theoretical geomechanics-marian ivan samizdat press
TRANSCRIPT
-
8/9/2019 Theoretical Geomechanics-Marian Ivan Samizdat Press
1/81
THEORETICALGEOMECHANICS
Marian IVANUniversity of Bucharest
Department of Geophysics
6 Traian VUIA str.70138 BUCHAREST o.p.37
ROMANIA
e-mail : [email protected]
Samizdat Press
-
8/9/2019 Theoretical Geomechanics-Marian Ivan Samizdat Press
2/81
2
In memoriam
ProfessorMarin DOROBANU
INTRODUCTION
These course notes present some theoretical problems related to the Mechanics of a Continuum Solid Body, of particular
importance 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 details
can be found into the excellent books of George Backus and Brian Kennett, also available at Samizdat Press. The linear
elasticity and the homogeneity of the continuum solid body are almost thoroughly assumed to be valid, but some elements of
Rheology 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 a
particular form, as a consequence of the simplified hypothesis previously assumed. It is verified that solution checks both the
corresponding equations and the boundary conditions. Based on the Uniqueness Theorem of the Linear Elasticity, it follows the
assumed particular solution is just the general solution of the problem. In all the cases discussed here, the assumed simplified
hypotheses allow one to obtain simple, analytical solutions. At a first glance, the importance of such solutions is minor with
respect 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 possibility
to shear his work.
-
8/9/2019 Theoretical Geomechanics-Marian Ivan Samizdat Press
3/81
3
CONTENTS
INTRODUCTION page
Chapter A: BASIC ELEMENTS. .... 5
A.1) The displacement vector. Lagrangean (material) and Eulerian (spatial) co-ordinates. ..5
A.2) Invariants of a tensor. Tensor deviator. ...5
A.3) Strain tensor. Stress tensor. Equation of motion / equilibrium. ..5
A.4) HOOKEs law. .....8
Chapter B: DEFORMATION OF A CYLINDRICAL BODY IN THE PRESENCE OF GRAVITY ....9
B.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. ....14
C.2) The model. Simplifying hypothesis. The planar deformation state. ...14
C.3) Equations of equilibrium. AIRYs potential. ..15
C.4) Boundary conditions. The final shape of the dam......16
Chapter D: KIRSCHs PROBLEM - the circular bore hole / tunnel..19
D.1) The model. ... ..19
D.2) The planar state of deformation in cylindrical co-ordinate system. ...20
D.3) The circle of MOHR. ..21
D.4) AIRYs potential in cylindrical co-ordinates. The bi-harmonic equation.. 21
D.5) The divergence of a tensor in cylindrical co-ordinates. ..22
D.6) The gradient of a vector and the strain tensor in cylindrical co-ordinates. ...23
D.7) The bi-harmonic equation cylindrical co-ordinates. ..24
D.8) The stress elements. Conditions at infinity for the stress elements. ...25
D.9) Strain and displacement vector. Conditions at infinity. .27
D.10) Boundary conditions for the stress elements on the wall of the circular cavity. ..28
D.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. 33
E.2) The model. ...35
E.3) The equations of equilibrium and strain tensor in spherical co-ordinates. .35
E.4) LAPLACE operator in spherical co-ordinates. LEGENDREs polynomials. .36
E.5) The displacement field. ...37
E.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.41
F.2) The planar state of a plate. The bending state..41
-
8/9/2019 Theoretical Geomechanics-Marian Ivan Samizdat Press
4/81
4
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. ......62
G.4) Integrals of displacement vector. ....62
G.5) Equation of motion in quasi-mean values. .....63
G.6) Quasi-mean value of the shell density. The differential equation. .....63
G.7) The buckling of a spherical shell. ......64
G.8) Load on the upper face. Stress on the lower surface of the shell. ...66
G.9) The differential equation of time dependent flexure. .....67
G.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. .69
b) MAXWELL (viscous-elastic) model. .70
c) BURGERS (general linear) model. .70
d) Remarks on the linear models. ...71
H.3) Non-linear models. ..73
H.4) Brittle. Creep. Empirical criteria. ...73
H.5) Empirical criteria for shear-faulting. TRESCA criterion. COULOMB-NAVIER criterion. .....74
H.6) Von MISES-HENCKY criterion for ductile flow (plasticity).. ..76
H.7) Rheological models. ...77
a) SAINT VENANT body (elastic-plastic material). .77
b) BINGHAM body (visco-plastic material). .77
Chapter I: THE ACCRETION WEDGE... 78
I.1) The model. 78
I.2) Equations of equilibrium. Yield condition. Stress field. ..78
I.3) Boundary conditions. Final results. ..79
References...81
-
8/9/2019 Theoretical Geomechanics-Marian Ivan Samizdat Press
5/81
5
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 t 0 , 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
-
8/9/2019 Theoretical Geomechanics-Marian Ivan Samizdat Press
6/81
6
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)
-
8/9/2019 Theoretical Geomechanics-Marian Ivan Samizdat Press
7/81
7
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 D l 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.
-
8/9/2019 Theoretical Geomechanics-Marian Ivan Samizdat Press
8/81
8
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.
-
8/9/2019 Theoretical Geomechanics-Marian Ivan Samizdat Press
9/81
9
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)
-
8/9/2019 Theoretical Geomechanics-Marian Ivan Samizdat Press
10/81
10
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
-
8/9/2019 Theoretical Geomechanics-Marian Ivan Samizdat Press
11/81
11
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
-
8/9/2019 Theoretical Geomechanics-Marian Ivan Samizdat Press
12/81
12
=
=
+ +
+ +
+
++
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
-
8/9/2019 Theoretical Geomechanics-Marian Ivan Samizdat Press
13/81
13
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.
-
8/9/2019 Theoretical Geomechanics-Marian Ivan Samizdat Press
14/81
14
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
23 1 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
-
8/9/2019 Theoretical Geomechanics-Marian Ivan Samizdat Press
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 3 310
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.
-
8/9/2019 Theoretical Geomechanics-Marian Ivan Samizdat Press
16/81
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
-
8/9/2019 Theoretical Geomechanics-Marian Ivan Samizdat Press
17/81
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
-
8/9/2019 Theoretical Geomechanics-Marian Ivan Samizdat Press
18/81
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)
-
8/9/2019 Theoretical Geomechanics-Marian Ivan Samizdat Press
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
rr a r bc
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
-
8/9/2019 Theoretical Geomechanics-Marian Ivan Samizdat Press
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 dn 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 dn n n n rn
n
A A
= + = ++
+
+
=
+ +
=
, ,cos sin
cos sin
2
0
2
1 1 2 2
1
2 2
1
(d70)
From (d70), it follows that
rrr
=lim 0 . (d71)
Also,
( ) ( ), ( ) sin cos sin cosr n n cn n n n rn
n
n d n 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)
From (d73), it follows that
r
r
=lim 0 . (d74)
-
8/9/2019 Theoretical Geomechanics-Marian Ivan Samizdat Press
27/81
-
8/9/2019 Theoretical Geomechanics-Marian Ivan Samizdat Press
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
u r
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
( )
( )
rrc
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
-
8/9/2019 Theoretical Geomechanics-Marian Ivan Samizdat Press
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
-
8/9/2019 Theoretical Geomechanics-Marian Ivan Samizdat Press
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.
-
8/9/2019 Theoretical Geomechanics-Marian Ivan Samizdat Press
31/81
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
-
8/9/2019 Theoretical Geomechanics-Marian Ivan Samizdat Press
32/81
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 .
-
8/9/2019 Theoretical Geomechanics-Marian Ivan Samizdat Press
33/81
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 gradt div=
+
. (e3)
The HOOKEs reversed law gives
=
+
1
1E E , (e4)
so
=
++
+E
1 11 , (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 iqq
q qij
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
-
8/9/2019 Theoretical Geomechanics-Marian Ivan Samizdat Press
34/81
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 grad
t 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 stress
tensor . Together with appropriate conditions (in tensions) on the boundary of the elastic body, they allow one to solve the
corresponding 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)
-
8/9/2019 Theoretical Geomechanics-Marian Ivan Samizdat Press
35/81
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 rr 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 3
c 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 is
div
rrr r r r r rrr rr r r
r
r r r
r
r
r
r r
r
r r r
r
r 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
-
8/9/2019 Theoretical Geomechanics-Marian Ivan Samizdat Press
36/81
36
grad r
r r r r
r
r r r
r
r r
r r r
r
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 r
r
r r
r r
r
r r r
r
r
r r r r
r
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 r
r
r r r
r
r
r
r
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
rr
r r
r rr
r
r
r
rr
r
r
r r
r
r
r
r r
+ + + =+ +
+ + + +
=
1 3
1 30
tan
tan tan
, (e38)
i.e.
rr
rrr
rr
rr
rr 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= =
+
+
1
22 1
2
2sinsin sin
sin
(e41)
Consider the LAPLACE equation
-
8/9/2019 Theoretical Geomechanics-Marian Ivan Samizdat Press
37/81
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
d
drr dR
dr
R
d
d
dY
d
Yk
2
=
=
sin
sin, (e45)
where kis 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
R n 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
R n 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 nn
Pn10
(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 10 ( ) , ( ) , , (e53)
where and are two unknown functions following to be obtained. Substituting (e53) into (e36) it follows that
( )
( ) ( )
rrr
rd 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
-
8/9/2019 Theoretical Geomechanics-Marian Ivan Samizdat Press
38/81
38
+ + =d d tg a/ / cos , (e56)where a is a constant remaining to be obtained. Hence
( )
( ) ( )
= = + = +
= + =
a r a r
rra d d r
cos / , ( ) cos / ,
cos / /
23 2 3 2
2
2
(e57)
By using HOOKEs law, it follows that
( ) ( )
( )[ ]
[ ]
rra r
rd 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. Because
dx
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
23 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
-
8/9/2019 Theoretical Geomechanics-Marian Ivan Samizdat Press
39/81
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
rC tg a r C tg a r
C tg a r
=
=
= +
22
2 1 2
2
2
1 2
2
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 the
points 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
r
t
re
e
e
Q
e
e
e
e
e
e
Q
e
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 to
a 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 2
21 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 minus
the hemisphere) is acting on the hemisphere with a total force equal to
-
8/9/2019 Theoretical Geomechanics-Marian Ivan Samizdat Press
40/81
40
rS
dArr r r
S
dAe e e
=
+
, (e78)
where dA is the surface element and the unit vectors are obtained with (e70). It follows that
( )[ ]
rdA
S
dr
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
2
2
( )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
2
30
u
uu
Q
u
ut
r
=
. (e87)
Of particular importance in real life are the components
333
2
3 53
1
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 a
certain domain) by assuming the principle of the superposition.
-
8/9/2019 Theoretical Geomechanics-Marian Ivan Samizdat Press
41/81
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.
-
8/9/2019 Theoretical Geomechanics-Marian Ivan Samizdat Press
42/81
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,
-
8/9/2019 Theoretical Geomechanics-Marian Ivan Samizdat Press
43/81
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
-
8/9/2019 Theoretical Geomechanics-Marian Ivan Samizdat Press
44/81
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
-
8/9/2019 Theoretical Geomechanics-Marian Ivan Samizdat Press
45/81
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
-
8/9/2019 Theoretical Geomechanics-Marian Ivan Samizdat Press
46/81
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,
-
8/9/2019 Theoretical Geomechanics-Marian Ivan Samizdat Press
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
hw
w
y
Hw
w
y
* *(f53)
Substituting eqs.(f51)-(f53) into eq.(f35) it follows the equation of Sophie GERMAIN:
D w gH j x y s x y
H jx y
sx y
jx y
sx y
q q
q q q q
* * ( , ) ( , )
( , ) ( , ),
( , ) ( , ),
= + +
+
+
3 3
2 1 1 12 2
2
, (f54)
where
Dh E H
= + =
+ ( )
( )
( )( )
22 3
3
1
1 1 2
3
12(f55)
represents the flexural rigidity of the plate.
So, obtaining the median surface asks someone to solve a bi-harmonic equation, with certain boundary con