Theory of Elasticity and Plasticity

한밭대학교 산업대학원

토목공학과

이  형  준

Contents

1. Introduction

2. Vector and tensor

3. Stress

4. Principal Stresses and Principal Axes

5. Analysis of Deformations

6. Velocity Fields and Campatibility Conditions

7. Constitutive Equations

1. Introduction

Mechanics :  the study of the motion(or equilibrium) of matter and forces that cause such motion(or equilibrium)

⇒ based on the concepts of time, space, force, energy and matter

→ needed for the study of all branches of physics, chemistry, biology and engineering

Definition of continuum

(1) Classical definition

three dimensions(space) + time → four dimensions

continuum ot matter : continuous distribution of matter in space

density → be measured by its mass

A material continuum is a material for witch the density of mass, momentum, and energy exist in the mathematical sense.

The mechanics of s such a material continuum is continuum mechanics

(2) Our definition

Defining lengths, areas, and volumes, the density, momentum, energy, stress, and strain can be defined at every point in the space , and if they are all continuous functions of spatial coordinates in

⇔  The material in  is a continuum.

The concept of stress

Consider a material  occupying a spatial region .

surface ,

small surface ,

force   : depend on the location and size of the area and

the orientation of the normal

: unit vector normal to  → outward

The limiting vector

⇒  is called traction or stress vector.

The collective entity of tractions in every orientation of the surface is called stress tensor.

2. Vectors and Tensors

2.1 Vectors

: Directed line segment with a given magnitude and a given direction in three dimensional space.

Notation : Ꭺ,  ℝ

○ Equal : Vectors have the same direction and same magnitude.

○ unit vector : a vector of magnitude of 1

○ zero vector : О,⍜ zero magnitude

○ Sum : parallelogram law                            B

A                       C

○ Multiple : yielded another vector

: same direction as  magnitude of k, k:real number

:

○ Subtraction : yoleds another vector

○  : unit vectors in the direction of positive  axexs.            every vectors in a three dimensional space.

: linear combination

: the component of

○ The magnitude of

If   ,

○ Scalar(Dot) product of and    ==> Scalar quantity

→ (magnitude of ) ․ (component of along )

○ Vector(Cross) product of and

==> yield another vector of

indirection normal to the plane containing the two vectors.

,     ,

2.2 Vector Equations

a particle on which the forces   act.

The condition of equilibrium

Equation of plane

p :constant

: unit normal vector of the plane

: vector represent a plane

⇒ a, b, c : component of the unit normal vector  of the plane

2.3 The summation of convention

○ Indicial notation

.

,    : constant

⇒       →

○ Dummy index   : is summed   →  summation

○ Free index   : is not summed

Example : Direction cosine

: angle between  and x axis

(i=12,3) : the component of unit vector on the coordinates

or

○ Line element with component   in the cartesian coordinate x,y,z

Length  :

○ Kronecker delta

○ Matrices and Determinants

m x n matrix

Transpose of A

Let, vector  : a row matrix ()

Determinant

--- the sum of all the products of the elements of the matrix

Permutation symbol.

The  identify

Kronecker delta

Differentiation

Let, be a function of n variables

2.4 Translation and Rotation of Coordinates.

- Translation

Consider two sets of rectangular Cartesian frame of reference O-xy and O'-x'y'

← the frame of reference O'-x'y' is obtained from O-xy by a shift of origin without a change in orientation.

If the coordinate of the new origin O' are (h,k) relative to O-xy

or

- Rotation

If the origin remains fixed, and the new axes are obtained by rotating Ox and Oy through an angle θ in the counterclockwise direction.

,

⇒  Orthogonal matrix

⇒  Kronecker delta

- Three-Dimensional Space

Consider two right handed rectangular Cartesian coordinate  and  with the same origin O

Let, x denote the position vector of a point P with components  and .

and  be unit vectors of the coordinates  and , respectively.

,

the vector x

A scalar product of both sides with

⇒

Define,

A dotr product of both sides with

,

The numbers  that represent the coordinate of the point P

: the cosine of the angle

between the axes  and

Three unit vectors along  from the edge of a cube with volume 1

The volume of a parallelepiped having any three vectors u,v,w as edge

= triple product u․(v×w)

→ The determinant of their components, if they are right-handed

Volume = u․(v×w) =

x1, x2, x3  and  x1', x2', x3', are right handed

2.5 Coordinate Transformation in General

A set of independent variables  specifies the coordinates of a point in a frame of reference.

A set equation

describes a transformation from  to a set of new variables

The inverse transformation

In order to ensure that such a reversible transformation exists and is in one-to-one correspondence in a certain region R of the variables ()

⇒ each set of numbers defines a unique set of number , for  in the region R

Sufficient Conditions

(1) The function  are single valued, are continuous,

and possess continuous first partial derivatives in the region R

(2) The Jacobian determinant J =  doses not vanish

at any point of the region R

Coordinate transformations with the properties (1) and (2) are called admissible transformations

Jacobian  = positive : right-hand set → right-hand set  ⇒ proper

= negative : right-hand set → reft-hand set  ⇒ improper

Our transformations : admissible and proper

Scalar and Vector Fields

A scalar quantity : such as temperature dependent on the position of a point in space, may be expressed as a function  in terms of the coordinates of the point. The function =constants.

⇒ Scalar field

An example of vector field is the velocity of a fluid particle based on position

Gradient of a Scalar Field

A scalar  is defined over a region of space.

possible to form the derivative of  with respect to each of the coordinates

→ three  : the components of a vector

→ called the gradient of .

→ Symbol  represent a vector operator

with the components

Generally, the gradient normal to the surface of a scalar field

⇒  is a scalar field and  is a vector whose direction is normal to

the surface  = constant.

The part is treated as an operator and a vector in its own right.

Divergence of a Vector

The dot product of the  operator with a vector V

⇒ scalar, at any point, it has only one value, and does not posses the three component that a vector would.

Curl of a Vector

The cross(vector)t product of the  operator with a vector V

⇒ in the expansion of the determinant, the operator  must precede .

If the partial derivatives of  and , and V are assumed to exist.

(1)  : Laplacian of

(2) ,   where  and  are scalar fields.

(3)

(4)

(5)

Vector and Index Notation

Vector Notation	Index Notation	Rank of Tensor
v (vector)		1
= u․v (dot, scala, or inner product)		0
w = u×v (cross or vector product)		1
(gradient of scalar field)		1
(vector gradient)		2
(divergence)		0
(curl)		1
(Laplacian)		1

Partial Derivatives

When only Cartesian coordinates are considered, the partial derivatives of any tensor field behave like the components of a Cartesian tensor.

Let consider two sets of Cartesian coordinates() and () related by

: constants

If  is a tensor, so that

then, on differentiating both sides,

Use a denote partial differentiating.

,    ,

3. Stress

3.1 Notations of Stress

Free body :

A body or a small element of the body is imagined to be cut from the material and free from the remaining part, with the external forces acting on that element. The internal interactions acting on the surfaces of the free body before cut are replaced by forces distributed over the whole surfaces of the free body after cut.

Plane :

A plane can be defined by its location and direction(direction of its normal)

Plane p is defined by its location at point Q, together with its direction defined by unit normal n.

Fig. Free body, cut plane and stress tensor

State of Stress at a Point :

A continuum D, having a point Po with coordinates  and enclosed by a small volume  ΔV

Two types of forces : body or volume forces and surface or contact forces

Body forces

resultant force R and a resultant G about

Considering the limit as  ΔV→0

1. A definite limiting force intensity F exists.

2. The limiting moment intensity is zero

where F represents the body force per unit volume.

Let,  components of F, be continuous function .

The total body force B

or

The resultant moment M of the body forces about the origin O

or

Surface forces and stress vector

Considering a plane ΔA with the unit vector v normal to the area  ΔA.

Let  and  be the resultant force and moment about  due to the action across area ΔA of the material from the positive side onto the negative side of the cut plane n.

In the limit as  ΔA→0

1. A definite limiting force intensity T exists.

2. The limiting moment intensity vanishes.

Hereby,  (the vector T with the fixed plane n) represents the stress vector at point   associated with the cut plane n.

Definition of the state of stress at a point

The state of stress at a point is defined as the totality of all stress vectors  at that point.

⇒ infinite number of cuts through a point → infinite number of values of ,  in general which are different from each other.

This infinite number of values  characterizes the state of stress(or the stress state) at a point. Fortunately, there is no need to know all the values of the stress vectors on the infinite numbers of plane containing the point. If the stress vectors  on three mutually perpendicular planes, are known, the stress vector on any plane containing this point can be found from equilibrium conditions at that point.

Notations of stress

Consider a continuum in a rectangular parallelepiped

Stress vector :  ← acting on Surface  with an outer normal vector pointing in the positive direction of -axis.

: normal stresses

: shear stresses

For Cartesian coordinates x, y, z

or

Sign Convection.

: positive

← If the outer normal of a surface element points in the positive direction

of the -axis, the vector representing the component of normal stress

acting on the surface element will point in the positive e -axis,

: positive (i≠j)

← shearing stress vectors points to the positive -axis if the outer normal

agree in sense with -axis.

3.2 The Laws of Motion

Continuum mechanics is founded on Newton's laws of motion.

Let rectangular Cartesian coordinate,

The space occupied by a material body at any time t be denoted B(t)

Let  be the position vector of a point with respect to the origin of the coordinate system.

Consider infinitesimal element of volume dv

: density of material,    : velocity at

The momentum over the domain B(t), ,  For over the domain B(t)

←  the linear momentum of the body in the configuration B(t)

The moment of momentum of the element about the origin, over the domain B(t)

←  the moment of momentum of the body

Newton's law :

1) The rate of change the linear momentum is equal to the total applied force F acting on the body.

2) The rate of change of moment of momentum is equal to the total applied torque L about the origin.

Types of external forces acting on the material bodies in the mechanics of continuum media.

1) Body force, acting on elements of volume of the body,

- Gravitational forces and electromagnetic forces

- The resultant force vector contributed by the body force

- In a gravitational field   :  ,    : gravitational acceleration

2) Surface force, or stresses, acting on the surface elements,

- Aerodynamic pressure and stress due to mechanical contact between

two bodies, or one part of a body on another.

- stress vector conceived in Euler and Cauchy's principle

- surface forces acting on the external surface : stress vector

Total forces acting upon the material occupying the region B.

: stress vector acting on the surface ,

whose outer normal vector is

Torque about the origin.

Equations of motion.

3.3 Cauchy's Formula

From the equations of motion, the stress vector  representing the action of material exterior to a surface element on the interior is equal in magnitude and opposite in direction to the stress vector  representing the action of interior material on the exterior across the same surface element.

Consider a small pillbox with two parallel surface of area  and thickness .

When,  → 0,  → small but finite,  the volume force, the linear momentum and its rate of change with time vanish.

Therefore,

: the stress vector is a function of the normal vector to a surface.

Knowing the components , the stress vector acting on any surface with outer normal vector , whose components are , This stress vector is denoted by  with components  given by Cauchy's formula :

Consider an infinitesimal tetrahedron formed by three surface parallel to the coordinate planes and one normal to the unit vector .

The area of the surface normal to  be , the areas of the other three surface are

= area of surface parallel to the

= area of surface parallel to the

= area of surface parallel to the

The volume of the tetrahedron,

h : the height of the vertex P from the base (

The forces in the positive direction of , acting on the three coordinate surface.

,  ,

ε's are inserted because the tractions act at point slightly different from P. If the stress field is continuous, then  are infinitesimal quantities.

The forces acting on the triangle normal to v has a component  in the positive  direction,

Body force :

The rate of change of linear momentum   :

: the component of acceleration in the direction of .

The first equation of motion

+  +

+  +   =

Dividing by ,  h→0,   →0

∴

Cauchy's formula assures us that the nine component of stress  are necessary and sufficient to define the traction across any surface element in a body. Hence, the stress state in a body is characterized by the set of quantities .     will be called a stress vector

Normal and Shearing Stress

On an element of surface with a unit outer normal (with component ),  there acts a traction (with components ). The component of  in the direction of  is the normal stress () acting on the surface element. Since the component of a vector in the direction of a unit vector is given by the scalar product of the two vectors.

On the other hand, since the vector  can be decomposed into two orthogonal components  and , where  denotes the shearing stress tangent to the surface. The magnitude of the shearing stress on a surface having the normal  is given by the equation

Fig. Normal and shearing stress

3.4 Equations of Equilibrium

Consider the static equilibrium state of an infinitesimal parallelpiped with surfaces parallel to the coordinate planes.

The stresses acting on the various surfaces :

,

The forces in the direction of , area of

on the right-hand side

on the left-hand side

The body force :

In nonuniform stress field, every stress component is a function of position.

The stress component  is a function of ,  at , and at

If   is a continuously differential function of , according to Taylor's theorem,

where .  If  is finite, then the last term can be eliminated because of a small compared with the other terms.

Equilibrium of the body in the  direction.

Dividing by

A similar equations of equilibrium of forces in the  and  direction can be obtained.

or

The equilibrium of an element requires also that the resultant moment vanish.

→ the stress vector is symmetric.

Considering the moment of all the forces about the  axis.

- Positive in counterclockwise direction.

On dividing through by , and passing to the limit as , , and

Equations of motion

by D'Alembert principle

A particle in motion may be considered in equilibrium if the negative of the product of the mass and acceleration of the particle is applied as an external force on the particle.  → inertial force

the mass :

the inertial force : ,      ; acceleration

, tec

or

3.5 Change of Stress Components in Transformation of Coordinates

The stress components  is defined with respect to a rectangular Cartesian system .

Consider the stress components in the new reference system.

: the direction cosine of the -axis with respect to the -axis.

The the force per unit area acting on  with unit outer normal vector  is a vector  with components.

If the normal vector  is chosen to be parallel to the axis ,

then  the stress vector   has components,

The component of vector in the direction of axis ,

= projection of  on the   axis

∴

In the matrix form,

or

3.6 Stress Components in Orthogonal Curvilinear Coordinate

Orthogonal curvilinear coordinates :  the boundary conditions are simplified

Ex) The flow in a circular cylindrical tube

The torsion of a circular shaft

In a set of cylindrical coordinate, , which are related to the rectangular Cartesian coordinates, .

,

The components of the stress tensor at point()

or

The local rectangular Cartesian frame of reference  at point () with the origin located at the point (), the -axis  in the direction of increasing , -axis  in the direction of increasing , and -axis parallel to .

The stress components by identifying  with ;

,  ,  ,  tec

Since the coordinate system  and  are both Cartesian, → apply the transformation law.

The direction cosine of the axes  relative to

By  and   ,  ,  etc

Homework #2.

Problem 2.1

The components of a stress tensor at a certain place in a body may be presented as a matrix.

,       i, j =1,2,3

1) What is the stress vector acting on the outer side(the side away from the origin) of the following plane, which passes through the place in question ?

2) What are the normal and tangential components of the stress vector to this plane ?.

Problem 2.2

With reference to the x-, y-, and z-coordinates, the state of stress at a certain point of a body is given by the following matrix.

Find the stress vector acting on a plane passing through the point and parallel to the plane  .

Problem 2.3

Does equilibrium exist for the following stress distribution in the absence of body force ?

,

,

Problem 2.4

If the state of stress at a point  is

Find the stress vector and the magnitude of the normal and shearing stress acting on the plane

Problem 2.5

Prove that if , then under the coordinate transformation given by Eq.   with

,

We have  ; that is, in a planar stress distribution, the sum of the two normal stresses in invariant.

Problem 2.6

Consider a vertical column of an isothermal atmosphere that obeys the gas law, , or , where  is the density of gas,  is the pressure,  is the gas constant, and  is the absolute temperature. This gas is subjected to a gravitational acceleration  so that the body force is  per unit volume, pointing to the ground.

If the pressure at the ground level  is , determine the relation between the pressure and the height  above the ground.

Problem 2.7

Consider a two-dimensional state of stress in a thin plate in which   . The equations of equilibrium acting in the plate in the absence of body force are

,

Show that these equations are satisfied identically  if  are derived from an arbitrary function  such that

,    ,

Problem 2.8

In a square plate in the region  the following stresses hold;

( are constant)

1) What must the shear stress  be in order to assure equilibrium ?. Assume that the material the plate is made of isotropic and obeys Hook's law.     2) What are the strains ?. Are they compatible?

3) What kind of boundary conditions are satisfied if the constant a=b=0, but c and d do not vanish?

4. Principal Stresses and Principal Axes

4.1 Introduction

Nine components of stress :

,

→  six independent

symmetric

Under rotation of coordinates, the matrix of stress components can be reduced to a diagonal of the form.

Principal axes : a set of coordinates axes with respect to which the stress

matrix is diagonal.

Principal stress : the corresponding stress components

Principal plane : the coordinate planes determined by the principal axes.

→ Each of the principal stress is a normal stress acting on a principal plane.

: Stress vector on normal to the plane

: There is no shear component.

Normal vector  of the surface S,

4.2 Plane State of Stress

Consider a simplified physical situation in which a thin membrane is stretched by forces acting on its edges and lying in its plane.

- the face  and   : free(unstressed)

- stress components :   on surface

→  through the membrane  because the membrane

is very thin

⇒   Plane state of stress(plane stress state) in the -plane.

Stress components in the symmetric matrix

:

Rotation of coordinates from  to

Stress components in the new frame of reference()

Direction cosine between the two systems of rectangular Cartesian coordinates

According to   or

The stress components in new coordinate.

=

=

Since,

,

⇒

, when

⇒ The directions of the , axes corresponding to the particular values of  given as above value.  →  called principal directions

The , axes  →  called principal axes

The stress    →  called principal stresses

If  ,  are principal axes, then

→  or  is either a maximum or minimum with respect to all choices of

On the other hand, find the angle  when  reaches its extreme value

→ on differentiating  with respect to , and setting the derivative equal to zero   :

The maximum value of

→ the maximum of the shear stress acting on all planes parallel to the z-axis.

4.3 Mohr's Circle for Plane Stress

A geometric representation of the principal stresses and maximum shear stress by Otto Mohr as shown in Figure.

The normal and shear stresses acting on the surface are plotted on a stress plane in which the abscissa represents the normal stress and the ordinate the shear.

○ Normal stress : tension is plotted as positive, compression as negative

○ Shear stress : special rule.

- positive on a face of element when it yields a clockwise

moment about the center point  of the element.

- negative - counterclockwise moment

○ Join the line  which intersect the -axis at .

○ Draw a circle through A and B

⇒ Mohr's circle

○ The stress acting on a surface whose normal makes an angle  with -axis in the counterclockwise direction.

- Draw a radius  that makes an angle  with the line

- The abscissa of  gives the normal stress on that surface

and the ordinate the shear

- The point  represents the stress acting on the surface

whose normal makes an angle  with -axis.

To prove that this construction is valid.

The abscissa of

,

The ordinate of

→ Difference in sign : negative - counterclockwise moment in Mohr's circle.

Maximum and minimum of with respect to all choices of , when

- Maximum

- Minimum

Maximum of

4.4 Mohr's Circle for Three-Dimensional Stress State

Let  be the principal stresses at a point. The components of stress vector acting on any section can be obtained by the tensor transformation law or Mohr's circle.

If the normal stress  and shear stress  acting any section are plotted on a plane, with and  as coordinates.

→ If , then  is the largest stress and  is the largest shear for all possible surface. The plane on which the largest shear acts is inclined  from the principal planes on which  and  act.

4.5 Principal Stresses

- The stress vector acting on a surface with normal  depend on the direction of .

- The angle between the stress vector and the normal  varies with the orientation of the surface.

- A surface so oriented that the stress vector is exactly normal to it.

← At least three mutually orthogonal surfaces that fulfill this requirement at any point in the body

⇒  Such surface is a principal plane, its normal a principal axes.

The value of normal stress acting on the principal plane is a principal stress.

Let,   : unit vector in the direction of a principal axis

: the corresponding principal stress

then, the stress vector acting on the surface normal to   has component

,

This same vector is given by the expression ,

Hence, on equating these two expression,

,

→ are to solved for

Since   is a unit vector which

⇒  Find a set of the nontrivial solutions

⇒ Eigenvalue problem

These exist three real valued principal stresses and a set of orthogonal principal axes. Whether the principal stresses are all positive, all negative, or mixed depends on whether the quadratic form  is positive definite, negative definite, or uncertain, respectively.

The system of equation has a nonvanishing solutions  if and only if the determinant of the coefficients vanishes.

⇒ cubic equation in , its roots are the principal stresses

For each value of the principal stress, a unit normal vector  can be determined.

→

⇒ characteristic equation

On the other hand,  if are the roots of the characteristic equation.

Since the principal stresses characterize the physical state of stress at a point, they are independent of any coordinates of reference. So,  are invariant with respect to the rotation of coordinates.

: invariants of the stress tensor with respect to the rotation of coordinates.

For a symmetric stress tensor, the three principal stresses are all real and the three principal plane are mutually orthogonal.

The proof that three principal plane are mutually orthogonal.

Let,  be a unit vector in the direction of the principal axes, with components

The solutions of equation  corresponding to the roots , respectively.

Multiplying the first equation by  and the second equation by , summing over , and subtracting the resulting equations.

On account of the symmetry condition,

⇒  The assumption of the roots  are all real.

When ,

,  ,

If ,  will be fixed and an infinite number of pairs of vector  orthogonal to  can be determined.

If , any set of orthogonal axes may be taken as the principal axes.

If the reference axes  are chosen to coincide with the principal axes, then the matrix of stress components become

<Example>

The stress tensor , at a point is given as following.

Find the principal stress and principal directions at this point.

(units of stress)

solution:

- Invariants

- Characteristic equation

,

⇒ three roots ,

Check,

Calculation of the principal direction

For

And,

,  ,

For

,  ,

For

,  ,

Orthogonality of , ,

,  ,

4.6 Shearing Stresses

On element of surface with a unit outer normal  (with components ) and acts a traction (with components ).

The component of  in the direction  is the normal stress acting on the surface element.

Let, this normal stress be denoted by ,

Fig. Normal and shearing stress

Since the vector  can be decomposed into two orthogonal components  and .  denotes the shearing stress tangent to the surface.

Let the principal axes be chosen as the coordinate axes, and let  be the principal stresses.

,  ,

Foe example, if ,

<Example>

The state of stress  , at a point is given as following.

(units of stress)

For a plane with unit normal

(1) The magnitude of the stress vector, , for plane

Thus the magnitude of the stress vector

(2) The normal and shear stress components,   and

,

4.7 Stress Deviation Tensor

The stress deviation tensor

where,

:  Kronecker delta

The separation of

= spherical or hydrostatic stress tensor + deviation stress tensor

↓                          ↓

(volume change)                (shape change)

= hydrostatic part    +  deviation

→ very important in describing the plastic behavior of metals.

To determine the principal stress deviations , the determinantal equation

⇒ cubic equation in ,

roots : the principal stresses deviation

or

⇒ characteristic equation

: the invariants of the stress deviation tensor

→ The first invariant of stress deviation tensor always vanishes

The invariants  are related to the invariants

4.8 Lame's Stress Ellipsoid

On element of surface with a unit outer normal  (with components ) and acts a traction (with components ).

Let the principal axes of the stress tensor be chosen as the coordinate axes , and let the principal stresses be written as , Then,

,  ,

Since  is unit vector,

The components of  satisfy the equation

→ the equation of an ellipsoid with reference to a sustem of rectangular coordinates with axes labeled , ,

→ This ellipsoid is the locus the end points of vector  issuing from a common center.

4.9 Pure Shear State

In an  coordinate system, let the stress state at a point  P  be . If some coordinate system  can be found such that

The state of stress at P is said to be pure shear state

A necessary and sufficient condition for a pure shear state is that

or

Proof

In order to that , one component of stress , which of the opposite sense to the other two.

Let fix the -axis and rotate -  and - axes. Since  and  are of opposite sign, a plane where normal stress is zero can be found.

→    This plane AB with its normal as axis 2.

Normal stresses on the 2-3'-x coordinate system,

From the condition of vanishing first invariant,

→   and  must be of opposite signs.

4.10 Octahedral Stress

An octahedral (stress) plane is a plane whose normal make equal angles with each of the principal axes of stress. Thus, the planes with normal   in the principal system.

→ Octahedral plane

In fig.,

Fig. Octahedral normal and shear stress

Fig. Octahedral planes in principal coordinate system

Referring to the principal stress axes, 1, 2, and 3, the stress tensor ,

The normal component of a stress vector at point O associated with any direction .

or

The normal stress on a face of the octahedron,

Note,

(1) The quantity is the mean normal stress(or hydrostatic pressure)

(2) For an isotropic material, gives rise only to the change of volume

without affecting the shape of the element.

(3) Magnitude of  on all the eight faces is the same.

The shear stress on a face of the octahedron, .

For the principal shear stresses,

where,  are the principal shear stresses.

⇒   change in shape, but dose not contribute to the change volume.

In terms of general nonprincipal stresses.

<Example>

The state of stress , at a point is given as following.

(units of stress)

(1) The octahedral normal and shearing stress

(2) The hydrostatic stress

(3) The stress deviator tensor,

⇒Since is a pure shear state,

Homework #3.  Due to 4 May 2007

Problem 3.1

Let , , , ,

, .

(1) What is the magnitude of the traction acting on whose normal vector is

?

(2) What are the three components ( in the x-, y-, and z-directions) of the stress vector on the surface?

(3) What is the normal stress acting on the surface?

(4) What is the resultant shear stress acting on the surface?

Problem 3.2  Stress concentration

Describe the boundary conditions for a plate with a circular hole subjected to a static uniform tensile loading with the normal stress  acting on the ends.

It is known that if this plate is made of a linear elastic material.

(1) Check the stress boundary conditions to see whether they are satisfied.

(2) Find the location of the point where the normal stress  is the maximum.

(3) Find the maximum shearing stress in the entire plate.

(4) Obtain the maximum principal stress in the plate.

Problem 3.3

The stress at a point in a body has the following components with respect to a set of rectangular Cartesian coordinates ;

(1) Find the values of the invariants , and the principal stresses.

(2) Find the deviation stress tensor, .

(3) Determine the invariants .

(4) Find the octahedral normal and shearing stress.

5. Analysis of Deformation

5.1 Deformation

The forces applied to a body cause the stress. In general, forces also give rise to movements and deformations in the body. When the relative position of any two points in a continuous body is changed, the body is said to be deformed or strained. If the distance between every pair of points in the body remains constant during motions of the body, then the body undergoes rigid body motion. The displacements of a rigid body motion consist of translations and rotations, translations and rotations are therefore called rigid body displacements.

Consider a string an initial length , if it is stretched to a length

Linear strain in

,     ,

,

Change in angle (

= measure dirtortion

= shear strain

If ,

If ,

⇒ In infinitesimal elongation, all of these strain measures are

approximately equal.

The selection of proper measures of strain is dictated basically by the stress-strain relationship(i.e., the constitutive equation of the material)

For example, in uniaxial stretching , the tensile stress , strain

: Young's modulus

→ Hook's law

⇒ Hookean material

The relationship for a Hookean material subjected to an infinitesimal shear strain.

: shear modulus or modulus of rigidity

Let us consider the mathematical description of deformation.

Let a body occupy a space . Referred to a rectangular Cartesian frame of reference, every particle in the body has a set of coordinates. When the body is deformed, every particle takes up a new position.

A particle P at a place with coordinates  is moved to new place Q with coordinates when the body moves and deforms.

The vector  is called the displacement vector of the particle.  The components of displacement vector are

, ,

The deformation of the body is known if  are known function of  .

⇒ Transformation(mapping) from  to

In continuum mechanics, the deformation is continuous. And , assume that the transformation is one to one.

⇒ the functions : unique and the unique inverse

The displacement vector , and its components

The displacement vector

: in the original position

: in the deformed position

5.2 The Strain

A rigid body motion induces no stress. Thus, the displacements themselves are not directly related to the stress.

To related deformation with stress,

⇒ Consider the stretching and distortion of the body

Consider three neighboring points  in the body.

Original position :

Deformed configuration :

The change in area and angles of the triangle is completely determined if we known the change in the length of the sides.

Consider an infinitesimal line element connecting point P, P', and Q, Q'

⇒ P and P' are deforemed to Q and Q', respectively

Original position :

Deformed position :

The square of the length  of  and   of in the original configuration and deformed position. .

Due to,   are a function of  and are a function of .

,

,

On introducing the Kronecker delta,

The difference between the square of the length

The strain tensor

⇒ Introduced by Green and St.-Vernant

Called Green's strain tensor

Refer to original position  : Largrangian

⇒ Introduced by Cauchy for infinitesimal strain

Almansi and Hamel for finite strain

Called Almansi's strain tensor

Refer to deformed position  : Eulerian

The tensor  and  are obviously symmetric,

,

implies and vice versa.

⇒ The necessary and sufficient condition that a deformation of a body be a rigid-body motion is that all components of the strain tensor , or  be zero throughout the body.

5.3 Strain Components in Terms of Displacements

Displacement vector  with components

,

The strain tensor

In unabridged notation,

for ,    for ,  and  for

○ Displacement vector  are considered function of  the position of point in the body in unstrained configuration.

⇒ Lagrangian strain tensor is evaluated

○ Displacement vector  are considered function of  the position of point in the body in strained configuration.

⇒ Eulerian strain tensor is evaluated

Cauchy's infinitesimal strain tensor

The first derivative of displacement  are very small, so that the squares and products of partial derivatives of  are negligible.

In unabridged notation,

for  and  for

,

,

,

⇒  called tensorial strain

In the case of infinitesimal displacement, the distinction the Lagrangian and Eulerian strain tensor disappear.

Notation for shear strain

,

,

,

⇒  inconvenience in the mathematical deal

called engineering notation

5.4 Geometric Interpretation of Infinitesimal Strain Components

Let  be a set of rectangular Cartesian coordinates. Consider a line element of length  parallel to the x-axis().

In this case,  , and  differs from only by a small quantity of the second order if we assume that the displacement  and the strain components  are infinitesimal.

⇒ represent extension or change of length per unit length

Let consider a small rectangular in the body with edges .

: represents the change of the angle

=  tan(change of angle )

In engineering usage,

⇒ called shearing strains or detrusions

5.5 Infinitesimal Rotation

Consider an infinitesimal displacement field ,

Form the Cartesian tensor

Which is antisymmetric.

Hence, the tensor  has only three independent components -

because  are zero.

Build the dual vector, from such an antisymmetric tensor.

or

⇒  be called dual(antisymmetric) tensor of a vector

and  be called the rotation vector and rotation tensor of

the displacement field , respectively.

The vanishing of the symmetric strain tensor  and  is a necessary and sufficient condition for a neighborhood of a particle to be moved like a rigid body.

A rigid-body motion consists of a translation and a rotation.

The translation is . What is the rotation?. In an infinitesimal displacement field,

- the strain tensor vanishing at a point

- the rotation is given by the tensor

Consider a point  in the neighborhood of . Let coordinates of  and  be  and , respectively.

The relative displacement of  with respect to  .

where, the first term of right side is the infinitesimal strain tensor, which is zero. The second term is the rotation tensor  ,

by  ,

⇒ The relative displacement is the vector product of   and

An infinitesimal rotation  about axis through

in the direction of  .

5.6 Finite Strain Components

When the strain components are not small, it is easy to give simple geometric interpretations for the components of the strain tensors.

Consider a set of rectangular Cartesian coordinates with respect to the strain components.

Let a line element before deformation be , with components

The extension  of this element

or

∴

when  is small compared to 1

<cf> Taylor series.

The component

Consider two line elements  and  that are at a right angle in the original state.

After deformation,

→   with components

→   with components

Forming the scalar product of the deformed element,

According to

,

∴

,

⇒  : the angle between the line element   and  after deformation.

The change of angle between the two line element.

⇒ exhibit the relationship  to the angles and   .

In Eulerian strain ; refer to deformed configuration.

,

If the deviation from a right angle between two elements in the original state,

In the infinitesimal strains,

,  ,  ,

5.7 Principal Strains ; Mohr's Circle

To exchange the word stress with strain.

Strain tensor, → symmetric

There are three principal

⇒ cubic equation in , its roots are the principal strains

For each value of the principal stress, a unit normal vector  can be determined.

⇒ characteristic equation

: invariants of the strain tensor

For a symmetric strain tensor, the three principal strains are all real and the three principal plane are mutually orthogonal.

Principal axis with direction cosine,  ,

,

i = 1,2,3,

Principal plane : a plane perpendicular to a principal axis.

If the reference axes  are chosen to coincide with the principal axes, then the matrix of strain components become

The principal shear strain.

,  ,

The engineering principal shear strain.

,  ,

The maximum shear strain

Strain Deviation Tensor

The strain deviation tensor

where,

:  Kronecker delta

To determine the principal strain deviations , the determinantal equation

⇒ cubic equation in ,

roots : the principal strain deviation

or

⇒ characteristic equation

: the invariants of the strain deviation tensor

Consider a unit cube with edges directed along the principal axes of strain 1,2 and 3. After deformation, the cube become rectangular parallelepiped with edges (), () and (), since there is no shear strain.

The relative change in volume

For small strain

⇒ called the dilatation or simply the volume change.

Cauchy's formulae for strains

The strain vector

The strain vector  for any fiber  can be decomposed into two components, one in the direction of the fiber , called the normal strain , and the other lying in the plane normal to the fiber, called the shear strain .

If the unit vector , normal to direction  has components(), the magnitude of the shear strain component

The resultant shear strain  at point for the fiber

Octahedral Strain

An octahedral plane is a plane whose normal make equal angles with each of the principal axes of strain. Thus, the planes with normal   in the principal system.

The normal strain on a face of the octahedron,

The shear strain on a face of the octahedron, .

In terms of general nonprincipal stresses.

<Example>

The strain tensor , at a point is given as following.

(1) Find the principal strain

- Invariants

- Characteristic equation

⇒ three roots ,

Check,

(2) the maximum shear strain

<Example>

The strain tensor , at a point is given as following.

(1) The deviatoric strain tensor

(2) Calculate the invariants

(3) Calculate the dilatation(volume change)

<Example>

The state of strain  , at a point is given as following.

For a plane with unit normal

(1) The normal strain

(2) The magnitude of the strain vector, , for plane

(3) The magnitude of the shear strain  , where unit vector

5.8 Infinitesimal Strain Components in Polar Coordinates

The strain components can be referred to a local rectangular frame of reference oriented in the direction of the curvilinear coordinates.

In polar coordinates,

strain components,  ←

tensor transformation law

Relations between the polar coordinates  and the rectangular coordinates

In a set of cylindrical coordinate, , which are related to the rectangular Cartesian coordinates, .

,

,

,

Derivative with respect to  and  in the Cartesian equations may be transformed into the derivative with respect to  and .

The components of the displacements vector ()

The strain components in polar coordinates

The direction cosines between the two sets of coordinate -

or or or	- 0	0	0 0 1

The tensor transformation law holds,

The strain-displacement relationship

,  ,

,

The strain components in local coordinates

,  ,

,

5.9 Direct Derivation of the Strain-Displacement Relations

in Polar Coordinates

The strain components can be derived directly from the geometric definition of the infinitesimal strain components.

Normal strain : the ratio of change of length per unit length

Shearing strain : one-half of the change of a right angle.

Consider first the displacement in the -direction, . (Fig.5.6(a))

A radial displacement of a circumferential element causes an elongation of that element.

A strain in the -direction,

element ab  : original length of

element a'b'  : displaced length of

The tangential strain due to this radial displacement

The tangential strain due to the tangential displacement

The total tangential strain,

The normal strain in the axial direction,

The shearing strain  is equal to one-half of the change of angle .

The change in the radial displacement in the -direction,

The change in the tangential displacement in the radial direction,

The change in slope of the line

⇒  These equations are the same as those derived by transformation law.

5.10 Other Strain Measures

The strain tensors are not the only ones suitable for the description of deformation.

The most natural ones : analysis of deformation on the change of the square of the distance between any two particles.

Using Pythagoras's theorem, the squre of the distance between two points  and  with coordinates referred to a rectangular Cartesian frame reference.

⇒  analysis on this equation

→ the result was a natural definition of strain tensor.

However, deformations does not have to be described this way.

For example, using the change of distance (instead of ) as starting point

or using the set of nine first derivatives of the displacement field();

called  "deformation gradients"

Separate the matrix  into a sum of symmetric part and antisymmetric part.

The symmetric part of deformation gradient matrix is the matrix of the infinitesimal strain

The antisymmetric part of deformation gradient matrix is the rotation tensor.

is called Lagrangian rotation tensor.

Other well known strain measures are Cauchy's strain tensor and Finger's strain tensor.

When the mapping is given as following,

Cauchy's strain tensor

,

Finger's strain tensor

,

Homework #4.  Due to 18 May 2007

Problem 4.1

Given the strain tensor  as

Find

(1) The principal strains and directions

(2) The maximum shear strain

(3) The octahedral strains

(4) The normal strain component  and the resultant shear strain component

for a fiber element with direction

(5) The deviatoric strain tensor  and its invariants   and

(6) The volume change per unit volume(dilatation)

(7) The strain invariants  and

Problem 4.2

The displacement components  at a point in a body are given by functional components

, ,

(1) Show that there is no rotation if the deformation are assumed to be small.

(2) Assuming large deformation, find the Lagrangian rotation and strain tensor,

and calculate the principal strains ,  and  for this case.

6. Velocity Fields and Compatibility Conditions

6.1 Velocity Field

For the study of fluid flow, we are generally concerned with velocity field. i.e., with the velocity of every particle in the body of the fluid.

The location of each fluid particle to a frame reference , the field of flow is described by the velocity vector field

The components of velocity vector is expressed by the functions

,  ,

or    by index notation.

For continuous flow, the continuous and differentiable functions

Study the relationship between velocities at neighboring points.

Let the particles  and  be located instantaneously at  and , respectively.

The difference in velocities at these two points is

The partial derivatives are evaluated at the particle .

Let define the rate-of-deformation tensor  and the spin tensor

Then,

: symmetric,

: antisymmetric,

→ only three independent element

there exist a vector  dual to ,

,   i.e.

The vector  is called the vorticity vector.

The analysis of the velocity field is very much like the analysis of an infinitesimal deformation field.

During infinitesimal interval of time , infinitesimal displacement is

About the infinitesimal strain field can be immediately extended correspondingly to the rate of change of strain with the word velocity replacing the word displacement.

6.2 The Compatibility Conditions

Suppose a set of two partial differential equations for one unknown function .

,

→ cannot be solved.  ↔  mutually inconsistent.

: unequal

When partial differential equations are given as follows,

,

The differential equations cannot be integrated unless the condition

⇒ Condition of integrability,  called equation of compatibility.

Consider a plane state of strain such as may exist in the solid propellant grain of a rock.

A set of strain

, , ,

Check the consistency of the equation.

Displacements  :   and

If the strain is small,

Differentiate the equations with x and y,

,

⇒

⇒  The equation of compatibility for a plane state of strain.

Two-dimensional velocity field of a fluid. The components of the rate-of-strain tensor may be measured.

To check the consistency,

is the components of the rate-of-strain tensor

6.3 Compatibility of Strain Components in Three Dimensions

In three dimensions, the components of strain with the displacement

or

How do we integrate the above differential equations to determine ?

⇒ six equations, three unknown function .

⇒ The system will be have a single-valued solution only if the function

satisfy the conditions of compatibility.

By differentiation

Interchanging subscript.

From these,

⇒ The equation of compatibility of St. Venant for infinitesimal strain.

⇒ 81 equations, only 6 essential(←symmetry of strain components)

For finite strains, a compatibility condition can be derived from Riemann's theorem by the fact that the deformed body remains in a Euclidean space.

Homework #5.  Due to 25 May 2007

Problem 5.1

The displacement fields in a unit circle are given as followings,

,

Is there any question of compatibility?

Problem 5.2

The displacement field in a unit circle is the followings. Is it compatible ?

,

,

Problem 5.3

In a two-dimensional, plane-strain field, the displacements are described by , whereas that long the z-axis, , is identically zero,  being a set of rectangular Cartesian coordinates.

(1) Express the strain components  ,  and  in terms of

(2) Derive the equation of compatibility for the strain system  ,  and

(3) Is the following strain system compatible ?

,  ,

where  are constants. All other strain components are zero.

Problem 5.4

Determine the relations among the constants , ,  ,  , ,  and  so that the following is possible state of strain.

7. Constitutive Equations

: Stress-Strain Relationships

7.1 Introduction

The solution of a solid mechanics problem at each instant of time must satisfy the following three conditions.

◎ Equation of equilibrium or of motion.

(Equilibrium equation) ← Statically(or dynamically) admissible.

-  or     at surface point

-  or  at interior point

or

◎ Conditions of geometry or the compatibility of strain and displacements

(Compatibility condition or Integrability condition)

← Kinematically admissible.

where,

or

◎ Material constitutive laws or stress-strain relations

or

			Body and Surface Force									Displacement
Equilibrium																		Compatibility
				Stresses									Strains
								Constitutive laws

7.2 Basic Assumption(Hypotheses)

In the majority of engineering applications of structural and geological materials such as metals, concrete, soils, rocks, and rubber under short-term loading, time independence of the stress-strain relation is a reasonable approximation and is made most often.

The constitutive models or stress-strain relations to be described in the forthcoming are based on the following two assumptions.

(1) Material behavior is time independent

- The rate sensitivity, creep and relaxation are not included

in such behavior

(2) Interaction between mechanical and thermal processes is neglected.

- Only material bodies under isothermal conditions are considered

7.3 Need for Elastic Material Models

The various elastic constitutive relationships are formulated to model the mechanical behavior of the class of real materials that can be described as elastic. The study of theses elastic model is needed fir two importanr reasons.

(1) Elastic models described well the behavior of many engineering materials at the working load. For example, the linear elastic models has been used successfully to describe the behavior of metal materials at stress levels below the elastic limit. Therefore, elastic constitutive relations are the basis for the theory of elasticity.

(2) The elastic constitutive models are needed in the theory of plasticity, which can be considered as a generalization of the theory of elasticity. For instance, elastic-plastic models are used extensively for metal materials at a overload stages.

7.4 Definitions

Elastic Material

A material body is deformed when subjected to applied forces. If upon the release of the applied forces the body recovers its original shape and size, then the material body is called elastic.

For such a material, the current state of stress depends only on the current state of deformation, that is, the stress is a function of strain.

: the elastic response function.

⇒ both reversible and path independent

→ strains are uniquely determined from the current state of stress

vice versa.

The elastic material defined above is usually termed Cauchy elastic material. Cauchy elastic material may generate energy under certain loading-unloading cycles. Clearly, in such cases it violates the laws of thermodynamics. Therefore, the term hyperelastic of Green elastic material is used to indicate that the elastic response function is further restricted by existence of an elastic strain energy function , which is in general a function of strain components .

⇒ This ensures that no energy can be generated through load cycles and thermodynamic laws are always satisfied.

Hyperelastic Model

The term hyperelastic model is used to describe the incremental elastic constitutive relations. These models are often used to describe the behavior of a class of materials in which the state of stress is generally a function of the current state of strain as well as of the stress path followed to reach that state.

stress rate(or increment) tensor

strain rate(or increment) tensor

elastic response function.

Material Symmetry Properties

If the mechanical behavior of the material is the same in certain directions, then the material is said to have material symmetry with respect to these directions. Material symmetry is exhibited to the form invariance of the constitutive relations under a group of transformations of the coordinate axes. If there is no material symmetry at all, the material is termed anisotropic.

(1) Orthotropic Material :

- Three orthogonal planes of material symmetry.

- such as wood material(normal, tangential and radial direction)

- Material symmetry is exhibited under any 180° rotation of the axes

(2) Transversely Isotropic Material :

- Rotational symmetry with respect to one of the coordinate axes.

- such as rolled plate(the material properties are the same in any direction

of the plane  parallel to the direction of rolling.

- Material symmetry is exhibited under any rotation of the axes().

- The plane  parallel to the direction of rolling is called

the plane of isotropic.

(3) Isotropic Material :

- The mechanical behavior is identical in all directions.

- Any plane is a plane of material symmetry, and any axis is an axis of

rotational symmetry.

- such as polycrystalline metals

- It is important to note that this isotropic is destroyed after plastic

deformations occur, since plastic deformations are physically anisotropic.

7.5 Isotropic Linear Elastic Stress-Strain Relations

(Generalized Hook's Law)

The most general form for linear elastic stress-strain relations for a Cauchy elastic material.

: components of initial stress tensor corresponding to the initial

strain free state (when all strain components  )

: tensor of material constants.

If it is assumed that the initial strain free state corresponds to an initial stress free state, that is,  .

Since both  and  are second-order tensor,  is a fourth-order tensor with 81() constants for such a tensor . However, since  and  are both symmetrical, one has the following symmetry conditions.

Hence, the maximum number of independent constants is reduced to 36.

For Green elastic material, the four subscripts of elastic constants can be considered as pair , and the order of the pairs can be interchanged, . The number of independent constants needed is reduced from 36 to 21.

If, in addition, having a plane of elastic symmetry, the number of independent constants needed is reduced from 21 to 13

For orthotropic symmetry, the number of independent constants is reduced to 9.

For transversely symmetry, the number of independent constants is reduced to 5.

7.5.1 Isotropic Linear Elastic Stress-Strain Relations

For an isotropic material, the elastic constants  must be the same for all directions. Thus the tensor  must be an isotropic fourth-order tensor.

The most general form for the isotropic tensor  is given.

,  and  are scalar constants.

←  ()

Since  must satisfy the symmetry conditions,

or

⇒ Constitutive law for an isotropic linear elastic material

For an isotropic linear elastic material, there are only two independent material constants,  and , which are called Lame's constants.

Conversely, strains  can be expressed in terms of stress in the constitutive law.

⇒ Constitutive law for an isotropic linear elastic material

The constant  and  are determined from experimental test results for some simple state of stresses and strains.

Simple tests are described as followings.

(1) Hydrostatic Compression Test

:

only nonzero components of stress

Bulk modulus  is defined for this case

as the ratio between the hydrostatic pressure

and the corresponding volume change .

, ,

,

(2) Simple Tension Test

and all other

,

(3) Simple Shear Test

and all other

(4) Uniaxial Strain Test

: Lateral movement restrained.

Constrained modulus,

and all other

The constitutive relations can be written in different forms by utilizing the relations between the elastic moduli. In particular, the following forms are used frequently in practice.

Relationship among  elastic moduli, E, G, K, ,  , M

),   , ,

, , ,

Decomposition of stress-strain Relations

A separations exists between the mean(hydrostatic or volumetric) and the deviatoric(shear) ( component(s in the isotropic linear elastic model.

The hydrostatic response can be derived directly from eqn.  and  ().

To derive the deviatoric response relations, using relation .

Substitute for ,   and .

∴

Combining these two equations,  and , we can write the total elastic strains  in terms of the hydrostatic and deviatoric stresses.

or

,    ()

Similarly,  the elastic stresses  can be expressed in terms of the volumetric and deviatoric strains.

Experimental results for real elastic materials have shown that the constants  and  are always positive, that is,

,

No practical experience with any existing material that will exhibit a negative value of . Thus actual values of  for most materials in practice are positive. .

The value   implies that  and  or elastic incompressibility.   Some rubberlike materials are almost incompressible and have a value of .

Concrete :

Steel    :

Condition for elastic materials

7.5.2 Isotropic Linear Elastic Stress-Strain Relations in Matrix Form

The stress-strain relationships discussed above can be conveniently expressed in matrix forms. These forms are suitable for use in solutions by numerical methods(e.g. finite element method). In the following matrix forms are given for various cases.

(1) Three-Dimensional Case :

The stress and strain components are defined by two vectors  and .

,      ,

: elastic constitutive or elastic moduli matrix.

or alternatively, substituting for  and  in terms of  and

Also, stress-strain relations can be written in matrix form as following.

: elastic compliance matrix

(2) Plane Stress Case :

The two dimensional plane stress case with

Example : thin plate

And,

It is to be noted that in the plane stress case, the strain component  is nonzero,  while shear strain components  and  are zero. The component  has the value.

⇒   is a linear function of  and

(3) Plane Strain Case :

The two dimensional plane strain conditions :

In elongated bodies of uniform cross sections subjected to uniform loading along their longitudinal axis(z-axis).

Example : tunnels, soil slopes, dams, and retaining walls.

And,

For this case, the stress components  , and the stress component  has the value.

(4) Axisymmetric Case :

Analysis of bodies of revolution under axisymmetric loading is similar to those of plane stress and plane strain conditions since it is also two-dimensional.

The nonzero stress components  ,  , , and  and the corresponding strains are   ,  , , and .

.

And,

7.6 Anisotropic, Orthotropic, and Transversely Isotropic Linear

(Green) Elastic Stress-Strain Relations

The strain energy density function  or complementary energy density function , may be represented as a power series expansion in the strain components , or the stress components  .

,

Expanding the function  in a polynomial expansion and retaining only second-order terms.

where,  denote constants.

Using Eq. , we can express the stress,

Finally, choosing (i.e. assuming that the initial strain free state corresponds to an initial stress free state) and taking

and clearly

the general form of the linear elastic stress-strain relation can be written as,

It is possible to insist that  , so that

For Green elastic material, the four subscripts of elastic constants can be considered as pair , and the order of the pairs can be interchanged, . The number of independent constants needed is reduced from 36 to 21.

The matrix forms of the linear stress-strain relations for anisotropic, orthotropic, and transversely isotropic elastic materials are given below.

< Example >

The elastic complementary energy density  is defined as,

Find an expression for  in terms of the stress invariants   and  for an isotropic linear elastic materials.

<Solution>

Substituting for

where , substituting for

which can be reduced to

, ,

or, in terms of

7.6.1 Anisotropic Linear Elastic Stress-Strain Relations

The general form of the linear elastic stress-strain relation

In matrix form with 21 constants

, ,  ,,  ‥‥

( 11→1, 22→2, 33→3, 12→4, 23→5, 31→6 )

7.6.2 Orthotropic Material(Nine Constants)

An orthotropic material possesses material elastic symmetry with respect to three mutually perpendicular axes. Taking the coordinate axes  and  perpendicular to the three planes of material symmetry and requiring that the elastic properties be unchanged under 180° rotations about these axes, certain relations can be found among the constants. The axes  and  in this case are called the principal directions(axes) of the material. In general, it can be shown material elastic symmetry under the transformation  of the coordinate axes requires that the forth-order tensor  satisfy the conditions;

where  is the symmetry transformation tensor containing the directional cosines, of the rotated axes  with respect to the original axes .

For example, consider first the material symmetry under 180° rotations about the -axis for the orthotropic material.

In this case, constants  gives the following relations.

,

,

so that

In addition,

In similar manner, similar relations can be obtained by considering rotation about the - and -axis.

These symmetries in the subscripts reduce the number of independent elastic constants to 12 only. Consideration of symmetry of   for a Green elastic material further reduces to number to only nine constants.

If engineering definition for the elastic moduli is used, an alternative form can be written as following.

where,  Young's moduli in directions ,  and , respectively.

shear moduli for plane that are parallel to the

coordinate planes ,  and ,  respectively.

For example,  the shear modulus  characterizes the

shear strain  produced by the shear stress .

Poisson's ratio which characterizes the

compressive strain in the  direction produced by a

tensile stress in the  direction. For instance,

characterizes the compressive strain in the  direction

produced by a tensile stress in the  direction.

Because of the symmetry requirements for Green elastic material,

The elastic constitutive or elastic moduli matrix  contains 12 elastic moduli, but only nine of these moduli are independent because the symmetry requirements.

7.6.3 Transversely Isotropic Material(Five Constants)

In this case, the material exhibit a rotational elastic symmetry about one of the coordinate axes. Let axis  be the axis of elastic symmetry. The plane of isotropy is the  plane(as shown Fig.)

The transversely isotropic material by simply imposing the additional requirement that Eq.  must also be satisfied for any rotational transformation,  about the -axis of the form.

where  is the angle of rotation about the axis of elastic symmetry(-axis).

This additional symmetry on the material reduces the number of independent elastic constants from nine to five only.

If engineering definition for the elastic moduli is used, an alternative form can be written as following.

where,  Young's moduli in the plane of isotropy and perpendicular

to it, respectively.

shear modulus for the plane of isotropy

shear modulus for a plane normal to the plane of isotropy.

Poisson's ratio which characterizes the transverse strain reduction

in the plane of isotropy due to a tensile stress in a direction

normal to it

Poisson's ratio which characterizes the transverse strain reduction

in the plane of isotropy due to a tensile stress in the same plane.

Note that the five independent constants are chosen to be [is not independent]

For an isotropic linear elastic material(), the same results are derived.

< Example >

The stress tensor  at a given point in a transversely isotropic material is given below. the material elastic moduli refered to material principal axes ,  and  are

,

,

(a) The components of the strain tensor  at the given point

(b) The corresponding value of the strain energy density

in-lb/in3(units of energy or work per unit volume)

Homework #6.  Due to 4 June 2007

Problem 6.1

Using Eq.  and the stress-strain relations of an isotropic linear elastic material, show that the equations of equilibrium , can be written in the following form(these equations are known as Navier's displacement equations)

in which  and  are poisson's ratio and shear modulus, respectively.

Problem 6.2

For an isotropic linear elastic material, the stress components,  , at a point are given by

The material constants,  and

Determine the following

(1) The strain deviatoric tensor, , at the given point

(2) The value of the strain energy density  and the complementary energy

density , corresponding to the given state of stress.

(3) The principal strains ,  and  at the same point.

Problem 6.3

For an orthotropic linear elastic material, the state of stress at a certain point refered to material axes of symmetry is given by

The material constants,

,    ,

,    ,

,    ,

(1) Find the strain components, , at the given point

(2) Find principal stress ,  and   and

the principal strain ,  and  at the point.

(3) Calculate the strain energy density , corresponding to the given state of

stress.

Problem 6.4

Prove the following relations between the elastic moduli,  and

7.7 Nonlinear Elastic Stress-Strain Relations

Many engineering materials exhibit nonlinear mechanical behavior, even when sustaining small deformations. Examples are materials such as concrete, rocks, and soils. This type material( or physical) nonlinearity had led to the development of various nonlinear elastic constitutive laws.

Many different approaches are used in the formulation of nonlinear elastic constitutive relations. For examples, some approach is based on an assumed series expansion  of the elastic energy functions  or  in terms of higher orders of the strain or stress components( or their invariants). Other approach, the constitutive relations of linear elastic models are generalized(modified) to represent nonlinear material behavior. Only isotropic materials are considered in the following discussion.

7.7.1 Isotropic Nonlinear Elastic Stress-Strain Relations Based on

Function  or

For an isotropic elastic material. the strain energy density function  can be expressed in terms of any three independent invariants of the strain tensor .

Choosing the three inveriants,  defined below,

can be written as

whrer the invariants  are given by

Then, we have

or, substituting for   and carrying out the differentiation, we get

where,

From above Eq., the function (material strain functions)are related by the three equations;

It should be noted that the choice of the three independent strain invariants appearing in Eqs is arbitrary. Instead, one may use the invarinats , or the invariants  of the strain deviator tensor , or even mixed invariants such as  . The particular advantage of the choice here is the separation of functions  in a simple convenient manner.

As an example of a typical constitutive equation. consider  to be written as a polynomial expansion of invarinats .  If the terms in  from the second to the forth order in strain are retained. one can express .

where,  are constants.

when these functions are substituted in Eq., we can obtain a 'cube' stress-strain relation. It remains only to determine the nine constants .

In a similar procedure, one may obtain different constitutive laws from Eq. by expanding the function  in terms of the stress invariants ( or ). Thus if we choose the following stress invariants(similar to those defined for strains) ;

the constitutive laws is

where,

and the constraints on the material stress function are given by the three relations.

It is to be emphasized that the behavior of isotropic models described in Eqs. is reversible and path independent, as in the linear elastic models, since the state of strain(stress) is uniquely determined by the current values of stresses(strains) without regard to the history of loading.

< Example >

The behavior of an isotropic nonlinear elastic material is described by the following assumed polynomial expression for the function .

where a, b, c are material constants and are the stress invariants .

(a) Derive the stress-strain relations for this material in terms of the constants a, b, and c

(b) Plot the stress-strain relation in simple tension( and all other

components are zero).

Find an expression for the tangent Young's modulus  , defined as

in simple tension, in terms of the stress .

What is the value of the initial tangent modulus, (at ) ?

(c) Show that the constitutive equations are reduced to those of the isotropic

linear elastic material for c=0.

Find the relations between the constants a, b, and the elastic moduli        for this case.

<Solution>

(a)

, ,

let, (),  ,

and

or

or

(b) For simple tension,

,

, ,

For  and

The tangent Young's modulus  , defined as

The initial tangent modulus

(c) For c=0,

The strain component for the simple tension case is calculated

With  ,  ,

Combining these results,

These equations can be solved for a and b in terms of  and  .

Or, utilizing the relations among the elastic moduli,

→   The energy density is composed of a volumetric energy term() and an energy of shear or distortion(), which are independent of each other.

7.7.2 Isotropic Nonlinear Elastic Stress-Strain Relations Based on

Modification of the Linear Elastic Models

The linear elastic stress-strain relations are isotropic and reversible. A simple extension of these relations with the elastic constants replaced by scalar functions associated with either the stress and/or the strain invariants has the property of isotropy and reversibility also. For instance, scalar functions associated with the stress include the values of the three principal ,  or equally well the three independent invariants . Therefore, different scalar functions such as  associated with the stress invariants, or associated with the strain invariants, may be employed to describe various nonlinear elastic constitutive models. The nonlinear stress-strain relations for each of these models reduce to the linear forms when the scalar function are taken to be constants.

As a first example, consider the linear elastic form modified by replacing the reciprocal of Young' modulus  with a scalar function of the invariants  labeled as . Thus one has

Poisson's ratio  also may be replaced by a function of the stress invariants. The above Eqs. are nonlinear stress-strain relations for isotropic elastic material which reduce to the linear forms when  is constant (). They represent elastic (reversible) behavior because the state of strain is determined uniquely by the current state of stress without regard to the history of the loading.

There is, of course,  a neat and logical separation between the mean response and the deviatoric or shear response of the material, exactly as for the linear elastic material. Specifically, one can write as