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ADAMA UNIVERSITY ADAMA UNIVERSITY SCHOOL OF ENGINEERING AND INFORMATION TECHNOLOGY SCHOOL OF ENGINEERING AND INFORMATION TECHNOLOGY

MECHANICAL AND VEHICLE ENGINEERING DEPARTMENT MECHANICAL AND VEHICLE ENGINEERING DEPARTMENT

POST GRADUATE PROGRAMPOST GRADUATE PROGRAMMANUFACTURING ENGINEERINGMANUFACTURING ENGINEERING

TERM PAPER PRESENTATION ON

STRESSSTRESS -- STRAIN RELATIONSTRAIN RELATION

PRESENTER TADELE BELAY PRESENTER TADELE BELAY



Descriptions1. Description of Stress-Strain Relation for Elastic Deformation

1.1 Mohr¶s Circle Equations

1.2 Strain

1.3 Small strain deformations

1.4 The strain Tensor

1.5 Elastic Deformation

1.6 Strain Energy

2. Description of Stress-Strain Relation for Plastic Deformation

2.1 Plastic Work

2.2 Effective Stress

2.3 Effective Strain

2.4 Flow Rules

Review Exercises



Introduction:

Stress is the intensity of force and strain is a measure of the amount of

deformation. Experiences shows that, all solid materials can be deformed when

subjected to external load.

It is further found that up to certain limiting loads a solid will recover its

original dimensions when the load is removed.

The recovery of the original dimensions of a deformed body when the load is

removed is known as Elastic Behavior .



For most materials, as long as the load does not exceed the elastic limit, the

deformation is proportional to the load. This relationship is known as Hookes

law.

It is more frequently stated as stress is directly proportional to the strain.

Hookes law requires that the load deformation relationship should be linear.

However, it does not necessarily follow that all materials which behave

elastically will have a linear stress-strain relationship. Rubber is an example of

a material with a non linear stress-strain relationship that still satisfies the

definition of an elastic material.



1. Description of Stress-Strain Relation for Elastic Deformation

Elastic deformations in metals are quite small and require very sensitive

instruments for their measurement.

Ultrasensitive instruments have shown that the elastic limits of the metals aremuch lower than the values usually measured in engineering tests of

materials.



Average strain which is the ratio of change

in length to the original length can berelated by the following representation;

As shown on fig. , the external load P is

balanced by the internal resisting force

where W is the stress normal to the cutting

plane and A is the cross sectional area of the

bar. The equilibrium equation is



Stress

There are nine components of stress as shown in Figure.

A normal stress component is one in which the force is acting normal to the

plane. It may be tensile or compressive.

A shear stress component is one in which the force acts parallel to the plane.

Stress components are defined with two subscripts. The first denotes the normal

to the plane on which the force acts and the second is the direction of the force.



For example,

xx is a tensile stress in the x -direction.

Shear stress acting on the x -plane in the y -

direction is denoted xy

.

Repeated subscripts (e.g., xx , yy , zz)

indicate normal stresses.

Mixed subscripts (e.g., zx , xy , yz) denote

shear stresses



A state of stress in tensor notation is expressed as;

The use of the opposite convention should cause no problem because i j = ji

where i and j are iterated over x, y, and z.

Except where tensor notation is required, it is simpler to use a single subscript for a

normal stress and denote a shear stress by .

For example, x xx and xy xy.

In the same manner,

Pairs of shear stresses with the same subscripts in reverse order are always equal

(e.g., i j = ji ).



Stress transformation;

The general equation for transforming the stresses from one set of axes

(e.g., n, m, p) to another set of axes (e.g., i, j, k ) is

Here, the term im is the cosine of the angle between the i and m axes

and the term jn is the cosine of the angle between the j and n axes. This

is often written as



Consider transforming stresses from the x , y , z axis system to the x , y , z

system shown in Figure



Stresses represented by the same subscripts represent the principal stresses,

where we can easily represent by a single subscript like; Wx, Wy, and Wz. They act

perpendicular to a plane. The magnitudes of the principal stresses, p, are theroots of;

where I 1, I 2, and I 3 are called the invariants of the stress tensor. They are

The first invariant I 1 = íp/ 3 where p is the pressure. I 1, I 2, and I 3 are

independent of the orientation of the axes. Expressed in terms of the principal

stresses, they are;



In the special cases where two of the three shear stress terms vanish (e.g., yx =

zx = 0), the stress z normal to the xy plane is a principal stress and the other

two principal stresses lie in the xy plane. This is illustrated in Figure

For these conditions xz = y z = 0, yz = zx = 0, x x = y y = cos J, and xy =í yx = sin J. Substituting these relations into equations results in;

From trigonometric general relations we have;

By substituting the above values we will obtain the following relations;



If is set to zero, equation can be re-arranged like the following;

The principal stresses can be calculated as;



Depending on the above equation, the following important facts will figure out;

The maximum and minimum values of normal stress on the oblique plane

through point O occur when the shear stress is zero.

The maximum and minimum values of both normal stress and shear stress

occur at angles which are 90O apart.

The maximum shear stress occurs at an angle halfway between the maximum

and minimum normal stresses.

The variation of normal stress and shear stress occurs in the form of a sine

wave, with a period of J =180o. These relationships are valid for any state of

stress.



Mohr¶s circle (three dimensional) for various states of stress for uniaxial

tension and compression

Mohr¶s circles(3D) for various states of stress. Biaxial tension and triaxial

tension (unequal)



Mohr¶s circles (3D) for various states of stress. Uniaxial tension plus biaxialcompression.



For the limiting case of equal triaxial tension (hydrostatic tension) Mohrs circle

reduces to a point, and there are no shear stresses acting on any plane in the body.

The effectiveness of biaxial and triaxial tension stresses in reducing the shear stresses

results in a considerable decrease in the ductility of the material because plastic

deformation is produced by shear stresses. Thus, brittle fracture is invariably associated

with triaxial stresses developed at a notch or stress raiser.

However fig e shows that, if compressive stresses are applied lateral to a tensile stress,

the maximum shear stress is larger than for the case of either uni-axial tension or

compression. Because of the high value of shear stress relative to the applied tensile

stress the material has an excellent opportunity to deform plastically without fracturing

under this state of stress.

For example: greater ductility is obtained in extrusion through a die than in simple

uniaxial tension because the reaction of the metal with the die will produce lateral

compressive stresses.



STRAIN

The displacement of points in a continuum may result from rigid body

translation, rotation and deformation. The deformation of a solid may be

made up of dilatation, change in volume, or distortion, change in shape.Two types of strains are found;

�Engineering Strain or Nominal strain

It is denoted by e;

True strain or logarithmic strain

It is represented by

The true strain and engineering strains are almost equal when they are small.

Look by expanding the above equation;



True strains are more convenient than engineering strains due to:

yTrue strains for an equivalent amount of tensile and compressive

deformation are equal except for sign.

yTrue strains are additive.

yThe volume strain is the sum of the three normal strains.

Small s

train deforma

tions

Small deformations are the province of elastic theory, while larger

deformations are treated in the disciplines of plasticity and hydrodynamics.

The equations developed in this section are basically geometrical;



Considering the above figure, a solid body in fixed coordinates at x, y, and z.

When the body displaces from Q to Q¶, the new coordinates will be: x+u, y+v,

z+w. The components of the displacement are u, v, and w.

In the undeformed state points A and B are separated by a distance dx. Sincedisplacement u, in this one dimensional case, is a function of x, B is displaced

slightly more thanA since it is further from the fixed end.

The normal strain is given by



The normal strain, ex, is defined as The shear strains are associated

with the angles between AD and

A¶D¶ and between AB and A¶B¶.For small deformations



To generalize this to three dimensions,

each of the components of the

displacement will be linearly related to

each of the three initial coordinates of the point.

Where the above strain coefficients

which can be identified as principal

stress is described as;

For the angular distortion of the axis;



The total shear strain is the sum of these two angles,

In general, displacement components such as exy, eyx « etc., produce both

shear strain and rigid body rotation.



The total shear strain is the sum of these two angles,

Similarly,

This definition of shear strain,K, is equivalent to the simple shear measured ina torsion of shear test.

The strain Tensor

If tensor shear strains are defined as;

Small shear strains from a tensor,



Because small strains form a tensor, they can be transformed from one set

of axes to another in a way identical to the transformation of stresses.

Mohr¶s circle relations can be used.

It must be remembered, however, that

and that the transformations hold only for small strains. If yz = zx = 0,

And

The principal strains can be found from the Mohr¶s circle equations for strains,



Strains on other planes are given by;

Elastic Deformation

Up to now discussion of stress and strain has been perfectly general and

applicable to any continuum.

Hooke¶s law states that;

Where E is the modulus of elasticity in tension or compression. While a tensile

force in the x direction produces an extension along that axis, it also produces

a contraction in the transverse y and z directions.



The transverse strain has been found by experience to be a constant fraction of

the strain in the longitudinal direction. This is known as Poisson¶s ratio, and

denoted by the symbol v.

Only the absolute value of v is used in calculat ions. F or most metal s the value

of v ar e cl ose t o 0.33.

To develop the stress-strain relations for a three dimenstional state of stress,

consider a unit cube subjected normal stresses and shearing stresses .

Because the elastic stresses are small and the material is isotropic, we can

assume that normal stress Wx does not produce shear strain on the x,y or z

planes and that a shear stress Xxy does not produce normal strains on the x, y or

z planes.



Applying the principle of superposition to determine the strain produced by

more than one stress component. For example, the stress Wx produces a normal

strain Ix and two transverse strains Iy=-v Ix

and Iz

= -v Ix



Hooke¶s laws can be expressed as

The shearing stresses acting on the unit cube produce shearing strains;

The proportionality constant G is the modulus of elasticity in shear, or the

modulus of rigidity. Values of G are usually determined from a torsion test.

Still another elastic constant is the bulk modulus or the volumetric modulus

of elasticity K . The bulk modulus is the ratio of the hydrostatic pressure to

the dilatation that it produces;



Where ±p is the hydrostatic pressure, ( is the volumetric strain and F is the

compressibility

Many useful relationships can be derived between the elastic constants E, G, v

and K. For example:

Another important relationship is the expression relating E, G and v. This

equation is usually developed in a first course in strength of materials;



Typical room-temperature values of elastic constants for isentropic materials

Modulus of elasticity = E

Modulus of rigidity (shear modulus) = G

Poisson¶s ratio = v



Strain Energy

The elastic strain energy U is the energy expended by the action of external forces

in deforming an elastic body. Essentially all the work performed during elastic

deformation is stored as elastic energy, and this energy is recovered on the release

of the applied forces.

Energy or work is equal to a force multiplied by the distance over which it acts.

In the deformation of an elastic body, the force and deformation increase linearly

from initial values of zero so that the average energy is equal to one half of their

product. This is also equal to the area under the load deformation curve;



For an elemental cube that is subjected to only a tensile stress along the x-axis,

the elastic strain energy is given by

The above equation describes the total elastic energy absorbed by the

element. SinceA

dx is the volume of the element, the strain energy per unitvolume or strain energy density Uo is given by;

Note that the lateral strains which accompany deformation in simple tension

do not enter into the expression for strain energy because forces do not exist

in the direction of the lateral strains.



By the same type of reasoning, the strain energy per unit volume of an

element subjected to pure shear is given by;

The elastic strain energy for a general three dimensional stress distribution

may be obtained by superposition.

Or in tensor notation

Substituting the Hooke¶s law in the above equation for strain;



Stress-strain relationship in plastic deformation

With elastic deformation, a body returns to its original shape when the stress is

removed and the stress and strain under elastic loading are related through

Hooke¶s laws. Any stress will cause some strain.

In contrast, no plastic deformation occurs until the stress reaches the yield

st r eng th. For ductile metals large amounts of plastic deformation can occur under

continually increasing stress.

In plastic deformation, the deformation is an irreversible process, strain

hardening usually occurs as a result of the deformation, and anisotropy may

occur with continued deformation.



Elastic deformation depends only on the initial and final state of stress and

strain, where as plastic deformation also depends on the stress-strain path and

may be inhomogeneous.

In elastic deformation usually small strains are involved, and the principal axes

for stress and strain coincide.

There are two theories of plasticity that are commonly used:

The flow or incremental theory as expressed by the Levy-Mises

equations

The deformation theory in which proportional loading is assumed



PLASTIC WORK

The differential amount of plastic work per volume associated with tensile

strain d of a bar of leng th l 0, subjected t o a for ce act ing on an ar ea is;

In the most general case where there are three normal stresses and three shear

stresses the plastic work per volume is

In term of principal stresses and strains,



It is useful to define an eff ect ive st r ess, , for a yield criterion such that

yield ing occurs when the magnitude of r eaches a crit ical value.

EFFECTIVE STRESS

This can also be expresses as;

For the Tresca criterion,



EFFECTIVE STR AIN

The eff ect ive st r ain, , is defined such that the incr emental plast ic work per

volume is

The von Mises effective strain may be expressed as



FLOW RULES ( LEVY-MISES EQUATION)

The strains that result from elastic deformation are described by Hooke¶s law.

There are similar relations for plastic deformation, called the f l ow r ules.

Stress-strain relationships that describe the path of plastic deformation of a

material are called flow rules.

In the most general form the flow rule may be written

where f is the f unct ion of i j that describes yield ing ( i.e., the yield criterion.) It

is related to what has been called the plastic potential. For the von Mises

criterion, differentiation results in ;



Conclusion

Elastic Deformation Plastic Deformation

Hooke¶s law is considered; Flow rules is applied



Exercise 1:

Consider a stress state with Wx=70MPa, Wy=35MPa, Xxy= 20MPa,

Wz=Xzx=Xyz=0. Find the principal stresses.

Solution:

But,

Formula:

Hence;

Ans:



Exercise 2

Determine the principal stresses for the state of stress

MPa

Solution:

Formula:

Analysis:

ANS



Exercise 3

Using the data¶s given on Exercise 1, Wx=70MPa, Wy=35MPa, Xxy= 20MPa,

Wz=Xzx=Xyz=0 and the principal stresses W1=79.1MPa, W2=25.9MPa and W3=0.Considering E = 61GPa and v = 0.3, calculate I1 and I2 by using Mohr¶s circle

equations

Solution:

Analysis:



Now using Mohr¶s circle



Exercise 4



Exercise 5

The state of stress is given by Wx

= 25P and Wy=5P plus shearing stresses X

xy. On

a plane at 45o counterclockwise to a the plane on which Wx acts the state of

stress is 50MPa tension and 5MPa shear. Determine the values of Wx,W

y,X

xy.

Solution:

Formula:



R eferences

Metal Forming, Mechanics and Metallurgy, Third Edition, William F. Hosford andRobert M. Caddell, Cambridge University

Metal Working Science and Engineering, Edward Mielnik, Part ± I

Mechanical Metallurgy, SI Metric Edition, George E. Dieter, Mc-Graw Hill

Company

Applied Metal Forming Including FEM Analysis, Henry S. Valberg, Norwegian

University of Science and Technology

Modeling of Metal Forming and Machining Processes by Finite Element and Soft

Computing Methods Engineering Materials, Prakash M. Dixit and Uday S. Dixi

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