Theory of Elasticity and Plasticity

한밭대학교 산업대학원

토목공학과

이  형  준

			Body and Surface Force									Displacement
Equilibrium																		Compatibility
				Stresses									Strains
								Constitutive laws

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 of matter : continuous distribution of matter in space

density of the mass distribution at the point P

→ 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.

Elementary Topics through Which Basic Ideas Evolved

Some elementary topics that are simple and useful and that have been fundamental in the history of mechanics.

- Newton's laws of motion, equations of equilibrium, free-body diagrams,

the analysis of a truss, a beam, a block, a plate and a shell,

and the classical beam theory

(1) Newton's Laws of Motion

Newton's laws are stated with respect to material particles in a three-dimensional space the obeys Euclidean geometry.

A material particle has a unique, positive measure, the mass of particle.   The location of the particle can be described with respect to a rectangular Cartesian frame of reference.

Consider a particle of mass m, Let the position, velocity, and acceleration of this particle be denoted by the vector x(or [eqn: vec{x}]), v(or [eqn: vec{v}]) and a(or [eqn: vec{a}]), respectively

[eqn: vec{v} `=` {d vec{x}} over {dt} ```,```` vec{a} `=` {d vec{v}} over {dt} ````]

Let F(or [eqn: vec{F}]) be the total force on the particle.

- Newton's first law :  F = 0 → v = constant

- Newton's second law :  F ≠ 0 →  [eqn: {d} over {dt} m vec{v} `=` vec{F} ``,```or````` vec{F} ``=`m` vec{a``}]

- An equation of equilibrium of two forces

[eqn: vec{F} ``+(-m``` vec{a} )``=`0]

→   [eqn: (-m``` vec{a} )`] : is called the inertial force

→ states that the sum of the external forces acting on a particle  and the

inertial force vanishes  : the inertial force balances the external force

The Newtonian equation of motion stated in this way is called D'Alembert's principle

Consider a system of particles that interact with each other. Every particle is influenced by all the other particles in the system.

Let [eqn: vec{F} _{IJ}] denote the forces of interaction exerted by particle number J on particle number I and [eqn: vec{F _{JI}}] , that of I on particle J

- Newton's third law

[eqn: vec{F _{IJ}} ``=`- vec{F _{JI}} ```````or``` vec{F _{IJ}} ```+`` vec{F _{JI}} ````=`0]

→  If I=J  [eqn: vec{F _{II}} ```=`0] ( I not summed)

.The force [eqn: vec{F _{I}}] that acts on the I-th particle consists of an external forces [eqn: vec{F _{I} ^{{} ^{}} `} ` ^{(e)}], such as gravity, and an internal force that is the resultant of mutual interaction between particles.

[eqn: vec{F _{I}} ```=` vec{F} ` _{I} ` ^{(e)} ``+`` sum _{J=1} ^{K} vec{F _{IJ}} ``]

[eqn: {d} over {dt} m _{I} ` vec{`v} ` _{I} ````=` vec{F} ` _{I} ` ^{(e)} ``+`` sum _{J=1} ^{K} vec{F _{IJ}} ```````````(I=`1,2`,`....,K)`]

(2) Equilibrium

A special motion is equilibrium, i.e. one in which there is no acceleration for any particle of the system.

- At equilibrium

[eqn: ` vec{F} ` _{I} ` ^{(e)} ``+`` sum _{J=1} ^{K} vec{F _{IJ}} `````=`0``````````(I=`1,2`,`....,K)`]

Summing over I from 1 to K

[eqn: sum _{I=1} ^{K} ` vec{F} ` _{I} ` ^{(e)} ``+`` sum _{I=1} ^{K} sum _{J=1} ^{K} vec{F _{IJ}} `````=`0```````````]

In last sum, whenever [eqn: vec{F} _{IJ}] appear, [eqn: vec{F _{JI}}] appears also → they add up to zero

[eqn: sum _{I=1} ^{K} ` vec{F} ` _{I} ` ^{(e)} ````=`0```````````]

===≫ For a body in equilibrium, the summation of all external

forces action on the body is zero.

Let us consider the tendency of a body to rotate. If a body pivoted at a point O and is acted no by a force [eqn: vec{F _{I}}],, then the moment of the force about O that tends to cause the body to rotate about O is given by the vector product [eqn: vec{r} _{I} `` TIMES ``` vec{F _{I}}]., where [eqn: vec{r _{I}} ``] is a radius vector from O to any point on the line of action of the force [eqn: vec{F _{I}}].

Summing over I from 1 to K

[eqn: sum _{I=1} ^{K} ` vec{r} ` _{I} `` TIMES ``` vec{F} ` _{I} ` ^{(e)} ````=`0```````````]

===≫ For a body in equilibrium, the summation of the moments of

all external forces action on the body about any point is zero.

(3) Use of Free-Body Diagram in the Analysis of Problems

If a machine is in equilibrium, every part of it is equilibrium.

A diagram of the part with all the external forces acting on it clearly indicated is called a free-body diagram. The method we use is therefore called the free-body method.

Example 1. Analysis of Truss

○ Compute the reactions at the two supports

→ Consider the whole truss as a free body

The conditions of equilibrium are ;

- Summation of vertical forces is zero, [eqn: sum _{} ^{} V``=`0]

[eqn: W``-``R _{L} ``-`R _{R} ``=`0]

- Summation of moments about the point a is zero, [eqn: sum _{} ^{} M _{a} ``=`0]

[eqn: W`` BULLET 3L``-`R _{R} ` BULLET 6L``=`0]

⇒  [eqn: R _{L} ``=`R _{R} ``=`W//2]

○ Compute the force(tension or compression) in the member ab and ac

→ Consider a free body ; cut through ab and ac

The conditions of equilibrium are ;

- Summing all the forces in vertical direction, [eqn: sum _{} ^{} V``=`0]

[eqn: `` {W} over {2} ``+F _{ab} ``sin theta ````=`0]

→  [eqn: theta `=`45 ^{o} `,`sin theta `=` sqrt {2} /2,````````` -> ``F _{ab} `=-W/ sqrt {2} `]

- Summing all the forces in horizontal direction, [eqn: sum _{} ^{} H``=`0]

[eqn: ``F _{ab} ``co theta ``+`F _{ac} ``=`0]

→  [eqn: theta `=`45 ^{o} `,`cos theta `=` sqrt {2} /2,````F _{ab} `=W/ sqrt {2} ````` -> ``F _{ac} `=W/2]

○ Compute the force in the member df, ef and eg

→ Consider a free body ; cut passing through f' and g'

The conditions of equilibrium are ;

- Summation of all the horizontal force vanish, [eqn: sum _{} ^{} H``=`0]

[eqn: ``F _{df} ``+`H _{ef} ``+``F _{eg} ``=`0]

- Summation of all the vertical force vanish, [eqn: sum _{} ^{} V``=`0]

[eqn: `` {W} over {2} ``+V _{ef} ````=`0]

- Sum of moments about the point e vanish, [eqn: sum _{} ^{} M _{e} ``=`0]

[eqn: `` {W} over {2} BULLET 2L``+F _{df} BULLET L+F _{ef} BULLET 0+F _{eg} BULLET 0```=`0]

→ [eqn: ``F _{df} `=-W]

- Sum of moments about the point f vanish, [eqn: sum _{} ^{} M _{f} ``=`0]

[eqn: `` {W} over {2} BULLET 3L``-F _{eg} BULLET L``=`0]

→ [eqn: ``F _{eg} `=3W/2]

→  [eqn: `V _{ef} `=-W/2``,`````H _{ef} `=-W/2]

A similar calculation can be done for other members of the truss.

Example 2. A Simply Supported Beam

○ Compute the reactions and

the bending moments in the beam,

The conditions of equilibrium are ;

- Sum of all forces in the horizontal

direction vanish  [eqn: sum _{} ^{} H``=`0]

[eqn: H``=`0]

- Sum of all forces in the vertical direction vanish  [eqn: sum _{} ^{} V``=`0]

[eqn: S``=`` {W} over {2}]

- Sum of the moments of all forces about the left end support vanish

[eqn: M=S``x``=( {W} over {2} )`x]

Example 3.  Stresses in a Block

○ Consider a block of solid material

compressed by a load W acting

on the end.

- The stress on a plane z=0

[eqn: sigma ``=`- {W} over {A}]

- Let us make a cut with a plane that

is inclined at an angle [eqn: theta] to the xy-plane

Let [eqn: vec{nu }] be the unit normal vector of

the inclined plane, and [eqn: vec{z}] be a unit

vector in the direction of the z-axis.

[eqn: vec{nu } ` BULLET ` vec{z} ``=`cos( vec{nu } `,` vec{z} `)``=`cos` theta]

Consider the upper half of the block as a free body,

- The balance of forces requires that the resultant force acting on the plane [eqn: vec{nu }] is exactly equal to -W

[eqn: N``=``-W`cos` theta `,```S``=``-W`sin` theta]

- The cross-sectional area of the block cut by the plane [eqn: vec{nu }] is [eqn: A/cos theta]

- The normal and shear stress

[eqn: sigma _{n} ``=``-`cos ^{2} ` theta `W/A`,`````` tau ``=``-sin` theta `cos theta ``W/A]

The normal [eqn: sigma _{n}] and shear stress [eqn: tau] vary with angle [eqn: theta]

→ The normal [eqn: sigma _{n}] is a maximum when [eqn: theta `=`0], whereas the shear stress [eqn: tau]

reaches a maximum when [eqn: theta =45 DEG], and [eqn: tau _{max} ``=` {1} over {2} W/A]

Example 5.  A Pressurized Spherical Shell

○ Consider the wall an inflated balloon

Let the inner radius of the shell be [eqn: r _{i}] and the outer radius be [eqn: r _{o}], the thickness of wall be [eqn: h=`r _{o} `-`r _{i}] and the internal pressure [eqn: p _{i}] acting on the inner wall.

The resultant pressure force acting on a hemisphere is [eqn: pi `r _{i} ^{2} p _{i}]. The area of the wall on the diametrical plane is [eqn: pi `r _{o} ^{2} ``-` pi r _{i} ^{2}] .

The normal stress in the wall of the shell is not uniform, To calculate the average tensile stress in the wall <[eqn: sigma]>

The balance of the forces in equilibrium requires,

[eqn: pi `(r _{o} ^{2} ``-`r _{i} ^{2} )`< sigma >``=` pi `r` _{i} ^{2} `p _{i}]

or   [eqn: < sigma >`=``p _{i} {r _{i} ^{2}} over {r _{o} ^{2} ``-`r _{i} ^{2}} ``=`p _{i} {r _{i} ^{2}} over {h`(r _{o} ^{} ``+`r _{i} ^{} )}]

This is useful formula that is valid for thick-walled, as well as thin walled, spherical shells.

If a pressure [eqn: p _{o}] acts on the outside of the wall.

The resulting normal stress in the wall will be

[eqn: < sigma >`=``- {r _{o} ^{2} ```p _{o}} over {h`(r _{o} ^{} ``+`r _{i} ^{} )}]

If the shell is subjected to both an internal and an external pressure, and the wall of the sphere is very thin, then [eqn: r _{o} `-r _{i} `=`h``;`r _{o} `` CONG `r _{i} ``=`r]

[eqn: < sigma >`=``- {r`p _{i}} over {2h} ``-` {r``p _{o}} over {2h}]

(4) Simple Beam Theory

Consider a prismatic beam of a uniform, isotropic Hookean material with a rectangular cross section subjected to a pair of bending moments of magnitude M acting on its end.

Consider two neighboring cross sections A nad B that are perpendicular to the plane y=0 when the beam is unloaded. When the beam is bent into a circular arc, the two  planes A and B are deformed into plane A' and B' that remain normal to the arc. That A' and B' are planes is because of symmetry. That they are perpendicular to the centroidal arc is also because of symmetry.      Let the radius of curvature of the centroidal are be R.

When the cross sections A' and B' are bent to a relative angle of [eqn: d theta], the centroidal arc length is [eqn: Rd theta]; whereas a line at a distance y above the centriodal line will have a length [eqn: (R`+`y)`d theta].

- The strain

[eqn: epsilon _{xx} `=` {(R+y)d theta ``-`R`d theta } over {R`d theta } `=` {yd theta } over {R`d theta } `=` {y} over {R`}]

In response to the strain [eqn: epsilon _{xx} `], there will be a stress [eqn: sigma _{xx} `]

The assumption that [eqn: sigma _{xx} `] is the only nonvanishing component of stress, whereas [eqn: sigma _{yy} =` sigma _{zz} = tau _{xy} = tau _{yz} = tau _{zx} =0].

Then according to Hooke's law,

[eqn: sigma _{xx} `=`E` epsilon _{xx} ``=`E`` {y} over {R}]

Since, only pure bending moments act on the beam, the resultant axial force must vanish.

[eqn: int _{A} ^{} {sigma _{xx} `dA} ``=0]

A : the cross section

dA : an element of area in the cross section

[eqn: int _{A} ^{} {sigma _{xx} `dA} ``= int _{A} ^{} {E`` {y} over {R} dA} `= {E} over {R} int _{A} ^{} {y`dA} ``=`0]   →  [eqn: int _{A} ^{} {y`dA} ``=`0]

→ Which says that the origin(y=0) must be the centroid of the cross section. This is explains our original choice of the centroid as origin.

The centroidal plane y=0 is unstressed during bending. Material particles on it are not strained in the axial direction. This plane is called neutral surface of the beam.

The resultant moment of the bending stress [eqn: sigma _{xx} `] about the z-axis must be equal to the external moment M.

[eqn: M``=`` int _{A} ^{} {y` sigma _{xx} `dA}]   →  [eqn: M``=``` {E} over {R} int _{A} ^{} {y ^{2} `dA}]

area moment of inertia of the cross section : [eqn: I``=``` int _{A} ^{} {y ^{2} `dA}]

[eqn: M``=``` {E`I} over {R} ,``` {M} over {EI} ``=` {1} over {R} ``=` phi ,````` sigma _{xx} ``=` sigma _{o} ` {y} over {c} ``,``` sigma _{o} ``=`` {M`c} over {I}]

← [eqn: c`=`h/2`,```````I``=`bh ^{3} /12]

(5) Deflection of Beams

Consider the beam under a lateral load and calculate the deflection of a beam.

Let the beam deflection curve be [eqn: y(x)]. When [eqn: y(x)] is small, its curvature can be approximated by [eqn: d ^{2} y/dx ^{2}].

[eqn: {d ^{2} y} over {dx ^{2} `} ``=` {1} over {E`I} `M`(x)]

The beam deflection [eqn: y(x)] can be obtained by solving this equation with approximate boundary conditions,

- Simple supported end : deflection and moment vanish at x=0 and x=l

[eqn: y``=``0`,````` {d ^{2} `y} over {d`x ^{2}} ``=``0]

- Clamped end : deflection and slope vanish at x=0 and x=l

[eqn: y``=``0`,````` {d`y} over {d`x ^{}} ``=``0]

- Free end : moment and shear specified

[eqn: E`I``` {d ^{2} `y} over {d`x ^{2}} ``=``M``,`````````E`I``` {d ^{3} `y} over {d`x ^{3}} ``=``S``]

The bending moment M, the transverse shear S, and the lateral load per unit length w are related by the equations.

[eqn: {d`M} over {d`x} ``=``S``,`````````` {d`S} over {d`x} ``=```w]

If the curvature is small(so that the preceding analysis can be valid), but the slope is finite, then we should use the exact expression for 1/R, which leads to the following equation in place.

[eqn: {d ^{2} y} over {dx ^{2} `} LEFT [ 1``+`` LEFT ( {d`y} over {d`x} RIGHT ) ^{2} RIGHT ] ^{-3/2} ``=` {1} over {E`I} `M`(x)]

→ Geometric Nonlinearity : Large deflection

As an example, consider the small deflection of a cantilever beam clamped at the left end and subjected to a constant bending moment.

The following equation can be integrated,

[eqn: y(x)`=` {M} over {E`I} `` {x ^{2}} over {2} ``+`A`x``+``B]

A, B : arbitrary constant

The boundary conditions [eqn: y`=`d`y/d`x`=`0] at x=0   → A = B = 0

In this special case, the boundary conditions on the free end are also satisfied because M = constant.

Can we, however, satisfy all the boundary conditions in general?. After all, our beam has two ends with two conditions each, so that we have four boundary conditions, whereas our differential equations is only of the second order.
Are we going to have a sufficient number of arbitrary constants to satisfy all boundary conditions?. A further reflection tells us, however, that for a general loading, the differential equation must be obtained.

Thus the general equation must be

[eqn: w```=` {d`S} over {d`x} ``=```` {d ^{2} `M} over {d`x ^{2}} ``=`` {d ^{2}} over {dx ^{2}} LEFT ( E`I`` {d ^{2} `y} over {`d`x ^{2}} RIGHT )]

which is a fourth-order differential equation, able to handle four boundary conditions.

In general case a uniform beam,

[eqn: E`I`` {d ^{4} `y} over {d`x ^{4}} ``=``w`(x)]

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.

form a right-handed system

,     ,

※  Homework # 1

Problem 2.1, 2.3, 2.5, 2.7, 2.9, 2.11

2.2 Vector Equations

A particle on which the forces   act.

The condition of equilibrium

Equation of plane

the vector r  represents a plane if n  is a unit vector and p is a constant

r · n = p

→ The scalar product r · n  represents the scalar projection of r on n

p :constant

n  : unit normal vector

of the plane

r  : vector represent a plane

⇒ a, b, c : component of the unit normal vector n  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  : Square matrix

--- A number : the sum of all the products of the elements of the matrix

Permutation symbol.

vanishes whenever the values of any two indices coincide;

when the subscripts permute as 1,2,3;  otherwise

The  identify

Kronecker delta

Differentiation

Let, be a function of n variables

※  Homework # 1 - continued

Problem 2.16, 2.19, 2.21

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 dot 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.

,    ,

※  Homework # 1 - continued

Problem 2.31, 2.34, 2.36, 2.37