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Hindawi Publishing CorporationISRNMechanical EngineeringVolume 2013, Article ID 640958, 11 pageshttp://dx.doi.org/10.1155/2013/640958
Research ArticleSpringback Analysis in Sheet Metal Forming UsingModified Ludwik Stress-Strain Relation
Sanjay Kumar Patel, Radha Krishna Lal, J. P. Dwivedi, and V. P. Singh
Department of Mechanical Engineering, Indian Institute of Technology, Banaras Hindu University, Varanasi 221005, India
Correspondence should be addressed to Radha Krishna Lal; radhakrishna773@gmail.com
Received 18 June 2013; Accepted 17 July 2013
Academic Editors: C. C. Huang and A. Z. Sahin
Copyright Β© 2013 Sanjay Kumar Patel et al. This is an open access article distributed under the Creative Commons AttributionLicense, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properlycited.
This paper deals with the springback analysis in sheet metal forming using modified Ludwik stress-strain relation. Using thedeformation theory of plasticity, formulation of the problem and spring back ratio is derived using modified Ludwik stress strainrelationship with Tresca and von Mises yielding criteraia. The results have been representing the effect of different value of π/πΈ orππ/πΈ ratio, different values of Strain hardening index (π), Poissonβs ratio (]), and thickness on spring back ratio (π
π/π π). The main
aim of this paper is to study the effects of the thickness, π/πΈ ratio, π and Poissonβs ratio in spring back ratio.
1. Introduction
In sheet metal working, sheets are deformed to cylindricaland helical shapes by plastic bending with the help of a punchand die set. When such bending is properly done, the insidecontour of the section matches the surface contour of the dieduring the forming operation. However, after the release ofthe applied loads, the contour assumes a different shape thanthat of the die because of the release of elastic stresses in themetal. This elastic distortion is commonly called springback.Springback complicates tool design in that the die must bedesigned to compensate for it. Consequently, it is desirable tohave a method of quantitatively predicting the magnitude ofspringback as a function of the properties of the material andgeometry involved in the forming operation.
In bending, the springback is ameasure of elastic recoveryof radius on removal of the applied bending moment orload after the bending section is beyond elastic limit. Whiledesigning the die set, the springback factor should be takeninto account to avoidmismatch while assembling formed dif-ferent sections.
Initially, springback studies are with sheet bending oper-ations. Sachs [1], Schroeder [2], Gardiner [3], Singh andJohnson [4], and others studied the springback consideredbending of sheets of different shapes and depicted springbackas a function of material thickness, length, and width of the
sheets taken. Their studies were limited to V- and U-shapeddies for applying bending loads and they predicted thespringback as a measure of change in the curvature dis-tribution. Huth [5], Nadai [6], and Upadhyay [7] have alldone a number of excellent work on the elasto-plastic torsionof bars with rectangular sections, but their work has beenlimited to monotonically increasing loads only. Dwivediet al. [8, 9] analytically predicted the residual angle of twistand torque relation, and so forth, for bars of elastic strain-hardening materials with narrow rectangular sections. Thiswork, however, has the limitation that it is valid for thinrectangular strips only. Dwivedi et al. [10, 11] dealt with thetorsional springback of square-section bars of linear andnonlinear work-hardening materials. Dwivedi et al. [12] alsodealt with the torsional springback of L-shaped section barsof nonlinear work-hardening materials.
An accurate analysis of springback has been made in thepast on sheet bending and tube bending operations throughexperiment [13β18]. Torsional springback in thin tubes withnonlinear work hardening analysis was made by Choubeyet al. [19β21].
In the following, approximation equations are derived inan attempt to provide a quantitative method with practicalutility for predicting the springback behavior in bent sectionof sheet metal as a function of die radius, sheet thickness, andstress-strain characteristics of the material.
2 ISRNMechanical Engineering
2. Basic Equations and Formulation
In this paper, springback prediction approach using themod-ified Ludwik stress-strain relation was studied. The assump-tion of narrow beam of a wide sheet can significantly alterthemagnitude of the predicted results in sheet metal bendingbecause this ignorance of the transverse stresses are presentduring forming. In this paper, it is assumed that the bendsection is a wide sheet of an elastic plastic strain hardeningmaterial in an effort to obtain more accurate expression forpredicting the springback behavior. In this method, use ofapplied moment and curvature relation during the formationof a wide sheet of metal around a portion of cylindricaldie was done to derive springback relationship. From thebending equation, we directly relate the bendingmoment andcurvature during the forming as follows:
π
πΌ=
π
π¦=
πΈ
π ,
π
πΈπΌ=
1
π ,
π = πΆ β1
π .
(1)
From this equation, the springback is related to theapplied bending moment and curvature during the forma-tion. The modified Ludwik stress-strain relation (Figure 2) isgiven by
π =
[[[
[
πΈπ π β€π
πΈ
π(πΈπ
π)
π
π β©ΎΜπ
πΈ
]]]
]
, (2)
where π = ππ, π = πΈπ is the elastic part and π = π
π(πΈπ/π
π)π
is the plastic part, and the equation gives the total stress.
3. Assumption for Derivation ofSpringback Behaviour
(1) Pure bending of the sheet to cylindrical surface.(2) The cross section dimensions of the sheet are such so
that the width to thickness ratio is large.(3) The stress-strain characteristics of material are the
same in tension and compression.(4) The natural surface is always in the centre of the sheet,
and plane sections remain plane during bending.(5) The cross section dimensions of the sheet do not
change significantly in bending.(6) The radius of bending is large compared to the thick-
ness of the sheet so radial stresses are assumed to benegligible.
(7) The circumferential strains are sufficiently small sothat the conventional strain and the true strain areapproximately equivalent.
(8) The transverse strain is zero at any point in the sheet.(9) The circumferential strain for any fiber does not vary
along the bent section.
AM
B
CO
Mmax
Curvature (1/R) 1/Rf 1/Ro
Slope =dME/d(1/R)
Figure 1: Schematic representation of curvature versus applied mo-ment during bending.
2.0
2.0
1.5
1.0
0.5
00
4.0Eπ/πo
π/π
0
3510
n = 2
β
πΌ = 3/7
Figure 2: Empirical stress-strain curve for elastic plastic material inmodified ludwik stress-strain relation.
4. Derivation of Springback Equation forTresca Yield Criteria
Figure 1 shows the schematic plot of applied bendingmomentversus curvature during the formation of awide sheet ofmetalaround a portion of a cylindrical die at point A, the materialyields and plastic deformation continuous until the insidesurface of the material conforms to the inside surface of thematerial conforms to die at point Bwhen the appliedmomentis released elastic springback occurs from B to C.The changein curvature due to this elastic springback is (1/π
π) β (1/π
π)
From Figure 3,
1
π π
β1
π π
=πmax
πππΈ/π (1/π )
. (3)
Figure 3 shows a section of bent sheet and the coordinatesystem adopted.Then calculation ofmaximum applied bend-ing moment by balancing the external and internal momentsis as follows:
πmax = β«
+π‘/2
βπ‘/2
ππ₯π¦ππ¦ = 2β«
π‘/2
0
ππ₯π¦ππ¦. (4)
ISRNMechanical Engineering 3
y
tz
x R
x = Circumferential directiony = Radial directionz = Transverse direction
Figure 3: Direction of principal stresses and curvature during bend-ing.
Before proceeding, it is desirable to attempt to character-ize the stress-strain behaviour of material in a simple tension.From the modified Ludwik stress-strain relation, we get
π =
[[[[
[
πΈπ π β€π
πΈ
π(πΈπ
π)
π
πβ©ΎΜπ
πΈ
]]]]
]
. (5)
This is the equation for springback for Modified Ludwikstress-strain relation.
In elastic region,
π = πΈπ. (6)
In plastic region, where π = ππ,
π = ππ(
πΈ
ππ
)
π
ππ
. (7)
Let
ππ(
πΈ
ππ
)
π
= πΎ. (8)
Then
π = πΎππ
. (9)
At yield point, the stress in elastic and plastic region is equal;so we can write
πΈππ= πΎππ
π,
ππ= (
πΎ
πΈ)
1/(1βπ)
.
(10)
From (10), stress in yield point is
ππ= πΎππ
π= πΎ(
πΎ
πΈ)
π/(1βπ)
. (11)
It will be assumed for stresses less thanππthat the stress-strain
behaviour of the material in a simple tension test is elastic (6)and that yielding occurs at π
πand (7) is applicable for stresses
greater than ππ.
For combined states of stress, the relationship betweenprincipal stresses and strain for the elastic.
By Hookeβs law,
ππ₯=
1
πΈ(ππ₯β ] (π
π¦+ ππ§)) ,
ππ¦=
1
πΈ(ππ¦β ] (π
π₯+ ππ§)) ,
ππ§=
1
πΈ(ππ§β ] (π
π¦+ ππ₯)) .
(12)
From assumptions that the radius of bending is largecompared to thickness of the sheet so radial stresses canassume to be negligible, and the transverse strain is zero atany point in the sheet,
ππ¦= ππ§= πΏπ§= 0. (13)
And the circumferential strain for any fiber does not varyalong the bent section.
So,
ππ₯= πΏπ₯. (14)
Then, from putting the value of (12) in (13),
ππ§=
1
πΈ(ππ§β ] (π
π¦+ ππ₯)) ,
0 =1
πΈ(ππ§β ] (π
π¦+ ππ₯)) ,
1
πΈ(ππ§β ] (π
π₯)) = 0,
ππ§= ]ππ₯.
(15)
From maximum shear stress theory of failure (Tresca yieldcriteria),
ππ
2=
ππ₯β ππ§
2,
ππ= ππ₯β ]ππ₯,
ππ= ππ₯(1 β ]) .
(16)
Substituting (16) in (11),
ππ= ππ₯(1 β ]) = πΎ(
πΎ
πΈ)
π/(1βπ)
,
ππ₯=
πΎ(πΎ/πΈ)1/(1βπ)
(1 β ]).
(17)
Then,
ππ₯=
1
πΈ(ππ₯β ] (π
π¦+ ππ§)) ,
ππ₯=
1
πΈ(ππ₯β ]2ππ₯) ,
ππ₯= (
1
πΈ)ππ₯(1 β ]2) .
(18)
4 ISRNMechanical Engineering
So in yield point, the circumferential strain is
ππ₯π
= (1
πΈ)ππ(1 β ]2) . (19)
Putting yield point stress value from (11), we get
ππ₯π
=πΎ(πΎ/πΈ)
π/(1βπ)
πΈ (1 β ])(1 β ]2) ,
ππ₯π
= (πΎ
πΈ)
1/(1βπ)
(1 + ]) .
(20)
This is the approximate value of circumferential strain at elas-tic plastic interface.Then in elastic region, the circumferentialstress is
ππ₯=
πΈ
(1 β ]2)ππ₯. (21)
Here, ππ₯= π¦/π
π.
Then,
ππ₯=
πΈ
(1 β ]2)(
π¦
π π
) , (22)
for 0 β€ ππ₯β€ (πΎ/πΈ)
1/(1βπ)
(1 + ]).From the deformation theory of plasticity,
πΏπ₯=
1
πΎ1/π(π2
π₯+ π2
π¦+ π2
π§β ππ₯ππ¦β ππ¦ππ§β ππ§ππ₯)(1βπ)/2π
Γ (ππ₯β
ππ¦
2β
ππ§
2) ,
πΏπ¦=
1
πΎ1/π(π2
π₯+ π2
π¦+ π2
π§β ππ₯ππ¦β ππ¦ππ§β ππ§ππ₯)(1βπ)/2π
Γ (ππ¦β
ππ₯
2β
ππ§
2) ,
πΏπ§=
1
πΎ1/π(π2
π₯+ π2
π¦+ π2
π§β ππ₯ππ¦β ππ¦ππ§β ππ§ππ₯)(1βπ)/2π
Γ (ππ§β
ππ₯
2β
ππ¦
2) .
(23)
From assumptions that the radius of bending is largecompared to thickness of the sheet so radial stresses canassume to be negligible and the transverse strain is zero atany point in the sheet, π
π¦= ππ§= πΏπ§= 0.
And the circumferential strain for any fiber does not varyalong the bent section. So, π
π₯= πΏπ₯.
Put this value in (23). We get
πΏπ§=
1
πΎ1/π(π2
π₯+ π2
π¦+ π2
π§β ππ₯ππ¦β ππ¦ππ§β ππ§ππ₯)(1βπ)/2π
Γ (ππ§β
ππ₯
2β
ππ¦
2) ,
0 =1
πΎ1/π(π2
π₯+ π2
π§β ππ§ππ₯)(1βπ)/2π
(ππ§β
ππ₯
2) .
(24)
So,
0 = (ππ§β
ππ₯
2) ,
ππ§=
ππ₯
2.
(25)
Putting the value of ππ§into (23), we found that
πΏπ₯=
1
πΎ1/π(π2
π₯+ π2
π¦+ π2
π§β ππ₯ππ¦β ππ¦ππ§β ππ§ππ₯)(1βπ)/2π
Γ (ππ₯β
ππ¦
2β
ππ§
2) ,
πΏπ₯=
1
πΎ1/π(π2
π₯+
π2
π₯
4β
π2
π₯
2)
(1βπ)/2π
(ππ₯β
ππ₯
4) ,
πΏπ₯=
1
πΎ1/π(
3π2
π₯
4)
(1βπ)/2π
(3ππ₯
4) ,
πΏπ₯=
1
πΎ1/π(3
4)
(1+π)/2π
ππ₯
1/π
,
ππ₯=
πΎ
(3/4)(1+π)/2
πΏπ
π₯.
(26)
From initial assumption that the circumferential strainare sufficiently small so that the conventional strain and truestrain are approximately equivalent,
ππ₯=
π¦
π π
= πΏπ₯. (27)
Putting the value of πΏπ₯into (26), we get
ππ₯=
πΎ
(3/4)(1+π)/2
(π¦
π π
)
π
, (28)
where ππ₯is valid for plastic region; that is,
(πΎ
πΈ)
1/(1βπ)
(1 + ]) β€ ππ₯β€
π‘
2. (29)
From (4), we found maximum bending moment is
πmax = 2 [β«
π‘/2
0
ππ₯π¦ππ¦] , (30)
where we can split the limit into elastic or plastic part, from0 to π
πππ₯elastic region and π
πππ₯to (π‘/2) plastic region, where
π πππ₯
is calculated by
π πππ₯
= π π(πΎ
πΈ)
1/(1βπ)
(1 + ]) . (31)
ISRNMechanical Engineering 5
Now,
πmax = 2 [β«
π πππ₯
0
ππ₯(elastic)π¦ππ¦ + β«
π‘/2
π πππ₯
ππ₯(plastic)π¦ππ¦]
= 2 [β«
π πππ₯
0
(πΈ
1 β ]2)π¦2
ππ¦
+β«
π‘/2
π πππ₯
πΎ
(3/4)(1+π)/2
(π¦
π π
)
π
π¦ππ¦]
= 2 [β«
π πππ₯
0
(πΈ
1 β ]2)π¦2
ππ¦
+ β«
π‘/2
π πππ₯
πΎ
(3/4)(1+π)/2
(1
π π
)
π
π¦π+1
ππ¦]
= 2(1
π π
)[(πΈ
1 β ]2)
π¦3
3]
π πππ₯
0
+ 2[πΎ
(3/4)(1+π)/2
(1
π π
)
π
π¦π+2
π + 2]
π‘/2
π πππ₯
.
(32)
Put the value of π πππ₯
in above equation.Then
= 2[πΈπ 2
π(πΎ/πΈ)
3/(1βπ)
3 (1 β ])(1 + ])2]
+ 2[πΎ(π‘/2)
π+2
(3/4)(1+π)/2
π ππ(π + 2)
βπΎπ 2
π(πΎ/πΈ)
(π+2)/(1βπ)
(1 + ])π+2
(3/4)(1+π)/2
(π + 2)
] .
(33)
From elementary plate theory [17] applied moment perunit width during elastic unloading,
ππΈ
πΌ=
ππ₯
π¦,
ππΈ=
2 β πΈ β (π‘/2)3
3 (1 β ]2) β π .
(34)
The slope of elastic recovery in elastic unloading is
πππΈ
π (1/π )=
2 β πΈ β (π‘/2)3
3 (1 β ]2). (35)
Putting this into (3), we get
1
π π
β1
π π
=πmax
πππΈ/π (1/π )
π π
π π
= 1 β π π(
πmaxπππΈ/π (1/π )
) ,
π π
π π
= 1 β π π{(
2π π
π‘)
3
(πΎ
πΈ)
3/(1βπ)
(1 + ])3
+
3πΎ (1 β ]2) (2π π/π‘)1βπ
πΈ(3/4)(1+π)/2
(π + 2)
β3((2π
π)/π‘)3
(πΎ/πΈ)3/(1βπ)
(1 + ])π+3 (1 β ])
(3/4)(1+π)/2
(π + 2)
} ,
π π
π π
= 1 β
3πΎ (1 β ]2)
πΈ(3/4)(1+π)/2
(π + 2)
(2π π
π‘)
1βπ
+ [(2π π
π‘) (
πΎ
πΈ)
1/(1βπ)
(1 + ])]3
Γ [3(1 + ])π (1 β ])
(3/4)(1+π)/2
(π + 2)
β 1] ,
π π
π π
= 1 β
3 (1 β ]2)
(3/4)(1+π)/2
(π + 2)
(πΈ
ππ
)
πβ1
(2π π
π‘)
1βπ
+ [(2π π
π‘) (
ππ
πΈ)
1
(1 + ])]3
Γ [3(1 + ])π (1 β ])
(3/4)(1+π)/2
(π + 2)
β 1] .
(36)
This is the equation for springback ratio for modi-fied Ludwik stress-strain relationship using maximum shearstress theory.
5. Derivation of Springback Equation forVon Mises Yielding
From maximum shear stress theory of failure (von Misesyielding criteria),
π2
π=
1
2{(ππ₯β ππ¦)2
+ (ππ¦β ππ§)2
+ (ππ§β ππ₯)2
} ,
π2
π=
1
2{π2
π₯+ π2
π§+ π2
π§β 2ππ§ππ₯+ π2
π₯} ,
ππ= ππ₯(1 + ]2 β ])
1/2
,
(37)
by using (13) and (14).
6 ISRNMechanical Engineering
Substituting (15) in (11), at yield point,
ππ= πΎ(
πΎ
πΈ)
π/(1βπ)
= πππ₯
(1 + ]2 β ])1/2
,
πππ₯
(1 + ]2 β ])1/2
= πΎ(πΎ
πΈ)
π/(1βπ)
,
ππ₯π
=πΎ(πΎ/πΈ)
π/(1βπ)
(1 β ] + ]2)1/2
.
(38)
Now,
ππ₯=
1
πΈ(ππ₯β ] (π
π¦+ ππ§)) ,
ππ₯=
1
πΈ(ππ₯β ]2ππ₯) ,
ππ₯=
1
πΈππ₯(1 β ]2) .
(39)
So, in yield point, the circumferential strain is
ππ₯π
=1
πΈπππ₯
(1 β ]2) . (40)
Putting yield point stress value from (38) in (40), we get
ππ₯π
=πΎ(πΎ/πΈ)
π/(1βπ)
πΈ(1 β π + π2)1/2
(1 β ]2) ,
ππ₯π
= (πΎ
πΈ)
1/(1βπ) (1 β ]2)
(1 β ] + ]2)1/2
.
(41)
This is the approximate value of circumferential strain atelastic plastic interface.
Then in elastic region, the circumferential stress is
ππ₯=
πΈ
(1 β ]2)ππ₯. (42)
Here, ππ₯= π¦/π
π.
Putting value of circumferential strain, we get
ππ₯=
πΈ
(1 β ]2)(
π¦
π π
) , (43)
for 0 β€ ππ₯β€ (πΎ/πΈ)
1/(1βπ)
((1 β ]2)/(1 β ] + ]2)1/2
).From the deformation theory of plasticity [16],
πΏπ₯=
1
πΎ1/π(π2
π₯+ π2
π¦+ π2
π§β ππ₯ππ¦β ππ¦ππ§β ππ§ππ₯)(1βπ)/2π
Γ (ππ₯β
ππ¦
2β
ππ§
2) ,
πΏπ¦=
1
πΎ1/π(π2
π₯+ π2
π¦+ π2
π§β ππ₯ππ¦β ππ¦ππ§β ππ§ππ₯)(1βπ)/2π
Γ (ππ¦β
ππ₯
2β
ππ§
2) ,
πΏπ§=
1
πΎ1/π(π2
π₯+ π2
π¦+ π2
π§β ππ₯ππ¦β ππ¦ππ§β ππ§ππ₯)(1βπ)/2π
Γ (ππ§β
ππ₯
2β
ππ¦
2) .
(44)
From assumptions that the radius of bending is largecompared to thickness of the sheet so radial stresses canassume to be negligible and the transverse strain is zero at anypoint in the sheet,
ππ¦= ππ§= πΏπ§= 0. (45)
And the circumferential strain for any fiber does not varyalong the bent section.
So,
ππ₯= πΏπ₯. (46)
Put these values in (44). We get
πΏπ§=
1
πΎ1/π(π2
π₯+ π2
π¦+ π2
π§β ππ₯ππ¦β ππ¦ππ§β ππ§ππ₯)(1βπ)/2π
Γ (ππ§β
ππ₯
2β
ππ¦
2) ,
0 =1
πΎ1/π(π2
π₯+ π2
π§β ππ§ππ₯)(1βπ)/2π
(ππ§β
ππ₯
2) .
(47)
So,
0 = (ππ§β
ππ₯
2) ,
ππ§=
ππ₯
2.
(48)
Putting the value of ππ§into (44), we get
πΏπ₯=
1
πΎ1/π(π2
π₯+ π2
π¦+ π2
π§β ππ₯ππ¦β ππ¦ππ§β ππ§ππ₯)(1βπ)/2π
Γ (ππ₯β
ππ¦
2β
ππ§
2) ,
πΏπ₯=
1
πΎ1/π(π2
π₯+
π2
π₯
4β
π2
π₯
2)
(1βπ)/2π
(ππ₯β
ππ₯
4) ,
πΏπ₯=
1
πΎ1/π(
3π2
π₯
4)
(1βπ)/2π
(3ππ₯
4) ,
πΏπ₯=
1
πΎ1/π(3
4)
(1+π)/2π
π1/π
π₯,
ππ₯=
πΎ
(3/4)(1+π)/π
πΏπ
π₯.
(49)
From initial assumption that the circumferential strainare sufficiently small so that the conventional strain and truestrain are approximately equivalent,
ππ₯=
π¦
π π
= πΏπ₯. (50)
Putting the value of πΏπ₯into (49), we get
ππ₯=
πΎ
(3/4)(1+π)/π
(π¦
π π
)
π
, (51)
ISRNMechanical Engineering 7
where ππ₯is valid for plastic region; that is,
for (πΎ
πΈ)
1/(1βπ) (1 β ]2)
(1 β ] + ]2)β€ ππ₯β€
π‘
2. (52)
From (4), maximum bending moment is
πmax = 2 [β«
π‘/2
0
ππ₯π¦ππ¦] . (53)
We can split the limit into elastic or plastic part from 0 to π‘/2
to 0 to π πππ₯
elastic region and π πππ₯
to π‘/2 plastic region whereπ πππ₯
is calculated by
π πππ₯
= (πΎ
πΈ)
1/(1βπ)
π π
(1 β ]2)
(1 β ] + ]2)1/2
,
πmax = 2 [β«
π πππ₯
0
ππ₯(elastic)π¦ππ¦ + β«
π‘/2
π πππ₯
ππ₯(plastic)π¦ππ¦]
= 2 [β«
π πππ₯
0
πΈ
1 β ]2π¦2
(1
π π
)ππ¦
+ β«
π‘/2
π πππ₯
πΎ
(3/4)(1+π)/2
(π¦
π π
)
π
π¦ππ¦]
= 2 [β«
π πππ₯
0
(πΈ
1 β ]2)π¦2
(1
π π
)ππ¦
+ β«
π‘/2
π πππ₯
πΎ
(3/4)(1+π)/2
(1
π π
)
π
π¦π+1
ππ¦]
= 2(1
π π
)[(πΈ
1 β ]2)
π¦3
3]
π πππ₯
0
+ 2[πΎ
(3/4)(1+π)/2
(1
π π
)
π
π¦π+2
π + 2]
π‘/2
π πππ₯
.
(54)
Putting the value of π πππ₯
, we get
π πππ₯
= (πΎ
πΈ)
1/(1βπ)
π π
(1 β ]2)
(1 β ] + ]2)1/2
. (55)
Then
πmax = 2[[
[
πΈπ 3
π(πΎ/πΈ)
3/(1βπ)
((1 β ]2)3
/(1 β ] + ]2)3/2
)
3 (1 β ]2) π π
]]
]
+ 2[πΎ(π‘/2)
π+2
(3/4)(1+π)/2
π ππ(π + 2)
βπΎπ (π+2)
π(πΎ/πΈ)
(π+2)/(1βπ)
(1 β ]2)π+2
(3/4)(1+π)/2
π ππ(π + 2) (1 β ] + ]2)
(π+2)/2
] .
(56)
From elementary plate theory [17] applied moment perunit width during elastic unloading,
ππΈ=
2 β πΈ β (π‘/2)3
3 (1 β V2) β π . (57)
Now, the slope of elastic recovery in elastic unloading is
πππΈ
π (1/π )=
2 β πΈ β (π‘/2)3
3 (1 β ]2). (58)
Putting this into (3), we get
1
π 0
β1
π π
=πmax
πππΈ/π (1/π )
,
π π
π π
= 1 β π π
πmaxπππΈ/π (1/π )
,
π π
π π
= 1 β
3πΎ (1 β ]2)
πΈ (π + 2) (3/4)(π+1)/2
(2π π
π‘)
1βπ
+ [(2π π
π‘) (
πΎ
πΈ)
1/(1βπ)
]
3
Γ [
[
3(1 β ]2)π+3
(3/4)(π+1)/2
(π + 2) (1 β ] + ]2)(π+2)/2
β
(1 β ]2)3
(1 β ] + ]2)3/2
]
]
,
π π
π π
= 1 β
3 (1 β ]2)
(π + 2) (3/4)(π+1)/2
(ππ
πΈ)
1βπ
(2π π
π‘)
1βπ
+ [(2π π
π‘) (
ππ
πΈ)]
3
Γ [
[
3(1 β ]2)π+3
(3/4)(π+1)/2
(π + 2) (1 β ] + ]2)(π+2)/2
β
(1 β ]2)3
(1 β ] + ]2)3/2
]
]
.
(59)
This is the equation for springback ratio for modified Ludwikstress-strain relationship using von Mises yield criteria.
8 ISRNMechanical Engineering
20 30 40 50 60 70 80Ro/t
1
0.9
0.8
0.7
0.6
0.5
0.4
Ro/R
f
For n = 0.1
Y/E = 5.5 Γ 10β4
Y/E = 1.5 Γ 10β3
Y/E = 2.4 Γ 10β3
Figure 4: Springback ratio with π π/π‘ with different values of π
π/πΈ
or π/πΈ at π = 0.1, ] = 0.33.
20 30 40 50 60 70 80Ro/t
1
0.9
0.8
0.7
0.6
0.5
0.4
Ro/R
f
For n = 0.2
Y/E = 5.5 Γ 10β4
Y/E = 1.5 Γ 10β3
Y/E = 2.4 Γ 10β3
Figure 5: Springback ratio with π π/π‘ with different values of π
π/πΈ
or π/πΈ at π = 0.2, ] = 0.33.
6. Results and Discussion
For the sheet metal bending, the springback is calculatedfrom derived equation (using modified Ludwik stress-strainrelationship with Tresca and von Mises yield criteria).
π π
π π
= 1 β
3 (1 β ]2)
(3/4)(1+π)/2
(π + 2)
(πΈ
ππ
)
πβ1
(2π π
π‘)
1βπ
+ [(2π π
π‘) (
ππ
πΈ)
1
(1 + ])]3
Γ [3(1 + ])π (1 β ])
(3/4)(1+π)/2
(π + 2)
β 1] ,
π π
π π
= 1 β
3 (1 β ]2)
(π + 2) (3/4)(π+1)/2
(ππ
πΈ)
1βπ
(2π π
π‘)
1βπ
+ [(2π π
π‘) (
ππ
πΈ)]
3
20 30 40 50 60 70 80Ro/t
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
Ro/R
f
For n = 0.3
Y/E = 5.5 Γ 10β4
Y/E = 1.5 Γ 10β3
Y/E = 2.4 Γ 10β3
Figure 6: Springback ratio with π π/π‘ with different values of π
π/πΈ
or π/πΈ at π = 0.2, ] = 0.33.
20 30 40 50 60 70 80Ro/t
10.90.80.70.60.50.40.30.2
Ro/R
f
For n = 0.4
Y/E = 5.5 Γ 10β4
Y/E = 1.5 Γ 10β3
Y/E = 2.4 Γ 10β3
Figure 7: Springback ratio with π π/π‘ with different values of π
π/πΈ
or π/πΈ at π = 0.4, ] = 0.33.
Γ [3(1 β ]2)
π+3
(3/4)(π+1)/2
(π + 2) (1 β ] + ]2)(π+2)/2
β
(1 β ]2)3
(1 β ] + ]2)3/2
]
]
.
(60)
Here, from (8),
πΎ = [πΈ(πΈ
ππ
)
πβ1
] . (61)
Then, πΎ/πΈ = (ππ/πΈ)1βπ.
We can see that both equations have a last term that isvery negligible because the π
π/πΈ or π/πΈ = 5.5 Γ 10
β4 and itsthree times will be 10
β12; therefore, we have to neglect the lastterm of both equations.
Then we write,
π π
π π
= 1 β
3 (1 β ]2)
(3/4)(1+π)/2
(π + 2)
(πΈ
π)
πβ1
(2π π
π‘)
1βπ
, (62)
π π
π π
= 1 β
3 (1 β ]2)
(π + 2) (3/4)(π+1)/2
(π
πΈ)
1βπ
(2π π
π‘)
1βπ
. (63)
ISRNMechanical Engineering 9
20 30 40 50 60 70 80Ro/t
1
0.9
0.95
0.8
0.85
0.7
0.75
0.65
Ro/R
f
n = 0.4
n = 0.3
n = 0.2
n = 0.1
Y/E = 5.5 Γ 10β4
Figure 8: Springback ratio with π π/π‘ with different values of π at
π/πΈ = 5.5 Γ 10β4, ] = 0.33.
20 30 40 50 60 70 80Ro/t
0.9
0.80.85
0.70.75
0.650.6
0.5
0.4
0.55
0.45
Ro/R
f
n = 0.4
n = 0.3
n = 0.2
n = 0.1
Y/E = 1.5 Γ 10β3
Figure 9: Springback ratio with π π/π‘ with different values of π at
π/πΈ = 1.5 Γ 10β3, ] = 0.33.
Here, we can see that both (62) and (63) are the same if we canapply Tresca or vonMises, getting same result but last term isdifferent. The values of last term are much less; therefore, weneglect this term.
Therefore, we can say that if we apply Tresca or vonMisesyield criteria, then we are getting same result. This equationis dependent on the π
π/π‘, ππ/πΈ, strain hardening coefficient
π, and ] is Possionβs ratio.Figure 4 (π = 0.1), Figure 5 (π = 0.2), Figure 6 (π = 0.3),
and Figure 7 (π = 0.4) show the variation in springbackratio along with change in π
π/π‘ with different π
π/πΈ ratio. The
higher value of π (strain hardening coefficient) material tocloser to elasto-ideal plastic material, and thickness is onemm, and π
π/πΈ value is 5.5Γ10
β4
, 1.522Γ10β3, and 2.4Γ10
β3
(for different material c10100 copper, 1100al, and 1065steel).From the figure, we come to know that as the with increasingratio of π
π/π‘ is decrease the springback ratio and with
decreasing ππ/πΈ springback ratio is increases.
Figures 8, 9, and 10 show the variation in springback ratioalong with π
π/π‘ with the different value of π with fixed value
20 30 40 50 60 70 80Ro/t
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
Ro/R
f
n = 0.4n = 0.3
n = 0.2
n = 0.1
Y/E = 2.4 Γ 10β3
Figure 10: Springback ratio with π π/π‘ with different values of π at
π/πΈ = 2.4 Γ 10β3, ] = 0.33.
20 30 40 50 60 70 80Ro/t
0.9
0.8
0.85
0.7
0.75
0.65
Ro/R
f
οΏ½ = 0.25
οΏ½ = 0.35
οΏ½ = 0.45
n = 0.3 and Y/E == 5.5 Γ 10β4
Figure 11: Springback ratio with π π/π‘ with different values of ] at
π/πΈ = 5.5 Γ 10β4.
of ππ/πΈ = 5.5 Γ 10
β4, ππ/πΈ = 1.522 Γ 10
β3, and ππ/πΈ = 2.4 Γ
10β3, here we take the values of π = 0.1, 0.2, 0.3, and 0.4. From
the figure, we come to know that with increasing value of π,springback ratio is decreasing.
Figures 11, 12, and 13 show the variation in springbackalongwithπ
π/π‘with different value of Possionβs ratio ] = 0.25,
] = 0.35, and ] = 0.45 with fixed ππ/πΈ = 5.5 Γ 10
β4,ππ/πΈ = 1.522 Γ 10
β3, ππ/πΈ = 2.4 Γ 10
β3, and π = 0.3. Fromthe figure, we come to know that with increasing value of ],springback ratio is increasing.
Figures 14, 15, and 16 show the variation in springbackalong with π‘ with different value of π = 0.1, 0.2, 0.3, and0.4 and where π
π/πΈ = 5.5 Γ 10
β4, ππ/πΈ = 1.522 Γ 10
β3,ππ/πΈ = 2.4 Γ 10
β3, ] = 0.33, and π π
= 40 is fixed. Fromthe figure, we come to know that the springback ratio rapidlyincreases in range of 0 to 2mm thickness, then the increasein springback ratio is stable.
7. Conclusions
Based on the results presented in chapter of results and dis-cussion, following conclusions about the springback analysisin sheet metal forming using nonlinear constitutive equation,the following applies.
10 ISRNMechanical Engineering
20 30 40 50 60 70 80Ro/t
0.8
0.70.75
0.60.65
0.50.55
0.40.45
0.35
Ro/R
f
οΏ½ = 0.25
οΏ½ = 0.35
οΏ½ = 0.45
n = 0.3 and Y/E = 1.5 Γ 10β3
Figure 12: Springback ratio with π π/π‘ with different values of ] at
π/πΈ = 1.5 Γ 10β3.
20 30 40 50 60 70 80Ro/t
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
Ro/R
f
οΏ½ = 0.25
οΏ½ = 0.35
οΏ½ = 0.45
n = 0.3 and Y/E = 2.4 Γ 10β3
Figure 13: Springback ratio with π π/π‘ with different values of ] at
π/πΈ = 2.4 Γ 10β3.
0 0.5 1 1.5 2 2.5 3 3.5 4 54.5t
0.9
0.95
1
0.8
0.85
0.7
0.75
Ro/R
f
n = 0.4
n = 0.3
n = 0.2
n = 0.1
οΏ½ = 0.33 and Y/E = 5.5 Γ 10β4
Figure 14: Springback ratio with π‘ for different values of π at π/πΈ =
5.5 Γ 10β4.
(i) The theoretical analysis for the sheetmetal under purebending has been done, and it was found that theprediction of springback is quite successfull.
(ii) As expected, elastic recovery is found to be more withdecreasing values of work hardening coefficient. At
0 0.5 1 1.5 2 2.5 3 3.5 4 54.5t
0.90.95
1
0.80.85
0.5
0.70.75
0.60.65
0.55
Ro/R
f
n = 0.4
n = 0.3
n = 0.2
n = 0.1
οΏ½ = 0.33 and Y/E = 1.5 Γ 10β3
Figure 15: Springback ratio with π‘ for different values of π at π/πΈ =
1.5 Γ 10β3.
0 0.5 1 1.5 2 2.5 3 3.5 4 54.5t
0.9
1
0.8
0.5
0.4
0.7
0.6Ro/R
f
n = 0.4
n = 0.3
n = 0.2
n = 0.1
οΏ½ = 0.33 and Y/E = 2.4 Γ 10β3
Figure 16: Springback ratio with π‘ for different values of π at π/πΈ =
2.4 Γ 10β3.
lower values of n, the material will approach to anelasto-ideally plastic behavior.
(iii) Springback ratio increases with increasing thickness.(iv) Springback ratio increases with decreasing ratio of
yield point stress to Youngβs modulus of elasticity(v) Springback ratio is increasing with increasing Pos-
sionβs ratio.(vi) Springback ratio for Tresca and von Mises yield cri-
teria, we say that π π/π‘ is less than 20, there will be
slightly change in that is negligible and for high valueof π π/π‘ or more than 20, differences in springback
ratio is increases.
Nomenclature
π π: Radius of die
π π: Radius of final product after removing load
πmax: Maximum applied bending moment per unit widthπ¦: Distance from midsectionπ‘: Thicknessπ: True stressπ: True strainπΈ: Modulus of elasticity
ISRNMechanical Engineering 11
]: Poissonβs ratioπ: Strain hardening indexπΌ: Empirical constantπΎ: Constantπ, ππ: Yield point stress
ππ: Strain at yield point
πΏ: Strain in plastic region.
References
[1] G. Sachs, Principles and Methods of Sheet Metal Fabricating,Reinhold, New York, NY, USA, 1951.
[2] W. Schroeder, βMechanics of sheetmetal bending,βTransactionsof the ASME, vol. 36, pp. 138β145, 1943.
[3] F. J. Gardiner, βThe springback of metals,β Transactions of theASEME, vol. 49, pp. 1β9, 1958.
[4] A. N. Singh and W. Johnson, βSpringback after cylindricallybending metal strips,β in Proceedings of the Dr Karunesh Memo-rial International Conference, NewDelhi, India, December 1979.
[5] J. H. Huth, βA note on plastic torsion,β Journal of Applied Me-chanics, vol. 22, pp. 432β434, 1955.
[6] A. Nadai,Theory of Flow and Fracture of Solids, vol. 1, McGraw-Hill, 1950.
[7] P. C. Upadhyay, Elasto-plastic torsion [M.Tech. thesis], Mechan-ical Engineering Department, I.I.T. Kanpur, 1970.
[8] J. P. Dwivedi, A. N. Singh, S. Ram, and N. K. Das Talukder,βSpringback analysis in torsion of rectangular strips,β Interna-tional Journal of Mechanical Sciences, vol. 28, no. 8, pp. 505β515,1986.
[9] J. P. Dwivedi, P. K. Sarkar, R. S. Ram. S., N. K. D. Talukder, andA. N. Singh, βExperimental aspects of torsional springback inrectangular strips,β Journal of the Institution of Engineers, vol.67, no. 4, pp. 70β73, 1987.
[10] J. P. Dwivedi, A. K. Shukla, and P. C. Upadhyay, βTorsionalspringback of square section bars of linear wor hardeningmaterials,βComputers and Structures, vol. 45, no. 3, pp. 421β429,1972.
[11] J. P.Dwivedi, P. C.Upadhyay, andN.K.DasTalukder, βTorsionalspringback in square section bars of nonlinear work-hardeningmaterials,β International Journal of Mechanical Sciences, vol. 32,no. 10, pp. 863β876, 1990.
[12] J. P. Dwivedi, P. C. Upadhyay, and N. K. D. Talukder, βSpring-back analysis of torsion of L-sectioned bars of work-hardeningmaterials,βComputers and Structures, vol. 43, no. 5, pp. 815β822,1992.
[13] Z. T. Zhang and S. J. Hu, βStress and residual stress distributionsin plane strain bending,β International Journal of MechanicalSciences, vol. 40, no. 6, pp. 533β543, 1998.
[14] T. Kuwabara, βAdvances in experiments on metal sheets andtubes in support of constitutive modeling and forming simula-tions,β International Journal of Plasticity, vol. 23, no. 3, pp. 385β419, 2007.
[15] H. K. Yi, D.W. Kim, C. J. van Tyne, and Y. H. Moon, βAnalyticalprediction of springback based on residual differential strainduring sheet metal bending,β Journal of Mechanical EngineeringScience, vol. 222, no. 2, pp. 117β129, 2008.
[16] A. El Megharbel, G. A. El Nasser, and A. El Domiaty, βBendingof tube and sectionmade of strain-hardeningmaterials,β JournalofMaterials Processing Technology, vol. 203, no. 1β3, pp. 372β380,2008.
[17] E. Da-xin, H. Hau-hui, L. Xiao-yi, and N. Ru-xin, βSpring-backdeformation in tube bending,β International Journal ofMinerals,Metallurgy and Materials, vol. 16, no. 2, pp. 177β183, 2009.
[18] D. E. Daxin E and Y. Liu, βSpringback and time-dependentspringback of 1Cr18Ni9Ti stainless steel tubes under bending,βMaterials and Design, vol. 31, no. 3, pp. 1256β1261, 2010.
[19] V. K. Choubey, M. Gangwar, and J. P. Dwivedi, βTorsionalspringback analysis in thin tubes with non-linear work harden-ing,β Journal of Mechanical Engineering, vol. 7, no. 1, pp. 15β34,2011.
[20] V. K. Choubey, M. Gangwar, and J. P. Dwivedi, βSpringbackanalysis of thin tubes,β Journal of Mechanical Engineering, vol.7, no. 2, 2011.
[21] V. K. Choubey, M. Gangwar, J. P. Dwivedi, and N. K. dasTalukder, βSpringback analysis of thin tubes with arbitrarystress-strain curves,β Journal of Mechanical Engineering, vol. 8,no. 1, 2011.
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