Appendix C Stress tensor

Appendix C Stress tensor V o l u m e a n d s u r f a c e forces in a n i d e a l fluid In studying propagation of acoustic waves in an ideal fluid, it was shown that there are two types of forces: volume forces and surface forces. The classical example of the former is the gravitational force. For instance, in the case of an elementary volume AV

this

force is equal to F(p) = f(p)AV =

p(p)g(p)AV,

(C-1)

where f(p) is the vector, characterizing the density of volume forces, f(p) = Here

p(p)

p(p) g(p)

(C-2)

is the mass density and g(p) is the gravitational field, caused by all masses,

except Am(p): =

(c-3)

Of course, masses of the Earth are main sources of the gravitational field. The surface forces may arise differently. Consider an elementary plane surface,

da(p),

da(p)

and at

inside an ideal fluid, Fig. C.la. A medium, situated at the right side of

its vicinity, acts on the medium, located at the left side of this elements, with the force F(p) The vector t surface

t(p)da

(C-4)

is the density of surface forces, and in an ideal fluid it is normal to the

da: t-

563

A P P E N D I X C. S T R E S S T E N S O R

564

{a}

(b} F -

n

F

{c}

F

{d]

t

F

:-

V

--"

F

da

(f)

{e} n

Figure C.I:

da

n

(a) Orientation of surface forces inside an ideal fluid (b) Orientation of

surface forces inside an elastic medium (c) Traction vector t (d) Arbitrary volume of elastic medium (e) Tractions at opposite sides of a disk where

n

is the unit vector, normal to the surface, and

P

is the pressure.

The

first remarkable feature of surface forces is the fact that they act only in the direction perpendicular to the surface. In other words, the tangential components of these forces are absent. This means that an action (pull or push) in the direction, tangential to the element da, does not have any influence on the ideal fluid, located on the other side of this surface. In accordance with Newton's third law a medium, situated at the left side of da, also exerts a force across this element, and it is equal to

F = - t (p)da

(C-6)

Thus, in the vicinity of any point p of the surface element there are two surface forces

APPENDIX C. STRESS TENSOR

565

with equal magnitude and opposite directions. They are applied at two points, located at different sides of the surface and infinitely close to each other. The second outstanding feature of these forces is independence of their magnitude on an orientation of the surface

da. A change of the direction of the unit vector n does not make an influence on IF]. As is well known, this allows us to characterize a force distribution in the ideal fluid by the single scalar function only, which is called the pressure.

S u r f a c e forces i n s i d e a n e l a s t i c m e d i u m Completely different behavior of surface forces is observed in an elastic medium. First of all, both the normal and tangential components of the force are transmitted through the surface. This means that, in general, the force F. acting on some element da, can be arbitrary oriented with respect to the normal

n,

Fig. C.lb. As in the case of an

ideal fluid, media located at both sides of the surface act on each other with forces F(p) and - F(p),

respectively. The second feature of these forces in an elastic medium is

the fact that a change of an orientation of the element

da results in a change of the

force, exerted across it. We can imagine infinite number of orientations of the element

da and, correspondingly, an unlimited number of different forces, acting at the same point of an elastic medium.

Because of this the following question arises.

How can

we characterize such a distribution of forces? We attempt to find one quantity, which will allow us to determine the force density,

t,

acting on the element

da, regardless

of its orientation. We have already performed a similar task then we studied studying scalar fields, (Part I). In principle, at each point there is always an infinite number of the directional derivatives of such field, and, in order to calculate them, the gradient of the scalar field was introduced. It turns out that a behavior of forces t

as a function

of an orientation of the elementary surface at the same point is also described by single quantity, which is called the stress tensor. Before we demonstrate this fact, let us make some comments about volume and surface forces. a. In the absence of external forces a body is not deformed and its atoms are in a stable equilibrium. Correspondingly, forces of interaction are equal to zero. Because of a deformation a relative position of atoms changes and the internal forces arise. They try to return atoms to their original position. b. The volume force, acting on elementary mass, Am(p), is caused by masses inside and outside an elastic body. Also these forces may have electric or magnetic origin.

566

APPENDIX

The traction vector

C. S T R E S S

TENSOR

t

Let us take a small element of the surface da inside an elastic body. We consider the force, t r a n s m i t t e d through the element da and caused by a medium, which is situated at a certain side of da.

In order to specie' this portion of the body, we draw the normal

n toward it, (da = da n).

In other words, a direction of n defines a medium, which

produces the surface force. As in the case of an ideal fluid, the density of the surface force is defined by eq. C-4 t=

F da'

and it implies that the force F is uniformly distributed over the element da. The vector t is called the traction across this surface at the point p, Fig. C.lc. The dimension of t is [t] - kg m -1 see -2, and, by definition, t components along the Cartesian coordinate axes are tx - t . i - t cos(t, i), Here cos(t,i),

ty - t . j - t cos(t,j),

tz - t . k - t cos(t, k)

(C-7)

cos(t, j), cos(t, k) are directional cosines of the vector t.

It is a simple m a t t e r to find the normal and tangential components of the traction with respect to the plane element

For instance, the scalar component along the

da.

normal n is

(c-8)

tn (p) -- t ( p ) - n - t ( p ) c o s ( t , n), where

t is the traction magnitude. If tn(p)

is negative, it is called the pressure. In

the opposite case, tn > 0, this component is called the tension. For instance, when the fluid is at rest, directions of the vector

t

and the normal

n

are exactly opposite to

each other. In an elastic medium the traction can be at any angle to the normal n, Fig. C.lc. E q u a t i o n s of e q u i l i b r i u m in i n t e g r a l f o r m In order to understand a distribution of internal forces it is very useful to consider the case when an elastic body is in a state of static equilibrium. The latter is provided by a system of external forces. This means that all particles of the body are at rest and, in

APPENDIX C. STRESS TENSOR

567

particular, wave propagation is absent. Consider an arbitrary volume V of the elastic medium, surrounded by the surface S, Fig. C.ld. Since the body is in equilibrium the resultant external force,

F,

and the resultant torque,

M,

have to be equal to zero,

(Appendix A): F - 0

and

M - 0

(C-9)

Earlier we pointed out that the force F consists of the external surface and volume forces. For instance, the former is caused by elements of the medium, located at the external side of the surface S. side of S.

They act on the neighboring elements near the internal

As in the case of the ideal fluid we will use the concept of the density of

volume forces, f, and the traction,

t.

Therefore, the elementary volume, dV, and

the elementary surface, da, are subjected to tile action of forces: d F - fdV

and

d F - tda

(C- 10)

As we already know, such presentation means that the volume and surface forces are uniformly distributed over dV

and

da, respectively. Now we are prepared to write

down conditions of an equilibrium when both translation and rotation are absent. Making use of eqs. C-9 and the principle of superposition we obtain

fdI" + / t I"

da - 0

(C-11)

S

and

f

(r x f) d V + / ( r

V

x t) da - 0

(C-12)

S

Here r is the radius-vector drawn from an arbitrary chosen origin to any element of the volume V or the surface S (Appendix A). The first equality shows that the volume V does not experience translation, while the second one guarantees that this body is not involved in rotation. In both cases it is assumed that at the initial instant the body was at rest. The two equations represent conditions of equilibrium in integral form, since the volume V may have arbitrary dimensions. Because our purpose is to find out relationships between surface forces on the vicinity of any point p inside an elastic body, we replace eqs. C-11-C-12 by their differential form. This task can be solved at least by two ways, related to each other.

The first

approach is based on an assumption that the volume I ,~ is very small. Correspondingly,

APPENDIX C. STRESS TENSOR

568

points of the surface S are close to the point p, located at the middle of the volume V. This allows us to expand components of the traction t at points of the surface in the Taylor series around the point p. Also we assume that these components linearly change within the volume V. For this reason, terms of the series, which contain the second and higher order derivatives, are discarded. Then it turns out that after an integration over S, eq. C-11, it becomes possible to express the first condition of an equilibrium in terms of the traction t and the density of volume forces f, at the point p. The same equation of an equilibrium with respect to a translation can be obtained slightly differently, and the second approach follows from the Gauss divergence theorem, (Part I):

f div M d I ~ - f M . d a I"

where d a =da n,

and

n

S

is the unit vector, directed outward the volume

orientation is in agreement with a direction of the traction

t.

V.

This

It emphasizes the fact

that a medium, surrounding the volume, generates a force, acting on V. In other words, these forces are external. In the Cartesian system of coordinates we have n

-

nxi + nuj + n~k

(C-13)

and n~, n u and n~ are directional cosines of the normal n with respect to coordinate axes.

Vectors

X,

Y,

Z

In order to obtain the differential form of eq. C-11 it is very useful to introduce three vectors:

X, Y and Z.

By definition

X - Xxi + Xyj + Xzk,

Y - }~i + }~j + }~zk,

Z - Zxi + Zyj + Zzk

(C-14)

There vectors obey the following rule. The dot product of each vector and the normal n of the elementary surface da gives the corresponding component of the traction t on the coordinate axes

tz = X . n , For instance, the dot product

X-n

ty = Y - n ,

t~ = Z . n

defines the projection of the vector

(C-15) X

on the

normal n, and it is equal to the x-component of the traction. As follows from eq. C-15

t - ( X - n ) i + ( Y - n ) j + (Z. n) k

APPENDIX C. STRESS TENSOR,

569

or

(C-16)

t - X~i + }~ + Z.k,

where X~, }~ and Z~ are projections of vectors X,

Y

and Z on the normal n.

Respectively, the normal component of the traction can be written as t~=t.n or

(C-17)

t~ - X~nx + l'~ny + )~nz Let us rewrite eqs. C-15 in the form

t~ - X~nz + Xyny + Xznz,

ty - } ~nx + ~'yny + lznz,

tz

= Z~n~ + Zynu +

Zznz (C-18)

or in the compact form

I Xz Xy Xz I ~ ly }z n

tHere

(C-19)

tx, ty and tz are the Cartesian components of the traction, acting on the plane

element with the normal n. Equations C-18 or C-19 can be treated as a transformation of the normal n into the vector t. However, they have much more important meaning and in order to understand it we consider three special orientations of the elementary surface, First, suppose that this element is perpendicular to the x-axis, nz-

1.

da, at the point p.

(n = i), that is

n v = nz = 0

Then, as follows from eq. C-18

tx(p) - X~(p),

ty(p) - ]'~(p).

tz(p) - Zx(p)

(C-20)

Comparison with eq. C-19 shows that the first column of the matrix characterizes the traction t(p), when the element

da is normal to the x-axis. At the same time, Xx,

1~, Zx are components of the vector t"

t=Xxi+

~xj+ Zxk

APPENDL~i C. STRESS TENSOR

570

In the second case the element da(p) is normal to the y-axis, and correspondingly nx - O.

nz = 0

ny - 1,

Then we have

tz - Xy,

ty - Yy,

tz -

Zy

and t(p) - Xyi + )~j + Zuk We see that the second column of the matrix defines the traction

t

at the same point,

when the element da is perpendicular to the y-axis. In a similar manner we find that the last column represents the vector t. if the clement da(p) is normal to the z-axis and t(p) - X z i +

~zj + Zzk

Cauchy formulas Thus, the matrix, (eq.

C-19), contains information about the traction

mutually perpendicular positions of the clement, da(p). the normal

n

t

for three

It is essential, that each time

and one of the unit vectors of the Cartesian svstem coincide. Assume

that components of vectors vector

t

X,

Y

and

Z are given. In other words, we know the

for three orientations of the element

da, corresponding to the coordinate

planes, (Part I). Then, an importance of eqs. C-15 or C-18 becomes clear. In fact, they allow us to calculate the traction t at the same point for any orientation of the element

da(p) and these relationships are called Cauchy formulas. One can say that we have solved our main task and found out that the matrix, given by eq. C-19, is the desired quantity, which completely describes the traction t for an arbitrary orientation of the surface element da. T h e first c o n d i t i o n of a n e q u i l i b r i u m in t h e d i f f e r e n t i a l f o r m To understand better some properties of this matrix we should return to conditions of an equilibrium and obtain their differential form. It is natural to start from eq. C-11. First, consider this equation for the x-component of vectors f and t.

/ f~d~" + J t~da - O V

S

It is clear that (C-21)

APPENDIX C. STRESS TENSOR

571

Substitution of the first equality of the set. C-15 into eq. C-21 yields

V

S

The integrand of the surface integral is represented as the flux of the vector X through

da, and, therefore, we can make use of the Gauss theorem. As was already mentioned, this was one of the reasons for introduction of vectors X, Y, Z. Correspondingly, in place of eq. C-22 we obtain the element

f L d i + / divX dI - O I"

I"

or

(fx + divX) dI" - 0

Since this equality takes place regardless of dimensions and shape of the volume V, we conclude that the integrand is also equal to zero L + divX - 0

(C-23)

By analogy, applying the same approach to components fy, ty and fz, tz, we have f~ + divY = 0,

f~ + divZ = 0

(C-24)

Thus, eqs. C-23-C-24 represent the differential form of eq. C-11, and they show that an elementary volume around some point p does not experience a translation. It is obvious that the left hand side of these equations describe the resultant force, acting on the unit volume. Also it may be proper to notice the following. By definition, the divergence is a sum of the first derivatives, for instance

OX~

OXy

div X - - - ~ - x + W

+

OXz Oz

or

div X - lim ~ X - n da

A I; --+ 0

Respectively, a calculation of divergence implies that the elementary volume AV

has

to be so small that functions X, Y and Z change almost linearly inside of it. At the

APPENDIX C. STRESS TENSOR

572

same time the density of volume forces remains the same. Of course, with a decrease of the volume a variation of each component of these vectors also tends to zero. As we know (Part I), an equilibrium does not take place instantly and it is always preceded by the dynamic stage. Suppose that at some instant the constant external forces are applied to the surface

S,

surrounding an elastic body. At tile same moment a wave begins

to propagate through the volume and ultimately it provides an equilibrium of each its portion. For illustration consider two examples of an elementary volume inside a body. Example one

Suppose that a volume is a very thin disk with elementary surfaces

da(p~) and da(p2), da (p~) - da (p2) - da Its lateral surface is so small that one can neglect the forces acting on it. At the same time, we assume that forces, exerted on surfaces

da(p~) and da(p2) are distributed

uniformly over them. Therefore, a distribution of these forces is characterized by the tractions t(pi)

and

t(p2).

Suppose that the wave approaches to the face da(p~) of

the disk and produces its expansion. Then, the traction surrounding medium, as well as the normal law the traction

n(pl).

t(pl)

is directed towards the

In accordance with Newton's third

t(p2) has the opposite direction on this volume, Fig. C.le. This means

that vector components of t(p~) and t(p2) in the direction, which is either normal or tangential to the disk, are also opposite to each other. In particular, in the state of an equilibrium t(pl) -- - t(p2), provided that we can neglect the volume forces. If tile wave produces a compression of an elementary disk, a direction of tractions is given in Fig. C.lf. It is essential that such an orientation of the traction at opposite faces of an elementary volume is always observed. Example two

Now consider an elementary parallelepiped, shown in Fig. C.2a.

The sides of this volume are equal to Ax, Ay and Az, and the middle point p has coordinates x, y, z.

As in the first example, because of the wave, the volume, AV,

is subjected to an action of forces, caused by a deformation of the surrounding medium. These surface forces are uniformly distributed over each face of AI:, but they may have different magnitudes and directions at different faces. First assume that the wave moves along the x-axis and produces a compression. Therefore. the vector component of the traction Ax -

2

'

y'

APPENDIX C. STRESS TENSOR

573

{b)

(o) Y t~

t:

,: 4 t V

t"/ y~"

u Xy ,

//

J

,~

t ~;

t:

4

t__

0

3

~X

~X

{cJ)

{c) z A

~ x y / / ~ ,,n

,r

y t

~X

Figure C.2: (a,b) Tractions on faces of an elementary parallelepiped (c) Derivation of Cauchy formulas is directed along the x-axis. When this wave reaches the opposite face, the force acts on a medium, which is in front of the volume, AI:. As follows from Newton's third law, the traction, caused by this medium, ./_~X

tx(x + - ~ ,

y, z)

has an opposite direction. If the wave is accompanied by tangential components of the traction, ty and tz, then, applying the same law, we find that t~(x

~x 2 ' y' z),

~x ty(x + --~-, y, z)

574

A P P E N D I X C. S T R E S S T E N S O R

and Ax 2 ' y' z)

tz(X

Ax tz(X + - ~ ,

and

y, z)

also have opposite directions. The same behavior of vector components of the traction is observed at other faces of the volume component of vectors X, Y

V, Fig. C.2a. At the same time each scalar

and Z has the same sign at the opposite faces. In fact,

by definition, we have G(x

Ax 2 ' y' z) = X - n - - i . X ( x

Ax 2 ' y' z) - - X x ( x

Ax 2 ' y' z)

(C-25)

For instance, in the case of the compressional wave, the scalar component tx(X - ~ , y, z) is positive. Therefore, we conclude that .'~X

Xx(x -

2 ' y' z) < O

In the opposite face we have Ax Ax - K ' y' z) - i. x ( ~ + - V '

t=(x +

Ax

~' ~) - x = ( ~ + --K' y' z)

(c-26)

In accordance with Newton's third law, the component tx is negative. Correspondingly, as on the back face:

x~(~ + - y ,

y, z) < 0

It is a simple matter to demonstrate that all other scalar components of vectors X, Y and Z do not change sign at opposite faces of the elementary volume. F l u x of t h e v e c t o r

X

Next we derive again eq. C-23 in more explicit way. With this purpose let us calculate the flux of the vector X through the closed surface, surrounding the volume AV, Fig. C.2a. Our goal is to simplify eq. C-22. when this volume is very small. It is clear that the flux through both faces, perpendicular to the x-axis, is Ax x ~ ( ~ + - { - , y, ~) - x ~ ( ~

2 ' g' z) A y A z

or

OX=(x,y,z)

Ox

OX~

Ax Ay Az-

Ox A I :

APPENDIX C. STRESS TENSOR

575

The pair of faces, normal to the y-axis, gives

[~u

~Y z)] Ax ~ z

y + -~-, z) - X,(x, y - - ~--,

or

OXy A x A y A z

OXy

0y Finally the flux through opposite faces, normal to the z-axis, is equal to Az Xz[X~(~, y, ~ + 7 - ) -

Xz(~, y, ~

Az ~-)]~~y

or

OXz At" Oz Thus, the total flux is

n da

_(ox \-~

OXz)

+ - ~ + - ~ 1 ZV

As before, assuming that the density of volume forces is constant inside AV, we again obtain eq. C-23. The same approach gives the flux of vectors Y and Z:

j Y . n d a - (o,.. o,; -~-x +--~-y + - - ~ - z / A l "

and

Z.n da-

(az,

OZz]

--O-~x+--~-y +---~-z ] A V,

and, correspondingly, eqs. C-24. Now it is appropriate to make several comments. 1. The last three equations allow us to express the flux through a surface, surrounding an elementary volume, in terms of the first derivatives of scalar components of X, Y and Z at the middle point p.

In other words, eqs. C-23 C-24 establish relationships

between these components and the density of the volume force around the same point p. 2. We use values of functions at all faces of an elementary volume, but in the limit obtain formulas, which characterize a behavior of vectors X.

Y,

Z and

f at one

point, p. 3. As was mentioned earlier we assume that each component of these vectors linearly changes between opposite faces. This implies that a difference between values of any component at the middle point and at a face is directly proportional to the distance,

(Ax/2, Ay/2 or Az/2), i.e., values of each scalar component, for instance, X u, differ only slightly at the opposite faces. opposite directions.

However, corresponding vector components have

A P P E N D I X C. S T R E S S T E N S O R

576 T h e s e c o n d c o n d i t i o n of a n e q u i l i b r i u m

We continue a study of vectors X, Y and Z, and with this purpose in mind consider eq. C-12. Its left hand side describes the resultant moment, and in order to provide an equilibrium, it has to be equal to zero. Respectively, each its component also vanishes. For instance, in the case of the x-component we have

/(yfz-

zfy)d~" + / ( g t z -

V

zty)da = O

(C-27)

S

To obtain its differential form we make use of eqs. C-15. and it gives

(yfz - z G ) d i " + f ( y Z - z Y ) - n da - 0 V

(C-28)

S

Applying again the Gauss's theorem we replace the surface integral by a volume integral, and eq. C-28 becomes

[yfz - zf~ + div(y Z - z Y)] d l ' - 0

(C-29)

V

By analogy with the first condition of an equilibrium, we take into account that eq. C-29 is valid for an arbitrary volume. This means that integrand is equal to zero, too:

Yfz - zfy + div(y Z -

z Y)-

(C-30)

0

This is the differential form of eq. C-28. and it shows that the torque is zero. It is a relationship between components

f~,

fz

x-component of the and vectors

Y

and

Z in the vicinity of any point. This equality contains extremely important information about scalar components of vectors

Y

and

Z.

To describe these new features we

perform some simplifications in eq. C-30 As is well known from vector analysis, div (y Z) - Z grad y + y div Z Since grad y - j

and grad z -

and

div ( z Y ) -

Y grad z + z div Y

k. instead of eq. C-30 we have

Y fz - z fy + Z . j + y div Z - Y

k-

z div Y -

or

Yfz - zfy + Z~ - ~'~ + y div Z - z div Y = 0

0

(C-31)

A P P E N D I X C. S T R E S S T E N S O R

577

Now, making use of the first condition of an equilibrium, eqs. C-23-C-24. we discover that Zy - } z

(C-32)

In the same manner, considering the y and z components of the resultant torque:

j

(z fx - x f z ) d V + / ( z

V

t, - x tz)da - 0

S

and

(xf u -

yfx)dl;

V

+

f (x t u -

g t.)da

-

O,

S

we see that

Zx - Xz

and

]}~: - Xy

(C-33)

In essence eqs. C-32-C-33 represent the second condition of an equilibrium of an elementary volume when its dimensions tend to zero. Thus, from both conditions of an equilibrium we found out that some elements of the matrix, eq. C-19, are equal to each other:

Xy(p) - Yx(p).

Xz(p) - Z~(p).

Iz(p) - Zy(p)

(C-34)

Taking into account an importance of these equalities, let us discuss them in some details. With this purpose consider an elementary cube, (Ax = Ay = Az) and its crosssection in the plane X O Y ,

Fig. C.2b. First, we pay attention to tangential components

of vectors X and Y, which act on faces 1-2 and 2 3. Applying again the Taylor series we have

Ay X,(x..~ + Z - " ~) - X~(p) +

and

~"]:(x + - i f . y. z) - ~'~(p) -~

ax (p) y Oy

Ox

2

(C-35)

2

The traction components tx and tu, associated with X~ and

t~, try to rotate the

cube in opposite directions. As follows from eqs. C-35, in the limit, when the volume

APPENDIX C. STRESS TENSOR

578

becomes infinitely small, their resultant torque vanishes, if

Xy(p) - Y~(p).

However, at

faces 1-2 and 2-3 of the elementary volume, we may have"

Ax The same components at the opposite faces of tile cube, 1-4 and 4-3, also form torques. As before, they have opposite directions and in the limit, when Ax -4 0 we again obtain

Xy(p)

that X,

Y

and

is equal to ~ ( p ) .

Similarly, studying all tangential components of vectors

Z, we again arrive at eqs. C-34. This consideration also shows that in a

state of an equilibrium these components are not usually equal at opposite faces. For example, /-.~Z

Now it is proper to make several comments. 1. The set of equalities C-34 describes relationships between tangential components of a traction at point p. They act on elements of coordinate planes, which are equal to

da~ - dydz, 2.

As was demonstrated, eqs.

day - dxdz,

daz - dxdy

C-34 remain valid, regardless of the volume force

density, f. 3. If we assume at the beginning that eqs. C-34 take place, then the second condition of an equilibrium is not independent and it follows from the first one. This approach is very useful, and it will be used later in deriving equations of motion. 4. In general, an equilibrium of an elementary volume depends on both the volume and surface forces. For instance, if the former can be neglected, the first condition is greatly simplified and we obtain divX-

O.

divY = O,

divZ = 0

(C-36)

Stress tensor As was already pointed out the matrix

I X~ Xy Xz I ZxZ

Zz

(C-37)

APPENDIX C. STRESS TENSOR

579

transforms the unit vector n of an arbitrary surface element into the traction, eq. C-19. This matrix is the tensor, and the relationship between

t

and

t(p) ,

n remains

linear in new system of Cartesian coordinates, obtained by a rotation from the old one. Correspondingly, elements of the tensor in the new system can be calculated, applying formulas, derived in the previous Appendix B. By definition, nine scalar elements of the tensor, eq.

C-37, are called stresses, and the)" allow us to find forces, acting on any

element da. Its diagonal elements X~.

Zz

~ ~ and

are called the normal stresses, since they characterize forces, which are perpendicular to corresponding coordinate planes.

The other elements are shear stresses, and it is

understandable, because they define tangential components of forces, exerted on the same coordinate planes. In accordance with eqs. C-34 the stress tensor is symmetrical, and. therefore, it, is defined by six elements only. There are different notations for tensor elements and one of them is given above, eq. C-37. It clearly shows the meaning of each element. For instance, Xy describes the force at the point p, directed along the x-axis and applied to the surface element da(p), which is perpendicular to the y-axis. The second notation uses one letter only for all elements, and it has a form

Xx - Tll

Xy - T12

Xz - T13

I~; -- T21

l ; - T22

1~, - T23

Z. - Tal

Z~ - 7a2

Z= - Taa

(C-38)

and in place of eq. C-37 we have

/

rl~

T~2 Tt3 /

7:21 T22 T23 T31

T32

(C-39)

733

Respectively, the first index defines the component of the force, while the second characterizes a direction of the normal to the surface element. For instance, the

Ta2 describes

z-component of the traction, which acts on the surface element, perpendicular to

the y-axis. It is obvious that Z21 - Z12 ,

Z13 - Z31

and

T32 = T2a

APPENDIX C. STRESS TENSOR

580 Also the stress tensor is sometimes written as

I Pzx Pzy Pxz I Pyx Pyy Pyz Pzx Pzy Pzz

(C-40)

Comparison with eq. C-37 easily defines the meaning of each element. Finally, in order to emphasize a difference between the normal and shear stresses the following notations are used, too:

and

Pxz = ax, Pxy -- Txy,

ay, Pxz = 7xz,

pyy =

Pzz -- az Pyz = Wyz

Therefore, the stress tensor is

I ~x Txy Txz I Tyx (~y Tyz Tzx Tzy ~z

(C-41)

In general, the latter is a function of a point and represents the example of a tensor field. Cauchy formulas and an equilibrium Earlier we obtained Cauchy formulas by simply introducing vectors X, Y and Z, eqs. C-15. It is also very fruitful to derive them, eqs. C-18, proceeding from the integral form of the first condition of an equilibrium. In other words, we again demonstrate, that the stress tensor, given at some point p, allows us to determine the traction, t(p), acting on the elementary surface,

da(p),

arbitrary oriented with respect to coordinate planes.

Solving this task it is convenient to deal with two different elementary volumes. C a s e one:

two-dimensional model

Consider an elementary volume inside an

elastic medium, which has a shape of the wedge,

Fig. C.2c. Before we use the first

condition of an equilibrium it is proper to notice the following. With a decrease of the wedge volume the surface forces decrease proportional to the area of its faces, that is as a square of linear dimensions. At the same time the volume force, for instance, the gravitational one, decays more rapidly; as a cube, since it is directly proportional to mass. For this reason we can neglect this force, that is [f[dl" <<

Itlda

(C-42)

A P P E N D I X C. S T R E S S T E N S O R

581

Also we assume that the volume width,

Ay,

is very small, and forces, acting on two

faces, perpendicular to the y-axis, cancel each other. This simplifies the first condition of an equilibrium, too, eq. C-11, and it can be written as

t(pa)AyAl + t ( p 2 ) A y A z + t ( p ~ ) A x A y -- O,

(C-43)

where +

Respectively, for the x and z-components we have (C-44)

tzn(p3)A1 + tx(p2)Az + tx(p,)J.-kx -- 0 and

tz~(p3)A1 + tz(p2)Az + tz(pl)Ax -- 0

By definition, t~ (Pl) -- - X ( P l ) "

and

k - - X : (p,)

t~(p2) - - X ( p ~ ) . i -

-X~(p2),

since at both faces the normal has a direction, opposite to the corresponding unit vector. Then, the first equation of the set C-44 becomes Az Ax t z n ( P 3 ) - X x ( P 2 ) ~ l + Xz(p2) As is seen from Fig. C.2d Az Al

z

COS

,~ --

are directional cosines of the normal n.

7"/x.

Ax Al

"-- C O S

--

//z

Thus, we have

t x , ( p 3 ) - Xx(p2)n~ + Xz(pl)nz It is clear that tz (p~) - - Z (Pl)" k - - Zz (pl)

and

O

t~(p2) - - Z ( p 2 ) - i -

-Zz(p2)

(C-45)

APPENDIX C. STRESS TENSOR

582 Therefore, the second equation of the set C-44 gives

tz~(p3) - Z~(p2)n~+Zz(p,)nz

(C-46)

Here txn(P3) and tzn(P3) are the z- and the x components of the traction t(p3) at the elementary plane with the normal n. As follows from eqs. C-45-C-46, they describe relationships between stresses at different points pl, p2 and P3. However, with a decrease of the wedge volume, all faces approach to the same point

p.

In the limit these stresses characterize forces, exerted

on three elementary surfaces, which have a common point p. Comparison with Cauchy formulas, eqs. C-18, shows that eqs. C-45-C-46 represent their special case, when an influence of forces, acting on faces, perpendicular to the y-axis, can be neglected. Until now we found the Cartesian components, tzn

and

tzn , in terms of the stress

tensor:

(X, Xz) Zx Zz It is also a simple m a t t e r to determine the normal and shear components of the traction t at the same point P3. As is seen from Fig. C.2d

tnn(Pa) -- tzn cos fl + t zn cos ct,

t sn (P3) -- -tzn cos c~ + t zn cos fl

or

tnn(P3) -- txnnz + tznnz, Here tnn and

tsn(P3) -- --tznnz + tznnz

(C-47)

tsn are the normal and shear components of the traction at the point

p3. Substitution of eqs. C-45-C-46 into the set C-47 yields

tun(P3) -- nx(Xxnx + Xznz) q- rtz(Zzrtx + Zzrtz) or

tnn(P3)-- n~Xx 2 + nxnzXz + n~nzZ~ + n2zZz Similarly, for the shear component we have t

.(v3)

-

+

-

+

(C-48)

APPENDIX

583

C. S T R E S S T E N S O R

or 2

2

(C-49)

tsn(P3) -- n x Z z + n z n z Z z - n z n z X z - n z X z

As we know, from the second condition of an equilibrium it follows that Assume that the system of coordinates,

x and

the new axis x' is directed along the normal the elementary surface around elements,

Xx,(p3)

and

p3.

Zx,(p3)

Then,

z, is rotated about the n,

while the

t~n(p3)

and

Zx -

Xz.

y-axis, and

z' axis is tangential to

ts~(p3)

represent the stress

in the new system of coordinates.

Therefore, eqs.

C-48-C-49 perform a transformation of two elements of the stress tensor, caused by a rotation of the coordinate system. Considering the surface element, normal to the old one, (Ay, AI), we can determine the stress Zz,. As concerns Xz,, it is equal to Zx,. It is proper to notice that the same result, eqs. C-48-C-49, follows from expressions derived in the Appendix B. C a s e two:

three-dimensional model

Next we study an equilibrium of a

tetrahedron, shown in Fig. C.3a. Three of its faces coincide with corresponding elements of the coordinate planes. The areas of all plane faces of this body are related to each other in the following way: davy - da cos(k, n),

da~. - da cos(j n)

and dayz = da cos(i, n)

Here da is the area of the face with the normal n. Our goal is to determine Cartesian components of the traction

t

at this oblique

element of the closed surface. Applying again the first condition of an equilibrium in the integral form, we see that t da + t(p~)dayz + t(p2)daxz + t(p3)da~y = 0

(C-50)

Considering the Cartesian components of this equality and making use of the set C-15, we again arrive at the Cauchy formulas, eqs. C-18. At the same time, eq. C-50 does not relate to each other tensor elements, but it establishes a relationship between the traction

t

at an oblique surface element and the stress tensor. In addition note the

following: a. Earlier we derived Cauchy formulas without the use of the condition of an equilibrium, and, therefore, the volume forces.

584

A P P E N D L ' i C. S T R E S S T E N S O R

{o)

z

(b) d ~ V-

F 't ! S

B

{c) n

A x

x

Figure C.3: (a) Cauchy formula for three dimensional cases (b) Equilibrium of bar (c) Illustration of eq. C-54

b. As was demonstrated in the first case, Cauchy formulas permit us to find elements of the stress tensor in the new Cartesian svstem of coordinates. Certainly, the same is correct in the three-dimensional case, (Appendix B)

S t r e s s behavior and a n e q u i l i b r i u m Next we proceed from Cauchy formulas and illustrate a behavior of the traction components in an elastic medium. With this purpose in mind consider two examples, assuming that a body is in an equilibrium. Example one

Suppose that the stress tensor at some point of a medium is Xx-

}~-

Zz - - P

and

}] = Zx - X y -

0

(C-51)

APPENDIX

C. S T R E S S

585

TENSOR

By definition, for an arbitrary oriented element d a with the normal n we have tx - X . n - - P c o s ( i , n ) - - P

n~

ty - Y - n = - P c o s ( j , n ) - - P

ny

tz -

nz

Z .n -

-P

cos(k, n) -

-P

Respectively, the traction t, acting on this surface element, is t-

-P

n,

(C-52)

that is the vector t has a direction, which is opposite to the normal n. is equal to the pressure P.

Its magnitude

As we know, such a behavior is observed in the ideal fluid

when an equilibrium takes place. Example two that two forces, F

Consider an elastic bar, oriented along the and

-F,

x-axis, and assume

applied at bar ends, provide an equilibrium, Fig. C.3b.

Because of these forces an extension occurs and internal forces arise. In order to find their distribution we mentally draw a cross- section S in any place of the bar. Its portions, A and B, are located at both sides of this surface. Inasmuch as the bar is an equilibrium, parts A and B are at rest, too. Therefore, the internal force, acting on S and caused by the portion A, is equal to F.

In other words, the resultant force, exerted on B, is

equal to zero. Otherwise, it would be in a state of motion. Changing a position of the cross-section S and bearing in mind that the force is distributed uniformly over it, we conclude that F is the same at all points of the bar. Besides, this force is perpendicular to the section S, that is the traction has only the normal component, equal to F tx - Xx = -S

(C-53)

As was shown, this stress element provides an equilibrium, while the others are equal to zero:

T

l

ax

0 0 /

0

0 0

0

0 0

Here it is proper to note that an influence of volume forces is ignored, and only surface forces are able to sustain this state of the bar. In the same manner we can consider the

586

.4PPENDI.,"t" C. S T R E S S

internal force, acting on the boundary S of the portion ,4.

TENSOR

It is clear that this force

differs from F by a sign only (Newton's third law). Now we take an arbitrary oriented surface element inside the bar, Fig.

C.3b.

In

accordance with Cauchy formulas, eq. C-18, the traction has only the component along the x-axis, and it is equal to tx - X~ cos(n, i),

(C-54)

while ty - tz - 0 at all points of the bar. In particular, if the normal n is perpendicular to the x-axis, the traction is equal to zero. In general, there are both the normal and tangential components of the traction. Indeed, as is seen from Fig. C.3c t ~ - tx cos . 3 - Xx cos 2 3 -

and

(c-s5)

X xn~

tsn - - t x sin/3 - - X x sin/3 cos 3

Thus, the component tnn gradually decreases with an increase of the angle /3, while the shear component has a maximum, when 3 - 7r/4, and it is equal to

tsn = Xx 2

(c-s6)

Stress equations of motion As is well known, a motion of an elastic body can be represented as a superposition of a translation and a rotation around its center of mass (Appendix A). Of course, there is also a deformation, and this phenomenon will be studied later. Equation describing translation is M a o - F,

(C-57)

Here 5I is the total mass of the body, a0 is an acceleration of the center of mass, and F is the resultant of external forces. By definition, we have

and

Mao-/padt ~

F-/fdV+/t dS

V

V

S

Correspondingly, the first equation of motion in the integral form is

f p a d," - f f dv + f t dS, V

~"

S

(C-58)

APPENDIX C. STRESS TENSOR

587

where a is the acceleration of an elementary mass,

p dV. To obtain the differential

form of this equation, consider, as in the case of an equilibrium, any component of this equality. For instance, the x-component is equal to:

f p axdl" - / fx dI'+ / X . n dS t"

I"

(C-59)

S

Replacing the last integral by the volume one. we have

(p

ax

Ix -

div X)dV - 0

(C-60)

Inasmuch as eq. C-60 is valid for any arbitrary volume, we conclude that the integrand is equal to zero, too: p ax = fx + div X

(C-61)

p a y - fy + div Y

(C-62)

p az - fz + div Z

(C-63)

By analogy we obtain

and

The last three equations are the differential form of eq. C-58. For example, eq. C-62 shows that an acceleration of an elementary volume,

AV, along the y-axis is defined

by the volume force fy dV and the resultant of surface forces" divY dl .... In fact, by a definition of the divergence, this product can be replaced as Y.dS, S

and the integral describes the total surface force caused by a surrounding medium. In most cases, discussed in this monograph, an influence of volume forces can be neglected. Therefore, in place of eq. C-58 we have

/padV-/tdS

(C-64)

APPENDIX C. STRESS TENSOR

588

As follows from eqs. C-61-C-63 in the Cartesian system of coordinates

OX~ OXy OXz P a ~ - f ~ + - - - ~ x + - ~ y + Oz

(c-65)

P aY - fY + -~-x + -~y + O---z OZ~

OZ~

OZz

Here

ax

~-

02Sx Ot2 ,

02% ay = Ot2,

az

02Sz Ot2

and

s =sxi+syj+szk is the displacement of the center of mass which in general takes place due to a translation of the volume, as a rigid body, and its deformation. Usually the set C-65 is called the stress equations. It may be proper to notice that they contain too many unknowns and this shortcoming will be removed later. Now we consider several examples, illustrating a stress behavior of a moving body. Example one

Suppose that at some instant t - 0 the constant force is applied to

one end of the bar and it is directed along the x-axis, Fig. C.4a. As is well known, (Part I), at the beginning we observe waves, propagating between bar ends, and its different elements move with different velocities. It is essential that within this time interval a deformation changes. Then, after some time an influence of waves becomes negligible, and it happens due to an attenuation. Correspondingly, each elementary mass starts to move with the same acceleration, and we can apply the second Newton's law to any part of the bar. It is obvious that with a decrease of its length the first time interval becomes smaller and in the limit it tends to zero. Let us mentally draw the cross-section S of the bar and consider a portion A, Fig. C.4a. Since it moves with the same acceleration as the whole bar, the internal force, Fi, acting at points of S is defined from the equality

p 1 S - p lAS

or

F~ -

F

(C-66)

589

A P P E N D I X C. S T R E S S T E N S O R (b}

(a}

r

I

P

L-'"

--B-. . . .

~.-"

0

{d}

[c)

! 0

*x

~X

0

Figure C.4: (a) Distribution of internal forces in the bar when F =const (b) Rotation of the bar with the constant acceleration (c) Translation of elementary volume (d) Rotation of elementary volume Thus, the internal force linearly decreases towards the front end of bar. Applying the principle of superposition we can determine a distribution of the internal forces, when both ends of the bar are under action of forces. Example two a,

Consider a bar rotating around its end with the constant acceleration

Fig. C.4b. As follows from the second equation of motion the force, normal to the

bar, is defined as r x Ft=I a,

(C-67)

Here I is the moment of inertia and r is the distance from an elementary mass, Am, to the axis of rotation. In this case, eq. C-67 is simplified and we obtain Ft r -

Am r2~

or

Ft - Am rc~

(C-68)

A P P E N D I X C. S T R E S S T E N S O R

590

Thus, the shearing force linearly decreases with the distance r. Of course, there is also the centripetal force, directed along the bar. E x a m p l e t h r e e . T r a n s l a t i o n of a n e l e m e n t a r y v o l u m e

Consider an elementary

volume of an elastic medium and assume that the wave propagates along the

x-axis,

Fig. C.4c. In addition, we suppose that the force, associated with wave, has only the

x- Ax/2, the left portion of the volume starts to move, while the other part remains at rest. Of course, during this motion we also observe a deformation. When the wave approaches the front face, x + A x / 2 , the force F~ begins to act on a medium in front of the volume. In accordance with the third Newton's law, the force, caused by the surrounding medium and acting on this face: component F~. As soon as the wave reached the face with the coordinate:

Fx(x + --2-, y, z, t) has the same magnitude but opposite direction. Thus, the total force, exerted on the opposite faces of the volume, is .~x Ax Fx(x - -~-, y, z, t) + f x ( x + --~, y, z, t)

(C-69)

Inasmuch as the x-component of the traction is related to the vector X as

tz = X . n, the sum in C-69 can be represented in the form

Ax

~-,y,z,t)

]

AyAz i

(c-7o)

Here Xx is the normal stress. Now it is appropriate to point out that at the opposite faces the forces have opposite directions but stress values differ only slightly from each other. Because of this we assume that the function Xx changes linearly between these faces, the difference in C-70 is written as

i,

y, z, t)

Ox

(c-71)

where Xx is the stress value at the middle point of the volume. The total force also includes the volume one

OXx

02Sx

0----~-t- fx -- P c3t2

APPENDIX

C. S T R E S S

TENSOR

591

Geometry of the wave, propagating along the x-axis, allows us to assume that vectors Y

and

Z

are absent.

At the same time stresses

X u and

Xz

at opposite faces,

perpendicular to the Y and Z-axes, are the same OXy Oy

--

OXz Oz

--0

In such a case the set of equations of motion, C-65, is reduced to the single equation. If at each instant of time Ax ~x X ~ ( x - - ~ , y, z, t) - X x ( x + - ~ - , y, z, t)

and

fz - 0,

then a deformation of the volume does not change, and it experiences a translation with the constant velocity.

In a more general case. when these stresses are different, the

volume is involved in a more complicated motion, including a vibration around its center of mass. E x a m p l e four. R o t a t i o n of e l e m e n t a r y v o l u m e

Now we investigate rotation

of an elementary volume and, for simplicity, assume that it has a shape of the cube, Fig. C.4d. Its sides are equal to h. Since we are interested in rotation, an influence of normal stresses are not considered. As before, assume that the wave propagates along the x-axis, but unlike the first example, it produces the force, perpendicular to the x-axis. At the beginning, consider an action of its vector component Fy. At points of the back face of the cube we have h Fy(x - -~, y, z, t) - tyh2j - ( Y - n)h2j or

h h Fy(x - -~, y, z, t) - - ) x ( X - -~, y, z, t)h2j

(C-72)

As soon as the wave front passes this face. the left portion of the volume begins to move along the y-axis, but the other portion is at rest. Such motion causes a deformation of the volume. Besides the normal and tangential components of the force appear at faces, perpendicular to the y-axis. When the wave approaches the face, x + h/2,

the medium

in front of the volume acts with the force Fy

-

h ty(x + -~, y, z, t)h'~j

or

h Fy - ~'~(x + -~, y, z, t)h2j

(C-73)

A P P E N D I X C. S T R E S S T E N S O R

592

It is essential that forces, given by eqs. C-72-C-73, have opposite directions, but their magnitudes differ only slightly from each other. Forces, acting on faces, normal to the y-axis, display the same behavior, and we have: h h F=(x, y + -~, z, t) - Xy(x, y + -~, z, t)h 2 i

h

(C-74)

h

F=(x, y - -~, z, t) - - X y ( x , y - -~, z, t)h 2 i

and

It is clear that all four forces may cause a rotation of the volume about the z-axis, and it is described by the equation

Mz = Iaz Here Mz

is the z-component of the torque,

(C-75) I

is moment of inertia and c~z is the

component of the angular acceleration. As is seen from Fig. C.4d, the forces, acting at opposite faces of the volume, produce the torques in the same direction. For this reason, evaluating the total torque, we have to add their magnitudes. It is important to emphasize that torques due to forces, oriented along the x and y axes, have opposite directions. Otherwise we would not be able to observe an equilibrium. As follows from eqs. C-72-C-73 the magnitude of the first pair of torques is equal to

h

h

1111 -- }'~(x + - ~ , y , z , t ) + } ~ ( x - -~,g,z,t)

] -~, h3

(c-7o)

since the level arm is hi2. In the same manner the torque magnitude of tile second pair is

M2 -

h

h

] h3

Xu(x + -~, z, t ) + Xy(x, g - -~, z, t) -~

(c-77)

Next we expand stresses at each face of the cube in the Tavlor series around the middle point of the volume, p(x, y, z). Discarding terms of the order h 3 and higher, we obtain

M1 -

and

[

M2 -

I~ (p, t) +

02}~(P't) h---8] h 3 Ox 2

Xy(p,t) +

02Xu(p, t) h 2 ] h3 Oy 2 8

(C-78)

if h - + O

APPENDIX

C. S T R E S S

593

TENSOR

It is proper to notice that the Taylor series of the magnitude of each torque contains the term with the first derivatives. However, fortunately they are absent in expressions for M1 and M2. As follows from eqs. C-78 the magnitude of the resultant torque around the z-axis is ]tl -- d]"I1 -- ,112 --

(} "x -- X u) +

Ox 2

Oy 2

In accordance with eq. A-51 the moment of inertia of the cube is equal to (C-80)

I-

Alh2 = ph5 6 6 that is it has the same order with respect to h,

as the second term in eq.

C-79.

Substitution of eqs. C-77-C-80 into eq. C-75 yields }~ (p, t) - X u (p, t) +

( 02yx Ox 2

02Xy) i)y 2

h 2 _ ph 2 8 - --60 z

(C-81 )

Here }~ and X u and their derivatives are taken at the point, p. Since the acceleration can not be infinitely large, we conclude that

X (p, t) -

t)

(c-s2)

By analogy, considering a rotation about the x and y axes it follows that and

X z - Z~

}z - Zy

(C-83)

The set of equations C-82-C-83 is very important result, because it shows that, as in the case of an equilibrium, the stress tensor is also symmetrical in the dynamic stage. Besides from eq. C-81 it follows that the acceleration C~z is defined by the difference of second derivatives of stress components and naturally, it is independent on h. In fact, we have ozz-

3( ~

02}~: Ox 2

02"yY) Oy 2

(C-84)

In the same manner, considering a rotation of the volume around two other axes, we can obtain expressions for C~x and av. Let us make some comments. a. During a propagation of the wave through a given elementary volume, eq. C-75 is applied to its portion, located behind the wave front b. If the shearing forces are such that 02:~ ~ Ox ~

02Xy Oy 2 '

the elementary volume rotates around tile z-axis with the constant velocity w, (h ~ 0).

APPENDIX C. STRESS TENSOR

594

Relationship between the second and first equations of motion Next we pay attention to the second equation of motion, which describes rotation. As is well known, (Appendix A), this equation has the form 0L

= M

Ot Here M equal to

(C-85)

is the resultant torque, caused by the external forces and, by definition, it is

g - f(r •

f)d~ + J r

dS

•

(C-86)

and r is the radius-vector, characterizing a position of the point of a volume with respect to the center of mass. The left hand side of eq. C-85 describes the rate of a change of the total angular momentum L,

L /

rzpvdI"

V

Respectively 0---~ =

r • p a dI',

(C-87)

I"

since

0 (r•215 Ot Thus, the second equation of motion in the integral form is

f pr•

dl'-

t"

f r• ~"

dI'+

f r•

dS

(C-88)

5;

It turns out that this equation is not independent, but it follows from the first equation of a motion. To demonstrate this very important fact, we consider at the beginning the z-component of eq. C-85, which is equal to

/ p ( y a z - z % ) d l ' - / ( y f z - ~,fy)dl'+ / ( y tz - z ty)dS V

V

S

(C-89)

APPENDIXC. STRESSTENSOR

595

It is proper to notice that the same approach was used in the case of an equilibrium. Taking into account eqs. C-61-C-63 in place of eq. C-89 we have

f [y(fz+div Z) -z(f,+div

Y)]dI" -

f (yfz-zf,)dl'+ f y(Z.n)dS- ~'z(Y.n)dS t"

V

S

S

or

f(y

div Z - z

div Y)dl" - / d i v ( g Z ) d l " - f div(zY)dI"

V

S

(C-90)

S

As was shown earlier div y Z - y

div Z +

Zy

and

div zY - z div Y + }z

Therefore, eq. C-89 is greatly simplified, and we obtain

(Zy -

I" ) d l ' -

0

(C-91)

I"

Considering two other components of eq. C-85, we have

L(Xz- Zx)di:- 0

and

f ( X ~ - ~x)di" - 0

(C-92)

ov

l"

These equalities can be interpreted in two different ways. First of all, we earlier proved that the stress tensor is symmetrical and, therefore, the left hand side of eqs. C-91-C-92 is equal to zero. This means that the second equation of motion follows from the first one. In other words, all information about both types of motion contains in eqs. C-61C63. Certainly, it is important result, which greatly simplifies a study of wave fields. At the same time in some cases, when a translation is absent, the use of eq.

C-85can

be

more preferable. Also, eqs. C-91-C-92 may serve as another proof of the stress tensor symmetry. This follows from the fact that these equalities are valid for any volume, and, correspondingly, integrands are also equal to zero, that is