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Theory of metal forming

Mechanical metallurgy is concerned with the response of metals to forces or loads. The objectivemay be to convert the cast ingot into a more useful shape, such as a flat plate, and hence it is

necessary to know the conditions of temperature and rate of loading which minimizes the forceneeded to do the job.The material is considered to be homogenous whose mechanical behaviour is precisely defined onthe basis of a very few material constants; otherwise understanding the response of materials toforces will involve many disciplines.

1.1Elastic and Plastic behaviour

All solid materials can be deformed when subject to external load. Upto certain limiting load, asolid will recover its original dimensions when the load is removed and this is the elastic behaviour.The limiting load beyond which the material no longer behaves elastically is the elastic limit. Onexceeding this limit, the solid will experience a permanent set or deformation when the load is

removed and the body is said to have undergone plastic deformation.

For most materials, as long as the load does not exceed the elastic limit, the deformation is proportional to the load, according to Hook’s law, i.e., stress is proportional to strain.

Elastic deformation in metals is quite small and requires very sensitive instruments for their measurement.

1.2Average Stress and Strain

Assume two gage marks put on the surface of the bar in its unstrained condition and Lo is the gage

length. A load P is applied to one end of the bar giving a slight increase in the length and decreasein diameter.The average linear strain e = δ/Lo = L-Lo/LoStrain is a dimensionless quantity.

In the free body diagram of the bar, the external load P is balanced by the internal resisting force ∫σdA, where σ is the stress normal to the plane of c/s area A.

P = ∫σ dA= σ∫dA = σA

So σ = P/A

This equation gives average stress over the area A.The inherent anisotropy between grains of a polycrystalline material rules out uniformity of stressover a body of macroscopic size. The presence of more than one phase also gives rise tononuniformity of stress on a microscopic scale.Below the elastic limit the Hook’s law can be considered to be valid, so that the average stress is proportional to average strain, σ / e = E = constantThe constant E is the modulus of elasticity or Young’s modulus.

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1.3Tensile deformation of a ductile material

The basic data on the mechanical properties of a ductile metal are obtained from a tension test, inwhich a suitably designed specimen is subjected to increasing axial load until it fractures. The load

and elongation are measured at frequent intervals during the test and are expressed as average stressand strain, as in figure. The figure shows a typical stress-strain curve for a metal such as Al or Cu.The initial linear portion of the curve OA is the elastic region within which Hook’s law is obeyed.Point A is the elastic limit, defined as the greatest stress that the metal can withstand withoutexperiencing a permanent strain when the load is removed. The determination of elastic limit istedious and it is often replaced by the proportional limit A’ – the stress at which the stress-straincurve deviates from linearity.

For engineering purposes the limit of usable elastic behaviour is described by the yield strength, point B. the yield strength is defines as the stress that will produce a small amount of plasticdeformation( say 0.2%). In fig. permanent strain or offset is OC.

Plastic deformation begins when the elastic limit is exceeded. As the plastic deformation of thespecimen increases, the metal becomes stronger (strain hardening) so that the load required tofurther extend the specimen increases. Eventually the load reaches a maximum value, and the stressat this point is the ultimate tensile strength.For a ductile material the diameter of specimen decreases rapidly beyond the maximum load, sothat the load required to continue deformation drops off until the specimen fractures.

Ductile Vs Brittle behaviour – depending on ability of the material to undergo plastic deformation. A brittle material fractures at the elastic limit – but white C.I shows some measure of plasticity before fracture. Adequate ductility is an important engineering consideration, because it allows thematerial to redistribute localized stresses. With brittle materials, localized stresses continue to buildup when there is no local yielding. Finally crack forms at one or more points of stress concentrationand spreads rapidly over the section. Even in the absence of stress concentration, the brittle materialwill still fracture suddenly, because the yield stress and tensile strength are identical. Tungsten – brittle at room temp., but ductile at elevated temperature

Steel – ductile at room temperature, but brittle due to hydrogen embrittlement

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Fig. typical stress strain curve for (a) brittle material (b) brittle material with slight ductility.

Fig. typical stress strain curve of ductile material

1.4What constitutes failure?

1. Excessive elastic deformation2. yielding or excessive plastic deformation3. Fracture

Two general types of excessive deformation may occur(1) excessive deflection under conditions of

stable equilibrium, such as the deflection of beams under gradual load; (2) sudden deflection or buckling, under conditions of unstable equilibrium.Excessive elastic deformation of a machine part may mean failure of the machine; for example, ashaft which is too flexible can cause rapid wear of bearings – or excessive deflection of closelymating parts can result in interference and damage. The sudden buckling type failure may result incase of slender beams when axial load exceeds Euler load. Failures due to excessive elasticdeformation are controlled by Elastic modulus, not by the strength of the material. Stiffness of amember can be effectively increased by changing the shape or the dimension.

A’

A

B

σmax

fracture

CO

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Yielding or excessive plastic deformation occurs when the elastic limit of the material has beenexceeded. Yielding may prevent the proper functioning of the part. Failure by plastic deformation iscontrolled by yield strength of the material for uniaxial loading. For more complex loading, yieldstrength is still the parameter but used with a failure criterion.

At higher temperatures the material constantly deforms at constant stress in a time dependentyielding known as creep.

The formation of a crack which can result in complete disruption of the continuity of the member constitutes fracture. Metals fail by fracture in three general ways: (1) sudden brittle fracture (2)fatigue or progressive fracture (3) delayed fracture

Fatigue failures occur in parts which are subjected to alternating or fluctuating stresses. A minutecrack starts at a localized spot(notch) and gradually spreads over the c/s until the member breaks.Fatigue failure occurs without any visible sign of yielding at average stresses below the tensilestrength. Design for fatigue failure is based on empirical relationship.

One common type of delayed fracture is stress-rupture failure, which occurs when metals have been loaded statically at elevated temperature for a long period of time. Depending on the stressand the temperature there may be no yielding prior to failure.

All engineering materials show variability in property. further, uncertainty exists regarding themagnitude of applied load. Accidental loads of high magnitude are also possible. In order to provide a margin of safety, it is necessary that the allowable, safe stresses be smaller than that producing failure. The safe stress is called working stress and is considered as either the yieldstrength or tensile strength divided by a number called factor of safety.

1.5Concept of stress and types of stresses

Stress is defined as the internal resistance of a body to an external applied force per unit area.Figure represents a body in equilibrium under the action of external forces P1,P2, P3, …P5.There are two kinds of external forces on a body – surface forces (hydrostatic pressure or pressureexerted by one body on another) and body forces (forces distributed over the volume of the bodysuch as gravitational, magnetic, or inertia forces).In general the forces will not be uniformly distributed over any cross section of the body. To obtainthe stress at some point O in a plane, a part of the body is removed and replaced by a system of external forces that will retain each point in the other part of the body in its original position as before the removal a part. See figure.

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We then take an area ∆A surrounding point O and note that a force ∆P acts on this area. If the area∆A is continuously reduced to zero, the limiting value of the ratio ∆P/∆A is the stress at the point O

on the given plane of the body.σ =

∆

∆

→∆A

P

A

lim0

The stress will be in the direction of the resultant force P and will be inclined at an angle to ∆A. thestress will be different on ay other plane passing through point O.It is inconvenient to use a stress which is inclined at some arbitrary angle to the area over which itacts. The total stress can be resolved into a normal component σ, perpendicular to ∆A, and ashearing stress τ, lying in the plane of the area considered. Considering the figure, the force Pmakes an angle θ with the normal to the plane of area A. also the plane containing normal z andforce P makes angle φ with y-axis.The normal stress is given by σ = (P/A) cosθThe shear stress in the plane acts along the line OC and has magnitude τ = (P/A) sinθThe shear stress may be further resolved into two components parallel to x and y directions lying inthe planeX direction τ = (P/A) sinθsinφY direction τ = (P/A) sinθcosφTherefore a given plane may have one normal stress and two shear stresses acting on it.

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Fig. resolution of total stress into components

1.6Concept of strain and types of strain

The average linear strain was defined as the ratio of change in length to the original length of samedimensions.

ε = δ/Lo = ∆L/Lo = (L-Lo)/Lowhere ε is average linear strain and δ is deformation.By analogy with definition of stress at a point, the strain at a point is the ratio of the deformation tothe gauge length as gauge length approaches zero. Not only will the elastic deformation of a body result in a change in length of a linear element inthe body but it may also result in a change in the initial angle between any two lines. The angular change in a right angle is known as shear strain. Figure shows pure shear of one face of a cube.The angle at A which was originally 90 degrees is decreased by a small amount θ by the shear stress. The shear strain γ is equal to the displacement a divided by the distance between the planes,h. the ratio a/h is also the tangent of the angle through which the element has rotated. For smallangles, γ = a/h = tanθ = θ

Fig.shear strain

θ

φ

P

C

x

y

z

h

a

θ

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1.7Stresses on an inclined plane

Consider a plane ABC inclined to the three Cartesian coordinate axes x1,x2,x3 and intersecting

them at the points A,B,C respectively. Thus ABCP forms a trapezoid. The stress components actingon the three perpendicular faces of this trapezoid, which lie in coordinate planes are thecomponents of the stress tensor or the stress state at the point P and are completely known.We have to determine the normal and shear stress on the inclined plane ABC.Let S1, S2, S3 be the magnitudes of the three stress components in the direction of the coordinateaxes, on the inclined plane ABC. Let As be the area of the triangular face ABC and A1, A2, A3 bethe area of the triangular faces BPC, CPA, APB respectively.The trapezoid ABCP would be in equilibrium if the resultant forces acting along the three axes arezero. Therefore, we can write the following equations.

S1 As - { σ11. A1 + σ21 .A2 + σ31 .A3} =0

S2 As – { σ12 .A2 + σ22. A2 + σ32 A3} = 0S3 As - { σ13 A3 + σ23 .A3 + σ33 . A3} =0

Now, dividing the above equations by As and substituting n1=A1/As; n2=A2/As; n3=A3/As

S1 = σ11 . n1 + σ21 n2 + σ31 . n3

S2 = σ12 .n1 + σ22 . n2 + σ32 . n3

S3 = σ13 . n1 + σ23 . n2 + σ33 . n3

Because σij = σji, we can write

Si = σij nj (i,j=1,2,3)

The stress normal to the plane is obtained by projecting the three components S1, S2, S3 on to thenormal.

σn = S1 n1 + S2 n2 + S3 n3= Si ni

Based on the previous equation for Si,σn = σij ni nj

The resultant of the stresses S1 ,S2, S3 acting on the plane ABC is obtained as

S2 = S12 + S22 + S32

The shear stress component σs acting on the plane ABC can be determined from the followingequation

σs2 = S2 – σn2

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Example

The stress state at a point is given by the stress tensor 5 3 2

T= 3 10 4 T =

246

3104

532

2 4 6Determine the normal and shear components on a plane which is equally inclined to the three axes.

The unit of stresses is N/mm2.

The direction cosines of the normal to the plane equally inclined to three axes are 1/√3, 1/√3, 1/√3.The normal stress σn is given by

σn = σij ni nj

= (5+3+2+3+10+4+2+4+6)/3 = 13 N/mm2

The stress components on ABC plane, S1,S2,S3 are given byS1 = σ11 n1 + σ21 n2 + σ31 n3

= 5/√3 + 3/√3 + 2/√3 = 10/√3 N/mm2Similarly S2 = 17/√3 N/mm2

S3 = 12/√3 N/mm2

Hence S2 = (100+289+144)/3 = 533/3

Therefore the shear stress is given byσs = (533/3 – 169)1/2

= 2.94 N/mm

2

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1.8State of stress at a point – in three dimensions

The general three dimensional state of stress consists of three unequal principal stresses acting at a point. This is called a triaxial state of stress. If two of the three principal stresses are equal, the stateof stress is known as cylindrical, while if all three principal stresses are equal, the state of stress issaid to be hydrostatic or spherical.

Determination of principal stresses for a three dimensional state of stress acting on an arbitraryCartesian coordinate system:

Figure represents an elemental free body, with diagonal plane JKL of area A. the plane JKL isassumed to be a principal plane cutting through the unit cube. σ is the principal stress acting normalto the plane JKL. Let l,m,n be the direction cosines of σ (i.e. cosines of angle between σ and the

x,y,z axes). Since the free body must be in equilibrium, the forces acting on each of its faces must balance. The components of σ along each of the axes are Sx, Sy and Sz.

Sx = σl Sy = σm Sz = σnArea KOL = Al Area JOK = Am Area JOL = An

Taking the summation of forces in x direction results inσAl - σxAl – τyxAm – τzxAn = 0

K

J

L

z

y

x

τzy

τzx

σz

σx

σy

τxy

τxz

τyx

τyz

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which reduces to(σ – σx) l – τyx m – τzx n = 0

Summing the forces in other two directions results in- τxy l +(σ – σy) m – τzy n = 0- τxz l – τyz m +(σ – σz) n = 0

the above three equations are homogenous linear equations in terms of l,m,n.the only solution can be obtained by setting the determinant of the coefficient of l,m,n equal tozero, since l,m,n can not all be zero.

| σ – σx -τyx -τzx || -τxy σ-σy -τxz | = 0| -τxz -τyz σ-σz|

Solution of the determinant results in a cubic equation in σ

σ3 – (σxx + σyy + σzz)σ2 + (σxxσyy + σyyσzz + σzzσxx – τxy2 – τyz

2 – τzx2)σ – (σxxσyyσzz + 2τxyτyzτzx – σxxτyz

2 – σyyτzx

2 – σzzτxy2) = 0

(i.e)σ3 – I σ2 + II σ –III = 0

The three roots are three principal stresses σ1, σ2, σ3.To determine the direction, with respect to the original x,y,z axes, in which the principal stressesact, it is necessary to substitute σ1, σ2, σ3 each in turn into the three equations of force balance.The resulting equations must be solved simultaneously for l,m,n with the help of relation,l2+m2+n2 =1. since there is only one set of principal stresses, the above equation must remain the sameirrespective of the changes in the co-ordinate system, because it has to have the same roots.With the change of co-ordinate system, the different stress components change, however thecoefficients I, II, III do not change. Hence these coefficients I, II, III are called the first, second andthird invariants of stress tensor.

If we choose the principal directions (direction of no shear) as the direction of co-ordinate axes,then the stress invariants take the formI = σ1 + σ2 + σ3II = -(σ1σ2 + σ2σ3 + σ3σ1)III = σ1σ2σ3

It is often possible apriori to establish from physical conditions, the direction of one of the principalaxes. Let it be z, so thatσ3 = σzz and τxz = τyz = 0the non-zero components of the stress are σxx, σyy, σzz and τxy, therefore

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σ1 = (σxx + σyy)/2 + {[(σxx – σyy)/2]2}1/2 + τxy2 σ2 = (σxx + σyy)/2 - {[(σxx – σyy)/2]2}1/2 + τxy2

σ3 = σzz

1.9Planes of maximum shear

Let S be the total stress before resolution into normal and shear components and acting on a plane,with l,m,n as the direction cosines with respect to the three principal axes.

S2 = Sx2 + Sy2 +Sz2 = σ12l2 + σ2

2m2 + σ32n2

The normal stress σ acting on this plane is given byσ = Sxl + Sym + Sz n = σ1l2 + σ2m2 + σ3n2

Therefore the shearing stress acting on the same plane is given byτ2 = S2 – σ2 = σ1

2l2 + σ22m2 + σ3

2n2 - (σ1l2 + σ2m2 + σ3n2)2 which reduces to

τ2 = (σ1 – σ2)2l2m2 + (σ1 – σ3)2l2n2 + (σ2 – σ3)2m2n2

Values of τ for the three particular sets of direction cosines listed below are of interest because they bisect the angle between the two of the three principal axes. Therefore they are the maximumshearing stresses or the principal shearing stresses.

l m n τ0 ±√1/2 ±√1/2 τ1 = (σ2-σ3)/2±√1/2 0 ±√1/2 τ2 = (σ1-σ3)/2±√1/2 ±√1/2 0 τ3 = (σ1 –σ2)/2

since according to convention, σ1 is algebraically greatest principal normal stress and σ3 is thealgebraically smallest principal stress, τ2 has the largest value of shear stress and it is calledmaximum shear stress τmax.The maximum shearing stress is of importance in theories of yielding and metal formingoperations.

1.10Mean Stress and Stress Deviator

The first invariant of the stress is the sum of the normal stresses, i.e.,I = σxx + σyy + σzz

The mean stress which is also known as hydrostatic stress is σm = I/3 = (σxx+σyy+σzz)/3

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The hydrostatic stress is regarded as a special stress tensor asσm 0 00 σm 00 0 σm

In this tensor, all the normal components are equal and the shear components vanish.When the mean stress components are subtracted from the stress tensor components, a stressdeviator tensor is obtained.

Sij = σxx σxy σxz σm 0 0

σyx σyy σyz - 0 σm 0

σzx σzy σzz 0 0 σm

the principal directions are the same for the deviator tensor.In terms of principal stresses the deviator tensor is

Sij = σ1 – σm 0 00 σ2 – σm 00 0 σ3 – σm

Example

The stress state at a point is given by10 4 0

T= 4 12 00 0 0

Determine the principal stresses. The stresses are in N/mm2

The principal stresses are the roots of the following equation| 10-σ 4 0 ||4 12-σ 0 | = 0|0 0 -σ|

The above equation may be simplified to,-σ{(10-σ)(12-σ) -16} = 0

Or σ[ σ2 - 22σ +104} = 0

From the above equation, it is clear that one of the principal stresses is zero, σ3=0. for σ2,σ1 we equate the second factor to zero.

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σ2 - 22σ +104 = 0so, σ1,σ2 = [22± (222 – 416)1/2]/2

= 11± √17 N/mm2

Example

Determine the principle stresses for the following stress state. The stresses are in N/mm2 .0 10 0

T= 10 0 00 0 0

The principal stresses are given by the following equation,

|-σ 10 0||10 -σ 0| = 0|0 0 -σ|

OR -σ( σ2 – 100) = 0

One of the principal stresses in zero, σ3 = 0. The remaining two principal stresses, σ1,σ2 are10 N/mm2 and -10N/mm2 respectively.

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1.11Strain at a point

ex = du/dx ; ey = dv/dy; ez = dw/dz

γγxy = du/dy + dv/dxγγxz = du/dz + dw/dxγγyz = dv/dz + dw/dy

thus six terms are necessary to completely describe the state of strain at a point.

In complete analogy with stress, it is possible to define a system of coordinate axes along whichthere is no shear strains. These axes are the principal strain axes. An element oriented along one of these principal axes will undergo pure extension or contraction.By analogy with the expression for principal stress, we can derive an expression for principalstrain. These equations may be obtained much more easily by replacing σ,τ in the stress equation by

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e and γ/2. for example, the linear strain on any plane in a two dimensional situation can beexpressed by

eθ = ex cos2θ + ey sin2θ+ γxy sinθ cosθthe principal strains are then the roots of the following equation

e3 – (ex +ey +ez) e2 +[exey +eyez +exez – ¼(γxy2 +γyz2 +γxz2)]e – [exeyez + ¼ (ex γyz2 +

ey γxz2

+ ez γxy2

)– ¼ (γxy γxz γyz)] = 0 by the same analogy the principal shearing strains can be obtained as

γ1 = e2 –e3γ2 = γmax = e1 –e3γ3 = e1 – e2

The strain at a point can be divided into spherical component (em) and a strain deviator.em = (e1+e2+e3)/3

the strain deviator in this case is e1 0 0 em 0 00 e2 0 - 0 em 0

0 0 e3 0 0 em

1.12Forming Properties of Metals and Alloys

Most of the metal forming processes require a combination of material properties for their successful operation. For example, for deep drawing of sheet metal, one of the requirements is thatthe material should be ductile. However, lead is quite ductile at room temperature but lead sheet isnot that suitable for drawing into cups in the way steel cups are drawn. The material properties thatare of importance for metal forming are:

1. Yield strength / Flow stress2. Ductility3. Strain hardening4. Strain rate sensitivity5. Effect of temperature on yield strength and ductility6. Effect of hydrostatic pressure on yield strength and ductility7. Instability and fracture strength.

1.13Testing for Yield Strength / Flow Stress

For the determination of yield strength or flow stress, one or more of the following three basic testsare conducted

1. Tension test2. Compression Test3. Torsion test

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For loading up to point A there is a linear relationship between the stress and strain. Point A iscalled proportionality limit. Thereafter the stress strain relation is nonlinear; up to point B thedeformation is elastic and on unloading the specimen regains its original dimensions. Beyond pointB the metal yields, it suffers plastic deformation. Most of the strain after this point is plastic strainwhich is not recovered upon unloading. The value of the stress at the point B is called upper yield

strength and point B is the upper yield point.E

B CA

Proportionality limit

Figure 1 A typical stress-strain curve for mild steel

With further increase in strain, the stress may fall a bit to a lower level; this is due to the formationof Luder Bands. With increase in tension, localized plastic flow takes place in a narrow band withthe boundary planes inclined at an angle to the axis of specimen. However due to strain hardeningof the material in the band, the load again increases until another Luder band appears in the test piece. This goes on until the whole specimen is full of Luder Bands.

Up to point C the stress is oscillating between two narrow limits. This occurs in alloys havinginterstitial solid solution structure like low carbon steel. The stress at the point C is called the lower yield strength.

However with further increase in strain beyond point C, the load or stress again starts increasing.The stress at the point C is the flow stress at yield point, however the data given in materialstandards generally refer to the upper yield point.After point C the stress-strain curve moves upwards, however, with further deformation, its slopegradually decreases to zero at the point E, the highest point on the curve. After E, the curve goesdown. Before point E, increase in strain increases the load on the specimen due to strain hardening.Even after point E, strain hardening is there; but at some point the area of c/s of the test piece startsdecreasing and a neck forms, with the result, the force that the test piece can bear decreasescontinuously with further deformation. After some elongation in the neck, the specimen fractures atthe point F. since we have defined the stress as the force divided by the original area of c/s, the

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stress value thus calculated also decreases after E. the stress at the point E is known as ultimatetensile strength.

1.14True stress – True strain

The stress-strain curve would appear very much different if we use true stress on the ordinate andtrue strain on the abscissa. The true stress and true strain are defined below.

True stress = Force applied on the specimen / current area of cross section

Ludwik(1909) first proposed the concept of true strain. For true strain let us take that a specimen of length l is elongated by dl. The differential strain dε is

dε = dl /l

The true strain in the specimen which has been elongated from the initial length lo to the finallength l is obtained by integrating the above equation.True strain ε = lo∫l (dl/l ) = loge(l/lo)

= loge (1+ dl/lo)= loge(1+E), where E is the engineering strain = dl/l

The typical stress-strain curve redrawn on the true stress – true strain axes would like the oneshown in figure. There are standard specifications for the shape and dimension of the test specimen,which should be adapted in order to get meaningful results. Besides, the following factors should be noted.

1. temperature at which the test is conducted2. strain rate during the test

T r u e s t r e s s

True strain

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3. accuracy of load measuring instrument4. accuracy of instrument which measures elongation.

1.15Strain Hardening

Let us again consider the tension test curve. In the figure shown, the test piece is loaded beyond theyield point up to a point P. the test piece is then unloaded. The elastic deformation recovers via theunloading curve PR which is more or less parallel to AO. Out of the total strain OS correspondingto the point P, the part RS is the elastic recovery. The line PR depicts the elastic recovery. The partOR which is not recovered is the plastic strain suffered by the test specimen. Now, if we reload the same test piece, it nearly follows the line RP. There is however somedeviation due to hysteresis, which is very small and yielding now, occurs at the point P. Further loading of the test piece beyond P gives the same stress-strain curve as we would have obtained if

there were no unloading. This shows that after suffering a plastic strain represented by OR, theyield strength of the metal has increased from point B to point P. This is called strain hardening or work hardening.

σot P

B

O R

σoc Q

1.16Bauschinger Effect

Now consider another experiment for which the test curve is shown in fig. in this test the specimenhas been loaded in tension up to a point P beyond the yield point B(yield strength = σot), thenunloaded to the point R and again reloaded in the reverse direction, i.e., compressed.

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ε ε ε

(a)Rigid ideal plastic (b) ideal plastic with elastic region (c) strain hardening

Yielding Criteria for ductile materialsThe problem of deducing mathematical relationships for predicting the conditions at which plasticyielding begins when a material is subjected to a complex state of stress is an importantconsideration in the field of plasticity. In uniaxial loading, plastic flow begins at the yield stress,and it is to be expected that yielding under a situation of combined stress is related to some particular combination of the principal stresses.

1.18Maximum shear stress theory

The maximum shear stress theory, some times called the Tresca, Coulomb, or Guest yield criterion,states that yielding will occur when the maximum shear stress reaches a critical value equal to theshearing yield stress in a uinaxial tension test. The maximum shear stress is given by

τmax = ½ (σ1 – σ3)where σ1 is the algebraically largest and σ3 is the algebraically smallest principal stress.

For uniaxial tension, σ1=σo, σ2=σ3=0. where σo is the yield strength in simple tension. Thereforethe shearing yield stress for simple tension τo is equal to one half of the tensile yield stress.

τ = σo/2therefore we can write, τmax = ½ (σ1-σ3) = τo = σo/2

or σ1 – σ3 = σo

Also, σ1 – σ3 = 2k, where k is the yield stress for pure shear.

The maximum shear stress theory is in good agreement with experimental results, being slightly onthe safer side, and is widely used by designers for ductile materials. It has replaced older and lessaccurate maximum stress theory, Rankine’s theory.Prager and Hodge(1951) have pointed out that in certain plasticity problems, the simple relations asabove can not be used as the yielding conditions, since it is not known which of these three principal stresses is the largest.

1.19Von Mises, or Distortion energy theory

A somewhat better fit with experimental results is provided by the yield criterion given byσo = 1/√2 [(σ1-σ2)2 + (σ2-σ3)2 + (σ3-σ1)2]1/2

according to this criterion, yielding will occur when the differences between the principal stressesexpressed by the right hand side of the equation exceed the yield stress in uniaxial tension, σo.The development of this yield criterion is associated with the names of Von Moises, Hencky,Maxwell and Huber.

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A number of attempts have been made to provide physical meaning to the Von Mises yieldcriterion. One commonly accepted concept is that this yield criterion expresses the strain energy of distortion.On the basis of distortion energy concept, yielding will occur when the strain energy of distortion per unit volume exceeds the strain energy of distortion per unit volume of a specimen strained to

the yield stress in uniaxial tension or compression.Another common physical interpretation is that it represents the critical value of the octahedralshear stress.

The total elastic strain energy per unit volume can be divided into two components: the strainenergy of distortion, Uo’, and the stain energy of volume change, Uo’’.As hydrostatic stress has no effect on stress state(yielding), only the stress deviator can causedistortion. The strain energy of distortion will be based on stress deviator. It represents the strainenergy associated with changing the shape of specimen and neglects the strain energy associatedwith volume change.It can be shown that, for uniaxial stress state, Uo’ = [(1+ν)/3E] σo2

For the condition of pure shear, as in torsion, τ=σ.σ1=σo σ2=0 σ3= -σo the strain energy of distortion for this state of stress can be shown to be

Uo’ = {(1+ν)/E}σ2 = [(1+ν)/E]τ2

The ratio between the strain energy of distortion at yielding for uniaxial stress and that for pureshear can be obtained by

{(1+ν)/3E}σo2 = [(1+ν)/E]τo

2

τo = (1/√3) σo = 0.577 σo

thus by distortion energy theory, yield by shear should be(i.e., yield occurs at) 0.577 times thetensile yield strength.

0.6 Distorsion energy

0.5

τxy/σo

0.3 Maximum shear stress

0.00.1 0.5 1.0

σx/σo

Comparison of max shear stress theory and distortion energy theory (by GI Taylor and H Quinney)

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For plane stress condition, the distortion energy theory of yielding can be expressed mathematically by

σ12 + σ22 – σ1σ2 = σo2

this is the equation of an ellipse whose major semiaxis is √2 σo and whose minor semiaxis is √(2/3)σo.

σo

Distortion energy theoryσo

√2σo

σ1

√2/3σo

Maximum shear stress theory

A convenient way of comparing the yielding criteria for a two dimensional state of stress is with a plot as in figure. Note that the maximum shear stress theory and the distortion energy theory predictthe same yield stress for conditions of uniaxial stress and balanced biaxial stress (σ1 = σ2). It can

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be shown that for this state of stress the shear stress law predicts a yield stress which is 15% lower than the value given by distortion energy criterion.

1.20Octahedral shear stress and shear strain

The octahedral stresses are a particular set of stress functions which are important in the theory of plasticity. They are the stresses acting on the faces of a three dimensional octahedron which has thegeometric property that the faces of the planes make equal angles with each of the three principaldirections of stress. For such a geometric body, the angle between the normal to one of the facesand the nearest principal axis is 54˚44”, and the cosine of this angle is 1/√3.

The stresses acting on each face of the octahedron can be resolved into a normal octahedral stressσoct and an octahedral shear stress τoct., lying in the octahedral plane.The normal octahedral stress is equal to the hydrostatic component of the total stress.

σoct = 1/3 (σ1 + σ2 + σ3) = σ”The octahedral shear stress is given by

τoct = 1/3 [(σ1 – σ2)

2

+ (σ2 – σ3)

2

+ (σ3 – σ1)

2

]

1/2

= √2σo/3

or σo = = 1/√2 [(σ1-σ2)2 + (σ2-σ3)2 + (σ3-σ1)2]1/2

In a sense the octahedral theory can be considered to be the stress equivalent of the distortionenergy theory. According to this theory, the octahedral shear stress corresponding to yielding inuniaxial tension is given by τoct = (√2) σo/3 = 0.471 σo

Octahedral strains are referred to the same 3D octahedron as the octahedral stresses. The octahedrallinear strain is given by εoct = 1/3(ε1 + ε2 + ε3)

Octahedral shear strain is given by γoct = 2/3[(ε1 – ε2)2 + (ε2 – ε3)2 + (ε3 – ε1)2]1/2

1.21Invariants of Stresses and Strains

It is frequently useful to simplify the representation of a complex state of stress or strain by meansof invariant functions of stress and strain. If the plastic stress strain curve is plotted in terms of invariants of stress and strain, approximately the same curve will be obtained regardless of the stateof stress. For example, the flow curves obtained in a uniaxial tension test and a biaxial torsion testof a thin tube with internal pressure will coincide when the curves are plotted in terms of invariantsof stress and strain functions. Nadai(1937) has shown that the octahedral shear stress and shear strain are invariant functionswhich describe the flow curve independent of the type of test. Other frequently used invariantfunctions are the effective or significant stress and strain.With the principal directions as co-ordinate axes,Effective or significant stress σ` = (√2)/2 [(σ1-σ2)2 + (σ2-σ3)2 + (σ3-σ1)2]1/2

Effective or significant strain ε` = (√2)/3 [(ε1-ε2)2 + (ε2-ε3)2 + (ε3-ε1)2]1/2

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The effective stress and strain are related to the octagonal shearing stress and strain asτoct = (√2 / 3) σ` and γoct = √2 ε`

1.22Slip-field theory (Hencky’s plastic section method)

Consider a volume element in plane strain within a plastic region of a body. Figure represents a 2Dstate of stress with respect to arbitrary Cartesian co-ordinates. It is possible to determine the principle planes such that the shear stresses vanish. The principal stresses are simply functions of spherical component of stress σ”, and the shearing stress k.

y σ2 y ασy τxy σ” σ”

σx xτxy

σ1 σ1=σ”+k σ1

σx τxy τxyβ

σy σ2=σ”-k σ” σ”

Figure: 2D state of stress in plane starin

The maximum shear stress will occur on planes 45˚ to the direction of principal stresses. Thus, thecritical shear stress k will first reach its value on these planes. In the figure shown, it is seen that themaximum shear stress occurs in two orthogonal directions, designated α and β. These lines of maximum shear stress are known as slip lines. The slip lines have the property that the shear strainis a maximum and the linear strain is zero in a direction tangent to them.

By comparing the figures, it is seen that the principal stresses have a direction 45˚ to the slip lines.The value of principal stresses can be determined if σ” is known since

σ1 = σ” + k σ2 = σ” – k

if σ” is constant throughout the region, slip lines will be straight lines.

However, if the slip lines curve by an angle φ, the following relationship holds:σ” + 2kφ = constant along α lineσ” -2kφ = constant along β line

The slip lines at the free surface must make an angle of 45˚ with the surface, since there can be noresultant tangential force at a free surface. Since there is no resultant normal stress at a free surface,

σ1=0 and therefore σ” = -k and

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σ2 = -2k. The transverse principal stress is compressive with a value of 2k.

σ1=0

Free surface

k k

σ2 σ2=2k k k

σ1Figure: slip line field at the free surface

1.23Deformation of an ideal plastic metal by a flat punch

Friction is considered negligible. Plastic deformation will first start at the corners of the punh andwill result in a slip line field as shown in figure.

Consider point M. since this is a free surface, normal stress is zero and σ” = k The equation of this slip line may be written as σ” + 2kφ = k. There is no change in the value of σ”until we reach the point N, where the slip line deviates from a straight line.In going from N to Q, the slip line turns through an angle of φ = -π/2 so that its equation at point Qis σ” -2k(π/2) = k Since there is no further change in φ in going to point R, the principal stress normal to the surfaceat R is σ1R = σ” + k = (k + 2kπ/2) + k Or σ1R = 2k(1+π/2)And σ2R = 2kπ/2

The pressure is uniform over the face of the punch and is equal to σ1 = 2k(1+π/2)Since k = σo/√3

σ1 = σmax = 2σo/√3 [1+π/2] = 3σo thus the theory predicts that full scale plastic flow, with the resulting indentation will occur whenthe stress across the face of the punch reaches three times the yield strength in tension.

σ1R σ1=0

R M

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O N

Slip line field produced by indentation of a punch1.24Construction of slip lines

Slip lines may be constructed in the body which is undergoing plastic deformation or is at the yield point, provided the normal and shear stresses are completely known on its entire boundary. Theconstruction gives an orthogonal network of slip lines covering the cross section of the deforming body. We can any point inside the body through this network. The application of Hencky’s stressequations as we go along the α and β lines of the network gives the state of stress at the desired point in the body. In this way we get the exact solution of the problem.

Suppose the boundary is a free line (A free line is one which is not a slip line). in this case the

directions of α and β lines may be determined from the stress prescribed on the boundary and thesolution may be extended into the deforming zone limited by the intersecting slip lines emanatingfrom the end points of the free line.

Directions of slip lines

For construction of slip lines, first of all we have to determine the direction of β and α lines on the boundary points from the stress prescribed on it. Four types of stress conditions may beencountered on the boundary. They are

1. stress free surface not contacting the tool2. frictionless interface between the tool and the deforming material3. interface with coulomb friction condition4. interface with sticking friction condition

(i) stress free surfacein this case there is no shear or normal stress acting on the surface. Since there is no shear stress,the plane tangent to the surface is a principal plane and normal to it is one of the principal stresses.The other two principal stresses lie in the tangential plane. Let σ1,σ2,σ3 be the principal stresses.Let us take σ3 along the normal to the tangential plane. Let σ2 be the middle stress, then

σ2 = (σ1 + σ3)/2

since σ3 = 0, σ2 = σ1/2this shows that plastic deformation is caused by by σ1 which may be compressive or tensile.

If σ1 is compressive(i.e) σ2 = σ1/ = -K and σ3=0

If σ1 is tensile(i.e) σ1= 2K and hence σ2 = K and hydrostatic pressure P = -K

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σ3=0 σ3=0

σ1 90 σ1 σ1 90 σ1

β-line α-lineσ2 σ2

α-line β-line

yielding in compression yielding in tension

(ii) frictionless interfaceOn a frictionless interface, there is no shear stress. Thus normal to interface at the desired

point is the direction of principal stress. Let σ3 be the principal stress, not equal to zero. In real caseσ3 can only be compressive. Let σ2 be the middle stress, then

σ2 = (σ1+σ3)/2

σ3 σ3

σ1 σ1 σ1 σ190 90

β α

α β

(a) σ1,σ3 compressive (b) σ1 tensile, σ3 compressive

(iii)Interface with coulomb frictionLet σ3 be the stress normal to the surface. The frictional shear stress is equal to μσ3. let one of theslip lines be inclined at θ with the free surface. Figure shows the Mohr’s stress circle. Take radiusCE such that EF = μσ3 and angle ECG = 2θ. Then,

EF = K cos2θ = μσ3Hence θ = ½ cos-1(μσ3/K)

σ3

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μσ3 2θ σ2

θ μσ3

σ3surface with coulomb friction

(iv) interface with sticking frictionThe sticking friction occurs when the shear stress at the interface reaches the value K = yieldstrength of the material in shear. This is also the maximum shear stress.Then cos 2θ = 1 or θ =0,90. therefore one of the slip lines is parallel to the surface and the other is

perpendicular to it. Fricton occurs only when σ3 is compressive. The directions of α and β lines areshown in figure.

Construction of slip lines when two slip lines are given

Let OA be an α slip line and OB be a β slip line intersecting α line at O. the normal stress,hydrostatic pressure and shear stress(K) are known at all the points on OA and OB. Either of theslip line may be part of the interface between the tool and deforming body on which there issticking friction (magnitude of friction = K).

In order to construct the slip lines in the region bounded by two slip lines, we take a number of points on the two lines, such as (0,1), (0,2), (0,3) etc. on the βline and (1,), (2,), (3,0) etc. on theαline. OX and OY are the reference co-ordinates. Let us denote the angle of inclination of αlinewith respect to OX at (1,), (2,0), (3,0) etc, by θ10,θ20,θ30 and so on. Similarly the inclination of αlineat the points (0,1), (0,2), (0,3) are denoted by θ01,θ02,θ03 and so on.the values of hydrostatic pressureare also denoted similarly.

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Let us take that the αline from point (0,1) and βline from point (1,0) intersect at the point(1,1). LetP11 be the pressure and θ11 be the inclination of αline at the point (1,1). For the segment (0,1) to(1,1) on the αline, we have

P01 +2Kθ01 = P11 + 2Kθ11 For the segment (1,0) to (1,1) of βline, P10 - 2Kθ10 = P11 – 2Kθ11

Based on the above two equations,4Kθ11 = P01 – P10 + 2K(θ01 + θ10)

And by adding the two equations, we get2P11 = P01 +P10 + 2K(θ01 – θ10)

Thus at point (1,1) we know the complete stress stste.To find the location of the point (1,1):The αline segment (0,1) to (1,1) as well as the βline segment (1,0) to (1,1) may be approximated bystraight segments. We may draw a line segment with the mean of angles of αline at (0,1) and at(1,1) i.e., (θ01 +θ11)/2. similarly we may draw a line segment with the mean of the angles of βline at

(1,0) and at (1,1) i.e., π/2 + (θ10 + θ11)/2.The intersection of these two points gives the location of point (1,1).Proceeding this way we may know the stress and location of (1,1),(2,2),(3,3), etc. bounded by theslip lines. The slip line from the extreme points A and B intersect at D. the region OADBO is theregion of the solution. Beyond this region the solution can not be extended.

2.1Friction in metal working

Friction condition between deforming tools, work piece in metal working are of great importance toa number of factors, viz., mode of deformation, properties of finished specimen, resulting surfaceroughness, etc. A stumbling block in correlating theory and practice is the certainty attached to the

magnitude of magnitude of interfacial friction. The friction stress is measured in units of force per unit area of contact. The surface area of contact is a boundary of the deformed material. Themechanics of friction is a complex phenomenon and depends on a number of variables such asnature of materials, size, and shape of specimen, velocity of deformation, temperature andatmospheric conditions.

Coulomb’s Friction Law

By Coulomb’s law, F = μ N or τ=μpThe coefficient of friction μ is taken as constant for a given material and die and is said to beindependent of velocity, applied load and area of contact.This relations describes the conditions well until the product μp becomes higher than the yieldstress in pure shear of the material. The material will then stick to the tool and yielding takes placein the interior of the material. Then the above law does not apply.

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Constant friction factor Some investigators prefer to assume a constant shear stress irrespective of the pressure between thedie and the material. It is assumed that the d=shear stress is proportional to the strength of thematerial. When one body is fully plastic,

τ = m σ/√3where m is the friction factor and is constant for a given die, work & constant surface , temperatureconditions.If a lubricant is brought into picture, three models are generally used to describe the effect such ashydrodynamic lubrication, boundary lubrication or hydrostatic lubrication. In the latter, well knownin metal forming, the lubricant is trapped in numerous small pockets formed in the surface of thespecimen when the tool and the specimen meet. The pressure in these is sufficiently high totransmit the necessary normal stresses.

Composite friction

It is well known that in plastic deformation of metals, the surface of the work piece is distorted andtakes on an impression of the tool surface. Hence, friction in plastic deformation is different fromsliding friction in machine parts. However high relative velocities between the work piece and thetool surfaces combined with high interface pressure and severe deformation mode will cause breakdown of surface film and will allow new surfaces to come in contact with the tool surface andhence facilitate the intimate contact essential for adhesion. For such a case the metal undergoingdeformation does not necessarily slip along the tool surface, though it is not ruled out. For such acomposite mechanism of friction, Deryagin suggested that it is a function of sliding and adhesion.The shear equation is

τ = μ(P + Aφ)where, the first term denotes sliding friction and the second term denotes adhesion. The load between the two surfaces is supported by the minute asperities. Under light loads thecontact may be elastic, but as the load increases, some of the asperities undergo plastic deformationand adhere to each other resulting in micro-welds. The strength of the bond depends on interfacialconditions, time, and temperature. Adhesion friction constitutes a major drawback in production because it affects the die life and surface finish of the product.By the well known theory of adhesion of Bowden and Tabor, the friction force becomes the forcethat is required to shear these junctions in order to slide the two surfaces over each other.Hence the coefficient of friction is defined as μ F/N = τ/σSince the asperity is surrounded by a large mass of material, the normal stress on the asperity isanalogous to the hardness of the material, thus, τ = μH.

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Friction in high speed forming

Coefficient of friction generally decreases with increasing velocity between the work piece and thetool, but at the same time increased velocity results in increased temperature and the resultinglubricant breakdown. However extremely high velocities may generate sufficient temperature tomelt a thin layer of metal at the work piece and deforming tool interface to act as a lubricant and is

usually known as flow zone. This layer of very high strain increases in thickness along the face of the deforming tool and in the direction of flow. This layer may be conveniently considered as a Newtonian fluid for which,

τ = η δu/δy(i.e), friction stress is proportional to viscosity and velocity gradient.It has been shown that thickness of flow zone increases parabolically, in the direction of flow.

2.2Lubrication in metal working

Much attention has been paid to produce low values of coefficient of friction. In metal working,however, elimination of the possibility of damage caused by metal transference from work piece to

the tool is important. Tool life can be prolonged both by reducing friction and by preventingmetallic contact with work piece, by using lubricants. Apart from increasing the force required,frictional stress may cause inhomogeneity as well as surface cracks in the worked part.To produce a bright surface it is necessary to sacrifice some lubrication efficiency. In rolling of flatstrips, the roll will slip if the friction is too low and it is necessary to use a poor lubricant to obtainthe greatest reduction per pass.Principle of lubricationMost metal working lubricants have been developed empirically, by trial and error. Lubricants inuse vary from gases to solid, but majority are fluids. Additives are compounded in oils or emulsionsto improve boundary lubrication under severe conditions of temperature and pressure.

An ideal metal working lubricant should perform all of the following functions:1. provide boundary and hydrodynamic lubrication at high pressure2. minimize and restrict surface temperature3. reduce wear rate at tool-work interface4. dissipate contaminants such as particles, dirt, and scale from the work surfaces5. dissipate the heat generated in metal working6. prevent metal adhesion and pick-up between the tools and work piece7. assist in maintaining the desired surface finish and metallurgical characteristics of the

finished product

The most common lubricants are oils, chemical compounds, soap solutions and phosphate andoxalate coatings. Certain elements are also added to the fluids to act as anti-welding agents. Nowthe developments in the field make use of chemical reactions between fluids and surfaces to formlow shear strength layers such as fluorides and chlorides. A thin layer of materials such as Teflon between the two surfaces is also helpful.For wire drawing of steels chemically deposited copper coatings are used. Small electric current of the order of a few amperes, reduce wear when applied to sliding surfaces. It appears that the current

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produces additional chemical compounds of low shear strength in lubricating oil containingadditives.For elevated temperatures and some cold operations, various solid lubricants are used such asgraphite, MoS2, phosphate, glasses and saw dust. A recent development in reducing friction is theuse of high frequency vibration between the die and the work piece.

In most metal working operations, the pressure involved are too high and the speed too low to permit full hydrodynamic lubrication and the best that can be obtained is a boundary lubrication or intermittent hydrodynamic lubrication. All metal working lubricants therefore contain constituentshaving boundary lubrication properties.

Lubrication mechanism

Most metal working processes are considered to be operating in the boundary regime of lubrication.Under high loads lubrication film is thinned to near molecular dimensions, rendering itdiscontinuous.There are three types of lubrication mechanisms.

In the ideal thick film lubrication (full fluid or hydrodynamic lubrication) the film is continuous,the true area of contact is zero and the friction is a function of viscosity of lubricant undergoingshear in the contact region. With this mechanism, coefficient of friction is low and wear is nil. Thismechanism is common in bearing system.Most metal deformation processes are carried out predominantly in boundary or mixed film regime.This system does consider the presence of any apparently thin film separating the surfaces, but atthe same time admits some asperity interactions. In this case, the film where present is extremelythin and discontinuous. Friction in boundary lubrication is comparatively high and coefficient of friction of 0.1 and above can be tolerated without damage.The term solid lubrication is used to describe the reduction of friction and wear through the use of inorganic solids having low shear strength. The mechanism involves complete separation of surfaces by the soild lubricants. Ex. : graphite and MoS2.

2.3Forging machines

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The upper die and ram are lifted by rolls gripping the board. When the ram is released it falls by itsown weight. In the board hammer, the energy supplied to the blow is equal to the potential energydue to the weight of the ram and the height of the fall. They range from 400lb – 35in fall to 7500lb-75in fall.

Greater forging capacity in the range 1000 to 10000 lb is available with steam hammer. Since thefalling ram is accelerated by steam pressure, the energy supplied to the blow is related to the kineticenergy of the ram.

W = Mv2/2gStriking velocity in excess of 25 ft/sec may be obtained.

Presses are rated by the load developed at the bottom of the stroke. Loads of 100 to 7000 tons arecommercially available. Hydraulic presses are developed for 500 to 18000 tons.

Horizontal presses are very useful for high production forging of symmetrical shapes from bar stock. The machine is basically a double acting press with dies to firmly grip the work aroung thecircumference and forming tools which upset the metal. Bolts, ears and rivets are typical partsmade.Rotary swaging

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Swaging is the process of extending the length at the cost of reducing the cross section

Rotary swaging subjects the a bar or tube to a series of blows from two dies which rotate aroundthe stock so that it is hammered from all the sides. Two dies backed up by rollers are rotated with aspindle which rotates in a cage containing a number of hardened steel rollers. When the hammers

contact the rollers, the des are forces together but when the spindle rotates the hammer to a position between the rollers, the dies fly open owing to centrifugal force. Thus the work is subjected toseveral thousand blows per minute.

In the absence of friction when dies are well lubricated, uniaxial compressive force required tocause yielding is P = σo AThe compressive stress p induced by uniaxial force P is given by

Pp=4Ph / πD2hoWhere h is height of cylinder at any instant and ho is original height of cylinder.

2.4Forging defects1. if the deformation during forging is limited to the surface llayers, as with light rapid

hammer blows, the dendritic ingot structure will not be broken down at the interior of theforging. Incomplete forging penetration can be readily detected by macroetching a c/s.

2. the examination of deep etch disk for segregation, dendritic structure, cracks is a standardquality control procedure for large forgings.

3. surface cracking can occur as a result of excessive working of surface at too low atemperature or as a result of hot shortness. Cracking at the flash is another surface defect in

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closed die forging, since crack generally penetrates into the body of the forging when theflash is trimmed off. Flash cracking can be avoided by increasing the flash thickness or byrelocating the flash to a less critical region.

4. cold shut in closed die forging is a discontinuity produced when two surfaces of metal foldagainst each other without welding completely. Ex. The flash of a forging operation may

press into the metal surface during a subsequent operation.5. in upsetting of bar stock, buckling of the bar should be prevented. Unsupported lengthshould be no greater than 2 or 3 times the diameter.

6. secondary tensile stresses can develop and produce cracking. Internal cracks develop as aresult of circumferential tensile stresses.

7. presence of flow lines or fiber structure.8. residual stresses in forgings are a result of inhomogenous deformation and are generally

quite small because deformation is carried out in hot working range.9. special precautions must be observed during cooling of large forgings. They are subjected to

cracks or flakes at the center of c/s.

ROLLINGRolling constitutes a group of processes in which deformation or change in shape of product is

brought about by compression between rotating cylindrical rolls, disc type rolls or dies.

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1. Longitudinal rolling processesProduce strips, sheets, flats, bars of circular/ non-circular cross section, angle sections, beams,channels, wide flange beams, rails, etc.

2. Transverse rolling or cross rollingProduce rolled gears, axles, rolled discs(pulley), seamless tubes, pipes, etc., thread rolling

3. Lateral rolling or axial rollingProduce spherical balls, finned tubes

4. Die rollingIt is a combination of rolling and forging processes. Instead of pressing the entire componentsimultaneously between the dies (as in forging) a rotating die presses the metal into a stationary diesuccessively in each rotation. Adapted, to do noiseless riveting.

5. Forge rollingIn this process, die like impressions are made on the surfaces of rolls. The component is rolled between them. Produce stepped shafts, components.

LONGITUDINAL ROLLING OF STRIPS AND SHAFTS

Process parametersThe ingot or billet is passed between two rolls mounted on a rigid frame or stand. The gap betweenthe rolls is adjusted to less than the height of the incoming billet and hence it gets rolled to areduced height while it increases in length and width.

Let h1, b1, v1 and h2, b2, v2 be the length, breadth and width of billet at the entrance and exit of the rolls. Then,

h1b1v1 = h2b2v2

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In cold rolling of thin sheets, the rigidity of roll frame is particularly important.h2 = hg + Δf + Δr + Δs

however in hot rolling the elastic deformation of rolls and elastic recovery of strip are very smallcompared to the reduction and hence may be neglected.

h/b < 2h/b >2

Non-uniformity of deformation in transverse section

Set gap Δf+Δr Δs

Rolled sheet thickness

Initial thickness of sheet

Rolling load curve

Elastic deformationcurve for rolling standand rolls

R o l l i n g l o a d

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Entry condition and maximum reductionFigure shows the forces acting on the rolls at the entry side. The radial pressure of the rolls pushesit away from the rolls while the frictional force tries to pull into the roll gap. Let α be the angle of contact. The billet will enter the roll gap if there is a net positive force in the direction of rolling.

Therefore the condition for the billet to enter the roll gap isμ P cos α >= P sinα

Where P is the force exerted by the rolls, μ is the coefficient of friction and α is the contact angle.In the limiting case for the maximum value of α, the two sides of the above equation may beequated giving

αmax = tan-1 μ

the maximum possible reduction is Δhmax = D(1-cosαmax), D is diameter of rolls.If we desire to icrease the reduction , we need to increase the coefficient of friction or diameter or both

In cold rolling, the μ is quite small because the rolls are ground and well lubricated. Μ varies 0.05

to 0.1In hot rolling the μ for steel may be determined from

Μ = k1 . k2. k3(1.05 – 0.0005t)Where t is the temperature of the billet in °C andk 1 coefficient which takes into account the material of the rolls.k2 coefficient for influence of rolling speedk3 – coefficient to take care of material being rolled; its value is 1 for M.S, 1-1.3 for alloy steels

α

P μP

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The area of contact between the rolls and the deforming work material affects the load on the rolls.

Curved area of contact = R.α bmProjected area of contact = L.bmWhere bm = (b1 + b2)/2L is the length of contact projected on the middle plane of rolling

L2 = (2R – Δh/2) Δh/2

It is observed that the speed of the billet on the entry side is less than the speed of the rolls, so thefrictional stress on the billet is directed in the direction of rolling. On the exit side the billet speed ishigher than the roll surface speed and the frictional stress is directed against the metal flow.

Between the entry and exit, there is a section called neutral section where the roll speed and the

billet speed are the same. Thus the deformation is divided into(i) leading zone which extends from the neutral section to exit(ii) lagging zone which extends from entry to neutral section

2 6 10 14 18rolling speed m/s

k2

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let vr – roll surface speedv1 – billet speed at entry sidev2 – billet speed at the exit sidevγ – billet speed at neutral section

% forward slip on exit side = [(v2 – vγ) / vγ] x 100% backward slip on entry side = [(vγ – v1)/vγ] x 100

v2

v1

γ

α

L

Neutral section

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Conical portion of die

Relief angle

D i e

a n g

l e

Die casting land

WCnib

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