Dr.R.Narayanasamy - Anisotropy of sheet metals.

Anisotropy in sheet metals
By
Dr. R. Narayanasamy,
B.E.,M.Tech.,M.Engg.,Ph.D.,(D.Sc.),
Professor,
Department of Production Engineering,
National Institute of Technology,
Tiruchirappalli- 620 015 ,
Tamil Nadu, India.
9/22/2015 1

Yield Locus
• For a biaxial plane stress condition the
Von Mises yield criterion can be expressed
mathematically as:
(Plot of this eqn is the Yield Locus)
is the equation of an ellipse. Major axis is
and Minor axis is
9/22/2015 2

Yield Locus cont…
• For maximum shear stress condition – Yield
locus falls inside of Von Mises Yield ellipse.
• (Uniaxial and balanced biaxial stress)
predicts the same yield stress.
• For pure shear there occurs a greatest
divergence
• Yield stress by Von Mises criterion 15.5% >
Yield stress by maximum shear stress
criterion.
9/22/2015 3

Comparison of yield criteria for plane
stress (Fig 1)
9/22/2015 4

• Change in property values with direction of measurement.
• Properties and amount of change depends on:
Structural origin of anisotropy
Intensity with which it is developed
Properties are:
Yield Strength (YS)
Ultimate Tensile Strength (UTS)
Percentage elongation
Reduction in area
Uniform and total ductility
True stress at fracture
Notched bar energy absorption.
Anisotropy in Yielding
9/22/2015 5

• Yield stress & ultimate tensile stress (under
tension) are sensitive to structural anisotropy.
• Other properties depend on fibering of
particles and weak interfaces.
• The intersection of (σ2 /σ1) = 1 (load path &
yield locus) has physical significance.
9/22/2015 6

Imagine
• A sheet is compressed through thickness
direction and yields at σy(3) . A hydrostatic
tension (in all three directions) is applied to
the yielding specimen.
• The magnitude of hydrostatic tension is
σh = σy(3) (It does change the stress state but it
will not yield).
9/22/2015 7

σ1
σ1 σ2
σ2
σ3
σ3
-σ3
-σ3
σ1
σ1 σ2
σ2
9/22/2015 8

• By adding compressive & tensile hydrostatic
stresses σh = σy(3) balanced biaxial tension can
be produced (responsible for yielding of sheet
metal).
• Yielding on (σ2 /σ1) = 1 load path is controlled
by through thickness compression yield stress.
• Yield condition is given by:
σ1 =σ2 = σy(3)
9/22/2015 9

• σy(3) changes independent of σy(1,2) (yield stress
in direction 1 & 2 respectively).
• Anisotropy is described by
– Normal anisotropy (&) Planar anisotropy (which
is related with crystallographic textures in rolled
sheets)
– The texture is symmetrical around the normal
sheet.
– Anisotropy has effect on yield locus.
9/22/2015 10

9/22/2015 11

• When σy(3) > σy(1,2) yielding is postponed to
above uniaxial yield level. (This is texture
hardening).
• When σy(3) < σy(1,2) the result is texture softening.
• Empirical equation for anisotropic yield locus:
This equation follows from Hill’s general
analysis of yielding in plastically anisotropic
material.
9/22/2015 12

• Yielding criteria is considered if the material is
isotropic.
• Material is no longer isotropic due to
appreciable plastic deformation.
• Most fabricated metal shapes have anisotropic
properties, so that it is likely that the tubular
specimen used for basic studies of yield
criteria incorporate some degree of anisotropy.
9/22/2015 13

Anisotropy in Yielding cont…
• Von-Mises’ criterion would not be valid for
highly oriented cold-rolled sheet or a fiber-
reinforced composite material.
• Hill’s formulated yield criterion for anisotropic
material having orthotropic symmetry.
• where F,G,…..N are constants defining degree
of anisotropy.
9/22/2015 14

• For principal axes of orthotropic symmetry
• If X is yield stress in 1 direction, Y is yield
stress in the 2 direction and Z is the yield
stress in the 3 direction, then we can evaluate
the constant as
9/22/2015 15

• Lubahn and Felgar give detailed plasticity
calculation for anisotropic behavoir
9/22/2015 16

Yield locus for textured titanium –
alloy sheet (Fig 2)
9/22/2015 17
ε = 0.002

• On a plane-stress yield locus as in fig 1,
anisotropic yielding results in distortion of the
yield locus.
• The yield locus is highly textured titanium alloy
sheet (fig 2).
• The experimentally determined curve is
non-symmetric when compared with ideal
isotropic curve.
9/22/2015 18

• An important aspect of yield anisotropy is
texture hardening.
• Consider a highly textured sheet that is
fabricated into a thin-wall pressure vessel, so
that the thickness stress is negligible.
9/22/2015 19

or
• We can assume that the yield stress in the
plane of the sheet are equal, i.e. X=Y.
•
9/22/2015 20

• However the yield stress in the thickness
direction of the sheet, Z, which is a difficult
property to measure.
• This problem can be circumvented by measuring
R value, (R = the ratio of the width strain to
thickness strain.)
The yield locus equation is written as:
9/22/2015 21

• The yield locus can be written as
• High through thickness yield stress Z results in
low-thickness strain and high value of R.
• The extent of strength is seen in fig. 1 from
the texture effect for spherical vessel
• The resistance to yielding (plastic
deformation) increases with increased R.
9/22/2015 22

Normal anisotropy9/22/2015 23

Anisotropy – Dr.R.Sowerby Notes
• One of the simplest functions capable of
description of initial anisotropy is the
quadratic form
f=
1
2
𝐶𝑖𝑗𝑘𝑙σ𝑖𝑗σ 𝑘𝑙 − σ̅2
= 0 ………….(1)
• If for example, 𝐶𝑖𝑗𝑘𝑙 , is expressed by the
following fourth order isotropic tensor
9/22/2015 24

Anisotropy cont…
• 𝐶𝑖𝑗𝑘𝑙 =
3
2
[𝛅𝑖𝑘 𝛅𝑗𝑙 + 𝛅𝑖𝑙 𝛅𝑗𝑘] -𝛅𝑖𝑗 𝛅 𝑘𝑙 ,
where the 𝛅𝑖𝑘 etc. are the Kvoneeker delta,
then (1) reduces to the Von-Mises yield
criterion.
NB: The 𝐶𝑖𝑗𝑘𝑙 represents 81 coefficients but
condition of symmetry reduces the independent
coefficients.
9/22/2015 25

Anisotropy cont…
• Note from (1) by flow rule
• dϵ𝑖𝑗 = 𝑑𝛌𝐶𝑖𝑗𝑘𝑙σ 𝑘𝑙
since σ 𝑘𝑙=σ𝑙𝑘 𝐶𝑖𝑗𝑘𝑙=𝐶𝑖𝑗𝑙𝑘 … … . . (𝑎)
dϵ𝑖𝑗=dϵ𝑗𝑖 𝐶𝑖𝑗𝑘𝑙=𝐶𝑖𝑗𝑙𝑘 ……….(b)
• Plastic work increment
𝑑𝑊 𝑝
= σ𝑖𝑗dϵ𝑖𝑗= 𝑑𝛌𝐶𝑖𝑗𝑘𝑙σ𝑖𝑗σ 𝑘𝑙
9/22/2015 26

Anisotropy cont…
• Since 𝑑𝑊 𝑝
is a scalar quantity the ijkl all
dummy suffices and are interchangeable.
• So 𝐶𝑖𝑗𝑘𝑙=𝐶𝑙𝑘𝑖𝑗 …………..(c)
• From (a)-(c), 𝐶𝑖𝑗𝑘𝑙 = 𝐶𝑖𝑗𝑙𝑘= 𝐶𝑗𝑖𝑘𝑙 = 𝐶 𝑘𝑙𝑖𝑗 ,
• Then the 81 independent coefficients, 𝐶𝑖𝑗𝑘𝑙
can be reduced to 21.
9/22/2015 27

Anisotropy cont…
• If it is now assumed that there exists three
orthogonal principal axes of anisotropy
such that:
i. The direct strain increments are independent
of shear stress,
ii. The shear strain increment are independent
of direct stresses,
iii. The shear strain increment depend on the
corresponding shear stresses only;
9/22/2015 28

Anisotropy cont…
• Then by invoking the plastic incompressibility
assumption the number of independent
coefficients are reduced to 6.
• With 6 independent coefficient the yield
condition can be expressed in the following
form, due to Hill
• 2𝑓(σ𝑖𝑗)𝝣 𝐹(σ 𝑦𝑦 − σ 𝑧𝑧)2
+𝐺(σ 𝑧𝑧 − σ 𝑥𝑥)2
+
𝐻(σ 𝑥𝑥 − σ 𝑦𝑦)2
+2𝐿σ 𝑦𝑧
2
+ 2𝑀σ 𝑧𝑥
2
+2𝑁σ 𝑥𝑦
2
= 1
…………..(2)9/22/2015 29

Anisotropy cont…
• The coefficients F, G, H etc. characterize the
current state of anisotropy.
when L=M=N=3F=3G=3H,
• The expression (2) reduces to the Von-Mises
yield criterion if F is equated to 1
2𝑦2 with y
the yield stress in uniaxial tension.
• If f in (2) is taken as plastic potential then the
plastic strain increments, referred to the
plastic axes of anisotropy are
9/22/2015 30

Anisotropy cont…
• 𝑑ϵ 𝑥𝑥 = 𝛌[𝐻(σ 𝑥𝑥-σ 𝑦𝑦)+G(σ 𝑥𝑥-σ 𝑧𝑧)]
• 𝑑ϵ 𝑦𝑦 = 𝛌[𝐹(σ 𝑦𝑦-σ 𝑧𝑧)+H(σ 𝑦𝑦-σ 𝑥𝑥)]
• 𝑑ϵ 𝑧𝑧 = 𝛌[𝐺(σ 𝑧𝑧-σ 𝑥𝑥)+F(σ 𝑧𝑧-σ 𝑦𝑦)]
• 𝑑ϵ 𝑦𝑧 = 𝛌𝐿σ 𝑦𝑧
• 𝑑ϵ 𝑧𝑥 = 𝛌𝑀σ 𝑧𝑥
• 𝑑ϵ 𝑥𝑦 = 𝛌𝑁σ 𝑥𝑦
9/22/2015 31

Some theoretical consideration of the
effect of Anisotropy
• We have already met Hill’s mode of an Anisotropy
Yield criterion :-
2𝑓(σ𝑖𝑗)𝝣 𝐹(σ 𝑦𝑦 − σ 𝑧𝑧)2
+𝐺(σ 𝑧𝑧 − σ 𝑥𝑥)2
+
𝐻(σ 𝑥𝑥 − σ 𝑦𝑦)2
+2𝐿σ 𝑦𝑧
2
+ 2𝑀σ 𝑧𝑥
2
+2𝑁σ 𝑥𝑦
2
= 1
………….…(18)
• If we consider only principal stresses acting in the
principal direction of Anisotropy then (18) reduces
to
2𝑓(σ𝑖𝑗) = 𝐹(σ22 − σ33)2
+𝐺(σ33 − σ11)2
+
𝐻(σ11 − σ22)2
=1 …………........…(19)
9/22/2015 32

effect of Anisotropy cont…
• Applying the flow rule to either (19) or (18)
(The form is unchanged for the direct strain
increments)
We obtain
• 𝑑𝜖11 = 𝑑𝛌[H(σ11-σ22)+G(σ11-σ33)] ……(a)
• 𝑑𝜖22 = 𝑑𝛌[F(σ22-σ33)+H(σ22-σ11)] ……(b)
• 𝑑𝜖33 = 𝑑𝛌[G(σ33-σ11)+F(σ33-σ22)] ……(c)
………………………………….(20)
9/22/2015 33

• Imagine preforming and tensile test in the ①
direction (i.e. in the X- direction of anisotropy)
• Then σ22= σ33=0
Hence
𝑑ϵ11: 𝑑ϵ22:𝑑ϵ33=(G+H):-H:-G ………(21)
• The ratio 𝐻
𝐺 is referred to as the 𝑅 𝑥value .
(the suffix ‘x’ implying the x- direction of
anisotropy)
9/22/2015 34

• It is the ratio of the width strain increment to
the thickness strain increment.
• In an isotropic material it would have the
value of unity.
• It is usual to refer to the integrated form of
(21) so that
ϵ22
ϵ33
=𝑅 𝑥
9/22/2015 35

• (in practice however the R value could change
during continuous loading)
• Similarly for tensile test in the Y-direction of
anisotropy.
• 𝑑ϵ22: 𝑑ϵ11:𝑑ϵ33=(F+H):-H:-F ………(22)
And
𝑑ϵ11
𝑑ϵ33
= 𝑅 𝑦 = (
ϵ11
ϵ22
if the integrated form of
the strain is employed)
• In general 𝑅 𝑥 ≠ 𝑅 𝑦 …………..(23)
9/22/2015 36

Plane Stress System
• This is appropriate to sheet metal forming
operation and σ33(say) is taken as zero.
• Equation (19) and (20) still hold of course with
σ33=0 .
• The form of equation (21) and (22) are unchanged.
• It is customary to identify the “Rolling Direction“
“Transverse Direction” and “Through Thickness
Direction” of a sheet metal of material, as the
principal direction of Anisotropy.
9/22/2015 37

Plane Stress System cont…
• Consequently a Tensile test conducted on a
specimen cut from the plane of the sheet.
• But in the Rolling Direction is usually designated
as “R” value of 𝑅0.
• The suffix 0 indicates 00
to the rolling direction.
• Similarly the 𝑅90would be the “R” value for a
tensile specimen cut at 90° to the rolling
direction.
• Again, in general, 𝑅0 ≠ 𝑅90
9/22/2015 38

Planar Anisotropy
• If by performing experiments on the sheet it is
found that the measured R value is unchanged
with orientation , the sheet is said to have
Planar Anisotropy.
• (if R=1 then the sheet is of course isotropic)
9/22/2015 39

Planar Anisotropy cont…
• Digression
• It can be shown that the R value at any angle
α (measured anticlockwise from rolling
direction) is:-
• 𝑅 α =
𝐻+ 2𝑁−𝐹−𝐺−4𝐻 𝑠𝑖𝑛2α𝑐𝑜𝑠2α
𝐹𝑠𝑖𝑛2α+𝐺𝑐𝑜𝑠2α
……….(24)
• Planar isotropy occurs (according to this
theory)
• When N=F+2H=G+2H (NB: F=G)
9/22/2015 40

Planar Anisotropy cont…
• In which case
• R(α) always equals 𝐻
𝐹 (or 𝐹
𝐺) …….(25)
• Technological definition of planar anisotropy
• You will some times see this defined as:-
• Δ𝑅 =
𝑅0+𝑅90−𝑅45
2
………………………..(26)
9/22/2015 41

Normal Anisotropy
• Already mentioned that a constant R value,
independent of rotation, implies Planar
Anisotropy.
• However for R≠1 the material is still
Anisotropic.
• A high R value(R>1) implies that the material
has a high resistance to thinning (A high
through thickness strength)
• The reverse is true for R<1.
9/22/2015 42

Normal Anisotropy
• Consequently the R value(for a sheet with
planar isotropy) is regarded as a measure of
the Normal Anisotropy.
• A technological measure of Normal Anisotropy
is often given as
• 𝑅͞=
𝑅0+2𝑅45+ 𝑅90
4
…………………….(27)
9/22/2015 43

Normal Anisotropy cont…
• The recognition of the role that the R value can play
in certain metal forming processes has lead to an
increased activity in the role of “Texture
Hardening”.
• The problem lies in first identifying the role ‘R’
might play and then suitably processing the sheet to
produce the desired Texture.
• For example an increased in ‘R’ value tends to
increase the deep drawing capabilities of sheet.
• All evidence up to now would suggest a similar
trend for Stretch Forming Operation.
9/22/2015 44

Influence of the Anisotropic Parameters in the
Shape of the Yield Surface for the Plane Stress Case
• With σ33 = 0 equation (19) reduces to
2𝑓(σ𝑖𝑗) = 𝐺 + 𝐻 σ11
2
− 2𝐻σ11σ22 +
𝐻 + 𝐹 σ22
2
= 1 ……………(23)
• By altering the ratio of the parameters H,G& F
you can easily recognize influence.
• This has on the shape of the yield surface ,
plotted in σ11, σ22 space.
9/22/2015 45

Representative Stress and
Representative Strain Increment
• As with the isotropic case expression for
representative stress and representative strain
increment can be formulated.
• Written in terms of principal components and
directed in the principal axes of Anisotropy, they
are:-
• σ =
3
2
𝐹(σ22−σ33)2+𝐺(σ33−σ11)2+𝐻(σ11−σ22)2
(𝐹+𝐺+𝐻)
1
2
…(29)
9/22/2015 46

Representative Stress and
Representative Strain Increment cont…
• 𝑑ϵ =
2
3
F + G + H
1
2
𝐹
𝐺𝑑ϵ22−𝐻𝑑ϵ33
𝑄
2
+ 𝐺
𝐻𝑑ϵ33−𝐹𝑑ϵ11
𝑄
2
+
𝐻
𝐹𝑑ϵ11−𝐺𝑑ϵ22
𝑄
2
………..(30)
Where Q= (FG+GH+HF)
When F=G=H (29)and (30) reduces to
the Isotropic expression.
9/22/2015 47

Verification of the σ̅ and dϵ̅ expression
for Hill’s Model of Anisotropy
• From (18) and the flow rule
𝑑ϵ 𝑥𝑥 = 𝑑𝛌 𝐻 σ 𝑥𝑥 − σ 𝑦𝑦 + σ 𝑥𝑥 − σ 𝑧𝑧 ……(a)
𝑑ϵ 𝑦𝑦 = 𝑑𝛌 𝐻 σ 𝑦𝑦 − σ 𝑧𝑧 + σ 𝑦𝑦 − σ 𝑥𝑥 …(b)
𝑑ϵ 𝑧𝑧 = 𝑑𝛌 𝐻 σ 𝑧𝑧 − σ 𝑥𝑥 + σ 𝑧𝑧 − σ 𝑥𝑥 ……(c)
𝑑ϵ 𝑥𝑦 = 𝑑𝛌Nσ 𝑥𝑦 ………… ….. (d)
ϵ 𝑦𝑧 = 𝑑𝛌Lσ 𝑦𝑧 ……………....(e)
dϵ 𝑧𝑥 = 𝑑𝛌Mσ 𝑧𝑥 ……………….(f)
……………………...(31)
9/22/2015 48

for Hill’s Model of Anisotropy cont…
• From (a),(b) and (c) of (31) we may obtain
• 𝐺𝑑ϵ 𝑦𝑦 − H𝑑ϵ 𝑧𝑧 = 𝑄 σ 𝑦𝑦 − σ 𝑧𝑧 𝑑𝛌…….(a)
• H𝑑ϵ 𝑧𝑧 − 𝐹𝑑ϵ 𝑥𝑥 = 𝑄 σ 𝑧𝑧 − σ 𝑥𝑥 𝑑𝛌…….(b)
• 𝐹𝑑ϵ 𝑥𝑥 − 𝐺𝑑ϵ 𝑦𝑦 = 𝑄 σ 𝑥𝑥 − σ 𝑦𝑦 𝑑𝛌…….(c)
………………………….(32)
where Q=FG+GH+HF
9/22/2015 49

• Square the equation (32)(a),(b) and (c) in turn
and multiply by F,G and H respectively add to
the sum
2
𝐿
𝑑ϵ 𝑦𝑧
2
+
2
𝑀
𝑑ϵ 𝑧𝑥
2
+
2
𝑁
𝑑ϵ 𝑥𝑦
2
On L.H.S.
Note(the addition to R.H.S through 31(d),(e)&(f))
9/22/2015 50

Verification of the σ̅ and d ϵ̅ expression
• Using equation (18) we obtain
𝐹
𝑄2
(Gdϵ 𝑦𝑦 − Hdϵ 𝑧𝑧)2+
𝐺
𝑄2
(Hdϵ 𝑧𝑧 − 𝐹dϵ 𝑥𝑥)2+
𝐻
𝑄2
(Fdϵ 𝑥𝑥 − 𝐺dϵ 𝑦𝑦)2+
2
𝐿
𝑑ϵ 𝑦𝑧
2
+
2
𝑀
𝑑ϵ 𝑧𝑥
2
+
2
𝑁
𝑑ϵ 𝑥𝑦
2
= 𝑑𝛌 2
…………….(33)
9/22/2015 51

• Now choose a generalized (representative)
yield stress, σ̅ , and generalized
(representative) plastic strain increment dϵ̅
such that an increment of plastic work 𝑑𝑊 𝑃
is
given by
• 𝑑𝑊 𝑃
= σ𝑑ϵ = σ𝑖𝑗 𝑑ϵ𝑖𝑗 = σ𝑖𝑗 𝑑𝛌
𝑑𝑓
𝑑σ 𝑖𝑗
= 2𝑓𝑑𝛌
• And hence from (18) =𝑑𝛌…………………(34)
9/22/2015 52

• The choice of one representative quantity is
Arbitrary.
• Select
σ̅ =
3
2
𝐹 σ 𝑦𝑦 − σ 𝑧𝑧
2
+ 𝐺 σ 𝑧𝑧 − σ 𝑥𝑥
2
… … + 2𝐿σ 𝑦𝑦
2
… . .
𝐹 + 𝐺 + 𝐻
1
2
1
2
……………(35)
9/22/2015 53

• Hence from (33),(34) and (35)
• 𝑑ϵ =
2
3
( 𝐹 + 𝐺 +
9/22/2015 54

• Note the assumption for σ̅ ~ reducing to the
expression for the isotropic case when F=G=H.
It then follows that, so to will dϵ̅ .
• Note also that the equation (29) and (30) are a
particular form of (35) and (36) where σ11 etc.
where principal stresses in the principal
direction of
9/22/2015 55

• We have already seen that conducting tensile
test and measuring the resulting strain
increments (or total strain) from the number
of tensile tests.
• We can obtain a ratio of the anisotropic
coefficient.
• See for example equation (21),(22) and (25).
9/22/2015 56

• Similarly the coefficient are also related to the
tensile stress properties for example using
equation(18) or (19).
• And measure of the yield strength in the
X(1),Y(2)and Z(3) direction we have
9/22/2015 57

•
1
σ 𝑥𝑥
2 = 𝐻 + 𝐺……(a)
•
1
σ 𝑦𝑦
2 = 𝐻 + 𝐹……(b)
•
1
σ 𝑧𝑧
2 = 𝐹 + 𝐺……(c) ……………………(37)
• Here σ 𝑥𝑥 etc. are the measured yield
strengths.
9/22/2015 58

• Hence if we know the coefficient(or their ratio)
then we are a liberty to predict the yield
strength in other direction if we know it in one
direction.
• For example planar isotropy (with F=G)
requires σ 𝑥𝑥 − σ 𝑦𝑦.
• Thus knowing σ 𝑥𝑥 then σ 𝑧𝑧 is given by
•
σ 𝑧𝑧
2
σ 𝑥𝑥
2 =
𝐻+𝐺
𝐹+𝐺
=
1+𝑅
2
…………………….(38)
• 𝑅 = 𝐻 𝐹
9/22/2015 59

• Already mentioned that altering the
coefficient s H,G and F will alter the shape of
the yield locus .
• You can verify this for yourselves using
equations (19) or (29), since one can establish
a Geometric representation of the yield
surface when plotted in principal stresses
space.
9/22/2015 60

• The plot can be further simplified by
considering the plane stress case (e.g. σ33=0)
• For example using equation (29)(with σ33=0)
and setting H=0, F=G . The yield surface when
plotted in σ11~σ22 is a circle.
• Putting F=G, but H>F elongates the yield
surface along the axis bisecting the σ11~σ22
axis (i.e. along a 45° line) .
9/22/2015 61

Verification of the σ̅ and dϵ̅expression
• The yield surface is still symmetric about this
45° line with F=G .
• If F≠G then the surface is not symmetric about
this 45° line.
• Note also the influence of F,G and H on a
Plane Strain loading line.
• Consider the case of 𝑑ϵ22 = 0, from (20) or
(31 (b)) (withσ33=0), then σ22 is given by
σ22=
𝐻σ11
𝐹+𝐻
9/22/2015 62

• With H=0 then 𝑑ϵ22 = 0 along a lineσ22=0
• If H>>F then 𝑑ϵ22 = 0 along a lineσ22≃σ11
• i.e. almost balanced biaxial tension.
• A high R value has already mentioned
indicates a high through thickness strength in
theory one should be able to obtain a
measure of σ22 by conducting a balanced
biaxial test(i.e. σ11=σ22).
9/22/2015 63

• For Planar Anisotropyσ 𝑥𝑥 ≠ σ 𝑦𝑦
• So
σ 𝑦𝑦
2
σ 𝑥𝑥
2 =
𝐻+𝐹
𝐻+𝐺
=
1+ 𝐹
𝐻
1+ 𝐺
𝐻
=
1+1
𝑅90
1+1
𝑅0
……….(39)
• Where for example 𝑅0and 𝑅90 would be the R
values in the Rolling and Transverse directions
of a rolled sheet and σ 𝑥𝑥 and σ 𝑦𝑦 are the
rolling strengths of tensile specimen cut in the
rolling and transverse directions.
9/22/2015 64

• The expression for the yield strength at any angle
measured α anticlockwise to the rolling direction is
given by
σα
2
=
1
𝐹 sin α 2+𝐺 cos α 2+𝐻+(2𝑁−𝐹−𝐺−4𝐻) sin α 2 cos α 2
…………………………..(40)
• For the derivation of equation(40)(likewise
equation 24)~ which defines the R value at any
angle α to the rolling direction.
9/22/2015 65

• Therefore equal biaxial stresses σ in the plane of
the sheet is the same as doing a compression (or
tension) test (i.e.in the 3 directions).
• Hence looking at the ellipse on diagram the
intersection the ellipse makes with the 45°
direction (distance OC) measures the through
thickness strength.
• This can be greater or less than OA or OB
depending on the values of H,G and F.
9/22/2015 66

Analytical approach to show
improvement in drawability with R value
• Assumption
• Material has planar
isotropy i.e. F=G
• Material does not strain
harden.
• Flange does not deform
under condition of plane
strain; no thickening in Z
direction i.e. 𝑑ϵ 𝑧𝑧 = 0
9/22/2015 67

Analytical approach to show improvement
in drawability with R value cont..
• From assumption (iii) using equation (20) with
Z=3, 𝜃=2, r=1; then
• 𝑑ϵ 𝑧𝑧 = 0 = 𝑑𝛌 𝐺 σ 𝑧𝑧 − σ 𝑟𝑟 + 𝐹 σ 𝑧𝑧 − σ 𝜃𝜃
• σ 𝑧𝑧 =
σ 𝑟𝑟−σ
𝜃𝜃
2
• N.B. F=G ………(1)
9/22/2015 68

• Using the yield function and inserting this
value of σ 𝑧𝑧 from (1)
• σ 𝑟𝑟 − σ 𝜃𝜃 =
2
𝐹+2𝐹
1
2
= 𝐴(say)……..(2)
• Now the radial equilibrium condition for an
element in the flange is
•
𝑑σ 𝑟𝑟
𝑑𝑟
+
σ 𝑟𝑟−σ
𝜃𝜃
𝑟
= 0…………………(3)
9/22/2015 69

• Using the value (2) in (3) and noting the
material does not harden (assumption(ii))
gives
• σ 𝑟𝑟 = −𝐴 ln 𝑟 + 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 …………..(4)
• Now σ 𝑟𝑟 = 0 at 𝑟 = 𝑟0
i.e. outside of Flange
• σ 𝑟𝑟 = 𝐴 ln 𝑟0
𝑟
9/22/2015 70

• At inside of flange at 𝑟 = 𝑟𝑖 then
• σ 𝑟𝑟 𝑟=𝑟 𝑖
= 𝐴𝑙𝑛 𝑟0
𝑟 𝑖………………….(5)
• In the absence of friction the punch load (P)
must be very nearly equal to
• σ 𝑟𝑟 𝑟=𝑟 𝑖
∗ π2𝑟𝑖 ∗ 𝑡
(t= material thickness)
• From equation (5)
• 𝑃 = 𝐴𝑙𝑛 𝑟0
𝑟 𝑖 ∗ π2𝑟𝑖 ∗ 𝑡………………..(6)
9/22/2015 71

• Now consider the
element in the cup wall
• σ 𝑟𝑟 ≃ 0
• 𝑑ϵ 𝜃𝜃 ≃0
9/22/2015 72

• 𝑑ϵ 𝜃𝜃 = 𝑑𝛌 𝐹 σ 𝜃𝜃 − 0 + 𝐻 σ 𝜃𝜃 − σ 𝑧𝑧 = 0
• σ 𝜃𝜃 =
𝐻
𝐻+𝐹
σ 𝑧𝑧
• Substitute this in the yield function with σ 𝑟𝑟 =
0 gives
• σ 𝑧𝑧 =
𝐹+𝐻
𝐹 2𝐻+𝐹
1
2
…………………(7)
9/22/2015 73

• The drawing load (P) must also equal to
2π𝑟𝑖 𝑡 ∗ σ 𝑧𝑧………………………..(8)
• Using (7) and (8)
• 𝑃 = 2π𝑟𝑖 𝑡 ∗
𝐹+𝐻
𝐹 2𝐻+𝐹
1
2
………………(9)
• Equation (9) and (6) (the assumption that t is
the same in each of these equations neglects
the possibility of the material thinning as it
behaves over the profile radius).
9/22/2015 74

• Hence the equation(9) and(6)
𝐴𝑙𝑛 𝑟0
𝑟 𝑖 ∗ π2𝑟𝑖 ∗ 𝑡 = 2π𝑟𝑖 𝑡 ∗
𝐹+𝐻
𝐹 2𝐻+𝐹
1
2
...(10)
• Using the value of A from (2) in (10)
• 𝑙𝑛 𝑟0
𝑟 𝑖 =
𝐹+𝐻
2𝐹
1
2
9/22/2015 75

• Defining R as 𝐻
𝐹
• 𝑙𝑛 𝑟0
𝑟 𝑖 =
1+𝑅
2
1
2
………………(11)
• Equation (11) states that the bigger the R
value the bigger the ratio 𝑟0
𝑟 𝑖(which is the
L.D.R)
• There is a limit, as found in practice to how
increasing R effects the L.D.R. in actual fact
the relationship is not
Limiting Draw Ratio in deep drawing.
9/22/2015 76

Dr.R.Narayanasamy - Anisotropy of sheet metals.

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