kinematics of continua - tamu...

KINEMATICS OF CONTINUA

Introduction Deformation of a continuum Configurations of a continuum Deformation mapping Descriptions of motionMaterial time derivative Velocity and acceleration Transformation relations and invariants Decomposition of displacement gradient tensor Compatibility conditions for solid continua An introduction to fluid mechanics Decomposition of velocity gradient tensor

This part of the course is devoted to the study of geometric changes in a continuous medium that is in equilibrium.

The study of geometric changes in a continuum without regard to the forces causing the changes is known as kinematics.

INTRODUCTION

V 0

m( , t) limV

xρ∆ →

∆≡

∆

In a continuous medium, any property of the medium, for example density, can be defined at every point of the medium. This is possible only if the medium contains no gaps between points:

Consider a body of known geometry in a three-dimensional Euclidean space. The body of matter may be viewed as a set of particles, each particle representing a large collection of molecules with a continuous distribution of matter in space and time. Under external stimuli, the body will undergo macroscopic geometric changes, which are termed deformations.

The geometric changes are accompanied by stresses that are induced in the body. If the applied loads are time dependent, the deformation of the body will be a function of time; that is, the geometry of the body of matter will change with time.

DEFORMATION OF A CONTINUUM

Configurations of a ContinuumIf the loads are applied slowly so that the deformation is dependent only on the loads, the body will occupy a sequence of geometrical regions. The region occupied by the continuum at a given time t is termed a configuration.

Suppose that the continuum initially occupies a configuration , in which a particle X occupies position X, referred to a reference frame of right-handed, rectangular Cartesian axes ( ) at a fixed origin O with orthonormal basis vectors , as shown in the figure.

0C

1 2 3, ,X X X

1 2 3,( ),E E E

CONFIGURATIONS OF A CONTINUUM

Reference configuration

Deformed configuration

Particle X, occupying position x at time

Deformed configuration

Particle X, occupying position x at time

x ,X3 3

0R C C 1Cx X, ( 0)

2 2t t x X

1 1t t x X

x ,X2 2

x ,X1 1

2CO

E eˆ ˆ,1 1

E e2 2ˆ ˆ,

E e3 3ˆ ˆ,

CONFIGURATIONS OF A CONTINUUM

Note that X (lightface roman letter) is the name of the particle that occupies location X (boldface letter) in configuration and therefore (X1, X2, X3) are called the material coordinates. After the application of some external stimuli (e.g., loads), the continuum changes its geometric shape and thus assumes a new configuration

, called the current or deformed configuration. Particle X now occupies position x in the deformed configuration .

C

0C

CThe mapping is called the deformation mapping of the body. The deformation mappingtakes the position vector X from the reference configuration and places the same point in the deformed configuration as .

0: C Cχ( , )tχ X

( , )tx χ X

DEFORMATION MAPPING

E2ˆ

3 3x , X

2 2x , X

1 1x , X

κO

•

0κ

( ),tX

Reference configuration,

X

X Xx

u Current configuration, C

Particle X occupying position X

Particle X occupying position x

0R C C

E eˆ ˆ,1 1

E e3 3ˆ ˆ,

DEFORMATION MAPPING

The inverse mapping takes the position vector x from the deformed configuration back to the reference configuration . It is not always possible to construct the inverse mapping from a known deformation mapping.

10: C C χ

DESCRIPTIONS OF MOTON

1( , )tX χ x

Material Description (Lagrangian description)In the material description, the motion of the body is referred to a reference configuration , which is often chosen to be the initial configuration. Thus, in the material description, also known as the Lagrangian description, the current coordinates x in are expressed in terms of the reference coordinates :

RC

C0CX

0( , ), ( , )t x χ X X χ X

( ( ), ) ( , )t t x X X

The variation of a typical variable over the body is described with respect to the material coordinates X and time t:

x( , )t

Spatial Description (Eulerian description)In the spatial description, also known as the Eulerian description, the motion is referred to the currentconfiguration occupied by the body, and is described with respect to the current position occupied by material particle X:

The coordinates x are termed the spatial coordinates.

x( , )tx C

1( , ), ( , ) ( , )t t t x X X x χ x

DESCRIPTIONS OF MOTON

MATERIAL TIME DERIVTIVE

x( , )tHowever, when is known in the spatial description, its time derivative for a given particle, known as the material derivative, is

( , ) ( , ) ( , )

( , )

i

i

ii

DxD t t tDt t Dt x

v tt x t

x x x

v x

When a function is known in the material description, , its total time derivative, D/Dt, is simply the partial derivative with respect to timebecause the material coordinates X do not change with time:

( , )t X

Xfixed

( , )( , ) ( , )D tt tDt t t

X

X X

x( , )t

VELOCITY AND ACCELERATION

The material time derivative of a function of current position and time is nothing but its total time derivative, which by the chain rule of differentiation is

( , ) ( , ) ( , )i

i

dxd t t tdt t dt x

x x x

where and v is the velocity vector( )tx xd Ddt Dt

x xv

The acceleration a of the particle x is defined by the total time derivative of the velocity v:

( , ) DtDt t

v va x v v

AN EXAMPLE

Suppose that the motion of a continuous medium is described by the mapping

1 2 1 2 1 2 3 3e e eˆ ˆ ˆ( , ) ( ) ( )t X At X X At X X χ X x

and that the temperature T in the continuum in the spatial description is given by

1 1 2 2 1 2( , ) ( )T t c x c tx x tx x

Determine (a) the inverse of the mapping , (b) the velocity components, and (c) the total time derivatives of T in the two descriptions.

χ

Problem statement:

Problem solution: From the given mapping have

1 1 2 2 2 1 3 3

1 1

2 2

3 3

1 01 0

0 0 1

, ,

or

x X AtX x X AtX x X

x At Xx At Xx X

Clearly, the mapping is linear. Therefore, polygons are mapped into polygons. In particular, a unit square is mapped into a square that is rotated in a clockwisedirection.

At

At

1 1,x X

x , X2 2

Initial (undeformed)body is shown in dotted lines

Deformed body

1 0.

1 0.

(1)

AN EXAMPLE (continued)


(a) The inverse mapping can be determined, when possible, by expressing in terms of . In the present case, it is possible to invert the relations in Eq. (1) and obtain

1 1

2 22 22 2

3 3

1 01 1 0

10 0 1

( )

X At xX At x

A tX A t x

(2)

(b) The velocity vector is given by

1 21 21 2 1 2 2 1, ,Dx Dxv v v AX v AX

Dt Dt v E E (3)

1 2 3( , , )x x x 1 2 3( , , )X X X

1.0

0.5

0.0

−0.5

−1.00.0 0.4 0.8 1.2 1.6 2.0

x , X1 1

2 2x , X

0 25A .

4 0t .

3 0t .

2 0t .

1 0t .

0 0t .

(c) The time rate of change of temperature of a materialparticle in the body is simply

On the other hand, the time rate of change of temperature at point x, which is now occupied by particle X in the spatial description, is


1 22 1 fixed

( , ) ( , ) ( )D T t T t AtX A XDt t

X

X X

2 1 2

1 2

1

2 1

( , )

( )

ii

D T TT t v x v v tDt t x

AtX A X

x

(4)

(5)

E2ˆ

3 3x , X

2 2x , X

1 1x , X

O

•( ),tX

Reference configuration,

X

X Xx

u Current configuration, C

0C

E eˆ ˆ,1 1

E e3 3ˆ ˆ,

KINEMATICS OF DEFORMATION OF SOLID CONTINUA

Displacement vector u x X ( )i i iu x X ( , ) ( , )( , ) ( , ) , ( , ) D t tt t tDt t

x X Xu X x X X X uv

One of the key quantities in deformation analysis is thedeformation gradient, denoted F, which providesthe relationship between a material line before deformation and the line (consisting of the same material as ) after deformation. It is defined as follows:

DEFORMATION GRADIENT

dXxd

dX

0 0

T TTT,d d d

χ xx F X X F F x xX X

e Eˆ ,ˆ iiJ i J iJ

J

xF FX

F

We note that F is a mapping from the undeformed body to deformed body; it is not a tensor. The rectangular Cartesian component form is

1 T T,d d d

XX F x x F F Xx

AN EXAMPLE

Consider the uniform deformation of a square block ofside 2 units and initially centered at $\bf X=(0,0)$. If the deformation is defined by the mapping

(a) sketch the deformation, (b) determine the deformation gradient F, and (c) compute the displacements.

1 2 1 2 2 3 33 5 0 5 4e e eˆ( ) ( . ˆ). ( ˆ)X X X X χ X

Problem statement:

Solution:(a) From the given mapping, we have in matrix form, we have

1 1 2 2 2 3 33 5 0 5 4. . , ,x X X x X x X


(−1,−1) (1,−1)

(−1,1) (1,1)

X1 , x1

X2 , x2

(2,3)

(3,5)

(4,3)

(5,5)(3.5,4)

( )Xχ

1 1ˆ ˆ,E e

2 2ˆ ˆ,E e

F E2ˆ( )

1 1ˆ ˆ( )F E e

1 ˆ( )F A

• •

• •

• •

•

•

• •

2.0 3.0 4.0 5.0(0,0)

2.0

3.0

4.0

5.0

°

°

°

°

2ˆ ˆ=A e

1 2 1 2

1 1 2 31 1 4 31 1 5 5

1 1 3 5

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

X X x x

1 1 2

2 2 3 3

1 1 2

2 2 3 3

3 5 0 54

1 5 0 54

x . X . X ,x X , x X .X . x . x ,X x , X x .

1 1 1

1 2 3

2 2 2

1 2 3

3 3 3

1 2 3

1 1 1

1 2 3

1 2 2 2

1 2 3

3 3 3

1 2 3

1 0 0 5 0 00 0 1 0 0 00 0 0 0 1 0

. . .[ ] . . . ,

. . .

[ ]

x x xX X Xx x xFX X Xx x xX X X

X X Xx x xX X XFx x xX X Xx x x

1 0 0 5 0 00 0 1 0 0 00 0 0 0 1 0

. . .

. . .

. . .

(a)

(b)

2 2 313 5 0 5 4 0ˆ ˆ ˆ. . X u x X e e e

GREEN-LAGRANGE STRAIN TENSOR

XQ

Q

P Q_

XP

xP P_

xQ

uQ

uP

0C

dx

dX C

(time t = 0)

(time t)

1 1x , X

2 2x , X

3 3x , X

(X)χ

Strain is a measure of geometric changes in a solid body. The elementary mechanics definition of the change in the length of a line element divided by the original length cannot be used in multi-dimensions because only the square of the length of a line element (in 2-D or 3-D)can be found. Thus, we have

where C is called the right Cauchy-Green deformation tensor.

2

2

T

( ) ,( )

( )

dS d dds d d

d dd d

X Xx xX F F XX C X

Td d d x F X X F


2 2 2( ) ( )ds dS d d X E XDefine the Green-Lagrange strain tensor E as

1 1 10 02 2 2

10 0 0 02

T T

T T

( ) ( )

( ) ( ) ( )

E F F I C I I u I u I

u u u u

where E is

12

JI K KIJ

J I I J

uu u uEX X X X

The rectangular Cartesian component form of E is

2 2 2

31 1 211

1 1 1 1

2 2 2

32 1 222

2 2 2 2

2

3 133

3 3

12

12

12

,uu u uEX X X X

uu u uEX X X X

u u uEX X

2 2

32

3 3

3 31 2 1 1 2 212

2 1 1 2 1 2 1 2

3 3 31 1 1 2 213

3 1 1 3 1 3 1 3

223

3

12

12

12

uX X

u uu u u u u uEX X X X X X X X

u u uu u u u uEX X X X X X X X

uEX

3 3 31 1 2 2

2 2 3 2 3 2 3

u u uu u u uX X X X X X X

The rectangular Cartesian components in explicit form are given by


1X

2X

3X

E1ˆ

E2ˆ

E3ˆ

11E

X1−face

X3−faceX2−face

21E

31E22E

32E

12E

33E

23E13E

th thStrain in the direction onthe faceIJE I J

Deformed body

TRANSFORMATION RELATIONSAND INVARIANTS

11 2 32, ( ),ii ii jj ij ijJ E J E E E E J E

The components of a tensor are not invariant, that is, the components depend on the coordinate system chosen. However, certain combinations of the components remain invariant under coordinate transformations. For example the trace (i.e., sum of the diagonal elements) of a second-order tensor is an invariant quantity. There are many invariants. The principal invariants of strain tensor are:

The strain transformation equations are the same as those for any second-order tensor:

T[ ] [ ][ ][ ]ij im jn mnE E E L E L

AN EXAMPLE

For the deformation mapping given in an earlier example,

determine the Cartesian components of the right Cauchy-Green deformation tensor C and the Green-Lagrange strain tensor E.

1 1 2 2 2 3 3

1 1 2 2 2 3 3

3 5 0 5 41 5 0 5 4

x . X . X , x X , x X .X . x . x , X x , X x .

Problem statement:

Solution: We have

1 0 0 0 1 0 0 5 0 1 0 0 50 00 5 1 0 0 0 0 1 0 0 0 5 1 25 00 0 0 0 1 0 0 0 0 1 0 0 0 00 1

0 0 0 50 01 1 0 5 0 25 02 2

0 0 0 00 0

T

. . . . .[ ] [ ] [ ] . . . . . .

. . . . . .

. .[ ] [ ] [ ] . .

. .

C F F

E C I

INFITESIMAL STRAIN TENSORIf E is of the order in , then we mean

If terms of the order can be omitted, then

can be approximated as

O( ) 0 u

0O( ) asI

J

uX

2O( )12

JI K KIJ

J I I J

uu u uEX X X X

2

0 0

1 02

12

TE

O( ) as

( )

JIIJ

J I

uuEX X

ε u u

, the infinitesimal strain tensor

EXERCISES ON KINEMATICS1. If the deformation mapping of a body is given by

where A and B are constants, determine (a) thedisplacement components in the material description,(b) the displacement components in the spatialdescription, and (c) the components of theGreen-Lagrange strain tensor.

1 2 1 2 1 2 3 3e e eˆ ˆ ˆ( ) ( ) ( )X AX X BX X χ X

B

A

EXERCISES ON KINEMATICSConsider a unit square block of material of thickness h(into the plane of the paper), as shown in figure below.If the block is subjected to a loading that deforms the square block into the shape shown (with no change in the thickness),

(a) determine the deformation mapping, assuming that it is a complete polynomial in X1 and X2 up to the term X1X2, (b) compute the components of the right Cauchy--Green deformation tensor C and Green-Lagrange strain tensor E at the point X=(1,1,0), and (c) compute the principal strainsand directions at X=(1,1,0) for .

3

2 2x , X

1 1x , X

1

100

1

2.

EXERCISES ON KINEMATICSGiven the following displacement vector in a material description using a cylindrical coordinate system

where A, B, and C are constants, determine the infinitesimal strains. Here ( ) denote the material coordinates.

e e eˆ ˆ ˆsinr zAr Brz C u

, ,r z

3.

The two-dimensional displacement field in a bodyis given by

where c1 and c2 are constants. Find the linear and nonlinear Green-Lagrange strains.

2 3 2 31 1 1 2 1 2 2 2 2

3 2 3 22 2 2 2 2 2 1 1 2

2 3

3 1 322 4 2

( ) ,

( )

u X X X c c c X X

u X c c X X c X X

X

X

4.

DECOMPOSITION OF DISPLACEMENT GRADIENT TENSOR

The displacement gradient tensor, , as the sum of symmetric e and skew-symmetric tensors:

0u∇

( )T T T1 10 0 0 0 02 2( ) ( )u u u u u e ∇ = ∇ + ∇ + ∇ − ∇ ≡ + Ω

T T1 10 0 0 02 2( ) , ( )e u u u u = ∇ + ∇ = ∇ − ∇ Ω

For example, we haveu e e e2 2

1 3 1 2 3 2 1 2 3

1 3 1 3

2 3 2

1 1 1

1 2 3

2 2 2

1 2 3

3 3 32 1

1 2

3

3

ˆ ˆ ˆ( ) ( )

( ) ( )2 , 0, 2 ,

0, 2 , 2 ,

, , 0

)

.

( ( )

u u uX X Xu u uX X Xu u u

X X X X X X

X X X X

X X X

X XX X X

X

For symmetric and skew-symmetric parts are

u2

T0 1

1

1 3

2 3

1 3 2 3

1 3 1 3 2

2 3 2 3

1 3 2 2 3

1 3

1

1

1

2

2 3

1 3 2 2 3

( ) 00 ( )

( ) ( ) 04( ) 0 ( )

1 0 4( ) ( )2 ( ) ( ) 0

0 0 ( )1 0 0 (

22

2 222

2 222

2)

2 ( 2) ( ) 0

X XX X

X X X XX X X X X

X X X X

XX

XX X X X X

X X XX X

X X

X

XXX X X

DECOMPOSITION OF DISPLACEMENT GRADIENT TENSOR

COMPATIBILITY CONDITIONS FOR SOLID CONTINUA

The task of computing strains from a given displacement field is a straightforward exercise. However, sometimes we face the problem of finding the displacements from a given strain field. This is not as straightforward because there are six independent partial differential equations (i.e., strain-displacement relations) for only three unknowndisplacements, which would in general over-determine the solution. We will find some conditions, known as Saint-Venant's compatibility equations, that will ensure the computation of a unique displacement field from a given strain field. The derivation is presented for infinitesimal strains.

COMPATIBILITY CONDITIONS FOR 2-D SOLID CONTINUA

We begin with infinitesimal strains in two dimensions. We have the following three strain-displacement relations:

If the given data are compatible, any two of the three equations should yield the same displacement components. For example, consider the following infinitesimal strain field:

In terms of the displacement components u1 and u2, we have

1 2 1 211 22 12

1 2 2 1

2, ,u u u uX X X X

11 22 12( , , )

11 22 12 1 20, X X


Integration of the first two equations gives

On substitution into the shear strain, we obtain

which cannot be satisfied.

If is specified as, it would be possible to determine f and g, and then u1 and u2. Thus, not all arbitrarily specified strain fields are compatible.

1 2 1 21 2

1 2 2 1

0 0 2, ,u u u u X XX X X X

1 2 2 1( ), ( )u f X u g X

1 22 1

2df dg X XdX dX

12ε 12 1 1 2 2c X c X


The compatibility of a given strain field can be established as follows. Differentiate the first equation with respect to X2 twice, the second equation with respect to X1 twice, and the third equation with respect to X1 and X2 each, and obtain

Using the first two equations in the third equation, we arrive at the following relation among the three strains:Equation (3.7.4) is called the strain compatibility condition for 2-D:

3 2 3 2 3 3 21 11 2 22 1 2 12

2 2 2 2 2 21 2 2 2 1 1 2 1 1 2 1 2

2, ,u u u uX X X X X X X X X X X X

2 2 211 22 122 22 1 1 2

2X X X X

AN EXAMPLE

Given the following two-dimensional, infinitesimal strain field:

where c1, c2, and c3 are constants, determine if the strainfield is compatible.

Using the 2-D compatibility condition, we obtain

Thus the strain field is not compatible, unless

2 2 3 211 1 1 1 2 22 2 1 12 3 1 2

13

, ,c X X X c X c X X

2 2 211 22 12

1 1 2 1 3 12 22 1 1 2

2 2 2 4c X c X c XX X X X

1 2 32 0c c c

Problem Statement:

Solution:


The six strain compatibility conditions for 3-D solid continua for infinitesimal deformation are as follows:

2 22 2 2 233 1311 22 12 11

2 2 2 22 1 1 2 3 1 1 3

2 2 2 22 2 233 23 23 1322 11 12

2 2 23 2 2 3 2 3 1 1 2 1 3

2 2 22 213 23 3322 122

1 3 2 1 2 2 3 1

2 2

2

, ,

, ,

,

X X X X X X X X

X X X X X X X X X X X

X X X X X X X X

2 2213 2312

22 3 2 3 1 3X X X X X X

In index notation they can be expressed as (out of 81 only 6 are distinctly different)

2 22 2ij jnmn im

i j m n j n i mX X X X X X X X

AN EXAMPLE

Given the following three-dimensional, infinitesimal strain field:

determine if the strain field is compatible.

Using the first compatibility condition, we obtain

Thus the strain field is not compatible.

22 3

222

211 1 22 12 1

3 23 3 33

, ,,

,, aconstan, t

X X X XX cX

2 2 211 22 122 22 1 1 2

2 0 0 0X X X X

Problem Statement:

Solution:

2 2233 1311

2 23 1 1 3

2 0 0 2X X X X

AN INTRODUCTION TO FLUID MECHANICS

Fluid mechanics is concerned with the motion of gases and liquids and their interaction with the surroundings. At the outset one must note that fluid mechanics is seldom concerned with fluids alone but rather with effects of fluids on the surroundings with which they come in contact. A fluid state of matter is characterized by the relative mobility of its molecules. The intermolecular forces are weaker in liquids and extremely small in gases. The stress in a fluid is proportional to the gradient of velocity, and the proportionality parameter is known as the viscosity. It is a measure of the intermolecular forces exerted as layers of fluid attempt to slide past one another.

DESCRIPTION OF FLUID CONTINUAThe flight of birds in the air and the motion of fish in the water can be understood by the principles of fluid mechanics. Such understanding helps us design airplanes and ships. The formation of tornadoes, hurricanes, and thunderstorms can also be explained with the help of the equations of fluid mechanics, which are expressed using the spatial description. That is, in fluid mechanics we are interested in the spatial location which the fluid is instantly occupies, and we do not keep track of where the fluid particles come from and where they go. Also, there is no concept of displacement in fluid mechanics. Thus, much of the preceding discussion is not relevant for study of fluid flows.

TYPES OF FLUIDS AND FLOWSFluids are classified based on characteristics of the fluid properties or the basic nature of the flow. An inviscid fluid is one where the viscosity is assumed to be zero. An incompressible fluid is one with constant density, and an incompressible flow is one in which density variations compared to a reference density are negligible. An inviscid and incompressible fluid is termed an ideal or a perfect fluid. A real fluid is one with finite viscosity, and it may or may not be incompressible. An ideal fluid is one for which . When the stress is linearly related to the strain rate, the fluid is said to be Newtonian. A non-Newtonian fluid is one which does not obey a linear stress-strain rate relation. A non-Newtonian constitutive relation can be of algebraic power-law, differential, or integral type.

0constant and

VARIOUS TYPES OF FLOWS

Uniform flow occurs when the convective term is negligible. The velocity field does not change with space in the direction of uniform flow. A uniform flow can be unsteady. In a nonuniform flow, velocity field is a function of position in all directions.

v v⋅∇

In a laminar flow fluid particles move smoothly; at high velocities, random motion, called turbulent motion, can occur.

Potential flows are irrotational flows of ideal fluids. Potential flows allow use of the velocity potential or stress function to write the governing equations in terms of a single variable.

v 0∇ × =

When the curl of the velocity field in a flow is zero, then the flow is called irrotational (i.e., the fluid does not rotate but only translates):

When the divergence of the velocity field in a flow is zero, then the fluid is called incompressible (i.e., the volume does not change but only its shape changes): 0v∇ ⋅ =

The flow is said to be steady if the velocity field is not a function of time: ; when , the flow is unsteady.

( / )v 0t∂ ∂ = ( / )v 0t∂ ∂ ≠

There is no concept of displacement in fluid mechanics. A pathline is the line traced out by a given fluid particle as it moves one point to another (the pathline is a Lagrangian concept).

VARIOUS TYPES OF FLOWS (continued)

DECOMPOSITION OF VELOCITY GRADIENT

In fluid mechanics, the velocity vector is the variable of interest. Similar to the displacementgradient tensor, we can write the velocity gradient tensor as the sum of symmetric D andskew-symmetric W tensors:

where D is called the rate of deformation tensor and Wis called the vorticity tensor or spin tensor:

The Green-Lagrange strain tensor E and the rate of deformation D are related (if and when needed):

( , )v x t

( )TL v≡ ∇

( )T T T1 12 2( ) ( )L v v v v v D W ≡ ∇ = ∇ + ∇ + ∇ − ∇ ≡ +

T T1 12 2( ) , ( )D v v W v v = ∇ + ∇ = ∇ − ∇

. .T T 1orE F D F D F E F− −= ⋅ ⋅ = ⋅ ⋅

EXERCISES ON KINEMATICS OF FLUID CONTINUA

1 1 12 2

1 1 12

, ,

, ,

zr r rrr r rz

z zrz zz

vv v v vD DvDr r r r z r

v vv vD D Dr r z r zv

1. Show that the components of rate of deformationtensor D in the cylindrical coordinate system are:

2. Show that the components of the vorticity tensorW in the cylindrical coordinate system are:

1 1 12 2

1 1 02

,

, zr rr r rz zr

zz z zzrr

vv v v vW W W Wr r r z r

v vW W W W Wr z

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