linearelasticity and the homogeneity of the continuum solid body are almost thoroughly assumed to be valid, but some elements ofRheology are also presented.

In most cases, the semi-inverse method is used to solve the problems. According to it, the solution is supposed to be of aparticular form, as a consequence of the simplified hypothesis previously assumed. It is verified that solution checks both thecorresponding equations and the boundary conditions. Based on the Uniqueness Theorem of the Linear Elasticity, it follows theassumed particular solution is just the general solution of the problem. In all the cases discussed here, the assumed simplifiedhypotheses allow one to obtain simple, analytical solutions. At a first glance, the importance of such solutions is minor withrespect to the real cases, where mainly the non-homogeneity of the medium plays a great role. However, the analytical solutions

are the basis for deriving finite element algorithms, allowing one to model satisfactory the complex real cases. Such examplesare also presented.

The author will kindly appreciate any critical remarks on these notes, being very grateful to Samizdat Press for the possibilityto shear his work.


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CONTENTS

INTRODUCTION page

Chapter A: BASIC ELEMENTS. .... 5

A.1) The displacement vector. Lagrangean (material) and Eulerian (spatial) co-ordinates. ..5A.2) Invariants of a tensor. Tensor deviator. ...5A.3) Strain tensor. Stress tensor. Equation of motion / equilibrium. ..5A.4) HOOKEs law. .....8

Chapter B: DEFORMATION OF A CYLINDRICAL BODY IN THE PRESENCE OF GRAVITY ....9B.1) The model .9

B.2) Equations of equilibrium. Boundary conditions. Simplifying hypothesis. ......9B.3) The final shape of the body. ...12

Chapter C: LVYs PROBLEM - the triangular dam. 14

C.1) The SAINT-VENANTs equations. ....14C.2) The model. Simplifying hypothesis. The planar deformation state. ...14C.3) Equations of equilibrium. AIRYs potential. ..15C.4) Boundary conditions. The final shape of the dam......16

Chapter D: KIRSCHs PROBLEM - the circular bore hole / tunnel..19

D.1) The model. ... ..19D.2) The planar state of deformation in cylindrical co-ordinate system. ...20D.3) The circle of MOHR. ..21D.4) AIRYs potential in cylindrical co-ordinates. The bi-harmonic equation..21D.5) The divergence of a tensor in cylindrical co-ordinates. ..22D.6) The gradient of a vector and the strain tensor in cylindrical co-ordinates. ...23D.7) The bi-harmonic equation cylindrical co-ordinates. ..24D.8) The stress elements. Conditions at infinity for the stress elements. ...25D.9) Strain and displacement vector. Conditions at infinity. .27D.10) Boundary conditions for the stress elements on the wall of the circular cavity. ..28D.11) The final shape of the wall. .31

Chapter E: BOUSSINESQs PROBLEM - concentrated load acting on an elastic semi-space....33

E.1) The equations of BELTRAMI and MITCHELL. 33E.2) The model. ...35E.3) The equations of equilibrium and strain tensor in spherical co-ordinates. .35E.4) LAPLACE operator in spherical co-ordinates. LEGENDREs polynomials. .36E.5) The displacement field. ...37E.6) Boundary conditions for the stress elements. The final solution. ...39

Chapter F: ELEMENTS OF THIN PLATE THEORY....41

F.1) The model of a thin elastic plane plate.41F.2) The planar state of a plate. The bending state..41


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F.3) Loads acting on the plate..42F.4.) Odd and even functions for the planar state and for the bending state. .42F.5) Mean value of a function. Equilibrium equations for thin plates. ..43F.6) Thin plate in the bending state. ...45F.7) BERNOULLIs hypothesis. .....45F.8) HOOKEs law for a thin plate. ....46F.9) The infinite, 1-dimensional (1-D) plate. The flexure of the lithosphere. ......47F.10) Exterior forces on the lateral surface of the plate. Buckling. ....48F.11) The buckling of a simply leaning thin plate. ....50F.12) The infinite extended 1-D plate. ....51F.13) FOURIER transforms. Properties. .....51F.14) Solution of the flexure equation by using FOURIER transforms. .....52F.15) Finite plates. .....55

a) Significance of ij and ij for the bending state. ...55

b) The rectangular plate. Boundary conditions. LVYs solution. ...55F.16) Vibrations of a plate laying on a viscous substratum. ..57

a) The differential equation. ...57b) The rectangular plate with 3 embedded sides. ...58

Chapter G: THE SPHERICAL SHELL. 60

G.1) The model. BERNOULLIs hypothesis. Displacement vector and strain tensor. ..60

G.2) Quasi-mean values. Equation of motion. ...61G.3) Integrals of the stress elements. Quasi-moments. ......62G.4) Integrals of displacement vector. ....62G.5) Equation of motion in quasi-mean values. .....63G.6) Quasi-mean value of the shell density. The differential equation. .....63G.7) The buckling of a spherical shell. ......64G.8) Load on the upper face. Stress on the lower surface of the shell. ...66G.9) The differential equation of time dependent flexure. .....67G.10) Spherical effects with respect to the plane plate. ......68

Chapter H: ELEMENTS OF RHEOLOGY...69

H.1) Introduction. ....69

H.2) Linear models. .69a) KELVIN-VOIGT (strong viscous) model. .69b) MAXWELL (viscous-elastic) model. .70c) BURGERS (general linear) model. .70d) Remarks on the linear models. ...71

H.3) Non-linear models. ..73H.4) Brittle. Creep. Empirical criteria. ...73H.5) Empirical criteria for shear-faulting. TRESCA criterion. COULOMB-NAVIER criterion. .....74H.6) Von MISES-HENCKY criterion for ductile flow (plasticity).. ..76H.7) Rheological models. ...77

a) SAINT VENANT body (elastic-plastic material). .77b) BINGHAM body (visco-plastic material). .77

Chapter I: THE ACCRETION WEDGE... 78

I.1) The model. 78I.2) Equations of equilibrium. Yield condition. Stress field. ..78I.3) Boundary conditions. Final results. ..79

References...81


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A) BASIC ELEMENTS.

A.1) The displacement vector. Lagrangean (material) and Eulerian (spatial) co-ordinates.

Consider an arbitrary material point inside a continuum body, subject to a deformation process. At the initial time t0 , that

point has the position vector denoted by X

, with respect to the origin of a co-ordinate system (Fig.A1). At a time t t 0 , the

new position vector is to be x

. The difference x X

represents the displacement vector. Taking into account that the

components ( )x x x1 2 3, , of the vector x

are all functions of the components ( )X X X1 2 3, , of X

, the displacement

vector can be written as

( )U U X X X t

=

1 2 3, , , . (a1)

This represents a Lagrangean (material)description of the deformation process. Here, ( )X X X1 2 3, , are representing the

Lagrangean (material) co-ordinates. Alternately, the components ( )X X X1 2 3, , of X

can be seen as functions of the

components ( )x x x1 2 3, , of the vector x

. Consequently, the displacement vector can be written as

( )u u x x x t

=

1 2 3, , , . (a2)

This represents a Eulerian (spatial)description of the deformation process, where ( )x x x1 2 3, , are the Eulerian (spatial)co-ordinates. A basic supposition assumed thoroughly in that notes is that the deformation process is a continuous one, i.e. all

the components of u

or U

are continuous functions together their derivatives with respect to both their spatial co-ordinates or

to time.

The Lagrangean co-ordinates are usual in the Solid Mechanics, while the Eulerian co-ordinates are commonly used in Fluid

Mechanics. However, in the Linear Elasticity, the distinction between these two kinds of co-ordinates is not important, as it will

be seen in the next chapters. More details on such aspects can be found in (Aki and Richards 1980; Ranalli 1987).

Fig.A.1. The continuum deformed body and the displacement vector.

A.2)Invariants of a tensor. Tensor deviator.

A second order tensor represents mainly a 3x3 matrix. The elements of the tensor are changing according to a certain rule

with respect to a change of the co-ordinate system. Such a change with respect to a rotation will be discussed later. For

simplicity, only symmetric tensors will be considered. A symmetric tensor is equal to its transpose

T Tt= (a3)

(or Tij Tji= ). The superscript t shows the transposed tensor (matrix).Let the components of the tensor be real numbers


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T

T T T

T T T

T T T

=

11 12 13

12 22 23

13 23 33

, (a4)

The scalar and the vector u

are representing the eigen-value and the eigen-vector respectively of that tensor if

T u u u

=

, 0 . (a5)

Its easy to see that the eigen-values are not changing with respect to a rotation of the co-ordinates system.

Suppose now that the eigen-value and the components of the eigen-vector u

are complex numbers. By taking the complex

conjugate (denoted by an asterisk) into (a5), it follows

Tu u* *

=

(a6)

Taking into account the symmetry of the tensor, the next inner product is evaluated into two different ways

<

>=<

>=

T u u u u u,* , *

2

, (a7)

and

<

>=<

>=<

>=<

>=

T T Tu u utu u u u u u, * , * , * , * * *

2

(a8)

From (a7) and (a8) it follows that the eigen-values (and the components of the eigen-vectors) of a symmetric tensor are real

numbers.

Eq.(a5) can be written as

T u u

=

1 0 0, , (a9)

where

1denotes the unit tensor. From (a9) it follows that the next determinant vanishes

T T T

T T T

T T T

11 12 13

12 22 23

13 23 33

0

=

(a10)

Hence the eigen-values are the roots of the third degree equation

+ + = 3 12

2 3 0I I I , (a11)

where

( )

( )

I T T T

I T T T T T T T T T

I T T T T T T

tr T

T

1 11 22 33

2 11 22 22 33 33 11 122 232 132

3 11 22 33 12 23 13

= + + =

= + +

= + + =

,

,

... det

(a12)

With respect to a rotation of the co-ordinates system, the elements of the tensor are generally changing. Because the quantities

defined by (a12) can also be expressed as functions of the roots of (a11), it follows their values are not changing with respect to

a rotation. They represent the main invariants of the tensor. The first invariant is the trace of the tensor, while the third one

represent just its determinant.

The tensor defined by

()

*

T T trT=

1

3 1, (a13)


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represents the tensor deviator, having its trace equal to zero. Elementary computations show its second invariant is

( ) ( ) ( )I T T T T T T T T T21

611 22

222 33

233 11

26

122

232

132* = + + + + +

(a14)

That invariant is especially important to define constitutive equations for plasticity.

A.3)Strain tensor. Stress tensor. Equation of motion / equilibrium.

By using the spatial co-ordinates, the strain tensor is defined as

=

+

1

2grad u grad t u . (a15)

where grad denotes the gradient. In Cartesian co-ordinates, that symmetric tensor has the elements

( )iju i

xj

uj

xiu i, j uj i

= +

= +

1

2

1

2, . (a16)

According to the CAUCHYs hypotheses there are two kinds of forces acting at an arbitrary point placed inside a body or on

its boundary. The first ones are represented by the mass forces, characterised by a mass density b

. For the problems

discussed in that book, such mass forces are ignored. Or, they are represented by the gravity, when b

is just the gravitational

acceleration g

. Suppose now a mechanical state of tension (stress) is present inside the deformed body, e.g. as a result of the

action of a pair forces T . An arbitrary cross section is considered through a certain point of the body, dividing it into a part

denoted by Dl at left and a pert Dr at the right respectively (Fig.A.2). A surface element dS is considered on the boundary

of Dl , having the outer pointing normal vector denoted by n

. The material points of the boundary of Dr are acting on dS

by an elementary force df

. It follows (e.g. Aki and Richards 1980; Ranalli 1987) that the next relation is valid

df

dSn

=

, (a.17)

where the tensor represents the CAUCHY stress tensor, spatial co-ordinates being used. According to the Principle ofthe Kinetic Momentum Balance, stress is a symmetric tensor. It can be shown too that the Principle of Impulse Balance leads to

the next equation of motion /equilibrium

div bd u

dt +

=

2

2, (a18)

That equation is valid at an arbitrary point inside the body , where is the density and d udt

2

2

represents the acceleration. By

projecting eq.(a18) on the co-ordinates system axes, three scalar equations are obtained.

Fig.A.2. An imaginary cross section through the deformed body.


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A.4)HOOKEs law.

Neglecting the initial stress (in most cases), it is further assumed a linear relation between the stress and strain tensors, i.e.

= H , (a19)or

ij ijkl klH = , (a20)

where H is a fourth-order tensor. Eq.(a19) represents HOOKEs law. In the usual cases discussed here, an elastic,homogeneous, isotropic medium is considered. Then eq.(a19) takes the particular form

= +tr 1 2 . (a21)Here, tr denotes the trace of the tensor, 1 is the unit tensor(matrix) and , are the elastic coefficients of LAM. Hence

( )

( )

( )

11 11 22 332

11

22 11 22 332

22

33 11 22 33 2 33

12 212

12 13 312

13 23 322

23

= + + +

= + + +

= + + +

= = = = = =

,

,

, ,

, (a.22)

Alternately, HOOKEs law ( a21) can be reversed to give

=+

1

1E E

tr , (a23)

where the modulus of YOUNG is

E =+

+

3 2(a24)

and the transverse contraction coefficient of POISSON is

=

+2( ). (a25)

By reversing (a24) and (a25), it follows

( )( ) ( )

=+

=+1 1 21

2 1E E, . (a26)

The parameter defined by

=+

= 3 2

3 3 1 2

E

( ) (a27)

represents the incompressibility or bulk modulus. For (theoretical) incompressible rocks, that modulus approaches infinity.

Other constitutive equations will be discussed in relation to the rheological bodies.


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B) DEFORMATION OF A CYLINDRICAL BODY IN THE PRESENCE OF

GRAVITY

B.1) The model.

An elastic homogeneous isotropic body is considered (Fig.B1). Its initial shape is a right, vertical, very thin cylinder of

radius equal to r and height equal to H . The base of the body is placed on the horizontal, absolutely rigid, plane x Ox1 2 .

The deformation of the body due to its own weight follows to be studied and the final shape of the body into the final

equilibrium state will be found. The approximations of the linear theory are assumed and the variation of the density is ignored.The problem is solved by following the next steps:

i) - the equations of equilibrium are used, the unknowns here being the components of the stress tensor ; these equationsare processed according to the simplifying hypothesis of the problem;

ii) - by using the reversed HOOKEs law, the equations of equilibrium are processed in order to have only the components of

the strain tensor as unknowns;

iii) - by using the definition of the strain tensor, the components of the displacement vector

u are obtained and the final

shape of the body is found.

Fig.B.1. (a) A vertical cylinder lying on a rigid plane; (b) The final shape of a vertical cross section (solid line) with respect to

the initial shape (dashed line). [NO SCALE]

B.2) The equations of equilibrium. Boundary conditions. Simplifying hypothesis.

A simplified approach can be derived by using cylindrical co-ordinates. However, the problem here is an introductory one. So

these co-ordinates will be used later, in relation to other problems. The equations of equilibrium in Cartesian co-ordinates are

111 12 2 13 30

12 1 22 2 23 30

131 23 2 33 30

, , ,

, , ,

, , ,

+ + =

+ + =

+ + =g

(b1)

Here, is the density and g is the gravitational acceleration. The forces acting upon the body are the reaction force of thehorizontal plane and the gravity of the cylinder. The boundary conditions are:

-on the lateral surface of the cylinder:

n for x H x x

=

0 3 0 1 2, [ , ], , (b2)

-on the upper base of the cylinder

n for x H x x

=

= 0 3 1 2, , , , (b3)

Here, is the disc of radius equal to r , having the centre at the origin of the co-ordinate system and the boundary denoted by. The outer pointing normal at the lateral surface of the body is a linear combination with variable coefficients of thehorizontal unit vectors, i.e.

n C x x C x xe e

=

+

1 1 21

2 1 22

( , ) ( , ) (b4)


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For x H x x3 0 1 2 [ , ], , eq.(b2) becomes

C x x C x xe e e e e e1 1 2 11 1 12 2 13 3 2 1 2 12 1 22 2 23 3 0( , )[ ] ( , )[ ]

+

+

+

+

+

=

(b5)

The outer pointing normal at the upper base of the body is the unit vector

e3 . For x H x x3 1 2= , , , eq. (b3) gives

13 1 23 2 33 3

0

+

+

=

e e e (b6)

Eq. (b5) is satisfied if the stress tensor has the form

=

0 0 0

0 0 0

0 033

(b7)

on the lateral surface of the body.

Because the cylinder is a very thin one, the stress at its inner points is approximately the same one to the stress on the lateral

surface. So, it is assumed that eq.(b7) holds inside the whole volume of the body. It follows eqs.(b1a)-(b1b) are identical

verified. From eq. (b1c) it follows that

333

33 1 2 30

xg x x x H= = =, ( , , ) (b8)

The problem represented by eq.(b8) has the next immediate solution

33 1 2 3 3 ( , , ) ( )x x x g x H= (b9)The reversed HOOKEs law is

( )[ ]

( )[ ]( )[ ]

111

111 11 22 33

22 1 1 22 11 22 33

331

133 11 22 33

121

12 231

23 131

13

= + + +

= + + +

= + + +

=+

=+

=+

E

E

E

E E E

( )

( )

( )

, ,

(b10)

Because

iji

xj

j

xi

u u

= +

1

2, (b11)

eqs. (b10) lead to

1

13

2

23

3

33

1

2

2

10 2

3

3

20 3

1

1

30

u u u

u u u u u u

x

g

EH x

x

g

EH x

x

g

Ex H

x x x x x x

= = =

+ = + = + =

( ) , ( ) , ( )

, ,

(b12)

By integrating eq.(b12) it follows that


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1 3 1 1 2 3

2 3 2 2 1 3

332

23 3 1 2

12

21

0 23

32

2 31

13

1

u

u

u

g

EH x x f x x

g

EH x x f x x

g

E

xHx f x x

f

x

f

x

f

x

f

x

g

E xf

x

f

x

g

E x

= +

= +

= +

+ = + = + =

( ) ( , ) ,

( ) ( , ) ,

( ) ( , ) ,

, ,

(b13)

Hence the displacement field is found if the unknown functions f f f1 2 3, , are finally obtained. Differentiating (b13e) with

respect to x1and (b13f) with respect to x2 and adding the results, it follows

22

3

1 2 3

1

2

2

10

f

x x x

f

x

f

x+ +

= . (b14)

So, using eq.(b13d) it follows

23

1 2 0

f

x x = (b15)From eq.(b15) it follows that

f x x h x h x3 1 2 1 1 2 2( , ) ( ) ( )= + , (b16)

where h h1 2, are two unknown functions, following to be found. Eqs.(b13e) and (b13f) give

f x xx

g

Ex

dh x

dx

f x x

x

g

Ex

dh x

dx

2 1 3

32

2 2

2

1 2 3

31

1 1

1

( , ) ( ),

( , ) ( )= = (b17)

The left side of eq.(b17a) is represented by a function depending on x x1 3, only, while the right side is a function of x2 .

Hence both sides are equal to a constant, i.e.

f x xx

adh

dx

g

Ex a2 1 3

32

2

22 2

( , ),= = + (b18)

It follows that

f x x a x g x h xg

Ex a x b2 1 3 2 3 2 1 2 2

2 22

2 2 2( , ) ( ) , ( )= + = + +

(b19)

In a similar manner, eq.(b17b) gives

f x x a x g x h xg

Ex a x b1 2 3 1 3 1 2 1 1

2 12

1 1 1( , ) ( ) , ( )= + = + +

(b20)

From eq. (b13d) it follows thatdg x

dx

dg x

dxK1 2

2

2 1

1

( ) ( )= = (b21)

where K is a constant. Then

g x Kx C g x Kx C1 2 2 1 2 1 1 2( ) , ( )= + = + (b22)

For simplicity, material co-ordinates are used to obtain the final expression of the displacement field


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=

=

+ +

+ +

+

++

u

u

u

u

Xg

E

H X X

H X X

XHX X X

a X KX

a X KX

a X a X

C

C

b b

( )

( )

( )

( )

1

2

3

3 1

3 2

32

23

2 12

22

1 3 2

2 3 1

1 1 2 2

2

1

1 2

(b23)

The first term in eq.(b23) is the true displacement, the last one is a translation while the second term is the rigid rotation

aa

K

X2

1

(b24)

B.3) The final shape of the body.

a) The final shape of theupperbaseConsider an arbitrary point of co-ordinates equal to ( , , )X X X H1 2 3 = . In the initial stage, it is placed on the upper base ofthe cylinder. Finally, the co-ordinates of the point are

xx

x

XX

H

g

E

X XH

1

2

3

1

2

00

2 12

22

2

2

=

+

+

( )

(b25)

Hence

x X

x X

xg

E

X Xg

E

H H

1 1

2 2

32 1

2

2

2

2

2

==

= + +

( )

(b26)

From eq.(b26) it follows that

x x X X r12

22

12

22 2+ = + = . (b27)

Hence the circle representing the contour of the upper base remains a circle of the same radius. The plane of the circle is

moving downward by a quantity equal to g H r E( ) / ( )2 2 2 . The surface of the disc representing the upper base of thebody is no longer a plane one. It becomes a rotational parabolic surface having the equation

xg

Ex x

g

EH H3

2 12

22

2

2= + +

( ) (b28)

b) The final shape of the lower base

Consider now an arbitrary point initially placed on the lower base of the body. The point has the co-ordinates equal to

( , , )X X X1 2 3 0= . The final co-ordinate of the point are

x

x

x

X

Xg

E

HX

HX

X X

1

2

3

1

2

0

1

2

12

22 2

=

+

+

( ) /

(b29)

Hence


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x gH E X

x gH E X

xg

EX X

1 1 1

2 1 2

32 1

222

= += +

= +

( / )

( / )

( )

(b30)

From eq.(b30) it follows that

x x gH E X X gH E r12 22 1 2 12 22 1 2 2+ = + + = +( / ) ( ) ( / ) , (31)

i.e. the circle representing the contour of the lower base remains a circle. The new radius is increased by a quantity equal to

gH E/ . The initial horizontal plane of the circle is uplifted by a quantity equal to gr E2 2/ ( ) . The surface of the discrepresenting the lower base becomes a rotational paraboloid having the equation

xg

E gH Ex x3

2 1 212

22=

++

( / )( ) (b32)

c) The final shape of the lateral surface

Consider now a point initially placed on a generatrix line of the cylinder. Because of the cylindrical symmetry of the problem,

the point having the initial co-ordinates equal to ( , , )X X r X1 0 2 3= = is considered. Finally, that point has the positioncharacterised by the co-ordinates

x

x rg

EH X r

x X

g

E

X

HX

g

E r

1 0

2 3

3 332

2 3 2

2

=

= +

= +

+

( ) (b33)

From (b33), it follows that the generatrix remains into the initial vertical plane. Its shape is changed from a straight line

segment to a convex parabolic segment, having the equation

( ) ( )xE

gx r

E

gx r H

g

EH

g

Er3

2 22 1

22 1

2

2

2

2= + +

/ / (b34)

OBSERVATION. On the lower base of the cylinder it is acting the reaction force of the rigid plane, equal to the weight of the

body. When the surface of the base is decreasing, approaching the paraboloid of eq.(b32), the normal unit effort (equal to the

weight divided by the contact area) is increasing. At a certain moment, its magnitude will exceed a yielding value of the

material. Then, HOOKEs law, valid in the elastic domain, will be no longer appropriate here.


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C) LVYs PROBLEM - THE TRIANGULAR DAM

C.1) The SAINT-VENANT s equations.

Differentiating a certain element of the strain tensor

ij i, j j iu u = + , / 2 , (c1)it follows, for example, that

( ) ( )( ) ( ) ( )

11 22 2211 11 22 2 2 11 1122 2 211

1 2 12 2 1 12 1 2 2 1 122

1212

, , , , , , , ,

, , , , , , , ,

+ = + = +

= + = + =

u u u u

u u u u(c2)

Hence

11 22 22 112

1212, , , + = (c3)Also,

22 33 33 222

23 23, , , + = (c4)3311 11 33

23131, , , + = (c5)

In a similar way, it follows that

( )

( )

( )

12 3 231 31 2 2 22 31

231 31 2 12 3 3 3312

31 2 12 3 231 1 11 23

, , , , ,

, , , , ,

, , , , ,

+ =

+ =

+ =

(c6)-(c8)

The above equations (c3)-(c8) represent the SAINT-VENANTs equations of compatibility.

C.2) The model . Simplifying hypothesis. The planar deformation state.

A horizontal dam of infinite length is considered. The cross-section is represented by a rectangular triangle OAB (Fig.C1).The length of the base is AB=l and the height is OA=h. On OA catheter is acting the hydrostatic pressure of a liquid (water)

having the specific weight equal to . As a result, the dam is deformed. The dam is represented by an elastic homogeneous,isotropic material. Its specific weight is equal to and its elastic constants are E and.

Fig.C1. A vertical cross section through the dam. N.Hs. is the free surface of the water, acting on OA side by a pressure linearly

increasing with depth.

Because the shape of the dam, the displacement vector has the components like


15/81

15

1 1 1 2

2 2 1 2

30

u u

u u

u

x x

x x

=

=

=

( , )

( , ) (c9)

It follows the strain tensor components are like

( )( )

( )

11 11 11 1 2

12

1

2 1 2 2 1 12 1 2

131

2 1 3 3 10

22 2 2 22 1 2

231

2 2 3 3 20

33 3 30

= =

= + =

= + =

= =

= + =

= =

, ( , )

, , ( , )

, ,

, ( , )

, ,

,

u

u u

u u

u

u u

u

x x

x x

x x

(c10)

Hence the strain matrix is

( )

=

=

x x1 211 12

0

12 220

0 0 0

, , (c11)

It corresponds to aplanar state of the strain (the plane here being 1-2).

The components of the stress tensor are

( )

( )

( ) ( ) ( ) ( )

11 11 222

11

122

12

132

130

22 11 22 2 22

232

230

33 11 222

33 11 22 2 11 22 11 22

= + +

=

= =

= + += =

= + + = + =+

+ = +( )

(c12)

Hence the stress matrix is

( )

( )

=

=

+

x x1 2

11 120

12 220

0 011 22

, (c13)

Because the component 33 of the stress has a non-zero value, eq.(c13) shows that the stress state corresponding to a planar state

of the strain is not generally a planar one too.

C.3) Equations of equilibrium. AIRYs potential.

The only body force acting on the dam is its weight. The equations of equilibrium are

111 12 20

12 1 22 20

, ,

, ,

+ + =

+ =

(c14)

Because the presence of , eqs.(c14) represent a non-homogeneous system. In the beginning, the homogeneous system issolved, i.e.


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16

111 12 20

12 1 22 2 0

, ,

, ,

+ =

+ =

(c15)

Using an unknown function , the first equation of (c15) is verified for

112

121

= =

x x

, (c16)

In the same way, the second equation of (c15) is verified for

122

121 = = x x, (c17)

It follows that

x x1 2

0+ = (c18)

i.e. the unknown functions are

= =

A Ax x2 1

, (c19)

The unknown function A A x x= ( , )1 2 represents the AIRYs potential. It allows one to obtain the next expressions for

the components of the stress tensor when the body force are absent:

11 22 12 12 22 11 = = =, , , , ,A A A (c20)From (c13), the trace of the stress tensor can be written using LAPLACEs operator in 1-2 co-ordinates

( )tr A = + + = +( ) ( ) *1 11 22 1 (c21)The components of the strain tensor are obtained using the reversed HOOKEs law

111

22 221

11 121

12

=+

=

+

=

+E E E

A A A A A,* , ,

* , , . (c22)

Using (c22) and (c3) it follows

,*

,,

*

,,2222

221111

112

1212A A A A A

+

= , (c23)

i.e.

( ) * *1 0 = A (c24)

Because < 05. , it follows that AIRYs potential is a solution of the bi-harmonic equation

* * A = 0 (c25)Because the trace of a tensor is an invariant, eq.(c25) holds too in the general case of the orthogonal curvilinear co-ordinates.

However, eq.(c20) has to be modified.

C.4) Boundary conditions. The final shape of the dam.

On the side OA of the dam is acting the hydrostatic pressure. It follows that

( )

=

2 1 2e ex (c26)On the side OB of the dam is acting the negligible atmospheric pressure. It follows that

=

n 0 (c27)where the outer pointing normal at the dam is

=

+

n e esin cos 1 2 (c28)

On the side OA, for x h x1 0 2 0 =[ , ], , it follows that


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17

120

22 1

=

=

,

x(c29)

On the side OB it follows for x h x x1 0 2 1 =[ , ], tan that

12 110

22 120

=

=

tan ,

tan(c30)

Eqs.(c29)-(c30) represent 4 boundary conditions, suggesting a solution of the bi-harmonic equation (c25) which depends on 4

unknown coefficients denoted by a b c d, , , , i.e.

A x xa

xb

x xc

x xd

x( , )1 26 1

3

2 12

22

1 22

6 23= + + + (c31)

Using (c20), the solution of the homogeneous system is

11 1 2

12 1 2

22 1 2

= +

= +

= +

cx dx

bx cx

ax bx

( ) (c32)

A particular solution of the non-homogeneous system (c14) is

11 22 0

12 2

= ==

x

(c33)

It follows the general solution of (c14) is

11 1 2

12 1 2 2

22 1 2

= +

= +

= +

cx dx

bx cx x

ax bx

( ) . (c34)

Replacing (c34) into (c29)-(c30) it follows that

+ = =

+ = = + + = =+ + + + = =

( ) , [ , ] ,

, [ , ] ,( ) tan ( ) , [ , ] , tan

( ) tan , [ , ] , tan

bx cx x for x h x

ax bx x for x h xcx dx bx cx x for x h x x

bx cx x ax bx for x h x x

1 2 2 0 1 0 2 0

1 2 1 1 0 2 01 2 1 2 2 0 1 0 2 1

1 2 2 1 2 0 1 0 2 1

(c35)

It follows that

a

b

c

d

= =

= +

=

0

2

2 3

/ tan

/ tan / tan

(c36)

and

11 1 2

12 2

22 1

= +

=

=

Ax Bx

Cx

x

(c37)

where

A h l B h l h l C h l= = = 2 2 2 3 3 2 2/ , / / , / (c38)Hence

11 11 1 1 2 2,u C x C x= = + , (c39)where


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18

C A A E C B E1 1 2 12= + = ( )[ ( )] / , ( ) / (c40)

It follows that

1 1 12 2 2 1 2 1 2u C x C x x f x= + +/ ( ) (c41)

where the unknown function f1 follows to be found. In the same manner,

2 3 1 2 4 22 2 2 1u C x x C x f x= + +/ ( ) (c42)

But

( )12 12 1 2 2 1 12 2 1 1 2 3 2 2 1 1 12 1 2 = + = + + + = + = +, , ( ( ) ( )u u C x f x C x f x E E Cx (c43)Hence

C x f x K

C x f x K

2 1 2 1

3 2 1 2

+ =

+ =

( )

( )

, (c44)

where K is an arbitrary constant. It follows

f x C x Kx K f x C x Kx K1 2 3 22 2 2 1 2 1 2 1

2 2 1 2( ) / , ( ) / = + = + + (c45)Hence, the displacement field is

1 1 1

2 22 1 2 3

2 12

2 22 1

2 2 12 2 3 1 2 4 2

2 2 1 2uu

C x C x x C C E x Kx K

C x C x x C x Kx K

= + + + +

= + + + +

/ [ ( ) / ] /

/ /

(c46)

The last terms into (c46) represent a rigid roto-translation.

It should be outlined that the above boundary conditions on stress values on the sides OA and OB are not complete ones. As a

result, the unknown constants C C3 4, are present in (c46). Boundary conditions on stress values (or displacements) on the

side AB are required in order to obtain an unique solution of the problem

For example, consider the case when the points A and B are fixed ones. It follows

( ) ( )

( ) [ ]{ }

1 1 12 2 2 2 1 2 3 2 1 2 2 2

2 22

12 2 3 2 1 2 2 3 2 1 2 1

u

u

C x h C x h x C E x l x

C h x C x x x h l C h C E l x h

= + + +

= + + + +

/ [ ( )C / ] /

/ / ( )C / / ( )

(c47)

An arbitrary point placed initially on the side AB has the initial co-ordinates ( )X h X1 2= ; . Its final position is

x X X X h C C E X l X

x X X X X C hX l X l

u

u

1 1 1 1 2 32 1 2 2 2

2 2 2 1 2 2 3 2 2

= + = + + +

= + = +

( , ) [ ( ) / ] ( ) /

( , ) ( ) /

(c48)

Elementary computations show that

CE

CE

h

l3

2 1 11 2

2

2+

+=

+

( )( ) ( )

(c49)

If

C E C3

2 1

0+

+

0 , where R is the radius of the bore hole.It follows to solve (d43) by using (d44) in order to derive the AIRYs potential. The stress components will be obtained from

(d27), imposing the boundary conditions on the wall of the bore hole. The components of the strain will be derived by using the

HOOKEs reversed law. The displacement vector will be obtained from the definition of strain elements, allowing one to find

the final shape of the deformed bore hole wall.

Consider the FOURIER expansion of the AIRYs potential, having the coefficients equal to functions of r

[ ]A A A Br r n r n n r nn

( , ) ( ) ( ) cos ( )sin = + +=

0

1

(d45)

It follows

A A A Br n

nn

n

n

= + +

=

01

cos

sin , (d46)

2

2 01

AA A B

rn

nn

n

n

= + +

=

cos sin , (d47)

( )

2

22

1

AA Bn n n n n

n

= +=

cos sin . (d48)

Hence

A A A A A B B Br

n

rn

r

n

rn

n

= + +

+ +

=

0

1 2

2

1 2

21

cos

sin (d49)

and

A A A A A A A

B B B B B

r

n

r

n

r

n n

rn

n

r

n

r

n

r

n n

rn

= + + +

++

+

=

+ + +

++

+

02 2 2 1

2

2 2 1

3

4 4 2

41

2 2 2 1

2

2 2 1

3

4 4 2

4

cos

sin

. (d50)

By using (d50), the bi-harmonic (d43) is verified if

0

0A = , (d51)and if the functions A B, are the solutions of the next differential equation

+ +

++

+

=2 2 2 1

2

2 2 1

3

4 4 2

40

r

n

r

n

r

n n

r. (d52)

Because0A is a function of r only, eq.(d51) is

1 1 0 0r

d

drr

d

dr r

d

drr

d

dr

A

= . (d53)


25/81

25

Hence

rd

dr r

d

drr

d

dra

d

drr

d

dra r r b r

A A1 00

00 0

=

= +, ln . (d54)

But

r rdrr

r const.ln ln=

+2

2

1

2(d55)

Hence

rd

dr

ar r

br c

A0 02

2 1

2

0

2

20=

+ +ln (d56)

Finally, denoting again the constants, it follows

0 02

02

0 0A r a r r b r c r d( ) ln ln= + + + . (d57)In order to solve eq.(52), a solution of the form

= rm (d58)is considered. Substituting (d58) in (d52), it follows that the exponent m is the solution of the algebraic equation

m m m m m m m n m m n m n n( )( )( ) ( )( ) ( ) ( ) ( ) + + + + + =1 2 3 2 1 2 2 2 1 1 2 2 1 4 4 2 0 ,(d59)

having the rootsm n m n m n m n1 2 2 3 4 2= = + = = +, , , . (d60)

Hence the AIRYs potential is

A r a r r b r c r d an rn bn r

n cn rn d n r

n n

n

n rn

n rn

n rn

n rn n

n

( , ) ln ln cos

sin

= + + + + + + + + +

=

+ + + + + +

=

02

02

0 02 2

1

2 2

1

,(d61)

where the unknown coefficients ai bi ci d i j j j j i j, , , , , , , , , , , ... , , , ... = =0 1 2 1 2 follows to be obtained.

D.8) The stress elements. Conditions at infinity for the stress elements.

By using (d61) and (d27), it follows

( )

= = + +

+ + + + + + +

=

+ + + + + + +

=

,ln

( )( ) ( ) ( )( ) ( ) cos

( )( ) ( ) ( )( ) ( ) sin

rra r b

c

r

an n n rn bn n n r

n cn n n rn d n n n r

n n

n

n n n rn

n n n rn

n n n rn

n n n rn n

n

A 0 2 3 2 002

2 1 1 2 2 1 1 2

1

2 1 1 2 2 1 1 2

1

(d62)

At great distances from the cylindrical cavity, the elastic perturbation has to vanish, i.e.

r=lim 0 . (d63)

It follows

a b b an n n

bn n n

0 0 2 0 2 2 2 2 0 0 1 2

0 3 4

= + + = = = =

= = =

, , , , , ...

, , ,...

(d64)

Hence the AIRYs potential is


26/81

26

( ) ( )

( )

A r c r d b r cn n n n rn

n

d n n n n rn

n

( , ) ln cos sin

cos sin

= + + + + + +

=

+ +

=

0 0 1 12

1

1

(d65)

and

( )

( )

= + + =

+ + +

=

cr

n n cn n n n r n

n

n n d n n n n rn

n

02

2 1

1

1 2

1

( )( ) cos sin

( ) cos sin

. (d66)

From (d65), it follows that

( )

( )

,( ) cos sin

( ) cos sin

rc

rb n cn n n n r

n

n

n dn n n n rn

n

A = + + + + + +

=

+ +

=

01 1 2

1

1

1

1

, (d67)

( ) ( ), sin cos sin cos A n cn n n n rn

n

n dn n n n rn

n

= + +

=

+ +

=

2

1 1

, (d68)

( ) ( ), cos sin cos sin A n cn n n n rn

n

n d n n n n rn

n

= + +

=

+

=

2 2

1

2

1

. (d69)

Hence

( )

( )

rr

r

r r

c

r

b

r

n n cn n n n rn

n

n n d n n n n rn

n

A A

= + = ++

+

+

=

+ +

=

, ,cos sin

cos sin

2

0

2

1 1 2 2

1

2 2

1

(d70)


rrr

=lim 0 . (d71)

Also,

( ) ( ), ( ) sin cos sin cosr n n cn n n n rn

n

n dn n n n rn

n

A = +

=

+

=

2 1

1

2 1

1

. (d72)

Hence

( ) ( )

rr

r r

n n cn n n n rn

n

n n d n n n n rn

n

A A

= +

=

=

+

=

, ,

( ) sin cos sin cos

2

2

1

2 2

1

(d73)


r

r

=lim 0 . (d74)


27/81

27

D.9) Strain and displacement vector. Conditions at infinity.

From (d39) and HOOKEs reversed law it follows that

( )

= +

=+

+

11

11 1E

trE rr

( ) (d75)

i.e.

rr E rr

u r

r

=

=

1 2

1, (d76)

Hence

( )

( )( )

E u r

r

c

r

b

rn n n cn n n n r

n

n

n n d n n n n rn

n

1 20

1

1

21 1 1 2

12

2

1

11 2

1

=

+

+ + +

+

=

+ +

=

( ) ( ) cos sin

cos sin

(d77)

Integrating (d77) it follows

( ) ( ) ( )

( )

Eu

r

c

rb r n n c

nn

nn r n

n

n d n n n n rn

n

1 2

0

1

1

1 12

12 1

2

1

1

1

1

=

+ + + + +

+ +

=

+

+

=

+

ln cos sin

cos sin ( )

(d78)

From

r

u r

=lim 0 , (d79)

it follows that

b1 1 0 0+ = , ( ) . (d80)Hence

( ) ( )

( )

Eu r

c

rn n cn n n n r

n

n

n dn n n n rn

n

1 20

1

12

12 1

2

1

1

1

1

=

+ + +

+ +

=

+

+

=

cos sin

cos sin

. (d81)

Finally

( )( )

( )

u rE

c

rn cn n n n r

n

n

n d n n n n rn

n

=+

+ + + +

=

+ + =

1 0 2 4 1

2

1

1

cos sin

cos sin

. (d82)

In the same way,

=

= +1 2

1E rr

u

r

u r

r. (d83)

It follows

E u Eu r r r rr

1 2 1 2 1=

+

. (d84)Substituting (d82), (d66) and (d70) into (d84) and integrating with respect to , it follows after some computations that


28/81

28

( )( )

( )

uE

r n cn n n n rn

n

n d n n n n rn

n

=+

+ + +

=

+

=

11 4 4 1

2

1

1

( ) ( ) sin cos

sin cos

, (d85)

Form the condition

r

u

=lim 0 , (d86)

it follows the unknown function = ( )r is subject to the condition

r

r

=lim ( ) 0 . (d87)

But

r E r

ur

r

u

r

u

r

=

+= +

1 1

2, (d88)

or

21

rr

E u r u ru

r

=+

+

. (d89)

Substituting (d73), (d82) and (d85) into (d89), it follows after some computations that

rd

dr

= , (d90)

i.e. = Cr . From (d87) it follows that ( )r 0 and, finally,

( )( ) ( )uE

n cn n n n rn

n

n dn n n n rn

n

=

+ + +

=

+

=

1 4 4 12

1

1

sin cos sin cos

(d91)

D.10) Boundary conditions for the stress elements on the wall of the circular cavity.

Using the previous results, the final expressions of the plane elements of the stress are equal to

( )

( )

rr

c

rn n cn n n n r

n

n

n n d n n n n rn

n

= + +

=

+ +

=

02

2 2

1

2 2

1

cos sin

cos sin

(d92)

( )

( )

= + + =

+ + +

=

c

rn n cn n n n r

n

n

n n d n n n n rn

n

02

2 1

1

1 2

1

( )( ) cos sin

( ) cos sin

(d93)

and

( ) ( )r n n cn n n n rn

n

n n d n n n n rn

n =

=

+

=

( ) sin cos sin cos2

1

2 2

1

(d94)

The cavity wall has the outer normal (with respect to the rock domain) equal to er and radius equal to r R= . In the case

of a bore hole, let p be the difference between the mud pressure and the pressure of the fluid contained by the porous rock


29/81

29

(usually, because the atmospheric pressure is negligible, it follows in the case of a tunnel that p = 0 ). It follows the final

stressf satisfies the next boundary condition

fr

pr

for r Re e

=

= , (d92)

Hence

rr r R rrp

r r R r

= =

==

0

0 . (d93)

With no loss of generality, it can be assumed that the stress at infinity is along its main axes, i.e.120

0 = . Hence

( ) ( )

( ) ( )

c

Rn n cn n n n R

n

n

n n d n n n n Rn

n

p

n n cn n n n Rn

n

n n d n n n n Rn

n

02

2 2

2

2 2

1

110

220

2

110

220

22

2

2

2 2

1

110

220

22

+ +

=

+

+

=

= +

+

=

+ +

+

=

=

cos sin cos sin

cos

sin cos sin cos

sin

(d94)

Hence

c

R

p

d

02

110

220

2

1 1 0

= +

= =

, (d95)

and

n ncn

R nn n

d n

R nn n n n

R nn n n

R nn

n

2 2 22

2 2 22

2

110

220

22

+ + +

+

+ +

+ +

+

=

=

cos sin

cos

(d96)

+ +

+

+

+ +

+

=

=

n n cnR n

n n d n

R nn n n n

R nn n n

R nn

n

2 22

2 22

2

110

220

22

sin cos

sin

. (d97)

It follows that


30/81

30

c

Rp

c

R

d

R

c

R

d

R

02

110

220

2

4 22

6 24

110

220

2

2

2

2 6

2

4

110

220

2

= +

+ =

=

. (d98

and

( ) ( ) , , ,...

( ) ( ) , , ,...

( ) ( ) , , ,...

( ) ( ) , , ,...

n ncn

R nn n

d n

R nn

n n n

R nn n n

R nn

n ncn

R nn n

d n

R nn

n n n

R nn n n

R nn

2 2 22

0 3 4

2 2 22

0 2 3

2 22

0 3 4

2 22

0 2 3

+ + ++

= =

+ + ++

= =

+ ++

= =

+ + + = =

(d99)

Hence

c p R

c R cn n

dR

d n n

n n n

0110

220

2

2

2110

220

2

2 0 3 4

2 110 220

4

40 3 4

0 0 2 3

= +

+

=

= =

= = =

= = =

, , , , ...

, , , , ...

, , , , ...

. (d100)

By using (d11-d13) for120

0 = , the expressions of the final stress are equal to

rrf

pR

r

R

r

R

r

=+

+

+

+

+

110

220

2

110

220

22

110

220

2

2

22

110

220

2

22

3

2 110

220

4

42

cos

cos cos

(d101)

and

f p

R

r

R

r=

+

+

++

110

220

2

110

220

22 11

0220

2

2

2

3

2 110

220

4

42cos cos (d102)

rf R

r

R

r

=

+

110

220

22

110

220

2

22

3

2 110

220

4

42sin sin sin (d103)

In real cases, the value of the stress at infinity are positive ones for compression.


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31

D.11) The final shape of the wall.

Consider again the case when the direction of the horizontal axes of the co-ordinate system is along the corresponding eigen

vectors of the initial stress0 . In that case, 12

00 = . Using the above results, it follows the displacement vector for the

points initially placed on the wall of the circular cavity is

ur

r RE

R p

u r RE

R

( , ) ( ) cos

( , ) ( ) sin

= =+ +

+ +

= = +

1 110

220

23 4 11

0220

22

13 4 11

0220

22

. (d104)

Consider an arbitrary point on the wall of the bore hole. In the initial state, it has the polar co-ordinates ( )r R,= . Itsposition vector with respect to the centre of the circle is

=

+

X e eR cos sin 1 2 (d105)

Using (d6) the position vector in the final stage is

=

+

+

=

+

++ +

+ +

+

+

+

X e e e e

e e

e e

u r R ru R R

ER p

ER

( , ) ( , ) cos sin

( ) cos cos sin

( ) sin sin cos

1 2

1 110

220

23 4 11

0220

22

1 2

13 4 11

0220

22

1 2

. (d106)

Hence

x RE

R p

ER

x R

E

R p

ER

11 11

0220

23 4 11

0220

22

13 4 11

0220

22

21 11

0220

2

3 4 110

220

2

2

13 4 11

0220

22

= ++ +

+ +

++

= ++ +

+ +

+

cos ( ) cos cos

( ) sin sin

sin ( ) cos sin

( ) sin cos

. (d107)

After elementary computations, it follows


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32

xE

p

xE

p

X

X

1 11 11

0220

23 4 11

0220

2 1

2 11 11

0220

23 4 11

0220

2 2

= ++ +

+ +

= ++ +

+

( )

( )

. (d108)

Taking into account that12 22 2X X R+ = , it follows the final shape of the cavity is an ellipse of equation

x

a

x

b

12

222

21+ = , (d109)

where the semi-axes of the ellipse are equal to

a RE

p

b RE

p

= ++ +

+ +

= + + + +

11 11

0220

23 4 11

0220

2

11 110 220

23 4 11

0 220

2

( )

( )

. (d110)

Fig.D2. The shape of the borehole (tunnel) in the initial state (the circle) and in the final state (the ellipse), corresponding to a

compressive stress.

In real cases, the initial stress0 is usually a compressive one, i.e. (see Fig.D2)

110

1 220

3 = = S S, (d111)where the maximum compressive stress S1 and the minimum compressive stress S3 have positive values 0 3 1 S S . Itfollows here that the major semi-axis a corresponds to the minimum stress and a b .


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33

E) BOUSSINESQS PROBLEM - the concentrated force acting on the elastic semi-

space

E.1) The equations of BELTRAMI and MITCHELL.

It follows to obtain the partial derivative equations for the stress tensor in the particular case of an elastic, homogeneous,isotropic media. In the beginning, the next symmetric tensor is evaluated

S grad b grad t b=

+

( ) ( )

, (e1)

Using the equilibrium equation, it follows that

div b =

(e2)

Hence

S grad div grad t div= +

. (e3)

The HOOKEs reversed law gives

=

+

1

1E E , (e4)

so

=

++

+E

1 1 1 , (e5)where

= tr ,

= =

trE

1 2 . (e6)

But

div f grad f ( )1 = . (e7)Hence

divE

div grad

= + + +1 1 (e8)Expression (e1) becomes

SE

grad div grad t div grad grad grad t grad=+

+

++

+

1 1

( ) ( ) ( ) ( ) , (e9)

Because the tensor grad grad( ) is a symmetric one, it follows that

SE

grad div grad t div grad grad=+

+

++12

1 ( ) ( ) ( ) . (e10)

The tensor in the first parenthesis of (e10) has the ij - component equal to

[ ]

grad div grad t div

ij

div i, j div j i iq qj jq qi

i,qqj q iqj j qqi q jqi i, j j i qq q q ij

ijtr

ij ij Eij

ij Egrad grad ij

u u u u u u u

( ) ( ) ( ) ( ) , , ,

, , , , , , ,

,, ( )

+

= + = + =

+ + += +

+

=

+

=

+

=

+

1

2

1

22

1 2 1 2

(e11)

Hence

grad div grad t divE

grad grad( ) ( ) ( )

+ = +

1 2

(e12)

Using (e12) and (e10), the expression (e1) becomes


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34

SE

grad grad=+

++11

1 (e13)Applying the 3-dimensional LAPLACE operator in (e4), it follows

=+

1

1E E (e14)Replacing (e14) into (e13) gives

S grad grad= + + + 1

1

11 , (e15)i.e.

+

++

=

+

1

1

11 grad grad grad b gradt b( ) ( ) (e16)

But

tr tr

tr grad grad

tr

tr grad b gradt

b tr grad b div b

( ) ( ) ,

( ) ,

( ) ,

( ) ( ) ( ) ( )

= =

=

=

+

=

=

1 3

2 2

(e17)

Applying the trace operator in (e16) and using eqs.(e17) gives

++

+

= 1

1

3

12

div b( ) (e18)

It follows the trace of the stress tensor verifies the relation

= +

11

div b( ) (e19)

Replacing (e19) into (e16) gives

+

+

+=

+

1

1

11div b grad grad grad b grad t b( ) ( ) ( ) (e20)

Eqs.(e19)-(e20) represents the equations of BELTRAMI and MITCHELL, having as unknowns only the elements of the stresstensor . Together with appropriate conditions (in tensions) on the boundary of the elastic body, they allow one to solve thecorresponding linear static problem. Particular cases.a) Suppose that

div b( )

= 0 , (e21)

i.e. a vector potential

exists having the property that

b rot

=

, (e22)From (e19) it follows that the traces of both stress and strain tensors are harmonic functions = = 0 , (e23)

b) Suppose that

b grad

= , where = 0 , (e24)it follows that

div b( )

= = 0 , (e25)i.e. eq. (e23) is verified and eq.(20) gives

+

+

= 1

1

2grad grad grad grad( ) (e26)


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35

Applying the LAPLACE operator to eq.(e26), it follows that the stress tensor is the solution of the i-harmonic equation

= 0 . (e27)In most real cases, the volume forces are neglected (or they are represented only by the weight of the body, satisfying eq.(24). Itfollows the elastic linear problem involve solving harmonic and bi-harmonic equations.

E.2) The model.

A co-ordinate system having the third vertical axis positive downward is used. The semi-space x3 0 is represented by an

elastic, homogeneous, isotropic medium having the elastic coefficients and (or E and respectively). In the origin of theco-ordinate system is acting a vertical force having the magnitude equal to P . It follows to find the stress and thedisplacements. Spherical co-ordinates will be used (Fig.E1):

x r x r x r1 2 3= = =sin cos , sin sin , cos (e28)

Fig.E1 The spherical co-ordinate system and a vertical force of magnitude equal to P acting at the origin.

Because symmetry, the displacement vector has the components like

r r r ru u u u u= = =( , ) , , ( , ) 0 (e29)It follows the components of both stress and strain tensors are functions of r and only.

E.3) The equations of equilibrium and strain tensor in spherical co-ordinates.

The LAM differential parameters for spherical co-ordinates are1 1 2 3h h hr r= = =, , sin (e30)

The generalised curvilinear co-ordinates are1 2 3c c cr= = =, , (e31)

The unit vectors of the axis are

1 2 3n e n e n er

=

=

=

, , (e32)

By using, for example, (IVAN, 1996), the divergence of a symmetric tensorT in spherical co-ordinates isdiv

rrr r r r r rrr rr r r

rr r r

rr

rr r

rr r r

rr r

T

T T T T T T Te

T T T T T T Te

T T T T Te

=

+ + + +

+

+

+ + + + +

+

+ + + +

1 1 2

1 1 3

1 1 3 2

sin tan

sin tan tan

sin tan

(e33)

In the same way, the gradient of a vector in spherical co-ordinates is


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grad rr r r r

rr r r

rr r

r r rr

r r r

r r r r

vv

e ev v

e ev v

e e

ve e

v ve e

v ve e

v

e e

v

e e

v

=

+

+

+

+ +

+

+

+

+

1 1

1 1

1 1

sin

sin tan

sin

+ +

r

r r

v v

e etan

(e34)

The components of the strain tensor are

rrr

r r rr

r r

r rr

r r rr

r

r r r rr

r r

u u u u

u u u u u

u u u u u u

= = +

= +

= +

= +

= + +

, ,

sin, ,

sin tan,

sin tan

1

2

1

1

2

1 1

1

2

1 1 1

(e35)

In the particular case of eqs.(e29), it follows

rrr

r r rr

r r r

rr

rr

r r

u u u u

u u u u

= = +

=

= + = = +

, ,

, ,tan

1

2

10

10

(e36)

Hence

r = = 0 , (e37)By using (e33), the equilibrium equations in the absence of the volume forces are

rrr r

r rrr

rr

rrr

rr r

rr

rr r

+ + + =+ +

+ + + +

=

1 3

1 30

tan

tan tan

, (e38)

i.e.

rr rr r r r

rr r r r

3 2 2

3 2 2

sin sin sin ,

sin sin cos

+

=

+

=

(e39)

E.4) LAPLACE operator in spherical co-ordinates. LEGENDREs polynomials.

Using, for example, (IVAN, 1996) the gradient of a scalar function in spherical co-ordinates is

grad ff

r r r

f

r

fe e e=

+

+

1 1

sin . (e40)

Also, the LAPLACE operator is

f div grad f r r

rf

r

f f= =

+

+

12

2 12

2sinsin sin

sin

(e41)

Consider the LAPLACE equation


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37

f= 0 (e42)for the particular case when the unknown function has the form ( )f f r= , . Equation (e42) becomes

r

rf

r

f2 1 0

+

=

sinsin . (e43)

By using the method of separation of the variables, a solution for eq.(43) has the form

( )f r R r Y, ( ) ( ) = , (e44)where R andY are two unknown functions. Eq.(e43) becomes

ddr

r dRdr

R

dd

dYd

Yk

2

=

=

sin

sin, (e45)

where k is a constant. From (e45) it follows that

rd R

drr

dR

drkR2

2

22 0+ = (e46)

In a general case, the function R can be developed in power series. Lets look for a particular solution having the form

Rn r rn( ) = where n is a natural number. It follows from (e46) that

k n n= +( )1 (e47)

Then the particular solution of eq. (e46) can be expressed with the aid of two arbitrary constants as

Rn r A n rn Bn r

n( ) /= + (e48)Because f has to approach finite values for r , it has to take A n = 0 .The second relation (e45) gives

sin cos ( ) sin

d Y

d

dY

dn n Y

2

21 0+ + + = (e49)

By performing the substitution z = cos eq.(e49) becomesd

dzz

dY

dzn n Y1 2 1 0

+ + =( ) , (e50)

The solution of (e50) is represented by the LEGENDRE polynomials denoted by Pn z n( ) , , , , ...= 0 1 2 So,Po z P z z P z z( ) , ( ) , ( ) ( ) / = = = 1 1 2 3

2 1 2 (e51)Because the trace of the stress tensor is a solution of the harmonic equation (e23), it follows that the general solution for thattrace in the case of the BOUSSINESQ problem is

=+

=

an

r nnPn1

0(cos ) . (e52)

According to eq.(e6), a similar solution exists for the trace of strain tensor.

E.5) The displacement field.

We look for a displacement field having the form

r r ru u u= = =1 1

0 ( ) , ( ) , , (e53)

where and are two unknown functions following to be obtained. Substituting (e53) into (e36) it follows that

( )

( ) ( )

rr r r d d r r

d d r tg r

= = =

= + = = +

/ , / / ,

/ / , , / /

2 2 2 2 0

2 0 2(e54)

It follows that the trace of the strain tensor is

( ) = = + + = + +tr rr d d tg r/ / / 2 (e55)

A comparison of (e55) to (e52) shows that


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38

+ + =d d tg a/ / cos , (e56)where a is a constant remaining to be obtained. Hence

( )

( ) ( )

= = + = +

= + =

a r a r

rr a d d r

cos / , ( ) cos / ,

cos / /

2 3 2 3 2 2

2

(e57)

By using HOOKEs law, it follows that

( ) ( )

( )[ ]

[ ]

rr a r r d d r

a d d r

a d d r

= =

= + +

= +

cos / , / / ,

cos / / ,

( ) cos / /

2 2 2 2

2 2

2 2 2

(e58)

Substituting (e58) into the first equilibrium equation (e39) and using (e56), it follows after elementary computations that

d

d

d

da

sin ( / ) sin

= +2 2 . (e59)

Hence

sin ( / ) cos ( / ) ( / ) sin ,

( / ) /

sin ( / )sin

d

d

ab b

aa

d

d

b aa

= + + = + + +

= +

+ +

22 2

22 2 2

2 22

(e60)

where b is an integration constant. Becausedx

xtg

xC

sinln = +2 , (e61)

where C is a new integration constant, it follows that

= +

+ +ba

tg a C2

22

2( / ) ln ( / ) cos . (e62)

For / 4 , the logarithmic term into (e62) leads to infinite radial displacements. That can be avoided by taking

ba

+ =

2

2 0( / ) . (e63)

Hence

= + + = +a Cd

da( / ) cos , ( / ) sin2 2 . (e64)

and

( )[ ]rr a C r = + 3 4 2 2cos / . (e65)Substituting eq.(e64) into eq.(e56), it follows that

( )

d

d tga C

d

d

aC

aC D

Da a

C

+ = + +

= +

= + + +

= + + + +

( / ) cos ,

sin ( / ) sin sin ,

sin ( / ) cos cos

( / ) ( / )sin cos

3

2 3 2

43 2

43

23 2

(e66)

Hence

= +

+ + +D a a

C ctg( / ) /

sin( / )sin

3 4

23 . (e67)

Because the tangential displacements has to be finite ones for 0 , it follows that


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39

=

+ + = + +

= + +

Ca

C tga

d

d

C a

cos

sin( / )sin ( / )sin ,

cos ( / )( / ) cos

1

23

2 23

2 2 2 23

(e68)

By substituting eqs.(e64) and (e68) into eq.(e58) it follows that

r C tg a r C tg a r

C tg a r

=

=

= +

22

2 1 22

2

1 22

2

sin / , cos / ,

cos / (e69)

Substituting eqs.(e69) into the second equilibrium equation (e39) it can be seen that the last one is identically verified.

E.6) Boundary conditions for stress elements. The final solutions.

Eqs.(e65) and (e69) contain the unknown coefficients a and C. These constants follow to be obtained taking into accountthat the force P concentrated in the origin of the co-ordinate system is acting on the elastic semi-space. It can be seen that thepoints of the horizontal plane x3 0= have the co-latitude = / 2 .The unit vectors of the spherical co-ordinate system are related to the same vectors of the rectangular co-ordinate system by

rt

re

e

e

Qe

e

e

e

e

e

Qe

e

e

=

=

1

2

3

1

2

3

, , (e70)

where the orthogonal matrix is

Q =

sin cos sin sin cos

cos cos cos sin sin

sin cos

0. (e71)

The outer pointing unit vector normal to the elastic semi-space is equal to

=

+

3e e ercos sin . (e72)The resulting exterior force acting on the elastic semi-space is vanishing for all the points of the horizontal plane = / 2 ,excepting the origin, i.e.

=

3 0e . (e73)

Substituting (e72) into (e73) for = / 2 gives

= =0 0,

r. (e74)

By using (e69), the first eq.(e74) becomes an identity and the second one leads toa C= 2 . (e75)

So, eqs.(e65) and (e69) give

( )[ ]rr C r = + 2 3 4 2cos / , (e76)

r C tg r C tg r

= =

22

2 22

1 2cos / , cos / . (e77)

Consider an elastic hemisphere having the centre at the origin of the co-ordinate system. The curved surface S of the

hemisphere has the outer pointing normal equal to

re . On that surface, the rest of the elastic body (i.e. the semi-space minusthe hemisphere) is acting on the hemisphere with a total force equal to


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rS

dA rr r rS

dAe e e

=

+

, (e78)

where dA is the surface element and the unit vectors are obtained with (e70). It follows that

( )[ ]

r dA

S

d r r d

C tg d C

e e

e e

=

= + +

= +

0

2

0

22

4 3 42

2

0

23 4 3

/sin

cos cos sin cos sin/ ( )

(e79)

Because the hemisphere is into an equilibrium state, it follows that

rS

dA Pe e

+

=

3 0 , (e80)

Hence

CP

= +4 ( )

. (e81)

Finally, the non-zero components of the stress tensor are equal to

( )[ ]rrP r = + + 2 3 4

2( )

cos / , (e82)

rP

tg r

=+2 2

2( )

cos / , (e83)

= +

Ptg r

41 2

22

( )cos / . (e84)

The non-zero components of the displacement vector are

[ ]rP

r

P tg r

u

u

=+

+

= + +

42 2 1

4 23

( )( / ) cos / ,

( )( / )sin /

(e85)

Using eq.(e71), the components of the stress tensor into the Cartesian base can be obtained as

11 12 13

22 23

33

0

0

0 0

=

trr r

rQ Q (e86)

Also, the components of the displacement vector into the Cartesian base are

1

23 0

u

uu Q

u

ut

r

=

. (e87)

Of particular importance in real life are the components

333

23 5

31

22 1

1 2

3

= =

+ +

Pz r

E

P

r

z

ru/ , ( ) . (e88)

The BOUSSINESQ problem has a great importance in Geomechanics, in relation to the computation of a building foundation.The above solution derived for a concentrated vertical force can be used in the case of arbitrary vertical forces (spread on acertain domain) by assuming the principle of the superposition.


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41

F) ELEMENTS OF THIN PLATES THEORY

F.1) The model of a thin elastic plane plate.

The thin plane plate is a cylindrical body having an arbitrary horizontal cross section in the initially non-deformed state. Its

height, denoted by H h= 2 , is much smaller than the other dimensions (usually, around 7-10 times). The material of the plateis an elastic, homogenous, isotropic one, having the constants denoted by E and , respectively and . The third axis ofthe co-ordinate system, denoted by Oz , is a vertical one, positive downward. Let be x x x y1 2= =, . In the initial, non-

deformed state, the plate is a plane horizontal one. The upper face has the equation z h= and the down face has z h=respectively. The plane of equation z = 0 represents the median plane. After the plate is deformed, it becomes a mediansurface.

F.2) The planar state of a plate. The bending state.

Two particular situation for a deformed plate are considered (Fig.F1).

Fig.F1. (a) The non-deformed plate; (b) The plate into a planar state; (c) The plate into a bending state. Here, the median plane

is denoted by m.p. while m.s. is the median surface.

In the first case, the horizontal components of the displacement vector are symmetrical ones with respect to the median plane,

being even functions with respect to the z -variable, while the vertical component of the displacement vector is an anti-symmetrical one (odd function with respect to z ), i.e.

kx y z

kx y z k

x y z x y z

u u

u u

( , , ) ( , , ), ,

( , , ) ( , , ).

= =

=

1 2

3 3

(f1)

From eq.(f1b) it follows that

3

0 0u x y( , , ) = , (f2)

i.e. the material points placed initially into the median plane have no vertical displacement as a consequence of the deformation

of the plate (the median plane holds a horizontal plane one). This kind of deformation represents theplanar state of the plate.

In the second case, the horizontal components of the displacement vector are anti-symmetrical ones (odd functions with respect

to z ), while the vertical component of the displacement vector is a symmetrical one with respect to the median plane (evenfunction with respect to z ), i.e.

kx y z

kx y z k

x y z x y z

u u

u u

( , , ) ( , , ), ,

( , , ) ( , , ).

= =

= 1 2

3 3

(f3)

By deformation, the material points initially placed into the median plane are displaced on the vertical direction, having no

horizontal movement. The median plane becomes a median surface. This state represents the bending state of a plate.

F.3) Loads acting on the plate.


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42

For simplicity, the forces acting on the lateral surface of the plate are neglected. On the upper surface of the plate, having the

equation z h= and the outward pointing normal vector

=

n e3 , it is acting the surface forces s

x yq q

=

( , ) . In the

same way, on the down surface, having the equation z h= and the normal vector

=

n e3 , it is acting the surface force

j jx yq q

=

( , ) . Hence

=

= e q s for z h3 , , (f4)

i.e.

13 1 23 2 33 3 ( , , ) ( , ) , ( , , ) ( , ) , ( , , ) ( , )x y hs

x y x y hs

x y x y hs

x yq q q = = = (f5)Also

=

=e qj for z h3 , , (f6)i.e.

13 1 23 2 33 3 ( , , ) ( , ) , ( , , ) ( , ) , ( , , ) ( , )x y hj

x y x y hj

x y x y hj

x yq q q= = = . (f7)We shell see that the above presented deformation states are compatible only to certain distributions of volume or surface

forces applied to the plate.

F.4) Odd and even functions for the planar state and for the bending state.

Let f f x y z= ( , , ) be a function of three variables, supposed to be smooth enough. It can be seen that

f x y zf x y z f x y z f x y z f x y z

( , , )( , , ) ( , , ) ( , , ) ( , , )

=+

+

2 2(f8)

Let

f x y zf x y z f x y z+ = + ( , , ) ( , , ) ( , , )

2, (f9)

f x y zf x y z f x y z =

( , , )

( , , ) ( , , )

2(f10)

It follows that

f x y z f x y z f x y z f x y z+ = + = ( , , ) ( , , ) , ( , , ) ( , , ) . (f11)

The function f+ represents the even part of f (with respect to z -variable), while the function f represents its odd part. Itfollows that

f f f f f f +

+

= +

= +

=

+

=, , 0 , (f12)

i.e. the even part of the even part is equal to the even part too. A similar relation holds for the odd part. The even part of an odd

part (and the odd part of an even part) is vanishing. For k x x x y= = =1 2 1 2, , , , it follows that

[ ]fk xk

f x y zxk

f x y zxk

f x y z f k x y z+

=

+ = +

=

+

,( , , ) ( , , ) ( , , ) , ( , , )

1

2(f13)

Hence, the partial derivative (with respect to a horizontal co-ordinate) of the even part of a certain function is equal to the even

part of that derivative. This property holds for the odd part. So,

( ) ( )fk

fk fk

fk+

=

+

=

,, ,

,, (f14)

Also,


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43

[ ] [ ] ( )fz

f x y z f x y z f x y z f x y z f + = +

= =

,( , , ) ( , , ) , ( , , ) , ( , , ) ,

3

1

2

1

23 3 3

(f15)

Hence the partial derivative of the even part with respect to the vertical co-ordinate is equal to the odd part of the partial

derivative of the function itself with respect to same co-ordinate. In the same way it follows that

( )f f =

+

,,

33 . (f16)

By using the HOOKEs law and the definition of the strain, it can be resumed that the planar state and the bending state are

characterised by the next components:

-for theplanar state: - the displacement vector:1 2 3+ + u u u, , ,

- strain tensor:11 12 13 22 23 33+ + + + + , , , , , , ,

- stress tensor:11 12 13 22 23 33+ + + + , , , , , .

-for the bending state: - the displacement vector:1 2 3 +

u u u, , ,

- strain tensor:11 12 13 22 23 33 + + , , , , , , ,

- stress tensor:

11 12 13 22 23 33

+ +

, , , , , .

F5) Mean value of a function. Equilibrium equations for thin plates.

Let f f x y z= ( , , )an integrable function with respect to z -variable. The mean value of fcomputed on the thickness of theplate is denoted by

f x yh

f x y z

h

h

( , ) ( , , )dz=

+

1

2. (f17)

For an odd function f, its mean value vanishes, i.e.

f x y =( , ) 0 , (f18)It follows that

f f f f f f = + + = + + = + (f19)For a function C C x y= ( , ) depending only on the horizontal co-ordinates, eq.(f17) leads to

C C zC= =, 0 . (f20)Differentiating eq.(f17) with respect to the horizontal co-ordinates, it follows that:

( )fk f k k, , , ,= = 1 2 . (f21)For the vertical derivative, it follows that

[ ]fh

fz

dzh

h

h

f x y h f x y hh

f x y z h, ( , , ) ( , , ) ( , , )3 12

12

1= =

+ = = . (f22)

Consider the function zf x y z( , , ) . It follows that

( ) ( )zf zf zf zf + = = +, . (f23)Its mean value is

( )zf z f f zf zf zf zf = + + = + + = = +( ) . (f24)Hence


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44

zfh

zf

zh

h

dzh z

zf f

h

h

dzh

zfhh

hfdz

h

h

f x y h f x y hf x y f x y z h f

, ( ) ( )

( , , ) ( , , )( , ) ( , , )

31

2

1

2

1

2

1

2

2

=

+=

+=

+

+=

+ = + =

. (f25)

Neglecting the volume forces (the weight of the plate itself, for example), the equilibrium equations for the planar state are

111

122

133

0

121

222

233

0

131

232

333

0

+

+

+

+

=

+

+

+

+

=

+

+

+

=

, , ,

, , ,

, , ,

. (f26)

Let the mean values of the stress components be denoted by

ij ij h ijx y z

h

h

i j = =

+=

1

21 2 3( , , )dz , , , , (f27)

Applying the mean value operator to eqs. (f26) gives

( ) ( ) [ ]

( ) ( ) [ ]

11 1 12 21

2 13 130

12 1 22 2

1

2 23 230

, ,( , , ) ( , , )

, ,( , , ) ( , , )

+ + =

+ + =

hx y h x y h

hx y h x y h

, (f28)

Using eqs.(f5) and (f7) it follows

( ) ( )

( ) ( )

11 1 12 2

1

2 1 10

12 1 22 2

1

2 2 20

, ,( , ) ( , )

, ,( , ) ( , )

+ + +

=

+ + +

=

h

jx y

sx y

h

jx y

sx y

q q

q q. (f29)

Let

ij z ij hz

ijx y z

h

h

i j = =

+=

1

21 2 3( , , )dz , , , . (f30)

Multiplying eqs.(f26) by z and using again the mean value operator, it follows

( ) ( )13 1 23 233 33

2 330 , ,

( , , ) ( , , )+ +

+ =

x y h x y h. (f31)

Hence, using again eqs.(f5) and (f7),

( ) ( )13 1 23 23 3

2 330 , ,

( , ) ( , )+ +

=

jx y

jx yq q

, (f32)

So, the equilibrium of a thin plate into the planar state leads to eqs.(f29) and (f32).

Consider the weight of the plate, the equation of equilibrium for the plate into the bending state are

111

122

133

0

121

222

233

0

131

232

333

0

+

+

+

=

+

+

+

=

+

+

+

+

+ =

, , ,

, , ,

, , ,g

. (f33)

Here, the density of the plate is and the acceleration of gravity is g .Proceeding in a similar manner to above, it follows the equations of equilibrium for the thin plate into a bending state are


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45

( ) ( )

( ) ( )

( ) ( )

13 1 23 2

1

2 3 30

11 1 12 2

1

2 1 1 130

12 1 22 2

1

2 2 2 230

, ,( , ) ( , ) ,

, ,( , ) ( , )

, ,( , ) ( , )

+ + +

+ =

+ +

=

+ +

=

h

jx y

sx y g

jx y

sx y

jx y

sx y

q q

q q

q q

. (f34)

Eq.(f34b) is differentiated with respect to x and eq.(f34c) is differentiated with respect to y. The results are substituted in

eq.(f34a). Hence

( ) ( ) ( )11 11 2 12 12 22 221

2 3 3

1

2 1 1 1

1

2 2 2 20

, , , ( , ) ( , )

( , ) ( , ),

( , ) ( , ),

+ + + +

+

+

+

=

h

jx y

sx y g

jx y

sx y

jx y

sx y

q q

q q q q

. (f35)

F.6) Thin plane plate in the bending state.

The component1u of the displacement vector is developed in power series with respect to z -variable. It follows

1 1 0 1 0u u ux y z x y z z x y( , , ) ( , , ) ( , , ) ...= + +

(f36)

For the bending state, the first term in eq.(f36) vanishes. Because the thickness of the plate is a small one, only the second term

is kept. So the horizontal components of the displacement vector are

11 0 2

2 0uu

uu

x y z zz

x y x y z zz

x y( , , ) ( , , ) , ( , , ) ( , , )= =

. (f37)

The vertical displacement of the points placed into the median plane is denoted by

w w x y x yu= =( , ) ( , , )3 0 , (f38)i.e. the mean surface has the equation z w x y= ( , ) . Here, w represents the arrow of the plate.

F7) BERNOULLIs hypothesis.

According to BERNOULLI, it is assumed that an arbitrary material segment of the plate, initially perpendicular on the

median plane in the non-deformed state, rests perpendicular on the mean surface in the deformed state. Let A x y z( , , ) a

certain point of the plate (not placed in the mean plane) and A x y0 0( , , ) its projection on the median plane. So, the segment

A A0

is perpendicular on the median plane. After deformation, the material point A is moving at the point having the co-

ordinates ( )M x x y z y x y z z x y zu u u+ + +1 2 3( , , ), ( , , ), ( , , ) , while the point A0 is moving at the point

( )M x x y y x y w x yu u0 1 0 2 0+ +( , , ), ( , , ), ( , ) . The horizontal displacements of the points placed in the median planeare vanishing. So, writing the vector along the line, it follows that

( )M M x y z x y z z x y z w x yu u u0 1 2 3 = + ( , , ), ( , , ), ( , , ) ( , ) (f39)

Expanding in power series the even function3u with respect to z -variable, it follows

32

u x y z w x y z x y( , , ) ( , ) ( , ) ...= + + (f40)Using eqs.(f37) and (f40), it follows that


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46

M M zz

x y zz

x y z z x y

zz

x yz

x y z x y

u u

u u

01 0 2 0 3

1 0 2 0 1 2

= + +

= + +

( , , ), ( , , ), ( , ) ...

( , , ), ( , , ), ( , ) ...

(f41)

Hence the direction is

M M

M M

z x y z x y z x y

z zz x y

u u

u u

0

0

1

0

2

0 1

2

12

22

1 22

=+ +

+

+ + +

|| ||

( , , ), ( , , ), ( , ) ...

( , ) ...

(f42)

The normal vector on the median surface is

=

+

+n

w

x

w

y

w

x

w

y

, ,1

2 2

1 . (f43)

Comparing eq.(f42) to eq.(f43), the supplemental hypothesis of BERNOULLI is satisfied if

1 0 2 0 3u u uz x y wx z x y wy x y z w x y( , , ) , ( , , ) , ( , , ) ( , )= = = , (f44)

i.e., by using eq.(f37), the displacement vector has the horizontal components equal to

1 2 3u u ux y z zw

xx y z z

w

yx y z w x y( , , ) , ( , , ) , ( , , ) ( , )= = =

. (f45)

From eq. (f45), the components of the strain tensor are

11

2

2 12

2

22

2

2 330

= = = =z

w

xz

w

x yz

w

y, , , . (f46)

The LAPLACE operator in horizontal components is denoted by

* = +

2

2

2

2x y. (f47)

Hence the trace of the strain tensor is

= = + + = tr z w11 22 33 *

. (f48)

F8) HOOKEs law for a thin plate.

By using eqs.(f46) and (f48) it follows that

112

112

2

2

= + = +

tr z w

w

x

*. (f49)

But

zh

z

h

h

dzh2 1

2

22

3=

+= (f50)

Hence

11 11

2

32

2

2

2

122

2

2 = = +

= +

z

hw

w

x

Hw

w

x

* *. (f51)

In the same way,


47/81

47

12 122

2

3

2 2

6

2

= = = zh w

x y

H w

x y

, (f52)

22 22

2

32

2

2

2

122

2

2 = = +

= +

z

h

marian ivan - theoretical geomechanics

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