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Linear Elasticity

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Linear Elasticity
Chapter 5
Linear Elasticity
1
Introduction
The simplest mechanical test consists of placing a standardized specimen with its ends in
the grips of a tensile testing machine and then applying load under controlled conditions.
Uniaxial loading conditions are thus approximately obtained. A force balance on a small
element of the specimen yields the longitudinal (true) stress as
F
A
where F is the applied force and A is the (instantaneous) cross sectional area of the specimen.
Alternatively, if the initial cross sectional area A0 is used, one obtains the engineering stress
=
e =
F
A0
For loading in the elastic regime, for most engineering materials e .
Likewise, the true strain is dened as
=
Z l dl
while the engineering strain is given by
e =
l0
l
= ln( )
l
l0
Z l dl
l0 l0
=
l
l0
l0
Again, for loading in the elastic regime, for most engineering materials e .
Linear elastic behavior in the tension test is well described by Hooke's law, namely
= E
where E is the modulus of elasticity or Young's modulus. For most materials, this is a large
number of the order of 1011 Pa.
1
Values of E can be readily determined by measuring the speed of propagation of longitudinal elastic waves in the material. Ultrasonic waves are induced by a piezoelectric device on
the surface on the specimen and their rate of propagation accurately measured. The velocity
of the longitudinal wave is given by
v
u
u
vL = t
E (1 )
(1 + )(1 2 )
Transverse wave propagation rates are also easily measured by ultrasonic techniques and
the corresponding relationship is
s
G
where G is the modulus of elasticity in shear or shear modulus.
The shear modulus is involved in the description of linear elastic behavior under shear
loading, as encountered, for instance during torsion testing of thin walled pipes. In this case,
if the shear stress is and the shear strain , Hooke's law is
vT =
= G
2
Generalized Hooke's Law
The statement that the component of stress at a given point inside a linear elastic medium
are linear homogeneous functions of the strain components at the point is known as the
generalized Hooke's law. Mathematically, this implies that
ij = Dijkl kl
where ij and ij are, respectively the stress and strain tensor components. The quantity
Dijkl is the tensor of elastic constants and it characterizes the elastic properties of the
medium. Since the stress tensor is symmetric, the elastic constants tensor consists of 36
components.
The elastic strain energy W is dened as the symmetric quadratic form
1
1
W = ij kl = Dijkl ij kl
2
2
and has the property that ij = @[email protected] . Because of the symmetry of W , the actual number
of elastic constants in the most general case is 21. This number is further reduced in special
cases that are of much interest in applications. For instance, for isotropic materials (elastic
properties the same in all directions) the number of elastic constants is 2. For orthotropic
materials (characterized by three mutually perpendicular planes of symmetry) the number
of constants is 9. If the material exhibits symmetry with respect to only one plane, the
number of constants is 13.
2
3
Stress-Strain Relations for Isotropic Elastic Solids
The generalized Hooke's law for isotropic solids is
= 3K
ij0 = 2G0ij
where K and G are the elastic constants bulk modulus and shear modulus, respectively and
the primes denote the stress and strain deviators.
Combination of the above with the denition of stress and strain deviation tensors yields
the following commonly used forms of Hooke's law; for stress, in terms of strain
ij = Æij + 2Gij
and for strain, in terms of stress
1+
ij
Æ
E
E ij
The constants and G are called Lame's constants, while E is Young's modulus and is
Poisson's ratio. Any of the above elastic constants can be expressed in terms of the others
and only two are independent. Values of the above elastic constants for a wide variety of
engineering materials are readily available in handbooks.
For an isotropic elastic solid in a rectangular Cartesian system of coordinates, the constitutive equations of behavior then become
1
xx = [xx (yy + zz )]
E
1
yy = [yy (xx + zz )]
E
1
zz = [zz (xx + yy )]
E
ij =
1+
1
xy = xy
E
2G
1+
1
yz =
yz = yz
E
2G
1+
1
zx =
zx = zx
E
2G
xy =
4
Stress-Strain Relations for Anisotropic Elastic Solids
It is conventional in studying elastic deformation of anisotropic materials to relabel the six
stress and strain components as follows:
11 = 1
3
22 = 2
33 = 3
23 = 4
13 = 5
12 = 6
11 = 1
22 = 2
33 = 3
1
23 = 4
2
1
13 = 5
2
1
12 = 6
2
With the new notation and using the summation convention, Hooke's law becomes
i = Cij j
or equivalently
i = Sij j
where Cij and Sij are, respectively the elastic stiness and compliance matrices. Depending
on the symmetries existing in the material, only a few components of the above matrices
are nonzero. For instance, for single crystals with cubic structure only C11, C12 and C44
are nonzero. Values of the components of the above matrices for a variety of anisotropic
materials are readily available in handbooks.
Following are the stress-strain relationships for linear elastic materials with two forms of
anisotropy. The general relatiosnhip for a fully anisotropic material is
0
B
B
B
B
B
B
B
B
@
1
2
3
4
5
6
1 0
C
B
C
B
C
B
C
B
C
=B
C
B
C
B
C
A B
@
C11
C21
C31
C41
C51
C61
C12
C22
C32
C42
C52
C62
C13
C23
C33
C43
C53
C63
C14
C24
C34
C44
C54
C64
C15
C25
C35
C45
C55
C65
C16
C26
C36
C46
C56
C66
10
C
B
C
B
C
B
C
B
C
B
C
B
C
B
C
AB
@
1
2
3
4
5
6
1
C
C
C
C
C
C
C
C
A
Since the stiness matrix Cij is symmetric only 21 independent constants are required to
fully specify the relationship.
4
Another common example is the orthotropic material (containing two orthogonal planes
of material property symmetry). In this case
0
B
B
B
B
B
B
B
B
@
1
2
3
4
5
6
1 0
C11
C
B
C21
C
B
C
B
C
B
C31
C
=B
C
B
C
B 0
C
A B
@ 0
0
C12 C13
C22 C23
C32 C33
0 0 C44
0 0 0
0 0 0
0
0
0
0
0
0
0
0
0
C55
0 C66
0
0
0
10
C
B
C
B
C
B
C
B
C
B
C
B
C
B
C
AB
@
1
2
3
4
5
6
1
C
C
C
C
C
C
C
C
A
And only nine independent constants are required to fully specify the relationship.
5
Formulation of Linear Elastic Problems
For steady state conditions, the governing equations of the isotropic linear elastic solid are,
the equilibrium equations
@ij
+ Xi = ij;j + Xi = 0
@xj
the stress-strain relations
ij = kk Æij + 2Gij
and the small displacement, strain-displacement relations
1 @u @u
ij = ( i + j )
2 @xj @xi
Admissible strain elds are those that satisfy the equations of compatibility. The three
equilibrium equations together with the six stress-strain relations constitute a set of nine
equations for the nine unknowns ui ; ij . One can show this system is complete, yields an
unique solution under suitable boundary conditions and the resulting strain satises the
compatibility relations.
The above system of equations must be solved in each particular case subject to appropriate boundary conditions. There are two fundamental boundary value problems of
elasticity:
Problem 1: Determine the stress and strain elds inside the elastic body subject to
specied values of the displacement ui at its boundary.
Problem 2: Determine the stress and strain elds inside the elastic body subject to a
specied values of the surface tractions Ti at its boundary.
5
Because of the nature of the boundary conditions associated with the two fundamental
problems above, it is convenient to produce formulations of the elasticity problem involving
only displacements or stresses.
Specically., combining the strain-displacement relationships with Hooke's law and subsequent introduction of the result into the equilibrium equations yields the Navier equations
Gui;jj + ( + G)uj;ji + Xi = 0
The above is then a set of three equations for the three unknowns ui .
Alternatively, combination of the compatibility equations, Hooke's law and the equilibrium equations yields the Beltrami-Michell equations
r2ij + 1 +1 kk;ij = 1 Xk;k Æij (Xi;j + Xj;i)
An interesting special case of the equilibrium equations is obtained if the body force
vanishes. In this case, one can readily show that the two invariants, the dilatation e = satises
r2e = 0
and the mean stress = 31 satises
r2 = 0
Furthermore
r4ui = 0
r4ij = 0
r4ij = 0
i.e. the dilatation and the mean stress are harmonic functions while the displacement vector
components, the strain tensor components and the stress tensor components are all biharmonic functions.
6
Governing Equations in a Rectangular Cartesian Coordinate System
For the sake of convenience, the equations of equilibrium, the constitutive equations of
material behavior and the compatibility relationships governing the deformation behavior of
linear elastic solids are now provided in a rectangular Cartesian coordinate system.
6
The equilibrium equations are
@xx @xy @xz
+
+
+ Xx = 0
@x
@y
@z
@yx @yy @yz
+
+
+ Xy = 0
@x
@y
@z
@zx @zy @zz
+
+
+ Xz = 0
@x
@y
@z
The constitutive equations (Hooke's law) are
1
[
(yy + zz )]
E xx
1
yy = [yy (xx + zz )]
E
1
zz = [zz (xx + yy )]
E
xx =
1
2G xy
1
yz = yz
2G
1
zx = zx
2G
The equations of compatibility are
xy =
@ 2xx @
= (
@[email protected] @x
@ 2 yy @
= (
@[email protected] @y
@ 2 zz
@
= (
@[email protected] @z
@yz @zx @xy
+
+
)
@x
@y
@z
@zx @xy @yz
+
+
)
@y
@z
@x
@xy @yz @zx
+
+
)
@z
@x
@y
@ 2 xy @ 2 xx @ 2 yy
= 2 + 2
@[email protected]
@y
@x
2
2
@
@
@ 2 zz
2 yz = yy
+
@[email protected]
@z 2
@y2
@ 2
@ 2
@ 2 xx
2 zx = zz
+
@[email protected] @x2
@z 2
2
7
Finally, the strain tensor components ij are related to the components of the displacement vector (u; v; w) as follows
@u
@x
@v
yy =
@y
@w
zz =
@z
xx =
and
1 @u @v
xy = ( + )
2 @y @x
1 @u @w
xz = ( + )
2 @z @x
1 @v @w
yz = ( + )
2 @z @y
7
Strain Energy Function
An important quantity in continuum mechanics is the energy associated with deformation.
The specic strain energy of the linear elastic material W was dened above as
1
W = Cijkl ij kl
2
It has the properties
@W
= ij
@ij
and
@W
= ij
@ij
It can be shown that, as long as the strain energy function exists and is positive denite, the
two fundamental boundary value problems of elasticity have unique solutions. Non-unique
solutions may occur if W is not positive denite (e.g. buckling, inelastic deformation) or
when the linear equations break down (e.g. nite deformation, forces with memory).
8
8
Torsion of Elastic Bars
Saint-Venant produced solutions for the torsion problem of a long bar (aligned with the z
direction), under small twist conditions by assuming the vector components of displacement
to be given by
u = zy
v = zx
w = (x; y)
where is the angle of twist per unit length of the bar and (x; y) is the warping function.
It can be shown that the warping function satises Laplace's equation
r2 = 0
subject to the condition
@
= y cos(x; n) x cos(y; n)
@n
at the lateral boundary of the bar and is therefore an harmonic function.
Alternatively, Prandtl proposed, in analogy with the stream function of hydrodynamics,
the use of the stress function (x; y) dened by
@
xz =
@y
@
yz =
@x
It can be shown that the equations of elasticity for the torsion problem in terms of the stream
function are equivalent to the problem of solving
r2
= 2G
subject to
=0
on the boundary of the bar.
For instance for the bar of elliptical cross section (axes a and b), a suitable expression for
the stress function is
x
y
= k[( )2 + ( )2 ) 1]
a
b
With the above
can readily obtain closed form expressions for the shearing stresses zx,
q one
2 + 2 , the torque Mt = 2 R R dxdy , angle of twist and the warpage w.
zy and z = zx
zy
9
9
Bending of Beams
Consider a cantilever beam of length L, breadth b = 1 and height h = 0:1 (moment of inertia
I ), loaded by a concentrated force per unit length P at its free end. Since b >> h a twodimensional model may be a reasonable approximation. Let the x-axis be aligned with the
beam axis and the y-axis with the beam height, let also the origin of coordinates be located
at the free end of the beam and disregard body forces. Let the stress function (x; y) be
dened by
xx =
@ 2
@y2
yy =
@ 2
@x2
and
xy =
@ 2
@[email protected]
Then, for mechanical equilibrium, the stress function must satisfy
@ 4
@ 4
@ 4
+
2
+
=0
@x4
@[email protected] @y4
A common method for nding solutions to the above is by taking polynomials of various
degrees. For instance, one can show that by assuming the stress function to have the form
of a fourth degree polynomial in x and y, the solution of the stress function equation and
subsequent substitutions taking into account the boundary conditions yield the distributions
of stresses and beam displacements; these are
xx =
P xy
I
yy = 0
xy =
u=
P x2 y
2EI
P h2
(
2I 4
y2)
P y3 P y3
P L2
+
+(
6EI 6GI
2EI
10
P h2
)y
8GI
and
P xy2 P x3 P L2x P L3
+
+
2EI
6EI 2EI 3EI
Note that the v displacement above for y = 0 (the beam's neutral axis) coincides with
the result obtained using elementary strength of materials methods by double integration of
the beam deection curve equation.
Consider now the three dimensional situation of a cantilever beam (length L, moment of
inertia of the cross section I ), loaded at one end with transverse force P . Select a rectangular
Cartesian system of coordinates with the z -axis aligned with the axis of the beam and the
x y plane coincident with its cross-section. The equilibrium equations are
v=
@zx
=0
@z
@yz
=0
@z
and
@xz @yz P x
+
+
=0
@x
@y
I
It can be shown that in order to satisfy the compatibility requirement (expressed in terms
of the stress tensor - Beltrami-Michell equations), the above problem can be represented by
the following equivalent one in terms of the stress function , namely, determine the function
(x; y) satisfying
@2
Py
@2
+
= r2 =
2
2
@x
@y
1+ I
df
dy
subject to the condition = 0 on the boundary of the cross section of the beam, where f (y)
is a conveniently dened but arbitrary function introduced by Timoshenko.
Specically, for the beam of circular cross section (radius r), a suitable form for f (y) is
P 2
(r
2I
becomes
f (y) =
with this, the above equation for
r2
=
y2)
1 + 2 P y
1+ I
with the solution
=
1 + 2 P 2 2
(x + y
8(1 + ) I
11
r2 )y
From this, the stress components are found to be
xz =
P x2
(3 + 2 ) P 2
+ f (y) =
(r
2I
8(1 + ) I
@
@y
yz =
x2
1 2 2
y)
3 + 2
@
(1 + 2 ) P xy
=
@x
4(1 + ) I
and, from elementary beam theory,
P (L z )x
I
zz =
10
Bending of Rectangular Plates
Starting from the equation of mechanical equilibrium Lagrange rst obtained the following
expression for the small displacements of a simply supported, large rectangular at plate of
width a, breadth b and thickness h, originally on the x y plane, as a consequence of a
normal pressure f (x; y) distributed over the surface of the plate, namely
@ 4w
@ 4w
@ 4w f
+
2
+
=
@x4
@[email protected] @y4 D
Here w(x; y) is the local plate deection and D is the exural rigidity of the plate dened by
Eh3
12(1 2)
and is equivalent to the stiness EI in the case of beams.
The above equilibrium equation must now be solved subject to the boundary conditions
for simply supported edges, i.e.
D=
w=
at x = 0 and x = a, and
@ 2w
=0
@x2
@ 2w
=0
@y2
at y = 0 and y = b. This is now a standard characteristic value problem.
Assuming the distributed load to be a reasonably well behaved function, it can be expressed in the form of a double Fourier series as
w=
f (x; y) =
1 X
1
X
m=1 n=1
amn sin(
12
mx
ny
) sin(
)
a
b
with the Fourier coeÆcients
mx
ny
4 Z aZ b
f (x; y) sin(
) sin(
)dxdy
amn =
ab 0 0
a
b
The plate deection is then obtained by superposition as
w=
1 X
1
1 X
amn
ny
mx
) sin(
)
2
2 2 sin(
m
n
4
D m=1 n=1 ( a2 + b2 )
a
b
Specically, for the plate loaded by a uniform pressure p0 one readily obtains
amn =
and the deection becomes
w=
16p0
2 mn
1 X
1 sin( mx ) sin( ny )
16p0 X
a
b
6 D m=1 n=1 mn( ma22 + nb22 )2
Note that in this case, the Fourier coeÆcients amn are non zero only for odd values of the
indices m and n.
11
Potential Methods
Helmholtz' theorem states that any reasonable vector eld u can be expressed in general in
terms of potentials, i.e.
u = r + r where, here is the scalar potential and
directly related to the dilation e by
is the vector potential. The scalar potential is
e = ; ii
while the vector potential relates to the rotation vector !i by
2!i = i ; jj
A simple example of the use of potentials in elasticity is Lame's strain potential function
dened by
2Gui =
@
@xi
13
It can be shown that in the absence of body forces the potential is an harmonic function, i.e.
r2 = 0
and the stress tensor components are given by
ij = ;ij
Since harmonic functions have long been studied and are well known, they can be readily
used to obtain solutions to many practical problems. For instance, for a hollow sphere (inner
radius a, outer radius b) subjected to inner pressure p and outer pressure q, one can use the
potential function
=
C
+ DR2
R
where C and D are constants and R2 = x2 + y2 + z 2 to determine the stress at (R; 0; 0). The
values of the constants C and D are determined by incorporating the boundary conditions
to yield
RR = p
(b=R)3 1
(b=a)3 1
q
1 (a=R)3
1 (a=b)3
and
p (b=R)3 + 2
= (
)
2 (b=a)3 1
q (a=R)3 + 2
(
)
2 1 (a=b)3
Many other interesting problems can be solved using the method of potentials.
12
Two-Dimensional Problems in Rectangular Cartesian Coordinates
In Cartesian coordinates (x; y), the Airy stress function (x; y) is implicitly dened by the
equations
@ 2
= xx V
@y2
@ 2
= xy
@[email protected]
@ 2
= yy V
@x2
14
where the potential V is implicitly given by
rV
= X
where X is the body force vector.
If X = 0 then
@ 4
@ 4
@ 4
+
2
+
=0
@x4
@[email protected] @y4
Hence, the stress function is a biharmonic function.
13
Two-Dimensional Problems in Polar Coordinates
The equations of equilibrium in two-dimensional polar coordinates (r; ), where the body
force acts only along the r direction are
@r 1 @r r +
+
+ Xr = 0
@r r @
r
1 @ @r 2r
+
+
=0
r @
@r
r
One can show that the appropriate form of the stress function equation required to satisfy
compatibility is
(
@2 1 @
1 @ 2 @ 2 1 @ 1 @ 2
+
+
)(
+
+
)=0
@r2 r @r r2 @2 @r2 r @r r2 @2
where
rr =
1 @
r @r
and
@ 2
@r2
An important special case is obtained when the stress distribution is symmetrical about
an axis. The stress function equation in this case becomes
=
d4 2 d3 1 d2 1 d
+
+
=0
dr4 r dr3 r2 dr2 r3 dr
One can show that a fairly general solution of the above is of the form
= A log r + Br2 log r + Cr2 + D
15
13.1
Pressurized Cylindrical Tube
Consider a long cylindrical tube subjected to uniform pressures at its inner and outer surfaces
( p at r = a and q at r = b). Axial symmetry can be assumed and body forces may be
neglected. The equilibrium equation becomes
drr rr +
=0
dr
r
Moreover, Hooke's law reduce to
rr =
1
(
E rr
) =
du
dr
=
1
(
E rr ) =
u
r
and
where u is the radial displacement.
It can be shown that an appropriate form for the stress function that yields physically
meaningful, single-valued displacement functions in this case is
= A log r + Cr2 + D
Taking derivatives of this and substituting the stated boundary conditions yields expressions for the unknown constants A and C and the stress eld is given by
rr =
(b=r)2
1 @
= p
r @r
(b=a)2
=
@ 2
(b=r)2 + 1
=
p
@r2
(b=a)2 1
1
1
q
q
1 (a=r)2
1 (a=b)2
1 + (a=r)2
1 (a=b)2
And the displacement is given by
u=
1 (a2 p b2 q)r 1 + (p q)a2b2
+
E
b2 a2
E (b2 a2 )r
In this particular case, the problem can also be solved without involving the stress
function by solving the equidimensional dierential equation that results from substituting
Hooke's law and the strain-displacement relationships into the equilibrium equation.
16
13.2
Bending of a Curved Bar
For the bending of a curved bar (inner curvature radius a, outer curvature radius b), by a
force P acting towards the origin at one end while the other end is clamped, one has
= (d1 r3 +
c01 0
+ d1 r log r) sin r
where
P
d1 =
2N
2 b2
P
a
c01 =
2N
P
(a2 + b2 )
d01 =
N
where N = a2
13.3
b2 + (a2 + b2 ) log(b=a).
Rotating Disks
Consider the problem of a solid disk of material with density , radius b, of uniform thickness
and rotating about its center with angular velocity !. If the disk thickness is small compared
to its radius, both, radial and tangential stresses can be regarded approximately constant
though the thickness Moreover, because of the symmetry, the stress components can be
regarded as functions of r only and the shear stresses vanish. The equilibrium equations
reduce to a single one, namely
@r r d
+
+ Xr = (rr ) + !2 r2 = 0
@r
r
dr
where Xr = !2 r.
It can be shown that a stress function F dened by the relationships
F = rr
and
dF
= !2 r2
dr
produces stresses satisfying the equilibrium equation.
The strain tensor components in this case are
r =
du 1
= (
dr E r
17
)
u 1
= ( r )
r E where u is the radial displacement and Hooke's law has been used.
By eliminating u from the above equations and using Hooke's law, the following dierential equation for F is obtained
=
d2 F
dF
+r
2
dr
dr
The general solution is of the form
r2
F + (3 + )!2 r3 = 0
C1 3 + 2 3
! r
r
8
Particular solutions to specic problems can now be obtained by introducing boundary
conditions. For instance, for a solid disk, the stresses are nite at the center and r vanishes
at its outer radius r = b. The resulting stresses are then
3+ 2 2 2
r =
! (b r )
8
F = Cr +
3 + 2 2 1 + 3 2 2
! b
! r
8
8
Alternatively, for a disk with a hole at the center (radius a), since r = 0 at both r = a
and r = b, the stresses become
=
r =
3+ 2 2 2
! (b + a
8
a2 b2
r2
r2 )
3 + 2 2 2 a2 b2 1 + 3 2
! (b + a + 2
r)
8
r
3+
Note that as the size of the hole approaches zero, the maximum tangential stress does not
converge to the value obtained for the solid disk. This is the eect of stress concentration
associated with the presence of the hole.
=
13.4
Plate with a Hole under Uniaxial Tension
Consider a large plate with a circular hole (radius a) in the middle and under uniform
uniaxial tension S along the x direction.
Far away from the hole (for radii b >> a), the stresses are given by
S
(r )jr=b = (1 + cos(2))
2
18
and
S
sin(2)
2
where is the angle between the position vector and the positive x axis measured clockwise.
It can be shown that the above stresses can be obtained from the stress function
r =
= f (r) cos(2)
by using the equations
r =
1 @
r @r
=
@ 2
@r2
and
@ 1 @
(
)
@r r @
By requiring the stress function to satisfy
@2 1 @
1 @ 2 @ 2 1 @ 1 @ 2 ( 2+
+ 2 2 )( 2 +
+
)=0
@r r @r r @ @r
r @r r2 @2
the compatibility condition is automatically satised. Introducing the assumed relationship
for yields
@2 1 @
4 @ 2 f 1 @f 4f
)(
+
)=0
( 2+
@r r @r r2 @r2 r @r r2
This can be readily solved yielding the following general solution for the stress function
C
= (AR2 + Br4 + 2 + D) cos(2)
r
Finally, introducing the far-eld stress conditions (at r = b) and the condition of zero external
forces at the hole boundary yields
S
a
S
a
a
r = (1 ( )2 ) + (1 + 3( )4 4( )2 ) cos(2)
2
r
2
r
r
r =
S
a
= (1 + ( )2 )
2
r
r =
S
a
(1 + 3( )4 ) cos(2)
2
r
S
a
a
(1 3( )4 + 2( )2 ) sin(2)
2
r
r
19
14
Thermodynamics of Elastic Bodies
Thermodynamics accounts for the eects of thermal phenomena on the response of materials
to mechanical loads. The rst law of thermodynamics states that energy is conserved. The
second law states that entropy is always created. The combined statement of the rst and
second laws of thermodynamics as applied to solid bodies is
1
dU = T dS + ij dij
where U is the specic internal energy or energy per unit mass and S is the specic entropy
or entropy per unit mass.
The Helmholtz and Gibbs free energy functions are often useful in solving practical
problems. They are dened respectively as
F =U
TS
and
G =U
TS
1
ij ij
The equations of thermodynamics can be used to obtain expressions of the condition
required for thermodynamic equilibrium. Two equivalent statements of the condition are
A system is in thermodynamic equilibrium if for all possible variations of the state
taking place at constant energy, the entropy change is negative, i.e. (S )jU < 0, the
entropy is maximal.
A system is in thermodynamic equilibrium if for all possible variations of the state
taking place at constant entropy, the energy change is non-negative, i.e. (U )jS 0,
the energy is minimal (or positive denite).
The laws of thermodynamics impose denite restrictions on the mathematical form of
constitutive equations. In the case of linear elastic bodies, combination of Hooke's law with
the denition of strain energy yields
dW = ij dij = ede + 2Gij dij
For a point in a body going from zero stress to a particular stressed state, the total strain
energy becomes
W=
Z
dW = e2 + Gij ij
2
20
or, in terms of the second invariant of the strain deviation tensor, J2 = 12 0ij 0ij ,
1
W = Ke2 + 2GJ2
2
Since, thermodynamics requires the energy to be positive denite, necessarily
E>0
and
1
1<<
2
The strain energy function must be modied when the body is subjected to thermal loads
in addition to mechanical ones since there is thermal expansion. The appropriate form of
the strain energy in this case is
1 0
ij kl
W = ij (T T0 )ij + Cijkl
2
where ij is a tensor of thermal expansion coeÆcients and T0 is a reference temperature.
With this, the stress becomes
ij = Cijkl kl ij (T T0 )
As in the isothermal case, the number of elastic constants is signicantly reduced for an
isotropic body and the expressions for stress, strain and energy become
E
ij = eÆij + 2Gij
(T T0 )Æij
1 2
1+
ij =
sÆij +
+ (T T0 )Æij
E
E ij
and
2 2(1 + )
e
(T T0 )e]
W = G[ij ij +
1 2
1 2
where is the coeÆcient of thermal expansion, e = ii and s = ii .
With the above, expressions for the various thermodynamic functions applicable to
isotropic linear elastic solid can be derived. For instance, the internal energy is given by
11
E
U = [ Ke2 + 2GJ2 +
T e + (Cv )je =0 (T T0 )]
2
1 2 0
and for the Gibbs free energy
ZT ZT
1 2 1+
dT 0
G= [ ij ij (T T0 ) + dT (Cp )j =0 0 ]
2E
2E
T
T0
T0
Moreover, various equations connecting the thermal and mechanical properties of the
elastic solid are also readily derived. One example is the formula
T @ij @ij
Cp Cv =
@T @T
ij
ij
21
15
15.1
Variational and Finite Element Methods
Introduction
The nite element method is the most widely used numerical technique for the solution of
complex problems in solid mechanics. The method is based on rst subdividing the domain of
interest into a collection of small volumes called nite elements which are connected at nodal
locations. The solution is approximated inside the elements from the values at the nodes
using special functions called shape functions and the result is introduced into the variational
(virtual power) formulation of the problem. The result is a set of algebraic equations that
upon solution yield the values of the desired unknown at the nodal locations.
15.2
Variational Methods
Recall that the boundary value problem consisting of nding the function u(x) such that
d2 u
= u00 = f
dx2
subject to the boundary conditions u(0) = u(1) = 0 is equivalent to the variational problem
consisting of determining the function u(x) such that
a(u; v) = (f; v)
where
a(u; v) =
and
Z1
0
u00 dx =
(f; v) =
Z1
0
Z1
0
u0 v0 dx
fvdx
and also to the problem of nding the function u(x) that minimizes the functional
1
I (u) = a(u; v) (f; v)
2
where v is a reasonably well behaved but rather arbitrary function.
When one looks for approximate solutions to the integrated forms of the problem by
the Galerkin method or by the Ritz method, one seeks for solutions in a nite dimensional
subspace of the actual solution space.
In the Ritz method, one searches for an approximation to the function that minimizes
the functional. The method starts by introducing a set of basis functions fi (x)gNi=1 and
expresses the approximate solution as the trial function uN u as
uN =
N
X
i=1
ai i
22
The functional obtained using this approximation then becomes
1
I (uN ) = a(uN ; v) (f; v)
2
Now, the function v being arbitrary, is selected as the test function
v = vN = uN =
N
X
i=1
ai i
and is substituted into the functional expression to give
N
N
N
X
X
1 X
I (uN ) = a( ai i ; ai i ) (f; ai i )
2 i=1
i=1
i=1
where
N
X
a(
i=1
ai i ;
N
X
i=1
ai i ) =
and
(f;
N
X
i=1
ai i ) =
Z1
0
(ai
Z1 X
N
0
f(
i=1
N d
X
i
i=1
dx
)2 dx
ai i )dx
Finally, the unknown coeÆcients ai ; i = 1; 2; :::; N are determined by solving the system
of algebraic equations obtained from the extremum conditions. i.e.
@I
@I
@I
=
= :::: =
=0
@a1 @a2
@aN
which in matrix notation are simply written as
Ka = F
In the Galerkin method, one seeks directly for an approximate solution of a(u; v) =
(f; v). Like in the Ritz method, in the Galerkin approach one starts by introducing a set
of basis functions fj (x)gNj=1 satisfying the stated boundary conditions and expresses the
approximate solution as the trial function uN u as
uN =
N
X
j =1
aj j
However, the test functions are selected so as to be identical to the basis functions, i.e.
fvigNi=1 = fi(x)gNi=1 .
23
The resulting discrete form of the problem is then
N
X
i=1
aj a(j ; i ) = (f; i )
for all i = 1; 2; :::; N . Here
a(j ; i ) =
Z 1 dj di
and
(f; i ) =
dx dx
0
Z1
0
dx
fi dx
for i = 1; 2; :::; N
Using matrix notation, the above is simply written as
Ka = F
Solving the algebraic problem yields the values of the coeÆcients ai ; i = 1; 2; :::; N .
The system of equations obtained with the Galerkin method is identical to the one obtained using the Ritz method. However, the Galerkin method is of more general applicability
since it works even when the problem of minimizing the functional has no solution. The nite
element method is obtained by implementing the Galerkin (or Ritz) procedure with a very
particular choice of basis functions.
15.3
The Finite Element Method
The nite element method is a numerical procedure designed to nd approximate solutions
of boundary value problems. The method is a direct implementation of the Galerkin (or
Ritz) procedure in which the chosen basis functions are nite element basis functions (also
called sometimes nite element shape functions). These functions are of very simple form
(piecewise polynomials of low order are most common) and their most distinctive feature is
that they are nonzero only in a small subregion of the computational domain, i.e. they have
a local character. Finite element basis functions are then said to posses local or compact
support.
This is in direct contrast with the classic Galerkin or Ritz approaches where basis functions valid over the entire computational domain are used. Another important feature of the
nite element basis functions is that the basis function associated with a particular element
partially overlap with those of immediately adjacent elements.
Improved approximation accuracy is readily obtained with the nite element method by
simply increasing the number of subdomains (i.e. by decreasing their size) without increasing
the complexity of the polynomials used to represent the solution inside each individual
subregion.
24
By implementing the Galerkin process on individual subregions another powerful feature
of the method emerges. Since the coordinate functions of adjacent subregions overlap partially with each other, the resulting system of algebraic equations can be constructed and
assembled together by the computer just as it is created. This feature makes it a trivial
process to determine increasingly better approximations by renement of the initial subdivision. Moreover, the matrix associated with the resulting system has a sparse structure and
eÆcient solution methods are applicable. This very desirable feature is absent in the classic
implementation of the Galerkin or Ritz methods and constitutes an important disadvantage
in those cases.
The simplest description of the fundamental characteristics of nite elements is obtained
for one dimensional systems. Finite element characteristics for two and three dimensional
elements are based on the same ideas but involve additional complexities of description that
distract from the basic points. In this section basic characteristics of selected one dimensional
elements are described. The implementation of the Ritz-Galerkin nite element method is
also covered.
In sum, a nite element is a triple of entities (K; PK ; ) where
K is a simple geometrical object, such as a line segment in one dimension, a triangle
in two dimensions and a parallelepiped in three dimensions,
PK is a nite dimensional linear space of basis functions dened on K , and
is a set of degrees of freedom.
A nite element is then the totality of ingredients associated with the denition of the global
approximating function inside the subdomain. Since the method produces approximations to
global functions from functions having only local support, it is key to be able to distinguish
and to relate the global and local descriptions. Therefore, one needs to carefully specify both
local and global parameters as well as their relationship.
15.4
Global and Local Coordinates and Basis Functions for Unidimensional Systems
Local parameters are specied at the element level. For instance, consider the one dimensional case. A simple subdivision of a given domain x 2 [a; b] is readily obtained by introducing a collection of N contiguous subdomains [x1 = a; x2 ]; [x2 ; x3 ]; :::; [xi ; xi+1 ]; :::; [xN ; xN +1 =
b]. The points x1 ; x2 ; x3 ; :::; xi ; :::; xN ; xN +1 are called nodes and each contiguous pair of
them, [xi ; xi+1 ], represents the boundaries of the element that spans the space hi between
the nodes. In this case, each element has two nodes one located at each end of the element.
The location of all these points is given with reference to a global system of coordinates. If
nodes are numbered based on their position is terms of global coordinates one obtains global
node numbers.
25
In the local description of the element bounded by nodes [xi ; xi+1 ], the positions measured
with respect to the global system of coordinates are transformed into coordinates [1 ; 2 ] =
[ 1; +1] referred to a system of coordinates with its origin in the center of the element, i.e.
[xi ; xi+1 ] ! [1 ; 2 ]. The aÆne or linear transformation is simplest, i.e.
2x xi xi+1
(x) =
hi
The inverse transformation is
h + xi + xi+1
x( ) = i
2
If for any given element, node numbers are assigned based on the local coordinates one
obtains local node numbers. For instance, the node with global number i is the same as
local node 2 for element i and also as local node 1 for element i + 1. Since in nite element
work on uses basis functions with local support and thus implements the Galerkin method at
the element level to subsequently assemble the resulting equations, the distinction between
local and global representations is quite important.
In the nite element method one starts by using simple interpolation functions as basis
functions at the element level and then proceeds to represent the solution in the entire
domain by collecting the contributions associated with each element. The basis functions
involved at the element level are called local basis functions while the ones that produce the
solution in the entire domain are called global basis functions. Clearly, the two sets of basis
functions are closely related.
The role of the local basis functions is to generate the value of the approximated quantity
inside the element from the values at the nodes by interpolation. Assuming again a one dimensional system and the simple subdivision introduced above, consider an arbitrary interior
element e where the index i denotes the global node number. If the values of the required
solution at xi and xi+1 are u(xi ) = ui and u(xi+1) = ui+1 , respectively, the approximate
values of u(x), u(x)e inside the element are calculated by simple interpolation as follows
i
i
ue (x) = ui i1 + ui+1 i2
i
Here, the local nite element basis functions for element e , i1 and i2 are dened as
x
x
i1 = i+1
hi
i
x xi
hi
where hi = xi+1 xi is the element size and all the positions are measured in the global
coordinate system. Note that the indices 1 and 2 on the local basis functions refer to the
local node numbers for the element.
i2 =
26
For simplicity, assume all nite elements are of equal size (uniformly spaced mesh) so
that h1 = h2 = ::: = hi = h. Now, the approximate solution on the entire domain uh (x) is
represented as a linear combination of the nite element basis functions, i.e.
uh (x) = u11 + u2 2 + ::: + ui i + uN N =
N
X
i=1
ui i
However, note that here, i ; i = 1; 2; :::; N are global nite element basis functions.
There is a simple relationship between local and global nite element basis functions,
specically, in element 1,
1 = 11
in element 2
2 =
in element ei ,
i =
(
(
12 if x 2 e1
21 if x 2 e2
i2
i1
if x 2 ei
if x 2 ei
1
1
and in element eN
N = N2
Using the introduced notation the Galerkin nite element method equations can be expressed as
N
X
i=1
uj a(j ; i ) = (f; i )
for all i = 1; 2; :::; N . Here
a(j ; i ) =
Z 1 dj di
and
(f; i ) =
dx dx
0
Z1
0
dx
fi dx
for i = 1; 2; :::; N .
Using matrix notation, the above is simply written as
Ku = F
27
Solving the algebraic problem yields the values of the nodal values ui ; i = 1; 2; :::; N . The
matrix K is called the nite element stiness matrix, the colum vector F is the force vector
and the column vector u is the vector of unknown nodal values of u.
The same thing can be done to express the Ritz nite element method equations using
the new notation. Specically, the functional I (uh ) is given by
N
N
N
X
X
1 X
I (uh ) = a( ui i ; ui i ) (f; ui i )
2 i=1
i=1
i=1
where
N
X
a(
i=1
ui i ;
N
X
i=1
ui i ) =
and
(f;
N
X
i=1
ai i ) =
Z1X
N
di
(
0 i=1
Z1 X
N
0
f(
i=1
ui
dx
)2 dx
ui i )dx
Finally, introduction of the extremum conditions
@I
@I
@I
=
= ::: =
@u1 @u2
@uN
yields the system
Ku = F
which is identical to the one obtained using Galerkin's method.
15.5
Finite Elements in Elastic Stress Analysis
Consider an elastic body being deformed by applied loads while constrained from rigid body
motion. The stress principle of Euler and Cauchy states that when the body is in mechanical
equilibrium the volume forces and surfaces at any small subdomain inside the body balance
each other, i.e.
Z
fdv +
Z
tds = 0
where f is the volume force vector eld, t is the stress vector eld and and are the
volume and surface area of the subdomain.
Furthermore, Cauchy's theorem states that related to the stress vector t there is a stress
tensor called the Cauchy stress tensor.
28
The displacement vector eld u described the deformation everywhere inside the loaded
body. However, it is convenient to introduce the strain tensor associated to the displacement
in the small deformation regime characteristic of elastic phenomena
1 @u @u
ij = (
+ ) = rs u
2 @xj @xi
where ij are the components of the strain tensor and rs is called the symmetric gradient
operator.
It is convenient in solid mechanics to state the condition of mechanical equilibrium in
variational form. In this representation, the deformation state corresponding to mechanical
equilibrium of the loaded body is the one that yields the minimum value of the functional
Z 1
Z
I = [ : f:u] + g:uds
V 2
1
P
where : = ik ik ik This is just the expression of the principle of minimum energy. A
closely related variational statement is the principle of virtual work.
All materials satisfy the above equation as a condition for mechanical equilibrium. However, depending on the material constitution, the response to a given load varies. This
variation is described by means of constitutive equations of mechanical behavior. The constitutive equation for elastic behavior is Hooke's law
1+
trI
E
E
where E is the elastic modulus, is Poisson's ratio and tr = ii where the summation
convention is used. Sometimes, the shorthand notation = C is used instead.
Te commonly used displacement formulation of the nite element method as it is applied
in solid mechanics is readily obtained by expressing the variational principle in terms of the
displacement, i.e.
Z 1
Z
I = [ rs u : C rs u f:u]dv + g:uds
V 2
1
Consider now as an illustration a linear elastic bar loaded by a distributed internal body
force f (x) constrained at the end x = 0, attached to a rigid wall by a spring of spring
constant kl at x = L and under imposed axial tractions Tds = cu(x) due to a distributed
spring acting along its length. The goal is to determine the longitudinal displacement of the
bar at equilibrium. If the cross sectional area of the bar is A(x) the condition of mechanical
equilibrium is obtained by performing a force balance on a small element x of the bar and
then taking the limit as jDeltax ! 0. The result is
=
d
du
(AE ) + cu = f
dx
dx
29
The principle of virtual work is obtained by multiplying the equilibrium equation by a
reasonable function v(x) and then integrating from x = 0 to x = L while using integration
by parts the result is, after minor rearrangement
ZL
ZL
du dv
+ cuv)dx = fvdx + (F v)x=L (F v)jx=0
dx dx
0
0
The function v is called the virtual displacement and dv=dx is the virtual strain. Introducing the notation
(AE
a(u; v) =
and
(f; v) =
ZL
0
ZL
0
(AE
du dv
+ cuv)dx
dx dx
fvdx + (F v)jx=L
(F v)jx=0
the principle of virtual work becomes
a(u; v) = (f; v)
Using the above notation one denes the elastic strain energy U as
1
U = a(u; u)
2
As in the general formulation of the nite element methodology, the function u that satisfyies the dierential mechanical equilibrium equations subject to given boundary conditions
is the same that solves the variational problem expressed by the principle of virtual work
and it is also the same that minimizes the energy functional.
30
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