matriz tensor

8/11/2019 Matriz Tensor

1/18

Stress, Strain, and Elasticity, 2006

Rick Aster

February 1, 2006

In seismology, a fundamental goal is to understand the vector equation ofmotion

F= ma (1)

for force, F, mass, m, and acceleration, a, and its many consequences in con-tinuous elastic and nearly elastic media.

In a continuous medium of density , we can write an equation of forcebalance at point xand time, t, as

f(x, t) = 2u(x, t)

t2 (2)

or, in index form as

fi(x, t) =2ui(x, t)

t2 (3)

where f is a force density(a force per unit volume) and the displacement of the

media from its equilibrium state in space and time isu

(x

, t).Stressis the continuum analog of pressure. It has units of force per area, andthe total force on a real or imagined surface within a volume can be calculatedby integrating the stress over that surface. Stress can thus be thought of as asurface force ofcontact force, in contrast with a volume force, such as gravity,where a total force calculation would involve integrating over a volume.

The tractionat some point x with a medium is the resolved force per unitarea on a surface defined by the normal n

(n) = limS0

f

S (4)

where S is a small surface element defined by n. Note that and n will not,in general, be parallel; the traction will typically have a nonzero normal and a

nonzero tangential component.A complete equation of motion with both surface and body force densities,

fb, is thus

ftotal=

S

(n)dS+

V

fb(n) dV =m2u

t2 . (5)

1


2/18

Figure 1: Traction Continuity

Now, consider a thin wafer of material. As its thickness decreases to zero,the volume, and the volume integrals above, go to zero, but the surface area andthe surface integral remain finite. It is thus clear that the integrated surfaceforce must go to zero for a thin waver of any orientation (otherwise we wouldhave a finite force applied to a zero mass!). Thus, the traction on each side mustbe equal and opposite, i.e.

(n) = (n) (6)

everywhere within a connected medium. Traction is thus a continuous quantity

in such a medium; it cannot have discontinuities.The stress tensor is simply a matrix of traction components

ij =

11 12 1321 22 23

31 32 33

=

(1)1

(2)1

(3)1

(1)2

(2)2

(3)2

(1)3

(2)3

(3)3

(7)

where (j)i is the i

th component of traction on the surface of an infinitesimalcube defined by the outward normal in the j th direction.

2


3/18

Figure 2: Traction Component Decomposition

To see how the stress tensor can be easily used to find the traction on anarbitrary surface, consider a right angle tetrahedron acted upon only by surfaceforces, with three of its faces, dSi, lying in the three coordinate planes.

As the volume of the tetrahedron in Figure 2 goes to zero, so must the netforce (otherwise we would have a finite force operating on a zero mass and a

resulting infinite acceleration). Thus

limV0

(n)dS+ (n1)dS1+ (n2)dS2+ (n3)dS3 = 0 (8)

or

(n) =(n1)dS1 (n2)dS2 (n3)dS3

dS . (9)

The components of n are just the projections onto the coordinate axes, orequivalantly the correspondin area ratios of the sizes of the right angle tetrahe-dron

n= (n n1, n n2, n n3) = (dS1/dS, dS2/dS, dS3/dS) (10)

so that the traction on dS is

(n) = (n1)n n1 (n2)n n2 (n3)n n3

= (n1)n n1+ (n2)n n2+ (n3)n n3

= (ni)(n ni) =ijnj .

3


4/18

Figure 3: Traction Torques

We thus see that traction for an arbitrary surface defined by the normaln is given by the matrix multiplication of and n; the stress tensor is themathematical machine that transforms one vector (the normal) into another(the traction).

The diagonal terms of the stress tensor define the normal stressesin the co-ordinate system in which is expressed. The off-diagonal terms define the shear

stresses. Normal stresses act parallel to nwhile shear stresses act perpendicularto n. In a viscosity-free liquid, where the flow of the material acts to eliminateshear tractions, the stress tensor is diagonal, and defines the pressure. Indeed,the stress tensor in such a medium has the very simple form:

ij = P ij (11)

wherePis the pressure and is the Kronecker delta. The minus sign arisesbecause of our outward normal convention; tractions that push inward are neg-ative (beaware that this convention is sometimes reversed in engineering appli-cations). Our convention is convenient, as we shall see, because positive stressesproduce positive strains.

Next, consider a small stressed rectangular prism (dx dy in length/width)

of in the (x, y) plane.The stresses on the faces in the (x, z) and (y, z) planes are approximately

given by the values of the stress tensor at O, ij , Because the cube is small, wecan approximate the stresses on the other faces at x = dx and y = dy using the

4


5/18

first-order approximation

xi+ xi= xi+ xix

dx (12)

and

yi+ yi = yi + yi

y dy (13)

The net torque, Tz (counterclockwise) on the prism in Figure 3 about z at Ofor a prism of thicknessdz is a sum of terms consisting of contact forces (stresstimes area), times their corresponding perpendicular moment arms (there areonly six terms instead of eight because xy acting on the x face and yxoperating on the y face both have associated lever arms of zero length andhence contribute zero torque)

Tz = (xy + xy) dy dz dx(yx + yx) dx dz dy1/2(xx+ xx) dy dz dy(14)

+1/2(xx) dy dz dy+ 1/2(yy + yy) dx dz dx 1/2(yy) dx dz dx

where the 1/2 factors arise because the normal forces operate on lever arms of1/2dy or 1/2dx.

To keep the prism in equilibrium (and from undergoing an infinite angularacceleration as its dimensions and volume, V = dx dy dz, go to zero), thenet torque must go to zero faster than the volume and moment of inertia, sothat

0 = limV0

Tz

V = (xy+ xy) (yx+ yx) 1/2(xx+ xx) dy/dx (15)

+1/2(xx) dy/dx + 1/2(yy+yy) dx/dy 1/2(yy) dx/dy

Asdx, dy, anddz all go to 0, each of the ij terms go to zero due to (13). Theremainingii terms cancel, giving

0 = limV0

V =xy yx . (16)

We could have just as easily have performed this derivaton in the ( x, z) or(y, z) planes, which would have given the results

0 = xz zx (17)

and0 = yz zy . (18)

Because of these symmetries, the stress tensor is a symmetric tensor, andthus has only 6 independent elements.

As stress is the first tensor object that we will encounter in this course, thisis a good time to review tensor properties. A tensor is an ob ject that specifiessome physical or abstract condition (e.g., stress, strain, conductivity), and can

5


6/18

be expressed in an arbitrary coordinate system. Tensors are just linear operatorsconverting one object (such as a vector) to another. Recall that a vector is an

object specifying a magnitude and direction. To represent this quantity in a newcoordinate system, we simply multiply by the appropriate matrix of directioncosines

A= ei ej (19)

where the ej are the original coordinate axes and the e

i are the new coordinateaxes. The coordinate transformation for an arbitrary vector is thus

v = Av (20)

To see how to express a tensor such as stress in a new coordinate system, considertraction and unit normal vectors in two coordinate systems

= A (21)

n = An (22)

As A is an orthogonal matrix, the inverse transformation is

= AT (23)

n= ATn (24)

In the primed coordinate system, we thus have

=n = A= An= AATn (25)

Thus, the stress tensor in the new coordinate system, which is the tensor relating

n

and

, is given by the transformation

= AAT (26)

Because it is symmetric, the stress tensor is diagonalizable, meaning there isalways a coordinate system where the off-diagonal elements are all zero. In thiscoordinate system the stress tensor can be written as

= = iji = VTV (27)

where the eigenvalues i are normal principal stresses, and the transformingmatrix V consists of columns of eigenvectors corresponding to each eigenvaluein the eigenvalue-eigenvector decomposition

= VVT

. (28)

Thus, the coordinate system where is diagonalized is defined by the columnsofV, and the tractions in each of these directions (called the principal stressdirections) are ivi.

6


7/18

To examine how shear and normal tractions vary with coordinate system,consider a biaxial stress

=

xx 0 00 yy 0

0 0 0

. (29)

A clockwise rotation of the coordinate system about the z axis by an angle is characterized by the rotation matrix of direction cosines between coordinateaxes, ei and ej

Aij = e

i ej =

cos sin 0sin cos 0

0 0 1

. (30)

so that the stress tensor in the new coordinate system is

= AAT =

cos sin 0sin cos 0

0 0 1

xx 0 00 yy 0

0 0 0

(31)

cos sin 0 sin cos 0

0 0 1

=

xxcos

2 + yysin2 (xx yy)sin cos 0

(xx yy)sin cos yycos2 + xxsin2 0

0 0 0

= xx

xy

xz

yx

yy

yzzx

zy

zz

.

Ifxx and yy are of opposite sign, there will be two angles where xx= 0(no normal traction on the surface defined by the x direction), when

xxcos2 + yysin

2 = 0 (32)

or

tan2 = xx/yy , (33)

which has a real solution when xxyy 0 of

= arctanxx/yy . (34)Similarly ifxxyy 0, then there are two angles for which

yy = 0 (no normaltraction on the surface defined by the x direction), whenxxyy 0

= arctanyy/xx . (35)

7


8/18

Figure 4: A Simple Stress Coordinate System Stress Rotation

We can thus rotate the stress tensor so that two of the four faces have onlyshear tractions. For example, if xx = yy , we can get all diagonal termsof the stress tensor to be zero through a 45 coordinate system rotation, asillustrated below. In this case, the pre- and post-rotation stress tensors are just

=

P 0 00 P 0

0 0 0

(36)

and

=

0 P 0P 0 0

0 0 0

. (37)

It is very important to realize that the physical state of stress has not changed,only the numbers and coordinate system that we chose to express it in!

There is, as you would expect, an important link between stresses and fault-ing, as tractions provide the force that makes faults slip. In general, faultmovement will be more likely to occur near planes that resolve maximum shearstress, although other factors such as fluid pressure or other processes acting to

8


9/18

reduce fault stress, as well as the long-term structural evolution and rotationof fault zones can produce active fault systems that are oriented at appreciable

angles from these maximum shear-traction planes.The mean stressis the average of the stress diagonal

M= (xx+ yy + zz)/3 =ii/3 . (38)

Mis invariant with respect to the coordinate system, as we can see by notingthat

ij =AikklATlj =AikklAjl = AikVkllmVjmAjm (39)

whereV is the matrix of eigenvectors that rotates into the diagonal matrix ofeigenvalues, n, . The double rotation effected by A and V can be expressedas a single rotation by the orthogonal matrix

Bil= AikVkl (40)

so thatij =BillmBjm = BillnnmnBjm . (41)

The trace of the stress tensor in any coordinate system thus the sum of theeigenvalues

ii= nBinBin=n

n . (42)

The deviatoric stress tensor is

Dij =ij Mij (43)

Stresses are usually measured in the MKS units of Newtons per square meter,or Pascals (Pa), but you may frequently see it expressed in bars, were 1 bar = 0.1MPa. Within the Earth, where the mean stress at shallow depths is essentiallydue to the lithostatic load

P = g

z0

(z)dz , (44)

P is typically much greater than the off-diagonal deviatoric stress tensor ele-ments. Nevertheless, the deviatoric stresses are frequently of the most interestin seismology, as they drive faulting and deformation that is not purely volu-metric. Deviatoric stresses can range up to 100s of MPa (kbars) in the crust,but appear to be much lower on most active faults. Lithostatic stresses, on theother hand, reach up approximately 20 GPa in the outermost approximately600 km of the Earth where earthquakes occur.

Note that any matrix, K, can be written as the sum of a symmetric part, Sand an antisymmetric part, A

Kij = 1/2(Kij+ Kji) + 1/2(Kij Kji) Sij+ Aij (45)

9


10/18

It is instructive to examine the geometric significance of antisymmetric andsymmetric matrix operators. An antisymmetric tensor can be closely related to

a vector. Consider an arbitrary antisymmetric tensor

Aij =

0 a12 a13a12 0 a23

a13 a23 0

(46)

Left multiplication of an arbitrary vector xby A gives

vi= Aijxj =

a12x2+ a13x3a12x1+ a23x3

a13x1+ a32x2

=

w2x3 w3x2w3x1 w1x3

w1x2 w2x1

w x (47)

where

w= a23a13

a12

(48)Thus, multiplication by an antisymmetric tensor is equivalent to a cross-productmultiplication by w, which introduces a 90 rotation and a scaling by |w| sin (where is the angle between wand x).

A symmetric tensor operator, S, applied to an arbitrary vector,x has anothergeometric interpretation which can be seen by looking at the dot product of theinput and output vectors of the operation, which is

C(x) =xi(Sijxj) = s11x21+s22x

22+s33x

23+2(s12x1x2+s13x1x3+s23x2x3) . (49)

The derivatives of the scalar function Cwith respect to xi are

C

xi= 2sijxj (50)

Thus, the gradient ofC, which is everywhere normal to the surface where C issome constant (say c0), is proportional to the result of the tensor operation onx. the C= c0 surface is thus described by the equation

c0= xiSijxj =s11x21+ s22x

22+ s33x

23+ 2(s12x1x2+ s13x1x3+ s23x2x3). (51)

If we select the principal coordinate system, where S is diagonalized, then

c0= xiSijxj =s11x21+ s22x

22+ s33x

23 (52)

which is the equation for a quadric surface. If all of the eigenvalues of S arepositive or negative, andc0 has the same sign as the eigenvalues, then this is theequation of an ellipsoid with semiaxes of length

c1/i. A symmetric tensor

operation thus scales and rotates the set of all unit vectors to trace out a quadricsurface in 3-space.

10


11/18

Figure 5: Strain

We are now prepared to introduce the deformational counterpart of stress,strain. Simply put, elastic strain (which we are most interested in seismology)is a normalized measure the elastic (reversible) deformation of a medium fromequilibrium. If we have a deformation field, ui(x), the first order linearizationfor this deformation near a point xis

ui(x+ x) = ui(x) + ui(x)xj

xj =ui(x) + ui (53)

The quantities of interest, the products of thexj and the spatial derivatives ofthe displacement field

ui=ui(x)

xjxj , (54)

can be written as a sum of symmetric and an antisymmetric matrices

ui = 1/2

ui(x)

xj+

uj(x)

xi

xj

+1/2ui(x)

xj

uj(x)

xi

xj (ij+ Wij)xj . (55)

First, lets consider the antisymmetric component; from our previous discus-

11


12/18

Figure 6: Rotating a Vector

sion, it is clear that left multiplication ofx by W is equivalent to

1/2

u3(x)x2

u2(x)x3

u1(x)x3

u3(x)x1

u2(x)x1

u1(x)x2

x w x (56)

For the first-order strain expansion (53) to be valid, wand xmust besufficiently small. To see what the cross product corresponds to in this case,consider the arbitrary clockwise rotation of a vector r about some axis n(Figure6).

We have

r =ON+ N V + V Q= n(n r) + (r n(n r))cos + (r n)sin (57)

= r cos + n(n r)(1 cos ) + (r n)sin

If the rotation is small, then cos 1 and sin , so that

r (r n) + r (58)

The change in the coordinate vector to the particle originally at r is thus

r= r r= (r n) = (n r) (59)

We thus see that, if the terms in w are small, then left multiplication by wcorresponds to a counterclockwise rotation about the w axis by an angle |w|.As pure rotations of materials do not release or store elastic energy, we do not

12


13/18

Figure 7: Basic strains and rotations: a) uxx

> 0; b) uxx

< 0; c) uxy

> 0; d)uxy

= uyx

=>0; e) uxy

= uyx

=>0

usually worry about them in seismology (just as we do not generally concernourselves with rigid translations). We are principally interested in the symmetriccomponent, the strain tensor

ij = 1/2

ui(x)

xj+

uj(x)

xi

. (60)

First, note that the elements of are dimensionless. The diagonal elementsspecify changes in lengths (contractions or extensions or longitudinal strains),

while the off-diagonal elements specify displacements perpendicular to the di-rections of interest, and are thus shear strains.

Figure 7 shows the basic types of strains. In the longitudinal strains, themedium is either extended or compressed along one axis. It thus undergoes achange in both volume and shape. Under simple shear, the shape changes (butnot, to first order, the volume), but this deformation requires an antisymmet-

13


14/18

ric (rotational) component of the strain tensor. If we remove the rotationalcomponent , we are left with pure shear.

It is important and useful to differentiate between changes in volume andchanges in shape. The trace of the strain tensor is the dilatation

= ii=ux

x +

uyy

+ uz

z = u (61)

which is just the divergence of the displacement field. To see the physicalsignificance of , consider the first-order change in volume that occurs when amaterial of volumeV =dx dy dz is strained to a volume V + V

V + V =

1 +

uxx

1 +

uyy

1 +

uzz

dx dy dz (62)

1 +

ux

x +

uy

y +

uz

z

dx dy dz= V + V .

We thus see that V /V is the volumetric strain.The relationship between stress and strain, called the constitutive equation

is fundamental to seismology. In one dimension, this relationship for an elasticsystem is given by Hookes law

fx= k x (63)

wherefx is the force applied to the system, k is the spring constant, and x isthe resulting displacement from equilibrium. Hookes law is linear; a doubling offorce produces a doubling of displacement, and so forth. If we look at a cartoon-ish relationship between displacement and force in the simple spring system, inFigure 8 we see that there is a range of forces and corresponding strains for

which we expect linear behavior, but more complicated behavior outside of thisrange. If the force gets too large, the metal in the spring will exceed its elasticlimit, and become permanently deformed. On the other hand, if the magnitudeof a negative force is large enough to push coils together, it suddenly becomesmuch harder to deform it, as the spring constant of a hollow metal cylinder,rather than a coil spring, will become appropriate. The slope of the curve is ameasure of the stiffness of the system, so there is a large slope increase whenthis occurs. In between these extremes, we see that there is a linear relationshipbetween force and displacement. To understand the relationship between stressand strain in elastic media, we must generalize Hookes law for continuous elas-tic media. In seismology, strains are commonly small (displacements of micronsand wavelengths of km), which makes much of seismology a linear science (thisis fortunate, especially considering the many other complications that we mustdeal with!). The most general linear relationship between stress and strain is

ij = cijklkl (64)

The elements of c which relate stress and strain are called elastic moduli. Asstrain is dimensionless, the moduli have units of stress (Pa). Although c has 34 =

14


15/18

Figure 8: Linear and Nonlinear Elasticity in the Spring System

15


16/18

81 elements, it is easy to see that the actual number must be considerably less,as both the stress and strain tensors are symmetric. This, and thermodynamic

considerations that we wont elaborate on here, reduce the maximum physicallypossible number of independent elements in c to 21 (still a daunting number).Fortunately, much of seismology can be done using the simplest possible c forsolid media, where there are only two independent moduli

cijkl = ijkl+ (ikjl+ iljk) (65)

where the moduli, and are called the Lame parameters. Thus, for thissimplest case of an isotropic medium, where elastic properties vary in no waywith coordinate system, the constitutive equation is

ij = (ijkl+ (ikjl + iljk))kl (66)

=kkij+ 2ij =ij+ 2ij

=

+ 2xx 2xy 2xz2yx + 2yy 2yz

2zx 2zy + 2zz

has no common name (well see its physical interpretation later). is animportant quantity in seismology called the rigidity, it is the ratio of pure shearstrain due to an applied shear stress component in an isotropic medium

= ij2ij

(67)

wherei =j .The bulk modulus or incompressibility describes the volume stiffness of

an isotropic medium. For a medium under isotropic stress characterized by apressure,P, we have

ij = P ij = cijklkl (68)

=ij+ 2ij .

Evaluating the trace of the stress tensor for isotropic media by summing thethree equations for the diagonal stress tensor terms gives

3P = 3 + 2 = (3 + 2). (69)

The bulk modulus is defined as the ratio of pressure to volume change

=

P

= +

2

3 . (70)

In an ideal (zero viscosity) fluid, mu = 0, thus, the Lame parameter is theincompressibility of an ideal fluid.

Consider a stretching experiment where tension is applied to an isotropicmedium along a principal axis (say x).

16


17/18

Figure 9: A uniaxial stress test

The stress state is

=

xx 0 00 0 0

0 0 0

. (71)

Using the isotropic cijkl (66) gives the stress-strain relationships

xx= + 2xx (72)

yy = zz = 0 = + 2yy = + 2zz (73)

xy = yx = xz = zx = yz = zy = 0 (74)

= 2xy= 2yx = 2xz = 2zx = 2yz = 2zy .

These expressions imply

yy = zz (75)

(as would be expected from symmetry), so that

(xx+ 2yy) + 2yy = 0 (76)

or

yy =

2( + )xx xx (77)

where, the negative of the dimensionless ratio of radial to longitudinal strainisPoissons Ratio.

The ratio of longitudinal stress to strain in the xdirection in the stretchingexperiment is found by substituting xx foryy and zz in (72)

xx= (xx 2xx) + 2xx (78)

17


18/18

to obtain the expression for Youngs Modulus

xxxx

= (3 + 2) +

E . (79)

Note that all of these moduli ( , , , , E ) are positive and have units ofstress (except for the dimensionless ). For many Earth materials . If wetake = , we have what is called a Poisson solid, where = 0.25, E= 5/2,and = 5/3.

18

matriz tensor

Documents