CHAPTER 3 Revised

8/4/2019 CHAPTER 3 Revised

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Manufacture Engineering Processes;By Ali M Alsamhan

CHAPTER THREE: Principle of Metal Forming Theory

1

CHAPTER THREE: Principles of Metal Forming Theory

3.1 Experimental stress-strain flow curve.3.2 Nominal and true stresses and strains.3.3 Volume constancy phenomena in metal forming processes.3.4 Plastic tensile instability and necking condition.3.5 Analytical stress-strain flow curves.3.6 Yielding criteria.3.7 Plane strain and plane stress conditions.3.8 Work and energy method application in metal forming processes.3.9 Determination of forces/torque required in metal forming processes.3.10 Illustrated examples.


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3

Thread -10

1 (25.4 mm)2.25(57.15mm)

2(50.8 mm)Gauge length.

R1/8 (3.18mm)

1.5(38.1mm) 2(50.8mm)

8 (203.2 mm) Gauge length

9 (228.6 mm)

3(76.2mm)

t

Fig. 3.2 Commonly used tension specimens for metallic materials;

a) reduced-section round with threaded ends.b) reduced-section flat with vise-grip ends.

(a)

(b)

Power law equations are commonly used to model strain-hardening behavior in analytical stress.

Fig. 3.2 shows the two commonly used tension test specimens, for metallic materials.

3.2 Nominal and true stresses and strains

Engineering stresses and strains are commonly used for small deformations (elastic deformation),

which is commonly used in structural engineering design. True stresses and true strains are

commonly used when large strains are involved like what is happening in metal forming.

Engineering stress and strain are calculated as follows:

oo

o

alno

o

alno

l

l

l

ll

lengthgaugeoriginal

lengthgaugeofextensionestraingEngineerin

A

P

areationcrossoriginal

loadappliedstressgEngineerin

min

minsec

Eq. 3.1 and 3.2

As the test specimen is loaded, it elongates and contracts along the lateral or traverse direction and

produce lateral strains and forms necking. The true stresses are calculated by dividing the load, P ,

by the current or instantaneous cross-section area, Ac ,at the instant of measuring the load, P, which

gives average stress value (distributed along the neck) and expressed as follows:


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4

AP/ Eq 3.3

True strain prior to necking, is obtained by referring small incremental change in length to the

instantaneous length, l. The true strain is calculated as follows:

o

l

l l

l

l

dlln

0

Eq 3.4

True strain also called logarithmic strain, incremental strain or natural strain.

True strain is given the symbol while engineering strain is given the symbol of e. The

relationship between true and engineering strains can be drawn as follows:

)1ln()1ln()ln()1(1 eel

le

l

l

l

l

l

lle

oooo

o

Eq. 3.5

The advantages of using true strain are given below:

1. True strain has the same numerical value in tension and in compression loadings (withnegative sign in compression), which is not the case for engineering strain. For example,

consider a test specimen the gauge length of which is elongated from 10 to 20 mm or

compressed from 20 to 10 mm, the true and engineering strains for both cases are given as

follows:

)(21

102010

)(210

1020

)(693.0)2

1ln()

20

10ln(

)(693.0)10

20ln()ln(

specimenloadingncompressioe

specimenloadingtensionl

le

specimenloadingncompressio

specimenloadingtensionl

l

C

o

T

C

o

T

2. True strain is additive, if done in successive loading. For example, if a specimen had agauge length, lo ,and was elongated to, l1 ,then to, l2 , the total true strain is given as

follows:

)ln(lnln 2

1

21

2121

oo l

l

l

l

l

l

While engineering strain is given by


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5

1l

2l

3l

11 ll 11 ll

11 ll

331 ll

Fig. 3.3 Parallelepiped of metal before and after deformation.

2121

2

21

1

12

2

1

1 ,,

eeel

lle

l

lle

l

lle

o

o

o

o

For example, consider the following loading condition;

Incremental loading step Length (mm)

0 50 (Gauge length)

1 55

2 60.5

3 66.55

True and engineering strains are calculated as follows:

totaltotal eeeeee

eee

and

331.050/)5055.66(3.0

1.05.60/05.65.60/)5.6055.66(,1.055/5.555/)555.60(,1.050/5

2859.00953.00953.00953.0)(2859.0)50/55.66ln(

0953.0)5.60/55.66ln(,0953.0)55/5.60ln(,0953.0)50/55ln(

30322110

322110

32211030

322110

3.3 Volume constancy phenomena in metal forming processes

Based on experimental evidence it was found for all metals, the volume of material is constant

during plastic deformation, which is not the case for elastic deformation (however this volume is

very small so the change could be neglected). This can be expressed as follows:

0d

dVEq 3.6

Consider a parallepiped of metal which has initial edge lengths of; l1 , l2 and l3 and final length

(after deformation) 333222111 ,, lllllllll finalfinalfinal ,( see Fig. 3.3).


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6

Since the volume is constant during plastic deformation, then;

321

332211

332211321..

)).().((1)).().((..

lll

llllllorlllllllll

Eq 3.7

Equation 3.7 can be written as follows:

000)1ln()1ln()1ln(

1)1).(1).(1(1)1)(1)(1(

321321

321

3

3

2

2

1

1

d

dVeee

eeeorl

l

l

l

l

l

Volume constancy can be expressed by: 2211 lAlAorlAlAAl ffoo Eq 3.8

Hence, true strain can be expressed also as: )ln()ln(2

1

1

2

21A

A

l

l Eq 3.9

The relationship between nominal and true stress can be drawn using volume constancy principles

and given as follows;

)1(. minminmin el

l

A

A

A

A

A

P

A

Palno

o

c

alno

c

o

alno

c

o

oc

Eq 3.10

3.4 Plastic tensile instability and necking conditions

Plastic tensile instability starts after yielding point, just before ultimate load. During this period,

the increase in load is associated with increased strain. At ultimate load, the specimen elongated

without any increase in load. At this point the material starts behave unstable. This deformation is

called Instability condition, under tension load. At this point necking occurs at the weakest points

and the deformation changes from being uniform distribution to local necking. However, the

change in load becomes zero.

0,,,0/

d

dA

d

dA

d

dPthenAPhoweverddP Eq 3.11


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7

Imperfection area

Diffuse neck

Homogeneous area

P

P

inst

instSlope=1

inst

d

d

Fig. 3.4 Graphical plot of necking strain Eq. 3.13.

For volume constancy:

d

dl

l

A

d

dA

d

dlA

d

dAl

d

dAl

d

dlA

d

Ald

d

dV 0

)(0 Eq 3.12

Substituting Eq. 3.12 into Eq.3.12;

)();(,0)( yinstabilitatd

dd

l

dlnote

d

dl

l

A

d

dA

Eq 3.13

Where; u is the ultimate tensile stress.

This result indicates that instability occurs when the slope of the stress-strain flow curve (rate of

work hardening) is equal to the magnitude of the existing stress (u in current case). Instability

loading condition will lead a necking formation and finally specimen fracture.

In metal forming processes, the maximum deformation can be obtained for ductile metals

subjected to tension loading without failure through the strain at the instability (instcritical true

strain).

The critical true stain, inst at necking can be obtained graphically by plotting dd / versus on

the stress-strain flow curve as shown in Fig. 3.4.


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3.5 Analytical stress-strain flow curves.

Numerical modeling of strain hardening behavior may be achieved using one of the following

equations:

1. nK Eq. 3.142. mnK Eq. 3.153.

o

o

n formK

)ln( Eq. 3.16

Where Krefers to the strength coefficient, n; the strain hardening exponent, m and m; strain rate

sensitivity indices, and o reference strain at which strain rate hardening is negligible [3.2]. Note,

the strain rate can be defined as the instantaneous deformation velocity divided by the

instantaneous length or height of tested specimen ( hv/ ), while true strain refers to the

instantaneous force divided by the instantaneous area of tested specimen.

The most commonly used analytical model is the power law Eq. 3.13, which do not include the

strain rate (idealized stress-strain flow curve) and called Lidwik-Hollomon equation. Using

equations 3.13 and 3.14, for instability condition;

nn

KnKknKd

d nnnn

1

11 Eq. 3.17

This means that necking occurs when = n [3.3]. From Eq. 3.17, it can be concluded that true

strain at instability is equal to the strain hardening exponent (n). This means that (n) is a measure

of the ability of the metal to undergo plastic deformation without failure.

Table 3.1, shows K and n values for two different metals 1100-O aluminum and 18-8 stainless

steel. It is clear from the table that n value for stainless steel is higher than that of aluminum. This

also means that stainless steel has more ability for elongation (just before instability) than

aluminum has.

Table 3.1 Material properties of 1100-O Aluminum and 18-8 stainless-steel

Property 1100-O Aluminum 18-8 Stainless steel

Yield strength, y 24 Mpa (3.5 ksi) 275 Mpa (40.1 ksi)

Ultimate strength, u 48 Mpa (7.0 ksi) 725 Mpa (105.6 ksi)

Elongation 45% 55%

(n or at necking 0.2 0.51

Therefore, it is important to be able to predict the stress and strain at the on-set of instability, by

adjusting process parameters, to avoid failure.


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It is worth noting that in metal forming processes the state of stress or loading conditions are more

complex than the tensile loading condition. For example, in tension load testing, yielding takes

place when yeild 1 (where is the stress along loading direction) at which plastic deformation

initiated. However, in metal forming processes the deformation takes place under more complex

state of stresses i.e. 0,0 32 . Hence, a criterion is required to predict the yielding under

this complex state of stresses.

Furthermore, it worth noting that under compressive state of stresses, necking phonemina at

instability will not occur, and the limits of deformation are set by fracture.

Hint:

To understand the roles of (n) and (K) values for starin-hardening metal behavior, consider the following cases:

1. Strain-hardening exponent (n) role:

0,

00,

..

12

12

d

dhenceand

andtherforeand

eindeformatiomoreFor

2. Strength coefficient (K)role;The strength coefficient is considered as a magnification or scale for the strength of a given material. Consider the

following example:

Slope=d /d

More slope flow curves

Less slope flow curves

This means, for more deformation the stresses willincrease and the material becomes more stronger,this is what is called strain-hardening behavior.

As it can be seen from Eq. 3.17; d/d=K n n-1;The slope is proportional to the strain hardeningexponent (n).

Example 3.1:Given two materials A and B,KA=150 and KB=150 N/mm

2 orMpa, nA=0.5 and

nB=0.25 and for given true strain =1 for bothcases;

d /d)A=150x0.5x1=75 N/mm2

d /d)B=150x0.25x1=37.5 N/mm2

then d/d)A > d/d)B or slope)A > slope)B

Example 3.2:Given two materials having the same strain hardeningexponent (n) and subjected to the same deformation(i.e. same true strain).KA=100 N/mm

2 and KB=50 N/mm

2, nA=nB=0.5 , and

A = B =1.0. Then the true stress for both cases willbe;

A=100(1)0.5

=100N/mm2 and B=50(1)0.5

=50

N/mm2

==> AB .

Material A ; n

Material B ; n


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10

3.6 Yielding criteria

Assumptions:

1. Metals are homogeneous, continuous and isotropic.2. Same yielding strength in tensile and compression loading (i.e. ductile metals).3. The volume is constant during plastic deformation, VV/ and the sum of the

plastic strain increments is zero i.e. ( 0321 ddd )

4. Strain rate and temperature effects are not considered.Two yielding criteria are commonly used to predict when yielding starts for complex state of

stresses, namely; Von Mises and Tresca criterions.

i)Von Mises criterion states that yielding will occur when the value of the root mean square of the

principal stresses reaches a critical value ,given as

1

2/12

13

2

32

2

21 ])()()[( C or Eq. 3.18a

2/1

212

2

13

2

32

2

21 ])()()[( CCwhereC Eq. 3.18b

This equation can be written in general form using normal and shear stresses as follows:

3

222222)(6)()()( Czxyzxyxzzyyx Eq. 3.19

These constants can be determined, by considering special cases such as the case of tensile test,

where 00 321 andy . By substituting these values in equation 3.18b, we obtain;

2

2

2

1 22 yC Re-substitute in Eq 3.18a and 3.18b to obtain;

yyCC 2)2(])()()[(2/122/1

21

2/12

13

2

32

2

21 Eq. 3.18

In metal forming processes, complex state of stress exists i.e. 0,0,0 321 and and not the

case of simple tension test. An expression of the effective stresses acting in this case is given as

follows;


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11

321

2/12

13

2

32

2

21 ])()()[(2

1 wherey Eq. 3.19

The associated effective strain and strain rate can be expressed as follows:

2/12

3

2

2

2

1 )](3

2[ Eq. 3.20

2/12

3

2

2

2

1 )](3

2[

Eq. 3.21

ii) Tresca criterion (also called maximum shear stress criterion)

In Tresca criterion, the effective stress is expressed as follows;

32131 whereor yyMinMax Eq. 3.22

The effective strain is expressed as follows;

3,2,1 andiwhereddMaxi

For the three principal directions Eq. 3.23

Hence, the analytical flow stress-strain curve is converted from nK to nK )(

using the effective stress and strain terms.

Example 3.3:

For the following state of loading, specify when yielding condition started, using Von Mises and

Tresca criterions;

a) 2321 /100,15,0,87 mmNy .

b) 2321 /100,50,0,50 mmNy .

Solution:

Tresca;

a) y 87087minmax No yielding condition exist according Tresca criteria.b) y 100)50(50minmax Yielding condition exists according to Tresca

criteria.

Von Mises;

a) 728715;15150;87087 133221


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12

y 55.80])72(2/1)15(2/1)87(2/1[2/1222 No yielding condition according to

Von Mises criteria.

b) 1005050;50500;50050 133221 y 6.86])100(2/1)50(2/1)50(2/1[

2/1222 No yielding condition according to

Von Mises criteria.

Note: The Von-Mises criterion is most commonly used in metal forming processes to predict the

initial yielding conditions.

Example 3.4:

Drive the effective stress and strain in simple tension test?

Solution:

Assuming the loading direction is the 1st

axis, then 0,0 321 .

Using Eq. 3.19

1

2/12

12

1

2

1

22

1 )2(2

1])0()00()0[(

2

1

For circular cross-section tension specimen 321 ;0 , and for volume

constancy 1323213121321 2

1220220

Using Eq. 3.20

1

2/12

1

2

1

2

1

2/12

3

2

2

2

1 )]4

1

4

1(

3

2[)](

3

2[ .

Hence, for the general case; the effective stress and the effective strain will be used for modeling

the material behavior, (i.e. nn KK )( ).


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13

Load

3

2

1

In general, strains can be expresses

using Hook Law:

))((1

))((

1

))((1

2133

3122

3211

vE

vE

vE

Eq.3.24

Where E Young Modulus and v

Poissons ratio.

For Plan strain conditions:

)(0 2122 v

Ifv equal to 0.5 then

2/)(0 2122 Eq.3.25

Fig. 3.5 Plane strain condition in metal working.

3.7 Plane strain and plane stress conditions.

There are two important general cases commonly used in metal forming to analyze the processes;

Plane strain and Plane stress conditions.

In plane strain conditions; all stress components exist (i.e. 0,0,0 321 ) and one of the

strains (and the two related shear strains) are equal to zero (i.e. 0,0,0 321 ), see Fig. 3.5.

Using plane strain condition, Von Misses effective stresses, volume constancy principle, and

effective strain equation (as shown below), it is possible to simplify the calculations of the

effective strain and stress for different metal forming processes (see Fig. 3.6):


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14

Fig. 3.6 Effective stresses and strains for different complex state of stresses of metal formingprocesses [3.4].


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15

.)](3

2[

0

;)(2

1

;])()()[(2

1

2/12

3

2

2

2

1

321

312

2/12

13

2

32

2

21

and

y

Eq. 3.26

In plane stress condition, a biaxial state of stress exists ( 0,0 21 ) while the stress in the third

direction (and its associated shear stresses) is equal to zero ( 03 ), (see Fig. 3.7).

For plane stress condition Eq 3.24 will be reduced to;

))((1

)(1

)(1

213

122

211

vE

vE

vE

Eq. 3.27

Plane stress condition exists in sheet metal forming processes, i.e. stretching process.

3.8 Work and energy method application in metal forming processes.

Work of deformation is an important property in metal forming, and usually used to calculate the

required external force and power in metal forming processes. This can be achieved by equating

the work of the external forces and the internal energy of deformation.

The work of the external force can be expressed by simple equation; W=F.l. The incremental form

of this equation can be written as;

dlFdW . Eq. 3.28

Dividing the incremental equation by volume gives;

Fig. 3.7 Plane stress condition.


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16

21 dd

Fig. 3.8 Work of the deformation.

l

dl

A

F

dV

dW Eq. 3.29

This can be rewritten using the effective stress and effective strain as follows;

ddw Eq. 3.30

Where dw; is the incremental work per unit volume. The right side of Eq. 3.30 represents the area

under the flow stress-strain curve, (see Fig. 3.8).

The total work per unit volume can be obtained

by integrating Eq. 3.30 from1 to 2 , (see Fig.

3.8) as shown below;

dw 2

1

Eq. 3.31

Then the total work for the entire volume will be;

dVdWV

2

1

Eq. 3.32a

or

dVW 2

1

Eq. 3.32

Also, the work per unit volume (shown in Fig. 3.9); can be approximated by multiplying the mean

yielding stressm

, and the total strain along the deformation, as follows

)( 12 mw

Eq. 3.33


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17

and

Equations 3.32 and 3.33 can be combined as follows;

2

112

2

112

11

dKd nm Eq. 3.34

Then the total work can be expressed as follows;

)(1

1

1

1

2

nn

n

KVW Eq. 3.35a

If the initial strain is zero, equation (3.35a) can re-written as follows;

1

1

n

n

KVW Eq. 3.35

Note, the total work obtained in equation (3.35) is ideal work, i.e. material is homogeneous and the

friction is neglected. Hence, the total work will be the ideal work plus work due to non-

homogeneity (redundant work) plus the work due to friction, see Fig. 3.10.

Fig. 3.9 Work of the deformation.

21

w w

21

m


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18

Initial work piece Homogeneous deformation Friction+redundant

Force

Force

Fig. 3.10 Work of the deformation ( redundantFrictionogeneoustotal WWW ho m ).

Example 3.5

The following data were obtained in a tensile test using circular cross-section speciam having a

wire gaug diameter of 15 mm of mild-steel material and wire gaug length of 50 mm;

P

KN52.05 51.85 57.43 61.32 64.8 67.59 69.98 72.28 73.77 74.86 76.16 79.35 80.74 80.55 78.95 68.69 58.33

L

mm

51.18 51.59 52.37 53.16 53.92 54.71 55.5 56.29 57.05 57.83 58.62 61.95 68.78 71.12 71.52 72.31 72.64

a) Plot the experimental stress-strain curve in terms of ;

Nominal stress and strain. True stress and strain.

b) What is the value ofKand n ?

c) The plastic work required to streatch the speciamen at maximum load ?

Solution:

a) The nominal stress nominal is calculated through dividing the load (Newton) by the initial cross-section area using Equation (3.1) ( 22 7.176)15(*4/*4/ mmdA oo ) for each given

data. The nomimal or engineering strain is calculated using Equation (3.2), (o

o

l

lle

).

The true stress is calculated by dividing each load value P by the actual or instantaneous

cross-sectional area A at the stage of the test. Since the volume is constant during plastic

deformation (3

88360.50*7.176. mmLAV ooo ), then LALALA oo /88368836 .


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19

And the true stress will be calculated asA

P up to the maximum load (80.74 KN). The true

strain or logarithmic strain is calculated from Equation 3.4 ( )./ln( oll ).

Nominal and true stress -strain curves for Example 3.5

0.00

100.00

200.00

300.00

400.00

500.00

600.00

700.00

0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40

(Engineering / True strains )

(Nominal

Engineering stress -strain curve

True stress -strain curve

b) The strain hardening exponent n (equal to the true strain at the maximum load (80.74 KN)) isequal to 0.14.

c) The K value for this material can be calualted from the flow stress-strain equation 3.14(

n

K

), and given as follows;

214.02

/73.827)14.0(/56.628 mmNKKmmN

.

d) The plastic work required to streatch the speciamen to the maximum load is calculated from:

)87.4.89.4871(

.031.4871890)14.0(14.1

73.827*8836

1.

1..

14.0

KJmN

mmNn

KLA

n

KVwVW

n

oo

n

Problems

A Cylinderical specimen is compressed to 1/3 of its height, given the initial height and

diameter as ho=20 mm andDo=30 mm. Calculate;

a) The final diameterD ?

b) True stress at the end of deformation?

c) Nominal stress at the end of deformation?

d) Total ideal work?


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20

The following data were reported from a tension test:Load (N) 11500 16400 17000 20800 20600

Elongation (mm) 0.5 5.0 15.0 21.5 25.5

The speciamen has a wire gauge length of 50.5 mm and wire gauge diameter of 7 mm.

Determine:

a) The cross-section area at amximum load?

b) The true stress at maximum load?

c) The ultimate tensile strength?

d) The strain hardening exponent n and strength coefficient K?

e) The ideal plastic work required to stretch the specimen to instability?

Estimate the plastic work necessary to stretch a tensile specimen to instability. The initial cross

section area is 40.8 mm2 and the initial length is 50.8 mm. The material follows

= 1200 0.35

MPa. Determine the ultimate tensile strength (UTS)?

The following data were reported in a tensile test having rectangular cross-section specimen:

Initial strip length lo: 60.0 mm

Initial strip thickness to: 5.0 mm

Initial strip width wo : 20.0 mm

The material flow: nK Determine the following:

i) The cross-section area at maximumload.ii) The true stress at maximum load.iii) Effective strain at maximum load.iv) The ultimate strength.v) The value ofKand n.vi) The plastic work necessary to stretch the specimen to instability.

The following data were reported from tension test having rectangular cross-sectional area:

Load(N) 11500 16400 17000 20800 20700

Length(mm) lf 63.0 68.0 75.0 80.0 85.0Thickness (mm) tf 4.8 4.6 4.3 4.1 3.9

Width (mm) wf 19.84 19.18 18.6 18.29 18.1


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Initial strip length lo: 60.0 mm

Initial strip thickness to: 5.0 mmInitial strip width wo : 20.0 mm

The material flow: nK

Determine the following:

vii) The strip width at maximumload.viii)The true stress at maximum load.ix) Effective strain at maximum load.x) The ultimate strength.xi) The value ofKand n.xii) The plastic work necessary to stretch the specimen to instability.

Load(N) 11500 16400 17000 20800

Length(mm) lf 61.0 65.5 76.0 82.0

Thickness (mm)tf

4.8 4.6 4.3 4.0

Width (mm) wf 19.8 19.7 19.5 wf

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CHAPTER 3 Revised